diff --git a/.gitignore b/.gitignore index 4b3a9dd..6f725ea 100644 --- a/.gitignore +++ b/.gitignore @@ -25,3 +25,4 @@ build/* *.swp *.swo +**/.DS_Store diff --git a/README.md b/README.md index d0e3978..1a10a97 100644 --- a/README.md +++ b/README.md @@ -30,6 +30,18 @@ The [milp-evolve](milp-evolve/) folder includes the following material to open s } ``` +### OptiMind: Teaching LLMs to Think Like Optimization Experts +The [optimind](optimind/) folder includes the following material to open source the paper, evaluation code, and data: + +```latex +@article{chen2025optimind, + title={OptiMind: Teaching LLMs to Think Like Optimization Experts}, + author={Zhang, Xinzhi and Chen, Zeyi and Zope, Humishka and Barbalho, Hugo and Mellou, Konstantina and Molinaro, Marco and Kulkarni, Janardhan and Menache, Ishai and Li, Sirui}, + journal={arXiv preprint arXiv:2509.22979}, + year={2025} +} +``` + ## Responsible AI Considerations diff --git a/optimind/LICENSE b/optimind/LICENSE new file mode 100644 index 0000000..269a897 --- /dev/null +++ b/optimind/LICENSE @@ -0,0 +1,21 @@ +MIT License + +Copyright (c) 2025 Microsoft + +Permission is hereby granted, free of charge, to any person obtaining a copy +of this software and associated documentation files (the "Software"), to deal +in the Software without restriction, including without limitation the rights +to use, copy, modify, merge, publish, distribute, sublicense, and/or sell +copies of the Software, and to permit persons to whom the Software is +furnished to do so, subject to the following conditions: + +The above copyright notice and this permission notice shall be included in all +copies or substantial portions of the Software. + +THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR +IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, +FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE +AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER +LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, +OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE +SOFTWARE. diff --git a/optimind/README.md b/optimind/README.md new file mode 100644 index 0000000..f3362ce --- /dev/null +++ b/optimind/README.md @@ -0,0 +1,96 @@ +# OptiMind: Teaching LLMs to Think Like Optimization Experts + +[![arXiv](https://img.shields.io/badge/arXiv-2509.22979-b31b1b.svg)](https://arxiv.org/abs/2509.22979) +[![Hugging Face](https://img.shields.io/badge/%F0%9F%A4%97%20Hugging%20Face-OptiMind--SFT-blue)](https://huggingface.co/microsoft/OptiMind-SFT) +[![License: MIT](https://img.shields.io/badge/License-MIT-yellow.svg)](https://mit-license.org/) + +This repository contains the official evaluation code and cleaned datasets for the paper **"OptiMind: Teaching LLMs to Think Like Optimization Experts"**. + +OptiMind is a comprehensive framework designed to enhance Large Language Models (LLMs) for formulating Mixed-Integer Linear Programming (MILP) problems. By systematically integrating optimization domain expertise into both training (via data cleaning) and inference (via class-specific error hints), OptiMind significantly improves formulation accuracy over base models and other open-source baselines. + +## Overview + +The repository provides: +1. **Evaluation Scripts**: The exact code used in our paper to evaluate models using single-turn generation, majority voting, and multi-turn self-correction with solver feedback. +2. **Cleaned Benchmarks**: Expert-verified versions of IndustryOR, Mamo-Complex, and OptMATH, where we fixed missing data, ambiguities, and incorrect ground truths. +3. **Expert Hints**: The library of error-analysis hints used to guide the model during inference. + +## Usage + +### 1. Install Enviornment +We recommend using `uv` for fast package management. Please follow the steps below to set up the environment. + +```bash +curl -LsSf https://astral.sh/uv/install.sh | sh + +uv venv .sglang --python 3.12 --seed +source .sglang/bin/activate +uv pip install "sglang" --prerelease=allow +uv pip install pandas numpy matplotlib json_repair nest_asyncio gurobipy hf_transfer azure-identity pyarrow termcolor +``` + +**Note:** Make sure you have a valid Gurobi license to execute the solver-based evaluation and self-correction loops. + +### 2. Model +Our fine-tuned model is available on Hugging Face: [microsoft/OptiMind-SFT](https://huggingface.co/microsoft/OptiMind-SFT). + +### 3. Running Evaluations +First, clone the repository and navigate to the project directory: + +```bash +git clone [https://github.com/microsoft/OptiGuide.git](https://github.com/microsoft/OptiGuide.git) +cd OptiMind +``` + +We provide shell scripts to automate the evaluation pipeline, which includes running experiments across multiple random seeds, temperatures, and majority-voting configurations. Before running the evaluation scripts, you must update the file paths in `submit_eval.sh` and `submit_eval_gptoss.sh` to match your local environment. + +For `gpt-oss-20b` and its fine-tuned variants (including [our fine-tuned model](https://huggingface.co/microsoft/OptiMind-SFT)), use `submit_eval_gptoss.sh`: +``` +bash submit_eval_gptoss.sh +``` +For other open-source models (e.g., Qwen, Llama), use `submit_eval.sh` that excludes the `--gpt-oss` and `--reasoning` flags. +``` +bash submit_eval.sh +``` +The evaluation result will be automatically saved in `eval_results//`. To aggregate the mean and stand deviation across the runs, run +``` +python calculate_avg_results.py +``` + + + + + +## Data and Benchmarks +We provide our rigorously cleaned versions of three challenging benchmarks in the `/data` folder. These files are provided in CSV format with question and answer columns. + + +- `data/optimind_cleaned_classified_industryor.csv`: Cleaned and classified IndustryOR benchmark. +- `data/optimind_cleaned_classified_mamo_complex.csv`: Cleaned and classified Mamo-Complex benchmark. +- `data/optimind_cleaned_classified_optmath.csv`: Cleaned and classified OptMATH benchmark. +- `data/hints.csv`: The dictionary mapping problem classes to specific error summaries and hints. + +### Dataset Cleaning Comparisons + +In the `data/comparison/` folder, we provide detailed HTML summary tables that transparently document every modification made to the original benchmarks. These files allow you to align the original instances with our cleaned versions and verify the specific fixes (e.g., addressing missing parameters, ambiguity, or wrong ground truths). + +- `data/comparison/industryOR_original_vs_ours.html`: Comparison table aligning the original [IndustryOR](https://huggingface.co/datasets/CardinalOperations/IndustryOR) instances with our cleaned versions. +- `data/comparison/OptMATH_original_vs_ours.html`: Comparison table aligning the [OptMATH](https://github.com/optsuite/OptMATH/blob/main/benchmark/OptMATH_Bench.json) instances with our cleaned versions. +- `data/comparison/compare_SIRL_Ours.html`: A comparison between our cleaned IndustryOR set and the cleaned version from [SIRL](https://github.com/Cardinal-Operations/SIRL/blob/main/test_data/IndustryOR_fixedV2.json) (identifying residual issues in the latter). + +Each row in these HTML tables contains the Problem Index, Original Problem, Original Answer, Updated Problem, Updated Answer, and a "How did we fix it" description. + + +## License +This project is licensed under the [MIT Licence](https://mit-license.org/). + +## Citation +If you find our work, code, or datasets useful, please cite our paper: +``` +@article{chen2025optimind, + title={OptiMind: Teaching LLMs to Think Like Optimization Experts}, + author={Chen, Zeyi and Zhang, Xinzhi and Zope, Humishka and Barbalho, Hugo and Mellou, Konstantina and Molinaro, Marco and Kulkarni, Janardhan and Menache, Ishai and Li, Sirui}, + journal={arXiv preprint arXiv:2509.22979}, + year={2025} +} +``` \ No newline at end of file diff --git a/optimind/SECURITY.md b/optimind/SECURITY.md new file mode 100644 index 0000000..e751608 --- /dev/null +++ b/optimind/SECURITY.md @@ -0,0 +1,14 @@ + + +## Security + +Microsoft takes the security of our software products and services seriously, which +includes all source code repositories in our GitHub organizations. + +**Please do not report security vulnerabilities through public GitHub issues.** + +For security reporting information, locations, contact information, and policies, +please review the latest guidance for Microsoft repositories at +[https://aka.ms/SECURITY.md](https://aka.ms/SECURITY.md). + + \ No newline at end of file diff --git a/optimind/calculate_avg_results.py b/optimind/calculate_avg_results.py new file mode 100644 index 0000000..5399313 --- /dev/null +++ b/optimind/calculate_avg_results.py @@ -0,0 +1,93 @@ +from pathlib import Path +import json +import re +import sys +import math + +def average_turn_accuracies(root_dir: str, turns=None, use_new: bool = False): + """ + Compute average accuracies and std dev for turn_0..turn_4 over subdirectories named 'xxx_j'. + Returns (averages_dict, stdevs_dict, counts_dict). + + If use_new is True, read from stats_new.json instead of stats.json. + """ + if turns is None: + turns = [f"turn_{i}" for i in range(5)] # turn_0 .. turn_4 + + subdir_pat = re.compile(r".+_\d+$") + sums = {t: 0.0 for t in turns} + sumsq = {t: 0.0 for t in turns} + counts = {t: 0 for t in turns} + + stats_filename = "stats_new.json" if use_new else "stats.json" + + for d in Path(root_dir).iterdir(): + if not d.is_dir() or not subdir_pat.match(d.name): + continue + stats_path = d / stats_filename + if not stats_path.exists(): + continue + try: + data = json.loads(stats_path.read_text()) + except Exception: + continue + acc = data.get("accuracy_per_turn", {}) + for t in turns: + v = acc.get(t) + if isinstance(v, (int, float)): + v = float(v) + sums[t] += v + sumsq[t] += v * v + counts[t] += 1 + + averages = {t: (sums[t] / counts[t] if counts[t] else None) for t in turns} + stdevs = {} + for t in turns: + n = counts[t] + if n > 1: + # sample standard deviation (ddof=1) + var = (sumsq[t] - (sums[t] ** 2) / n) / (n - 1) + # guard tiny negative from FP error + stdevs[t] = math.sqrt(var) if var > 0 else 0.0 + elif n == 1: + stdevs[t] = 0.0 + else: + stdevs[t] = None + + return averages, stdevs, counts + +def main(): + prog = Path(sys.argv[0]).name + args = sys.argv[1:] + + if not args: + print(f"Usage: {prog} [--new] ") + sys.exit(1) + + use_new = False + if "--new" in args: + use_new = True + args.remove("--new") + + if not args: + print(f"Usage: {prog} [--new] ") + sys.exit(1) + + root_dir = args[0] + averages, stdevs, counts = average_turn_accuracies(root_dir, use_new=use_new) + + # Print sorted by turn index + for t in sorted(averages, key=lambda k: int(k.split('_')[1])): + avg = averages[t] + std = stdevs[t] + n = counts[t] + + if avg is None: + print(f"{t}: None (std=None) (n={n})") + else: + avg *= 100.0 + std_str = "None" if std is None else f"{std * 100.0:.6f}" + print(f"{t}: {avg:.6f} ± {std_str} (n={n})") + +if __name__ == "__main__": + main() diff --git a/optimind/data/comparisons/industryOR_original_vs_ours.html b/optimind/data/comparisons/industryOR_original_vs_ours.html new file mode 100644 index 0000000..1c7cd1a --- /dev/null +++ b/optimind/data/comparisons/industryOR_original_vs_ours.html @@ -0,0 +1,3161 @@ + + + + + Comparison: IndustryOR Original VS Ours + + + +

Comparison: IndustryOR Original VS Ours

+

File 1: .\OptiMind-data\data\industryOR\industryOR_original_from_json.xlsx

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File 2: .\OptiMind-data\data\industryOR\industryOR_ours.xlsx

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Generated: 2025-11-14 17:59:52

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RowDescription 1Value 1Description 2Value 2Description DiffValue DiffFixFix Type
1A factory produces two types of food, I and II, and currently has 50 skilled workers. It is known that one skilled worker can produce $10 \ \mathrm{kg} / \ \mathrm{h}$ of food I or $6 \ \mathrm{kg} / \ \mathrm{h}$ of food II. According to contract bookings, the weekly demand for these two foods will rise sharply, as shown in Table 1-11. Therefore, the factory has decided to train 50 new workers by the end of the 8th week. It is known that a worker works $40 \ \mathrm{h}$ per week, and a skilled worker can train up to three new workers in two weeks (during the training period, both the skilled worker and the trainees do not participate in production). The weekly wage of a skilled worker is 360 yuan, the weekly wage of a trainee during the training period is 120 yuan, and after training, the wage is 240 yuan per week, with the same production efficiency as skilled workers. During the transition period of training, many skilled workers are willing to work overtime, and the factory has decided to arrange some workers to work $60 \ \mathrm{h}$ per week, with a weekly wage of 540 yuan. If the booked food cannot be delivered on time, the compensation fee for each week of delay per $ \ \mathrm{kg}$ is 0.5 yuan for food I and 0.6 yuan for food II. Under these conditions, how should the factory make comprehensive arrangements to minimize the total cost? + +Table 1-11 + +| Week | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | +|------|---|---|---|---|---|---|---|---| +| I | 10000 | 10000 | 12000 | 12000 | 16000 | 16000 | 20000 | 20000 | +| II | 6000 | 7200 | 8400 | 10800 | 10800 | 12000 | 12000 | 12000 |219816A factory produces two types of food, I and II, and currently has 50 skilled workers. It is known that one skilled worker can produce $10 \ \mathrm{kg} / \ \mathrm{h}$ of food I or $6 \ \mathrm{kg} / \ \mathrm{h}$ of food II. According to contract bookings, the weekly demand for these two foods will rise sharply, as shown in Table 1-11. Therefore, the factory has decided to train 50 new workers by the end of the 8th week. It is known that a worker works $40 \ \mathrm{h}$ per week, and a skilled worker can train up to three new workers in two weeks (during the training period, both the skilled worker and the trainees do not participate in production). The weekly wage of a skilled worker is 360 yuan, the weekly wage of a trainee during the training period is 120 yuan, and after training, the wage is 240 yuan per week, with the same production efficiency as skilled workers. During the transition period of training, many skilled workers are willing to work overtime, and the factory has decided to arrange some workers to work $60 \ \mathrm{h}$ per week, with a weekly wage of 540 yuan. If the booked food cannot be delivered on time, the compensation fee for each week of delay per $ \ \mathrm{kg}$ is 0.5 yuan for food I and 0.6 yuan for food II. Under these conditions, how should the factory make comprehensive arrangements to minimize the total cost? + +Table 1-11 + +| Week | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | +|------|---|---|---|---|---|---|---|---| +| I | 10000 | 10000 | 12000 | 12000 | 16000 | 16000 | 20000 | 20000 | +| II | 6000 | 7200 | 8400 | 10800 | 10800 | 12000 | 12000 | 12000 |219816No differencesNo differencesok
2A fighter jet is a crucial combat tool, but to make a fighter jet effective, enough pilots are required. Therefore, a portion of the produced fighter jets, besides those used directly in combat, must be allocated for training pilots. Given that the annual production of fighter jets is \(a_j(j=1, 2)\), with \(a_1=10\) and \(a_2=15\), furthermore, each fighter jet can train 5 pilots per year. How should the annual production of fighter jets be allocated to maximize their contribution to air defense over \(n\) years?125Each year $t=1,\dots ,n$ two production lines deliver $a_1=10$ and $a_2=15$ new fighter jets (25 total). $n=10$. Decide how many of that year's 25 aircraft, $x_t$, enter combat immediately and how many, $y_t=25-x_t$, become training platforms. A training jet produces five newly qualified pilots who are available at the start of the next year; every combat jet must be matched with one trained pilot to be operational, and training jets can be reassigned to combat in later years. Starting with no aircraft or pilots, choose integer sequences $\{x_t,y_t\}_{t=1}^n$ to maximise the cumulative number of operational combat jet-years $\sum_{t=1}^{n} x_t$, subject to annual pilot-availability and fleet-balance constraints.1350A Each year $t=1,\dots ,n$ two production lines deliver $a_1=10$ and $a_2=15$ new fighter jet is a crucial jets (25 total). $n=10$. Decide how many of that year's 25 aircraft, $x_t$, enter combat tool, but to make a fighter immediately and how many, $y_t=25-x_t$, become training platforms. A training jet effective, enough produces five newly qualified pilots who are required. Therefore, a portion available at the start of the produced fighter jets, besides those used directly in combat, next year; every combat jet must be allocated for training pilots. Given that the annual production of fighter jets is \(a_j(j=1, 2)\), matched with \(a_1=10\) one trained pilot to be operational, and \(a_2=15\), furthermore, each fighter jet can train 5 pilots per year. How should the annual production of fighter training jets can be allocated reassigned to combat in later years. Starting with no aircraft or pilots, choose integer sequences $\{x_t,y_t\}_{t=1}^n$ to maximize their contribution maximise the cumulative number of operational combat jet-years $\sum_{t=1}^{n} x_t$, subject to air defense over \(n\) years? annual pilot-availability and fleet-balance constraints.125 1350The original problem was not well defined, in particular it lacked a clear objective functionambiguous
3A company specializing in foldable tables needs to create an optimal production and human resources plan for a six-month period (January to June) to maximize its total net profit. The plan must detail monthly in-house production levels, outsourcing quantities, and workforce management (hiring/firing). + +**Initial Conditions (at the start of January):** +- Initial Workforce: 1,000 employees +- Initial Inventory: 15,000 units + +**Revenue and Cost Structure:** +- **Sales Price:** 300 Yuan per unit sold. +- **Raw Material Cost:** 90 Yuan per unit, applicable *only* to units produced in-house. +- **Outsourcing Cost:** 200 Yuan per unit for finished tables acquired from a third-party supplier. This is an all-inclusive cost. +- **Inventory Holding Cost:** 15 Yuan per unit for any inventory held at the end of a month. +- **Backorder Cost:** 35 Yuan per unit for any unfulfilled demand (stockout) carried over to the next month. + +**Labor and Production Parameters:** +- **Labor Requirement:** Each in-house unit requires 5 labor hours to produce. +- **Regular Labor:** Each worker provides 160 regular working hours per month (8 hours/day * 20 days/month). The company pays a regular wage of 30 Yuan/hour for these 160 hours, regardless of full utilization. +- **Overtime Labor:** Workers can perform overtime. Total overtime hours per month for the entire workforce cannot exceed 20 hours per worker. The overtime wage is 40 Yuan/hour. +- **Workforce Management:** The company can hire or fire workers each month. The cost to hire a new worker is 5,000 Yuan, and the cost to fire a worker is 8,000 Yuan. + +**Demand and Fulfillment Logic:** +- Unfulfilled demand from one month is back-ordered and must be met in subsequent months. +- The company fulfills orders (both current demand and backorders) using available inventory from the previous month, current in-house production, and outsourced units. + +**Terminal Condition (at the end of June):** +- The ending inventory must be at least 10,000 units. +- All backorders must be cleared (i.e., ending backorders must be zero). + +**Forecasted Demand:** +| Month | January | February | March | April | May | June | +|:---:|:---:|:---:|:---:|:---:|:---:|:---:| +| Demand Forecast | 20,000 | 40,000 | 42,000 | 35,000 | 19,000 | 18,500 | + +Based on this information, formulate the optimal six-month operational plan.10349920A company specializing in foldable tables needs to create an optimal production and human resources plan for a six-month period (January to June) to maximize its total net profit. The plan must detail monthly in-house production levels, outsourcing quantities, and workforce management (hiring/firing). + +**Initial Conditions (at the start of January):** +- Initial Workforce: 1,000 employees +- Initial Inventory: 15,000 units + +**Revenue and Cost Structure:** +- **Sales Price:** 300 Yuan per unit sold. +- **Raw Material Cost:** 90 Yuan per unit, applicable *only* to units produced in-house. +- **Outsourcing Cost:** 200 Yuan per unit for finished tables acquired from a third-party supplier. This is an all-inclusive cost. +- **Inventory Holding Cost:** 15 Yuan per unit for any inventory held at the end of a month. +- **Backorder Cost:** 35 Yuan per unit for any unfulfilled demand (stockout) carried over to the next month. + +**Labor and Production Parameters:** +- **Labor Requirement:** Each in-house unit requires 5 labor hours to produce. +- **Regular Labor:** Each worker provides 160 regular working hours per month (8 hours/day * 20 days/month). The company pays a regular wage of 30 Yuan/hour for these 160 hours, regardless of full utilization. +- **Overtime Labor:** Workers can perform overtime. Total overtime hours per month for the entire workforce cannot exceed 20 hours per worker. The overtime wage is 40 Yuan/hour. +- **Workforce Management:** The company can hire or fire workers each month. The cost to hire a new worker is 5,000 Yuan, and the cost to fire a worker is 8,000 Yuan. + +**Demand and Fulfillment Logic:** +- Unfulfilled demand from one month is back-ordered and must be met in subsequent months. +- The company fulfills orders (both current demand and backorders) using available inventory from the previous month, current in-house production, and outsourced units. + +**Terminal Condition (at the end of June):** +- The ending inventory must be at least 10,000 units. +- All backorders must be cleared (i.e., ending backorders must be zero). + +**Forecasted Demand:** +| Month | January | February | March | April | May | June | +|:---:|:---:|:---:|:---:|:---:|:---:|:---:| +| Demand Forecast | 20,000 | 40,000 | 42,000 | 35,000 | 19,000 | 18,500 | + +Based on this information, formulate the optimal six-month operational plan.10349920No differencesNo differencesok
4A farmer needs to decide how many cows, sheep, and chickens to raise in order to achieve maximum profit. The farmer can sell cows, sheep, and chickens for $500, $200, and $8 each, respectively. The feed costs for each cow, sheep, and chicken are $100, $80, and $5, respectively. The profit is the difference between the selling price and the feed cost. Each cow, sheep, and chicken produces 10, 5, and 3 units of manure per day, respectively. Due to the limited time the farm staff has for cleaning the farm each day, they can handle up to 800 units of manure. Additionally, because of the limited farm size, the farmer can raise at most 50 chickens. Furthermore, the farmer must have at least 10 cows to meet customer demand. The farmer must also raise at least 20 sheep. Finally, the total number of animals cannot exceed 100.30400A farmer needs to decide how many cows, sheep, and chickens to raise in order to achieve maximum profit. The farmer can sell cows, sheep, and chickens for $500, $200, and $8 each, respectively. The feed costs for each cow, sheep, and chicken are $100, $80, and $5, respectively. The profit is the difference between the selling price and the feed cost. Each cow, sheep, and chicken produces 10, 5, and 3 units of manure per day, respectively. Due to the limited time the farm staff has for cleaning the farm each day, they can handle up to 800 units of manure. Additionally, because of the limited farm size, the farmer can raise at most 50 chickens. Furthermore, the farmer must have at least 10 cows to meet customer demand. The farmer must also raise at least 20 sheep. Finally, the total number of animals cannot exceed 100.30400No differencesNo differencesok
5Mary is planning her dinner tonight. Every 100 grams of okra contains 3.2 grams of fiber, every 100 grams of carrots contains 2.7 grams of fiber, every 100 grams of celery contains 1.6 grams of fiber, and every 100 grams of cabbage contains 2 grams of fiber. How many grams of each type of food should Mary buy to maximize her fiber intake? + +She is considering choosing one among salmon, beef, and pork as a protein source. + +She also considers choosing at least two kinds of vegetables among okra, carrots, celery, and cabbage. + +The price of salmon is $4 per 100 grams, beef is $3.6 per 100 grams, pork is $1.8 per 100 grams. The price of okra is $2.6 per 100 grams, carrots are $1.2 per 100 grams, celery is $1.6 per 100 grams, and cabbage is $2.3 per 100 grams. Mary has a budget of $15 for this meal. + +The total food intake should be 600 grams.1600Mary is planning her dinner tonight. Every 100 grams of okra contains 3.2 grams of fiber, every 100 grams of carrots contains 2.7 grams of fiber, every 100 grams of celery contains 1.6 grams of fiber, and every 100 grams of cabbage contains 2 grams of fiber. How many grams of each type of food should Mary buy to maximize her fiber intake? + +She is considering choosing one among salmon, beef, and pork as a protein source. For the chosen protein she must take at least one gram of it. + +She also considers choosing at least two kinds of vegetables among okra, carrots, celery, and cabbage. For each of the selected vegetables, she must take at least one gram. + +The price of salmon is $4 per 100 grams, beef is $3.6 per 100 grams, pork is $1.8 per 100 grams. The price of okra is $2.6 per 100 grams, carrots are $1.2 per 100 grams, celery is $1.6 per 100 grams, and cabbage is $2.3 per 100 grams. Mary has a budget of $15 for this meal. + +The total food intake should be 600 grams.18.95657143Mary is planning her dinner tonight. Every 100 grams of okra contains 3.2 grams of fiber, every 100 grams of carrots contains 2.7 grams of fiber, every 100 grams of celery contains 1.6 grams of fiber, and every 100 grams of cabbage contains 2 grams of fiber. How many grams of each type of food should Mary buy to maximize her fiber intake? + +She is considering choosing one among salmon, beef, and pork as a protein source. For the chosen protein she must take at least one gram of it. + +She also considers choosing at least two kinds of vegetables among okra, carrots, celery, and cabbage. For each of the selected vegetables, she must take at least one gram. + +The price of salmon is $4 per 100 grams, beef is $3.6 per 100 grams, pork is $1.8 per 100 grams. The price of okra is $2.6 per 100 grams, carrots are $1.2 per 100 grams, celery is $1.6 per 100 grams, and cabbage is $2.3 per 100 grams. Mary has a budget of $15 for this meal.1600 18.95657143Incorrect ground truth: One could never have 1600 grams of fibres with at most 600 g of total food intake. incorrect ground truth
6The contract reservations for the next year for products I, II, and III of a certain factory in each quarter are shown in Table 1-10. + +Table 1-10 +| Product | 1 | 2 | 3 | 4 | +|---------|------|------|------|------| +| I | 1500 | 1000 | 2000 | 1200 | +| II | 1500 | 1500 | 1200 | 1500 | +| III | 1000 | 2000 | 1500 | 2500 | + +At the beginning of the first quarter, there is no inventory for these three products, and it is required to have 150 units in stock for each product by the end of the fourth quarter. It is known that the factory has 15,000 production hours per quarter, and each unit of products I, II, and III requires 2, 4, and 3 hours respectively. Due to a change in equipment, product I cannot be produced in the second quarter. It is stipulated that if the products cannot be delivered on time, a compensation of 20 yuan per unit per quarter delay is required for products I and II, while for product III, the compensation is 10 yuan. Additionally, for products produced but not delivered in the current quarter, the inventory cost is 5 yuan per unit per quarter. How should the factory schedule production to minimize the total cost of compensation and inventory?10755The contract reservations for the next year for products I, II, and III of a certain factory in each quarter are shown in Table 1-10. + +Table 1-10 +| Product | 1 | 2 | 3 | 4 | +|---------|------|------|------|------| +| I | 1500 | 1000 | 2000 | 1200 | +| II | 1500 | 1500 | 1200 | 1500 | +| III | 1000 | 2000 | 1500 | 2500 | + +At the beginning of the first quarter, there is no inventory for these three products, and it is required to have 150 units in stock for each product by the end of the fourth quarter. It is known that the factory has 15,000 production hours per quarter, and each unit of products I, II, and III requires 2, 4, and 3 hours respectively. Due to a change in equipment, product I cannot be produced in the second quarter. It is stipulated that if the products cannot be delivered on time, a compensation of 20 yuan per unit per quarter delay is required for products I and II, while for product III, the compensation is 10 yuan. Additionally, for products produced but not delivered in the current quarter, the inventory cost is 5 yuan per unit per quarter. How should the factory schedule production to minimize the total cost of compensation and inventory?10755No differencesNo differencesok
7An Italian transportation company needs to move some empty containers from its 6 warehouses (located in Verona, Perugia, Rome, Pescara, Taranto, and Lamezia) to major national ports (Genoa, Venice, Ancona, Naples, Bari). The container inventory at the warehouses is as follows: + +| | Empty Containers | +|:---:|:---:| +| Verona | 10 | +| Perugia | 12 | +| Rome | 20 | +| Pescara | 24 | +| Taranto | 18 | +| Lamezia | 40 | + +The demand at the ports is as follows: + +| | Container Demand | +|:---:|:---:| +| Genoa | 20 | +| Venice | 15 | +| Ancona | 25 | +| Naples | 33 | +| Bari | 21 | + +The transport is carried out by a fleet of trucks. The cost to transport each container is proportional to the distance traveled by the trucks, with a rate of 30 euros per kilometer. Each truck can carry up to 2 containers. The distances are as follows: + +| | Genoa | Venice | Ancona | Naples | Bari | +|:---:|:---:|:---:|:---:|:---:|:---:| +| Verona | $290 \mathrm{~km}$ | $115 \mathrm{~km}$ | $355 \mathrm{~km}$ | $715 \mathrm{~km}$ | $810 \mathrm{~km}$ | +| Perugia | $380 \mathrm{~km}$ | $340 \mathrm{~km}$ | $165 \mathrm{~km}$ | $380 \mathrm{~km}$ | $610 \mathrm{~km}$ | +| Rome | $505 \mathrm{~km}$ | $530 \mathrm{~km}$ | $285 \mathrm{~km}$ | $220 \mathrm{~km}$ | $450 \mathrm{~km}$ | +| Pescara | $655 \mathrm{~km}$ | $450 \mathrm{~km}$ | $155 \mathrm{~km}$ | $240 \mathrm{~km}$ | $315 \mathrm{~km}$ | +| Taranto | $1010 \mathrm{~km}$ | $840 \mathrm{~km}$ | $550 \mathrm{~km}$ | $305 \mathrm{~km}$ | $95 \mathrm{~km}$ | +| Lamezia | $1072 \mathrm{~km}$ | $1097 \mathrm{~km}$ | $747 \mathrm{~km}$ | $372 \mathrm{~km}$ | $333 \mathrm{~km}$ | + +Write a mathematical program to find the minimum cost transportation policy and solve it using COPTPY.904590An Italian transportation company needs to move some empty containers from its 6 warehouses (located in Verona, Perugia, Rome, Pescara, Taranto, and Lamezia) to major national ports (Genoa, Venice, Ancona, Naples, Bari). The container inventory at the warehouses is as follows: + +| | Empty Containers | +|:---:|:---:| +| Verona | 10 | +| Perugia | 12 | +| Rome | 20 | +| Pescara | 24 | +| Taranto | 18 | +| Lamezia | 40 | + +The demand at the ports is as follows: + +| | Container Demand | +|:---:|:---:| +| Genoa | 20 | +| Venice | 15 | +| Ancona | 25 | +| Naples | 33 | +| Bari | 21 | + +The transport is carried out by a fleet of trucks. The cost to transport each container is proportional to the distance traveled by the trucks, with a rate of 30 euros per kilometer. Each truck can carry up to 2 containers. The distances are as follows: + +| | Genoa | Venice | Ancona | Naples | Bari | +|:---:|:---:|:---:|:---:|:---:|:---:| +| Verona | $290 \mathrm{~km}$ | $115 \mathrm{~km}$ | $355 \mathrm{~km}$ | $715 \mathrm{~km}$ | $810 \mathrm{~km}$ | +| Perugia | $380 \mathrm{~km}$ | $340 \mathrm{~km}$ | $165 \mathrm{~km}$ | $380 \mathrm{~km}$ | $610 \mathrm{~km}$ | +| Rome | $505 \mathrm{~km}$ | $530 \mathrm{~km}$ | $285 \mathrm{~km}$ | $220 \mathrm{~km}$ | $450 \mathrm{~km}$ | +| Pescara | $655 \mathrm{~km}$ | $450 \mathrm{~km}$ | $155 \mathrm{~km}$ | $240 \mathrm{~km}$ | $315 \mathrm{~km}$ | +| Taranto | $1010 \mathrm{~km}$ | $840 \mathrm{~km}$ | $550 \mathrm{~km}$ | $305 \mathrm{~km}$ | $95 \mathrm{~km}$ | +| Lamezia | $1072 \mathrm{~km}$ | $1097 \mathrm{~km}$ | $747 \mathrm{~km}$ | $372 \mathrm{~km}$ | $333 \mathrm{~km}$ | + +Write a mathematical program to find the minimum cost transportation policy and solve it.904590| Taranto | $1010 \mathrm{~km}$ | $840 \mathrm{~km}$ | $550 \mathrm{~km}$ | $305 \mathrm{~km}$ | $95 \mathrm{~km}$ | +| Lamezia | $1072 \mathrm{~km}$ | $1097 \mathrm{~km}$ | $747 \mathrm{~km}$ | $372 \mathrm{~km}$ | $333 \mathrm{~km}$ | + +Write a mathematical program to find the minimum cost transportation policy and solve it using COPTPY. it.No differencesRemoved mention to specific solver/library ("using COPTPY")specific solver
8Now, we need to determine 4 out of 5 workers to complete one of the four tasks respectively. Due to each worker's different technical specialties, the time required for them to complete each task varies. The hours required by each worker to complete each task are shown in Table 5-2. + +Table 5-2 +| Worker | $A$ | $B$ | $C$ | $D$ | +|--------|-----|-----|-----|-----| +| I | 9 | 4 | 3 | 7 | +| II | 4 | 6 | 5 | 6 | +| III | 5 | 4 | 7 | 5 | +| IV | 7 | 5 | 2 | 3 | +| V | 10 | 6 | 7 | 4 | + +Try to find a job assignment plan that minimizes the total working hours.14Now, we need to determine 4 out of 5 workers to complete one of the four tasks respectively. Due to each worker's different technical specialties, the time required for them to complete each task varies. The hours required by each worker to complete each task are shown in Table 5-2. + +Table 5-2 +| Worker | $A$ | $B$ | $C$ | $D$ | +|--------|-----|-----|-----|-----| +| I | 9 | 4 | 3 | 7 | +| II | 4 | 6 | 5 | 6 | +| III | 5 | 4 | 7 | 5 | +| IV | 7 | 5 | 2 | 3 | +| V | 10 | 6 | 7 | 4 | + +Try to find a job assignment plan that minimizes the total working hours.14No differencesNo differencesok
9Haus Toys can manufacture and sell toy trucks, toy airplanes, toy boats, and toy trains. The profit for each truck sold is $5, each airplane $10, each boat $8, and each train $7. How many types of toys should Haus Toys manufacture to maximize profits? + +There are 890 units of wood available. Each truck requires 12 units, each airplane 20 units, each boat 15 units, and each train 10 units. + +There are 500 units of steel available. Each airplane requires 3 units, each boat 5 units, each train 4 units, and each truck 6 units. + +If Haus Toys manufactures trucks, they will not manufacture trains. + +However, if they manufacture boats, they will also manufacture airplanes. + +The number of toy boats manufactured cannot exceed the number of toy trains manufactured.623Haus Toys can manufacture and sell toy trucks, toy airplanes, toy boats, and toy trains. The profit for each truck sold is $5, each airplane $10, each boat $8, and each train $7. How many types of toys should Haus Toys manufacture to maximize profits? + +There are 890 units of wood available. Each truck requires 12 units, each airplane 20 units, each boat 15 units, and each train 10 units. + +There are 500 units of steel available. Each airplane requires 3 units, each boat 5 units, each train 4 units, and each truck 6 units. + +If Haus Toys manufactures trucks, they will not manufacture trains. + +However, if they manufacture boats, they will also manufacture airplanes. + +The number of toy boats manufactured cannot exceed the number of toy trains manufactured.623No differencesNo differencesok
10A convenience supermarket is planning to open several chain stores in a newly built residential area in the northwest suburb of the city. For shopping convenience, the distance from any residential area to one of the chain stores should not exceed $800 \mathrm{~m}$. Table 5-1 shows the new residential areas and the residential areas within a radius of $800 \mathrm{~m}$ from each of them. Question: What is the minimum number of chain stores the supermarket needs to build among the mentioned residential areas, and in which residential areas should they be built? + +| Area Code | Residential Areas within $800 \mathrm{~m}$ Radius | +|-----------|---------------------------------------------------| +| A | A, C, E, G, H, I | +| B | B, H, I | +| C | A, C, G, H, I | +| D | D, J | +| E | A, E, G | +| F | F, J, K | +| G | A, C, E, G | +| H | A, B, C, H, I | +| I | A, B, C, H, I | +| J | D, F, J, K, L | +| K | F, J, K, L | +| L | J, K, L |3A convenience supermarket is planning to open several chain stores in a newly built residential area in the northwest suburb of the city. For shopping convenience, the distance from any residential area to one of the chain stores should not exceed $800 \mathrm{~m}$. Table 5-1 shows the new residential areas and the residential areas within a radius of $800 \mathrm{~m}$ from each of them. Question: What is the minimum number of chain stores the supermarket needs to build among the mentioned residential areas, and in which residential areas should they be built? + +| Area Code | Residential Areas within $800 \mathrm{~m}$ Radius | +|-----------|---------------------------------------------------| +| A | A, C, E, G, H, I | +| B | B, H, I | +| C | A, C, G, H, I | +| D | D, J | +| E | A, E, G | +| F | F, J, K | +| G | A, C, E, G | +| H | A, B, C, H, I | +| I | A, B, C, H, I | +| J | D, F, J, K, L | +| K | F, J, K, L | +| L | J, K, L |3No differencesNo differencesok
11A company produces two types of small motorcycles, where type A is entirely manufactured by the company, and type B is assembled from imported parts. The production, assembly, and inspection time required for each unit of these two products are shown in Table 3.2. + +Table 3.2 + +| Type | Process | | | Selling Price <br> (Yuan/unit) | +| :---: | :---: | :---: | :---: | :---: | +| | Manufacturing | Assembly | Inspection | | +| Type A (hours/unit) | 20 | 5 | 3 | 650 | +| Type B (hours/unit) | 0 | 7 | 6 | 725 | +| Max production capacity per week (hours) | 120 | 80 | 40 | | +| Production cost per hour (Yuan) | 12 | 8 | 10 | | + +If the company's operational goals and targets are as follows: + +$p_{1}$ : The total profit per week should be at least 3000 yuan; + +$p_{2}$ : At least 5 units of type A motorcycles should be produced per week; + +$p_{3}$ : Minimize the idle time of each process as much as possible. The weight coefficients of the three processes are proportional to their hourly costs, and overtime is not allowed. + +Try to establish a model for this problem.4136A company produces two types of small motorcycles, where type A is entirely manufactured by the company, and type B is assembled from imported parts. The production, assembly, and inspection time required for each unit of these two products are shown in Table 3.2. + +Table 3.2 + +| Type | Process | | | Selling Price <br> (Yuan/unit) | +| :---: | :---: | :---: | :---: | :---: | +| | Manufacturing | Assembly | Inspection | | +| Type A (hours/unit) | 20 | 5 | 3 | 650 | +| Type B (hours/unit) | 0 | 7 | 6 | 725 | +| Max production capacity per week (hours) | 120 | 80 | 40 | | +| Production cost per hour (Yuan) | 12 | 8 | 10 | | + +If the company's operational goals and targets are as follows: + +$p_{1}$ : The total profit per week should be at least 3000 yuan; + +$p_{2}$ : At least 5 units of type A motorcycles should be produced per week; + +$p_{3}$ : Minimize the idle time of each process as much as possible. The weight coefficients of the three processes are their hourly costs, and overtime is not allowed. + +Try to establish a model for this problem.272$p_{2}$ : At least 5 units of type A motorcycles should be produced per week; + +$p_{3}$ : Minimize the idle time of each process as much as possible. The weight coefficients of the three processes are proportional to their hourly costs, and overtime is not allowed. + +Try to establish a model for this problem.4136 272Incorrect ground truth. Also, we clarified the exact weights in the objective function by removing the words "proportional to"incorrect ground truth, ambiguous
12Red Star Plastics Factory produces six distinct types of plastic containers. Each container type is characterized by a specific volume, market demand, and unit variable production cost, as detailed in Table 5-11. + +**Table 5-11: Container Data** +| Container Type (Code) | 1 | 2 | 3 | 4 | 5 | 6 | +| :------------------------------ | :--- | :--- | :--- | :--- | :--- | :---- | +| Volume ($\text{cm}^3$) | 1500 | 2500 | 4000 | 6000 | 9000 | 12000 | +| Market Demand (units) | 500 | 550 | 700 | 900 | 400 | 300 | +| Unit Variable Production Cost (Yuan/unit) | 5 | 8 | 10 | 12 | 16 | 18 | + +The production of any container type necessitates the use of its dedicated specialized equipment. If the decision is made to **activate** the production equipment for a particular container type (i.e., if the production quantity of that type is greater than zero), a fixed setup cost of 1200 Yuan is incurred for that specific equipment. + +Should the production quantity of a certain container type be insufficient to meet its direct demand, the factory has the option to utilize other container types with **larger or equal volume** as substitutes to fulfill this unmet demand. For instance, type 2 containers (volume 2500 $\text{cm}^3$) can be used to satisfy the demand for type 1 containers (requiring a volume of 1500 $\text{cm}^3$), but type 1 containers cannot be used for type 2 demand. In this problem, the container type codes are pre-sorted in ascending order of their volumes. + +**Question:** +How should the factory organize its production? The objective is to develop a production plan that minimizes the total cost—comprising the sum of variable production costs for all containers produced and the fixed costs for all activated equipment—while ensuring that the demand for all container types is fully met.43200Red Star Plastics Factory produces six distinct types of plastic containers. Each container type is characterized by a specific volume, market demand, and unit variable production cost, as detailed in Table 5-11. + +**Table 5-11: Container Data** +| Container Type (Code) | 1 | 2 | 3 | 4 | 5 | 6 | +| :------------------------------ | :--- | :--- | :--- | :--- | :--- | :---- | +| Volume ($\text{cm}^3$) | 1500 | 2500 | 4000 | 6000 | 9000 | 12000 | +| Market Demand (units) | 500 | 550 | 700 | 900 | 400 | 300 | +| Unit Variable Production Cost (Yuan/unit) | 5 | 8 | 10 | 12 | 16 | 18 | + +The production of any container type necessitates the use of its dedicated specialized equipment. If the decision is made to **activate** the production equipment for a particular container type (i.e., if the production quantity of that type is greater than zero), a fixed setup cost of 1200 Yuan is incurred for that specific equipment. + +Should the production quantity of a certain container type be insufficient to meet its direct demand, the factory has the option to utilize other container types with **larger or equal volume** as substitutes to fulfill this unmet demand. For instance, type 2 containers (volume 2500 $\text{cm}^3$) can be used to satisfy the demand for type 1 containers (requiring a volume of 1500 $\text{cm}^3$), but type 1 containers cannot be used for type 2 demand. In this problem, the container type codes are pre-sorted in ascending order of their volumes. + +**Question:** +How should the factory organize its production? The objective is to develop a production plan that minimizes the total cost—comprising the sum of variable production costs for all containers produced and the fixed costs for all activated equipment—while ensuring that the demand for all container types is fully met.43200No differencesNo differencesok
13Tom and Jerry just bought a farm in Sunshine Valley, and they are considering using it to plant corn, wheat, soybeans, and sorghum. The profit per acre for planting corn is $1500, the profit per acre for planting wheat is $1200, the profit per acre for planting soybeans is $1800, and the profit per acre for planting sorghum is $1600. To maximize their profit, how many acres of land should they allocate to each crop? Tom and Jerry’s farm has a total area of 100 acres. + +The land area used for planting corn must be at least twice the land area used for planting wheat. + +The land area used for planting soybeans must be at least half the land area used for planting sorghum. + +The land area used for planting wheat must be three times the land area used for planting sorghum.180000Tom and Jerry just bought a farm in Sunshine Valley, and they are considering using it to plant corn, wheat, soybeans, and sorghum. The profit per acre for planting corn is $1500, the profit per acre for planting wheat is $1200, the profit per acre for planting soybeans is $1800, and the profit per acre for planting sorghum is $1600. To maximize their profit, how many acres of land should they allocate to each crop? Tom and Jerry’s farm has a total area of 100 acres. + +The land area used for planting corn must be at least twice the land area used for planting wheat. + +The land area used for planting soybeans must be at least half the land area used for planting sorghum. + +The land area used for planting wheat must be three times the land area used for planting sorghum.180000No differencesNo differencesok
14Mary is planning tonight's dinner. She wants to choose a combination of protein and vegetables to maximize her protein intake for the meal. Her protein options are chicken, salmon, and tofu, which can be bought in any quantity. + +- Chicken: 23g protein, $3.00 cost, per 100g. +- Salmon: 20g protein, $5.00 cost, per 100g. +- Tofu: 8g protein, $1.50 cost, per 100g. + +She also wants to choose from a list of five vegetables, sold in 100g packs. She must select at least three different types of vegetables. + +- Broccoli (100g pack): 2.8g protein, $1.20 cost. +- Carrots (100g pack): 0.9g protein, $0.80 cost. +- Spinach (100g pack): 2.9g protein, $1.50 cost. +- Bell Pepper (100g pack): 1.0g protein, $1.00 cost. +- Mushrooms (100g pack): 3.1g protein, $2.00 cost. + +Mary has two main constraints: +1. Her total budget is $20. +2. The total weight of all food must not exceed 800 grams. + +How should Mary choose her ingredients to get the maximum possible amount of protein?123.8Mary is planning tonight's dinner. She wants to choose a combination of protein and vegetables to maximize her protein intake for the meal. Her protein options are chicken, salmon, and tofu, which can be bought in any quantity. + +- Chicken: 23g protein, $3.00 cost, per 100g. +- Salmon: 20g protein, $5.00 cost, per 100g. +- Tofu: 8g protein, $1.50 cost, per 100g. + +She also wants to choose from a list of five vegetables, sold in 100g packs. She must select at least three different types of vegetables. + +- Broccoli (100g pack): 2.8g protein, $1.20 cost. +- Carrots (100g pack): 0.9g protein, $0.80 cost. +- Spinach (100g pack): 2.9g protein, $1.50 cost. +- Bell Pepper (100g pack): 1.0g protein, $1.00 cost. +- Mushrooms (100g pack): 3.1g protein, $2.00 cost. + +Mary has two main constraints: +1. Her total budget is $20. +2. The total weight of all food must not exceed 800 grams. + +How should Mary choose her ingredients to get the maximum possible amount of protein?123.8No differencesNo differencesok
15A certain factory needs to use a special tool over $n$ planning stages. At stage $j$, $r_j$ specialized tools are needed. At the end of this stage, all tools used within this stage must be sent for repair before they can be reused. There are two repair methods: one is slow repair, which is cheaper (costs $b$ per tool) but takes longer ($p$ stages to return); the other is fast repair, which costs $c$ per tool $(c > b)$ and is faster, requiring only $q$ stages to return $(q < p)$. If the repaired tools cannot meet the needs, new ones must be purchased, with a cost of $a$ per new tool $(a > c)$. This special tool will no longer be used after $n$ stages. Determine an optimal plan for purchasing and repairing the tools to minimize the cost spent on tools during the planning period.\n\nn = 10 # number of stages\nr = [3, 5, 2, 4, 6, 5, 4, 3, 2, 1] # tool requirements per stage, indexing starts at 1\na = 10 # cost of buying a new tool\nb = 1 # cost of slow repair\nc = 3 # cost of fast repair\np = 3 # slow repair duration\nq = 1 # fast repair duration36A certain factory needs to use a special tool over $n$ planning stages. At stage $j$, $r_j$ specialized tools are needed. At the end of this stage, all tools used within this stage must be sent for repair before they can be reused. There are two repair methods: one is slow repair, which is cheaper (costs $b$ per tool) but takes longer ($p$ stages to return, e.g. if a tool goes to repair after stage 1, it will return at stage 1+p); the other is fast repair, which costs $c$ per tool $(c > b)$ and is faster, requiring only $q$ stages to return $(q < p)$. If the repaired tools cannot meet the needs, new ones must be purchased, with a cost of $a$ per new tool $(a > c)$. This special tool will no longer be used after $n$ stages. Determine an optimal plan for purchasing and repairing the tools to minimize the cost spent on tools during the planning period.\n\nn = 10 # number of stages\nr = [3, 5, 2, 4, 6, 5, 4, 3, 2, 1] # tool requirements per stage, indexing starts at 1\na = 10 # cost of buying a new tool\nb = 1 # cost of slow repair\nc = 3 # cost of fast repair\np = 3 # slow repair duration\nq = 1 # fast repair duration134A certain factory needs to use a special tool over $n$ planning stages. At stage $j$, $r_j$ specialized tools are needed. At the end of this stage, all tools used within this stage must be sent for repair before they can be reused. There are two repair methods: one is slow repair, which is cheaper (costs $b$ per tool) but takes longer ($p$ stages to return); return, e.g. if a tool goes to repair after stage 1, it will return at stage 1+p); the other is fast repair, which costs $c$ per tool $(c > b)$ and is faster, requiring only $q$ stages to return $(q < p)$. If the repaired tools cannot meet the needs, new ones must be purchased, with a cost of $a$ per new tool $(a > c)$. This special tool will no longer be used after $n$ stages. Determine an optimal plan for purchasing and repairing the tools to minimize the cost spent on tools during the planning period.\n\nn = 10 # number of stages\nr = [3, 5, 2, 4, 6, 5, 4, 3, 2, 1] # tool requirements per stage, indexing starts at 1\na = 10 # cost of buying a new tool\nb = 1 # cost of slow repair\nc = 3 # cost of fast repair\np = 3 # slow repair duration\nq = 1 # fast repair duration36 134Ambiguity about fixing a tool at time 1, say, whether the tool is available at time p or at time p+1, incorrect ground truthambiguous, incorrect ground truth
16A store plans to formulate the purchasing and sales plan for a certain product for the first quarter of next year. It is known that the warehouse capacity of the store can store up to 500 units of the product, and there are 200 units in stock at the end of this year. The store purchases goods once at the beginning of each month. The purchasing and selling prices of the product in each month are shown in Table 1.3. + +Table 1.3 + +| Month | 1 | 2 | 3 | +| :---: | :---: | :---: | :---: | +| Purchasing Price (Yuan) | 8 | 6 | 9 | +| Selling Price (Yuan) | 9 | 8 | 10 | + +Now, determine how many units should be purchased and sold each month to maximize the total profit, and express this problem as a linear programming model.4100A store plans to formulate the purchasing and sales plan for a certain product for the first quarter of next year. It is known that the warehouse capacity of the store can store up to 500 units of the product, and there are 200 units in stock at the end of this year. The store purchases goods once at the beginning of each month. The purchasing and selling prices of the product in each month are shown in Table 1.3. + +Table 1.3 + +| Month | 1 | 2 | 3 | +| :---: | :---: | :---: | :---: | +| Purchasing Price (Yuan) | 8 | 6 | 9 | +| Selling Price (Yuan) | 9 | 8 | 10 | + +Now, determine how many units should be purchased and sold each month to maximize the total profit, and express this problem as a linear programming model.4100No differencesNo differencesok
17Certain strategic bomber groups are tasked with destroying enemy military targets. It is known that the target has four key parts, and destroying at least two of them will suffice. + +Resources and constraints: + +Bomb stockpile: A maximum of 28 heavy bombs and 12 light bombs can be used. +Fuel limit: Total fuel consumption must not exceed 10,000 liters. +Fuel consumption rules: Same as before (depend on bomb type, flight distance, and takeoff/landing). + +Table 1-17 +| Key Part | Distance from Airport (km) | Probability of Destruction per Heavy Bomb | Probability of Destruction per Light Bomb | +|----------|----------------------------|-----------------------------------------|------------------------------------------| +| | | | | +| 1 | 450 | 0.03 | 0.08 | +| 2 | 480 | 0.10 | 0.11 | +| 3 | 540 | 0.05 | 0.12 | +| 4 | 600 | 0.05 | 0.09 | + +How should the bombing plan be determined to maximize the probability of success? What is the maximum probability of success?0.5765Certain strategic bomber groups are tasked with destroying enemy military targets. It is known that the target has four key parts, and destroying at least two of them will suffice. + + Resources and constraints: + + Bomb stockpile: A maximum of 28 heavy bombs and 12 light bombs can be used. + Fuel limit: Total fuel consumption must not exceed 10,000 liters. + Fuel consumption rules: Same as before (depend on bomb type, flight distance, and takeoff/landing). + + Table 1-17 + | Key Part | Distance from Airport (km) | Probability of Destruction per Heavy Bomb | Probability of Destruction per Light Bomb | + |----------|----------------------------|-----------------------------------------|------------------------------------------| + | | | | | + | 1 | 450 | 0.03 | 0.08 | + | 2 | 480 | 0.10 | 0.11 | + | 3 | 540 | 0.05 | 0.12 | + | 4 | 600 | 0.05 | 0.09 | + + How should the bombing plan be determined to maximize the probability of success? What is the maximum probability of success?0.5765This involves the product of probabilities, so it is a non-linear problem.non-linear problem
18A textile factory produces two types of fabrics: one for clothing and the other for curtains. The factory operates two shifts, with a weekly production time set at 110 hours. Both types of fabrics are produced at a rate of 1000 meters per hour. Assuming that up to 70,000 meters of curtain fabric can be sold per week, with a profit of 2.5 yuan per meter, and up to 45,000 meters of clothing fabric can be sold per week, with a profit of 1.5 yuan per meter, the factory has the following objectives in formulating its production plan: + +$p_{1}$ : The weekly production time must fully utilize 110 hours; + +$p_{2}$ : Overtime should not exceed 10 hours per week; + +$p_{3}$ : At least 70,000 meters of curtain fabric and 45,000 meters of clothing fabric must be sold per week; + +$p_{4}$ : Minimize overtime as much as possible. + +Formulate a model for this problem.227500A textile factory produces two types of fabrics: one for clothing and the other for curtains. The factory operates two shifts, with a weekly production time set at 110 hours. Both types of fabrics are produced at a rate of 1000 meters per hour. Assuming that up to 70,000 meters of curtain fabric can be sold per week, with a profit of 2.5 yuan per meter, and up to 45,000 meters of clothing fabric can be sold per week, with a profit of 1.5 yuan per meter, the factory has the following objectives in formulating its production plan: + +$p_{1}$ : The weekly production time must fully utilize 110 hours; + +$p_{2}$ : Overtime should not exceed 10 hours per week; + +$p_{3}$ : At least 70,000 meters of curtain fabric and 45,000 meters of clothing fabric must be sold per week; + +$p_{4}$ : Minimize overtime as much as possible. + +Formulate a model for this problem.5No differences227500 5Incorrect ground truthincorrect ground truth
19A furniture store can choose to order chairs from three different manufacturers: A, B, and C. The cost of ordering each chair from manufacturer A is $50, from manufacturer B is $45, and from manufacturer C is $40. The store needs to minimize the total cost of the order. + +Additionally, each order from manufacturer A will include 15 chairs, while each order from manufacturers B and C will include 10 chairs. The number of orders must be an integer. The store needs to order at least 100 chairs. + +Each order from manufacturer A will include 15 chairs, while each order from manufacturers B and C will include 10 chairs. The store needs to order at most 500 chairs. + +If the store decides to order chairs from manufacturer A, it must also order at least 10 chairs from manufacturer B. + +Furthermore, if the store decides to order chairs from manufacturer B, it must also order chairs from manufacturer C.4000A furniture store can choose to order chairs from three different manufacturers: A, B, and C. The cost of ordering each chair from manufacturer A is $50, from manufacturer B is $45, and from manufacturer C is $40. The store needs to minimize the total cost of the order. + +Additionally, each order from manufacturer A will include 15 chairs, while each order from manufacturers B and C will include 10 chairs. The number of orders must be an integer. The store needs to order at least 100 chairs. + +Each order from manufacturer A will include 15 chairs, while each order from manufacturers B and C will include 10 chairs. The store needs to order at most 500 chairs. + +If the store decides to order chairs from manufacturer A, it must also order at least 10 chairs from manufacturer B. + +Furthermore, if the store decides to order chairs from manufacturer B, it must also order chairs from manufacturer C.4000No differencesNo differencesok
20Bright Future Toys wants to build and sell robots, model cars, building blocks, and dolls. The profit for each robot sold is $15, for each model car sold is $8, for each set of building blocks sold is $12, and for each doll sold is $5. How many types of toys should Bright Future Toys manufacture to maximize profit? +There are 1200 units of plastic available. Each robot requires 30 units of plastic, each model car requires 10 units of plastic, each set of building blocks requires 20 units of plastic, and each doll requires 15 units of plastic. + +There are 800 units of electronic components available. Each robot requires 8 units of electronic components, each model car requires 5 units of electronic components, each set of building blocks requires 3 units of electronic components, and each doll requires 2 units of electronic components. + +If Bright Future Toys manufactures robots, they will not manufacture dolls. + +However, if they manufacture model cars, they will also manufacture building blocks. + +The number of dolls manufactured cannot exceed the number of model cars manufactured.956Bright Future Toys wants to build and sell robots, model cars, building blocks, and dolls. The profit for each robot sold is $15, for each model car sold is $8, for each set of building blocks sold is $12, and for each doll sold is $5. How many types of toys should Bright Future Toys manufacture to maximize profit? +There are 1200 units of plastic available. Each robot requires 30 units of plastic, each model car requires 10 units of plastic, each set of building blocks requires 20 units of plastic, and each doll requires 15 units of plastic. + +There are 800 units of electronic components available. Each robot requires 8 units of electronic components, each model car requires 5 units of electronic components, each set of building blocks requires 3 units of electronic components, and each doll requires 2 units of electronic components. + +If Bright Future Toys manufactures robots, they will not manufacture dolls. + +However, if they manufacture model cars, they will also manufacture building blocks. + +The number of dolls manufactured cannot exceed the number of model cars manufactured.956No differencesNo differencesok
21A restaurant needs to order dining tables from three different suppliers, A, B, and C. The cost of ordering each dining table from Supplier A is $120, from Supplier B is $110, and from Supplier C is $100. The restaurant needs to minimize the total cost of the order. + +Additionally, each order from Supplier A will include 20 tables, while each order from Suppliers B and C will include 15 tables. The number of orders must be an integer. The restaurant needs to order at least 150 tables. + +Each order from Supplier A will include 20 tables, and each order from Suppliers B and C will include 15 tables. The restaurant needs to order no more than 600 tables. + +If the restaurant decides to order tables from Supplier A, it must also order at least 30 tables from Supplier B. + +Additionally, if the restaurant decides to order tables from Supplier B, it must also order tables from Supplier C.1000A restaurant needs to order dining tables from three different suppliers, A, B, and C. The cost of ordering each dining table from Supplier A is $120, from Supplier B is $110, and from Supplier C is $100. The restaurant needs to minimize the total cost of the order. + +Additionally, each order from Supplier A will include 20 tables, while each order from Suppliers B and C will include 15 tables. The number of orders must be an integer. The restaurant needs to order at least 150 tables. + +Each order from Supplier A will include 20 tables, and each order from Suppliers B and C will include 15 tables. The restaurant needs to order no more than 600 tables. + +If the restaurant decides to order tables from Supplier A, it must also order at least 30 tables from Supplier B. + +Additionally, if the restaurant decides to order tables from Supplier B, it must also order tables from Supplier C.15000No differences1000 15000Incorrect ground truthincorrect ground truth
22A company plans to produce 3 types of products $A_{1}, A_{2}, A_{3}$. It can produce for 22 days in a month. The following table gives the maximum demand (unit $=100 \mathrm{~kg}$), price ($\$ / 100 \mathrm{Kg}$), production cost (per 100Kg product), and production quota (the maximum number of 100kg units that can be produced in one day if all production lines are devoted to this product). + +| Product | $A_{1}$ | $A_{2}$ | $A_{3}$ | +| :---: | :---: | :---: | :---: | +| Maximum Demand | 5300 | 4500 | 5400 | +| Selling Price | $124$ | $109$ | $115$ | +| Production Cost | $73.30$ | $52.90$ | $65.40$ | +| Production Quota | 500 | 450 | 550 | + +The fixed activation cost of the production line is as follows: + +| Product | $A_{1}$ | $A_{2}$ | $A_{3}$ | +| :---: | :---: | :---: | :---: | +| Activation Cost | $170000$ | $150000$ | $100000$ | + +Minimum production batch: + +$$ +\begin{array}{c|ccc} +Product & A_{1} & A_{2} & A_{3} \\ +\hline +Minimum Batch & 20 & 20 & 16 +\end{array} +$$ + +Please formulate an operations research model to determine a production plan that maximizes total revenue while accommodating fixed activation costs and minimum production batch constraints.7890A company plans to produce 3 types of products $A_{1}, A_{2}, A_{3}$. It can produce for 22 days in a month. The following table gives the maximum demand (unit $=100 \mathrm{~kg}$), price ($\$ / 100 \mathrm{Kg}$), production cost (per 100Kg product), and production quota (the maximum number of 100kg units that can be produced in one day if all production lines are devoted to this product). + +| Product | $A_{1}$ | $A_{2}$ | $A_{3}$ | +| :---: | :---: | :---: | :---: | +| Maximum Demand | 5300 | 4500 | 5400 | +| Selling Price | $124$ | $109$ | $115$ | +| Production Cost | $73.30$ | $52.90$ | $65.40$ | +| Production Quota | 500 | 450 | 550 | + +The fixed activation cost of the production line is as follows: + +| Product | $A_{1}$ | $A_{2}$ | $A_{3}$ | +| :---: | :---: | :---: | :---: | +| Activation Cost | $170000$ | $150000$ | $100000$ | + +Minimum production batch: + +$$ +\begin{array}{c|ccc} +Product & A_{1} & A_{2} & A_{3} \\ +\hline +Minimum Batch & 20 & 20 & 16 +\end{array} +$$ + +Please formulate an operations research model to determine a production plan that maximizes total revenue while accommodating fixed activation costs and minimum production batch constraints.270290No differences7890 270290Incorrect ground truthincorrect ground truth
23Hongdou Clothing Factory uses three special equipment to produce shirts, short-sleeved shirts, and casual clothes respectively. It is known that the labor, material usage, selling price, and variable cost of each of the above products are as shown in Table 5-10. + +Table 5-10 + +| Product Name | Labor per unit | Material per unit | Selling Price | Variable Cost | +|--------------|----------------|------------------|---------------|---------------| +| Shirt | 3 | 4 | 120 | 60 | +| Short-sleeve | 2 | 3 | 80 | 40 | +| Casual Cloth | 6 | 6 | 180 | 80 | + +It is known that the available labor per week is 1500 units, the available material is 1600 units, and the weekly fixed costs for the three special equipment for producing shirts, short-sleeved shirts, and casual clothes are 2000, 1500, and 1000 respectively. Design a weekly production plan for the factory to maximize its profit.21500Hongdou Clothing Factory uses three special equipment to produce shirts, short-sleeved shirts, and casual clothes respectively. It is known that the labor, material usage, selling price, and variable cost of each of the above products are as shown in Table 5-10. + +Table 5-10 + +| Product Name | Labor per unit | Material per unit | Selling Price | Variable Cost | +|--------------|----------------|------------------|---------------|---------------| +| Shirt | 3 | 4 | 120 | 60 | +| Short-sleeve | 2 | 3 | 80 | 40 | +| Casual Cloth | 6 | 6 | 180 | 80 | + +It is known that the available labor per week is 1500 units, the available material is 1600 units, and the weekly fixed costs for the three special equipment for producing shirts, short-sleeved shirts, and casual clothes are 2000, 1500, and 1000 respectively. Design a weekly production plan for the factory to maximize its profit.24000No differences21500 24000Incorrect ground truthincorrect ground truth
24A manufacturing company needs to transport 1800 units of product from the warehouse to three different sales points. The company has four transportation options to choose from: truck, van, motorcycle, and electric vehicle. Since the van and electric vehicle both consume a lot of energy, the company wants to choose only one of these two options. Each trip with a truck generates 100 units of pollution, a van generates 50 units of pollution, a motorcycle generates 10 units of pollution, and an electric vehicle generates 0 units of pollution. The total pollution generated from all trips cannot exceed 2000 units. At least 10 trips must use a truck. Trucks, vans, motorcycles, and electric vehicles can transport 100 units, 80 units, 40 units, and 60 units of product per trip, respectively. The company needs to ensure that the total amount of transported product is at least 1800 units.1000A manufacturing company needs to transport 1800 units of product from the warehouse to three different sales points. The company has four transportation options to choose from: truck, van, motorcycle, and electric vehicle. Since the van and electric vehicle both consume a lot of energy, the company wants to choose only one of these two options. Each trip with a truck generates 100 units of pollution, a van generates 50 units of pollution, a motorcycle generates 10 units of pollution, and an electric vehicle generates 0 units of pollution. The total pollution generated from all trips cannot exceed 2000 units. At least 10 trips must use a truck. Trucks, vans, motorcycles, and electric vehicles can transport 100 units, 80 units, 40 units, and 60 units of product per trip, respectively. The company needs to ensure that the total amount of transported product is at least 1800 units. Return the minimized pollution in units while meeting all constraints.1000A manufacturing company needs to transport 1800 units of product from the warehouse to three different sales points. The company has four transportation options to choose from: truck, van, motorcycle, and electric vehicle. Since the van and electric vehicle both consume a lot of energy, the company wants to choose only one of these two options. Each trip with a truck generates 100 units of pollution, a van generates 50 units of pollution, a motorcycle generates 10 units of pollution, and an electric vehicle generates 0 units of pollution. The total pollution generated from all trips cannot exceed 2000 units. At least 10 trips must use a truck. Trucks, vans, motorcycles, and electric vehicles can transport 100 units, 80 units, 40 units, and 60 units of product per trip, respectively. The company needs to ensure that the total amount of transported product is at least 1800 units. Return the minimized pollution in units while meeting all constraints.No differencesThe problem did not state the objective (minimize pollution), we added itmissing objective
25An investor plans to invest 100,000 yuan, with two investment options to choose from. The first investment guarantees a return of 0.7 yuan for every 1 yuan invested after one year. The second investment guarantees a return of 2 yuan for every 1 yuan invested after two years, but the investment time must be in multiples of two years. In order to maximize the investor's earnings by the end of the third year, how should the investments be made? Formulate this as a linear programming problem.510000An investor plans to invest 100,000 yuan, with two investment options to choose from. The first investment guarantees a return of 0.7 yuan for every 1 yuan invested after one year. The second investment guarantees a return of 2 yuan for every 1 yuan invested after two years, but the investment time must be in multiples of two years. In order to maximize the investor's earnings by the end of the third year, how should the investments be made? Formulate this as a linear programming problem.510000No differencesNo differencesok
26The number of salespeople required at a 24-hour convenience store in different time periods is as follows: 2:00-6:00 - 10 people, 6:00-10:00 - 15 people, 10:00-14:00 - 25 people, 14:00-18:00 - 20 people, 18:00-22:00 - 18 people, 22:00-2:00 - 12 people. Salespeople start their shifts at 2:00, 6:00, 10:00, 14:00, 18:00, and 22:00, working continuously for 8 hours. Determine the minimum number of salespeople needed to meet the requirements.53The number of salespeople required at a 24-hour convenience store in different time periods is as follows: 2:00-6:00 - 10 people, 6:00-10:00 - 15 people, 10:00-14:00 - 25 people, 14:00-18:00 - 20 people, 18:00-22:00 - 18 people, 22:00-2:00 - 12 people. Salespeople start their shifts at 2:00, 6:00, 10:00, 14:00, 18:00, and 22:00, working continuously for 8 hours. Determine the minimum number of salespeople needed to meet the requirements.53No differencesNo differencesok
27A factory produces three types of products: I, II, and III. Each product needs to go through two processing procedures, A and B. The factory has two pieces of equipment that can complete process A, denoted as A1 and A2; it has three pieces of equipment that complete process B, denoted as B1, B2, and B3. Product I can be processed on any equipment for A and B; Product II can be processed on any A equipment but only on B1 for process B; Product III can only be processed on A2 and B2. Given the unit processing time on various machines, raw material costs, product sale prices, effective machine hours, and the costs of operating the machines at full capacity as shown in Table 1-4, the task is to arrange the optimal production plan to maximize the factory's profit. + +Table 1-4 +| Equipment | Product I | Product II | Product III | Effective Machine Hours | Operating Costs at Full Capacity (Yuan) | +|------------|-----------|------------|-------------|--------------------------|------------------------------------------| +| A1 | 5 | 10 | | 6000 | 300 | +| A2 | 7 | 9 | 12 | 10000 | 321 | +| B1 | 6 | 8 | | 4000 | 250 | +| B2 | 4 | | 11 | 7000 | 783 | +| B3 | 7 | | | 4000 | 200 | +| Raw Material Cost (Yuan/Unit) | 0.25 | 0.35 | 0.50 | | | +| Unit Price (Yuan/Unit) | 1.25 | 2.00 | 2.80 | | |1146.4142A factory produces three types of products: I, II, and III. Each product needs to go through two processing procedures, A and B. The factory has two pieces of equipment that can complete process A, denoted as A1 and A2; it has three pieces of equipment that complete process B, denoted as B1, B2, and B3. Product I can be processed on any equipment for A and B; Product II can be processed on any A equipment but only on B1 for process B; Product III can only be processed on A2 and B2. Given the unit processing time on various machines, raw material costs, product sale prices, effective machine hours, and the costs of operating the machines at full capacity as shown in Table 1-4, the task is to arrange the optimal production plan to maximize the factory's profit. + +Table 1-4 +| Equipment | Product I | Product II | Product III | Effective Machine Hours | Operating Costs at Full Capacity (Yuan) | +|------------|-----------|------------|-------------|--------------------------|------------------------------------------| +| A1 | 5 | 10 | | 6000 | 300 | +| A2 | 7 | 9 | 12 | 10000 | 321 | +| B1 | 6 | 8 | | 4000 | 250 | +| B2 | 4 | | 11 | 7000 | 783 | +| B3 | 7 | | | 4000 | 200 | +| Raw Material Cost (Yuan/Unit) | 0.25 | 0.35 | 0.50 | | | +| Unit Price (Yuan/Unit) | 1.25 | 2.00 | 2.80 | | |1146.4142No differencesNo differencesok
28Someone has a fund of 300,000 yuan and has the following investment projects in the next three years: +(1) Investment can be made at the beginning of each year within three years, with an annual profit of 20% of the investment amount, and the principal and interest can be used for investment in the following year; +(2) Investment is only allowed at the beginning of the first year, and it can be recovered at the end of the second year, with the total principal and interest amounting to 150% of the investment amount, but the investment limit is no more than 150,000 yuan; +(3) Investment is allowed at the beginning of the second year within three years, and it can be recovered at the end of the third year, with the total principal and interest amounting to 160% of the investment amount, and the investment limit is 200,000 yuan; +(4) Investment is allowed at the beginning of the third year within three years, and it can be recovered in one year with a profit of 40%, and the investment limit is 100,000 yuan. +Chapter One: Linear Programming and Simplex Method +Try to determine an investment plan for this person that maximizes the principal and interest at the end of the third year.580000Someone has a fund of 300,000 yuan and has the following investment projects in the next three years: +(1) Investment can be made at the beginning of each year within three years, with an annual profit of 20% of the investment amount, and the principal and interest can be used for investment in the following year; +(2) Investment is only allowed at the beginning of the first year, and it can be recovered at the end of the second year, with the total principal and interest amounting to 150% of the investment amount, but the investment limit is no more than 150,000 yuan; +(3) Investment is allowed at the beginning of the second year within three years, and it can be recovered at the end of the third year, with the total principal and interest amounting to 160% of the investment amount, and the investment limit is 200,000 yuan; +(4) Investment is allowed at the beginning of the third year within three years, and it can be recovered in one year with a profit of 40%, and the investment limit is 100,000 yuan. +Chapter One: Linear Programming and Simplex Method +Try to determine an investment plan for this person that maximizes the principal and interest at the end of the third year.580000No differencesNo differencesok
29Jieli Company needs to recruit three types of professionals to work in the two regional branches located in Donghai City and Nanjiang City. The demand for different professionals in these regional branches is shown in Table 4-3. After assessing the situation of the applicants, the company has categorized them into 6 types. Table 4-4 lists the specialties each type of person can handle, the specialty they prefer, and the city they prefer to work in. The company's personnel arrangement considers the following three priorities: +$p_1$: All three types of professionals needed are fully met; +$p_2$: 8000 recruited personnel meet their preferred specialty; +$p_3$: 8000 recruited personnel meet their preferred city. +Try to establish a mathematical model for goal planning accordingly. + +Table 4-3 +| Branch Location | Specialty | Demand | +|-----------------|-----------|--------| +| Donghai City | 1 | 1000 | +| Donghai City | 2 | 2000 | +| Nanjiang City | 3 | 1500 | +| Nanjiang City | 1 | 2000 | +| Nanjiang City | 2 | 1000 | +| Nanjiang City | 3 | 1000 | + +Table 4-4 + +| Type | Number of People | Suitable Specialty | Preferred Specialty | Preferred City | +|------|------------------|--------------------|---------------------|----------------| +| 1 | 1500 | 1,2 | 1 | Donghai | +| 2 | 1500 | 2,3 | 2 | Donghai | +| 3 | 1500 | 1,3 | 1 | Nanjiang | +| 4 | 1500 | 1,3 | 3 | Nanjiang | +| 5 | 1500 | 2,3 | 3 | Donghai | +| 6 | 1500 | 3 | 3 | Nanjiang |11500Jieli Company needs to recruit three types of professionals to work in the two regional branches located in Donghai City and Nanjiang City. The demand for different professionals in these regional branches is shown in Table 4-3. After assessing the situation of the applicants, the company has categorized them into 6 types. Table 4-4 lists the specialties each type of person can handle, the specialty they prefer, and the city they prefer to work in. The company's personnel arrangement considers the following three priorities: +$p_1$: All three types of professionals needed are fully met; +$p_2$: 4000 recruited personnel meet their preferred specialty; +$p_3$: 4000 recruited personnel meet their preferred city. +Formulate a plan to minimize the total number of people that need to move from one city to another to meet these priorities. Return the minimized objective value. + +Table 4-3 +| Branch Location | Specialty | Demand | +|-----------------|-----------|--------| +| Donghai City | 1 | 1000 | +| Donghai City | 2 | 2000 | +| Donghai City | 3 | 1500 | +| Nanjiang City | 1 | 2000 | +| Nanjiang City | 2 | 1000 | +| Nanjiang City | 3 | 1000 | + +Table 4-4 + +| Type | Number of People | Suitable Specialty | Preferred Specialty | Preferred City | +|------|------------------|--------------------|---------------------|----------------| +| 1 | 1500 | 1,2 | 1 | Donghai | +| 2 | 1500 | 2,3 | 2 | Donghai | +| 3 | 1500 | 1,3 | 1 | Nanjiang | +| 4 | 1500 | 1,3 | 3 | Nanjiang | +| 5 | 1500 | 2,3 | 3 | Donghai | +| 6 | 1500 | 3 | 3 | Nanjiang |2000Jieli Company needs to recruit three types of professionals to work in the two regional branches located in Donghai City and Nanjiang City. The demand for different professionals in these regional branches is shown in Table 4-3. After assessing the situation of the applicants, the company has categorized them into 6 types. Table 4-4 lists the specialties each type of person can handle, the specialty they prefer, and the city they prefer to work in. The company's personnel arrangement considers the following three priorities: +$p_1$: All three types of professionals needed are fully met; +$p_2$: 8000 4000 recruited personnel meet their preferred specialty; +$p_3$: 8000 4000 recruited personnel meet their preferred city. + Try to establish Formulate a mathematical model for goal planning accordingly. plan to minimize the total number of people that need to move from one city to another to meet these priorities. Return the minimized objective value. + +Table 4-3 +| Branch Location | Specialty | Demand | +|-----------------|-----------|--------| +| Donghai City | 1 | 1000 | +| Donghai City | 2 | 2000 | +| Nanjiang Donghai City | 3 | 1500 | +| Nanjiang City | 1 | 2000 | +| Nanjiang City | 2 | 1000 | +| Nanjiang City | 3 | 1000 |11500 2000The original description missed a clear objective function; we added it and changed problem data to make the problem feasible. We also fixed a wrong city name.underspecified
30Suppose a certain animal needs at least $700 \mathrm{~g}$ of protein, $30 \mathrm{~g}$ of minerals, and $100 \mathrm{mg}$ of vitamins daily. There are 5 types of feed available, and the nutritional content and price per gram of each type of feed are shown in Table 1-5: +Try to formulate a linear programming model that meets the animal's growth needs while minimizing the cost of selecting the feed. +Table 1-6 +| Feed | Protein (g) | Minerals (g) | Vitamins (mg) | Price (¥/kg) | Feed | Protein (g) | Minerals (g) | Vitamins (mg) | Price (¥/kg) | +|------|-------------|--------------|---------------|--------------|------|-------------|--------------|---------------|--------------| +| 1 | 3 | 1 | 0.5 | 0.2 | 4 | 6 | 2 | 2 | 0.3 | +| 2 | 2 | 0.5 | 1 | 0.7 | 5 | 18 | 0.5 | 0.8 | 0.8 | +| 3 | 1 | 0.2 | 0.2 | 0.4 | | | | | |32.43589744Suppose a certain animal needs at least $700 \mathrm{~g}$ of protein, $30 \mathrm{~g}$ of minerals, and $100 \mathrm{mg}$ of vitamins daily. There are 5 types of feed available, and the nutritional content and price per kilogram of each type of feed are shown in Table 1-5: +Try to formulate a linear programming model that meets the animal's growth needs while minimizing the cost of selecting the feed. +Table 1-6 +| Feed | Protein (g) | Minerals (g) | Vitamins (mg) | Price (¥/kg) | Feed | Protein (g) | Minerals (g) | Vitamins (mg) | Price (¥/kg) | +|------|-------------|--------------|---------------|--------------|------|-------------|--------------|---------------|--------------| +| 1 | 3 | 1 | 0.5 | 0.2 | 4 | 6 | 2 | 2 | 0.3 | +| 2 | 2 | 0.5 | 1 | 0.7 | 5 | 18 | 0.5 | 0.8 | 0.8 | +| 3 | 1 | 0.2 | 0.2 | 0.4 | | | | | |32.43589744Suppose a certain animal needs at least $700 \mathrm{~g}$ of protein, $30 \mathrm{~g}$ of minerals, and $100 \mathrm{mg}$ of vitamins daily. There are 5 types of feed available, and the nutritional content and price per gram kilogram of each type of feed are shown in Table 1-5: +Try to formulate a linear programming model that meets the animal's growth needs while minimizing the cost of selecting the feed. +Table 1-6 +| Feed | Protein (g) | Minerals (g) | Vitamins (mg) | Price (¥/kg) | Feed | Protein (g) | Minerals (g) | Vitamins (mg) | Price (¥/kg) |No differencesThe statement was inconsistent: the text said "price per gram" but the table gave price per kilogram. We made it consistent (price per kilogram)inconsistent
31A factory produces three types of products: I, II, and III. Each product must undergo two processing stages, A and B. The factory has two types of equipment to complete stage A (A1, A2) and three types of equipment to complete stage B (B1, B2, B3). + +The production rules are as follows: +- Product I can be processed on any type of A equipment (A1 or A2) and any type of B equipment (B1, B2, or B3). +- Product II can be processed on any type of A equipment (A1 or A2), but for stage B, it can only be processed on B1 equipment. +- Product III can only be processed on A2 equipment for stage A and B2 equipment for stage B. + +The detailed data for processing time per piece, costs, sales price, and machine availability is provided in the table below. The objective is to determine the optimal production plan to maximize the factory's total profit. + +Data Table +| Equipment | Product I | Product II | Product III | Effective Machine Hours | Full - load Equipment Cost (Yuan) | Processing Cost per Machine Hour (Yuan/hour) | +| :--- | :--- | :--- | :--- | :--- | :--- | :--- | +| A1 | 5 | 10 | - | 6000 | 300 | 0.05 | +| A2 | 7 | 9 | 12 | 10000 | 321 | 0.03 | +| B1 | 6 | 8 | - | 4000 | 250 | 0.06 | +| B2 | 4 | - | 11 | 7000 | 783 | 0.11 | +| B3 | 7 | - | - | 4000 | 200 | 0.05 | +| Raw Material Cost (Yuan/piece) | 0.25 | 0.35 | 0.5 | - | - | - | +| Unit Price (Yuan/piece) | 1.25 | 2 | 2.8 | - | - | - |1146.39A factory produces three types of products: I, II, and III. Each product must undergo two processing stages, A and B. The factory has two types of equipment to complete stage A (A1, A2) and three types of equipment to complete stage B (B1, B2, B3). + +The production rules are as follows: +- Product I can be processed on any type of A equipment (A1 or A2) and any type of B equipment (B1, B2, or B3). +- Product II can be processed on any type of A equipment (A1 or A2), but for stage B, it can only be processed on B1 equipment. +- Product III can only be processed on A2 equipment for stage A and B2 equipment for stage B. + +The detailed data for processing time per piece, costs, sales price, and machine availability is provided in the table below. The objective is to determine the optimal production plan to maximize the factory's total profit. + +Data Table +| Equipment | Product I | Product II | Product III | Effective Machine Hours | Full - load Equipment Cost (Yuan) | Processing Cost per Machine Hour (Yuan/hour) | +| :--- | :--- | :--- | :--- | :--- | :--- | :--- | +| A1 | 5 | 10 | - | 6000 | 300 | 0.05 | +| A2 | 7 | 9 | 12 | 10000 | 321 | 0.03 | +| B1 | 6 | 8 | - | 4000 | 250 | 0.06 | +| B2 | 4 | - | 11 | 7000 | 783 | 0.11 | +| B3 | 7 | - | - | 4000 | 200 | 0.05 | +| Raw Material Cost (Yuan/piece) | 0.25 | 0.35 | 0.5 | - | - | - | +| Unit Price (Yuan/piece) | 1.25 | 2 | 2.8 | - | - | - |1190.38No differences1146.39 1190.38Incorrect ground truthincorrect ground truth
32A product consists of three components produced by four workshops, each with a limited number of production hours. Table 1.4 below provides the production rates of the three components. The objective is to determine the number of hours each workshop should allocate to each component to maximize the number of completed products. Formulate this problem as a linear programming problem. + +Table 1.4 + +| Workshop | Production Capacity (hours) | Production Rate (units/hour) | | | +| :------: | :-------------------------: | :--------------------------: | - | - | +| | | Component 1 | Component 2 | Component 3 | +| A | 100 | 10 | 15 | 5 | +| B | 150 | 15 | 10 | 5 | +| C | 80 | 20 | 5 | 10 | +| D | 200 | 10 | 15 | 20 |2924A product consists of three components produced by four workshops, each with a limited number of production hours. Table 1.4 below provides the production rates of the three components. The objective is to determine the number of hours each workshop should allocate to each component to maximize the number of completed products. Formulate this problem. + +Table 1.4 + +| Workshop | Production Capacity (hours) | Production Rate (units/hour) | | | +| :------: | :-------------------------: | :--------------------------: | - | - | +| | | Component 1 | Component 2 | Component 3 | +| A | 100 | 10 | 15 | 5 | +| B | 150 | 15 | 10 | 5 | +| C | 80 | 20 | 5 | 10 | +| D | 200 | 10 | 15 | 20 |2924A product consists of three components produced by four workshops, each with a limited number of production hours. Table 1.4 below provides the production rates of the three components. The objective is to determine the number of hours each workshop should allocate to each component to maximize the number of completed products. Formulate this problem as a linear programming problem. + +Table 1.4No differencesThe statement explicitly said "linear programming", but the ground truth is integral. Removed "linear programming" from the descriptioninconsistent
33A wealthy noble passed away, leaving the following inheritance: + +- A painting by Caillebotte: $25000 +- A bust of Diocletian: $5000 +- A Yuan dynasty Chinese vase: $20000 +- A 911 Porsche: $40000 +- Three diamonds: each $12000 +- A Louis XV sofa: $3000 +- Two very precious Jack Russell racing dogs: each $3000 (will stipulates they must not be separated) +- A sculpture from 200 AD: $10000 +- A sailing boat: $15000 +- A Harley Davidson motorcycle: $10000 +- A piece of furniture once belonging to Cavour: $13000, + +which must be shared between two sons. How to formulate a mathematical program and solve it using COPTPY to minimize the difference in value between the two parts?1000A wealthy noble passed away, leaving the following inheritance: + +- A painting by Caillebotte: $25000 +- A bust of Diocletian: $5000 +- A Yuan dynasty Chinese vase: $20000 +- A 911 Porsche: $40000 +- Three diamonds: each $12000 +- A Louis XV sofa: $3000 +- Two very precious Jack Russell racing dogs: each $3000 (will stipulates they must not be separated) +- A sculpture from 200 AD: $10000 +- A sailing boat: $15000 +- A Harley Davidson motorcycle: $10000 +- A piece of furniture once belonging to Cavour: $13000, + +which must be shared between two sons. How to formulate a mathematical program and solve it to minimize the difference in value between the two parts?1000- A Harley Davidson motorcycle: $10000 +- A piece of furniture once belonging to Cavour: $13000, + +which must be shared between two sons. How to formulate a mathematical program and solve it using COPTPY to minimize the difference in value between the two parts?No differencesRemoved mention to specific solver/library ("using COPTPY")specific solver
34The current problem faced by the company is how to use the fewest number of containers to pack the currently needed goods for transportation, while considering the weight of the goods, specific packaging requirements, and inventory limitations. Professional modeling and analysis are needed for a batch of goods’ transportation strategy to ensure maximum utilization of the limited container space. + +The company currently has a batch to be transported, with each container able to hold a maximum of 60 tons of goods and each container used must load at least 18 tons of goods. The goods to be loaded include five types: A, B, C, D, and E, with quantities of 120, 90, 300, 90, and 120 respectively. The weights are 0.5 tons for A, 1 ton for B, 0.4 tons for C, 0.6 tons for D, and 0.65 tons for E. Additionally, to meet specific usage requirements, every time A goods are loaded, at least 1 unit of C must also be loaded, but loading C alone does not require simultaneously loading A; and considering the demand limitation for D goods, each container must load at least 12 units of D. + +Establish an operations research model so that the company can use the fewest number of containers to pack this batch of goods.7The current problem faced by the company is how to use the fewest number of containers to pack the currently needed goods for transportation, while considering the weight of the goods, specific packaging requirements, and inventory limitations. Professional modeling and analysis are needed for a batch of goods’ transportation strategy to ensure maximum utilization of the limited container space. + +The company currently has a batch to be transported, with each container able to hold a maximum of 60 tons of goods and each container used must load at least 18 tons of goods. The goods to be loaded include five types: A, B, C, D, and E, with quantities of 120, 90, 300, 90, and 120 respectively. The weights are 0.5 tons for A, 1 ton for B, 0.4 tons for C, 0.6 tons for D, and 0.65 tons for E. Additionally, to meet specific usage requirements, every time A goods are loaded, at least 1 unit of C must also be loaded, but loading C alone does not require simultaneously loading A; and considering the demand limitation for D goods, each container must load at least 12 units of D. + +Establish an operations research model so that the company can use the fewest number of containers to pack this batch of goods.7No differencesNo differencesok
35A fabric dyeing plant has 3 dyeing vats. Each batch of fabric must be dyed in sequence in each vat: first, the second, and third vats. The plant must color five batches of fabric of different sizes. The time required in hours to dye batch $i$ in vat $j$ is given in the following matrix: + +$$ +\left(\begin{array}{ccc} +3 & 1 & 1 \\ +2 & 1.5 & 1 \\ +3 & 1.2 & 1.3 \\ +2 & 2 & 2 \\ +2.1 & 2 & 3 +\end{array}\right) +$$ + +Schedule the dyeing operations in the vats to minimize the completion time of the last batch.14.1A fabric dyeing plant has 3 dyeing vats. Each batch of fabric must be dyed in sequence in each vat: first, the second, and third vats. The plant must color five batches of fabric of different sizes. The time required in hours to dye batch $i$ in vat $j$ is given in the following matrix: + +$$ +\left(\begin{array}{ccc} +3 & 1 & 1 \\ +2 & 1.5 & 1 \\ +3 & 1.2 & 1.3 \\ +2 & 2 & 2 \\ +2.1 & 2 & 3 +\end{array}\right) +$$ + +Schedule the dyeing operations in the vats to minimize the completion time of the last batch.14.1No differencesNo differencesok
36The Vehicle Routing Problem (VRP) was first proposed by Dantzig and Ramser in 1959. It is a classic combinatorial optimization problem. The basic VRP can be described as follows: in a certain area, there is a number of customers and a distribution center or depot. Customers are generally located at different positions, and each has a specific demand for goods. The distribution center needs to dispatch a fleet of vehicles and design appropriate delivery routes to fulfill the demands of all customers. The objective of VRP is to optimize a certain benefit metric while satisfying all customer demands. The benefit metric is usually presented as an objective function, which varies according to the company's requirements. Common objective functions include minimizing the total distance traveled by vehicles, minimizing the total delivery time, or minimizing the number of vehicles used. In addition to satisfying customer demands, VRP often needs to consider various other constraints, leading to several variants. For example, if the vehicle's load cannot exceed its maximum capacity, the problem becomes the Capacitated Vehicle Routing Problem (CVRP). If each customer's delivery must be made within a specific time frame, the problem becomes the Vehicle Routing Problem with Time Windows (VRPTW). + +The Vehicle Routing Problem with Time Windows (VRPTW) is a classic variant of the VRP. There are many real-world applications of VRPTW, as customer locations often have service time windows. For instance, some logistics centers need to stock parcels during off-peak hours, and large supermarkets need to replenish goods outside of business hours. Real-time delivery services like food delivery also require strict delivery time windows. Time windows can be categorized as hard or soft. A Hard Time Window (HTW) means that a vehicle must arrive at the delivery point within or before the time window; late arrivals are not permitted. If a vehicle arrives early, it must wait until the time window opens to begin service. This is common in scenarios like supermarket restocking and logistics center inbound operations. A Soft Time Window (STW) means that a vehicle is not strictly required to arrive within the time window, but it is encouraged to do so. A penalty is incurred for early or late arrivals. This is applicable in scenarios such as meal delivery, school bus services, and industrial deliveries. + +The Vehicle Routing Problem with Hard Time Windows (VRPHTW) can be described as follows: within a region, there is a set of customer locations and a central depot. Vehicles must start from the depot and return to the depot, following continuous paths. Each customer must be served by exactly one vehicle, and vehicles have a limited capacity. Each customer has a specific service time window, and service is only accepted within this window. A vehicle can arrive at a customer location early and wait for the time window to open, or it can arrive within the time window to provide service. Service can only begin within the time window, and the service duration is known. The distribution center must arrange an optimal delivery plan to both complete the delivery tasks and minimize travel costs. Because VRPHTW does not allow for delays, it, like the VRP, primarily emphasizes the minimization of travel costs along the routes. + + Now we consider a major enterprise logistics provider, 'Global Logistics', is responsible for providing precise material delivery services for multiple high-end office buildings and shops in a city's central business district (CBD). Due to traffic control in the CBD and the specific receiving requirements of the customers, the delivery task is highly challenging. + +**Specific Requirements:** + +1. **Delivery Task**: There are 20 customers requiring delivery service on the day, and the demands of all customers must be met. +2. **Vehicle Constraints**: The company can use at most 5 trucks, and the capacity of each truck is 200 units. +3. **Capacity Constraint**: The total demand of all customers on a single route must not exceed the truck's maximum capacity (200 units). +4. **Time Window Constraint**: Each customer has a strict 'hard time window.' Service must begin within this specified time window. Early arrivals must wait, and late arrivals are not permitted. +5. **Service Time**: Due to the complex handover procedures at customer sites, a fixed service time of 90 minutes is required for unloading, handover, and paperwork at each customer location. +6. **Optimization Objective**: While satisfying all constraints, the company's objective is to **minimize the total distance traveled by all vehicles** to reduce operational costs. + +**Data Details:** + +* **Central Depot (Depot 0)**: + * Coordinates: (40, 50) + * Operating Time Window: [0, 1236] (minutes) +* **Customer Locations (Customers 1-20)**: The coordinates, demand, service time window, and service duration for each customer are shown in the table below. + +| Customer ID | Coordinates (X, Y) | Demand (units) | Time Window (minutes) | Service Duration (minutes) | +| :--- | :--- | :--- |:--- | :--- | +| 1 | (45, 68) | 10 | [912, 967] | 90 | +| 2 | (45, 70) | 30 | [825, 870] | 90 | +| 3 | (42, 66) | 10 | [65, 146] | 90 | +| 4 | (42, 68) | 10 | [727, 782] | 90 | +| 5 | (42, 65) | 10 | [15, 67] | 90 | +| 6 | (40, 69) | 20 | [621, 702] | 90 | +| 7 | (40, 66) | 20 | [170, 225] | 90 | +| 8 | (38, 68) | 20 | [255, 324] | 90 | +| 9 | (38, 70) | 10 | [534, 605] | 90 | +| 10 | (35, 66) | 10 | [357, 410] | 90 | +| 11 | (35, 69) | 10 | [448, 505] | 90 | +| 12 | (25, 85) | 20 | [652, 721] | 90 | +| 13 | (22, 75) | 30 | [30, 92] | 90 | +| 14 | (22, 85) | 10 | [567, 620] | 90 | +| 15 | (20, 80) | 40 | [384, 429] | 90 | +| 16 | (20, 85) | 40 | [475, 528] | 90 | +| 17 | (18, 75) | 20 | [99, 148] | 90 | +| 18 | (15, 75) | 20 | [179, 254] | 90 | +| 19 | (15, 80) | 10 | [278, 345] | 90 | +| 20 | (30, 50) | 10 | [10, 73] | 90 | + +Now, please provide an operations research model for this VRPHTW.175.37The Vehicle Routing Problem (VRP) was first proposed by Dantzig and Ramser in 1959. It is a classic combinatorial optimization problem. The basic VRP can be described as follows: in a certain area, there is a number of customers and a distribution center or depot. Customers are generally located at different positions, and each has a specific demand for goods. The distribution center needs to dispatch a fleet of vehicles and design appropriate delivery routes to fulfill the demands of all customers. The objective of VRP is to optimize a certain benefit metric while satisfying all customer demands. The benefit metric is usually presented as an objective function, which varies according to the company's requirements. Common objective functions include minimizing the total distance traveled by vehicles, minimizing the total delivery time, or minimizing the number of vehicles used. In addition to satisfying customer demands, VRP often needs to consider various other constraints, leading to several variants. For example, if the vehicle's load cannot exceed its maximum capacity, the problem becomes the Capacitated Vehicle Routing Problem (CVRP). If each customer's delivery must be made within a specific time frame, the problem becomes the Vehicle Routing Problem with Time Windows (VRPTW). + +The Vehicle Routing Problem with Time Windows (VRPTW) is a classic variant of the VRP. There are many real-world applications of VRPTW, as customer locations often have service time windows. For instance, some logistics centers need to stock parcels during off-peak hours, and large supermarkets need to replenish goods outside of business hours. Real-time delivery services like food delivery also require strict delivery time windows. Time windows can be categorized as hard or soft. A Hard Time Window (HTW) means that a vehicle must arrive at the delivery point within or before the time window; late arrivals are not permitted. If a vehicle arrives early, it must wait until the time window opens to begin service. This is common in scenarios like supermarket restocking and logistics center inbound operations. A Soft Time Window (STW) means that a vehicle is not strictly required to arrive within the time window, but it is encouraged to do so. A penalty is incurred for early or late arrivals. This is applicable in scenarios such as meal delivery, school bus services, and industrial deliveries. + +The Vehicle Routing Problem with Hard Time Windows (VRPHTW) can be described as follows: within a region, there is a set of customer locations and a central depot. Vehicles must start from the depot and return to the depot, following continuous paths. Each customer must be served by exactly one vehicle, and vehicles have a limited capacity. Each customer has a specific service time window, and service is only accepted within this window. A vehicle can arrive at a customer location early and wait for the time window to open, or it can arrive within the time window to provide service. Service can only begin within the time window, and the service duration is known. The distribution center must arrange an optimal delivery plan to both complete the delivery tasks and minimize travel costs. Because VRPHTW does not allow for delays, it, like the VRP, primarily emphasizes the minimization of travel costs along the routes. + + Now we consider a major enterprise logistics provider, 'Global Logistics', is responsible for providing precise material delivery services for multiple high-end office buildings and shops in a city's central business district (CBD). Due to traffic control in the CBD and the specific receiving requirements of the customers, the delivery task is highly challenging. + +**Specific Requirements:** + +1. **Delivery Task**: There are 20 customers requiring delivery service on the day, and the demands of all customers must be met. +2. **Vehicle Constraints**: The company can use at most 5 trucks, and the capacity of each truck is 200 units. +3. **Capacity Constraint**: The total demand of all customers on a single route must not exceed the truck's maximum capacity (200 units). +4. **Time Window Constraint**: Each customer has a strict 'hard time window.' Service must begin within this specified time window. Early arrivals must wait, and late arrivals are not permitted. +5. **Service Time**: Due to the complex handover procedures at customer sites, a fixed service time of 90 minutes is required for unloading, handover, and paperwork at each customer location. +6. **Optimization Objective**: While satisfying all constraints, the company's objective is to **minimize the total distance traveled by all vehicles** to reduce operational costs. + +**Data Details:** + +* **Central Depot (Depot 0)**: + * Coordinates: (40, 50) + * Operating Time Window: [0, 1236] (minutes) +* **Customer Locations (Customers 1-20)**: The coordinates, demand, service time window, and service duration for each customer are shown in the table below. + +| Customer ID | Coordinates (X, Y) | Demand (units) | Time Window (minutes) | Service Duration (minutes) | +| :--- | :--- | :--- |:--- | :--- | +| 1 | (45, 68) | 10 | [912, 967] | 90 | +| 2 | (45, 70) | 30 | [825, 870] | 90 | +| 3 | (42, 66) | 10 | [65, 146] | 90 | +| 4 | (42, 68) | 10 | [727, 782] | 90 | +| 5 | (42, 65) | 10 | [15, 67] | 90 | +| 6 | (40, 69) | 20 | [621, 702] | 90 | +| 7 | (40, 66) | 20 | [170, 225] | 90 | +| 8 | (38, 68) | 20 | [255, 324] | 90 | +| 9 | (38, 70) | 10 | [534, 605] | 90 | +| 10 | (35, 66) | 10 | [357, 410] | 90 | +| 11 | (35, 69) | 10 | [448, 505] | 90 | +| 12 | (25, 85) | 20 | [652, 721] | 90 | +| 13 | (22, 75) | 30 | [30, 92] | 90 | +| 14 | (22, 85) | 10 | [567, 620] | 90 | +| 15 | (20, 80) | 40 | [384, 429] | 90 | +| 16 | (20, 85) | 40 | [475, 528] | 90 | +| 17 | (18, 75) | 20 | [99, 148] | 90 | +| 18 | (15, 75) | 20 | [179, 254] | 90 | +| 19 | (15, 80) | 10 | [278, 345] | 90 | +| 20 | (30, 50) | 10 | [10, 73] | 90 | + +Now, please provide an operations research model for this VRPHTW.175.37No differencesNo differencesok
37A factory produces two types of microcomputers, A and B. Each type of microcomputer requires the same two production processes. The processing time, profit from sales, and the maximum weekly processing capacity for each type are shown in Table 3.1. + +Table 3.1 + +| Process | Model | | Maximum Weekly Processing Capacity | +| :---: | :---: | :---: | :---: | +| | $\\mathrm{A}$ | $\\mathrm{B}$ | | +| I (hours / unit) | 4 | 6 | 150 | +| II (hours / unit) | 3 | 2 | 70 | +| Profit ($ per unit) | 300 | 450 | | + +The expected values for the factory's operational goals are as follows: + +$p_{1}$: The total weekly profit must not be less than $10,000. + +$p_{2}$: Due to contractual requirements, at least 10 units of Model A and at least 15 units of Model B must be produced per week. + +$p_{3}$: The weekly production time for Process I should be exactly 150 hours, and the production time for Process II should be fully utilized, with potential overtime if necessary. + +Try to establish the mathematical programming model for this problem.11250A factory produces two types of microcomputers, A and B. Each type of microcomputer requires the same two production processes. The processing time, profit from sales, and the maximum weekly processing capacity for each type are shown in Table 3.1. + +Table 3.1 + +| Process | Model | | Maximum Weekly Processing Capacity | +| :---: | :---: | :---: | :---: | +| | $\\mathrm{A}$ | $\\mathrm{B}$ | | +| I (hours / unit) | 4 | 6 | 150 | +| II (hours / unit) | 3 | 2 | 70 | +| Profit ($ per unit) | 300 | 450 | | + +The expected values for the factory's operational goals are as follows: + +$p_{1}$: The total weekly profit must not be less than $10,000. + +$p_{2}$: Due to contractual requirements, at least 10 units of Model A and at least 15 units of Model B must be produced per week. + +$p_{3}$: The weekly production time for Process I should be exactly 150 hours, and the production time for Process II should be fully utilized, with potential overtime if necessary. + +Try to establish the mathematical programming model for this problem in oder to maximize total profit.11250$p_{3}$: The weekly production time for Process I should be exactly 150 hours, and the production time for Process II should be fully utilized, with potential overtime if necessary. + +Try to establish the mathematical programming model for this problem. problem in oder to maximize total profit.No differencesClarified the objective for the problemambiguous
38There are three different products to be processed on three machine tools. Each product must first be processed on machine 1, then sequentially on machines 2 and 3. The order of processing the three products on each machine should remain the same. Assuming $t_{ij}$ represents the time to process the $i$-th product on the $j$-th machine, how should the schedule be arranged to minimize the total processing cycle for the three products? The timetable is as follows: +| Product | Machine 1 | Machine 2 | Machine 3 | +|---------|-----------|-----------|-----------| +| Product 1 | 2 | 3 | 1 | +| Product 2 | 4 | 2 | 3 | +| Product 3 | 3 | 5 | 2 |14There are three different products to be processed on three machine tools. Each product must first be processed on machine 1, then sequentially on machines 2 and 3. The order of processing the three products on each machine should remain the same. Assuming $t_{ij}$ represents the time to process the $i$-th product on the $j$-th machine, how should the schedule be arranged to minimize the total processing cycle for the three products? The timetable is as follows: +| Product | Machine 1 | Machine 2 | Machine 3 | +|---------|-----------|-----------|-----------| +| Product 1 | 2 | 3 | 1 | +| Product 2 | 4 | 2 | 3 | +| Product 3 | 3 | 5 | 2 |14No differencesNo differencesok
39A company plans to transport goods between the city and the suburb and needs to choose the most environmentally friendly transportation method. The company can choose from the following three methods: motorcycle, small truck, and large truck. Each motorcycle trip produces 40 units of pollution, each small truck trip produces 70 units of pollution, and each large truck trip produces 100 units of pollution. The company's goal is to minimize total pollution. + +The company can only choose two out of these three transportation methods. + +Due to certain road restrictions, the number of motorcycle trips cannot exceed 8. + +Each motorcycle trip can transport 10 units of products, each small truck trip can transport 20 units of products, and each large truck trip can transport 50 units of products. The company needs to transport at least 300 units of products. + +The total number of trips must be less than or equal to 20.600A company plans to transport goods between the city and the suburb and needs to choose the most environmentally friendly transportation method. The company can choose from the following three methods: motorcycle, small truck, and large truck. Each motorcycle trip produces 40 units of pollution, each small truck trip produces 70 units of pollution, and each large truck trip produces 100 units of pollution. The company's goal is to minimize total pollution. + +The company can only choose two out of these three transportation methods. + +Due to certain road restrictions, the number of motorcycle trips cannot exceed 8. + +Each motorcycle trip can transport 10 units of products, each small truck trip can transport 20 units of products, and each large truck trip can transport 50 units of products. The company needs to transport at least 300 units of products. + +The total number of trips must be less than or equal to 20.600No differencesNo differencesok
40The independent country of Carelland mainly exports four commodities: steel, engines, electronic components, and plastic. Carelland's Minister of Finance (i.e., Minister of Economy) wants to maximize exports and minimize imports. The unit prices of steel, engines, electronics, and plastic on the world market are, in local currency (Klunz), 500, 1500, 300, 1200 respectively. Producing 1 unit of steel requires 0.02 units of engines, 0.01 units of plastic, 250 Klunz of other imported goods, and 6 person-months of labor. Producing 1 unit of engines requires 0.8 units of steel, 0.15 units of electronic components, 0.11 units of plastic, 300 Klunz of imported goods, and 1 person-year. One unit of electronics requires: 0.01 units of steel, 0.01 units of engines, 0.05 units of plastic, 50 Klunz of imported goods, and 6 person-months of labor. One unit of plastic requires: 0.03 units of engines, 0.2 units of steel, 0.05 units of electronic components, 300 Klunz of imported goods, and 2 person-years. Engine production is limited to 650000 units, and plastic production is limited to 60000 units. The total available labor force per year is 830000 person-months. Write a mathematical program to maximize domestic GDP and solve the problem using AMPL.33288066.63The independent country of Carelland mainly exports four commodities: steel, engines, electronic components, and plastic. Carelland's Minister of Finance (i.e., Minister of Economy) wants to maximize exports and minimize imports. The unit prices of steel, engines, electronics, and plastic on the world market are, in local currency (Klunz), 500, 1500, 300, 1200 respectively. Producing 1 unit of steel requires 0.02 units of engines, 0.01 units of plastic, 250 Klunz of other imported goods, and 6 person-months of labor. Producing 1 unit of engines requires 0.8 units of steel, 0.15 units of electronic components, 0.11 units of plastic, 300 Klunz of imported goods, and 1 person-year. One unit of electronics requires: 0.01 units of steel, 0.01 units of engines, 0.05 units of plastic, 50 Klunz of imported goods, and 6 person-months of labor. One unit of plastic requires: 0.03 units of engines, 0.2 units of steel, 0.05 units of electronic components, 300 Klunz of imported goods, and 2 person-years. Engine production is limited to 650000 units, and plastic production is limited to 60000 units. The total available labor force per year is 830000 person-months. Write a mathematical program to maximize domestic GDP and solve the problem.36288567The independent country of Carelland mainly exports four commodities: steel, engines, electronic components, and plastic. Carelland's Minister of Finance (i.e., Minister of Economy) wants to maximize exports and minimize imports. The unit prices of steel, engines, electronics, and plastic on the world market are, in local currency (Klunz), 500, 1500, 300, 1200 respectively. Producing 1 unit of steel requires 0.02 units of engines, 0.01 units of plastic, 250 Klunz of other imported goods, and 6 person-months of labor. Producing 1 unit of engines requires 0.8 units of steel, 0.15 units of electronic components, 0.11 units of plastic, 300 Klunz of imported goods, and 1 person-year. One unit of electronics requires: 0.01 units of steel, 0.01 units of engines, 0.05 units of plastic, 50 Klunz of imported goods, and 6 person-months of labor. One unit of plastic requires: 0.03 units of engines, 0.2 units of steel, 0.05 units of electronic components, 300 Klunz of imported goods, and 2 person-years. Engine production is limited to 650000 units, and plastic production is limited to 60000 units. The total available labor force per year is 830000 person-months. Write a mathematical program to maximize domestic GDP and solve the problem using AMPL. problem.33288066.63 36288567 The ground truth uses the incorrect objective that only maximizes the total produce * price but not the adjusted produce after using some produce for other materials. Also, removed mention to specific solver/library ("using AMPL")incorrect ground truth, specific solver
41A person has a fund of 500,000 yuan and the following investment projects available in the next three years: + +(1) Investment can be made at the beginning of each year within three years, and the annual profit is 20% of the investment amount. + +(2) Investment is only allowed at the beginning of the first year, and can be recovered at the end of the second year, with the total principal and interest being 150% of the investment amount. However, this type of investment is limited to no more than 120,000 yuan. + +(3) Investment at the beginning of the second year, recoverable at the end of the second year, with the total principal and interest being 160% of the investment amount. This type of investment is limited to 150,000 yuan. + +(4) Investment is allowed at the beginning of the third year, recoverable in one year, with a profit of 40%, and the investment limit is 100,000 yuan. + +Determine an investment plan for the person that maximizes the total principal and interest by the end of the third year.96.464A person has a fund of 500,000 yuan and the following investment projects available in the next three years: + +(1) Investment can be made at the beginning of each year within three years, and the annual profit is 20% of the investment amount. + +(2) Investment is only allowed at the beginning of the first year, and can be recovered at the end of the second year, with the total principal and interest being 150% of the investment amount. However, this type of investment is limited to no more than 120,000 yuan. + +(3) Investment at the beginning of the second year, recoverable at the end of the second year, with the total principal and interest being 160% of the investment amount. This type of investment is limited to 150,000 yuan. + +(4) Investment is allowed at the beginning of the third year, recoverable in one year, with a profit of 40%, and the investment limit is 100,000 yuan. + +Determine an investment plan for the person that maximizes the total principal and interest by the end of the third year.964640No differences96.464 964640Incorrect ground truthincorrect ground truth
42Two steel furnaces at a steel plant each use two methods of steelmaking simultaneously. The first method takes $a=2$ hours per furnace and costs $m=50$ in fuel expenses; the second method takes $b=3$ hours per furnace and costs $n=70$ in fuel expenses. Assuming each furnace produces $k=10$ tons of steel regardless of the method used, and that at least $d=30$ tons of steel must be produced within $c=12$ hours, how should these two methods be allocated to minimize fuel expenses? Formulate this problem as a linear programming model.150Two steel furnaces at a steel plant each use two methods of steelmaking simultaneously. The first method takes $a=2$ hours per furnace and costs $m=50$ in fuel expenses; the second method takes $b=3$ hours per furnace and costs $n=70$ in fuel expenses. Assuming each furnace produces $k=10$ tons of steel regardless of the method used, and that at least $d=30$ tons of steel must be produced within $c=12$ hours, how should these two methods be allocated to minimize fuel expenses? Formulate this problem as a linear programming model.150No differencesNo differencesok
43A production base needs to extract raw materials from warehouses A and B every day for production. The required raw materials are: at least 240 pieces of raw material A, at least 80 kg of raw material B, and at least 120 tons of raw material C. It is known that: Each truck from warehouse A can transport back to the production base 4 pieces of raw material A, 2 kg of raw material B, 6 tons of raw material C, with a freight cost of 200 yuan per truck; each truck from warehouse B can transport back to the production base 7 pieces of raw material A, 2 kg of raw material B, 2 tons of raw material C per day, with a freight cost of 160 yuan per truck. Question: In order to meet production needs, how many trucks should be dispatched daily from warehouse A and warehouse B to minimize the total freight cost?6800A production base needs to extract raw materials from warehouses A and B every day for production. The required raw materials are: at least 240 pieces of raw material A, at least 80 kg of raw material B, and at least 120 tons of raw material C. It is known that: Each truck from warehouse A can transport back to the production base 4 pieces of raw material A, 2 kg of raw material B, 6 tons of raw material C, with a freight cost of 200 yuan per truck; each truck from warehouse B can transport back to the production base 7 pieces of raw material A, 2 kg of raw material B, 2 tons of raw material C per day, with a freight cost of 160 yuan per truck. Question: In order to meet production needs, how many trucks should be dispatched daily from warehouse A and warehouse B to minimize the total freight cost?6800No differencesNo differencesok
44Given that there are $m=2$ production points for a certain type of material, where the output at the $i$-th point $(i=1,2)$ is $a_i$, $a_1 = 100$, and $a_2 = 150$. This material is to be shipped to $n=2$ demand points, where the demand at the $j$-th point $(j=1, 2)$ is $b_j$, $b_1 = 80$, and $b_2 = 120$. It is known that $\sum_i a_i \geqslant \sum_j b_j$. It is also known that when shipping from production points to demand points, it must pass through one of the $p=2$ intermediate marshaling stations. If the $k$-th $(k=1, 2)$ intermediate marshaling station is used, a fixed cost $f_k$ is incurred regardless of the transshipment volume, where $f_1 = 10$ and $f_2 = 15$. The $k$-th intermediate marshaling station has a maximum transshipment capacity limitation $q_k$, where $q_1 = 100$ and $q_2 = 100$. Let $c_{i k}$ and $c'_{k j}$ denote the unit transportation cost from $i$ to $k$ and from $k$ to $j$, respectively, where $c_{11}=2$, $c_{12}=3$, $c_{21}=4$, $c_{22}=1$, $c'_{11}=3$, $c'_{12}=2$, $c'_{21}=1$, and $c'_{22}=4$. Try to determine a transportation plan for this material that minimizes the total cost.685Given that there are $m=2$ production points for a certain type of material, where the output at the $i$-th point $(i=1,2)$ is $a_i$, $a_1 = 100$, and $a_2 = 150$. This material is to be shipped to $n=2$ demand points, where the demand at the $j$-th point $(j=1, 2)$ is $b_j$, $b_1 = 80$, and $b_2 = 120$. It is known that $\sum_i a_i \geqslant \sum_j b_j$. It is also known that when shipping from production points to demand points, it must pass through one of the $p=2$ intermediate marshaling stations. If the $k$-th $(k=1, 2)$ intermediate marshaling station is used, a fixed cost $f_k$ is incurred regardless of the transshipment volume, where $f_1 = 10$ and $f_2 = 15$. The $k$-th intermediate marshaling station has a maximum transshipment capacity limitation $q_k$, where $q_1 = 100$ and $q_2 = 100$. Let $c_{i k}$ and $c'_{k j}$ denote the unit transportation cost from $i$ to $k$ and from $k$ to $j$, respectively, where $c_{11}=2$, $c_{12}=3$, $c_{21}=4$, $c_{22}=1$, $c'_{11}=3$, $c'_{12}=2$, $c'_{21}=1$, and $c'_{22}=4$. Try to determine a transportation plan for this material that minimizes the total cost.685No differencesNo differencesok
45A factory produces three types of products, A, B, and C. Each unit of product A requires 1 hour for technical preparation, 10 hours of direct labor, and 3 kg of materials. Each unit of product B requires 2 hours for technical preparation, 4 hours of labor, and 2 kg of materials. Each unit of product C requires 1 hour for technical preparation, 5 hours of labor, and 1 kg of materials. The available technical preparation time is 100 hours, labor time is 700 hours, and materials are 400 kg. The company offers larger discounts for bulk purchases, as detailed in Table 1-22. Determine the company's production plan to maximize profit. +Table 1-22 +| Product A | | Product B | | Product C | | +|:---------------|:---------:|:---------------|:---------:|:---------------|:---------:| +| Sales Volume (pieces) | Profit (yuan) | Sales Volume (pieces) | Profit (yuan) | Sales Volume (pieces) | Profit (yuan) | +| 0 ~ 40 | 10 | 0 ~ 50 | 6 | 0 ~ 100 | 5 | +| 40 ~ 100 | 9 | 50 ~ 100 | 4 | Above 100 | 4 | +| 100 ~ 150 | 8 | Above 100 | 3 | | | +| Above 150 | 7 | | | | |734A factory produces three types of products, A, B, and C. Each unit of product A requires 1 hour for technical preparation, 10 hours of direct labor, and 3 kg of materials. Each unit of product B requires 2 hours for technical preparation, 4 hours of labor, and 2 kg of materials. Each unit of product C requires 1 hour for technical preparation, 5 hours of labor, and 1 kg of materials. The available technical preparation time is 100 hours, labor time is 700 hours, and materials are 400 kg. The company offers larger discounts for bulk purchases, as detailed in Table 1-22. Determine the company's production plan to maximize profit. +Table 1-22 +| Product A | | Product B | | Product C | | +|:---------------|:---------:|:---------------|:---------:|:---------------|:---------:| +| Sales Volume (pieces) | Profit (yuan) | Sales Volume (pieces) | Profit (yuan) | Sales Volume (pieces) | Profit (yuan) | +| 0 ~ 40 | 10 | 0 ~ 50 | 6 | 0 ~ 100 | 5 | +| 40 ~ 100 | 9 | 50 ~ 100 | 4 | Above 100 | 4 | +| 100 ~ 150 | 8 | Above 100 | 3 | | | +| Above 150 | 7 | | | | |712No differences734 712Incorrect ground truthincorrect ground truth
46A university computer lab hires 4 undergraduates (designated 1, 2, 3, and 4) and 2 graduate students (designated 5 and 6) for duty answering questions. The maximum duty hours from Monday to Friday and the hourly wage for each person are shown in Table 5-9. + +Table 5-9 +| Student ID | Wage (CNY/h) | Monday | Tuesday | Wednesday | Thursday | Friday | +|------------|--------------|--------|---------|-----------|----------|--------| +| 1 | 10.0 | 6 | 0 | 6 | 0 | 7 | +| 2 | 10.0 | 0 | 6 | 0 | 6 | 0 | +| 3 | 9.9 | 4 | 8 | 3 | 0 | 5 | +| 4 | 9.8 | 5 | 5 | 6 | 0 | 4 | +| 5 | 10.8 | 3 | 0 | 4 | 8 | 0 | +| 6 | 11.3 | 0 | 6 | 0 | 6 | 3 | + +The lab operates from 8:00 AM to 10:00 PM, and there must be one and only one student on duty during open hours. It is also required that each undergraduate must work at least 8 hours per week, and each graduate student must work at least 7 hours per week. Additionally, supplement the following requirements: each student can work no more than 2 shifts per week, and no more than 3 students can be scheduled for duty each day. Based on these conditions, establish a new mathematical model.472.3A university computer lab hires 4 undergraduates (designated 1, 2, 3, and 4) and 2 graduate students (designated 5 and 6) for duty answering questions. The maximum duty hours from Monday to Friday and the hourly wage for each person are shown in Table 5-9. + +Table 5-9 +Student ID | Wage (CNY/h) | Monday | Tuesday | Wednesday | Thursday | Friday +1 | 10.0 | 6 | 0 | 6 | 0 | 7 +2 | 10.0 | 0 | 6 | 0 | 6 | 7 +3 | 9.9 | 4 | 8 | 4 | 0 | 5 +4 | 9.8 | 5 | 5 | 6 | 0 | 4 +5 | 10.8 | 4 | 0 | 4 | 8 | 0 +6 | 11.3 | 5 | 6 | 0 | 6 | 3 + +The lab operates from 8:00 AM to 10:00 PM, and there must be one and only one student on duty during open hours. It is also required that each undergraduate must work at least 8 hours per week, and each graduate student must work at least 7 hours per week. Additionally, each student can work no more than 2 shifts per week, and no more than 3 students can be scheduled for duty each day. + +Based on these conditions, establish a mathematical model to determine the work schedule that satisfies all requirements.717.9A university computer lab hires 4 undergraduates (designated 1, 2, 3, and 4) and 2 graduate students (designated 5 and 6) for duty answering questions. The maximum duty hours from Monday to Friday and the hourly wage for each person are shown in Table 5-9. + +Table 5-9 + | Student ID | Wage (CNY/h) | Monday | Tuesday | Wednesday | Thursday | Friday | + |------------|--------------|--------|---------|-----------|----------|--------| + | +1 | 10.0 | 6 | 0 | 6 | 0 | 7 | + | +2 | 10.0 | 0 | 6 | 0 | 6 | 0 | + | 7 +3 | 9.9 | 4 | 8 | 3 4 | 0 | 5 | + | +4 | 9.8 | 5 | 5 | 6 | 0 | 4 | + | +5 | 10.8 | 3 4 | 0 | 4 | 8 | 0 | + | +6 | 11.3 | 0 5 | 6 | 0 | 6 | 3 | + +The lab operates from 8:00 AM to 10:00 PM, and there must be one and only one student on duty during open hours. It is also required that each undergraduate must work at least 8 hours per week, and each graduate student must work at least 7 hours per week. Additionally, supplement the following requirements: each student can work no more than 2 shifts per week, and no more than 3 students can be scheduled for duty each day. + +Based on these conditions, establish a new mathematical model. model to determine the work schedule that satisfies all requirements.472.3 717.9The original instance was infeasible and had a ground truth of 472.3; we updated the data to make it feasible and updated the ground truth accordinglyinfeasible
47A certain farm has 100 hectares of land and 15,000 yuan in funds for production development. The labor force situation on the farm is 3,500 person-days in autumn and winter, and 4,000 person-days in spring and summer. If the labor force itself is not fully utilized, they can work externally, earning 2.1 yuan/person-day in spring and summer and 1.8 yuan/person-day in autumn and winter. + +The farm cultivates three types of crops: soybeans, corn, and wheat, and also raises dairy cows and chickens. Crop cultivation requires no specialized investment, but raising animals involves an investment of 400 yuan per dairy cow and 3 yuan per chicken. Raising dairy cows requires allocating 1.5 hectares of land per cow to grow feed, and involves 100 person-days in autumn and winter, and 50 person-days in spring and summer per cow. The annual net income is 400 yuan per dairy cow. Raising chickens does not use land, requires 0.6 person-days in autumn and winter, and 0.3 person-days in spring and summer per chicken. Annual net income is 2 yuan per chicken. The current chicken coop can accommodate up to 3,000 chickens, and the cow barn can accommodate up to 32 dairy cows. The labor and income requirements for the three types of crops per year are shown in Table 1-9. + +Table 1-9 +| Item | Soybean | Corn | Wheat | +|----------------|---------|------|-------| +| Person-days (Autumn/Winter) | 20 | 35 | 10 | +| Person-days (Spring/Summer) | 50 | 75 | 40 | +| Annual Net Income (Yuan/hectare) | 175 | 300 | 120 | + +Determine the farm's operating plan to maximize annual net income. Please note that Labor days are calculated in whole days, fractions are not allowed.20241.61538A certain farm has 100 hectares of land and 15,000 yuan in funds for production development. The labor force situation on the farm is 3,500 person-days in autumn and winter, and 4,000 person-days in spring and summer. If the labor force itself is not fully utilized, they can work externally, earning 2.1 yuan/person-day in spring and summer and 1.8 yuan/person-day in autumn and winter. + +The farm cultivates three types of crops: soybeans, corn, and wheat, and also raises dairy cows and chickens. Crop cultivation requires no specialized investment, but raising animals involves an investment of 400 yuan per dairy cow and 3 yuan per chicken. Raising dairy cows requires allocating 1.5 hectares of land per cow to grow feed, and involves 100 person-days in autumn and winter, and 50 person-days in spring and summer per cow. The annual net income is 400 yuan per dairy cow. Raising chickens does not use land, requires 0.6 person-days in autumn and winter, and 0.3 person-days in spring and summer per chicken. Annual net income is 2 yuan per chicken. The current chicken coop can accommodate up to 3,000 chickens, and the cow barn can accommodate up to 32 dairy cows. The labor and income requirements for the three types of crops per year are shown in Table 1-9. + +Table 1-9 +| Item | Soybean | Corn | Wheat | +|----------------|---------|------|-------| +| Person-days (Autumn/Winter) | 20 | 35 | 10 | +| Person-days (Spring/Summer) | 50 | 75 | 40 | +| Annual Net Income (Yuan/hectare) | 175 | 300 | 120 | + +Determine the farm's operating plan to maximize annual net income. Please note that workers can only work externally for full days, fractions are not allowed. It is not possible to change the crop and animal raising plans from season to season.20241.8| Person-days (Spring/Summer) | 50 | 75 | 40 | +| Annual Net Income (Yuan/hectare) | 175 | 300 | 120 | + +Determine the farm's operating plan to maximize annual net income. Please note that Labor days are calculated in whole workers can only work externally for full days, fractions are not allowed. It is not possible to change the crop and animal raising plans from season to season.20241.61538 20241.8Ambiguous, added clarifications.ambiguous
48A factory produces two models of microcomputers, A and B. Each model requires the same two processes. The processing time, sales profit, and the factory’s maximum weekly processing capacity for each model are shown in Table 3.1. + +Table 3.1 + +| Process | Model | | Maximum Weekly Processing Capacity | +| :---: | :---: | :---: | :---: | +| | $A$ | $B$ | | +| I (hours/unit) | 4 | 6 | 150 | +| II (hours/unit) | 3 | 2 | 70 | +| Profit (yuan/unit) | 300 | 450 | | + +Given the factory's business goals: + +$p_{1}$: The total weekly profit should not be less than 10,000 yuan; + +$p_{2}$: Due to contract requirements, at least 10 units of model A and at least 15 units of model B must be produced each week; + +$p_{3}$: The processing time for Process I should be exactly 150 hours per week, and the processing time for Process II should ideally be fully utilized, with potential for appropriate overtime; + +$p_{4}$: If products are produced during overtime in Process II, the profit per unit is reduced by 20 yuan for model A and 25 yuan for model B, and the maximum overtime for Process II is 30 hours per week. Formulate the mathematical model for this problem.11250A factory produces two models of microcomputers, A and B. Each model requires the same two processes. The processing time, sales profit, and the factory’s maximum weekly processing capacity for each model are shown in Table 3.1. + +Table 3.1 + +| Process | Model | | Maximum Weekly Processing Capacity | +| :---: | :---: | :---: | :---: | +| | $A$ | $B$ | | +| I (hours/unit) | 4 | 6 | 150 | +| II (hours/unit) | 3 | 2 | 70 | +| Profit (yuan/unit) | 300 | 450 | | + +Given the factory's business goals: + +$p_{1}$: The total weekly profit should not be less than 10,000 yuan; + +$p_{2}$: Due to contract requirements, at least 10 units of model A and at least 15 units of model B must be produced each week; + +$p_{3}$: The processing time for Process I should be exactly 150 hours per week, and the processing time for Process II should ideally be fully utilized, with potential for appropriate overtime; + +$p_{4}$: If products are produced during overtime in Process II, the profit per unit is reduced by 20 yuan for model A and 25 yuan for model B, and the maximum overtime for Process II is 30 hours per week. Formulate the mathematical model for this problem.11250No differencesNo differencesok
49A factory needs to rent a warehouse to store materials in the next 4 months. The required warehouse area for each month is listed in Table 1-14. + +Table 1-14 +| Month | 1 | 2 | 3 | 4 | +|-------|------|------|------|------| +| Required warehouse area (?) | 1500 | 1000 | 2000 | 1200 | + +The longer the rental contract period, the greater the discount on warehouse rental fees. The specific data is listed in Table 1-15. + +Table 1-15 +| Contract rental period (months) | 1 | 2 | 3 | 4 | +|---------------------------------|----|----|----|----| +| Rental fees for warehouse area during the contract period (yuan)56000A factory must rent warehouse space to cover storage needs over the next four months. The required storage areas are: +Month 1: 1500 m² +Month 2: 1000 m² +Month 3: 2000 m² +Month 4: 1200 m² + +Warehouse space can be rented via contracts of fixed duration. A contract of length k months (k ? {1, 2, 3, 4}) may start at the beginning of any month t provided it ends no later than Month 4 (i.e., t + k ? 1 ? 4). A contract starting in month t covers months t through t + k ? 1. The rental fee is charged per square meter per month and depends on the contract length as follows: +1-month contract: 22 yuan per m² per month +2-month contract: 21 yuan per m² per month +3-month contract: 20 yuan per m² per month +4-month contract: 19 yuan per m² per month + +Additional rules and assumptions: + +You may sign any number of contracts. + +Rented area is divisible (you may rent any nonnegative real number of m²). + +Supply is unlimited at the listed rates. + +In each month, the total active rented area must be at least the required area for that month. + +You pay for the entire area specified in each contract for every month it is active, even if some capacity is unused. + +Your task is to choose the start times, durations, and areas of contracts to minimize the total rental cost over the four-month horizon while satisfying the monthly area requirements.113000A factory needs to must rent a warehouse space to store materials in cover storage needs over the next 4 four months. The required warehouse area for each month is listed in Table 1-14. storage areas are: + Month 1: 1500 m² + Month 2: 1000 m² + Month 3: 2000 m² + Month 4: 1200 m² + + Table 1-14 + | Warehouse space can be rented via contracts of fixed duration. A contract of length k months (k ? {1, 2, 3, 4}) may start at the beginning of any month t provided it ends no later than Month | 1 | 2 | 3 | 4 | + |-------|------|------|------|------| + | Required warehouse area (?) | 1500 | 1000 | 2000 | 1200 | (i.e., t + k ? 1 ? 4). A contract starting in month t covers months t through t + k ? 1. The rental fee is charged per square meter per month and depends on the contract length as follows: + 1-month contract: 22 yuan per m² per month + 2-month contract: 21 yuan per m² per month + 3-month contract: 20 yuan per m² per month + 4-month contract: 19 yuan per m² per month + + The longer the rental contract period, the greater the discount on warehouse rental fees. The specific data is listed in Table 1-15. Additional rules and assumptions: + + Table 1-15 + | Contract rental period (months) | 1 | 2 | 3 | 4 | + |---------------------------------|----|----|----|----| + | Rental fees for warehouse You may sign any number of contracts. + + Rented area is divisible (you may rent any nonnegative real number of m²). + + Supply is unlimited at the listed rates. + + In each month, the total active rented area during must be at least the required area for that month. + + You pay for the entire area specified in each contract period (yuan) for every month it is active, even if some capacity is unused. + + Your task is to choose the start times, durations, and areas of contracts to minimize the total rental cost over the four-month horizon while satisfying the monthly area requirements.56000 113000Incomplete problem statement and data. We completed itunderspecified
50A store has formulated a purchase and sales plan for a certain product from July to December. It is known that the warehouse capacity must not exceed 500 units, with 200 units in stock at the end of June. Thereafter, purchases are made at the beginning of each month. Assume the purchase and selling prices of this product for each month are shown in Table 1-21. How much should be purchased and sold each month to maximize the total revenue? + +Table 1-21 +| Month | 7 | 8 | 9 | 10 | 11 | 12 | +|-------|----|----|----|----|----|----| +| Buy | 28 | 24 | 25 | 27 | 23 | 23 | +| Sell | 29 | 24 | 26 | 28 | 22 | 25 |9100A store has formulated a purchase and sales plan for a certain product from July to December. It is known that the warehouse capacity must not exceed 500 units, with 200 units in stock at the end of June. Thereafter, purchases are made at the beginning of each month. Assume the purchase and selling prices of this product for each month are shown in Table 1-21. How much should be purchased and sold each month to maximize the total revenue? + +Table 1-21 +| Month | 7 | 8 | 9 | 10 | 11 | 12 | +|-------|----|----|----|----|----|----| +| Buy | 28 | 24 | 25 | 27 | 23 | 23 | +| Sell | 29 | 24 | 26 | 28 | 22 | 25 |9100No differencesNo differencesok
51The number of nurses required in each time period over 24 hours at a certain hospital is as follows: 2:00-6:00 - 10 people, 6:00-10:00 - 15 people, 10:00-14:00 - 25 people, 14:00-18:00 - 20 people, 18:00-22:00 - 18 people, 22:00-2:00 - 12 people. Nurses start shifts in 6 batches at 2:00, 6:00, 10:00, 14:00, 18:00, and 22:00 and work continuously for 8 hours. Please determine: If the hospital can hire contract nurses with the same working hours as regular nurses, and if the pay for regular nurses is 10 yuan/hour and for contract nurses is 15 yuan/hour, should the hospital hire contract nurses and if so, how many?4240The number of nurses required in each time period over 24 hours at a certain hospital is as follows: 2:00-6:00 - 10 people, 6:00-10:00 - 15 people, 10:00-14:00 - 25 people, 14:00-18:00 - 20 people, 18:00-22:00 - 18 people, 22:00-2:00 - 12 people. Nurses start shifts in 6 batches at 2:00, 6:00, 10:00, 14:00, 18:00, and 22:00 and work continuously for 8 hours. Please determine: If the hospital can hire contract nurses with the same working hours as regular nurses, and if the pay for regular nurses is 10 yuan/hour and for contract nurses is 15 yuan/hour, should the hospital hire contract nurses and if so, how many?4240No differencesNo differencesok
52For a certain 24-hour bus service, the number of drivers and crew members required during different time periods each day is shown in Table 1-2: +Table 1-2 +\begin{tabular}{|c|c|c||c|c|c|} +\hline Shift & Time & Required number & Shift & Time & Required number \\ +\hline 1 & $6: 00 \sim 10: 00$ & 60 & 4 & $18 ; 00 \sim 22 ; 00$ & 50 \\ +\hline 2 & $10 ; 00 \sim 14 ; 00$ & 70 & 5 & $22 ; 00 \sim 2 ; 00$ & 20 \\ +\hline 3 & $14 ; 00 \sim 18 ; 00$ & 60 & 6 & $2: 00 \sim 6 ; 00$ & 30 \\ +\hline +\end{tabular} + +Assuming that drivers and crew members start their shifts at the beginning of each time period and work continuously for 8 hours, determine the minimum number of drivers and crew members needed for this bus route. Formulate the linear programming model for this problem.150For a certain 24-hour bus service, the number of drivers and crew members required during different time periods each day is shown in Table 1-2: +Table 1-2 +\begin{tabular}{|c|c|c||c|c|c|} +\hline Shift & Time & Required number & Shift & Time & Required number \\ +\hline 1 & $6: 00 \sim 10: 00$ & 60 & 4 & $18 ; 00 \sim 22 ; 00$ & 50 \\ +\hline 2 & $10 ; 00 \sim 14 ; 00$ & 70 & 5 & $22 ; 00 \sim 2 ; 00$ & 20 \\ +\hline 3 & $14 ; 00 \sim 18 ; 00$ & 60 & 6 & $2: 00 \sim 6 ; 00$ & 30 \\ +\hline +\end{tabular} + +Assuming that drivers and crew members start their shifts at the beginning of each time period and work continuously for 8 hours, determine the minimum number of drivers and crew members needed for this bus route. Formulate the linear programming model for this problem.150No differencesNo differencesok
53The Zhang family has 6 children: Harry, Hermione, Ron, Fred, George, and Ginny. The cost of taking Harry is $1200, Hermione is $1650, Ron is $750, Fred is $800, George is $800, and Ginny is $1500. Which children should the couple take to minimize the total cost of taking the children? They can take up to four children on the upcoming trip. + +Ginny is the youngest, so the Zhang family will definitely take her. + +If the couple takes Harry, they will not take Fred because Harry does not get along with him. + +If the couple takes Harry, they will not take George because Harry does not get along with him. + +If they take George, they must also take Fred. + +If they take George, they must also take Hermione. + +Even though it will cost them a lot of money, the Zhang family has decided to take at least three children.3050The Zhang family has 6 children: Harry, Hermione, Ron, Fred, George, and Ginny. The cost of taking Harry is $1200, Hermione is $1650, Ron is $750, Fred is $800, George is $800, and Ginny is $1500. Which children should the couple take to minimize the total cost of taking the children? They can take up to four children on the upcoming trip. + +Ginny is the youngest, so the Zhang family will definitely take her. + +If the couple takes Harry, they will not take Fred because Harry does not get along with him. + +If the couple takes Harry, they will not take George because Harry does not get along with him. + +If they take George, they must also take Fred. + +If they take George, they must also take Hermione. + +Even though it will cost them a lot of money, the Zhang family has decided to take at least three children.3050No differencesNo differencesok
54Given that a certain factory plans to produce three types of products, I, II, and III, each product needs to be processed on equipment $A, B, C$ as shown in Table 2-3: + +Table 2-3 +| Equipment Code | I | II | III | Effective Monthly Equipment Hours | +|----------------|----|----|-----|----------------------------------| +| A | 8 | 2 | 10 | 300 | +| B | 10 | 5 | 8 | 400 | +| C | 2 | 13 | 10 | 420 | +| Unit Product Profit (per thousand yuan) | 3 | 2 | 2.9 | | + +How can the equipment capacity be fully utilized to maximize production profit?135.2666667Given that a certain factory plans to produce three types of products, I, II, and III, each product needs to be processed on equipment $A, B, C$ as shown in Table 2-3: + +Table 2-3 +| Equipment Code | I | II | III | Effective Monthly Equipment Hours | +|----------------|----|----|-----|----------------------------------| +| A | 8 | 2 | 10 | 300 | +| B | 10 | 5 | 8 | 400 | +| C | 2 | 13 | 10 | 420 | +| Unit Product Profit (per thousand yuan) | 3 | 2 | 2.9 | | + +How can the equipment capacity be fully utilized to maximize production profit? The quantity of each product must be an integer.134.5| C | 2 | 13 | 10 | 420 | +| Unit Product Profit (per thousand yuan) | 3 | 2 | 2.9 | | + +How can the equipment capacity be fully utilized to maximize production profit? The quantity of each product must be an integer.135.2666667 134.5Ambiguity: The original ground truth hints to producing fractional products, but it could also be interpreted as only whole products are allowed. Clarified that the quantity of the products must be integers.integral/fractional
55A master's student in Operations Research at a certain university is required to select two courses in mathematics, two in operations research, and two in computer science from a total of seven courses: Calculus, Operations Research, Data Structures, Management Statistics, Computer Simulation, Computer Programming, and Forecasting. Some courses belong to only one category: Calculus falls under Mathematics, Computer Programming under Computer Science. However, some courses fall under multiple categories: Operations Research can be considered both Operations Research and Mathematics, Data Structures both Computer Science and Mathematics, Management Statistics both Mathematics and Operations Research, Computer Simulation both Computer Science and Operations Research, and Forecasting both Operations Research and Mathematics. Courses that fall under multiple categories can fulfill the requirement of both categories simultaneously. Additionally, some courses have prerequisites: Computer Simulation or Data Structures requires Computer Programming first, Management Statistics requires Calculus first, and Forecasting requires Management Statistics first. The question is: What is the minimum number of courses a master's student must take, and which specific courses, to meet the above requirements?4A master's student in Operations Research at a certain university is required to select two courses in mathematics, two in operations research, and two in computer science from a total of seven courses: Calculus, Operations Research, Data Structures, Management Statistics, Computer Simulation, Computer Programming, and Forecasting. Some courses belong to only one category: Calculus falls under Mathematics, Computer Programming under Computer Science. However, some courses fall under multiple categories: Operations Research can be considered both Operations Research and Mathematics, Data Structures both Computer Science and Mathematics, Management Statistics both Mathematics and Operations Research, Computer Simulation both Computer Science and Operations Research, and Forecasting both Operations Research and Mathematics. Courses that fall under multiple categories can fulfill the requirement of both categories simultaneously. Additionally, some courses have prerequisites: Computer Simulation or Data Structures requires Computer Programming first, Management Statistics requires Calculus first, and Forecasting requires Management Statistics first. The question is: What is the minimum number of courses a master's student must take, and which specific courses, to meet the above requirements?4No differencesNo differencesok
56A trading company specializes in the wholesale business of certain grains. The company currently has a warehouse with a capacity of 5000 dan. On January 1, the company has 1000 dan of grain in stock and 20,000 yuan in funds. The estimated grain prices for the first quarter are shown in Table 1-8. + +Table 1-8 +| Month | Purchase Price (yuan/dan) | Selling Price (yuan/dan) | +|-------|---------------------------|--------------------------| +| 1 | 2.85 | 3.10 | +| 2 | 3.05 | 3.25 | +| 3 | 2.90 | 2.95 | + +The purchased grains will be delivered in the same month but can only be sold in the next month, and payment is required upon delivery. The company hopes to have an inventory of 2000 dan at the end of the quarter. What purchasing and selling strategy should be adopted to maximize the total profit over the three months?1360.45A trading company specializes in the wholesale business of certain grains. The company currently has a warehouse with a capacity of 5000 dan. On January 1, the company has 1000 dan of grain in stock and 20,000 yuan in funds. The estimated grain prices for the first quarter are shown in Table 1-8. + +Table 1-8 +| Month | Purchase Price (yuan/dan) | Selling Price (yuan/dan) | +|-------|---------------------------|--------------------------| +| 1 | 2.85 | 3.10 | +| 2 | 3.05 | 3.25 | +| 3 | 2.90 | 2.95 | + +The purchased grains will be delivered in the same month but can only be sold in the next month, and payment is required upon delivery. The company hopes to have an inventory of 2000 dan at the end of the quarter. What purchasing and selling strategy should be adopted to maximize the total profit over the three months?-700No differences1360.45 -700Incorrect ground truthincorrect ground truth
57Assuming a paper mill receives three orders for rolls of paper, with length and width requirements as shown in Table 1.2. + +Table 1.2 + +| Order Number | Width (meters) | Length (meters) | +| :---: | :---: | :---: | +| 1 | 0.5 | 1000 | +| 2 | 0.7 | 3000 | +| 3 | 0.9 | 2000 | + +The mill produces rolls of paper with standard widths of 1 meter and 2 meters. Assuming the length of the rolls is unlimited and can be spliced to reach the required length, how should the rolls be cut to minimize the area of waste?600Assuming a paper mill receives three orders for rolls of paper, with length and width requirements as shown in Table 1.2. + +Table 1.2 + +| Order Number | Width (meters) | Length (meters) | +| :---: | :---: | :---: | +| 1 | 0.5 | 1000 | +| 2 | 0.7 | 3000 | +| 3 | 0.9 | 2000 | + +The mill produces rolls of paper with standard widths of 1 meter and 2 meters. Assuming the length of the rolls is unlimited and can be spliced to reach the required length, how should the rolls be cut to minimize the area of waste?600No differencesNo differencesok
58Vicky and David have just bought a farm in the Yarra Valley, and they are considering using it to grow apples, pears, oranges, and lemons. The profit for growing one acre of apples is $2000, for one acre of pears is $1800, for one acre of oranges is $2200, and for one acre of lemons is $3000. To achieve maximum profit, how many acres of land should they use to grow each type of fruit? Vicky and David have just bought a farm in the Yarra Valley with a total area of 120 acres. + +The land used to grow apples should be at least twice the land used to grow pears. + +The land used to grow apples should be at least three times the land used to grow lemons. + +The land used to grow oranges must be twice the land used to grow lemons. + +Vicky and David are unwilling to grow more than two types of fruit.264000Vicky and David have just bought a farm in the Yarra Valley, and they are considering using it to grow apples, pears, oranges, and lemons. The profit for growing one acre of apples is $2000, for one acre of pears is $1800, for one acre of oranges is $2200, and for one acre of lemons is $3000. To achieve maximum profit, how many acres of land should they use to grow each type of fruit? Vicky and David have just bought a farm in the Yarra Valley with a total area of 120 acres. + +The land used to grow apples should be at least twice the land used to grow pears. + +The land used to grow apples should be at least three times the land used to grow lemons. + +The land used to grow oranges must be twice the land used to grow lemons if lemons are grown. If no lemons are grown, then we do not have this constraint. + +Vicky and David are unwilling to grow more than two types of fruit.264000The land used to grow apples should be at least three times the land used to grow lemons. + +The land used to grow oranges must be twice the land used to grow lemons. lemons if lemons are grown. If no lemons are grown, then we do not have this constraint. + +Vicky and David are unwilling to grow more than two types of fruit.No differencesUnclear description: "oranges must be twice the land used to grow lemons" should only be enforced if lemons are grown. if there's no lemons, oranges quantity can be anything. If prefer to always enforce it instead, then we would need to change the ground truth, as it would be incorrect in that case.ambiguous
59A candy factory uses raw materials A, B, and C to process three different brands of candies, A, B, and C. It is known that the content of A, B, and C in each brand of candy, the cost of raw materials, the monthly limit of each raw material, and the unit processing fee and selling price of the three brands of candies are shown in Table 1-7. + +Table 1-7 + +| Item | A | B | C | Raw Material Cost (Yuan/kg) | Monthly Limit (kg) | +|:----------------|:---------------|:---------------|:---------------|:-----------------------------|:-------------------| +| A | ? 60% | ? 15% | | 2.00 | 2000 | +| B | | | | 1.50 | 2500 | +| C | ? 20% | ? 60% | ? 50% | 1.00 | 1200 | +| Processing Fee (Yuan/kg) | 0.50 | 0.40 | 0.30 | | | +| Selling Price (Yuan/kg) | 3.40 | 2.85 | 2.25 | | | + +How many kilograms of each of the three brands of candies should the factory produce each month to maximize the profit?6160A candy factory uses raw materials A, B, and C to process three different brands of candies, A, B, and C. It is known that the content of A, B, and C in each brand of candy, the cost of raw materials, the monthly limit of each raw material, and the unit processing fee and selling price of the three brands of candies are shown in Table 1-7. + +Table 1-7 + +| Item | A | B | C | Raw Material Cost (Yuan/kg) | Monthly Limit (kg) | +|:----------------|:---------------|:---------------|:---------------|:-----------------------------|:-------------------| +| A | ? 60% | ? 15% | | 2.00 | 2000 | +| B | | | | 1.50 | 2500 | +| C | ? 20% | ? 60% | ? 50% | 1.00 | 1200 | +| Processing Fee (Yuan/kg) | 0.50 | 0.40 | 0.30 | | | +| Selling Price (Yuan/kg) | 3.40 | 2.85 | 2.25 | | | + +How many kilograms of each of the three brands of candies should the factory produce each month to maximize the profit?6160No differencesNo differencesok
60A traveling salesman must visit 7 customers at 7 different locations, with the (symmetric) distance matrix as follows: + +| | 1 | 2 | 3 | 4 | 5 | 6 | 7 | +| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | +| 1 | - | 86 | 49 | 57 | 31 | 69 | 50 | +| 2 | | - | 68 | 79 | 93 | 24 | 5 | +| 3 | | | - | 16 | 7 | 72 | 67 | +| 4 | | | | - | 90 | 69 | 1 | +| 5 | | | | | - | 86 | 59 | +| 6 | | | | | | - | 81 | + +Formulate a mathematical program to determine the visiting order starting and ending at location 1 to minimize the travel distance, and solve it using COPTPY.153A traveling salesman must visit 7 customers at 7 different locations, with the (symmetric) distance matrix as follows: + +| | 1 | 2 | 3 | 4 | 5 | 6 | 7 | +| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | +| 1 | - | 86 | 49 | 57 | 31 | 69 | 50 | +| 2 | | - | 68 | 79 | 93 | 24 | 5 | +| 3 | | | - | 16 | 7 | 72 | 67 | +| 4 | | | | - | 90 | 69 | 1 | +| 5 | | | | | - | 86 | 59 | +| 6 | | | | | | - | 81 | + +Formulate a mathematical program to determine the visiting order starting and ending at location 1 to minimize the travel distance.153| 5 | | | | | - | 86 | 59 | +| 6 | | | | | | - | 81 | + +Formulate a mathematical program to determine the visiting order starting and ending at location 1 to minimize the travel distance, and solve it using COPTPY. distance.No differencesRemoved mention to specific solver/library ("using COPTPY")specific solver
61A product can be processed on any one of the four devices: A, B, C, or D. The preparation completion costs when each device is enabled, the unit production cost for the product, and the maximum processing capacity of each device are shown in Table 5-7. If 2000 units of the product need to be produced, how can the total cost be minimized? Try to establish a mathematical model. + +Table 5-7 +| Device | Prep Completion Cost (Yuan) | Unit Production Cost (Yuan/Unit) | Maximum Processing Capacity (Units) | +|--------|------------------------------|----------------------------------|------------------------------------| +| A | 1000 | 20 | 900 | +| B | 920 | 24 | 1000 | +| C | 800 | 16 | 1200 | +| D | 700 | 28 | 1600 |37000A product can be processed on any one of the four devices: A, B, C, or D. The preparation completion costs when each device is enabled, the unit production cost for the product, and the maximum processing capacity of each device are shown in Table 5-7. If 2000 units of the product need to be produced, how can the total cost be minimized? Try to establish a mathematical model. + +Table 5-7 +| Device | Prep Completion Cost (Yuan) | Unit Production Cost (Yuan/Unit) | Maximum Processing Capacity (Units) | +|--------|------------------------------|----------------------------------|------------------------------------| +| A | 1000 | 20 | 900 | +| B | 920 | 24 | 1000 | +| C | 800 | 16 | 1200 | +| D | 700 | 28 | 1600 |37000No differencesNo differencesok
62The Zhang family is deciding to invest in several different restaurants. The annual revenue of Restaurant A is $15,000, Restaurant B is $40,000, Restaurant C is $30,000, and Restaurant D is $50,000. They need to decide whether to purchase each restaurant, with each restaurant being able to be purchased only once. Help them decide which restaurants to buy to maximize their annual income. +The cost of Restaurant A is 1.6 million, Restaurant B is 2.5 million, Restaurant C is 1.8 million, and Restaurant D is 3 million. The Zhang family's investment budget is 6 million. + +If they purchase Restaurant D, then they cannot purchase Restaurant A.90000The Zhang family is deciding to invest in several different restaurants. The annual revenue of Restaurant A is $15,000, Restaurant B is $40,000, Restaurant C is $30,000, and Restaurant D is $50,000. They need to decide whether to purchase each restaurant, with each restaurant being able to be purchased only once. Help them decide which restaurants to buy to maximize their annual income. +The cost of Restaurant A is 1.6 million, Restaurant B is 2.5 million, Restaurant C is 1.8 million, and Restaurant D is 3 million. The Zhang family's investment budget is 6 million. + +If they purchase Restaurant D, then they cannot purchase Restaurant A.90000No differencesNo differencesok
63A farmer needs to transport 1000 units of fresh produce from the farm to a nearby market. The farmer has three transportation options: a horse, a bicycle, and a handcart. Since both the bicycle and handcart are very physically demanding, the farmer wants to choose only one of these two transportation methods. The horse generates 80 units of pollution per trip, the bicycle generates 0 units of pollution, and the handcart generates 0 units of pollution. The total amount of pollution generated by all trips must not exceed 1000 units. At least 8 trips must be made using the horse. The horse, bicycle, and handcart can carry 55 units, 30 units, and 40 units of produce per trip respectively. The farmer needs to ensure that the total amount of transported produce is at least 1000 units.640A farmer needs to transport 1000 units of fresh produce from the farm to a nearby market. The farmer has three transportation options: a horse, a bicycle, and a handcart. Since both the bicycle and handcart are very physically demanding, the farmer wants to choose only one of these two transportation methods. The horse generates 80 units of pollution per trip, the bicycle generates 0 units of pollution, and the handcart generates 0 units of pollution. The total amount of pollution generated by all trips must not exceed 1000 units. At least 8 trips must be made using the horse. The horse, bicycle, and handcart can carry 55 units, 30 units, and 40 units of produce per trip respectively. The farmer needs to ensure that the total amount of transported produce is at least 1000 units while minimizing the total amount of pollution. What is the minimum amount of pollution that the farmer can achieve?640A farmer needs to transport 1000 units of fresh produce from the farm to a nearby market. The farmer has three transportation options: a horse, a bicycle, and a handcart. Since both the bicycle and handcart are very physically demanding, the farmer wants to choose only one of these two transportation methods. The horse generates 80 units of pollution per trip, the bicycle generates 0 units of pollution, and the handcart generates 0 units of pollution. The total amount of pollution generated by all trips must not exceed 1000 units. At least 8 trips must be made using the horse. The horse, bicycle, and handcart can carry 55 units, 30 units, and 40 units of produce per trip respectively. The farmer needs to ensure that the total amount of transported produce is at least 1000 units. units while minimizing the total amount of pollution. What is the minimum amount of pollution that the farmer can achieve?No differencesThe objective should be to minimize the total pollution, but it is not stated in the problem descriptionunderspecified
64A company needs to decide whether to hire some of the five candidates to join their R&D team. The salary requirements for candidates F, G, H, I, and J are $12,000, $15,000, $18,000, $5,000, and $10,000 respectively. The company wants to minimize the total amount paid to candidates without exceeding the budget. + +The company's budget is $40,000 and they wish to hire a maximum of 4 new employees. + +The skill levels of the candidates are as follows: +Candidate F: Level 2 +Candidate G: Level 3 +Candidate H: Level 4 +Candidate I: Level 1 +Candidate J: Level 2 + +The company needs to ensure that the total skill level of the hired employees is at least 8. + +The project management experience years of each candidate are as follows: +Candidate F: 1 year +Candidate G: 2 years +Candidate H: 2 years +Candidate I: 5 years +Candidate J: 4 years + +They hope the total project management experience of the team is at least 8 years. + +Due to the similar technical background of candidates G and J, the company can choose at most one of them.38000A company needs to decide whether to hire some of the five candidates to join their R&D team. The salary requirements for candidates F, G, H, I, and J are $12,000, $15,000, $18,000, $5,000, and $10,000 respectively. The company wants to minimize the total amount paid to candidates without exceeding the budget. + +The company's budget is $40,000 and they wish to hire a maximum of 4 new employees. + +The skill levels of the candidates are as follows: +Candidate F: Level 2 +Candidate G: Level 3 +Candidate H: Level 4 +Candidate I: Level 1 +Candidate J: Level 2 + +The company needs to ensure that the total skill level of the hired employees is at least 8. + +The project management experience years of each candidate are as follows: +Candidate F: 1 year +Candidate G: 2 years +Candidate H: 2 years +Candidate I: 5 years +Candidate J: 4 years + +They hope the total project management experience of the team is at least 8 years. + +Due to the similar technical background of candidates G and J, the company can choose at most one of them.38000No differencesNo differencesok
65A company produces two types of products: microwave ovens and water heaters, which are manufactured in both workshops A and B. It is known that apart from the purchased parts, the production of one microwave oven requires 2 hours of processing in workshop A and 1 hour of assembly in workshop B. The production of one water heater requires 1 hour of processing in workshop A and 3 hours of assembly in workshop B. After production, both products need inspection, sales, and other procedures. The inspection and sales cost for each microwave oven is 30 yuan, and for each water heater is 50 yuan. Workshop A has 250 hours of available production time per month, with each hour costing 80 yuan; workshop B has 150 hours of available production time per month, with each hour costing 20 yuan. It is estimated that an average of 80 microwave ovens and 50 water heaters can be sold per month next year. Based on these actual conditions, the company has established the following monthly plan constraints: + +1. Inspection and sales costs should not exceed 5500 yuan per month; +2. At least 80 microwave ovens should be sold per month; +3. The production hours of both workshops A and B should be fully utilized; +4. Overtime in workshop A should not exceed 20 hours; +5. At least 50 water heaters should be sold per month. + +Try to determine the monthly production plan for the company.30500A company produces two types of products: microwave ovens and water heaters, which are manufactured in both workshops A and B. It is known that apart from the purchased parts, the production of one microwave oven requires 2 hours of processing in workshop A and 1 hour of assembly in workshop B. The production of one water heater requires 1 hour of processing in workshop A and 3 hours of assembly in workshop B. After production, both products need inspection, sales, and other procedures. The inspection and sales cost for each microwave oven is 30 yuan, and for each water heater is 50 yuan. Workshop A has 250 hours of available production time per month, with each hour costing 80 yuan; workshop B has 150 hours of available production time per month, with each hour costing 20 yuan. It is estimated that an average of 80 microwave ovens and 50 water heaters can be sold per month next year. Based on these actual conditions, the company has established the following monthly plan constraints: + +1. Inspection and sales costs should not exceed 5500 yuan per month; +2. At least 80 microwave ovens should be sold per month; +3. The production hours of both workshops A and B should be fully utilized, and overtime for workshop A and B are allowed. +4. Overtime in workshop A should not exceed 20 hours; we do not have upper limit on workshop B's overtime. +5. At least 50 water heaters should be sold per month. + +Try to determine the monthly production plan for the company.305001. Inspection and sales costs should not exceed 5500 yuan per month; +2. At least 80 microwave ovens should be sold per month; +3. The production hours of both workshops A and B should be fully utilized; utilized, and overtime for workshop A and B are allowed. +4. Overtime in workshop A should not exceed 20 hours; we do not have upper limit on workshop B's overtime. +5. At least 50 water heaters should be sold per month. + +Try to determine the monthly production plan for the company.No differencesAmbiguity: Clarification needed on overtime being allowed and unlimited for Bambiguous
66A toy company manufactures three types of tabletop golf toys, each requiring different manufacturing techniques. The high-end type requires 17 hours of manufacturing labor, 8 hours of inspection, and yields a profit of 300 yuan per unit. The mid-range type requires 10 hours of labor, 4 hours of inspection, and yields a profit of 200 yuan per unit. The low-end type requires 2 hours of labor, 2 hours of inspection, and yields a profit of 100 yuan per unit. Available labor hours are 1000, and available inspection hours are 500. Additionally, market forecasts indicate a demand of no more than 50 units for the high-end type, no more than 80 units for the mid-range type, and no more than 150 units for the low-end type. Determine the production plan for the company to maximize profit.25000A toy company manufactures three types of tabletop golf toys, each requiring different manufacturing techniques. The high-end type requires 17 hours of manufacturing labor, 8 hours of inspection, and yields a profit of 300 yuan per unit. The mid-range type requires 10 hours of labor, 4 hours of inspection, and yields a profit of 200 yuan per unit. The low-end type requires 2 hours of labor, 2 hours of inspection, and yields a profit of 100 yuan per unit. Available labor hours are 1000, and available inspection hours are 500. Additionally, market forecasts indicate a demand of no more than 50 units for the high-end type, no more than 80 units for the mid-range type, and no more than 150 units for the low-end type. Determine the production plan for the company to maximize profit.25000No differencesNo differencesok
67The market demand for products I and II is as follows: Product I requires 10,000 units per month from January to April, 30,000 units per month from May to September, and 100,000 units per month from October to December. Product II requires 15,000 units per month from March to September and 50,000 units per month during other months. The cost of producing these two products at a certain factory is as follows: Product I costs 5 yuan per unit to produce from January to May, and 4.50 yuan per unit from June to December; Product II costs 8 yuan per unit to produce from January to May, and 7 yuan per unit from June to December. The factory's combined production capacity for both products should not exceed 120,000 units per month. Product I has a volume of 0.2 cubic meters per unit, Product II has a volume of 0.4 cubic meters per unit, and the factory's warehouse capacity is 15,000 cubic meters. If the factory's warehouse space is insufficient, external warehouse space can be rented. Using the factory’s own warehouse costs 1 yuan per cubic meter per month, while renting an external warehouse increases this cost to 1.5 yuan per cubic meter per month. Given that the initial inventory of both products at the beginning of July is zero, how should production be scheduled from July to December to minimize the total production and inventory costs while meeting market demand?3160500The market demand for products I and II is as follows: Product I requires 10,000 units per month from January to April, 30,000 units per month from May to September, and 100,000 units per month from October to December. Product II requires 15,000 units per month from March to September and 50,000 units per month during other months. The cost of producing these two products at a certain factory is as follows: Product I costs 5 yuan per unit to produce from January to May, and 4.50 yuan per unit from June to December; Product II costs 8 yuan per unit to produce from January to May, and 7 yuan per unit from June to December. The factory's combined production capacity for both products should not exceed 120,000 units per month. Product I has a volume of 0.2 cubic meters per unit, Product II has a volume of 0.4 cubic meters per unit, and the factory's warehouse capacity is 15,000 cubic meters. If the factory's warehouse space is insufficient, external warehouse space can be rented. Using the factory’s own warehouse costs 1 yuan per cubic meter per month, while renting an external warehouse increases this cost to 1.5 yuan per cubic meter per month. Given that the initial inventory of both products at the beginning of July is zero, how should production be scheduled from July to December to minimize the total production and inventory costs while meeting market demand?3160500No differencesNo differencesok
68There are two coal yards A and B, each receiving no less than 80 tons and 100 tons of coal per month, respectively. They are responsible for supplying coal to three residential areas, which need 55 tons, 75 tons, and 50 tons of coal per month, respectively. Coal yard A is located 10 kilometers, 5 kilometers, and 6 kilometers from these three residential areas. Coal yard B is located 4 kilometers, 8 kilometers, and 15 kilometers from these three residential areas. How should these two coal yards distribute coal to the three residential areas to minimize the ton-kilometers of transportation?1030There are two coal yards A and B, each receiving no less than 80 tons and 100 tons of coal per month, respectively. They are responsible for supplying coal to three residential areas, which need 55 tons, 75 tons, and 50 tons of coal per month, respectively. Coal yard A is located 10 kilometers, 5 kilometers, and 6 kilometers from these three residential areas. Coal yard B is located 4 kilometers, 8 kilometers, and 15 kilometers from these three residential areas. How should these two coal yards distribute coal to the three residential areas to minimize the ton-kilometers of transportation?1030No differencesNo differencesok
69A steel reinforcement workshop produces a batch of steel bars (with the same diameter), consisting of 90 pieces of 3 meters in length and 60 pieces of 4 meters in length. It is known that each piece of raw steel bar used is 10 meters in length. How can the raw material be cut most efficiently? Establish a linear programming model for this problem.20A steel reinforcement workshop produces a batch of steel bars (with the same diameter), consisting of 90 pieces of 3 meters in length and 60 pieces of 4 meters in length. It is known that each piece of raw steel bar used is 10 meters in length. How can the raw material be cut most efficiently? Establish a linear programming model for this problem.53No differences20 53Incorrect ground truth, impossible to use 20 steel bars (200 meters in total) to obtain 90 pieces of 3 meters + 60 pieces of 4 metersincorrect ground truth
70The famous Traveling Salesman Problem (TSP) in operations research can be described as follows: A traveling salesman departs from a certain city, visits two other cities to sell merchandise, and must visit each city exactly once before returning to the original starting city. The distances between the cities are provided in the table below. +| City | 1 | 2 | 3 | 4 | +| ---- | ------ | ------ | ------ | ------ | +| 1 | 0 | 10 | 20 | 12 | +| 2 | 10 | 0 | 5 | 10 | +| 3 | 20 | 5 | 0 | 8 | +| 4 | 15 | 12 | 8 | 0 | + +What route should the salesman choose to travel in order to minimize the total distance? Try to formulate an integer programming model for this problem.35The famous Traveling Salesman Problem (TSP) in operations research can be described as follows: A traveling salesman departs from a certain city, and must visit each city exactly once before returning to the original starting city. The distances between the cities are provided in the table below (the entry at row i and column j represents the cost of going from city i to city j) +| City | 1 | 2 | 3 | 4 | +| ---- | ------ | ------ | ------ | ------ | +| 1 | 0 | 10 | 20 | 12 | +| 2 | 10 | 0 | 5 | 10 | +| 3 | 20 | 5 | 0 | 8 | +| 4 | 15 | 12 | 8 | 0 | + +What route should the salesman choose to travel in order to minimize the total distance? Try to formulate an integer programming model for this problem.35The famous Traveling Salesman Problem (TSP) in operations research can be described as follows: A traveling salesman departs from a certain city, visits two other cities to sell merchandise, and must visit each city exactly once before returning to the original starting city. The distances between the cities are provided in the table below. below (the entry at row i and column j represents the cost of going from city i to city j) +| City | 1 | 2 | 3 | 4 | +| ---- | ------ | ------ | ------ | ------ | +| 1 | 0 | 10 | 20 | 12 |No differencesAmbiguous: distances not symmetric but not clear the from/to); we specified this. Also, it is not clear the role of "visits two other cities", so we removed the sentenceambiguous
71Consider assigning $n=2$ factories to $n$ locations. The transportation volume between factory $i$ and factory $j$ is $d_{ij}$, and the unit transportation cost from location $p$ to location $q$ is $c_{pq}$. The specific values are shown in the following table: Table 1.1 + +| | Transportation volume to Location 1 | Transportation volume to Location 2 | Transportation cost to Location 1 | Transportation cost to Location 2 | +| :----: | :---------------------------------: | :---------------------------------: | :-------------------------------: | :-------------------------------: | +| Factory 1 | 10 | 20 | 5 | 8 | +| Factory 2 | 30 | 40 | 6 | 7 | + +In order to minimize the total transportation cost, formulate this problem as an integer model.330Consider assigning $n=2$ factories to $n$ locations. The transportation volume between factory $i$ and factory $j$ is $d_{ij}$, and the unit transportation cost from location $p$ to location $q$ is $c_{pq}$. The specific values are shown in the following table: Table 1.1 + +| | Transportation volume to Location 1 | Transportation volume to Location 2 | Transportation cost to Location 1 | Transportation cost to Location 2 | +| :----: | :---------------------------------: | :---------------------------------: | :-------------------------------: | :-------------------------------: | +| Factory 1 | 10 | 20 | 5 | 8 | +| Factory 2 | 30 | 40 | 6 | 7 | + +In order to minimize the total transportation cost, formulate this problem as an integer model.330No differencesNo differencesok
72The Li family plans to invest their retirement fund in commercial real estate. The annual income from Property 1 is $12,500, Property 2 is $35,000, Property 3 is $23,000, and Property 4 is $100,000. The decision to be made is whether to buy each property or not, rather than how many to buy, as there is only one of each property available. Help them decide which properties to purchase to maximize their annual income. + +The cost of Property 1 is $1.5 million, Property 2 is $2.1 million, Property 3 is $2.3 million, and Property 4 is $4.2 million. The Li family's budget is $7 million. + +If they purchase Property 4, they cannot purchase Property 3.135000The Li family plans to invest their retirement fund in commercial real estate. The annual income from Property 1 is $12,500, Property 2 is $35,000, Property 3 is $23,000, and Property 4 is $100,000. The decision to be made is whether to buy each property or not, rather than how many to buy, as there is only one of each property available. Help them decide which properties to purchase to maximize their annual income. + +The cost of Property 1 is $1.5 million, Property 2 is $2.1 million, Property 3 is $2.3 million, and Property 4 is $4.2 million. The Li family's budget is $7 million. + +If they purchase Property 4, they cannot purchase Property 3.135000No differencesNo differencesok
73The Li family has 5 children: Alice, Bob, Charlie, Diana, and Ella. The cost to take Alice is $1000, Bob is $900, Charlie is $600, Diana is $500, and Ella is $700. Which children should the couple take to minimize the total cost of taking the children? + +They can take up to 3 children on the upcoming trip. + +Bob is the youngest, so the Li family will definitely take him. + +If the couple takes Alice, they will not take Diana because Alice does not get along with her. + +If the couple takes Bob, they will not take Charlie because Bob does not get along with him. + +If they take Charlie, they must also take Diana. + +If they take Diana, they must also take Ella. + +Despite the cost, the Li family has decided to take at least two children.1200The Li family has 5 children: Alice, Bob, Charlie, Diana, and Ella. The cost to take Alice is $1000, Bob is $900, Charlie is $600, Diana is $500, and Ella is $700. Which children should the couple take to minimize the total cost of taking the children? + +They can take up to 3 children on the upcoming trip. + +Bob is the youngest, so the Li family will definitely take him. + +If the couple takes Alice, they will not take Diana because Alice does not get along with her. + +If the couple takes Bob, they will not take Charlie because Bob does not get along with him. + +If they take Charlie, they must also take Diana. + +If they take Diana, they must also take Ella. + +Despite the cost, the Li family has decided to take at least two children.1600No differences1200 1600Incorrect ground truthincorrect ground truth
74A project includes the following 7 activities, with their durations (in days) as follows: $A(4), B(3), C(5), D(2), E(10), F(10), G(1)$. The precedence relationships are also given as: $A \rightarrow G, D ; E, G \rightarrow F; D, F \rightarrow C ; F \rightarrow B$. The cost of work per day is 1000 Euros; additionally, a special machine must be rented from the start of activity $A$ to the end of activity $B$, costing 5000 Euros per day. Formulate this as a linear programming problem and solve it using COPTPY.230000A project includes the following 7 activities, with their durations (in days) as follows: $A(4), B(3), C(5), D(2), E(10), F(10), G(1)$. The precedence relationships are also given as: $A \rightarrow G, D ; E, G \rightarrow F; D, F \rightarrow C ; F \rightarrow B$. The cost of work per day is 1000 Euros; additionally, a special machine must be rented from the start of activity $A$ to the end of activity $B$, costing 5000 Euros per day. Formulate this as a linear programming problem to minimize cost and complete all activities.115000A project includes the following 7 activities, with their durations (in days) as follows: $A(4), B(3), C(5), D(2), E(10), F(10), G(1)$. The precedence relationships are also given as: $A \rightarrow G, D ; E, G \rightarrow F; D, F \rightarrow C ; F \rightarrow B$. The cost of work per day is 1000 Euros; additionally, a special machine must be rented from the start of activity $A$ to the end of activity $B$, costing 5000 Euros per day. Formulate this as a linear programming problem to minimize cost and solve it using COPTPY. complete all activities.230000 115000Incorrect ground truthincorrect ground truth
75There are $\mathrm{A}$ and $\mathrm{B}$ two products, both requiring two successive chemical reaction processes. Each unit of product $\mathrm{A}$ needs 2 hours for the first process and 3 hours for the second process. Each unit of product $\mathrm{B}$ needs 3 hours for the first process and 4 hours for the second process. Available time for the first process is 16 hours, and available time for the second process is 24 hours. + +For each unit of product $\mathrm{B}$ produced, 2 units of by-product $\mathrm{C}$ are generated simultaneously, requiring no additional cost. By-product $\mathrm{C}$ can be sold up to 5 units, and the rest must be disposed of at a cost of 2 yuan per unit. + +Each unit of product $\mathrm{A}$ sold yields a profit of 4 yuan, each unit of product $\mathrm{B}$ yields a profit of 10 yuan, and each unit of by-product $\mathrm{C}$ sold yields a profit of 3 yuan. + +In order to maximize total profit, establish the linear programming model for this problem.57There are $\mathrm{A}$ and $\mathrm{B}$ two products, both requiring two successive chemical reaction processes. Each unit of product $\mathrm{A}$ needs 2 hours for the first process and 3 hours for the second process. Each unit of product $\mathrm{B}$ needs 3 hours for the first process and 4 hours for the second process. Available time for the first process is 16 hours, and available time for the second process is 24 hours. + +For each unit of product $\mathrm{B}$ produced, 2 units of by-product $\mathrm{C}$ are generated simultaneously, requiring no additional cost. By-product $\mathrm{C}$ can be sold up to 5 units, and the rest must be disposed of at a cost of 2 yuan per unit. + +Each unit of product $\mathrm{A}$ sold yields a profit of 4 yuan, each unit of product $\mathrm{B}$ yields a profit of 10 yuan, and each unit of by-product $\mathrm{C}$ sold yields a profit of 3 yuan. + +In order to maximize total profit, establish the linear programming model for this problem.57No differencesNo differencesok
76A timber storage and transport company has a large warehouse for storing and transporting timber for sale. Due to seasonal price fluctuations, the company purchases timber at the beginning of each quarter, with part of it being sold within the quarter and part being stored for future sales. It is known that the maximum storage capacity of the company’s warehouse is 200,000 m³, and the storage cost is $(a+b u)$ yuan/m³, where $a=70$, $b=100$, and $u$ is the storage time (in quarters). The purchase and sale prices for each quarter and the estimated maximum sales volumes are shown in Table 1-18. + +Table 1-18 +| Quarter | Purchase Price (10,000 yuan/10,000 m²) | Sale Price (10,000 yuan/10,000 m²) | Estimated Maximum Sales Volume (10,000 m³) | +|---------|----------------------------------------|------------------------------------|---------------------------------------------| +| Winter | 410 | 425 | 100 | +| Spring | 430 | 440 | 140 | +| Summer | 460 | 465 | 200 | +| Autumn | 450 | 455 | 160 | + +Since timber is not suitable for long-term storage, all inventory should be sold by the end of autumn. Try to establish a linear programming model for this problem to maximize the company's annual profit.4700A timber storage and transport company has a large warehouse for storing and transporting timber for sale. Due to seasonal price fluctuations, the company purchases timber at the beginning of each quarter, with part of it being sold within the quarter and part being stored for future sales. It is known that the maximum storage capacity of the company's warehouse is 200,000 m³, and the storage cost is $(a+b u)$ yuan/m³, where $a=70$, $b=100$, and $u$ is the storage time (in quarters). The purchase and sale prices for each quarter and the estimated maximum sales volumes are shown in Table 1-18. + +Table 1-18 +| Quarter | Purchase Price (10,000 yuan/10,000 m²) | Sale Price (10,000 yuan/10,000 m²) | Estimated Maximum Sales Volume (10,000 m³) | +|---------|----------------------------------------|------------------------------------|---------------------------------------------| +| Winter | 410 | 425 | 100 | +| Spring | 430 | 440 | 140 | +| Summer | 460 | 465 | 200 | +| Autumn | 450 | 455 | 160 | + +Since timber is not suitable for long-term storage, all inventory should be sold by the end of autumn. Try to establish a linear programming model for this problem to maximize the company's annual profit. Return your answer in the unit of 10000 yuan.4700A timber storage and transport company has a large warehouse for storing and transporting timber for sale. Due to seasonal price fluctuations, the company purchases timber at the beginning of each quarter, with part of it being sold within the quarter and part being stored for future sales. It is known that the maximum storage capacity of the company’s company's warehouse is 200,000 m³, and the storage cost is $(a+b u)$ yuan/m³, where $a=70$, $b=100$, and $u$ is the storage time (in quarters). The purchase and sale prices for each quarter and the estimated maximum sales volumes are shown in Table 1-18. + +Table 1-18 +| Quarter | Purchase Price (10,000 yuan/10,000 m²) | Sale Price (10,000 yuan/10,000 m²) | Estimated Maximum Sales Volume (10,000 m³) | +| Summer | 460 | 465 | 200 | +| Autumn | 450 | 455 | 160 | + +Since timber is not suitable for long-term storage, all inventory should be sold by the end of autumn. Try to establish a linear programming model for this problem to maximize the company's annual profit. Return your answer in the unit of 10000 yuan.No differencesTo match the ground truth, the answer has to be reported in 10,000 yuans, we added this to the statementunderspecified
77There are 10 different parts, and they can all be processed on machine \( A \), machine \( B \), or machine \( C \). The unit processing costs are shown in Table 5-6. Additionally, as long as any part is processed on the aforementioned machines, a one-time setup cost will be incurred regardless of whether one or multiple types of parts are processed, with the respective costs being \( d_A = 100 \), \( d_B = 135 \), and \( d_C = 200 \) yuan. If the requirements are: + +1. One piece of each of the aforementioned 10 types of parts needs to be processed; +2. If the 1st part is processed on machine \( A \), then the 2nd part must be processed on machine \( B \) or \( C \); conversely, if the 1st part is processed on machine \( B \) or \( C \), then the 2nd part must be processed on machine \( A \); +3. Parts 3, 4, and 5 must be processed on machines A, B, and C respectively; +4. The number of parts processed on machine \( C \) should not exceed 3 types. + +Try to establish an integer programming mathematical model for this problem with the objective of minimizing the total cost. + +Table 5-6 +| Machine/Part | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | +|--------------|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----| +| A | $10$ | $20$ | $30$ | $40$ | $50$ | $60$ | $70$ | $80$ | $90$ | $100$ | +| B | $15$ | $25$ | $35$ | $45$ | $55$ | $65$ | $75$ | $85$ | $95$ | $105$ | +| C | $20$ | $30$ | $40$ | $50$ | $60$ | $70$ | $80$ | $90$ | $100$ | $110$ |1000There are 10 different parts, and they can all be processed on machine \( A \), machine \( B \), or machine \( C \). The unit processing costs are shown in Table 5-6. Additionally, as long as any part is processed on the aforementioned machines, a one-time setup cost will be incurred regardless of whether one or multiple types of parts are processed, with the respective costs being \( d_A = 100 \), \( d_B = 135 \), and \( d_C = 200 \) yuan. If the requirements are: + +1. One piece of each of the aforementioned 10 types of parts needs to be processed; +2. If the 1st part is processed on machine \( A \), then the 2nd part must be processed on machine \( B \) or \( C \); conversely, if the 1st part is processed on machine \( B \) or \( C \), then the 2nd part must be processed on machine \( A \); +3. Parts 3, 4, and 5 must be processed on machines A, B, and C respectively; +4. The number of parts processed on machine \( C \) should not exceed 3 types. + +Try to establish an integer programming mathematical model for this problem with the objective of minimizing the total cost. + +Table 5-6 +| Machine/Part | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | +|--------------|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----| +| A | $10$ | $20$ | $30$ | $40$ | $50$ | $60$ | $70$ | $80$ | $90$ | $100$ | +| B | $15$ | $25$ | $35$ | $45$ | $55$ | $65$ | $75$ | $85$ | $95$ | $105$ | +| C | $20$ | $30$ | $40$ | $50$ | $60$ | $70$ | $80$ | $90$ | $100$ | $110$ |1005No differences1000 1005Incorrect ground truthincorrect ground truth
78A shoe store employs 5 full-time sales clerks and 4 part-time sales clerks. Their working hours and wage conditions are shown in Table 3.3. + +Table 3.3 + +| | Monthly Working Hours | Sales Volume (Pairs/Hour) | Wage (Yuan/Hour) | Overtime Pay (Yuan/Hour) | +| :---: | :---: | :---: | :---: | :---: | +| Full-time | 160 | 5 | 1 | 1.5 | +| Part-time | 80 | 2 | 0.6 | 0.7 | + +Each pair of shoes sold earns a profit of 0.3 yuan. The store has set the following goals: + +$p_{1}$: Achieve monthly sales of 5500 pairs; + +$p_{2}$: Ensure full employment of all sales clerks; + +$p_{3}$: Minimize overtime hours. + +Try to establish a model for this problem.400A shoe store employs 5 full-time sales clerks and 4 part-time sales clerks. Their working hours and wage conditions are shown in Table 3.3. + +Table 3.3 + +| | Monthly Working Hours | Sales Volume (Pairs/Hour) | Wage (Yuan/Hour) | Overtime Pay (Yuan/Hour) | +| :---: | :---: | :---: | :---: | :---: | +| Full-time | 160 | 5 | 1 | 1.5 | +| Part-time | 80 | 2 | 0.6 | 0.7 | + +Each pair of shoes sold earns a profit of 0.3 yuan. The store has set the following goals: + +$p_{1}$: Achieve monthly sales of 5500 pairs; + +$p_{2}$: Ensure full employment of all sales clerks; + +$p_{3}$: Minimize overtime hours. + +Try to establish a model for this problem.172No differences400 172Incorrect ground truthincorrect ground truth
79A furniture factory needs to decide how many tables, chairs, and bookshelves to produce in order to maximize its profit. The factory can sell each table for $200, each chair for $50, and each bookshelf for $150. The manufacturing costs for each table, chair, and bookshelf are $120, $20, and $90 respectively. The profit is the difference between the selling price and the manufacturing cost. Each table, chair, and bookshelf occupy 5, 2, and 3 square meters of warehouse space respectively. Due to limited warehouse space, the total space cannot exceed 500 square meters. In addition, due to market demand, the factory needs to produce at least 10 tables and 20 bookshelves. Finally, the total number of items produced by the factory cannot exceed 200.9800A furniture factory needs to decide how many tables, chairs, and bookshelves to produce in order to maximize its profit. The factory can sell each table for $200, each chair for $50, and each bookshelf for $150. The manufacturing costs for each table, chair, and bookshelf are $120, $20, and $90 respectively. The profit is the difference between the selling price and the manufacturing cost. Each table, chair, and bookshelf occupy 5, 2, and 3 square meters of warehouse space respectively. Due to limited warehouse space, the total space cannot exceed 500 square meters. In addition, due to market demand, the factory needs to produce at least 10 tables and 20 bookshelves. Finally, the total number of items produced by the factory cannot exceed 200.9800No differencesNo differencesok
80A company requires skilled workers and laborers for three tasks. The first task can be completed by one skilled worker alone, or by a group of one skilled worker and two laborers. The second task can be done by one skilled worker or one laborer alone. The third task can be completed by a group of five laborers, or by one skilled worker leading three laborers. The weekly wages for skilled workers and laborers are 100 yuan and 80 yuan respectively. They work 48 hours per week, but their actual effective working hours are 42 hours and 36 hours respectively. To complete these tasks, the company needs a total effective working time of 10,000 hours for the first task, 20,000 hours for the second task, and 30,000 hours for the third task per week. The number of workers that can be recruited is limited to a maximum of 400 skilled workers and 800 laborers. Establish a mathematical model to determine how many skilled workers and laborers should be hired in order to minimize the total wage expenditure.45960A company requires skilled workers and laborers for three tasks. The first task can be completed by one skilled worker alone, or by a group of one skilled worker and two laborers. The second task can be done by one skilled worker or one laborer alone. The third task can be completed by a group of five laborers, or by one skilled worker leading three laborers. The weekly wages for skilled workers and laborers are 100 yuan and 80 yuan respectively. They work 48 hours per week, but their actual effective working hours are 42 hours and 36 hours respectively. To complete these tasks, the company needs a total effective working time of 8400 hours for the first task, 10800 hours for the second task, and 18000 hours for the third task per week. The number of workers that can be recruited is limited to a maximum of 400 skilled workers and 800 laborers. Establish a mathematical model to determine how many skilled workers and laborers should be hired in order to minimize the total wage expenditure.84000A company requires skilled workers and laborers for three tasks. The first task can be completed by one skilled worker alone, or by a group of one skilled worker and two laborers. The second task can be done by one skilled worker or one laborer alone. The third task can be completed by a group of five laborers, or by one skilled worker leading three laborers. The weekly wages for skilled workers and laborers are 100 yuan and 80 yuan respectively. They work 48 hours per week, but their actual effective working hours are 42 hours and 36 hours respectively. To complete these tasks, the company needs a total effective working time of 10,000 8400 hours for the first task, 20,000 10800 hours for the second task, and 30,000 18000 hours for the third task per week. The number of workers that can be recruited is limited to a maximum of 400 skilled workers and 800 laborers. Establish a mathematical model to determine how many skilled workers and laborers should be hired in order to minimize the total wage expenditure.45960 84000The original instance was infeasible, we updated the data to make it feasibleinfeasible
81On Danzig Street, vehicles can park on both sides of the street. Mr. Edmonds, who lives at No. 1, is organizing a party with about 30 participants, and they will arrive in 15 cars. The length of the i-th car is ?_i, in meters, as follows: + +| i | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | +|----|----|-----|----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----| +| ?_i | 4 | 4.5 | 5 | 4.1 | 2.4 | 5.2 | 3.7 | 3.5 | 3.2 | 4.5 | 2.3 | 3.3 | 3.8 | 4.6 | 3 | + +In order to avoid disturbing the neighbors, Mr. Edmonds wants to arrange parking on both sides of the street so that the total length of the street occupied by his friends' vehicles is minimized. Please provide a mathematical programming formulation and solve this problem using AMPL. +How does the program change if the cars on one side of the street cannot occupy more than 30 meters?28.6On Danzig Street, vehicles can park on both sides of the street. Mr. Edmonds, who lives at No. 1, is organizing a party with about 30 participants, and they will arrive in 15 cars. The length of the i-th car is ?_i, in meters, as follows: + +| i | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | +|----|----|-----|----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----| +| ?_i | 4 | 4.5 | 5 | 4.1 | 2.4 | 5.2 | 3.7 | 3.5 | 3.2 | 4.5 | 2.3 | 3.3 | 3.8 | 4.6 | 3 | + +In order to avoid disturbing the neighbors, Mr. Edmonds wants to arrange parking on both sides of the street so that the total length of the street occupied by his friends' vehicles is minimized. Please provide a mathematical programming formulation and solve this problem. +How does the program change if the cars on one side of the street cannot occupy more than 30 meters?28.6|----|----|-----|----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----| +| ?_i | 4 | 4.5 | 5 | 4.1 | 2.4 | 5.2 | 3.7 | 3.5 | 3.2 | 4.5 | 2.3 | 3.3 | 3.8 | 4.6 | 3 | + +In order to avoid disturbing the neighbors, Mr. Edmonds wants to arrange parking on both sides of the street so that the total length of the street occupied by his friends' vehicles is minimized. Please provide a mathematical programming formulation and solve this problem using AMPL. problem. +How does the program change if the cars on one side of the street cannot occupy more than 30 meters?No differencesRemoved mention to specific solver/library ("using AMPL")specific solver
82Changjiang Comprehensive Shopping Mall has 5000 m² of space for lease and plans to attract the following 5 types of stores as tenants. The table below shows the area occupied by each type of store for one shop, the minimum and maximum number of shops for each type within the mall, and the expected annual profit (in ten thousand yuan) per store for different numbers of stores. Each store pays 20% of its annual profit as rent to the mall. Question: How many of each type of store should the mall lease to maximize total rental income? + +Table 5-12 + +| Code | Store Type | Area per Shop / m² | Min | Max | 1 Store | 2 Stores | 3 Stores | +|------|------------|--------------------|-----|-----|---------|----------|----------| +| 1 | Jewelry | 250 | 1 | 3 | 9 | 8 | 7 | +| 2 | Shoes & Hats | 350 | 1 | 2 | 10 | 9 | - | +| 3 | General Merchandise | 800 | 1 | 3 | 27 | 21 | 20 | +| 4 | Bookstore | 400 | 0 | 2 | 16 | 10 | - | +| 5 | Catering | 500 | 1 | 3 | 17 | 15 | 12 |28Changjiang Comprehensive Shopping Mall has 5000 m² of space for lease and plans to attract the following 5 types of stores as tenants. The table below shows the area occupied by each type of store for one shop, the minimum and maximum number of shops for each type within the mall, and the expected annual profit (in ten thousand yuan) per store for different numbers of stores. Each store pays 20% of its annual profit as rent to the mall. Question: How many of each type of store should the mall lease to maximize total rental income? + +Table 5-12 + +| Code | Store Type | Area per Shop / m² | Min | Max | 1 Store | 2 Stores | 3 Stores | +|------|------------|--------------------|-----|-----|---------|----------|----------| +| 1 | Jewelry | 250 | 1 | 3 | 9 | 8 | 7 | +| 2 | Shoes & Hats | 350 | 1 | 2 | 10 | 9 | - | +| 3 | General Merchandise | 800 | 1 | 3 | 27 | 21 | 20 | +| 4 | Bookstore | 400 | 0 | 2 | 16 | 10 | - | +| 5 | Catering | 500 | 1 | 3 | 17 | 15 | 12 |28No differencesNo differencesok
83A certain restaurant operates around the clock, and the number of waiters needed in 24 hours is shown in Table 1.1. + +Table 1.1 + +| Time | Minimum Number of Waiters Needed | Time | Minimum Number of Waiters Needed | +|:-----------:|:-------------------------------:|:-----------:|:-------------------------------:| +| $2 \sim 6$ | 4 | $14 \sim 18$| 7 | +| $6 \sim 10$ | 8 | $18 \sim 22$| 12 | +| $10 \sim 14$| 10 | $22 \sim 2$ | 4 | + +Each waiter works continuously for 8 hours a day. The goal is to find the minimum number of waiters that meet the above conditions and represent this problem as a linear programming model.26A certain restaurant operates around the clock, and the number of waiters needed in 24 hours is shown in Table 1.1. + +Table 1.1 + +| Time | Minimum Number of Waiters Needed | Time | Minimum Number of Waiters Needed | +|:-----------:|:-------------------------------:|:-----------:|:-------------------------------:| +| $2 \sim 6$ | 4 | $14 \sim 18$| 7 | +| $6 \sim 10$ | 8 | $18 \sim 22$| 12 | +| $10 \sim 14$| 10 | $22 \sim 2$ | 4 | + +Each waiter works continuously for 8 hours a day. The goal is to find the minimum number of waiters that meet the above conditions and represent this problem as a linear programming model.26No differencesNo differencesok
84A company hopes to recruit new employees for its team. The salary requirements for candidates A, B, C, D, and E are $8100, $20000, $21000, $3000, and $8000 respectively. They need to decide whether to hire each candidate. The team wants to minimize the total amount paid to the candidates. + +They hope to hire a maximum of 3 new employees. + +The team has a limited budget of $35,000. They need to ensure that the total payment to the selected candidates does not exceed the budget. + +The qualifications of the five candidates are as follows: +Candidate A: Bachelor's degree; +Candidate B: Master's degree; +Candidate C: Doctoral degree; +Candidate D: No degree; +Candidate E: No degree. +They will select at least one candidate with a Master's or Doctoral degree. + +The work experience of the five candidates is as follows: +Candidate A: 3 years of work experience; +Candidate B: 10 years of work experience; +Candidate C: 4 years of work experience; +Candidate D: 3 years of work experience; +Candidate E: 7 years of work experience. +They hope the total work experience of the selected candidates is no less than 12 years. + +Due to the equivalent professional skills of candidates A and E, the company will choose at most one from the two. + +They will hire at least 2 new employees.23000A company hopes to recruit new employees for its team. The salary requirements for candidates A, B, C, D, and E are $8100, $20000, $21000, $3000, and $8000 respectively. They need to decide whether to hire each candidate. The team wants to minimize the total amount paid to the candidates. + +They hope to hire a maximum of 3 new employees. + +The team has a limited budget of $35,000. They need to ensure that the total payment to the selected candidates does not exceed the budget. + +The qualifications of the five candidates are as follows: +Candidate A: Bachelor's degree; +Candidate B: Master's degree; +Candidate C: Doctoral degree; +Candidate D: No degree; +Candidate E: No degree. +They will select at least one candidate with a Master's or Doctoral degree. + +The work experience of the five candidates is as follows: +Candidate A: 3 years of work experience; +Candidate B: 10 years of work experience; +Candidate C: 4 years of work experience; +Candidate D: 3 years of work experience; +Candidate E: 7 years of work experience. +They hope the total work experience of the selected candidates is no less than 12 years. + +Due to the equivalent professional skills of candidates A and E, the company will choose at most one from the two. + +They will hire at least 2 new employees.23000No differencesNo differencesok
85A company is producing two products (X and Y). The resources required for the production of X and Y are divided into two parts: machine time for automated processing and craftsman time for manual finishing. The table below shows the number of minutes required for each product: + +| Item | Machine Time (minutes) | Craftsman Time (minutes) | +| :---: | :---: | :---: | +| X | 13 | 20 | +| Y | 19 | 29 | + +The company has 40 hours of machine time available in the next working week, but only 35 hours of craftsman time. The cost of machine time is £10 per hour, and the cost of craftsman time is £2 per hour. Idle time for machines and craftsmen incurs no cost. For each product produced (all products produced will be sold), the revenue for product X is £20, and the revenue for product Y is £30. The company has a specific contract that requires 10 units of product X to be produced for a customer each week. Formulate a model for this problem.1866.37A company is producing two products (X and Y). The resources required for the production of X and Y are divided into two parts: machine time for automated processing and craftsman time for manual finishing. The table below shows the number of minutes required for each product: + +| Item | Machine Time (minutes) | Craftsman Time (minutes) | +| :---: | :---: | :---: | +| X | 13 | 20 | +| Y | 19 | 29 | + +The company has 40 hours of machine time available in the next working week, but only 35 hours of craftsman time. The cost of machine time is £10 per hour, and the cost of craftsman time is £2 per hour. Idle time for machines and craftsmen incurs no cost. For each product produced (all products produced will be sold), the revenue for product X is £20, and the revenue for product Y is £30. Products can only be produced in whole units. The company has a specific contract that requires 10 units of product X to be produced for a customer each week. Formulate a model for this problem.1861.466667| X | 13 | 20 | +| Y | 19 | 29 | + +The company has 40 hours of machine time available in the next working week, but only 35 hours of craftsman time. The cost of machine time is £10 per hour, and the cost of craftsman time is £2 per hour. Idle time for machines and craftsmen incurs no cost. For each product produced (all products produced will be sold), the revenue for product X is £20, and the revenue for product Y is £30. Products can only be produced in whole units. The company has a specific contract that requires 10 units of product X to be produced for a customer each week. Formulate a model for this problem.1866.37 1861.466667Ambiguity: The original ground truth hints to producing fractional products, but it could also be interpreted as only whole products are allowed. Clarified that the quantity of the products must be integers.integer/fractional
86Healthy Pet Foods Company produces two types of dog food: Meaties and Yummies. Each pack of Meaties contains 2 pounds of grains and 3 pounds of meat; each pack of Yummies contains 3 pounds of grains and 1.5 pounds of meat. The company believes it can sell any quantity of dog food that it can produce. Meaties sell for $2.80 per pack, and Yummies sell for $2.00 per pack. The company's production is subject to several constraints. First, a maximum of 400,000 pounds of grains can be purchased each month at a price of $0.20 per pound of grains. A maximum of 300,000 pounds of meat can be purchased each month at a price of $0.50 per pound of meat. Additionally, a special machine is required to produce Meaties, with a monthly capacity of 90,000 packs. The variable costs for mixing and packaging dog food are $0.25 per pack (Meaties) and $0.20 per pack (Yummies). Detailed information is provided in Table B-1. + +**Table B-1 Healthy Pet Foods Data** + +| | Meaties | Yummies | +|--------------------|--------------|------------| +| Price per pack | $2.80 | $2.00 | +| Raw materials | | | +| - Grains | 2.0 lbs | 3.0 lbs | +| - Meat | 3.0 lbs | 1.5 lbs | +| Variable cost | $0.25/pack | $0.20/pack | +| Resources | | | +| Meaties capacity | 90,000 packs/month | | +| Monthly available grains | 400,000 lbs | | +| Monthly available meat | 300,000 lbs | | + +Assume you are the manager of the dog food department at Healthy Pet Foods Company. Your salary is based on the department's profit, so you will try to maximize profit. How should you operate the department to maximize both the profit and your salary?77500Healthy Pet Foods Company produces two types of dog food: Meaties and Yummies. Each pack of Meaties contains 2 pounds of grains and 3 pounds of meat; each pack of Yummies contains 3 pounds of grains and 1.5 pounds of meat. The company believes it can sell any quantity of dog food that it can produce. Meaties sell for $2.80 per pack, and Yummies sell for $2.00 per pack. The company's production is subject to several constraints. First, a maximum of 400,000 pounds of grains can be purchased each month at a price of $0.20 per pound of grains. A maximum of 300,000 pounds of meat can be purchased each month at a price of $0.50 per pound of meat. Additionally, a special machine is required to produce Meaties, with a monthly capacity of 90,000 packs. The variable costs for mixing and packaging dog food are $0.25 per pack (Meaties) and $0.20 per pack (Yummies). Detailed information is provided in Table B-1. + +**Table B-1 Healthy Pet Foods Data** + +| | Meaties | Yummies | +|--------------------|--------------|------------| +| Price per pack | $2.80 | $2.00 | +| Raw materials | | | +| - Grains | 2.0 lbs | 3.0 lbs | +| - Meat | 3.0 lbs | 1.5 lbs | +| Variable cost | $0.25/pack | $0.20/pack | +| Resources | | | +| Meaties capacity | 90,000 packs/month | | +| Monthly available grains | 400,000 lbs | | +| Monthly available meat | 300,000 lbs | | + +Assume you are the manager of the dog food department at Healthy Pet Foods Company. Your salary is based on the department's profit, so you will try to maximize profit. How should you operate the department to maximize both the profit and your salary?77500No differencesNo differencesok
87A transportation company has two types of trucks, Type A and Type B. Type A trucks have 20 cubic meters of refrigerated capacity and 40 cubic meters of non-refrigerated capacity. In contrast, Type B trucks have the same total capacity, but the capacities for refrigerated and non-refrigerated cargo are equal. A grocer needs to rent trucks to transport 3000 cubic meters of refrigerated cargo and 4000 cubic meters of non-refrigerated cargo. The rental cost per kilometer for Type A trucks is £30, while the rental cost per kilometer for Type B trucks is £40. How many of each type of truck should the grocer rent to minimize the total cost? + +Try to formulate a model for this problem.4170A transportation company has two types of trucks, Type A and Type B. Type A trucks have 20 cubic meters of refrigerated capacity and 40 cubic meters of non-refrigerated capacity. In contrast, Type B trucks have the same total capacity, but the capacities for refrigerated and non-refrigerated cargo are equal. A grocer needs to rent trucks to transport 3000 cubic meters of refrigerated cargo and 4000 cubic meters of non-refrigerated cargo. The rental cost per kilometer for Type A trucks is £30, while the rental cost per kilometer for Type B trucks is £40. How many of each type of truck should the grocer rent to minimize the total cost? + +Try to formulate a model for this problem.4170No differencesNo differencesok
88A company uses two machines (Machine 1 and Machine 2) to produce two types of products (liquid fertilizer and solid fertilizer). To produce one unit of liquid fertilizer, it takes 50 minutes on Machine 1 and 30 minutes on Machine 2. To produce one unit of solid fertilizer, it takes 24 minutes on Machine 1 and 33 minutes on Machine 2. At the beginning of the week, there are 30 units of liquid fertilizer and 90 units of solid fertilizer in inventory. The available processing time for Machine 1 this week is expected to be 40 hours, and for Machine 2 it is expected to be 35 hours. The demand for liquid fertilizer this week is estimated at 75 units, and for solid fertilizer at 95 units. The company's policy is to maximize the total number of units of liquid fertilizer and solid fertilizer in inventory at the end of the week. + +Formulate a model for this problem.1.25A company uses two machines (Machine 1 and Machine 2) to produce two types of products (liquid fertilizer and solid fertilizer). To produce one unit of liquid fertilizer, it takes 50 minutes on Machine 1 and 30 minutes on Machine 2. To produce one unit of solid fertilizer, it takes 24 minutes on Machine 1 and 33 minutes on Machine 2. Fertilizers must be produced in whole units, and fractional amounts are not allowed. At the beginning of the week, there are 30 units of liquid fertilizer and 90 units of solid fertilizer in inventory. The available processing time for Machine 1 this week is expected to be 40 hours, and for Machine 2 it is expected to be 35 hours. The demand for liquid fertilizer this week is estimated at 75 units, and for solid fertilizer at 95 units. The company's policy is to maximize the total number of units of liquid fertilizer and solid fertilizer in inventory at the end of the week. + +Formulate a model for this problem.1A company uses two machines (Machine 1 and Machine 2) to produce two types of products (liquid fertilizer and solid fertilizer). To produce one unit of liquid fertilizer, it takes 50 minutes on Machine 1 and 30 minutes on Machine 2. To produce one unit of solid fertilizer, it takes 24 minutes on Machine 1 and 33 minutes on Machine 2. Fertilizers must be produced in whole units, and fractional amounts are not allowed. At the beginning of the week, there are 30 units of liquid fertilizer and 90 units of solid fertilizer in inventory. The available processing time for Machine 1 this week is expected to be 40 hours, and for Machine 2 it is expected to be 35 hours. The demand for liquid fertilizer this week is estimated at 75 units, and for solid fertilizer at 95 units. The company's policy is to maximize the total number of units of liquid fertilizer and solid fertilizer in inventory at the end of the week. + +Formulate a model for this problem.1.25 1It's unclear whether integer or fractional units should be used, the ground truth implied fractional, but the description talks about units that could hint to integer. We clarified the problem description to ensure there is no ambiguity.integral/fractional
89A company produces product A and product B. Each unit of product A sold generates a profit of £30, while each unit of product B sold generates a profit of £10. The company can allocate a maximum of 40 hours per week for production. Producing one unit of product A requires 6 hours, while producing one unit of product B requires 3 hours. Market demand requires that the quantity of product B produced must be at least three times the quantity of product A. The storage space occupied by product A is four times that of product B, and a maximum of four units of product A can be stored per week. + +Formulate a model for this problem.146.667A company produces product A and product B. Each unit of product A sold generates a profit of £30, while each unit of product B sold generates a profit of £10. The company can allocate a maximum of 40 hours per week for production. Producing one unit of product A requires 6 hours, while producing one unit of product B requires 3 hours, and products can only be produced in whole units. Market demand requires that the quantity of product B produced must be at least three times the quantity of product A. The storage space occupied by product A is four times that of product B. The storage space's capacity is such that it can store 4 units of product A when only product A is stored. + +Formulate a model for this problem.140A company produces product A and product B. Each unit of product A sold generates a profit of £30, while each unit of product B sold generates a profit of £10. The company can allocate a maximum of 40 hours per week for production. Producing one unit of product A requires 6 hours, while producing one unit of product B requires 3 hours. hours, and products can only be produced in whole units. Market demand requires that the quantity of product B produced must be at least three times the quantity of product A. The storage space occupied by product A is four times that of product B, and a maximum of four B. The storage space's capacity is such that it can store 4 units of product A can be stored per week. when only product A is stored. + +Formulate a model for this problem.146.667 140The constraint regarding maximum storage was unclear, we clarified it. Also, it was unclear whether integer or fractional should be used (the ground truth hints to fractional, but talking about units may hint to integer), we clarified it. ambiguous, integral/fractional
90A store wants to clear out 200 shirts and 100 pairs of pants from last season. They decide to introduce two promotional packages, A and B. Package A includes one shirt and two pairs of pants, priced at £30. Package B includes three shirts and one pair of pants, priced at £50. The store does not want to sell fewer than 20 A packages and 10 B packages. How many of each package do they need to sell to maximize the revenue from the promotion? + +Try to establish a model for this problem.4000A store wants to clear out 200 shirts and 100 pairs of pants from last season. They decide to introduce two promotional packages, A and B. Package A includes one shirt and two pairs of pants, priced at £30. Package B includes three shirts and one pair of pants, priced at £50. The store does not want to sell fewer than 20 A packages and 10 B packages. How many of each package do they need to sell to maximize the revenue from the promotion? + +Try to establish a model for this problem.3600No differences4000 3600Incorrect ground truthincorrect ground truth
91A company produces two products (A and B), with a profit of £3 and £5 per unit sold, respectively. Each product must be assembled on a specific machine, requiring 12 minutes of assembly time per unit for product A and 25 minutes per unit for product B. The company's estimated effective machine working time per week is only 30 hours (due to maintenance or malfunctions). Technical constraints mean that for every five units of product A produced, at least two units of product B must be produced. + +Try to formulate a model for this problem.6200A company produces two products (A and B), with a profit of £3 and £5 per unit sold, respectively. Each product must be assembled on a specific machine, requiring 12 minutes of assembly time per unit for product A and 25 minutes per unit for product B. The company's estimated effective machine working time per week is only 30 hours (due to maintenance or malfunctions). Technical constraints mean that for every five units of product A produced, at least two units of product B must be produced. + +Try to formulate a model for this problem.408No differences6200 408Incorrect ground truthincorrect ground truth
92A school is preparing a trip for 400 students. The transportation company has 10 buses with 50 seats each and 8 minibuses with 40 seats each, but only 9 drivers are available. The rental cost for a bus is £800, and the rental cost for a minibus is £600. Calculate how many of each type of bus should be used to achieve the lowest cost. + +Try to formulate a model for this problem.408.9A school is preparing a trip for 400 students. The transportation company has 10 buses with 50 seats each and 8 minibuses with 40 seats each, but only 9 drivers are available. The rental cost for a bus is £800, and the rental cost for a minibus is £600. Calculate how many of each type of bus should be used to achieve the lowest cost. + +Try to formulate a model for this problem.6200No differences408.9 6200Incorrect ground truthincorrect ground truth
93A dairy processing plant uses milk to produce two dairy products, \( A_{1} \) and \( A_{2} \). One barrel of milk can be processed into 3 kg of \( A_{1} \) in 12 hours on Type A equipment or into 4 kg of \( A_{2} \) in 8 hours on Type B equipment. According to market demand, all produced \( A_{1} \) and \( A_{2} \) can be sold. The profit is 24 yuan per kilogram of \( A_{1} \) and 16 yuan per kilogram of \( A_{2} \). The processing plant can get a daily supply of 50 barrels of milk, with a total of 480 hours of labor time available from regular workers each day. The Type A equipment can process up to 100 kg of \( A_{1} \) per day, while the processing capacity of Type B equipment is not limited. Formulate a production plan for the plant to maximize daily profit.3360A dairy processing plant uses milk to produce two dairy products, \( A_{1} \) and \( A_{2} \). One barrel of milk can be processed into 3 kg of \( A_{1} \) in 12 hours on Type A equipment or into 4 kg of \( A_{2} \) in 8 hours on Type B equipment. According to market demand, all produced \( A_{1} \) and \( A_{2} \) can be sold. The profit is 24 yuan per kilogram of \( A_{1} \) and 16 yuan per kilogram of \( A_{2} \). The processing plant can get a daily supply of 50 barrels of milk, with a total of 480 hours of labor time available from regular workers each day. The Type A equipment can process up to 100 kg of \( A_{1} \) per day, while the processing capacity of Type B equipment is not limited. Formulate a production plan for the plant to maximize daily profit.3360No differencesNo differencesok
94A company blends two types of crude oil (A and B) to produce two types of gasoline (Type I and Type II). The minimum proportion of crude oil A in gasoline Types I and II is 50% and 60%, respectively. The selling prices are 4800 yuan/t and 5600 yuan/t, respectively. The company has current inventories of 500 t of crude oil A and 1000 t of crude oil B, and they can purchase up to 1500 t of crude oil A from the market. The market price for crude oil A is: 10,000 yuan/t for purchases up to 500 t; 8,000 yuan/t for the portion exceeding 500 t but not exceeding 1000 t; 6,000 yuan/t for the portion exceeding 1000 t. How should the company plan its purchasing and processing of crude oil?4800000A company blends two types of crude oil (A and B) to produce two types of gasoline (Type I and Type II). The minimum proportion of crude oil A in gasoline Types I and II is 50% and 60%, respectively. The selling prices are 4800 yuan/t and 5600 yuan/t, respectively. The company has current inventories of 500 t of crude oil A and 1000 t of crude oil B, and they can purchase up to 1500 t of crude oil A from the market. The market price for crude oil A is: 10,000 yuan/t for purchases up to 500 t; 8,000 yuan/t for the portion exceeding 500 t but not exceeding 1000 t; 6,000 yuan/t for the portion exceeding 1000 t. How should the company plan its purchasing and processing of crude oil? Return the maximized profit in yuan.5000000A company blends two types of crude oil (A and B) to produce two types of gasoline (Type I and Type II). The minimum proportion of crude oil A in gasoline Types I and II is 50% and 60%, respectively. The selling prices are 4800 yuan/t and 5600 yuan/t, respectively. The company has current inventories of 500 t of crude oil A and 1000 t of crude oil B, and they can purchase up to 1500 t of crude oil A from the market. The market price for crude oil A is: 10,000 yuan/t for purchases up to 500 t; 8,000 yuan/t for the portion exceeding 500 t but not exceeding 1000 t; 6,000 yuan/t for the portion exceeding 1000 t. How should the company plan its purchasing and processing of crude oil? Return the maximized profit in yuan.4800000 5000000Incorrect ground truthincorrect ground truth
95A beverage factory produces a kind of beverage to meet market demand. According to market forecasts, the sales department of the factory has determined the demand for the beverage for the next 4 weeks. The planning department, based on the actual situation of the factory, has provided the production capacity and production cost for the next 4 weeks, as shown in Table 1. When there is a surplus of beverages after meeting the demand each week, a storage cost of 0.2 thousand yuan per week per thousand boxes of beverages needs to be paid. How should the production plan be arranged to minimize the total cost (the sum of production cost and storage cost) over the four weeks while meeting the weekly market demand? + +Table 1 Beverage Production and Demand Data: + +\begin{tabular}{c|c|c|c} +\hline +Week & Demand/1000 boxes & Production Capacity/1000 boxes & Cost per 1000 boxes/1000 yuan \\ +\hline +1 & 15 & 30 & 5.0 \\ +\hline +2 & 25 & 40 & 5.1 \\ +\hline +3 & 35 & 45 & 5.4 \\ +\hline +4 & 25 & 20 & 5.5 \\ +\hline +Total & 100 & 135 & \\ +\hline +\end{tabular}528A beverage factory produces a kind of beverage to meet market demand. According to market forecasts, the sales department of the factory has determined the demand for the beverage for the next 4 weeks. The planning department, based on the actual situation of the factory, has provided the production capacity and production cost for the next 4 weeks, as shown in Table 1. When there is a surplus of beverages after meeting the demand each week, a storage cost of 0.2 thousand yuan per week per thousand boxes of beverages needs to be paid. How should the production plan be arranged to minimize the total cost (the sum of production cost and storage cost) over the four weeks while meeting the weekly market demand? + +Table 1 Beverage Production and Demand Data: + +\begin{tabular}{c|c|c|c} +\hline +Week & Demand/1000 boxes & Production Capacity/1000 boxes & Cost per 1000 boxes/1000 yuan \\ +\hline +1 & 15 & 30 & 5.0 \\ +\hline +2 & 25 & 40 & 5.1 \\ +\hline +3 & 35 & 45 & 5.4 \\ +\hline +4 & 25 & 20 & 5.5 \\ +\hline +Total & 100 & 135 & \\ +\hline +\end{tabular}528No differencesNo differencesok
96A steel pipe retailer sources raw steel pipes from a steel pipe factory, cuts the pipes according to customer requirements, and sells them. The raw steel pipes obtained from the factory are all 1850 mm in length. A customer now needs 15 pieces of 290 mm, 28 pieces of 315 mm, 21 pieces of 350 mm, and 30 pieces of 455 mm steel pipes. To simplify the production process, it is required that no more than 4 types of cutting patterns are used. The most frequently used cutting pattern incurs an additional cost of 1/10 of the value of a raw steel pipe, the second most frequent incurs an additional cost of 2/10, and so on. Moreover, the number of cuts for each pattern cannot be too many (a single raw steel pipe can produce up to 5 products). Additionally, to minimize waste, the leftover material for each cutting pattern should not exceed 100 mm. How should the material be cut to minimize total cost, and what is the total cost in this case?19.6A steel pipe retailer sources raw steel pipes from a steel pipe factory, cuts the pipes according to customer requirements, and sells them. The raw steel pipes obtained from the factory are all 1850 mm in length. A customer now needs 15 pieces of 290 mm, 28 pieces of 315 mm, 21 pieces of 350 mm, and 30 pieces of 455 mm steel pipes. To simplify the production process, it is required that no more than 4 types of cutting patterns are used. The most frequently used cutting pattern incurs an additional cost of 1/10 of the value of a raw steel pipe, the second most frequent incurs an additional cost of 2/10, and so on. Moreover, the number of cuts for each pattern cannot be too many (a single raw steel pipe can produce up to 5 products). Additionally, to minimize waste, the leftover material for each cutting pattern should not exceed 100 mm. How should the material be cut to minimize total cost, and what is the total cost in this case?21.5No differences19.6 21.5Incorrect ground truth: Even with perfect ranking, 19.6 is impossible. With Y total raw pipes and the surcharge schedule 0.1,0.2,0.3,0.4, the extra cost is at least 0.1Y (the most-used pattern alone contributes 0.1 per pipe, others add more). +So if Y=19, total cost ≥19+0.1⋅19=20.9. Hence 19.6 can’t be optimal.incorrect ground truth
97A company mixes four types of liquid raw materials with different sulfur contents (denoted as A, B, C, and D, respectively) to produce two products (denoted as \( \mathrm{A} \) and \( \mathrm{B} \)). According to the production process requirements, raw materials A, B, and D must first be mixed in a mixing tank, and then the mixed liquid is further mixed with raw material C to produce \( \mathrm{A} \) and \( \mathrm{B} \). The sulfur contents of raw materials A, B, C, and D are \( 3\%, 1\%, 2\%, 1\% \) respectively, and their purchase prices are 6, 16, 10, 15 (thousand yuan per ton) respectively. The sulfur content of products \( \mathrm{A} \) and \( \mathrm{B} \) must not exceed \( 2.5\% \) and \( 1.5\% \) respectively, and their selling prices are 9.15 (thousand yuan per ton). According to market information, there is no limit to the supply of raw materials A, B, and C, but the supply of raw material D is limited to a maximum of 50 tons. The market demand for products \( \mathrm{A} \) and \( \mathrm{B} \) is 100 tons and 200 tons respectively. How should the production be arranged?449.999999A company mixes four types of liquid raw materials with different sulfur contents (denoted as A, B, C, and D, respectively) to produce two products (denoted as \( \mathrm{A} \) and \( \mathrm{B} \)). According to the production process requirements, raw materials A, B, and D must first be mixed in a mixing tank, and then the mixed liquid is further mixed with raw material C to produce \( \mathrm{A} \) and \( \mathrm{B} \). The sulfur contents of raw materials A, B, C, and D are \( 3\%, 1\%, 2\%, 1\% \) respectively, and their purchase prices are 6, 16, 10, 15 (thousand yuan per ton) respectively. The sulfur content of products \( \mathrm{A} \) and \( \mathrm{B} \) must not exceed \( 2.5\% \) and \( 1.5\% \) respectively, and their selling prices are 9, 15 (thousand yuan per ton) respectively. According to market information, there is no limit to the supply of raw materials A, B, and C, but the supply of raw material D is limited to a maximum of 50 tons. The market demand for products \( \mathrm{A} \) and \( \mathrm{B} \) is 100 tons and 200 tons respectively. How should the production be arranged to maximize the total profit?450A company mixes four types of liquid raw materials with different sulfur contents (denoted as A, B, C, and D, respectively) to produce two products (denoted as \( \mathrm{A} \) and \( \mathrm{B} \)). According to the production process requirements, raw materials A, B, and D must first be mixed in a mixing tank, and then the mixed liquid is further mixed with raw material C to produce \( \mathrm{A} \) and \( \mathrm{B} \). The sulfur contents of raw materials A, B, C, and D are \( 3\%, 1\%, 2\%, 1\% \) respectively, and their purchase prices are 6, 16, 10, 15 (thousand yuan per ton) respectively. The sulfur content of products \( \mathrm{A} \) and \( \mathrm{B} \) must not exceed \( 2.5\% \) and \( 1.5\% \) respectively, and their selling prices are 9.15 9, 15 (thousand yuan per ton). ton) respectively. According to market information, there is no limit to the supply of raw materials A, B, and C, but the supply of raw material D is limited to a maximum of 50 tons. The market demand for products \( \mathrm{A} \) and \( \mathrm{B} \) is 100 tons and 200 tons respectively. How should the production be arranged? arranged to maximize the total profit?449.999999 450Fixed issue with price informationambiguous
98A company uses steel and aluminum as raw materials to produce two products (A and B). A single unit of product A requires 6 kg of steel, 8 kg of aluminum, 11 hours of labor, and yields a profit of 5000 yuan (excluding worker overtime pay). A single unit of product B requires 12 kg of steel, 20 kg of aluminum, 24 hours of labor, and yields a profit of 11000 yuan (excluding worker overtime pay). The company currently has 200 kg of steel, 300 kg of aluminum, and 300 hours of labor available. If workers need to work overtime, the overtime pay is 100 yuan per hour. Please develop a production plan to maximize the company's profit and minimize worker overtime.137500A company uses steel and aluminum as raw materials to produce two products (A and B). A single unit of product A requires 6 kg of steel, 8 kg of aluminum, 11 hours of labor, and yields a profit of 5000 yuan (excluding worker overtime pay). A single unit of product B requires 12 kg of steel, 20 kg of aluminum, 24 hours of labor, and yields a profit of 11000 yuan (excluding worker overtime pay). Products can only be produced in whole units. The company currently has 200 kg of steel, 300 kg of aluminum, and 300 hours of labor available. If workers need to work overtime, the overtime pay is 100 yuan per hour. Please develop a production plan to maximize the company's overall profit taking into account worker overtime.165900A company uses steel and aluminum as raw materials to produce two products (A and B). A single unit of product A requires 6 kg of steel, 8 kg of aluminum, 11 hours of labor, and yields a profit of 5000 yuan (excluding worker overtime pay). A single unit of product B requires 12 kg of steel, 20 kg of aluminum, 24 hours of labor, and yields a profit of 11000 yuan (excluding worker overtime pay). Products can only be produced in whole units. The company currently has 200 kg of steel, 300 kg of aluminum, and 300 hours of labor available. If workers need to work overtime, the overtime pay is 100 yuan per hour. Please develop a production plan to maximize the company's overall profit and minimize taking into account worker overtime.137500 165900Original desription was ambiguous mentioning two objectives (max profit, min overtime), we clarified. Original ground truth was incorrect. Also, it's unclear whether integer or fractional units should be used, we clarified.ambiguous, incorrect ground truth, integral/fractional
99An electronic system is composed of 3 types of components. The system operates normally if all three components function properly. By installing one or more spare parts for any of the components, the reliability of the components can be improved. The system's operational reliability is the product of the reliabilities of each component, and the reliability of each component is a function of the number of spare parts installed. The first half of the table below shows the function relationship between the number of spare parts and the reliability of a specific component. The prices and weights of the 3 types of components are shown in rows 8 to 9 of the table. Given that the total budget for all spare parts is limited to 150 yuan, and the weight limit is 20 kg, how should spare parts be installed to maximize the system's operational reliability? + +\begin{table}[h] +\centering +\begin{tabular}{|c|c|c|c|} +\hline +\textbf{Component Number} & \textbf{1} & \textbf{2} & \textbf{3} \\ \hline +\textbf{Number of Spares} & & & \\ \hline +0 & 0.5 & 0.6 & 0.7 \\ \hline +1 & 0.6 & 0.75 & 0.9 \\ \hline +2 & 0.7 & 0.95 & 1.0 \\ \hline +3 & 0.8 & 1.0 & 1.0 \\ \hline +4 & 0.9 & 1.0 & 1.0 \\ \hline +5 & 1.0 & 1.0 & 1.0 \\ \hline +\textbf{Unit Price (yuan)} & 20 & 30 & 40 \\ \hline +\textbf{Unit Weight (kg)} & 2 & 4 & 6 \\ \hline +\end{tabular} +\caption{Spare Component Data Table} +\end{table}0.6075An electronic system is composed of 3 types of components. The system operates normally if all three components function properly. By installing one or more spare parts for any of the components, the reliability of the components can be improved. The system's operational reliability is the product of the reliabilities of each component, and the reliability of each component is a function of the number of spare parts installed. The first half of the table below shows the function relationship between the number of spare parts and the reliability of a specific component. The prices and weights of the 3 types of components are shown in rows 8 to 9 of the table. Given that the total budget for all spare parts is limited to 150 yuan, and the weight limit is 20 kg, how should spare parts be installed to maximize the system's operational reliability? + +\begin{table}[h] +\centering +\begin{tabular}{|c|c|c|c|} +\hline +\textbf{Component Number} & \textbf{1} & \textbf{2} & \textbf{3} \\ \hline +\textbf{Number of Spares} & & & \\ \hline +0 & 0.5 & 0.6 & 0.7 \\ \hline +1 & 0.6 & 0.75 & 0.9 \\ \hline +2 & 0.7 & 0.95 & 1.0 \\ \hline +3 & 0.8 & 1.0 & 1.0 \\ \hline +4 & 0.9 & 1.0 & 1.0 \\ \hline +5 & 1.0 & 1.0 & 1.0 \\ \hline +\textbf{Unit Price (yuan)} & 20 & 30 & 40 \\ \hline +\textbf{Unit Weight (kg)} & 2 & 4 & 6 \\ \hline +\end{tabular} +\caption{Spare Component Data Table} +\end{table}0.6075No differencesNo differencesok
100In network communication services, bandwidth plays an important role. Below is a bandwidth communication table between several communication nodes, showing the bandwidth between any two nodes. If two nodes cannot be directly connected, the corresponding bandwidth is $0$. It is required to establish a link between node $A$ and node $E$ that must pass through service node $C$ (without loops). The bandwidth of this link is defined as the minimum bandwidth value on the link. Please propose a reasonable link arrangement to maximize the bandwidth of this link and find out the maximum bandwidth. + +\begin{table}[h] + \centering + \begin{tabular}{|c|c|c|c|c|c|} + \hline + & A & B & C & D & E \\ + \hline + A & 0 & 90 & 85 & 0 & 65 \\ + \hline + B & 95 & 0 & 70 & 65 & 34 \\ + \hline + C & 60 & 0 & 0 & 88 & 80 \\ + \hline + D & 67 & 30 & 25 & 0 & 84 \\ + \hline + E & 0 & 51 & 0 & 56 & 0 \\ + \hline + \end{tabular} +\end{table}84In network communication services, bandwidth plays an important role. Below is a bandwidth communication table between several communication nodes, showing the bandwidth between any two nodes. If two nodes cannot be directly connected, the corresponding bandwidth is $0$. It is required to establish a link between node $A$ and node $E$ that must pass through service node $C$ (without loops). The bandwidth of this link is defined as the minimum bandwidth value on the link. Please propose a reasonable link arrangement to maximize the bandwidth of this link and find out the maximum bandwidth. + +\begin{table}[h] + \centering + \begin{tabular}{|c|c|c|c|c|c|} + \hline + & A & B & C & D & E \\ + \hline + A & 0 & 90 & 85 & 0 & 65 \\ + \hline + B & 95 & 0 & 70 & 65 & 34 \\ + \hline + C & 60 & 0 & 0 & 88 & 80 \\ + \hline + D & 67 & 30 & 25 & 0 & 84 \\ + \hline + E & 0 & 51 & 0 & 56 & 0 \\ + \hline + \end{tabular} +\end{table}84No differencesNo differencesok
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RowDescription 1Value 1Description 2Value 2Description DiffValue DiffFixFix Type
45A factory produces three types of products, A, B, and C. Each unit of product A requires 1 hour for technical preparation, 10 hours of direct labor, and 3 kg of materials. Each unit of product B requires 2 hours for technical preparation, 4 hours of labor, and 2 kg of materials. Each unit of product C requires 1 hour for technical preparation, 5 hours of labor, and 1 kg of materials. The available technical preparation time is 100 hours, labor time is 700 hours, and materials are 400 kg. The company offers larger discounts for bulk purchases, as detailed in Table 1-22. Determine the company's production plan to maximize profit. +Table 1-22 +| Product A | | Product B | | Product C | | +|:---------------|:---------:|:---------------|:---------:|:---------------|:---------:| +| Sales Volume (pieces) | Profit (yuan) | Sales Volume (pieces) | Profit (yuan) | Sales Volume (pieces) | Profit (yuan) | +| 0 ~ 40 | 10 | 0 ~ 50 | 6 | 0 ~ 100 | 5 | +| 40 ~ 100 | 9 | 50 ~ 100 | 4 | Above 100 | 4 | +| 100 ~ 150 | 8 | Above 100 | 3 | | | +| Above 150 | 7 | | | | |700A factory produces three types of products, A, B, and C. Each unit of product A requires 1 hour for technical preparation, 10 hours of direct labor, and 3 kg of materials. Each unit of product B requires 2 hours for technical preparation, 4 hours of labor, and 2 kg of materials. Each unit of product C requires 1 hour for technical preparation, 5 hours of labor, and 1 kg of materials. The available technical preparation time is 100 hours, labor time is 700 hours, and materials are 400 kg. The company offers larger discounts for bulk purchases, as detailed in Table 1-22. Determine the company's production plan to maximize profit. +Table 1-22 +| Product A | | Product B | | Product C | | +|:---------------|:---------:|:---------------|:---------:|:---------------|:---------:| +| Sales Volume (pieces) | Profit (yuan) | Sales Volume (pieces) | Profit (yuan) | Sales Volume (pieces) | Profit (yuan) | +| 0 ~ 40 | 10 | 0 ~ 50 | 6 | 0 ~ 100 | 5 | +| 40 ~ 100 | 9 | 50 ~ 100 | 4 | Above 100 | 4 | +| 100 ~ 150 | 8 | Above 100 | 3 | | | +| Above 150 | 7 | | | | |712No differences700 712Incorrect ground truth. Solution that achieves our value: Product A: 62 units (40 units at profit 10, 22 units at profit 9), Product B: 19 units (at profit 6), Product C: 0 unitsincorrect ground truth
70The famous Traveling Salesman Problem (TSP) in operations research can be described as follows: A traveling salesman departs from a certain city, visits two other cities to sell merchandise, and must visit each city exactly once before returning to the original starting city. The distances between the cities are provided in the table below. +| City | 1 | 2 | 3 | 4 | +| ---- | ------ | ------ | ------ | ------ | +| 1 | 0 | 10 | 20 | 12 | +| 2 | 10 | 0 | 5 | 10 | +| 3 | 20 | 5 | 0 | 8 | +| 4 | 15 | 12 | 8 | 0 | + +What route should the salesman choose to travel in order to minimize the total distance? Try to formulate an integer programming model for this problem.35The famous Traveling Salesman Problem (TSP) in operations research can be described as follows: A traveling salesman departs from a certain city, and must visit each city exactly once before returning to the original starting city. The distances between the cities are provided in the table below (the entry at row i and column j represents the cost of going from city i to city j) +| City | 1 | 2 | 3 | 4 | +| ---- | ------ | ------ | ------ | ------ | +| 1 | 0 | 10 | 20 | 12 | +| 2 | 10 | 0 | 5 | 10 | +| 3 | 20 | 5 | 0 | 8 | +| 4 | 15 | 12 | 8 | 0 | + +What route should the salesman choose to travel in order to minimize the total distance? Try to formulate an integer programming model for this problem.35The famous Traveling Salesman Problem (TSP) in operations research can be described as follows: A traveling salesman departs from a certain city, visits two other cities to sell merchandise, and must visit each city exactly once before returning to the original starting city. The distances between the cities are provided in the table below. below (the entry at row i and column j represents the cost of going from city i to city j) +| City | 1 | 2 | 3 | 4 | +| ---- | ------ | ------ | ------ | ------ | +| 1 | 0 | 10 | 20 | 12 |No differencesAmbiguous description: The problem says "departs from a certain city, and must visit each city exactly once before returning to the original starting city", that implies visiting 3 cities instead of 4 for TSP. Also it's not clear in the table the "from city" and "to city" for the distances. We clarified both.ambiguity
76A timber storage and transport company has a large warehouse for storing and transporting timber for sale. Due to seasonal price fluctuations, the company purchases timber at the beginning of each quarter, with part of it being sold within the quarter and part being stored for future sales. It is known that the maximum storage capacity of the company’s warehouse is 200,000 m³, and the storage cost is $(a+b u)$ yuan/m³, where $a=70$, $b=100$, and $u$ is the storage time (in quarters). The purchase and sale prices for each quarter and the estimated maximum sales volumes are shown in Table 1-18. + +Table 1-18 +| Quarter | Purchase Price (10,000 yuan/10,000 m²) | Sale Price (10,000 yuan/10,000 m²) | Estimated Maximum Sales Volume (10,000 m³) | +|---------|----------------------------------------|------------------------------------|---------------------------------------------| +| Winter | 410 | 425 | 100 | +| Spring | 430 | 440 | 140 | +| Summer | 460 | 465 | 200 | +| Autumn | 450 | 455 | 160 | + +Since timber is not suitable for long-term storage, all inventory should be sold by the end of autumn. Try to establish a linear programming model for this problem to maximize the company's annual profit.4700A timber storage and transport company has a large warehouse for storing and transporting timber for sale. Due to seasonal price fluctuations, the company purchases timber at the beginning of each quarter, with part of it being sold within the quarter and part being stored for future sales. It is known that the maximum storage capacity of the company's warehouse is 200,000 m³, and the storage cost is $(a+b u)$ yuan/m³, where $a=70$, $b=100$, and $u$ is the storage time (in quarters). The purchase and sale prices for each quarter and the estimated maximum sales volumes are shown in Table 1-18. + +Table 1-18 +| Quarter | Purchase Price (10,000 yuan/10,000 m²) | Sale Price (10,000 yuan/10,000 m²) | Estimated Maximum Sales Volume (10,000 m³) | +|---------|----------------------------------------|------------------------------------|---------------------------------------------| +| Winter | 410 | 425 | 100 | +| Spring | 430 | 440 | 140 | +| Summer | 460 | 465 | 200 | +| Autumn | 450 | 455 | 160 | + +Since timber is not suitable for long-term storage, all inventory should be sold by the end of autumn. Try to establish a linear programming model for this problem to maximize the company's annual profit. Return your answer in the unit of 10000 yuan.4700A timber storage and transport company has a large warehouse for storing and transporting timber for sale. Due to seasonal price fluctuations, the company purchases timber at the beginning of each quarter, with part of it being sold within the quarter and part being stored for future sales. It is known that the maximum storage capacity of the company’s company's warehouse is 200,000 m³, and the storage cost is $(a+b u)$ yuan/m³, where $a=70$, $b=100$, and $u$ is the storage time (in quarters). The purchase and sale prices for each quarter and the estimated maximum sales volumes are shown in Table 1-18. + +Table 1-18 +| Quarter | Purchase Price (10,000 yuan/10,000 m²) | Sale Price (10,000 yuan/10,000 m²) | Estimated Maximum Sales Volume (10,000 m³) | +| Summer | 460 | 465 | 200 | +| Autumn | 450 | 455 | 160 | + +Since timber is not suitable for long-term storage, all inventory should be sold by the end of autumn. Try to establish a linear programming model for this problem to maximize the company's annual profit. Return your answer in the unit of 10000 yuan.No differencesThe problem asks to maximize annual profit, but it doesn't specify the unit. One could return it in yuan, while the ground truth expects unit of 10000 yuan.ambiguity
91A company produces two products (A and B), with a profit of £3 and £5 per unit sold, respectively. Each product must be assembled on a specific machine, requiring 12 minutes of assembly time per unit for product A and 25 minutes per unit for product B. The company's estimated effective machine working time per week is only 30 hours (due to maintenance or malfunctions). Technical constraints mean that for every five units of product A produced, at least two units of product B must be produced. + +Try to formulate a model for this problem.408.9A company produces two products (A and B), with a profit of £3 and £5 per unit sold, respectively. Each product must be assembled on a specific machine, requiring 12 minutes of assembly time per unit for product A and 25 minutes per unit for product B. The company's estimated effective machine working time per week is only 30 hours (due to maintenance or malfunctions). Technical constraints mean that for every five units of product A produced, at least two units of product B must be produced. + +Try to formulate a model for this problem.408No differences408.9 408Incorrect ground truth. incorrect ground truth
98A company uses steel and aluminum as raw materials to produce two products (A and B). A single unit of product A requires 6 kg of steel, 8 kg of aluminum, 11 hours of labor, and yields a profit of 5000 yuan (excluding worker overtime pay). A single unit of product B requires 12 kg of steel, 20 kg of aluminum, 24 hours of labor, and yields a profit of 11000 yuan (excluding worker overtime pay). The company currently has 200 kg of steel, 300 kg of aluminum, and 300 hours of labor available. If workers need to work overtime, the overtime pay is 100 yuan per hour. Please develop a production plan to maximize the company's profit with minimial worker overtime.137500A company uses steel and aluminum as raw materials to produce two products (A and B). A single unit of product A requires 6 kg of steel, 8 kg of aluminum, 11 hours of labor, and yields a profit of 5000 yuan (excluding worker overtime pay). A single unit of product B requires 12 kg of steel, 20 kg of aluminum, 24 hours of labor, and yields a profit of 11000 yuan (excluding worker overtime pay). Products can only be produced in whole units. The company currently has 200 kg of steel, 300 kg of aluminum, and 300 hours of labor available. If workers need to work overtime, the overtime pay is 100 yuan per hour. Please develop a production plan to maximize the company's overall profit taking into account worker overtime.165900A company uses steel and aluminum as raw materials to produce two products (A and B). A single unit of product A requires 6 kg of steel, 8 kg of aluminum, 11 hours of labor, and yields a profit of 5000 yuan (excluding worker overtime pay). A single unit of product B requires 12 kg of steel, 20 kg of aluminum, 24 hours of labor, and yields a profit of 11000 yuan (excluding worker overtime pay). Products can only be produced in whole units. The company currently has 200 kg of steel, 300 kg of aluminum, and 300 hours of labor available. If workers need to work overtime, the overtime pay is 100 yuan per hour. Please develop a production plan to maximize the company's overall profit with minimial taking into account worker overtime.137500 165900Incorrect ground truth. Can produce A=15, B=9, overtime=81incorrect ground truth
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File 1: C:\Users\kmellou\OneDrive - Microsoft\OptiMind-data\data\optmath\optmath_original_xinzhi_excel.xlsx

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Generated: 2025-11-19 18:17:49

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1The manufacturing facility produces custom components for two jobs, Job 0 and Job 1, each consisting of a sequence of operations that must be performed in a specific order. The goal is to schedule these operations to minimize the total completion time (makespan) while satisfying all operational constraints. Job 0 has five operations with processing times: Operation 0 takes 4 units, Operation 1 takes 1 unit, Operation 2 takes 6 units, Operation 3 takes 6 units, and Operation 4 takes 8 units. Job 1 has four operations with processing times: Operation 0 takes 9 units, Operation 1 takes 1 unit, Operation 2 takes 4 units, and Operation 3 takes 2 units. + +Precedence constraints ensure that operations within each job are performed in sequence with specific gaps. For Job 0, Operation 1 must start at least 4 units after Operation 0 starts, Operation 2 must start at least 1 unit after Operation 1 starts, Operation 3 must start at least 6 units after Operation 2 starts, and Operation 4 must start at least 6 units after Operation 3 starts. For Job 1, Operation 1 must start at least 9 units after Operation 0 starts, Operation 2 must start at least 1 unit after Operation 1 starts, and Operation 3 must start at least 4 units after Operation 2 starts. + +Machine capacity constraints ensure that operations assigned to the same machine do not overlap. Binary variables determine the order of operations on shared machines. For example, if Operation 1 of Job 0 and Operation 3 of Job 0 are on the same machine, one must complete at least 6 units before the other starts. Similarly, if Operation 1 of Job 0 and Operation 2 of Job 1 are on the same machine, one must complete at least 4 units before the other starts. These constraints apply to all operation pairs on shared machines, ensuring no overlap and maintaining required time gaps. The large constant of 100,000 is used in these constraints to enforce the sequencing logic by ensuring that the constraints are only active when the binary variable is set to 1. + +The makespan must be at least as large as the completion time of every operation. Specifically, the makespan must be at least 4 units after Operation 0 of Job 0 starts, 1 unit after Operation 1 of Job 0 starts, 6 units after Operation 2 of Job 0 starts, 6 units after Operation 3 of Job 0 starts, 8 units after Operation 4 of Job 0 starts, 9 units after Operation 0 of Job 1 starts, 1 unit after Operation 1 of Job 1 starts, 4 units after Operation 2 of Job 1 starts, and 2 units after Operation 3 of Job 1 starts. + +The objective is to determine the start times for all operations and the sequence of operations on shared machines to minimize the makespan while respecting all precedence constraints, machine capacity constraints, and the makespan definition. Binary variables are used to enforce the correct sequencing of operations on shared machines, with a large constant of 100,000 ensuring the constraints are properly applied. The start times for all operations and the makespan must be non-negative. The binary variables are explicitly defined for all relevant operation pairs on shared machines, and the start times and makespan are continuous variables.25The manufacturing facility produces custom components for two jobs, Job 0 and Job 1, each consisting of a sequence of operations that must be performed in a specific order. The goal is to schedule these operations to minimize the total completion time (makespan) while satisfying all operational constraints. Job 0 has five operations with processing times: Operation 0 takes 4 units, Operation 1 takes 1 unit, Operation 2 takes 6 units, Operation 3 takes 6 units, and Operation 4 takes 8 units. Job 1 has four operations with processing times: Operation 0 takes 9 units, Operation 1 takes 1 unit, Operation 2 takes 4 units, and Operation 3 takes 2 units. + +Precedence constraints ensure that operations within each job are performed in sequence with specific gaps. For Job 0, Operation 1 must start at least 4 units after Operation 0 starts, Operation 2 must start at least 1 unit after Operation 1 starts, Operation 3 must start at least 6 units after Operation 2 starts, and Operation 4 must start at least 6 units after Operation 3 starts. For Job 1, Operation 1 must start at least 9 units after Operation 0 starts, Operation 2 must start at least 1 unit after Operation 1 starts, and Operation 3 must start at least 4 units after Operation 2 starts. + +Machine capacity constraints ensure that operations assigned to the same machine do not overlap. In particular, there are 3 machines (M0, M1, M2). Operation 1 of Job 0, Operation 3 of Job 0, and Operation 2 of Job 1 are on machine M0. Operation 0of Job 0, Operation 2 of Job 0, and Operation 3 of Job 1 are on machine M1. Finally, Operation 4 of Job 0, Operation 0 of Job 1, and Operation 1 of Job 1 are on machine M2. Binary variables determine the order of operations on shared machines. Constraints apply to all operation pairs on shared machines, ensuring no overlap and maintaining required time gaps (one operation must finish before another can start). The large constant of 100,000 is used in these constraints to enforce the sequencing logic by ensuring that the constraints are only active when the binary variable is set to 1. + +The makespan must be at least as large as the completion time of every operation. Specifically, the makespan must be at least 4 units after Operation 0 of Job 0 starts, 1 unit after Operation 1 of Job 0 starts, 6 units after Operation 2 of Job 0 starts, 6 units after Operation 3 of Job 0 starts, 8 units after Operation 4 of Job 0 starts, 9 units after Operation 0 of Job 1 starts, 1 unit after Operation 1 of Job 1 starts, 4 units after Operation 2 of Job 1 starts, and 2 units after Operation 3 of Job 1 starts. + +The objective is to determine the start times for all operations and the sequence of operations on shared machines to minimize the makespan while respecting all precedence constraints, machine capacity constraints, and the makespan definition. Binary variables are used to enforce the correct sequencing of operations on shared machines, with a large constant of 100,000 ensuring the constraints are properly applied. The start times for all operations and the makespan must be non-negative. The binary variables are explicitly defined for all relevant operation pairs on shared machines, and the start times and makespan are continuous variables.25Precedence constraints ensure that operations within each job are performed in sequence with specific gaps. For Job 0, Operation 1 must start at least 4 units after Operation 0 starts, Operation 2 must start at least 1 unit after Operation 1 starts, Operation 3 must start at least 6 units after Operation 2 starts, and Operation 4 must start at least 6 units after Operation 3 starts. For Job 1, Operation 1 must start at least 9 units after Operation 0 starts, Operation 2 must start at least 1 unit after Operation 1 starts, and Operation 3 must start at least 4 units after Operation 2 starts. + +Machine capacity constraints ensure that operations assigned to the same machine do not overlap. Binary variables determine the order In particular, there are 3 machines (M0, M1, M2). Operation 1 of operations on shared machines. For example, if Job 0, Operation 3 of Job 0, and Operation 2 of Job 1 are on machine M0. Operation 0of Job 0, Operation 2 of Job 0 0, and Operation 3 of Job 0 1 are on the same machine, one must complete at least 6 units before the other starts. Similarly, if machine M1. Finally, Operation 1 4 of Job 0, Operation 0 of Job 1, and Operation 2 1 of Job 1 are on machine M2. Binary variables determine the same machine, one must complete at least 4 units before the other starts. These constraints order of operations on shared machines. Constraints apply to all operation pairs on shared machines, ensuring no overlap and maintaining required time gaps. gaps (one operation must finish before another can start). The large constant of 100,000 is used in these constraints to enforce the sequencing logic by ensuring that the constraints are only active when the binary variable is set to 1. + +The makespan must be at least as large as the completion time of every operation. Specifically, the makespan must be at least 4 units after Operation 0 of Job 0 starts, 1 unit after Operation 1 of Job 0 starts, 6 units after Operation 2 of Job 0 starts, 6 units after Operation 3 of Job 0 starts, 8 units after Operation 4 of Job 0 starts, 9 units after Operation 0 of Job 1 starts, 1 unit after Operation 1 of Job 1 starts, 4 units after Operation 2 of Job 1 starts, and 2 units after Operation 3 of Job 1 starts.No differencesMissing data (e.g., shared machines partially specified)missing data
2The problem involves scheduling the landing times of three aircraft to minimize the total penalty costs associated with deviations from their earliest and latest possible landing times while ensuring safe separation between aircraft and adhering to their respective time windows. Each aircraft has an earliest and latest possible landing time, and penalty costs for landing early or late. Aircraft 0 has an earliest landing time of 201 minutes, a latest landing time of 261 minutes, an early penalty cost of \$37 per minute, and a late penalty cost of \$73 per minute. Aircraft 1 has an earliest landing time of 234 minutes, a latest landing time of 294 minutes, an early penalty cost of \$99 per minute, and a late penalty cost of \$84 per minute. Aircraft 2 has an earliest landing time of 29 minutes, a latest landing time of 89 minutes, an early penalty cost of \$90 per minute, and a late penalty cost of \$56 per minute. + +The separation requirements are enforced using binary decision variables that determine the order in which the aircraft land. If Aircraft 0 lands before Aircraft 1, there must be at least 11 minutes of separation; if Aircraft 1 lands before Aircraft 0, the separation must be at least 27 minutes. If Aircraft 0 lands before Aircraft 2, there must be at least 9 minutes of separation; if Aircraft 2 lands before Aircraft 0, the separation must be at least 232 minutes. If Aircraft 1 lands before Aircraft 2, there must be at least 9 minutes of separation; if Aircraft 2 lands before Aircraft 1, the separation must be at least 265 minutes. + +The landing time windows must be respected: Aircraft 0 must land between 201 and 261 minutes, Aircraft 1 between 234 and 294 minutes, and Aircraft 2 between 29 and 89 minutes. The penalties for early and late landings are calculated based on the difference between the actual landing time and the earliest or latest possible landing time, multiplied by the respective penalty costs. The objective is to determine the optimal landing times for all three aircraft to minimize the total penalty costs while ensuring all separation and time window constraints are satisfied. The total penalty cost is the sum of the early and late penalties for all aircraft. The order in which the aircraft land affects the separation requirements and landing times, so the schedule must balance these costs to achieve the lowest total penalty. Binary decision variables are used to determine the order of aircraft landings, ensuring that the separation constraints are enforced correctly. The constraints also ensure that the landing times are within the specified windows and that the penalties for early and late landings are calculated accurately based on the deviations from the earliest and latest possible landing times.0The problem involves scheduling the landing times of three aircraft to minimize the total penalty costs associated with deviations from their target landing times while ensuring safe separation between aircraft and adhering to their respective time windows. Each aircraft has an earliest and latest possible landing time, and penalty costs for landing earlier or later than their target landing time. Aircraft 0 has a target landing time of 231 minutes, an earliest landing time of 201 minutes, a latest landing time of 261 minutes, an early penalty cost of \$37 per minute, and a late penalty cost of \$73 per minute. Aircraft 1 has a target landing time of 264 minutes, an earliest landing time of 234 minutes, a latest landing time of 294 minutes, an early penalty cost of \$99 per minute, and a late penalty cost of \$84 per minute. Aircraft 2 has a target landing time of 59 minutes, an earliest landing time of 29 minutes, a latest landing time of 89 minutes, an early penalty cost of \$90 per minute, and a late penalty cost of \$56 per minute. + +The separation requirements are enforced using binary decision variables that determine the order in which the aircraft land. If Aircraft 0 lands before Aircraft 1, there must be at least 11 minutes of separation; if Aircraft 1 lands before Aircraft 0, the separation must be at least 27 minutes. If Aircraft 0 lands before Aircraft 2, there must be at least 9 minutes of separation; if Aircraft 2 lands before Aircraft 0, the separation must be at least 232 minutes. If Aircraft 1 lands before Aircraft 2, there must be at least 9 minutes of separation; if Aircraft 2 lands before Aircraft 1, the separation must be at least 265 minutes. + +The landing time windows must be respected: Aircraft 0 must land between 201 and 261 minutes, Aircraft 1 between 234 and 294 minutes, and Aircraft 2 between 29 and 89 minutes. The penalties for early and late landings are calculated based on the difference between the actual landing time and the target landing time, multiplied by the respective penalty costs. The objective is to determine the optimal landing times for all three aircraft to minimize the total penalty costs while ensuring all separation and time window constraints are satisfied. The total penalty cost is the sum of the early and late penalties for all aircraft. The order in which the aircraft land affects the separation requirements and landing times, so the schedule must balance these costs to achieve the lowest total penalty. Binary decision variables are used to determine the order of aircraft landings, ensuring that the separation constraints are enforced correctly. The constraints also ensure that the landing times are within the specified windows and that the penalties for early and late landings are calculated accurately based on the deviations from the target landing times.7410The problem involves scheduling the landing times of three aircraft to minimize the total penalty costs associated with deviations from their earliest and latest possible target landing times while ensuring safe separation between aircraft and adhering to their respective time windows. Each aircraft has an earliest and latest possible landing time, and penalty costs for landing early earlier or late. later than their target landing time. Aircraft 0 has a target landing time of 231 minutes, an earliest landing time of 201 minutes, a latest landing time of 261 minutes, an early penalty cost of \$37 per minute, and a late penalty cost of \$73 per minute. Aircraft 1 has a target landing time of 264 minutes, an earliest landing time of 234 minutes, a latest landing time of 294 minutes, an early penalty cost of \$99 per minute, and a late penalty cost of \$84 per minute. Aircraft 2 has a target landing time of 59 minutes, an earliest landing time of 29 minutes, a latest landing time of 89 minutes, an early penalty cost of \$90 per minute, and a late penalty cost of \$56 per minute. + +The separation requirements are enforced using binary decision variables that determine the order in which the aircraft land. If Aircraft 0 lands before Aircraft 1, there must be at least 11 minutes of separation; if Aircraft 1 lands before Aircraft 0, the separation must be at least 27 minutes. If Aircraft 0 lands before Aircraft 2, there must be at least 9 minutes of separation; if Aircraft 2 lands before Aircraft 0, the separation must be at least 232 minutes. If Aircraft 1 lands before Aircraft 2, there must be at least 9 minutes of separation; if Aircraft 2 lands before Aircraft 1, the separation must be at least 265 minutes. + +The landing time windows must be respected: Aircraft 0 must land between 201 and 261 minutes, Aircraft 1 between 234 and 294 minutes, and Aircraft 2 between 29 and 89 minutes. The penalties for early and late landings are calculated based on the difference between the actual landing time and the earliest or latest possible target landing time, multiplied by the respective penalty costs. The objective is to determine the optimal landing times for all three aircraft to minimize the total penalty costs while ensuring all separation and time window constraints are satisfied. The total penalty cost is the sum of the early and late penalties for all aircraft. The order in which the aircraft land affects the separation requirements and landing times, so the schedule must balance these costs to achieve the lowest total penalty. Binary decision variables are used to determine the order of aircraft landings, ensuring that the separation constraints are enforced correctly. The constraints also ensure that the landing times are within the specified windows and that the penalties for early and late landings are calculated accurately based on the deviations from the earliest and latest possible target landing times.7410Ambiguous & missing data: Part of the statement said that the aicrafts *have* to land within windows, but part of the statement said that they can land outside the windows by paying penalty. The description is missing information about the target landing times of each aircraft. Updated ground truth accordingly.ambiguous, missing data
3A logistics company must deliver packages to four cities: City 0, City 1, City 2, and City 3, minimizing the total delivery cost while ensuring each city is visited exactly once and the route forms a continuous loop without subtours. The delivery costs between cities are as follows: City 0 to City 1 costs 616 units, City 0 to City 2 costs 567 units, City 0 to City 3 costs 143 units, City 1 to City 0 costs 388 units, City 1 to City 2 costs 947 units, City 1 to City 3 costs 628 units, City 2 to City 0 costs 495 units, City 2 to City 1 costs 650 units, City 2 to City 3 costs 604 units, City 3 to City 0 costs 77 units, City 3 to City 1 costs 630 units, and City 3 to City 2 costs 23 units. The company uses binary decision variables to indicate whether a route is used and assigns position variables to each city to prevent subtours. The position variable for City 0 is fixed at 0, while the position variables for City 1, City 2, and City 3 must be integers between 0 and 3. The objective is to minimize the total delivery cost, which is the sum of the costs of all selected routes, including the term 0 times the position variable for City 0. Constraints ensure each city is visited and departed exactly once. Subtour elimination is enforced by ensuring that if a route from City i to City j is used, the position of City j must be greater than the position of City i, except for City 0, which is fixed as the starting point. These constraints are implemented through specific relationships between the route variables and the position variables, guaranteeing the delivery route forms a single loop without smaller loops. The constraints also explicitly bound the position variables for City 1, City 2, and City 3 to be between 0 and 3, and ensure that the position variables are correctly related to the route variables to eliminate subtours.1204A logistics company must deliver packages to four cities: City 0, City 1, City 2, and City 3, minimizing the total delivery cost while ensuring each city is visited exactly once and the route forms a continuous loop without subtours. The delivery costs between cities are as follows: City 0 to City 1 costs 616 units, City 0 to City 2 costs 567 units, City 0 to City 3 costs 143 units, City 1 to City 0 costs 388 units, City 1 to City 2 costs 947 units, City 1 to City 3 costs 628 units, City 2 to City 0 costs 495 units, City 2 to City 1 costs 650 units, City 2 to City 3 costs 604 units, City 3 to City 0 costs 77 units, City 3 to City 1 costs 630 units, and City 3 to City 2 costs 23 units. The company uses binary decision variables to indicate whether a route is used and assigns position variables to each city to prevent subtours. The position variable for City 0 is fixed at 0, while the position variables for City 1, City 2, and City 3 must be integers between 0 and 3. The objective is to minimize the total delivery cost, which is the sum of the costs of all selected routes, including the term 0 times the position variable for City 0. Constraints ensure each city is visited and departed exactly once. Subtour elimination is enforced by ensuring that if a route from City i to City j is used, the position of City j must be greater than the position of City i, except for City 0, which is fixed as the starting point. These constraints are implemented through specific relationships between the route variables and the position variables, guaranteeing the delivery route forms a single loop without smaller loops. The constraints also explicitly bound the position variables for City 1, City 2, and City 3 to be between 0 and 3, and ensure that the position variables are correctly related to the route variables to eliminate subtours.1204No differencesNo differencesok
4The problem involves scheduling operations for three jobs (Job 0, Job 1, and Job 2) on shared machines to minimize the total completion time (makespan). Each job consists of a sequence of operations with specific processing times: Job 0 has two operations (Operation 0 takes 9 units of time, Operation 1 takes 8 units), Job 1 has three operations (Operation 0 takes 8 units, Operation 1 takes 8 units, Operation 2 takes 4 units), and Job 2 has four operations (Operation 0 takes 1 unit, Operation 1 takes 6 units, Operation 2 takes 4 units, Operation 3 takes 2 units). Operations within a job must follow a strict order, meaning each operation cannot start until the previous one in the same job has finished. For example, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 has completed after 9 units of time, Operation 1 of Job 1 cannot start until Operation 0 of Job 1 has completed after 8 units of time, and Operation 2 of Job 1 cannot start until Operation 1 of Job 1 has completed after 8 units of time. Similarly, Operation 1 of Job 2 cannot start until Operation 0 of Job 2 has completed after 1 unit of time, Operation 2 of Job 2 cannot start until Operation 1 of Job 2 has completed after 6 units of time, and Operation 3 of Job 2 cannot start until Operation 2 of Job 2 has completed after 4 units of time. + +Certain operations from different jobs share the same machine and cannot overlap in time. For instance, Operation 1 of Job 0 and Operation 1 of Job 2 are assigned to the same machine, so one must finish before the other starts. Similarly, Operation 0 of Job 0 and Operation 1 of Job 1 share a machine, and their schedules must not overlap. This applies to all pairs of operations sharing the same machine, such as Operation 0 of Job 0 and Operation 0 of Job 2, Operation 1 of Job 1 and Operation 0 of Job 2, Operation 0 of Job 1 and Operation 2 of Job 2, Operation 0 of Job 1 and Operation 3 of Job 2, Operation 2 of Job 1 and Operation 2 of Job 2, Operation 2 of Job 1 and Operation 3 of Job 2, and Operation 2 of Job 2 and Operation 3 of Job 2. + +To enforce these constraints, binary variables are used to determine the order of operations on shared machines. A large constant (100000) is used in the constraints to ensure that operations on the same machine do not overlap. For example, if Operation 1 of Job 0 and Operation 1 of Job 2 are on the same machine, the constraints ensure that one operation starts only after the other has finished, using the binary variable to decide the order. The binary variables are restricted to values of 0 or 1, and the start times of operations are continuous and non-negative. + +The makespan must be at least as large as the completion time of every operation. For example, it must be at least 9 units of time (completion of Operation 0 of Job 0), 8 units (completion of Operation 1 of Job 0), 8 units (completion of Operation 0 of Job 1), 8 units (completion of Operation 1 of Job 1), 4 units (completion of Operation 2 of Job 1), 1 unit (completion of Operation 0 of Job 2), 6 units (completion of Operation 1 of Job 2), 4 units (completion of Operation 2 of Job 2), and 2 units (completion of Operation 3 of Job 2). + +The objective is to determine the start times for each operation on each machine such that all precedence and machine capacity constraints are satisfied, and the total completion time (makespan) is minimized. This involves deciding the order of operations on shared machines and ensuring no overlaps, while respecting the processing times and sequence constraints of each job.22The problem involves scheduling operations for three jobs (Job 0, Job 1, and Job 2) on shared machines to minimize the total completion time (makespan). Each job consists of a sequence of operations with specific processing times: Job 0 has two operations (Operation 0 takes 9 units of time, Operation 1 takes 8 units), Job 1 has three operations (Operation 0 takes 8 units, Operation 1 takes 8 units, Operation 2 takes 4 units), and Job 2 has four operations (Operation 0 takes 1 unit, Operation 1 takes 6 units, Operation 2 takes 4 units, Operation 3 takes 2 units). Operations within a job must follow a strict order, meaning each operation cannot start until the previous one in the same job has finished. For example, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 has completed after 9 units of time, Operation 1 of Job 1 cannot start until Operation 0 of Job 1 has completed after 8 units of time, and Operation 2 of Job 1 cannot start until Operation 1 of Job 1 has completed after 8 units of time. Similarly, Operation 1 of Job 2 cannot start until Operation 0 of Job 2 has completed after 1 unit of time, Operation 2 of Job 2 cannot start until Operation 1 of Job 2 has completed after 6 units of time, and Operation 3 of Job 2 cannot start until Operation 2 of Job 2 has completed after 4 units of time. + +Certain operations from different jobs share the same machine and cannot overlap in time: Operation 1 of Job 0 and Operation 1 of Job 2 are assigned to the same machine, so one must finish before the other starts. Similarly, Operation 0 of Job 0 and Operation 1 of Job 1 share a machine, and their schedules must not overlap. This also applies to the following additional pairs of operations: Operation 0 of Job 0 and Operation 0 of Job 2, Operation 1 of Job 1 and Operation 0 of Job 2, Operation 0 of Job 1 and Operation 2 of Job 2, Operation 0 of Job 1 and Operation 3 of Job 2, Operation 2 of Job 1 and Operation 2 of Job 2, Operation 2 of Job 1 and Operation 3 of Job 2, and Operation 2 of Job 2 and Operation 3 of Job 2. + +To enforce these constraints, binary variables are used to determine the order of operations on shared machines. A large constant (1000) is used in the constraints to ensure that operations on the same machine do not overlap. The binary variables are restricted to values of 0 or 1, and the start times of operations are continuous and non-negative. + +The makespan must be at least as large as the completion time of every operation. For example, it must be at least 9 units of time (completion of Operation 0 of Job 0), 8 units (completion of Operation 1 of Job 0), 8 units (completion of Operation 0 of Job 1), 8 units (completion of Operation 1 of Job 1), 4 units (completion of Operation 2 of Job 1), 1 unit (completion of Operation 0 of Job 2), 6 units (completion of Operation 1 of Job 2), 4 units (completion of Operation 2 of Job 2), and 2 units (completion of Operation 3 of Job 2). + +The objective is to determine the start times for each operation on each machine such that all precedence and machine capacity constraints are satisfied, and the total completion time (makespan) is minimized. This involves deciding the order of operations on shared machines and ensuring no overlaps, while respecting the processing times and sequence constraints of each job.22The problem involves scheduling operations for three jobs (Job 0, Job 1, and Job 2) on shared machines to minimize the total completion time (makespan). Each job consists of a sequence of operations with specific processing times: Job 0 has two operations (Operation 0 takes 9 units of time, Operation 1 takes 8 units), Job 1 has three operations (Operation 0 takes 8 units, Operation 1 takes 8 units, Operation 2 takes 4 units), and Job 2 has four operations (Operation 0 takes 1 unit, Operation 1 takes 6 units, Operation 2 takes 4 units, Operation 3 takes 2 units). Operations within a job must follow a strict order, meaning each operation cannot start until the previous one in the same job has finished. For example, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 has completed after 9 units of time, Operation 1 of Job 1 cannot start until Operation 0 of Job 1 has completed after 8 units of time, and Operation 2 of Job 1 cannot start until Operation 1 of Job 1 has completed after 8 units of time. Similarly, Operation 1 of Job 2 cannot start until Operation 0 of Job 2 has completed after 1 unit of time, Operation 2 of Job 2 cannot start until Operation 1 of Job 2 has completed after 6 units of time, and Operation 3 of Job 2 cannot start until Operation 2 of Job 2 has completed after 4 units of time. + +Certain operations from different jobs share the same machine and cannot overlap in time. For instance, time: Operation 1 of Job 0 and Operation 1 of Job 2 are assigned to the same machine, so one must finish before the other starts. Similarly, Operation 0 of Job 0 and Operation 1 of Job 1 share a machine, and their schedules must not overlap. This also applies to all the following additional pairs of operations sharing the same machine, such as operations: Operation 0 of Job 0 and Operation 0 of Job 2, Operation 1 of Job 1 and Operation 0 of Job 2, Operation 0 of Job 1 and Operation 2 of Job 2, Operation 0 of Job 1 and Operation 3 of Job 2, Operation 2 of Job 1 and Operation 2 of Job 2, Operation 2 of Job 1 and Operation 3 of Job 2, and Operation 2 of Job 2 and Operation 3 of Job 2. + +To enforce these constraints, binary variables are used to determine the order of operations on shared machines. A large constant (100000) (1000) is used in the constraints to ensure that operations on the same machine do not overlap. For example, if Operation 1 of Job 0 and Operation 1 of Job 2 are on the same machine, the constraints ensure that one operation starts only after the other has finished, using the binary variable to decide the order. The binary variables are restricted to values of 0 or 1, and the start times of operations are continuous and non-negative. + +The makespan must be at least as large as the completion time of every operation. For example, it must be at least 9 units of time (completion of Operation 0 of Job 0), 8 units (completion of Operation 1 of Job 0), 8 units (completion of Operation 0 of Job 1), 8 units (completion of Operation 1 of Job 1), 4 units (completion of Operation 2 of Job 1), 1 unit (completion of Operation 0 of Job 2), 6 units (completion of Operation 1 of Job 2), 4 units (completion of Operation 2 of Job 2), and 2 units (completion of Operation 3 of Job 2).No differencesOriginal sentences like "For instance, Operation 1 of Job 0 and Operation 1 of Job 2 are assigned to the same machine" indicated that the description was incomplete (i.e., there could be other operations sharing machines). We made the statement clearer without adding/removing dataambiguous
5Imagine you are managing a supply chain network for a company that operates across three key locations: **Node 0**, **Node 1**, and **Node 2**. Each location has specific supply and demand requirements, and your goal is to optimize the flow of goods between these nodes while minimizing the total cost of establishing connections and shipping products. + +#### Supply and Demand: +- **Node 0** has a net supply of **236 units**, meaning it can provide this amount to other nodes. +- **Node 1** has a net demand of **29 units**, requiring this amount to be fulfilled from other nodes. +- **Node 2** has a net demand of **207 units**, which must also be met through shipments from other nodes. + +#### Shipping Costs and Capacities: +To move goods between nodes, you must decide whether to establish a connection (which incurs a fixed cost) and how much to ship (which incurs a variable cost per unit). The costs and capacities for each possible connection are as follows: + +1. **Node 0 to Node 1**: + - Fixed cost to establish the connection: **$3,532** + - Variable shipping cost per unit: **$45** + - Maximum shipping capacity: **773 units** + +2. **Node 0 to Node 2**: + - Fixed cost: **$3,895** + - Variable shipping cost: **$49** + - Maximum capacity: **973 units** + +3. **Node 1 to Node 0**: + - Fixed cost: **$3,485** + - Variable shipping cost: **$11** + - Maximum capacity: **785 units** + +4. **Node 1 to Node 2**: + - Fixed cost: **$1,845** + - Variable shipping cost: **$40** + - Maximum capacity: **787 units** + +5. **Node 2 to Node 0**: + - Fixed cost: **$1,291** + - Variable shipping cost: **$13** + - Maximum capacity: **602 units** + +6. **Node 2 to Node 1**: + - Fixed cost: **$1,818** + - Variable shipping cost: **$47** + - Maximum capacity: **766 units** + +#### Key Decisions: +- For each connection between nodes, you must decide whether to establish it (a binary decision: yes or no). +- For each established connection, you must determine the amount of goods to ship, ensuring it does not exceed the connection's capacity. + +#### Objective: +Your goal is to minimize the total cost, which includes: +1. The fixed costs of establishing the necessary connections. +2. The variable shipping costs based on the amount of goods transported. + +#### Constraints: +1. **Capacity Constraints**: The amount of goods shipped between any two nodes cannot exceed the maximum capacity of that connection. For example, if you decide to ship goods from **Node 0 to Node 1**, the shipment must be **773 units or less**. +2. **Flow Balance Constraints**: The total inflow and outflow of goods at each node must match its supply or demand. For example: + - At **Node 0**, the total inflow from **Node 1** and **Node 2** minus the outflow to **Node 1** and **Node 2** must equal **236 units**. + - At **Node 1**, the total inflow from **Node 0** and **Node 2** minus the outflow to **Node 0** and **Node 2** must equal **-29 units** (indicating a net demand). + - At **Node 2**, the total inflow from **Node 0** and **Node 1** minus the outflow to **Node 0** and **Node 1** must equal **-207 units** (indicating a net demand). + +#### Summary: +Your task is to determine the most cost-effective way to establish connections and ship goods between **Node 0**, **Node 1**, and **Node 2** to meet the supply and demand requirements while minimizing the total cost. This involves balancing fixed and variable costs, adhering to capacity limits, and ensuring the flow of goods aligns with the needs of each node.7364Imagine you are managing a supply chain network for a company that operates across three key locations: **Node 0**, **Node 1**, and **Node 2**. Each location has specific supply and demand requirements, and your goal is to optimize the flow of goods between these nodes while minimizing the total cost of establishing connections and shipping products. + +#### Supply and Demand: +- **Node 0** has a net supply of **236 units**, meaning it can provide this amount to other nodes. +- **Node 1** has a net demand of **29 units**, requiring this amount to be fulfilled from other nodes. +- **Node 2** has a net demand of **207 units**, which must also be met through shipments from other nodes. + +#### Shipping Costs and Capacities: +To move goods between nodes, you must decide whether to establish a connection (which incurs a fixed cost) and how much to ship (which incurs a variable cost per unit). The costs and capacities for each possible connection are as follows: + +1. **Node 0 to Node 1**: + - Fixed cost to establish the connection: **$3,532** + - Variable shipping cost per unit: **$45** + - Maximum shipping capacity: **773 units** + +2. **Node 0 to Node 2**: + - Fixed cost: **$3,895** + - Variable shipping cost: **$49** + - Maximum capacity: **973 units** + +3. **Node 1 to Node 0**: + - Fixed cost: **$3,485** + - Variable shipping cost: **$11** + - Maximum capacity: **785 units** + +4. **Node 1 to Node 2**: + - Fixed cost: **$1,845** + - Variable shipping cost: **$40** + - Maximum capacity: **787 units** + +5. **Node 2 to Node 0**: + - Fixed cost: **$1,291** + - Variable shipping cost: **$13** + - Maximum capacity: **602 units** + +6. **Node 2 to Node 1**: + - Fixed cost: **$1,818** + - Variable shipping cost: **$47** + - Maximum capacity: **766 units** + +#### Key Decisions: +- For each connection between nodes, you must decide whether to establish it (a binary decision: yes or no). +- For each established connection, you must determine the amount of goods to ship, ensuring it does not exceed the connection's capacity. + +#### Objective: +Your goal is to minimize the total cost, which includes: +1. The fixed costs of establishing the necessary connections. +2. The variable shipping costs based on the amount of goods transported. + +#### Constraints: +1. **Capacity Constraints**: The amount of goods shipped between any two nodes cannot exceed the maximum capacity of that connection. For example, if you decide to ship goods from **Node 0 to Node 1**, the shipment must be **773 units or less**. +2. **Flow Balance Constraints**: The total inflow and outflow of goods at each node must match its supply or demand. For example: + - At **Node 0**, the total inflow from **Node 1** and **Node 2** minus the outflow to **Node 1** and **Node 2** must equal **236 units**. + - At **Node 1**, the total inflow from **Node 0** and **Node 2** minus the outflow to **Node 0** and **Node 2** must equal **-29 units** (indicating a net demand). + - At **Node 2**, the total inflow from **Node 0** and **Node 1** minus the outflow to **Node 0** and **Node 1** must equal **-207 units** (indicating a net demand). + +#### Summary: +Your task is to determine the most cost-effective way to establish connections and ship goods between **Node 0**, **Node 1**, and **Node 2** to meet the supply and demand requirements while minimizing the total cost. This involves balancing fixed and variable costs, adhering to capacity limits, and ensuring the flow of goods aligns with the needs of each node.18640No differences7364 18640Ground truth was wrong (just the variable costs are more than the initially provided ground truth)incorrect ground truth
6The problem involves scheduling the landing times of four aircraft (aircraft_0, aircraft_1, aircraft_2, aircraft_3) at an airport to minimize the total penalty costs associated with deviations from their earliest possible landing times while adhering to separation and timing constraints. Each aircraft has an earliest and latest possible landing time, and penalty costs for landing early or late. Aircraft_0 has an earliest time of 0 minutes, latest time of 37 minutes, early penalty of \$61 per minute, and late penalty of \$82 per minute. Aircraft_1 has an earliest time of 7 minutes, latest time of 67 minutes, early penalty of \$35 per minute, and late penalty of \$34 per minute. Aircraft_2 has an earliest time of 225 minutes, latest time of 285 minutes, early penalty of \$74 per minute, and late penalty of \$41 per minute. Aircraft_3 has an earliest time of 119 minutes, latest time of 179 minutes, early penalty of \$93 per minute, and late penalty of \$56 per minute. + +The landing order between any two aircraft must be explicitly determined using binary decision variables, ensuring one lands before the other. Separation constraints are enforced based on these landing orders, with coefficients modifying the required separation time depending on the landing sequence. For example, if aircraft_0 lands before aircraft_1, the separation constraint is adjusted by coefficients such as 2 and 30 depending on the order. Similar adjustments apply to other pairs: aircraft_0 and aircraft_2, aircraft_0 and aircraft_3, aircraft_1 and aircraft_2, aircraft_1 and aircraft_3, and aircraft_2 and aircraft_3. These constraints ensure that the separation between consecutive landings is maintained according to the specific coefficients provided in the LP data. + +Each aircraft must land within its specified earliest and latest possible landing times. Early deviations are calculated as the difference between the earliest time and the actual landing time if the aircraft lands early, and late deviations are calculated as the difference between the actual landing time and the earliest time if the aircraft lands late. The total penalty cost is the sum of the early and late penalties for all aircraft, weighted by their respective penalty rates. The objective is to determine the optimal landing times for all four aircraft to minimize the total penalty costs while ensuring all separation and timing constraints are satisfied.0The problem involves scheduling the landing times of four aircraft (aircraft_0, aircraft_1, aircraft_2, aircraft_3) at an airport to minimize the total penalty costs associated with deviations from their earliest possible landing times while adhering to separation and timing constraints. Each aircraft has an earliest and latest possible landing time, and penalty costs for landing early or late. +Aircraft_0 has an earliest time of 0 minutes, latest time of 37 minutes, early penalty of \$61 per minute, and late penalty of \$82 per minute. +Aircraft_1 has an earliest time of 7 minutes, latest time of 67 minutes, early penalty of \$35 per minute, and late penalty of \$34 per minute. +Aircraft_2 has an earliest time of 225 minutes, latest time of 285 minutes, early penalty of \$74 per minute, and late penalty of \$41 per minute. +Aircraft_3 has an earliest time of 119 minutes, latest time of 179 minutes, early penalty of \$93 per minute, and late penalty of \$56 per minute. + +The landing order between any two aircraft must be explicitly determined using binary decision variables, ensuring one lands before the other. Separation constraints are enforced based on these landing orders, with two different separation times for each unordered pair {i,j} depending on which aircraft lands first. The required separation times (in minutes) are: + +• Pair (0,1): + – If aircraft_0 lands before aircraft_1, separation = 30 + – If aircraft_1 lands before aircraft_0, separation = 2 + +• Pair (0,2): + – If aircraft_0 lands before aircraft_2, separation = 30 + – If aircraft_2 lands before aircraft_0, separation = 2 + +• Pair (0,3): + – If aircraft_0 lands before aircraft_3, separation = 30 + – If aircraft_3 lands before aircraft_0, separation = 2 + +• Pair (1,2): + – If aircraft_1 lands before aircraft_2, separation = 20 + – If aircraft_2 lands before aircraft_1, separation = 3 + +• Pair (1,3): + – If aircraft_1 lands before aircraft_3, separation = 25 + – If aircraft_3 lands before aircraft_1, separation = 4 + +• Pair (2,3): + – If aircraft_2 lands before aircraft_3, separation = 15 + – If aircraft_3 lands before aircraft_2, separation = 2 + +These constraints ensure that the separation between consecutive landings is maintained according to the specific coefficients provided above. + +Each aircraft should land within its specified earliest and latest possible landing times. Early deviations are calculated as the difference between the earliest time and the actual landing time if the aircraft lands early, and late deviations are calculated as the difference between the actual landing time and the earliest time if the aircraft lands late. The total penalty cost is the sum of the early and late penalties for all aircraft. The objective is to determine the optimal landing times for all four aircraft to minimize the total penalty costs while ensuring all separation and timing constraints are satisfied.738The problem involves scheduling the landing times of four aircraft (aircraft_0, aircraft_1, aircraft_2, aircraft_3) at an airport to minimize the total penalty costs associated with deviations from their earliest possible landing times while adhering to separation and timing constraints. Each aircraft has an earliest and latest possible landing time, and penalty costs for landing early or late. +Aircraft_0 has an earliest time of 0 minutes, latest time of 37 minutes, early penalty of \$61 per minute, and late penalty of \$82 per minute. +Aircraft_1 has an earliest time of 7 minutes, latest time of 67 minutes, early penalty of \$35 per minute, and late penalty of \$34 per minute. +Aircraft_2 has an earliest time of 225 minutes, latest time of 285 minutes, early penalty of \$74 per minute, and late penalty of \$41 per minute. +Aircraft_3 has an earliest time of 119 minutes, latest time of 179 minutes, early penalty of \$93 per minute, and late penalty of \$56 per minute. + +The landing order between any two aircraft must be explicitly determined using binary decision variables, ensuring one lands before the other. Separation constraints are enforced based on these landing orders, with coefficients modifying the required two different separation time times for each unordered pair {i,j} depending on the landing sequence. For example, if aircraft_0 which aircraft lands before aircraft_1, the separation constraint is adjusted by coefficients such as 2 and 30 depending on the order. Similar adjustments apply to other pairs: aircraft_0 and aircraft_2, aircraft_0 and aircraft_3, aircraft_1 and aircraft_2, aircraft_1 and aircraft_3, and aircraft_2 and aircraft_3. These constraints ensure that the first. The required separation between consecutive landings is maintained according to the specific coefficients provided in the LP data. times (in minutes) are: + + • Pair (0,1): +  – If aircraft_0 lands before aircraft_1, separation = 30 +  – If aircraft_1 lands before aircraft_0, separation = 2 + + • Pair (0,2): +  – If aircraft_0 lands before aircraft_2, separation = 30 +  – If aircraft_2 lands before aircraft_0, separation = 2 + + • Pair (0,3): +  – If aircraft_0 lands before aircraft_3, separation = 30 +  – If aircraft_3 lands before aircraft_0, separation = 2 + + • Pair (1,2): +  – If aircraft_1 lands before aircraft_2, separation = 20 +  – If aircraft_2 lands before aircraft_1, separation = 3 + + • Pair (1,3): +  – If aircraft_1 lands before aircraft_3, separation = 25 +  – If aircraft_3 lands before aircraft_1, separation = 4 + + • Pair (2,3): +  – If aircraft_2 lands before aircraft_3, separation = 15 +  – If aircraft_3 lands before aircraft_2, separation = 2 + + These constraints ensure that the separation between consecutive landings is maintained according to the specific coefficients provided above. + +Each aircraft must should land within its specified earliest and latest possible landing times. Early deviations are calculated as the difference between the earliest time and the actual landing time if the aircraft lands early, and late deviations are calculated as the difference between the actual landing time and the earliest time if the aircraft lands late. The total penalty cost is the sum of the early and late penalties for all aircraft, weighted by their respective penalty rates. aircraft. The objective is to determine the optimal landing times for all four aircraft to minimize the total penalty costs while ensuring all separation and timing constraints are satisfied.738Missing data about the required separation between landingsmissing data
7The problem involves scheduling the landings of six aircraft to minimize the total penalty costs associated with deviations from their target landing times. Each aircraft has a target landing time, an earliest possible landing time, and a latest possible landing time. Penalties are incurred for landing early or late, with specific penalty rates for each aircraft: aircraft_0 has an early penalty rate of 72 and a late penalty rate of 59, aircraft_1 has an early penalty rate of 81 and a late penalty rate of 40, aircraft_2 has an early penalty rate of 46 and a late penalty rate of 34, aircraft_3 has an early penalty rate of 78 and a late penalty rate of 24, aircraft_4 has an early penalty rate of 40 and a late penalty rate of 80, and aircraft_5 has an early penalty rate of 60 and a late penalty rate of 14. The landing times must respect minimum separation times between consecutive landings, which vary depending on the pair of aircraft. The separation constraints are enforced using binary decision variables that determine the order in which the aircraft land. The objective is to minimize the sum of early and late landing penalties, weighted by the respective penalty rates for each aircraft. The landing times must fall within the specified time windows for each aircraft: aircraft_0 must land between 221 and 281, aircraft_1 between 179 and 239, aircraft_2 between 268 and 300, aircraft_3 between 0 and 36, aircraft_4 between 131 and 191, and aircraft_5 between 265 and 300. The order of landings must ensure that the separation constraints are satisfied, with specific separation times for each pair of aircraft. The binary decision variables ensure that the separation constraints are met by determining the sequence of landings. The problem requires determining the exact landing times for each aircraft within their respective time windows, ensuring that the separation requirements are met, and minimizing the total penalty costs. The binary variables are used to enforce the separation constraints, ensuring that if one aircraft lands before another, the required separation time is maintained. The constraints also ensure that the landing times fall within the specified time windows and that the penalties for early or late landings are correctly calculated based on the deviation from the target landing times. The binary variables are explicitly defined as binary (0 or 1) to determine the sequence of landings and enforce the separation constraints.34The problem involves scheduling the landings of six aircraft to minimize the total penalty costs associated with deviations from their target landing times. Each aircraft has a target landing time, an earliest possible landing time, and a latest possible landing time. + +• Target landing times (minutes): +– aircraft_0: 244 +– aircraft_1: 192 +– aircraft_2: 293 +– aircraft_3: 28 +– aircraft_4: 150 +– aircraft_5: 287 + +Penalties are incurred for landing early or late, with specific penalty rates for each aircraft: aircraft_0 has an early penalty rate of 72 and a late penalty rate of 59, aircraft_1 has an early penalty rate of 81 and a late penalty rate of 40, aircraft_2 has an early penalty rate of 46 and a late penalty rate of 34, aircraft_3 has an early penalty rate of 78 and a late penalty rate of 24, aircraft_4 has an early penalty rate of 40 and a late penalty rate of 80, and aircraft_5 has an early penalty rate of 60 and a late penalty rate of 14. + +The landing times must respect minimum separation times between consecutive landings, which vary depending on the pair of aircraft. The required minimum separation times s_ij (in minutes, where i lands before j) are: + +j → 0 1 2 3 4 5 +i ↓ +0 0 29 24 31 26 25 +1 29 0 35 29 28 29 +2 24 35 0 29 26 27 +3 31 29 29 0 29 24 +4 26 28 26 29 0 25 +5 25 29 27 24 25 0 + +The separation constraints are enforced using binary decision variables that determine the order in which the aircraft land. The objective is to minimize the sum of early and late landing penalties, weighted by the respective penalty rates for each aircraft. The landing times must fall within the specified time windows for each aircraft: aircraft_0 must land between 221 and 281, aircraft_1 between 179 and 239, aircraft_2 between 268 and 300, aircraft_3 between 0 and 36, aircraft_4 between 131 and 191, and aircraft_5 between 265 and 300. The order of landings must ensure that the separation constraints are satisfied, with the specific separation times shown above for each pair of aircraft. The binary decision variables ensure that the separation constraints are met by determining the sequence of landings. + +The problem requires determining the exact landing times for each aircraft within their respective time windows, ensuring that the separation requirements are met, and minimizing the total penalty costs. The binary variables are used to enforce the separation constraints, ensuring that if one aircraft lands before another, the required separation time is maintained. The constraints also ensure that the landing times fall within the specified time windows and that the penalties for early or late landings are correctly calculated based on the deviation from the target landing times. The binary variables are explicitly defined as binary (0 or 1) to determine the sequence of landings and enforce the separation constraints.1078The problem involves scheduling the landings of six aircraft to minimize the total penalty costs associated with deviations from their target landing times. Each aircraft has a target landing time, an earliest possible landing time, and a latest possible landing time. + + • Target landing times (minutes): + – aircraft_0: 244 + – aircraft_1: 192 + – aircraft_2: 293 + – aircraft_3: 28 + – aircraft_4: 150 + – aircraft_5: 287 + +Penalties are incurred for landing early or late, with specific penalty rates for each aircraft: aircraft_0 has an early penalty rate of 72 and a late penalty rate of 59, aircraft_1 has an early penalty rate of 81 and a late penalty rate of 40, aircraft_2 has an early penalty rate of 46 and a late penalty rate of 34, aircraft_3 has an early penalty rate of 78 and a late penalty rate of 24, aircraft_4 has an early penalty rate of 40 and a late penalty rate of 80, and aircraft_5 has an early penalty rate of 60 and a late penalty rate of 14. + +The landing times must respect minimum separation times between consecutive landings, which vary depending on the pair of aircraft. The required minimum separation times s_ij (in minutes, where i lands before j) are: + + j → 0 1 2 3 4 5 + i ↓ + 0 0 29 24 31 26 25 + 1 29 0 35 29 28 29 + 2 24 35 0 29 26 27 + 3 31 29 29 0 29 24 + 4 26 28 26 29 0 25 + 5 25 29 27 24 25 0 + + The separation constraints are enforced using binary decision variables that determine the order in which the aircraft land. The objective is to minimize the sum of early and late landing penalties, weighted by the respective penalty rates for each aircraft. The landing times must fall within the specified time windows for each aircraft: aircraft_0 must land between 221 and 281, aircraft_1 between 179 and 239, aircraft_2 between 268 and 300, aircraft_3 between 0 and 36, aircraft_4 between 131 and 191, and aircraft_5 between 265 and 300. The order of landings must ensure that the separation constraints are satisfied, with the specific separation times shown above for each pair of aircraft. The binary decision variables ensure that the separation constraints are met by determining the sequence of landings. + +The problem requires determining the exact landing times for each aircraft within their respective time windows, ensuring that the separation requirements are met, and minimizing the total penalty costs. The binary variables are used to enforce the separation constraints, ensuring that if one aircraft lands before another, the required separation time is maintained. The constraints also ensure that the landing times fall within the specified time windows and that the penalties for early or late landings are correctly calculated based on the deviation from the target landing times. The binary variables are explicitly defined as binary (0 or 1) to determine the sequence of landings and enforce the separation constraints.34 1078Missing data about target landing times and pairwise separation values.missing data
8The manufacturing facility produces custom orders for three jobs: Job 0, Job 1, and Job 2. Each job consists of a sequence of operations that must be performed in a specific order, with each operation requiring a specific duration. The facility has limited machines, and some operations from different jobs must share the same machine. The goal is to schedule all operations to minimize the total completion time (makespan). + +Job 0 has two operations: Operation 0 and Operation 1. Operation 1 cannot start until Operation 0 has finished, with at least 8 hours between their start times. The makespan must be at least 8 hours after the start of Operation 0 and at least 5 hours after the start of Operation 1. + +Job 1 has five operations: Operation 0, Operation 1, Operation 2, Operation 3, and Operation 4. Operation 1 cannot start until Operation 0 has finished, with at least 7 hours between their start times. Operation 2 cannot start until Operation 1 has finished, with at least 9 hours between their start times. Operation 3 cannot start until Operation 2 has finished, with at least 5 hours between their start times. Operation 4 cannot start until Operation 3 has finished, with at least 6 hours between their start times. The makespan must be at least 7 hours after the start of Operation 0, 9 hours after the start of Operation 1, 5 hours after the start of Operation 2, 6 hours after the start of Operation 3, and 2 hours after the start of Operation 4. + +Job 2 has five operations: Operation 0, Operation 1, Operation 2, Operation 3, and Operation 4. Operation 1 cannot start until Operation 0 has finished, with at least 9 hours between their start times. Operation 2 cannot start until Operation 1 has finished, with at least 5 hours between their start times. Operation 3 cannot start until Operation 2 has finished, with at least 8 hours between their start times. Operation 4 cannot start until Operation 3 has finished, with at least 3 hours between their start times. The makespan must be at least 9 hours after the start of Operation 0, 5 hours after the start of Operation 1, 8 hours after the start of Operation 2, 3 hours after the start of Operation 3, and 1 hour after the start of Operation 4. + +Certain operations from different jobs share machines and must be scheduled without overlap. The following pairs of operations share machines: Operation 0 of Job 1 and Operation 2 of Job 1; Operation 1 of Job 0 and Operation 3 of Job 2; Operation 0 of Job 0 and Operation 1 of Job 2; Operation 0 of Job 0 and Operation 2 of Job 2; Operation 1 of Job 2 and Operation 2 of Job 2; Operation 4 of Job 1 and Operation 0 of Job 2. The scheduling system must determine the order of these operations to avoid conflicts, using binary variables to enforce the order. A large constant (100000) is used in the constraints to ensure that the order of operations sharing machines is respected. + +The objective is to determine the start times for all operations such that all precedence constraints within each job are satisfied, no two operations sharing the same machine overlap in time, and the total completion time (makespan) is minimized. The makespan must be at least as large as the completion time of every operation. All start times must be non-negative. The binary variables are used to enforce the order of operations sharing machines, and the large constant ensures that the constraints are properly applied. The start times and the makespan are general variables, and the objective is explicitly to minimize the makespan.29The manufacturing facility produces custom orders for three jobs: Job 0, Job 1, and Job 2. Each job consists of a sequence of operations that must be performed in a specific order, with each operation requiring a specific duration. The facility has limited machines, and some operations from different jobs must share the same machine. The goal is to schedule all operations to minimize the total completion time (makespan). + +Job 0 has two operations: Operation 0 and Operation 1. Operation 1 cannot start until Operation 0 has finished, with at least 8 hours between their start times. The makespan must be at least 8 hours after the start of Operation 0 and at least 5 hours after the start of Operation 1. + +Job 1 has five operations: Operation 0, Operation 1, Operation 2, Operation 3, and Operation 4. Operation 1 cannot start until Operation 0 has finished, with at least 7 hours between their start times. Operation 2 cannot start until Operation 1 has finished, with at least 9 hours between their start times. Operation 3 cannot start until Operation 2 has finished, with at least 5 hours between their start times. Operation 4 cannot start until Operation 3 has finished, with at least 6 hours between their start times. The makespan must be at least 7 hours after the start of Operation 0, 9 hours after the start of Operation 1, 5 hours after the start of Operation 2, 6 hours after the start of Operation 3, and 2 hours after the start of Operation 4. + +Job 2 has five operations: Operation 0, Operation 1, Operation 2, Operation 3, and Operation 4. Operation 1 cannot start until Operation 0 has finished, with at least 9 hours between their start times. Operation 2 cannot start until Operation 1 has finished, with at least 5 hours between their start times. Operation 3 cannot start until Operation 2 has finished, with at least 8 hours between their start times. Operation 4 cannot start until Operation 3 has finished, with at least 3 hours between their start times. The makespan must be at least 9 hours after the start of Operation 0, 5 hours after the start of Operation 1, 8 hours after the start of Operation 2, 3 hours after the start of Operation 3, and 1 hour after the start of Operation 4. + +Certain operations from different jobs share machines and must be scheduled without overlap. The following pairs of operations share machines: Operation 0 of Job 1 and Operation 2 of Job 1; Operation 1 of Job 0 and Operation 3 of Job 2; Operation 0 of Job 0 and Operation 1 of Job 2; Operation 0 of Job 0 and Operation 2 of Job 2; Operation 1 of Job 2 and Operation 2 of Job 2; Operation 4 of Job 1 and Operation 0 of Job 2. The scheduling system must determine the order of these operations to avoid conflicts, using binary variables to enforce the order. A large constant (100000) is used in the constraints to ensure that the order of operations sharing machines is respected. + +The objective is to determine the start times for all operations such that all precedence constraints within each job are satisfied, no two operations sharing the same machine overlap in time, and the total completion time (makespan) is minimized. The makespan must be at least as large as the completion time of every operation. All start times must be non-negative. The binary variables are used to enforce the order of operations sharing machines, and the large constant ensures that the constraints are properly applied. The start times and the makespan are general variables, and the objective is explicitly to minimize the makespan.29No differencesNo differencesok
9Determine the optimal sequence of six cities (City 0, City 1, City 2, City 3, City 4, and City 5) to minimize the total travel distance while ensuring each city is visited exactly once. The distances between the cities are as follows: From City 0 to City 1 (722 units), City 2 (758 units), City 3 (688 units), City 4 (897 units), and City 5 (100 units); from City 1 to City 0 (746 units), City 2 (76 units), City 3 (49 units), City 4 (86 units), and City 5 (385 units); from City 2 to City 0 (479 units), City 1 (154 units), City 3 (11 units), City 4 (374 units), and City 5 (350 units); from City 3 to City 0 (645 units), City 1 (842 units), City 2 (117 units), City 4 (619 units), and City 5 (535 units); from City 4 to City 0 (278 units), City 1 (829 units), City 2 (499 units), City 3 (268 units), and City 5 (973 units); from City 5 to City 0 (951 units), City 1 (144 units), City 2 (810 units), City 3 (943 units), and City 4 (989 units). Each city must be visited exactly once, and each city must be departed exactly once. To prevent subtours, the sequence of cities must be continuous, and the position of each city in the sequence must be between 0 and 5. If a route is taken from city i to city j, the position of city i must be less than the position of city j. The decision variables include selecting whether to include each route (binary) and assigning a position to each city in the sequence (integer). The position variables u[city_0] to u[city_5] must satisfy the constraints that ensure the sequence is continuous and the position of each city is between 0 and 5. The goal is to minimize the total distance traveled while satisfying all constraints, including the subtour elimination constraints that link the route variables and the position variables. The constraints R12 to R41 ensure that if a route is taken from city i to city j, the position of city i must be less than the position of city j, and the position variables u[city_1] to u[city_5] are bounded between 0 and 5. The route variables are binary, and the position variables are general (integer) variables.1062Determine the optimal sequence of six cities (City 0, City 1, City 2, City 3, City 4, and City 5) to minimize the total travel distance while ensuring each city is visited exactly once and return to the origin city. The distances between the cities are as follows: From City 0 to City 1 (722 units), City 2 (758 units), City 3 (688 units), City 4 (897 units), and City 5 (100 units); from City 1 to City 0 (746 units), City 2 (76 units), City 3 (49 units), City 4 (86 units), and City 5 (385 units); from City 2 to City 0 (479 units), City 1 (154 units), City 3 (11 units), City 4 (374 units), and City 5 (350 units); from City 3 to City 0 (645 units), City 1 (842 units), City 2 (117 units), City 4 (619 units), and City 5 (535 units); from City 4 to City 0 (278 units), City 1 (829 units), City 2 (499 units), City 3 (268 units), and City 5 (973 units); from City 5 to City 0 (951 units), City 1 (144 units), City 2 (810 units), City 3 (943 units), and City 4 (989 units). The task is to find the shortest possible route that visits each city exactly once, and returns to the origin city. What is the minimal total distance travelled?1062Determine the optimal sequence of six cities (City 0, City 1, City 2, City 3, City 4, and City 5) to minimize the total travel distance while ensuring each city is visited exactly once. once and return to the origin city. The distances between the cities are as follows: From City 0 to City 1 (722 units), City 2 (758 units), City 3 (688 units), City 4 (897 units), and City 5 (100 units); from City 1 to City 0 (746 units), City 2 (76 units), City 3 (49 units), City 4 (86 units), and City 5 (385 units); from City 2 to City 0 (479 units), City 1 (154 units), City 3 (11 units), City 4 (374 units), and City 5 (350 units); from City 3 to City 0 (645 units), City 1 (842 units), City 2 (117 units), City 4 (619 units), and City 5 (535 units); from City 4 to City 0 (278 units), City 1 (829 units), City 2 (499 units), City 3 (268 units), and City 5 (973 units); from City 5 to City 0 (951 units), City 1 (144 units), City 2 (810 units), City 3 (943 units), and City 4 (989 units). Each city must be visited exactly once, and each city must be departed exactly once. To prevent subtours, the sequence of cities must be continuous, and the position of each city in the sequence must be between 0 and 5. If a route The task is taken from city i to city j, the position of city i must be less than find the position of city j. The decision variables include selecting whether to include each shortest possible route (binary) and assigning a position to each city in the sequence (integer). The position variables u[city_0 to u[city_5] must satisfy the constraints] that ensure the sequence is continuous and the position of visits each city is between 0 exactly once, and 5. The goal is returns to minimize the total distance traveled while satisfying all constraints, including the subtour elimination constraints that link the route variables and the position variables. The constraints R12 to R41 ensure that if a route origin city. What is taken from city i to city j, the position of city i must be less than the position of city j, and the position variables u[city_1 to u[city_5] are bounded between 0 and 5. The route variables are binary, and] the position variables are general (integer) variables. minimal total distance travelled?No differencesIt mentions "ensuring each city is visited exactly once", while we need to return to the origin city. Added clarification, and also removed redundant content that is part of the solution.ambiguous
10Minimize the total cost, which includes fixed costs for establishing transit lines, operational costs based on the frequency of each line, and penalty costs for any unmet passenger demand. The transit network consists of seven potential lines (L0 to L6), each with specific fixed costs, operational costs per trip, and frequency bounds. The fixed costs are 2960 units for L0, 3834 for L1, 2998 for L2, 4562 for L3, 3636 for L4, 1972 for L5, and 1570 for L6. The operational costs per trip are 166 units for L0, 154 for L1, 133 for L2, 473 for L3, 381 for L4, 459 for L5, and 242 for L6. The frequency of each line must be between 2 and 10 trips per unit time if the line is selected. The total vehicle usage across all lines must not exceed the available fleet of 35 vehicles. Vehicle usage per trip is 66 units for L0, 113 for L1, 142 for L2, 179 for L3, 89 for L4, 106 for L5, and 129 for L6. + +For each origin-destination (OD) pair, the total capacity provided by the selected lines must meet or exceed the demand. If not, the unmet demand incurs a penalty. The demand for OD pair 4-3 is 19 passengers with a penalty of 945 units, OD pair 5-2 is 42 passengers with a penalty of 649 units, OD pair 0-1 is 33 passengers with a penalty of 605 units, OD pair 7-4 is 28 passengers with a penalty of 747 units, OD pair 1-4 is 30 passengers with a penalty of 799 units, OD pair 3-7 is 31 passengers with a penalty of 960 units, OD pair 6-3 is 27 passengers with a penalty of 558 units, OD pair 2-5 is 13 passengers with a penalty of 530 units, OD pair 4-7 is 27 passengers with a penalty of 630 units, OD pair 1-3 is 42 passengers with a penalty of 528 units, OD pair 7-1 is 42 passengers with a penalty of 695 units, OD pair 6-1 is 31 passengers with a penalty of 814 units, OD pair 5-1 is 35 passengers with a penalty of 652 units, OD pair 3-1 is 39 passengers with a penalty of 748 units, OD pair 6-2 is 20 passengers with a penalty of 786 units, OD pair 5-3 is 34 passengers with a penalty of 599 units, OD pair 2-1 is 39 passengers with a penalty of 948 units, and OD pair 6-5 is 39 passengers with a penalty of 577 units. + +A station can only be designated as a transfer station if at least two lines pass through it. For example, station 0 can only be a transfer station if line L5 is selected, station 1 can only be a transfer station if at least two of lines L3, L4, or L6 are selected, station 2 can only be a transfer station if line L0 is selected, station 3 can only be a transfer station if line L1 is selected, station 4 can only be a transfer station if line L2 is selected, station 5 can only be a transfer station if at least two of lines L2, L3, L4, L5, or L6 are selected, and stations 6 and 7 cannot be transfer stations. The binary variables y[0] to y[7] indicate whether a station is a transfer station, and constraints ensure that y[n] can only be 1 if at least two lines pass through station n. The total frequency of lines serving each OD pair must meet a minimum threshold to ensure adequate service quality. Each selected line must form a continuous route, ensuring that the sequence of stations is logically connected. Frequencies and unmet demands must be non-negative, and decisions about line selection and transfer stations must be binary (yes/no). The constraints also ensure that the frequency of a line is between 2 and 10 if the line is selected, and the total vehicle usage across all lines does not exceed 35 units.4029430Minimize the total cost, which includes fixed costs for establishing transit lines, operational costs based on the frequency of each line, and penalty costs for any unmet passenger demand. The transit network consists of seven potential lines (L0 to L6), each with specific fixed costs, operational costs per trip, and frequency bounds. The fixed costs are 2960 units for L0, 3834 for L1, 2998 for L2, 4562 for L3, 3636 for L4, 1972 for L5, and 1570 for L6. The operational costs per trip are 166 units for L0, 154 for L1, 133 for L2, 473 for L3, 381 for L4, 459 for L5, and 242 for L6. The frequency of each line must be between 2 and 10 trips per unit time if the line is selected. The total vehicle usage across all lines must not exceed the available fleet of 35 vehicles. Vehicle usage per trip is 66 units for L0, 113 for L1, 142 for L2, 179 for L3, 89 for L4, 106 for L5, and 129 for L6. + + For each origin-destination (OD) pair, the total capacity provided by the selected lines must meet or exceed the demand. If not, the unmet demand incurs a penalty. The demand for OD pair 4-3 is 19 passengers with a penalty of 945 units, OD pair 5-2 is 42 passengers with a penalty of 649 units, OD pair 0-1 is 33 passengers with a penalty of 605 units, OD pair 7-4 is 28 passengers with a penalty of 747 units, OD pair 1-4 is 30 passengers with a penalty of 799 units, OD pair 3-7 is 31 passengers with a penalty of 960 units, OD pair 6-3 is 27 passengers with a penalty of 558 units, OD pair 2-5 is 13 passengers with a penalty of 530 units, OD pair 4-7 is 27 passengers with a penalty of 630 units, OD pair 1-3 is 42 passengers with a penalty of 528 units, OD pair 7-1 is 42 passengers with a penalty of 695 units, OD pair 6-1 is 31 passengers with a penalty of 814 units, OD pair 5-1 is 35 passengers with a penalty of 652 units, OD pair 3-1 is 39 passengers with a penalty of 748 units, OD pair 6-2 is 20 passengers with a penalty of 786 units, OD pair 5-3 is 34 passengers with a penalty of 599 units, OD pair 2-1 is 39 passengers with a penalty of 948 units, and OD pair 6-5 is 39 passengers with a penalty of 577 units. + + A station can only be designated as a transfer station if at least two lines pass through it. For example, station 0 can only be a transfer station if line L5 is selected, station 1 can only be a transfer station if at least two of lines L3, L4, or L6 are selected, station 2 can only be a transfer station if line L0 is selected, station 3 can only be a transfer station if line L1 is selected, station 4 can only be a transfer station if line L2 is selected, station 5 can only be a transfer station if at least two of lines L2, L3, L4, L5, or L6 are selected, and stations 6 and 7 cannot be transfer stations. The binary variables y[0 to y[7] indicate whether a station is a transfer station, and constraints ensure that y[n] can only be 1 if at least two lines pass through station n. The total frequency of lines serving each OD pair must meet a minimum threshold to ensure adequate service quality. Each selected line must form a continuous route, ensuring that the sequence of stations is logically connected. Frequencies and unmet demands must be non-negative, and decisions about line selection and transfer stations must be binary (yes/no). The constraints also ensure that the frequency of a line is between 2 and 10 if the line is selected, and the total vehicle usage across all lines does not exceed 35 units.]402943Very underspecified problem. For example, it is missing station-line/OD-line incidence and other information. underspecified
11A manufacturing facility produces custom orders for two jobs, Job 0 and Job 1. Each job consists of a sequence of operations that must be performed in a specific order, and some operations require the use of shared machinery. The goal is to schedule these operations to minimize the total production time, or makespan, ensuring all orders are completed as quickly as possible. + +Job 0 has three operations: Operation 0, Operation 1, and Operation 2. Operation 1 must start at least 8 units of time after Operation 0 completes, and Operation 2 must start at least 1 unit of time after Operation 1 completes. Job 1 has six operations: Operation 0, Operation 1, Operation 2, Operation 3, Operation 4, and Operation 5. Operation 1 must start at least 5 units of time after Operation 0 completes, Operation 2 must start at least 5 units of time after Operation 1 completes, Operation 3 must start at least 7 units of time after Operation 2 completes, Operation 4 must start at least 4 units of time after Operation 3 completes, and Operation 5 must start at least 2 units of time after Operation 4 completes. + +Some operations from both jobs require the use of the same machine, and only one operation can be processed on a machine at any given time. The following pairs of operations share the same machine: Operation 3 and Operation 4 of Job 1, Operation 0 and Operation 1 of Job 0, Operation 0 of Job 0 and Operation 2 of Job 1, and Operation 1 of Job 0 and Operation 2 of Job 1. For each pair of operations sharing a machine, a decision must be made about which operation is processed first. This decision is represented by a binary variable, where a value of 1 indicates that the first operation in the pair is processed before the second, and a value of 0 indicates the opposite. + +The primary objective is to minimize the total production time, or makespan, which is defined as the time at which the last operation across all jobs is completed. The makespan must be at least as large as the completion time of every operation in both jobs. Specifically, the makespan must be at least 8 units of time after Operation 0 of Job 0 completes, at least 1 unit of time after Operation 1 of Job 0 completes, at least 4 units of time after Operation 2 of Job 0 completes, at least 5 units of time after Operation 0 of Job 1 completes, at least 5 units of time after Operation 1 of Job 1 completes, at least 7 units of time after Operation 2 of Job 1 completes, at least 4 units of time after Operation 3 of Job 1 completes, at least 2 units of time after Operation 4 of Job 1 completes, and at least 9 units of time after Operation 5 of Job 1 completes. + +All operation start times must be non-negative, meaning no operation can start before time zero. For each pair of operations sharing a machine, the following must hold: if Operation A is scheduled before Operation B, then Operation B cannot start until Operation A has been completed, and conversely, if Operation B is scheduled before Operation A, then Operation A cannot start until Operation B has been completed. These constraints are enforced using a large constant (100,000) to ensure that the scheduling logic is maintained. The specific values derived from this constant, such as 99996, 99992, and 99999, are used in the constraints to enforce the sequencing logic. + +The task is to determine the optimal schedule for all operations in both jobs, ensuring that all precedence and machine capacity constraints are satisfied, and that the total production time (makespan) is minimized. This involves deciding the order of operations on shared machines and ensuring that all operations are completed as quickly as possible.32A manufacturing facility produces custom orders for two jobs, Job 0 and Job 1. Each job consists of a sequence of operations that must be performed in a specific order, and some operations require the use of shared machinery. The goal is to schedule these operations to minimize the total production time, or makespan, ensuring all orders are completed as quickly as possible. + +Job 0 has three operations: Operation 0, Operation 1, and Operation 2. Operation 1 must start at least 8 units of time after Operation 0 completes, and Operation 2 must start at least 1 unit of time after Operation 1 completes. Job 1 has six operations: Operation 0, Operation 1, Operation 2, Operation 3, Operation 4, and Operation 5. Operation 1 must start at least 5 units of time after Operation 0 completes, Operation 2 must start at least 5 units of time after Operation 1 completes, Operation 3 must start at least 7 units of time after Operation 2 completes, Operation 4 must start at least 4 units of time after Operation 3 completes, and Operation 5 must start at least 2 units of time after Operation 4 completes. + +Some operations from both jobs require the use of the same machine, and only one operation can be processed on a machine at any given time. The following pairs of operations share the same machine: Operation 3 and Operation 4 of Job 1, Operation 0 and Operation 1 of Job 0, Operation 0 of Job 0 and Operation 2 of Job 1, and Operation 1 of Job 0 and Operation 2 of Job 1. For each pair of operations sharing a machine, a decision must be made about which operation is processed first. This decision is represented by a binary variable, where a value of 1 indicates that the first operation in the pair is processed before the second, and a value of 0 indicates the opposite. + +The primary objective is to minimize the total production time, or makespan, which is defined as the time at which the last operation across all jobs is completed. The makespan must be at least as large as the completion time of every operation in both jobs. Specifically, the makespan must be at least 8 units of time after Operation 0 of Job 0 completes, at least 1 unit of time after Operation 1 of Job 0 completes, at least 4 units of time after Operation 2 of Job 0 completes, at least 5 units of time after Operation 0 of Job 1 completes, at least 5 units of time after Operation 1 of Job 1 completes, at least 7 units of time after Operation 2 of Job 1 completes, at least 4 units of time after Operation 3 of Job 1 completes, at least 2 units of time after Operation 4 of Job 1 completes, and at least 9 units of time after Operation 5 of Job 1 completes. + +All operation start times must be non-negative, meaning no operation can start before time zero. For each pair of operations sharing a machine, the following must hold: if Operation A is scheduled before Operation B, then Operation B cannot start until Operation A has been completed, and conversely, if Operation B is scheduled before Operation A, then Operation A cannot start until Operation B has been completed. These constraints are enforced using a large constant (100,000) to ensure that the scheduling logic is maintained. The specific values derived from this constant, such as 99996, 99992, and 99999, are used in the constraints to enforce the sequencing logic. + +The task is to determine the optimal schedule for all operations in both jobs, ensuring that all precedence and machine capacity constraints are satisfied, and that the total production time (makespan) is minimized. This involves deciding the order of operations on shared machines and ensuring that all operations are completed as quickly as possible.32No differencesNo differencesok
12The problem involves scheduling the landing times of three aircraft (Aircraft 0, Aircraft 1, and Aircraft 2) to minimize the total penalty costs associated with landing early or late. Each aircraft has a target landing time, and the actual landing time must fall within a specified earliest and latest possible landing time window. Aircraft 0 must land between 24 and 84 minutes, Aircraft 1 between 180 and 240 minutes, and Aircraft 2 between 215 and 275 minutes. The penalties for landing early are \$21, \$16, and \$49 per minute for Aircraft 0, 1, and 2, respectively. The penalties for landing late are \$57, \$34, and \$45 per minute for Aircraft 0, 1, and 2, respectively. The target landing times are defined such that Aircraft 0 must land no earlier than 54 minutes, Aircraft 1 no earlier than 210 minutes, and Aircraft 2 no earlier than 245 minutes. + +To ensure safe landings, separation times between aircraft must be maintained based on the order in which they land. Binary variables determine the order of landings between any two aircraft. If Aircraft 0 lands before Aircraft 1, there must be at least 5 minutes between their landings; if Aircraft 1 lands before Aircraft 0, the separation must be at least 96 minutes. If Aircraft 0 lands before Aircraft 2, the separation must be at least 4 minutes; if Aircraft 2 lands before Aircraft 0, the separation must be at least 131 minutes. If Aircraft 1 lands before Aircraft 2, the separation must be at least 5 minutes; if Aircraft 2 lands before Aircraft 1, the separation must be at least 25 minutes. Additionally, if Aircraft 0 lands before Aircraft 1, the separation must be at least 216 minutes; if Aircraft 1 lands before Aircraft 0, the separation must be at least 2 minutes. If Aircraft 0 lands before Aircraft 2, the separation must be at least 251 minutes; if Aircraft 2 lands before Aircraft 0, the separation must be at least 1 minute. If Aircraft 1 lands before Aircraft 2, the separation must be at least 95 minutes; if Aircraft 2 lands before Aircraft 1, the separation must be at least 3 minutes. + +The objective is to determine the landing times for each aircraft that minimize the total penalty costs, which are calculated as the sum of the early and late penalties for all aircraft. The early penalty for each aircraft is the number of minutes it lands before its target time multiplied by its early penalty rate, and the late penalty is the number of minutes it lands after its target time multiplied by its late penalty rate. The solution must ensure that all aircraft land within their respective time windows and that the separation times between landings are respected based on the order in which the aircraft land. The decision variables include the actual landing times for each aircraft and binary variables that determine the order of landings between any two aircraft. The goal is to find the optimal landing schedule that minimizes the total penalty costs while satisfying all constraints.0The problem involves scheduling the landing times of three aircraft (Aircraft 0, Aircraft 1, and Aircraft 2) to minimize the total penalty costs associated with landing early or late. Each aircraft has a target landing time, and the actual landing time must fall within a specified earliest and latest possible landing time window. Aircraft 0 must land between 24 and 84 minutes, Aircraft 1 between 180 and 240 minutes, and Aircraft 2 between 215 and 275 minutes. The penalties for landing early are \$21, \$16, and \$49 per minute for Aircraft 0, 1, and 2, respectively. The penalties for landing late are \$57, \$34, and \$45 per minute for Aircraft 0, 1, and 2, respectively. The target landing times are defined such that Aircraft 0 should land at 54 minutes, Aircraft 1 should land at 210 minutes, and Aircraft 2 should land at 245 minutes. + +To ensure safe landings, separation times between aircraft must be maintained based on the order in which they land. Binary variables determine the order of landings between any two aircraft. If Aircraft 0 lands before Aircraft 1, there must be at least 5 minutes between their landings; if Aircraft 1 lands before Aircraft 0, the separation must be at least 96 minutes. If Aircraft 0 lands before Aircraft 2, the separation must be at least 4 minutes; if Aircraft 2 lands before Aircraft 0, the separation must be at least 131 minutes. If Aircraft 1 lands before Aircraft 2, the separation must be at least 5 minutes; if Aircraft 2 lands before Aircraft 1, the separation must be at least 25 minutes. + +The objective is to determine the landing times for each aircraft that minimize the total penalty costs, which are calculated as the sum of the early and late penalties for all aircraft. The early penalty for each aircraft is the number of minutes it lands before its target time multiplied by its early penalty rate, and the late penalty is the number of minutes it lands after its target time multiplied by its late penalty rate. The solution must ensure that all aircraft land within their respective time windows and that the separation times between landings are respected based on the order in which the aircraft land. The decision variables include the actual landing times for each aircraft and binary variables that determine the order of landings between any two aircraft. The goal is to find the optimal landing schedule that minimizes the total penalty costs while satisfying all constraints.0The problem involves scheduling the landing times of three aircraft (Aircraft 0, Aircraft 1, and Aircraft 2) to minimize the total penalty costs associated with landing early or late. Each aircraft has a target landing time, and the actual landing time must fall within a specified earliest and latest possible landing time window. Aircraft 0 must land between 24 and 84 minutes, Aircraft 1 between 180 and 240 minutes, and Aircraft 2 between 215 and 275 minutes. The penalties for landing early are \$21, \$16, and \$49 per minute for Aircraft 0, 1, and 2, respectively. The penalties for landing late are \$57, \$34, and \$45 per minute for Aircraft 0, 1, and 2, respectively. The target landing times are defined such that Aircraft 0 must should land no earlier than at 54 minutes, Aircraft 1 no earlier than should land at 210 minutes, and Aircraft 2 no earlier than should land at 245 minutes. + +To ensure safe landings, separation times between aircraft must be maintained based on the order in which they land. Binary variables determine the order of landings between any two aircraft. If Aircraft 0 lands before Aircraft 1, there must be at least 5 minutes between their landings; if Aircraft 1 lands before Aircraft 0, the separation must be at least 96 minutes. If Aircraft 0 lands before Aircraft 2, the separation must be at least 4 minutes; if Aircraft 2 lands before Aircraft 0, the separation must be at least 131 minutes. If Aircraft 1 lands before Aircraft 2, the separation must be at least 5 minutes; if Aircraft 2 lands before Aircraft 1, the separation must be at least 25 minutes. Additionally, if Aircraft 0 lands before Aircraft 1, the separation must be at least 216 minutes; if Aircraft 1 lands before Aircraft 0, the separation must be at least 2 minutes. If Aircraft 0 lands before Aircraft 2, the separation must be at least 251 minutes; if Aircraft 2 lands before Aircraft 0, the separation must be at least 1 minute. If Aircraft 1 lands before Aircraft 2, the separation must be at least 95 minutes; if Aircraft 2 lands before Aircraft 1, the separation must be at least 3 minutes. + +The objective is to determine the landing times for each aircraft that minimize the total penalty costs, which are calculated as the sum of the early and late penalties for all aircraft. The early penalty for each aircraft is the number of minutes it lands before its target time multiplied by its early penalty rate, and the late penalty is the number of minutes it lands after its target time multiplied by its late penalty rate. The solution must ensure that all aircraft land within their respective time windows and that the separation times between landings are respected based on the order in which the aircraft land. The decision variables include the actual landing times for each aircraft and binary variables that determine the order of landings between any two aircraft. The goal is to find the optimal landing schedule that minimizes the total penalty costs while satisfying all constraints.No differencesAmbiguity in the description: It said "must land no earlier than" the target times, although that is the role of the time windows, the target times are for calculating the early/late penalties. Also, 2 conflicting values are provided for each separation time, deleted the second one.ambiguous
13In a manufacturing facility, you are tasked with scheduling the production of two distinct jobs, Job 0 and Job 1, across a set of machines. Each job consists of a sequence of operations that must be performed in a specific order, and each operation requires a certain amount of time to complete. The goal is to minimize the total production time, known as the makespan, while ensuring that all operations are scheduled without overlapping on the same machine and that the sequence of operations within each job is respected. + +Job 0 consists of two operations: Operation 0 takes 5 units of time to complete, and Operation 1 can only start after Operation 0 has finished. The makespan must be at least 5 units of time after the start of Operation 0 and at least 2 units of time after the start of Operation 1. + +Job 1 consists of five operations: Operation 0 takes 9 units of time to complete, Operation 1 takes 9 units of time and can only start after Operation 0 has finished, Operation 2 takes 7 units of time and can only start after Operation 1 has finished, Operation 3 takes 2 units of time and can only start after Operation 2 has finished, and Operation 4 takes 8 units of time and can only start after Operation 3 has finished. The makespan must be at least 9 units of time after the start of Operation 0, at least 9 units of time after the start of Operation 1, at least 7 units of time after the start of Operation 2, at least 2 units of time after the start of Operation 3, and at least 8 units of time after the start of Operation 4. + +Certain operations from Job 1 must be performed on the same machine, and their scheduling must avoid overlaps. Specifically, Operation 0 and Operation 4 of Job 1 cannot overlap on the same machine. If Operation 0 starts first, Operation 4 must start at least 8 units of time after Operation 0 finishes. Conversely, if Operation 4 starts first, Operation 0 must start at least 9 units of time after Operation 4 finishes. Similarly, Operation 2 and Operation 3 of Job 1 cannot overlap on the same machine. If Operation 2 starts first, Operation 3 must start at least 2 units of time after Operation 2 finishes. Conversely, if Operation 3 starts first, Operation 2 must start at least 7 units of time after Operation 3 finishes. + +The primary objective is to minimize the makespan, which is the total time taken to complete all operations across both jobs. The makespan must be at least as long as the completion time of the last operation in either job. All operations must start at or after time 0, and the order of operations within each job must be strictly followed. The scheduling of operations on shared machines must respect the no-overlap constraints described above. Binary variables are used to enforce the no-overlap constraints on the shared machines, ensuring that the operations do not overlap and the sequence constraints are satisfied.35In a manufacturing facility, you are tasked with scheduling the production of two distinct jobs, Job 0 and Job 1, across a set of machines. Each job consists of a sequence of operations that must be performed in a specific order, and each operation requires a certain amount of time to complete. The goal is to minimize the total production time, known as the makespan, while ensuring that all operations are scheduled without overlapping on the same machine and that the sequence of operations within each job is respected. + +Job 0 consists of two operations: Operation 0 takes 5 units of time to complete, and Operation 1 can only start after Operation 0 has finished. The makespan must be at least 5 units of time after the start of Operation 0 and at least 2 units of time after the start of Operation 1. + +Job 1 consists of five operations: Operation 0 takes 9 units of time to complete, Operation 1 takes 9 units of time and can only start after Operation 0 has finished, Operation 2 takes 7 units of time and can only start after Operation 1 has finished, Operation 3 takes 2 units of time and can only start after Operation 2 has finished, and Operation 4 takes 8 units of time and can only start after Operation 3 has finished. The makespan must be at least 9 units of time after the start of Operation 0, at least 9 units of time after the start of Operation 1, at least 7 units of time after the start of Operation 2, at least 2 units of time after the start of Operation 3, and at least 8 units of time after the start of Operation 4. + +Certain operations must be performed on the same machine, and their scheduling must avoid overlaps. + +The primary objective is to minimize the makespan, which is the total time taken to complete all operations across both jobs. The makespan must be at least as long as the completion time of the last operation in either job. All operations must start at or after time 0, and the order of operations within each job must be strictly followed. The scheduling of operations on shared machines must respect the no-overlap constraints described above. Binary variables are used to enforce the no-overlap constraints on the shared machines, ensuring that the operations do not overlap and the sequence constraints are satisfied. + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 2 + - Operation 1 of Job 1: Machine 2 + - Operation 2 of Job 1: Machine 2 + - Operation 3 of Job 1: Machine 1 + - Operation 4 of Job 1: Machine 137Job 1 consists of five operations: Operation 0 takes 9 units of time to complete, Operation 1 takes 9 units of time and can only start after Operation 0 has finished, Operation 2 takes 7 units of time and can only start after Operation 1 has finished, Operation 3 takes 2 units of time and can only start after Operation 2 has finished, and Operation 4 takes 8 units of time and can only start after Operation 3 has finished. The makespan must be at least 9 units of time after the start of Operation 0, at least 9 units of time after the start of Operation 1, at least 7 units of time after the start of Operation 2, at least 2 units of time after the start of Operation 3, and at least 8 units of time after the start of Operation 4. + +Certain operations from Job 1 must be performed on the same machine, and their scheduling must avoid overlaps. Specifically, Operation 0 and Operation 4 of Job 1 cannot overlap on the same machine. If Operation 0 starts first, Operation 4 must start at least 8 units of time after Operation 0 finishes. Conversely, if Operation 4 starts first, Operation 0 must start at least 9 units of time after Operation 4 finishes. Similarly, Operation 2 and Operation 3 of Job 1 cannot overlap on the same machine. If Operation 2 starts first, Operation 3 must start at least 2 units of time after Operation 2 finishes. Conversely, if Operation 3 starts first, Operation 2 must start at least 7 units of time after Operation 3 finishes. + +The primary objective is to minimize the makespan, which is the total time taken to complete all operations across both jobs. The makespan must be at least as long as the completion time of the last operation in either job. All operations must start at or after time 0, and the order of operations within each job must be strictly followed. The scheduling of operations on shared machines must respect the no-overlap constraints described above. Binary variables are used to enforce the no-overlap constraints on the shared machines, ensuring that the operations do not overlap and the sequence constraints are satisfied. + + + Additional information: + - There are 3 machines, numbered from 0 to 2 + - The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 2 + - Operation 1 of Job 1: Machine 2 + - Operation 2 of Job 1: Machine 2 + - Operation 3 of Job 1: Machine 1 + - Operation 4 of Job 1: Machine 135 37It was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
14There are two aircraft, **aircraft_0** and **aircraft_1**, that need to be scheduled for landing. The goal is to minimize the total penalty costs associated with deviations from their target landing times. **aircraft_0** must land between 0 and 30 units of time, while **aircraft_1** must land between 128 and 188 units of time. The target landing time for **aircraft_0** is 0 units, and for **aircraft_1**, it is 158 units. Penalties are incurred for deviations from the target times: **aircraft_0** incurs a penalty of 42 units per unit time for landing early and 32 units per unit time for landing late, while **aircraft_1** incurs a penalty of 62 units per unit time for landing early and 55 units per unit time for landing late. + +To ensure safety, a minimum separation time must be maintained between the landings of the two aircraft. If **aircraft_0** lands before **aircraft_1**, there must be at least 3 units of separation time. If **aircraft_1** lands before **aircraft_0**, there must be at least 98 units of separation time. The order of landings is determined by binary variables **AircraftOrder[aircraft_0,aircraft_1]** and **AircraftOrder[aircraft_1,aircraft_0]**, which are mutually exclusive and must sum to 1. The landing times must respect the separation requirements based on the landing order, and each aircraft must land within its specified time window. + +The early and late deviations for each aircraft are calculated based on the difference between the actual landing time and the target time. For **aircraft_0**, the early deviation is non-negative, and the late deviation is non-negative. For **aircraft_1**, the early deviation is calculated as the difference between the target time and the actual landing time, and the late deviation is calculated as the difference between the actual landing time and the target time. The objective is to minimize the total penalty costs, which are calculated as the sum of the penalties for early and late deviations for both aircraft.0There are two aircraft, **aircraft_0** and **aircraft_1**, that need to be scheduled for landing. The goal is to minimize the total penalty costs associated with deviations from their target landing times. **aircraft_0** must land between 0 and 30 units of time, while **aircraft_1** must land between 128 and 188 units of time. The target landing time for **aircraft_0** is 0 units, and for **aircraft_1**, it is 158 units. Penalties are incurred for deviations from the target times: **aircraft_0** incurs a penalty of 42 units per unit time for landing early and 32 units per unit time for landing late, while **aircraft_1** incurs a penalty of 62 units per unit time for landing early and 55 units per unit time for landing late. + +To ensure safety, a minimum separation time must be maintained between the landings of the two aircraft. If **aircraft_0** lands before **aircraft_1**, there must be at least 3 units of separation time. If **aircraft_1** lands before **aircraft_0**, there must be at least 98 units of separation time. The order of landings is determined by binary variables **AircraftOrder[aircraft_0,aircraft_1]** and **AircraftOrder[aircraft_1,aircraft_0]**, which are mutually exclusive and must sum to 1. The landing times must respect the separation requirements based on the landing order, and each aircraft must land within its specified time window. + +The early and late deviations for each aircraft are calculated based on the difference between the actual landing time and the target time. For **aircraft_0**, the early deviation is non-negative, and the late deviation is non-negative. For **aircraft_1**, the early deviation is calculated as the difference between the target time and the actual landing time, and the late deviation is calculated as the difference between the actual landing time and the target time. The objective is to minimize the total penalty costs, which are calculated as the sum of the penalties for early and late deviations for both aircraft.0No differencesNo differencesok
15The manufacturing facility produces custom parts for two jobs, Job 0 and Job 1, each with a sequence of operations that must follow specific precedence constraints. Job 0 has two operations: Operation 0 must start at least 2 hours before Operation 1 begins. Job 1 has five operations: Operation 1 must start at least 4 hours after Operation 0 begins, Operation 2 must start at least 2 hours after Operation 1 completes, Operation 3 must start at least 1 hour after Operation 2 completes, and Operation 4 must start at least 3 hours after Operation 3 completes. + +Operations from different jobs share machines: Operation 0 of Job 0 and Operation 0 of Job 1 share a machine, Operation 0 of Job 0 and Operation 1 of Job 1 share a machine, Operation 0 of Job 1 and Operation 1 of Job 1 share a machine, and Operation 2 of Job 1 and Operation 4 of Job 1 share a machine. To prevent overlapping, binary variables determine the order of operations on shared machines. If Operation A is scheduled before Operation B on a shared machine, the binary variable is 1; otherwise, it is 0. + +The objective is to minimize the makespan, the total completion time for all operations. The makespan must be at least as large as the completion time of the last operation in either job. Constraints ensure that operations within the same job follow their precedence requirements and that operations sharing a machine do not overlap. For example, if Operation 0 of Job 0 is scheduled before Operation 0 of Job 1, Operation 0 of Job 1 must start at least 4 hours after Operation 0 of Job 0 begins. Similarly, if Operation 2 of Job 1 is scheduled before Operation 4 of Job 1, Operation 4 of Job 1 must start at least 7 hours after Operation 2 of Job 1 begins. + +The makespan accounts for the completion times of all operations: Operation 0 of Job 0 contributes at least 2 hours, Operation 1 of Job 0 contributes at least 4 hours, Operation 0 of Job 1 contributes at least 4 hours, Operation 1 of Job 1 contributes at least 2 hours, Operation 2 of Job 1 contributes at least 1 hour, Operation 3 of Job 1 contributes at least 3 hours, and Operation 4 of Job 1 contributes at least 7 hours. All operation start times must be non-negative, and the binary variables determining the order of operations on shared machines must be either 0 or 1. The goal is to determine the optimal schedule for all operations across both jobs, ensuring all constraints are met while minimizing the makespan.17The manufacturing facility produces custom parts for two jobs, Job 0 and Job 1, each with a sequence of operations that must follow specific precedence constraints. Job 0 has two operations: Operation 0 with duration 2, and Operation 1 with duration 4. Job 1 has five operations: Operation 0 with duration 4, Operation 1 with duration 2, Operation 2 with duration 1, Operation 3 with duration 3, and Operation 4 with duration 7. For each job, all of its operations need to be processed in order, and one operation can only start once the previous operation has been completed. In particular, for Job 0, Operation 1 must start at least 2 hours after Operation 0 begins. For Job 1, Operation 1 must start at least 4 hours after Operation 0 begins, Operation 2 must start at least 2 hours after Operation 1 begins, Operation 3 must start at least 1 hour after Operation 2 begins, and Operation 4 must start at least 3 hours after Operation 3 begins. + +There are 3 machines (M0, M1, M2). Operations from different jobs share machines: Operation 0 of Job 0, Operation 0 of Job 1, Operation 1 of Job 0, and Operation 1 of Job 1 share machine M0. Operation 2 of Job 1 and Operation 4 of Job 1 share machine M1. Operation 3 of Job 1 is on machine M2. To prevent overlapping, binary variables determine the order of operations on shared machines. If Operation A is scheduled before Operation B on a shared machine, the binary variable is 1; otherwise, it is 0. + +The objective is to minimize the makespan, the total completion time for all operations. The makespan must be at least as large as the completion time of the last operation in either job. Constraints ensure that operations within the same job follow their precedence requirements and that operations sharing a machine do not overlap. + +The goal is to determine the optimal schedule for all operations across both jobs, ensuring all constraints are met while minimizing the makespan.17The manufacturing facility produces custom parts for two jobs, Job 0 and Job 1, each with a sequence of operations that must follow specific precedence constraints. Job 0 has two operations: Operation 0 with duration 2, and Operation 1 with duration 4. Job 1 has five operations: Operation 0 with duration 4, Operation 1 with duration 2, Operation 2 with duration 1, Operation 3 with duration 3, and Operation 4 with duration 7. For each job, all of its operations need to be processed in order, and one operation can only start once the previous operation has been completed. In particular, for Job 0, Operation 1 must start at least 2 hours before after Operation 1 0 begins. For Job 1 has five operations: 1, Operation 1 must start at least 4 hours after Operation 0 begins, Operation 2 must start at least 2 hours after Operation 1 completes, begins, Operation 3 must start at least 1 hour after Operation 2 completes, begins, and Operation 4 must start at least 3 hours after Operation 3 completes. begins. + + There are 3 machines (M0, M1, M2). Operations from different jobs share machines: Operation 0 of Job 0 and 0, Operation 0 of Job 1 share a machine, 1, Operation 0 1 of Job 0 0, and Operation 1 of Job 1 share a machine, machine M0. Operation 0 2 of Job 1 and Operation 1 4 of Job 1 share a machine, and Operation 2 of Job 1 and machine M1. Operation 4 3 of Job 1 share a machine. is on machine M2. To prevent overlapping, binary variables determine the order of operations on shared machines. If Operation A is scheduled before Operation B on a shared machine, the binary variable is 1; otherwise, it is 0. + +The objective is to minimize the makespan, the total completion time for all operations. The makespan must be at least as large as the completion time of the last operation in either job. Constraints ensure that operations within the same job follow their precedence requirements and that operations sharing a machine do not overlap. For example, if Operation 0 of Job 0 is scheduled before Operation 0 of Job 1, Operation 0 of Job 1 must start at least 4 hours after Operation 0 of Job 0 begins. Similarly, if Operation 2 of Job 1 is scheduled before Operation 4 of Job 1, Operation 4 of Job 1 must start at least 7 hours after Operation 2 of Job 1 begins. + +The makespan accounts for the completion times of all operations: Operation 0 of Job 0 contributes at least 2 hours, Operation 1 of Job 0 contributes at least 4 hours, Operation 0 of Job 1 contributes at least 4 hours, Operation 1 of Job 1 contributes at least 2 hours, Operation 2 of Job 1 contributes at least 1 hour, Operation 3 of Job 1 contributes at least 3 hours, and Operation 4 of Job 1 contributes at least 7 hours. All operation start times must be non-negative, and the binary variables determining the order of operations on shared machines must be either 0 or 1. The goal is to determine the optimal schedule for all operations across both jobs, ensuring all constraints are met while minimizing the makespan.No differencesMissing data (e.g., shared machines partially specified)missing data
16The problem involves scheduling four jobs (Job 0, Job 1, Job 2, and Job 3) on a set of machines to minimize the total completion time (makespan). Each job consists of a sequence of operations, and each operation must be processed on a specific machine with a given processing time. Job 0 has two operations: Operation 0 takes 2 units of time, and Operation 1 takes 3 units of time. Job 1 has five operations: Operation 0 takes 8 units of time, Operation 1 takes 7 units of time, Operation 2 takes 4 units of time, Operation 3 takes 3 units of time, and Operation 4 takes 2 units of time. Job 2 has five operations: Operation 0 takes 8 units of time, Operation 1 takes 3 units of time, Operation 2 takes 7 units of time, Operation 3 takes 8 units of time, and Operation 4 takes 4 units of time. Job 3 has seven operations: Operation 0 takes 6 units of time, Operation 1 takes 4 units of time, Operation 2 takes 7 units of time, Operation 3 takes 6 units of time, Operation 4 takes 2 units of time, Operation 5 takes 8 units of time, and Operation 6 takes 9 units of time. + +Each operation in a job must start only after the previous operation in the same job has completed. For example, in Job 1, Operation 1 cannot start until Operation 0 has finished, and in Job 3, Operation 6 cannot start until Operation 5 has finished. No two operations can be processed on the same machine at the same time. For example, if Operation 3 of Job 1 and Operation 4 of Job 2 are assigned to the same machine, they cannot overlap, and the start time of one must be at least 4 units after the other, depending on the order in which they are scheduled. Similarly, Operation 1 of Job 1 and Operation 2 of Job 3 cannot overlap on the same machine, and the start time of one must be at least 7 units after the other. + +The total completion time (makespan) must be at least as large as the completion time of every operation. For example, the makespan must be at least 2 units to account for the completion of Operation 0 in Job 0, and it must also be at least 9 units to account for the completion of Operation 6 in Job 3. The objective is to minimize the makespan, which involves determining the start times for each operation on each machine while respecting the precedence and machine capacity constraints. All start times must be non-negative, and the order of operations on each machine is determined by binary decision variables, which decide whether one operation starts before another on the same machine. A large constant (100000) is used in the machine capacity constraints to ensure that the constraints are only active when the binary variable is 1 or 0. + +The binary decision variables, such as \(X_{(1,3)(2,4)}\), \(X_{(1,3)(3,0)}\), and others, determine the order of operations on the same machine. For example, \(X_{(1,3)(2,4)}\) decides whether Operation 3 of Job 1 starts before Operation 4 of Job 2 on the same machine. The constraints involving these binary variables ensure that the start times of operations on the same machine do not overlap. The start times \(S_{(i,j)}\) and the makespan \(C_{\text{max}}\) are general variables that can take non-negative real values. The precedence constraints between operations of different jobs on the same machine, such as those involving \(S_{(1,3)}\), \(S_{(2,4)}\), and \(S_{(3,0)}\), are explicitly enforced to ensure that the operations are scheduled in a feasible order.42The problem involves scheduling four jobs (Job 0, Job 1, Job 2, and Job 3) on a set of machines to minimize the total completion time (makespan). Each job consists of a sequence of operations, and each operation must be processed on a specific machine with a given processing time. Job 0 has two operations: Operation 0 takes 2 units of time, and Operation 1 takes 3 units of time. Job 1 has five operations: Operation 0 takes 8 units of time, Operation 1 takes 7 units of time, Operation 2 takes 4 units of time, Operation 3 takes 3 units of time, and Operation 4 takes 2 units of time. Job 2 has five operations: Operation 0 takes 8 units of time, Operation 1 takes 3 units of time, Operation 2 takes 7 units of time, Operation 3 takes 8 units of time, and Operation 4 takes 4 units of time. Job 3 has seven operations: Operation 0 takes 6 units of time, Operation 1 takes 4 units of time, Operation 2 takes 7 units of time, Operation 3 takes 6 units of time, Operation 4 takes 2 units of time, Operation 5 takes 8 units of time, and Operation 6 takes 9 units of time. + +Each operation in a job must start only after the previous operation in the same job has completed. For example, in Job 1, Operation 1 cannot start until Operation 0 has finished, and in Job 3, Operation 6 cannot start until Operation 5 has finished. No two operations can be processed on the same machine at the same time. For example, if Operation 3 of Job 1 and Operation 4 of Job 2 are assigned to the same machine, they cannot overlap, and the start time of one must be at least 4 units after the other, depending on the order in which they are scheduled. Similarly, Operation 1 of Job 1 and Operation 2 of Job 3 cannot overlap on the same machine, and the start time of one must be at least 7 units after the other. + +The total completion time (makespan) must be at least as large as the completion time of every operation. For example, the makespan must be at least 2 units to account for the completion of Operation 0 in Job 0, and it must also be at least 9 units to account for the completion of Operation 6 in Job 3. The objective is to minimize the makespan, which involves determining the start times for each operation on each machine while respecting the precedence and machine capacity constraints. All start times must be non-negative, and the order of operations on each machine is determined by binary decision variables, which decide whether one operation starts before another on the same machine. A large constant (1000) is used in the machine capacity constraints to ensure that the constraints are only active when the binary variable is 1 or 0. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3} +- For each job j in J the operations are indexed from 0 to (n_j - 1). We denote operation (j,k) where k is the operation index in job j. +- Machines: M = {0, 1, 2} + +2) Machine assignment for every operation +- Job 0: + - (0,0) -> Machine 0 + - (0,1) -> Machine 1 + +- Job 1: + - (1,0) -> Machine 0 + - (1,1) -> Machine 2 + - (1,2) -> Machine 1 + - (1,3) -> Machine 0 + - (1,4) -> Machine 2 + +- Job 2: + - (2,0) -> Machine 2 + - (2,1) -> Machine 1 + - (2,2) -> Machine 0 + - (2,3) -> Machine 2 + - (2,4) -> Machine 1 + +- Job 3: + - (3,0) -> Machine 1 + - (3,1) -> Machine 0 + - (3,2) -> Machine 2 + - (3,3) -> Machine 1 + - (3,4) -> Machine 0 + - (3,5) -> Machine 2 + - (3,6) -> Machine 1 + +3) Processing times (duration) for every operation (repeated from above for clarity) +- Job 0: p[(0,0)] = 2, p[(0,1)] = 3 +- Job 1: p[(1,0)] = 8, p[(1,1)] = 7, p[(1,2)] = 4, p[(1,3)] = 3, p[(1,4)] = 2 +- Job 2: p[(2,0)] = 8, p[(2,1)] = 3, p[(2,2)] = 7, p[(2,3)] = 8, p[(2,4)] = 4 +- Job 3: p[(3,0)] = 6, p[(3,1)] = 4, p[(3,2)] = 7, p[(3,3)] = 6, p[(3,4)] = 2, p[(3,5)] = 8, p[(3,6)] = 946Each operation in a job must start only after the previous operation in the same job has completed. For example, in Job 1, Operation 1 cannot start until Operation 0 has finished, and in Job 3, Operation 6 cannot start until Operation 5 has finished. No two operations can be processed on the same machine at the same time. For example, if Operation 3 of Job 1 and Operation 4 of Job 2 are assigned to the same machine, they cannot overlap, and the start time of one must be at least 4 units after the other, depending on the order in which they are scheduled. Similarly, Operation 1 of Job 1 and Operation 2 of Job 3 cannot overlap on the same machine, and the start time of one must be at least 7 units after the other. + +The total completion time (makespan) must be at least as large as the completion time of every operation. For example, the makespan must be at least 2 units to account for the completion of Operation 0 in Job 0, and it must also be at least 9 units to account for the completion of Operation 6 in Job 3. The objective is to minimize the makespan, which involves determining the start times for each operation on each machine while respecting the precedence and machine capacity constraints. All start times must be non-negative, and the order of operations on each machine is determined by binary decision variables, which decide whether one operation starts before another on the same machine. A large constant (100000) (1000) is used in the machine capacity constraints to ensure that the constraints are only active when the binary variable is 1 or 0. + + The binary decision variables, such as \(X_{(1,3)(2,4)}\), \(X_{(1,3)(3,0)}\), ADDITIONAL INPUT DATA: + + 1) Sets and others, determine indices + - Jobs: J = {0, 1, 2, 3} + - For each job j in J the order of operations on are indexed from 0 to (n_j - 1). We denote operation (j,k) where k is the same machine. For example, \(X_{(1,3)(2,4)}\) decides whether Operation 3 of operation index in job j. + - Machines: M = {0, 1, 2} + + 2) Machine assignment for every operation + - Job 0: + - (0,0) -> Machine 0 + - (0,1) -> Machine 1 + + starts before Operation 4 of - Job 1: + - (1,0) -> Machine 0 + - (1,1) -> Machine 2 + on the same machine. The constraints involving these binary variables ensure that the start times of operations on the same machine do not overlap. The start - (1,2) -> Machine 1 + - (1,3) -> Machine 0 + - (1,4) -> Machine 2 + + - Job 2: + - (2,0) -> Machine 2 + - (2,1) -> Machine 1 + - (2,2) -> Machine 0 + - (2,3) -> Machine 2 + - (2,4) -> Machine 1 + + - Job 3: + - (3,0) -> Machine 1 + - (3,1) -> Machine 0 + - (3,2) -> Machine 2 + - (3,3) -> Machine 1 + - (3,4) -> Machine 0 + - (3,5) -> Machine 2 + - (3,6) -> Machine 1 + + 3) Processing times \(S_{(i,j)}\) and the makespan \(C_{\text{max}}\) are general variables that can take non-negative real values. The precedence constraints between operations of different jobs on the same machine, such as those involving \(S_{(1,3)}\), \(S_{(2,4)}\), and \(S_{(3,0)}\), are explicitly enforced to ensure that the operations are scheduled in a feasible order. (duration) for every operation (repeated from above for clarity) + - Job 0: p[(0,0) = 2, p[(0,1)] = 3] + - Job 1: p[(1,0) = 8, p[(1,1)] = 7, p[(1,2)] = 4, p[(1,3)] = 3, p[(1,4)] = 2] + - Job 2: p[(2,0) = 8, p[(2,1)] = 3, p[(2,2)] = 7, p[(2,3)] = 8, p[(2,4)] = 4] + - Job 3: p[(3,0) = 6, p[(3,1)] = 4, p[(3,2)] = 7, p[(3,3)] = 6, p[(3,4)] = 2, p[(3,5)] = 8, p[(3,6)] = 9]42 46It was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
17The problem involves scheduling the landing times of six aircraft at a busy airport to minimize the total penalty costs associated with deviations from their target landing times while ensuring safe separation between landings. Each aircraft has a target landing time, earliest and latest possible landing times, and penalty costs for landing early or late. The landing times must fall within the specified bounds for each aircraft: Aircraft 0 between 8 and 68 minutes, Aircraft 1 between 46 and 106 minutes, Aircraft 2 between 90 and 150 minutes, Aircraft 3 between 16 and 76 minutes, Aircraft 4 between 0 and 57 minutes, and Aircraft 5 between 39 and 99 minutes. The penalties are calculated based on the deviation from the target landing times, with specific costs per minute for early or late landings: 55 for early landing of Aircraft 0, 94 for Aircraft 1, 61 for Aircraft 2, 64 for Aircraft 3, 74 for Aircraft 4, and 84 for Aircraft 5; and 38 for late landing of Aircraft 0, 81 for Aircraft 1, 71 for Aircraft 2, 80 for Aircraft 3, 53 for Aircraft 4, and 84 for Aircraft 5. The binary decision variables ensure that for each pair of aircraft, one lands before the other, and the separation times are enforced based on the landing order. The separation constraints include specific times such as 22 minutes between Aircraft 0 and 1, 22 minutes between Aircraft 0 and 2, 5 minutes between Aircraft 0 and 3, 3 minutes between Aircraft 0 and 4, 3 minutes between Aircraft 0 and 5, 4 minutes between Aircraft 1 and 2, 3 minutes between Aircraft 1 and 3, 1 minute between Aircraft 1 and 4, 5 minutes between Aircraft 1 and 5, 2 minutes between Aircraft 2 and 3, 1 minute between Aircraft 2 and 4, 5 minutes between Aircraft 2 and 5, 5 minutes between Aircraft 3 and 4, 3 minutes between Aircraft 3 and 5, and 5 minutes between Aircraft 4 and 5. The goal is to determine the actual landing times for all six aircraft such that the total penalty cost is minimized, and all separation and landing time constraints are satisfied. The binary decision variables `AircraftOrder[i,j]` are used to determine the landing order, and the constraints involving `Early` and `Late` variables ensure that the landing times are within the acceptable deviations from the target times. The target landing times for each aircraft are implied by the constraints involving `Early` and `Late` variables, and the `AircraftOrder` variables are binary, ensuring that one aircraft lands before the other in each pair. The relationship between the separation constraints and the binary decision variables is essential for understanding the problem structure.0The problem involves scheduling the landing times of six aircraft at a busy airport to minimize the total penalty costs associated with deviations from their target landing times while ensuring safe separation between landings. Each aircraft has a target landing time, earliest and latest possible landing times, and penalty costs for landing early or late. The landing times must fall within the specified bounds for each aircraft: Aircraft 0 between 8 and 68 minutes, Aircraft 1 between 46 and 106 minutes, Aircraft 2 between 90 and 150 minutes, Aircraft 3 between 16 and 76 minutes, Aircraft 4 between 0 and 57 minutes, and Aircraft 5 between 39 and 99 minutes. The target landing times are 38 for Aircraft 0, 76 for Aircraft 1, 120 for Aircraft 2, 46 for Aircraft 3, 28 for Aircraft 4, and 69 for Aircraft 5. The penalties are calculated based on the deviation from the target landing times, with specific costs per minute for early or late landings: 55 for early landing of Aircraft 0, 94 for Aircraft 1, 61 for Aircraft 2, 64 for Aircraft 3, 74 for Aircraft 4, and 84 for Aircraft 5; and 38 for late landing of Aircraft 0, 81 for Aircraft 1, 71 for Aircraft 2, 80 for Aircraft 3, 53 for Aircraft 4, and 84 for Aircraft 5. The binary decision variables ensure that for each pair of aircraft, one lands before the other, and the separation times are enforced based on the landing order. The separation constraints include specific times such as 22 minutes between Aircraft 0 and 1, 22 minutes between Aircraft 0 and 2, 5 minutes between Aircraft 0 and 3, 3 minutes between Aircraft 0 and 4, 3 minutes between Aircraft 0 and 5, 4 minutes between Aircraft 1 and 2, 3 minutes between Aircraft 1 and 3, 1 minute between Aircraft 1 and 4, 5 minutes between Aircraft 1 and 5, 2 minutes between Aircraft 2 and 3, 1 minute between Aircraft 2 and 4, 5 minutes between Aircraft 2 and 5, 5 minutes between Aircraft 3 and 4, 3 minutes between Aircraft 3 and 5, and 5 minutes between Aircraft 4 and 5. The goal is to determine the actual landing times for all six aircraft such that the total penalty cost is minimized, and all separation and landing time constraints are satisfied. 0The problem involves scheduling the landing times of six aircraft at a busy airport to minimize the total penalty costs associated with deviations from their target landing times while ensuring safe separation between landings. Each aircraft has a target landing time, earliest and latest possible landing times, and penalty costs for landing early or late. The landing times must fall within the specified bounds for each aircraft: Aircraft 0 between 8 and 68 minutes, Aircraft 1 between 46 and 106 minutes, Aircraft 2 between 90 and 150 minutes, Aircraft 3 between 16 and 76 minutes, Aircraft 4 between 0 and 57 minutes, and Aircraft 5 between 39 and 99 minutes. The target landing times are 38 for Aircraft 0, 76 for Aircraft 1, 120 for Aircraft 2, 46 for Aircraft 3, 28 for Aircraft 4, and 69 for Aircraft 5. The penalties are calculated based on the deviation from the target landing times, with specific costs per minute for early or late landings: 55 for early landing of Aircraft 0, 94 for Aircraft 1, 61 for Aircraft 2, 64 for Aircraft 3, 74 for Aircraft 4, and 84 for Aircraft 5; and 38 for late landing of Aircraft 0, 81 for Aircraft 1, 71 for Aircraft 2, 80 for Aircraft 3, 53 for Aircraft 4, and 84 for Aircraft 5. The binary decision variables ensure that for each pair of aircraft, one lands before the other, and the separation times are enforced based on the landing order. The separation constraints include specific times such as 22 minutes between Aircraft 0 and 1, 22 minutes between Aircraft 0 and 2, 5 minutes between Aircraft 0 and 3, 3 minutes between Aircraft 0 and 4, 3 minutes between Aircraft 0 and 5, 4 minutes between Aircraft 1 and 2, 3 minutes between Aircraft 1 and 3, 1 minute between Aircraft 1 and 4, 5 minutes between Aircraft 1 and 5, 2 minutes between Aircraft 2 and 3, 1 minute between Aircraft 2 and 4, 5 minutes between Aircraft 2 and 5, 5 minutes between Aircraft 3 and 4, 3 minutes between Aircraft 3 and 5, and 5 minutes between Aircraft 4 and 5. The goal is to determine the actual landing times for all six aircraft such that the total penalty cost is minimized, and all separation and landing time constraints are satisfied. The binary decision variables `AircraftOrder[i,j` are used to determine the landing order, and the constraints involving `Early` and `Late` variables ensure that the landing times are within the acceptable deviations from the target times. The target landing times for each aircraft are implied by the constraints involving `Early` and `Late` variables, and the `AircraftOrder` variables are binary, ensuring that one aircraft lands before the other in each pair. The relationship between the separation constraints and the binary decision variables is essential for understanding the problem structure.]No differencesAdded missing data for target times, deleted redundant text that was describing details of the formulation (solution) instead of problem descriptionmissing data
18The problem involves scheduling the landing times of three aircraft (aircraft_0, aircraft_1, and aircraft_2) to minimize the total penalty costs associated with deviations from their target landing times. Each aircraft has specific time windows within which it must land: aircraft_0 must land between 19 and 79 minutes, aircraft_1 between 58 and 118 minutes, and aircraft_2 between 205 and 265 minutes. The penalties for landing early or late are as follows: aircraft_0 incurs a penalty of 58 units per minute early and 57 units per minute late, aircraft_1 incurs 25 units per minute early and 61 units per minute late, and aircraft_2 incurs 37 units per minute early and 70 units per minute late. + +The landing sequence between any two aircraft is determined by binary variables, where a value of 1 indicates that one aircraft lands before the other. The constraints ensure that the landing times respect the sequence and separation requirements. For example, if aircraft_0 lands before aircraft_1, there must be at least 4 minutes of separation, but if aircraft_1 lands before aircraft_0, the required separation is 21 minutes. Similarly, if aircraft_0 lands before aircraft_2, the separation must be at least 12 minutes, but if aircraft_2 lands before aircraft_0, the separation must be at least 126 minutes. For aircraft_1 and aircraft_2, the separation is at least 13 minutes if aircraft_1 lands first, and at least 87 minutes if aircraft_2 lands first. + +The objective is to minimize the total penalty costs, which include the penalties for early and late landings for all three aircraft. The constraints ensure that each aircraft lands within its specified time window, the landing sequence respects the separation requirements, and the binary variables enforce the correct sequence between aircraft pairs. The penalties for early or late landings are calculated based on the deviation from the target landing times, which are implicitly defined by the constraints on the landing times and the penalties. Specifically, the constraints ensure that the landing time of aircraft_0 plus the early penalty for aircraft_0 is at least 49 minutes, and the landing time of aircraft_0 minus the late penalty for aircraft_0 is at most 49 minutes. Similarly, the landing time of aircraft_1 plus the early penalty for aircraft_1 is at least 88 minutes, and the landing time of aircraft_1 minus the late penalty for aircraft_1 is at most 88 minutes. For aircraft_2, the landing time plus the early penalty is at least 235 minutes, and the landing time minus the late penalty is at most 235 minutes. These constraints ensure that the penalties are correctly calculated based on the deviation from the target landing times.0The problem involves scheduling the landing times of three aircraft (aircraft_0, aircraft_1, and aircraft_2) to minimize the total penalty costs associated with deviations from their target landing times. Each aircraft has specific time windows within which it must land: aircraft_0 must land between 19 and 79 minutes, aircraft_1 between 58 and 118 minutes, and aircraft_2 between 205 and 265 minutes. The target times are 49 for aircraft_0, 88 for aircraft_1, and 235 for aircraft_2. The penalties for landing early or late are as follows: aircraft_0 incurs a penalty of 58 units per minute early and 57 units per minute late, aircraft_1 incurs 25 units per minute early and 61 units per minute late, and aircraft_2 incurs 37 units per minute early and 70 units per minute late. + +The landing sequence between any two aircraft is determined by binary variables, where a value of 1 indicates that one aircraft lands before the other. The constraints ensure that the landing times respect the sequence and separation requirements. For example, if aircraft_0 lands before aircraft_1, there must be at least 4 minutes of separation, but if aircraft_1 lands before aircraft_0, the required separation is 21 minutes. Similarly, if aircraft_0 lands before aircraft_2, the separation must be at least 12 minutes, but if aircraft_2 lands before aircraft_0, the separation must be at least 126 minutes. For aircraft_1 and aircraft_2, the separation is at least 13 minutes if aircraft_1 lands first, and at least 87 minutes if aircraft_2 lands first. + +The objective is to minimize the total penalty costs, which include the penalties for early and late landings for all three aircraft. The constraints ensure that each aircraft lands within its specified time window, the landing sequence respects the separation requirements, and the binary variables enforce the correct sequence between aircraft pairs. The penalties for early or late landings are calculated based on the deviation from the target landing times.0The problem involves scheduling the landing times of three aircraft (aircraft_0, aircraft_1, and aircraft_2) to minimize the total penalty costs associated with deviations from their target landing times. Each aircraft has specific time windows within which it must land: aircraft_0 must land between 19 and 79 minutes, aircraft_1 between 58 and 118 minutes, and aircraft_2 between 205 and 265 minutes. The target times are 49 for aircraft_0, 88 for aircraft_1, and 235 for aircraft_2. The penalties for landing early or late are as follows: aircraft_0 incurs a penalty of 58 units per minute early and 57 units per minute late, aircraft_1 incurs 25 units per minute early and 61 units per minute late, and aircraft_2 incurs 37 units per minute early and 70 units per minute late. + +The landing sequence between any two aircraft is determined by binary variables, where a value of 1 indicates that one aircraft lands before the other. The constraints ensure that the landing times respect the sequence and separation requirements. For example, if aircraft_0 lands before aircraft_1, there must be at least 4 minutes of separation, but if aircraft_1 lands before aircraft_0, the required separation is 21 minutes. Similarly, if aircraft_0 lands before aircraft_2, the separation must be at least 12 minutes, but if aircraft_2 lands before aircraft_0, the separation must be at least 126 minutes. For aircraft_1 and aircraft_2, the separation is at least 13 minutes if aircraft_1 lands first, and at least 87 minutes if aircraft_2 lands first. + +The objective is to minimize the total penalty costs, which include the penalties for early and late landings for all three aircraft. The constraints ensure that each aircraft lands within its specified time window, the landing sequence respects the separation requirements, and the binary variables enforce the correct sequence between aircraft pairs. The penalties for early or late landings are calculated based on the deviation from the target landing times, which are implicitly defined by the constraints on the landing times and the penalties. Specifically, the constraints ensure that the landing time of aircraft_0 plus the early penalty for aircraft_0 is at least 49 minutes, and the landing time of aircraft_0 minus the late penalty for aircraft_0 is at most 49 minutes. Similarly, the landing time of aircraft_1 plus the early penalty for aircraft_1 is at least 88 minutes, and the landing time of aircraft_1 minus the late penalty for aircraft_1 is at most 88 minutes. For aircraft_2, the landing time plus the early penalty is at least 235 minutes, and the landing time minus the late penalty is at most 235 minutes. These constraints ensure that the penalties are correctly calculated based on the deviation from the target landing times.No differencesAdded missing data on target times, and removed the last sentences that are unnecessary and seem to conflate the minutes early/late with the early/late penalties - these values are different due to per-minute cost.ambiguous, missing data
19The problem involves scheduling the landing times of five aircraft to minimize the total penalty costs associated with deviations from their target landing times, while ensuring that each aircraft lands within its specified time window and that the required separation times between consecutive landings are maintained. Each aircraft has a target landing time, an earliest possible landing time, and a latest possible landing time. Aircraft 0 must land between 178 and 238, aircraft 1 between 35 and 95, aircraft 2 between 266 and 300, aircraft 3 between 0 and 40, and aircraft 4 between 69 and 129. Penalties are incurred for landing early or late, with specific costs per unit time for each aircraft: aircraft 0 has an early penalty of 62 and a late penalty of 82, aircraft 1 has an early penalty of 10 and a late penalty of 65, aircraft 2 has an early penalty of 89 and a late penalty of 72, aircraft 3 has an early penalty of 81 and a late penalty of 69, and aircraft 4 has an early penalty of 50 and a late penalty of 42. The separation requirements between aircraft depend on the order in which they land, and these requirements are enforced using binary decision variables that determine the landing sequence. The separation times are as follows: between aircraft 0 and 1, the required separation is 203 if 0 lands before 1 and 3 if 1 lands before 0; between aircraft 0 and 2, the required separation is 2 if 0 lands before 2 and 28 if 2 lands before 0; between aircraft 0 and 3, the required separation is 4 if 0 lands before 3 and 238 if 3 lands before 0; between aircraft 0 and 4, the required separation is 5 if 0 lands before 4 and 169 if 4 lands before 0; between aircraft 1 and 2, the required separation is 3 if 1 lands before 2 and 171 if 2 lands before 1; between aircraft 1 and 3, the required separation is 2 if 1 lands before 3 and 95 if 3 lands before 1; between aircraft 1 and 4, the required separation is 4 if 1 lands before 4 and 26 if 4 lands before 1; between aircraft 2 and 3, the required separation is 5 if 2 lands before 3 and 300 if 3 lands before 2; between aircraft 2 and 4, the required separation is 2 if 2 lands before 4 and 231 if 4 lands before 2; between aircraft 3 and 4, the required separation is 5 if 3 lands before 4 and 29 if 4 lands before 3. The objective is to minimize the sum of the penalties for early and late landings across all aircraft. The constraints include ensuring that each aircraft lands within its time window, enforcing the separation times based on the landing order, and calculating the penalties for deviations from the target landing times. The binary decision variables ensure that only one of the two possible orders between any pair of aircraft is chosen, and the separation constraints are applied accordingly. The landing times must also satisfy bounds that are specific to each aircraft. The penalties are calculated based on the difference between the actual landing time and the target landing time, with separate variables for early and late deviations. The goal is to find the optimal landing times and sequence that minimize the total penalty costs while satisfying all constraints.0The problem involves scheduling the landing times of five aircraft to minimize the total penalty costs associated with deviations from their target landing times, while ensuring that each aircraft lands within its specified time window and that the required separation times between consecutive landings are maintained. Each aircraft has a target landing time: Aircraft 0's target is 200, aircraft 1's target is 65, aircraft 2's target is 283, aircraft 3's target is 20, aircraft 4's target is 99. Each aircraft also has an earliest possible landing time, and a latest possible landing time. Aircraft 0 must land between 178 and 238, aircraft 1 between 35 and 95, aircraft 2 between 266 and 300, aircraft 3 between 0 and 40, and aircraft 4 between 69 and 129. Penalties are incurred for landing early or late, with specific costs per unit time for each aircraft: aircraft 0 has an early penalty of 62 and a late penalty of 82, aircraft 1 has an early penalty of 10 and a late penalty of 65, aircraft 2 has an early penalty of 89 and a late penalty of 72, aircraft 3 has an early penalty of 81 and a late penalty of 69, and aircraft 4 has an early penalty of 50 and a late penalty of 42. The separation requirements between aircraft depend on the order in which they land, and these requirements are enforced using binary decision variables that determine the landing sequence. The separation times are as follows: between aircraft 0 and 1, the required separation is 203 if 0 lands before 1 and 3 if 1 lands before 0; between aircraft 0 and 2, the required separation is 2 if 0 lands before 2 and 28 if 2 lands before 0; between aircraft 0 and 3, the required separation is 4 if 0 lands before 3 and 238 if 3 lands before 0; between aircraft 0 and 4, the required separation is 5 if 0 lands before 4 and 169 if 4 lands before 0; between aircraft 1 and 2, the required separation is 3 if 1 lands before 2 and 171 if 2 lands before 1; between aircraft 1 and 3, the required separation is 2 if 1 lands before 3 and 95 if 3 lands before 1; between aircraft 1 and 4, the required separation is 4 if 1 lands before 4 and 26 if 4 lands before 1; between aircraft 2 and 3, the required separation is 5 if 2 lands before 3 and 300 if 3 lands before 2; between aircraft 2 and 4, the required separation is 2 if 2 lands before 4 and 231 if 4 lands before 2; between aircraft 3 and 4, the required separation is 5 if 3 lands before 4 and 29 if 4 lands before 3. The objective is to minimize the sum of the penalties for early and late landings across all aircraft. The constraints include ensuring that each aircraft lands within its time window, enforcing the separation times based on the landing order, and calculating the penalties for deviations from the target landing times. The binary decision variables ensure that only one of the two possible orders between any pair of aircraft is chosen, and the separation constraints are applied accordingly. The landing times must also satisfy bounds that are specific to each aircraft. The penalties are calculated based on the difference between the actual landing time and the target landing time, with separate variables for early and late deviations. The goal is to find the optimal landing times and sequence that minimize the total penalty costs while satisfying all constraints.9410The problem involves scheduling the landing times of five aircraft to minimize the total penalty costs associated with deviations from their target landing times, while ensuring that each aircraft lands within its specified time window and that the required separation times between consecutive landings are maintained. Each aircraft has a target landing time, time: Aircraft 0's target is 200, aircraft 1's target is 65, aircraft 2's target is 283, aircraft 3's target is 20, aircraft 4's target is 99. Each aircraft also has an earliest possible landing time, and a latest possible landing time. Aircraft 0 must land between 178 and 238, aircraft 1 between 35 and 95, aircraft 2 between 266 and 300, aircraft 3 between 0 and 40, and aircraft 4 between 69 and 129. Penalties are incurred for landing early or late, with specific costs per unit time for each aircraft: aircraft 0 has an early penalty of 62 and a late penalty of 82, aircraft 1 has an early penalty of 10 and a late penalty of 65, aircraft 2 has an early penalty of 89 and a late penalty of 72, aircraft 3 has an early penalty of 81 and a late penalty of 69, and aircraft 4 has an early penalty of 50 and a late penalty of 42. The separation requirements between aircraft depend on the order in which they land, and these requirements are enforced using binary decision variables that determine the landing sequence. The separation times are as follows: between aircraft 0 and 1, the required separation is 203 if 0 lands before 1 and 3 if 1 lands before 0; between aircraft 0 and 2, the required separation is 2 if 0 lands before 2 and 28 if 2 lands before 0; between aircraft 0 and 3, the required separation is 4 if 0 lands before 3 and 238 if 3 lands before 0; between aircraft 0 and 4, the required separation is 5 if 0 lands before 4 and 169 if 4 lands before 0; between aircraft 1 and 2, the required separation is 3 if 1 lands before 2 and 171 if 2 lands before 1; between aircraft 1 and 3, the required separation is 2 if 1 lands before 3 and 95 if 3 lands before 1; between aircraft 1 and 4, the required separation is 4 if 1 lands before 4 and 26 if 4 lands before 1; between aircraft 2 and 3, the required separation is 5 if 2 lands before 3 and 300 if 3 lands before 2; between aircraft 2 and 4, the required separation is 2 if 2 lands before 4 and 231 if 4 lands before 2; between aircraft 3 and 4, the required separation is 5 if 3 lands before 4 and 29 if 4 lands before 3. The objective is to minimize the sum of the penalties for early and late landings across all aircraft. The constraints include ensuring that each aircraft lands within its time window, enforcing the separation times based on the landing order, and calculating the penalties for deviations from the target landing times. The binary decision variables ensure that only one of the two possible orders between any pair of aircraft is chosen, and the separation constraints are applied accordingly. The landing times must also satisfy bounds that are specific to each aircraft. The penalties are calculated based on the difference between the actual landing time and the target landing time, with separate variables for early and late deviations. The goal is to find the optimal landing times and sequence that minimize the total penalty costs while satisfying all constraints.9410Missing data about target landing timesmissing data
20There are six jobs, each with a specific sequence of operations that must be performed on shared machines to complete their tasks. The goal is to schedule these operations to minimize the total time (makespan) required to complete all jobs. Each job has a predefined sequence of operations with specific processing times. Job 0 has two operations: the first takes 1 unit of time, and the second takes 6 units. Job 1 has two operations: the first takes 4 units, and the second takes 2 units. Job 2 has three operations: the first takes 9 units, the second takes 2 units, and the third takes 2 units. Job 3 has four operations: the first takes 5 units, the second takes 7 units, the third takes 6 units, and the fourth takes 3 units. Job 4 has four operations: the first takes 2 units, the second takes 9 units, the third takes 8 units, and the fourth takes 7 units. Job 5 has two operations: the first takes 4 units, and the second takes 8 units. + +Operations from different jobs may share the same machine, and the order in which they are processed must be determined. For example, the second operation of Job 0 and the fourth operation of Job 3 share a machine. If Job 0’s second operation starts first, it must finish before Job 3’s fourth operation begins, or vice versa. Similarly, the second operation of Job 1 and the second operation of Job 2 share a machine. If Job 1’s second operation starts first, it must finish before Job 2’s second operation begins, or vice versa. + +Each job’s operations must be performed in a specific order. For Job 0, the second operation cannot start until the first operation is completed. For Job 1, the second operation cannot start until the first operation is completed. For Job 2, the second operation cannot start until the first operation is completed, and the third operation cannot start until the second operation is completed. For Job 3, the second operation cannot start until the first operation is completed, the third operation cannot start until the second operation is completed, and the fourth operation cannot start until the third operation is completed. For Job 4, the second operation cannot start until the first operation is completed, the third operation cannot start until the second operation is completed, and the fourth operation cannot start until the third operation is completed. For Job 5, the second operation cannot start until the first operation is completed. + +The objective is to minimize the total time (makespan) required to complete all operations for all jobs. The makespan is determined by the latest completion time among all operations. The constraints include precedence constraints, which ensure that each operation starts only after its preceding operation in the same job’s sequence is completed. For example, Job 0’s second operation must start at least 1 unit of time after the first operation starts. Job 1’s second operation must start at least 4 units of time after the first operation starts. Job 2’s second operation must start at least 9 units of time after the first operation starts, and the third operation must start at least 2 units of time after the second operation starts. Job 3’s second operation must start at least 5 units of time after the first operation starts, the third operation must start at least 7 units of time after the second operation starts, and the fourth operation must start at least 6 units of time after the third operation starts. Job 4’s second operation must start at least 2 units of time after the first operation starts, the third operation must start at least 9 units of time after the second operation starts, and the fourth operation must start at least 8 units of time after the third operation starts. Job 5’s second operation must start at least 4 units of time after the first operation starts. + +Machine capacity constraints ensure that operations sharing the same machine do not overlap. For example, if Job 0’s second operation starts first on a shared machine, it must finish before Job 3’s fourth operation begins, or vice versa. A large constant (100,000) is used to enforce the order of operations on shared machines. Non-negativity constraints ensure that all operation start times are non-negative. Binary decision variables determine the order of operations on shared machines. + +The processing times for each operation are fixed and provided. The precedence constraints specify the minimum time gaps between operations. The machine capacity constraints ensure that operations on shared machines do not overlap. The decisions to be made include determining the start time for each operation for all jobs, deciding the order of operations on shared machines using binary variables, and ensuring that all constraints are satisfied while minimizing the total makespan. The objective is to minimize the total time (makespan) required to complete all operations for all jobs, ensuring that all precedence and machine capacity constraints are met.29There are six jobs, each with a specific sequence of operations that must be performed on shared machines to complete their tasks. The goal is to schedule these operations to minimize the total time (makespan) required to complete all jobs. Each job has a predefined sequence of operations with specific processing times. Job 0 has two operations: the first takes 1 unit of time, and the second takes 6 units. Job 1 has two operations: the first takes 4 units, and the second takes 2 units. Job 2 has three operations: the first takes 9 units, the second takes 2 units, and the third takes 2 units. Job 3 has four operations: the first takes 5 units, the second takes 7 units, the third takes 6 units, and the fourth takes 3 units. Job 4 has four operations: the first takes 2 units, the second takes 9 units, the third takes 8 units, and the fourth takes 7 units. Job 5 has two operations: the first takes 4 units, and the second takes 8 units. + +Operations from different jobs may share the same machine, and the order in which they are processed must be determined. For example, the second operation of Job 0 and the fourth operation of Job 3 share a machine. If Job 0’s second operation starts first, it must finish before Job 3’s fourth operation begins, or vice versa. Similarly, the second operation of Job 1 and the second operation of Job 2 share a machine. If Job 1’s second operation starts first, it must finish before Job 2’s second operation begins, or vice versa. + +Each job’s operations must be performed in a specific order. For Job 0, the second operation cannot start until the first operation is completed. For Job 1, the second operation cannot start until the first operation is completed. For Job 2, the second operation cannot start until the first operation is completed, and the third operation cannot start until the second operation is completed. For Job 3, the second operation cannot start until the first operation is completed, the third operation cannot start until the second operation is completed, and the fourth operation cannot start until the third operation is completed. For Job 4, the second operation cannot start until the first operation is completed, the third operation cannot start until the second operation is completed, and the fourth operation cannot start until the third operation is completed. For Job 5, the second operation cannot start until the first operation is completed. + +The objective is to minimize the total time (makespan) required to complete all operations for all jobs. The makespan is determined by the latest completion time among all operations. The constraints include precedence constraints, which ensure that each operation starts only after its preceding operation in the same job’s sequence is completed. For example, Job 0’s second operation must start at least 1 unit of time after the first operation starts. Job 1’s second operation must start at least 4 units of time after the first operation starts. Job 2’s second operation must start at least 9 units of time after the first operation starts, and the third operation must start at least 2 units of time after the second operation starts. Job 3’s second operation must start at least 5 units of time after the first operation starts, the third operation must start at least 7 units of time after the second operation starts, and the fourth operation must start at least 6 units of time after the third operation starts. Job 4’s second operation must start at least 2 units of time after the first operation starts, the third operation must start at least 9 units of time after the second operation starts, and the fourth operation must start at least 8 units of time after the third operation starts. Job 5’s second operation must start at least 4 units of time after the first operation starts. + +Machine capacity constraints ensure that operations sharing the same machine do not overlap. For example, if Job 0’s second operation starts first on a shared machine, it must finish before Job 3’s fourth operation begins, or vice versa. A large constant (100,000) is used to enforce the order of operations on shared machines. Non-negativity constraints ensure that all operation start times are non-negative. Binary decision variables determine the order of operations on shared machines. + +The processing times for each operation are fixed and provided. The precedence constraints specify the minimum time gaps between operations. The machine capacity constraints ensure that operations on shared machines do not overlap. The decisions to be made include determining the start time for each operation for all jobs, deciding the order of operations on shared machines using binary variables, and ensuring that all constraints are satisfied while minimizing the total makespan. The objective is to minimize the total time (makespan) required to complete all operations for all jobs, ensuring that all precedence and machine capacity constraints are met.26No differences29 26Incorrect ground truthincorrect ground truth
21The manufacturing facility produces custom parts for two jobs, Job 0 and Job 1, each consisting of operations performed on shared machines. The goal is to minimize the total completion time (makespan) by scheduling operations efficiently while respecting constraints. + +Job 0 has three operations: Operation 0 takes 2 hours, Operation 1 takes 8 hours, and Operation 2 takes 5 hours. Job 1 has two operations: Operation 0 takes 1 hour, and Operation 1 takes 7 hours. + +For Job 0, Operation 1 cannot start until Operation 0 has finished, with at least 2 hours between the finish of Operation 0 and the start of Operation 1. Operation 2 cannot start until Operation 1 has finished, with at least 8 hours between the finish of Operation 1 and the start of Operation 2. + +For Job 1, Operation 1 cannot start until Operation 0 has finished, with at least 1 hour between the finish of Operation 0 and the start of Operation 1. + +Operations on shared machines cannot overlap. If Operation 0 of Job 1 is scheduled before Operation 1 of Job 1, there must be at least 1 hour between the finish of Operation 0 and the start of Operation 1. If Operation 1 of Job 1 is scheduled before Operation 0 of Job 1, there must be at least 7 hours between the finish of Operation 1 and the start of Operation 0. + +For Job 0, if Operation 0 is scheduled before Operation 1, there must be at least 8 hours between the finish of Operation 0 and the start of Operation 1. If Operation 0 is scheduled before Operation 2, there must be at least 5 hours between the finish of Operation 0 and the start of Operation 2. If Operation 1 is scheduled before Operation 2, there must be at least 5 hours between the finish of Operation 1 and the start of Operation 2. + +The makespan must be at least as large as the completion time of every operation: Operation 0 of Job 0 contributes at least 2 hours, Operation 1 of Job 0 contributes at least 8 hours, Operation 2 of Job 0 contributes at least 5 hours, Operation 0 of Job 1 contributes at least 1 hour, and Operation 1 of Job 1 contributes at least 7 hours. + +Binary variables determine the order of operations on shared machines, ensuring no overlap. The objective is to determine the start times for all operations and their processing order on shared machines to minimize the makespan while satisfying all constraints.15The manufacturing facility produces custom parts for two jobs, Job 0 and Job 1, each consisting of operations performed on machines. The goal is to minimize the total completion time (makespan) by scheduling operations efficiently while respecting constraints. +Job 0 has three operations: Operation 0 takes 2 hours, Operation 1 takes 8 hours, and Operation 2 takes 5 hours. Job 1 has two operations: Operation 0 takes 1 hour, and Operation 1 takes 7 hours. +The makespan must be at least as large as the completion time of every operation: Operation 0 of Job 0 takes 2 hours, Operation 1 of Job 0 takes 8 hours, Operation 2 of Job 0 takes 5 hours, Operation 0 of Job 1 takes 1 hour, and Operation 1 of Job 1 takes 7 hours. Each machine can only process one operation at a time, and the operations of a job must be processes sequentially. For example, operation 1 of job 0 can only be processed after the completion of operation 0 of job 0. +The objective is to determine the start times for all operations and their processing order on shared machines to minimize the makespan while satisfying all constraints. + +Additional Information: + +There are 3 machines: M1, M2, M3 +Assignment of operations to machines: +Job 0: +Operation 0 → Machine M1 +Operation 1 → Machine M2 +Operation 2 → Machine M3 + +Job 1: +Operation 0 → Machine M2 +Operation 1 → Machine M115The manufacturing facility produces custom parts for two jobs, Job 0 and Job 1, each consisting of operations performed on shared machines. The goal is to minimize the total completion time (makespan) by scheduling operations efficiently while respecting constraints. +Job 0 has three operations: Operation 0 takes 2 hours, Operation 1 takes 8 hours, and Operation 2 takes 5 hours. Job 1 has two operations: Operation 0 takes 1 hour, and Operation 1 takes 7 hours. + The makespan must be at least as large as the completion time of every operation: Operation 0 of Job 0 takes 2 hours, Operation 1 of Job 0 takes 8 hours, Operation 2 of Job 0 takes 5 hours, Operation 0 of Job 1 takes 1 hour, and Operation 1 of Job 1 takes 7 hours. Each machine can only process one operation at a time, and the operations of a job must be processes sequentially. For example, operation 1 of job 0 can only be processed after the completion of operation 0 of job 0. + The objective is to determine the start times for all operations and their processing order on shared machines to minimize the makespan while satisfying all constraints. + + For Job 0, Operation 1 cannot start until Operation 0 has finished, with at least 2 hours between the finish of Operation 0 and the start of Operation 1. Operation 2 cannot start until Operation 1 has finished, with at least 8 hours between the finish of Operation 1 and the start of Operation 2. + + For Job 1, Operation 1 cannot start until Operation 0 has finished, with at least 1 hour between the finish of Operation 0 and the start of Operation 1. + + Operations on shared machines cannot overlap. If Operation 0 of Job 1 is scheduled before Operation 1 of Job 1, there must be at least 1 hour between the finish of Operation 0 and the start of Operation 1. If Operation 1 of Job 1 is scheduled before Operation 0 of Job 1, there must be at least 7 hours between the finish of Operation 1 and the start of Operation 0. + + For Job 0, if Operation 0 is scheduled before Operation 1, there must be at least 8 hours between the finish of Operation 0 and the start of Operation 1. If Operation 0 is scheduled before Operation 2, there must be at least 5 hours between the finish of Operation 0 and the start of Operation 2. If Operation 1 is scheduled before Operation 2, there must be at least 5 hours between the finish of Operation 1 and the start of Operation 2. Additional Information: + + The makespan must be at least as large as the completion time of every operation: Operation 0 There are 3 machines: M1, M2, M3 + Assignment of operations to machines: +Job 0 contributes at least 2 hours, 0: +Operation 1 of Job 0 contributes at least 8 hours, Operation 2 of Job 0 contributes at least 5 hours, → Machine M1 +Operation 0 of Job 1 contributes at least 1 hour, and → Machine M2 +Operation 1 of Job 1 contributes at least 7 hours. 2 → Machine M3 + + Binary variables determine the order of operations on shared machines, ensuring no overlap. The objective is to determine the start times for all operations and their processing order on shared machines to minimize the makespan while satisfying all constraints. Job 1: + Operation 0 → Machine M2 + Operation 1 → Machine M1No differencesIt was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
22The problem involves scheduling operations for three jobs (Job 0, Job 1, and Job 2) on a set of machines in a manufacturing facility to minimize the total completion time (makespan). Each job consists of a sequence of operations with specific processing times. Job 0 has 5 operations with processing times of 7, 9, 8, 3, and 9 hours. Job 1 has 6 operations with processing times of 1, 3, 4, 1, 2, and 9 hours. Job 2 has 7 operations with processing times of 9, 3, 1, 3, 3, 3, and 5 hours. Each operation must start only after the previous operation in the same job has been completed. For example, Operation 1 of Job 0 cannot start until Operation 0 has been completed (7 hours), and Operation 1 of Job 1 cannot start until Operation 0 has been completed (1 hour). Operations from different jobs that require the same machine must be scheduled in a specific order to avoid overlapping. Binary decision variables determine the order of operations on shared machines. For instance, if Operation 0 of Job 0 is scheduled before Operation 0 of Job 1, the binary variable is set to 1; otherwise, it is set to 0. A large constant (100,000) is used in the constraints to ensure they are only active when necessary. The makespan must be at least as long as the completion time of the last operation across all jobs. The start times for all operations must be non-negative, and the objective is to minimize the makespan while satisfying all precedence and machine capacity constraints. Specifically, the constraints include precedence constraints within each job, machine capacity constraints to prevent overlapping operations on the same machine, and makespan constraints to ensure the makespan is at least as large as the completion time of every operation. The binary variables are used to enforce the order of operations on shared machines, and the large constant ensures the constraints are only active when necessary. The start times and the makespan are continuous variables, and the binary variables are restricted to 0 or 1. The machine capacity constraints are formulated such that for each pair of operations assigned to the same machine, the start time of one operation plus its processing time must be less than or equal to the start time of the other operation plus the large constant multiplied by the complement of the binary variable. This ensures that the operations do not overlap on the same machine. The makespan constraints ensure that the makespan is at least as large as the completion time of every operation. The precedence constraints ensure that operations within the same job are scheduled in the correct order. The binary variables are used to determine the order of operations on shared machines, and the large constant is used to make the constraints active only when necessary. The start times for all operations must be non-negative, and the objective is to minimize the makespan while satisfying all constraints.36The problem involves scheduling operations for three jobs (Job 0, Job 1, and Job 2) on a set of machines in a manufacturing facility to minimize the total completion time (makespan). Each job consists of a sequence of operations with specific processing times. Job 0 has 5 operations with processing times of 7, 9, 8, 3, and 9 hours. Job 1 has 6 operations with processing times of 1, 3, 4, 1, 2, and 9 hours. Job 2 has 7 operations with processing times of 9, 3, 1, 3, 3, 3, and 5 hours. Each operation must start only after the previous operation in the same job has been completed. For example, Operation 1 of Job 0 cannot start until Operation 0 has been completed (7 hours), and Operation 1 of Job 1 cannot start until Operation 0 has been completed (1 hour). Operations from different jobs that require the same machine must be scheduled in a specific order to avoid overlapping. Binary decision variables determine the order of operations on shared machines. For instance, if Operation 0 of Job 0 is scheduled before Operation 0 of Job 1, the binary variable is set to 1; otherwise, it is set to 0. A large constant (1000) is used in the constraints to ensure they are only active when necessary. The makespan must be at least as long as the completion time of the last operation across all jobs. The start times for all operations must be non-negative, and the objective is to minimize the makespan while satisfying all precedence and machine capacity constraints. Specifically, the constraints include precedence constraints within each job, machine capacity constraints to prevent overlapping operations on the same machine, and makespan constraints to ensure the makespan is at least as large as the completion time of every operation. The binary variables are used to enforce the order of operations on shared machines, and the large constant ensures the constraints are only active when necessary. The start times for all operations must be non-negative, and the objective is to minimize the makespan while satisfying all constraints. + +ADDITIONAL INPUT DATA (explicit, needed to implement Gurobi model): + +1) Sets and indices +- Jobs: J = {0, 1, 2} +- Operations per job (indexed by k starting at 0): + - Job 0: K0 = {0, 1, 2, 3, 4} + - Job 1: K1 = {0, 1, 2, 3, 4, 5} + - Job 2: K2 = {0, 1, 2, 3, 4, 5, 6} + +2) Machines (explicit list) +- Machines: M = {M0, M1, M2, M3} + +3) Machine assignment for every operation (explicit mapping) +- Job 0 operations: + - (0,0) -> M0 + - (0,1) -> M1 + - (0,2) -> M2 + - (0,3) -> M0 + - (0,4) -> M3 +- Job 1 operations: + - (1,0) -> M1 + - (1,1) -> M2 + - (1,2) -> M0 + - (1,3) -> M3 + - (1,4) -> M1 + - (1,5) -> M2 +- Job 2 operations: + - (2,0) -> M2 + - (2,1) -> M0 + - (2,2) -> M1 + - (2,3) -> M3 + - (2,4) -> M2 + - (2,5) -> M1 + - (2,6) -> M3 + +4) Processing times (repeated here explicitly per operation) +- Job 0 durations: + - p(0,0) = 7 + - p(0,1) = 9 + - p(0,2) = 8 + - p(0,3) = 3 + - p(0,4) = 9 +- Job 1 durations: + - p(1,0) = 1 + - p(1,1) = 3 + - p(1,2) = 4 + - p(1,3) = 1 + - p(1,4) = 2 + - p(1,5) = 9 +- Job 2 durations: + - p(2,0) = 9 + - p(2,1) = 3 + - p(2,2) = 1 + - p(2,3) = 3 + - p(2,4) = 3 + - p(2,5) = 3 + - p(2,6) = 541The problem involves scheduling operations for three jobs (Job 0, Job 1, and Job 2) on a set of machines in a manufacturing facility to minimize the total completion time (makespan). Each job consists of a sequence of operations with specific processing times. Job 0 has 5 operations with processing times of 7, 9, 8, 3, and 9 hours. Job 1 has 6 operations with processing times of 1, 3, 4, 1, 2, and 9 hours. Job 2 has 7 operations with processing times of 9, 3, 1, 3, 3, 3, and 5 hours. Each operation must start only after the previous operation in the same job has been completed. For example, Operation 1 of Job 0 cannot start until Operation 0 has been completed (7 hours), and Operation 1 of Job 1 cannot start until Operation 0 has been completed (1 hour). Operations from different jobs that require the same machine must be scheduled in a specific order to avoid overlapping. Binary decision variables determine the order of operations on shared machines. For instance, if Operation 0 of Job 0 is scheduled before Operation 0 of Job 1, the binary variable is set to 1; otherwise, it is set to 0. A large constant (100,000) (1000) is used in the constraints to ensure they are only active when necessary. The makespan must be at least as long as the completion time of the last operation across all jobs. The start times for all operations must be non-negative, and the objective is to minimize the makespan while satisfying all precedence and machine capacity constraints. Specifically, the constraints include precedence constraints within each job, machine capacity constraints to prevent overlapping operations on the same machine, and makespan constraints to ensure the makespan is at least as large as the completion time of every operation. The binary variables are used to enforce the order of operations on shared machines, and the large constant ensures the constraints are only active when necessary. The start times and the makespan are continuous variables, and the binary variables are restricted to 0 or 1. The machine capacity constraints are formulated such that for each pair of operations assigned to the same machine, the start time of one operation plus its processing time must be less than or equal to the start time of the other operation plus the large constant multiplied by the complement of the binary variable. This ensures that the operations do not overlap on the same machine. The makespan constraints ensure that the makespan is at least as large as the completion time of every operation. The precedence constraints ensure that operations within the same job are scheduled in the correct order. The binary variables are used to determine the order of operations on shared machines, and the large constant is used to make the constraints active only when necessary. The start times for all operations must be non-negative, and the objective is to minimize the makespan while satisfying all constraints. + + ADDITIONAL INPUT DATA (explicit, needed to implement Gurobi model): + + 1) Sets and indices + - Jobs: J = {0, 1, 2} + - Operations per job (indexed by k starting at 0): + - Job 0: K0 = {0, 1, 2, 3, 4} + - Job 1: K1 = {0, 1, 2, 3, 4, 5} + - Job 2: K2 = {0, 1, 2, 3, 4, 5, 6} + + 2) Machines (explicit list) + - Machines: M = {M0, M1, M2, M3} + + 3) Machine assignment for every operation (explicit mapping) + - Job 0 operations: + - (0,0) -> M0 + - (0,1) -> M1 + - (0,2) -> M2 + - (0,3) -> M0 + - (0,4) -> M3 + - Job 1 operations: + - (1,0) -> M1 + - (1,1) -> M2 + - (1,2) -> M0 + - (1,3) -> M3 + - (1,4) -> M1 + - (1,5) -> M2 + - Job 2 operations: + - (2,0) -> M2 + - (2,1) -> M0 + - (2,2) -> M1 + - (2,3) -> M3 + - (2,4) -> M2 + - (2,5) -> M1 + - (2,6) -> M3 + + 4) Processing times (repeated here explicitly per operation) + - Job 0 durations: + - p(0,0) = 7 + - p(0,1) = 9 + - p(0,2) = 8 + - p(0,3) = 3 + - p(0,4) = 9 + - Job 1 durations: + - p(1,0) = 1 + - p(1,1) = 3 + - p(1,2) = 4 + - p(1,3) = 1 + - p(1,4) = 2 + - p(1,5) = 9 + - Job 2 durations: + - p(2,0) = 9 + - p(2,1) = 3 + - p(2,2) = 1 + - p(2,3) = 3 + - p(2,4) = 3 + - p(2,5) = 3 + - p(2,6) = 536 41While the statement required to avoid overlap of operations on the same machines, it did not state how many machines are available or the assignment of operations to machines.missing data
23The problem involves scheduling four jobs (Job 0, Job 1, Job 2, and Job 3) across multiple machines to minimize the total completion time (makespan). Each job consists of a sequence of operations that must be performed in a specific order, and each operation is assigned to a particular machine. The processing times for each operation are as follows: Job 0 has Operation 0 (6 units), Operation 1 (8 units), and Operation 2 (9 units); Job 1 has Operation 0 (3 units), Operation 1 (8 units), Operation 2 (5 units), Operation 3 (5 units), and Operation 4 (2 units); Job 2 has Operation 0 (8 units), Operation 1 (1 unit), and Operation 2 (6 units); Job 3 has Operation 0 (1 unit), Operation 1 (9 units), Operation 2 (4 units), and Operation 3 (5 units). + +Each machine can process only one operation at a time. If two operations are assigned to the same machine, they cannot overlap in time. The order of operations on the same machine is determined by binary decision variables. For example, if Operation 0 of Job 0 and Operation 3 of Job 1 are on the same machine, the binary variable determines whether Operation 0 of Job 0 is processed before Operation 3 of Job 1 or vice versa. Similar constraints apply to other operation pairs on the same machine, such as Operation 2 of Job 1 and Operation 4 of Job 1, Operation 2 of Job 0 and Operation 0 of Job 2, Operation 2 of Job 0 and Operation 3 of Job 3, Operation 0 of Job 2 and Operation 3 of Job 3, Operation 0 of Job 1 and Operation 1 of Job 2, Operation 0 of Job 1 and Operation 2 of Job 3, and Operation 1 of Job 2 and Operation 2 of Job 3. + +The objective is to minimize the makespan, which must be at least as large as the completion time of every operation. For example, the completion time of Operation 0 of Job 0 is at least 6 units, so the makespan must be at least 6 units. Similarly, the completion time of Operation 1 of Job 0 is at least 8 units, so the makespan must be at least 8 units. This applies to all operations, with the final operation of Job 3 requiring at least 5 units. + +Constraints include precedence constraints within jobs, ensuring that each operation starts only after the previous operation in the same job has completed. For example, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 has completed, which takes at least 6 units. Machine capacity constraints ensure that operations assigned to the same machine do not overlap, with binary variables determining the order of operations on the same machine. The makespan must be at least as large as the completion time of every operation, and all start times for operations must be non-negative. Binary variables must be either 0 or 1, determining the order of operations on the same machine. + +The goal is to find the optimal schedule that meets all these constraints while minimizing the total time required to complete all jobs.28The problem involves scheduling four jobs (Job 0, Job 1, Job 2, and Job 3) across multiple machines to minimize the total completion time (makespan). Each job consists of a sequence of operations that must be performed in a specific order, and each operation is assigned to a particular machine. The processing times for each operation are as follows: Job 0 has Operation 0 (6 units), Operation 1 (8 units), and Operation 2 (9 units); Job 1 has Operation 0 (3 units), Operation 1 (8 units), Operation 2 (5 units), Operation 3 (5 units), and Operation 4 (2 units); Job 2 has Operation 0 (8 units), Operation 1 (1 unit), and Operation 2 (6 units); Job 3 has Operation 0 (1 unit), Operation 1 (9 units), Operation 2 (4 units), and Operation 3 (5 units). + +Each machine can process only one operation at a time. If two operations are assigned to the same machine, they cannot overlap in time. The order of operations on the same machine is determined by binary decision variables. For example, if Operation 0 of Job 0 and Operation 3 of Job 1 are on the same machine, the binary variable determines whether Operation 0 of Job 0 is processed before Operation 3 of Job 1 or vice versa. Similar constraints apply to other operation pairs on the same machine, such as Operation 2 of Job 1 and Operation 4 of Job 1, Operation 2 of Job 0 and Operation 0 of Job 2, Operation 2 of Job 0 and Operation 3 of Job 3, Operation 0 of Job 2 and Operation 3 of Job 3, Operation 0 of Job 1 and Operation 1 of Job 2, Operation 0 of Job 1 and Operation 2 of Job 3, and Operation 1 of Job 2 and Operation 2 of Job 3. + +The objective is to minimize the makespan, which must be at least as large as the completion time of every operation. For example, the completion time of Operation 0 of Job 0 is at least 6 units, so the makespan must be at least 6 units. Similarly, the completion time of Operation 1 of Job 0 is at least 8 units, so the makespan must be at least 8 units. This applies to all operations, with the final operation of Job 3 requiring at least 5 units. + +Constraints include precedence constraints within jobs, ensuring that each operation starts only after the previous operation in the same job has completed. For example, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 has completed, which takes at least 6 units. Machine capacity constraints ensure that operations assigned to the same machine do not overlap, with binary variables determining the order of operations on the same machine. The makespan must be at least as large as the completion time of every operation, and all start times for operations must be non-negative. Binary variables must be either 0 or 1, determining the order of operations on the same machine. + +The goal is to find the optimal schedule that meets all these constraints while minimizing the total time required to complete all jobs. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3} +- For each job j in J, operations are indexed sequentially starting at 0. We denote operation (j,k) for job j and operation index k. +- Machines: M = {0, 1, 2} # machines are indexed 0..2 + +2) Machine assignment for every operation (explicit mapping) +- Job 0: Operation (0,0) -> Machine 0 +- Job 0: Operation (0,1) -> Machine 1 +- Job 0: Operation (0,2) -> Machine 2 + +- Job 1: Operation (1,0) -> Machine 0 +- Job 1: Operation (1,1) -> Machine 1 +- Job 1: Operation (1,2) -> Machine 2 +- Job 1: Operation (1,3) -> Machine 1 +- Job 1: Operation (1,4) -> Machine 2 + +- Job 2: Operation (2,0) -> Machine 2 +- Job 2: Operation (2,1) -> Machine 0 +- Job 2: Operation (2,2) -> Machine 1 + +- Job 3: Operation (3,0) -> Machine 0 +- Job 3: Operation (3,1) -> Machine 2 +- Job 3: Operation (3,2) -> Machine 1 +- Job 3: Operation (3,3) -> Machine 2 + +3) Processing durations (explicit list matching the original description) +- Job 0: durations = [6, 8, 9] +- Job 1: durations = [3, 8, 5, 5, 2] +- Job 2: durations = [8, 1, 6] +- Job 3: durations = [1, 9, 4, 5]38Constraints include precedence constraints within jobs, ensuring that each operation starts only after the previous operation in the same job has completed. For example, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 has completed, which takes at least 6 units. Machine capacity constraints ensure that operations assigned to the same machine do not overlap, with binary variables determining the order of operations on the same machine. The makespan must be at least as large as the completion time of every operation, and all start times for operations must be non-negative. Binary variables must be either 0 or 1, determining the order of operations on the same machine. + +The goal is to find the optimal schedule that meets all these constraints while minimizing the total time required to complete all jobs. + + ADDITIONAL INPUT DATA: + + 1) Sets and indices + - Jobs: J = {0, 1, 2, 3} + - For each job j in J, operations are indexed sequentially starting at 0. We denote operation (j,k) for job j and operation index k. + - Machines: M = {0, 1, 2} # machines are indexed 0..2 + + 2) Machine assignment for every operation (explicit mapping) + - Job 0: Operation (0,0) -> Machine 0 + - Job 0: Operation (0,1) -> Machine 1 + - Job 0: Operation (0,2) -> Machine 2 + + - Job 1: Operation (1,0) -> Machine 0 + - Job 1: Operation (1,1) -> Machine 1 + - Job 1: Operation (1,2) -> Machine 2 + - Job 1: Operation (1,3) -> Machine 1 + - Job 1: Operation (1,4) -> Machine 2 + + - Job 2: Operation (2,0) -> Machine 2 + - Job 2: Operation (2,1) -> Machine 0 + - Job 2: Operation (2,2) -> Machine 1 + + - Job 3: Operation (3,0) -> Machine 0 + - Job 3: Operation (3,1) -> Machine 2 + - Job 3: Operation (3,2) -> Machine 1 + - Job 3: Operation (3,3) -> Machine 2 + + 3) Processing durations (explicit list matching the original description) + - Job 0: durations = [6, 8, 9] + - Job 1: durations = [3, 8, 5, 5, 2] + - Job 2: durations = [8, 1, 6] + - Job 3: durations = [1, 9, 4, 5]28 38While the statement required to aoid overlap of operations on the same machines, it did not state how many machines are available or the assignment of operations to machines. missing data
24Imagine you are managing a supply chain network for a company that operates across six key locations (nodes). Each location has specific supply and demand requirements, and your goal is to optimize the flow of goods between these locations while minimizing the total cost, which includes both fixed costs for establishing connections and variable costs for shipping goods. + +#### Supply and Demand: +- **Location 0** has a demand of 116 units. +- **Location 1** has a demand of 134 units. +- **Location 2** has no net supply or demand (balance is 0). +- **Location 3** has a demand of 141 units. +- **Location 4** has a demand of 8 units. +- **Location 5** has a demand of 101 units. + +#### Costs and Capacities: +To establish a connection between any two locations, there is a fixed cost. Additionally, shipping goods between locations incurs a variable cost per unit. Each connection also has a maximum capacity, limiting the amount of goods that can be shipped. + +Here are the specific costs and capacities for each possible connection: + +- **Fixed Costs (for establishing connections):** + - From Location 0: To Location 1 costs 1972, to Location 2 costs 2030, to Location 3 costs 2048, to Location 4 costs 1951, to Location 5 costs 2043. + - From Location 1: To Location 0 costs 1993, to Location 2 costs 2029, to Location 3 costs 1975, to Location 4 costs 1967, to Location 5 costs 2026. + - From Location 2: To Location 0 costs 2042, to Location 1 costs 1973, to Location 3 costs 2001, to Location 4 costs 2034, to Location 5 costs 2035. + - From Location 3: To Location 0 costs 2007, to Location 1 costs 1968, to Location 2 costs 2004, to Location 4 costs 1996, to Location 5 costs 1969. + - From Location 4: To Location 0 costs 1973, to Location 1 costs 1955, to Location 2 costs 2038, to Location 3 costs 1953, to Location 5 costs 2000. + - From Location 5: To Location 0 costs 1999, to Location 1 costs 2031, to Location 2 costs 2011, to Location 3 costs 1965, to Location 4 costs 1984. + +- **Variable Costs (per unit shipped):** + - From Location 0: To Location 1 costs 21, to Location 2 costs 19, to Location 3 costs 20, to Location 4 costs 21, to Location 5 costs 21. + - From Location 1: To Location 0 costs 21, to Location 2 costs 19, to Location 3 costs 21, to Location 4 costs 19, to Location 5 costs 20. + - From Location 2: To Location 0 costs 20, to Location 1 costs 20, to Location 3 costs 21, to Location 4 costs 21, to Location 5 costs 20. + - From Location 3: To Location 0 costs 19, to Location 1 costs 19, to Location 2 costs 20, to Location 4 costs 19, to Location 5 costs 20. + - From Location 4: To Location 0 costs 20, to Location 1 costs 19, to Location 2 costs 20, to Location 3 costs 20, to Location 5 costs 19. + - From Location 5: To Location 0 costs 21, to Location 1 costs 20, to Location 2 costs 20, to Location 3 costs 19, to Location 4 costs 21. + +- **Maximum Capacities (units that can be shipped):** + - From Location 0: To Location 1 can handle up to 137 units, to Location 2 up to 154 units, to Location 3 up to 189 units, to Location 4 up to 188 units, to Location 5 up to 247 units. + - From Location 1: To Location 0 up to 163 units, to Location 2 up to 197 units, to Location 3 up to 242 units, to Location 4 up to 191 units, to Location 5 up to 135 units. + - From Location 2: To Location 0 up to 226 units, to Location 1 up to 155 units, to Location 3 up to 182 units, to Location 4 up to 189 units, to Location 5 up to 160 units. + - From Location 3: To Location 0 up to 179 units, to Location 1 up to 232 units, to Location 2 up to 210 units, to Location 4 up to 146 units, to Location 5 up to 241 units. + - From Location 4: To Location 0 up to 238 units, to Location 1 up to 164 units, to Location 2 up to 238 units, to Location 3 up to 218 units, to Location 5 up to 195 units. + - From Location 5: To Location 0 up to 237 units, to Location 1 up to 191 units, to Location 2 up to 165 units, to Location 3 up to 133 units, to Location 4 up to 191 units. + +#### Objective: +Your task is to determine which connections to establish and how much to ship between locations to meet all demand requirements while minimizing the total cost. The total cost includes both the fixed costs of establishing connections and the variable costs of shipping goods. + +#### Constraints: +1. **Capacity Constraints:** The amount of goods shipped between any two locations cannot exceed the maximum capacity of that connection. +2. **Flow Balance:** The total amount of goods entering a location minus the total amount leaving must equal the net supply or demand at that location. +3. **Binary Decisions:** For each connection, you must decide whether to establish it (yes or no), which incurs the fixed cost. + +By carefully analyzing these factors, you aim to find the most cost-effective way to manage the supply chain network, ensuring all locations receive the goods they need without exceeding capacity limits or incurring unnecessary costs.13033Imagine you are managing a supply chain network for a company that operates across six key locations (nodes). Each location has specific supply and demand requirements, and your goal is to optimize the flow of goods between these locations while minimizing the total cost, which includes both fixed costs for establishing connections and variable costs for shipping goods. + +#### Supply and Demand: +- **Location 0** has a demand of 116 units. +- **Location 1** has a demand of 134 units. +- **Location 2** has no net supply or demand (balance is 0). +- **Location 3** has a demand of 141 units. +- **Location 4** has a demand of 8 units. +- **Location 5** has a demand of 101 units. + +#### Costs and Capacities: +To establish a connection between any two locations, there is a fixed cost. Additionally, shipping goods between locations incurs a variable cost per unit. Each connection also has a maximum capacity, limiting the amount of goods that can be shipped. + +Here are the specific costs and capacities for each possible connection: + +- **Fixed Costs (for establishing connections):** + - From Location 0: To Location 1 costs 1972, to Location 2 costs 2030, to Location 3 costs 2048, to Location 4 costs 1951, to Location 5 costs 2043. + - From Location 1: To Location 0 costs 1993, to Location 2 costs 2029, to Location 3 costs 1975, to Location 4 costs 1967, to Location 5 costs 2026. + - From Location 2: To Location 0 costs 2042, to Location 1 costs 1973, to Location 3 costs 2001, to Location 4 costs 2034, to Location 5 costs 2035. + - From Location 3: To Location 0 costs 2007, to Location 1 costs 1968, to Location 2 costs 2004, to Location 4 costs 1996, to Location 5 costs 1969. + - From Location 4: To Location 0 costs 1973, to Location 1 costs 1955, to Location 2 costs 2038, to Location 3 costs 1953, to Location 5 costs 2000. + - From Location 5: To Location 0 costs 1999, to Location 1 costs 2031, to Location 2 costs 2011, to Location 3 costs 1965, to Location 4 costs 1984. + +- **Variable Costs (per unit shipped):** + - From Location 0: To Location 1 costs 21, to Location 2 costs 19, to Location 3 costs 20, to Location 4 costs 21, to Location 5 costs 21. + - From Location 1: To Location 0 costs 21, to Location 2 costs 19, to Location 3 costs 21, to Location 4 costs 19, to Location 5 costs 20. + - From Location 2: To Location 0 costs 20, to Location 1 costs 20, to Location 3 costs 21, to Location 4 costs 21, to Location 5 costs 20. + - From Location 3: To Location 0 costs 19, to Location 1 costs 19, to Location 2 costs 20, to Location 4 costs 19, to Location 5 costs 20. + - From Location 4: To Location 0 costs 20, to Location 1 costs 19, to Location 2 costs 20, to Location 3 costs 20, to Location 5 costs 19. + - From Location 5: To Location 0 costs 21, to Location 1 costs 20, to Location 2 costs 20, to Location 3 costs 19, to Location 4 costs 21. + +- **Maximum Capacities (units that can be shipped):** + - From Location 0: To Location 1 can handle up to 137 units, to Location 2 up to 154 units, to Location 3 up to 189 units, to Location 4 up to 188 units, to Location 5 up to 247 units. + - From Location 1: To Location 0 up to 163 units, to Location 2 up to 197 units, to Location 3 up to 242 units, to Location 4 up to 191 units, to Location 5 up to 135 units. + - From Location 2: To Location 0 up to 226 units, to Location 1 up to 155 units, to Location 3 up to 182 units, to Location 4 up to 189 units, to Location 5 up to 160 units. + - From Location 3: To Location 0 up to 179 units, to Location 1 up to 232 units, to Location 2 up to 210 units, to Location 4 up to 146 units, to Location 5 up to 241 units. + - From Location 4: To Location 0 up to 238 units, to Location 1 up to 164 units, to Location 2 up to 238 units, to Location 3 up to 218 units, to Location 5 up to 195 units. + - From Location 5: To Location 0 up to 237 units, to Location 1 up to 191 units, to Location 2 up to 165 units, to Location 3 up to 133 units, to Location 4 up to 191 units. + +#### Objective: +Your task is to determine which connections to establish and how much to ship between locations to meet all demand requirements while minimizing the total cost. The total cost includes both the fixed costs of establishing connections and the variable costs of shipping goods. + +#### Constraints: +1. **Capacity Constraints:** The amount of goods shipped between any two locations cannot exceed the maximum capacity of that connection. +2. **Flow Balance:** The total amount of goods entering a location minus the total amount leaving must equal the net supply or demand at that location. +3. **Binary Decisions:** For each connection, you must decide whether to establish it (yes or no), which incurs the fixed cost. + +By carefully analyzing these factors, you aim to find the most cost-effective way to manage the supply chain network, ensuring all locations receive the goods they need without exceeding capacity limits or incurring unnecessary costs.13033No differencesNo differencesok
25You are tasked with optimizing the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 33 distinct areas, each with a specific population size that needs to be covered by cell tower signals. The goal is to determine which cell towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + +For each of the 29 available cell tower locations (tower_0 to tower_28), you must decide whether to build the tower (a binary decision: build or not build). For each of the 33 regions (region_0 to region_32), you must decide whether the region will be covered by at least one active cell tower (a binary decision: cover or not cover). + +The objective is to maximize the total population covered by the deployed cell towers. The population sizes for the regions are as follows: Region 0: 4,901; Region 1: 2,727; Region 2: 2,994; Region 3: 2,984; Region 4: 3,693; Region 5: 2,615; Region 6: 3,354; Region 7: 4,965; Region 8: 4,223; Region 9: 4,738; Region 10: 2,345; Region 11: 3,915; Region 12: 4,068; Region 13: 2,613; Region 14: 3,140; Region 15: 4,894; Region 16: 3,503; Region 17: 3,733; Region 18: 4,275; Region 19: 3,795; Region 20: 3,803; Region 21: 3,497; Region 22: 2,403; Region 23: 3,441; Region 24: 3,116; Region 25: 2,091; Region 26: 4,494; Region 27: 3,593; Region 28: 2,090; Region 29: 3,397; Region 30: 3,973; Region 31: 2,362; Region 32: 2,068. + +Each region can only be considered covered if at least one of the cell towers that can serve it is built. For example, Region 0 can be covered if any of the following towers are built: Tower 1, Tower 2, Tower 3, Tower 4, Tower 5, Tower 6, Tower 9, Tower 12, Tower 14, Tower 15, Tower 16, Tower 17, Tower 19, or Tower 26. Similar constraints apply to all 33 regions, with each region having a specific set of towers that can provide coverage. + +The total cost of building the selected cell towers must not exceed the available budget of 2,996 units. The costs for building each tower are as follows: Tower 0: 227; Tower 1: 355; Tower 2: 287; Tower 3: 487; Tower 4: 497; Tower 5: 412; Tower 6: 495; Tower 7: 294; Tower 8: 212; Tower 9: 290; Tower 10: 427; Tower 11: 399; Tower 12: 361; Tower 13: 253; Tower 14: 350; Tower 15: 314; Tower 16: 234; Tower 17: 203; Tower 18: 257; Tower 19: 248; Tower 20: 406; Tower 21: 264; Tower 22: 277; Tower 23: 419; Tower 24: 426; Tower 25: 297; Tower 26: 474; Tower 27: 333; Tower 28: 491. + +Your task is to select which cell towers to build and which regions to cover, ensuring that the total population covered is maximized while staying within the budget. Each region can only be covered if at least one of its associated towers is built, and the total cost of building the selected towers must not exceed 2,996 units. The solution must balance the trade-off between coverage and cost to achieve the best possible outcome.113803You are tasked with optimizing the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 33 distinct areas, each with a specific population size that needs to be covered by cell tower signals. The goal is to determine which cell towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + +For each of the 29 available cell tower locations (tower_0 to tower_28), you must decide whether to build the tower (a binary decision: build or not build). For each of the 33 regions (region_0 to region_32), you must decide whether the region will be covered by at least one active cell tower (a binary decision: cover or not cover). + +The objective is to maximize the total population covered by the deployed cell towers. The population sizes for the regions are as follows: Region 0: 4,901; Region 1: 2,727; Region 2: 2,994; Region 3: 2,984; Region 4: 3,693; Region 5: 2,615; Region 6: 3,354; Region 7: 4,965; Region 8: 4,223; Region 9: 4,738; Region 10: 2,345; Region 11: 3,915; Region 12: 4,068; Region 13: 2,613; Region 14: 3,140; Region 15: 4,894; Region 16: 3,503; Region 17: 3,733; Region 18: 4,275; Region 19: 3,795; Region 20: 3,803; Region 21: 3,497; Region 22: 2,403; Region 23: 3,441; Region 24: 3,116; Region 25: 2,091; Region 26: 4,494; Region 27: 3,593; Region 28: 2,090; Region 29: 3,397; Region 30: 3,973; Region 31: 2,362; Region 32: 2,068. + +Each region can only be considered covered if at least one of the cell towers that can serve it is built. - Region 0 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 7, Tower 9, Tower 16, Tower 17, Tower 20, Tower 22, Tower 24, Tower 25, Tower 26, Tower 27. + - Region 1 can be covered by towers Tower 0, Tower 1, Tower 15, Tower 19, Tower 22, Tower 23, Tower 25, Tower 26, Tower 27, Tower 28. + - Region 2 can be covered by towers Tower 1, Tower 3, Tower 5, Tower 6, Tower 9, Tower 10, Tower 15, Tower 16, Tower 18, Tower 19, Tower 20, Tower 21, Tower 23, Tower 25. + - Region 3 can be covered by towers Tower 2, Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 11, Tower 13, Tower 16, Tower 17, Tower 18. + - Region 4 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 4, Tower 7, Tower 8, Tower 9, Tower 13, Tower 15, Tower 18, Tower 19, Tower 25, Tower 26, Tower 27. + - Region 5 can be covered by towers Tower 1, Tower 2, Tower 5, Tower 6, Tower 11, Tower 13, Tower 16, Tower 18, Tower 20, Tower 21, Tower 23, Tower 25, Tower 26. + - Region 6 can be covered by towers Tower 1, Tower 2, Tower 5, Tower 6, Tower 7, Tower 10, Tower 11, Tower 14, Tower 25, Tower 26, Tower 27. + - Region 7 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 9, Tower 12, Tower 15, Tower 16, Tower 19, Tower 21, Tower 22, Tower 26, Tower 27. + - Region 8 can be covered by towers Tower 1, Tower 4, Tower 6, Tower 9, Tower 12, Tower 14, Tower 20, Tower 21, Tower 22, Tower 24, Tower 25, Tower 26. + - Region 9 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 6, Tower 7, Tower 9, Tower 10, Tower 13, Tower 15, Tower 16, Tower 19, Tower 20, Tower 22, Tower 27. + - Region 10 can be covered by towers Tower 3, Tower 5, Tower 7, Tower 11, Tower 13, Tower 15, Tower 18, Tower 19, Tower 20, Tower 21, Tower 27. + - Region 11 can be covered by towers Tower 0, Tower 1, Tower 4, Tower 9, Tower 12, Tower 13, Tower 17, Tower 20, Tower 21, Tower 22, Tower 24, Tower 26, Tower 27. + - Region 12 can be covered by towers Tower 3, Tower 4, Tower 6, Tower 8, Tower 10, Tower 13, Tower 15, Tower 16, Tower 19, Tower 20, Tower 23, Tower 24, Tower 26, Tower 28. + - Region 13 can be covered by towers Tower 0, Tower 3, Tower 8, Tower 9, Tower 11, Tower 14, Tower 17, Tower 20, Tower 21, Tower 22, Tower 24, Tower 25, Tower 26, Tower 27. + - Region 14 can be covered by towers Tower 0, Tower 2, Tower 4, Tower 12, Tower 13, Tower 15, Tower 17, Tower 18, Tower 22, Tower 24, Tower 25. + - Region 15 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 7, Tower 12, Tower 13, Tower 16, Tower 19, Tower 21, Tower 24, Tower 25, Tower 26, Tower 28. + - Region 16 can be covered by towers Tower 1, Tower 2, Tower 9, Tower 11, Tower 18, Tower 19, Tower 22, Tower 23, Tower 24, Tower 25, Tower 26, Tower 27, Tower 28. + - Region 17 can be covered by towers Tower 5, Tower 7, Tower 10, Tower 11, Tower 15, Tower 16, Tower 17, Tower 19, Tower 20, Tower 25. + - Region 18 can be covered by towers Tower 6, Tower 7, Tower 9, Tower 10, Tower 17, Tower 19, Tower 20, Tower 21, Tower 22, Tower 24, Tower 25, Tower 27. + - Region 19 can be covered by towers Tower 0, Tower 2, Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 12, Tower 15, Tower 21, Tower 22, Tower 23, Tower 25, Tower 27. + - Region 20 can be covered by towers Tower 1, Tower 2, Tower 5, Tower 6, Tower 10, Tower 12, Tower 14, Tower 17, Tower 19, Tower 25, Tower 26. + - Region 21 can be covered by towers Tower 3, Tower 7, Tower 9, Tower 10, Tower 12, Tower 14, Tower 16, Tower 20, Tower 22, Tower 24, Tower 25. + - Region 22 can be covered by towers Tower 0, Tower 2, Tower 3, Tower 7, Tower 14, Tower 16, Tower 17, Tower 22, Tower 26, Tower 28. + - Region 23 can be covered by towers Tower 0, Tower 3, Tower 5, Tower 6, Tower 7, Tower 8, Tower 9, Tower 13, Tower 14, Tower 16, Tower 17, Tower 20, Tower 23, Tower 25. + - Region 24 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 6, Tower 8, Tower 9, Tower 10, Tower 14, Tower 20, Tower 22, Tower 25, Tower 26. + - Region 25 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 6, Tower 7, Tower 9, Tower 13, Tower 14, Tower 18, Tower 20, Tower 21, Tower 22, Tower 23, Tower 26, Tower 27. + - Region 26 can be covered by towers Tower 0, Tower 2, Tower 4, Tower 5, Tower 7, Tower 10, Tower 16, Tower 17, Tower 18, Tower 22, Tower 23, Tower 25, Tower 26. + - Region 27 can be covered by towers Tower 0, Tower 6, Tower 8, Tower 11, Tower 12, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 23, Tower 24, Tower 26, Tower 28. + - Region 28 can be covered by towers Tower 0, Tower 2, Tower 4, Tower 5, Tower 6, Tower 12, Tower 15, Tower 16, Tower 18, Tower 19, Tower 21, Tower 24. + - Region 29 can be covered by towers Tower 1, Tower 4, Tower 7, Tower 10, Tower 12, Tower 13, Tower 15, Tower 18, Tower 27, Tower 28. + - Region 30 can be covered by towers Tower 1, Tower 2, Tower 4, Tower 8, Tower 12, Tower 13, Tower 17, Tower 19, Tower 21, Tower 22, Tower 23, Tower 25, Tower 27, Tower 28. + - Region 31 can be covered by towers Tower 0, Tower 6, Tower 8, Tower 9, Tower 12, Tower 16, Tower 17, Tower 18, Tower 22, Tower 27, Tower 28. + - Region 32 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 5, Tower 6, Tower 7, Tower 9, Tower 12, Tower 15, Tower 17, Tower 23, Tower 26. +The total cost of building the selected cell towers must not exceed the available budget of 2,996 units. The costs for building each tower are as follows: Tower 0: 227; Tower 1: 355; Tower 2: 287; Tower 3: 487; Tower 4: 497; Tower 5: 412; Tower 6: 495; Tower 7: 294; Tower 8: 212; Tower 9: 290; Tower 10: 427; Tower 11: 399; Tower 12: 361; Tower 13: 253; Tower 14: 350; Tower 15: 314; Tower 16: 234; Tower 17: 203; Tower 18: 257; Tower 19: 248; Tower 20: 406; Tower 21: 264; Tower 22: 277; Tower 23: 419; Tower 24: 426; Tower 25: 297; Tower 26: 474; Tower 27: 333; Tower 28: 491. + +Your task is to select which cell towers to build and which regions to cover, ensuring that the total population covered is maximized while staying within the budget. Each region can only be covered if at least one of its associated towers is built, and the total cost of building the selected towers must not exceed 2,996 units. The solution must balance the trade-off between coverage and cost to achieve the best possible outcome.113803The objective is to maximize the total population covered by the deployed cell towers. The population sizes for the regions are as follows: Region 0: 4,901; Region 1: 2,727; Region 2: 2,994; Region 3: 2,984; Region 4: 3,693; Region 5: 2,615; Region 6: 3,354; Region 7: 4,965; Region 8: 4,223; Region 9: 4,738; Region 10: 2,345; Region 11: 3,915; Region 12: 4,068; Region 13: 2,613; Region 14: 3,140; Region 15: 4,894; Region 16: 3,503; Region 17: 3,733; Region 18: 4,275; Region 19: 3,795; Region 20: 3,803; Region 21: 3,497; Region 22: 2,403; Region 23: 3,441; Region 24: 3,116; Region 25: 2,091; Region 26: 4,494; Region 27: 3,593; Region 28: 2,090; Region 29: 3,397; Region 30: 3,973; Region 31: 2,362; Region 32: 2,068. + +Each region can only be considered covered if at least one of the cell towers that can serve it is built. For example, - Region 0 can be covered if any of the following by towers Tower 0, Tower 3, Tower 4, Tower 7, Tower 9, Tower 16, Tower 17, Tower 20, Tower 22, Tower 24, Tower 25, Tower 26, Tower 27. + - Region 1 can be covered by towers Tower 0, Tower 1, Tower 15, Tower 19, Tower 22, Tower 23, Tower 25, Tower 26, Tower 27, Tower 28. + - Region 2 can be covered by towers Tower 1, Tower 3, Tower 5, Tower 6, Tower 9, Tower 10, Tower 15, Tower 16, Tower 18, Tower 19, Tower 20, Tower 21, Tower 23, Tower 25. + - Region 3 can be covered by towers Tower 2, Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 11, Tower 13, Tower 16, Tower 17, Tower 18. + - Region 4 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 4, Tower 7, Tower 8, Tower 9, Tower 13, Tower 15, Tower 18, Tower 19, Tower 25, Tower 26, Tower 27. + - Region 5 can be covered by towers Tower 1, Tower 2, Tower 5, Tower 6, Tower 11, Tower 13, Tower 16, Tower 18, Tower 20, Tower 21, Tower 23, Tower 25, Tower 26. + - Region 6 can be covered by towers Tower 1, Tower 2, Tower 5, Tower 6, Tower 7, Tower 10, Tower 11, Tower 14, Tower 25, Tower 26, Tower 27. + - Region 7 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 9, Tower 12, Tower 15, Tower 16, Tower 19, Tower 21, Tower 22, Tower 26, Tower 27. + - Region 8 can be covered by towers Tower 1, Tower 4, Tower 6, Tower 9, Tower 12, Tower 14, Tower 20, Tower 21, Tower 22, Tower 24, Tower 25, Tower 26. + - Region 9 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 6, Tower 7, Tower 9, Tower 10, Tower 13, Tower 15, Tower 16, Tower 19, Tower 20, Tower 22, Tower 27. + - Region 10 can be covered by towers Tower 3, Tower 5, Tower 7, Tower 11, Tower 13, Tower 15, Tower 18, Tower 19, Tower 20, Tower 21, Tower 27. + - Region 11 can be covered by towers Tower 0, Tower 1, Tower 4, Tower 9, Tower 12, Tower 13, Tower 17, Tower 20, Tower 21, Tower 22, Tower 24, Tower 26, Tower 27. + - Region 12 can be covered by towers Tower 3, Tower 4, Tower 6, Tower 8, Tower 10, Tower 13, Tower 15, Tower 16, Tower 19, Tower 20, Tower 23, Tower 24, Tower 26, Tower 28. + - Region 13 can be covered by towers Tower 0, Tower 3, Tower 8, Tower 9, Tower 11, Tower 14, Tower 17, Tower 20, Tower 21, Tower 22, Tower 24, Tower 25, Tower 26, Tower 27. + - Region 14 can be covered by towers Tower 0, Tower 2, Tower 4, Tower 12, Tower 13, Tower 15, Tower 17, Tower 18, Tower 22, Tower 24, Tower 25. + - Region 15 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 7, Tower 12, Tower 13, Tower 16, Tower 19, Tower 21, Tower 24, Tower 25, Tower 26, Tower 28. + - Region 16 can be covered by towers Tower 1, Tower 2, Tower 9, Tower 11, Tower 18, Tower 19, Tower 22, Tower 23, Tower 24, Tower 25, Tower 26, Tower 27, Tower 28. + - Region 17 can be covered by towers Tower 5, Tower 7, Tower 10, Tower 11, Tower 15, Tower 16, Tower 17, Tower 19, Tower 20, Tower 25. + - Region 18 can be covered by towers Tower 6, Tower 7, Tower 9, Tower 10, Tower 17, Tower 19, Tower 20, Tower 21, Tower 22, Tower 24, Tower 25, Tower 27. + - Region 19 can be covered by towers Tower 0, Tower 2, Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 12, Tower 15, Tower 21, Tower 22, Tower 23, Tower 25, Tower 27. + - Region 20 can be covered by towers Tower 1, Tower 2, Tower 5, Tower 6, Tower 10, Tower 12, Tower 14, Tower 17, Tower 19, Tower 25, Tower 26. + - Region 21 can be covered by towers are built: Tower 3, Tower 7, Tower 9, Tower 10, Tower 12, Tower 14, Tower 16, Tower 20, Tower 22, Tower 24, Tower 25. + - Region 22 can be covered by towers Tower 0, Tower 2, Tower 3, Tower 7, Tower 14, Tower 16, Tower 17, Tower 22, Tower 26, Tower 28. + - Region 23 can be covered by towers Tower 0, Tower 3, Tower 5, Tower 6, Tower 7, Tower 8, Tower 9, Tower 13, Tower 14, Tower 16, Tower 17, Tower 20, Tower 23, Tower 25. + - Region 24 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 6, Tower 8, Tower 9, Tower 10, Tower 14, Tower 20, Tower 22, Tower 25, Tower 26. + - Region 25 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 6, Tower 7, Tower 9, Tower 13, Tower 14, Tower 18, Tower 20, Tower 21, Tower 22, Tower 23, Tower 26, Tower 27. + - Region 26 can be covered by towers Tower 0, Tower 2, Tower 4, Tower 5, Tower 7, Tower 10, Tower 16, Tower 17, Tower 18, Tower 22, Tower 23, Tower 25, Tower 26. + - Region 27 can be covered by towers Tower 0, Tower 6, Tower 9, 8, Tower 11, Tower 12, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 23, Tower 24, Tower 26, Tower 28. + - Region 28 can be covered by towers Tower 0, Tower 2, Tower 4, Tower 5, Tower 6, Tower 12, Tower 15, Tower 16, Tower 18, Tower 19, Tower 21, Tower 24. + - Region 29 can be covered by towers Tower 1, Tower 4, Tower 7, Tower 10, Tower 12, Tower 13, Tower 15, Tower 18, Tower 27, Tower 28. + - Region 30 can be covered by towers Tower 1, Tower 2, Tower 4, Tower 8, Tower 12, Tower 13, Tower 17, Tower 19, or Tower 26. Similar constraints apply to all 33 regions, with each region having a specific set of 21, Tower 22, Tower 23, Tower 25, Tower 27, Tower 28. + - Region 31 can be covered by towers that Tower 0, Tower 6, Tower 8, Tower 9, Tower 12, Tower 16, Tower 17, Tower 18, Tower 22, Tower 27, Tower 28. + - Region 32 can provide coverage. be covered by towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 5, Tower 6, Tower 7, Tower 9, Tower 12, Tower 15, Tower 17, Tower 23, Tower 26. +The total cost of building the selected cell towers must not exceed the available budget of 2,996 units. The costs for building each tower are as follows: Tower 0: 227; Tower 1: 355; Tower 2: 287; Tower 3: 487; Tower 4: 497; Tower 5: 412; Tower 6: 495; Tower 7: 294; Tower 8: 212; Tower 9: 290; Tower 10: 427; Tower 11: 399; Tower 12: 361; Tower 13: 253; Tower 14: 350; Tower 15: 314; Tower 16: 234; Tower 17: 203; Tower 18: 257; Tower 19: 248; Tower 20: 406; Tower 21: 264; Tower 22: 277; Tower 23: 419; Tower 24: 426; Tower 25: 297; Tower 26: 474; Tower 27: 333; Tower 28: 491. + +Your task is to select which cell towers to build and which regions to cover, ensuring that the total population covered is maximized while staying within the budget. Each region can only be covered if at least one of its associated towers is built, and the total cost of building the selected towers must not exceed 2,996 units. The solution must balance the trade-off between coverage and cost to achieve the best possible outcome.No differencesMissing data about which towers cover which regionsmissing data
26You are tasked with managing the distribution of resources across a network of nine cities. Each city has a specific supply or demand for resources, and your goal is to ensure that all cities meet their requirements while minimizing the total transportation cost. The transportation costs vary depending on the route taken between cities, and each route has a maximum capacity limit for resource shipments. + +#### City Requirements: +- **City 0** has a net demand of **41 units**. +- **City 1** has a net demand of **23 units**. +- **City 2** has a net supply of **15 units** (indicated by a negative demand of -15). +- **City 3** has a net demand of **2 units**. +- **City 4** has a net supply of **13 units** (indicated by a negative demand of -13). +- **City 5** has a net supply of **4 units** (indicated by a negative demand of -4). +- **City 6** has a net supply of **7 units** (indicated by a negative demand of -7). +- **City 7** has a net supply of **2 units** (indicated by a negative demand of -2). +- **City 8** has a net supply of **25 units** (indicated by a negative demand of -25). + +#### Transportation Costs and Capacities: +The cost of shipping resources between cities varies, and each route has a maximum capacity. Below are the costs and capacities for each possible route: + +- **From City 0**: + - To City 1: Cost = 7, Capacity = 35 + - To City 2: Cost = 9, Capacity = 60 + - To City 3: Cost = 8, Capacity = 48 + - To City 4: Cost = 8, Capacity = 58 + - To City 5: Cost = 10, Capacity = 39 + - To City 6: Cost = 8, Capacity = 60 + - To City 7: Cost = 6, Capacity = 54 + - To City 8: Cost = 9, Capacity = 31 + +- **From City 1**: + - To City 0: Cost = 10, Capacity = 39 + - To City 2: Cost = 8, Capacity = 31 + - To City 3: Cost = 9, Capacity = 56 + - To City 4: Cost = 8, Capacity = 60 + - To City 5: Cost = 9, Capacity = 37 + - To City 6: Cost = 10, Capacity = 54 + - To City 7: Cost = 9, Capacity = 39 + - To City 8: Cost = 7, Capacity = 50 + +- **From City 2**: + - To City 0: Cost = 10, Capacity = 31 + - To City 1: Cost = 6, Capacity = 35 + - To City 3: Cost = 7, Capacity = 50 + - To City 4: Cost = 7, Capacity = 53 + - To City 5: Cost = 9, Capacity = 36 + - To City 6: Cost = 10, Capacity = 48 + - To City 7: Cost = 8, Capacity = 46 + - To City 8: Cost = 10, Capacity = 60 + +- **From City 3**: + - To City 0: Cost = 7, Capacity = 50 + - To City 1: Cost = 6, Capacity = 38 + - To City 2: Cost = 7, Capacity = 39 + - To City 4: Cost = 10, Capacity = 47 + - To City 5: Cost = 10, Capacity = 37 + - To City 6: Cost = 6, Capacity = 54 + - To City 7: Cost = 7, Capacity = 44 + - To City 8: Cost = 10, Capacity = 55 + +- **From City 4**: + - To City 0: Cost = 9, Capacity = 31 + - To City 1: Cost = 8, Capacity = 31 + - To City 2: Cost = 7, Capacity = 30 + - To City 3: Cost = 7, Capacity = 40 + - To City 5: Cost = 6, Capacity = 60 + - To City 6: Cost = 6, Capacity = 46 + - To City 7: Cost = 7, Capacity = 31 + - To City 8: Cost = 9, Capacity = 49 + +- **From City 5**: + - To City 0: Cost = 7, Capacity = 34 + - To City 1: Cost = 7, Capacity = 30 + - To City 2: Cost = 6, Capacity = 51 + - To City 3: Cost = 8, Capacity = 32 + - To City 4: Cost = 9, Capacity = 33 + - To City 6: Cost = 7, Capacity = 47 + - To City 7: Cost = 7, Capacity = 42 + - To City 8: Cost = 8, Capacity = 43 + +- **From City 6**: + - To City 0: Cost = 8, Capacity = 55 + - To City 1: Cost = 7, Capacity = 48 + - To City 2: Cost = 7, Capacity = 35 + - To City 3: Cost = 7, Capacity = 31 + - To City 4: Cost = 8, Capacity = 43 + - To City 5: Cost = 8, Capacity = 56 + - To City 7: Cost = 6, Capacity = 41 + - To City 8: Cost = 6, Capacity = 55 + +- **From City 7**: + - To City 0: Cost = 7, Capacity = 31 + - To City 1: Cost = 6, Capacity = 31 + - To City 2: Cost = 9, Capacity = 38 + - To City 3: Cost = 8, Capacity = 34 + - To City 4: Cost = 7, Capacity = 31 + - To City 5: Cost = 8, Capacity = 40 + - To City 6: Cost = 9, Capacity = 39 + - To City 8: Cost = 6, Capacity = 42 + +- **From City 8**: + - To City 0: Cost = 8, Capacity = 48 + - To City 1: Cost = 8, Capacity = 60 + - To City 2: Cost = 8, Capacity = 36 + - To City 3: Cost = 10, Capacity = 47 + - To City 4: Cost = 8, Capacity = 32 + - To City 5: Cost = 7, Capacity = 36 + - To City 6: Cost = 6, Capacity = 57 + - To City 7: Cost = 10, Capacity = 35 + +#### Objective: +Your task is to determine the optimal amount of resources to ship between cities to meet all demands and supplies while minimizing the total transportation cost. The shipments must respect the capacity limits of each route, and the total flow of resources into and out of each city must balance according to their respective requirements. + +#### Constraints: +1. **Flow Balance**: For each city, the total resources shipped into the city plus its supply must equal the total resources shipped out of the city plus its demand. +2. **Capacity Limits**: The amount of resources shipped on any route cannot exceed the maximum capacity for that route. +3. **Non-Negativity**: All shipments must be non-negative. + +By carefully planning the shipments, you aim to achieve the lowest possible total cost while ensuring all cities receive the resources they need.493You are tasked with managing the distribution of resources across a network of nine cities. Each city has a specific supply or demand for resources, and your goal is to ensure that all cities meet their requirements while minimizing the total transportation cost. The transportation costs vary depending on the route taken between cities, and each route has a maximum capacity limit for resource shipments. + +#### City Requirements: +- **City 0** has a net demand of **41 units**. +- **City 1** has a net demand of **23 units**. +- **City 2** has a net supply of **15 units** (indicated by a negative demand of -15). +- **City 3** has a net demand of **2 units**. +- **City 4** has a net supply of **13 units** (indicated by a negative demand of -13). +- **City 5** has a net supply of **4 units** (indicated by a negative demand of -4). +- **City 6** has a net supply of **7 units** (indicated by a negative demand of -7). +- **City 7** has a net supply of **2 units** (indicated by a negative demand of -2). +- **City 8** has a net supply of **25 units** (indicated by a negative demand of -25). + +#### Transportation Costs and Capacities: +The cost of shipping resources between cities varies, and each route has a maximum capacity. Below are the costs and capacities for each possible route: + +- **From City 0**: + - To City 1: Cost = 7, Capacity = 35 + - To City 2: Cost = 9, Capacity = 60 + - To City 3: Cost = 8, Capacity = 48 + - To City 4: Cost = 8, Capacity = 58 + - To City 5: Cost = 10, Capacity = 39 + - To City 6: Cost = 8, Capacity = 60 + - To City 7: Cost = 6, Capacity = 54 + - To City 8: Cost = 9, Capacity = 31 + +- **From City 1**: + - To City 0: Cost = 10, Capacity = 39 + - To City 2: Cost = 8, Capacity = 31 + - To City 3: Cost = 9, Capacity = 56 + - To City 4: Cost = 8, Capacity = 60 + - To City 5: Cost = 9, Capacity = 37 + - To City 6: Cost = 10, Capacity = 54 + - To City 7: Cost = 9, Capacity = 39 + - To City 8: Cost = 7, Capacity = 50 + +- **From City 2**: + - To City 0: Cost = 10, Capacity = 31 + - To City 1: Cost = 6, Capacity = 35 + - To City 3: Cost = 7, Capacity = 50 + - To City 4: Cost = 7, Capacity = 53 + - To City 5: Cost = 9, Capacity = 36 + - To City 6: Cost = 10, Capacity = 48 + - To City 7: Cost = 8, Capacity = 46 + - To City 8: Cost = 10, Capacity = 60 + +- **From City 3**: + - To City 0: Cost = 7, Capacity = 50 + - To City 1: Cost = 6, Capacity = 38 + - To City 2: Cost = 7, Capacity = 39 + - To City 4: Cost = 10, Capacity = 47 + - To City 5: Cost = 10, Capacity = 37 + - To City 6: Cost = 6, Capacity = 54 + - To City 7: Cost = 7, Capacity = 44 + - To City 8: Cost = 10, Capacity = 55 + +- **From City 4**: + - To City 0: Cost = 9, Capacity = 31 + - To City 1: Cost = 8, Capacity = 31 + - To City 2: Cost = 7, Capacity = 30 + - To City 3: Cost = 7, Capacity = 40 + - To City 5: Cost = 6, Capacity = 60 + - To City 6: Cost = 6, Capacity = 46 + - To City 7: Cost = 7, Capacity = 31 + - To City 8: Cost = 9, Capacity = 49 + +- **From City 5**: + - To City 0: Cost = 7, Capacity = 34 + - To City 1: Cost = 7, Capacity = 30 + - To City 2: Cost = 6, Capacity = 51 + - To City 3: Cost = 8, Capacity = 32 + - To City 4: Cost = 9, Capacity = 33 + - To City 6: Cost = 7, Capacity = 47 + - To City 7: Cost = 7, Capacity = 42 + - To City 8: Cost = 8, Capacity = 43 + +- **From City 6**: + - To City 0: Cost = 8, Capacity = 55 + - To City 1: Cost = 7, Capacity = 48 + - To City 2: Cost = 7, Capacity = 35 + - To City 3: Cost = 7, Capacity = 31 + - To City 4: Cost = 8, Capacity = 43 + - To City 5: Cost = 8, Capacity = 56 + - To City 7: Cost = 6, Capacity = 41 + - To City 8: Cost = 6, Capacity = 55 + +- **From City 7**: + - To City 0: Cost = 7, Capacity = 31 + - To City 1: Cost = 6, Capacity = 31 + - To City 2: Cost = 9, Capacity = 38 + - To City 3: Cost = 8, Capacity = 34 + - To City 4: Cost = 7, Capacity = 31 + - To City 5: Cost = 8, Capacity = 40 + - To City 6: Cost = 9, Capacity = 39 + - To City 8: Cost = 6, Capacity = 42 + +- **From City 8**: + - To City 0: Cost = 8, Capacity = 48 + - To City 1: Cost = 8, Capacity = 60 + - To City 2: Cost = 8, Capacity = 36 + - To City 3: Cost = 10, Capacity = 47 + - To City 4: Cost = 8, Capacity = 32 + - To City 5: Cost = 7, Capacity = 36 + - To City 6: Cost = 6, Capacity = 57 + - To City 7: Cost = 10, Capacity = 35 + +#### Objective: +Your task is to determine the optimal amount of resources to ship between cities to meet all demands and supplies while minimizing the total transportation cost. The shipments must respect the capacity limits of each route, and the total flow of resources into and out of each city must balance according to their respective requirements. + +#### Constraints: +1. **Flow Balance**: For each city, the total resources shipped into the city plus its supply must equal the total resources shipped out of the city plus its demand. +2. **Capacity Limits**: The amount of resources shipped on any route cannot exceed the maximum capacity for that route. +3. **Non-Negativity**: All shipments must be non-negative. + +By carefully planning the shipments, you aim to achieve the lowest possible total cost while ensuring all cities receive the resources they need.493No differencesNo differencesok
27Imagine you are managing the distribution of goods across a network of eight cities. Each city has a specific supply or demand for goods, and your goal is to ensure that all cities receive the goods they need while minimizing the total transportation costs. The transportation costs vary depending on the route taken between cities, and each route has a maximum capacity for the amount of goods that can be shipped. + +#### **Cities and Their Requirements:** +- **City 0** has a net demand of **86 units** of goods. +- **City 1** has a net demand of **19 units** of goods. +- **City 2** has a net supply of **23 units** of goods (meaning it can send out 23 units). +- **City 3** has a net supply of **37 units** of goods. +- **City 4** has a net supply of **24 units** of goods. +- **City 5** has a balanced supply and demand, meaning it neither supplies nor demands goods. +- **City 6** has a net demand of **4 units** of goods. +- **City 7** has a net supply of **25 units** of goods. + +#### **Transportation Costs and Capacities:** +The cost of shipping goods between cities varies, and each route has a maximum capacity. Below are the costs and capacities for each possible route: + +- **From City 0:** + - To City 1: Cost = **9**, Capacity = **44** + - To City 2: Cost = **6**, Capacity = **47** + - To City 3: Cost = **8**, Capacity = **60** + - To City 4: Cost = **7**, Capacity = **40** + - To City 5: Cost = **7**, Capacity = **39** + - To City 6: Cost = **10**, Capacity = **40** + - To City 7: Cost = **7**, Capacity = **52** + +- **From City 1:** + - To City 0: Cost = **6**, Capacity = **32** + - To City 2: Cost = **7**, Capacity = **59** + - To City 3: Cost = **6**, Capacity = **58** + - To City 4: Cost = **8**, Capacity = **58** + - To City 5: Cost = **7**, Capacity = **51** + - To City 6: Cost = **6**, Capacity = **31** + - To City 7: Cost = **6**, Capacity = **34** + +- **From City 2:** + - To City 0: Cost = **10**, Capacity = **59** + - To City 1: Cost = **10**, Capacity = **49** + - To City 3: Cost = **9**, Capacity = **51** + - To City 4: Cost = **9**, Capacity = **39** + - To City 5: Cost = **10**, Capacity = **34** + - To City 6: Cost = **6**, Capacity = **47** + - To City 7: Cost = **6**, Capacity = **55** + +- **From City 3:** + - To City 0: Cost = **9**, Capacity = **60** + - To City 1: Cost = **7**, Capacity = **34** + - To City 2: Cost = **7**, Capacity = **30** + - To City 4: Cost = **7**, Capacity = **32** + - To City 5: Cost = **9**, Capacity = **43** + - To City 6: Cost = **7**, Capacity = **30** + - To City 7: Cost = **10**, Capacity = **43** + +- **From City 4:** + - To City 0: Cost = **9**, Capacity = **57** + - To City 1: Cost = **9**, Capacity = **38** + - To City 2: Cost = **6**, Capacity = **54** + - To City 3: Cost = **10**, Capacity = **46** + - To City 5: Cost = **6**, Capacity = **30** + - To City 6: Cost = **9**, Capacity = **48** + - To City 7: Cost = **7**, Capacity = **43** + +- **From City 5:** + - To City 0: Cost = **9**, Capacity = **44** + - To City 1: Cost = **6**, Capacity = **35** + - To City 2: Cost = **8**, Capacity = **32** + - To City 3: Cost = **7**, Capacity = **58** + - To City 4: Cost = **8**, Capacity = **56** + - To City 6: Cost = **6**, Capacity = **48** + - To City 7: Cost = **8**, Capacity = **49** + +- **From City 6:** + - To City 0: Cost = **9**, Capacity = **52** + - To City 1: Cost = **10**, Capacity = **48** + - To City 2: Cost = **7**, Capacity = **33** + - To City 3: Cost = **8**, Capacity = **39** + - To City 4: Cost = **10**, Capacity = **36** + - To City 5: Cost = **10**, Capacity = **47** + - To City 7: Cost = **9**, Capacity = **49** + +- **From City 7:** + - To City 0: Cost = **6**, Capacity = **38** + - To City 1: Cost = **8**, Capacity = **44** + - To City 2: Cost = **10**, Capacity = **33** + - To City 3: Cost = **6**, Capacity = **30** + - To City 4: Cost = **10**, Capacity = **60** + - To City 5: Cost = **7**, Capacity = **39** + - To City 6: Cost = **8**, Capacity = **55** + +#### **Objective:** +Your task is to determine the optimal amount of goods to ship between each pair of cities to meet all demands and supplies while minimizing the total transportation cost. The total cost is calculated by summing the cost of shipping goods across all routes, weighted by the amount shipped on each route. + +#### **Constraints:** +1. **Flow Balance:** For each city, the total amount of goods shipped into the city plus its own supply must equal the total amount shipped out of the city plus its demand. +2. **Capacity Limits:** The amount of goods shipped on any route cannot exceed the maximum capacity of that route. +3. **Non-Negativity:** The amount of goods shipped on any route must be zero or positive. + +By carefully planning the shipments, you aim to ensure that all cities receive the goods they need while keeping the total transportation costs as low as possible.875Imagine you are managing the distribution of goods across a network of eight cities. Each city has a specific supply or demand for goods, and your goal is to ensure that all cities receive the goods they need while minimizing the total transportation costs. The transportation costs vary depending on the route taken between cities, and each route has a maximum capacity for the amount of goods that can be shipped. + +#### **Cities and Their Requirements:** +- **City 0** has a net demand of **86 units** of goods. +- **City 1** has a net demand of **19 units** of goods. +- **City 2** has a net supply of **23 units** of goods (meaning it can send out 23 units). +- **City 3** has a net supply of **37 units** of goods. +- **City 4** has a net supply of **24 units** of goods. +- **City 5** has a balanced supply and demand, meaning it neither supplies nor demands goods. +- **City 6** has a net demand of **4 units** of goods. +- **City 7** has a net supply of **25 units** of goods. + +#### **Transportation Costs and Capacities:** +The cost of shipping goods between cities varies, and each route has a maximum capacity. Below are the costs and capacities for each possible route: + +- **From City 0:** + - To City 1: Cost = **9**, Capacity = **44** + - To City 2: Cost = **6**, Capacity = **47** + - To City 3: Cost = **8**, Capacity = **60** + - To City 4: Cost = **7**, Capacity = **40** + - To City 5: Cost = **7**, Capacity = **39** + - To City 6: Cost = **10**, Capacity = **40** + - To City 7: Cost = **7**, Capacity = **52** + +- **From City 1:** + - To City 0: Cost = **6**, Capacity = **32** + - To City 2: Cost = **7**, Capacity = **59** + - To City 3: Cost = **6**, Capacity = **58** + - To City 4: Cost = **8**, Capacity = **58** + - To City 5: Cost = **7**, Capacity = **51** + - To City 6: Cost = **6**, Capacity = **31** + - To City 7: Cost = **6**, Capacity = **34** + +- **From City 2:** + - To City 0: Cost = **10**, Capacity = **59** + - To City 1: Cost = **10**, Capacity = **49** + - To City 3: Cost = **9**, Capacity = **51** + - To City 4: Cost = **9**, Capacity = **39** + - To City 5: Cost = **10**, Capacity = **34** + - To City 6: Cost = **6**, Capacity = **47** + - To City 7: Cost = **6**, Capacity = **55** + +- **From City 3:** + - To City 0: Cost = **9**, Capacity = **60** + - To City 1: Cost = **7**, Capacity = **34** + - To City 2: Cost = **7**, Capacity = **30** + - To City 4: Cost = **7**, Capacity = **32** + - To City 5: Cost = **9**, Capacity = **43** + - To City 6: Cost = **7**, Capacity = **30** + - To City 7: Cost = **10**, Capacity = **43** + +- **From City 4:** + - To City 0: Cost = **9**, Capacity = **57** + - To City 1: Cost = **9**, Capacity = **38** + - To City 2: Cost = **6**, Capacity = **54** + - To City 3: Cost = **10**, Capacity = **46** + - To City 5: Cost = **6**, Capacity = **30** + - To City 6: Cost = **9**, Capacity = **48** + - To City 7: Cost = **7**, Capacity = **43** + +- **From City 5:** + - To City 0: Cost = **9**, Capacity = **44** + - To City 1: Cost = **6**, Capacity = **35** + - To City 2: Cost = **8**, Capacity = **32** + - To City 3: Cost = **7**, Capacity = **58** + - To City 4: Cost = **8**, Capacity = **56** + - To City 6: Cost = **6**, Capacity = **48** + - To City 7: Cost = **8**, Capacity = **49** + +- **From City 6:** + - To City 0: Cost = **9**, Capacity = **52** + - To City 1: Cost = **10**, Capacity = **48** + - To City 2: Cost = **7**, Capacity = **33** + - To City 3: Cost = **8**, Capacity = **39** + - To City 4: Cost = **10**, Capacity = **36** + - To City 5: Cost = **10**, Capacity = **47** + - To City 7: Cost = **9**, Capacity = **49** + +- **From City 7:** + - To City 0: Cost = **6**, Capacity = **38** + - To City 1: Cost = **8**, Capacity = **44** + - To City 2: Cost = **10**, Capacity = **33** + - To City 3: Cost = **6**, Capacity = **30** + - To City 4: Cost = **10**, Capacity = **60** + - To City 5: Cost = **7**, Capacity = **39** + - To City 6: Cost = **8**, Capacity = **55** + +#### **Objective:** +Your task is to determine the optimal amount of goods to ship between each pair of cities to meet all demands and supplies while minimizing the total transportation cost. The total cost is calculated by summing the cost of shipping goods across all routes, weighted by the amount shipped on each route. + +#### **Constraints:** +1. **Flow Balance:** For each city, the total amount of goods shipped into the city plus its own supply must equal the total amount shipped out of the city plus its demand. +2. **Capacity Limits:** The amount of goods shipped on any route cannot exceed the maximum capacity of that route. +3. **Non-Negativity:** The amount of goods shipped on any route must be zero or positive. + +By carefully planning the shipments, you aim to ensure that all cities receive the goods they need while keeping the total transportation costs as low as possible.875No differencesNo differencesok
28Assign seven staff members (Person 0 to Person 6) to three projects (Project 0, Project 1, and Project 2) such that each staff member is assigned to exactly one project. Each project has specific skill requirements: Project 0 requires a skill level of 8 for Skill 0 and 8 for Skill 1, Project 1 requires a skill level of 7 for Skill 0 and 9 for Skill 1, and Project 2 requires a skill level of 9 for Skill 0 and 7 for Skill 1. Each staff member has specific skill levels: Person 0 has Skill 0 level 5 and Skill 1 level 7, Person 1 has Skill 0 level 6 and Skill 1 level 7, Person 2 has Skill 0 level 5 and Skill 1 level 7, Person 3 has Skill 0 level 6 and Skill 1 level 7, Person 4 has Skill 0 level 5 and Skill 1 level 6, Person 5 has Skill 0 level 5 and Skill 1 level 6, and Person 6 has Skill 0 level 5 and Skill 1 level 7. The attained skill level for each project is the sum of the skill levels of the assigned staff members. The skill shortage for each project is the difference between the required skill level and the attained skill level. The goal is to minimize the maximum skill shortage across all projects and skills. Constraints include ensuring that each staff member is assigned to exactly one project, and that the skill shortages for Project 0, Skill 0 must be less than or equal to -8, for Project 0, Skill 1 must be less than or equal to -8, for Project 1, Skill 0 must be less than or equal to -7, for Project 1, Skill 1 must be less than or equal to -9, for Project 2, Skill 0 must be less than or equal to -9, and for Project 2, Skill 1 must be less than or equal to -7.0Assign seven staff members (Person 0 to Person 6) to three projects (Project 0, Project 1, and Project 2) such that each staff member is assigned to exactly one project. Each project has specific skill requirements: Project 0 requires a skill level of 8 for Skill 0 and 8 for Skill 1, Project 1 requires a skill level of 7 for Skill 0 and 9 for Skill 1, and Project 2 requires a skill level of 9 for Skill 0 and 7 for Skill 1. Each staff member has specific skill levels: Person 0 has Skill 0 level 5 and Skill 1 level 7, Person 1 has Skill 0 level 6 and Skill 1 level 7, Person 2 has Skill 0 level 5 and Skill 1 level 7, Person 3 has Skill 0 level 6 and Skill 1 level 7, Person 4 has Skill 0 level 5 and Skill 1 level 6, Person 5 has Skill 0 level 5 and Skill 1 level 6, and Person 6 has Skill 0 level 5 and Skill 1 level 7. The attained skill level for each project is the sum of the skill levels of the assigned staff members. The skill shortage for each project is the difference between the required skill level and the attained skill level. The goal is to minimize the maximum skill shortage across all projects and skills. Constraints include ensuring that each staff member is assigned to exactly one project, and that the skill shortages for Project 0, Skill 0 must be less than or equal to -8, for Project 0, Skill 1 must be less than or equal to -8, for Project 1, Skill 0 must be less than or equal to -7, for Project 1, Skill 1 must be less than or equal to -9, for Project 2, Skill 0 must be less than or equal to -9, and for Project 2, Skill 1 must be less than or equal to -7.0No differencesNo differencesok
29The problem involves scheduling six jobs, labeled Job 0 through Job 5, in a manufacturing facility. Each job consists of two operations that must be performed in a specific sequence. The goal is to minimize the total completion time, known as the makespan, by determining the start times for all operations and the order of operations on shared machines. Each operation has a processing time of 1 unit of time. Operations within the same job must follow a strict sequence, meaning the second operation cannot start until the first operation has completed. Certain operations are assigned to the same machine, and these operations cannot overlap in time. To enforce this, binary decision variables are used to determine the order of operations on shared machines. For example, if Operation 1 of Job 1 is scheduled before Operation 1 of Job 3 on the same machine, the start time of Operation 1 of Job 3 must be at least 1 unit of time after the completion of Operation 1 of Job 1. The makespan is defined as the maximum completion time across all operations, and it must be at least as large as the completion time of every operation. The constraints include precedence constraints within jobs, ensuring that each operation starts only after the previous operation in the same job has completed; machine capacity constraints, ensuring that operations assigned to the same machine do not overlap; and non-negativity constraints, ensuring that all start times are non-negative. The objective is to minimize the makespan, ensuring that all jobs are completed as quickly as possible. The binary decision variables, such as those determining the order of Operation 1 of Job 1 and Operation 1 of Job 3, are critical in enforcing the machine capacity constraints. The numerical parameters include a large constant (100000) used in the constraints involving binary variables to ensure the correct sequencing of operations on shared machines. The start times and the makespan are general variables that can take on any real value within the bounds defined by the constraints.5The problem involves scheduling six jobs, labeled Job 0 through Job 5, in a manufacturing facility. Each job consists of two operations that must be performed in a specific sequence. The goal is to minimize the total completion time, known as the makespan, by determining the start times for all operations and the order of operations on shared machines. Each operation has a processing time of 1 unit of time. Operations within the same job must follow a strict sequence, meaning the second operation cannot start until the first operation has completed. Certain operations are assigned to the same machine, and these operations cannot overlap in time. To enforce this, binary decision variables are used to determine the order of operations on shared machines. For example, if Operation 1 of Job 1 is scheduled before Operation 1 of Job 3 on the same machine, the start time of Operation 1 of Job 3 must be at least 1 unit of time after the completion of Operation 1 of Job 1. The makespan is defined as the maximum completion time across all operations, and it must be at least as large as the completion time of every operation. The constraints include precedence constraints within jobs, ensuring that each operation starts only after the previous operation in the same job has completed; machine capacity constraints, ensuring that operations assigned to the same machine do not overlap; and non-negativity constraints, ensuring that all start times are non-negative. The objective is to minimize the makespan, ensuring that all jobs are completed as quickly as possible. The binary decision variables, such as those determining the order of Operation 1 of Job 1 and Operation 1 of Job 3, are critical in enforcing the machine capacity constraints. The numerical parameters include a large constant (1000) used in the constraints involving binary variables to ensure the correct sequencing of operations on shared machines. The start times and the makespan are general variables that can take on any real value within the bounds defined by the constraints. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3, 4, 5} +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Machines: M = {A, B, C} + +2) Machine assignment for every operation (explicit) +- Operation (0,1) -> Machine A +- Operation (0,2) -> Machine B +- Operation (1,1) -> Machine A +- Operation (1,2) -> Machine C +- Operation (2,1) -> Machine B +- Operation (2,2) -> Machine C +- Operation (3,1) -> Machine A +- Operation (3,2) -> Machine B +- Operation (4,1) -> Machine C +- Operation (4,2) -> Machine B +- Operation (5,1) -> Machine A +- Operation (5,2) -> Machine C5The problem involves scheduling six jobs, labeled Job 0 through Job 5, in a manufacturing facility. Each job consists of two operations that must be performed in a specific sequence. The goal is to minimize the total completion time, known as the makespan, by determining the start times for all operations and the order of operations on shared machines. Each operation has a processing time of 1 unit of time. Operations within the same job must follow a strict sequence, meaning the second operation cannot start until the first operation has completed. Certain operations are assigned to the same machine, and these operations cannot overlap in time. To enforce this, binary decision variables are used to determine the order of operations on shared machines. For example, if Operation 1 of Job 1 is scheduled before Operation 1 of Job 3 on the same machine, the start time of Operation 1 of Job 3 must be at least 1 unit of time after the completion of Operation 1 of Job 1. The makespan is defined as the maximum completion time across all operations, and it must be at least as large as the completion time of every operation. The constraints include precedence constraints within jobs, ensuring that each operation starts only after the previous operation in the same job has completed; machine capacity constraints, ensuring that operations assigned to the same machine do not overlap; and non-negativity constraints, ensuring that all start times are non-negative. The objective is to minimize the makespan, ensuring that all jobs are completed as quickly as possible. The binary decision variables, such as those determining the order of Operation 1 of Job 1 and Operation 1 of Job 3, are critical in enforcing the machine capacity constraints. The numerical parameters include a large constant (100000) (1000) used in the constraints involving binary variables to ensure the correct sequencing of operations on shared machines. The start times and the makespan are general variables that can take on any real value within the bounds defined by the constraints. + + ADDITIONAL INPUT DATA: + + 1) Sets and indices + - Jobs: J = {0, 1, 2, 3, 4, 5} + - For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). + - Machines: M = {A, B, C} + + 2) Machine assignment for every operation (explicit) + - Operation (0,1) -> Machine A + - Operation (0,2) -> Machine B + - Operation (1,1) -> Machine A + - Operation (1,2) -> Machine C + - Operation (2,1) -> Machine B + - Operation (2,2) -> Machine C + - Operation (3,1) -> Machine A + - Operation (3,2) -> Machine B + - Operation (4,1) -> Machine C + - Operation (4,2) -> Machine B + - Operation (5,1) -> Machine A + - Operation (5,2) -> Machine CNo differencesIt was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
30You are tasked with planning the most efficient route for a delivery service that must visit seven cities: City 0, City 1, City 2, City 3, City 4, City 5, and City 6. The goal is to minimize the total travel cost while ensuring that each city is visited exactly once and that the delivery route forms a complete loop, starting and ending at the same city. The travel costs between each pair of cities are as follows: From City 0 to City 1 costs 285, to City 2 costs 275, to City 3 costs 303, to City 4 costs 282, to City 5 costs 308, and to City 6 costs 290. From City 1 to City 0 costs 293, to City 2 costs 291, to City 3 costs 319, to City 4 costs 295, to City 5 costs 294, and to City 6 costs 324. From City 2 to City 0 costs 309, to City 1 costs 325, to City 3 costs 313, to City 4 costs 292, to City 5 costs 275, and to City 6 costs 320. From City 3 to City 0 costs 307, to City 1 costs 286, to City 2 costs 305, to City 4 costs 297, to City 5 costs 299, and to City 6 costs 278. From City 4 to City 0 costs 320, to City 1 costs 313, to City 2 costs 278, to City 3 costs 300, to City 5 costs 318, and to City 6 costs 325. From City 5 to City 0 costs 281, to City 1 costs 291, to City 2 costs 305, to City 3 costs 285, to City 4 costs 298, and to City 6 costs 292. From City 6 to City 0 costs 323, to City 1 costs 296, to City 2 costs 307, to City 3 costs 295, to City 4 costs 293, and to City 5 costs 301. Each city must be visited by exactly one incoming route and must have exactly one outgoing route. To ensure the route forms a single loop and does not break into smaller disconnected cycles, the position of each city in the tour is tracked. The position of each city must be an integer between 0 and 6, and if a route from City i to City j is taken, the position of City i must be at least one less than the position of City j. The objective is to minimize the total travel cost across all routes while ensuring all constraints are satisfied.1987You are tasked with planning the most efficient route for a delivery service that must visit seven cities: City 0, City 1, City 2, City 3, City 4, City 5, and City 6. The goal is to minimize the total travel cost while ensuring that each city is visited exactly once and that the delivery route forms a complete loop, starting and ending at the same city. The travel costs between each pair of cities are as follows: From City 0 to City 1 costs 285, to City 2 costs 275, to City 3 costs 303, to City 4 costs 282, to City 5 costs 308, and to City 6 costs 290. From City 1 to City 0 costs 293, to City 2 costs 291, to City 3 costs 319, to City 4 costs 295, to City 5 costs 294, and to City 6 costs 324. From City 2 to City 0 costs 309, to City 1 costs 325, to City 3 costs 313, to City 4 costs 292, to City 5 costs 275, and to City 6 costs 320. From City 3 to City 0 costs 307, to City 1 costs 286, to City 2 costs 305, to City 4 costs 297, to City 5 costs 299, and to City 6 costs 278. From City 4 to City 0 costs 320, to City 1 costs 313, to City 2 costs 278, to City 3 costs 300, to City 5 costs 318, and to City 6 costs 325. From City 5 to City 0 costs 281, to City 1 costs 291, to City 2 costs 305, to City 3 costs 285, to City 4 costs 298, and to City 6 costs 292. From City 6 to City 0 costs 323, to City 1 costs 296, to City 2 costs 307, to City 3 costs 295, to City 4 costs 293, and to City 5 costs 301. The task is to find the shortest possible route that visits each city exactly once, and return to the origin city. What is the minimal total distance travelled?1987You are tasked with planning the most efficient route for a delivery service that must visit seven cities: City 0, City 1, City 2, City 3, City 4, City 5, and City 6. The goal is to minimize the total travel cost while ensuring that each city is visited exactly once and that the delivery route forms a complete loop, starting and ending at the same city. The travel costs between each pair of cities are as follows: From City 0 to City 1 costs 285, to City 2 costs 275, to City 3 costs 303, to City 4 costs 282, to City 5 costs 308, and to City 6 costs 290. From City 1 to City 0 costs 293, to City 2 costs 291, to City 3 costs 319, to City 4 costs 295, to City 5 costs 294, and to City 6 costs 324. From City 2 to City 0 costs 309, to City 1 costs 325, to City 3 costs 313, to City 4 costs 292, to City 5 costs 275, and to City 6 costs 320. From City 3 to City 0 costs 307, to City 1 costs 286, to City 2 costs 305, to City 4 costs 297, to City 5 costs 299, and to City 6 costs 278. From City 4 to City 0 costs 320, to City 1 costs 313, to City 2 costs 278, to City 3 costs 300, to City 5 costs 318, and to City 6 costs 325. From City 5 to City 0 costs 281, to City 1 costs 291, to City 2 costs 305, to City 3 costs 285, to City 4 costs 298, and to City 6 costs 292. From City 6 to City 0 costs 323, to City 1 costs 296, to City 2 costs 307, to City 3 costs 295, to City 4 costs 293, and to City 5 costs 301. Each city must be visited by exactly one incoming route and must have exactly one outgoing route. To ensure The task is to find the shortest possible route forms a single loop and does not break into smaller disconnected cycles, the position of each city in the tour is tracked. The position of that visits each city must be an integer between 0 and 6, exactly once, and if a route from City i return to City j is taken, the position of City i must be at least one less than the position of City j. The objective origin city. What is to minimize the minimal total travel cost across all routes while ensuring all constraints are satisfied. distance travelled?No differencesWe removed information that was not the description of the problem, but describing the variables/constraints of a possible model. Also, this additional information could lead to a wrong model (MTZ constraint on the return to the first node)ambiguous
31There are six jobs, each consisting of two operations that must be performed in a specific order. Each operation takes 1 unit of time to complete. Operations are assigned to shared machines, and each machine can handle only one operation at a time. The goal is to schedule the operations to minimize the total completion time, known as the makespan, which is the latest completion time of all operations. + +For each job, the second operation cannot start until the first operation is completed. If two operations are assigned to the same machine, they must be scheduled in a specific order, determined by binary decision variables. These variables indicate whether one operation starts before or after the other on the same machine. A large constant, 100,000, is used to enforce the sequencing logic in the constraints. + +The start times of all operations must be non-negative, and the makespan must be at least as large as the completion time of every operation. The objective is to determine the start times for all operations and the order in which they are processed on shared machines, ensuring that all precedence constraints are satisfied, overlaps on shared machines are avoided, and the makespan is minimized.5There are six jobs, each consisting of two operations that must be performed in a specific order. Each operation takes 1 unit of time to complete. Operations are assigned to shared machines, and each machine can handle only one operation at a time. The goal is to schedule the operations to minimize the total completion time, known as the makespan, which is the latest completion time of all operations. + +For each job, the second operation cannot start until the first operation is completed. If two operations are assigned to the same machine, they must be scheduled in a specific order, determined by binary decision variables. These variables indicate whether one operation starts before or after the other on the same machine. A large constant, 1000, is used to enforce the sequencing logic in the constraints. + +The start times of all operations must be non-negative, and the makespan must be at least as large as the completion time of every operation. The objective is to determine the start times for all operations and the order in which they are processed on shared machines, ensuring that all precedence constraints are satisfied, overlaps on shared machines are avoided, and the makespan is minimized. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {1, 2, 3, 4, 5, 6} +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Machines: M = {1, 2, 3} + +2) Machine assignment for every operation (which machine executes the operation) +- Operation (1,1) -> Machine 1 +- Operation (1,2) -> Machine 2 +- Operation (2,1) -> Machine 1 +- Operation (2,2) -> Machine 3 +- Operation (3,1) -> Machine 2 +- Operation (3,2) -> Machine 3 +- Operation (4,1) -> Machine 1 +- Operation (4,2) -> Machine 2 +- Operation (5,1) -> Machine 2 +- Operation (5,2) -> Machine 3 +- Operation (6,1) -> Machine 1 +- Operation (6,2) -> Machine 3 + +3) Operation durations +- All operations have processing time p[j,k] = 1 for every job j and operation k. + +Notes about index conventions and mapping +- Each operation (j,k) will be treated as a single task in the model. For pairwise sequencing constraints we consider only pairs of operations that share the same machine.5There are six jobs, each consisting of two operations that must be performed in a specific order. Each operation takes 1 unit of time to complete. Operations are assigned to shared machines, and each machine can handle only one operation at a time. The goal is to schedule the operations to minimize the total completion time, known as the makespan, which is the latest completion time of all operations. + +For each job, the second operation cannot start until the first operation is completed. If two operations are assigned to the same machine, they must be scheduled in a specific order, determined by binary decision variables. These variables indicate whether one operation starts before or after the other on the same machine. A large constant, 100,000, 1000, is used to enforce the sequencing logic in the constraints. + +The start times of all operations must be non-negative, and the makespan must be at least as large as the completion time of every operation. The objective is to determine the start times for all operations and the order in which they are processed on shared machines, ensuring that all precedence constraints are satisfied, overlaps on shared machines are avoided, and the makespan is minimized. + + ADDITIONAL INPUT DATA: + + 1) Sets and indices + - Jobs: J = {1, 2, 3, 4, 5, 6} + - For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). + - Machines: M = {1, 2, 3} + + 2) Machine assignment for every operation (which machine executes the operation) + - Operation (1,1) -> Machine 1 + - Operation (1,2) -> Machine 2 + - Operation (2,1) -> Machine 1 + - Operation (2,2) -> Machine 3 + - Operation (3,1) -> Machine 2 + - Operation (3,2) -> Machine 3 + - Operation (4,1) -> Machine 1 + - Operation (4,2) -> Machine 2 + - Operation (5,1) -> Machine 2 + - Operation (5,2) -> Machine 3 + - Operation (6,1) -> Machine 1 + - Operation (6,2) -> Machine 3 + + 3) Operation durations + - All operations have processing time p[j,k = 1 for every job j and operation k.] + + Notes about index conventions and mapping + - Each operation (j,k) will be treated as a single task in the model. For pairwise sequencing constraints we consider only pairs of operations that share the same machine.No differencesWhile the statement required to avoid overlap of operations on the same machines, it did not state how many machines are available or the assignment of operations to machines. missing data
32There are five jobs, each consisting of two operations that must be performed in a specific order. The first operation of each job must finish before the second operation of the same job can start. All operations are assigned to shared machines, and no two operations on the same machine can overlap in time. The goal is to schedule the operations such that the total completion time, or makespan, is minimized. + +Each operation has a processing time of 1 unit. For each job, the start time of the second operation must be at least 1 unit after the start time of the first operation. For every pair of operations assigned to the same machine, one operation must finish before the other can start, enforced by binary decision variables that determine the order of operations on the machine. A large constant (100,000) is used in the constraints to ensure that operations do not overlap on the same machine. The makespan must be at least as large as the completion time of every operation. + +The decision variables include the start time of each operation and the binary variables that determine the order of operations on shared machines. The objective is to minimize the makespan while satisfying all precedence and machine constraints.3There are five jobs, each consisting of two operations that must be performed in a specific order. The first operation of each job must finish before the second operation of the same job can start. All operations are assigned to shared machines, and no two operations on the same machine can overlap in time. The goal is to schedule the operations such that the total completion time, or makespan, is minimized. + +Each operation has a processing time of 1 unit. For each job, the start time of the second operation must be at least 1 unit after the start time of the first operation. For every pair of operations assigned to the same machine, one operation must finish before the other can start, enforced by binary decision variables that determine the order of operations on the machine. A large constant (1000) is used in the constraints to ensure that operations do not overlap on the same machine. The makespan must be at least as large as the completion time of every operation. + +The decision variables include the start time of each operation and the binary variables that determine the order of operations on shared machines. The objective is to minimize the makespan while satisfying all precedence and machine constraints. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {1, 2, 3, 4, 5} +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Machines: M = {1, 2} + +2) Machine assignment for every operation (which machine executes the operation) +- Operation (1,1) -> Machine 1 +- Operation (1,2) -> Machine 2 +- Operation (2,1) -> Machine 2 +- Operation (2,2) -> Machine 1 +- Operation (3,1) -> Machine 1 +- Operation (3,2) -> Machine 2 +- Operation (4,1) -> Machine 2 +- Operation (4,2) -> Machine 1 +- Operation (5,1) -> Machine 1 +- Operation (5,2) -> Machine 2 + +3) Processing times +- All operations have processing time p_{j,k} = 1 for all jobs j and operations k.5There are five jobs, each consisting of two operations that must be performed in a specific order. The first operation of each job must finish before the second operation of the same job can start. All operations are assigned to shared machines, and no two operations on the same machine can overlap in time. The goal is to schedule the operations such that the total completion time, or makespan, is minimized. + +Each operation has a processing time of 1 unit. For each job, the start time of the second operation must be at least 1 unit after the start time of the first operation. For every pair of operations assigned to the same machine, one operation must finish before the other can start, enforced by binary decision variables that determine the order of operations on the machine. A large constant (100,000) (1000) is used in the constraints to ensure that operations do not overlap on the same machine. The makespan must be at least as large as the completion time of every operation. + +The decision variables include the start time of each operation and the binary variables that determine the order of operations on shared machines. The objective is to minimize the makespan while satisfying all precedence and machine constraints. + + ADDITIONAL INPUT DATA: + + 1) Sets and indices + - Jobs: J = {1, 2, 3, 4, 5} + - For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). + - Machines: M = {1, 2} + + 2) Machine assignment for every operation (which machine executes the operation) + - Operation (1,1) -> Machine 1 + - Operation (1,2) -> Machine 2 + - Operation (2,1) -> Machine 2 + - Operation (2,2) -> Machine 1 + - Operation (3,1) -> Machine 1 + - Operation (3,2) -> Machine 2 + - Operation (4,1) -> Machine 2 + - Operation (4,2) -> Machine 1 + - Operation (5,1) -> Machine 1 + - Operation (5,2) -> Machine 2 + + 3) Processing times + - All operations have processing time p_{j,k} = 1 for all jobs j and operations k.3 5It was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
33You are tasked with managing the logistics of a supply chain network involving seven key facilities (nodes) that produce and distribute goods. Each facility has specific supply and demand requirements, and the goal is to minimize the total cost of establishing connections between facilities and transporting goods while ensuring all demands are met. + +Facility 0 has a net demand of 106 units, Facility 1 has a net demand of 144 units, Facility 2 has no net supply or demand, Facility 3 has no net supply or demand, Facility 4 has a net demand of 202 units, Facility 5 has a net demand of 40 units, and Facility 6 has a net demand of 8 units. + +Establishing a connection between any two facilities incurs a fixed cost, and transporting goods between them incurs a variable cost per unit. Each connection has a maximum capacity for goods that can be transported. The fixed costs, variable costs, and maximum capacities for each possible connection are as follows: + +- Fixed Costs (for establishing connections): + - Facility 0 to Facility 1: 1985 units + - Facility 0 to Facility 2: 1968 units + - Facility 0 to Facility 3: 2043 units + - Facility 0 to Facility 4: 2035 units + - Facility 0 to Facility 5: 2043 units + - Facility 0 to Facility 6: 1987 units + - Facility 1 to Facility 0: 2035 units + - Facility 1 to Facility 2: 2044 units + - Facility 1 to Facility 3: 1985 units + - Facility 1 to Facility 4: 2047 units + - Facility 1 to Facility 5: 2003 units + - Facility 1 to Facility 6: 2005 units + - Facility 2 to Facility 0: 2000 units + - Facility 2 to Facility 1: 1991 units + - Facility 2 to Facility 3: 2005 units + - Facility 2 to Facility 4: 1969 units + - Facility 2 to Facility 5: 2010 units + - Facility 2 to Facility 6: 1965 units + - Facility 3 to Facility 0: 1962 units + - Facility 3 to Facility 1: 2024 units + - Facility 3 to Facility 2: 2025 units + - Facility 3 to Facility 4: 2043 units + - Facility 3 to Facility 5: 2034 units + - Facility 3 to Facility 6: 2017 units + - Facility 4 to Facility 0: 1974 units + - Facility 4 to Facility 1: 2012 units + - Facility 4 to Facility 2: 2009 units + - Facility 4 to Facility 3: 1993 units + - Facility 4 to Facility 5: 1991 units + - Facility 4 to Facility 6: 2040 units + - Facility 5 to Facility 0: 2044 units + - Facility 5 to Facility 1: 2022 units + - Facility 5 to Facility 2: 2025 units + - Facility 5 to Facility 3: 1955 units + - Facility 5 to Facility 4: 2001 units + - Facility 5 to Facility 6: 2029 units + - Facility 6 to Facility 0: 1968 units + - Facility 6 to Facility 1: 1956 units + - Facility 6 to Facility 2: 1986 units + - Facility 6 to Facility 3: 2030 units + - Facility 6 to Facility 4: 2001 units + - Facility 6 to Facility 5: 1961 units + +- Variable Costs (per unit transported): + - Facility 0 to Facility 1: 19 units + - Facility 0 to Facility 2: 19 units + - Facility 0 to Facility 3: 19 units + - Facility 0 to Facility 4: 19 units + - Facility 0 to Facility 5: 21 units + - Facility 0 to Facility 6: 21 units + - Facility 1 to Facility 0: 21 units + - Facility 1 to Facility 2: 19 units + - Facility 1 to Facility 3: 20 units + - Facility 1 to Facility 4: 20 units + - Facility 1 to Facility 5: 21 units + - Facility 1 to Facility 6: 21 units + - Facility 2 to Facility 0: 21 units + - Facility 2 to Facility 1: 19 units + - Facility 2 to Facility 3: 21 units + - Facility 2 to Facility 4: 21 units + - Facility 2 to Facility 5: 21 units + - Facility 2 to Facility 6: 20 units + - Facility 3 to Facility 0: 21 units + - Facility 3 to Facility 1: 19 units + - Facility 3 to Facility 2: 21 units + - Facility 3 to Facility 4: 19 units + - Facility 3 to Facility 5: 19 units + - Facility 3 to Facility 6: 20 units + - Facility 4 to Facility 0: 21 units + - Facility 4 to Facility 1: 21 units + - Facility 4 to Facility 2: 21 units + - Facility 4 to Facility 3: 20 units + - Facility 4 to Facility 5: 20 units + - Facility 4 to Facility 6: 19 units + - Facility 5 to Facility 0: 21 units + - Facility 5 to Facility 1: 19 units + - Facility 5 to Facility 2: 20 units + - Facility 5 to Facility 3: 20 units + - Facility 5 to Facility 4: 20 units + - Facility 5 to Facility 6: 21 units + - Facility 6 to Facility 0: 19 units + - Facility 6 to Facility 1: 21 units + - Facility 6 to Facility 2: 19 units + - Facility 6 to Facility 3: 20 units + - Facility 6 to Facility 4: 21 units + - Facility 6 to Facility 5: 21 units + +- Maximum Capacities (for goods transported): + - Facility 0 to Facility 1: 153 units + - Facility 0 to Facility 2: 202 units + - Facility 0 to Facility 3: 185 units + - Facility 0 to Facility 4: 134 units + - Facility 0 to Facility 5: 248 units + - Facility 0 to Facility 6: 170 units + - Facility 1 to Facility 0: 214 units + - Facility 1 to Facility 2: 184 units + - Facility 1 to Facility 3: 193 units + - Facility 1 to Facility 4: 181 units + - Facility 1 to Facility 5: 153 units + - Facility 1 to Facility 6: 139 units + - Facility 2 to Facility 0: 217 units + - Facility 2 to Facility 1: 238 units + - Facility 2 to Facility 3: 231 units + - Facility 2 to Facility 4: 130 units + - Facility 2 to Facility 5: 203 units + - Facility 2 to Facility 6: 232 units + - Facility 3 to Facility 0: 147 units + - Facility 3 to Facility 1: 222 units + - Facility 3 to Facility 2: 129 units + - Facility 3 to Facility 4: 206 units + - Facility 3 to Facility 5: 202 units + - Facility 3 to Facility 6: 134 units + - Facility 4 to Facility 0: 183 units + - Facility 4 to Facility 1: 210 units + - Facility 4 to Facility 2: 238 units + - Facility 4 to Facility 3: 193 units + - Facility 4 to Facility 5: 157 units + - Facility 4 to Facility 6: 137 units + - Facility 5 to Facility 0: 134 units + - Facility 5 to Facility 1: 128 units + - Facility 5 to Facility 2: 241 units + - Facility 5 to Facility 3: 143 units + - Facility 5 to Facility 4: 198 units + - Facility 5 to Facility 6: 213 units + - Facility 6 to Facility 0: 213 units + - Facility 6 to Facility 1: 206 units + - Facility 6 to Facility 2: 184 units + - Facility 6 to Facility 3: 231 units + - Facility 6 to Facility 4: 207 units + - Facility 6 to Facility 5: 165 units + +Decide whether to establish a connection between each pair of facilities, where establishing a connection incurs a fixed cost. Determine the amount of goods to transport between connected facilities, ensuring that the total flow does not exceed the connection's capacity. The total incoming goods minus the total outgoing goods must equal the facility's net demand for each facility. The decision to establish a connection is binary. The goal is to minimize the total cost, which includes both the fixed costs of establishing connections and the variable costs of transporting goods.13022You are tasked with managing the logistics of a supply chain network involving seven key facilities (nodes) that produce and distribute goods. Each facility has specific supply and demand requirements, and the goal is to minimize the total cost of establishing connections between facilities and transporting goods while ensuring all demands are met. + +Facility 0 has a net demand of 106 units, Facility 1 has a net demand of 144 units, Facility 2 has a supply of 250 units, Facility 3 has a supply of 250 units, Facility 4 has a net demand of 202 units, Facility 5 has a net demand of 40 units, and Facility 6 has a net demand of 8 units. + +Establishing a connection between any two facilities incurs a fixed cost, and transporting goods between them incurs a variable cost per unit. Each connection has a maximum capacity for goods that can be transported. The fixed costs, variable costs, and maximum capacities for each possible connection are as follows: + +- Fixed Costs (for establishing connections): + - Facility 0 to Facility 1: 1985 units + - Facility 0 to Facility 2: 1968 units + - Facility 0 to Facility 3: 2043 units + - Facility 0 to Facility 4: 2035 units + - Facility 0 to Facility 5: 2043 units + - Facility 0 to Facility 6: 1987 units + - Facility 1 to Facility 0: 2035 units + - Facility 1 to Facility 2: 2044 units + - Facility 1 to Facility 3: 1985 units + - Facility 1 to Facility 4: 2047 units + - Facility 1 to Facility 5: 2003 units + - Facility 1 to Facility 6: 2005 units + - Facility 2 to Facility 0: 2000 units + - Facility 2 to Facility 1: 1991 units + - Facility 2 to Facility 3: 2005 units + - Facility 2 to Facility 4: 1969 units + - Facility 2 to Facility 5: 2010 units + - Facility 2 to Facility 6: 1965 units + - Facility 3 to Facility 0: 1962 units + - Facility 3 to Facility 1: 2024 units + - Facility 3 to Facility 2: 2025 units + - Facility 3 to Facility 4: 2043 units + - Facility 3 to Facility 5: 2034 units + - Facility 3 to Facility 6: 2017 units + - Facility 4 to Facility 0: 1974 units + - Facility 4 to Facility 1: 2012 units + - Facility 4 to Facility 2: 2009 units + - Facility 4 to Facility 3: 1993 units + - Facility 4 to Facility 5: 1991 units + - Facility 4 to Facility 6: 2040 units + - Facility 5 to Facility 0: 2044 units + - Facility 5 to Facility 1: 2022 units + - Facility 5 to Facility 2: 2025 units + - Facility 5 to Facility 3: 1955 units + - Facility 5 to Facility 4: 2001 units + - Facility 5 to Facility 6: 2029 units + - Facility 6 to Facility 0: 1968 units + - Facility 6 to Facility 1: 1956 units + - Facility 6 to Facility 2: 1986 units + - Facility 6 to Facility 3: 2030 units + - Facility 6 to Facility 4: 2001 units + - Facility 6 to Facility 5: 1961 units + +- Variable Costs (per unit transported): + - Facility 0 to Facility 1: 19 units + - Facility 0 to Facility 2: 19 units + - Facility 0 to Facility 3: 19 units + - Facility 0 to Facility 4: 19 units + - Facility 0 to Facility 5: 21 units + - Facility 0 to Facility 6: 21 units + - Facility 1 to Facility 0: 21 units + - Facility 1 to Facility 2: 19 units + - Facility 1 to Facility 3: 20 units + - Facility 1 to Facility 4: 20 units + - Facility 1 to Facility 5: 21 units + - Facility 1 to Facility 6: 21 units + - Facility 2 to Facility 0: 21 units + - Facility 2 to Facility 1: 19 units + - Facility 2 to Facility 3: 21 units + - Facility 2 to Facility 4: 21 units + - Facility 2 to Facility 5: 21 units + - Facility 2 to Facility 6: 20 units + - Facility 3 to Facility 0: 21 units + - Facility 3 to Facility 1: 19 units + - Facility 3 to Facility 2: 21 units + - Facility 3 to Facility 4: 19 units + - Facility 3 to Facility 5: 19 units + - Facility 3 to Facility 6: 20 units + - Facility 4 to Facility 0: 21 units + - Facility 4 to Facility 1: 21 units + - Facility 4 to Facility 2: 21 units + - Facility 4 to Facility 3: 20 units + - Facility 4 to Facility 5: 20 units + - Facility 4 to Facility 6: 19 units + - Facility 5 to Facility 0: 21 units + - Facility 5 to Facility 1: 19 units + - Facility 5 to Facility 2: 20 units + - Facility 5 to Facility 3: 20 units + - Facility 5 to Facility 4: 20 units + - Facility 5 to Facility 6: 21 units + - Facility 6 to Facility 0: 19 units + - Facility 6 to Facility 1: 21 units + - Facility 6 to Facility 2: 19 units + - Facility 6 to Facility 3: 20 units + - Facility 6 to Facility 4: 21 units + - Facility 6 to Facility 5: 21 units + +- Maximum Capacities (for goods transported): + - Facility 0 to Facility 1: 153 units + - Facility 0 to Facility 2: 202 units + - Facility 0 to Facility 3: 185 units + - Facility 0 to Facility 4: 134 units + - Facility 0 to Facility 5: 248 units + - Facility 0 to Facility 6: 170 units + - Facility 1 to Facility 0: 214 units + - Facility 1 to Facility 2: 184 units + - Facility 1 to Facility 3: 193 units + - Facility 1 to Facility 4: 181 units + - Facility 1 to Facility 5: 153 units + - Facility 1 to Facility 6: 139 units + - Facility 2 to Facility 0: 217 units + - Facility 2 to Facility 1: 238 units + - Facility 2 to Facility 3: 231 units + - Facility 2 to Facility 4: 130 units + - Facility 2 to Facility 5: 203 units + - Facility 2 to Facility 6: 232 units + - Facility 3 to Facility 0: 147 units + - Facility 3 to Facility 1: 222 units + - Facility 3 to Facility 2: 129 units + - Facility 3 to Facility 4: 206 units + - Facility 3 to Facility 5: 202 units + - Facility 3 to Facility 6: 134 units + - Facility 4 to Facility 0: 183 units + - Facility 4 to Facility 1: 210 units + - Facility 4 to Facility 2: 238 units + - Facility 4 to Facility 3: 193 units + - Facility 4 to Facility 5: 157 units + - Facility 4 to Facility 6: 137 units + - Facility 5 to Facility 0: 134 units + - Facility 5 to Facility 1: 128 units + - Facility 5 to Facility 2: 241 units + - Facility 5 to Facility 3: 143 units + - Facility 5 to Facility 4: 198 units + - Facility 5 to Facility 6: 213 units + - Facility 6 to Facility 0: 213 units + - Facility 6 to Facility 1: 206 units + - Facility 6 to Facility 2: 184 units + - Facility 6 to Facility 3: 231 units + - Facility 6 to Facility 4: 207 units + - Facility 6 to Facility 5: 165 units + +Decide whether to establish a connection between each pair of facilities, where establishing a connection incurs a fixed cost. Determine the amount of goods to transport between connected facilities, ensuring that the total flow does not exceed the connection's capacity. The total incoming goods minus the total outgoing goods must equal the facility's net demand for each facility. The decision to establish a connection is binary. The goal is to minimize the total cost, which includes both the fixed costs of establishing connections and the variable costs of transporting goods.19805You are tasked with managing the logistics of a supply chain network involving seven key facilities (nodes) that produce and distribute goods. Each facility has specific supply and demand requirements, and the goal is to minimize the total cost of establishing connections between facilities and transporting goods while ensuring all demands are met. + +Facility 0 has a net demand of 106 units, Facility 1 has a net demand of 144 units, Facility 2 has no net a supply or demand, of 250 units, Facility 3 has no net a supply or demand, of 250 units, Facility 4 has a net demand of 202 units, Facility 5 has a net demand of 40 units, and Facility 6 has a net demand of 8 units. + +Establishing a connection between any two facilities incurs a fixed cost, and transporting goods between them incurs a variable cost per unit. Each connection has a maximum capacity for goods that can be transported. The fixed costs, variable costs, and maximum capacities for each possible connection are as follows:13022 19805The original description was infeasible, since no node had supplyinfeasible
34A logistics company must deliver packages to six cities: City 0, City 1, City 2, City 3, City 4, and City 5. Each city must be visited exactly once, and each city must be departed exactly once, ensuring the route forms a continuous loop. The goal is to minimize the total delivery cost, which is the sum of the costs of traveling between the cities. The costs between cities are as follows: + +- From City 0: To City 1 costs 321, to City 2 costs 294, to City 3 costs 323, to City 4 costs 286, and to City 5 costs 325. +- From City 1: To City 0 costs 290, to City 2 costs 319, to City 3 costs 321, to City 4 costs 280, and to City 5 costs 324. +- From City 2: To City 0 costs 304, to City 1 costs 317, to City 3 costs 322, to City 4 costs 313, and to City 5 costs 300. +- From City 3: To City 0 costs 312, to City 1 costs 324, to City 2 costs 280, to City 4 costs 318, and to City 5 costs 284. +- From City 4: To City 0 costs 289, to City 1 costs 279, to City 2 costs 278, to City 3 costs 301, and to City 5 costs 324. +- From City 5: To City 0 costs 325, to City 1 costs 309, to City 2 costs 304, to City 3 costs 305, and to City 4 costs 310. + +To prevent smaller loops that do not include all cities, the company uses position variables for each city. These variables represent the order in which the cities are visited. The position variables for City 1 to City 5 must be between 0 and 5, while the position variable for City 0 has no specified bounds. Additionally, constraints ensure that if a route is taken from one city to another, the position of the first city must be less than the position of the second city, adjusted by a factor to eliminate subtours. The objective is to minimize the total delivery cost, including the sum of all route costs, while satisfying all constraints.1766A logistics company must deliver packages to six cities: City 0, City 1, City 2, City 3, City 4, and City 5. Each city must be visited exactly once, and each city must be departed exactly once, ensuring the route forms a continuous loop, and finally return to the origin city. The goal is to minimize the total delivery cost, which is the sum of the costs of traveling between the cities. The costs between cities are as follows: + +- From City 0: To City 1 costs 321, to City 2 costs 294, to City 3 costs 323, to City 4 costs 286, and to City 5 costs 325. +- From City 1: To City 0 costs 290, to City 2 costs 319, to City 3 costs 321, to City 4 costs 280, and to City 5 costs 324. +- From City 2: To City 0 costs 304, to City 1 costs 317, to City 3 costs 322, to City 4 costs 313, and to City 5 costs 300. +- From City 3: To City 0 costs 312, to City 1 costs 324, to City 2 costs 280, to City 4 costs 318, and to City 5 costs 284. +- From City 4: To City 0 costs 289, to City 1 costs 279, to City 2 costs 278, to City 3 costs 301, and to City 5 costs 324. +- From City 5: To City 0 costs 325, to City 1 costs 309, to City 2 costs 304, to City 3 costs 305, and to City 4 costs 310. +The task is to find the shortest possible route that visits each city exactly once, and return to the origin city. What is the minimal total distance travelled?1766A logistics company must deliver packages to six cities: City 0, City 1, City 2, City 3, City 4, and City 5. Each city must be visited exactly once, and each city must be departed exactly once, ensuring the route forms a continuous loop. loop, and finally return to the origin city. The goal is to minimize the total delivery cost, which is the sum of the costs of traveling between the cities. The costs between cities are as follows: + +- From City 0: To City 1 costs 321, to City 2 costs 294, to City 3 costs 323, to City 4 costs 286, and to City 5 costs 325. +- From City 1: To City 0 costs 290, to City 2 costs 319, to City 3 costs 321, to City 4 costs 280, and to City 5 costs 324. +- From City 3: To City 0 costs 312, to City 1 costs 324, to City 2 costs 280, to City 4 costs 318, and to City 5 costs 284. +- From City 4: To City 0 costs 289, to City 1 costs 279, to City 2 costs 278, to City 3 costs 301, and to City 5 costs 324. +- From City 5: To City 0 costs 325, to City 1 costs 309, to City 2 costs 304, to City 3 costs 305, and to City 4 costs 310. + To prevent smaller loops that do not include all cities, the company uses position variables for each city. These variables represent the order in which the cities are visited. The position variables for City 1 task is to City 5 must be between 0 and 5, while find the position variable for City 0 has no specified bounds. Additionally, constraints ensure that if a shortest possible route is taken from one that visits each city exactly once, and return to another, the position of the first city must be less than the position of the second city, adjusted by a factor to eliminate subtours. The objective origin city. What is to minimize the minimal total delivery cost, including the sum of all route costs, while satisfying all constraints. distance travelled?No differencesWe removed information that was not the description of the problem, but describing the variables/constraints of a possible model. Also, this additional information could lead to a wrong model (MTZ constraint on the return to the first node)ambiguous
35Maximize the total population covered by deploying cell towers across 35 distinct areas, ensuring that the total cost of building the towers does not exceed the available budget of 2,697 units. The population of each area is as follows: Area 0 has 3,870 people, Area 1 has 4,020 people, Area 2 has 2,440 people, Area 3 has 4,263 people, Area 4 has 2,727 people, Area 5 has 3,650 people, Area 6 has 3,155 people, Area 7 has 4,805 people, Area 8 has 2,213 people, Area 9 has 3,635 people, Area 10 has 3,262 people, Area 11 has 4,015 people, Area 12 has 4,991 people, Area 13 has 2,657 people, Area 14 has 2,150 people, Area 15 has 3,655 people, Area 16 has 2,141 people, Area 17 has 2,108 people, Area 18 has 4,200 people, Area 19 has 3,137 people, Area 20 has 4,255 people, Area 21 has 3,452 people, Area 22 has 2,155 people, Area 23 has 3,142 people, Area 24 has 4,909 people, Area 25 has 2,865 people, Area 26 has 3,204 people, Area 27 has 4,244 people, Area 28 has 4,975 people, Area 29 has 2,131 people, Area 30 has 4,710 people, Area 31 has 4,721 people, Area 32 has 4,061 people, Area 33 has 3,162 people, and Area 34 has 3,377 people. + +For each of the 27 potential cell tower locations, decide whether to build the tower, where building a tower incurs a specific cost: Tower 0 costs 233 units, Tower 1 costs 216 units, Tower 2 costs 366 units, Tower 3 costs 264 units, Tower 4 costs 413 units, Tower 5 costs 332 units, Tower 6 costs 414 units, Tower 7 costs 233 units, Tower 8 costs 432 units, Tower 9 costs 461 units, Tower 10 costs 465 units, Tower 11 costs 248 units, Tower 12 costs 392 units, Tower 13 costs 244 units, Tower 14 costs 416 units, Tower 15 costs 399 units, Tower 16 costs 448 units, Tower 17 costs 212 units, Tower 18 costs 216 units, Tower 19 costs 224 units, Tower 20 costs 489 units, Tower 21 costs 288 units, Tower 22 costs 256 units, Tower 23 costs 308 units, Tower 24 costs 325 units, Tower 25 costs 359 units, and Tower 26 costs 339 units. + +Each area must be covered by at least one of the built towers. For example, Area 0 can be covered by Towers 0, 3, 4, 5, 8, 11, 12, 13, 14, 15, 17, 19, 21, 22, 25, or 26; Area 1 can be covered by Towers 0, 4, 5, 7, 9, 12, 13, 17, 18, 19, 20, 21, 23, or 26; and similar coverage relationships apply to all 35 areas. The decisions to build towers and cover areas are binary, meaning you cannot partially build a tower or partially cover an area. The goal is to maximize the total population covered while ensuring the total cost of building the selected towers does not exceed the budget.122457"Maximize the total population covered by deploying cell towers across 35 distinct areas, ensuring that the total cost of building the towers does not exceed the available budget of 697 units. The population of each area is as follows: Area 0 has 3,870 people, Area 1 has 4,020 people, Area 2 has 2,440 people, Area 3 has 4,263 people, Area 4 has 2,727 people, Area 5 has 3,650 people, Area 6 has 3,155 people, Area 7 has 4,805 people, Area 8 has 2,213 people, Area 9 has 3,635 people, Area 10 has 3,262 people, Area 11 has 4,015 people, Area 12 has 4,991 people, Area 13 has 2,657 people, Area 14 has 2,150 people, Area 15 has 3,655 people, Area 16 has 2,141 people, Area 17 has 2,108 people, Area 18 has 4,200 people, Area 19 has 3,137 people, Area 20 has 4,255 people, Area 21 has 3,452 people, Area 22 has 2,155 people, Area 23 has 3,142 people, Area 24 has 4,909 people, Area 25 has 2,865 people, Area 26 has 3,204 people, Area 27 has 4,244 people, Area 28 has 4,975 people, Area 29 has 2,131 people, Area 30 has 4,710 people, Area 31 has 4,721 people, Area 32 has 4,061 people, Area 33 has 3,162 people, and Area 34 has 3,377 people. + +For each of the 27 potential cell tower locations, decide whether to build the tower, where building a tower incurs a specific cost: Tower 0 costs 233 units, Tower 1 costs 216 units, Tower 2 costs 366 units, Tower 3 costs 264 units, Tower 4 costs 413 units, Tower 5 costs 332 units, Tower 6 costs 414 units, Tower 7 costs 233 units, Tower 8 costs 432 units, Tower 9 costs 461 units, Tower 10 costs 465 units, Tower 11 costs 248 units, Tower 12 costs 392 units, Tower 13 costs 244 units, Tower 14 costs 416 units, Tower 15 costs 399 units, Tower 16 costs 448 units, Tower 17 costs 212 units, Tower 18 costs 216 units, Tower 19 costs 224 units, Tower 20 costs 489 units, Tower 21 costs 288 units, Tower 22 costs 256 units, Tower 23 costs 308 units, Tower 24 costs 325 units, Tower 25 costs 359 units, and Tower 26 costs 339 units. + +Each area must be covered by at least one of the built towers + - Area 0 can be covered if any of the towers 20, 3, 0, 23, 8, 7, 21, 4, 25, 17, 2, or 13 are built. + - Area 1 can be covered if any of the towers 1, 0, 2, 6, 7, 16, 19, 25, 17, 23, 13, or 22 are built. + - Area 2 can be covered if any of the towers 14, 18, 8, 0, 5, 13, 10, 24, 4, 6, 20, or 3 are built. + - Area 3 can be covered if any of the towers 2, 12, 3, 11, 23, 19, 8, 1, 14, 17, 24, or 25 are built. + - Area 4 can be covered if any of the towers 2, 17, 9, 20, 19, 11, 18, 6, 26, 1, 7, or 24 are built. + - Area 5 can be covered if any of the towers 2, 7, 3, 12, 8, 14, 20, 11, 5, 19, 17, or 6 are built. + - Area 6 can be covered if any of the towers 21, 8, 22, 26, 20, 2, 19, 5, 17, 7, 24, or 14 are built. + - Area 7 can be covered if any of the towers 12, 8, 20, 22, 17, 7, 10, 1, 21, 19, 24, or 26 are built. + - Area 8 can be covered if any of the towers 8, 2, 6, 18, 22, 10, 24, 15, 12, 14, 4, or 26 are built. + - Area 9 can be covered if any of the towers 4, 7, 23, 17, 24, 8, 18, 13, 20, 12, 11, or 25 are built. + - Area 10 can be covered if any of the towers 4, 16, 15, 2, 1, 3, 26, 5, 13, 23, 12, or 25 are built. + - Area 11 can be covered if any of the towers 19, 14, 16, 8, 17, 0, 3, 22, 23, 10, 20, or 9 are built. + - Area 12 can be covered if any of the towers 13, 5, 14, 0, 8, 16, 25, 21, 3, 9, 19, or 6 are built. + - Area 13 can be covered if any of the towers 4, 11, 24, 5, 17, 16, 0, 19, 10, 15, 20, or 3 are built. + - Area 14 can be covered if any of the towers 11, 25, 9, 7, 1, 23, 18, 2, 19, 15, 20, or 4 are built. + - Area 15 can be covered if any of the towers 4, 21, 15, 17, 5, 8, 16, 19, 13, 6, 23, or 9 are built. + - Area 16 can be covered if any of the towers 12, 21, 20, 11, 14, 16, 22, 3, 7, 18, 2, or 10 are built. + - Area 17 can be covered if any of the towers 0, 18, 17, 7, 25, 23, 26, 2, 1, 21, 19, or 22 are built. + - Area 18 can be covered if any of the towers 10, 2, 16, 7, 8, 21, 15, 6, 17, 4, 20, or 23 are built. + - Area 19 can be covered if any of the towers 25, 15, 13, 6, 3, 22, 24, 11, 20, 18, 14, or 1 are built. + - Area 20 can be covered if any of the towers 21, 20, 25, 3, 1, 12, 10, 23, 7, 6, 17, or 14 are built. + - Area 21 can be covered if any of the towers 4, 13, 5, 8, 14, 7, 2, 22, 17, 3, 1, or 0 are built. + - Area 22 can be covered if any of the towers 2, 24, 7, 5, 13, 15, 21, 6, 12, 1, 23, or 18 are built. + - Area 23 can be covered if any of the towers 0, 12, 8, 14, 9, 13, 17, 15, 4, 6, 22, or 20 are built. + - Area 24 can be covered if any of the towers 1, 18, 23, 17, 26, 10, 22, 20, 25, 15, 16, or 5 are built. + - Area 25 can be covered if any of the towers 1, 16, 2, 5, 24, 19, 22, 7, 12, 3, 21, or 26 are built. + - Area 26 can be covered if any of the towers 19, 2, 13, 21, 18, 22, 16, 10, 8, 6, 26, or 7 are built. + - Area 27 can be covered if any of the towers 8, 12, 4, 21, 20, 9, 14, 10, 2, 0, 22, or 3 are built. + - Area 28 can be covered if any of the towers 2, 17, 6, 16, 8, 4, 11, 26, 7, 20, 9, or 5 are built. + - Area 29 can be covered if any of the towers 14, 17, 22, 9, 19, 20, 16, 0, 25, 23, 3, or 4 are built. + - Area 30 can be covered if any of the towers 8, 3, 25, 23, 17, 4, 26, 9, 6, 10, 18, or 20 are built. + - Area 31 can be covered if any of the towers 16, 15, 8, 1, 2, 20, 13, 24, 23, 0, 10, or 4 are built. + - Area 32 can be covered if any of the towers 20, 8, 5, 23, 14, 17, 13, 21, 0, 3, 2, or 4 are built. + - Area 33 can be covered if any of the towers 17, 1, 11, 18, 26, 4, 13, 21, 25, 9, 24, or 23 are built. + - Area 34 can be covered if any of the towers 11, 6, 21, 7, 24, 3, 26, 17, 13, 4, 23, or 5 are built.. + The decisions to build towers and cover areas are binary, meaning you cannot partially build a tower or partially cover an area. The goal is to maximize the total population covered while ensuring the total cost of building the selected towers does not exceed the budget."117736Maximize "Maximize the total population covered by deploying cell towers across 35 distinct areas, ensuring that the total cost of building the towers does not exceed the available budget of 2,697 697 units. The population of each area is as follows: Area 0 has 3,870 people, Area 1 has 4,020 people, Area 2 has 2,440 people, Area 3 has 4,263 people, Area 4 has 2,727 people, Area 5 has 3,650 people, Area 6 has 3,155 people, Area 7 has 4,805 people, Area 8 has 2,213 people, Area 9 has 3,635 people, Area 10 has 3,262 people, Area 11 has 4,015 people, Area 12 has 4,991 people, Area 13 has 2,657 people, Area 14 has 2,150 people, Area 15 has 3,655 people, Area 16 has 2,141 people, Area 17 has 2,108 people, Area 18 has 4,200 people, Area 19 has 3,137 people, Area 20 has 4,255 people, Area 21 has 3,452 people, Area 22 has 2,155 people, Area 23 has 3,142 people, Area 24 has 4,909 people, Area 25 has 2,865 people, Area 26 has 3,204 people, Area 27 has 4,244 people, Area 28 has 4,975 people, Area 29 has 2,131 people, Area 30 has 4,710 people, Area 31 has 4,721 people, Area 32 has 4,061 people, Area 33 has 3,162 people, and Area 34 has 3,377 people. + +For each of the 27 potential cell tower locations, decide whether to build the tower, where building a tower incurs a specific cost: Tower 0 costs 233 units, Tower 1 costs 216 units, Tower 2 costs 366 units, Tower 3 costs 264 units, Tower 4 costs 413 units, Tower 5 costs 332 units, Tower 6 costs 414 units, Tower 7 costs 233 units, Tower 8 costs 432 units, Tower 9 costs 461 units, Tower 10 costs 465 units, Tower 11 costs 248 units, Tower 12 costs 392 units, Tower 13 costs 244 units, Tower 14 costs 416 units, Tower 15 costs 399 units, Tower 16 costs 448 units, Tower 17 costs 212 units, Tower 18 costs 216 units, Tower 19 costs 224 units, Tower 20 costs 489 units, Tower 21 costs 288 units, Tower 22 costs 256 units, Tower 23 costs 308 units, Tower 24 costs 325 units, Tower 25 costs 359 units, and Tower 26 costs 339 units. + +Each area must be covered by at least one of the built towers. For example, towers + - Area 0 can be covered if any of the towers 20, 3, 0, 23, 8, 7, 21, 4, 25, 17, 2, or 13 are built. + - Area 1 can be covered if any of the towers 1, 0, 2, 6, 7, 16, 19, 25, 17, 23, 13, or 22 are built. + - Area 2 can be covered if any of the towers 14, 18, 8, 0, 5, 13, 10, 24, 4, 6, 20, or 3 are built. + - Area 3 can be covered if any of the towers 2, 12, 3, 11, 23, 19, 8, 1, 14, 17, 24, or 25 are built. + - Area 4 can be covered if any of the towers 2, 17, 9, 20, 19, 11, 18, 6, 26, 1, 7, or 24 are built. + - Area 5 can be covered if any of the towers 2, 7, 3, 12, 8, 14, 20, 11, 5, 19, 17, or 6 are built. + - Area 6 can be covered if any of the towers 21, 8, 22, 26, 20, 2, 19, 5, 17, 7, 24, or 14 are built. + - Area 7 can be covered if any of the towers 12, 8, 20, 22, 17, 7, 10, 1, 21, 19, 24, or 26 are built. + - Area 8 can be covered if any of the towers 8, 2, 6, 18, 22, 10, 24, 15, 12, 14, 4, or 26 are built. + - Area 9 can be covered if any of the towers 4, 7, 23, 17, 24, 8, 18, 13, 20, 12, 11, or 25 are built. + - Area 10 can be covered if any of the towers 4, 16, 15, 2, 1, 3, 26, 5, 13, 23, 12, or 25 are built. + - Area 11 can be covered if any of the towers 19, 14, 16, 8, 17, 0, 3, 22, 23, 10, 20, or 9 are built. + - Area 12 can be covered if any of the towers 13, 5, 14, 0, 8, 16, 25, 21, 3, 9, 19, or 6 are built. + - Area 13 can be covered if any of the towers 4, 11, 24, 5, 17, 16, 0, 19, 10, 15, 20, or 3 are built. + - Area 14 can be covered if any of the towers 11, 25, 9, 7, 1, 23, 18, 2, 19, 15, 20, or 4 are built. + - Area 15 can be covered if any of the towers 4, 21, 15, 17, 5, 8, 16, 19, 13, 6, 23, or 9 are built. + - Area 16 can be covered if any of the towers 12, 21, 20, 11, 14, 16, 22, 3, 7, 18, 2, or 10 are built. + - Area 17 can be covered if any of the towers 0, 18, 17, 7, 25, 23, 26, 2, 1, 21, 19, or 22 are built. + - Area 18 can be covered if any of the towers 10, 2, 16, 7, 8, 21, 15, 6, 17, 4, 20, or 23 are built. + - Area 19 can be covered if any of the towers 25, 15, 13, 6, 3, 22, 24, 11, 20, 18, 14, or 1 are built. + - Area 0 20 can be covered by Towers 0, if any of the towers 21, 20, 25, 3, 1, 12, 10, 23, 7, 6, 17, or 14 are built. + - Area 21 can be covered if any of the towers 4, 13, 5, 8, 11, 12, 14, 7, 2, 22, 17, 3, 1, or 0 are built. + - Area 22 can be covered if any of the towers 2, 24, 7, 5, 13, 15, 21, 6, 12, 1, 23, or 18 are built. + - Area 23 can be covered if any of the towers 0, 12, 8, 14, 9, 13, 17, 15, 4, 6, 22, or 20 are built. + - Area 24 can be covered if any of the towers 1, 18, 23, 17, 26, 10, 22, 20, 25, 15, 16, or 5 are built. + - Area 25 can be covered if any of the towers 1, 16, 2, 5, 24, 19, 22, 7, 12, 3, 21, or 26 are built. + - Area 26 can be covered if any of the towers 19, 2, 13, 21, 18, 22, 25, 16, 10, 8, 6, 26, or 26; 7 are built. + - Area 1 27 can be covered by Towers if any of the towers 8, 12, 4, 21, 20, 9, 14, 10, 2, 0, 22, or 3 are built. + - Area 28 can be covered if any of the towers 2, 17, 6, 16, 8, 4, 5, 11, 26, 7, 20, 9, 12, 13, or 5 are built. + - Area 29 can be covered if any of the towers 14, 17, 18, 22, 9, 19, 20, 16, 0, 25, 23, 3, or 4 are built. + - Area 30 can be covered if any of the towers 8, 3, 25, 23, 17, 4, 26, 9, 6, 10, 18, or 20 are built. + - Area 31 can be covered if any of the towers 16, 15, 8, 1, 2, 20, 13, 24, 23, 0, 10, or 4 are built. + - Area 32 can be covered if any of the towers 20, 8, 5, 23, 14, 17, 13, 21, 0, 3, 2, or 4 are built. + - Area 33 can be covered if any of the towers 17, 1, 11, 18, 26, 4, 13, 21, 25, 9, 24, or 23 are built. + - Area 34 can be covered if any of the towers 11, 6, 21, 7, 24, 3, 26, 17, 13, 4, 23, or 26; and similar coverage relationships apply to all 35 areas. 5 are built.. + The decisions to build towers and cover areas are binary, meaning you cannot partially build a tower or partially cover an area. The goal is to maximize the total population covered while ensuring the total cost of building the selected towers does not exceed the budget. budget."122457 117736Missing data about which towers cover which regionsmissing data
36The problem involves scheduling 7 jobs, each consisting of 2 operations, on a set of machines to minimize the total production time, or makespan. Each operation has a processing time of 1 unit of time. Operations within the same job must follow a specific sequence: operation 0 must be completed before operation 1 can begin. Some operations from different jobs are assigned to the same machine, and the machine can only process one operation at a time. The order in which operations are scheduled on a shared machine is determined by binary decision variables, which take a value of 1 if one operation is scheduled before another and 0 otherwise. A large constant of 100,000 is used to enforce that operations on the same machine do not overlap in time. The makespan must be at least as large as the completion time of every operation, ensuring it accounts for the longest-running job. Start times for all operations must be non-negative, and the binary variables must be either 0 or 1. Specific machine assignments include operations (4,1), (5,0), and (6,0) on one machine; (0,1) and (3,0) on another; (0,0), (2,0), and (4,0) on another; and (1,0), (1,1), (2,1), (3,1), and (5,1) on another. The objective is to determine the start times for all operations and the order in which operations are processed on shared machines to minimize the makespan while respecting the sequence of operations within each job and ensuring no overlaps on shared machines.3The problem involves scheduling 7 jobs, each consisting of 2 operations, on a set of machines to minimize the total production time, or makespan. Each operation has a processing time of 1 unit of time. Operations within the same job must follow a specific sequence: operation 0 must be completed before operation 1 can begin. Some operations from different jobs are assigned to the same machine, and the machine can only process one operation at a time. The order in which operations are scheduled on a shared machine is determined by binary decision variables, which take a value of 1 if one operation is scheduled before another and 0 otherwise. A large constant of 100 is used to enforce that operations on the same machine do not overlap in time. The makespan must be at least as large as the completion time of every operation, ensuring it accounts for the longest-running job. Start times for all operations must be non-negative, and the binary variables must be either 0 or 1. The objective is to determine the start times for all operations and the order in which operations are processed on shared machines to minimize the makespan while respecting the sequence of operations within each job and ensuring no overlaps on shared machines. + + +Additional information: +- There are 4 machines, numbered from 0 to 3 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 3 + - Operation 1 of Job 1: Machine 3 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 3 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 3 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 0 + - Operation 0 of Job 5: Machine 0 + - Operation 1 of Job 5: Machine 3 + - Operation 0 of Job 6: Machine 0 + - Operation 1 of Job 6: Machine 15The problem involves scheduling 7 jobs, each consisting of 2 operations, on a set of machines to minimize the total production time, or makespan. Each operation has a processing time of 1 unit of time. Operations within the same job must follow a specific sequence: operation 0 must be completed before operation 1 can begin. Some operations from different jobs are assigned to the same machine, and the machine can only process one operation at a time. The order in which operations are scheduled on a shared machine is determined by binary decision variables, which take a value of 1 if one operation is scheduled before another and 0 otherwise. A large constant of 100,000 100 is used to enforce that operations on the same machine do not overlap in time. The makespan must be at least as large as the completion time of every operation, ensuring it accounts for the longest-running job. Start times for all operations must be non-negative, and the binary variables must be either 0 or 1. Specific machine assignments include operations (4,1), (5,0), and (6,0) on one machine; (0,1) and (3,0) on another; (0,0), (2,0), and (4,0) on another; and (1,0), (1,1), (2,1), (3,1), and (5,1) on another. The objective is to determine the start times for all operations and the order in which operations are processed on shared machines to minimize the makespan while respecting the sequence of operations within each job and ensuring no overlaps on shared machines. + + + Additional information: + - There are 4 machines, numbered from 0 to 3 + - The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 3 + - Operation 1 of Job 1: Machine 3 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 3 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 3 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 0 + - Operation 0 of Job 5: Machine 0 + - Operation 1 of Job 5: Machine 3 + - Operation 0 of Job 6: Machine 0 + - Operation 1 of Job 6: Machine 13 5It was missing the number of machines and there was only a partial assignment of jobs to machinesmissing data
37The problem involves managing the distribution of a critical resource across seven interconnected cities: City 0, City 1, City 2, City 3, City 4, City 5, and City 6. The goal is to minimize the total cost of transporting resources between cities while ensuring that all demands are met and all operational constraints are satisfied. The cost of transporting resources from City 0 to City 1 is 2 units, from City 0 to City 2 is 1 unit, from City 0 to City 3 is 1 unit, from City 0 to City 4 is 3 units, from City 0 to City 5 is 3 units, and from City 0 to City 6 is 3 units. The cost of transporting resources from City 1 to City 0 is 2 units, from City 1 to City 2 is 1 unit, from City 1 to City 3 is 1 unit, from City 1 to City 4 is 1 unit, from City 1 to City 5 is 1 unit, and from City 1 to City 6 is 3 units. The cost of transporting resources from City 2 to City 0 is 2 units, from City 2 to City 1 is 2 units, from City 2 to City 3 is 3 units, from City 2 to City 4 is 2 units, from City 2 to City 5 is 1 unit, and from City 2 to City 6 is 3 units. The cost of transporting resources from City 3 to City 0 is 2 units, from City 3 to City 1 is 2 units, from City 3 to City 2 is 3 units, from City 3 to City 4 is 2 units, from City 3 to City 5 is 3 units, and from City 3 to City 6 is 1 unit. The cost of transporting resources from City 4 to City 0 is 1 unit, from City 4 to City 1 is 3 units, from City 4 to City 2 is 2 units, from City 4 to City 3 is 2 units, from City 4 to City 5 is 3 units, and from City 4 to City 6 is 2 units. The cost of transporting resources from City 5 to City 0 is 3 units, from City 5 to City 1 is 1 unit, from City 5 to City 2 is 1 unit, from City 5 to City 3 is 1 unit, from City 5 to City 4 is 2 units, and from City 5 to City 6 is 1 unit. The cost of transporting resources from City 6 to City 0 is 3 units, from City 6 to City 1 is 2 units, from City 6 to City 2 is 2 units, from City 6 to City 3 is 2 units, from City 6 to City 4 is 3 units, and from City 6 to City 5 is 1 unit. + +Each city must maintain a balance between incoming and outgoing resources. City 0 must ensure that the total resources it sends out are balanced by the resources it receives from other cities. City 1 has a net demand of 1 unit, City 2 has a net surplus of 2 units, City 3 has a net demand of 2 units, City 4 has a net surplus of 2 units, City 5 has a net demand of 1 unit, and City 6 has no net demand or surplus. The route from City 0 to City 1 can handle up to 7 units of resources, from City 0 to City 2 up to 8 units, from City 0 to City 3 up to 9 units, from City 0 to City 4 up to 9 units, from City 0 to City 5 up to 9 units, and from City 0 to City 6 up to 8 units. The route from City 1 to City 0 can handle up to 8 units, from City 1 to City 2 up to 8 units, from City 1 to City 3 up to 8 units, from City 1 to City 4 up to 8 units, from City 1 to City 5 up to 7 units, and from City 1 to City 6 up to 8 units. The route from City 2 to City 0 can handle up to 8 units, from City 2 to City 1 up to 8 units, from City 2 to City 3 up to 8 units, from City 2 to City 4 up to 8 units, from City 2 to City 5 up to 7 units, and from City 2 to City 6 up to 9 units. The route from City 3 to City 0 can handle up to 8 units, from City 3 to City 1 up to 9 units, from City 3 to City 2 up to 8 units, from City 3 to City 4 up to 9 units, from City 3 to City 5 up to 9 units, and from City 3 to City 6 up to 8 units. The route from City 4 to City 0 can handle up to 7 units, from City 4 to City 1 up to 7 units, from City 4 to City 2 up to 9 units, from City 4 to City 3 up to 7 units, from City 4 to City 5 up to 8 units, and from City 4 to City 6 up to 7 units. The route from City 5 to City 0 can handle up to 9 units, from City 5 to City 1 up to 8 units, from City 5 to City 2 up to 7 units, from City 5 to City 3 up to 9 units, from City 5 to City 4 up to 9 units, and from City 5 to City 6 up to 9 units. The route from City 6 to City 0 can handle up to 7 units, from City 6 to City 1 up to 9 units, from City 6 to City 2 up to 9 units, from City 6 to City 3 up to 9 units, from City 6 to City 4 up to 9 units, and from City 6 to City 5 up to 7 units. + +Each city has a maximum capacity for handling incoming resources. City 0 can receive up to 16 units, City 1 up to 15 units, City 2 up to 12 units, City 3 up to 15 units, City 4 up to 15 units, City 5 up to 15 units, and City 6 up to 17 units. The amount of resources transported between any two cities cannot be negative. The objective is to determine the optimal amount of resources to transport between cities to minimize the total transportation cost while ensuring that all cities meet their resource demands or surpluses, no transportation route exceeds its capacity, and no city exceeds its capacity for receiving resources.7The problem involves managing the distribution of a critical resource across seven interconnected cities: City 0, City 1, City 2, City 3, City 4, City 5, and City 6. The goal is to minimize the total cost of transporting resources between cities while ensuring that all demands are met and all operational constraints are satisfied. The cost of transporting resources from City 0 to City 1 is 2 units, from City 0 to City 2 is 1 unit, from City 0 to City 3 is 1 unit, from City 0 to City 4 is 3 units, from City 0 to City 5 is 3 units, and from City 0 to City 6 is 3 units. The cost of transporting resources from City 1 to City 0 is 2 units, from City 1 to City 2 is 1 unit, from City 1 to City 3 is 1 unit, from City 1 to City 4 is 1 unit, from City 1 to City 5 is 1 unit, and from City 1 to City 6 is 3 units. The cost of transporting resources from City 2 to City 0 is 2 units, from City 2 to City 1 is 2 units, from City 2 to City 3 is 3 units, from City 2 to City 4 is 2 units, from City 2 to City 5 is 1 unit, and from City 2 to City 6 is 3 units. The cost of transporting resources from City 3 to City 0 is 2 units, from City 3 to City 1 is 2 units, from City 3 to City 2 is 3 units, from City 3 to City 4 is 2 units, from City 3 to City 5 is 3 units, and from City 3 to City 6 is 1 unit. The cost of transporting resources from City 4 to City 0 is 1 unit, from City 4 to City 1 is 3 units, from City 4 to City 2 is 2 units, from City 4 to City 3 is 2 units, from City 4 to City 5 is 3 units, and from City 4 to City 6 is 2 units. The cost of transporting resources from City 5 to City 0 is 3 units, from City 5 to City 1 is 1 unit, from City 5 to City 2 is 1 unit, from City 5 to City 3 is 1 unit, from City 5 to City 4 is 2 units, and from City 5 to City 6 is 1 unit. The cost of transporting resources from City 6 to City 0 is 3 units, from City 6 to City 1 is 2 units, from City 6 to City 2 is 2 units, from City 6 to City 3 is 2 units, from City 6 to City 4 is 3 units, and from City 6 to City 5 is 1 unit. + +Each city must maintain a balance between incoming and outgoing resources. City 0 must ensure that the total resources it sends out are balanced by the resources it receives from other cities. City 1 has a net demand of 1 unit, City 2 has a net surplus of 2 units, City 3 has a net demand of 2 units, City 4 has a net surplus of 2 units, City 5 has a net demand of 1 unit, and City 6 has no net demand or surplus. The route from City 0 to City 1 can handle up to 7 units of resources, from City 0 to City 2 up to 8 units, from City 0 to City 3 up to 9 units, from City 0 to City 4 up to 9 units, from City 0 to City 5 up to 9 units, and from City 0 to City 6 up to 8 units. The route from City 1 to City 0 can handle up to 8 units, from City 1 to City 2 up to 8 units, from City 1 to City 3 up to 8 units, from City 1 to City 4 up to 8 units, from City 1 to City 5 up to 7 units, and from City 1 to City 6 up to 8 units. The route from City 2 to City 0 can handle up to 8 units, from City 2 to City 1 up to 8 units, from City 2 to City 3 up to 8 units, from City 2 to City 4 up to 8 units, from City 2 to City 5 up to 7 units, and from City 2 to City 6 up to 9 units. The route from City 3 to City 0 can handle up to 8 units, from City 3 to City 1 up to 9 units, from City 3 to City 2 up to 8 units, from City 3 to City 4 up to 9 units, from City 3 to City 5 up to 9 units, and from City 3 to City 6 up to 8 units. The route from City 4 to City 0 can handle up to 7 units, from City 4 to City 1 up to 7 units, from City 4 to City 2 up to 9 units, from City 4 to City 3 up to 7 units, from City 4 to City 5 up to 8 units, and from City 4 to City 6 up to 7 units. The route from City 5 to City 0 can handle up to 9 units, from City 5 to City 1 up to 8 units, from City 5 to City 2 up to 7 units, from City 5 to City 3 up to 9 units, from City 5 to City 4 up to 9 units, and from City 5 to City 6 up to 9 units. The route from City 6 to City 0 can handle up to 7 units, from City 6 to City 1 up to 9 units, from City 6 to City 2 up to 9 units, from City 6 to City 3 up to 9 units, from City 6 to City 4 up to 9 units, and from City 6 to City 5 up to 7 units. + +Each city has a maximum capacity for handling incoming resources. City 0 can receive up to 16 units, City 1 up to 15 units, City 2 up to 12 units, City 3 up to 15 units, City 4 up to 15 units, City 5 up to 15 units, and City 6 up to 17 units. The amount of resources transported between any two cities cannot be negative. The objective is to determine the optimal amount of resources to transport between cities to minimize the total transportation cost while ensuring that all cities meet their resource demands or surpluses, no transportation route exceeds its capacity, and no city exceeds its capacity for receiving resources.7No differencesNo differencesok
38The manufacturing facility produces three products over a six-month period, aiming to minimize total costs, including production, setup, and inventory holding costs, while meeting customer demand and adhering to resource constraints. Product 0 has cumulative demands of 75, 141, 241, 315, 386, and 469 units over the six months. Product 1 has cumulative demands of 65, 151, 228, 286, 338, and 390 units. Product 2 has cumulative demands of 57, 143, 193, 263, 363, and 428 units. Production costs per unit for Product 0 are $42, $47, $41, $40, $43, and $47 across the six months. For Product 1, they are $43, $50, $41, $50, $49, and $46. For Product 2, they are $42, $49, $40, $47, $49, and $45. Setup costs for Product 0 are $1,282, $1,493, $1,006, $1,286, $1,305, and $1,360. For Product 1, they are $1,402, $1,498, $1,296, $1,366, $1,376, and $1,172. For Product 2, they are $1,362, $1,094, $1,283, $1,279, $1,430, and $1,146. Inventory holding costs per unit for Product 0 are $4, $5, $4, $5, $4, and $4. For Product 1, they are $5, $5, $5, $4, $4, and $4. For Product 2, they are $5, $5, $4, $5, $5, and $4. The facility has monthly resource capacities of 1,173, 1,102, 1,018, 1,077, 1,190, and 1,132 units. Each unit of Product 0 consumes 1.778 resource units, Product 1 consumes 1.502, and Product 2 consumes 1.556. If a product is produced in a month, a setup cost is incurred, and production quantities are limited by setup decisions: Product 0 can produce up to 469, 394, 328, 228, 154, and 83 units; Product 1 up to 390, 325, 239, 162, 104, and 52 units; and Product 2 up to 428, 371, 285, 235, 165, and 65 units. Inventory at the end of each month must satisfy demand, and no initial or final inventory is allowed. The objective is to determine optimal production quantities, setup decisions, and inventory levels to minimize total costs while meeting demand and resource constraints.67501The manufacturing facility produces three products over a six-month period, aiming to minimize total costs, including production, setup, and inventory holding costs, while meeting customer demand and adhering to resource constraints. Product 0 has cumulative demands of 75, 141, 241, 315, 386, and 469 units over the six months. Product 1 has cumulative demands of 65, 151, 228, 286, 338, and 390 units. Product 2 has cumulative demands of 57, 143, 193, 263, 363, and 428 units. Production costs per unit for Product 0 are $42, $47, $41, $40, $43, and $47 across the six months. For Product 1, they are $43, $50, $41, $50, $49, and $46. For Product 2, they are $42, $49, $40, $47, $49, and $45. Setup costs for Product 0 are $1,282, $1,493, $1,006, $1,286, $1,305, and $1,360. For Product 1, they are $1,402, $1,498, $1,296, $1,366, $1,376, and $1,172. For Product 2, they are $1,362, $1,094, $1,283, $1,279, $1,430, and $1,146. Inventory holding costs per unit for Product 0 are $4, $5, $4, $5, $4, and $4. For Product 1, they are $5, $5, $5, $4, $4, and $4. For Product 2, they are $5, $5, $4, $5, $5, and $4. The facility has monthly resource capacities of 1,173, 1,102, 1,018, 1,077, 1,190, and 1,132 units. Each unit of Product 0 consumes 1.778 resource units, Product 1 consumes 1.502, and Product 2 consumes 1.556. If a product is produced in a month, a setup cost is incurred, and production quantities are limited by setup decisions: Product 0 can produce up to 469, 394, 328, 228, 154, and 83 units; Product 1 up to 390, 325, 239, 162, 104, and 52 units; and Product 2 up to 428, 371, 285, 235, 165, and 65 units. Inventory at the end of each month must satisfy demand, and no initial or final inventory is allowed. The objective is to determine optimal production quantities, setup decisions, and inventory levels to minimize total costs while meeting demand and resource constraints.67501No differencesNo differencesok
39There are five clients, labeled Client 0 to Client 4, each requiring two operations to be performed in sequence. Each operation takes exactly 1 unit of time to complete. The operations for each client must follow a strict sequence: the second operation cannot start until the first operation is completed, with a minimum gap of 1 unit of time between them. Operations from different clients may need to be performed on the same machine, and no two operations can overlap on the same machine. To enforce this, binary decision variables determine the order of operations on shared machines. The goal is to minimize the total production time, known as the makespan, which is the time when the last operation across all clients is completed. The makespan must be at least as long as the completion time of every operation. Start times for all operations are continuous variables, and the binary variables ensure that operations on the same machine do not overlap. A large number (100,000) is used in the constraints to enforce the scheduling logic, ensuring that one operation finishes before another starts on the same machine. The problem involves determining the optimal start times for all operations and the sequence of operations on shared machines to minimize the total production time while satisfying all constraints.4There are five clients, labeled Client 0 to Client 4, each requiring two operations to be performed in sequence. Each operation takes exactly 1 unit of time to complete. The operations for each client must follow a strict sequence: the second operation cannot start until the first operation is completed, with a minimum gap of 1 unit of time between them. Operations from different clients may need to be performed on the same machine, and no two operations can overlap on the same machine. To enforce this, binary decision variables determine the order of operations on shared machines. The goal is to minimize the total production time, known as the makespan, which is the time when the last operation across all clients is completed. The makespan must be at least as long as the completion time of every operation. Start times for all operations are continuous variables, and the binary variables ensure that operations on the same machine do not overlap. A large number (1000) is used in the constraints to enforce the scheduling logic, ensuring that one operation finishes before another starts on the same machine. The problem involves determining the optimal start times for all operations and the sequence of operations on shared machines to minimize the total production time while satisfying all constraints. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3, 4} (these correspond to Client 0..Client 4) +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Machines: M = {0, 1} (two machines, indexed 0 and 1) + +2) Machine assignment for every operation (explicit): +- Operation (0,1) -> Machine 0 +- Operation (0,2) -> Machine 1 +- Operation (1,1) -> Machine 0 +- Operation (1,2) -> Machine 1 +- Operation (2,1) -> Machine 0 +- Operation (2,2) -> Machine 1 +- Operation (3,1) -> Machine 0 +- Operation (3,2) -> Machine 1 +- Operation (4,1) -> Machine 0 +- Operation (4,2) -> Machine 1 + +3) Operation durations (time units): +- All operations have duration p_{j,k} = 1 for every job j and operation k7There are five clients, labeled Client 0 to Client 4, each requiring two operations to be performed in sequence. Each operation takes exactly 1 unit of time to complete. The operations for each client must follow a strict sequence: the second operation cannot start until the first operation is completed, with a minimum gap of 1 unit of time between them. Operations from different clients may need to be performed on the same machine, and no two operations can overlap on the same machine. To enforce this, binary decision variables determine the order of operations on shared machines. The goal is to minimize the total production time, known as the makespan, which is the time when the last operation across all clients is completed. The makespan must be at least as long as the completion time of every operation. Start times for all operations are continuous variables, and the binary variables ensure that operations on the same machine do not overlap. A large number (100,000) (1000) is used in the constraints to enforce the scheduling logic, ensuring that one operation finishes before another starts on the same machine. The problem involves determining the optimal start times for all operations and the sequence of operations on shared machines to minimize the total production time while satisfying all constraints. + + ADDITIONAL INPUT DATA: + + 1) Sets and indices + - Jobs: J = {0, 1, 2, 3, 4} (these correspond to Client 0..Client 4) + - For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). + - Machines: M = {0, 1} (two machines, indexed 0 and 1) + + 2) Machine assignment for every operation (explicit): + - Operation (0,1) -> Machine 0 + - Operation (0,2) -> Machine 1 + - Operation (1,1) -> Machine 0 + - Operation (1,2) -> Machine 1 + - Operation (2,1) -> Machine 0 + - Operation (2,2) -> Machine 1 + - Operation (3,1) -> Machine 0 + - Operation (3,2) -> Machine 1 + - Operation (4,1) -> Machine 0 + - Operation (4,2) -> Machine 1 + + 3) Operation durations (time units): + - All operations have duration p_{j,k} = 1 for every job j and operation k4 7It was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
40You are responsible for managing the distribution of resources across a network of eight cities. Each city has a specific supply of resources and a demand that must be met. The goal is to minimize the total transportation cost while ensuring that all cities receive the resources they need, without exceeding the capacity of the transportation links or the storage capacity of the cities. + +City 0 has a net demand of 2 units, City 1 has a balanced supply and demand, City 2 has a net supply of 1 unit, City 3 has a net demand of 2 units, City 4 has a net supply of 2 units, City 5 has a balanced supply and demand, City 6 has a net demand of 1 unit, and City 7 has a net supply of 2 units. + +The cost of transporting resources from City 0 to City 1 is 1 unit, to City 2 is 1 unit, to City 3 is 1 unit, to City 4 is 1 unit, to City 5 is 3 units, to City 6 is 3 units, and to City 7 is 3 units. From City 1 to City 0 is 1 unit, to City 2 is 2 units, to City 3 is 2 units, to City 4 is 2 units, to City 5 is 3 units, to City 6 is 2 units, and to City 7 is 1 unit. From City 2 to City 0 is 2 units, to City 1 is 3 units, to City 3 is 3 units, to City 4 is 2 units, to City 5 is 2 units, to City 6 is 3 units, and to City 7 is 3 units. From City 3 to City 0 is 2 units, to City 1 is 3 units, to City 2 is 1 unit, to City 4 is 3 units, to City 5 is 1 unit, to City 6 is 3 units, and to City 7 is 3 units. From City 4 to City 0 is 3 units, to City 1 is 1 unit, to City 2 is 3 units, to City 3 is 1 unit, to City 5 is 1 unit, to City 6 is 2 units, and to City 7 is 1 unit. From City 5 to City 0 is 1 unit, to City 1 is 1 unit, to City 2 is 3 units, to City 3 is 2 units, to City 4 is 1 unit, to City 6 is 3 units, and to City 7 is 3 units. From City 6 to City 0 is 2 units, to City 1 is 3 units, to City 2 is 3 units, to City 3 is 2 units, to City 4 is 3 units, to City 5 is 1 unit, and to City 7 is 1 unit. From City 7 to City 0 is 2 units, to City 1 is 1 unit, to City 2 is 2 units, to City 3 is 2 units, to City 4 is 1 unit, to City 5 is 1 unit, and to City 6 is 1 unit. + +The maximum capacity for transporting resources from City 0 to City 1 is 7 units, to City 2 is 9 units, to City 3 is 7 units, to City 4 is 8 units, to City 5 is 9 units, to City 6 is 7 units, and to City 7 is 7 units. From City 1 to City 0 is 8 units, to City 2 is 8 units, to City 3 is 8 units, to City 4 is 7 units, to City 5 is 9 units, to City 6 is 8 units, and to City 7 is 9 units. From City 2 to City 0 is 7 units, to City 1 is 9 units, to City 3 is 9 units, to City 4 is 7 units, to City 5 is 9 units, to City 6 is 8 units, and to City 7 is 9 units. From City 3 to City 0 is 8 units, to City 1 is 8 units, to City 2 is 9 units, to City 4 is 7 units, to City 5 is 9 units, to City 6 is 7 units, and to City 7 is 8 units. From City 4 to City 0 is 7 units, to City 1 is 7 units, to City 2 is 7 units, to City 3 is 9 units, to City 5 is 8 units, to City 6 is 7 units, and to City 7 is 7 units. From City 5 to City 0 is 9 units, to City 1 is 9 units, to City 2 is 9 units, to City 3 is 9 units, to City 4 is 7 units, to City 6 is 7 units, and to City 7 is 7 units. From City 6 to City 0 is 8 units, to City 1 is 9 units, to City 2 is 7 units, to City 3 is 9 units, to City 4 is 9 units, to City 5 is 7 units, and to City 7 is 9 units. From City 7 to City 0 is 9 units, to City 1 is 7 units, to City 2 is 8 units, to City 3 is 8 units, to City 4 is 8 units, to City 5 is 9 units, and to City 6 is 8 units. + +The maximum storage capacity for City 0 is 16 units, City 1 is 16 units, City 2 is 18 units, City 3 is 12 units, City 4 is 14 units, City 5 is 16 units, City 6 is 12 units, and City 7 is 14 units. + +Your task is to determine the optimal amount of resources to transport between cities to minimize the total transportation cost while ensuring that each city's net demand or supply is satisfied, the transportation links are not overloaded, and the storage capacities of the cities are not exceeded.7You are responsible for managing the distribution of resources across a network of eight cities. Each city has a specific supply of resources and a demand that must be met. The goal is to minimize the total transportation cost while ensuring that all cities receive the resources they need, without exceeding the capacity of the transportation links or the storage capacity of the cities. + +City 0 has a net demand of 2 units, City 1 has a balanced supply and demand, City 2 has a net supply of 1 unit, City 3 has a net demand of 2 units, City 4 has a net supply of 2 units, City 5 has a balanced supply and demand, City 6 has a net demand of 1 unit, and City 7 has a net supply of 2 units. + +The cost of transporting resources from City 0 to City 1 is 1 unit, to City 2 is 1 unit, to City 3 is 1 unit, to City 4 is 1 unit, to City 5 is 3 units, to City 6 is 3 units, and to City 7 is 3 units. From City 1 to City 0 is 1 unit, to City 2 is 2 units, to City 3 is 2 units, to City 4 is 2 units, to City 5 is 3 units, to City 6 is 2 units, and to City 7 is 1 unit. From City 2 to City 0 is 2 units, to City 1 is 3 units, to City 3 is 3 units, to City 4 is 2 units, to City 5 is 2 units, to City 6 is 3 units, and to City 7 is 3 units. From City 3 to City 0 is 2 units, to City 1 is 3 units, to City 2 is 1 unit, to City 4 is 3 units, to City 5 is 1 unit, to City 6 is 3 units, and to City 7 is 3 units. From City 4 to City 0 is 3 units, to City 1 is 1 unit, to City 2 is 3 units, to City 3 is 1 unit, to City 5 is 1 unit, to City 6 is 2 units, and to City 7 is 1 unit. From City 5 to City 0 is 1 unit, to City 1 is 1 unit, to City 2 is 3 units, to City 3 is 2 units, to City 4 is 1 unit, to City 6 is 3 units, and to City 7 is 3 units. From City 6 to City 0 is 2 units, to City 1 is 3 units, to City 2 is 3 units, to City 3 is 2 units, to City 4 is 3 units, to City 5 is 1 unit, and to City 7 is 1 unit. From City 7 to City 0 is 2 units, to City 1 is 1 unit, to City 2 is 2 units, to City 3 is 2 units, to City 4 is 1 unit, to City 5 is 1 unit, and to City 6 is 1 unit. + +The maximum capacity for transporting resources from City 0 to City 1 is 7 units, to City 2 is 9 units, to City 3 is 7 units, to City 4 is 8 units, to City 5 is 9 units, to City 6 is 7 units, and to City 7 is 7 units. From City 1 to City 0 is 8 units, to City 2 is 8 units, to City 3 is 8 units, to City 4 is 7 units, to City 5 is 9 units, to City 6 is 8 units, and to City 7 is 9 units. From City 2 to City 0 is 7 units, to City 1 is 9 units, to City 3 is 9 units, to City 4 is 7 units, to City 5 is 9 units, to City 6 is 8 units, and to City 7 is 9 units. From City 3 to City 0 is 8 units, to City 1 is 8 units, to City 2 is 9 units, to City 4 is 7 units, to City 5 is 9 units, to City 6 is 7 units, and to City 7 is 8 units. From City 4 to City 0 is 7 units, to City 1 is 7 units, to City 2 is 7 units, to City 3 is 9 units, to City 5 is 8 units, to City 6 is 7 units, and to City 7 is 7 units. From City 5 to City 0 is 9 units, to City 1 is 9 units, to City 2 is 9 units, to City 3 is 9 units, to City 4 is 7 units, to City 6 is 7 units, and to City 7 is 7 units. From City 6 to City 0 is 8 units, to City 1 is 9 units, to City 2 is 7 units, to City 3 is 9 units, to City 4 is 9 units, to City 5 is 7 units, and to City 7 is 9 units. From City 7 to City 0 is 9 units, to City 1 is 7 units, to City 2 is 8 units, to City 3 is 8 units, to City 4 is 8 units, to City 5 is 9 units, and to City 6 is 8 units. + +The maximum storage capacity for City 0 is 16 units, City 1 is 16 units, City 2 is 18 units, City 3 is 12 units, City 4 is 14 units, City 5 is 16 units, City 6 is 12 units, and City 7 is 14 units. + +Your task is to determine the optimal amount of resources to transport between cities to minimize the total transportation cost while ensuring that each city's net demand or supply is satisfied, the transportation links are not overloaded, and the storage capacities of the cities are not exceeded.7No differencesNo differencesok
41Minimize the total penalty costs for landing five aircraft, where penalties are incurred for landing earlier or later than the target times. The target landing times are 89, 63, 64, 97, and 85 for aircraft 0, 1, 2, 3, and 4, respectively. Each aircraft must land within its specified time window: aircraft 0 between 60 and 100, aircraft 1 between 60 and 93, aircraft 2 between 60 and 94, aircraft 3 between 67 and 100, and aircraft 4 between 60 and 100. The penalty costs per unit time for landing early are 19, 10, 14, 12, and 19 for aircraft 0, 1, 2, 3, and 4, respectively. The penalty costs per unit time for landing late are 14, 13, 15, 15, and 14 for aircraft 0, 1, 2, 3, and 4, respectively. + +To ensure safety, a minimum separation time must be maintained between consecutive landings. The required separation times between aircraft pairs are as follows: between aircraft 0 and 1, 7 units if aircraft 0 lands first or 8 units if aircraft 1 lands first; between aircraft 0 and 2, 6 units if aircraft 0 lands first or 7 units if aircraft 2 lands first; between aircraft 0 and 3, 5 units if aircraft 0 lands first or 8 units if aircraft 3 lands first; between aircraft 0 and 4, 5 units if aircraft 0 lands first or 7 units if aircraft 4 lands first; between aircraft 1 and 2, 5 units if aircraft 1 lands first or 6 units if aircraft 2 lands first; between aircraft 1 and 3, 5 units if aircraft 1 lands first or 7 units if aircraft 3 lands first; between aircraft 1 and 4, 6 units if aircraft 1 lands first or 5 units if aircraft 4 lands first; between aircraft 2 and 3, 5 units if aircraft 2 lands first or 6 units if aircraft 3 lands first; between aircraft 2 and 4, 5 units if aircraft 2 lands first or 7 units if aircraft 4 lands first; between aircraft 3 and 4, 6 units if aircraft 3 lands first or 6 units if aircraft 4 lands first. + +The order in which aircraft land must be determined, and for each pair of aircraft, one must land before the other. The separation time must be enforced based on the landing order. The actual landing times for each aircraft must be determined, and the penalties for landing early or late must be calculated based on the deviations from the target landing times. The objective is to minimize the total penalty costs while ensuring that all constraints are satisfied.87Minimize the total penalty costs for landing five aircraft, where penalties are incurred for landing earlier or later than the target times. The target landing times are 89, 63, 64, 97, and 85 for aircraft 0, 1, 2, 3, and 4, respectively. Each aircraft must land within its specified time window: aircraft 0 between 60 and 100, aircraft 1 between 60 and 93, aircraft 2 between 60 and 94, aircraft 3 between 67 and 100, and aircraft 4 between 60 and 100. The penalty costs per unit time for landing early are 19, 10, 14, 12, and 19 for aircraft 0, 1, 2, 3, and 4, respectively. The penalty costs per unit time for landing late are 14, 13, 15, 15, and 14 for aircraft 0, 1, 2, 3, and 4, respectively. + +To ensure safety, a minimum separation time must be maintained between consecutive landings. The required separation times between aircraft pairs are as follows: between aircraft 0 and 1, 7 units if aircraft 0 lands first or 8 units if aircraft 1 lands first; between aircraft 0 and 2, 6 units if aircraft 0 lands first or 7 units if aircraft 2 lands first; between aircraft 0 and 3, 5 units if aircraft 0 lands first or 8 units if aircraft 3 lands first; between aircraft 0 and 4, 5 units if aircraft 0 lands first or 7 units if aircraft 4 lands first; between aircraft 1 and 2, 5 units if aircraft 1 lands first or 6 units if aircraft 2 lands first; between aircraft 1 and 3, 5 units if aircraft 1 lands first or 7 units if aircraft 3 lands first; between aircraft 1 and 4, 6 units if aircraft 1 lands first or 5 units if aircraft 4 lands first; between aircraft 2 and 3, 5 units if aircraft 2 lands first or 6 units if aircraft 3 lands first; between aircraft 2 and 4, 5 units if aircraft 2 lands first or 7 units if aircraft 4 lands first; between aircraft 3 and 4, 6 units if aircraft 3 lands first or 6 units if aircraft 4 lands first. + +The order in which aircraft land must be determined, and for each pair of aircraft, one must land before the other. The separation time must be enforced based on the landing order. The actual landing times for each aircraft must be determined, and the penalties for landing early or late must be calculated based on the deviations from the target landing times. The objective is to minimize the total penalty costs while ensuring that all constraints are satisfied.87No differencesNo differencesok
42Facility Location for Maximum Dispersion + +You are tasked with selecting a set of facilities from a network of five potential locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall dispersion between the selected facilities. Dispersion is measured by the weighted distances between pairs of facilities, where higher weights indicate greater desirability for separation. The goal is to ensure that the selected facilities are as spread out as possible, given the constraints. + +#### Key Decisions: +1. **Facility Selection**: Decide which facilities to open. Each facility can either be selected (1) or not selected (0). Specifically, you must select exactly 2 facilities out of the 5 available. +2. **Pairwise Dispersion**: Determine whether a pair of facilities contributes to the dispersion score. A pair contributes only if both facilities in the pair are selected. + +#### Objective: +Maximize the total dispersion score, which is the sum of the weighted distances between all pairs of selected facilities. The weights for each pair are as follows: +- Node 0 and Node 1: 22 +- Node 0 and Node 2: 19 +- Node 0 and Node 3: 24 +- Node 0 and Node 4: 22 +- Node 1 and Node 0: 19 +- Node 1 and Node 2: 23 +- Node 1 and Node 3: 16 +- Node 1 and Node 4: 25 +- Node 2 and Node 0: 19 +- Node 2 and Node 1: 16 +- Node 2 and Node 3: 19 +- Node 2 and Node 4: 21 +- Node 3 and Node 0: 25 +- Node 3 and Node 1: 19 +- Node 3 and Node 2: 21 +- Node 3 and Node 4: 17 +- Node 4 and Node 0: 19 +- Node 4 and Node 1: 22 +- Node 4 and Node 2: 19 +- Node 4 and Node 3: 21 + +#### Constraints: +1. **Facility Selection Limit**: Exactly 2 facilities must be selected from the 5 available. This means the sum of the selection decisions for all facilities must equal 2. +2. **Pairwise Contribution Rules**: + - A pair of facilities can only contribute to the dispersion score if both facilities in the pair are selected. + - If either facility in a pair is not selected, the pair cannot contribute to the dispersion score. + - If both facilities in a pair are selected, the pair must contribute to the dispersion score. + +#### Numerical Parameters: +- The weights for each pair of facilities are explicitly provided in the objective section above. +- The total number of facilities to select is fixed at 2. + +#### Summary: +Your task is to select 2 facilities out of 5 such that the total dispersion score, calculated as the sum of the weighted distances between all selected pairs, is maximized. The selection must adhere to the constraints that ensure only valid pairs contribute to the dispersion score and that exactly 2 facilities are chosen.49Facility Location for Maximum Dispersion + +You are tasked with selecting a set of facilities from a network of five potential locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall dispersion between the selected facilities. Dispersion is measured by the weighted distances between pairs of facilities, where higher weights indicate greater desirability for separation. The goal is to ensure that the selected facilities are as spread out as possible, given the constraints. + +#### Key Decisions: +1. **Facility Selection**: Decide which facilities to open. Each facility can either be selected (1) or not selected (0). Specifically, you must select exactly 2 facilities out of the 5 available. +2. **Pairwise Dispersion**: Determine whether a pair of facilities contributes to the dispersion score. A pair contributes only if both facilities in the pair are selected. + +#### Objective: +Maximize the total dispersion score, which is the sum of the weighted distances between all pairs of selected facilities. The weights for each pair are as follows: +- Node 0 and Node 1: 22 +- Node 0 and Node 2: 19 +- Node 0 and Node 3: 24 +- Node 0 and Node 4: 22 +- Node 1 and Node 0: 19 +- Node 1 and Node 2: 23 +- Node 1 and Node 3: 16 +- Node 1 and Node 4: 25 +- Node 2 and Node 0: 19 +- Node 2 and Node 1: 16 +- Node 2 and Node 3: 19 +- Node 2 and Node 4: 21 +- Node 3 and Node 0: 25 +- Node 3 and Node 1: 19 +- Node 3 and Node 2: 21 +- Node 3 and Node 4: 17 +- Node 4 and Node 0: 19 +- Node 4 and Node 1: 22 +- Node 4 and Node 2: 19 +- Node 4 and Node 3: 21 + +#### Constraints: +1. **Facility Selection Limit**: Exactly 2 facilities must be selected from the 5 available. This means the sum of the selection decisions for all facilities must equal 2. +2. **Pairwise Contribution Rules**: + - A pair of facilities can only contribute to the dispersion score if both facilities in the pair are selected. + - If either facility in a pair is not selected, the pair cannot contribute to the dispersion score. + - If both facilities in a pair are selected, the pair must contribute to the dispersion score. + +#### Numerical Parameters: +- The weights for each pair of facilities are explicitly provided in the objective section above. +- The total number of facilities to select is fixed at 2. + +#### Summary: +Your task is to select 2 facilities out of 5 such that the total dispersion score, calculated as the sum of the weighted distances between all selected pairs, is maximized. The selection must adhere to the constraints that ensure only valid pairs contribute to the dispersion score and that exactly 2 facilities are chosen.49No differencesNo differencesok
43The problem involves scheduling production tasks for five jobs (Job 0 through Job 4) in a manufacturing facility. Each job consists of a sequence of operations that must be performed in a specific order, with each operation taking exactly 1 unit of time to complete. The facility has a limited number of machines, and certain operations from different jobs may require the same machine. The goal is to minimize the total production time, known as the makespan, while ensuring that all operational constraints are satisfied. + +Each operation within a job must start only after the previous operation in the same job is completed. For example, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 is finished. Operations that require the same machine cannot overlap in time. To enforce this, binary decision variables are used to determine the order of operations on each machine. For instance, if Operation 0 of Job 0 is scheduled before Operation 0 of Job 1 on Machine 1, the binary variable \( X_{(0,0),(1,0)} \) is set to 1; otherwise, it is set to 0. A large constant (100000) is used in the constraints to ensure that the sequencing is properly enforced. + +The makespan must be at least as long as the completion time of the last operation across all jobs. For example, if Operation 1 of Job 4 finishes at time 10, the makespan must be at least 10. All operation start times must be non-negative, and the binary decision variables can only take values of 0 or 1. + +The objective is to determine the start times for all operations and the sequence of operations on shared machines such that the total production time is minimized. This must be done while ensuring that all precedence constraints, machine capacity constraints, and non-negativity constraints are satisfied. The processing time for each operation is fixed at 1 unit of time, and the binary decision variables are used to enforce the order of operations on shared machines.5The problem involves scheduling production tasks for five jobs (Job 0 through Job 4) in a manufacturing facility. Each job consists of a sequence of 2 operations that must be performed in sequential order, with each operation taking exactly 1 unit of time to complete. The facility has a limited number of machines, and certain operations from different jobs may require the same machine. The goal is to minimize the total production time, known as the makespan, while ensuring that all operational constraints are satisfied. + +Each operation within a job must start only after the previous operation in the same job is completed. For example, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 is finished. Operations that require the same machine cannot overlap in time. To enforce this, binary decision variables are used to determine the order of operations on each machine. For instance, if Operation 0 of Job 0 is scheduled before Operation 0 of Job 1 on Machine 1, the binary variable \( X_{(0,0),(1,0)} \) is set to 1; otherwise, it is set to 0. A large constant (100) is used in the constraints to ensure that the sequencing is properly enforced. + +The makespan must be at least as long as the completion time of the last operation across all jobs. For example, if Operation 1 of Job 4 finishes at time 10, the makespan must be at least 10. All operation start times must be non-negative, and the binary decision variables can only take values of 0 or 1. + +The objective is to determine the start times for all operations and the sequence of operations on shared machines such that the total production time is minimized. This must be done while ensuring that all precedence constraints, machine capacity constraints, and non-negativity constraints are satisfied. The processing time for each operation is fixed at 1 unit of time, and the binary decision variables are used to enforce the order of operations on shared machines. + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 1 + - Operation 1 of Job 0: Machine 0 + - Operation 0 of Job 1: Machine 2 + - Operation 1 of Job 1: Machine 0 + - Operation 0 of Job 2: Machine 1 + - Operation 1 of Job 2: Machine 0 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 14The problem involves scheduling production tasks for five jobs (Job 0 through Job 4) in a manufacturing facility. Each job consists of a sequence of 2 operations that must be performed in a specific sequential order, with each operation taking exactly 1 unit of time to complete. The facility has a limited number of machines, and certain operations from different jobs may require the same machine. The goal is to minimize the total production time, known as the makespan, while ensuring that all operational constraints are satisfied. + +Each operation within a job must start only after the previous operation in the same job is completed. For example, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 is finished. Operations that require the same machine cannot overlap in time. To enforce this, binary decision variables are used to determine the order of operations on each machine. For instance, if Operation 0 of Job 0 is scheduled before Operation 0 of Job 1 on Machine 1, the binary variable \( X_{(0,0),(1,0)} \) is set to 1; otherwise, it is set to 0. A large constant (100000) (100) is used in the constraints to ensure that the sequencing is properly enforced. + +The makespan must be at least as long as the completion time of the last operation across all jobs. For example, if Operation 1 of Job 4 finishes at time 10, the makespan must be at least 10. All operation start times must be non-negative, and the binary decision variables can only take values of 0 or 1. + +The objective is to determine the start times for all operations and the sequence of operations on shared machines such that the total production time is minimized. This must be done while ensuring that all precedence constraints, machine capacity constraints, and non-negativity constraints are satisfied. The processing time for each operation is fixed at 1 unit of time, and the binary decision variables are used to enforce the order of operations on shared machines. + + + Additional information: + - There are 3 machines, numbered from 0 to 2 + - The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 1 + - Operation 1 of Job 0: Machine 0 + - Operation 0 of Job 1: Machine 2 + - Operation 1 of Job 1: Machine 0 + - Operation 0 of Job 2: Machine 1 + - Operation 1 of Job 2: Machine 0 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 15 4It was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
44Managing Resource Distribution Across Cities + +Imagine you are responsible for managing the distribution of a critical resource (e.g., medical supplies, food, or energy) across seven interconnected cities. Each city has a specific supply of the resource, a demand that must be met, and limited capacity to store or process incoming resources. Your goal is to minimize the total cost of transporting the resource between cities while ensuring that all demands are satisfied and no capacity limits are exceeded. + +#### City Details: +- **City 0** has a net demand of 2 units (it requires 2 more units than it currently has). +- **City 1** has a net demand of 2 units. +- **City 2** has a net supply of 1 unit (it has 1 more unit than it needs). +- **City 3** has a net supply of 2 units. +- **City 4** is balanced, with no net demand or supply. +- **City 5** has a net supply of 1 unit. +- **City 6** is balanced, with no net demand or supply. + +#### Transportation Costs: +The cost of transporting resources between cities varies depending on the route. Below are the costs per unit transported: +- From **City 0**: + - To City 1: 1 unit cost + - To City 2: 2 units cost + - To City 3: 3 units cost + - To City 4: 2 units cost + - To City 5: 3 units cost + - To City 6: 3 units cost +- From **City 1**: + - To City 0: 2 units cost + - To City 2: 3 units cost + - To City 3: 2 units cost + - To City 4: 2 units cost + - To City 5: 3 units cost + - To City 6: 3 units cost +- From **City 2**: + - To City 0: 2 units cost + - To City 1: 1 unit cost + - To City 3: 3 units cost + - To City 4: 2 units cost + - To City 5: 3 units cost + - To City 6: 2 units cost +- From **City 3**: + - To City 0: 2 units cost + - To City 1: 2 units cost + - To City 2: 2 units cost + - To City 4: 3 units cost + - To City 5: 1 unit cost + - To City 6: 1 unit cost +- From **City 4**: + - To City 0: 1 unit cost + - To City 1: 1 unit cost + - To City 2: 1 unit cost + - To City 3: 1 unit cost + - To City 5: 1 unit cost + - To City 6: 1 unit cost +- From **City 5**: + - To City 0: 3 units cost + - To City 1: 2 units cost + - To City 2: 3 units cost + - To City 3: 2 units cost + - To City 4: 3 units cost + - To City 6: 3 units cost +- From **City 6**: + - To City 0: 3 units cost + - To City 1: 3 units cost + - To City 2: 2 units cost + - To City 3: 3 units cost + - To City 4: 3 units cost + - To City 5: 3 units cost + +#### Transportation Constraints: +Each route between cities has a maximum capacity for resource transportation: +- **City 0**: + - To City 1: Maximum 8 units + - To City 2: Maximum 7 units + - To City 3: Maximum 7 units + - To City 4: Maximum 8 units + - To City 5: Maximum 7 units + - To City 6: Maximum 8 units +- **City 1**: + - To City 0: Maximum 8 units + - To City 2: Maximum 9 units + - To City 3: Maximum 8 units + - To City 4: Maximum 7 units + - To City 5: Maximum 7 units + - To City 6: Maximum 8 units +- **City 2**: + - To City 0: Maximum 9 units + - To City 1: Maximum 9 units + - To City 3: Maximum 7 units + - To City 4: Maximum 9 units + - To City 5: Maximum 9 units + - To City 6: Maximum 9 units +- **City 3**: + - To City 0: Maximum 9 units + - To City 1: Maximum 7 units + - To City 2: Maximum 7 units + - To City 4: Maximum 9 units + - To City 5: Maximum 8 units + - To City 6: Maximum 9 units +- **City 4**: + - To City 0: Maximum 8 units + - To City 1: Maximum 9 units + - To City 2: Maximum 9 units + - To City 3: Maximum 9 units + - To City 5: Maximum 8 units + - To City 6: Maximum 8 units +- **City 5**: + - To City 0: Maximum 8 units + - To City 1: Maximum 7 units + - To City 2: Maximum 9 units + - To City 3: Maximum 8 units + - To City 4: Maximum 8 units + - To City 6: Maximum 8 units +- **City 6**: + - To City 0: Maximum 9 units + - To City 1: Maximum 8 units + - To City 2: Maximum 8 units + - To City 3: Maximum 7 units + - To City 4: Maximum 9 units + - To City 5: Maximum 8 units + +#### City Capacity Constraints: +Each city has a maximum capacity for incoming resources: +- **City 0**: Maximum 18 units +- **City 1**: Maximum 13 units +- **City 2**: Maximum 14 units +- **City 3**: Maximum 16 units +- **City 4**: Maximum 15 units +- **City 5**: Maximum 18 units +- **City 6**: Maximum 14 units + +#### Objective: +Your task is to determine the optimal amount of resources to transport between cities to meet all demands while minimizing the total transportation cost. Ensure that: +1. All net demands and supplies are balanced. +2. No transportation route exceeds its capacity. +3. No city exceeds its capacity for incoming resources. +4. All transportation amounts are non-negative. + +By solving this problem, you will ensure efficient resource distribution across the cities at the lowest possible cost.9Managing Resource Distribution Across Cities + +Imagine you are responsible for managing the distribution of a critical resource (e.g., medical supplies, food, or energy) across seven interconnected cities. Each city has a specific supply of the resource, a demand that must be met, and limited capacity to store or process incoming resources. Your goal is to minimize the total cost of transporting the resource between cities while ensuring that all demands are satisfied and no capacity limits are exceeded. + +#### City Details: +- **City 0** has a net demand of 2 units (it requires 2 more units than it currently has). +- **City 1** has a net demand of 2 units. +- **City 2** has a net supply of 1 unit (it has 1 more unit than it needs). +- **City 3** has a net supply of 2 units. +- **City 4** is balanced, with no net demand or supply. +- **City 5** has a net supply of 1 unit. +- **City 6** is balanced, with no net demand or supply. + +#### Transportation Costs: +The cost of transporting resources between cities varies depending on the route. Below are the costs per unit transported: +- From **City 0**: + - To City 1: 1 unit cost + - To City 2: 2 units cost + - To City 3: 3 units cost + - To City 4: 2 units cost + - To City 5: 3 units cost + - To City 6: 3 units cost +- From **City 1**: + - To City 0: 2 units cost + - To City 2: 3 units cost + - To City 3: 2 units cost + - To City 4: 2 units cost + - To City 5: 3 units cost + - To City 6: 3 units cost +- From **City 2**: + - To City 0: 2 units cost + - To City 1: 1 unit cost + - To City 3: 3 units cost + - To City 4: 2 units cost + - To City 5: 3 units cost + - To City 6: 2 units cost +- From **City 3**: + - To City 0: 2 units cost + - To City 1: 2 units cost + - To City 2: 2 units cost + - To City 4: 3 units cost + - To City 5: 1 unit cost + - To City 6: 1 unit cost +- From **City 4**: + - To City 0: 1 unit cost + - To City 1: 1 unit cost + - To City 2: 1 unit cost + - To City 3: 1 unit cost + - To City 5: 1 unit cost + - To City 6: 1 unit cost +- From **City 5**: + - To City 0: 3 units cost + - To City 1: 2 units cost + - To City 2: 3 units cost + - To City 3: 2 units cost + - To City 4: 3 units cost + - To City 6: 3 units cost +- From **City 6**: + - To City 0: 3 units cost + - To City 1: 3 units cost + - To City 2: 2 units cost + - To City 3: 3 units cost + - To City 4: 3 units cost + - To City 5: 3 units cost + +#### Transportation Constraints: +Each route between cities has a maximum capacity for resource transportation: +- **City 0**: + - To City 1: Maximum 8 units + - To City 2: Maximum 7 units + - To City 3: Maximum 7 units + - To City 4: Maximum 8 units + - To City 5: Maximum 7 units + - To City 6: Maximum 8 units +- **City 1**: + - To City 0: Maximum 8 units + - To City 2: Maximum 9 units + - To City 3: Maximum 8 units + - To City 4: Maximum 7 units + - To City 5: Maximum 7 units + - To City 6: Maximum 8 units +- **City 2**: + - To City 0: Maximum 9 units + - To City 1: Maximum 9 units + - To City 3: Maximum 7 units + - To City 4: Maximum 9 units + - To City 5: Maximum 9 units + - To City 6: Maximum 9 units +- **City 3**: + - To City 0: Maximum 9 units + - To City 1: Maximum 7 units + - To City 2: Maximum 7 units + - To City 4: Maximum 9 units + - To City 5: Maximum 8 units + - To City 6: Maximum 9 units +- **City 4**: + - To City 0: Maximum 8 units + - To City 1: Maximum 9 units + - To City 2: Maximum 9 units + - To City 3: Maximum 9 units + - To City 5: Maximum 8 units + - To City 6: Maximum 8 units +- **City 5**: + - To City 0: Maximum 8 units + - To City 1: Maximum 7 units + - To City 2: Maximum 9 units + - To City 3: Maximum 8 units + - To City 4: Maximum 8 units + - To City 6: Maximum 8 units +- **City 6**: + - To City 0: Maximum 9 units + - To City 1: Maximum 8 units + - To City 2: Maximum 8 units + - To City 3: Maximum 7 units + - To City 4: Maximum 9 units + - To City 5: Maximum 8 units + +#### City Capacity Constraints: +Each city has a maximum capacity for incoming resources: +- **City 0**: Maximum 18 units +- **City 1**: Maximum 13 units +- **City 2**: Maximum 14 units +- **City 3**: Maximum 16 units +- **City 4**: Maximum 15 units +- **City 5**: Maximum 18 units +- **City 6**: Maximum 14 units + +#### Objective: +Your task is to determine the optimal amount of resources to transport between cities to meet all demands while minimizing the total transportation cost. Ensure that: +1. All net demands and supplies are balanced. +2. No transportation route exceeds its capacity. +3. No city exceeds its capacity for incoming resources. +4. All transportation amounts are non-negative. + +By solving this problem, you will ensure efficient resource distribution across the cities at the lowest possible cost.7No differences9 7Incorrect ground truthincorrect ground truth
45You are responsible for assigning 8 employees to 7 projects to minimize the total cost while meeting all project requirements and respecting employee availability and capacity limits. Each employee has a fixed total number of hours they must contribute, and each project requires a specific number of hours to be completed. The decision variables are binary, indicating whether an employee is assigned to a project or not. + +Employee 0 must contribute exactly 1 hour in total, Employee 1 must contribute exactly 18 hours, Employee 2 must contribute exactly 15 hours, Employee 3 must contribute exactly 19 hours, Employee 4 must contribute exactly 18 hours, Employee 5 must contribute exactly 17 hours, Employee 6 must contribute exactly 15 hours, and Employee 7 must contribute exactly 19 hours. + +Project 0 requires exactly 15 hours, Project 1 requires exactly 19 hours, Project 2 requires exactly 16 hours, Project 3 requires exactly 20 hours, Project 4 requires exactly 17 hours, Project 5 requires exactly 16 hours, and Project 6 requires exactly 19 hours. + +Each employee has a maximum number of hours they can contribute to each project: Employee 0 can contribute up to 8 hours to Project 0, 9 hours to Project 1, 5 hours to Project 2, 10 hours to Project 3, 7 hours to Project 4, 10 hours to Project 5, and 10 hours to Project 6. Employee 1 can contribute up to 8 hours to Project 0, 10 hours to Project 1, 6 hours to Project 2, 6 hours to Project 3, 6 hours to Project 4, 8 hours to Project 5, and 5 hours to Project 6. Employee 2 can contribute up to 5 hours to Project 0, 7 hours to Project 1, 9 hours to Project 2, 7 hours to Project 3, 8 hours to Project 4, 9 hours to Project 5, and 7 hours to Project 6. Employee 3 can contribute up to 5 hours to Project 0, 6 hours to Project 1, 7 hours to Project 2, 7 hours to Project 3, 6 hours to Project 4, 8 hours to Project 5, and 9 hours to Project 6. Employee 4 can contribute up to 7 hours to Project 0, 10 hours to Project 1, 7 hours to Project 2, 6 hours to Project 3, 9 hours to Project 4, 5 hours to Project 5, and 8 hours to Project 6. Employee 5 can contribute up to 6 hours to Project 0, 10 hours to Project 1, 5 hours to Project 2, 9 hours to Project 3, 9 hours to Project 4, 6 hours to Project 5, and 10 hours to Project 6. Employee 6 can contribute up to 7 hours to Project 0, 8 hours to Project 1, 10 hours to Project 2, 5 hours to Project 3, 10 hours to Project 4, 10 hours to Project 5, and 6 hours to Project 6. Employee 7 can contribute up to 10 hours to Project 0, 7 hours to Project 1, 6 hours to Project 2, 9 hours to Project 3, 10 hours to Project 4, 5 hours to Project 5, and 7 hours to Project 6. + +The cost of assigning Employee 0 to Project 0 is 29, to Project 1 is 30, to Project 2 is 32, to Project 3 is 34, to Project 4 is 26, to Project 5 is 35, and to Project 6 is 27. The cost of assigning Employee 1 to Project 0 is 32, to Project 1 is 25, to Project 2 is 35, to Project 3 is 25, to Project 4 is 28, to Project 5 is 26, and to Project 6 is 33. The cost of assigning Employee 2 to Project 0 is 31, to Project 1 is 29, to Project 2 is 35, to Project 3 is 29, to Project 4 is 32, to Project 5 is 25, and to Project 6 is 27. The cost of assigning Employee 3 to Project 0 is 30, to Project 1 is 29, to Project 2 is 28, to Project 3 is 35, to Project 4 is 29, to Project 5 is 27, and to Project 6 is 28. The cost of assigning Employee 4 to Project 0 is 33, to Project 1 is 28, to Project 2 is 26, to Project 3 is 31, to Project 4 is 26, to Project 5 is 29, and to Project 6 is 32. The cost of assigning Employee 5 to Project 0 is 35, to Project 1 is 34, to Project 2 is 29, to Project 3 is 32, to Project 4 is 33, to Project 5 is 31, and to Project 6 is 28. The cost of assigning Employee 6 to Project 0 is 27, to Project 1 is 30, to Project 2 is 27, to Project 3 is 30, to Project 4 is 32, to Project 5 is 32, and to Project 6 is 26. The cost of assigning Employee 7 to Project 0 is 30, to Project 1 is 27, to Project 2 is 31, to Project 3 is 32, to Project 4 is 34, to Project 5 is 32, and to Project 6 is 26. + +The objective is to assign employees to projects such that all project requirements are met, no employee exceeds their total available hours, no employee exceeds their maximum contribution limit for any specific project, and the total cost of all assignments is minimized.3320You are responsible for assigning 8 employees to 7 projects to minimize the total cost while meeting all project requirements and respecting employee availability and capacity limits. Each employee has a fixed total number of hours they must contribute, and each project requires a specific number of hours to be completed. The decision variables are binary, indicating whether an employee is assigned to a project or not. + +Employee 0 must contribute exactly 1 hour in total, Employee 1 must contribute exactly 18 hours, Employee 2 must contribute exactly 15 hours, Employee 3 must contribute exactly 19 hours, Employee 4 must contribute exactly 18 hours, Employee 5 must contribute exactly 17 hours, Employee 6 must contribute exactly 15 hours, and Employee 7 must contribute exactly 19 hours. + +Project 0 requires exactly 15 hours, Project 1 requires exactly 19 hours, Project 2 requires exactly 16 hours, Project 3 requires exactly 20 hours, Project 4 requires exactly 17 hours, Project 5 requires exactly 16 hours, and Project 6 requires exactly 19 hours. + +Each employee has a maximum number of hours they can contribute to each project: Employee 0 can contribute up to 8 hours to Project 0, 9 hours to Project 1, 5 hours to Project 2, 10 hours to Project 3, 7 hours to Project 4, 10 hours to Project 5, and 10 hours to Project 6. Employee 1 can contribute up to 8 hours to Project 0, 10 hours to Project 1, 6 hours to Project 2, 6 hours to Project 3, 6 hours to Project 4, 8 hours to Project 5, and 5 hours to Project 6. Employee 2 can contribute up to 5 hours to Project 0, 7 hours to Project 1, 9 hours to Project 2, 7 hours to Project 3, 8 hours to Project 4, 9 hours to Project 5, and 7 hours to Project 6. Employee 3 can contribute up to 5 hours to Project 0, 6 hours to Project 1, 7 hours to Project 2, 7 hours to Project 3, 6 hours to Project 4, 8 hours to Project 5, and 9 hours to Project 6. Employee 4 can contribute up to 7 hours to Project 0, 10 hours to Project 1, 7 hours to Project 2, 6 hours to Project 3, 9 hours to Project 4, 5 hours to Project 5, and 8 hours to Project 6. Employee 5 can contribute up to 6 hours to Project 0, 10 hours to Project 1, 5 hours to Project 2, 9 hours to Project 3, 9 hours to Project 4, 6 hours to Project 5, and 10 hours to Project 6. Employee 6 can contribute up to 7 hours to Project 0, 8 hours to Project 1, 10 hours to Project 2, 5 hours to Project 3, 10 hours to Project 4, 10 hours to Project 5, and 6 hours to Project 6. Employee 7 can contribute up to 10 hours to Project 0, 7 hours to Project 1, 6 hours to Project 2, 9 hours to Project 3, 10 hours to Project 4, 5 hours to Project 5, and 7 hours to Project 6. + +The cost of assigning Employee 0 to Project 0 is 29, to Project 1 is 30, to Project 2 is 32, to Project 3 is 34, to Project 4 is 26, to Project 5 is 35, and to Project 6 is 27. The cost of assigning Employee 1 to Project 0 is 32, to Project 1 is 25, to Project 2 is 35, to Project 3 is 25, to Project 4 is 28, to Project 5 is 26, and to Project 6 is 33. The cost of assigning Employee 2 to Project 0 is 31, to Project 1 is 29, to Project 2 is 35, to Project 3 is 29, to Project 4 is 32, to Project 5 is 25, and to Project 6 is 27. The cost of assigning Employee 3 to Project 0 is 30, to Project 1 is 29, to Project 2 is 28, to Project 3 is 35, to Project 4 is 29, to Project 5 is 27, and to Project 6 is 28. The cost of assigning Employee 4 to Project 0 is 33, to Project 1 is 28, to Project 2 is 26, to Project 3 is 31, to Project 4 is 26, to Project 5 is 29, and to Project 6 is 32. The cost of assigning Employee 5 to Project 0 is 35, to Project 1 is 34, to Project 2 is 29, to Project 3 is 32, to Project 4 is 33, to Project 5 is 31, and to Project 6 is 28. The cost of assigning Employee 6 to Project 0 is 27, to Project 1 is 30, to Project 2 is 27, to Project 3 is 30, to Project 4 is 32, to Project 5 is 32, and to Project 6 is 26. The cost of assigning Employee 7 to Project 0 is 30, to Project 1 is 27, to Project 2 is 31, to Project 3 is 32, to Project 4 is 34, to Project 5 is 32, and to Project 6 is 26. + +The objective is to assign employees to projects such that all project requirements are met, no employee exceeds their total available hours, no employee exceeds their maximum contribution limit for any specific project, and the total cost of all assignments is minimized.3320No differencesNo differencesok
46You are tasked with optimizing the placement of facilities across a network of five locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall benefit derived from their interactions. Each location has the potential to host a facility, but due to budget constraints, you can only select **exactly 3 locations** to establish facilities. The goal is to maximize the total benefit generated by the interactions between the selected facilities. + +#### Key Details: +1. **Facility Selection**: You must choose exactly 3 out of the 5 locations to establish facilities. Each location can either be selected (1) or not selected (0). +2. **Interaction Benefits**: The benefit of interaction between any two facilities depends on the specific pair of locations. For example: + - If facilities are placed at Node 0 and Node 1, the interaction benefit is **16**. + - If facilities are placed at Node 0 and Node 2, the interaction benefit is **20**. + - Similarly, the interaction benefits for all other pairs are provided in the table below. + +3. **Interaction Rules**: + - An interaction benefit is only counted if **both facilities in the pair are selected**. For example, if Node 0 and Node 1 are both selected, the interaction benefit of 16 is added to the total. + - If only one facility in a pair is selected, no interaction benefit is counted for that pair. + +4. **Interaction Benefit Table**: + | Pair of Nodes | Interaction Benefit | + |---------------|---------------------| + | Node 0 & Node 1 | 16 | + | Node 0 & Node 2 | 20 | + | Node 0 & Node 3 | 18 | + | Node 0 & Node 4 | 22 | + | Node 1 & Node 0 | 25 | + | Node 1 & Node 2 | 24 | + | Node 1 & Node 3 | 17 | + | Node 1 & Node 4 | 16 | + | Node 2 & Node 0 | 19 | + | Node 2 & Node 1 | 22 | + | Node 2 & Node 3 | 18 | + | Node 2 & Node 4 | 23 | + | Node 3 & Node 0 | 21 | + | Node 3 & Node 1 | 21 | + | Node 3 & Node 2 | 25 | + | Node 3 & Node 4 | 17 | + | Node 4 & Node 0 | 20 | + | Node 4 & Node 1 | 19 | + | Node 4 & Node 2 | 22 | + | Node 4 & Node 3 | 16 | + +5. **Constraints**: + - **Facility Selection Constraint**: Exactly 3 facilities must be selected out of the 5 locations. + - **Interaction Constraints**: For each pair of locations, the interaction benefit is only counted if both facilities in the pair are selected. This is enforced by ensuring that: + - If either facility in a pair is not selected, the interaction benefit for that pair is not counted. + - If both facilities in a pair are selected, the interaction benefit is fully counted. + +6. **Objective**: + - Maximize the total interaction benefit across all selected facility pairs. + +#### Example: +Suppose you select facilities at Node 0, Node 1, and Node 2. The total interaction benefit would be calculated as follows: +- Node 0 & Node 1: 16 +- Node 0 & Node 2: 20 +- Node 1 & Node 2: 24 +Total Benefit = 16 + 20 + 24 = **60** + +Your task is to determine the optimal selection of 3 facilities that maximizes the total interaction benefit, considering all the constraints and interaction rules provided.127You are tasked with optimizing the placement of facilities across a network of five locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall benefit derived from their interactions. Each location has the potential to host a facility, but due to budget constraints, you can only select **exactly 3 locations** to establish facilities. The goal is to maximize the total benefit generated by the interactions between the selected facilities. + +#### Key Details: +1. **Facility Selection**: You must choose exactly 3 out of the 5 locations to establish facilities. Each location can either be selected (1) or not selected (0). +2. **Interaction Benefits**: The benefit of interaction between any two facilities depends on the specific pair of locations. For example: + - If facilities are placed at Node 0 and Node 1, the interaction benefit is **16**. + - If facilities are placed at Node 0 and Node 2, the interaction benefit is **20**. + - Similarly, the interaction benefits for all other pairs are provided in the table below. + +3. **Interaction Rules**: + - An interaction benefit is only counted if **both facilities in the pair are selected**. For example, if Node 0 and Node 1 are both selected, the interaction benefit of 16 is added to the total. + - If only one facility in a pair is selected, no interaction benefit is counted for that pair. + +4. **Interaction Benefit Table**: + | Pair of Nodes | Interaction Benefit | + |---------------|---------------------| + | Node 0 & Node 1 | 16 | + | Node 0 & Node 2 | 20 | + | Node 0 & Node 3 | 18 | + | Node 0 & Node 4 | 22 | + | Node 1 & Node 0 | 25 | + | Node 1 & Node 2 | 24 | + | Node 1 & Node 3 | 17 | + | Node 1 & Node 4 | 16 | + | Node 2 & Node 0 | 19 | + | Node 2 & Node 1 | 22 | + | Node 2 & Node 3 | 18 | + | Node 2 & Node 4 | 23 | + | Node 3 & Node 0 | 21 | + | Node 3 & Node 1 | 21 | + | Node 3 & Node 2 | 25 | + | Node 3 & Node 4 | 17 | + | Node 4 & Node 0 | 20 | + | Node 4 & Node 1 | 19 | + | Node 4 & Node 2 | 22 | + | Node 4 & Node 3 | 16 | + +5. **Constraints**: + - **Facility Selection Constraint**: Exactly 3 facilities must be selected out of the 5 locations. + - **Interaction Constraints**: For each pair of locations, the interaction benefit is only counted if both facilities in the pair are selected. This is enforced by ensuring that: + - If either facility in a pair is not selected, the interaction benefit for that pair is not counted. + - If both facilities in a pair are selected, the interaction benefit is fully counted. + +6. **Objective**: + - Maximize the total interaction benefit across all selected facility pairs. + +#### Example: +Suppose you select facilities at Node 0, Node 1, and Node 2. The total interaction benefit would be calculated as follows: +- Node 0 & Node 1: 16 +- Node 0 & Node 2: 20 +- Node 1 & Node 2: 24 +Total Benefit = 16 + 20 + 24 = **60** + +Your task is to determine the optimal selection of 3 facilities that maximizes the total interaction benefit, considering all the constraints and interaction rules provided.127No differencesNo differencesok
47There are 5 jobs labeled Job 0 to Job 4, each requiring processing on 4 machines labeled Machine 0 to Machine 3. Each job has a specific processing time on each machine: Job 0 takes 5 hours on Machine 0, 9 hours on Machine 1, 7 hours on Machine 2, and 7 hours on Machine 3; Job 1 takes 10 hours on Machine 0, 8 hours on Machine 1, 8 hours on Machine 2, and 9 hours on Machine 3; Job 2 takes 6 hours on Machine 0, 8 hours on Machine 1, 9 hours on Machine 2, and 7 hours on Machine 3; Job 3 takes 7 hours on Machine 0, 7 hours on Machine 1, 8 hours on Machine 2, and 8 hours on Machine 3; Job 4 takes 9 hours on Machine 0, 5 hours on Machine 1, 5 hours on Machine 2, and 9 hours on Machine 3. Each job must be assigned to exactly one position in the schedule, and each position must be assigned exactly one job. A job cannot start on a machine until it has completed processing on the previous machine in the sequence. Additionally, a job in a later position in the schedule cannot start on a machine until the job in the previous position has completed processing on that machine. The objective is to minimize the total completion time, which includes the start time of the last job on the last machine (Machine 3) and the processing time of the last job on Machine 3. The start times for each job on each machine must be non-negative, and all constraints must be satisfied to avoid delays and resource conflicts.62There are 5 jobs labeled Job 0 to Job 4, each requiring processing on 4 machines labeled Machine 0 to Machine 3. Each job has a specific processing time on each machine: Job 0 takes 5 hours on Machine 0, 9 hours on Machine 1, 7 hours on Machine 2, and 7 hours on Machine 3; Job 1 takes 10 hours on Machine 0, 8 hours on Machine 1, 8 hours on Machine 2, and 9 hours on Machine 3; Job 2 takes 6 hours on Machine 0, 8 hours on Machine 1, 9 hours on Machine 2, and 7 hours on Machine 3; Job 3 takes 7 hours on Machine 0, 7 hours on Machine 1, 8 hours on Machine 2, and 8 hours on Machine 3; Job 4 takes 9 hours on Machine 0, 5 hours on Machine 1, 5 hours on Machine 2, and 9 hours on Machine 3. Each job must be assigned to exactly one position in the schedule, and each position must be assigned exactly one job. A job cannot start on a machine until it has completed processing on the previous machine in the sequence. Additionally, a job in a later position in the schedule cannot start on a machine until the job in the previous position has completed processing on that machine. The objective is to minimize the total completion time, which includes the start time of the last job on the last machine (Machine 3) and the processing time of the last job on Machine 3. The start times for each job on each machine must be non-negative, and all constraints must be satisfied to avoid delays and resource conflicts.62No differencesNo differencesok
48Minimize the total operational cost of a power generation system over six periods, consisting of Type 0 and Type 1 generators. The cost includes a fixed base cost of 80 units per generator per period, a power generation cost of 4 units per MW produced, and startup costs of 354 units for Type 0 generators and 347 units for Type 1 generators. The system must meet the following demand in each period: 201 MW in period 0, 191 MW in period 1, 182 MW in period 2, 190 MW in period 3, 194 MW in period 4, and 168 MW in period 5. Each Type 0 generator can produce between 34 MW and 127 MW, and no more than 4 Type 0 generators can operate in any period. Each Type 1 generator can produce between 35 MW and 129 MW, and no more than 1 Type 1 generator can operate in any period. The total maximum power output of all generators in each period must be at least 15% higher than the demand for that period, ensuring a reserve margin. For Type 0 generators, the number of operational generators in any period cannot exceed the number from the previous period plus the number started in the current period, and the same applies to Type 1 generators. The number of generators and startup decisions must be integers. The goal is to determine the optimal number of generators to operate, their power output levels, and startup decisions to minimize total costs while satisfying all constraints.5464Minimize the total operational cost of a power generation system over six periods, consisting of Type 0 and Type 1 generators. The cost includes a fixed base cost of 80 units per generator per period, a power generation cost of 4 units per MW produced, and startup costs of 354 units for Type 0 generators and 347 units for Type 1 generators. The system must meet the following demand in each period: 201 MW in period 0, 191 MW in period 1, 182 MW in period 2, 190 MW in period 3, 194 MW in period 4, and 168 MW in period 5. Each Type 0 generator can produce between 34 MW and 127 MW, and no more than 4 Type 0 generators can operate in any period. Each Type 1 generator can produce between 35 MW and 129 MW, and no more than 1 Type 1 generator can operate in any period. The total maximum power output of all generators in each period must be at least 15% higher than the demand for that period, ensuring a reserve margin. For Type 0 generators, the number of operational generators in any period cannot exceed the number from the previous period plus the number started in the current period, and the same applies to Type 1 generators. The number of generators and startup decisions must be integers. The goal is to determine the optimal number of generators to operate, their power output levels, and startup decisions to minimize total costs while satisfying all constraints.6165No differences5464 6165Incorrect ground truthincorrect ground truth
49You are tasked with managing the distribution of resources across a network of eight cities. Each city has a specific supply of resources and a demand that must be met. The goal is to minimize the total cost of transporting resources between cities while ensuring that all demands are satisfied and all constraints are respected. + +#### City Details: +- **City 0** has a net supply of 4 units. +- **City 1** has a net demand of 1 unit. +- **City 2** has a net demand of 2 units. +- **City 3** has a net supply of 1 unit. +- **City 4** has a net demand of 1 unit. +- **City 5** has a net demand of 1 unit. +- **City 6** and **City 7** are balanced, with no net supply or demand. + +#### Transportation Costs: +The cost of transporting resources between cities varies. Here are the costs for each possible route: +- From **City 0** to **City 1**: 3 units +- From **City 0** to **City 2**: 3 units +- From **City 0** to **City 3**: 3 units +- From **City 0** to **City 4**: 3 units +- From **City 0** to **City 5**: 1 unit +- From **City 0** to **City 6**: 3 units +- From **City 0** to **City 7**: 2 units +- From **City 1** to **City 0**: 3 units +- From **City 1** to **City 2**: 2 units +- From **City 1** to **City 3**: 2 units +- From **City 1** to **City 4**: 3 units +- From **City 1** to **City 5**: 2 units +- From **City 1** to **City 6**: 2 units +- From **City 1** to **City 7**: 1 unit +- From **City 2** to **City 0**: 1 unit +- From **City 2** to **City 1**: 1 unit +- From **City 2** to **City 3**: 3 units +- From **City 2** to **City 4**: 1 unit +- From **City 2** to **City 5**: 2 units +- From **City 2** to **City 6**: 2 units +- From **City 2** to **City 7**: 3 units +- From **City 3** to **City 0**: 1 unit +- From **City 3** to **City 1**: 3 units +- From **City 3** to **City 2**: 1 unit +- From **City 3** to **City 4**: 1 unit +- From **City 3** to **City 5**: 1 unit +- From **City 3** to **City 6**: 2 units +- From **City 3** to **City 7**: 3 units +- From **City 4** to **City 0**: 3 units +- From **City 4** to **City 1**: 2 units +- From **City 4** to **City 2**: 1 unit +- From **City 4** to **City 3**: 2 units +- From **City 4** to **City 5**: 3 units +- From **City 4** to **City 6**: 2 units +- From **City 4** to **City 7**: 3 units +- From **City 5** to **City 0**: 1 unit +- From **City 5** to **City 1**: 2 units +- From **City 5** to **City 2**: 3 units +- From **City 5** to **City 3**: 2 units +- From **City 5** to **City 4**: 1 unit +- From **City 5** to **City 6**: 3 units +- From **City 5** to **City 7**: 3 units +- From **City 6** to **City 0**: 3 units +- From **City 6** to **City 1**: 1 unit +- From **City 6** to **City 2**: 1 unit +- From **City 6** to **City 3**: 3 units +- From **City 6** to **City 4**: 2 units +- From **City 6** to **City 5**: 1 unit +- From **City 6** to **City 7**: 3 units +- From **City 7** to **City 0**: 3 units +- From **City 7** to **City 1**: 3 units +- From **City 7** to **City 2**: 1 unit +- From **City 7** to **City 3**: 2 units +- From **City 7** to **City 4**: 2 units +- From **City 7** to **City 5**: 1 unit +- From **City 7** to **City 6**: 3 units + +#### Constraints: +1. **Flow Balance**: The total resources entering a city must equal the total resources leaving that city, adjusted for its supply or demand. +2. **Link Capacity**: The amount of resources transported between any two cities cannot exceed the capacity of the link: + - **City 0** to **City 1**: 9 units + - **City 0** to **City 2**: 7 units + - **City 0** to **City 3**: 9 units + - **City 0** to **City 4**: 7 units + - **City 0** to **City 5**: 8 units + - **City 0** to **City 6**: 9 units + - **City 0** to **City 7**: 9 units + - **City 1** to **City 0**: 9 units + - **City 1** to **City 2**: 7 units + - **City 1** to **City 3**: 9 units + - **City 1** to **City 4**: 7 units + - **City 1** to **City 5**: 9 units + - **City 1** to **City 6**: 8 units + - **City 1** to **City 7**: 7 units + - **City 2** to **City 0**: 8 units + - **City 2** to **City 1**: 7 units + - **City 2** to **City 3**: 8 units + - **City 2** to **City 4**: 9 units + - **City 2** to **City 5**: 9 units + - **City 2** to **City 6**: 7 units + - **City 2** to **City 7**: 8 units + - **City 3** to **City 0**: 7 units + - **City 3** to **City 1**: 7 units + - **City 3** to **City 2**: 9 units + - **City 3** to **City 4**: 9 units + - **City 3** to **City 5**: 9 units + - **City 3** to **City 6**: 7 units + - **City 3** to **City 7**: 9 units + - **City 4** to **City 0**: 8 units + - **City 4** to **City 1**: 8 units + - **City 4** to **City 2**: 8 units + - **City 4** to **City 3**: 8 units + - **City 4** to **City 5**: 8 units + - **City 4** to **City 6**: 8 units + - **City 4** to **City 7**: 9 units + - **City 5** to **City 0**: 7 units + - **City 5** to **City 1**: 9 units + - **City 5** to **City 2**: 8 units + - **City 5** to **City 3**: 7 units + - **City 5** to **City 4**: 7 units + - **City 5** to **City 6**: 9 units + - **City 5** to **City 7**: 7 units + - **City 6** to **City 0**: 9 units + - **City 6** to **City 1**: 8 units + - **City 6** to **City 2**: 8 units + - **City 6** to **City 3**: 7 units + - **City 6** to **City 4**: 7 units + - **City 6** to **City 5**: 8 units + - **City 6** to **City 7**: 8 units + - **City 7** to **City 0**: 9 units + - **City 7** to **City 1**: 9 units + - **City 7** to **City 2**: 7 units + - **City 7** to **City 3**: 8 units + - **City 7** to **City 4**: 8 units + - **City 7** to **City 5**: 8 units + - **City 7** to **City 6**: 8 units + +3. **City Capacity**: The total resources entering a city cannot exceed its capacity: + - **City 0**: 19 units + - **City 1**: 17 units + - **City 2**: 15 units + - **City 3**: 16 units + - **City 4**: 14 units + - **City 5**: 15 units + - **City 6**: 14 units + - **City 7**: 13 units + +#### Objective: +Your objective is to determine the optimal amount of resources to transport between each pair of cities to minimize the total transportation cost while ensuring that all demands are met and all constraints are satisfied. + +This problem requires careful planning to balance the supply and demand across the network, respecting the capacity limits of both the transportation links and the cities themselves. The solution will provide the most cost-effective way to distribute resources across the cities.7You are tasked with managing the distribution of resources across a network of eight cities. Each city has a specific supply of resources and a demand that must be met. The goal is to minimize the total cost of transporting resources between cities while ensuring that all demands are satisfied and all constraints are respected. + +#### City Details: +- **City 0** has a net supply of 4 units. +- **City 1** has a net demand of 1 unit. +- **City 2** has a net demand of 2 units. +- **City 3** has a net supply of 1 unit. +- **City 4** has a net demand of 1 unit. +- **City 5** has a net demand of 1 unit. +- **City 6** and **City 7** are balanced, with no net supply or demand. + +#### Transportation Costs: +The cost of transporting resources between cities varies. Here are the costs for each possible route: +- From **City 0** to **City 1**: 3 units +- From **City 0** to **City 2**: 3 units +- From **City 0** to **City 3**: 3 units +- From **City 0** to **City 4**: 3 units +- From **City 0** to **City 5**: 1 unit +- From **City 0** to **City 6**: 3 units +- From **City 0** to **City 7**: 2 units +- From **City 1** to **City 0**: 3 units +- From **City 1** to **City 2**: 2 units +- From **City 1** to **City 3**: 2 units +- From **City 1** to **City 4**: 3 units +- From **City 1** to **City 5**: 2 units +- From **City 1** to **City 6**: 2 units +- From **City 1** to **City 7**: 1 unit +- From **City 2** to **City 0**: 1 unit +- From **City 2** to **City 1**: 1 unit +- From **City 2** to **City 3**: 3 units +- From **City 2** to **City 4**: 1 unit +- From **City 2** to **City 5**: 2 units +- From **City 2** to **City 6**: 2 units +- From **City 2** to **City 7**: 3 units +- From **City 3** to **City 0**: 1 unit +- From **City 3** to **City 1**: 3 units +- From **City 3** to **City 2**: 1 unit +- From **City 3** to **City 4**: 1 unit +- From **City 3** to **City 5**: 1 unit +- From **City 3** to **City 6**: 2 units +- From **City 3** to **City 7**: 3 units +- From **City 4** to **City 0**: 3 units +- From **City 4** to **City 1**: 2 units +- From **City 4** to **City 2**: 1 unit +- From **City 4** to **City 3**: 2 units +- From **City 4** to **City 5**: 3 units +- From **City 4** to **City 6**: 2 units +- From **City 4** to **City 7**: 3 units +- From **City 5** to **City 0**: 1 unit +- From **City 5** to **City 1**: 2 units +- From **City 5** to **City 2**: 3 units +- From **City 5** to **City 3**: 2 units +- From **City 5** to **City 4**: 1 unit +- From **City 5** to **City 6**: 3 units +- From **City 5** to **City 7**: 3 units +- From **City 6** to **City 0**: 3 units +- From **City 6** to **City 1**: 1 unit +- From **City 6** to **City 2**: 1 unit +- From **City 6** to **City 3**: 3 units +- From **City 6** to **City 4**: 2 units +- From **City 6** to **City 5**: 1 unit +- From **City 6** to **City 7**: 3 units +- From **City 7** to **City 0**: 3 units +- From **City 7** to **City 1**: 3 units +- From **City 7** to **City 2**: 1 unit +- From **City 7** to **City 3**: 2 units +- From **City 7** to **City 4**: 2 units +- From **City 7** to **City 5**: 1 unit +- From **City 7** to **City 6**: 3 units + +#### Constraints: +1. **Flow Balance**: The total resources entering a city must equal the total resources leaving that city, adjusted for its supply or demand. +2. **Link Capacity**: The amount of resources transported between any two cities cannot exceed the capacity of the link: + - **City 0** to **City 1**: 9 units + - **City 0** to **City 2**: 7 units + - **City 0** to **City 3**: 9 units + - **City 0** to **City 4**: 7 units + - **City 0** to **City 5**: 8 units + - **City 0** to **City 6**: 9 units + - **City 0** to **City 7**: 9 units + - **City 1** to **City 0**: 9 units + - **City 1** to **City 2**: 7 units + - **City 1** to **City 3**: 9 units + - **City 1** to **City 4**: 7 units + - **City 1** to **City 5**: 9 units + - **City 1** to **City 6**: 8 units + - **City 1** to **City 7**: 7 units + - **City 2** to **City 0**: 8 units + - **City 2** to **City 1**: 7 units + - **City 2** to **City 3**: 8 units + - **City 2** to **City 4**: 9 units + - **City 2** to **City 5**: 9 units + - **City 2** to **City 6**: 7 units + - **City 2** to **City 7**: 8 units + - **City 3** to **City 0**: 7 units + - **City 3** to **City 1**: 7 units + - **City 3** to **City 2**: 9 units + - **City 3** to **City 4**: 9 units + - **City 3** to **City 5**: 9 units + - **City 3** to **City 6**: 7 units + - **City 3** to **City 7**: 9 units + - **City 4** to **City 0**: 8 units + - **City 4** to **City 1**: 8 units + - **City 4** to **City 2**: 8 units + - **City 4** to **City 3**: 8 units + - **City 4** to **City 5**: 8 units + - **City 4** to **City 6**: 8 units + - **City 4** to **City 7**: 9 units + - **City 5** to **City 0**: 7 units + - **City 5** to **City 1**: 9 units + - **City 5** to **City 2**: 8 units + - **City 5** to **City 3**: 7 units + - **City 5** to **City 4**: 7 units + - **City 5** to **City 6**: 9 units + - **City 5** to **City 7**: 7 units + - **City 6** to **City 0**: 9 units + - **City 6** to **City 1**: 8 units + - **City 6** to **City 2**: 8 units + - **City 6** to **City 3**: 7 units + - **City 6** to **City 4**: 7 units + - **City 6** to **City 5**: 8 units + - **City 6** to **City 7**: 8 units + - **City 7** to **City 0**: 9 units + - **City 7** to **City 1**: 9 units + - **City 7** to **City 2**: 7 units + - **City 7** to **City 3**: 8 units + - **City 7** to **City 4**: 8 units + - **City 7** to **City 5**: 8 units + - **City 7** to **City 6**: 8 units + +3. **City Capacity**: The total resources entering a city cannot exceed its capacity: + - **City 0**: 19 units + - **City 1**: 17 units + - **City 2**: 15 units + - **City 3**: 16 units + - **City 4**: 14 units + - **City 5**: 15 units + - **City 6**: 14 units + - **City 7**: 13 units + +#### Objective: +Your objective is to determine the optimal amount of resources to transport between each pair of cities to minimize the total transportation cost while ensuring that all demands are met and all constraints are satisfied. + +This problem requires careful planning to balance the supply and demand across the network, respecting the capacity limits of both the transportation links and the cities themselves. The solution will provide the most cost-effective way to distribute resources across the cities.10No differences7 10Incorrect ground truthincorrect ground truth
50The problem involves determining the most cost-effective way to transport medical supplies across a network of 10 hospitals (node_0 to node_9) while minimizing total transportation costs. Each hospital is interconnected, and the cost of transporting supplies between any two hospitals varies. The decision variables are binary, representing whether a route between two hospitals is used (1) or not (0). The objective is to minimize the total cost of transporting supplies, calculated as the sum of the costs of all activated routes. + +The constraints ensure that the supply balance is maintained across all hospitals. Specifically, node_0 must have a net supply of +1, indicating it is the source of supplies, while node_9 must have a net supply of -1, indicating it is the destination. All intermediate hospitals (node_1 to node_8) must have a net supply of 0, meaning they neither gain nor lose supplies overall. The flow balance constraints ensure that the total flow into each hospital equals the total flow out of it, with the exception of node_0 and node_9, which have specific net supply requirements. All decision variables are binary, meaning routes are either fully used or not used at all. The goal is to find the set of routes that satisfies these constraints while minimizing the total transportation cost.20The problem involves determining the most cost-effective way to transport medical supplies across a network of 10 hospitals (node_0 to node_9) while minimizing total transportation costs. Each hospital is interconnected, and the cost of transporting supplies between any two hospitals varies. The decision variables are binary, representing whether a route between two hospitals is used (1) or not (0). The objective is to minimize the total cost of transporting supplies, calculated as the sum of the costs of all activated routes. + +The constraints ensure that the supply balance is maintained across all hospitals. Specifically, node_0 must have a net supply of +1, indicating it is the source of supplies, while node_9 must have a net supply of -1, indicating it is the destination. All intermediate hospitals (node_1 to node_8) must have a net supply of 0, meaning they neither gain nor lose supplies overall. The flow balance constraints ensure that the total flow into each hospital equals the total flow out of it, with the exception of node_0 and node_9, which have specific net supply requirements. All decision variables are binary, meaning routes are either fully used or not used at all. The goal is to find the set of routes that satisfies these constraints while minimizing the total transportation cost. + +Cost matrix c_(i,j) for i,j ∈ {0,…,9} + 0 1 2 3 4 5 6 7 8 9 + ------------------------------------------ +0 | 0 12 23 31 22 44 37 55 28 60 +1 | 12 0 9 17 35 27 31 48 19 58 +2 | 23 9 0 8 14 26 22 33 12 45 +3 | 31 17 8 0 11 16 29 38 21 40 +4 | 22 35 14 11 0 13 18 27 17 32 +5 | 44 27 26 16 13 0 7 21 29 26 +6 | 37 31 22 29 18 7 0 10 16 24 +7 | 55 48 33 38 27 21 10 0 14 18 +8 | 28 19 12 21 17 29 16 14 0 20 +9 | 60 58 45 40 32 26 24 18 20 048The problem involves determining the most cost-effective way to transport medical supplies across a network of 10 hospitals (node_0 to node_9) while minimizing total transportation costs. Each hospital is interconnected, and the cost of transporting supplies between any two hospitals varies. The decision variables are binary, representing whether a route between two hospitals is used (1) or not (0). The objective is to minimize the total cost of transporting supplies, calculated as the sum of the costs of all activated routes. + +The constraints ensure that the supply balance is maintained across all hospitals. Specifically, node_0 must have a net supply of +1, indicating it is the source of supplies, while node_9 must have a net supply of -1, indicating it is the destination. All intermediate hospitals (node_1 to node_8) must have a net supply of 0, meaning they neither gain nor lose supplies overall. The flow balance constraints ensure that the total flow into each hospital equals the total flow out of it, with the exception of node_0 and node_9, which have specific net supply requirements. All decision variables are binary, meaning routes are either fully used or not used at all. The goal is to find the set of routes that satisfies these constraints while minimizing the total transportation cost. + + Cost matrix c_(i,j) for i,j ∈ {0,…,9} + 0 1 2 3 4 5 6 7 8 9 + ------------------------------------------ + 0 | 0 12 23 31 22 44 37 55 28 60 + 1 | 12 0 9 17 35 27 31 48 19 58 + 2 | 23 9 0 8 14 26 22 33 12 45 + 3 | 31 17 8 0 11 16 29 38 21 40 + 4 | 22 35 14 11 0 13 18 27 17 32 + 5 | 44 27 26 16 13 0 7 21 29 26 + 6 | 37 31 22 29 18 7 0 10 16 24 + 7 | 55 48 33 38 27 21 10 0 14 18 + 8 | 28 19 12 21 17 29 16 14 0 20 + 9 | 60 58 45 40 32 26 24 18 20 020 48Missing data about costsmissing data
51There are 7 jobs, each with 2 operations that must be performed in sequence. Each operation has a processing time of 1 unit of time. The facility has multiple machines, and some operations from different jobs must be performed on the same machine. Operations sharing the same machine cannot overlap in time. The order of operations on shared machines is determined by binary decision variables, which can only take values of 0 or 1. The goal is to minimize the total completion time, known as the makespan, which is the latest completion time among all operations. + +Each operation in a job must start only after the previous operation in the same job is completed. Operations that share the same machine must be scheduled such that one starts only after the other is completed, as determined by the binary variables. The makespan must be at least as large as the completion time of every operation. All operation start times must be non-negative. The binary variables enforce the order of operations on shared machines, ensuring no overlap. The processing time for all operations is 1 unit of time. The makespan is determined by the latest completion time among all operations, and the objective is to minimize this makespan.3There are 7 jobs, each with 2 operations that must be performed in sequence. Each operation has a processing time of 1 unit of time. The facility has multiple machines, and some operations from different jobs must be performed on the same machine. Operations sharing the same machine cannot overlap in time. The order of operations on shared machines is determined by binary decision variables, which can only take values of 0 or 1. The goal is to minimize the total completion time, known as the makespan, which is the latest completion time among all operations. + +Each operation in a job must start only after the previous operation in the same job is completed. Operations that share the same machine must be scheduled such that one starts only after the other is completed, as determined by the binary variables. The makespan must be at least as large as the completion time of every operation. All operation start times must be non-negative. The binary variables enforce the order of operations on shared machines, ensuring no overlap. The processing time for all operations is 1 unit of time. The makespan is determined by the latest completion time among all operations, and the objective is to minimize this makespan. + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 1 + - Operation 1 of Job 0: Machine 0 + - Operation 0 of Job 1: Machine 2 + - Operation 1 of Job 1: Machine 0 + - Operation 0 of Job 2: Machine 1 + - Operation 1 of Job 2: Machine 0 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 1 + - Operation 0 of Job 5: Machine 1 + - Operation 1 of Job 5: Machine 0 + - Operation 0 of Job 6: Machine 0 + - Operation 1 of Job 6: Machine 25There are 7 jobs, each with 2 operations that must be performed in sequence. Each operation has a processing time of 1 unit of time. The facility has multiple machines, and some operations from different jobs must be performed on the same machine. Operations sharing the same machine cannot overlap in time. The order of operations on shared machines is determined by binary decision variables, which can only take values of 0 or 1. The goal is to minimize the total completion time, known as the makespan, which is the latest completion time among all operations. + +Each operation in a job must start only after the previous operation in the same job is completed. Operations that share the same machine must be scheduled such that one starts only after the other is completed, as determined by the binary variables. The makespan must be at least as large as the completion time of every operation. All operation start times must be non-negative. The binary variables enforce the order of operations on shared machines, ensuring no overlap. The processing time for all operations is 1 unit of time. The makespan is determined by the latest completion time among all operations, and the objective is to minimize this makespan. + + + Additional information: + - There are 3 machines, numbered from 0 to 2 + - The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 1 + - Operation 1 of Job 0: Machine 0 + - Operation 0 of Job 1: Machine 2 + - Operation 1 of Job 1: Machine 0 + - Operation 0 of Job 2: Machine 1 + - Operation 1 of Job 2: Machine 0 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 1 + - Operation 0 of Job 5: Machine 1 + - Operation 1 of Job 5: Machine 0 + - Operation 0 of Job 6: Machine 0 + - Operation 1 of Job 6: Machine 23 5It was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
52Minimize the total penalty costs associated with landing five aircraft either earlier or later than their target times, while ensuring safe separation between consecutive landings. Each aircraft has a target landing time, earliest and latest possible landing times, and penalties for landing early or late. The penalties are \$15, \$10, \$17, \$15, and \$15 per minute early, and \$11, \$16, \$17, \$15, and \$12 per minute late for Aircraft 0, 1, 2, 3, and 4, respectively. The target landing times are 71, 95, 73, 80, and 78 minutes, with earliest landing times of 60, 65, 60, 60, and 60 minutes, and latest landing times of 100 minutes for all aircraft. + +To ensure safety, specific separation times must be maintained between consecutive landings. If Aircraft 0 lands before Aircraft 1, there must be at least 7 minutes between their landings. If Aircraft 0 lands before Aircraft 2, there must be at least 5 minutes between their landings. If Aircraft 0 lands before Aircraft 3, there must be at least 5 minutes between their landings. If Aircraft 0 lands before Aircraft 4, there must be at least 6 minutes between their landings. If Aircraft 1 lands before Aircraft 2, there must be at least 5 minutes between their landings. If Aircraft 1 lands before Aircraft 3, there must be at least 8 minutes between their landings. If Aircraft 1 lands before Aircraft 4, there must be at least 6 minutes between their landings. If Aircraft 2 lands before Aircraft 3, there must be at least 7 minutes between their landings. If Aircraft 2 lands before Aircraft 4, there must be at least 7 minutes between their landings. If Aircraft 3 lands before Aircraft 4, there must be at least 7 minutes between their landings. + +The landing time for each aircraft must be between its earliest and latest possible landing times. The separation times between consecutive landings must be enforced based on the order in which the aircraft land. The penalties for landing early or late are applied based on the deviation from the target landing times. The objective is to determine the actual landing times for all five aircraft such that the total penalty cost is minimized, each aircraft lands within its specified earliest and latest landing times, and the required separation times between consecutive landings are strictly maintained. Binary decision variables are used to determine the order of landings between aircraft pairs, ensuring proper separation based on the landing order. Constraints enforce the relationship between landing times and the binary variables, ensuring proper separation based on the landing order. The binary variables are restricted to 0 or 1.153Minimize the total penalty costs associated with landing five aircraft either earlier or later than their target times, while ensuring safe separation between consecutive landings. Each aircraft has a target landing time, earliest and latest possible landing times, and penalties for landing early or late. The penalties are \$15, \$10, \$17, \$15, and \$15 per minute early, and \$11, \$16, \$17, \$15, and \$12 per minute late for Aircraft 0, 1, 2, 3, and 4, respectively. The target landing times are 71, 95, 73, 80, and 78 minutes, with earliest landing times of 60, 65, 60, 60, and 60 minutes, and latest landing times of 100 minutes for all aircraft. + +To ensure safety, specific separation times must be maintained between consecutive landings: there must be at least 7 minutes between Aircraft 0 and Aircraft 1, 5 minutes between Aircraft 0 and Aircraft 2, 5 minutes between Aircraft 0 and Aircraft 3, 6 minutes between Aircraft 0 and Aircraft 4, 5 minutes between Aircraft 1 and Aircraft 2, 8 minutes between Aircraft 1 and Aircraft 3, 6 minutes between Aircraft 1 and Aircraft 4, 7 minutes between Aircraft 2 and Aircraft 3, 7 minutes between Aircraft 2 and Aircraft 4, and 7 minutes between Aircraft 3 and Aircraft 4. + +The landing time for each aircraft must be between its earliest and latest possible landing times. The separation times between consecutive landings must be enforced based on the order in which the aircraft land. The penalties for landing early or late are applied based on the deviation from the target landing times. The objective is to determine the actual landing times for all five aircraft such that the total penalty cost is minimized, each aircraft lands within its specified earliest and latest landing times, and the required separation times between consecutive landings are strictly maintained. Binary decision variables are used to determine the order of landings between aircraft pairs, ensuring proper separation based on the landing order. Constraints enforce the relationship between landing times and the binary variables, ensuring proper separation based on the landing order. The binary variables are restricted to 0 or 1.153Minimize the total penalty costs associated with landing five aircraft either earlier or later than their target times, while ensuring safe separation between consecutive landings. Each aircraft has a target landing time, earliest and latest possible landing times, and penalties for landing early or late. The penalties are \$15, \$10, \$17, \$15, and \$15 per minute early, and \$11, \$16, \$17, \$15, and \$12 per minute late for Aircraft 0, 1, 2, 3, and 4, respectively. The target landing times are 71, 95, 73, 80, and 78 minutes, with earliest landing times of 60, 65, 60, 60, and 60 minutes, and latest landing times of 100 minutes for all aircraft. + +To ensure safety, specific separation times must be maintained between consecutive landings. If Aircraft 0 lands before Aircraft 1, landings: there must be at least 7 minutes between their landings. If Aircraft 0 lands before and Aircraft 1, 5 minutes between Aircraft 0 and Aircraft 2, there must be at least 5 minutes between their landings. If Aircraft 0 lands before and Aircraft 3, there must be at least 5 6 minutes between their landings. If Aircraft 0 lands before and Aircraft 4, there must be at least 6 5 minutes between their landings. If Aircraft 1 lands before and Aircraft 2, there must be at least 5 8 minutes between their landings. If Aircraft 1 lands before and Aircraft 3, there must be at least 8 6 minutes between their landings. If Aircraft 1 lands before and Aircraft 4, there must be at least 6 7 minutes between their landings. If Aircraft 2 lands before and Aircraft 3, there must be at least 7 minutes between their landings. If Aircraft 2 lands before and Aircraft 4, there must be at least and 7 minutes between their landings. If Aircraft 3 lands before and Aircraft 4, there must be at least 7 minutes between their landings. 4. + +The landing time for each aircraft must be between its earliest and latest possible landing times. The separation times between consecutive landings must be enforced based on the order in which the aircraft land. The penalties for landing early or late are applied based on the deviation from the target landing times. The objective is to determine the actual landing times for all five aircraft such that the total penalty cost is minimized, each aircraft lands within its specified earliest and latest landing times, and the required separation times between consecutive landings are strictly maintained. Binary decision variables are used to determine the order of landings between aircraft pairs, ensuring proper separation based on the landing order. Constraints enforce the relationship between landing times and the binary variables, ensuring proper separation based on the landing order. The binary variables are restricted to 0 or 1.No differencesThe separation times were stated only for one direction per aircraft pair (e.g., if Aircraft 0 lands before Aircraft 1, but not for the opposite), we adjusted the description to refer to both directions.missing data
53You are managing a manufacturing facility that produces three types of products (Product 0, Product 1, and Product 2) using two machines (Machine 0 and Machine 1) over three time periods (Period 0, Period 1, and Period 2). Your goal is to minimize the total operational costs, which include production setup costs, machine startup costs, inventory holding costs, and backlogging costs, while meeting customer demand for each product in each period. + +For each product, machine, and time period, decide whether to produce the product and how much to produce. For each machine and time period, decide whether to start up the machine to begin production. For each product and time period, decide how much inventory to carry over to the next period or how much demand to backlog if production falls short. + +Minimize the total cost, which includes production setup costs of 145.4242073380383 per production decision, machine startup costs of 79.18604813454679 per startup decision, inventory holding costs of 0.9008900951620358 for Product 0, 1.068925395295676 for Product 1, and 1.066244796883431 for Product 2 per unit per period, and backlogging costs of 6.734178682720012 for Product 0, 7.442225721133133 for Product 1, and 7.104627209109436 for Product 2 per unit per period. + +For each product and time period, the total production across machines, plus any inventory carried over from the previous period, minus any backlog from the previous period, must equal the demand for that period plus the current inventory minus the current backlog. The demands for each product and period are as follows: Product 0: Period 0 = 20.33410585875594, Period 1 = 22.57961652322203, Period 2 = 15.89714240515548; Product 1: Period 0 = 24.05647618194615, Period 1 = 20.36522893983323, Period 2 = 17.47466293074195; Product 2: Period 0 = 16.36553930950429, Period 1 = 22.7831030020558, Period 2 = 19.7805033393852. + +For each machine and time period, the production quantity multiplied by -71.79935969683524 for Machine 0 or -73.09691843304661 for Machine 1, plus the startup time multiplied by 6.342821301312698 for Machine 0 or 6.171760083826499 for Machine 1, plus the amount produced must be less than or equal to zero. Each machine can only produce one product at a time in any given period, meaning the sum of production decisions across all products for a given machine and period must be less than or equal to 1. + +If a machine is producing a product in a period but was not producing it in the previous period, a startup decision must be made. Production quantities, inventory levels, and backlogs must be non-negative. Production and startup decisions are binary (either 0 or 1).919.3047225564642You are managing a manufacturing facility that produces three types of products (Product 0, Product 1, and Product 2) using two machines (Machine 0 and Machine 1) over three time periods (Period 0, Period 1, and Period 2). Your goal is to minimize the total operational costs, which include production setup costs, machine startup costs, inventory holding costs, and backlogging costs, while meeting customer demand for each product in each period. + +For each product, machine, and time period, decide whether to produce the product and how much to produce. For each machine and time period, decide whether to start up the machine to begin production. For each product and time period, decide how much inventory to carry over to the next period or how much demand to backlog if production falls short. + +Minimize the total cost, which includes production setup costs of 145.4242073380383 per production decision, machine startup costs of 79.18604813454679 per startup decision, inventory holding costs of 0.9008900951620358 for Product 0, 1.068925395295676 for Product 1, and 1.066244796883431 for Product 2 per unit per period, and backlogging costs of 6.734178682720012 for Product 0, 7.442225721133133 for Product 1, and 7.104627209109436 for Product 2 per unit per period. + +For each product and time period, the total production across machines, plus any inventory carried over from the previous period, minus any backlog from the previous period, must equal the demand for that period plus the current inventory minus the current backlog. The demands for each product and period are as follows: Product 0: Period 0 = 20.33410585875594, Period 1 = 22.57961652322203, Period 2 = 15.89714240515548; Product 1: Period 0 = 24.05647618194615, Period 1 = 20.36522893983323, Period 2 = 17.47466293074195; Product 2: Period 0 = 16.36553930950429, Period 1 = 22.7831030020558, Period 2 = 19.7805033393852. + +For each machine and time period, the production quantity multiplied by -71.79935969683524 for Machine 0 or -73.09691843304661 for Machine 1, plus the startup time multiplied by 6.342821301312698 for Machine 0 or 6.171760083826499 for Machine 1, plus the amount produced must be less than or equal to zero. Each machine can only produce one product at a time in any given period, meaning the sum of production decisions across all products for a given machine and period must be less than or equal to 1. + +If a machine is producing a product in a period but was not producing it in the previous period, a startup decision must be made. Production quantities, inventory levels, and backlogs must be non-negative. Production and startup decisions are binary (either 0 or 1).919.3047225564642No differencesNo differencesok
54There are 5 jobs, labeled Job 0 to Job 4, each consisting of 2 operations. Each operation must be performed on a specific machine, and the order of operations within a job is fixed. For example, Operation 1 of Job 0 must be completed before Operation 2 of Job 0 can begin. The processing time for each operation is 1 unit of time. The facility has a limited number of machines, and some operations from different jobs must be performed on the same machine. To avoid conflicts, no two operations can be scheduled on the same machine at the same time. This requires determining the order in which operations are processed on each machine. For example, Operations from Job 1 and Job 2 both require Machine 1, and you must decide whether to process Job 1's operation before Job 2's, or vice versa. Similarly, Operations from Job 4 and Job 0 must be sequenced on Machine 2. Binary variables are used to indicate whether one operation is scheduled before another on the same machine, ensuring that the scheduling constraints are satisfied. The primary objective is to minimize the makespan, which is the total time required to complete all jobs, determined by the latest completion time among all operations. For example, if the last operation finishes at time 10, the makespan is 10. Constraints include precedence constraints, ensuring each operation within a job starts only after the previous operation in the same job has finished; machine capacity constraints, ensuring operations sharing the same machine do not overlap in time; makespan definition, ensuring the makespan is at least as large as the completion time of every operation; non-negativity constraints, ensuring all operation start times are non-negative; and binary variables, determining the order of operations on shared machines. The processing time for each operation is 1 unit of time, and binary variables are used with a large constant (100,000) to enforce sequencing constraints. For example, if a binary variable is 1, then one operation must finish before another starts on the same machine. The task is to determine the start times for all operations across all jobs, ensuring operations within the same job are performed in the correct sequence, operations sharing the same machine do not overlap, and the total completion time is minimized.4There are 5 jobs, labeled Job 0 to Job 4, each consisting of 2 operations. Each operation must be performed on a specific machine, and the order of operations within a job is fixed. For example, Operation 1 of Job 0 must be completed before Operation 2 of Job 0 can begin. The processing time for each operation is 1 unit of time. The facility has a limited number of machines, and some operations from different jobs must be performed on the same machine. To avoid conflicts, no two operations can be scheduled on the same machine at the same time. This requires determining the order in which operations are processed on each machine. For example, Operations from Job 1 and Job 2 both require Machine 1, and you must decide whether to process Job 1's operation before Job 2's, or vice versa. Similarly, Operations from Job 4 and Job 0 must be sequenced on Machine 2. Binary variables are used to indicate whether one operation is scheduled before another on the same machine, ensuring that the scheduling constraints are satisfied. The primary objective is to minimize the makespan, which is the total time required to complete all jobs, determined by the latest completion time among all operations. For example, if the last operation finishes at time 10, the makespan is 10. Constraints include precedence constraints, ensuring each operation within a job starts only after the previous operation in the same job has finished; machine capacity constraints, ensuring operations sharing the same machine do not overlap in time; makespan definition, ensuring the makespan is at least as large as the completion time of every operation; non-negativity constraints, ensuring all operation start times are non-negative; and binary variables, determining the order of operations on shared machines. The processing time for each operation is 1 unit of time, and binary variables are used with a large constant (1000) to enforce sequencing constraints. For example, if a binary variable is 1, then one operation must finish before another starts on the same machine. The task is to determine the start times for all operations across all jobs, ensuring operations within the same job are performed in the correct sequence, operations sharing the same machine do not overlap, and the total completion time is minimized. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3, 4} +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Machines: M = {1, 2, 3} + +2) Processing times (in time units) +- All operations have processing time p_{j,k} = 1 for every job j and operation k. + +3) Machine assignment for every operation (explicit mapping) +- Operation (0,1) -> Machine 2 +- Operation (0,2) -> Machine 2 +- Operation (1,1) -> Machine 1 +- Operation (1,2) -> Machine 3 +- Operation (2,1) -> Machine 1 +- Operation (2,2) -> Machine 3 +- Operation (3,1) -> Machine 3 +- Operation (3,2) -> Machine 1 +- Operation (4,1) -> Machine 2 +- Operation (4,2) -> Machine 14There are 5 jobs, labeled Job 0 to Job 4, each consisting of 2 operations. Each operation must be performed on a specific machine, and the order of operations within a job is fixed. For example, Operation 1 of Job 0 must be completed before Operation 2 of Job 0 can begin. The processing time for each operation is 1 unit of time. The facility has a limited number of machines, and some operations from different jobs must be performed on the same machine. To avoid conflicts, no two operations can be scheduled on the same machine at the same time. This requires determining the order in which operations are processed on each machine. For example, Operations from Job 1 and Job 2 both require Machine 1, and you must decide whether to process Job 1's operation before Job 2's, or vice versa. Similarly, Operations from Job 4 and Job 0 must be sequenced on Machine 2. Binary variables are used to indicate whether one operation is scheduled before another on the same machine, ensuring that the scheduling constraints are satisfied. The primary objective is to minimize the makespan, which is the total time required to complete all jobs, determined by the latest completion time among all operations. For example, if the last operation finishes at time 10, the makespan is 10. Constraints include precedence constraints, ensuring each operation within a job starts only after the previous operation in the same job has finished; machine capacity constraints, ensuring operations sharing the same machine do not overlap in time; makespan definition, ensuring the makespan is at least as large as the completion time of every operation; non-negativity constraints, ensuring all operation start times are non-negative; and binary variables, determining the order of operations on shared machines. The processing time for each operation is 1 unit of time, and binary variables are used with a large constant (100,000) (1000) to enforce sequencing constraints. For example, if a binary variable is 1, then one operation must finish before another starts on the same machine. The task is to determine the start times for all operations across all jobs, ensuring operations within the same job are performed in the correct sequence, operations sharing the same machine do not overlap, and the total completion time is minimized. + + ADDITIONAL INPUT DATA: + + 1) Sets and indices + - Jobs: J = {0, 1, 2, 3, 4} + - For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). + - Machines: M = {1, 2, 3} + + 2) Processing times (in time units) + - All operations have processing time p_{j,k} = 1 for every job j and operation k. + + 3) Machine assignment for every operation (explicit mapping) + - Operation (0,1) -> Machine 2 + - Operation (0,2) -> Machine 2 + - Operation (1,1) -> Machine 1 + - Operation (1,2) -> Machine 3 + - Operation (2,1) -> Machine 1 + - Operation (2,2) -> Machine 3 + - Operation (3,1) -> Machine 3 + - Operation (3,2) -> Machine 1 + - Operation (4,1) -> Machine 2 + - Operation (4,2) -> Machine 1No differencesIt was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
55Cell Tower Coverage Planning + +You are tasked with planning the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 35 distinct areas, each with a specific population size that needs to be served. Your goal is to determine which cell towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + +#### Key Decisions: +1. **Build or Not Build**: For each of the 26 potential cell tower locations, you must decide whether to build the tower (a binary decision: build or not build). +2. **Coverage Assignment**: For each of the 35 areas, you must determine whether it will be covered by the deployed towers (a binary decision: covered or not covered). + +#### Objective: +Maximize the total population covered by the deployed cell towers. The population sizes for the 35 areas are as follows: +- Area 0: 3,215 people +- Area 1: 2,395 people +- Area 2: 4,466 people +- Area 3: 2,520 people +- Area 4: 2,192 people +- Area 5: 4,625 people +- Area 6: 4,796 people +- Area 7: 2,973 people +- Area 8: 2,034 people +- Area 9: 2,754 people +- Area 10: 2,143 people +- Area 11: 2,473 people +- Area 12: 3,099 people +- Area 13: 4,136 people +- Area 14: 2,154 people +- Area 15: 3,411 people +- Area 16: 3,230 people +- Area 17: 2,155 people +- Area 18: 4,907 people +- Area 19: 2,665 people +- Area 20: 4,708 people +- Area 21: 4,494 people +- Area 22: 2,381 people +- Area 23: 3,077 people +- Area 24: 2,699 people +- Area 25: 3,320 people +- Area 26: 4,931 people +- Area 27: 4,645 people +- Area 28: 2,346 people +- Area 29: 4,393 people +- Area 30: 4,051 people +- Area 31: 4,507 people +- Area 32: 4,610 people +- Area 33: 3,412 people +- Area 34: 3,059 people + +#### Constraints: +1. **Coverage Requirements**: + - Each area can only be considered covered if at least one of the cell towers assigned to it is built. For example: + - Area 0 can be covered if any of the towers 1, 5, 10, 11, 13, 15, 18, 19, 20, 22, 23, or 24 are built. + - Area 1 can be covered if any of the towers 3, 4, 6, 11, 12, 14, 15, 17, 18, 19, 21, 23, or 24 are built. + - Similar constraints apply to all 35 areas, with specific combinations of towers required for coverage. + +2. **Budget Constraint**: + - The total cost of building the selected towers must not exceed the available budget of 2,894 units. The costs for building each tower are as follows: + - Tower 0: 411 units + - Tower 1: 294 units + - Tower 2: 376 units + - Tower 3: 447 units + - Tower 4: 414 units + - Tower 5: 498 units + - Tower 6: 437 units + - Tower 7: 272 units + - Tower 8: 381 units + - Tower 9: 323 units + - Tower 10: 356 units + - Tower 11: 200 units + - Tower 12: 297 units + - Tower 13: 428 units + - Tower 14: 427 units + - Tower 15: 484 units + - Tower 16: 494 units + - Tower 17: 436 units + - Tower 18: 218 units + - Tower 19: 428 units + - Tower 20: 399 units + - Tower 21: 484 units + - Tower 22: 367 units + - Tower 23: 204 units + - Tower 24: 289 units + - Tower 25: 283 units + +3. **Binary Decisions**: + - Each tower can either be built (1) or not built (0). + - Each area can either be covered (1) or not covered (0). + +#### Summary: +Your task is to select which towers to build and which areas to cover, ensuring that the total population covered is maximized while staying within the budget. The coverage of each area depends on the specific combination of towers built, and the total cost of building the selected towers must not exceed 2,894 units.118976Cell Tower Coverage Planning + +You are tasked with planning the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 35 distinct areas, each with a specific population size that needs to be served. Your goal is to determine which cell towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + +#### Key Decisions: +1. **Build or Not Build**: For each of the 26 potential cell tower locations, you must decide whether to build the tower (a binary decision: build or not build). +2. **Coverage Assignment**: For each of the 35 areas, you must determine whether it will be covered by the deployed towers (a binary decision: covered or not covered). + +#### Objective: +Maximize the total population covered by the deployed cell towers. The population sizes for the 35 areas are as follows: +- Area 0: 3,215 people +- Area 1: 2,395 people +- Area 2: 4,466 people +- Area 3: 2,520 people +- Area 4: 2,192 people +- Area 5: 4,625 people +- Area 6: 4,796 people +- Area 7: 2,973 people +- Area 8: 2,034 people +- Area 9: 2,754 people +- Area 10: 2,143 people +- Area 11: 2,473 people +- Area 12: 3,099 people +- Area 13: 4,136 people +- Area 14: 2,154 people +- Area 15: 3,411 people +- Area 16: 3,230 people +- Area 17: 2,155 people +- Area 18: 4,907 people +- Area 19: 2,665 people +- Area 20: 4,708 people +- Area 21: 4,494 people +- Area 22: 2,381 people +- Area 23: 3,077 people +- Area 24: 2,699 people +- Area 25: 3,320 people +- Area 26: 4,931 people +- Area 27: 4,645 people +- Area 28: 2,346 people +- Area 29: 4,393 people +- Area 30: 4,051 people +- Area 31: 4,507 people +- Area 32: 4,610 people +- Area 33: 3,412 people +- Area 34: 3,059 people + +#### Constraints: +1. **Coverage Requirements**: + - Each area can only be considered covered if at least one of the cell towers assigned to it is built. For example: + - Area 0 can be covered if any of the towers 1, 5, 10, 11, 13, 15, 18, 19, 20, 22, 23, or 24 are built. + - Area 1 can be covered if any of the towers 3, 4, 6, 11, 12, 14, 15, 17, 18, 19, 21, 23, or 24 are built. + - Area 2 can be covered if any of the towers 23, 14, 0, 10, 15, 6, 21, 11, 20, 25, 9, or 17 are built. + - Area 3 can be covered if any of the towers 17, 11, 19, 2, 7, 18, 6, 16, 8, 1, 0, or 23 are built. + - Area 4 can be covered if any of the towers 12, 11, 19, 9, 6, 23, 3, 24, 15, 8, 22, or 20 are built. + - Area 5 can be covered if any of the towers 24, 5, 13, 17, 22, 21, 15, 19, 1, 4, 0, or 6 are built. + - Area 6 can be covered if any of the towers 3, 5, 19, 24, 0, 6, 20, 15, 18, 17, 25, or 2 are built. + - Area 7 can be covered if any of the towers 1, 16, 5, 7, 20, 11, 12, 19, 8, 14, 21, or 15 are built. + - Area 8 can be covered if any of the towers 7, 20, 15, 14, 4, 10, 1, 0, 22, 3, 11, or 9 are built. + - Area 9 can be covered if any of the towers 19, 5, 8, 13, 0, 6, 4, 22, 20, 12, 24, or 2 are built. + - Area 10 can be covered if any of the towers 2, 1, 3, 24, 6, 10, 14, 17, 11, 7, 23, or 8 are built. + - Area 11 can be covered if any of the towers 24, 6, 15, 18, 16, 9, 11, 17, 0, 2, 10, or 12 are built. + - Area 12 can be covered if any of the towers 18, 11, 1, 10, 8, 16, 9, 13, 19, 7, 20, or 6 are built. + - Area 13 can be covered if any of the towers 7, 21, 6, 10, 2, 16, 9, 4, 15, 0, 17, or 18 are built. + - Area 14 can be covered if any of the towers 11, 13, 16, 10, 1, 24, 4, 8, 0, 25, 7, or 19 are built. + - Area 15 can be covered if any of the towers 7, 11, 1, 24, 15, 14, 21, 12, 9, 0, 16, or 2 are built. + - Area 16 can be covered if any of the towers 7, 10, 9, 0, 13, 15, 12, 2, 1, 25, 20, or 6 are built. + - Area 17 can be covered if any of the towers 24, 8, 21, 2, 0, 18, 25, 4, 23, 7, 10, or 5 are built. + - Area 18 can be covered if any of the towers 11, 19, 14, 0, 18, 24, 5, 23, 2, 10, 7, or 8 are built. + - Area 19 can be covered if any of the towers 4, 20, 25, 17, 14, 12, 8, 7, 10, 22, 21, or 23 are built. + - Area 20 can be covered if any of the towers 21, 19, 25, 20, 4, 6, 10, 3, 2, 14, 8, or 15 are built. + - Area 21 can be covered if any of the towers 16, 25, 4, 11, 3, 18, 5, 19, 9, 2, 22, or 1 are built. + - Area 22 can be covered if any of the towers 23, 18, 20, 12, 21, 17, 10, 25, 2, 9, 4, or 6 are built. + - Area 23 can be covered if any of the towers 14, 25, 20, 18, 11, 23, 7, 13, 2, 4, 19, or 12 are built. + - Area 24 can be covered if any of the towers 0, 5, 16, 24, 7, 23, 14, 6, 25, 10, 19, or 22 are built. + - Area 25 can be covered if any of the towers 5, 20, 3, 23, 10, 2, 6, 19, 13, 25, 9, or 15 are built. + - Area 26 can be covered if any of the towers 18, 2, 5, 22, 3, 12, 6, 15, 9, 10, 21, or 1 are built. + - Area 27 can be covered if any of the towers 13, 10, 23, 7, 9, 3, 18, 17, 21, 25, 2, or 14 are built. + - Area 28 can be covered if any of the towers 11, 17, 0, 10, 22, 21, 2, 3, 14, 25, 9, or 6 are built. + - Area 29 can be covered if any of the towers 1, 16, 4, 14, 3, 19, 2, 18, 9, 7, 0, or 13 are built. + - Area 30 can be covered if any of the towers 2, 6, 14, 25, 15, 21, 13, 22, 23, 12, 11, or 9 are built. + - Area 31 can be covered if any of the towers 8, 6, 11, 23, 16, 1, 13, 18, 14, 2, 20, or 19 are built. + - Area 32 can be covered if any of the towers 19, 25, 16, 24, 1, 4, 22, 11, 13, 15, 21, or 12 are built. + - Area 33 can be covered if any of the towers 11, 2, 9, 3, 4, 21, 25, 5, 12, 6, 19, or 7 are built. + - Area 34 can be covered if any of the towers 10, 19, 0, 6, 18, 21, 12, 25, 14, 13, 11, or 20 are built. + +1. **Budget Constraint**: + - The total cost of building the selected towers must not exceed the available budget of 894 units. The costs for building each tower are as follows: + - Tower 0: 411 units + - Tower 1: 294 units + - Tower 2: 376 units + - Tower 3: 447 units + - Tower 4: 414 units + - Tower 5: 498 units + - Tower 6: 437 units + - Tower 7: 272 units + - Tower 8: 381 units + - Tower 9: 323 units + - Tower 10: 356 units + - Tower 11: 200 units + - Tower 12: 297 units + - Tower 13: 428 units + - Tower 14: 427 units + - Tower 15: 484 units + - Tower 16: 494 units + - Tower 17: 436 units + - Tower 18: 218 units + - Tower 19: 428 units + - Tower 20: 399 units + - Tower 21: 484 units + - Tower 22: 367 units + - Tower 23: 204 units + - Tower 24: 289 units + - Tower 25: 283 units + +3. **Binary Decisions**: + - Each tower can either be built (1) or not built (0). + - Each area can either be covered (1) or not covered (0). + +#### Summary: +Your task is to select which towers to build and which areas to cover, ensuring that the total population covered is maximized while staying within the budget. The coverage of each area depends on the specific combination of towers built, and the total cost of building the selected towers must not exceed 894 units.116311- Each area can only be considered covered if at least one of the cell towers assigned to it is built. For example: + - Area 0 can be covered if any of the towers 1, 5, 10, 11, 13, 15, 18, 19, 20, 22, 23, or 24 are built. + - Area 1 can be covered if any of the towers 3, 4, 6, 11, 12, 14, 15, 17, 18, 19, 21, 23, or 24 are built. + - Similar constraints apply to all 35 areas, with specific combinations of towers required for coverage. + + 2. Area 2 can be covered if any of the towers 23, 14, 0, 10, 15, 6, 21, 11, 20, 25, 9, or 17 are built. + - Area 3 can be covered if any of the towers 17, 11, 19, 2, 7, 18, 6, 16, 8, 1, 0, or 23 are built. + - Area 4 can be covered if any of the towers 12, 11, 19, 9, 6, 23, 3, 24, 15, 8, 22, or 20 are built. + - Area 5 can be covered if any of the towers 24, 5, 13, 17, 22, 21, 15, 19, 1, 4, 0, or 6 are built. + - Area 6 can be covered if any of the towers 3, 5, 19, 24, 0, 6, 20, 15, 18, 17, 25, or 2 are built. + - Area 7 can be covered if any of the towers 1, 16, 5, 7, 20, 11, 12, 19, 8, 14, 21, or 15 are built. + - Area 8 can be covered if any of the towers 7, 20, 15, 14, 4, 10, 1, 0, 22, 3, 11, or 9 are built. + - Area 9 can be covered if any of the towers 19, 5, 8, 13, 0, 6, 4, 22, 20, 12, 24, or 2 are built. + - Area 10 can be covered if any of the towers 2, 1, 3, 24, 6, 10, 14, 17, 11, 7, 23, or 8 are built. + - Area 11 can be covered if any of the towers 24, 6, 15, 18, 16, 9, 11, 17, 0, 2, 10, or 12 are built. + - Area 12 can be covered if any of the towers 18, 11, 1, 10, 8, 16, 9, 13, 19, 7, 20, or 6 are built. + - Area 13 can be covered if any of the towers 7, 21, 6, 10, 2, 16, 9, 4, 15, 0, 17, or 18 are built. + - Area 14 can be covered if any of the towers 11, 13, 16, 10, 1, 24, 4, 8, 0, 25, 7, or 19 are built. + - Area 15 can be covered if any of the towers 7, 11, 1, 24, 15, 14, 21, 12, 9, 0, 16, or 2 are built. + - Area 16 can be covered if any of the towers 7, 10, 9, 0, 13, 15, 12, 2, 1, 25, 20, or 6 are built. + - Area 17 can be covered if any of the towers 24, 8, 21, 2, 0, 18, 25, 4, 23, 7, 10, or 5 are built. + - Area 18 can be covered if any of the towers 11, 19, 14, 0, 18, 24, 5, 23, 2, 10, 7, or 8 are built. + - Area 19 can be covered if any of the towers 4, 20, 25, 17, 14, 12, 8, 7, 10, 22, 21, or 23 are built. + - Area 20 can be covered if any of the towers 21, 19, 25, 20, 4, 6, 10, 3, 2, 14, 8, or 15 are built. + - Area 21 can be covered if any of the towers 16, 25, 4, 11, 3, 18, 5, 19, 9, 2, 22, or 1 are built. + - Area 22 can be covered if any of the towers 23, 18, 20, 12, 21, 17, 10, 25, 2, 9, 4, or 6 are built. + - Area 23 can be covered if any of the towers 14, 25, 20, 18, 11, 23, 7, 13, 2, 4, 19, or 12 are built. + - Area 24 can be covered if any of the towers 0, 5, 16, 24, 7, 23, 14, 6, 25, 10, 19, or 22 are built. + - Area 25 can be covered if any of the towers 5, 20, 3, 23, 10, 2, 6, 19, 13, 25, 9, or 15 are built. + - Area 26 can be covered if any of the towers 18, 2, 5, 22, 3, 12, 6, 15, 9, 10, 21, or 1 are built. + - Area 27 can be covered if any of the towers 13, 10, 23, 7, 9, 3, 18, 17, 21, 25, 2, or 14 are built. + - Area 28 can be covered if any of the towers 11, 17, 0, 10, 22, 21, 2, 3, 14, 25, 9, or 6 are built. + - Area 29 can be covered if any of the towers 1, 16, 4, 14, 3, 19, 2, 18, 9, 7, 0, or 13 are built. + - Area 30 can be covered if any of the towers 2, 6, 14, 25, 15, 21, 13, 22, 23, 12, 11, or 9 are built. + - Area 31 can be covered if any of the towers 8, 6, 11, 23, 16, 1, 13, 18, 14, 2, 20, or 19 are built. + - Area 32 can be covered if any of the towers 19, 25, 16, 24, 1, 4, 22, 11, 13, 15, 21, or 12 are built. + - Area 33 can be covered if any of the towers 11, 2, 9, 3, 4, 21, 25, 5, 12, 6, 19, or 7 are built. + - Area 34 can be covered if any of the towers 10, 19, 0, 6, 18, 21, 12, 25, 14, 13, 11, or 20 are built. + + 1. **Budget Constraint**: + - The total cost of building the selected towers must not exceed the available budget of 2,894 894 units. The costs for building each tower are as follows: + - Tower 0: 411 units + - Tower 1: 294 units + - Tower 2: 376 units + - Each area can either be covered (1) or not covered (0). + +#### Summary: +Your task is to select which towers to build and which areas to cover, ensuring that the total population covered is maximized while staying within the budget. The coverage of each area depends on the specific combination of towers built, and the total cost of building the selected towers must not exceed 2,894 894 units.118976 116311Missing data about which towers cover which regionsmissing data
56You are responsible for managing the distribution of resources across a network of nine cities. Each city has a specific supply or demand for resources, and your goal is to minimize the total transportation cost while ensuring that all cities receive the resources they need. The transportation costs vary depending on the route taken between cities, and there are capacity limits on how much can be transported between any two cities. + +City 0 has a net demand of 63 units of resources. City 1 has a net demand of 2 units of resources. City 2 has a net demand of 5 units of resources. City 3 has a net demand of 16 units of resources. City 4 has a net supply of 13 units of resources. City 5 has a net supply of 13 units of resources. City 6 has a net supply of 26 units of resources. City 7 has a net supply of 19 units of resources. City 8 has a net supply of 15 units of resources. + +The cost of transporting resources from City 0 to City 1 is 6, with a capacity of 48. From City 0 to City 2, the cost is 7, with a capacity of 37. From City 0 to City 3, the cost is 10, with a capacity of 57. From City 0 to City 4, the cost is 9, with a capacity of 41. From City 0 to City 5, the cost is 9, with a capacity of 40. From City 0 to City 6, the cost is 10, with a capacity of 48. From City 0 to City 7, the cost is 9, with a capacity of 38. From City 0 to City 8, the cost is 8, with a capacity of 44. + +The cost of transporting resources from City 1 to City 0 is 10, with a capacity of 44. From City 1 to City 2, the cost is 8, with a capacity of 58. From City 1 to City 3, the cost is 9, with a capacity of 57. From City 1 to City 4, the cost is 6, with a capacity of 34. From City 1 to City 5, the cost is 8, with a capacity of 49. From City 1 to City 6, the cost is 8, with a capacity of 57. From City 1 to City 7, the cost is 8, with a capacity of 46. From City 1 to City 8, the cost is 6, with a capacity of 59. + +The cost of transporting resources from City 2 to City 0 is 8, with a capacity of 31. From City 2 to City 1, the cost is 8, with a capacity of 43. From City 2 to City 3, the cost is 7, with a capacity of 43. From City 2 to City 4, the cost is 6, with a capacity of 46. From City 2 to City 5, the cost is 6, with a capacity of 45. From City 2 to City 6, the cost is 8, with a capacity of 41. From City 2 to City 7, the cost is 8, with a capacity of 59. From City 2 to City 8, the cost is 10, with a capacity of 40. + +The cost of transporting resources from City 3 to City 0 is 9, with a capacity of 44. From City 3 to City 1, the cost is 10, with a capacity of 46. From City 3 to City 2, the cost is 10, with a capacity of 32. From City 3 to City 4, the cost is 7, with a capacity of 56. From City 3 to City 5, the cost is 9, with a capacity of 42. From City 3 to City 6, the cost is 6, with a capacity of 52. From City 3 to City 7, the cost is 8, with a capacity of 51. From City 3 to City 8, the cost is 7, with a capacity of 39. + +The cost of transporting resources from City 4 to City 0 is 9, with a capacity of 49. From City 4 to City 1, the cost is 10, with a capacity of 31. From City 4 to City 2, the cost is 7, with a capacity of 47. From City 4 to City 3, the cost is 7, with a capacity of 60. From City 4 to City 5, the cost is 10, with a capacity of 46. From City 4 to City 6, the cost is 7, with a capacity of 32. From City 4 to City 7, the cost is 8, with a capacity of 42. From City 4 to City 8, the cost is 7, with a capacity of 45. + +The cost of transporting resources from City 5 to City 0 is 8, with a capacity of 31. From City 5 to City 1, the cost is 8, with a capacity of 51. From City 5 to City 2, the cost is 10, with a capacity of 53. From City 5 to City 3, the cost is 7, with a capacity of 47. From City 5 to City 4, the cost is 8, with a capacity of 36. From City 5 to City 6, the cost is 6, with a capacity of 30. From City 5 to City 7, the cost is 9, with a capacity of 57. From City 5 to City 8, the cost is 9, with a capacity of 40. + +The cost of transporting resources from City 6 to City 0 is 10, with a capacity of 57. From City 6 to City 1, the cost is 8, with a capacity of 36. From City 6 to City 2, the cost is 10, with a capacity of 56. From City 6 to City 3, the cost is 9, with a capacity of 51. From City 6 to City 4, the cost is 7, with a capacity of 48. From City 6 to City 5, the cost is 7, with a capacity of 56. From City 6 to City 7, the cost is 7, with a capacity of 60. From City 6 to City 8, the cost is 6, with a capacity of 56. + +The cost of transporting resources from City 7 to City 0 is 10, with a capacity of 37. From City 7 to City 1, the cost is 10, with a capacity of 36. From City 7 to City 2, the cost is 8, with a capacity of 48. From City 7 to City 3, the cost is 7, with a capacity of 60. From City 7 to City 4, the cost is 9, with a capacity of 38. From City 7 to City 5, the cost is 7, with a capacity of 43. From City 7 to City 6, the cost is 9, with a capacity of 60. From City 7 to City 8, the cost is 9, with a capacity of 53. + +The cost of transporting resources from City 8 to City 0 is 7, with a capacity of 54. From City 8 to City 1, the cost is 8, with a capacity of 42. From City 8 to City 2, the cost is 8, with a capacity of 36. From City 8 to City 3, the cost is 10, with a capacity of 46. From City 8 to City 4, the cost is 8, with a capacity of 58. From City 8 to City 5, the cost is 7, with a capacity of 54. From City 8 to City 6, the cost is 10, with a capacity of 56. From City 8 to City 7, the cost is 10, with a capacity of 34. + +Your task is to determine the optimal amount of resources to transport between each pair of cities to meet all demands and supplies while minimizing the total transportation cost. The transportation must respect the capacity limits on each route, and the amount of resources transported cannot be negative. For each city, the total amount of resources sent out plus its supply must equal the total amount received plus its demand.714You are responsible for managing the distribution of resources across a network of nine cities. Each city has a specific supply or demand for resources, and your goal is to minimize the total transportation cost while ensuring that all cities receive the resources they need. The transportation costs vary depending on the route taken between cities, and there are capacity limits on how much can be transported between any two cities. + +City 0 has a net demand of 63 units of resources. City 1 has a net demand of 2 units of resources. City 2 has a net demand of 5 units of resources. City 3 has a net demand of 16 units of resources. City 4 has a net supply of 13 units of resources. City 5 has a net supply of 13 units of resources. City 6 has a net supply of 26 units of resources. City 7 has a net supply of 19 units of resources. City 8 has a net supply of 15 units of resources. + +The cost of transporting resources from City 0 to City 1 is 6, with a capacity of 48. From City 0 to City 2, the cost is 7, with a capacity of 37. From City 0 to City 3, the cost is 10, with a capacity of 57. From City 0 to City 4, the cost is 9, with a capacity of 41. From City 0 to City 5, the cost is 9, with a capacity of 40. From City 0 to City 6, the cost is 10, with a capacity of 48. From City 0 to City 7, the cost is 9, with a capacity of 38. From City 0 to City 8, the cost is 8, with a capacity of 44. + +The cost of transporting resources from City 1 to City 0 is 10, with a capacity of 44. From City 1 to City 2, the cost is 8, with a capacity of 58. From City 1 to City 3, the cost is 9, with a capacity of 57. From City 1 to City 4, the cost is 6, with a capacity of 34. From City 1 to City 5, the cost is 8, with a capacity of 49. From City 1 to City 6, the cost is 8, with a capacity of 57. From City 1 to City 7, the cost is 8, with a capacity of 46. From City 1 to City 8, the cost is 6, with a capacity of 59. + +The cost of transporting resources from City 2 to City 0 is 8, with a capacity of 31. From City 2 to City 1, the cost is 8, with a capacity of 43. From City 2 to City 3, the cost is 7, with a capacity of 43. From City 2 to City 4, the cost is 6, with a capacity of 46. From City 2 to City 5, the cost is 6, with a capacity of 45. From City 2 to City 6, the cost is 8, with a capacity of 41. From City 2 to City 7, the cost is 8, with a capacity of 59. From City 2 to City 8, the cost is 10, with a capacity of 40. + +The cost of transporting resources from City 3 to City 0 is 9, with a capacity of 44. From City 3 to City 1, the cost is 10, with a capacity of 46. From City 3 to City 2, the cost is 10, with a capacity of 32. From City 3 to City 4, the cost is 7, with a capacity of 56. From City 3 to City 5, the cost is 9, with a capacity of 42. From City 3 to City 6, the cost is 6, with a capacity of 52. From City 3 to City 7, the cost is 8, with a capacity of 51. From City 3 to City 8, the cost is 7, with a capacity of 39. + +The cost of transporting resources from City 4 to City 0 is 9, with a capacity of 49. From City 4 to City 1, the cost is 10, with a capacity of 31. From City 4 to City 2, the cost is 7, with a capacity of 47. From City 4 to City 3, the cost is 7, with a capacity of 60. From City 4 to City 5, the cost is 10, with a capacity of 46. From City 4 to City 6, the cost is 7, with a capacity of 32. From City 4 to City 7, the cost is 8, with a capacity of 42. From City 4 to City 8, the cost is 7, with a capacity of 45. + +The cost of transporting resources from City 5 to City 0 is 8, with a capacity of 31. From City 5 to City 1, the cost is 8, with a capacity of 51. From City 5 to City 2, the cost is 10, with a capacity of 53. From City 5 to City 3, the cost is 7, with a capacity of 47. From City 5 to City 4, the cost is 8, with a capacity of 36. From City 5 to City 6, the cost is 6, with a capacity of 30. From City 5 to City 7, the cost is 9, with a capacity of 57. From City 5 to City 8, the cost is 9, with a capacity of 40. + +The cost of transporting resources from City 6 to City 0 is 10, with a capacity of 57. From City 6 to City 1, the cost is 8, with a capacity of 36. From City 6 to City 2, the cost is 10, with a capacity of 56. From City 6 to City 3, the cost is 9, with a capacity of 51. From City 6 to City 4, the cost is 7, with a capacity of 48. From City 6 to City 5, the cost is 7, with a capacity of 56. From City 6 to City 7, the cost is 7, with a capacity of 60. From City 6 to City 8, the cost is 6, with a capacity of 56. + +The cost of transporting resources from City 7 to City 0 is 10, with a capacity of 37. From City 7 to City 1, the cost is 10, with a capacity of 36. From City 7 to City 2, the cost is 8, with a capacity of 48. From City 7 to City 3, the cost is 7, with a capacity of 60. From City 7 to City 4, the cost is 9, with a capacity of 38. From City 7 to City 5, the cost is 7, with a capacity of 43. From City 7 to City 6, the cost is 9, with a capacity of 60. From City 7 to City 8, the cost is 9, with a capacity of 53. + +The cost of transporting resources from City 8 to City 0 is 7, with a capacity of 54. From City 8 to City 1, the cost is 8, with a capacity of 42. From City 8 to City 2, the cost is 8, with a capacity of 36. From City 8 to City 3, the cost is 10, with a capacity of 46. From City 8 to City 4, the cost is 8, with a capacity of 58. From City 8 to City 5, the cost is 7, with a capacity of 54. From City 8 to City 6, the cost is 10, with a capacity of 56. From City 8 to City 7, the cost is 10, with a capacity of 34. + +Your task is to determine the optimal amount of resources to transport between each pair of cities to meet all demands and supplies while minimizing the total transportation cost. The transportation must respect the capacity limits on each route, and the amount of resources transported cannot be negative. For each city, the total amount of resources sent out plus its supply must equal the total amount received plus its demand.714No differencesNo differencesok
57Job Shop Scheduling in a Manufacturing Facility + +You are managing a manufacturing facility that produces custom parts for various clients. The facility has multiple machines, each capable of performing specific operations. There are seven jobs (Job 0 to Job 6) that need to be processed, and each job consists of a sequence of operations that must be performed in a specific order. Each operation is assigned to a particular machine, and no two operations can be processed on the same machine simultaneously. + +#### Objective: +Your goal is to minimize the **makespan**, which is the total time required to complete all jobs. This ensures that all jobs are finished as quickly as possible, allowing the facility to meet client deadlines and optimize resource utilization. + +#### Decisions to Be Made: +1. **Start Times**: Determine the start time for each operation of every job. +2. **Operation Order**: Decide the order in which operations are processed on each machine, ensuring no overlaps. + +#### Constraints: +1. **Precedence Constraints**: + - Each operation in a job must start only after the previous operation in the same job has been completed. For example: + - Operation 1 of Job 0 must start at least 1 time unit after Operation 0 of Job 0 has finished. + - Operation 1 of Job 1 must start at least 1 time unit after Operation 0 of Job 1 has finished. + - This pattern applies to all jobs and their respective operations. + +2. **Machine Capacity Constraints**: + - No two operations assigned to the same machine can overlap in time. For example: + - If Operation 1 of Job 3 and Operation 0 of Job 4 are assigned to the same machine, they must be scheduled such that one starts only after the other has finished. A binary decision variable determines the order of these operations. + - Similar constraints apply to all pairs of operations sharing the same machine, such as: + - Operation 1 of Job 3 and Operation 1 of Job 4. + - Operation 1 of Job 3 and Operation 1 of Job 6. + - Operation 0 of Job 4 and Operation 1 of Job 4. + - Operation 0 of Job 4 and Operation 1 of Job 6. + - Operation 1 of Job 4 and Operation 1 of Job 6. + - Operation 1 of Job 0 and Operation 0 of Job 1. + - Operation 1 of Job 0 and Operation 0 of Job 2. + - Operation 0 of Job 1 and Operation 0 of Job 2. + - Operation 0 of Job 0 and Operation 1 of Job 1. + - Operation 0 of Job 0 and Operation 1 of Job 2. + - Operation 0 of Job 0 and Operation 0 of Job 3. + - Operation 0 of Job 0 and Operation 1 of Job 5. + - Operation 0 of Job 0 and Operation 0 of Job 6. + - Operation 1 of Job 1 and Operation 1 of Job 2. + - Operation 1 of Job 1 and Operation 0 of Job 3. + - Operation 1 of Job 1 and Operation 1 of Job 5. + - Operation 1 of Job 1 and Operation 0 of Job 6. + - Operation 1 of Job 2 and Operation 0 of Job 3. + - Operation 1 of Job 2 and Operation 1 of Job 5. + - Operation 1 of Job 2 and Operation 0 of Job 6. + - Operation 0 of Job 3 and Operation 1 of Job 5. + - Operation 0 of Job 3 and Operation 0 of Job 6. + - Operation 1 of Job 5 and Operation 0 of Job 6. + +3. **Makespan Definition**: + - The makespan must be at least as large as the completion time of every operation. For example: + - The makespan must be at least 1 time unit after the completion of Operation 0 of Job 0. + - The makespan must be at least 1 time unit after the completion of Operation 1 of Job 0. + - This applies to all operations across all jobs. + +4. **Non-Negativity**: + - All start times must be non-negative, meaning no operation can start before time zero. + +5. **Binary Variables**: + - Binary variables are used to determine the order of operations on shared machines. For example: + - A binary variable decides whether Operation 1 of Job 3 is processed before or after Operation 0 of Job 4 on the same machine. + - Similar binary variables are used for all relevant pairs of operations sharing the same machine. + +#### Numerical Parameters: +- The processing time for each operation is implicitly defined by the constraints, with a minimum required gap of 1 time unit between consecutive operations of the same job. +- The large constant \( N = 100,000 \) is used in the machine capacity constraints to ensure that the binary variables correctly enforce the order of operations. + +#### Summary: +Your task is to schedule the start times of all operations across all jobs and machines in such a way that the total completion time (makespan) is minimized. You must ensure that all precedence and machine capacity constraints are satisfied, and that the makespan reflects the latest completion time of any operation. The use of binary variables is critical to determining the correct order of operations on shared machines.3Job Shop Scheduling in a Manufacturing Facility + +You are managing a manufacturing facility that produces custom parts for various clients. The facility has multiple machines, each capable of performing specific operations. There are seven jobs (Job 0 to Job 6) that need to be processed, and each job consists of 2 operations that must be performed in a specific order. All operations take 1 time unit. + +#### Objective: +Your goal is to minimize the **makespan**, which is the total time required to complete all jobs. This ensures that all jobs are finished as quickly as possible, allowing the facility to meet client deadlines and optimize resource utilization. + +#### Decisions to Be Made: +1. **Start Times**: Determine the start time for each operation of every job. +2. **Operation Order**: Decide the order in which operations are processed on each machine, ensuring no overlaps. + +#### Constraints: +1. **Precedence Constraints**: + - Each operation in a job must start only after the previous operation in the same job has been completed. For example: + - Operation 1 of Job 0 must start at least 1 time unit after Operation 0 of Job 0 has finished. + - Operation 1 of Job 1 must start at least 1 time unit after Operation 0 of Job 1 has finished. + - This pattern applies to all jobs and their respective operations. + +2. **Machine Capacity Constraints**: + - Certain additional pairs of operations also cannot overlap in time. These additional pairs are: + - Operation 1 of Job 3 and Operation 1 of Job 4. + - Operation 1 of Job 3 and Operation 1 of Job 6. + - Operation 0 of Job 4 and Operation 1 of Job 4. + - Operation 0 of Job 4 and Operation 1 of Job 6. + - Operation 1 of Job 4 and Operation 1 of Job 6. + - Operation 1 of Job 0 and Operation 0 of Job 1. + - Operation 1 of Job 0 and Operation 0 of Job 2. + - Operation 0 of Job 1 and Operation 0 of Job 2. + - Operation 0 of Job 0 and Operation 1 of Job 1. + - Operation 0 of Job 0 and Operation 1 of Job 2. + - Operation 0 of Job 0 and Operation 0 of Job 3. + - Operation 0 of Job 0 and Operation 1 of Job 5. + - Operation 0 of Job 0 and Operation 0 of Job 6. + - Operation 1 of Job 1 and Operation 1 of Job 2. + - Operation 1 of Job 1 and Operation 0 of Job 3. + - Operation 1 of Job 1 and Operation 1 of Job 5. + - Operation 1 of Job 1 and Operation 0 of Job 6. + - Operation 1 of Job 2 and Operation 0 of Job 3. + - Operation 1 of Job 2 and Operation 1 of Job 5. + - Operation 1 of Job 2 and Operation 0 of Job 6. + - Operation 0 of Job 3 and Operation 1 of Job 5. + - Operation 0 of Job 3 and Operation 0 of Job 6. + - Operation 1 of Job 5 and Operation 0 of Job 6. + +3. **Makespan Definition**: + - The makespan must be at least as large as the completion time of every operation + +4. **Non-Negativity**: + - All start times must be non-negative, meaning no operation can start before time zero. + +5. **Binary Variables**: + - Binary variables are used to determine the order of operations on shared machines. For example: + - A binary variable decides whether Operation 1 of Job 3 is processed before or after Operation 0 of Job 4 on the same machine. + - Similar binary variables are used for all relevant pairs of operations sharing the same machine. + +#### Numerical Parameters: +- The large constant \( N = 100 \) is used in the machine capacity constraints to ensure that the binary variables correctly enforce the order of operations. + +#### Summary: +Your task is to schedule the start times of all operations across all jobs and machines in such a way that the total completion time (makespan) is minimized. You must ensure that all precedence and machine capacity constraints are satisfied, and that the makespan reflects the latest completion time of any operation. The use of binary variables is critical to determining the correct order of operations on shared machines.6Job Shop Scheduling in a Manufacturing Facility + +You are managing a manufacturing facility that produces custom parts for various clients. The facility has multiple machines, each capable of performing specific operations. There are seven jobs (Job 0 to Job 6) that need to be processed, and each job consists of a sequence of 2 operations that must be performed in a specific order. Each operation is assigned to a particular machine, and no two All operations can be processed on the same machine simultaneously. take 1 time unit. + +#### Objective: +Your goal is to minimize the **makespan**, which is the total time required to complete all jobs. This ensures that all jobs are finished as quickly as possible, allowing the facility to meet client deadlines and optimize resource utilization. + - This pattern applies to all jobs and their respective operations. + +2. **Machine Capacity Constraints**: + - No two Certain additional pairs of operations assigned to the same machine can also cannot overlap in time. For example: + - If Operation 1 of Job 3 and Operation 0 of Job 4 are assigned to the same machine, they must be scheduled such that one starts only after the other has finished. A binary decision variable determines the order of these operations. + - Similar constraints apply to all These additional pairs of operations sharing the same machine, such as: are: + - Operation 1 of Job 3 and Operation 1 of Job 4. + - Operation 1 of Job 3 and Operation 1 of Job 6. + - Operation 0 of Job 4 and Operation 1 of Job 4. + - Operation 1 of Job 5 and Operation 0 of Job 6. + +3. **Makespan Definition**: + - The makespan must be at least as large as the completion time of every operation. For example: + - The makespan must be at least 1 time unit after the completion of Operation 0 of Job 0. + - The makespan must be at least 1 time unit after the completion of Operation 1 of Job 0. + - This applies to all operations across all jobs. operation + +4. **Non-Negativity**: + - All start times must be non-negative, meaning no operation can start before time zero. + - Similar binary variables are used for all relevant pairs of operations sharing the same machine. + +#### Numerical Parameters: +- The processing time for each operation is implicitly defined by the constraints, with a minimum required gap of 1 time unit between consecutive operations of the same job. + - The large constant \( N = 100,000 100 \) is used in the machine capacity constraints to ensure that the binary variables correctly enforce the order of operations. + +#### Summary: +Your task is to schedule the start times of all operations across all jobs and machines in such a way that the total completion time (makespan) is minimized. You must ensure that all precedence and machine capacity constraints are satisfied, and that the makespan reflects the latest completion time of any operation. The use of binary variables is critical to determining the correct order of operations on shared machines.3 6Information missing on how many operations per job, their duration, etcmissing data
58The problem involves scheduling operations for five jobs (Job 0 through Job 4) in a manufacturing facility. Each job consists of two operations that must be performed in sequence, with the second operation starting only after the first operation is completed. All operations have a processing time of 1 unit. Operations from different jobs may share the same machine, and each machine can handle only one operation at a time. The goal is to minimize the total time required to complete all jobs, known as the makespan. + +The start time of each operation must be determined, ensuring that the sequence of operations within each job is respected and that no two operations on the same machine overlap. Binary variables are used to determine the order of operations on shared machines. For example, if Operation 0 of Job 0 and Operation 0 of Job 2 are on the same machine, a binary variable indicates whether Operation 0 of Job 0 precedes Operation 0 of Job 2 or vice versa. A large constant (100,000) is used in the constraints to enforce the correct ordering of operations on shared machines. + +The makespan must be at least as large as the completion time of every operation. For example, if Operation 0 of Job 0 finishes at time 2, the makespan must be at least 2. The constraints include precedence constraints within jobs, ensuring that the start time of the second operation is at least 1 unit after the completion of the first operation. Machine capacity constraints ensure that two operations assigned to the same machine do not overlap in time. The solution will provide a schedule that ensures all jobs are completed as quickly as possible, respecting the constraints of machine availability and operation precedence.4The problem involves scheduling operations for five jobs (Job 0 through Job 4) in a manufacturing facility. Each job consists of two operations that must be performed in sequence, with the second operation starting only after the first operation is completed. All operations have a processing time of 1 unit. Operations from different jobs may share the same machine, and each machine can handle only one operation at a time. The goal is to minimize the total time required to complete all jobs, known as the makespan. + +The start time of each operation must be determined, ensuring that the sequence of operations within each job is respected and that no two operations on the same machine overlap. Binary variables are used to determine the order of operations on shared machines. For example, if Operation 0 of Job 0 and Operation 0 of Job 2 are on the same machine, a binary variable indicates whether Operation 0 of Job 0 precedes Operation 0 of Job 2 or vice versa. A large constant (100) is used in the constraints to enforce the correct ordering of operations on shared machines. + +The makespan must be at least as large as the completion time of every operation. For example, if Operation 0 of Job 0 finishes at time 2, the makespan must be at least 2. The constraints include precedence constraints within jobs, ensuring that the start time of the second operation is at least 1 unit after the completion of the first operation. Machine capacity constraints ensure that two operations assigned to the same machine do not overlap in time. The solution will provide a schedule that ensures all jobs are completed as quickly as possible, respecting the constraints of machine availability and operation precedence. + + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 0 + - Operation 0 of Job 1: Machine 1 + - Operation 1 of Job 1: Machine 1 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 0 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 0 + - Operation 0 of Job 4: Machine 1 + - Operation 1 of Job 4: Machine 15The problem involves scheduling operations for five jobs (Job 0 through Job 4) in a manufacturing facility. Each job consists of two operations that must be performed in sequence, with the second operation starting only after the first operation is completed. All operations have a processing time of 1 unit. Operations from different jobs may share the same machine, and each machine can handle only one operation at a time. The goal is to minimize the total time required to complete all jobs, known as the makespan. + +The start time of each operation must be determined, ensuring that the sequence of operations within each job is respected and that no two operations on the same machine overlap. Binary variables are used to determine the order of operations on shared machines. For example, if Operation 0 of Job 0 and Operation 0 of Job 2 are on the same machine, a binary variable indicates whether Operation 0 of Job 0 precedes Operation 0 of Job 2 or vice versa. A large constant (100,000) (100) is used in the constraints to enforce the correct ordering of operations on shared machines. + +The makespan must be at least as large as the completion time of every operation. For example, if Operation 0 of Job 0 finishes at time 2, the makespan must be at least 2. The constraints include precedence constraints within jobs, ensuring that the start time of the second operation is at least 1 unit after the completion of the first operation. Machine capacity constraints ensure that two operations assigned to the same machine do not overlap in time. The solution will provide a schedule that ensures all jobs are completed as quickly as possible, respecting the constraints of machine availability and operation precedence. + + + + Additional information: + - There are 3 machines, numbered from 0 to 2 + - The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 0 + - Operation 0 of Job 1: Machine 1 + - Operation 1 of Job 1: Machine 1 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 0 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 0 + - Operation 0 of Job 4: Machine 1 + - Operation 1 of Job 4: Machine 14 5It was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
59There are 5 jobs labeled as Job 0, Job 1, Job 2, Job 3, and Job 4, and 4 machines labeled as Machine 0, Machine 1, Machine 2, and Machine 3. Each job must be assigned to one of 5 schedule positions (Position 0 to Position 4) on each machine. The goal is to minimize the total completion time of all jobs, which includes the start time of the job in the last schedule position (Position 4) on the last machine (Machine 3) and the processing time of that job. Each job has specific processing times on each machine: Job 0 takes 9 units on Machine 0, 7 units on Machine 1, 5 units on Machine 2, and 6 units on Machine 3; Job 1 takes 7 units on Machine 0, 10 units on Machine 1, 10 units on Machine 2, and 8 units on Machine 3; Job 2 takes 6 units on Machine 0, 9 units on Machine 1, 9 units on Machine 2, and 8 units on Machine 3; Job 3 takes 10 units on Machine 0, 10 units on Machine 1, 7 units on Machine 2, and 8 units on Machine 3; Job 4 takes 6 units on Machine 0, 10 units on Machine 1, 5 units on Machine 2, and 10 units on Machine 3. Each schedule position on each machine must be assigned exactly one job, and each job must be assigned to exactly one schedule position across all machines. The start time of a job on the next machine must be at least equal to its completion time on the current machine. Additionally, the start time of a job in the next schedule position must be at least equal to the completion time of the job in the current position on the same machine. The processing times for the jobs in the last schedule position (Position 4) are: Job 0 takes 6 units, Job 1 takes 8 units, Job 2 takes 8 units, Job 3 takes 8 units, and Job 4 takes 10 units. The start time of the job in Position 4 on Machine 3 is a critical factor in the objective function. The decision variables include assigning each job to a specific schedule position on each machine and determining the start times of each job on each machine to ensure all constraints are satisfied. The objective is to minimize the total completion time of all jobs while adhering to the constraints of the manufacturing process.67There are 5 jobs labeled as Job 0, Job 1, Job 2, Job 3, and Job 4, and 4 machines labeled as Machine 0, Machine 1, Machine 2, and Machine 3. Each job must be assigned to one of 5 schedule positions (Position 0 to Position 4) on each machine. The goal is to minimize the total completion time of all jobs, which includes the start time of the job in the last schedule position (Position 4) on the last machine (Machine 3) and the processing time of that job. Each job has specific processing times on each machine: Job 0 takes 9 units on Machine 0, 7 units on Machine 1, 5 units on Machine 2, and 6 units on Machine 3; Job 1 takes 7 units on Machine 0, 10 units on Machine 1, 10 units on Machine 2, and 8 units on Machine 3; Job 2 takes 6 units on Machine 0, 9 units on Machine 1, 9 units on Machine 2, and 8 units on Machine 3; Job 3 takes 10 units on Machine 0, 10 units on Machine 1, 7 units on Machine 2, and 8 units on Machine 3; Job 4 takes 6 units on Machine 0, 10 units on Machine 1, 5 units on Machine 2, and 10 units on Machine 3. Each schedule position on each machine must be assigned exactly one job, and each job must be assigned to exactly one schedule position across all machines. The start time of a job on the next machine must be at least equal to its completion time on the current machine. Additionally, the start time of a job in the next schedule position must be at least equal to the completion time of the job in the current position on the same machine. The processing times for the jobs in the last schedule position (Position 4) are: Job 0 takes 6 units, Job 1 takes 8 units, Job 2 takes 8 units, Job 3 takes 8 units, and Job 4 takes 10 units. The start time of the job in Position 4 on Machine 3 is a critical factor in the objective function. The decision variables include assigning each job to a specific schedule position on each machine and determining the start times of each job on each machine to ensure all constraints are satisfied. The objective is to minimize the total completion time of all jobs while adhering to the constraints of the manufacturing process.67No differencesNo differencesok
60There are 7 jobs, each consisting of 2 operations that must be processed in sequence. Each operation takes exactly 1 unit of time to complete. The operations are assigned to specific machines, and no two operations on the same machine can overlap in time. For each pair of operations assigned to the same machine, a binary decision variable determines the order in which they are processed. If one operation is scheduled before the other, the corresponding binary variable is set to 1; otherwise, it is set to 0. A large constant value of 100,000 is used to enforce the order of operations on shared machines. The start time of each operation must be non-negative, and the start time of the second operation in each job must be at least 1 unit of time after the start time of the first operation in the same job. The total completion time, or makespan, is defined as the maximum completion time across all operations. The objective is to minimize the makespan by determining the optimal start times for all operations and the order of operations on shared machines, while ensuring that all constraints are satisfied.3There are 7 jobs, each consisting of 2 operations that must be processed in sequence. Each operation takes exactly 1 unit of time to complete. The operations are assigned to specific machines, and no two operations on the same machine can overlap in time. For each pair of operations assigned to the same machine, a binary decision variable determines the order in which they are processed. If one operation is scheduled before the other, the corresponding binary variable is set to 1; otherwise, it is set to 0. A large constant value of 100 is used to enforce the order of operations on shared machines. The start time of each operation must be non-negative, and the start time of the second operation in each job must be at least 1 unit of time after the start time of the first operation in the same job. The total completion time, or makespan, is defined as the maximum completion time across all operations. The objective is to minimize the makespan by determining the optimal start times for all operations and the order of operations on shared machines, while ensuring that all constraints are satisfied. + + + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 2 + - Operation 1 of Job 1: Machine 2 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 0 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 0 + - Operation 0 of Job 5: Machine 0 + - Operation 1 of Job 5: Machine 1 + - Operation 0 of Job 6: Machine 0 + - Operation 1 of Job 6: Machine 15There are 7 jobs, each consisting of 2 operations that must be processed in sequence. Each operation takes exactly 1 unit of time to complete. The operations are assigned to specific machines, and no two operations on the same machine can overlap in time. For each pair of operations assigned to the same machine, a binary decision variable determines the order in which they are processed. If one operation is scheduled before the other, the corresponding binary variable is set to 1; otherwise, it is set to 0. A large constant value of 100,000 100 is used to enforce the order of operations on shared machines. The start time of each operation must be non-negative, and the start time of the second operation in each job must be at least 1 unit of time after the start time of the first operation in the same job. The total completion time, or makespan, is defined as the maximum completion time across all operations. The objective is to minimize the makespan by determining the optimal start times for all operations and the order of operations on shared machines, while ensuring that all constraints are satisfied. + + + + + Additional information: + - There are 3 machines, numbered from 0 to 2 + - The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 2 + - Operation 1 of Job 1: Machine 2 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 0 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 0 + - Operation 0 of Job 5: Machine 0 + - Operation 1 of Job 5: Machine 1 + - Operation 0 of Job 6: Machine 0 + - Operation 1 of Job 6: Machine 13 5It was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
61Select two facilities from five potential locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the minimum distance between any two selected facilities. Each location can either be selected (1) or not selected (0). For every pair of locations, determine whether both are selected as facilities, represented by a binary decision variable. The distances between the locations are scaled by a factor of 1,000,000 and are as follows: Node 0 and Node 1: 19 units, Node 0 and Node 2: 21 units, Node 0 and Node 3: 31 units, Node 0 and Node 4: 35 units, Node 1 and Node 0: 29 units, Node 1 and Node 2: 13 units, Node 1 and Node 3: 46 units, Node 1 and Node 4: 37 units, Node 2 and Node 0: 50 units, Node 2 and Node 1: 22 units, Node 2 and Node 3: 49 units, Node 2 and Node 4: 33 units, Node 3 and Node 0: 18 units, Node 3 and Node 1: 20 units, Node 3 and Node 2: 46 units, Node 3 and Node 4: 33 units, Node 4 and Node 0: 18 units, Node 4 and Node 1: 26 units, Node 4 and Node 2: 39 units, Node 4 and Node 3: 43 units. The total number of selected facilities must equal two. For every pair of locations, the selection of both facilities in the pair is only possible if both individual facilities are selected, enforced through constraints linking the pairwise selection variables to the individual facility selection variables. All decision variables are binary, ensuring facilities are either fully selected or not selected at all. The objective is to maximize the minimum distance between any two selected facilities, ensuring geographical dispersion while adhering to the constraints on pairwise selection and total facility count.46Select two facilities from five potential locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the minimum distance between any two selected facilities. Each location can either be selected (1) or not selected (0). For every pair of locations, determine whether both are selected as facilities, represented by a binary decision variable. The distances between the locations are scaled by a factor of 1,000,000 and are as follows: Node 0 and Node 1: 19 units, Node 0 and Node 2: 21 units, Node 0 and Node 3: 31 units, Node 0 and Node 4: 35 units, Node 1 and Node 0: 29 units, Node 1 and Node 2: 13 units, Node 1 and Node 3: 46 units, Node 1 and Node 4: 37 units, Node 2 and Node 0: 50 units, Node 2 and Node 1: 22 units, Node 2 and Node 3: 49 units, Node 2 and Node 4: 33 units, Node 3 and Node 0: 18 units, Node 3 and Node 1: 20 units, Node 3 and Node 2: 46 units, Node 3 and Node 4: 33 units, Node 4 and Node 0: 18 units, Node 4 and Node 1: 26 units, Node 4 and Node 2: 39 units, Node 4 and Node 3: 43 units. The total number of selected facilities must equal two. For every pair of locations, the selection of both facilities in the pair is only possible if both individual facilities are selected, enforced through constraints linking the pairwise selection variables to the individual facility selection variables. All decision variables are binary, ensuring facilities are either fully selected or not selected at all. The objective is to maximize the minimum distance between any two selected facilities, ensuring geographical dispersion while adhering to the constraints on pairwise selection and total facility count.46No differencesNo differencesok
62The problem involves scheduling operations for six jobs (Job 0 through Job 5) in a manufacturing facility, where each job consists of two operations that must be performed in sequence. Each operation has a processing time of 1 unit of time. Operations from different jobs may share the same machine, and a machine can only process one operation at a time. The goal is to minimize the total time to complete all jobs, known as the makespan. + +For each job, the second operation cannot start until the first operation is completed, ensuring a sequence within each job. Binary variables are used to determine the order of operations on shared machines. If a binary variable is set to 1, it indicates that one operation is scheduled before another on the same machine; if set to 0, the order is reversed. A large constant (100,000) is used in the constraints to enforce this sequencing logic. + +The makespan must be at least as large as the completion time of every operation across all jobs. All start times for operations must be non-negative, and the binary variables must be either 0 or 1. The objective is to determine the start times for all operations and the sequence of operations on shared machines to minimize the makespan, ensuring efficient completion of all jobs. + +Constraints include ensuring that the start time of the second operation in each job is at least 1 unit of time after the start time of the first operation. For operations sharing the same machine, one operation must finish before the other starts, enforced by the binary variables and the large constant. The makespan must be at least as large as the completion time of every operation, and all start times must be non-negative. The binary variables must be either 0 or 1 to determine the correct sequence of operations on shared machines.6The problem involves scheduling operations for six jobs (Job 0 through Job 5) in a manufacturing facility, where each job consists of two operations that must be performed in sequence. Each operation has a processing time of 1 unit of time. Operations from different jobs may share the same machine, and a machine can only process one operation at a time. The goal is to minimize the total time to complete all jobs, known as the makespan. + +For each job, the second operation cannot start until the first operation is completed, ensuring a sequence within each job. Binary variables are used to determine the order of operations on shared machines. If a binary variable is set to 1, it indicates that one operation is scheduled before another on the same machine; if set to 0, the order is reversed. A large constant (1000) is used in the constraints to enforce this sequencing logic. + +The makespan must be at least as large as the completion time of every operation across all jobs. All start times for operations must be non-negative, and the binary variables must be either 0 or 1. The objective is to determine the start times for all operations and the sequence of operations on shared machines to minimize the makespan, ensuring efficient completion of all jobs. + +Constraints include ensuring that the start time of the second operation in each job is at least 1 unit of time after the start time of the first operation. For operations sharing the same machine, one operation must finish before the other starts, enforced by the binary variables and the large constant. The makespan must be at least as large as the completion time of every operation, and all start times must be non-negative. The binary variables must be either 0 or 1 to determine the correct sequence of operations on shared machines. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3, 4, 5} +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Operation identifiers: we'll flatten operations into a set O with elements o = (j,k) and also give each a unique integer id o_id = 0..11 for modeling convenience. +- Machines: M = {A, B, C} + +2) Machine assignment for every operation (explicit) +- Operation (0,1) -> Machine A +- Operation (0,2) -> Machine B +- Operation (1,1) -> Machine A +- Operation (1,2) -> Machine C +- Operation (2,1) -> Machine B +- Operation (2,2) -> Machine C +- Operation (3,1) -> Machine A +- Operation (3,2) -> Machine B +- Operation (4,1) -> Machine C +- Operation (4,2) -> Machine A +- Operation (5,1) -> Machine B +- Operation (5,2) -> Machine C + +3) Processing durations (time units) +All operations have processing time p(o) = 14The problem involves scheduling operations for six jobs (Job 0 through Job 5) in a manufacturing facility, where each job consists of two operations that must be performed in sequence. Each operation has a processing time of 1 unit of time. Operations from different jobs may share the same machine, and a machine can only process one operation at a time. The goal is to minimize the total time to complete all jobs, known as the makespan. + +For each job, the second operation cannot start until the first operation is completed, ensuring a sequence within each job. Binary variables are used to determine the order of operations on shared machines. If a binary variable is set to 1, it indicates that one operation is scheduled before another on the same machine; if set to 0, the order is reversed. A large constant (100,000) (1000) is used in the constraints to enforce this sequencing logic. + +The makespan must be at least as large as the completion time of every operation across all jobs. All start times for operations must be non-negative, and the binary variables must be either 0 or 1. The objective is to determine the start times for all operations and the sequence of operations on shared machines to minimize the makespan, ensuring efficient completion of all jobs. + +Constraints include ensuring that the start time of the second operation in each job is at least 1 unit of time after the start time of the first operation. For operations sharing the same machine, one operation must finish before the other starts, enforced by the binary variables and the large constant. The makespan must be at least as large as the completion time of every operation, and all start times must be non-negative. The binary variables must be either 0 or 1 to determine the correct sequence of operations on shared machines. + + ADDITIONAL INPUT DATA: + + 1) Sets and indices + - Jobs: J = {0, 1, 2, 3, 4, 5} + - For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). + - Operation identifiers: we'll flatten operations into a set O with elements o = (j,k) and also give each a unique integer id o_id = 0..11 for modeling convenience. + - Machines: M = {A, B, C} + + 2) Machine assignment for every operation (explicit) + - Operation (0,1) -> Machine A + - Operation (0,2) -> Machine B + - Operation (1,1) -> Machine A + - Operation (1,2) -> Machine C + - Operation (2,1) -> Machine B + - Operation (2,2) -> Machine C + - Operation (3,1) -> Machine A + - Operation (3,2) -> Machine B + - Operation (4,1) -> Machine C + - Operation (4,2) -> Machine A + - Operation (5,1) -> Machine B + - Operation (5,2) -> Machine C + + 3) Processing durations (time units) + All operations have processing time p(o) = 16 4While the statement required to aoid overlap of operations on the same machines, it did not state how many machines are available or the assignment of operations to machines. missing data
63In a manufacturing facility, six distinct jobs need to be processed on shared machines. Each job consists of two operations that must be performed in a specific order, with the second operation starting only after the first operation is completed. Each operation has a processing time of 1 unit. The goal is to minimize the total completion time, known as the makespan, by scheduling operations on the machines efficiently. + +Each operation is assigned to a specific machine, and some operations from different jobs share the same machine, creating potential conflicts. To resolve these conflicts, the scheduler must determine the order in which operations are processed on each machine. This is achieved using binary variables, where a value of 1 indicates that one operation is scheduled before another, and 0 indicates the opposite. A large constant of 100,000 is used in the constraints to ensure that only one of the two possible orders is active at any time, preventing overlapping operations on the same machine. + +The start time of each operation must be non-negative, and the makespan must be at least as large as the completion time of every operation. The precedence constraints ensure that for each job, the start time of the second operation is at least 1 unit of time after the start time of the first operation. The machine capacity constraints ensure that for every pair of operations assigned to the same machine, one operation must finish before the other starts, determined by the binary variables. + +The objective is to minimize the makespan by scheduling the operations in a way that respects the precedence constraints within each job, avoids overlapping operations on the same machine, and ensures all start times are non-negative. The solution must also accurately reflect the completion time of the last operation, ensuring the makespan is minimized.4In a manufacturing facility, six distinct jobs need to be processed on shared machines. Each job consists of two operations that must be performed in a specific order, with the second operation starting only after the first operation is completed. Each operation has a processing time of 1 unit. The goal is to minimize the total completion time, known as the makespan, by scheduling operations on the machines efficiently. + +Each operation is assigned to a specific machine, and some operations from different jobs share the same machine, creating potential conflicts. To resolve these conflicts, the scheduler must determine the order in which operations are processed on each machine. This is achieved using binary variables, where a value of 1 indicates that one operation is scheduled before another, and 0 indicates the opposite. A large constant of 1000 is used in the constraints to ensure that only one of the two possible orders is active at any time, preventing overlapping operations on the same machine. + +The start time of each operation must be non-negative, and the makespan must be at least as large as the completion time of every operation. The precedence constraints ensure that for each job, the start time of the second operation is at least 1 unit of time after the start time of the first operation. The machine capacity constraints ensure that for every pair of operations assigned to the same machine, one operation must finish before the other starts, determined by the binary variables. + +The objective is to minimize the makespan by scheduling the operations in a way that respects the precedence constraints within each job, avoids overlapping operations on the same machine, and ensures all start times are non-negative. The solution must also accurately reflect the completion time of the last operation, ensuring the makespan is minimized. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {1, 2, 3, 4, 5, 6} +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Machines: M = {1, 2, 3} + +2) Machine assignment for every operation (which machine executes the operation) +- Operation (1,1) -> Machine 1 +- Operation (1,2) -> Machine 2 +- Operation (2,1) -> Machine 1 +- Operation (2,2) -> Machine 3 +- Operation (3,1) -> Machine 2 +- Operation (3,2) -> Machine 3 +- Operation (4,1) -> Machine 1 +- Operation (4,2) -> Machine 2 +- Operation (5,1) -> Machine 2 +- Operation (5,2) -> Machine 3 +- Operation (6,1) -> Machine 3 +- Operation (6,2) -> Machine 1 + +3) Processing times (duration) for each operation +- All operations have processing time p_{j,k} = 14In a manufacturing facility, six distinct jobs need to be processed on shared machines. Each job consists of two operations that must be performed in a specific order, with the second operation starting only after the first operation is completed. Each operation has a processing time of 1 unit. The goal is to minimize the total completion time, known as the makespan, by scheduling operations on the machines efficiently. + +Each operation is assigned to a specific machine, and some operations from different jobs share the same machine, creating potential conflicts. To resolve these conflicts, the scheduler must determine the order in which operations are processed on each machine. This is achieved using binary variables, where a value of 1 indicates that one operation is scheduled before another, and 0 indicates the opposite. A large constant of 100,000 1000 is used in the constraints to ensure that only one of the two possible orders is active at any time, preventing overlapping operations on the same machine. + +The start time of each operation must be non-negative, and the makespan must be at least as large as the completion time of every operation. The precedence constraints ensure that for each job, the start time of the second operation is at least 1 unit of time after the start time of the first operation. The machine capacity constraints ensure that for every pair of operations assigned to the same machine, one operation must finish before the other starts, determined by the binary variables. + +The objective is to minimize the makespan by scheduling the operations in a way that respects the precedence constraints within each job, avoids overlapping operations on the same machine, and ensures all start times are non-negative. The solution must also accurately reflect the completion time of the last operation, ensuring the makespan is minimized. + + ADDITIONAL INPUT DATA: + + 1) Sets and indices + - Jobs: J = {1, 2, 3, 4, 5, 6} + - For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). + - Machines: M = {1, 2, 3} + + 2) Machine assignment for every operation (which machine executes the operation) + - Operation (1,1) -> Machine 1 + - Operation (1,2) -> Machine 2 + - Operation (2,1) -> Machine 1 + - Operation (2,2) -> Machine 3 + - Operation (3,1) -> Machine 2 + - Operation (3,2) -> Machine 3 + - Operation (4,1) -> Machine 1 + - Operation (4,2) -> Machine 2 + - Operation (5,1) -> Machine 2 + - Operation (5,2) -> Machine 3 + - Operation (6,1) -> Machine 3 + - Operation (6,2) -> Machine 1 + + 3) Processing times (duration) for each operation + - All operations have processing time p_{j,k} = 1No differencesIt was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
64The goal is to minimize the total travel cost for a delivery vehicle visiting six cities: City 0, City 1, City 2, City 3, City 4, and City 5. The vehicle must visit each city exactly once and return to the starting point, ensuring the route forms a single tour without subtours. The decision variables include binary choices for whether the vehicle travels directly from one city to another and general integer variables representing the sequence in which cities are visited. The travel costs between cities are as follows: from City 0 to City 1 costs 281, to City 2 costs 300, to City 3 costs 307, to City 4 costs 313, and to City 5 costs 308; from City 1 to City 0 costs 314, to City 2 costs 313, to City 3 costs 324, to City 4 costs 321, and to City 5 costs 323; from City 2 to City 0 costs 317, to City 1 costs 277, to City 3 costs 297, to City 4 costs 299, and to City 5 costs 308; from City 3 to City 0 costs 287, to City 1 costs 304, to City 2 costs 291, to City 4 costs 299, and to City 5 costs 320; from City 4 to City 0 costs 312, to City 1 costs 314, to City 2 costs 292, to City 3 costs 322, and to City 5 costs 275; from City 5 to City 0 costs 320, to City 1 costs 282, to City 2 costs 280, to City 3 costs 313, and to City 4 costs 321. The constraints include ensuring the vehicle arrives at and departs from each city exactly once, assigning sequence numbers to cities to prevent subtours, and ensuring sequence numbers are unique and range from 0 to 5. Subtour elimination is enforced by linking route decisions with sequence numbers, ensuring logical visit order and preventing invalid loops. The solution must respect all constraints to form a valid tour while minimizing total travel costs.1741The goal is to minimize the total travel cost for a delivery vehicle visiting six cities: City 0, City 1, City 2, City 3, City 4, and City 5. The vehicle must visit each city exactly once and return to the starting point, ensuring the route forms a single tour without subtours. The decision variables include binary choices for whether the vehicle travels directly from one city to another and general integer variables representing the sequence in which cities are visited. The travel costs between cities are as follows: from City 0 to City 1 costs 281, to City 2 costs 300, to City 3 costs 307, to City 4 costs 313, and to City 5 costs 308; from City 1 to City 0 costs 314, to City 2 costs 313, to City 3 costs 324, to City 4 costs 321, and to City 5 costs 323; from City 2 to City 0 costs 317, to City 1 costs 277, to City 3 costs 297, to City 4 costs 299, and to City 5 costs 308; from City 3 to City 0 costs 287, to City 1 costs 304, to City 2 costs 291, to City 4 costs 299, and to City 5 costs 320; from City 4 to City 0 costs 312, to City 1 costs 314, to City 2 costs 292, to City 3 costs 322, and to City 5 costs 275; from City 5 to City 0 costs 320, to City 1 costs 282, to City 2 costs 280, to City 3 costs 313, and to City 4 costs 321. The constraints include ensuring the vehicle arrives at and departs from each city exactly once, assigning sequence numbers to cities to prevent subtours, and ensuring sequence numbers are unique and range from 0 to 5. Subtour elimination is enforced by linking route decisions with sequence numbers, ensuring logical visit order and preventing invalid loops. The solution must respect all constraints to form a valid tour while minimizing total travel costs.1741No differencesNo differencesok
65Minimize the total operational costs, which include production setup costs of 145.2319299071881 per production run, machine startup costs of 65.79360965435782 per startup, inventory holding costs of 0.8652264753415143 for Product 0, 0.9341662877492469 for Product 1, and 1.07339072521956 for Product 2 per unit, and backlogging penalties of 7.607410261300208 for Product 0, 6.920344391340704 for Product 1, and 6.497781941350707 for Product 2 per unit. The decisions include whether to produce each product on each machine in each period, whether to start up a machine for a specific product in a given period, the exact amount of each product to produce on each machine in each period, how much of each product to keep in inventory at the end of each period, and how much to backlog if demand cannot be met. + +The constraints include meeting customer demand in each period, where the total production of each product across both machines, plus any inventory carried over from the previous period, minus any backlog from the previous period, must equal the demand for that period plus the current inventory minus the current backlog. The demands are 21.53845915883999 for Product 0 in Period 0, 22.99877365370049 in Period 1, and 20.41750946364395 in Period 2; 24.34292430654904 for Product 1 in Period 0, 18.69161579278047 in Period 1, and 16.96075239636408 in Period 2; and 15.44135005337596 for Product 2 in Period 0, 24.75403844533352 in Period 1, and 17.09921582696376 in Period 2. + +The total production quantity for each product on each machine in each period, plus any startup time required, must not exceed the machine's capacity. The startup time for Machine 0 is 6.900949355420833, and for Machine 1, it is 7.255877688483005. The capacity of Machine 0 is 73.62985121550206, and for Machine 1, it is 69.3191618930891. Each machine can only be used for one product at a time in each period, meaning the sum of production decisions for all products on a single machine in a given period cannot exceed 1. + +If a machine is used to produce a product in a period but was not used for that product in the previous period, a startup cost is incurred. For the first period, a startup is automatically required if production occurs. Production quantities, inventory levels, and backlogs must be non-negative, and production and startup decisions are binary (0 or 1).856.1453747955272Minimize the total operational costs, which include production setup costs of 145.2319299071881 per production run, machine startup costs of 65.79360965435782 per startup, inventory holding costs of 0.8652264753415143 for Product 0, 0.9341662877492469 for Product 1, and 1.07339072521956 for Product 2 per unit, and backlogging penalties of 7.607410261300208 for Product 0, 6.920344391340704 for Product 1, and 6.497781941350707 for Product 2 per unit. The decisions include whether to produce each product on each machine in each period, whether to start up a machine for a specific product in a given period, the exact amount of each product to produce on each machine in each period, how much of each product to keep in inventory at the end of each period, and how much to backlog if demand cannot be met. + +The constraints include meeting customer demand in each period, where the total production of each product across both machines, plus any inventory carried over from the previous period, minus any backlog from the previous period, must equal the demand for that period plus the current inventory minus the current backlog. The demands are 21.53845915883999 for Product 0 in Period 0, 22.99877365370049 in Period 1, and 20.41750946364395 in Period 2; 24.34292430654904 for Product 1 in Period 0, 18.69161579278047 in Period 1, and 16.96075239636408 in Period 2; and 15.44135005337596 for Product 2 in Period 0, 24.75403844533352 in Period 1, and 17.09921582696376 in Period 2. + +The total production quantity for each product on each machine in each period, plus any startup time required, must not exceed the machine's capacity. The startup time for Machine 0 is 6.900949355420833, and for Machine 1, it is 7.255877688483005. The capacity of Machine 0 is 73.62985121550206, and for Machine 1, it is 69.3191618930891. Each machine can only be used for one product at a time in each period, meaning the sum of production decisions for all products on a single machine in a given period cannot exceed 1. + +If a machine is used to produce a product in a period but was not used for that product in the previous period, a startup cost is incurred. For the first period, a startup is automatically required if production occurs. Production quantities, inventory levels, and backlogs must be non-negative, and production and startup decisions are binary (0 or 1).856.1453747955272No differencesNo differencesok
66A logistics company must deliver packages to seven cities: City 0, City 1, City 2, City 3, City 4, City 5, and City 6. The goal is to determine the most efficient route that minimizes the total delivery cost while ensuring each city is visited exactly once and the route forms a single continuous loop. The delivery costs between cities are as follows: + +- From City 0: To City 1 costs 277, to City 2 costs 301, to City 3 costs 299, to City 4 costs 301, to City 5 costs 280, and to City 6 costs 282. +- From City 1: To City 0 costs 311, to City 2 costs 291, to City 3 costs 291, to City 4 costs 310, to City 5 costs 292, and to City 6 costs 313. +- From City 2: To City 0 costs 315, to City 1 costs 310, to City 3 costs 285, to City 4 costs 320, to City 5 costs 278, and to City 6 costs 311. +- From City 3: To City 0 costs 290, to City 1 costs 289, to City 2 costs 313, to City 4 costs 289, to City 5 costs 320, and to City 6 costs 305. +- From City 4: To City 0 costs 277, to City 1 costs 313, to City 2 costs 325, to City 3 costs 300, to City 5 costs 300, and to City 6 costs 281. +- From City 5: To City 0 costs 281, to City 1 costs 301, to City 2 costs 311, to City 3 costs 292, to City 4 costs 299, and to City 6 costs 305. +- From City 6: To City 0 costs 280, to City 1 costs 291, to City 2 costs 311, to City 3 costs 298, to City 4 costs 275, and to City 5 costs 298. + +The problem requires that each city is visited exactly once and that each city is the starting point for exactly one trip. To ensure the route forms a single continuous loop without smaller disconnected loops, the company uses constraints involving the position of each city in the tour. Each city is assigned a position in the delivery sequence, ranging from 1 to 7, and these positions must satisfy constraints that prevent subtours. For example, if City 1 is visited before City 2, the position of City 1 must be less than that of City 2. The objective is to minimize the total delivery cost, which is the sum of the costs of all trips taken, while satisfying all constraints.1988A logistics company must deliver packages to seven cities: City 0, City 1, City 2, City 3, City 4, City 5, and City 6. The goal is to determine the most efficient route that minimizes the total delivery cost while ensuring each city is visited exactly once and the route forms a single continuous loop. The delivery costs between cities are as follows: + +- From City 0: To City 1 costs 277, to City 2 costs 301, to City 3 costs 299, to City 4 costs 301, to City 5 costs 280, and to City 6 costs 282. +- From City 1: To City 0 costs 311, to City 2 costs 291, to City 3 costs 291, to City 4 costs 310, to City 5 costs 292, and to City 6 costs 313. +- From City 2: To City 0 costs 315, to City 1 costs 310, to City 3 costs 285, to City 4 costs 320, to City 5 costs 278, and to City 6 costs 311. +- From City 3: To City 0 costs 290, to City 1 costs 289, to City 2 costs 313, to City 4 costs 289, to City 5 costs 320, and to City 6 costs 305. +- From City 4: To City 0 costs 277, to City 1 costs 313, to City 2 costs 325, to City 3 costs 300, to City 5 costs 300, and to City 6 costs 281. +- From City 5: To City 0 costs 281, to City 1 costs 301, to City 2 costs 311, to City 3 costs 292, to City 4 costs 299, and to City 6 costs 305. +- From City 6: To City 0 costs 280, to City 1 costs 291, to City 2 costs 311, to City 3 costs 298, to City 4 costs 275, and to City 5 costs 298. + +The problem requires that each city is visited exactly once and that each city is the starting point for exactly one trip. To ensure the route forms a single continuous loop without smaller disconnected loops, the company uses constraints involving the position of each city in the tour. Each city is assigned a position in the delivery sequence, ranging from 1 to 7, and these positions must satisfy constraints that prevent subtours. For example, if City 1 is visited before City 2, the position of City 1 must be less than that of City 2. The objective is to minimize the total delivery cost, which is the sum of the costs of all trips taken, while satisfying all constraints.1988No differencesNo differencesok
67Cell Tower Deployment for Regional Coverage + +You are tasked with deploying cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 34 distinct areas, each with a specific population size that needs to be covered by the cell towers. Your goal is to determine which towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + +#### Key Decisions: +1. **Tower Deployment**: Decide whether to build each of the 30 available cell towers. Each tower has a unique construction cost, and building it will enable coverage for certain areas. +2. **Area Coverage**: Determine which of the 34 areas will be covered by the deployed towers. An area is considered covered if at least one of the towers that can serve it is built. + +#### Objective: +Maximize the total population covered by the deployed cell towers. The population sizes for the 34 areas are as follows: +- Area 0: 3,328 people +- Area 1: 2,066 people +- Area 2: 2,261 people +- Area 3: 4,527 people +- Area 4: 4,205 people +- Area 5: 4,927 people +- Area 6: 2,941 people +- Area 7: 4,277 people +- Area 8: 3,276 people +- Area 9: 4,326 people +- Area 10: 2,852 people +- Area 11: 3,957 people +- Area 12: 2,068 people +- Area 13: 4,724 people +- Area 14: 3,226 people +- Area 15: 3,575 people +- Area 16: 2,922 people +- Area 17: 3,850 people +- Area 18: 4,898 people +- Area 19: 2,288 people +- Area 20: 4,726 people +- Area 21: 2,513 people +- Area 22: 2,336 people +- Area 23: 4,669 people +- Area 24: 4,626 people +- Area 25: 4,585 people +- Area 26: 4,367 people +- Area 27: 4,981 people +- Area 28: 4,049 people +- Area 29: 4,948 people +- Area 30: 2,182 people +- Area 31: 2,421 people +- Area 32: 3,061 people +- Area 33: 4,310 people + +#### Constraints: +1. **Coverage Requirements**: Each area must be covered by at least one of the towers that can serve it. For example: + - Area 0 can be covered by Towers 0, 3, 4, 7, 9, 16, 17, 20, 22, 24, 25, 26, or 27. + - Area 1 can be covered by Towers 0, 1, 15, 19, 22, 23, 25, 26, 27, or 29. + - Similar coverage relationships exist for all 34 areas, as detailed in the input data. + +2. **Budget Constraint**: The total cost of building the selected towers must not exceed the available budget of 3,375 units. The construction costs for the towers are: + - Tower 0: 248 units + - Tower 1: 226 units + - Tower 2: 454 units + - Tower 3: 466 units + - Tower 4: 241 units + - Tower 5: 201 units + - Tower 6: 415 units + - Tower 7: 486 units + - Tower 8: 485 units + - Tower 9: 492 units + - Tower 10: 249 units + - Tower 11: 486 units + - Tower 12: 417 units + - Tower 13: 327 units + - Tower 14: 257 units + - Tower 15: 486 units + - Tower 16: 438 units + - Tower 17: 266 units + - Tower 18: 473 units + - Tower 19: 268 units + - Tower 20: 293 units + - Tower 21: 379 units + - Tower 22: 476 units + - Tower 23: 482 units + - Tower 24: 367 units + - Tower 25: 239 units + - Tower 26: 368 units + - Tower 27: 492 units + - Tower 28: 454 units + - Tower 29: 319 units + +3. **Binary Decisions**: Each tower can either be built (1) or not built (0), and each area can either be covered (1) or not covered (0). + +#### Summary: +Your task is to select which towers to build and which areas to cover to maximize the total population served, ensuring that all coverage requirements are met and the total construction cost does not exceed the budget. The solution must account for the specific coverage capabilities of each tower and the population sizes of the areas.124268Cell Tower Deployment for Regional Coverage + + You are tasked with deploying cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 34 distinct areas, each with a specific population size that needs to be covered by the cell towers. Your goal is to determine which towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + + #### Key Decisions: + 1. **Tower Deployment**: Decide whether to build each of the 30 available cell towers. Each tower has a unique construction cost, and building it will enable coverage for certain areas. + 2. **Area Coverage**: Determine which of the 34 areas will be covered by the deployed towers. An area is considered covered if at least one of the towers that can serve it is built. + + #### Objective: + Maximize the total population covered by the deployed cell towers. The population sizes for the 34 areas are as follows: +- Area 0: 3,328 people +- Area 1: 2,066 people +- Area 2: 2,261 people +- Area 3: 4,527 people +- Area 4: 4,205 people +- Area 5: 4,927 people +- Area 6: 2,941 people +- Area 7: 4,277 people +- Area 8: 3,276 people +- Area 9: 4,326 people +- Area 10: 2,852 people +- Area 11: 3,957 people +- Area 12: 2,068 people +- Area 13: 4,724 people +- Area 14: 3,226 people +- Area 15: 3,575 people +- Area 16: 2,922 people +- Area 17: 3,850 people +- Area 18: 4,898 people +- Area 19: 2,288 people +- Area 20: 4,726 people +- Area 21: 2,513 people +- Area 22: 2,336 people +- Area 23: 4,669 people +- Area 24: 4,626 people +- Area 25: 4,585 people +- Area 26: 4,367 people +- Area 27: 4,981 people +- Area 28: 4,049 people +- Area 29: 4,948 people +- Area 30: 2,182 people +- Area 31: 2,421 people +- Area 32: 3,061 people +- Area 33: 4,310 people + + #### Constraints: + 1. **Coverage Requirements**: Each area must be covered by at least one of the towers that can serve it. For example: + - Area 0 can be covered by Towers 0, 3, 4, 7, 9, 16, 17, 20, 22, 24, 25, 26, or 27. + - Area 1 can be covered by Towers 0, 1, 15, 19, 22, 23, 25, 26, 27, or 29. + - Full coverage relationships for all areas: + - Area 0 can be covered by Towers Tower 0, Tower 3, Tower 4, Tower 7, Tower 9, Tower 16, Tower 17, Tower 20, Tower 22, Tower 24, Tower 25, Tower 26, Tower 27. + - Area 1 can be covered by Towers Tower 0, Tower 1, Tower 15, Tower 19, Tower 22, Tower 23, Tower 25, Tower 26, Tower 27, Tower 29. + - Area 2 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 7, Tower 8, Tower 13, Tower 17, Tower 18, Tower 21, Tower 23, Tower 25, Tower 28, Tower 29. + - Area 3 can be covered by Towers Tower 0, Tower 6, Tower 7, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 24, Tower 29. + - Area 4 can be covered by Towers Tower 0, Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 10, Tower 13, Tower 14, Tower 18, Tower 22, Tower 25, Tower 26. + - Area 5 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 8, Tower 11, Tower 12, Tower 14, Tower 17, Tower 19, Tower 25, Tower 26, Tower 29. + - Area 6 can be covered by Towers Tower 1, Tower 2, Tower 6, Tower 7, Tower 9, Tower 11, Tower 17, Tower 18, Tower 19, Tower 20, Tower 26, Tower 27, Tower 29. + - Area 7 can be covered by Towers Tower 3, Tower 5, Tower 7, Tower 8, Tower 11, Tower 12, Tower 14, Tower 20, Tower 27, Tower 29. + - Area 8 can be covered by Towers Tower 2, Tower 5, Tower 6, Tower 8, Tower 11, Tower 17, Tower 19, Tower 20, Tower 21, Tower 22, Tower 23, Tower 27. + - Area 9 can be covered by Towers Tower 1, Tower 5, Tower 7, Tower 8, Tower 10, Tower 12, Tower 14, Tower 17, Tower 19, Tower 20, Tower 21, Tower 22, Tower 24, Tower 26, Tower 29. + - Area 10 can be covered by Towers Tower 2, Tower 4, Tower 6, Tower 10, Tower 12, Tower 14, Tower 15, Tower 18, Tower 20, Tower 22, Tower 23, Tower 28. + - Area 11 can be covered by Towers Tower 4, Tower 7, Tower 8, Tower 11, Tower 12, Tower 13, Tower 17, Tower 18, Tower 22, Tower 23, Tower 26, Tower 27. + - Area 12 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 4, Tower 5, Tower 13, Tower 15, Tower 16, Tower 20, Tower 24, Tower 29. + - Area 13 can be covered by Towers Tower 0, Tower 2, Tower 3, Tower 8, Tower 10, Tower 12, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 23, Tower 28. + - Area 14 can be covered by Towers Tower 0, Tower 3, Tower 5, Tower 8, Tower 9, Tower 13, Tower 14, Tower 16, Tower 22, Tower 23, Tower 25, Tower 28. + - Area 15 can be covered by Towers Tower 0, Tower 3, Tower 4, Tower 5, Tower 6, Tower 10, Tower 11, Tower 15, Tower 16, Tower 17, Tower 19, Tower 20, Tower 24, Tower 28, Tower 29. + - Area 16 can be covered by Towers Tower 1, Tower 2, Tower 7, Tower 9, Tower 15, Tower 18, Tower 20, Tower 21, Tower 25, Tower 26, Tower 27, Tower 28. + - Area 17 can be covered by Towers Tower 4, Tower 5, Tower 6, Tower 8, Tower 13, Tower 15, Tower 16, Tower 17, Tower 18, Tower 19, Tower 21, Tower 24, Tower 28, Tower 29. + - Area 18 can be covered by Towers Tower 0, Tower 2, Tower 3, Tower 7, Tower 9, Tower 10, Tower 11, Tower 12, Tower 14, Tower 16, Tower 19, Tower 20, Tower 21, Tower 24, Tower 26. + - Area 19 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 7, Tower 8, Tower 10, Tower 16, Tower 20, Tower 22, Tower 23, Tower 24, Tower 26, Tower 27, Tower 29. + - Area 20 can be covered by Towers Tower 3, Tower 4, Tower 6, Tower 7, Tower 13, Tower 15, Tower 17, Tower 18, Tower 19, Tower 22, Tower 23, Tower 24, Tower 27, Tower 28, Tower 29. + - Area 21 can be covered by Towers Tower 1, Tower 3, Tower 10, Tower 12, Tower 13, Tower 14, Tower 20, Tower 21, Tower 23, Tower 25, Tower 26, Tower 29. + - Area 22 can be covered by Towers Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 13, Tower 14, Tower 17, Tower 25, Tower 28. + - Area 23 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 14, Tower 17, Tower 20, Tower 25, Tower 26, Tower 27, Tower 28. + - Area 24 can be covered by Towers Tower 0, Tower 1, Tower 5, Tower 6, Tower 12, Tower 13, Tower 15, Tower 21, Tower 24, Tower 27, Tower 29. + - Area 25 can be covered by Towers Tower 4, Tower 6, Tower 9, Tower 13, Tower 14, Tower 15, Tower 17, Tower 21, Tower 22, Tower 25, Tower 28, Tower 29. + - Area 26 can be covered by Towers Tower 1, Tower 6, Tower 10, Tower 15, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 26, Tower 27, Tower 28. + - Area 27 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 5, Tower 7, Tower 12, Tower 16, Tower 19, Tower 21, Tower 23, Tower 26, Tower 27, Tower 28, Tower 29. + - Area 28 can be covered by Towers Tower 1, Tower 2, Tower 6, Tower 7, Tower 8, Tower 10, Tower 13, Tower 16, Tower 18, Tower 19, Tower 21, Tower 22, Tower 27, Tower 28. + - Area 29 can be covered by Towers Tower 0, Tower 2, Tower 3, Tower 4, Tower 7, Tower 8, Tower 9, Tower 10, Tower 12, Tower 14, Tower 20, Tower 21, Tower 22, Tower 28, Tower 29. + - Area 30 can be covered by Towers Tower 2, Tower 4, Tower 6, Tower 7, Tower 8, Tower 9, Tower 11, Tower 16, Tower 17, Tower 24. + - Area 31 can be covered by Towers Tower 0, Tower 9, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 25, Tower 26, Tower 27. + - Area 32 can be covered by Towers Tower 3, Tower 4, Tower 6, Tower 8, Tower 9, Tower 10, Tower 17, Tower 18, Tower 20, Tower 23, Tower 25, Tower 26, Tower 27, Tower 28, Tower 29. + - Area 33 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 4, Tower 5, Tower 8, Tower 10, Tower 13, Tower 15, Tower 20, Tower 22, Tower 26, Tower 27, Tower 28. + + 2. **Budget Constraint**: The total cost of building the selected towers must not exceed the available budget of 675 units. The construction costs for the towers are: + - Tower 0: 248 units + - Tower 1: 226 units + - Tower 2: 454 units + - Tower 3: 466 units + - Tower 4: 241 units + - Tower 5: 201 units + - Tower 6: 415 units + - Tower 7: 486 units + - Tower 8: 485 units + - Tower 9: 492 units + - Tower 10: 249 units + - Tower 11: 486 units + - Tower 12: 417 units + - Tower 13: 327 units + - Tower 14: 257 units + - Tower 15: 486 units + - Tower 16: 438 units + - Tower 17: 266 units + - Tower 18: 473 units + - Tower 19: 268 units + - Tower 20: 293 units + - Tower 21: 379 units + - Tower 22: 476 units + - Tower 23: 482 units + - Tower 24: 367 units + - Tower 25: 239 units + - Tower 26: 368 units + - Tower 27: 492 units + - Tower 28: 454 units + - Tower 29: 319 units + + 3. **Binary Decisions**: Each tower can either be built (1) or not built (0), and each area can either be covered (1) or not covered (0). + + #### Summary: + Select which towers to build and which areas to cover to maximize the total population served, ensuring coverage requirements are met and the total construction cost does not exceed the budget. The solution must account for the specific coverage capabilities of each tower and the population sizes of the areas.107698Cell Tower Deployment for Regional Coverage + + You are tasked with deploying cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 34 distinct areas, each with a specific population size that needs to be covered by the cell towers. Your goal is to determine which towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + + #### Key Decisions: + 1. **Tower Deployment**: Decide whether to build each of the 30 available cell towers. Each tower has a unique construction cost, and building it will enable coverage for certain areas. + 2. **Area Coverage**: Determine which of the 34 areas will be covered by the deployed towers. An area is considered covered if at least one of the towers that can serve it is built. + + #### Objective: + Maximize the total population covered by the deployed cell towers. The population sizes for the 34 areas are as follows: +- Area 0: 3,328 people +- Area 1: 2,066 people +- Area 2: 2,261 people +- Area 32: 3,061 people +- Area 33: 4,310 people + + #### Constraints: + 1. **Coverage Requirements**: Each area must be covered by at least one of the towers that can serve it. For example: + - Area 0 can be covered by Towers 0, 3, 4, 7, 9, 16, 17, 20, 22, 24, 25, 26, or 27. + - Area 1 can be covered by Towers 0, 1, 15, 19, 22, 23, 25, 26, 27, or 29. + - Similar Full coverage relationships exist for all 34 areas, as detailed in the input data. areas: + - Area 0 can be covered by Towers Tower 0, Tower 3, Tower 4, Tower 7, Tower 9, Tower 16, Tower 17, Tower 20, Tower 22, Tower 24, Tower 25, Tower 26, Tower 27. + - Area 1 can be covered by Towers Tower 0, Tower 1, Tower 15, Tower 19, Tower 22, Tower 23, Tower 25, Tower 26, Tower 27, Tower 29. + - Area 2 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 7, Tower 8, Tower 13, Tower 17, Tower 18, Tower 21, Tower 23, Tower 25, Tower 28, Tower 29. + - Area 3 can be covered by Towers Tower 0, Tower 6, Tower 7, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 24, Tower 29. + - Area 4 can be covered by Towers Tower 0, Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 10, Tower 13, Tower 14, Tower 18, Tower 22, Tower 25, Tower 26. + - Area 5 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 8, Tower 11, Tower 12, Tower 14, Tower 17, Tower 19, Tower 25, Tower 26, Tower 29. + - Area 6 can be covered by Towers Tower 1, Tower 2, Tower 6, Tower 7, Tower 9, Tower 11, Tower 17, Tower 18, Tower 19, Tower 20, Tower 26, Tower 27, Tower 29. + - Area 7 can be covered by Towers Tower 3, Tower 5, Tower 7, Tower 8, Tower 11, Tower 12, Tower 14, Tower 20, Tower 27, Tower 29. + - Area 8 can be covered by Towers Tower 2, Tower 5, Tower 6, Tower 8, Tower 11, Tower 17, Tower 19, Tower 20, Tower 21, Tower 22, Tower 23, Tower 27. + - Area 9 can be covered by Towers Tower 1, Tower 5, Tower 7, Tower 8, Tower 10, Tower 12, Tower 14, Tower 17, Tower 19, Tower 20, Tower 21, Tower 22, Tower 24, Tower 26, Tower 29. + - Area 10 can be covered by Towers Tower 2, Tower 4, Tower 6, Tower 10, Tower 12, Tower 14, Tower 15, Tower 18, Tower 20, Tower 22, Tower 23, Tower 28. + - Area 11 can be covered by Towers Tower 4, Tower 7, Tower 8, Tower 11, Tower 12, Tower 13, Tower 17, Tower 18, Tower 22, Tower 23, Tower 26, Tower 27. + - Area 12 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 4, Tower 5, Tower 13, Tower 15, Tower 16, Tower 20, Tower 24, Tower 29. + - Area 13 can be covered by Towers Tower 0, Tower 2, Tower 3, Tower 8, Tower 10, Tower 12, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 23, Tower 28. + - Area 14 can be covered by Towers Tower 0, Tower 3, Tower 5, Tower 8, Tower 9, Tower 13, Tower 14, Tower 16, Tower 22, Tower 23, Tower 25, Tower 28. + - Area 15 can be covered by Towers Tower 0, Tower 3, Tower 4, Tower 5, Tower 6, Tower 10, Tower 11, Tower 15, Tower 16, Tower 17, Tower 19, Tower 20, Tower 24, Tower 28, Tower 29. + - Area 16 can be covered by Towers Tower 1, Tower 2, Tower 7, Tower 9, Tower 15, Tower 18, Tower 20, Tower 21, Tower 25, Tower 26, Tower 27, Tower 28. + - Area 17 can be covered by Towers Tower 4, Tower 5, Tower 6, Tower 8, Tower 13, Tower 15, Tower 16, Tower 17, Tower 18, Tower 19, Tower 21, Tower 24, Tower 28, Tower 29. + - Area 18 can be covered by Towers Tower 0, Tower 2, Tower 3, Tower 7, Tower 9, Tower 10, Tower 11, Tower 12, Tower 14, Tower 16, Tower 19, Tower 20, Tower 21, Tower 24, Tower 26. + - Area 19 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 7, Tower 8, Tower 10, Tower 16, Tower 20, Tower 22, Tower 23, Tower 24, Tower 26, Tower 27, Tower 29. + - Area 20 can be covered by Towers Tower 3, Tower 4, Tower 6, Tower 7, Tower 13, Tower 15, Tower 17, Tower 18, Tower 19, Tower 22, Tower 23, Tower 24, Tower 27, Tower 28, Tower 29. + - Area 21 can be covered by Towers Tower 1, Tower 3, Tower 10, Tower 12, Tower 13, Tower 14, Tower 20, Tower 21, Tower 23, Tower 25, Tower 26, Tower 29. + - Area 22 can be covered by Towers Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 13, Tower 14, Tower 17, Tower 25, Tower 28. + - Area 23 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 14, Tower 17, Tower 20, Tower 25, Tower 26, Tower 27, Tower 28. + - Area 24 can be covered by Towers Tower 0, Tower 1, Tower 5, Tower 6, Tower 12, Tower 13, Tower 15, Tower 21, Tower 24, Tower 27, Tower 29. + - Area 25 can be covered by Towers Tower 4, Tower 6, Tower 9, Tower 13, Tower 14, Tower 15, Tower 17, Tower 21, Tower 22, Tower 25, Tower 28, Tower 29. + - Area 26 can be covered by Towers Tower 1, Tower 6, Tower 10, Tower 15, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 26, Tower 27, Tower 28. + - Area 27 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 5, Tower 7, Tower 12, Tower 16, Tower 19, Tower 21, Tower 23, Tower 26, Tower 27, Tower 28, Tower 29. + - Area 28 can be covered by Towers Tower 1, Tower 2, Tower 6, Tower 7, Tower 8, Tower 10, Tower 13, Tower 16, Tower 18, Tower 19, Tower 21, Tower 22, Tower 27, Tower 28. + - Area 29 can be covered by Towers Tower 0, Tower 2, Tower 3, Tower 4, Tower 7, Tower 8, Tower 9, Tower 10, Tower 12, Tower 14, Tower 20, Tower 21, Tower 22, Tower 28, Tower 29. + - Area 30 can be covered by Towers Tower 2, Tower 4, Tower 6, Tower 7, Tower 8, Tower 9, Tower 11, Tower 16, Tower 17, Tower 24. + - Area 31 can be covered by Towers Tower 0, Tower 9, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 25, Tower 26, Tower 27. + - Area 32 can be covered by Towers Tower 3, Tower 4, Tower 6, Tower 8, Tower 9, Tower 10, Tower 17, Tower 18, Tower 20, Tower 23, Tower 25, Tower 26, Tower 27, Tower 28, Tower 29. + - Area 33 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 4, Tower 5, Tower 8, Tower 10, Tower 13, Tower 15, Tower 20, Tower 22, Tower 26, Tower 27, Tower 28. + + 2. **Budget Constraint**: The total cost of building the selected towers must not exceed the available budget of 3,375 675 units. The construction costs for the towers are: + - Tower 0: 248 units + - Tower 1: 226 units + - Tower 2: 454 units + - Tower 28: 454 units + - Tower 29: 319 units + + 3. **Binary Decisions**: Each tower can either be built (1) or not built (0), and each area can either be covered (1) or not covered (0). + + #### Summary: + Your task is to select Select which towers to build and which areas to cover to maximize the total population served, ensuring that all coverage requirements are met and the total construction cost does not exceed the budget. The solution must account for the specific coverage capabilities of each tower and the population sizes of the areas.124268 107698Missing data about which towers cover which regionsmissing data
68Minimize the total operational costs, which include production setup costs, machine startup costs, inventory holding costs, and backlogging costs, while meeting customer demand for each product in each period. The decisions involve determining whether to produce each product on each machine in each period, deciding whether to start up each machine for each product in each period, determining the exact quantity of each product to produce on each machine in each period, and deciding how much inventory to carry over or how much demand to backlog for each product in each period. The total cost includes production setup costs of 149.1695690337089 per production run, machine startup costs of 69.24269269143792 per machine startup, inventory holding costs of 1.080318154052025 for Product 0 and 1.143069947055348 for Product 1, and backlogging costs of 6.798895023015501 for Product 0 and 6.372651501311341 for Product 1. The constraints ensure that the total production of each product across all machines, plus any inventory carried over from the previous period, minus any backlog from the previous period, equals the demand for that product in the current period, plus any inventory carried forward, minus any backlogged demand. The demands are 22.92965094851524 units for Product 0 in Period 0, 19.03396834281949 units in Period 1, and 18.90167422155575 units in Period 2, and 19.80185049682801 units for Product 1 in Period 0, 24.22810173621122 units in Period 1, and 16.84556169689551 units in Period 2. The production quantity on each machine, plus the startup time, cannot exceed the machine's capacity, which is 76.7045484622153 for Machine 0 and 69.3695364951939 for Machine 1. Each machine can only produce one product at a time in any given period. If a machine is used for production in a period but was not used in the previous period, a startup cost is incurred. Production quantities, inventory levels, and backlogs must be non-negative, and production and startup decisions must be binary.564.4326342045792Minimize the total operational costs, which include production setup costs, machine startup costs, inventory holding costs, and backlogging costs, while meeting customer demand for each product in each period. The decisions involve determining whether to produce each product on each machine in each period, deciding whether to start up each machine for each product in each period, determining the exact quantity of each product to produce on each machine in each period, and deciding how much inventory to carry over or how much demand to backlog for each product in each period. The total cost includes production setup costs of 149.1695690337089 per production run, machine startup costs of 69.24269269143792 per machine startup, inventory holding costs of 1.080318154052025 for Product 0 and 1.143069947055348 for Product 1, and backlogging costs of 6.798895023015501 for Product 0 and 6.372651501311341 for Product 1. The constraints ensure that the total production of each product across all machines, plus any inventory carried over from the previous period, minus any backlog from the previous period, equals the demand for that product in the current period, plus any inventory carried forward, minus any backlogged demand. The demands are 22.92965094851524 units for Product 0 in Period 0, 19.03396834281949 units in Period 1, and 18.90167422155575 units in Period 2, and 19.80185049682801 units for Product 1 in Period 0, 24.22810173621122 units in Period 1, and 16.84556169689551 units in Period 2. The production quantity on each machine, plus the startup time, cannot exceed the machine's capacity, which is 76.7045484622153 for Machine 0 and 69.3695364951939 for Machine 1. Each machine can only produce one product at a time in any given period. If a machine is used for production in a period but was not used in the previous period, a startup cost is incurred. Production quantities, inventory levels, and backlogs must be non-negative, and production and startup decisions must be binary.564.4326342045792No differencesNo differencesok
69Minimize the total cost, which includes fixed costs for establishing selected transit lines, operational costs based on the frequency of each line, and penalty costs for any unmet passenger demand. The fixed costs for establishing lines L_0 to L_13 are 2048, 2064, 2058, 3000, 2107, 2583, 2569, 2081, 2887, 2900, 2922, 2106, 2140, and 2227 units, respectively. The operational costs per unit frequency for lines L_0 to L_13 are 242, 236, 200, 300, 286, 249, 230, 260, 293, 206, 296, 240, 217, and 295 units, respectively. The penalty costs for unmet demand for OD pairs OD_5_9, OD_10_2, OD_2_7, OD_7_12, OD_2_0, OD_7_5, OD_3_11, OD_0_3, OD_7_9, and OD_7_10 are 709, 755, 739, 722, 725, 796, 753, 709, 732, and 723 units, respectively. + +For each OD pair, the total capacity provided by the selected lines must meet or exceed the demand. If not, the unmet demand incurs a penalty. For OD pair OD_5_9, the combined capacity of lines L_1, L_2, L_5, L_7, L_8, L_9, L_11, L_12, and L_13 must be at least 24 passengers. For OD pair OD_10_2, the combined capacity of lines L_0 and L_5 must be at least 24 passengers. For OD pair OD_2_7, the combined capacity of lines L_0, L_4, L_5, L_7, L_9, L_10, L_12, and L_13 must be at least 23 passengers. For OD pair OD_7_12, the combined capacity of lines L_5, L_8, and L_11 must be at least 28 passengers. For OD pair OD_2_0, the combined capacity of lines L_4, L_6, L_8, L_12, and L_13 must be at least 25 passengers. For OD pair OD_7_5, the combined capacity of lines L_3 and L_11 must be at least 27 passengers. For OD pair OD_3_11, the combined capacity of lines L_3, L_4, L_7, L_8, and L_10 must be at least 20 passengers. For OD pair OD_0_3, the combined capacity of lines L_9 and L_12 must be at least 21 passengers. For OD pair OD_7_9, the combined capacity of lines L_0, L_6, L_11, and L_12 must be at least 22 passengers. For OD pair OD_7_10, the combined capacity of lines L_0, L_2, L_5, L_8, L_10, and L_11 must be at least 24 passengers. + +Each line’s frequency must be within a specified range if the line is selected. For lines L_0 to L_13, the frequency must be between 2 and 10 trips per unit time if the line is selected. The total vehicle usage across all lines cannot exceed the available fleet size of 70 vehicles. Vehicle usage is calculated based on the round-trip time and frequency of each line, with the total usage for lines L_0 to L_13 being 70, 71, 67, 78, 104, 75, 110, 96, 120, 65, 99, 61, 91, and 65 units, respectively. + +A station can only be designated as a transfer station if at least two transit lines pass through it. For station 0, lines L_2, L_4, L_8, or L_13 must be selected. For station 1, at least two of lines L_0, L_1, L_2, L_3, L_4, L_10, or L_12 must be selected. For station 2, lines L_5, L_12, or L_13 must be selected. For station 3, lines L_2, L_7, or L_13 must be selected. For station 4, lines L_4 or L_11 must be selected. For station 5, at least two of lines L_1, L_3, L_8, L_9, or L_11 must be selected. For station 6, line L_0 must be selected. For station 7, at least two of lines L_1, L_5, L_6, or L_13 must be selected. For station 8, at least two of lines L_1, L_2, L_5, or L_8 must be selected. For station 9, at least two of lines L_0, L_3, L_4, L_10, L_11, or L_13 must be selected. For station 10, at least two of lines L_2, L_7, L_10, or L_13 must be selected. For station 11, at least two of lines L_2, L_9, L_10, or L_11 must be selected. For station 12, at least two of lines L_0, L_2, L_3, L_6, or L_11 must be selected. + +The total frequency of lines serving each OD pair must meet a minimum threshold to ensure adequate service. For example, the combined frequency of lines serving OD_5_9 must be at least 2 trips per unit time. If a line is selected, it must use a continuous path of arcs (connections between stations) in the network, ensuring that each line operates as a coherent route.1753710Minimize the total cost, which includes fixed costs for establishing selected transit lines, operational costs based on the frequency of each line, and penalty costs for any unmet passenger demand. The fixed costs for establishing lines L_0 to L_13 are 2048, 2064, 2058, 3000, 2107, 2583, 2569, 2081, 2887, 2900, 2922, 2106, 2140, and 2227 units, respectively. The operational costs per unit frequency for lines L_0 to L_13 are 242, 236, 200, 300, 286, 249, 230, 260, 293, 206, 296, 240, 217, and 295 units, respectively. The penalty costs for unmet demand for OD pairs OD_5_9, OD_10_2, OD_2_7, OD_7_12, OD_2_0, OD_7_5, OD_3_11, OD_0_3, OD_7_9, and OD_7_10 are 709, 755, 739, 722, 725, 796, 753, 709, 732, and 723 units, respectively. + + For each OD pair, the total capacity provided by the selected lines must meet or exceed the demand. If not, the unmet demand incurs a penalty. For OD pair OD_5_9, the combined capacity of lines L_1, L_2, L_5, L_7, L_8, L_9, L_11, L_12, and L_13 must be at least 24 passengers. For OD pair OD_10_2, the combined capacity of lines L_0 and L_5 must be at least 24 passengers. For OD pair OD_2_7, the combined capacity of lines L_0, L_4, L_5, L_7, L_9, L_10, L_12, and L_13 must be at least 23 passengers. For OD pair OD_7_12, the combined capacity of lines L_5, L_8, and L_11 must be at least 28 passengers. For OD pair OD_2_0, the combined capacity of lines L_4, L_6, L_8, L_12, and L_13 must be at least 25 passengers. For OD pair OD_7_5, the combined capacity of lines L_3 and L_11 must be at least 27 passengers. For OD pair OD_3_11, the combined capacity of lines L_3, L_4, L_7, L_8, and L_10 must be at least 20 passengers. For OD pair OD_0_3, the combined capacity of lines L_9 and L_12 must be at least 21 passengers. For OD pair OD_7_9, the combined capacity of lines L_0, L_6, L_11, and L_12 must be at least 22 passengers. For OD pair OD_7_10, the combined capacity of lines L_0, L_2, L_5, L_8, L_10, and L_11 must be at least 24 passengers. + + Each line’s frequency must be within a specified range if the line is selected. For lines L_0 to L_13, the frequency must be between 2 and 10 trips per unit time if the line is selected. The total vehicle usage across all lines cannot exceed the available fleet size of 70 vehicles. Vehicle usage is calculated based on the round-trip time and frequency of each line, with the total usage for lines L_0 to L_13 being 70, 71, 67, 78, 104, 75, 110, 96, 120, 65, 99, 61, 91, and 65 units, respectively. + + A station can only be designated as a transfer station if at least two transit lines pass through it. For station 0, lines L_2, L_4, L_8, or L_13 must be selected. For station 1, at least two of lines L_0, L_1, L_2, L_3, L_4, L_10, or L_12 must be selected. For station 2, lines L_5, L_12, or L_13 must be selected. For station 3, lines L_2, L_7, or L_13 must be selected. For station 4, lines L_4 or L_11 must be selected. For station 5, at least two of lines L_1, L_3, L_8, L_9, or L_11 must be selected. For station 6, line L_0 must be selected. For station 7, at least two of lines L_1, L_5, L_6, or L_13 must be selected. For station 8, at least two of lines L_1, L_2, L_5, or L_8 must be selected. For station 9, at least two of lines L_0, L_3, L_4, L_10, L_11, or L_13 must be selected. For station 10, at least two of lines L_2, L_7, L_10, or L_13 must be selected. For station 11, at least two of lines L_2, L_9, L_10, or L_11 must be selected. For station 12, at least two of lines L_0, L_2, L_3, L_6, or L_11 must be selected. + + The total frequency of lines serving each OD pair must meet a minimum threshold to ensure adequate service. For example, the combined frequency of lines serving OD_5_9 must be at least 2 trips per unit time. If a line is selected, it must use a continuous path of arcs (connections between stations) in the network, ensuring that each line operates as a coherent route.175371Very underspecified problem.underspecified
70You are tasked with optimizing the placement of facilities across a network of five locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall dispersion between the selected facilities. Dispersion is measured by the weighted distances between pairs of facilities, where the weights represent the importance or benefit of separating specific pairs. Your goal is to select exactly three facilities out of the five possible locations while ensuring that the total dispersion benefit is maximized. + +#### Key Decisions: +- **Facility Selection:** Decide which three out of the five locations will host the facilities. Each location can either be selected (1) or not selected (0). +- **Pairwise Dispersion Activation:** Determine whether the dispersion benefit between a pair of selected facilities is activated. This is represented by a binary decision (1 if both facilities in the pair are selected, 0 otherwise). + +#### Objective: +Maximize the total dispersion benefit across all pairs of selected facilities. The dispersion benefit between each pair of locations is as follows: +- Node 0 and Node 1: 17 units +- Node 0 and Node 2: 18 units +- Node 0 and Node 3: 21 units +- Node 0 and Node 4: 22 units +- Node 1 and Node 0: 22 units +- Node 1 and Node 2: 15 units +- Node 1 and Node 3: 21 units +- Node 1 and Node 4: 21 units +- Node 2 and Node 0: 16 units +- Node 2 and Node 1: 21 units +- Node 2 and Node 3: 23 units +- Node 2 and Node 4: 17 units +- Node 3 and Node 0: 21 units +- Node 3 and Node 1: 22 units +- Node 3 and Node 2: 19 units +- Node 3 and Node 4: 24 units +- Node 4 and Node 0: 24 units +- Node 4 and Node 1: 16 units +- Node 4 and Node 2: 15 units +- Node 4 and Node 3: 23 units + +#### Constraints: +1. **Facility Selection Limit:** Exactly three facilities must be selected out of the five locations. This means the sum of selected facilities must equal three. +2. **Pairwise Dispersion Activation Rules:** + - The dispersion benefit between two locations can only be activated if both locations are selected. + - If either location in a pair is not selected, the dispersion benefit for that pair cannot be activated. + - If both locations in a pair are selected, the dispersion benefit must be activated. + +#### Numerical Parameters: +- The dispersion benefits between each pair of locations are explicitly provided in the objective description above. +- The total number of facilities to be selected is fixed at three. + +#### Practical Implications: +This problem ensures that the selected facilities are spread out in a way that maximizes their overall separation benefit, which could be critical in scenarios such as minimizing risk, optimizing coverage, or ensuring redundancy. The constraints ensure that the solution is both feasible and aligned with the operational requirements of selecting exactly three facilities. + +By solving this problem, you will determine the optimal set of three facilities and the corresponding pairwise dispersion benefits that maximize the total separation value across the network.135You are tasked with optimizing the placement of facilities across a network of five locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall dispersion between the selected facilities. Dispersion is measured by the weighted distances between pairs of facilities, where the weights represent the importance or benefit of separating specific pairs. Your goal is to select exactly three facilities out of the five possible locations while ensuring that the total dispersion benefit is maximized. + +#### Key Decisions: +- **Facility Selection:** Decide which three out of the five locations will host the facilities. Each location can either be selected (1) or not selected (0). +- **Pairwise Dispersion Activation:** Determine whether the dispersion benefit between a pair of selected facilities is activated. This is represented by a binary decision (1 if both facilities in the pair are selected, 0 otherwise). + +#### Objective: +Maximize the total dispersion benefit across all pairs of selected facilities. The dispersion benefit between each pair of locations is as follows: +- Node 0 and Node 1: 17 units +- Node 0 and Node 2: 18 units +- Node 0 and Node 3: 21 units +- Node 0 and Node 4: 22 units +- Node 1 and Node 0: 22 units +- Node 1 and Node 2: 15 units +- Node 1 and Node 3: 21 units +- Node 1 and Node 4: 21 units +- Node 2 and Node 0: 16 units +- Node 2 and Node 1: 21 units +- Node 2 and Node 3: 23 units +- Node 2 and Node 4: 17 units +- Node 3 and Node 0: 21 units +- Node 3 and Node 1: 22 units +- Node 3 and Node 2: 19 units +- Node 3 and Node 4: 24 units +- Node 4 and Node 0: 24 units +- Node 4 and Node 1: 16 units +- Node 4 and Node 2: 15 units +- Node 4 and Node 3: 23 units + +#### Constraints: +1. **Facility Selection Limit:** Exactly three facilities must be selected out of the five locations. This means the sum of selected facilities must equal three. +2. **Pairwise Dispersion Activation Rules:** + - The dispersion benefit between two locations can only be activated if both locations are selected. + - If either location in a pair is not selected, the dispersion benefit for that pair cannot be activated. + - If both locations in a pair are selected, the dispersion benefit must be activated. + +#### Numerical Parameters: +- The dispersion benefits between each pair of locations are explicitly provided in the objective description above. +- The total number of facilities to be selected is fixed at three. + +#### Practical Implications: +This problem ensures that the selected facilities are spread out in a way that maximizes their overall separation benefit, which could be critical in scenarios such as minimizing risk, optimizing coverage, or ensuring redundancy. The constraints ensure that the solution is both feasible and aligned with the operational requirements of selecting exactly three facilities. + +By solving this problem, you will determine the optimal set of three facilities and the corresponding pairwise dispersion benefits that maximize the total separation value across the network.135No differencesNo differencesok
71Inventory Management with Backlogging in a Retail Supply Chain + +You are managing the inventory and ordering process for a retail supply chain over a 19-week period. Your goal is to minimize the total costs associated with ordering, holding inventory, and backlogging unmet demand, while ensuring that customer demand is met as efficiently as possible. The costs and constraints are as follows: + +#### **Cost Components:** +1. **Ordering Costs**: Each time an order is placed in a given week, a fixed cost is incurred. These costs vary by week and are as follows: + - Week 1: \$930, Week 2: \$881, Week 3: \$925, Week 4: \$808, Week 5: \$711, Week 6: \$785, Week 7: \$997, Week 8: \$820, Week 9: \$843, Week 10: \$631, Week 11: \$726, Week 12: \$1,071, Week 13: \$741, Week 14: \$1,157, Week 15: \$916, Week 16: \$791, Week 17: \$1,055, Week 18: \$932, Week 19: \$794. + +2. **Variable Ordering Costs**: In addition to the fixed costs, there is a variable cost per unit ordered in each week. These costs are: + - Week 1: \$40, Week 2: \$26, Week 3: \$38, Week 4: \$20, Week 5: \$25, Week 6: \$22, Week 7: \$40, Week 8: \$39, Week 9: \$25, Week 10: \$26, Week 11: \$34, Week 12: \$25, Week 13: \$26, Week 14: \$27, Week 15: \$27, Week 16: \$35, Week 17: \$23, Week 18: \$26, Week 19: \$36. + +3. **Inventory Holding Costs**: Holding inventory at the end of each week incurs a cost per unit. These costs are: + - Week 1: \$19, Week 2: \$16, Week 3: \$24, Week 4: \$23, Week 5: \$24, Week 6: \$23, Week 7: \$20, Week 8: \$21, Week 9: \$21, Week 10: \$25, Week 11: \$15, Week 12: \$20, Week 13: \$24, Week 14: \$16, Week 15: \$16, Week 16: \$16, Week 17: \$19, Week 18: \$15, Week 19: \$20. + +4. **Backlogging Costs**: If demand cannot be met in a given week, the unmet demand is backlogged to the next week, incurring a cost per backlogged unit. These costs are: + - Week 1: \$55, Week 2: \$54, Week 3: \$71, Week 4: \$58, Week 5: \$51, Week 6: \$55, Week 7: \$45, Week 8: \$52, Week 9: \$60, Week 10: \$52, Week 11: \$52, Week 12: \$52, Week 13: \$76, Week 14: \$50, Week 15: \$47, Week 16: \$66, Week 17: \$70, Week 18: \$48, Week 19: \$69. + +#### **Demand Constraints:** +- The demand for each week must be met either by the inventory on hand, the units ordered in that week, or by backlogging to the next week. The demand for each week is as follows: + - Week 1: 495 units, Week 2: 316 units, Week 3: 375 units, Week 4: 467 units, Week 5: 418 units, Week 6: 398 units, Week 7: 533 units, Week 8: 525 units, Week 9: 361 units, Week 10: 452 units, Week 11: 470 units, Week 12: 378 units, Week 13: 302 units, Week 14: 463 units, Week 15: 438 units, Week 16: 383 units, Week 17: 377 units, Week 18: 429 units, Week 19: 324 units. + +#### **Inventory and Backlogging Rules:** +- The inventory at the start of Week 1 and the end of Week 19 must be zero. +- The backlogged demand at the start of Week 1 and the end of Week 19 must also be zero. +- The amount ordered in any week cannot exceed the total demand over the 19-week period (7,904 units) if an order is placed in that week. + +#### **Objective:** +Your task is to determine the optimal ordering schedule, inventory levels, and backlogged amounts for each week to minimize the total cost, which includes fixed ordering costs, variable ordering costs, inventory holding costs, and backlogging costs. + +#### **Key Decisions:** +1. **Order Placement**: Decide whether to place an order in each week (a binary decision: yes or no). +2. **Order Quantity**: Determine the number of units to order in each week. +3. **Inventory Levels**: Decide how much inventory to hold at the end of each week. +4. **Backlogged Demand**: Determine the amount of unmet demand to backlog to the next week. + +By carefully balancing these decisions, you aim to minimize the total cost while ensuring that all customer demand is met either immediately or through backlogging.252401Inventory Management with Backlogging in a Retail Supply Chain + +You are managing the inventory and ordering process for a retail supply chain over a 19-week period. Your goal is to minimize the total costs associated with ordering, holding inventory, and backlogging unmet demand, while ensuring that customer demand is met as efficiently as possible. The costs and constraints are as follows: + +#### **Cost Components:** +1. **Ordering Costs**: Each time an order is placed in a given week, a fixed cost is incurred. These costs vary by week and are as follows: + - Week 1: \$930, Week 2: \$881, Week 3: \$925, Week 4: \$808, Week 5: \$711, Week 6: \$785, Week 7: \$997, Week 8: \$820, Week 9: \$843, Week 10: \$631, Week 11: \$726, Week 12: \$1,071, Week 13: \$741, Week 14: \$1,157, Week 15: \$916, Week 16: \$791, Week 17: \$1,055, Week 18: \$932, Week 19: \$794. + +2. **Variable Ordering Costs**: In addition to the fixed costs, there is a variable cost per unit ordered in each week. These costs are: + - Week 1: \$40, Week 2: \$26, Week 3: \$38, Week 4: \$20, Week 5: \$25, Week 6: \$22, Week 7: \$40, Week 8: \$39, Week 9: \$25, Week 10: \$26, Week 11: \$34, Week 12: \$25, Week 13: \$26, Week 14: \$27, Week 15: \$27, Week 16: \$35, Week 17: \$23, Week 18: \$26, Week 19: \$36. + +3. **Inventory Holding Costs**: Holding inventory at the end of each week incurs a cost per unit. These costs are: + - Week 1: \$19, Week 2: \$16, Week 3: \$24, Week 4: \$23, Week 5: \$24, Week 6: \$23, Week 7: \$20, Week 8: \$21, Week 9: \$21, Week 10: \$25, Week 11: \$15, Week 12: \$20, Week 13: \$24, Week 14: \$16, Week 15: \$16, Week 16: \$16, Week 17: \$19, Week 18: \$15, Week 19: \$20. + +4. **Backlogging Costs**: If demand cannot be met in a given week, the unmet demand is backlogged to the next week, incurring a cost per backlogged unit. These costs are: + - Week 1: \$55, Week 2: \$54, Week 3: \$71, Week 4: \$58, Week 5: \$51, Week 6: \$55, Week 7: \$45, Week 8: \$52, Week 9: \$60, Week 10: \$52, Week 11: \$52, Week 12: \$52, Week 13: \$76, Week 14: \$50, Week 15: \$47, Week 16: \$66, Week 17: \$70, Week 18: \$48, Week 19: \$69. + +#### **Demand Constraints:** +- The demand for each week must be met either by the inventory on hand, the units ordered in that week, or by backlogging to the next week. The demand for each week is as follows: + - Week 1: 495 units, Week 2: 316 units, Week 3: 375 units, Week 4: 467 units, Week 5: 418 units, Week 6: 398 units, Week 7: 533 units, Week 8: 525 units, Week 9: 361 units, Week 10: 452 units, Week 11: 470 units, Week 12: 378 units, Week 13: 302 units, Week 14: 463 units, Week 15: 438 units, Week 16: 383 units, Week 17: 377 units, Week 18: 429 units, Week 19: 324 units. + +#### **Inventory and Backlogging Rules:** +- The inventory at the start of Week 1 and the end of Week 19 must be zero. +- The backlogged demand at the start of Week 1 and the end of Week 19 must also be zero. +- The amount ordered in any week cannot exceed the total demand over the 19-week period (7,904 units) if an order is placed in that week. + +#### **Objective:** +Your task is to determine the optimal ordering schedule, inventory levels, and backlogged amounts for each week to minimize the total cost, which includes fixed ordering costs, variable ordering costs, inventory holding costs, and backlogging costs. + +#### **Key Decisions:** +1. **Order Placement**: Decide whether to place an order in each week (a binary decision: yes or no). +2. **Order Quantity**: Determine the number of units to order in each week. +3. **Inventory Levels**: Decide how much inventory to hold at the end of each week. +4. **Backlogged Demand**: Determine the amount of unmet demand to backlog to the next week. + +By carefully balancing these decisions, you aim to minimize the total cost while ensuring that all customer demand is met either immediately or through backlogging.252401No differencesNo differencesok
72The problem involves assigning employees to shifts across two restaurants to minimize costs associated with unfulfilled positions and employee assignments. There are two restaurants, each with two shifts per day, and eight employees available for assignment. Each shift at each restaurant has specific staffing demands for a particular skill, and employees have varying availability and skill sets. Employees can only be assigned to shifts they are available for and to roles that match their skills. Each employee can be assigned to at most one shift across all restaurants. Assigning employees to shifts incurs specific costs: assigning Employee 0 to Skill 0 costs 1 unit, assigning Employee 3 to Skill 0 costs 2 units, and assigning Employee 4 to Skill 0 costs 3 units. If a restaurant cannot meet the demand for a specific skill in a shift, unfulfilled positions incur a penalty cost of 100 units per position. The goal is to minimize the total cost, which includes the penalty costs for unfulfilled positions and the costs of assigning employees to shifts. Constraints ensure that the total number of employees assigned to a shift at a restaurant, plus any unfulfilled positions, equals the demand for that shift. Employees can only be assigned to shifts they are available for and to roles that match their skills. Each employee can be assigned to at most one shift across all restaurants.16The problem involves assigning employees to shifts across two restaurants to minimize costs associated with unfulfilled positions and employee assignments. There are two restaurants, each with two shifts per day, and eight employees available for assignment. Each shift at each restaurant has specific staffing demands for a particular skill, and employees have varying availability and skill sets. + +Restaurant 0, Shift 0, Skill 0 requires 1 employee. Restaurant 0, Shift 0, Skill 1 requires 1 employee. Restaurant 0, Shift 1, Skill 0 requires 1 employee. Restaurant 1, Shift 0, Skill 0 requires 2 employees. Restaurant 1, Shift 1, Skill 1 requires 1 employee. + +Employees can only be assigned to shifts they are available for and to roles that match their skills. Employee 0 is available for Shift 0 but not for Shift 1. Employee 1 is available for Shift 0 but not for Shift 1. Employee 2 is available for Shifts 0 and 1. Employee 3 is available for Shift 1 but not for Shift 0. Employee 4 is available for Shifts 0 and 1. Employee 5 is available for Shift 1 but not for Shift 0. Employees 6 and 7 are available for Shifts 0 and 1. + +Each employee is qualified for specific skills. For instance, Employees 0, 6, 7 are not qualified for Skill 0 but qualified for Skill 1, while Employees 3 and 5 are not qualified for Skill 1 but qualified for Skill 0; Employees 1, 2, 4 are qualified by both Skill 0 and 1. Each employee can be assigned to at most one shift across all restaurants. Assigning employees to shifts incurs specific costs: assigning Employee 0 to Skill 1 costs 1 unit, assigning Employee 1 to Skill 0 costs 2 units and to Skill 1 costs 1 unit, assigning Employee 2 to Skill 0 costs 3 units and to Skill 1 costs 2 units, assigning Employee 3 to Skill 0 costs 2 units, assigning Employee 4 to Skill 0 costs 2 units and to Skill 1 costs 4 units, assigning Employee 5 to Skill 0 costs 3 units, assigning Employee 6 to Skill 1 costs 2 units, and assigning Employee 7 to Skill 1 costs 1 unit. If a restaurant cannot meet the demand for a specific skill in a shift, unfulfilled positions incur a penalty cost of 100 units per position. The goal is to minimize the total cost, which includes the penalty costs for unfulfilled positions and the costs of assigning employees to shifts. Employees can only be assigned to shifts they are available for and to roles that match their skills.11The problem involves assigning employees to shifts across two restaurants to minimize costs associated with unfulfilled positions and employee assignments. There are two restaurants, each with two shifts per day, and eight employees available for assignment. Each shift at each restaurant has specific staffing demands for a particular skill, and employees have varying availability and skill sets. + + Restaurant 0, Shift 0, Skill 0 requires 1 employee. Restaurant 0, Shift 0, Skill 1 requires 1 employee. Restaurant 0, Shift 1, Skill 0 requires 1 employee. Restaurant 1, Shift 0, Skill 0 requires 2 employees. Restaurant 1, Shift 1, Skill 1 requires 1 employee. + +Employees can only be assigned to shifts they are available for and to roles that match their skills. Employee 0 is available for Shift 0 but not for Shift 1. Employee 1 is available for Shift 0 but not for Shift 1. Employee 2 is available for Shifts 0 and 1. Employee 3 is available for Shift 1 but not for Shift 0. Employee 4 is available for Shifts 0 and 1. Employee 5 is available for Shift 1 but not for Shift 0. Employees 6 and 7 are available for Shifts 0 and 1. + + Each employee is qualified for specific skills. For instance, Employees 0, 6, 7 are not qualified for Skill 0 but qualified for Skill 1, while Employees 3 and 5 are not qualified for Skill 1 but qualified for Skill 0; Employees 1, 2, 4 are qualified by both Skill 0 and 1. Each employee can be assigned to at most one shift across all restaurants. Assigning employees to shifts incurs specific costs: assigning Employee 0 to Skill 1 costs 1 unit, assigning Employee 1 to Skill 0 costs 2 units and to Skill 1 costs 1 unit, assigning Employee 2 to Skill 0 costs 3 units and to Skill 1 costs 2 units, assigning Employee 3 to Skill 0 costs 2 units, and assigning Employee 4 to Skill 0 costs 2 units and to Skill 1 costs 4 units, assigning Employee 5 to Skill 0 costs 3 units. units, assigning Employee 6 to Skill 1 costs 2 units, and assigning Employee 7 to Skill 1 costs 1 unit. If a restaurant cannot meet the demand for a specific skill in a shift, unfulfilled positions incur a penalty cost of 100 units per position. The goal is to minimize the total cost, which includes the penalty costs for unfulfilled positions and the costs of assigning employees to shifts. Constraints ensure that the total number of employees assigned to a shift at a restaurant, plus any unfulfilled positions, equals the demand for that shift. Employees can only be assigned to shifts they are available for and to roles that match their skills. Each employee can be assigned to at most one shift across all restaurants.16 11Missing data about the requirements of the restaurants and the allowed shifts and skills of workersmissing data
73Minimize the total completion time of six jobs processed on three machines, where each job must be assigned to exactly one of six schedule positions and each position must have exactly one job. The processing times for each job on each machine are as follows: Job 0 takes 5, 5, and 7 hours on Machines 0, 1, and 2 respectively; Job 1 takes 5, 9, and 10 hours; Job 2 takes 6, 7, and 10 hours; Job 3 takes 8, 7, and 9 hours; Job 4 takes 9, 5, and 9 hours; and Job 5 takes 8, 8, and 7 hours. Each job must start on Machine 1 only after completing on Machine 0, and on Machine 2 only after completing on Machine 1. Additionally, the start time of a job in a later position must be at least the start time of the job in the previous position plus its processing time on the same machine. The objective is to minimize the total completion time, which includes the start time of the last job on the last machine plus its processing time. All start times must be non-negative, and job assignments to positions are binary (either assigned or not). The specific coefficients for the objective function are 5, 6, 10, 7, 8, and 8 for JobSchedule[0,5], JobSchedule[1,5], JobSchedule[2,5], JobSchedule[3,5], JobSchedule[4,5], and JobSchedule[5,5] respectively, with an additional term StartTime[5,3]. The constraints include ensuring one job per schedule position, one schedule position per job, machine precedence, and job precedence, with exact coefficients as detailed in the LP data.69Minimize the total completion time of six jobs processed on three machines, where each job must be assigned to exactly one of six schedule positions and each position must have exactly one job. The processing times for each job on each machine are as follows: Job 0 takes 5, 5, and 7 hours on Machines 0, 1, and 2 respectively; Job 1 takes 5, 9, and 10 hours; Job 2 takes 6, 7, and 10 hours; Job 3 takes 8, 7, and 9 hours; Job 4 takes 9, 5, and 9 hours; and Job 5 takes 8, 8, and 7 hours. Each job must start on Machine 1 only after completing on Machine 0, and on Machine 2 only after completing on Machine 1. Additionally, the start time of a job in a later position must be at least the start time of the job in the previous position plus its processing time on the same machine. The objective is to minimize the total completion time, which includes the start time of the last job on the last machine plus its processing time. All start times must be non-negative, and job assignments to positions are binary (either assigned or not). The objective is to minimize the weighted combination of the completion times of the jobs (on the last machine). The weights for Jobs 0 to 5 are, respectively, 5, 6, 10, 7, 8, and 8. The constraints include ensuring one job per schedule position, one schedule position per job, machine precedence, and job precedence.1746Minimize the total completion time of six jobs processed on three machines, where each job must be assigned to exactly one of six schedule positions and each position must have exactly one job. The processing times for each job on each machine are as follows: Job 0 takes 5, 5, and 7 hours on Machines 0, 1, and 2 respectively; Job 1 takes 5, 9, and 10 hours; Job 2 takes 6, 7, and 10 hours; Job 3 takes 8, 7, and 9 hours; Job 4 takes 9, 5, and 9 hours; and Job 5 takes 8, 8, and 7 hours. Each job must start on Machine 1 only after completing on Machine 0, and on Machine 2 only after completing on Machine 1. Additionally, the start time of a job in a later position must be at least the start time of the job in the previous position plus its processing time on the same machine. The objective is to minimize the total completion time, which includes the start time of the last job on the last machine plus its processing time. All start times must be non-negative, and job assignments to positions are binary (either assigned or not). The specific coefficients for the objective function are is to minimize the weighted combination of the completion times of the jobs (on the last machine). The weights for Jobs 0 to 5 are, respectively, 5, 6, 10, 7, 8, and 8 for JobSchedule[0,5, JobSchedule[1,5], JobSchedule[2,5], JobSchedule[3,5], JobSchedule[4,5], and JobSchedule[5,5] respectively, with an additional term StartTime[5,3].] 8. The constraints include ensuring one job per schedule position, one schedule position per job, machine precedence, and job precedence, with exact coefficients as detailed in the LP data. precedence.69 1746The statement had an ill-defined objective function, we clarified itambiguous
74Facility Dispersion in a Network + +You are tasked with selecting a set of facilities from a network of five locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall dispersion between the selected facilities. Dispersion is measured by the pairwise distances between the facilities, and your goal is to ensure that the selected facilities are as spread out as possible. + +#### Key Decisions: +1. **Facility Selection**: You must choose exactly 2 facilities out of the 5 available locations. Each location can either be selected (1) or not selected (0). +2. **Pairwise Dispersion Activation**: For every pair of facilities, a binary decision determines whether both facilities in the pair are selected. This ensures that dispersion is only calculated for pairs where both facilities are chosen. + +#### Objective: +Maximize the total dispersion across all selected facility pairs. The dispersion values between the facilities are as follows: +- Node 0 and Node 1: 24 units +- Node 0 and Node 2: 25 units +- Node 0 and Node 3: 18 units +- Node 0 and Node 4: 24 units +- Node 1 and Node 0: 22 units +- Node 1 and Node 2: 17 units +- Node 1 and Node 3: 22 units +- Node 1 and Node 4: 24 units +- Node 2 and Node 0: 18 units +- Node 2 and Node 1: 19 units +- Node 2 and Node 3: 25 units +- Node 2 and Node 4: 20 units +- Node 3 and Node 0: 16 units +- Node 3 and Node 1: 22 units +- Node 3 and Node 2: 16 units +- Node 3 and Node 4: 17 units +- Node 4 and Node 0: 22 units +- Node 4 and Node 1: 21 units +- Node 4 and Node 2: 19 units +- Node 4 and Node 3: 16 units + +#### Constraints: +1. **Facility Selection Limit**: Exactly 2 facilities must be selected from the 5 available locations. +2. **Pairwise Activation Rules**: + - For any pair of facilities, the pairwise dispersion can only be counted if both facilities in the pair are selected. + - If either facility in a pair is not selected, the pairwise dispersion for that pair cannot be counted. + - If both facilities in a pair are selected, the pairwise dispersion must be counted. + +#### Numerical Parameters: +- Total facilities to select: 2 +- Dispersion values between facilities are explicitly provided in the objective section above. + +#### Goal: +Determine which 2 facilities to select and which pairwise dispersions to activate to maximize the total dispersion across the network while adhering to the constraints.46Facility Dispersion in a Network + +You are tasked with selecting a set of facilities from a network of five locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall dispersion between the selected facilities. Dispersion is measured by the pairwise distances between the facilities, and your goal is to ensure that the selected facilities are as spread out as possible. + +#### Key Decisions: +1. **Facility Selection**: You must choose exactly 2 facilities out of the 5 available locations. Each location can either be selected (1) or not selected (0). +2. **Pairwise Dispersion Activation**: For every pair of facilities, a binary decision determines whether both facilities in the pair are selected. This ensures that dispersion is only calculated for pairs where both facilities are chosen. + +#### Objective: +Maximize the total dispersion across all selected facility pairs. The dispersion values between the facilities are as follows: +- Node 0 and Node 1: 24 units +- Node 0 and Node 2: 25 units +- Node 0 and Node 3: 18 units +- Node 0 and Node 4: 24 units +- Node 1 and Node 2: 17 units +- Node 1 and Node 3: 22 units +- Node 1 and Node 4: 24 units +- Node 2 and Node 3: 25 units +- Node 2 and Node 4: 20 units +- Node 3 and Node 4: 17 units + +#### Constraints: +1. **Facility Selection Limit**: Exactly 2 facilities must be selected from the 5 available locations. +2. **Pairwise Activation Rules**: + - For any pair of facilities, the pairwise dispersion can only be counted if both facilities in the pair are selected. + - If either facility in a pair is not selected, the pairwise dispersion for that pair cannot be counted. + - If both facilities in a pair are selected, the pairwise dispersion must be counted. + +#### Numerical Parameters: +- Total facilities to select: 2 +- Dispersion values between facilities are explicitly provided in the objective section above. + +#### Goal: +Determine which 2 facilities to select and which pairwise dispersions to activate to maximize the total dispersion across the network while adhering to the constraints.25- Node 0 and Node 2: 25 units +- Node 0 and Node 3: 18 units +- Node 0 and Node 4: 24 units + - Node 1 and Node 0: 22 units +- Node 1 and Node 2: 17 units +- Node 1 and Node 3: 22 units +- Node 1 and Node 4: 24 units + - Node 2 and Node 0: 18 units + - Node 2 and Node 1: 19 units +- Node 2 and Node 3: 25 units +- Node 2 and Node 4: 20 units + - Node 3 and Node 0: 16 units + - Node 3 and Node 1: 22 units + - Node 3 and Node 2: 16 units +- Node 3 and Node 4: 17 units + - Node 4 and Node 0: 22 units + - Node 4 and Node 1: 21 units + - Node 4 and Node 2: 19 units + - Node 4 and Node 3: 16 units + +#### Constraints: +1. **Facility Selection Limit**: Exactly 2 facilities must be selected from the 5 available locations.46 25Original statement had different dispersions between (0,1) and (1,0), which does not make sense in this context. We just kept one of themambiguous
75Minimize the total operational cost, which includes the base cost for active generators, the per MW cost for power output, and the startup cost for generators. The base cost is 79 units per generator for Type 0, 78 units for Type 1, and 78 units for Type 2. The per MW cost is 4 units for Type 0, 3 units for Type 1, and 4 units for Type 2. The startup cost is 355 units for Type 0, 352 units for Type 1, and 346 units for Type 2. The total power output from all generators must meet or exceed the demand in each period, with demands of 173 MW in period 0, 197 MW in period 1, 227 MW in period 2, 201 MW in period 3, and 196 MW in period 4. Each active generator must produce at least its minimum output and no more than its maximum output, with minimum outputs of 35 MW for all types and maximum outputs of 127 MW for Type 0, 129 MW for Type 1, and 128 MW for Type 2. The total maximum output capacity of all active generators must be at least 15% higher than the demand in each period. The number of active generators cannot exceed the available units for each type, with a maximum of 3 generators for Type 0, 1 generator for Type 1, and 5 generators for Type 2. The number of generators started in a period cannot exceed the difference between the number of active generators in that period and the previous period, plus any generators that were already started. The number of active generators and startups must be non-negative integers, and the power output must be non-negative.4373Minimize the total operational cost, which includes the base cost for active generators, the per MW cost for power output, and the startup cost for generators. The base cost is 79 units per generator for Type 0, 78 units for Type 1, and 78 units for Type 2. The per MW cost is 4 units for Type 0, 3 units for Type 1, and 4 units for Type 2. The startup cost is 355 units for Type 0, 352 units for Type 1, and 346 units for Type 2. The total power output from all generators must meet or exceed the demand in each period, with demands of 173 MW in period 0, 197 MW in period 1, 227 MW in period 2, 201 MW in period 3, and 196 MW in period 4. Each active generator must produce at least its minimum output and no more than its maximum output, with minimum outputs of 35 MW for all types and maximum outputs of 127 MW for Type 0, 129 MW for Type 1, and 128 MW for Type 2. The total maximum output capacity of all active generators must be at least 15% higher than the demand in each period. The number of active generators cannot exceed the available units for each type, with a maximum of 3 generators for Type 0, 1 generator for Type 1, and 5 generators for Type 2. Each generator of each type that is active in a period but was not active in the previous period needs to pay the appropriate startup cost. Whenever a generator is active, it needs to pay its base cost. The number of active generators and startups must be non-negative integers, and the power output must be non-negative.5233Minimize the total operational cost, which includes the base cost for active generators, the per MW cost for power output, and the startup cost for generators. The base cost is 79 units per generator for Type 0, 78 units for Type 1, and 78 units for Type 2. The per MW cost is 4 units for Type 0, 3 units for Type 1, and 4 units for Type 2. The startup cost is 355 units for Type 0, 352 units for Type 1, and 346 units for Type 2. The total power output from all generators must meet or exceed the demand in each period, with demands of 173 MW in period 0, 197 MW in period 1, 227 MW in period 2, 201 MW in period 3, and 196 MW in period 4. Each active generator must produce at least its minimum output and no more than its maximum output, with minimum outputs of 35 MW for all types and maximum outputs of 127 MW for Type 0, 129 MW for Type 1, and 128 MW for Type 2. The total maximum output capacity of all active generators must be at least 15% higher than the demand in each period. The number of active generators cannot exceed the available units for each type, with a maximum of 3 generators for Type 0, 1 generator for Type 1, and 5 generators for Type 2. The number Each generator of generators started each type that is active in a period cannot exceed the difference between the number of but was not active generators in that period and the previous period, plus any generators that were already started. period needs to pay the appropriate startup cost. Whenever a generator is active, it needs to pay its base cost. The number of active generators and startups must be non-negative integers, and the power output must be non-negative.4373 5233It was not clear the role of the "base cost", we clarifiedambiguous, incorrect ground truth
76Minimize the total operational costs, which include setup costs, startup costs, inventory holding costs, and backlogging costs, while meeting the demand for each item in each period. The setup cost is **136.4441474007699** per production run, and the startup cost is **72.08169936581939** per startup. The holding cost for Item 0 is **1.045681545748948** per unit, for Item 1 is **1.076278851472706** per unit, and for Item 2 is **1.141130030903227** per unit. The backlogging cost for Item 0 is **7.516688931886524** per unit, for Item 1 is **6.810219227144024** per unit, and for Item 2 is **7.989326755092092** per unit. + +The production system involves three items (Item 0, Item 1, and Item 2), two machines (Machine 0 and Machine 1), and three time periods (Period 0, Period 1, and Period 2). For each item, machine, and time period, decide whether to produce the item (binary decision) and how much to produce (continuous decision). Determine whether to start up a machine for a specific item in a given period (binary decision). Decide how much inventory to carry over or how much demand to backlog for each item in each period (continuous decisions). + +The total production, inventory, and backlog must balance the demand for each item in each period. For example, in Period 0, the demand for Item 0 is **15.13481645049243**, for Item 1 is **15.96509719647714**, and for Item 2 is **15.05799468831989**. In Period 1, the demand for Item 0 is **23.22348134905647**, for Item 1 is **24.89682877145792**, and for Item 2 is **20.28924491987942**. In Period 2, the demand for Item 0 is **19.81696900208589**, for Item 1 is **22.4702270586731**, and for Item 2 is **18.30123580755789**. + +The production quantity and startup time must not exceed the machine's capacity. Machine 0 has a capacity of **74.37564803880434**, and starting it up requires **6.598818345230923** units of time. Machine 1 has a capacity of **61.80119129480688**, and starting it up requires **6.132700881798776** units of time. Each machine can produce only one item at a time in any given period. For example, in Period 0, Machine 0 can produce either Item 0, Item 1, or Item 2, but not more than one. Similarly, Machine 1 can produce only one item at a time in any period. + +If a machine is used to produce an item in a period after being idle in the previous period, a startup cost is incurred. For example, if Machine 0 is used to produce Item 0 in Period 1 but was not used in Period 0, a startup cost of **72.08169936581939** is incurred. Production quantities, inventory levels, and backlogs must be non-negative. Production and startup decisions are binary (either 0 or 1). + +The goal is to determine the optimal production, startup, inventory, and backlog decisions for each item, machine, and time period to minimize the total cost while satisfying all constraints.889.1372720402694Minimize the total operational costs, which include setup costs, startup costs, inventory holding costs, and backlogging costs, while meeting the demand for each item in each period. The setup cost is **136.4441474007699** per production run, and the startup cost is **72.08169936581939** per startup. The holding cost for Item 0 is **1.045681545748948** per unit, for Item 1 is **1.076278851472706** per unit, and for Item 2 is **1.141130030903227** per unit. The backlogging cost for Item 0 is **7.516688931886524** per unit, for Item 1 is **6.810219227144024** per unit, and for Item 2 is **7.989326755092092** per unit. + +The production system involves three items (Item 0, Item 1, and Item 2), two machines (Machine 0 and Machine 1), and three time periods (Period 0, Period 1, and Period 2). For each item, machine, and time period, decide whether to produce the item (binary decision) and how much to produce (continuous decision). Determine whether to start up a machine for a specific item in a given period (binary decision). Decide how much inventory to carry over or how much demand to backlog for each item in each period (continuous decisions). + +The total production, inventory, and backlog must balance the demand for each item in each period. For example, in Period 0, the demand for Item 0 is **15.13481645049243**, for Item 1 is **15.96509719647714**, and for Item 2 is **15.05799468831989**. In Period 1, the demand for Item 0 is **23.22348134905647**, for Item 1 is **24.89682877145792**, and for Item 2 is **20.28924491987942**. In Period 2, the demand for Item 0 is **19.81696900208589**, for Item 1 is **22.4702270586731**, and for Item 2 is **18.30123580755789**. + +The production quantity and startup time must not exceed the machine's capacity. Machine 0 has a capacity of **74.37564803880434**, and starting it up requires **6.598818345230923** units of time. Machine 1 has a capacity of **61.80119129480688**, and starting it up requires **6.132700881798776** units of time. Each machine can produce only one item at a time in any given period. For example, in Period 0, Machine 0 can produce either Item 0, Item 1, or Item 2, but not more than one. Similarly, Machine 1 can produce only one item at a time in any period. + +If a machine is used to produce an item in a period after being idle in the previous period, a startup cost is incurred. For example, if Machine 0 is used to produce Item 0 in Period 1 but was not used in Period 0, a startup cost of **72.08169936581939** is incurred. Production quantities, inventory levels, and backlogs must be non-negative. Production and startup decisions are binary (either 0 or 1). + +The goal is to determine the optimal production, startup, inventory, and backlog decisions for each item, machine, and time period to minimize the total cost while satisfying all constraints.889.1372720402694No differencesNo differencesok
77In a manufacturing facility, six distinct jobs labeled Job 0 to Job 5 need to be processed on a set of machines. Each job consists of two operations that must be performed in a specific order, with each operation assigned to a specific machine. The goal is to schedule these operations to minimize the total completion time, known as the makespan, ensuring all jobs are completed as quickly as possible. + +Each operation has a processing time of 1 unit. For each job, the second operation cannot start until the first operation is completed. For example, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 is finished, and this applies to all jobs. + +If two operations are assigned to the same machine, they cannot overlap in time. The order in which they are processed is determined by a binary decision variable. For example, if Operation 1 of Job 0 and Operation 1 of Job 4 are assigned to the same machine, one must be scheduled before the other. The binary variable X_(0,1)_(4,1) determines the order: if X_(0,1)_(4,1) is 1, then Operation 1 of Job 0 is scheduled before Operation 1 of Job 4; if it is 0, then Operation 1 of Job 4 is scheduled before Operation 1 of Job 0. This logic applies to all pairs of operations on shared machines. + +The total completion time, C_max, must be at least as large as the completion time of every operation. For example, C_max must be greater than or equal to the completion time of Operation 0 of Job 0, which is S_(0,0) plus 1 unit. This applies to all operations across all jobs. + +The start time of every operation must be non-negative. For example, S_(0,0) must be greater than or equal to 0, and this applies to all operations. + +The binary variables X_(j,i)_(k,l) are used to enforce the order of operations on shared machines and can only take values of 0 or 1. + +The objective is to minimize the makespan, C_max, which represents the total time required to complete all jobs. This ensures that the manufacturing process is as efficient as possible, reducing idle time and maximizing throughput. All processing times are 1 unit, and the makespan must be at least 1 unit greater than the completion time of every operation.4In a manufacturing facility, six distinct jobs labeled Job 0 to Job 5 need to be processed on a set of machines. Each job consists of two operations that must be performed in a specific order, with each operation assigned to a specific machine. The goal is to schedule these operations to minimize the total completion time, known as the makespan, ensuring all jobs are completed as quickly as possible. + +Each operation has a processing time of 1 unit. For each job, the second operation cannot start until the first operation is completed. For example, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 is finished, and this applies to all jobs. + +If two operations are assigned to the same machine, they cannot overlap in time. The order in which they are processed is determined by a binary decision variable. For example, if Operation 1 of Job 0 and Operation 1 of Job 4 are assigned to the same machine, one must be scheduled before the other. The binary variable X_(0,1)_(4,1) determines the order: if X_(0,1)_(4,1) is 1, then Operation 1 of Job 0 is scheduled before Operation 1 of Job 4; if it is 0, then Operation 1 of Job 4 is scheduled before Operation 1 of Job 0. This logic applies to all pairs of operations on shared machines. + +The total completion time, C_max, must be at least as large as the completion time of every operation. For example, C_max must be greater than or equal to the completion time of Operation 0 of Job 0, which is S_(0,0) plus 1 unit. This applies to all operations across all jobs. + +The start time of every operation must be non-negative. For example, S_(0,0) must be greater than or equal to 0, and this applies to all operations. + +The binary variables X_(j,i)_(k,l) are used to enforce the order of operations on shared machines and can only take values of 0 or 1. + +The objective is to minimize the makespan, C_max, which represents the total time required to complete all jobs. This ensures that the manufacturing process is as efficient as possible, reducing idle time and maximizing throughput. All processing times are 1 unit, and the makespan must be at least 1 unit greater than the completion time of every operation. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3, 4, 5} +- For each job j in J, there are 2 operations indexed i = 0..1. We denote operation (j,i). +- Machines: M = {M0, M1, M2} + +2) Machine assignment for every operation (explicit listing) +- Operation (0,0) -> Machine M0 +- Operation (0,1) -> Machine M1 +- Operation (1,0) -> Machine M0 +- Operation (1,1) -> Machine M2 +- Operation (2,0) -> Machine M1 +- Operation (2,1) -> Machine M2 +- Operation (3,0) -> Machine M2 +- Operation (3,1) -> Machine M0 +- Operation (4,0) -> Machine M1 +- Operation (4,1) -> Machine M0 +- Operation (5,0) -> Machine M2 +- Operation (5,1) -> Machine M1 + +3) Operation durations +- All operations have processing time p_(j,i) = 1 for every job j and operation i.4The binary variables X_(j,i)_(k,l) are used to enforce the order of operations on shared machines and can only take values of 0 or 1. + +The objective is to minimize the makespan, C_max, which represents the total time required to complete all jobs. This ensures that the manufacturing process is as efficient as possible, reducing idle time and maximizing throughput. All processing times are 1 unit, and the makespan must be at least 1 unit greater than the completion time of every operation. + + ADDITIONAL INPUT DATA: + + 1) Sets and indices + - Jobs: J = {0, 1, 2, 3, 4, 5} + - For each job j in J, there are 2 operations indexed i = 0..1. We denote operation (j,i). + - Machines: M = {M0, M1, M2} + + 2) Machine assignment for every operation (explicit listing) + - Operation (0,0) -> Machine M0 + - Operation (0,1) -> Machine M1 + - Operation (1,0) -> Machine M0 + - Operation (1,1) -> Machine M2 + - Operation (2,0) -> Machine M1 + - Operation (2,1) -> Machine M2 + - Operation (3,0) -> Machine M2 + - Operation (3,1) -> Machine M0 + - Operation (4,0) -> Machine M1 + - Operation (4,1) -> Machine M0 + - Operation (5,0) -> Machine M2 + - Operation (5,1) -> Machine M1 + + 3) Operation durations + - All operations have processing time p_(j,i) = 1 for every job j and operation i.No differencesIt was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
78The problem involves designing a public transit network to minimize total costs while meeting passenger demand as much as possible. There are 11 potential transit lines (L_0 to L_10), each with a fixed cost for establishment and an operational cost per unit frequency. The fixed costs are: L_0 (\$2,478), L_1 (\$2,408), L_2 (\$2,443), L_3 (\$2,777), L_4 (\$2,152), L_5 (\$2,620), L_6 (\$2,359), L_7 (\$2,724), L_8 (\$2,247), L_9 (\$2,991), L_10 (\$2,941). The operational costs per unit frequency are: L_0 (\$297), L_1 (\$220), L_2 (\$288), L_3 (\$236), L_4 (\$212), L_5 (\$283), L_6 (\$206), L_7 (\$219), L_8 (\$242), L_9 (\$217), L_10 (\$227). Each line, if selected, must operate between 2 and 10 trips per hour. There are 13 stations, some of which can serve as transfer points if at least two lines pass through them. There are 12 origin-destination pairs with specific passenger demands and penalty costs for unsatisfied demand: OD_8_12 (29 passengers, \$702 penalty), OD_3_8 (20 passengers, \$751 penalty), OD_2_0 (28 passengers, \$740 penalty), OD_11_2 (28 passengers, \$722 penalty), OD_2_1 (30 passengers, \$785 penalty), OD_10_5 (26 passengers, \$722 penalty), OD_10_12 (26 passengers, \$735 penalty), OD_5_6 (26 passengers, \$773 penalty), OD_1_12 (25 passengers, \$770 penalty), OD_10_6 (27 passengers, \$757 penalty), OD_4_9 (26 passengers, \$747 penalty), OD_4_0 (27 passengers, \$757 penalty). The total number of vehicles available is 55, and the combined vehicle usage across all lines must not exceed this limit. The objective is to minimize the total cost, which includes fixed costs for establishing selected lines, operational costs based on line frequencies, and penalty costs for any unsatisfied demand. Constraints include ensuring that the total capacity provided by all lines serving each OD pair, plus any unsatisfied demand, meets or exceeds the passenger demand; enforcing frequency bounds for selected lines; limiting total vehicle usage; designating transfer stations only if at least two lines pass through them; and ensuring that the total frequency of lines serving each OD pair meets a minimum threshold.2373940The problem involves designing a public transit network to minimize total costs while meeting passenger demand as much as possible. There are 11 potential transit lines (L_0 to L_10), each with a fixed cost for establishment and an operational cost per unit frequency. The fixed costs are: L_0 (\$2,478), L_1 (\$2,408), L_2 (\$2,443), L_3 (\$2,777), L_4 (\$2,152), L_5 (\$2,620), L_6 (\$2,359), L_7 (\$2,724), L_8 (\$2,247), L_9 (\$2,991), L_10 (\$2,941). The operational costs per unit frequency are: L_0 (\$297), L_1 (\$220), L_2 (\$288), L_3 (\$236), L_4 (\$212), L_5 (\$283), L_6 (\$206), L_7 (\$219), L_8 (\$242), L_9 (\$217), L_10 (\$227). Each line, if selected, must operate between 2 and 10 trips per hour. There are 13 stations, some of which can serve as transfer points if at least two lines pass through them. There are 12 origin-destination pairs with specific passenger demands and penalty costs for unsatisfied demand: OD_8_12 (29 passengers, \$702 penalty), OD_3_8 (20 passengers, \$751 penalty), OD_2_0 (28 passengers, \$740 penalty), OD_11_2 (28 passengers, \$722 penalty), OD_2_1 (30 passengers, \$785 penalty), OD_10_5 (26 passengers, \$722 penalty), OD_10_12 (26 passengers, \$735 penalty), OD_5_6 (26 passengers, \$773 penalty), OD_1_12 (25 passengers, \$770 penalty), OD_10_6 (27 passengers, \$757 penalty), OD_4_9 (26 passengers, \$747 penalty), OD_4_0 (27 passengers, \$757 penalty). The total number of vehicles available is 55, and the combined vehicle usage across all lines must not exceed this limit. The objective is to minimize the total cost, which includes fixed costs for establishing selected lines, operational costs based on line frequencies, and penalty costs for any unsatisfied demand. Constraints include ensuring that the total capacity provided by all lines serving each OD pair, plus any unsatisfied demand, meets or exceeds the passenger demand; enforcing frequency bounds for selected lines; limiting total vehicle usage; designating transfer stations only if at least two lines pass through them; and ensuring that the total frequency of lines serving each OD pair meets a minimum threshold.237394Very underspecified problem.underspecified
79Minimize the total operational cost of a power generation system over six periods, consisting of Type 0 and Type 1 generators. The cost includes a base cost of 78 units per generator operational in a period, a variable cost of 4 units per megawatt of power generated, and startup costs of 347 units for Type 0 generators and 352 units for Type 1 generators when activated. The total power generated in each period must meet or exceed the demand: 172 MW in Period 0, 171 MW in Period 1, 166 MW in Period 2, 160 MW in Period 3, 195 MW in Period 4, and 181 MW in Period 5. Each generator has output limits: Type 0 generators must produce between 34 MW and 128 MW, and Type 1 generators must produce between 34 MW and 125 MW. The total maximum output capacity of all generators in each period must be at least 115% of the demand for that period, ensuring sufficient reserve capacity. The number of operational generators cannot exceed 4 for Type 0 and 5 for Type 1 in any period. For startup constraints, no more than 2 Type 0 generators can be started in Period 0, and no more than 1 Type 1 generator can be started in Period 0. The number of generators operational in a period cannot exceed the number operational in the previous period plus the number started in the current period. All decision variables must be non-negative integers, and power output must be non-negative.5116Minimize the total operational cost of a power generation system over six periods, consisting of Type 0 and Type 1 generators. The cost includes a base cost of 78 units per generator operational in a period, a variable cost of 4 units per megawatt of power generated, and startup costs of 347 units for Type 0 generators and 352 units for Type 1 generators when activated. The total power generated in each period must meet or exceed the demand: 172 MW in Period 0, 171 MW in Period 1, 166 MW in Period 2, 160 MW in Period 3, 195 MW in Period 4, and 181 MW in Period 5. Each generator has output limits: Type 0 generators must produce between 34 MW and 128 MW, and Type 1 generators must produce between 34 MW and 125 MW. The total maximum output capacity of all generators in each period must be at least 115% of the demand for that period, ensuring sufficient reserve capacity. The number of operational generators cannot exceed 4 for Type 0 and 5 for Type 1 in any period. For startup constraints, no more than 2 Type 0 generators can be started in Period 0, and no more than 1 Type 1 generator can be started in Period 0. The number of generators operational in a period cannot exceed the number operational in the previous period plus the number started in the current period. All decision variables must be non-negative integers, and power output must be non-negative.5810No differences5116 5810Incorrect ground truthincorrect ground truth
80Determine the number of generators of each type (Type 0 and Type 1) to operate, the power output for each generator type, and the number of startups in each of the five consecutive time periods to minimize the total cost of operation. The total cost includes a base cost of 80 units per Type 0 generator and 79 units per Type 1 generator for each period they are active, a power output cost of 3 units per MW for Type 0 generators and 4 units per MW for Type 1 generators, and a startup cost of 352 units for Type 0 generators and 346 units for Type 1 generators. The total power output from all generators must meet or exceed the demand in each period: 217 MW in period 0, 209 MW in period 1, 194 MW in period 2, 182 MW in period 3, and 164 MW in period 4. Each Type 0 generator must produce at least 34 MW and at most 129 MW when active, and each Type 1 generator must produce at least 34 MW and at most 127 MW when active. The combined maximum output of all active generators must meet or exceed specific capacity requirements: 249.55 MW in period 0, 240.35 MW in period 1, 223.1 MW in period 2, 209.3 MW in period 3, and 188.6 MW in period 4. No more than 1 Type 0 generator and 2 Type 1 generators can be active in any period. The number of startups for each generator type is constrained by the number of generators available and the number of generators active in the previous period. The variables representing the number of generators and startups are general integer variables.4014Determine the number of generators of each type (Type 0 and Type 1) to operate, the power output for each generator type, and the number of startups in each of the five consecutive time periods to minimize the total cost of operation. The total cost includes a base cost of 80 units per Type 0 generator and 79 units per Type 1 generator for each period they are active, a power output cost of 3 units per MW for Type 0 generators and 4 units per MW for Type 1 generators, and a startup cost of 352 units for Type 0 generators and 346 units for Type 1 generators. The total power output from all generators must meet or exceed the demand in each period: 217 MW in period 0, 209 MW in period 1, 194 MW in period 2, 182 MW in period 3, and 164 MW in period 4. Each Type 0 generator must produce at least 34 MW and at most 129 MW when active, and each Type 1 generator must produce at least 34 MW and at most 127 MW when active. The combined maximum output of all active generators must meet or exceed specific capacity requirements: 249.55 MW in period 0, 240.35 MW in period 1, 223.1 MW in period 2, 209.3 MW in period 3, and 188.6 MW in period 4. No more than 1 Type 0 generator and 2 Type 1 generators can be active in any period. The number of startups for each generator type is constrained by the number of generators available and the number of generators active in the previous period. The variables representing the number of generators and startups are general integer variables.4712No differences4014 4712Incorrect ground truthincorrect ground truth
81A manufacturing facility produces custom parts for seven jobs, labeled Job 0 through Job 6. Each job consists of two operations that must be performed in sequence, with the second operation starting only after the first is completed. Operations are assigned to specific machines, and some operations share the same machine, meaning they cannot be processed simultaneously. The goal is to schedule all operations to minimize the total completion time, known as the makespan. + +Each operation has a processing time of 1 unit. For example, Operation 0 of Job 0 takes 1 unit of time, and Operation 1 of Job 0 also takes 1 unit of time. This applies to all operations across all jobs. Operations sharing the same machine must be scheduled such that one finishes before the other starts. Binary decision variables determine the order of operations on shared machines. For instance, if Operation 1 of Job 0 and Operation 0 of Job 1 share a machine, a binary variable decides whether Operation 1 of Job 0 finishes before Operation 0 of Job 1 starts, or vice versa. + +The makespan must be at least as large as the completion time of every operation. For example, if Operation 1 of Job 0 finishes at time 2, the makespan must be at least 2. A large constant, 100,000, is used in the constraints to ensure that operations on the same machine do not overlap. The objective is to minimize the makespan, ensuring the facility operates as efficiently as possible while respecting all precedence and machine capacity constraints.3A manufacturing facility produces custom parts for seven jobs, labeled Job 0 through Job 6. Each job consists of two operations that must be performed in sequence, with the second operation starting only after the first is completed. Operations are assigned to specific machines, and some operations share the same machine, meaning they cannot be processed simultaneously. The goal is to schedule all operations to minimize the total completion time, known as the makespan. + +Each operation has a processing time of 1 unit. For example, Operation 0 of Job 0 takes 1 unit of time, and Operation 1 of Job 0 also takes 1 unit of time. This applies to all operations across all jobs. Operations sharing the same machine must be scheduled such that one finishes before the other starts. Binary decision variables determine the order of operations on shared machines. For instance, if Operation 1 of Job 0 and Operation 0 of Job 1 share a machine, a binary variable decides whether Operation 1 of Job 0 finishes before Operation 0 of Job 1 starts, or vice versa. + +The makespan must be at least as large as the completion time of every operation. For example, if Operation 1 of Job 0 finishes at time 2, the makespan must be at least 2. A large constant, 100, is used in the constraints to ensure that operations on the same machine do not overlap. The objective is to minimize the makespan, ensuring the facility operates as efficiently as possible while respecting all precedence and machine capacity constraints. + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 0 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 1 + - Operation 1 of Job 1: Machine 2 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 0 + - Operation 0 of Job 3: Machine 0 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 1 + - Operation 1 of Job 4: Machine 0 + - Operation 0 of Job 5: Machine 2 + - Operation 1 of Job 5: Machine 1 + - Operation 0 of Job 6: Machine 0 + - Operation 1 of Job 6: Machine 15Each operation has a processing time of 1 unit. For example, Operation 0 of Job 0 takes 1 unit of time, and Operation 1 of Job 0 also takes 1 unit of time. This applies to all operations across all jobs. Operations sharing the same machine must be scheduled such that one finishes before the other starts. Binary decision variables determine the order of operations on shared machines. For instance, if Operation 1 of Job 0 and Operation 0 of Job 1 share a machine, a binary variable decides whether Operation 1 of Job 0 finishes before Operation 0 of Job 1 starts, or vice versa. + +The makespan must be at least as large as the completion time of every operation. For example, if Operation 1 of Job 0 finishes at time 2, the makespan must be at least 2. A large constant, 100,000, 100, is used in the constraints to ensure that operations on the same machine do not overlap. The objective is to minimize the makespan, ensuring the facility operates as efficiently as possible while respecting all precedence and machine capacity constraints. + + Additional information: + - There are 3 machines, numbered from 0 to 2 + - The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 0 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 1 + - Operation 1 of Job 1: Machine 2 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 0 + - Operation 0 of Job 3: Machine 0 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 1 + - Operation 1 of Job 4: Machine 0 + - Operation 0 of Job 5: Machine 2 + - Operation 1 of Job 5: Machine 1 + - Operation 0 of Job 6: Machine 0 + - Operation 1 of Job 6: Machine 13 5It was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
82There are 5 jobs labeled Job 0 to Job 4 and 4 machines labeled Machine 0 to Machine 3. Each job must be processed on all machines in sequence, with specific processing times for each job on each machine. The goal is to assign each job to a unique schedule position (Position 0 to Position 4) and determine the start times for each job on each machine to minimize the total completion time. Each schedule position must be assigned to exactly one job, and each job must be assigned to exactly one schedule position. A job cannot start on a machine until it has finished processing on the previous machine, and a job in a later schedule position cannot start on a machine until the job in the previous position has finished processing on that machine. The objective function includes coefficients for the schedule variables in the last position: 6 for Job 0, 5 for Job 1, 10 for Job 2, 10 for Job 3, and 8 for Job 4, as well as the start time of the last job on the last machine. The processing times for each job on each machine are as follows: Job 0 takes 9, 9, 7, and 6 hours on Machines 0 to 3, respectively; Job 1 takes 6, 6, 7, and 5 hours; Job 2 takes 9, 10, 10, and 10 hours; Job 3 takes 6, 10, 6, and 10 hours; and Job 4 takes 10, 7, 6, and 8 hours. The JobSchedule variables are binary, indicating whether a job is assigned to a specific schedule position, and the StartTime variables represent the start times of jobs on machines, which must be non-negative.65There are 5 jobs labeled Job 0 to Job 4 and 4 machines labeled Machine 0 to Machine 3. Each job must be processed on all machines in sequence, with specific processing times for each job on each machine. The goal is to assign each job to a unique schedule position (Position 0 to Position 4) and determine the start times for each job on each machine to minimize the total completion time. Each schedule position must be assigned to exactly one job, and each job must be assigned to exactly one schedule position. A job cannot start on a machine until it has finished processing on the previous machine, and a job in a later schedule position cannot start on a machine until the job in the previous position has finished processing on that machine. The objective function includes coefficients for the schedule variables in the last position: 6 for Job 0, 5 for Job 1, 10 for Job 2, 10 for Job 3, and 8 for Job 4, as well as the start time of the last job on the last machine. The processing times for each job on each machine are as follows: Job 0 takes 9, 9, 7, and 6 hours on Machines 0 to 3, respectively; Job 1 takes 6, 6, 7, and 5 hours; Job 2 takes 9, 10, 10, and 10 hours; Job 3 takes 6, 10, 6, and 10 hours; and Job 4 takes 10, 7, 6, and 8 hours. The JobSchedule variables are binary, indicating whether a job is assigned to a specific schedule position, and the StartTime variables represent the start times of jobs on machines, which must be non-negative.65No differencesNo differencesok
83You are tasked with selecting exactly 2 out of 5 nodes (node_0, node_1, node_2, node_3, node_4) to host facilities, with the goal of maximizing the minimum distance between any two selected facilities. The decision variables are binary: each node can either be selected (1) or not selected (0), and for every pair of nodes, a binary variable indicates whether both nodes in the pair are selected. The distances between nodes are derived from the LP constraints, where the distance between node_0 and node_1 is 18 units (1,000,018 - 1,000,000), node_0 and node_2 is 36 units (1,000,036 - 1,000,000), node_0 and node_3 is 28 units (1,000,028 - 1,000,000), node_0 and node_4 is 20 units (1,000,020 - 1,000,000), node_1 and node_0 is 25 units (1,000,025 - 1,000,000), node_1 and node_2 is 37 units (1,000,037 - 1,000,000), node_1 and node_3 is 36 units (1,000,036 - 1,000,000), node_1 and node_4 is 17 units (1,000,017 - 1,000,000), node_2 and node_0 is 42 units (1,000,042 - 1,000,000), node_2 and node_1 is 42 units (1,000,042 - 1,000,000), node_2 and node_3 is 23 units (1,000,023 - 1,000,000), node_2 and node_4 is 46 units (1,000,046 - 1,000,000), node_3 and node_0 is 31 units (1,000,031 - 1,000,000), node_3 and node_1 is 36 units (1,000,036 - 1,000,000), node_3 and node_2 is 11 units (1,000,011 - 1,000,000), node_3 and node_4 is 50 units (1,000,050 - 1,000,000), node_4 and node_0 is 10 units (1,000,010 - 1,000,000), node_4 and node_1 is 29 units (1,000,029 - 1,000,000), node_4 and node_2 is 24 units (1,000,024 - 1,000,000), and node_4 and node_3 is 42 units (1,000,042 - 1,000,000). The constraints ensure that exactly 2 nodes are selected, and for each pair of nodes, the minimum distance between any two selected facilities must be at least as large as the distance between those two nodes if both nodes in the pair are selected. If either node in the pair is not selected, the constraint is relaxed using a large constant (1,000,000). Additionally, the selection of a pair of nodes is only valid if both nodes in the pair are selected, enforced by ensuring that if a pair is selected, both nodes in the pair must be selected, and if both nodes in a pair are selected, the pair must be selected. The solution must ensure that the facilities are optimally dispersed across the network.42You are tasked with selecting exactly 2 out of 5 nodes (node_0, node_1, node_2, node_3, node_4) to host facilities, with the goal of maximizing the minimum distance between any two selected facilities. The decision variables are binary: each node can either be selected (1) or not selected (0), and for every pair of nodes, a binary variable indicates whether both nodes in the pair are selected. The distances between nodes are derived from the LP constraints, where the distance between node_0 and node_1 is 18 units (1,000,018 - 1,000,000), node_0 and node_2 is 36 units (1,000,036 - 1,000,000), node_0 and node_3 is 28 units (1,000,028 - 1,000,000), node_0 and node_4 is 20 units (1,000,020 - 1,000,000), node_1 and node_0 is 25 units (1,000,025 - 1,000,000), node_1 and node_2 is 37 units (1,000,037 - 1,000,000), node_1 and node_3 is 36 units (1,000,036 - 1,000,000), node_1 and node_4 is 17 units (1,000,017 - 1,000,000), node_2 and node_0 is 42 units (1,000,042 - 1,000,000), node_2 and node_1 is 42 units (1,000,042 - 1,000,000), node_2 and node_3 is 23 units (1,000,023 - 1,000,000), node_2 and node_4 is 46 units (1,000,046 - 1,000,000), node_3 and node_0 is 31 units (1,000,031 - 1,000,000), node_3 and node_1 is 36 units (1,000,036 - 1,000,000), node_3 and node_2 is 11 units (1,000,011 - 1,000,000), node_3 and node_4 is 50 units (1,000,050 - 1,000,000), node_4 and node_0 is 10 units (1,000,010 - 1,000,000), node_4 and node_1 is 29 units (1,000,029 - 1,000,000), node_4 and node_2 is 24 units (1,000,024 - 1,000,000), and node_4 and node_3 is 42 units (1,000,042 - 1,000,000). The constraints ensure that exactly 2 nodes are selected, and for each pair of nodes, the minimum distance between any two selected facilities must be at least as large as the distance between those two nodes if both nodes in the pair are selected. If either node in the pair is not selected, the constraint is relaxed using a large constant (1,000,000). Additionally, the selection of a pair of nodes is only valid if both nodes in the pair are selected, enforced by ensuring that if a pair is selected, both nodes in the pair must be selected, and if both nodes in a pair are selected, the pair must be selected. The solution must ensure that the facilities are optimally dispersed across the network.42No differencesNo differencesok
84Cell Tower Coverage Planning + +You are tasked with planning the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 30 distinct areas, each with a specific population size that needs to be served. Your goal is to determine which cell towers to build and which areas to cover, ensuring that the total cost of building the towers does not exceed the available budget. + +#### Key Decisions: +1. **Build or Not Build a Tower**: For each of the 25 potential cell tower locations, you must decide whether to build the tower (a binary decision: build or not build). +2. **Coverage of Areas**: For each of the 30 areas, you must determine whether it will be covered by at least one of the built towers (a binary decision: covered or not covered). + +#### Objective: +Maximize the total population covered by the deployed cell towers. The population sizes for the 30 areas are as follows: +- Area 0: 2,882 people +- Area 1: 4,926 people +- Area 2: 3,930 people +- Area 3: 3,468 people +- Area 4: 3,415 people +- Area 5: 2,697 people +- Area 6: 2,746 people +- Area 7: 3,817 people +- Area 8: 4,806 people +- Area 9: 2,889 people +- Area 10: 2,276 people +- Area 11: 2,087 people +- Area 12: 4,452 people +- Area 13: 3,015 people +- Area 14: 2,305 people +- Area 15: 3,196 people +- Area 16: 4,854 people +- Area 17: 4,466 people +- Area 18: 2,154 people +- Area 19: 4,661 people +- Area 20: 2,587 people +- Area 21: 4,013 people +- Area 22: 3,519 people +- Area 23: 4,837 people +- Area 24: 2,465 people +- Area 25: 3,807 people +- Area 26: 4,498 people +- Area 27: 3,423 people +- Area 28: 2,832 people +- Area 29: 3,064 people + +#### Constraints: +1. **Coverage Requirements**: + - Each area must be covered by at least one of the built towers. For example: + - Area 0 can be covered by towers 3, 4, 6, 7, 8, 9, 10, 11, 13, 14, 15, 18, or 20. + - Area 1 can be covered by towers 0, 1, 3, 4, 5, 8, 11, 13, 15, 16, 20, 23, or 24. + - Similar coverage constraints apply to all 30 areas, with specific combinations of towers required for each area. + +2. **Budget Constraint**: + - The total cost of building the selected towers must not exceed the available budget of 2,501 units. The costs for building each tower are as follows: + - Tower 0: 453 units + - Tower 1: 208 units + - Tower 2: 223 units + - Tower 3: 336 units + - Tower 4: 473 units + - Tower 5: 274 units + - Tower 6: 257 units + - Tower 7: 436 units + - Tower 8: 440 units + - Tower 9: 242 units + - Tower 10: 272 units + - Tower 11: 277 units + - Tower 12: 232 units + - Tower 13: 292 units + - Tower 14: 369 units + - Tower 15: 318 units + - Tower 16: 475 units + - Tower 17: 238 units + - Tower 18: 440 units + - Tower 19: 496 units + - Tower 20: 275 units + - Tower 21: 240 units + - Tower 22: 291 units + - Tower 23: 320 units + - Tower 24: 462 units + +3. **Binary Decisions**: + - Each tower can either be built (1) or not built (0). + - Each area can either be covered (1) or not covered (0). + +#### Summary: +Your task is to select which towers to build and which areas to cover, ensuring that the total population covered is maximized while staying within the budget. The coverage of each area depends on the specific combination of towers built, and the total cost of building the selected towers must not exceed 2,501 units.104087Cell Tower Coverage Planning + +You are tasked with planning the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 30 distinct areas, each with a specific population size that needs to be served. Your goal is to determine which cell towers to build and which areas to cover, ensuring that the total cost of building the towers does not exceed the available budget. + +#### Key Decisions: +1. **Build or Not Build a Tower**: For each of the 25 potential cell tower locations, you must decide whether to build the tower (a binary decision: build or not build). +2. **Coverage of Areas**: For each of the 30 areas, you must determine whether it will be covered by at least one of the built towers (a binary decision: covered or not covered). + +#### Objective: +Maximize the total population covered by the deployed cell towers. The population sizes for the 30 areas are as follows: +- Area 0: 2,882 people +- Area 1: 4,926 people +- Area 2: 3,930 people +- Area 3: 3,468 people +- Area 4: 3,415 people +- Area 5: 2,697 people +- Area 6: 2,746 people +- Area 7: 3,817 people +- Area 8: 4,806 people +- Area 9: 2,889 people +- Area 10: 2,276 people +- Area 11: 2,087 people +- Area 12: 4,452 people +- Area 13: 3,015 people +- Area 14: 2,305 people +- Area 15: 3,196 people +- Area 16: 4,854 people +- Area 17: 4,466 people +- Area 18: 2,154 people +- Area 19: 4,661 people +- Area 20: 2,587 people +- Area 21: 4,013 people +- Area 22: 3,519 people +- Area 23: 4,837 people +- Area 24: 2,465 people +- Area 25: 3,807 people +- Area 26: 4,498 people +- Area 27: 3,423 people +- Area 28: 2,832 people +- Area 29: 3,064 people + +#### Constraints: +1. **Coverage Requirements**: + - Each area must be covered by at least one of the built towers. For example: + - Area 0 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 7, Tower 9, Tower 16, Tower 17, Tower 20, Tower 22, Tower 24. + - Area 1 can be covered by towers Tower 0, Tower 1, Tower 15, Tower 19, Tower 22, Tower 23. + - Area 2 can be covered by towers Tower 1, Tower 4, Tower 8, Tower 11, Tower 12, Tower 13, Tower 16, Tower 18, Tower 20, Tower 22, Tower 23, Tower 24. + - Area 3 can be covered by towers Tower 1, Tower 3, Tower 4, Tower 7, Tower 8, Tower 10, Tower 14, Tower 16, Tower 19, Tower 20, Tower 22, Tower 24. + - Area 4 can be covered by towers Tower 0, Tower 3, Tower 6, Tower 10, Tower 15, Tower 16, Tower 17, Tower 18, Tower 20, Tower 21, Tower 23. + - Area 5 can be covered by towers Tower 3, Tower 5, Tower 7, Tower 10, Tower 11, Tower 12, Tower 13, Tower 14, Tower 16, Tower 17, Tower 18, Tower 19, Tower 21, Tower 22, Tower 24. + - Area 6 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 6, Tower 8, Tower 12, Tower 13, Tower 14, Tower 15, Tower 17, Tower 20, Tower 23, Tower 24. + - Area 7 can be covered by towers Tower 2, Tower 3, Tower 4, Tower 5, Tower 9, Tower 12, Tower 14, Tower 16, Tower 17, Tower 20. + - Area 8 can be covered by towers Tower 0, Tower 2, Tower 7, Tower 12, Tower 13, Tower 14, Tower 17, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24. + - Area 9 can be covered by towers Tower 0, Tower 2, Tower 3, Tower 5, Tower 6, Tower 7, Tower 8, Tower 9, Tower 10, Tower 12, Tower 17, Tower 18, Tower 19, Tower 21, Tower 24. + - Area 10 can be covered by towers Tower 1, Tower 2, Tower 6, Tower 7, Tower 12, Tower 14, Tower 16, Tower 17, Tower 19, Tower 23, Tower 24. + - Area 11 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 7, Tower 12, Tower 13, Tower 15, Tower 16, Tower 17, Tower 20, Tower 22, Tower 23. + - Area 12 can be covered by towers Tower 2, Tower 4, Tower 5, Tower 8, Tower 10, Tower 11, Tower 13, Tower 14, Tower 17, Tower 19, Tower 22, Tower 23, Tower 24. + - Area 13 can be covered by towers Tower 4, Tower 5, Tower 6, Tower 8, Tower 9, Tower 11, Tower 12, Tower 15, Tower 17, Tower 19, Tower 20, Tower 22, Tower 23. + - Area 14 can be covered by towers Tower 4, Tower 8, Tower 10, Tower 11, Tower 12, Tower 18, Tower 19, Tower 20, Tower 23, Tower 24. + - Area 15 can be covered by towers Tower 0, Tower 2, Tower 3, Tower 4, Tower 6, Tower 7, Tower 8, Tower 9, Tower 11, Tower 13, Tower 14, Tower 15, Tower 18, Tower 22. + - Area 16 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 4, Tower 6, Tower 12, Tower 14, Tower 17, Tower 20, Tower 21, Tower 22. + - Area 17 can be covered by towers Tower 2, Tower 3, Tower 4, Tower 7, Tower 12, Tower 17, Tower 18, Tower 19, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24. + - Area 18 can be covered by towers Tower 1, Tower 3, Tower 4, Tower 5, Tower 7, Tower 8, Tower 9, Tower 11, Tower 12, Tower 15, Tower 18, Tower 19, Tower 20, Tower 22, Tower 23. + - Area 19 can be covered by towers Tower 1, Tower 3, Tower 8, Tower 9, Tower 11, Tower 14, Tower 16, Tower 18, Tower 19, Tower 23. + - Area 20 can be covered by towers Tower 1, Tower 3, Tower 4, Tower 5, Tower 6, Tower 7, Tower 11, Tower 12, Tower 13, Tower 18, Tower 20, Tower 21. + - Area 21 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 6, Tower 8, Tower 10, Tower 12, Tower 16, Tower 18, Tower 21, Tower 24. + - Area 22 can be covered by towers Tower 1, Tower 4, Tower 5, Tower 7, Tower 8, Tower 10, Tower 12, Tower 13, Tower 16, Tower 17, Tower 19, Tower 22, Tower 24. + - Area 23 can be covered by towers Tower 0, Tower 1, Tower 3, Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 12, Tower 14, Tower 15, Tower 17, Tower 21. + - Area 24 can be covered by towers Tower 1, Tower 2, Tower 3, Tower 5, Tower 6, Tower 7, Tower 8, Tower 9, Tower 10, Tower 13, Tower 14, Tower 16, Tower 17, Tower 18, Tower 21. + - Area 25 can be covered by towers Tower 1, Tower 2, Tower 6, Tower 7, Tower 9, Tower 10, Tower 11, Tower 12, Tower 13, Tower 14, Tower 17, Tower 18, Tower 19, Tower 24. + - Area 26 can be covered by towers Tower 1, Tower 2, Tower 4, Tower 6, Tower 9, Tower 12, Tower 15, Tower 16, Tower 18, Tower 19, Tower 20, Tower 22. + - Area 27 can be covered by towers Tower 0, Tower 1, Tower 4, Tower 8, Tower 9, Tower 10, Tower 13, Tower 15, Tower 17, Tower 19, Tower 23, Tower 24. + - Area 28 can be covered by towers Tower 0, Tower 1, Tower 3, Tower 5, Tower 11, Tower 12, Tower 13, Tower 16, Tower 18, Tower 19, Tower 20, Tower 21, Tower 22, Tower 24. + - Area 29 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 8, Tower 10, Tower 11, Tower 12, Tower 14, Tower 15, Tower 16, Tower 17, Tower 18, Tower 20, Tower 23. + +2. **Budget Constraint**: + - The total cost of building the selected towers must not exceed the available budget of 2,501 units. The costs for building each tower are as follows: + - Tower 0: 453 units + - Tower 1: 208 units + - Tower 2: 223 units + - Tower 3: 336 units + - Tower 4: 473 units + - Tower 5: 274 units + - Tower 6: 257 units + - Tower 7: 436 units + - Tower 8: 440 units + - Tower 9: 242 units + - Tower 10: 272 units + - Tower 11: 277 units + - Tower 12: 232 units + - Tower 13: 292 units + - Tower 14: 369 units + - Tower 15: 318 units + - Tower 16: 475 units + - Tower 17: 238 units + - Tower 18: 440 units + - Tower 19: 496 units + - Tower 20: 275 units + - Tower 21: 240 units + - Tower 22: 291 units + - Tower 23: 320 units + - Tower 24: 462 units + +3. **Binary Decisions**: + - Each tower can either be built (1) or not built (0). + - Each area can either be covered (1) or not covered (0). + +#### Summary: +Your task is to select which towers to build and which areas to cover, ensuring that the total population covered is maximized while staying within the budget. The coverage of each area depends on the specific combination of towers built, and the total cost of building the selected towers must not exceed 2,501 units.104087#### Constraints: +1. **Coverage Requirements**: + - Each area must be covered by at least one of the built towers. For example: + - Area 0 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 7, Tower 9, Tower 16, Tower 17, Tower 20, Tower 22, Tower 24. + - Area 1 can be covered by towers Tower 0, Tower 1, Tower 15, Tower 19, Tower 22, Tower 23. + - Area 2 can be covered by towers Tower 1, Tower 4, Tower 8, Tower 11, Tower 12, Tower 13, Tower 16, Tower 18, Tower 20, Tower 22, Tower 23, Tower 24. + - Area 3 can be covered by towers Tower 1, Tower 3, Tower 4, Tower 7, Tower 8, Tower 10, Tower 14, Tower 16, Tower 19, Tower 20, Tower 22, Tower 24. + - Area 4 can be covered by towers Tower 0, Tower 3, Tower 6, Tower 10, Tower 15, Tower 16, Tower 17, Tower 18, Tower 20, Tower 21, Tower 23. + - Area 5 can be covered by towers Tower 3, Tower 5, Tower 7, Tower 10, Tower 11, Tower 12, Tower 13, Tower 14, Tower 16, Tower 17, Tower 18, Tower 19, Tower 21, Tower 22, Tower 24. + - Area 6 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 6, Tower 8, Tower 12, Tower 13, Tower 14, Tower 15, Tower 17, Tower 20, Tower 23, Tower 24. + - Area 7 can be covered by towers Tower 2, Tower 3, Tower 4, Tower 5, Tower 9, Tower 12, Tower 14, Tower 16, Tower 17, Tower 20. + - Area 8 can be covered by towers Tower 0, Tower 2, Tower 7, Tower 12, Tower 13, Tower 14, Tower 17, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24. + - Area 9 can be covered by towers Tower 0, Tower 2, Tower 3, Tower 5, Tower 6, Tower 7, Tower 8, Tower 9, Tower 10, Tower 12, Tower 17, Tower 18, Tower 19, Tower 21, Tower 24. + - Area 10 can be covered by towers Tower 1, Tower 2, Tower 6, Tower 7, Tower 12, Tower 14, Tower 16, Tower 17, Tower 19, Tower 23, Tower 24. + - Area 11 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 7, Tower 12, Tower 13, Tower 15, Tower 16, Tower 17, Tower 20, Tower 22, Tower 23. + - Area 12 can be covered by towers Tower 2, Tower 4, Tower 5, Tower 8, Tower 10, Tower 11, Tower 13, Tower 14, Tower 17, Tower 19, Tower 22, Tower 23, Tower 24. + - Area 13 can be covered by towers Tower 4, Tower 5, Tower 6, Tower 8, Tower 9, Tower 11, Tower 12, Tower 15, Tower 17, Tower 19, Tower 20, Tower 22, Tower 23. + - Area 14 can be covered by towers Tower 4, Tower 8, Tower 10, Tower 11, Tower 12, Tower 18, or 20. Tower 19, Tower 20, Tower 23, Tower 24. + - Area 1 15 can be covered by towers Tower 0, Tower 2, Tower 3, Tower 4, Tower 6, Tower 7, Tower 8, Tower 9, Tower 11, Tower 13, Tower 14, Tower 15, Tower 18, Tower 22. + - Area 16 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 4, Tower 6, Tower 12, Tower 14, Tower 17, Tower 20, Tower 21, Tower 22. + - Area 17 can be covered by towers Tower 2, Tower 3, Tower 4, Tower 7, Tower 12, Tower 17, Tower 18, Tower 19, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24. + - Area 18 can be covered by towers Tower 1, Tower 3, Tower 4, Tower 5, Tower 7, Tower 8, Tower 9, Tower 11, Tower 12, Tower 15, Tower 18, Tower 19, Tower 20, Tower 22, Tower 23. + - Area 19 can be covered by towers Tower 1, Tower 3, Tower 8, Tower 9, Tower 11, Tower 14, Tower 16, Tower 18, Tower 19, Tower 23. + - Area 20 can be covered by towers Tower 1, Tower 3, Tower 4, Tower 5, Tower 6, Tower 7, Tower 11, Tower 12, Tower 13, Tower 18, Tower 20, Tower 21. + - Area 21 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 6, Tower 8, Tower 10, Tower 12, Tower 16, Tower 18, Tower 21, Tower 24. + - Area 22 can be covered by towers Tower 1, Tower 4, Tower 5, Tower 7, Tower 8, Tower 10, Tower 12, Tower 13, Tower 16, Tower 17, Tower 19, Tower 22, Tower 24. + - Area 23 can be covered by towers Tower 0, Tower 1, Tower 3, Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 12, Tower 14, Tower 15, Tower 17, Tower 21. + - Area 24 can be covered by towers Tower 1, Tower 2, Tower 3, Tower 5, Tower 6, Tower 7, Tower 8, Tower 9, Tower 10, Tower 13, Tower 14, Tower 16, Tower 17, Tower 18, Tower 21. + - Area 25 can be covered by towers Tower 1, Tower 2, Tower 6, Tower 7, Tower 9, Tower 10, Tower 11, Tower 12, Tower 13, Tower 14, Tower 17, Tower 18, Tower 19, Tower 24. + - Area 26 can be covered by towers Tower 1, Tower 2, Tower 4, Tower 6, Tower 9, Tower 12, Tower 15, Tower 16, Tower 18, Tower 19, Tower 20, Tower 22. + - Area 27 can be covered by towers Tower 0, Tower 1, Tower 4, Tower 8, Tower 9, Tower 10, Tower 13, Tower 15, Tower 17, Tower 19, Tower 23, or Tower 24. + - Similar coverage constraints apply to all 30 areas, with specific combinations of Area 28 can be covered by towers required for each area. Tower 0, Tower 1, Tower 3, Tower 5, Tower 11, Tower 12, Tower 13, Tower 16, Tower 18, Tower 19, Tower 20, Tower 21, Tower 22, Tower 24. + - Area 29 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 8, Tower 10, Tower 11, Tower 12, Tower 14, Tower 15, Tower 16, Tower 17, Tower 18, Tower 20, Tower 23. + +2. **Budget Constraint**: + - The total cost of building the selected towers must not exceed the available budget of 2,501 units. The costs for building each tower are as follows:No differencesMissing data about which towers cover which regionsmissing data
85Imagine you are managing the distribution of critical medical supplies across eight cities in a region. Each city has a specific supply or demand for these supplies, and your goal is to ensure that all cities receive the necessary resources while minimizing the total transportation cost. The transportation network connects these cities, and each route has a specific cost per unit of supply shipped, as well as a maximum capacity limit. + +#### Cities and Their Supply/Demand: +- **City 0** has a net demand of **34 units** (it needs 34 more units than it currently has). +- **City 1** has a surplus of **11 units** (it can supply 11 units to other cities). +- **City 2** has a surplus of **22 units**. +- **City 3** has a surplus of **10 units**. +- **City 4** has a surplus of **2 units**. +- **City 5** has a surplus of **9 units**. +- **City 6** has a surplus of **6 units**. +- **City 7** has a net demand of **26 units**. + +#### Transportation Costs and Capacities: +The cost of shipping one unit of supply between cities varies depending on the route. Below are the costs and maximum capacities for each route: + +- **From City 0**: + - To City 1: Cost = **9**, Capacity = **59** + - To City 2: Cost = **8**, Capacity = **59** + - To City 3: Cost = **8**, Capacity = **59** + - To City 4: Cost = **9**, Capacity = **43** + - To City 5: Cost = **10**, Capacity = **56** + - To City 6: Cost = **10**, Capacity = **40** + - To City 7: Cost = **9**, Capacity = **37** + +- **From City 1**: + - To City 0: Cost = **10**, Capacity = **36** + - To City 2: Cost = **9**, Capacity = **36** + - To City 3: Cost = **6**, Capacity = **40** + - To City 4: Cost = **9**, Capacity = **39** + - To City 5: Cost = **6**, Capacity = **57** + - To City 6: Cost = **6**, Capacity = **44** + - To City 7: Cost = **8**, Capacity = **56** + +- **From City 2**: + - To City 0: Cost = **6**, Capacity = **52** + - To City 1: Cost = **6**, Capacity = **40** + - To City 3: Cost = **6**, Capacity = **39** + - To City 4: Cost = **7**, Capacity = **31** + - To City 5: Cost = **9**, Capacity = **38** + - To City 6: Cost = **6**, Capacity = **44** + - To City 7: Cost = **10**, Capacity = **58** + +- **From City 3**: + - To City 0: Cost = **10**, Capacity = **30** + - To City 1: Cost = **10**, Capacity = **49** + - To City 2: Cost = **6**, Capacity = **52** + - To City 4: Cost = **10**, Capacity = **47** + - To City 5: Cost = **9**, Capacity = **47** + - To City 6: Cost = **6**, Capacity = **30** + - To City 7: Cost = **8**, Capacity = **34** + +- **From City 4**: + - To City 0: Cost = **6**, Capacity = **39** + - To City 1: Cost = **8**, Capacity = **45** + - To City 2: Cost = **9**, Capacity = **55** + - To City 3: Cost = **8**, Capacity = **36** + - To City 5: Cost = **6**, Capacity = **48** + - To City 6: Cost = **6**, Capacity = **35** + - To City 7: Cost = **10**, Capacity = **54** + +- **From City 5**: + - To City 0: Cost = **9**, Capacity = **43** + - To City 1: Cost = **7**, Capacity = **58** + - To City 2: Cost = **9**, Capacity = **49** + - To City 3: Cost = **6**, Capacity = **33** + - To City 4: Cost = **9**, Capacity = **59** + - To City 6: Cost = **10**, Capacity = **30** + - To City 7: Cost = **7**, Capacity = **59** + +- **From City 6**: + - To City 0: Cost = **10**, Capacity = **33** + - To City 1: Cost = **7**, Capacity = **37** + - To City 2: Cost = **6**, Capacity = **50** + - To City 3: Cost = **7**, Capacity = **34** + - To City 4: Cost = **9**, Capacity = **30** + - To City 5: Cost = **6**, Capacity = **56** + - To City 7: Cost = **10**, Capacity = **44** + +- **From City 7**: + - To City 0: Cost = **7**, Capacity = **46** + - To City 1: Cost = **7**, Capacity = **58** + - To City 2: Cost = **9**, Capacity = **60** + - To City 3: Cost = **7**, Capacity = **34** + - To City 4: Cost = **8**, Capacity = **39** + - To City 5: Cost = **6**, Capacity = **57** + - To City 6: Cost = **9**, Capacity = **42** + +#### Objective: +Your task is to determine the optimal amount of supplies to ship between cities to meet all demands and surpluses while minimizing the total transportation cost. The shipments must respect the capacity limits of each route, and no negative shipments are allowed. + +#### Constraints: +1. **Flow Balance**: For each city, the total amount of supplies shipped into the city must equal the total amount shipped out, adjusted for the city's supply or demand. +2. **Capacity Limits**: The amount of supplies shipped on any route cannot exceed the maximum capacity for that route. +3. **Non-Negativity**: All shipments must be non-negative. + +By solving this problem, you will ensure that all cities receive the necessary medical supplies at the lowest possible transportation cost.443Imagine you are managing the distribution of critical medical supplies across eight cities in a region. Each city has a specific supply or demand for these supplies, and your goal is to ensure that all cities receive the necessary resources while minimizing the total transportation cost. The transportation network connects these cities, and each route has a specific cost per unit of supply shipped, as well as a maximum capacity limit. + +#### Cities and Their Supply/Demand: +- **City 0** has a net demand of **34 units** (it needs 34 more units than it currently has). +- **City 1** has a surplus of **11 units** (it can supply 11 units to other cities). +- **City 2** has a surplus of **22 units**. +- **City 3** has a surplus of **10 units**. +- **City 4** has a surplus of **2 units**. +- **City 5** has a surplus of **9 units**. +- **City 6** has a surplus of **6 units**. +- **City 7** has a net demand of **26 units**. + +#### Transportation Costs and Capacities: +The cost of shipping one unit of supply between cities varies depending on the route. Below are the costs and maximum capacities for each route: + +- **From City 0**: + - To City 1: Cost = **9**, Capacity = **59** + - To City 2: Cost = **8**, Capacity = **59** + - To City 3: Cost = **8**, Capacity = **59** + - To City 4: Cost = **9**, Capacity = **43** + - To City 5: Cost = **10**, Capacity = **56** + - To City 6: Cost = **10**, Capacity = **40** + - To City 7: Cost = **9**, Capacity = **37** + +- **From City 1**: + - To City 0: Cost = **10**, Capacity = **36** + - To City 2: Cost = **9**, Capacity = **36** + - To City 3: Cost = **6**, Capacity = **40** + - To City 4: Cost = **9**, Capacity = **39** + - To City 5: Cost = **6**, Capacity = **57** + - To City 6: Cost = **6**, Capacity = **44** + - To City 7: Cost = **8**, Capacity = **56** + +- **From City 2**: + - To City 0: Cost = **6**, Capacity = **52** + - To City 1: Cost = **6**, Capacity = **40** + - To City 3: Cost = **6**, Capacity = **39** + - To City 4: Cost = **7**, Capacity = **31** + - To City 5: Cost = **9**, Capacity = **38** + - To City 6: Cost = **6**, Capacity = **44** + - To City 7: Cost = **10**, Capacity = **58** + +- **From City 3**: + - To City 0: Cost = **10**, Capacity = **30** + - To City 1: Cost = **10**, Capacity = **49** + - To City 2: Cost = **6**, Capacity = **52** + - To City 4: Cost = **10**, Capacity = **47** + - To City 5: Cost = **9**, Capacity = **47** + - To City 6: Cost = **6**, Capacity = **30** + - To City 7: Cost = **8**, Capacity = **34** + +- **From City 4**: + - To City 0: Cost = **6**, Capacity = **39** + - To City 1: Cost = **8**, Capacity = **45** + - To City 2: Cost = **9**, Capacity = **55** + - To City 3: Cost = **8**, Capacity = **36** + - To City 5: Cost = **6**, Capacity = **48** + - To City 6: Cost = **6**, Capacity = **35** + - To City 7: Cost = **10**, Capacity = **54** + +- **From City 5**: + - To City 0: Cost = **9**, Capacity = **43** + - To City 1: Cost = **7**, Capacity = **58** + - To City 2: Cost = **9**, Capacity = **49** + - To City 3: Cost = **6**, Capacity = **33** + - To City 4: Cost = **9**, Capacity = **59** + - To City 6: Cost = **10**, Capacity = **30** + - To City 7: Cost = **7**, Capacity = **59** + +- **From City 6**: + - To City 0: Cost = **10**, Capacity = **33** + - To City 1: Cost = **7**, Capacity = **37** + - To City 2: Cost = **6**, Capacity = **50** + - To City 3: Cost = **7**, Capacity = **34** + - To City 4: Cost = **9**, Capacity = **30** + - To City 5: Cost = **6**, Capacity = **56** + - To City 7: Cost = **10**, Capacity = **44** + +- **From City 7**: + - To City 0: Cost = **7**, Capacity = **46** + - To City 1: Cost = **7**, Capacity = **58** + - To City 2: Cost = **9**, Capacity = **60** + - To City 3: Cost = **7**, Capacity = **34** + - To City 4: Cost = **8**, Capacity = **39** + - To City 5: Cost = **6**, Capacity = **57** + - To City 6: Cost = **9**, Capacity = **42** + +#### Objective: +Your task is to determine the optimal amount of supplies to ship between cities to meet all demands and surpluses while minimizing the total transportation cost. The shipments must respect the capacity limits of each route, and no negative shipments are allowed. + +#### Constraints: +1. **Flow Balance**: For each city, the total amount of supplies shipped into the city must equal the total amount shipped out, adjusted for the city's supply or demand. +2. **Capacity Limits**: The amount of supplies shipped on any route cannot exceed the maximum capacity for that route. +3. **Non-Negativity**: All shipments must be non-negative. + +By solving this problem, you will ensure that all cities receive the necessary medical supplies at the lowest possible transportation cost.443No differencesNo differencesok
86You are managing the landing schedule for six aircraft at a busy airport. Each aircraft has a target landing time, but due to operational constraints, the actual landing time may vary. Your goal is to minimize the total penalty costs associated with landing aircraft either earlier or later than their target times, while ensuring safe separation between consecutive landings. + +Aircraft 0 has a target landing time of 86 units, can land as early as 60 units or as late as 100 units, and incurs a penalty of 12 units per time unit for landing early and 20 units per time unit for landing late. Aircraft 1 has a target landing time of 60 units, can land as early as 60 units or as late as 90 units, and incurs a penalty of 13 units per time unit for landing early and 16 units per time unit for landing late. Aircraft 2 has a target landing time of 71 units, can land as early as 60 units or as late as 100 units, and incurs a penalty of 16 units per time unit for landing early and 15 units per time unit for landing late. Aircraft 3 has a target landing time of 97 units, can land as early as 67 units or as late as 100 units, and incurs a penalty of 14 units per time unit for landing early and 13 units per time unit for landing late. Aircraft 4 has a target landing time of 89 units, can land as early as 60 units or as late as 100 units, and incurs a penalty of 15 units per time unit for landing early and 18 units per time unit for landing late. Aircraft 5 has a target landing time of 74 units, can land as early as 60 units or as late as 100 units, and incurs a penalty of 14 units per time unit for landing early and 19 units per time unit for landing late. + +To ensure safety, there must be a minimum separation time between the landings of any two aircraft. If Aircraft 0 lands before Aircraft 1, there must be at least 7 units of separation. If Aircraft 1 lands before Aircraft 0, the separation must be at least 5 units. If Aircraft 0 lands before Aircraft 2, there must be at least 7 units of separation. If Aircraft 2 lands before Aircraft 0, the separation must be at least 5 units. If Aircraft 0 lands before Aircraft 3, there must be at least 5 units of separation. If Aircraft 3 lands before Aircraft 0, the separation must be at least 8 units. If Aircraft 0 lands before Aircraft 4, there must be at least 5 units of separation. If Aircraft 4 lands before Aircraft 0, the separation must be at least 7 units. If Aircraft 0 lands before Aircraft 5, there must be at least 5 units of separation. If Aircraft 5 lands before Aircraft 0, the separation must be at least 7 units. If Aircraft 1 lands before Aircraft 2, there must be at least 7 units of separation. If Aircraft 2 lands before Aircraft 1, the separation must be at least 8 units. If Aircraft 1 lands before Aircraft 3, there must be at least 5 units of separation. If Aircraft 3 lands before Aircraft 1, the separation must be at least 8 units. If Aircraft 1 lands before Aircraft 4, there must be at least 8 units of separation. If Aircraft 4 lands before Aircraft 1, the separation must be at least 7 units. If Aircraft 1 lands before Aircraft 5, there must be at least 7 units of separation. If Aircraft 5 lands before Aircraft 1, the separation must be at least 5 units. If Aircraft 2 lands before Aircraft 3, there must be at least 8 units of separation. If Aircraft 3 lands before Aircraft 2, the separation must be at least 8 units. If Aircraft 2 lands before Aircraft 4, there must be at least 8 units of separation. If Aircraft 4 lands before Aircraft 2, the separation must be at least 6 units. If Aircraft 2 lands before Aircraft 5, there must be at least 6 units of separation. If Aircraft 5 lands before Aircraft 2, the separation must be at least 8 units. If Aircraft 3 lands before Aircraft 4, there must be at least 6 units of separation. If Aircraft 4 lands before Aircraft 3, the separation must be at least 8 units. If Aircraft 3 lands before Aircraft 5, there must be at least 5 units of separation. If Aircraft 5 lands before Aircraft 3, the separation must be at least 7 units. If Aircraft 4 lands before Aircraft 5, there must be at least 6 units of separation. If Aircraft 5 lands before Aircraft 4, the separation must be at least 7 units. + +Your task is to determine the actual landing times for all six aircraft such that each aircraft lands within its allowed time window, the separation requirements between consecutive landings are satisfied, and the total penalty cost, which is the sum of the penalties for landing early or late for all aircraft, is minimized. For each pair of aircraft, one must land before the other, and the separation time must be respected. The landing time for each aircraft must fall within its specified earliest and latest landing times. The penalties for landing early or late are calculated based on the difference between the actual landing time and the target landing time.72You are managing the landing schedule for six aircraft at a busy airport. Each aircraft has a target landing time, but due to operational constraints, the actual landing time may vary. Your goal is to minimize the total penalty costs associated with landing aircraft either earlier or later than their target times, while ensuring safe separation between consecutive landings. + +Aircraft 0 has a target landing time of 86 units, can land as early as 60 units or as late as 100 units, and incurs a penalty of 12 units per time unit for landing early and 20 units per time unit for landing late. Aircraft 1 has a target landing time of 60 units, can land as early as 60 units or as late as 90 units, and incurs a penalty of 13 units per time unit for landing early and 16 units per time unit for landing late. Aircraft 2 has a target landing time of 71 units, can land as early as 60 units or as late as 100 units, and incurs a penalty of 16 units per time unit for landing early and 15 units per time unit for landing late. Aircraft 3 has a target landing time of 97 units, can land as early as 67 units or as late as 100 units, and incurs a penalty of 14 units per time unit for landing early and 13 units per time unit for landing late. Aircraft 4 has a target landing time of 89 units, can land as early as 60 units or as late as 100 units, and incurs a penalty of 15 units per time unit for landing early and 18 units per time unit for landing late. Aircraft 5 has a target landing time of 74 units, can land as early as 60 units or as late as 100 units, and incurs a penalty of 14 units per time unit for landing early and 19 units per time unit for landing late. + +To ensure safety, there must be a minimum separation time between the landings of any two aircraft. If Aircraft 0 lands before Aircraft 1, there must be at least 7 units of separation. If Aircraft 1 lands before Aircraft 0, the separation must be at least 5 units. If Aircraft 0 lands before Aircraft 2, there must be at least 7 units of separation. If Aircraft 2 lands before Aircraft 0, the separation must be at least 5 units. If Aircraft 0 lands before Aircraft 3, there must be at least 5 units of separation. If Aircraft 3 lands before Aircraft 0, the separation must be at least 8 units. If Aircraft 0 lands before Aircraft 4, there must be at least 5 units of separation. If Aircraft 4 lands before Aircraft 0, the separation must be at least 7 units. If Aircraft 0 lands before Aircraft 5, there must be at least 5 units of separation. If Aircraft 5 lands before Aircraft 0, the separation must be at least 7 units. If Aircraft 1 lands before Aircraft 2, there must be at least 7 units of separation. If Aircraft 2 lands before Aircraft 1, the separation must be at least 8 units. If Aircraft 1 lands before Aircraft 3, there must be at least 5 units of separation. If Aircraft 3 lands before Aircraft 1, the separation must be at least 8 units. If Aircraft 1 lands before Aircraft 4, there must be at least 8 units of separation. If Aircraft 4 lands before Aircraft 1, the separation must be at least 7 units. If Aircraft 1 lands before Aircraft 5, there must be at least 7 units of separation. If Aircraft 5 lands before Aircraft 1, the separation must be at least 5 units. If Aircraft 2 lands before Aircraft 3, there must be at least 8 units of separation. If Aircraft 3 lands before Aircraft 2, the separation must be at least 8 units. If Aircraft 2 lands before Aircraft 4, there must be at least 8 units of separation. If Aircraft 4 lands before Aircraft 2, the separation must be at least 6 units. If Aircraft 2 lands before Aircraft 5, there must be at least 6 units of separation. If Aircraft 5 lands before Aircraft 2, the separation must be at least 8 units. If Aircraft 3 lands before Aircraft 4, there must be at least 6 units of separation. If Aircraft 4 lands before Aircraft 3, the separation must be at least 8 units. If Aircraft 3 lands before Aircraft 5, there must be at least 5 units of separation. If Aircraft 5 lands before Aircraft 3, the separation must be at least 7 units. If Aircraft 4 lands before Aircraft 5, there must be at least 6 units of separation. If Aircraft 5 lands before Aircraft 4, the separation must be at least 7 units. + +Your task is to determine the actual landing times for all six aircraft such that each aircraft lands within its allowed time window, the separation requirements between consecutive landings are satisfied, and the total penalty cost, which is the sum of the penalties for landing early or late for all aircraft, is minimized. For each pair of aircraft, one must land before the other, and the separation time must be respected. The landing time for each aircraft must fall within its specified earliest and latest landing times. The penalties for landing early or late are calculated based on the difference between the actual landing time and the target landing time.72No differencesNo differencesok
87In a manufacturing facility, six jobs labeled Job 0 through Job 5 must be scheduled, each consisting of two operations. The goal is to minimize the total production time (makespan), ensuring all jobs are completed efficiently. Each operation has a processing time of 1 unit of time. Operations within the same job must follow a sequence: the second operation cannot start until the first operation is completed. For example, Operation (0,1) of Job 0 cannot start until Operation (0,0) of Job 0 is finished, and this applies to all jobs. + +Certain operations share the same machine and cannot overlap in time. The machine assignments are inferred from the constraints, which specify pairwise relationships between operations. For example, Operation (0,0) of Job 0 and Operation (3,0) of Job 3 must be scheduled on the same machine without overlapping, and similarly for other operation pairs. The order of operations on shared machines is determined by binary decision variables. For instance, if Operation (0,0) of Job 0 is scheduled before Operation (3,0) of Job 3 on a shared machine, the corresponding binary variable is set to 1; otherwise, it is set to 0. + +The total production time (makespan) must be at least as large as the completion time of every operation. For example, if Operation (0,0) of Job 0 finishes at time 1 and Operation (0,1) of Job 0 finishes at time 2, the makespan must be at least 2. Similarly, if Operation (5,1) of Job 5 finishes at time 10, the makespan must be at least 10. + +To enforce the constraints, a large constant \( N = 100,000 \) is used in the machine capacity constraints to ensure that operations sharing the same machine do not overlap. The start times of all operations are continuous variables, and the binary variables determine the order of operations on shared machines. The objective is to minimize the makespan while respecting the precedence constraints within jobs, the machine capacity constraints, and the non-negativity and binary constraints on the decision variables.4In a manufacturing facility, six jobs labeled Job 0 through Job 5 must be scheduled, each consisting of two operations. The goal is to minimize the total production time (makespan), ensuring all jobs are completed efficiently. Each operation has a processing time of 1 unit of time. Operations within the same job must follow a sequence: the second operation cannot start until the first operation is completed. For example, Operation (0,1) of Job 0 cannot start until Operation (0,0) of Job 0 is finished, and this applies to all jobs. + +Certain operations share the same machine and cannot overlap in time. The machine assignments are listed explicitly below so a solver can be implemented directly. When two operations share the same machine, their order is decided by a binary variable which enforces a non-overlap constraint using a large constant N. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3, 4, 5} +- For each job j in J, there are 2 operations indexed k = 0..1. We denote operation (j,k) where k=0 is the first operation and k=1 is the second operation. +- Machines: M = {M0, M1, M2} + +2) Machine assignment for every operation (which machine executes the operation) +- Operation (0,0) -> M0 +- Operation (0,1) -> M1 +- Operation (1,0) -> M0 +- Operation (1,1) -> M2 +- Operation (2,0) -> M1 +- Operation (2,1) -> M2 +- Operation (3,0) -> M0 +- Operation (3,1) -> M1 +- Operation (4,0) -> M1 +- Operation (4,1) -> M2 +- Operation (5,0) -> M0 +- Operation (5,1) -> M2 + +3) Processing times (duration) for every operation (in time units) +- All operations have processing time p(j,k) = 1 for all j in J and k in {0,1}.5In a manufacturing facility, six jobs labeled Job 0 through Job 5 must be scheduled, each consisting of two operations. The goal is to minimize the total production time (makespan), ensuring all jobs are completed efficiently. Each operation has a processing time of 1 unit of time. Operations within the same job must follow a sequence: the second operation cannot start until the first operation is completed. For example, Operation (0,1) of Job 0 cannot start until Operation (0,0) of Job 0 is finished, and this applies to all jobs. + +Certain operations share the same machine and cannot overlap in time. The machine assignments are inferred from the constraints, which specify pairwise relationships between operations. For example, Operation (0,0) of Job 0 and Operation (3,0) of Job 3 must listed explicitly below so a solver can be scheduled on implemented directly. When two operations share the same machine without overlapping, and similarly for other operation pairs. The machine, their order of operations on shared machines is determined decided by binary decision variables. For instance, if Operation (0,0) of Job 0 is scheduled before Operation (3,0) of Job 3 on a shared machine, the corresponding binary variable is set to 1; otherwise, it is set to 0. which enforces a non-overlap constraint using a large constant N. + + The total production time (makespan) must be at least as large as the completion time of every operation. For example, if Operation (0,0) of Job 0 finishes at time 1 and Operation (0,1) of Job 0 finishes at time 2, the makespan must be at least 2. Similarly, if Operation (5,1) of Job 5 finishes at time 10, the makespan must be at least 10. ADDITIONAL INPUT DATA: + + To enforce the constraints, a large constant \( N 1) Sets and indices + - Jobs: J = 100,000 \) is used {0, 1, 2, 3, 4, 5} + - For each job j in the machine capacity constraints to ensure that operations sharing the same machine do not overlap. The start times of all operations J, there are continuous variables, and the binary variables determine the order of 2 operations on shared machines. The objective indexed k = 0..1. We denote operation (j,k) where k=0 is to minimize the makespan while respecting the precedence constraints within jobs, first operation and k=1 is the second operation. + - Machines: M = {M0, M1, M2} + + 2) Machine assignment for every operation (which machine capacity constraints, and executes the non-negativity operation) + - Operation (0,0) -> M0 + - Operation (0,1) -> M1 + - Operation (1,0) -> M0 + - Operation (1,1) -> M2 + - Operation (2,0) -> M1 + - Operation (2,1) -> M2 + - Operation (3,0) -> M0 + - Operation (3,1) -> M1 + - Operation (4,0) -> M1 + - Operation (4,1) -> M2 + - Operation (5,0) -> M0 + - Operation (5,1) -> M2 + + 3) Processing times (duration) for every operation (in time units) + - All operations have processing time p(j,k) = 1 for all j in J and binary constraints on the decision variables. k in {0,1}.4 5It was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
88Cell Tower Coverage Planning + +You are tasked with planning the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 32 distinct areas, each with a specific population size that needs to be covered by cell tower signals. Your goal is to determine which cell towers to build and which areas to cover, ensuring that the total cost of building the towers does not exceed the available budget. + +#### Key Details: +1. **Population Coverage**: Each area has a specific population size that can be covered if a cell tower is built in a location that serves that area. The populations of the areas range from 2,010 to 4,970 people, with the total population across all areas being a critical factor in your decision-making. + +2. **Cell Tower Options**: There are 29 potential cell tower locations, each with a unique construction cost. The costs range from 210 units for Tower 1 to 498 units for Tower 6. Building a tower at any of these locations will enable coverage for specific areas, depending on the tower's range and location. + +3. **Coverage Requirements**: For an area to be considered covered, at least one of the cell towers that can serve it must be built. For example, Area 0 can be covered if any of Towers 0, 2, 5, 6, 11, 18, 20, 21, 26, or 27 are built. Similar coverage relationships exist for all 32 areas. + +4. **Budget Constraint**: The total cost of building the selected cell towers must not exceed 3,226 units. This budget must be carefully allocated to maximize population coverage while staying within the financial limit. + +5. **Decision Variables**: + - **Build Decisions**: For each of the 29 cell tower locations, you must decide whether to build the tower (yes or no). + - **Coverage Decisions**: For each of the 32 areas, you must determine whether it will be covered (yes or no), based on the towers built. + +#### Objective: +Your primary objective is to maximize the total population covered by the cell towers. This means selecting a combination of towers to build that ensures the largest possible number of people across the 32 areas have access to cell service, while staying within the budget. + +#### Constraints: +1. **Coverage Logic**: An area can only be considered covered if at least one of the cell towers capable of serving it is built. For example, Area 1 can only be covered if at least one of Towers 0, 2, 3, 4, 5, 7, 10, 11, 15, 16, 17, 20, 23, or 28 is built. + +2. **Budget Limit**: The total cost of building the selected towers must not exceed 3,226 units. This requires careful selection of towers to balance coverage and cost. + +3. **Binary Decisions**: Both the decision to build a tower and the decision to cover an area are binary (yes or no). You cannot partially build a tower or partially cover an area. + +#### Numerical Parameters: +- **Population Sizes**: The populations of the 32 areas are as follows: + - Area 0: 2,285 + - Area 1: 3,023 + - Area 2: 3,557 + - Area 3: 3,938 + - Area 4: 3,689 + - Area 5: 2,426 + - Area 6: 4,697 + - Area 7: 3,214 + - Area 8: 2,580 + - Area 9: 2,010 + - Area 10: 3,180 + - Area 11: 4,083 + - Area 12: 2,787 + - Area 13: 3,095 + - Area 14: 2,898 + - Area 15: 4,374 + - Area 16: 4,970 + - Area 17: 3,689 + - Area 18: 4,612 + - Area 19: 3,232 + - Area 20: 4,177 + - Area 21: 2,211 + - Area 22: 3,123 + - Area 23: 4,194 + - Area 24: 3,774 + - Area 25: 3,391 + - Area 26: 3,959 + - Area 27: 3,010 + - Area 28: 3,213 + - Area 29: 3,106 + - Area 30: 2,652 + - Area 31: 3,441 + +- **Tower Costs**: The construction costs for the 29 towers are as follows: + - Tower 0: 323 units + - Tower 1: 210 units + - Tower 2: 360 units + - Tower 3: 480 units + - Tower 4: 375 units + - Tower 5: 491 units + - Tower 6: 498 units + - Tower 7: 267 units + - Tower 8: 413 units + - Tower 9: 417 units + - Tower 10: 291 units + - Tower 11: 388 units + - Tower 12: 284 units + - Tower 13: 463 units + - Tower 14: 309 units + - Tower 15: 382 units + - Tower 16: 476 units + - Tower 17: 325 units + - Tower 18: 428 units + - Tower 19: 265 units + - Tower 20: 323 units + - Tower 21: 282 units + - Tower 22: 441 units + - Tower 23: 467 units + - Tower 24: 491 units + - Tower 25: 232 units + - Tower 26: 396 units + - Tower 27: 370 units + - Tower 28: 309 units + +#### Summary: +Your task is to select which cell towers to build and which areas to cover, ensuring that the total population covered is maximized while staying within the budget of 3,226 units. The decisions must respect the coverage requirements for each area and the binary nature of the build and coverage decisions.108590Cell Tower Coverage Planning + +You are tasked with planning the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 32 distinct areas, each with a specific population size that needs to be covered by cell tower signals. Your goal is to determine which cell towers to build and which areas to cover, ensuring that the total cost of building the towers does not exceed the available budget. + +#### Key Details: +1. **Population Coverage**: Each area has a specific population size that can be covered if a cell tower is built in a location that serves that area. +2. **Cell Tower Options**: There are 29 potential cell tower locations, each with a unique construction cost. The costs range from 210 units for Tower 1 to 498 units for Tower 6. Building a tower at any of these locations will enable coverage for specific areas, depending on the tower's range and location. + +3. **Coverage Requirements**: For an area to be considered covered, at least one of the cell towers that can serve it must be built. Here is the list of which towers can cover which areas: + - Area 0 can be covered if any of the towers 20, 3, 0, 23, 8, 7, 25, 4, 27, 17, 2, or 13 are built. + - Area 1 can be covered if any of the towers 1, 0, 2, 6, 7, 16, 19, 27, 17, 25, 20, or 13 are built. + - Area 2 can be covered if any of the towers 7, 14, 18, 8, 0, 5, 22, 13, 10, 25, 4, or 6 are built. + - Area 3 can be covered if any of the towers 24, 10, 3, 2, 12, 26, 11, 22, 19, 8, 1, or 14 are built. + - Area 4 can be covered if any of the towers 17, 3, 12, 2, 28, 9, 20, 19, 11, 18, 6, or 25 are built. + - Area 5 can be covered if any of the towers 1, 21, 7, 24, 9, 2, 26, 3, 12, 8, 14, or 11 are built. + - Area 6 can be covered if any of the towers 5, 11, 27, 6, 21, 8, 22, 24, 20, 2, 28, or 17 are built. + - Area 7 can be covered if any of the towers 23, 7, 5, 14, 12, 8, 20, 17, 27, 10, 1, or 22 are built. + - Area 8 can be covered if any of the towers 26, 1, 25, 10, 12, 8, 2, 6, 18, 28, 21, or 15 are built. + - Area 9 can be covered if any of the towers 12, 20, 14, 4, 8, 25, 7, 17, 21, 24, 18, or 13 are built. + - Area 10 can be covered if any of the towers 28, 18, 12, 11, 7, 4, 16, 15, 2, 1, 3, or 23 are built. + - Area 11 can be covered if any of the towers 20, 5, 25, 21, 13, 19, 2, 12, 26, 23, 14, or 16 are built. + - Area 12 can be covered if any of the towers 8, 17, 0, 21, 23, 3, 25, 27, 28, 10, 24, or 9 are built. + - Area 13 can be covered if any of the towers 13, 5, 14, 0, 23, 24, 8, 16, 27, 21, 3, or 9 are built. + - Area 14 can be covered if any of the towers 26, 20, 16, 19, 6, 4, 11, 5, 17, 28, 0, or 10 are built. + - Area 15 can be covered if any of the towers 15, 0, 3, 11, 9, 7, 1, 23, 18, 2, 19, or 28 are built. + - Area 16 can be covered if any of the towers 26, 2, 24, 17, 28, 4, 23, 21, 15, 25, 5, or 8 are built. + - Area 17 can be covered if any of the towers 16, 27, 19, 13, 6, 17, 22, 24, 9, 12, 11, or 14 are built. + - Area 18 can be covered if any of the towers 28, 16, 14, 3, 7, 24, 2, 10, 0, 18, 17, or 23 are built. + - Area 19 can be covered if any of the towers 18, 7, 0, 2, 22, 20, 1, 27, 25, 24, 10, or 23 are built. + - Area 20 can be covered if any of the towers 16, 7, 8, 21, 15, 6, 17, 4, 18, 20, 24, or 27 are built. + - Area 21 can be covered if any of the towers 25, 15, 28, 13, 6, 3, 23, 21, 26, 11, 20, or 18 are built. + - Area 22 can be covered if any of the towers 14, 27, 23, 1, 21, 20, 26, 3, 25, 12, 10, or 24 are built. + - Area 23 can be covered if any of the towers 7, 6, 27, 17, 14, 4, 13, 5, 8, 24, 28, or 2 are built. + - Area 24 can be covered if any of the towers 14, 25, 17, 3, 1, 20, 26, 0, 2, 7, 5, or 13 are built. + - Area 25 can be covered if any of the towers 15, 28, 6, 12, 1, 5, 25, 0, 22, 8, 14, or 9 are built. + - Area 26 can be covered if any of the towers 13, 22, 23, 25, 17, 21, 27, 15, 4, 6, 9, or 19 are built. + - Area 27 can be covered if any of the towers 1, 18, 23, 17, 28, 26, 10, 24, 21, 27, 15, or 16 are built. + - Area 28 can be covered if any of the towers 27, 16, 5, 1, 28, 2, 26, 23, 19, 21, 7, or 12 are built. + - Area 29 can be covered if any of the towers 3, 18, 7, 27, 19, 1, 24, 2, 13, 25, 22, or 16 are built. + - Area 30 can be covered if any of the towers 10, 8, 6, 21, 22, 28, 7, 27, 12, 4, 9, or 14 are built. + - Area 31 can be covered if any of the towers 10, 24, 2, 0, 14, 19, 18, 3, 26, 17, 6, or 16 are built. +4. **Budget Constraint**: The total cost of building the selected cell towers must not exceed 3,226 units. This budget must be carefully allocated to maximize population coverage while staying within the financial limit. + +5. **Decision Variables**: + - **Build Decisions**: For each of the 29 cell tower locations, you must decide whether to build the tower (yes or no). + - **Coverage Decisions**: For each of the 32 areas, you must determine whether it will be covered (yes or no), based on the towers built. + +#### Objective: +Your primary objective is to maximize the total population covered by the cell towers. This means selecting a combination of towers to build that ensures the largest possible number of people across the 32 areas have access to cell service, while staying within the budget. + +#### Constraints: +1. **Coverage Logic**: An area can only be considered covered if at least one of the cell towers capable of serving it is built. For example, Area 1 can only be covered if at least one of Towers 0, 2, 3, 4, 5, 7, 10, 11, 15, 16, 17, 20, 23, or 28 is built. + +2. **Budget Limit**: The total cost of building the selected towers must not exceed 3,226 units. This requires careful selection of towers to balance coverage and cost. + +3. **Binary Decisions**: Both the decision to build a tower and the decision to cover an area are binary (yes or no). You cannot partially build a tower or partially cover an area. + +#### Numerical Parameters: +- **Population Sizes**: The populations of the 32 areas are as follows: + - Area 0: 2,285 + - Area 1: 3,023 + - Area 2: 3,557 + - Area 3: 3,938 + - Area 4: 3,689 + - Area 5: 2,426 + - Area 6: 4,697 + - Area 7: 3,214 + - Area 8: 2,580 + - Area 9: 2,010 + - Area 10: 3,180 + - Area 11: 4,083 + - Area 12: 2,787 + - Area 13: 3,095 + - Area 14: 2,898 + - Area 15: 4,374 + - Area 16: 4,970 + - Area 17: 3,689 + - Area 18: 4,612 + - Area 19: 3,232 + - Area 20: 4,177 + - Area 21: 2,211 + - Area 22: 3,123 + - Area 23: 4,194 + - Area 24: 3,774 + - Area 25: 3,391 + - Area 26: 3,959 + - Area 27: 3,010 + - Area 28: 3,213 + - Area 29: 3,106 + - Area 30: 2,652 + - Area 31: 3,441 + +- **Tower Costs**: The construction costs for the 29 towers are as follows: + - Tower 0: 323 units + - Tower 1: 210 units + - Tower 2: 360 units + - Tower 3: 480 units + - Tower 4: 375 units + - Tower 5: 491 units + - Tower 6: 498 units + - Tower 7: 267 units + - Tower 8: 413 units + - Tower 9: 417 units + - Tower 10: 291 units + - Tower 11: 388 units + - Tower 12: 284 units + - Tower 13: 463 units + - Tower 14: 309 units + - Tower 15: 382 units + - Tower 16: 476 units + - Tower 17: 325 units + - Tower 18: 428 units + - Tower 19: 265 units + - Tower 20: 323 units + - Tower 21: 282 units + - Tower 22: 441 units + - Tower 23: 467 units + - Tower 24: 491 units + - Tower 25: 232 units + - Tower 26: 396 units + - Tower 27: 370 units + - Tower 28: 309 units + +#### Summary: +Your task is to select which cell towers to build and which areas to cover, ensuring that the total population covered is maximized while staying within the budget of 3,226 units. The decisions must respect the coverage requirements for each area and the binary nature of the build and coverage decisions.108590You are tasked with planning the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 32 distinct areas, each with a specific population size that needs to be covered by cell tower signals. Your goal is to determine which cell towers to build and which areas to cover, ensuring that the total cost of building the towers does not exceed the available budget. + +#### Key Details: +1. **Population Coverage**: Each area has a specific population size that can be covered if a cell tower is built in a location that serves that area. The populations of the areas range from 2,010 to 4,970 people, with the total population across all areas being a critical factor in your decision-making. +2. **Cell Tower Options**: There are 29 potential cell tower locations, each with a unique construction cost. The costs range from 210 units for Tower 1 to 498 units for Tower 6. Building a tower at any of these locations will enable coverage for specific areas, depending on the tower's range and location. + +3. **Coverage Requirements**: For an area to be considered covered, at least one of the cell towers that can serve it must be built. For example, Here is the list of which towers can cover which areas: + - Area 0 can be covered if any of Towers the towers 20, 3, 0, 23, 8, 7, 25, 4, 27, 17, 2, or 13 are built. + - Area 1 can be covered if any of the towers 1, 0, 2, 6, 7, 16, 19, 27, 17, 25, 20, or 13 are built. + - Area 2 can be covered if any of the towers 7, 14, 18, 8, 0, 5, 22, 13, 10, 25, 4, or 6 are built. + - Area 3 can be covered if any of the towers 24, 10, 3, 2, 12, 26, 11, 22, 19, 8, 1, or 14 are built. + - Area 4 can be covered if any of the towers 17, 3, 12, 2, 28, 9, 20, 19, 11, 18, 6, or 25 are built. + - Area 5 can be covered if any of the towers 1, 21, 7, 24, 9, 2, 26, 3, 12, 8, 14, or 11 are built. + - Area 6 can be covered if any of the towers 5, 11, 27, 6, 21, 8, 22, 24, 20, 2, 28, or 17 are built. + - Area 7 can be covered if any of the towers 23, 7, 5, 14, 12, 8, 20, 17, 27, 10, 1, or 22 are built. + - Area 8 can be covered if any of the towers 26, 1, 25, 10, 12, 8, 2, 6, 18, 28, 21, or 15 are built. + - Area 9 can be covered if any of the towers 12, 20, 14, 4, 8, 25, 7, 17, 21, 24, 18, or 13 are built. + - Area 10 can be covered if any of the towers 28, 18, 12, 11, 7, 4, 16, 15, 2, 1, 3, or 23 are built. + - Area 11 can be covered if any of the towers 20, 5, 25, 21, 13, 19, 2, 12, 26, 23, 14, or 16 are built. + - Area 12 can be covered if any of the towers 8, 17, 0, 21, 23, 3, 25, 27, 28, 10, 24, or 9 are built. + - Area 13 can be covered if any of the towers 13, 5, 14, 0, 23, 24, 8, 16, 27, 21, 3, or 9 are built. + - Area 14 can be covered if any of the towers 26, 20, 16, 19, 6, 4, 11, 5, 17, 28, 0, or 10 are built. + - Area 15 can be covered if any of the towers 15, 0, 3, 11, 9, 7, 1, 23, 18, 2, 19, or 28 are built. + - Area 16 can be covered if any of the towers 26, 2, 24, 17, 28, 4, 23, 21, 15, 25, 5, or 8 are built. + - Area 17 can be covered if any of the towers 16, 27, 19, 13, 6, 17, 22, 24, 9, 12, 11, or 14 are built. + - Area 18 can be covered if any of the towers 28, 16, 14, 3, 7, 24, 2, 10, 0, 18, 17, or 23 are built. + - Area 19 can be covered if any of the towers 18, 7, 0, 2, 22, 20, 1, 27, 25, 24, 10, or 23 are built. + - Area 20 can be covered if any of the towers 16, 7, 8, 21, 15, 6, 17, 4, 18, 20, 24, or 27 are built. + - Area 21 can be covered if any of the towers 25, 15, 28, 13, 6, 3, 23, 21, 26, 11, 20, or 18 are built. + - Area 22 can be covered if any of the towers 14, 27, 23, 1, 21, 20, 26, 3, 25, 12, 10, or 24 are built. + - Area 23 can be covered if any of the towers 7, 6, 27, 17, 14, 4, 13, 5, 8, 24, 28, or 2 are built. + - Area 24 can be covered if any of the towers 14, 25, 17, 3, 1, 20, 26, 0, 2, 7, 5, or 13 are built. + - Area 25 can be covered if any of the towers 15, 28, 6, 12, 1, 5, 25, 0, 22, 8, 14, or 9 are built. + - Area 26 can be covered if any of the towers 13, 22, 23, 25, 17, 21, 27, 15, 4, 6, 9, or 19 are built. + - Area 27 can be covered if any of the towers 1, 18, 23, 17, 28, 26, 10, 24, 21, 27, 15, or 16 are built. + - Area 28 can be covered if any of the towers 27, 16, 5, 1, 28, 2, 26, 23, 19, 21, 7, or 12 are built. + - Area 29 can be covered if any of the towers 3, 18, 7, 27, 19, 1, 24, 2, 13, 25, 22, or 16 are built. + - Area 30 can be covered if any of the towers 10, 8, 6, 21, 22, 28, 7, 27, 12, 4, 9, or 14 are built. + - Area 31 can be covered if any of the towers 10, 24, 2, 0, 14, 19, 18, 3, 26, 17, 6, or 16 are built. Similar coverage relationships exist for all 32 areas. +4. **Budget Constraint**: The total cost of building the selected cell towers must not exceed 3,226 units. This budget must be carefully allocated to maximize population coverage while staying within the financial limit. + +5. **Decision Variables**:No differencesMissing data about tower coveragemissing data
89Assign six staff members to four projects such that each staff member is assigned to exactly one project. Each staff member has specific skill levels in three skills, and each project has required skill levels for each skill. The skill level attained for each project is determined by the highest skill level among the staff members assigned to that project. The skill shortage for each project is the difference between the required skill level and the attained skill level. The goal is to minimize the maximum skill shortage across all projects and skills. Staff members and their skill levels are as follows: Person 0 has Skill 0 (Level 5), Skill 1 (Level 5), Skill 2 (Level 5); Person 1 has Skill 0 (Level 6), Skill 1 (Level 5), Skill 2 (Level 7); Person 2 has Skill 0 (Level 6), Skill 1 (Level 5), Skill 2 (Level 7); Person 3 has Skill 0 (Level 5), Skill 1 (Level 5), Skill 2 (Level 6); Person 4 has Skill 0 (Level 7), Skill 1 (Level 5), Skill 2 (Level 6); Person 5 has Skill 0 (Level 6), Skill 1 (Level 7), Skill 2 (Level 5). Project skill requirements are as follows: Project 0 requires Skill 0 (Level 9), Skill 1 (Level 9), Skill 2 (Level 7); Project 1 requires Skill 0 (Level 7), Skill 1 (Level 7), Skill 2 (Level 8); Project 2 requires Skill 0 (Level 9), Skill 1 (Level 9), Skill 2 (Level 7); Project 3 requires Skill 0 (Level 7), Skill 1 (Level 8), Skill 2 (Level 9). The objective is to minimize the maximum skill shortage across all projects and skills.0Assign six staff members to four projects such that each staff member is assigned to exactly one project. Each staff member has specific skill levels in three skills, and each project has required skill levels for each skill. The skill level attained for each project is determined by the highest skill level among the staff members assigned to that project. The skill shortage for each project is the difference between the required skill level and the attained skill level. The goal is to minimize the maximum skill shortage across all projects and skills. Staff members and their skill levels are as follows: Person 0 has Skill 0 (Level 5), Skill 1 (Level 5), Skill 2 (Level 5); Person 1 has Skill 0 (Level 6), Skill 1 (Level 5), Skill 2 (Level 7); Person 2 has Skill 0 (Level 6), Skill 1 (Level 5), Skill 2 (Level 7); Person 3 has Skill 0 (Level 5), Skill 1 (Level 5), Skill 2 (Level 6); Person 4 has Skill 0 (Level 7), Skill 1 (Level 5), Skill 2 (Level 6); Person 5 has Skill 0 (Level 6), Skill 1 (Level 7), Skill 2 (Level 5). Project skill requirements are as follows: Project 0 requires Skill 0 (Level 9), Skill 1 (Level 9), Skill 2 (Level 7); Project 1 requires Skill 0 (Level 7), Skill 1 (Level 7), Skill 2 (Level 8); Project 2 requires Skill 0 (Level 9), Skill 1 (Level 9), Skill 2 (Level 7); Project 3 requires Skill 0 (Level 7), Skill 1 (Level 8), Skill 2 (Level 9). The objective is to minimize the maximum skill shortage across all projects and skills.0No differencesNo differencesok
90Facility Dispersion in a Network of Locations + +Imagine you are tasked with selecting a subset of facilities to open across a network of five locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall dispersion between the selected facilities. Dispersion is measured by the pairwise distances between the facilities, and your goal is to ensure that the selected facilities are as spread out as possible to optimize coverage and minimize redundancy. + +#### Key Details: +1. **Facility Selection**: You must select exactly **2 facilities** to open from the five available locations. Each location can either be chosen (1) or not chosen (0) to host a facility. +2. **Dispersion Values**: The pairwise distances (dispersion values) between the locations are as follows: + - Node 0 and Node 1: 20 units + - Node 0 and Node 2: 20 units + - Node 0 and Node 3: 18 units + - Node 0 and Node 4: 18 units + - Node 1 and Node 0: 22 units + - Node 1 and Node 2: 21 units + - Node 1 and Node 3: 16 units + - Node 1 and Node 4: 17 units + - Node 2 and Node 0: 16 units + - Node 2 and Node 1: 23 units + - Node 2 and Node 3: 15 units + - Node 2 and Node 4: 24 units + - Node 3 and Node 0: 23 units + - Node 3 and Node 1: 16 units + - Node 3 and Node 2: 23 units + - Node 3 and Node 4: 20 units + - Node 4 and Node 0: 19 units + - Node 4 and Node 1: 21 units + - Node 4 and Node 2: 15 units + - Node 4 and Node 3: 25 units + +3. **Objective**: Maximize the total dispersion between the selected facilities. This means you want to choose the two facilities such that the sum of their pairwise distances is as large as possible. + +4. **Constraints**: + - **Selection Constraint**: Exactly two facilities must be selected from the five locations. + - **Pairwise Activation**: For any two locations, the pairwise dispersion value is only counted if both locations are selected. For example, the dispersion between Node 0 and Node 1 (20 units) is only included in the total if both Node 0 and Node 1 are chosen. + - **Logical Relationships**: If a location is not selected, no pairwise dispersion involving that location can be included in the total. + +#### Example: +If you select Node 0 and Node 1, the total dispersion would be 20 units (from Node 0 to Node 1). If you select Node 2 and Node 4, the total dispersion would be 24 units (from Node 2 to Node 4). Your goal is to find the pair of locations that yields the highest possible total dispersion. + +#### Decision to Make: +- Which two facilities to open from the five available locations to maximize the total pairwise dispersion. + +#### Numerical Parameters: +- All pairwise dispersion values are provided above. +- The total number of facilities to select is fixed at 2. + +This problem requires balancing the selection of facilities to ensure they are as dispersed as possible, while adhering to the constraint of selecting exactly two locations. The solution will provide the optimal pair of facilities to open, maximizing the overall dispersion across the network.45Facility Dispersion in a Network of Locations + +Imagine you are tasked with selecting a subset of facilities to open across a network of five locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall dispersion between the selected facilities. Dispersion is measured by the pairwise distances between the facilities, and your goal is to ensure that the selected facilities are as spread out as possible to optimize coverage and minimize redundancy. + +#### Key Details: +1. **Facility Selection**: You must select exactly **2 facilities** to open from the five available locations. Each location can either be chosen (1) or not chosen (0) to host a facility. +2. **Dispersion Values**: The pairwise distances (dispersion values) between the locations are as follows: + - Node 0 and Node 1: 20 units + - Node 0 and Node 2: 20 units + - Node 0 and Node 3: 18 units + - Node 0 and Node 4: 18 units + - Node 1 and Node 0: 22 units + - Node 1 and Node 2: 21 units + - Node 1 and Node 3: 16 units + - Node 1 and Node 4: 17 units + - Node 2 and Node 0: 16 units + - Node 2 and Node 1: 23 units + - Node 2 and Node 3: 15 units + - Node 2 and Node 4: 24 units + - Node 3 and Node 0: 23 units + - Node 3 and Node 1: 16 units + - Node 3 and Node 2: 23 units + - Node 3 and Node 4: 20 units + - Node 4 and Node 0: 19 units + - Node 4 and Node 1: 21 units + - Node 4 and Node 2: 15 units + - Node 4 and Node 3: 25 units + +3. **Objective**: Maximize the total dispersion between the selected facilities. This means you want to choose the two facilities such that the sum of their pairwise distances is as large as possible. + +4. **Constraints**: + - **Selection Constraint**: Exactly two facilities must be selected from the five locations. + - **Pairwise Activation**: For any two locations, the pairwise dispersion value is only counted if both locations are selected. + - **Logical Relationships**: If a location is not selected, no pairwise dispersion involving that location can be included in the total. + +#### Decision to Make: +- Which two facilities to open from the five available locations to maximize the total pairwise dispersion. + +#### Numerical Parameters: +- All pairwise dispersion values are provided above. +- The total number of facilities to select is fixed at 2. + +This problem requires balancing the selection of facilities to ensure they are as dispersed as possible, while adhering to the constraint of selecting exactly two locations. The solution will provide the optimal pair of facilities to open, maximizing the overall dispersion across the network.454. **Constraints**: + - **Selection Constraint**: Exactly two facilities must be selected from the five locations. + - **Pairwise Activation**: For any two locations, the pairwise dispersion value is only counted if both locations are selected. For example, the dispersion between Node 0 and Node 1 (20 units) is only included in the total if both Node 0 and Node 1 are chosen. + - **Logical Relationships**: If a location is not selected, no pairwise dispersion involving that location can be included in the total. + + #### Example: + If you select Node 0 and Node 1, the total dispersion would be 20 units (from Node 0 to Node 1). If you select Node 2 and Node 4, the total dispersion would be 24 units (from Node 2 to Node 4). Your goal is to find the pair of locations that yields the highest possible total dispersion. + +#### Decision to Make: +- Which two facilities to open from the five available locations to maximize the total pairwise dispersion.No differencesThe example in the description was misleading: there are two distances from (0,1) and (1,0) that should be counted if these nodes are selected, but only one of them was counted in the exampleambiguous
91Select a set of facilities from five potential locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the minimum distance between any two selected facilities. The decision variables include binary choices for selecting each location and binary variables for selecting pairs of locations. The objective is to maximize the minimum distance (MinDistance) between any two selected facilities. Exactly two facilities must be selected. For each pair of locations, if both are selected, the distance between them must be at least MinDistance. If either or both locations in a pair are not selected, the distance constraint is relaxed using a large constant (1,000,000). The pairwise selection variables are symmetric, meaning selecting Node i and Node j is the same as selecting Node j and Node i. Additionally, the pairwise selection variables for the same node (e.g., Node 0 and Node 0) are set to 0. The distances between locations are as follows: Node 0 and Node 1 (41 units), Node 0 and Node 2 (31 units), Node 0 and Node 3 (43 units), Node 0 and Node 4 (10 units), Node 1 and Node 2 (34 units), Node 1 and Node 3 (46 units), Node 1 and Node 4 (12 units), Node 2 and Node 3 (41 units), Node 2 and Node 4 (24 units), and Node 3 and Node 4 (23 units). All decision variables are binary, and the constraints ensure that pairwise selections are consistent with individual selections.46Select a set of facilities from five potential locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the minimum distance between any two selected facilities. The decision variables include binary choices for selecting each location and binary variables for selecting pairs of locations. The objective is to maximize the minimum distance (MinDistance) between any two selected facilities. Exactly two facilities must be selected. For each pair of locations, if both are selected, the distance between them must be at least MinDistance. If either or both locations in a pair are not selected, the distance constraint is relaxed using a large constant (1,000,000). The pairwise selection variables are symmetric, meaning selecting Node i and Node j is the same as selecting Node j and Node i. Additionally, the pairwise selection variables for the same node (e.g., Node 0 and Node 0) are set to 0. The distances between locations are as follows: Node 0 and Node 1 (41 units), Node 0 and Node 2 (31 units), Node 0 and Node 3 (43 units), Node 0 and Node 4 (10 units), Node 1 and Node 2 (34 units), Node 1 and Node 3 (46 units), Node 1 and Node 4 (12 units), Node 2 and Node 3 (41 units), Node 2 and Node 4 (24 units), and Node 3 and Node 4 (23 units). All decision variables are binary, and the constraints ensure that pairwise selections are consistent with individual selections.46No differencesNo differencesok
92Managing Resource Distribution Across Cities + +You are responsible for managing the distribution of a critical resource (e.g., medical supplies, food, or energy) across eight interconnected cities. Each city has a specific supply or demand for the resource, and your goal is to ensure that all cities receive the required amount while minimizing the total transportation cost. The transportation costs vary depending on the routes between cities, and each route has a maximum capacity limit for resource shipments. + +#### City Requirements: +- **City 0** has a surplus of 51 units of the resource. +- **City 1** requires an additional 2 units to meet its demand. +- **City 2** requires an additional 7 units. +- **City 3** requires an additional 11 units. +- **City 4** requires an additional 9 units. +- **City 5** requires an additional 25 units. +- **City 6** has a surplus of 19 units. +- **City 7** requires an additional 16 units. + +#### Transportation Costs and Capacities: +The cost of transporting one unit of the resource between cities varies. Below are the costs and maximum capacities for each route: + +- **From City 0**: + - To City 1: Cost = 10, Capacity = 42 + - To City 2: Cost = 10, Capacity = 47 + - To City 3: Cost = 10, Capacity = 31 + - To City 4: Cost = 9, Capacity = 54 + - To City 5: Cost = 8, Capacity = 41 + - To City 6: Cost = 9, Capacity = 55 + - To City 7: Cost = 10, Capacity = 52 + +- **From City 1**: + - To City 0: Cost = 9, Capacity = 40 + - To City 2: Cost = 7, Capacity = 57 + - To City 3: Cost = 10, Capacity = 56 + - To City 4: Cost = 10, Capacity = 57 + - To City 5: Cost = 10, Capacity = 44 + - To City 6: Cost = 9, Capacity = 60 + - To City 7: Cost = 7, Capacity = 50 + +- **From City 2**: + - To City 0: Cost = 7, Capacity = 45 + - To City 1: Cost = 7, Capacity = 43 + - To City 3: Cost = 8, Capacity = 50 + - To City 4: Cost = 9, Capacity = 36 + - To City 5: Cost = 10, Capacity = 42 + - To City 6: Cost = 6, Capacity = 58 + - To City 7: Cost = 10, Capacity = 33 + +- **From City 3**: + - To City 0: Cost = 9, Capacity = 57 + - To City 1: Cost = 9, Capacity = 42 + - To City 2: Cost = 6, Capacity = 30 + - To City 4: Cost = 9, Capacity = 55 + - To City 5: Cost = 10, Capacity = 60 + - To City 6: Cost = 10, Capacity = 41 + - To City 7: Cost = 8, Capacity = 40 + +- **From City 4**: + - To City 0: Cost = 10, Capacity = 42 + - To City 1: Cost = 6, Capacity = 42 + - To City 2: Cost = 9, Capacity = 33 + - To City 3: Cost = 8, Capacity = 52 + - To City 5: Cost = 10, Capacity = 33 + - To City 6: Cost = 7, Capacity = 36 + - To City 7: Cost = 8, Capacity = 31 + +- **From City 5**: + - To City 0: Cost = 7, Capacity = 48 + - To City 1: Cost = 8, Capacity = 42 + - To City 2: Cost = 9, Capacity = 49 + - To City 3: Cost = 6, Capacity = 34 + - To City 4: Cost = 9, Capacity = 48 + - To City 6: Cost = 7, Capacity = 58 + - To City 7: Cost = 7, Capacity = 38 + +- **From City 6**: + - To City 0: Cost = 7, Capacity = 41 + - To City 1: Cost = 7, Capacity = 31 + - To City 2: Cost = 8, Capacity = 31 + - To City 3: Cost = 7, Capacity = 34 + - To City 4: Cost = 6, Capacity = 42 + - To City 5: Cost = 7, Capacity = 48 + - To City 7: Cost = 9, Capacity = 49 + +- **From City 7**: + - To City 0: Cost = 9, Capacity = 49 + - To City 1: Cost = 6, Capacity = 31 + - To City 2: Cost = 6, Capacity = 34 + - To City 3: Cost = 9, Capacity = 50 + - To City 4: Cost = 10, Capacity = 59 + - To City 5: Cost = 10, Capacity = 49 + - To City 6: Cost = 10, Capacity = 60 + +#### Objective: +Your task is to determine the optimal amount of resources to transport between cities to meet all demands while minimizing the total transportation cost. Ensure that: +1. The total amount of resources shipped into each city equals the total amount shipped out, adjusted for the city's supply or demand. +2. No shipment exceeds the capacity of its respective route. +3. All shipments are non-negative. + +By solving this problem, you will ensure efficient resource distribution across the cities at the lowest possible cost.541Managing Resource Distribution Across Cities + +You are responsible for managing the distribution of a critical resource (e.g., medical supplies, food, or energy) across eight interconnected cities. Each city has a specific supply or demand for the resource, and your goal is to ensure that all cities receive the required amount while minimizing the total transportation cost. The transportation costs vary depending on the routes between cities, and each route has a maximum capacity limit for resource shipments. + +#### City Requirements: +- **City 0** has a surplus of 51 units of the resource. +- **City 1** requires an additional 2 units to meet its demand. +- **City 2** requires an additional 7 units. +- **City 3** requires an additional 11 units. +- **City 4** requires an additional 9 units. +- **City 5** requires an additional 25 units. +- **City 6** has a surplus of 19 units. +- **City 7** requires an additional 16 units. + +#### Transportation Costs and Capacities: +The cost of transporting one unit of the resource between cities varies. Below are the costs and maximum capacities for each route: + +- **From City 0**: + - To City 1: Cost = 10, Capacity = 42 + - To City 2: Cost = 10, Capacity = 47 + - To City 3: Cost = 10, Capacity = 31 + - To City 4: Cost = 9, Capacity = 54 + - To City 5: Cost = 8, Capacity = 41 + - To City 6: Cost = 9, Capacity = 55 + - To City 7: Cost = 10, Capacity = 52 + +- **From City 1**: + - To City 0: Cost = 9, Capacity = 40 + - To City 2: Cost = 7, Capacity = 57 + - To City 3: Cost = 10, Capacity = 56 + - To City 4: Cost = 10, Capacity = 57 + - To City 5: Cost = 10, Capacity = 44 + - To City 6: Cost = 9, Capacity = 60 + - To City 7: Cost = 7, Capacity = 50 + +- **From City 2**: + - To City 0: Cost = 7, Capacity = 45 + - To City 1: Cost = 7, Capacity = 43 + - To City 3: Cost = 8, Capacity = 50 + - To City 4: Cost = 9, Capacity = 36 + - To City 5: Cost = 10, Capacity = 42 + - To City 6: Cost = 6, Capacity = 58 + - To City 7: Cost = 10, Capacity = 33 + +- **From City 3**: + - To City 0: Cost = 9, Capacity = 57 + - To City 1: Cost = 9, Capacity = 42 + - To City 2: Cost = 6, Capacity = 30 + - To City 4: Cost = 9, Capacity = 55 + - To City 5: Cost = 10, Capacity = 60 + - To City 6: Cost = 10, Capacity = 41 + - To City 7: Cost = 8, Capacity = 40 + +- **From City 4**: + - To City 0: Cost = 10, Capacity = 42 + - To City 1: Cost = 6, Capacity = 42 + - To City 2: Cost = 9, Capacity = 33 + - To City 3: Cost = 8, Capacity = 52 + - To City 5: Cost = 10, Capacity = 33 + - To City 6: Cost = 7, Capacity = 36 + - To City 7: Cost = 8, Capacity = 31 + +- **From City 5**: + - To City 0: Cost = 7, Capacity = 48 + - To City 1: Cost = 8, Capacity = 42 + - To City 2: Cost = 9, Capacity = 49 + - To City 3: Cost = 6, Capacity = 34 + - To City 4: Cost = 9, Capacity = 48 + - To City 6: Cost = 7, Capacity = 58 + - To City 7: Cost = 7, Capacity = 38 + +- **From City 6**: + - To City 0: Cost = 7, Capacity = 41 + - To City 1: Cost = 7, Capacity = 31 + - To City 2: Cost = 8, Capacity = 31 + - To City 3: Cost = 7, Capacity = 34 + - To City 4: Cost = 6, Capacity = 42 + - To City 5: Cost = 7, Capacity = 48 + - To City 7: Cost = 9, Capacity = 49 + +- **From City 7**: + - To City 0: Cost = 9, Capacity = 49 + - To City 1: Cost = 6, Capacity = 31 + - To City 2: Cost = 6, Capacity = 34 + - To City 3: Cost = 9, Capacity = 50 + - To City 4: Cost = 10, Capacity = 59 + - To City 5: Cost = 10, Capacity = 49 + - To City 6: Cost = 10, Capacity = 60 + +#### Objective: +Your task is to determine the optimal amount of resources to transport between cities to meet all demands while minimizing the total transportation cost. Ensure that: +1. The total amount of resources shipped into each city equals the total amount shipped out, adjusted for the city's supply or demand. +2. No shipment exceeds the capacity of its respective route. +3. All shipments are non-negative. + +By solving this problem, you will ensure efficient resource distribution across the cities at the lowest possible cost.584No differences541 584Incorrect ground truthincorrect ground truth
93The problem involves assigning employees to shifts across two restaurants to meet specific skill demands while minimizing costs. There are seven employees, each with specific availability and skill capabilities. Each shift at each restaurant requires one employee with Skill 0. Assigning an employee to a shift incurs a cost: some assignments cost 1 unit, while others cost 3 units. If a shift cannot be filled, a penalty of 100 units is incurred per unfulfilled position. + +Employees are only available for certain shifts, as specified by the availability constraints. For example, Employee 0 is available for Shift 0 but not for Shift 1 at Restaurant 1, and Employee 1 is available for Shift 0 but not for Shift 1 at Restaurant 1. Employees can only be assigned to roles that match their skills, and each employee can be assigned to at most one shift across all restaurants and roles. The total number of employees assigned to a shift plus any unfulfilled positions must equal the demand for that shift. + +The objective is to minimize the total cost, which includes the cost of assigning employees to roles based on their preferences and the penalty cost for any unfulfilled positions. The preference costs for assignments range from 1 to 3 units, and the penalty cost for unfulfilled positions is 100 units each. The demand for each shift is one employee with the required skill.4The problem involves assigning employees to shifts across two restaurants to meet specific skill demands while minimizing costs. There are seven employees, each with specific availability and skill capabilities. Each shift at each restaurant requires one employee with Skill 0. Assigning an employee to a shift incurs a cost: some assignments cost 1 unit, while others cost 3 units. If a shift cannot be filled, a penalty of 100 units is incurred per unfulfilled position. + +Employees are only available for certain shifts, as specified by the availability constraints. For example, Employee 0 is available for Shift 0 but not for Shift 1 at Restaurant 1, and Employee 1 is available for Shift 0 but not for Shift 1 at Restaurant 1. Employees can only be assigned to roles that match their skills, and each employee can be assigned to at most one shift across all restaurants and roles. The total number of employees assigned to a shift plus any unfulfilled positions must equal the demand for that shift. + +The objective is to minimize the total cost, which includes the cost of assigning employees to roles based on their preferences and the penalty cost for any unfulfilled positions. The preference costs for assignments range from 1 to 3 units, and the penalty cost for unfulfilled positions is 100 units each. The demand for each shift is one employee with the required skill.4No differencesNo differencesok
94Minimize the total travel distance for a delivery route across six cities, labeled City 0 through City 5, starting and ending at City 0. Each city must be visited exactly once, and the route must form a single loop without subtours. The distances between cities are as follows: +- From City 0: To City 1: 301 km, To City 2: 313 km, To City 3: 298 km, To City 4: 280 km, To City 5: 300 km. +- From City 1: To City 0: 309 km, To City 2: 317 km, To City 3: 317 km, To City 4: 279 km, To City 5: 321 km. +- From City 2: To City 0: 297 km, To City 1: 290 km, To City 3: 302 km, To City 4: 278 km, To City 5: 323 km. +- From City 3: To City 0: 292 km, To City 1: 305 km, To City 2: 321 km, To City 4: 292 km, To City 5: 321 km. +- From City 4: To City 0: 299 km, To City 1: 295 km, To City 2: 282 km, To City 3: 284 km, To City 5: 300 km. +- From City 5: To City 0: 318 km, To City 1: 317 km, To City 2: 304 km, To City 3: 278 km, To City 4: 284 km. + +Each city must be visited exactly once, and each city must be departed exactly once. To prevent subtours, auxiliary variables represent the position of each city in the tour, with positions ranging from 0 to 5. Subtour elimination constraints ensure that if a route segment from City i to City j is included, the position of City i must be less than the position of City j. Specifically, for each pair of cities i and j, the constraint 6 times the route segment from i to j plus the position of City i minus the position of City j must be less than or equal to 5. The decision variables are binary, indicating whether a route segment is included, and the auxiliary variables are general, representing the position of each city in the tour. The goal is to determine the sequence of cities that minimizes the total travel distance while satisfying all constraints.1741Minimize the total travel distance for a delivery route across six cities, labeled City 0 through City 5, starting and ending at City 0. Each city must be visited exactly once, and the route must form a single loop without subtours, and finally returning to the origin city. The distances between cities are as follows: +- From City 0: To City 1: 301 km, To City 2: 313 km, To City 3: 298 km, To City 4: 280 km, To City 5: 300 km. +- From City 1: To City 0: 309 km, To City 2: 317 km, To City 3: 317 km, To City 4: 279 km, To City 5: 321 km. +- From City 2: To City 0: 297 km, To City 1: 290 km, To City 3: 302 km, To City 4: 278 km, To City 5: 323 km. +- From City 3: To City 0: 292 km, To City 1: 305 km, To City 2: 321 km, To City 4: 292 km, To City 5: 321 km. +- From City 4: To City 0: 299 km, To City 1: 295 km, To City 2: 282 km, To City 3: 284 km, To City 5: 300 km. +- From City 5: To City 0: 318 km, To City 1: 317 km, To City 2: 304 km, To City 3: 278 km, To City 4: 284 km. + +The task is to find the shortest possible route that visits each city exactly once, and return to the origin city. What is the minimal total distance travelled?1741Minimize the total travel distance for a delivery route across six cities, labeled City 0 through City 5, starting and ending at City 0. Each city must be visited exactly once, and the route must form a single loop without subtours. subtours, and finally returning to the origin city. The distances between cities are as follows: +- From City 0: To City 1: 301 km, To City 2: 313 km, To City 3: 298 km, To City 4: 280 km, To City 5: 300 km. +- From City 1: To City 0: 309 km, To City 2: 317 km, To City 3: 317 km, To City 4: 279 km, To City 5: 321 km. +- From City 2: To City 0: 297 km, To City 1: 290 km, To City 3: 302 km, To City 4: 278 km, To City 5: 323 km. +- From City 4: To City 0: 299 km, To City 1: 295 km, To City 2: 282 km, To City 3: 284 km, To City 5: 300 km. +- From City 5: To City 0: 318 km, To City 1: 317 km, To City 2: 304 km, To City 3: 278 km, To City 4: 284 km. + + Each city must be visited exactly once, and each city must be departed exactly once. To prevent subtours, auxiliary variables represent the position of each city in the tour, with positions ranging from 0 to 5. Subtour elimination constraints ensure that if a route segment from City i to City j The task is included, the position of City i must be less than the position of City j. Specifically, for each pair of cities i and j, the constraint 6 times the route segment from i to j plus find the position of City i minus the position of City j must be less than or equal to 5. The decision variables are binary, indicating whether a shortest possible route segment is included, and the auxiliary variables are general, representing the position of that visits each city in the tour. The goal is exactly once, and return to determine the sequence of cities that minimizes origin city. What is the minimal total travel distance while satisfying all constraints. travelled?No differencesWe removed information that was not the description of the problem, but describing the variables/constraints of a possible model. Also, this additional information could lead to a wrong model (MTZ constraint on the return to the first node)ambiguous
95You are tasked with optimizing the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 35 distinct areas, each with a specific population size that needs to be covered by the cell towers. The goal is to determine which towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + +#### Key Decisions: +1. **Build or Not Build**: For each of the 29 potential tower locations, you must decide whether to build the tower (a binary decision: build or not build). +2. **Coverage**: For each of the 35 regions, you must determine whether it will be covered by at least one of the built towers (a binary decision: covered or not covered). + +#### Objective: +Maximize the total population covered by the deployed cell towers. The population sizes for each region are as follows: +- Region 0: 2,328 people +- Region 1: 2,364 people +- Region 2: 2,101 people +- Region 3: 4,451 people +- Region 4: 4,175 people +- Region 5: 3,684 people +- Region 6: 3,871 people +- Region 7: 4,018 people +- Region 8: 2,922 people +- Region 9: 3,610 people +- Region 10: 4,982 people +- Region 11: 3,296 people +- Region 12: 3,211 people +- Region 13: 3,875 people +- Region 14: 4,545 people +- Region 15: 4,382 people +- Region 16: 3,873 people +- Region 17: 3,481 people +- Region 18: 2,656 people +- Region 19: 3,621 people +- Region 20: 4,167 people +- Region 21: 2,027 people +- Region 22: 4,889 people +- Region 23: 3,684 people +- Region 24: 3,987 people +- Region 25: 2,597 people +- Region 26: 4,578 people +- Region 27: 2,159 people +- Region 28: 2,635 people +- Region 29: 2,699 people +- Region 30: 3,364 people +- Region 31: 3,164 people +- Region 32: 4,867 people +- Region 33: 3,938 people +- Region 34: 3,341 people + +#### Constraints: +1. **Coverage Requirement**: For a region to be considered covered, at least one of the towers that can serve it must be built. The specific towers that can cover each region are listed in the problem data. For example: + - Region 0 can be covered by towers 0, 1, 2, 4, 6, 9, 15, 18, 20, 21, 23, 24, 25, or 28. + - Region 1 can be covered by towers 0, 3, 4, 5, 6, 10, 13, 16, 17, 19, 20, 25, or 28. + - Similar constraints apply to all 35 regions. + +2. **Budget Constraint**: The total cost of building the selected towers must not exceed the available budget of 2,968 units. The cost of building each tower is as follows: + - Tower 0: 476 units + - Tower 1: 388 units + - Tower 2: 444 units + - Tower 3: 230 units + - Tower 4: 449 units + - Tower 5: 363 units + - Tower 6: 458 units + - Tower 7: 400 units + - Tower 8: 208 units + - Tower 9: 373 units + - Tower 10: 454 units + - Tower 11: 348 units + - Tower 12: 465 units + - Tower 13: 445 units + - Tower 14: 378 units + - Tower 15: 288 units + - Tower 16: 335 units + - Tower 17: 361 units + - Tower 18: 254 units + - Tower 19: 362 units + - Tower 20: 290 units + - Tower 21: 287 units + - Tower 22: 291 units + - Tower 23: 213 units + - Tower 24: 237 units + - Tower 25: 232 units + - Tower 26: 356 units + - Tower 27: 221 units + - Tower 28: 289 units + +#### Summary: +Your task is to select which towers to build and which regions to cover, ensuring that the total population covered is maximized while staying within the budget. The coverage of each region depends on the specific towers built, and the total cost of building the selected towers must not exceed 2,968 units. The goal is to make strategic decisions that balance coverage and cost to achieve the best possible outcome for the region.123542"You are tasked with optimizing the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 35 distinct areas, each with a specific population size that needs to be covered by the cell towers. The goal is to determine which towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + +#### Key Decisions: +1. **Build or Not Build**: For each of the 29 potential tower locations, you must decide whether to build the tower (a binary decision: build or not build). +2. **Coverage**: For each of the 35 regions, you must determine whether it will be covered by at least one of the built towers (a binary decision: covered or not covered). + +#### Objective: +Maximize the total population covered by the deployed cell towers. The population sizes for each region are as follows: +- Region 0: 2,328 people +- Region 1: 2,364 people +- Region 2: 2,101 people +- Region 3: 4,451 people +- Region 4: 4,175 people +- Region 5: 3,684 people +- Region 6: 3,871 people +- Region 7: 4,018 people +- Region 8: 2,922 people +- Region 9: 3,610 people +- Region 10: 4,982 people +- Region 11: 3,296 people +- Region 12: 3,211 people +- Region 13: 3,875 people +- Region 14: 4,545 people +- Region 15: 4,382 people +- Region 16: 3,873 people +- Region 17: 3,481 people +- Region 18: 2,656 people +- Region 19: 3,621 people +- Region 20: 4,167 people +- Region 21: 2,027 people +- Region 22: 4,889 people +- Region 23: 3,684 people +- Region 24: 3,987 people +- Region 25: 2,597 people +- Region 26: 4,578 people +- Region 27: 2,159 people +- Region 28: 2,635 people +- Region 29: 2,699 people +- Region 30: 3,364 people +- Region 31: 3,164 people +- Region 32: 4,867 people +- Region 33: 3,938 people +- Region 34: 3,341 people + +#### Constraints: +1. **Coverage Requirement**: For a region to be considered covered, at least one of the towers that can serve it must be built. The specific towers that can cover each region are listed in the problem data. +- Region 0 can be covered by Towers Tower 0, Tower 3, Tower 4, Tower 7, Tower 9, Tower 16, Tower 17, Tower 20, Tower 22, Tower 24, Tower 25, Tower 26, Tower 27. + - Region 1 can be covered by Towers Tower 0, Tower 1, Tower 15, Tower 19, Tower 22, Tower 23, Tower 25, Tower 26, Tower 27, Tower 29. + - Region 2 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 7, Tower 8, Tower 13, Tower 16, Tower 17, Tower 18, Tower 23, Tower 24, Tower 27, Tower 28. + - Region 3 can be covered by Towers Tower 0, Tower 6, Tower 7, Tower 13, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 24, Tower 28. + - Region 4 can be covered by Towers Tower 0, Tower 4, Tower 5, Tower 6, Tower 8, Tower 10, Tower 13, Tower 18, Tower 20, Tower 22, Tower 24, Tower 25, Tower 27. + - Region 5 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 8, Tower 11, Tower 12, Tower 14, Tower 17, Tower 19, Tower 25. + - Region 6 can be covered by Towers Tower 2, Tower 6, Tower 9, Tower 11, Tower 12, Tower 17, Tower 18, Tower 19, Tower 20, Tower 27. + - Region 7 can be covered by Towers Tower 2, Tower 3, Tower 7, Tower 8, Tower 9, Tower 12, Tower 14, Tower 21, Tower 24, Tower 27. + - Region 8 can be covered by Towers Tower 2, Tower 5, Tower 6, Tower 7, Tower 8, Tower 11, Tower 16, Tower 17, Tower 18, Tower 20, Tower 21, Tower 25, Tower 26, Tower 27, Tower 28. + - Region 9 can be covered by Towers Tower 1, Tower 7, Tower 8, Tower 10, Tower 12, Tower 17, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24, Tower 26, Tower 28. + - Region 10 can be covered by Towers Tower 6, Tower 10, Tower 12, Tower 14, Tower 15, Tower 18, Tower 20, Tower 22, Tower 23, Tower 28. + - Region 11 can be covered by Towers Tower 4, Tower 7, Tower 8, Tower 12, Tower 13, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 28. + - Region 12 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 4, Tower 5, Tower 7, Tower 13, Tower 15, Tower 16, Tower 20, Tower 24, Tower 27. + - Region 13 can be covered by Towers Tower 0, Tower 3, Tower 8, Tower 10, Tower 12, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 21, Tower 23, Tower 24. + - Region 14 can be covered by Towers Tower 0, Tower 3, Tower 5, Tower 8, Tower 9, Tower 13, Tower 14, Tower 16, Tower 21, Tower 23, Tower 24, Tower 27. + - Region 15 can be covered by Towers Tower 0, Tower 3, Tower 4, Tower 5, Tower 6, Tower 10, Tower 11, Tower 14, Tower 15, Tower 16, Tower 17, Tower 19, Tower 20, Tower 27, Tower 28. + - Region 16 can be covered by Towers Tower 1, Tower 2, Tower 7, Tower 9, Tower 15, Tower 18, Tower 19, Tower 20, Tower 24, Tower 25, Tower 26, Tower 28. + - Region 17 can be covered by Towers Tower 4, Tower 5, Tower 6, Tower 8, Tower 9, Tower 13, Tower 15, Tower 16, Tower 17, Tower 19, Tower 21, Tower 23, Tower 24, Tower 27. + - Region 18 can be covered by Towers Tower 0, Tower 2, Tower 3, Tower 7, Tower 10, Tower 11, Tower 14, Tower 16, Tower 20, Tower 21, Tower 25, Tower 27, Tower 28. + - Region 19 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 7, Tower 8, Tower 10, Tower 16, Tower 20, Tower 21, Tower 22, Tower 23, Tower 25, Tower 26, Tower 28. + - Region 20 can be covered by Towers Tower 3, Tower 4, Tower 6, Tower 7, Tower 10, Tower 13, Tower 15, Tower 16, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 27, Tower 28. + - Region 21 can be covered by Towers Tower 1, Tower 3, Tower 10, Tower 11, Tower 12, Tower 13, Tower 14, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24, Tower 27. + - Region 22 can be covered by Towers Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 13, Tower 14, Tower 17, Tower 24, Tower 27. + - Region 23 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 14, Tower 17, Tower 20, Tower 24, Tower 25, Tower 27, Tower 28. + - Region 24 can be covered by Towers Tower 0, Tower 1, Tower 5, Tower 6, Tower 12, Tower 13, Tower 15, Tower 20, Tower 23, Tower 26, Tower 28. + - Region 25 can be covered by Towers Tower 4, Tower 6, Tower 9, Tower 13, Tower 14, Tower 15, Tower 17, Tower 21, Tower 22, Tower 25, Tower 27, Tower 28. + - Region 26 can be covered by Towers Tower 1, Tower 10, Tower 15, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 26, Tower 27, Tower 28. + - Region 27 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 5, Tower 7, Tower 12, Tower 16, Tower 18, Tower 19, Tower 21, Tower 23, Tower 26, Tower 27, Tower 28. + - Region 28 can be covered by Towers Tower 1, Tower 2, Tower 6, Tower 7, Tower 8, Tower 10, Tower 13, Tower 16, Tower 18, Tower 19, Tower 21, Tower 22, Tower 26, Tower 28. + - Region 29 can be covered by Towers Tower 0, Tower 2, Tower 3, Tower 4, Tower 9, Tower 10, Tower 12, Tower 14, Tower 18, Tower 20, Tower 21, Tower 23. + - Region 30 can be covered by Towers Tower 2, Tower 4, Tower 6, Tower 7, Tower 8, Tower 9, Tower 11, Tower 16, Tower 17, Tower 23. + - Region 31 can be covered by Towers Tower 0, Tower 9, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 24, Tower 26, Tower 27. + - Region 32 can be covered by Towers Tower 3, Tower 4, Tower 6, Tower 8, Tower 9, Tower 10, Tower 15, Tower 17, Tower 18, Tower 20, Tower 23, Tower 25, Tower 26, Tower 27, Tower 28. + - Region 33 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 4, Tower 5, Tower 8, Tower 10, Tower 13, Tower 20, Tower 22, Tower 26, Tower 27, Tower 28. + - Region 34 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 11, Tower 13, Tower 14, Tower 17, Tower 20, Tower 21, Tower 22, Tower 24, Tower 25, Tower 27. + +2. **Budget Constraint**: The total cost of building the selected towers must not exceed the available budget of 2,968 units. The cost of building each tower is as follows: + - Tower 0: 476 units + - Tower 1: 388 units + - Tower 2: 444 units + - Tower 3: 230 units + - Tower 4: 449 units + - Tower 5: 363 units + - Tower 6: 458 units + - Tower 7: 400 units + - Tower 8: 208 units + - Tower 9: 373 units + - Tower 10: 454 units + - Tower 11: 348 units + - Tower 12: 465 units + - Tower 13: 445 units + - Tower 14: 378 units + - Tower 15: 288 units + - Tower 16: 335 units + - Tower 17: 361 units + - Tower 18: 254 units + - Tower 19: 362 units + - Tower 20: 290 units + - Tower 21: 287 units + - Tower 22: 291 units + - Tower 23: 213 units + - Tower 24: 237 units + - Tower 25: 232 units + - Tower 26: 356 units + - Tower 27: 221 units + - Tower 28: 289 units + +#### Summary: +Your task is to select which towers to build and which regions to cover, ensuring that the total population covered is maximized while staying within the budget. The coverage of each region depends on the specific towers built, and the total cost of building the selected towers must not exceed 2,968 units. The goal is to make strategic decisions that balance coverage and cost to achieve the best possible outcome for the region."123542You "You are tasked with optimizing the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 35 distinct areas, each with a specific population size that needs to be covered by the cell towers. The goal is to determine which towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + +#### Key Decisions: +1. **Build or Not Build**: For each of the 29 potential tower locations, you must decide whether to build the tower (a binary decision: build or not build). +- Region 34: 3,341 people + +#### Constraints: +1. **Coverage Requirement**: For a region to be considered covered, at least one of the towers that can serve it must be built. The specific towers that can cover each region are listed in the problem data. For example: +- Region 0 can be covered by towers Towers Tower 0, Tower 3, Tower 4, Tower 7, Tower 9, Tower 16, Tower 17, Tower 20, Tower 22, Tower 24, Tower 25, Tower 26, Tower 27. + - Region 1 can be covered by Towers Tower 0, Tower 1, Tower 15, Tower 19, Tower 22, Tower 23, Tower 25, Tower 26, Tower 27, Tower 29. + - Region 2 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 7, Tower 8, Tower 13, Tower 16, Tower 17, Tower 18, Tower 23, Tower 24, Tower 27, Tower 28. + - Region 3 can be covered by Towers Tower 0, Tower 6, Tower 7, Tower 13, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 24, Tower 28. + - Region 4 can be covered by Towers Tower 0, Tower 4, Tower 5, Tower 6, Tower 8, Tower 10, Tower 13, Tower 18, Tower 20, Tower 22, Tower 24, Tower 25, Tower 27. + - Region 5 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 8, Tower 11, Tower 12, Tower 14, Tower 17, Tower 19, Tower 25. + - Region 6 can be covered by Towers Tower 2, Tower 6, Tower 9, Tower 11, Tower 12, Tower 17, Tower 18, Tower 19, Tower 20, Tower 27. + - Region 7 can be covered by Towers Tower 2, Tower 3, Tower 7, Tower 8, Tower 9, Tower 12, Tower 14, Tower 21, Tower 24, Tower 27. + - Region 8 can be covered by Towers Tower 2, Tower 5, Tower 6, Tower 7, Tower 8, Tower 11, Tower 16, Tower 17, Tower 18, Tower 20, Tower 21, Tower 25, Tower 26, Tower 27, Tower 28. + - Region 9 can be covered by Towers Tower 1, Tower 7, Tower 8, Tower 10, Tower 12, Tower 17, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24, Tower 26, Tower 28. + - Region 10 can be covered by Towers Tower 6, Tower 10, Tower 12, Tower 14, Tower 15, Tower 18, Tower 20, Tower 22, Tower 23, Tower 28. + - Region 11 can be covered by Towers Tower 4, Tower 7, Tower 8, Tower 12, Tower 13, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 28. + - Region 12 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 4, Tower 5, Tower 7, Tower 13, Tower 15, Tower 16, Tower 20, Tower 24, Tower 27. + - Region 13 can be covered by Towers Tower 0, Tower 3, Tower 8, Tower 10, Tower 12, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 21, Tower 23, Tower 24. + - Region 14 can be covered by Towers Tower 0, Tower 3, Tower 5, Tower 8, Tower 9, Tower 13, Tower 14, Tower 16, Tower 21, Tower 23, Tower 24, Tower 27. + - Region 15 can be covered by Towers Tower 0, Tower 3, Tower 4, Tower 5, Tower 6, Tower 10, Tower 11, Tower 14, Tower 15, Tower 16, Tower 17, Tower 19, Tower 20, Tower 27, Tower 28. + - Region 16 can be covered by Towers Tower 1, Tower 2, Tower 7, Tower 9, Tower 15, Tower 18, Tower 19, Tower 20, Tower 24, Tower 25, or Tower 26, Tower 28. + - Region 1 17 can be covered by towers Towers Tower 4, Tower 5, Tower 6, Tower 8, Tower 9, Tower 13, Tower 15, Tower 16, Tower 17, Tower 19, Tower 21, Tower 23, Tower 24, Tower 27. + - Region 18 can be covered by Towers Tower 0, Tower 2, Tower 3, Tower 7, Tower 10, Tower 11, Tower 14, Tower 16, Tower 20, Tower 21, Tower 25, Tower 27, Tower 28. + - Region 19 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 7, Tower 8, Tower 10, Tower 16, Tower 20, Tower 21, Tower 22, Tower 23, Tower 25, Tower 26, Tower 28. + - Region 20 can be covered by Towers Tower 3, Tower 4, Tower 6, Tower 7, Tower 10, Tower 13, Tower 15, Tower 16, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 27, Tower 28. + - Region 21 can be covered by Towers Tower 1, Tower 3, Tower 10, Tower 11, Tower 12, Tower 13, Tower 14, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24, Tower 27. + - Region 22 can be covered by Towers Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 13, Tower 14, Tower 17, Tower 24, Tower 27. + - Region 23 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 14, Tower 17, Tower 20, Tower 24, Tower 25, Tower 27, Tower 28. + - Region 24 can be covered by Towers Tower 0, Tower 1, Tower 5, Tower 6, Tower 12, Tower 13, Tower 15, Tower 20, Tower 23, Tower 26, Tower 28. + - Region 25 can be covered by Towers Tower 4, Tower 6, Tower 9, Tower 13, Tower 14, Tower 15, Tower 17, Tower 21, Tower 22, Tower 25, Tower 27, Tower 28. + - Region 26 can be covered by Towers Tower 1, Tower 10, Tower 15, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 26, Tower 27, Tower 28. + - Region 27 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 5, Tower 7, Tower 12, Tower 16, Tower 18, Tower 19, Tower 21, Tower 23, Tower 26, Tower 27, Tower 28. + - Region 28 can be covered by Towers Tower 1, Tower 2, Tower 6, Tower 7, Tower 8, Tower 10, Tower 13, Tower 16, Tower 18, Tower 19, Tower 21, Tower 22, Tower 26, Tower 28. + - Region 29 can be covered by Towers Tower 0, Tower 2, Tower 3, Tower 4, Tower 9, Tower 10, Tower 12, Tower 14, Tower 18, Tower 20, Tower 21, Tower 23. + - Region 30 can be covered by Towers Tower 2, Tower 4, Tower 6, Tower 7, Tower 8, Tower 9, Tower 11, Tower 16, Tower 17, Tower 23. + - Region 31 can be covered by Towers Tower 0, Tower 9, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 24, Tower 26, Tower 27. + - Region 32 can be covered by Towers Tower 3, Tower 4, Tower 6, Tower 8, Tower 9, Tower 10, Tower 15, Tower 17, Tower 18, Tower 20, Tower 23, Tower 25, or Tower 26, Tower 27, Tower 28. + - Similar constraints apply to all 35 regions. Region 33 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 4, Tower 5, Tower 8, Tower 10, Tower 13, Tower 20, Tower 22, Tower 26, Tower 27, Tower 28. + - Region 34 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 11, Tower 13, Tower 14, Tower 17, Tower 20, Tower 21, Tower 22, Tower 24, Tower 25, Tower 27. + +2. **Budget Constraint**: The total cost of building the selected towers must not exceed the available budget of 2,968 units. The cost of building each tower is as follows: + - Tower 0: 476 units + - Tower 28: 289 units + +#### Summary: +Your task is to select which towers to build and which regions to cover, ensuring that the total population covered is maximized while staying within the budget. The coverage of each region depends on the specific towers built, and the total cost of building the selected towers must not exceed 2,968 units. The goal is to make strategic decisions that balance coverage and cost to achieve the best possible outcome for the region. region."No differencesMissing data about which towers cover which regionsmissing data
96Minimize the total production costs, which include setup costs, startup costs, inventory holding costs, and backlogging costs, while meeting customer demand for each product in each period. The setup cost is **121.0782541217011** per production run, and the startup cost is **69.65555018358044** per startup. The inventory holding cost for Product 0 is **1.196682091019819** per unit per period, and for Product 1, it is **1.146459407730801** per unit per period. The backlogging cost for Product 0 is **7.384397449671641** per unit per period, and for Product 1, it is **6.58642780982885** per unit per period. + +The production facility manufactures two types of products (Product 0 and Product 1) using two machines (Machine 0 and Machine 1) over four time periods (Period 0 to Period 3). The demand for Product 0 in Period 0 is **22.28694038228271**, in Period 1 is **15.40553617195959**, in Period 2 is **16.19622794603683**, and in Period 3 is **23.77133167016131**. The demand for Product 1 in Period 0 is **16.51020937307772**, in Period 1 is **15.41395287058592**, in Period 2 is **23.98409764196543**, and in Period 3 is **18.84683167137874**. + +Each machine has a limited capacity: Machine 0 has a capacity of **63.4339018881536**, and Machine 1 has a capacity of **67.51600987446864**. The startup time for Machine 0 is **7.917022863645049**, and for Machine 1, it is **7.682853596177669**. Each machine can produce only one product at a time in any given period. A machine can only be started up for a product if it was not producing that product in the previous period. + +The total production, inventory, and backlog must balance customer demand for each product in each period. Production quantities, inventory, and backlog must be non-negative. Production and startup decisions are binary (either 0 or 1). The goal is to determine the optimal production schedule, inventory levels, and backlog amounts to minimize the total cost while meeting all constraints and customer demand.763.2675651474576Minimize the total production costs, which include setup costs, startup costs, inventory holding costs, and backlogging costs, while meeting customer demand for each product in each period. The setup cost is **121.0782541217011** per production run, and the startup cost is **69.65555018358044** per startup. The inventory holding cost for Product 0 is **1.196682091019819** per unit per period, and for Product 1, it is **1.146459407730801** per unit per period. The backlogging cost for Product 0 is **7.384397449671641** per unit per period, and for Product 1, it is **6.58642780982885** per unit per period. + +The production facility manufactures two types of products (Product 0 and Product 1) using two machines (Machine 0 and Machine 1) over four time periods (Period 0 to Period 3). The demand for Product 0 in Period 0 is **22.28694038228271**, in Period 1 is **15.40553617195959**, in Period 2 is **16.19622794603683**, and in Period 3 is **23.77133167016131**. The demand for Product 1 in Period 0 is **16.51020937307772**, in Period 1 is **15.41395287058592**, in Period 2 is **23.98409764196543**, and in Period 3 is **18.84683167137874**. + +The total production quantity for each product on each machine in each period, plus any startup time required, must not exceed the machine's capacity. Each machine has a limited capacity: Machine 0 has a capacity of **63.4339018881536**, and Machine 1 has a capacity of **67.51600987446864**. The startup time for Machine 0 is **7.917022863645049**, and for Machine 1, it is **7.682853596177669**. Each machine can produce only one product at a time in any given period. A machine can only be started up for a product if it was not producing that product in the previous period. + +The total production, inventory, and backlog must balance customer demand for each product in each period. Production quantities, inventory, and backlog must be non-negative. Production and startup decisions are binary (either 0 or 1). The goal is to determine the optimal production schedule, inventory levels, and backlog amounts to minimize the total cost while meeting all constraints and customer demand.763.2675651474576The production facility manufactures two types of products (Product 0 and Product 1) using two machines (Machine 0 and Machine 1) over four time periods (Period 0 to Period 3). The demand for Product 0 in Period 0 is **22.28694038228271**, in Period 1 is **15.40553617195959**, in Period 2 is **16.19622794603683**, and in Period 3 is **23.77133167016131**. The demand for Product 1 in Period 0 is **16.51020937307772**, in Period 1 is **15.41395287058592**, in Period 2 is **23.98409764196543**, and in Period 3 is **18.84683167137874**. + + The total production quantity for each product on each machine in each period, plus any startup time required, must not exceed the machine's capacity. Each machine has a limited capacity: Machine 0 has a capacity of **63.4339018881536**, and Machine 1 has a capacity of **67.51600987446864**. The startup time for Machine 0 is **7.917022863645049**, and for Machine 1, it is **7.682853596177669**. Each machine can produce only one product at a time in any given period. A machine can only be started up for a product if it was not producing that product in the previous period. + +The total production, inventory, and backlog must balance customer demand for each product in each period. Production quantities, inventory, and backlog must be non-negative. Production and startup decisions are binary (either 0 or 1). The goal is to determine the optimal production schedule, inventory levels, and backlog amounts to minimize the total cost while meeting all constraints and customer demand.No differencesAmbiguity, clarified connection of start time with capacityambiguous
97Minimize the total completion time (makespan) required to process all jobs in a manufacturing facility. There are 7 jobs, each consisting of 2 operations that must be performed in sequence. Each operation has a processing time of 1 unit. Operations from different jobs may share the same machine, but a machine can only handle one operation at a time. To avoid overlaps, the order of operations on shared machines must be determined. + +The start time of each operation must be non-negative, and the start time of an operation in a job must be at least 1 unit after the completion of the previous operation in the same job. For operations sharing a machine, one operation must finish before the other starts. This is enforced using binary decision variables that determine the order of operations on shared machines. A large constant (100,000) is used in the constraints to ensure logical consistency when determining the order of operations. + +The makespan must be at least as large as the completion time of every operation across all jobs. The objective is to minimize the makespan while ensuring that all precedence and machine capacity constraints are satisfied. The decision variables include the start times of all operations, the binary variables determining the order of operations on shared machines, and the makespan itself.5Minimize the total completion time (makespan) required to process all jobs in a manufacturing facility. There are 7 jobs, each consisting of 2 operations that must be performed in sequence. Each operation has a processing time of 1 unit. Operations from different jobs may share the same machine, but a machine can only handle one operation at a time. To avoid overlaps, the order of operations on shared machines must be determined. + +The start time of each operation must be non-negative, and the start time of an operation in a job must be at least 1 unit after the completion of the previous operation in the same job. For operations sharing a machine, one operation must finish before the other starts. This is enforced using binary decision variables that determine the order of operations on shared machines. A large constant (100) is used in the constraints to ensure logical consistency when determining the order of operations. + +The makespan must be at least as large as the completion time of every operation across all jobs. The objective is to minimize the makespan while ensuring that all precedence and machine capacity constraints are satisfied. The decision variables include the start times of all operations, the binary variables determining the order of operations on shared machines, and the makespan itself. + + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 1 + - Operation 1 of Job 0: Machine 2 + - Operation 0 of Job 1: Machine 0 + - Operation 1 of Job 1: Machine 1 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 0 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 1 + - Operation 1 of Job 4: Machine 1 + - Operation 0 of Job 5: Machine 1 + - Operation 1 of Job 5: Machine 0 + - Operation 0 of Job 6: Machine 0 + - Operation 1 of Job 6: Machine 26Minimize the total completion time (makespan) required to process all jobs in a manufacturing facility. There are 7 jobs, each consisting of 2 operations that must be performed in sequence. Each operation has a processing time of 1 unit. Operations from different jobs may share the same machine, but a machine can only handle one operation at a time. To avoid overlaps, the order of operations on shared machines must be determined. + +The start time of each operation must be non-negative, and the start time of an operation in a job must be at least 1 unit after the completion of the previous operation in the same job. For operations sharing a machine, one operation must finish before the other starts. This is enforced using binary decision variables that determine the order of operations on shared machines. A large constant (100,000) (100) is used in the constraints to ensure logical consistency when determining the order of operations. + +The makespan must be at least as large as the completion time of every operation across all jobs. The objective is to minimize the makespan while ensuring that all precedence and machine capacity constraints are satisfied. The decision variables include the start times of all operations, the binary variables determining the order of operations on shared machines, and the makespan itself. + + + + Additional information: + - There are 3 machines, numbered from 0 to 2 + - The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 1 + - Operation 1 of Job 0: Machine 2 + - Operation 0 of Job 1: Machine 0 + - Operation 1 of Job 1: Machine 1 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 0 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 1 + - Operation 1 of Job 4: Machine 1 + - Operation 0 of Job 5: Machine 1 + - Operation 1 of Job 5: Machine 0 + - Operation 0 of Job 6: Machine 0 + - Operation 1 of Job 6: Machine 25 6It was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
98There are five jobs, labeled Job 0 through Job 4, each consisting of two operations that must be processed in sequence. Each operation takes exactly 1 unit of time to complete and is assigned to a specific machine. The goal is to schedule these operations to minimize the total completion time, known as the makespan, which is the maximum completion time across all operations. + +For each job, the second operation cannot start until the first operation is completed. This ensures that the operations within a job follow a strict sequence. Additionally, each machine can process only one operation at a time. If two operations are assigned to the same machine, they cannot overlap. A binary decision variable determines the order in which these operations are scheduled on the machine. If the variable is 1, the first operation is scheduled before the second; if it is 0, the order is reversed. + +The constraints ensure that the start time of the second operation in a job is at least 1 unit of time after the start time of the first operation. For operations assigned to the same machine, the start time of one operation must be at least 1 unit of time after the completion of the other operation. A large number (100,000) is used in the constraints to enforce this order without overlapping. + +The objective is to determine the start times for all operations and the order in which operations are processed on shared machines, such that the total makespan is minimized. This involves ensuring that all precedence constraints within jobs are satisfied, preventing overlapping operations on the same machine, and minimizing the latest completion time across all operations.3There are five jobs, labeled Job 0 through Job 4, each consisting of two operations that must be processed in sequence. Each operation takes exactly 1 unit of time to complete and is assigned to a specific machine. The goal is to schedule these operations to minimize the total completion time, known as the makespan, which is the maximum completion time across all operations. + +For each job, the second operation cannot start until the first operation is completed. This ensures that the operations within a job follow a strict sequence. Additionally, each machine can process only one operation at a time. If two operations are assigned to the same machine, they cannot overlap. A binary decision variable determines the order in which these operations are scheduled on the machine. If the variable is 1, the first operation is scheduled before the second; if it is 0, the order is reversed. + +The constraints ensure that the start time of the second operation in a job is at least 1 unit of time after the start time of the first operation. For operations assigned to the same machine, the start time of one operation must be at least 1 unit of time after the completion of the other operation. A large number (100) is used in the constraints to enforce this order without overlapping. + +The objective is to determine the start times for all operations and the order in which operations are processed on shared machines, such that the total makespan is minimized. This involves ensuring that all precedence constraints within jobs are satisfied, preventing overlapping operations on the same machine, and minimizing the latest completion time across all operations. + + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 1 + - Operation 1 of Job 0: Machine 2 + - Operation 0 of Job 1: Machine 0 + - Operation 1 of Job 1: Machine 1 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 0 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 1 + - Operation 1 of Job 4: Machine 15For each job, the second operation cannot start until the first operation is completed. This ensures that the operations within a job follow a strict sequence. Additionally, each machine can process only one operation at a time. If two operations are assigned to the same machine, they cannot overlap. A binary decision variable determines the order in which these operations are scheduled on the machine. If the variable is 1, the first operation is scheduled before the second; if it is 0, the order is reversed. + +The constraints ensure that the start time of the second operation in a job is at least 1 unit of time after the start time of the first operation. For operations assigned to the same machine, the start time of one operation must be at least 1 unit of time after the completion of the other operation. A large number (100,000) (100) is used in the constraints to enforce this order without overlapping. + +The objective is to determine the start times for all operations and the order in which operations are processed on shared machines, such that the total makespan is minimized. This involves ensuring that all precedence constraints within jobs are satisfied, preventing overlapping operations on the same machine, and minimizing the latest completion time across all operations. + + + + Additional information: + - There are 3 machines, numbered from 0 to 2 + - The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 1 + - Operation 1 of Job 0: Machine 2 + - Operation 0 of Job 1: Machine 0 + - Operation 1 of Job 1: Machine 1 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 0 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 1 + - Operation 1 of Job 4: Machine 13 5It was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
99Facility Dispersion in a Network + +Imagine you are tasked with selecting a set of facilities to maximize their dispersion across a network of five locations (nodes). The goal is to ensure that the selected facilities are as spread out as possible, which is measured by the pairwise distances between them. You must select exactly two facilities out of the five available locations. + +Each pair of locations has a specific distance value associated with it, representing how far apart they are. These distances are as follows: + +- Distance between Node 0 and Node 1: 24 units +- Distance between Node 0 and Node 2: 22 units +- Distance between Node 0 and Node 3: 23 units +- Distance between Node 0 and Node 4: 21 units +- Distance between Node 1 and Node 0: 20 units +- Distance between Node 1 and Node 2: 19 units +- Distance between Node 1 and Node 3: 19 units +- Distance between Node 1 and Node 4: 25 units +- Distance between Node 2 and Node 0: 17 units +- Distance between Node 2 and Node 1: 20 units +- Distance between Node 2 and Node 3: 18 units +- Distance between Node 2 and Node 4: 20 units +- Distance between Node 3 and Node 0: 25 units +- Distance between Node 3 and Node 1: 18 units +- Distance between Node 3 and Node 2: 21 units +- Distance between Node 3 and Node 4: 24 units +- Distance between Node 4 and Node 0: 21 units +- Distance between Node 4 and Node 1: 18 units +- Distance between Node 4 and Node 2: 15 units +- Distance between Node 4 and Node 3: 23 units + +Your objective is to maximize the total dispersion, which is the sum of the distances between all pairs of selected facilities. However, there are specific rules governing the selection process: + +1. **Selection Constraint**: Exactly two facilities must be selected from the five available locations. +2. **Pairwise Activation**: A distance between two locations is only counted if both locations are selected as facilities. For example, the distance between Node 0 and Node 1 (24 units) is only included in the total dispersion if both Node 0 and Node 1 are chosen. +3. **Logical Consistency**: If a pair of locations is selected, the distance between them must be fully accounted for in the total dispersion. Conversely, if either location in a pair is not selected, the distance between them must not contribute to the total dispersion. + +Your task is to determine which two locations to select as facilities to maximize the total dispersion while adhering to the constraints above. The solution must ensure that the selection of facilities is logically consistent and that the total dispersion is calculated accurately based on the pairwise distances provided.48Facility Dispersion in a Network + +Imagine you are tasked with selecting a set of facilities to maximize their dispersion across a network of five locations (nodes). The goal is to ensure that the selected facilities are as spread out as possible, which is measured by the pairwise distances between them. You must select exactly two facilities out of the five available locations. + +Each pair of locations has a specific distance value associated with it, representing how far apart they are. These distances are as follows: + +- Distance between Node 0 and Node 1: 24 units +- Distance between Node 0 and Node 2: 22 units +- Distance between Node 0 and Node 3: 23 units +- Distance between Node 0 and Node 4: 21 units +- Distance between Node 1 and Node 0: 20 units +- Distance between Node 1 and Node 2: 19 units +- Distance between Node 1 and Node 3: 19 units +- Distance between Node 1 and Node 4: 25 units +- Distance between Node 2 and Node 0: 17 units +- Distance between Node 2 and Node 1: 20 units +- Distance between Node 2 and Node 3: 18 units +- Distance between Node 2 and Node 4: 20 units +- Distance between Node 3 and Node 0: 25 units +- Distance between Node 3 and Node 1: 18 units +- Distance between Node 3 and Node 2: 21 units +- Distance between Node 3 and Node 4: 24 units +- Distance between Node 4 and Node 0: 21 units +- Distance between Node 4 and Node 1: 18 units +- Distance between Node 4 and Node 2: 15 units +- Distance between Node 4 and Node 3: 23 units + +Your objective is to maximize the total dispersion, which is the sum of the distances between all pairs of selected facilities. However, there are specific rules governing the selection process: + +1. **Selection Constraint**: Exactly two facilities must be selected from the five available locations. +2. **Pairwise Activation**: A distance between two locations is only counted if both locations are selected as facilities. +3. **Logical Consistency**: If a pair of locations is selected, the distance between them must be fully accounted for in the total dispersion. Conversely, if either location in a pair is not selected, the distance between them must not contribute to the total dispersion. + +Your task is to determine which two locations to select as facilities to maximize the total dispersion while adhering to the constraints above. The solution must ensure that the selection of facilities is logically consistent and that the total dispersion is calculated accurately based on the pairwise distances provided.48Your objective is to maximize the total dispersion, which is the sum of the distances between all pairs of selected facilities. However, there are specific rules governing the selection process: + +1. **Selection Constraint**: Exactly two facilities must be selected from the five available locations. +2. **Pairwise Activation**: A distance between two locations is only counted if both locations are selected as facilities. For example, the distance between Node 0 and Node 1 (24 units) is only included in the total dispersion if both Node 0 and Node 1 are chosen. +3. **Logical Consistency**: If a pair of locations is selected, the distance between them must be fully accounted for in the total dispersion. Conversely, if either location in a pair is not selected, the distance between them must not contribute to the total dispersion. + +Your task is to determine which two locations to select as facilities to maximize the total dispersion while adhering to the constraints above. The solution must ensure that the selection of facilities is logically consistent and that the total dispersion is calculated accurately based on the pairwise distances provided.No differencesThe example in the description was misleading: there are two distances from (0,1) and (1,0) that should be counted if these nodes are selected, but only one of them was counted in the exampleambiguous
100There are 5 jobs, each consisting of 2 operations that must be performed in sequence. Each operation has a processing time of 1 unit of time. Operations are assigned to machines, and each machine can process only one operation at a time. The goal is to minimize the total completion time (makespan) of all jobs by determining the start times of each operation and the order in which operations are processed on the machines. + +For each job, the second operation cannot start until the first operation has been completed. For example, Operation 1 of Job 0 must start at least 1 unit of time after Operation 0 of Job 0 starts. This applies to all jobs. Operations assigned to the same machine cannot overlap in time. For every pair of operations on the same machine, one must be processed before the other. If Operation A is processed before Operation B, then Operation B must start at least 1 unit of time after Operation A completes. Conversely, if Operation B is processed before Operation A, then Operation A must start at least 1 unit of time after Operation B completes. These constraints apply to all relevant pairs of operations. + +The makespan must be at least as large as the completion time of every operation. For example, the makespan must be at least 1 unit of time after the start of Operation 0 of Job 0, and this applies to all operations across all jobs. All start times must be non-negative, and the order variables, which determine the sequence of operations on the machines, must be binary (0 or 1). A large constant (100,000) is used in the constraints to ensure the correct order of operations is enforced. The objective is to minimize the makespan, ensuring all operations are scheduled efficiently without violating the constraints.5There are 5 jobs, each consisting of 2 operations that must be performed in sequence. Each operation has a processing time of 1 unit of time. Operations are assigned to machines, and each machine can process only one operation at a time. The goal is to minimize the total completion time (makespan) of all jobs by determining the start times of each operation and the order in which operations are processed on the machines. + +For each job, the second operation cannot start until the first operation has been completed. For example, Operation 1 of Job 0 must start at least 1 unit of time after Operation 0 of Job 0 starts. This applies to all jobs. Operations assigned to the same machine cannot overlap in time. For every pair of operations on the same machine, one must be processed before the other. If Operation A is processed before Operation B, then Operation B must start at least 1 unit of time after Operation A completes. Conversely, if Operation B is processed before Operation A, then Operation A must start at least 1 unit of time after Operation B completes. These constraints apply to all relevant pairs of operations. + +The makespan must be at least as large as the completion time of every operation. For example, the makespan must be at least 1 unit of time after the start of Operation 0 of Job 0, and this applies to all operations across all jobs. All start times must be non-negative, and the order variables, which determine the sequence of operations on the machines, must be binary (0 or 1). A large constant (100) is used in the constraints to ensure the correct order of operations is enforced. The objective is to minimize the makespan, ensuring all operations are scheduled efficiently without violating the constraints. + + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 2 + - Operation 1 of Job 1: Machine 2 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 0 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 05For each job, the second operation cannot start until the first operation has been completed. For example, Operation 1 of Job 0 must start at least 1 unit of time after Operation 0 of Job 0 starts. This applies to all jobs. Operations assigned to the same machine cannot overlap in time. For every pair of operations on the same machine, one must be processed before the other. If Operation A is processed before Operation B, then Operation B must start at least 1 unit of time after Operation A completes. Conversely, if Operation B is processed before Operation A, then Operation A must start at least 1 unit of time after Operation B completes. These constraints apply to all relevant pairs of operations. + +The makespan must be at least as large as the completion time of every operation. For example, the makespan must be at least 1 unit of time after the start of Operation 0 of Job 0, and this applies to all operations across all jobs. All start times must be non-negative, and the order variables, which determine the sequence of operations on the machines, must be binary (0 or 1). A large constant (100,000) (100) is used in the constraints to ensure the correct order of operations is enforced. The objective is to minimize the makespan, ensuring all operations are scheduled efficiently without violating the constraints. + + + + Additional information: + - There are 3 machines, numbered from 0 to 2 + - The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 2 + - Operation 1 of Job 1: Machine 2 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 0 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 0No differencesIt was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
101Facility Dispersion in a Network of Locations + +You are tasked with selecting a subset of facilities to open across a network of five locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall dispersion between the selected facilities. Dispersion is measured by the pairwise distances between the facilities, with higher distances indicating better dispersion. The goal is to ensure that the selected facilities are as spread out as possible within the network. + +#### Key Decisions: +1. **Facility Selection**: You must decide which facilities to open. Specifically, you need to select exactly 2 out of the 5 available locations to open facilities. Each location can either be selected (1) or not selected (0). +2. **Pairwise Dispersion Activation**: For each pair of selected facilities, you must determine whether their dispersion contributes to the overall objective. This is represented by a binary decision variable that activates only if both facilities in the pair are selected. + +#### Objective: +Maximize the total dispersion across all pairs of selected facilities. The dispersion values between each pair of locations are as follows: +- Node 0 and Node 1: 18 units +- Node 0 and Node 2: 24 units +- Node 0 and Node 3: 21 units +- Node 0 and Node 4: 24 units +- Node 1 and Node 0: 17 units +- Node 1 and Node 2: 21 units +- Node 1 and Node 3: 21 units +- Node 1 and Node 4: 17 units +- Node 2 and Node 0: 19 units +- Node 2 and Node 1: 15 units +- Node 2 and Node 3: 19 units +- Node 2 and Node 4: 22 units +- Node 3 and Node 0: 15 units +- Node 3 and Node 1: 17 units +- Node 3 and Node 2: 22 units +- Node 3 and Node 4: 16 units +- Node 4 and Node 0: 25 units +- Node 4 and Node 1: 24 units +- Node 4 and Node 2: 16 units +- Node 4 and Node 3: 23 units + +The objective is to maximize the sum of these dispersion values for all pairs of selected facilities. + +#### Constraints: +1. **Facility Selection Limit**: Exactly 2 facilities must be selected out of the 5 available locations. This ensures that the solution is neither too sparse nor too dense. +2. **Pairwise Dispersion Activation Rules**: + - For any pair of locations, their dispersion can only contribute to the objective if both facilities in the pair are selected. + - If either facility in a pair is not selected, their dispersion cannot be counted. + - If both facilities in a pair are selected, their dispersion must be counted. + +#### Numerical Parameters: +- The number of facilities to select: 2. +- The dispersion values between each pair of locations are explicitly provided above. + +#### Summary: +Your task is to select 2 out of 5 locations to open facilities such that the total dispersion between the selected facilities is maximized. The dispersion values between each pair of locations are given, and the solution must adhere to the constraints on facility selection and pairwise dispersion activation. The goal is to ensure that the selected facilities are as spread out as possible within the network.49Facility Dispersion in a Network of Locations + +You are tasked with selecting a subset of facilities to open across a network of five locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall dispersion between the selected facilities. Dispersion is measured by the pairwise distances between the facilities, with higher distances indicating better dispersion. The goal is to ensure that the selected facilities are as spread out as possible within the network. + +#### Key Decisions: +1. **Facility Selection**: You must decide which facilities to open. Specifically, you need to select exactly 2 out of the 5 available locations to open facilities. Each location can either be selected (1) or not selected (0). +2. **Pairwise Dispersion Activation**: For each pair of selected facilities, you must determine whether their dispersion contributes to the overall objective. This is represented by a binary decision variable that activates only if both facilities in the pair are selected. + +#### Objective: +Maximize the total dispersion across all pairs of selected facilities. The dispersion values between each pair of locations are as follows: +- Node 0 and Node 1: 18 units +- Node 0 and Node 2: 24 units +- Node 0 and Node 3: 21 units +- Node 0 and Node 4: 24 units +- Node 1 and Node 0: 17 units +- Node 1 and Node 2: 21 units +- Node 1 and Node 3: 21 units +- Node 1 and Node 4: 17 units +- Node 2 and Node 0: 19 units +- Node 2 and Node 1: 15 units +- Node 2 and Node 3: 19 units +- Node 2 and Node 4: 22 units +- Node 3 and Node 0: 15 units +- Node 3 and Node 1: 17 units +- Node 3 and Node 2: 22 units +- Node 3 and Node 4: 16 units +- Node 4 and Node 0: 25 units +- Node 4 and Node 1: 24 units +- Node 4 and Node 2: 16 units +- Node 4 and Node 3: 23 units + +The objective is to maximize the sum of these dispersion values for all pairs of selected facilities. + +#### Constraints: +1. **Facility Selection Limit**: Exactly 2 facilities must be selected out of the 5 available locations. This ensures that the solution is neither too sparse nor too dense. +2. **Pairwise Dispersion Activation Rules**: + - For any pair of locations, their dispersion can only contribute to the objective if both facilities in the pair are selected. + - If either facility in a pair is not selected, their dispersion cannot be counted. + - If both facilities in a pair are selected, their dispersion must be counted. + +#### Numerical Parameters: +- The number of facilities to select: 2. +- The dispersion values between each pair of locations are explicitly provided above. + +#### Summary: +Your task is to select 2 out of 5 locations to open facilities such that the total dispersion between the selected facilities is maximized. The dispersion values between each pair of locations are given, and the solution must adhere to the constraints on facility selection and pairwise dispersion activation. The goal is to ensure that the selected facilities are as spread out as possible within the network.49No differencesNo differencesok
102You are managing a factory that produces three types of products: **product_0**, **product_1**, and **product_2**, over a five-period planning horizon. The factory operates with three types of machines: **grinders**, **drills**, and **borers**, each with specific capacities and maintenance schedules. Your goal is to maximize the factory's profit by optimizing production, inventory, and sales decisions while adhering to operational constraints. + +The profit per unit sold is **$249** for **product_0**, **$222** for **product_1**, and **$249** for **product_2**. The holding cost for each unit in inventory is **$15** per period. + +In the first period, the production of each product must equal the sum of its sales and inventory for that period. For subsequent periods, the inventory from the previous period plus the current production must equal the sum of current sales and inventory. + +Each unit of **product_0** requires **2 hours** on grinders, **2 hours** on drills, and **1 hour** on borers. Each unit of **product_1** requires **1 hour** on grinders, **2 hours** on drills, and **1 hour** on borers. Each unit of **product_2** requires **1 hour** on grinders, **2 hours** on drills, and **2 hours** on borers. + +The total available grinding capacity is **480 hours** per period, except in period 3, where it is **480 hours**. The total available drilling capacity is **320 hours** per period, except in period 1, where it is reduced to **160 hours**. The total available boring capacity is **160 hours** per period, except in period 0, where it is **0 hours**. + +The maximum inventory capacity for each product is **80 units** per period. The maximum number of units that can be sold for each product in each period is as follows: **product_0**: **63** (period 0), **68** (period 1), **42** (period 2), **58** (period 3), **54** (period 4); **product_1**: **53** (period 0), **62** (period 1), **67** (period 2), **46** (period 3), **56** (period 4); **product_2**: **64** (period 0), **67** (period 1), **65** (period 2), **65** (period 3), **58** (period 4). + +At the end of the planning horizon (period 4), the inventory for each product must be exactly **20 units**. All production, inventory, and sales quantities must be non-negative. Machine downtime for maintenance is accounted for in the capacity constraints. + +Your task is to determine the optimal production, inventory, and sales plan that maximizes profit while satisfying all operational constraints.104245.5You are managing a factory that produces three types of products: **product_0**, **product_1**, and **product_2**, over a five-period planning horizon. The factory operates with three types of machines: **grinders**, **drills**, and **borers**, each with specific capacities and maintenance schedules. Your goal is to maximize the factory's profit by optimizing production, inventory, and sales decisions while adhering to operational constraints. + +The profit per unit sold is **$249** for **product_0**, **$222** for **product_1**, and **$249** for **product_2**. The holding cost for each unit in inventory is **$15** per period. + +In the first period, the production of each product must equal the sum of its sales and inventory for that period. For subsequent periods, the inventory from the previous period plus the current production must equal the sum of current sales and inventory. + +Each unit of **product_0** requires **2 hours** on grinders, **2 hours** on drills, and **1 hour** on borers. Each unit of **product_1** requires **1 hour** on grinders, **2 hours** on drills, and **1 hour** on borers. Each unit of **product_2** requires **1 hour** on grinders, **2 hours** on drills, and **2 hours** on borers. + +The total available grinding capacity is **480 hours** per period, except in period 3, where it is **480 hours**. The total available drilling capacity is **320 hours** per period, except in period 1, where it is reduced to **160 hours**. The total available boring capacity is **160 hours** per period, except in period 0, where it is **0 hours**. + +The maximum inventory capacity for each product is **80 units** per period. The maximum number of units that can be sold for each product in each period is as follows: **product_0**: **63** (period 0), **68** (period 1), **42** (period 2), **58** (period 3), **54** (period 4); **product_1**: **53** (period 0), **62** (period 1), **67** (period 2), **46** (period 3), **56** (period 4); **product_2**: **64** (period 0), **67** (period 1), **65** (period 2), **65** (period 3), **58** (period 4). + +At the end of the planning horizon (period 4), the inventory for each product must be exactly **20 units**. All production, inventory, and sales quantities must be non-negative. Machine downtime for maintenance is accounted for in the capacity constraints. + +Your task is to determine the optimal production, inventory, and sales plan that maximizes profit while satisfying all operational constraints.104148No differences104245.5 104148The production quantity, inventory etc should be integers, but the ground truth was fractionalinteger/fractional
103Cell Tower Coverage Planning + +You are tasked with planning the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 32 distinct areas, each with a specific population size that needs to be served. Your goal is to determine which cell towers to build and which areas to cover, ensuring that the total cost of building the towers does not exceed the available budget. + +#### Key Decisions: +1. **Build or Not Build**: For each of the 30 potential cell tower locations, you must decide whether to build the tower (yes or no). Building a tower incurs a specific cost, which varies by location. +2. **Coverage Assignment**: For each of the 32 areas, you must determine whether it will be covered by at least one of the built towers (yes or no). An area is considered covered if at least one of the towers that can serve it is built. + +#### Objective: +Maximize the total population covered by the deployed cell towers. The population sizes for the 32 areas are as follows: +- Area 0: 4,657 +- Area 1: 4,663 +- Area 2: 3,139 +- Area 3: 2,579 +- Area 4: 3,171 +- Area 5: 2,796 +- Area 6: 3,502 +- Area 7: 4,951 +- Area 8: 4,211 +- Area 9: 2,885 +- Area 10: 4,948 +- Area 11: 2,195 +- Area 12: 4,002 +- Area 13: 3,856 +- Area 14: 2,663 +- Area 15: 3,197 +- Area 16: 4,465 +- Area 17: 4,648 +- Area 18: 3,796 +- Area 19: 4,714 +- Area 20: 4,952 +- Area 21: 3,088 +- Area 22: 4,029 +- Area 23: 3,379 +- Area 24: 2,002 +- Area 25: 3,840 +- Area 26: 4,037 +- Area 27: 4,246 +- Area 28: 4,852 +- Area 29: 4,423 +- Area 30: 2,630 +- Area 31: 4,951 + +#### Constraints: +1. **Coverage Requirements**: Each area must be covered by at least one of the built towers. The specific towers that can cover each area are listed below: + - Area 0 can be covered by Towers 2, 3, 4, 5, 8, 9, 10, 13, 14, 15, 20, 21, 22, 23, 26, or 28. + - Area 1 can be covered by Towers 1, 3, 5, 6, 9, 14, 15, 17, 18, 20, 21, 25, or 27. + - (Similar detailed coverage relationships apply for all 32 areas, as specified in the LP data.) + +2. **Budget Constraint**: The total cost of building the selected towers must not exceed the available budget of 3,204 units. The costs for building each tower are as follows: + - Tower 0: 455 + - Tower 1: 288 + - Tower 2: 208 + - Tower 3: 441 + - Tower 4: 428 + - Tower 5: 424 + - Tower 6: 285 + - Tower 7: 414 + - Tower 8: 459 + - Tower 9: 488 + - Tower 10: 362 + - Tower 11: 233 + - Tower 12: 341 + - Tower 13: 251 + - Tower 14: 422 + - Tower 15: 261 + - Tower 16: 437 + - Tower 17: 384 + - Tower 18: 442 + - Tower 19: 310 + - Tower 20: 218 + - Tower 21: 491 + - Tower 22: 412 + - Tower 23: 243 + - Tower 24: 311 + - Tower 25: 247 + - Tower 26: 352 + - Tower 27: 318 + - Tower 28: 380 + - Tower 29: 375 + +3. **Binary Decisions**: Both the decision to build a tower and the decision to cover an area are binary (yes or no). You cannot partially build a tower or partially cover an area. + +#### Summary: +Your task is to select which cell towers to build and which areas to cover, ensuring that: +- Each area is served by at least one built tower. +- The total cost of building the towers does not exceed the budget of 3,204 units. +- The total population covered by the selected towers is maximized. + +This problem requires careful balancing of costs and coverage to achieve the best possible outcome within the given constraints.121467Cell Tower Coverage Planning + +You are tasked with planning the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 32 distinct areas, each with a specific population size that needs to be served. Your goal is to determine which cell towers to build and which areas to cover, ensuring that the total cost of building the towers does not exceed the available budget. + +#### Key Decisions: +1. **Build or Not Build**: For each of the 30 potential cell tower locations, you must decide whether to build the tower (yes or no). Building a tower incurs a specific cost, which varies by location. +2. **Coverage Assignment**: For each of the 32 areas, you must determine whether it will be covered by at least one of the built towers (yes or no). An area is considered covered if at least one of the towers that can serve it is built. + +#### Objective: +Maximize the total population covered by the deployed cell towers. The population sizes for the 32 areas are as follows: +- Area 0: 4,657 +- Area 1: 4,663 +- Area 2: 3,139 +- Area 3: 2,579 +- Area 4: 3,171 +- Area 5: 2,796 +- Area 6: 3,502 +- Area 7: 4,951 +- Area 8: 4,211 +- Area 9: 2,885 +- Area 10: 4,948 +- Area 11: 2,195 +- Area 12: 4,002 +- Area 13: 3,856 +- Area 14: 2,663 +- Area 15: 3,197 +- Area 16: 4,465 +- Area 17: 4,648 +- Area 18: 3,796 +- Area 19: 4,714 +- Area 20: 4,952 +- Area 21: 3,088 +- Area 22: 4,029 +- Area 23: 3,379 +- Area 24: 2,002 +- Area 25: 3,840 +- Area 26: 4,037 +- Area 27: 4,246 +- Area 28: 4,852 +- Area 29: 4,423 +- Area 30: 2,630 +- Area 31: 4,951 + +#### Constraints: +1. **Coverage Requirements**: Each area must be covered by at least one of the built towers. The specific towers that can cover each area are listed below: + - Area 0 can be covered by Towers 2, 3, 4, 5, 8, 9, 10, 13, 14, 15, 20, 21, 22, 23, 26, or 28. + - Area 1 can be covered by Towers 1, 3, 5, 6, 9, 14, 15, 17, 18, 20, 21, 25, or 27. + - Area 2 can be covered if any of the towers 20, 3, 0, 23, 8, 7, 24, 4, 28, 17, 2, or 18 are built. + - Area 3 can be covered if any of the towers 13, 1, 0, 2, 6, 7, 16, 19, 27, 17, 25, or 20 are built. + - Area 4 can be covered if any of the towers 13, 7, 14, 18, 8, 0, 5, 22, 29, 10, 25, or 4 are built. + - Area 5 can be covered if any of the towers 6, 24, 10, 3, 2, 12, 26, 11, 22, 19, 8, or 1 are built. + - Area 6 can be covered if any of the towers 23, 14, 17, 3, 12, 2, 27, 9, 20, 19, 11, or 18 are built. + - Area 7 can be covered if any of the towers 6, 22, 2, 1, 21, 7, 9, 27, 24, 3, 12, or 8 are built. + - Area 8 can be covered if any of the towers 14, 20, 26, 11, 5, 27, 24, 6, 21, 8, 2, or 25 are built. + - Area 9 can be covered if any of the towers 17, 23, 7, 5, 14, 12, 8, 20, 29, 27, 10, or 1 are built. + - Area 10 can be covered if any of the towers 7, 26, 1, 25, 10, 12, 8, 2, 6, 18, 28, or 21 are built. + - Area 11 can be covered if any of the towers 20, 15, 12, 29, 14, 4, 8, 24, 7, 17, 26, or 23 are built. + - Area 12 can be covered if any of the towers 23, 18, 13, 28, 12, 11, 7, 4, 16, 15, 2, or 1 are built. + - Area 13 can be covered if any of the towers 27, 3, 4, 20, 5, 21, 13, 19, 2, 12, 26, or 14 are built. + - Area 14 can be covered if any of the towers 16, 8, 17, 0, 21, 23, 3, 25, 27, 28, 10, or 24 are built. + - Area 15 can be covered if any of the towers 9, 13, 5, 14, 0, 23, 24, 8, 16, 27, 21, or 3 are built. + - Area 16 can be covered if any of the towers 27, 20, 9, 26, 28, 16, 19, 6, 4, 11, 5, or 17 are built. + - Area 17 can be covered if any of the towers 24, 16, 0, 19, 10, 15, 27, 3, 11, 9, 7, or 1 are built. + - Area 18 can be covered if any of the towers 7, 28, 18, 2, 26, 23, 15, 25, 17, 4, 20, or 24 are built. + - Area 19 can be covered if any of the towers 17, 5, 8, 16, 19, 13, 6, 29, 23, 9, 12, or 11 are built. + - Area 20 can be covered if any of the towers 14, 28, 16, 29, 3, 7, 24, 2, 10, 0, 18, or 17 are built. + - Area 21 can be covered if any of the towers 7, 18, 29, 0, 2, 22, 20, 1, 27, 25, 24, or 10 are built. + - Area 22 can be covered if any of the towers 2, 16, 7, 8, 21, 15, 6, 17, 4, 18, 20, or 24 are built. + - Area 23 can be covered if any of the towers 7, 25, 15, 28, 13, 6, 3, 23, 21, 26, 11, or 20 are built. + - Area 24 can be covered if any of the towers 13, 14, 27, 23, 1, 21, 20, 26, 3, 25, 12, or 10 are built. + - Area 25 can be covered if any of the towers 25, 27, 3, 7, 6, 29, 17, 14, 4, 13, 5, or 8 are built. + - Area 26 can be covered if any of the towers 14, 7, 27, 2, 29, 17, 3, 1, 20, 24, 0, or 26 are built. + - Area 27 can be covered if any of the towers 29, 24, 27, 7, 5, 13, 15, 23, 6, 12, 1, or 25 are built. + - Area 28 can be covered if any of the towers 12, 0, 29, 8, 25, 14, 9, 13, 17, 15, 4, or 6 are built. + - Area 29 can be covered if any of the towers 9, 6, 1, 18, 23, 17, 27, 10, 25, 21, 26, or 15 are built. + - Area 30 can be covered if any of the towers 16, 27, 29, 5, 1, 28, 2, 26, 23, 19, 21, or 7 are built. + - Area 31 can be covered if any of the towers 12, 3, 18, 7, 27, 19, 1, 24, 2, 13, 25, or 22 are built. +2. **Budget Constraint**: The total cost of building the selected towers must not exceed the available budget of 1,204 units. The costs for building each tower are as follows: + - Tower 0: 455 + - Tower 1: 288 + - Tower 2: 208 + - Tower 3: 441 + - Tower 4: 428 + - Tower 5: 424 + - Tower 6: 285 + - Tower 7: 414 + - Tower 8: 459 + - Tower 9: 488 + - Tower 10: 362 + - Tower 11: 233 + - Tower 12: 341 + - Tower 13: 251 + - Tower 14: 422 + - Tower 15: 261 + - Tower 16: 437 + - Tower 17: 384 + - Tower 18: 442 + - Tower 19: 310 + - Tower 20: 218 + - Tower 21: 491 + - Tower 22: 412 + - Tower 23: 243 + - Tower 24: 311 + - Tower 25: 247 + - Tower 26: 352 + - Tower 27: 318 + - Tower 28: 380 + - Tower 29: 375 + +3. **Binary Decisions**: Both the decision to build a tower and the decision to cover an area are binary (yes or no). You cannot partially build a tower or partially cover an area. + +#### Summary: +Your task is to select which cell towers to build and which areas to cover, ensuring that: +- Each area is served by at least one built tower. +- The total cost of building the towers does not exceed the budget of 1,204 units. +- The total population covered by the selected towers is maximized. + +This problem requires careful balancing of costs and coverage to achieve the best possible outcome within the given constraints.121467#### Constraints: +1. **Coverage Requirements**: Each area must be covered by at least one of the built towers. The specific towers that can cover each area are listed below: + - Area 0 can be covered by Towers 2, 3, 4, 5, 8, 9, 10, 13, 14, 15, 20, 21, 22, 23, 26, or 28. + - Area 1 can be covered by Towers 1, 3, 5, 6, 9, 14, 15, 17, 18, 20, 21, 25, or 27. + - (Similar detailed coverage relationships apply for all 32 areas, as specified in Area 2 can be covered if any of the towers 20, 3, 0, 23, 8, 7, 24, 4, 28, 17, 2, or 18 are built. + - Area 3 can be covered if any of the towers 13, 1, 0, 2, 6, 7, 16, 19, 27, 17, 25, or 20 are built. + - Area 4 can be covered if any of the towers 13, 7, 14, 18, 8, 0, 5, 22, 29, 10, 25, or 4 are built. + - Area 5 can be covered if any of the towers 6, 24, 10, 3, 2, 12, 26, 11, 22, 19, 8, or 1 are built. + - Area 6 can be covered if any of the towers 23, 14, 17, 3, 12, 2, 27, 9, 20, 19, 11, or 18 are built. + - Area 7 can be covered if any of the towers 6, 22, 2, 1, 21, 7, 9, 27, 24, 3, 12, or 8 are built. + - Area 8 can be covered if any of the towers 14, 20, 26, 11, 5, 27, 24, 6, 21, 8, 2, or 25 are built. + - Area 9 can be covered if any of the towers 17, 23, 7, 5, 14, 12, 8, 20, 29, 27, 10, or 1 are built. + - Area 10 can be covered if any of the towers 7, 26, 1, 25, 10, 12, 8, 2, 6, 18, 28, or 21 are built. + - Area 11 can be covered if any of the towers 20, 15, 12, 29, 14, 4, 8, 24, 7, 17, 26, or 23 are built. + - Area 12 can be covered if any of the towers 23, 18, 13, 28, 12, 11, 7, 4, 16, 15, 2, or 1 are built. + - Area 13 can be covered if any of the towers 27, 3, 4, 20, 5, 21, 13, 19, 2, 12, 26, or 14 are built. + - Area 14 can be covered if any of the towers 16, 8, 17, 0, 21, 23, 3, 25, 27, 28, 10, or 24 are built. + - Area 15 can be covered if any of the towers 9, 13, 5, 14, 0, 23, 24, 8, 16, 27, 21, or 3 are built. + - Area 16 can be covered if any of the towers 27, 20, 9, 26, 28, 16, 19, 6, 4, 11, 5, or 17 are built. + - Area 17 can be covered if any of the towers 24, 16, 0, 19, 10, 15, 27, 3, 11, 9, 7, or 1 are built. + - Area 18 can be covered if any of the towers 7, 28, 18, 2, 26, 23, 15, 25, 17, 4, 20, or 24 are built. + - Area 19 can be covered if any of the towers 17, 5, 8, 16, 19, 13, 6, 29, 23, 9, 12, or 11 are built. + - Area 20 can be covered if any of the towers 14, 28, 16, 29, 3, 7, 24, 2, 10, 0, 18, or 17 are built. + - Area 21 can be covered if any of the towers 7, 18, 29, 0, 2, 22, 20, 1, 27, 25, 24, or 10 are built. + - Area 22 can be covered if any of the towers 2, 16, 7, 8, 21, 15, 6, 17, 4, 18, 20, or 24 are built. + - Area 23 can be covered if any of the towers 7, 25, 15, 28, 13, 6, 3, 23, 21, 26, 11, or 20 are built. + - Area 24 can be covered if any of the towers 13, 14, 27, 23, 1, 21, 20, 26, 3, 25, 12, or 10 are built. + - Area 25 can be covered if any of the towers 25, 27, 3, 7, 6, 29, 17, 14, 4, 13, 5, or 8 are built. + - Area 26 can be covered if any of the towers 14, 7, 27, 2, 29, 17, 3, 1, 20, 24, 0, or 26 are built. + - Area 27 can be covered if any of the towers 29, 24, 27, 7, 5, 13, 15, 23, 6, 12, 1, or 25 are built. + - Area 28 can be covered if any of the towers 12, 0, 29, 8, 25, 14, 9, 13, 17, 15, 4, or 6 are built. + - Area 29 can be covered if any of the towers 9, 6, 1, 18, 23, 17, 27, 10, 25, 21, 26, or 15 are built. + - Area 30 can be covered if any of the LP data.) towers 16, 27, 29, 5, 1, 28, 2, 26, 23, 19, 21, or 7 are built. + - Area 31 can be covered if any of the towers 12, 3, 18, 7, 27, 19, 1, 24, 2, 13, 25, or 22 are built. +2. **Budget Constraint**: The total cost of building the selected towers must not exceed the available budget of 3,204 1,204 units. The costs for building each tower are as follows: + - Tower 0: 455 + - Tower 1: 288 + - Tower 2: 208 +#### Summary: +Your task is to select which cell towers to build and which areas to cover, ensuring that: +- Each area is served by at least one built tower. +- The total cost of building the towers does not exceed the budget of 3,204 1,204 units. +- The total population covered by the selected towers is maximized. + +This problem requires careful balancing of costs and coverage to achieve the best possible outcome within the given constraints.No differencesMissing data about which towers cover which areasmissing data
104In a manufacturing facility, you are tasked with scheduling a set of jobs across multiple machines to minimize the total completion time, known as the makespan. Each job consists of a sequence of operations that must be performed in a specific order, and each operation is assigned to a particular machine. The goal is to determine the start times for each operation on its assigned machine while respecting the following constraints: + +Each job has a predefined sequence of operations. For example, Job 0 has two operations: Operation 0 must be completed before Operation 1 can begin. Similarly, Job 1 has Operation 0 and Operation 1, with Operation 0 preceding Operation 1. This pattern applies to all jobs, ensuring that no operation starts before its predecessor in the same job has finished. Specifically, for Job 0, Operation 1 cannot start until Operation 0 has been completed, and for Job 1, Operation 1 cannot start until Operation 0 has been completed. This logic applies to all jobs, ensuring the correct sequence of operations. + +Each machine can process only one operation at a time. To prevent overlapping, the start times of operations assigned to the same machine must be carefully scheduled. For example, if Operation 0 of Job 0 and Operation 0 of Job 1 are assigned to the same machine, they cannot overlap. A binary decision variable determines the order in which these operations are processed. If Operation 0 of Job 0 is scheduled first, Operation 0 of Job 1 must start only after Operation 0 of Job 0 has finished, and vice versa. Similar constraints apply to all pairs of operations assigned to the same machine, ensuring no two operations overlap. A large constant (100,000) is used in the constraints to enforce the order of operations on shared machines. + +The makespan is defined as the total time required to complete all jobs. It must be at least as large as the completion time of the last operation across all jobs. For example, if Operation 1 of Job 6 finishes at time 100, the makespan must be at least 100. This ensures that the makespan accurately reflects the total time needed to complete all operations. + +The start time of every operation must be non-negative, meaning no operation can start before time zero. Binary variables are used to determine the order of operations on shared machines. These variables can only take values of 0 or 1, representing the scheduling decisions. The processing time for each operation is implicitly defined by the constraints. For example, the precedence constraints specify that Operation 1 of Job 0 cannot start until Operation 0 of Job 0 has been completed, with a required gap of at least 1 time unit. The binary variables determine the order of operations on shared machines. For instance, if the binary variable for Operation 0 of Job 0 and Operation 0 of Job 1 is 1, Operation 0 of Job 0 is scheduled before Operation 0 of Job 1 on the same machine. The makespan must be at least as large as the completion time of every operation. For example, if Operation 0 of Job 0 finishes at time 10, the makespan must be at least 10. + +The objective is to minimize the makespan, ensuring that all jobs are completed in the shortest possible time while respecting the precedence and machine capacity constraints. This problem requires determining the optimal start times for all operations and the sequence of operations on shared machines to achieve the minimum total completion time.3In a manufacturing facility, you are tasked with scheduling a set of jobs across multiple machines to minimize the total completion time, known as the makespan. Each job consists of a sequence of operations that must be performed in a specific order, and each operation is assigned to a particular machine. The goal is to determine the start times for each operation on its assigned machine while respecting the following constraints: + +Each job has a predefined sequence of operations. For example, Job 0 has two operations: Operation 0 must be completed before Operation 1 can begin. Similarly, Job 1 has Operation 0 and Operation 1, with Operation 0 preceding Operation 1. This pattern applies to all jobs, ensuring that no operation starts before its predecessor in the same job has finished. Specifically, for Job 0, Operation 1 cannot start until Operation 0 has been completed, and for Job 1, Operation 1 cannot start until Operation 0 has been completed. This logic applies to all jobs, ensuring the correct sequence of operations. + +Each machine can process only one operation at a time. To prevent overlapping, the start times of operations assigned to the same machine must be carefully scheduled. For example, if Operation 0 of Job 0 and Operation 0 of Job 1 are assigned to the same machine, they cannot overlap. A binary decision variable determines the order in which these operations are processed. If Operation 0 of Job 0 is scheduled first, Operation 0 of Job 1 must start only after Operation 0 of Job 0 has finished, and vice versa. Similar constraints apply to all pairs of operations assigned to the same machine, ensuring no two operations overlap. A large constant (1000) is used in the constraints to enforce the order of operations on shared machines. + +The makespan is defined as the total time required to complete all jobs. It must be at least as large as the completion time of the last operation across all jobs. For example, if Operation 1 of Job 6 finishes at time 100, the makespan must be at least 100. This ensures that the makespan accurately reflects the total time needed to complete all operations. + +The start time of every operation must be non-negative, meaning no operation can start before time zero. Binary variables are used to determine the order of operations on shared machines. These variables can only take values of 0 or 1, representing the scheduling decisions. The processing time for each operation is implicitly defined by the constraints. For example, the precedence constraints specify that Operation 1 of Job 0 cannot start until Operation 0 of Job 0 has been completed, with a required gap of at least 1 time unit. The binary variables determine the order of operations on shared machines. For instance, if the binary variable for Operation 0 of Job 0 and Operation 0 of Job 1 is 1, Operation 0 of Job 0 is scheduled before Operation 0 of Job 1 on the same machine. The makespan must be at least as large as the completion time of every operation. For example, if Operation 0 of Job 0 finishes at time 10, the makespan must be at least 10. + +The objective is to minimize the makespan, ensuring that all jobs are completed in the shortest possible time while respecting the precedence and machine capacity constraints. This problem requires determining the optimal start times for all operations and the sequence of operations on shared machines to achieve the minimum total completion time. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3} +- For each job j in J, there are 2 operations indexed k = 0..1. We denote operation (j,k). +- Machines: M = {0, 1} + +2) Machine assignment for every operation (which machine executes the operation) +- Operation (0,0) -> Machine 0 +- Operation (0,1) -> Machine 1 +- Operation (1,0) -> Machine 0 +- Operation (1,1) -> Machine 1 +- Operation (2,0) -> Machine 1 +- Operation (2,1) -> Machine 0 +- Operation (3,0) -> Machine 1 +- Operation (3,1) -> Machine 0 + +3) Processing times (duration) for every operation (in time units) +- p(0,0) = 3 +- p(0,1) = 2 +- p(1,0) = 2 +- p(1,1) = 4 +- p(2,0) = 4 +- p(2,1) = 3 +- p(3,0) = 1 +- p(3,1) = 211Each job has a predefined sequence of operations. For example, Job 0 has two operations: Operation 0 must be completed before Operation 1 can begin. Similarly, Job 1 has Operation 0 and Operation 1, with Operation 0 preceding Operation 1. This pattern applies to all jobs, ensuring that no operation starts before its predecessor in the same job has finished. Specifically, for Job 0, Operation 1 cannot start until Operation 0 has been completed, and for Job 1, Operation 1 cannot start until Operation 0 has been completed. This logic applies to all jobs, ensuring the correct sequence of operations. + +Each machine can process only one operation at a time. To prevent overlapping, the start times of operations assigned to the same machine must be carefully scheduled. For example, if Operation 0 of Job 0 and Operation 0 of Job 1 are assigned to the same machine, they cannot overlap. A binary decision variable determines the order in which these operations are processed. If Operation 0 of Job 0 is scheduled first, Operation 0 of Job 1 must start only after Operation 0 of Job 0 has finished, and vice versa. Similar constraints apply to all pairs of operations assigned to the same machine, ensuring no two operations overlap. A large constant (100,000) (1000) is used in the constraints to enforce the order of operations on shared machines. + +The makespan is defined as the total time required to complete all jobs. It must be at least as large as the completion time of the last operation across all jobs. For example, if Operation 1 of Job 6 finishes at time 100, the makespan must be at least 100. This ensures that the makespan accurately reflects the total time needed to complete all operations. + +The start time of every operation must be non-negative, meaning no operation can start before time zero. Binary variables are used to determine the order of operations on shared machines. These variables can only take values of 0 or 1, representing the scheduling decisions. The processing time for each operation is implicitly defined by the constraints. For example, the precedence constraints specify that Operation 1 of Job 0 cannot start until Operation 0 of Job 0 has been completed, with a required gap of at least 1 time unit. The binary variables determine the order of operations on shared machines. For instance, if the binary variable for Operation 0 of Job 0 and Operation 0 of Job 1 is 1, Operation 0 of Job 0 is scheduled before Operation 0 of Job 1 on the same machine. The makespan must be at least as large as the completion time of every operation. For example, if Operation 0 of Job 0 finishes at time 10, the makespan must be at least 10. + +The objective is to minimize the makespan, ensuring that all jobs are completed in the shortest possible time while respecting the precedence and machine capacity constraints. This problem requires determining the optimal start times for all operations and the sequence of operations on shared machines to achieve the minimum total completion time. + + ADDITIONAL INPUT DATA: + + 1) Sets and indices + - Jobs: J = {0, 1, 2, 3} + - For each job j in J, there are 2 operations indexed k = 0..1. We denote operation (j,k). + - Machines: M = {0, 1} + + 2) Machine assignment for every operation (which machine executes the operation) + - Operation (0,0) -> Machine 0 + - Operation (0,1) -> Machine 1 + - Operation (1,0) -> Machine 0 + - Operation (1,1) -> Machine 1 + - Operation (2,0) -> Machine 1 + - Operation (2,1) -> Machine 0 + - Operation (3,0) -> Machine 1 + - Operation (3,1) -> Machine 0 + + 3) Processing times (duration) for every operation (in time units) + - p(0,0) = 3 + - p(0,1) = 2 + - p(1,0) = 2 + - p(1,1) = 4 + - p(2,0) = 4 + - p(2,1) = 3 + - p(3,0) = 1 + - p(3,1) = 23 11It was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
105There are 7 jobs, labeled Job 0 through Job 6, each consisting of 2 operations that must be performed in sequence. Each operation takes 1 unit of time to complete. The facility has a limited number of machines, and some operations from different jobs must be performed on the same machine. To avoid conflicts, no two operations can be scheduled on the same machine at the same time. A binary decision variable determines the order in which these operations are scheduled. For example, if Operation 1 of Job 0 and Operation 0 of Job 3 are assigned to the same machine, the binary variable ensures that one operation starts at least 1 unit of time after the other. Within each job, operations must be performed in the specified order. For instance, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 has been completed. The makespan, defined as the total time required to complete all jobs, must be at least as large as the completion time of the last operation across all jobs. For every pair of operations that share the same machine, a binary variable determines the order in which they are scheduled. The objective is to minimize the makespan, ensuring that all jobs are completed in the shortest possible time. The start time of the second operation in each job must be at least 1 unit of time after the start time of the first operation. For every pair of operations assigned to the same machine, the schedule must ensure that they do not overlap. The makespan must be at least as large as the completion time of every operation. All start times must be non-negative. The binary variables that determine the order of operations on the same machine must be either 0 or 1. A large constant, 100,000, is used in the constraints to enforce the order of operations on shared machines. The processing time for every operation is 1 unit of time. The objective is to minimize the total time required to complete all jobs, ensuring that all operations are scheduled efficiently without violating any constraints.5There are 7 jobs, labeled Job 0 through Job 6, each consisting of 2 operations that must be performed in sequence. Each operation takes 1 unit of time to complete. The facility has a limited number of machines, and some operations from different jobs must be performed on the same machine. To avoid conflicts, no two operations can be scheduled on the same machine at the same time. A binary decision variable determines the order in which these operations are scheduled. The makespan, defined as the total time required to complete all jobs, must be at least as large as the completion time of the last operation across all jobs. For every pair of operations that share the same machine, a binary variable determines the order in which they are scheduled. The objective is to minimize the makespan, ensuring that all jobs are completed in the shortest possible time. The start time of the second operation in each job must be at least 1 unit of time after the start time of the first operation. For every pair of operations assigned to the same machine, the schedule must ensure that they do not overlap. The makespan must be at least as large as the completion time of every operation. All start times must be non-negative. The binary variables that determine the order of operations on the same machine must be either 0 or 1. A large constant, 100, is used in the constraints to enforce the order of operations on shared machines. The processing time for every operation is 1 unit of time. The objective is to minimize the total time required to complete all jobs, ensuring that all operations are scheduled efficiently without violating any constraints. + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 1 + - Operation 1 of Job 1: Machine 0 + - Operation 0 of Job 2: Machine 0 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 0 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 0 + - Operation 0 of Job 5: Machine 0 + - Operation 1 of Job 5: Machine 1 + - Operation 0 of Job 6: Machine 2 + - Operation 1 of Job 6: Machine 15There are 7 jobs, labeled Job 0 through Job 6, each consisting of 2 operations that must be performed in sequence. Each operation takes 1 unit of time to complete. The facility has a limited number of machines, and some operations from different jobs must be performed on the same machine. To avoid conflicts, no two operations can be scheduled on the same machine at the same time. A binary decision variable determines the order in which these operations are scheduled. For example, if Operation 1 of Job 0 and Operation 0 of Job 3 are assigned to the same machine, the binary variable ensures that one operation starts at least 1 unit of time after the other. Within each job, operations must be performed in the specified order. For instance, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 has been completed. The makespan, defined as the total time required to complete all jobs, must be at least as large as the completion time of the last operation across all jobs. For every pair of operations that share the same machine, a binary variable determines the order in which they are scheduled. The objective is to minimize the makespan, ensuring that all jobs are completed in the shortest possible time. The start time of the second operation in each job must be at least 1 unit of time after the start time of the first operation. For every pair of operations assigned to the same machine, the schedule must ensure that they do not overlap. The makespan must be at least as large as the completion time of every operation. All start times must be non-negative. The binary variables that determine the order of operations on the same machine must be either 0 or 1. A large constant, 100,000, 100, is used in the constraints to enforce the order of operations on shared machines. The processing time for every operation is 1 unit of time. The objective is to minimize the total time required to complete all jobs, ensuring that all operations are scheduled efficiently without violating any constraints. + + + Additional information: + - There are 3 machines, numbered from 0 to 2 + - The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 1 + - Operation 1 of Job 1: Machine 0 + - Operation 0 of Job 2: Machine 0 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 0 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 0 + - Operation 0 of Job 5: Machine 0 + - Operation 1 of Job 5: Machine 1 + - Operation 0 of Job 6: Machine 2 + - Operation 1 of Job 6: Machine 1No differencesIt was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
106The problem involves managing the distribution of a critical resource across seven interconnected cities to minimize the total transportation cost while meeting the demand or surplus of each city and respecting capacity constraints. Each city has a specific net demand or surplus: City 0 requires 3 additional units, City 1 requires 1 additional unit, City 2 has a surplus of 2 units, City 3 is balanced with no net demand, City 4 has a surplus of 2 units, City 5 has a surplus of 1 unit, and City 6 requires 1 additional unit. The cost of transporting resources between cities varies by route, with costs ranging from 1 to 3 units per resource. For example, transporting from City 0 to City 1 costs 1 unit, while transporting from City 0 to City 2 costs 2 units. Each city has a maximum capacity for receiving resources: City 0 can receive up to 18 units, City 1 up to 18 units, City 2 up to 12 units, City 3 up to 12 units, City 4 up to 14 units, City 5 up to 18 units, and City 6 up to 18 units. Additionally, each transportation link between cities has a maximum capacity, such as the link from City 0 to City 1, which can handle up to 9 units, and the link from City 0 to City 2, which can handle up to 8 units. The objective is to determine the optimal amount of resources to transport between cities to meet all demands while minimizing the total transportation cost. This requires ensuring that the net flow of resources into each city matches its demand or surplus, respecting the capacity limits of both the cities and the transportation links, and ensuring that no transportation link carries more resources than its maximum capacity. All transportation amounts must be non-negative.9The problem involves managing the distribution of a critical resource across seven interconnected cities to minimize the total transportation cost while meeting the demand or surplus of each city and respecting capacity constraints. + +• Net demand / surplus of every city +– City 0 requires 3 additional units +– City 1 requires 1 additional unit +– City 2 has a surplus of 2 units +– City 3 is balanced (0) +– City 4 has a surplus of 2 units +– City 5 has a surplus of 1 unit +– City 6 requires 1 additional unit + +• Maximum amount a city is allowed to receive +– City 0: 18 units +– City 1: 18 units +– City 2: 12 units +– City 3: 12 units +– City 4: 14 units +– City 5: 18 units +– City 6: 18 units + +• Transportation links (all links are bidirectional; each direction has the same +cost and capacity). +For every ordered pair (i, j) listed below, the first number is the cost per +unit shipped, the second number is the maximum capacity of that link. + +(0, 1): cost 1, capacity 9  (1, 0): cost 1, capacity 9 +(0, 2): cost 2, capacity 8  (2, 0): cost 2, capacity 8 +(0, 3): cost 2, capacity 10 (3, 0): cost 2, capacity 10 +(0, 4): cost 3, capacity 10 (4, 0): cost 3, capacity 10 +(0, 5): cost 3, capacity 12 (5, 0): cost 3, capacity 12 +(0, 6): cost 2, capacity 10 (6, 0): cost 2, capacity 10 + +(1, 2): cost 1, capacity 9  (2, 1): cost 1, capacity 9 +(1, 3): cost 2, capacity 9  (3, 1): cost 2, capacity 9 +(1, 4): cost 3, capacity 10 (4, 1): cost 3, capacity 10 +(1, 5): cost 2, capacity 11 (5, 1): cost 2, capacity 11 +(1, 6): cost 2, capacity 10 (6, 1): cost 2, capacity 10 + +(2, 3): cost 1, capacity 8  (3, 2): cost 1, capacity 8 +(2, 4): cost 1, capacity 7  (4, 2): cost 1, capacity 7 +(2, 5): cost 2, capacity 8  (5, 2): cost 2, capacity 8 +(2, 6): cost 3, capacity 9  (6, 2): cost 3, capacity 9 + +(3, 4): cost 2, capacity 8  (4, 3): cost 2, capacity 8 +(3, 5): cost 1, capacity 9  (5, 3): cost 1, capacity 9 +(3, 6): cost 2, capacity 9  (6, 3): cost 2, capacity 9 + +(4, 5): cost 1, capacity 7  (5, 4): cost 1, capacity 7 +(4, 6): cost 1, capacity 8  (6, 4): cost 1, capacity 8 + +(5, 6): cost 2, capacity 12 (6, 5): cost 2, capacity 12 + +The objective is to determine the optimal quantities to transport over every link so that all city demands are satisfied, no link or city capacity is violated, and the total transportation cost is minimized. All shipment amounts must be non-negative.10The problem involves managing the distribution of a critical resource across seven interconnected cities to minimize the total transportation cost while meeting the demand or surplus of each city and respecting capacity constraints. + + Each city has a specific net • Net demand or surplus: / surplus of every city + – City 0 requires 3 additional units, units + – City 1 requires 1 additional unit, unit + – City 2 has a surplus of 2 units, units + – City 3 is balanced with no net demand, (0) + – City 4 has a surplus of 2 units, units + – City 5 has a surplus of 1 unit, and unit + – City 6 requires 1 additional unit. The cost of transporting resources between cities varies by route, with costs ranging from 1 to 3 units per resource. For example, transporting from City 0 to City 1 costs 1 unit, while transporting from City 0 to City 2 costs 2 units. Each city has unit + + • Maximum amount a maximum capacity for receiving resources: City 0 can receive up city is allowed to receive + – City 0: 18 units, units + – City 1 up to 1: 18 units, units + – City 2 up to 2: 12 units, units + – City 3 up to 3: 12 units, units + – City 4 up to 4: 14 units, units + – City 5 up to 5: 18 units, and units + – City 6 up to 6: 18 units. Additionally, units + + • Transportation links (all links are bidirectional; each transportation link between cities direction has a maximum capacity, such as the link from City 0 to City 1, which can handle up to 9 units, same + cost and capacity). + For every ordered pair (i, j) listed below, the link from City 0 to City first number is the cost per + unit shipped, the second number is the maximum capacity of that link. + + (0, 1): cost 1, capacity 9  (1, 0): cost 1, capacity 9 + (0, 2): cost 2, which can handle up to capacity 8  (2, 0): cost 2, capacity 8 + units. (0, 3): cost 2, capacity 10 (3, 0): cost 2, capacity 10 + (0, 4): cost 3, capacity 10 (4, 0): cost 3, capacity 10 + (0, 5): cost 3, capacity 12 (5, 0): cost 3, capacity 12 + (0, 6): cost 2, capacity 10 (6, 0): cost 2, capacity 10 + + (1, 2): cost 1, capacity 9  (2, 1): cost 1, capacity 9 + (1, 3): cost 2, capacity 9  (3, 1): cost 2, capacity 9 + (1, 4): cost 3, capacity 10 (4, 1): cost 3, capacity 10 + (1, 5): cost 2, capacity 11 (5, 1): cost 2, capacity 11 + (1, 6): cost 2, capacity 10 (6, 1): cost 2, capacity 10 + + (2, 3): cost 1, capacity 8  (3, 2): cost 1, capacity 8 + (2, 4): cost 1, capacity 7  (4, 2): cost 1, capacity 7 + (2, 5): cost 2, capacity 8  (5, 2): cost 2, capacity 8 + (2, 6): cost 3, capacity 9  (6, 2): cost 3, capacity 9 + + (3, 4): cost 2, capacity 8  (4, 3): cost 2, capacity 8 + (3, 5): cost 1, capacity 9  (5, 3): cost 1, capacity 9 + (3, 6): cost 2, capacity 9  (6, 3): cost 2, capacity 9 + + (4, 5): cost 1, capacity 7  (5, 4): cost 1, capacity 7 + (4, 6): cost 1, capacity 8  (6, 4): cost 1, capacity 8 + + (5, 6): cost 2, capacity 12 (6, 5): cost 2, capacity 12 + +The objective is to determine the optimal amount of resources quantities to transport between cities to meet all demands while minimizing the total transportation cost. This requires ensuring over every link so that the net flow of resources into each all city matches its demand demands are satisfied, no link or surplus, respecting the city capacity limits of both the cities is violated, and the total transportation links, and ensuring that no transportation link carries more resources than its maximum capacity. cost is minimized. All transportation shipment amounts must be non-negative.9 10Missing some of the costs and capacities on the edges missing data
107Imagine you are managing the distribution of resources across nine interconnected cities. Each city has a specific supply or demand for resources, and your task is to ensure that the flow of resources between cities meets these requirements while minimizing the total transportation cost. The transportation costs vary depending on the route taken between cities, and each route has a maximum capacity limit that cannot be exceeded. + +#### City Requirements: +- **City 0** has a surplus of 69 units of resources. +- **City 1** has a deficit of 3 units. +- **City 2** has a surplus of 8 units. +- **City 3** has a deficit of 10 units. +- **City 4** has a deficit of 28 units. +- **City 5** has a deficit of 15 units. +- **City 6** has a surplus of 8 units. +- **City 7** has a deficit of 20 units. +- **City 8** has a deficit of 9 units. + +#### Transportation Costs: +The cost of transporting resources between cities varies as follows: +- From **City 0**: To City 1 costs 7, City 2 costs 6, City 3 costs 6, City 4 costs 10, City 5 costs 6, City 6 costs 8, City 7 costs 7, and City 8 costs 6. +- From **City 1**: To City 0 costs 10, City 2 costs 7, City 3 costs 9, City 4 costs 10, City 5 costs 10, City 6 costs 9, City 7 costs 8, and City 8 costs 6. +- From **City 2**: To City 0 costs 6, City 1 costs 6, City 3 costs 7, City 4 costs 8, City 5 costs 6, City 6 costs 7, City 7 costs 7, and City 8 costs 6. +- From **City 3**: To City 0 costs 8, City 1 costs 9, City 2 costs 6, City 4 costs 8, City 5 costs 8, City 6 costs 8, City 7 costs 10, and City 8 costs 10. +- From **City 4**: To City 0 costs 8, City 1 costs 7, City 2 costs 8, City 3 costs 7, City 5 costs 8, City 6 costs 7, City 7 costs 10, and City 8 costs 10. +- From **City 5**: To City 0 costs 10, City 1 costs 9, City 2 costs 6, City 3 costs 8, City 4 costs 10, City 6 costs 9, City 7 costs 8, and City 8 costs 9. +- From **City 6**: To City 0 costs 9, City 1 costs 7, City 2 costs 10, City 3 costs 7, City 4 costs 10, City 5 costs 10, City 7 costs 7, and City 8 costs 9. +- From **City 7**: To City 0 costs 10, City 1 costs 6, City 2 costs 6, City 3 costs 10, City 4 costs 9, City 5 costs 9, City 6 costs 10, and City 8 costs 10. +- From **City 8**: To City 0 costs 9, City 1 costs 10, City 2 costs 10, City 3 costs 10, City 4 costs 7, City 5 costs 9, City 6 costs 6, and City 7 costs 8. + +#### Capacity Constraints: +Each route between cities has a maximum capacity limit: +- **City 0**: To City 1 (59), City 2 (51), City 3 (49), City 4 (36), City 5 (38), City 6 (37), City 7 (55), City 8 (33). +- **City 1**: To City 0 (59), City 2 (36), City 3 (43), City 4 (38), City 5 (35), City 6 (41), City 7 (45), City 8 (47). +- **City 2**: To City 0 (47), City 1 (50), City 3 (58), City 4 (48), City 5 (53), City 6 (55), City 7 (50), City 8 (60). +- **City 3**: To City 0 (32), City 1 (34), City 2 (54), City 4 (31), City 5 (55), City 6 (45), City 7 (34), City 8 (49). +- **City 4**: To City 0 (59), City 1 (50), City 2 (57), City 3 (47), City 5 (33), City 6 (59), City 7 (56), City 8 (58). +- **City 5**: To City 0 (38), City 1 (34), City 2 (37), City 3 (40), City 4 (35), City 6 (46), City 7 (36), City 8 (59). +- **City 6**: To City 0 (51), City 1 (46), City 2 (54), City 3 (48), City 4 (35), City 5 (34), City 7 (39), City 8 (45). +- **City 7**: To City 0 (33), City 1 (55), City 2 (54), City 3 (35), City 4 (44), City 5 (56), City 6 (52), City 8 (35). +- **City 8**: To City 0 (59), City 1 (43), City 2 (56), City 3 (53), City 4 (53), City 5 (55), City 6 (38), City 7 (56). + +#### Objective: +Your goal is to determine the optimal flow of resources between cities to meet all supply and demand requirements while minimizing the total transportation cost. Ensure that the flow on each route does not exceed its capacity and that all resource flows are non-negative. + +This problem requires balancing the flow of resources across the network of cities, respecting the capacity limits of each route, and achieving the lowest possible transportation cost.709Imagine you are managing the distribution of resources across nine interconnected cities. Each city has a specific supply or demand for resources, and your task is to ensure that the flow of resources between cities meets these requirements while minimizing the total transportation cost. The transportation costs vary depending on the route taken between cities, and each route has a maximum capacity limit that cannot be exceeded. + +#### City Requirements: +- **City 0** has a surplus of 69 units of resources. +- **City 1** has a deficit of 3 units. +- **City 2** has a surplus of 8 units. +- **City 3** has a deficit of 10 units. +- **City 4** has a deficit of 28 units. +- **City 5** has a deficit of 15 units. +- **City 6** has a surplus of 8 units. +- **City 7** has a deficit of 20 units. +- **City 8** has a deficit of 9 units. + +#### Transportation Costs: +The cost of transporting resources between cities varies as follows: +- From **City 0**: To City 1 costs 7, City 2 costs 6, City 3 costs 6, City 4 costs 10, City 5 costs 6, City 6 costs 8, City 7 costs 7, and City 8 costs 6. +- From **City 1**: To City 0 costs 10, City 2 costs 7, City 3 costs 9, City 4 costs 10, City 5 costs 10, City 6 costs 9, City 7 costs 8, and City 8 costs 6. +- From **City 2**: To City 0 costs 6, City 1 costs 6, City 3 costs 7, City 4 costs 8, City 5 costs 6, City 6 costs 7, City 7 costs 7, and City 8 costs 6. +- From **City 3**: To City 0 costs 8, City 1 costs 9, City 2 costs 6, City 4 costs 8, City 5 costs 8, City 6 costs 8, City 7 costs 10, and City 8 costs 10. +- From **City 4**: To City 0 costs 8, City 1 costs 7, City 2 costs 8, City 3 costs 7, City 5 costs 8, City 6 costs 7, City 7 costs 10, and City 8 costs 10. +- From **City 5**: To City 0 costs 10, City 1 costs 9, City 2 costs 6, City 3 costs 8, City 4 costs 10, City 6 costs 9, City 7 costs 8, and City 8 costs 9. +- From **City 6**: To City 0 costs 9, City 1 costs 7, City 2 costs 10, City 3 costs 7, City 4 costs 10, City 5 costs 10, City 7 costs 7, and City 8 costs 9. +- From **City 7**: To City 0 costs 10, City 1 costs 6, City 2 costs 6, City 3 costs 10, City 4 costs 9, City 5 costs 9, City 6 costs 10, and City 8 costs 10. +- From **City 8**: To City 0 costs 9, City 1 costs 10, City 2 costs 10, City 3 costs 10, City 4 costs 7, City 5 costs 9, City 6 costs 6, and City 7 costs 8. + +#### Capacity Constraints: +Each route between cities has a maximum capacity limit: +- **City 0**: To City 1 (59), City 2 (51), City 3 (49), City 4 (36), City 5 (38), City 6 (37), City 7 (55), City 8 (33). +- **City 1**: To City 0 (59), City 2 (36), City 3 (43), City 4 (38), City 5 (35), City 6 (41), City 7 (45), City 8 (47). +- **City 2**: To City 0 (47), City 1 (50), City 3 (58), City 4 (48), City 5 (53), City 6 (55), City 7 (50), City 8 (60). +- **City 3**: To City 0 (32), City 1 (34), City 2 (54), City 4 (31), City 5 (55), City 6 (45), City 7 (34), City 8 (49). +- **City 4**: To City 0 (59), City 1 (50), City 2 (57), City 3 (47), City 5 (33), City 6 (59), City 7 (56), City 8 (58). +- **City 5**: To City 0 (38), City 1 (34), City 2 (37), City 3 (40), City 4 (35), City 6 (46), City 7 (36), City 8 (59). +- **City 6**: To City 0 (51), City 1 (46), City 2 (54), City 3 (48), City 4 (35), City 5 (34), City 7 (39), City 8 (45). +- **City 7**: To City 0 (33), City 1 (55), City 2 (54), City 3 (35), City 4 (44), City 5 (56), City 6 (52), City 8 (35). +- **City 8**: To City 0 (59), City 1 (43), City 2 (56), City 3 (53), City 4 (53), City 5 (55), City 6 (38), City 7 (56). + +#### Objective: +Your goal is to determine the optimal flow of resources between cities to meet all supply and demand requirements while minimizing the total transportation cost. Ensure that the flow on each route does not exceed its capacity and that all resource flows are non-negative. + +This problem requires balancing the flow of resources across the network of cities, respecting the capacity limits of each route, and achieving the lowest possible transportation cost.629No differences709 629Incorrect ground truthincorrect ground truth
108Minimize the total operational cost, which includes the base cost of operating generators, the variable cost based on power output, and the startup cost for generators. The base cost is 80 units per generator per period for Type 0 and Type 2 generators, and 78 units for Type 1 generators. The variable cost is 4 units per MW for Type 0 and Type 1 generators, and 3 units per MW for Type 2 generators. The startup cost is 353 units for Type 0 generators, 349 units for Type 1 generators, and 351 units for Type 2 generators. + +The total power output from all generators must meet or exceed the demand in each period, which is 163 MW in Period 0, 180 MW in Period 1, 208 MW in Period 2, 183 MW in Period 3, and 227 MW in Period 4. Each generator type has a minimum and maximum power output per generator: Type 0 generators must produce at least 35 MW and at most 126 MW, Type 1 generators must produce at least 35 MW and at most 130 MW, and Type 2 generators must produce at least 35 MW and at most 125 MW. + +The total maximum possible output from all generators must be at least 15% higher than the demand in each period. For example, in Period 0, the total maximum output must be at least 187.45 MW. The number of generators operating in each period cannot exceed the available capacity: up to 5 Type 0 generators, 2 Type 1 generators, and 1 Type 2 generator can be operated in any period. + +The number of generators started in a period is limited by the change in the number of generators operating from the previous period to the current period. For example, in Period 1, the number of Type 0 generators started cannot exceed the difference between the number of Type 0 generators operating in Period 1 and Period 0. This rule applies similarly for Type 1 and Type 2 generators across all periods. + +The goal is to determine the optimal number of generators to operate, the power output for each generator type, and the number of generators to start in each period to minimize the total operational cost while meeting all constraints.4436Minimize the total operational cost, which includes the base cost of operating generators, the variable cost based on power output, and the startup cost for generators. The base cost is 80 units per generator per period for Type 0 and Type 2 generators, and 78 units for Type 1 generators. The variable cost is 4 units per MW for Type 0 and Type 1 generators, and 3 units per MW for Type 2 generators. The startup cost is 353 units for Type 0 generators, 349 units for Type 1 generators, and 351 units for Type 2 generators. + +The total power output from all generators must meet or exceed the demand in each period, which is 163 MW in Period 0, 180 MW in Period 1, 208 MW in Period 2, 183 MW in Period 3, and 227 MW in Period 4. Each generator type has a minimum and maximum power output per generator: Type 0 generators must produce at least 35 MW and at most 126 MW, Type 1 generators must produce at least 35 MW and at most 130 MW, and Type 2 generators must produce at least 35 MW and at most 125 MW. + +The total maximum possible output from all generators must be at least 15% higher than the demand in each period. For example, in Period 0, the total maximum output must be at least 187.45 MW. The number of generators operating in each period cannot exceed the available capacity: up to 5 Type 0 generators, 2 Type 1 generators, and 1 Type 2 generator can be operated in any period. + +The number of generators started in a period is limited by the change in the number of generators operating from the previous period to the current period. For example, in Period 1, the number of Type 0 generators started cannot exceed the difference between the number of Type 0 generators operating in Period 1 and Period 0. This rule applies similarly for Type 1 and Type 2 generators across all periods. + +The goal is to determine the optimal number of generators to operate, the power output for each generator type, and the number of generators to start in each period to minimize the total operational cost while meeting all constraints.5136No differences4436 5136Incorrect ground truthincorrect ground truth
109A manufacturing facility processes six jobs (Job 0 through Job 5) on four machines (Machine 0 through Machine 3). Each job must be processed on all four machines in sequence, and the facility must determine the optimal schedule to minimize the total production time. Each job has a unique processing time on each machine, as follows: + +- **Job 0**: 5 hours on Machine 0, 8 hours on Machine 1, 8 hours on Machine 2, and 5 hours on Machine 3. +- **Job 1**: 7 hours on Machine 0, 6 hours on Machine 1, 7 hours on Machine 2, and 5 hours on Machine 3. +- **Job 2**: 10 hours on Machine 0, 10 hours on Machine 1, 8 hours on Machine 2, and 7 hours on Machine 3. +- **Job 3**: 7 hours on Machine 0, 10 hours on Machine 1, 5 hours on Machine 2, and 8 hours on Machine 3. +- **Job 4**: 9 hours on Machine 0, 9 hours on Machine 1, 5 hours on Machine 2, and 6 hours on Machine 3. +- **Job 5**: 6 hours on Machine 0, 9 hours on Machine 1, 6 hours on Machine 2, and 6 hours on Machine 3. + +The facility must adhere to the following constraints: + +1. **One Job per Schedule Position**: Each of the six schedule positions (Position 0 through Position 5) must be assigned exactly one job. No two positions can have the same job. + +2. **One Schedule Position per Job**: Each job must be assigned to exactly one schedule position. No job can be assigned to multiple positions. + +3. **Machine Precedence**: The processing of a job on Machine 1 cannot start until it has completed processing on Machine 0. Similarly, processing on Machine 2 cannot begin until the job has finished on Machine 1, and processing on Machine 3 cannot start until the job has completed on Machine 2. + +4. **Job Precedence**: The processing of a job in a later schedule position cannot start on any machine until the job in the previous position has completed its processing on that machine. For example, if Job 0 is in Position 0 and Job 1 is in Position 1, Job 1 cannot start on Machine 0 until Job 0 has finished its processing on Machine 0. + +The objective is to determine the optimal schedule for processing the six jobs across the four machines to minimize the total production time, ensuring that all constraints are satisfied. The total production time is defined as the time when the last job completes its processing on Machine 3, plus any additional processing time required for the last job in the schedule.68A manufacturing facility processes six jobs (Job 0 through Job 5) on four machines (Machine 0 through Machine 3). Each job must be processed on all four machines in sequence, and the facility must determine the optimal schedule to minimize the total production time. Each job has a unique processing time on each machine, as follows: + +- **Job 0**: 5 hours on Machine 0, 8 hours on Machine 1, 8 hours on Machine 2, and 5 hours on Machine 3. +- **Job 1**: 7 hours on Machine 0, 6 hours on Machine 1, 7 hours on Machine 2, and 5 hours on Machine 3. +- **Job 2**: 10 hours on Machine 0, 10 hours on Machine 1, 8 hours on Machine 2, and 7 hours on Machine 3. +- **Job 3**: 7 hours on Machine 0, 10 hours on Machine 1, 5 hours on Machine 2, and 8 hours on Machine 3. +- **Job 4**: 9 hours on Machine 0, 9 hours on Machine 1, 5 hours on Machine 2, and 6 hours on Machine 3. +- **Job 5**: 6 hours on Machine 0, 9 hours on Machine 1, 6 hours on Machine 2, and 6 hours on Machine 3. + +The facility must adhere to the following constraints: + +1. **One Job per Schedule Position**: Each of the six schedule positions (Position 0 through Position 5) must be assigned exactly one job. No two positions can have the same job. + +2. **One Schedule Position per Job**: Each job must be assigned to exactly one schedule position. No job can be assigned to multiple positions. + +3. **Machine Precedence**: The processing of a job on Machine 1 cannot start until it has completed processing on Machine 0. Similarly, processing on Machine 2 cannot begin until the job has finished on Machine 1, and processing on Machine 3 cannot start until the job has completed on Machine 2. + +4. **Job Precedence**: The processing of a job in a later schedule position cannot start on any machine until the job in the previous position has completed its processing on that machine. For example, if Job 0 is in Position 0 and Job 1 is in Position 1, Job 1 cannot start on Machine 0 until Job 0 has finished its processing on Machine 0. + +The objective is to determine the optimal schedule for processing the six jobs across the four machines to minimize the total production time, ensuring that all constraints are satisfied. The total production time is defined as the time when the last job completes its processing on Machine 3, plus any additional processing time required for the last job in the schedule.68No differencesNo differencesok
110The problem involves scheduling the landing times of five aircraft to minimize the total penalty costs associated with deviations from their target landing times while ensuring safe separation between landings and adhering to time windows. Each aircraft has a target landing time, earliest and latest possible landing times, and penalty costs for landing early or late. The aircraft details are as follows: Aircraft 0 has a target landing time of 83 minutes, earliest landing time of 60 minutes, latest landing time of 100 minutes, early penalty cost of \$16 per minute, and late penalty cost of \$16 per minute. Aircraft 1 has a target landing time of 79 minutes, earliest landing time of 60 minutes, latest landing time of 100 minutes, early penalty cost of \$16 per minute, and late penalty cost of \$10 per minute. Aircraft 2 has a target landing time of 75 minutes, earliest landing time of 60 minutes, latest landing time of 100 minutes, early penalty cost of \$16 per minute, and late penalty cost of \$13 per minute. Aircraft 3 has a target landing time of 95 minutes, earliest landing time of 65 minutes, latest landing time of 100 minutes, early penalty cost of \$19 per minute, and late penalty cost of \$19 per minute. Aircraft 4 has a target landing time of 98 minutes, earliest landing time of 68 minutes, latest landing time of 100 minutes, early penalty cost of \$14 per minute, and late penalty cost of \$15 per minute. + +To ensure safety, the following minimum separation times must be maintained between landings: 5 minutes between Aircraft 0 and Aircraft 1, 5 minutes between Aircraft 0 and Aircraft 2, 5 minutes between Aircraft 0 and Aircraft 3, 5 minutes between Aircraft 0 and Aircraft 4, 7 minutes between Aircraft 1 and Aircraft 2, 7 minutes between Aircraft 1 and Aircraft 3, 5 minutes between Aircraft 1 and Aircraft 4, 6 minutes between Aircraft 2 and Aircraft 3, 7 minutes between Aircraft 2 and Aircraft 4, and 5 minutes between Aircraft 3 and Aircraft 4. + +The landing times of each aircraft must fall within their specified earliest and latest possible landing times. The separation constraints must be respected, meaning that the landing times of any two aircraft must maintain the required separation based on their order of landing. The penalty for landing early is calculated as the number of minutes early multiplied by the early penalty cost, and the penalty for landing late is calculated as the number of minutes late multiplied by the late penalty cost. The objective is to determine the landing times for all five aircraft such that the total penalty cost is minimized, and all separation requirements and landing time windows are strictly adhered to. This involves deciding the exact landing time for each aircraft and the order in which they land, ensuring that the schedule is both safe and cost-effective.94The problem involves scheduling the landing times of five aircraft to minimize the total penalty costs associated with deviations from their target landing times while ensuring safe separation between landings and adhering to time windows. Each aircraft has a target landing time, earliest and latest possible landing times, and penalty costs for landing early or late. The aircraft details are as follows: Aircraft 0 has a target landing time of 83 minutes, earliest landing time of 60 minutes, latest landing time of 100 minutes, early penalty cost of \$16 per minute, and late penalty cost of \$16 per minute. Aircraft 1 has a target landing time of 79 minutes, earliest landing time of 60 minutes, latest landing time of 100 minutes, early penalty cost of \$16 per minute, and late penalty cost of \$10 per minute. Aircraft 2 has a target landing time of 75 minutes, earliest landing time of 60 minutes, latest landing time of 100 minutes, early penalty cost of \$16 per minute, and late penalty cost of \$13 per minute. Aircraft 3 has a target landing time of 95 minutes, earliest landing time of 65 minutes, latest landing time of 100 minutes, early penalty cost of \$19 per minute, and late penalty cost of \$19 per minute. Aircraft 4 has a target landing time of 98 minutes, earliest landing time of 68 minutes, latest landing time of 100 minutes, early penalty cost of \$14 per minute, and late penalty cost of \$15 per minute. + +To ensure safety, the following minimum separation times must be maintained between landings: 5 minutes between Aircraft 0 and Aircraft 1, 5 minutes between Aircraft 0 and Aircraft 2, 5 minutes between Aircraft 0 and Aircraft 3, 5 minutes between Aircraft 0 and Aircraft 4, 7 minutes between Aircraft 1 and Aircraft 2, 7 minutes between Aircraft 1 and Aircraft 3, 5 minutes between Aircraft 1 and Aircraft 4, 6 minutes between Aircraft 2 and Aircraft 3, 7 minutes between Aircraft 2 and Aircraft 4, and 5 minutes between Aircraft 3 and Aircraft 4. + +The landing times of each aircraft must fall within their specified earliest and latest possible landing times. The separation constraints must be respected, meaning that the landing times of any two aircraft must maintain the required separation based on their order of landing. The penalty for landing early is calculated as the number of minutes early multiplied by the early penalty cost, and the penalty for landing late is calculated as the number of minutes late multiplied by the late penalty cost. The objective is to determine the landing times for all five aircraft such that the total penalty cost is minimized, and all separation requirements and landing time windows are strictly adhered to. This involves deciding the exact landing time for each aircraft and the order in which they land, ensuring that the schedule is both safe and cost-effective.94No differencesNo differencesok
111The problem involves assigning employees to shifts across two restaurants to meet skill demands while minimizing costs. There are 2 restaurants, 6 shifts per restaurant, and 6 employees. Each employee has specific skills and availability for shifts. Employees have preference costs for certain skills, and unfulfilled positions incur a penalty cost of 100 units. The demand for skills varies by restaurant, shift, and skill type. For example, Restaurant 0, Shift 0 requires 2 employees with Skill 0, and Restaurant 1, Shift 0 requires 1 employee with Skill 0. Employees can only be assigned to shifts they are available for and to roles they are skilled in. Each employee can be assigned to at most one shift across all restaurants and skills. The objective is to minimize the total cost, which includes preference costs for employee assignments and penalty costs for unfulfilled positions. Constraints ensure that the total number of employees assigned to a specific skill in a shift, plus any unfulfilled positions, equals the demand for that skill. Employee availability and skill requirements are respected, and each employee is limited to one assignment. The preference costs for assignments are as follows: Employee 0 to Skill 0 costs 1 unit, Employee 1 to Skill 0 costs 2 units, and Employee 3 to Skill 0 costs 3 units. The penalty cost for unfulfilled positions is 100 units per position. The goal is to assign employees to shifts and roles in a way that meets all demands while minimizing the total cost.6The problem involves assigning employees to shifts across two restaurants to meet skill demands while minimizing costs. There are 2 restaurants, 6 shifts per restaurant, and 6 employees. Each employee has specific skills and availability for shifts. Employees have preference costs for certain skills, and unfulfilled positions incur a penalty cost of 100 units. The demand for skills varies by restaurant, shift, and skill type. For example, Restaurant 0, Shift 0 requires 2 employees with Skill 0, and Restaurant 1, Shift 0 requires 1 employee with Skill 0. Employees can only be assigned to shifts they are available for and to roles they are skilled in. Each employee can be assigned to at most one shift across all restaurants and skills. The objective is to minimize the total cost, which includes preference costs for employee assignments and penalty costs for unfulfilled positions. Constraints ensure that the total number of employees assigned to a specific skill in a shift, plus any unfulfilled positions, equals the demand for that skill. Employee availability and skill requirements are respected, and each employee is limited to one assignment. The preference costs for assignments are as follows: Employee 0 to Skill 0 costs 1 unit, Employee 1 to Skill 0 costs 2 units, and Employee 3 to Skill 0 costs 3 units. The penalty cost for unfulfilled positions is 100 units per position. The goal is to assign employees to shifts and roles in a way that meets all demands while minimizing the total cost.6No differencesNo differencesok
112You are responsible for managing the distribution of emergency medical supplies across eight cities in a region. Each city has a specific supply of medical resources and a demand that must be met to ensure adequate healthcare coverage. The goal is to minimize the total transportation cost while ensuring that all cities receive the necessary supplies and that no distribution routes exceed their capacity. + +### City Supply and Demand: +- **City 0** has a net demand of 1 unit of medical supplies. +- **City 1** has a balanced supply and demand (net demand of 0 units). +- **City 2** has a net supply of 1 unit of medical supplies. +- **City 3** has a net demand of 2 units of medical supplies. +- **City 4** has a balanced supply and demand (net demand of 0 units). +- **City 5** has a balanced supply and demand (net demand of 0 units). +- **City 6** has a net supply of 2 units of medical supplies. +- **City 7** has a balanced supply and demand (net demand of 0 units). + +### Transportation Costs: +The cost of transporting medical supplies between cities varies depending on the route. Below are the transportation costs per unit of supplies: + +- **From City 0**: To City 1 costs 3, to City 2 costs 2, to City 3 costs 2, to City 4 costs 2, to City 5 costs 3, to City 6 costs 3, and to City 7 costs 1. +- **From City 1**: To City 0 costs 1, to City 2 costs 2, to City 3 costs 3, to City 4 costs 1, to City 5 costs 2, to City 6 costs 1, and to City 7 costs 2. +- **From City 2**: To City 0 costs 2, to City 1 costs 2, to City 3 costs 3, to City 4 costs 3, to City 5 costs 2, to City 6 costs 1, and to City 7 costs 2. +- **From City 3**: To City 0 costs 1, to City 1 costs 2, to City 2 costs 1, to City 4 costs 3, to City 5 costs 3, to City 6 costs 2, and to City 7 costs 3. +- **From City 4**: To City 0 costs 3, to City 1 costs 2, to City 2 costs 1, to City 3 costs 1, to City 5 costs 3, to City 6 costs 2, and to City 7 costs 2. +- **From City 5**: To City 0 costs 1, to City 1 costs 2, to City 2 costs 1, to City 3 costs 2, to City 4 costs 1, to City 6 costs 2, and to City 7 costs 1. +- **From City 6**: To City 0 costs 2, to City 1 costs 3, to City 2 costs 1, to City 3 costs 1, to City 4 costs 1, to City 5 costs 1, and to City 7 costs 1. +- **From City 7**: To City 0 costs 1, to City 1 costs 1, to City 2 costs 3, to City 3 costs 1, to City 4 costs 2, to City 5 costs 3, and to City 6 costs 2. + +### Route Capacity Constraints: +Each route between cities has a maximum capacity for transporting medical supplies: + +- **From City 0**: To City 1 (7 units), to City 2 (7 units), to City 3 (7 units), to City 4 (7 units), to City 5 (8 units), to City 6 (8 units), and to City 7 (8 units). +- **From City 1**: To City 0 (8 units), to City 2 (7 units), to City 3 (8 units), to City 4 (8 units), to City 5 (7 units), to City 6 (7 units), and to City 7 (9 units). +- **From City 2**: To City 0 (8 units), to City 1 (7 units), to City 3 (7 units), to City 4 (7 units), to City 5 (7 units), to City 6 (9 units), and to City 7 (7 units). +- **From City 3**: To City 0 (7 units), to City 1 (7 units), to City 2 (9 units), to City 4 (8 units), to City 5 (7 units), to City 6 (7 units), and to City 7 (9 units). +- **From City 4**: To City 0 (9 units), to City 1 (7 units), to City 2 (8 units), to City 3 (9 units), to City 5 (7 units), to City 6 (7 units), and to City 7 (7 units). +- **From City 5**: To City 0 (7 units), to City 1 (8 units), to City 2 (9 units), to City 3 (9 units), to City 4 (8 units), to City 6 (9 units), and to City 7 (8 units). +- **From City 6**: To City 0 (9 units), to City 1 (8 units), to City 2 (7 units), to City 3 (8 units), to City 4 (8 units), to City 5 (7 units), and to City 7 (8 units). +- **From City 7**: To City 0 (9 units), to City 1 (8 units), to City 2 (7 units), to City 3 (9 units), to City 4 (9 units), to City 5 (8 units), and to City 6 (8 units). + +### City Capacity Constraints: +Each city has a maximum capacity for receiving medical supplies: + +- **City 0**: Can receive up to 19 units. +- **City 1**: Can receive up to 15 units. +- **City 2**: Can receive up to 15 units. +- **City 3**: Can receive up to 14 units. +- **City 4**: Can receive up to 15 units. +- **City 5**: Can receive up to 15 units. +- **City 6**: Can receive up to 14 units. +- **City 7**: Can receive up to 16 units. + +### Objective: +Your task is to determine the optimal distribution of medical supplies between the cities to minimize the total transportation cost while ensuring that all cities meet their supply and demand requirements, no route exceeds its capacity, and no city exceeds its receiving capacity.6You are responsible for managing the distribution of emergency medical supplies across eight cities in a region. Each city has a specific supply of medical resources and a demand that must be met to ensure adequate healthcare coverage. The goal is to minimize the total transportation cost while ensuring that all cities receive the necessary supplies and that no distribution routes exceed their capacity. + +### City Supply and Demand: +- **City 0** has a net demand of 1 unit of medical supplies. +- **City 1** has a balanced supply and demand (net demand of 0 units). +- **City 2** has a net supply of 1 unit of medical supplies. +- **City 3** has a net demand of 2 units of medical supplies. +- **City 4** has a balanced supply and demand (net demand of 0 units). +- **City 5** has a balanced supply and demand (net demand of 0 units). +- **City 6** has a net supply of 2 units of medical supplies. +- **City 7** has a balanced supply and demand (net demand of 0 units). + +### Transportation Costs: +The cost of transporting medical supplies between cities varies depending on the route. Below are the transportation costs per unit of supplies: + +- **From City 0**: To City 1 costs 3, to City 2 costs 2, to City 3 costs 2, to City 4 costs 2, to City 5 costs 3, to City 6 costs 3, and to City 7 costs 1. +- **From City 1**: To City 0 costs 1, to City 2 costs 2, to City 3 costs 3, to City 4 costs 1, to City 5 costs 2, to City 6 costs 1, and to City 7 costs 2. +- **From City 2**: To City 0 costs 2, to City 1 costs 2, to City 3 costs 3, to City 4 costs 3, to City 5 costs 2, to City 6 costs 1, and to City 7 costs 2. +- **From City 3**: To City 0 costs 1, to City 1 costs 2, to City 2 costs 1, to City 4 costs 3, to City 5 costs 3, to City 6 costs 2, and to City 7 costs 3. +- **From City 4**: To City 0 costs 3, to City 1 costs 2, to City 2 costs 1, to City 3 costs 1, to City 5 costs 3, to City 6 costs 2, and to City 7 costs 2. +- **From City 5**: To City 0 costs 1, to City 1 costs 2, to City 2 costs 1, to City 3 costs 2, to City 4 costs 1, to City 6 costs 2, and to City 7 costs 1. +- **From City 6**: To City 0 costs 2, to City 1 costs 3, to City 2 costs 1, to City 3 costs 1, to City 4 costs 1, to City 5 costs 1, and to City 7 costs 1. +- **From City 7**: To City 0 costs 1, to City 1 costs 1, to City 2 costs 3, to City 3 costs 1, to City 4 costs 2, to City 5 costs 3, and to City 6 costs 2. + +### Route Capacity Constraints: +Each route between cities has a maximum capacity for transporting medical supplies: + +- **From City 0**: To City 1 (7 units), to City 2 (7 units), to City 3 (7 units), to City 4 (7 units), to City 5 (8 units), to City 6 (8 units), and to City 7 (8 units). +- **From City 1**: To City 0 (8 units), to City 2 (7 units), to City 3 (8 units), to City 4 (8 units), to City 5 (7 units), to City 6 (7 units), and to City 7 (9 units). +- **From City 2**: To City 0 (8 units), to City 1 (7 units), to City 3 (7 units), to City 4 (7 units), to City 5 (7 units), to City 6 (9 units), and to City 7 (7 units). +- **From City 3**: To City 0 (7 units), to City 1 (7 units), to City 2 (9 units), to City 4 (8 units), to City 5 (7 units), to City 6 (7 units), and to City 7 (9 units). +- **From City 4**: To City 0 (9 units), to City 1 (7 units), to City 2 (8 units), to City 3 (9 units), to City 5 (7 units), to City 6 (7 units), and to City 7 (7 units). +- **From City 5**: To City 0 (7 units), to City 1 (8 units), to City 2 (9 units), to City 3 (9 units), to City 4 (8 units), to City 6 (9 units), and to City 7 (8 units). +- **From City 6**: To City 0 (9 units), to City 1 (8 units), to City 2 (7 units), to City 3 (8 units), to City 4 (8 units), to City 5 (7 units), and to City 7 (8 units). +- **From City 7**: To City 0 (9 units), to City 1 (8 units), to City 2 (7 units), to City 3 (9 units), to City 4 (9 units), to City 5 (8 units), and to City 6 (8 units). + +### City Capacity Constraints: +Each city has a maximum capacity for receiving medical supplies: + +- **City 0**: Can receive up to 19 units. +- **City 1**: Can receive up to 15 units. +- **City 2**: Can receive up to 15 units. +- **City 3**: Can receive up to 14 units. +- **City 4**: Can receive up to 15 units. +- **City 5**: Can receive up to 15 units. +- **City 6**: Can receive up to 14 units. +- **City 7**: Can receive up to 16 units. + +### Objective: +Your task is to determine the optimal distribution of medical supplies between the cities to minimize the total transportation cost while ensuring that all cities meet their supply and demand requirements, no route exceeds its capacity, and no city exceeds its receiving capacity.4No differences6 4Incorrect ground truthincorrect ground truth
113Imagine you are managing a power generation system responsible for supplying electricity to a region over five distinct periods. The system consists of three types of power generators, each with unique operational characteristics and costs. Your goal is to determine the optimal number of generators to operate, the amount of power each generator should produce, and when to start up additional generators to meet the region's electricity demand at the lowest possible cost. + +#### Key Decisions: +1. **Number of Generators to Operate**: For each generator type (Type 0, Type 1, Type 2) and each period (Period 0 to Period 4), decide how many generators should be active. +2. **Power Output**: Determine the amount of power (in MW) each active generator should produce during each period. +3. **Generator Startups**: Decide whether to start up additional generators in each period, incurring a startup cost. + +#### Objective: +Minimize the total cost of operating the power generation system, which includes: +- **Base Costs**: Each active generator incurs a fixed cost of 79 units per period, regardless of its power output. +- **Per-MW Costs**: Each generator also incurs a variable cost based on its power output: 3 units per MW for Type 0, 4 units per MW for Type 1, and 4 units per MW for Type 2. +- **Startup Costs**: Starting up a generator incurs a one-time cost of 355 units for Type 0 and 349 units for Types 1 and 2. + +#### Constraints: +1. **Demand Satisfaction**: The total power output across all generators must meet or exceed the region's electricity demand in each period. The demands are: + - Period 0: 167 MW + - Period 1: 212 MW + - Period 2: 224 MW + - Period 3: 221 MW + - Period 4: 189 MW + +2. **Generator Capacity Limits**: + - Each generator has a minimum and maximum power output: + - Type 0: Minimum 35 MW, Maximum 128 MW + - Type 1: Minimum 35 MW, Maximum 129 MW + - Type 2: Minimum 35 MW, Maximum 125 MW + - The power output of each generator must be within these limits based on whether it is active. + +3. **Reserve Margin**: To ensure reliability, the total maximum possible power output across all generators must be at least 15% higher than the demand in each period. This translates to: + - Period 0: 192.05 MW + - Period 1: 243.8 MW + - Period 2: 257.6 MW + - Period 3: 254.15 MW + - Period 4: 217.35 MW + +4. **Generator Availability**: + - The number of generators that can be active is limited: + - Type 0: Maximum 1 generator per period + - Type 1: Maximum 4 generators per period + - Type 2: Maximum 3 generators per period + +5. **Startup Logic**: + - If a generator is started in a period, it must have been inactive in the previous period. + - For Type 2 generators, up to 2 additional generators can be started in Period 0 without requiring a startup in the previous period. + +#### Numerical Parameters: +- **Base Costs**: 79 units per active generator per period. +- **Per-MW Costs**: 3 units for Type 0, 4 units for Types 1 and 2. +- **Startup Costs**: 355 units for Type 0, 349 units for Types 1 and 2. +- **Demands**: 167 MW (Period 0), 212 MW (Period 1), 224 MW (Period 2), 221 MW (Period 3), 189 MW (Period 4). +- **Capacity Limits**: + - Type 0: 35 MW (min), 128 MW (max) + - Type 1: 35 MW (min), 129 MW (max) + - Type 2: 35 MW (min), 125 MW (max) +- **Reserve Margins**: 192.05 MW (Period 0), 243.8 MW (Period 1), 257.6 MW (Period 2), 254.15 MW (Period 3), 217.35 MW (Period 4). + +Your task is to determine the optimal combination of active generators, their power outputs, and startup decisions to meet the region's electricity demand at the lowest total cost while adhering to all operational constraints.4904Imagine you are managing a power generation system responsible for supplying electricity to a region over five distinct periods. The system consists of three types of power generators, each with unique operational characteristics and costs. Your goal is to determine the optimal number of generators to operate, the amount of power each generator should produce, and when to start up additional generators to meet the region's electricity demand at the lowest possible cost. + +#### Key Decisions: +1. **Number of Generators to Operate**: For each generator type (Type 0, Type 1, Type 2) and each period (Period 0 to Period 4), decide how many generators should be active. +2. **Power Output**: Determine the amount of power (in MW) each active generator should produce during each period. +3. **Generator Startups**: Decide whether to start up additional generators in each period, incurring a startup cost. + +#### Objective: +Minimize the total cost of operating the power generation system, which includes: +- **Base Costs**: Each active generator incurs a fixed cost of 79 units per period, regardless of its power output. +- **Per-MW Costs**: Each generator also incurs a variable cost based on its power output: 3 units per MW for Type 0, 4 units per MW for Type 1, and 4 units per MW for Type 2. +- **Startup Costs**: Starting up a generator incurs a one-time cost of 355 units for Type 0 and 349 units for Types 1 and 2. + +#### Constraints: +1. **Demand Satisfaction**: The total power output across all generators must meet or exceed the region's electricity demand in each period. The demands are: + - Period 0: 167 MW + - Period 1: 212 MW + - Period 2: 224 MW + - Period 3: 221 MW + - Period 4: 189 MW + +2. **Generator Capacity Limits**: + - Each generator has a minimum and maximum power output: + - Type 0: Minimum 35 MW, Maximum 128 MW + - Type 1: Minimum 35 MW, Maximum 129 MW + - Type 2: Minimum 35 MW, Maximum 125 MW + - The power output of each generator must be within these limits based on whether it is active. + +3. **Reserve Margin**: To ensure reliability, the total maximum possible power output across all generators must be at least 15% higher than the demand in each period. This translates to: + - Period 0: 192.05 MW + - Period 1: 243.8 MW + - Period 2: 257.6 MW + - Period 3: 254.15 MW + - Period 4: 217.35 MW + +4. **Generator Availability**: + - The number of generators that can be active is limited: + - Type 0: Maximum 1 generator per period + - Type 1: Maximum 4 generators per period + - Type 2: Maximum 3 generators per period + +5. **Startup Logic**: + - If a generator is started in a period, it must have been inactive in the previous period. + - For Type 2 generators, up to 2 generators can be started in Period 0 without requiring a startup cost. + +#### Numerical Parameters: +- **Base Costs**: 79 units per active generator per period. +- **Per-MW Costs**: 3 units for Type 0, 4 units for Types 1 and 2. +- **Startup Costs**: 355 units for Type 0, 349 units for Types 1 and 2. +- **Demands**: 167 MW (Period 0), 212 MW (Period 1), 224 MW (Period 2), 221 MW (Period 3), 189 MW (Period 4). +- **Capacity Limits**: + - Type 0: 35 MW (min), 128 MW (max) + - Type 1: 35 MW (min), 129 MW (max) + - Type 2: 35 MW (min), 125 MW (max) +- **Reserve Margins**: 192.05 MW (Period 0), 243.8 MW (Period 1), 257.6 MW (Period 2), 254.15 MW (Period 3), 217.35 MW (Period 4). + +Your task is to determine the optimal combination of active generators, their power outputs, and startup decisions to meet the region's electricity demand at the lowest total cost while adhering to all operational constraints.49045. **Startup Logic**: + - If a generator is started in a period, it must have been inactive in the previous period. + - For Type 2 generators, up to 2 additional generators can be started in Period 0 without requiring a startup in the previous period. cost. + +#### Numerical Parameters: +- **Base Costs**: 79 units per active generator per period.No differencesConfusing when in period 0 the statement talks about "previous period"ambiguous
114Minimize the total cost, which includes production costs of 143.7514410203047 units per production decision, startup costs of 65.60228476581376 units per machine startup, inventory holding costs of 1.164609943740513 units for Product 0, 1.188814519462053 units for Product 1, and 0.8889238919759239 units for Product 2 per period, and backlogging costs of 6.772870896644705 units for Product 0, 7.291249122946056 units for Product 1, and 6.513011399630116 units for Product 2 per period. The production decisions are binary, indicating whether a product is produced on a specific machine in a given period, and the startup decisions are binary, indicating whether a machine is started up for a specific product in a given period. The production quantities must be non-negative, and the inventory and backlog levels must balance to meet the demand for each product in each period. The demand for Product 0 is 16.86823685495585 units in Period 0, 21.66307881187974 units in Period 1, and 22.77330720589323 units in Period 2. The demand for Product 1 is 23.97615061004427 units in Period 0, 19.60905873481086 units in Period 1, and 23.31252847872402 units in Period 2. The demand for Product 2 is 23.6195969696504 units in Period 0, 18.33370330296734 units in Period 1, and 22.95855759036257 units in Period 2. Each machine has a limited capacity, and the total production on a machine in a given period cannot exceed this capacity. For Machine 0, the capacity constraints are -71.6204295699619 times the production decision plus 6.477399095211152 times the startup decision plus the production quantity must be less than or equal to zero for each product and period. For Machine 1, the capacity constraints are -71.47086166745689 times the production decision plus 6.845777110793686 times the startup decision plus the production quantity must be less than or equal to zero for each product and period. Each machine can only produce one product at a time in a given period, and the production and startup decisions must follow logical constraints to ensure that a machine is only started up if it was not already producing the same product in the previous period. The production and startup decisions are binary, and the production quantities, inventory, and backlog levels must be non-negative.914.6636376994903Minimize the total cost, which includes production costs of 143.7514410203047 units per production decision, startup costs of 65.60228476581376 units per machine startup, inventory holding costs of 1.164609943740513 units for Product 0, 1.188814519462053 units for Product 1, and 0.8889238919759239 units for Product 2 per period, and backlogging costs of 6.772870896644705 units for Product 0, 7.291249122946056 units for Product 1, and 6.513011399630116 units for Product 2 per period. The production decisions are binary, indicating whether a product is produced on a specific machine in a given period, and the startup decisions are binary, indicating whether a machine is started up for a specific product in a given period. The production quantities must be non-negative, and the inventory and backlog levels must balance to meet the demand for each product in each period. The demand for Product 0 is 16.86823685495585 units in Period 0, 21.66307881187974 units in Period 1, and 22.77330720589323 units in Period 2. The demand for Product 1 is 23.97615061004427 units in Period 0, 19.60905873481086 units in Period 1, and 23.31252847872402 units in Period 2. The demand for Product 2 is 23.6195969696504 units in Period 0, 18.33370330296734 units in Period 1, and 22.95855759036257 units in Period 2. Each machine has a limited capacity, and the total production on a machine in a given period cannot exceed this capacity. For Machine 0, the capacity constraints are -71.6204295699619 times the production decision plus 6.477399095211152 times the startup decision plus the production quantity must be less than or equal to zero for each product and period. For Machine 1, the capacity constraints are -71.47086166745689 times the production decision plus 6.845777110793686 times the startup decision plus the production quantity must be less than or equal to zero for each product and period. Each machine can only produce one product at a time in a given period, and the production and startup decisions must follow logical constraints to ensure that a machine is only started up if it was not already producing the same product in the previous period. The production and startup decisions are binary, and the production quantities, inventory, and backlog levels must be non-negative.914.6636376994903No differencesNo differencesok
115In a manufacturing facility, the goal is to schedule the production of seven jobs, each consisting of two operations that must be performed in sequence, to minimize the total production time (makespan). Each operation has a processing time of 1 unit of time. Operations from different jobs may share machines, requiring careful scheduling to avoid overlaps. Binary variables determine the order of operations on shared machines, ensuring that one operation finishes before another starts. A large constant (100,000) is used to enforce this order in the constraints. + +The constraints include precedence constraints within each job, ensuring that the second operation cannot start until the first operation is completed. Machine capacity constraints prevent overlapping operations on shared machines, with specific pairs of operations (e.g., Operation 1 of Job 2 and Operation 0 of Job 4, Operation 1 of Job 2 and Operation 0 of Job 6, etc.) requiring explicit ordering. The makespan is defined as the maximum completion time across all operations, ensuring it is at least as long as the last operation's finish time. All start times for operations must be non-negative, and binary variables are restricted to 0 or 1. + +The objective is to minimize the makespan, ensuring all operations are completed efficiently while respecting machine usage and operation sequencing constraints. The problem involves 26 pairs of operations on shared machines, with binary variables determining their order.5In a manufacturing facility, the goal is to schedule the production of seven jobs, each consisting of two operations that must be performed in sequence, to minimize the total production time (makespan). Each operation has a processing time of 1 unit of time. Operations from different jobs may share machines, requiring careful scheduling to avoid overlaps. Binary variables determine the order of operations on shared machines, ensuring that one operation finishes before another starts. A large constant (100) is used to enforce this order in the constraints. + +The constraints include precedence constraints within each job, ensuring that the second operation cannot start until the first operation is completed. Machine capacity constraints prevent overlapping operations on shared machines, with specific pairs of operations requiring explicit ordering. The pairs of operations that cannot be overlapped are: Operation 1 of Job 2 and Operation 1 of Job 4, and Operation 0 of Job 2 and Operation 0 of Job 6. The makespan is defined as the maximum completion time across all operations, ensuring it is at least as long as the last operation's finish time. All start times for operations must be non-negative, and binary variables are restricted to 0 or 1. + +The objective is to minimize the makespan, ensuring all operations are completed efficiently while respecting machine usage and operation sequencing constraints.3In a manufacturing facility, the goal is to schedule the production of seven jobs, each consisting of two operations that must be performed in sequence, to minimize the total production time (makespan). Each operation has a processing time of 1 unit of time. Operations from different jobs may share machines, requiring careful scheduling to avoid overlaps. Binary variables determine the order of operations on shared machines, ensuring that one operation finishes before another starts. A large constant (100,000) (100) is used to enforce this order in the constraints. + +The constraints include precedence constraints within each job, ensuring that the second operation cannot start until the first operation is completed. Machine capacity constraints prevent overlapping operations on shared machines, with specific pairs of operations (e.g., requiring explicit ordering. The pairs of operations that cannot be overlapped are: Operation 1 of Job 2 and Operation 0 1 of Job 4, and Operation 1 0 of Job 2 and Operation 0 of Job 6, etc.) requiring explicit ordering. 6. The makespan is defined as the maximum completion time across all operations, ensuring it is at least as long as the last operation's finish time. All start times for operations must be non-negative, and binary variables are restricted to 0 or 1. + +The objective is to minimize the makespan, ensuring all operations are completed efficiently while respecting machine usage and operation sequencing constraints. The problem involves 26 pairs of operations on shared machines, with binary variables determining their order.5 3It states that there are 26 overlaps but only a few are provided, we omitted that sentence and clarified these are the overlaps.missing data
116The problem involves scheduling operations for six products, each requiring two operations to be performed in sequence. Each operation is assigned to a specific machine, and the goal is to minimize the total production time (makespan). Each operation has a processing time of 1 unit. The start time of each operation must be determined, and the order of operations on shared machines must be decided to avoid overlaps. Precedence constraints ensure that the second operation of a product cannot start until the first operation is completed. Machine capacity constraints ensure that operations assigned to the same machine do not overlap in time. The makespan must be at least as long as the completion time of the last operation across all products. Binary decision variables determine the order of operations on shared machines, and a large number (100,000) is used in the constraints to enforce this order without overlapping. The objective is to minimize the makespan by determining the start times of all operations and the order of operations on shared machines, ensuring all products are completed as quickly as possible.6"The problem involves scheduling operations for six products, each requiring two operations to be performed in sequence. Each operation is assigned to a specific machine, and the goal is to minimize the total production time (makespan). Each operation has a processing time of 1 unit. The start time of each operation must be determined, and the order of operations on shared machines must be decided to avoid overlaps. Precedence constraints ensure that the second operation of a product cannot start until the first operation is completed. Machine capacity constraints ensure that operations assigned to the same machine do not overlap in time. The makespan must be at least as long as the completion time of the last operation across all products. Binary decision variables determine the order of operations on shared machines, and a large number (100) is used in the constraints to enforce this order without overlapping. The objective is to minimize the makespan by determining the start times of all operations and the order of operations on shared machines, ensuring all products are completed as quickly as possible. + + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Product 0: Machine 2 + - Operation 1 of Product 0: Machine 1 + - Operation 0 of Product 1: Machine 2 + - Operation 1 of Product 1: Machine 2 + - Operation 0 of Product 2: Machine 2 + - Operation 1 of Product 2: Machine 1 + - Operation 0 of Product 3: Machine 1 + - Operation 1 of Product 3: Machine 0 + - Operation 0 of Product 4: Machine 2 + - Operation 1 of Product 4: Machine 0 + - Operation 0 of Product 5: Machine 0 + - Operation 1 of Product 5: Machine 1"5The "The problem involves scheduling operations for six products, each requiring two operations to be performed in sequence. Each operation is assigned to a specific machine, and the goal is to minimize the total production time (makespan). Each operation has a processing time of 1 unit. The start time of each operation must be determined, and the order of operations on shared machines must be decided to avoid overlaps. Precedence constraints ensure that the second operation of a product cannot start until the first operation is completed. Machine capacity constraints ensure that operations assigned to the same machine do not overlap in time. The makespan must be at least as long as the completion time of the last operation across all products. Binary decision variables determine the order of operations on shared machines, and a large number (100,000) (100) is used in the constraints to enforce this order without overlapping. The objective is to minimize the makespan by determining the start times of all operations and the order of operations on shared machines, ensuring all products are completed as quickly as possible. + + + + Additional information: + - There are 3 machines, numbered from 0 to 2 + - The assignment or operations to machines is the following: + - Operation 0 of Product 0: Machine 2 + - Operation 1 of Product 0: Machine 1 + - Operation 0 of Product 1: Machine 2 + - Operation 1 of Product 1: Machine 2 + - Operation 0 of Product 2: Machine 2 + - Operation 1 of Product 2: Machine 1 + - Operation 0 of Product 3: Machine 1 + - Operation 1 of Product 3: Machine 0 + - Operation 0 of Product 4: Machine 2 + - Operation 1 of Product 4: Machine 0 + - Operation 0 of Product 5: Machine 0 + - Operation 1 of Product 5: Machine 1"6 5While the statement required to aoid overlap of operations on the same machines, it did not state how many machines are available or the assignment of operations to machines. missing data
117There are 7 jobs, each consisting of 2 operations that must be performed in sequence. Each operation takes exactly 1 unit of time to complete. The operations are assigned to specific machines, and each machine can process only one operation at a time. The goal is to schedule the operations on the machines to minimize the total completion time, known as the makespan, while satisfying all constraints. + +For each job, the second operation cannot start until the first operation is completed. If two operations are assigned to the same machine, they cannot overlap in time. Binary decision variables are used to determine the order of operations on shared machines. If one operation is scheduled before another on the same machine, the corresponding binary variable is set to 1; otherwise, it is set to 0. A large constant (100000) is used in the constraints to ensure that the binary variables correctly enforce the order of operations on shared machines. + +The start time of each operation must be non-negative, and the makespan must be at least as large as the completion time of every operation. The objective is to determine the start times for all operations on their respective machines such that all precedence and machine capacity constraints are satisfied, and the total completion time is minimized.5There are 7 jobs, each consisting of 2 operations that must be performed in sequence. Each operation takes exactly 1 unit of time to complete. The operations are assigned to specific machines, and each machine can process only one operation at a time. The goal is to schedule the operations on the machines to minimize the total completion time, known as the makespan, while satisfying all constraints. + +For each job, the second operation cannot start until the first operation is completed. If two operations are assigned to the same machine, they cannot overlap in time. Binary decision variables are used to determine the order of operations on shared machines. If one operation is scheduled before another on the same machine, the corresponding binary variable is set to 1; otherwise, it is set to 0. A large constant (1000) is used in the constraints to ensure that the binary variables correctly enforce the order of operations on shared machines. + +The start time of each operation must be non-negative, and the makespan must be at least as large as the completion time of every operation. The objective is to determine the start times for all operations on their respective machines such that all precedence and machine capacity constraints are satisfied, and the total completion time is minimized. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {1, 2, 3, 4, 5, 6, 7} +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Machines: M = {A, B, C, D} + +2) Operation durations (all durations are in time units) +- Duration of every operation (j,k) is p_{j,k} = 1 for all j in J and k = 1,2. + +3) Machine assignment for every operation (explicit mapping) +- Operation (1,1) -> Machine A +- Operation (1,2) -> Machine B +- Operation (2,1) -> Machine A +- Operation (2,2) -> Machine B +- Operation (3,1) -> Machine C +- Operation (3,2) -> Machine D +- Operation (4,1) -> Machine B +- Operation (4,2) -> Machine C +- Operation (5,1) -> Machine D +- Operation (5,2) -> Machine A +- Operation (6,1) -> Machine B +- Operation (6,2) -> Machine D +- Operation (7,1) -> Machine C +- Operation (7,2) -> Machine A4There are 7 jobs, each consisting of 2 operations that must be performed in sequence. Each operation takes exactly 1 unit of time to complete. The operations are assigned to specific machines, and each machine can process only one operation at a time. The goal is to schedule the operations on the machines to minimize the total completion time, known as the makespan, while satisfying all constraints. + +For each job, the second operation cannot start until the first operation is completed. If two operations are assigned to the same machine, they cannot overlap in time. Binary decision variables are used to determine the order of operations on shared machines. If one operation is scheduled before another on the same machine, the corresponding binary variable is set to 1; otherwise, it is set to 0. A large constant (100000) (1000) is used in the constraints to ensure that the binary variables correctly enforce the order of operations on shared machines. + +The start time of each operation must be non-negative, and the makespan must be at least as large as the completion time of every operation. The objective is to determine the start times for all operations on their respective machines such that all precedence and machine capacity constraints are satisfied, and the total completion time is minimized. + + ADDITIONAL INPUT DATA: + + 1) Sets and indices + - Jobs: J = {1, 2, 3, 4, 5, 6, 7} + - For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). + - Machines: M = {A, B, C, D} + + 2) Operation durations (all durations are in time units) + - Duration of every operation (j,k) is p_{j,k} = 1 for all j in J and k = 1,2. + + 3) Machine assignment for every operation (explicit mapping) + - Operation (1,1) -> Machine A + - Operation (1,2) -> Machine B + - Operation (2,1) -> Machine A + - Operation (2,2) -> Machine B + - Operation (3,1) -> Machine C + - Operation (3,2) -> Machine D + - Operation (4,1) -> Machine B + - Operation (4,2) -> Machine C + - Operation (5,1) -> Machine D + - Operation (5,2) -> Machine A + - Operation (6,1) -> Machine B + - Operation (6,2) -> Machine D + - Operation (7,1) -> Machine C + - Operation (7,2) -> Machine A5 4While the statement required to aoid overlap of operations on the same machines, it did not state how many machines are available or the assignment of operations to machines. missing data
118Determine the production amounts, inventory levels, and setup decisions for two products, Product A and Product B, over a seven-period planning horizon to minimize total costs, including production, setup, and inventory holding costs, while meeting demand and adhering to resource constraints. Production costs vary by period: producing one unit of Product A costs 42 units in Period 0, 44 in Period 1, 40 in Period 2, 43 in Period 3, 45 in Period 4, 45 in Period 5, and 42 in Period 6. Producing one unit of Product B costs 46 units in Period 0, 44 in Period 1, 50 in Period 2, 40 in Period 3, 41 in Period 4, 41 in Period 5, and 43 in Period 6. Setup costs are fixed: setting up production for Product A costs 1032 units in Period 0, 1318 in Period 1, 1002 in Period 2, 1073 in Period 3, 1387 in Period 4, 1491 in Period 5, and 1063 in Period 6. For Product B, setup costs are 1260 units in Period 0, 1418 in Period 1, 1455 in Period 2, 1266 in Period 3, 1095 in Period 4, 1431 in Period 5, and 1420 in Period 6. Inventory holding costs are constant: holding one unit of Product A costs 5 units in all periods except Periods 5 and 6, where it costs 4 units. Holding one unit of Product B costs 5 units in Periods 0, 2, 4, 5, and 6, and 4 units in Periods 1 and 3. + +Demand must be satisfied in each period: cumulative demand for Product A is 67 units in Period 0, 160 in Period 1, 213 in Period 2, 309 in Period 3, 376 in Period 4, 464 in Period 5, and 564 in Period 6. For Product B, cumulative demand is 100 units in Period 0, 199 in Period 1, 250 in Period 2, 332 in Period 3, 394 in Period 4, 450 in Period 5, and 516 in Period 6. Resource capacity limits production: in Period 0, the combined production of Product A and Product B cannot exceed 1138 units of resource capacity, with Product A consuming 1.547487551721541 units and Product B consuming 1.91801512951683 units per unit produced. Capacity limits are 1019 units in Period 1, 1166 in Period 2, 1163 in Period 3, 1042 in Period 4, 1098 in Period 5, and 1166 in Period 6. + +Production of a product in a period can only occur if the setup for that product is activated in that period. For Product A, production in Period 0 cannot exceed 564 units if the setup is activated, 497 in Period 1, 404 in Period 2, 351 in Period 3, 255 in Period 4, 188 in Period 5, and 100 in Period 6. For Product B, production in Period 0 cannot exceed 516 units if the setup is activated, 416 in Period 1, 317 in Period 2, 266 in Period 3, 184 in Period 4, 122 in Period 5, and 66 in Period 6. + +Initial inventory for both products is zero at the start of Period 0, and all inventory must be depleted by the end of Period 6. Production amounts and inventory levels must be non-negative. The objective is to minimize the total cost, which includes production costs, setup costs, and inventory holding costs, while ensuring all constraints are satisfied.56408Determine the production amounts, inventory levels, and setup decisions for two products, Product A and Product B, over a seven-period planning horizon to minimize total costs, including production, setup, and inventory holding costs, while meeting demand and adhering to resource constraints. Production costs vary by period: producing one unit of Product A costs 42 units in Period 0, 44 in Period 1, 40 in Period 2, 43 in Period 3, 45 in Period 4, 45 in Period 5, and 42 in Period 6. Producing one unit of Product B costs 46 units in Period 0, 44 in Period 1, 50 in Period 2, 40 in Period 3, 41 in Period 4, 41 in Period 5, and 43 in Period 6. Setup costs are fixed: setting up production for Product A costs 1032 units in Period 0, 1318 in Period 1, 1002 in Period 2, 1073 in Period 3, 1387 in Period 4, 1491 in Period 5, and 1063 in Period 6. For Product B, setup costs are 1260 units in Period 0, 1418 in Period 1, 1455 in Period 2, 1266 in Period 3, 1095 in Period 4, 1431 in Period 5, and 1420 in Period 6. Inventory holding costs are constant: holding one unit of Product A costs 5 units in all periods except Periods 5 and 6, where it costs 4 units. Holding one unit of Product B costs 5 units in Periods 0, 2, 4, 5, and 6, and 4 units in Periods 1 and 3. + +Demand must be satisfied in each period: cumulative demand for Product A is 67 units in Period 0, 160 in Period 1, 213 in Period 2, 309 in Period 3, 376 in Period 4, 464 in Period 5, and 564 in Period 6. For Product B, cumulative demand is 100 units in Period 0, 199 in Period 1, 250 in Period 2, 332 in Period 3, 394 in Period 4, 450 in Period 5, and 516 in Period 6. Resource capacity limits production: in Period 0, the combined production of Product A and Product B cannot exceed 1138 units of resource capacity, with Product A consuming 1.547487551721541 units and Product B consuming 1.91801512951683 units per unit produced. Capacity limits are 1019 units in Period 1, 1166 in Period 2, 1163 in Period 3, 1042 in Period 4, 1098 in Period 5, and 1166 in Period 6. + +Production of a product in a period can only occur if the setup for that product is activated in that period. For Product A, production in Period 0 cannot exceed 564 units if the setup is activated, 497 in Period 1, 404 in Period 2, 351 in Period 3, 255 in Period 4, 188 in Period 5, and 100 in Period 6. For Product B, production in Period 0 cannot exceed 516 units if the setup is activated, 416 in Period 1, 317 in Period 2, 266 in Period 3, 184 in Period 4, 122 in Period 5, and 66 in Period 6. + +Initial inventory for both products is zero at the start of Period 0, and all inventory must be depleted by the end of Period 6. Production amounts and inventory levels must be non-negative. The objective is to minimize the total cost, which includes production costs, setup costs, and inventory holding costs, while ensuring all constraints are satisfied.56408No differencesNo differencesok
119Inventory Management for a Retail Supply Chain + +You are responsible for managing the inventory and ordering process for a retail supply chain over a 21-week period. The goal is to minimize the total costs associated with ordering and holding inventory while ensuring that customer demand is met each week. The costs include ordering costs, unit purchase costs, and inventory holding costs. Additionally, there are specific constraints on inventory levels and ordering decisions. + +#### Key Decisions: +1. **Ordering Decisions**: Each week, you must decide whether to place an order (a binary decision) and, if so, how much to order. Placing an order incurs a fixed cost, and the amount ordered cannot exceed a certain limit based on the total demand over the 21-week period. +2. **Inventory Levels**: You must manage the ending inventory each week, ensuring that it meets the flow balance constraints (i.e., the inventory carried over from the previous week plus the amount ordered equals the ending inventory plus the demand for that week). + +#### Objective: +Minimize the total cost, which includes: +- **Ordering Costs**: Fixed costs for placing an order in a given week. These costs vary by week and are as follows: 459 in week 1, 555 in week 2, 547 in week 3, 530 in week 4, 436 in week 5, 401 in week 6, 490 in week 7, 462 in week 8, 479 in week 9, 563 in week 10, 456 in week 11, 442 in week 12, 530 in week 13, 427 in week 14, 512 in week 15, 574 in week 16, 430 in week 17, 424 in week 18, 484 in week 19, 507 in week 20, and 446 in week 21. +- **Unit Purchase Costs**: Variable costs for each unit ordered. These costs also vary by week and are as follows: 49 in week 1, 48 in week 2, 55 in week 3, 44 in week 4, 51 in week 5, 48 in week 6, 53 in week 7, 43 in week 8, 42 in week 9, 52 in week 10, 45 in week 11, 42 in week 12, 43 in week 13, 42 in week 14, 59 in week 15, 57 in week 16, 57 in week 17, 44 in week 18, 57 in week 19, 60 in week 20, and 43 in week 21. +- **Holding Costs**: Costs for holding inventory at the end of each week. These costs are as follows: 23 in week 1, 28 in week 2, 28 in week 3, 25 in week 4, 30 in week 5, 25 in week 6, 28 in week 7, 22 in week 8, 30 in week 9, 30 in week 10, 23 in week 11, 28 in week 12, 27 in week 13, 22 in week 14, 22 in week 15, 23 in week 16, 20 in week 17, 28 in week 18, 30 in week 19, 30 in week 20, and 24 in week 21. + +#### Constraints: +1. **Flow Balance**: The ending inventory for each week must satisfy the following relationship: + - Ending inventory from the previous week + amount ordered = ending inventory for the current week + demand for the current week. + - The demand for each week is as follows: 150 in week 1, 132 in week 2, 135 in week 3, 178 in week 4, 133 in week 5, 173 in week 6, 161 in week 7, 195 in week 8, 110 in week 9, 118 in week 10, 131 in week 11, 198 in week 12, 118 in week 13, 120 in week 14, 129 in week 15, 148 in week 16, 189 in week 17, 178 in week 18, 145 in week 19, 161 in week 20, and 155 in week 21. + +2. **Ordering Limits**: The amount ordered in any week cannot exceed the total demand over the 21-week period (3157 units) if an order is placed. If no order is placed, the amount ordered must be zero. + +3. **Starting and Ending Inventory**: The starting inventory at the beginning of week 1 is zero, and the ending inventory at the end of week 21 must also be zero. + +#### Goal: +Your task is to determine the optimal ordering and inventory management plan that minimizes the total cost while meeting all constraints. This includes deciding when to place orders, how much to order each week, and how much inventory to carry over each week to meet demand without exceeding the budget or violating any constraints.165425Inventory Management for a Retail Supply Chain + +You are responsible for managing the inventory and ordering process for a retail supply chain over a 21-week period. The goal is to minimize the total costs associated with ordering and holding inventory while ensuring that customer demand is met each week. The costs include ordering costs, unit purchase costs, and inventory holding costs. Additionally, there are specific constraints on inventory levels and ordering decisions. + +#### Key Decisions: +1. **Ordering Decisions**: Each week, you must decide whether to place an order (a binary decision) and, if so, how much to order. Placing an order incurs a fixed cost, and the amount ordered cannot exceed a certain limit based on the total demand over the 21-week period. +2. **Inventory Levels**: You must manage the ending inventory each week, ensuring that it meets the flow balance constraints (i.e., the inventory carried over from the previous week plus the amount ordered equals the ending inventory plus the demand for that week). + +#### Objective: +Minimize the total cost, which includes: +- **Ordering Costs**: Fixed costs for placing an order in a given week. These costs vary by week and are as follows: 459 in week 1, 555 in week 2, 547 in week 3, 530 in week 4, 436 in week 5, 401 in week 6, 490 in week 7, 462 in week 8, 479 in week 9, 563 in week 10, 456 in week 11, 442 in week 12, 530 in week 13, 427 in week 14, 512 in week 15, 574 in week 16, 430 in week 17, 424 in week 18, 484 in week 19, 507 in week 20, and 446 in week 21. +- **Unit Purchase Costs**: Variable costs for each unit ordered. These costs also vary by week and are as follows: 49 in week 1, 48 in week 2, 55 in week 3, 44 in week 4, 51 in week 5, 48 in week 6, 53 in week 7, 43 in week 8, 42 in week 9, 52 in week 10, 45 in week 11, 42 in week 12, 43 in week 13, 42 in week 14, 59 in week 15, 57 in week 16, 57 in week 17, 44 in week 18, 57 in week 19, 60 in week 20, and 43 in week 21. +- **Holding Costs**: Costs for holding inventory at the end of each week. These costs are as follows: 23 in week 1, 28 in week 2, 28 in week 3, 25 in week 4, 30 in week 5, 25 in week 6, 28 in week 7, 22 in week 8, 30 in week 9, 30 in week 10, 23 in week 11, 28 in week 12, 27 in week 13, 22 in week 14, 22 in week 15, 23 in week 16, 20 in week 17, 28 in week 18, 30 in week 19, 30 in week 20, and 24 in week 21. + +#### Constraints: +1. **Flow Balance**: The ending inventory for each week must satisfy the following relationship: + - Ending inventory from the previous week + amount ordered = ending inventory for the current week + demand for the current week. + - The demand for each week is as follows: 150 in week 1, 132 in week 2, 135 in week 3, 178 in week 4, 133 in week 5, 173 in week 6, 161 in week 7, 195 in week 8, 110 in week 9, 118 in week 10, 131 in week 11, 198 in week 12, 118 in week 13, 120 in week 14, 129 in week 15, 148 in week 16, 189 in week 17, 178 in week 18, 145 in week 19, 161 in week 20, and 155 in week 21. + +2. **Ordering Limits**: The amount ordered in any week cannot exceed the total demand over the 21-week period (3157 units) if an order is placed. If no order is placed, the amount ordered must be zero. + +3. **Starting and Ending Inventory**: The starting inventory at the beginning of week 1 is zero, and the ending inventory at the end of week 21 must also be zero. + +#### Goal: +Your task is to determine the optimal ordering and inventory management plan that minimizes the total cost while meeting all constraints. This includes deciding when to place orders, how much to order each week, and how much inventory to carry over each week to meet demand without exceeding the budget or violating any constraints.165425No differencesNo differencesok
120You are tasked with managing the distribution of resources across eight interconnected cities to ensure each city meets its specific demand while minimizing the total transportation cost. Each city has a unique supply of resources, a specific demand, and a limited capacity to handle incoming shipments. Additionally, the transportation links between cities have varying costs and capacity constraints. + +City 0 has a net outflow of 5 units, meaning it is a supply node, and it can handle up to 13 units of incoming shipments. City 1 has a net inflow of 2 units, meaning it is a demand node, and it can accommodate up to 14 units of incoming shipments. City 2 is balanced, with no net demand or surplus, and can handle up to 17 units of incoming shipments. City 3 is also balanced, with no net demand or surplus, and can handle up to 12 units of incoming shipments. City 4 has a net outflow of 1 unit, meaning it is a supply node, and it can handle up to 16 units of incoming shipments. City 5 has a net inflow of 2 units, meaning it is a demand node, and it can accommodate up to 19 units of incoming shipments. City 6 is balanced, with no net demand or surplus, and can handle up to 16 units of incoming shipments. City 7 has a net inflow of 2 units, meaning it is a demand node, and it can accommodate up to 14 units of incoming shipments. + +The cost of transporting resources between cities varies, and each transportation link has a maximum capacity. From City 0: To City 1: Cost = 2, Capacity = 8; To City 2: Cost = 2, Capacity = 7; To City 3: Cost = 2, Capacity = 7; To City 4: Cost = 2, Capacity = 9; To City 5: Cost = 2, Capacity = 8; To City 6: Cost = 1, Capacity = 9; To City 7: Cost = 2, Capacity = 7. From City 1: To City 0: Cost = 1, Capacity = 8; To City 2: Cost = 1, Capacity = 8; To City 3: Cost = 3, Capacity = 7; To City 4: Cost = 2, Capacity = 8; To City 5: Cost = 1, Capacity = 7; To City 6: Cost = 3, Capacity = 9; To City 7: Cost = 1, Capacity = 9. From City 2: To City 0: Cost = 1, Capacity = 8; To City 1: Cost = 2, Capacity = 9; To City 3: Cost = 3, Capacity = 7; To City 4: Cost = 2, Capacity = 9; To City 5: Cost = 2, Capacity = 8; To City 6: Cost = 3, Capacity = 8; To City 7: Cost = 1, Capacity = 8. From City 3: To City 0: Cost = 3, Capacity = 7; To City 1: Cost = 3, Capacity = 7; To City 2: Cost = 3, Capacity = 7; To City 4: Cost = 2, Capacity = 7; To City 5: Cost = 3, Capacity = 7; To City 6: Cost = 2, Capacity = 9; To City 7: Cost = 3, Capacity = 9. From City 4: To City 0: Cost = 2, Capacity = 9; To City 1: Cost = 3, Capacity = 8; To City 2: Cost = 3, Capacity = 8; To City 3: Cost = 1, Capacity = 8; To City 5: Cost = 1, Capacity = 7; To City 6: Cost = 3, Capacity = 8; To City 7: Cost = 2, Capacity = 9. From City 5: To City 0: Cost = 1, Capacity = 7; To City 1: Cost = 3, Capacity = 7; To City 2: Cost = 3, Capacity = 7; To City 3: Cost = 3, Capacity = 9; To City 4: Cost = 2, Capacity = 8; To City 6: Cost = 2, Capacity = 9; To City 7: Cost = 3, Capacity = 8. From City 6: To City 0: Cost = 2, Capacity = 8; To City 1: Cost = 2, Capacity = 7; To City 2: Cost = 3, Capacity = 7; To City 3: Cost = 2, Capacity = 7; To City 4: Cost = 1, Capacity = 9; To City 5: Cost = 1, Capacity = 8; To City 7: Cost = 3, Capacity = 7. From City 7: To City 0: Cost = 2, Capacity = 8; To City 1: Cost = 2, Capacity = 8; To City 2: Cost = 2, Capacity = 7; To City 3: Cost = 3, Capacity = 9; To City 4: Cost = 1, Capacity = 7; To City 5: Cost = 2, Capacity = 8; To City 6: Cost = 3, Capacity = 7. + +Your goal is to determine the optimal amount of resources to transport between cities to meet all demands while minimizing the total transportation cost. Ensure that each city's net resource flow matches its demand or surplus, no transportation link exceeds its capacity, and no city exceeds its capacity to handle incoming shipments.11You are tasked with managing the distribution of resources across eight interconnected cities to ensure each city meets its specific demand while minimizing the total transportation cost. Each city has a unique supply of resources, a specific demand, and a limited capacity to handle incoming shipments. Additionally, the transportation links between cities have varying costs and capacity constraints. + +City 0 has a net outflow of 5 units, meaning it is a supply node, and it can handle up to 13 units of incoming shipments. City 1 has a net inflow of 2 units, meaning it is a demand node, and it can accommodate up to 14 units of incoming shipments. City 2 is balanced, with no net demand or surplus, and can handle up to 17 units of incoming shipments. City 3 is also balanced, with no net demand or surplus, and can handle up to 12 units of incoming shipments. City 4 has a net outflow of 1 unit, meaning it is a supply node, and it can handle up to 16 units of incoming shipments. City 5 has a net inflow of 2 units, meaning it is a demand node, and it can accommodate up to 19 units of incoming shipments. City 6 is balanced, with no net demand or surplus, and can handle up to 16 units of incoming shipments. City 7 has a net inflow of 2 units, meaning it is a demand node, and it can accommodate up to 14 units of incoming shipments. + +The cost of transporting resources between cities varies, and each transportation link has a maximum capacity. From City 0: To City 1: Cost = 2, Capacity = 8; To City 2: Cost = 2, Capacity = 7; To City 3: Cost = 2, Capacity = 7; To City 4: Cost = 2, Capacity = 9; To City 5: Cost = 2, Capacity = 8; To City 6: Cost = 1, Capacity = 9; To City 7: Cost = 2, Capacity = 7. From City 1: To City 0: Cost = 1, Capacity = 8; To City 2: Cost = 1, Capacity = 8; To City 3: Cost = 3, Capacity = 7; To City 4: Cost = 2, Capacity = 8; To City 5: Cost = 1, Capacity = 7; To City 6: Cost = 3, Capacity = 9; To City 7: Cost = 1, Capacity = 9. From City 2: To City 0: Cost = 1, Capacity = 8; To City 1: Cost = 2, Capacity = 9; To City 3: Cost = 3, Capacity = 7; To City 4: Cost = 2, Capacity = 9; To City 5: Cost = 2, Capacity = 8; To City 6: Cost = 3, Capacity = 8; To City 7: Cost = 1, Capacity = 8. From City 3: To City 0: Cost = 3, Capacity = 7; To City 1: Cost = 3, Capacity = 7; To City 2: Cost = 3, Capacity = 7; To City 4: Cost = 2, Capacity = 7; To City 5: Cost = 3, Capacity = 7; To City 6: Cost = 2, Capacity = 9; To City 7: Cost = 3, Capacity = 9. From City 4: To City 0: Cost = 2, Capacity = 9; To City 1: Cost = 3, Capacity = 8; To City 2: Cost = 3, Capacity = 8; To City 3: Cost = 1, Capacity = 8; To City 5: Cost = 1, Capacity = 7; To City 6: Cost = 3, Capacity = 8; To City 7: Cost = 2, Capacity = 9. From City 5: To City 0: Cost = 1, Capacity = 7; To City 1: Cost = 3, Capacity = 7; To City 2: Cost = 3, Capacity = 7; To City 3: Cost = 3, Capacity = 9; To City 4: Cost = 2, Capacity = 8; To City 6: Cost = 2, Capacity = 9; To City 7: Cost = 3, Capacity = 8. From City 6: To City 0: Cost = 2, Capacity = 8; To City 1: Cost = 2, Capacity = 7; To City 2: Cost = 3, Capacity = 7; To City 3: Cost = 2, Capacity = 7; To City 4: Cost = 1, Capacity = 9; To City 5: Cost = 1, Capacity = 8; To City 7: Cost = 3, Capacity = 7. From City 7: To City 0: Cost = 2, Capacity = 8; To City 1: Cost = 2, Capacity = 8; To City 2: Cost = 2, Capacity = 7; To City 3: Cost = 3, Capacity = 9; To City 4: Cost = 1, Capacity = 7; To City 5: Cost = 2, Capacity = 8; To City 6: Cost = 3, Capacity = 7. + +Your goal is to determine the optimal amount of resources to transport between cities to meet all demands while minimizing the total transportation cost. Ensure that each city's net resource flow matches its demand or surplus, no transportation link exceeds its capacity, and no city exceeds its capacity to handle incoming shipments.11No differencesNo differencesok
121The problem involves scheduling operations for five jobs (Job 0 through Job 4) in a manufacturing facility, where each job consists of two operations that must be performed in sequence. Each operation takes exactly 1 unit of time to complete. The facility has a limited number of machines, and multiple operations from different jobs may need to be assigned to the same machine. A machine can only handle one operation at a time, so the order of operations on shared machines must be determined to avoid overlaps. + +The goal is to minimize the total completion time, known as the makespan, which is the time at which the last operation across all jobs finishes. To achieve this, the start times of all operations must be scheduled such that: (1) for each job, the second operation starts only after the first operation is completed; (2) if two operations are assigned to the same machine, one must finish before the other starts; and (3) the makespan is at least as large as the completion time of every operation. + +The order of operations on shared machines is determined using binary decision variables, which indicate whether one operation is scheduled before another. A large constant (100000) is used in the constraints to enforce the sequencing of operations on shared machines. The start times of operations and the makespan are continuous variables, while the binary variables are restricted to values of 0 or 1. The objective is to minimize the makespan, ensuring all jobs are completed as quickly as possible.4The problem involves scheduling operations for five jobs (Job 0 through Job 4) in a manufacturing facility, where each job consists of two operations that must be performed in sequence. Each operation takes exactly 1 unit of time to complete. The facility has a limited number of machines, and multiple operations from different jobs may need to be assigned to the same machine. A machine can only handle one operation at a time, so the order of operations on shared machines must be determined to avoid overlaps. + +The goal is to minimize the total completion time, known as the makespan, which is the time at which the last operation across all jobs finishes. To achieve this, the start times of all operations must be scheduled such that: (1) for each job, the second operation starts only after the first operation is completed; (2) if two operations are assigned to the same machine, one must finish before the other starts; and (3) the makespan is at least as large as the completion time of every operation. + +The order of operations on shared machines is determined using binary decision variables, which indicate whether one operation is scheduled before another. A large constant (100) is used in the constraints to enforce the sequencing of operations on shared machines. The start times of operations and the makespan are continuous variables, while the binary variables are restricted to values of 0 or 1. The objective is to minimize the makespan, ensuring all jobs are completed as quickly as possible. + + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 1 + - Operation 1 of Job 1: Machine 0 + - Operation 0 of Job 2: Machine 0 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 0 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 04The goal is to minimize the total completion time, known as the makespan, which is the time at which the last operation across all jobs finishes. To achieve this, the start times of all operations must be scheduled such that: (1) for each job, the second operation starts only after the first operation is completed; (2) if two operations are assigned to the same machine, one must finish before the other starts; and (3) the makespan is at least as large as the completion time of every operation. + +The order of operations on shared machines is determined using binary decision variables, which indicate whether one operation is scheduled before another. A large constant (100000) (100) is used in the constraints to enforce the sequencing of operations on shared machines. The start times of operations and the makespan are continuous variables, while the binary variables are restricted to values of 0 or 1. The objective is to minimize the makespan, ensuring all jobs are completed as quickly as possible. + + + + Additional information: + - There are 3 machines, numbered from 0 to 2 + - The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 1 + - Operation 1 of Job 1: Machine 0 + - Operation 0 of Job 2: Machine 0 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 0 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 0No differencesIt was missing the number of machines. We added it, and also an explicit assignment of jobs to machinesmissing data
122There are 2 restaurants, 7 employees, and 2 shifts per day. Each shift requires employees with specific skills, and each employee has a set of skills they are qualified to perform. Employees have preferences for certain skills, and assigning them to tasks they prefer incurs lower costs. The cost of leaving a position unfulfilled is significantly higher, so the goal is to minimize both the preference costs and the number of unfulfilled positions. + +Restaurant 0, Shift 0, Skill 0 requires 1 employee. Restaurant 0, Shift 1, Skill 0 requires 1 employee. Restaurant 1, Shift 0, Skill 0 requires 2 employees. Restaurant 1, Shift 1, Skill 0 requires 1 employee. + +Employees have varying availability for shifts. For example, Employee 0 is available for Shift 0 but not for Shift 1. Employee 1 is unavailable for Shift 1. Employee 3 is unavailable for Shift 0. Each employee is qualified for specific skills. For instance, Employee 0 is qualified for Skill 0, while Employee 3 is not qualified for Skill 0. + +Assigning employees to tasks they prefer incurs lower costs. For example, assigning Employee 0 to Skill 0 costs 1 unit, and assigning Employee 1 to Skill 0 costs 2 units. Each unfulfilled position costs 100 units, which is significantly higher than preference costs. + +The goal is to assign employees to specific shifts and restaurants based on their skills and availability while determining the number of unfulfilled positions for each shift and restaurant. The total cost to minimize includes the cost of assigning employees to tasks they prefer and the penalty cost for unfulfilled positions. + +The total number of employees assigned to a shift at a restaurant, plus any unfulfilled positions, must equal the demand for that shift and skill. Employees can only be assigned to shifts they are available for and tasks they are qualified for. Each employee can be assigned to at most one shift across all restaurants and skills. + +Preference costs for assignments range from 1 to 3 units. The cost of an unfulfilled position is 100 units. Specific demands and constraints are as outlined above.9"There are 2 restaurants, 7 employees, and 2 shifts per day. Each shift requires employees with specific skills, and each employee has a set of 2 skills they are qualified to perform. Employees have preferences for certain skills, and assigning them to tasks they prefer incurs lower costs. The cost of leaving a position unfulfilled is significantly higher, so the goal is to minimize both the preference costs and the number of unfulfilled positions. + +Restaurant 0, Shift 0, Skill 0 requires 1 employee. Restaurant 0, Shift 1, Skill 0 requires 1 employee. Restaurant 1, Shift 0, Skill 0 requires 2 employees. Restaurant 1, Shift 1, Skill 0 requires 1 employee. + +Employees have varying availability for shifts. Employee 0 is available for Shift 0 but not for Shift 1. Employee 1 is available for Shift 0 but not for Shift 1. Employee 2 is available for Shifts 0 and 1. Employee 3 is available for Shift 1 but not for Shift 0. Employee 4 is available for Shifts 0 and 1. Employee 5 is available for Shift 1 but not for Shift 0. Employee 6 is available for Shifts 0 and 1. Each employee is qualified for specific skills. For instance, Employee 0 is not qualified for Skill 0 but qualified for Skill 1, while Employees 3 and 5 are not qualified for Skill 1 but qualified for Skill 0; Employees 1, 2, 4 and 6 are qualified by both Skill 0 and 1. +Assigning employees to tasks they prefer incurs lower costs. In particular, assigning Employee 1 to Skill 0 costs 2 units. +Assigning Employee 2 to Skill 0 costs 0 units. Assigning Employee 3 to Skill 0 costs 2 units. +Assigning Employee 4 to Skill 0 costs 1 unit. +Assigning Employee 5 to Skill 0 costs 3 units. +Assigning Employee 6 to Skill 0 costs 2 units. Each unfulfilled position costs 100 units, which is significantly higher than preference costs. + +The goal is to assign employees to specific shifts and restaurants based on their skills and availability while determining the number of unfulfilled positions for each shift and restaurant. The total cost to minimize includes the cost of assigning employees to tasks they prefer and the penalty cost for unfulfilled positions. + +The total number of employees assigned to a shift at a restaurant, plus any unfulfilled positions, must equal the demand for that shift and skill. Employees can only be assigned to shifts they are available for and tasks they are qualified for. Each employee can be assigned to at most one shift across all restaurants and skills."7There "There are 2 restaurants, 7 employees, and 2 shifts per day. Each shift requires employees with specific skills, and each employee has a set of 2 skills they are qualified to perform. Employees have preferences for certain skills, and assigning them to tasks they prefer incurs lower costs. The cost of leaving a position unfulfilled is significantly higher, so the goal is to minimize both the preference costs and the number of unfulfilled positions. + +Restaurant 0, Shift 0, Skill 0 requires 1 employee. Restaurant 0, Shift 1, Skill 0 requires 1 employee. Restaurant 1, Shift 0, Skill 0 requires 2 employees. Restaurant 1, Shift 1, Skill 0 requires 1 employee. + +Employees have varying availability for shifts. For example, Employee 0 is available for Shift 0 but not for Shift 1. Employee 1 is unavailable available for Shift 0 but not for Shift 1. Employee 2 is available for Shifts 0 and 1. Employee 3 is unavailable available for Shift 1 but not for Shift 0. Employee 4 is available for Shifts 0 and 1. Employee 5 is available for Shift 1 but not for Shift 0. Employee 6 is available for Shifts 0 and 1. Each employee is qualified for specific skills. For instance, Employee 0 is not qualified for Skill 0, 0 but qualified for Skill 1, while Employee Employees 3 is and 5 are not qualified for Skill 0. 1 but qualified for Skill 0; Employees 1, 2, 4 and 6 are qualified by both Skill 0 and 1. +Assigning employees to tasks they prefer incurs lower costs. For example, In particular, assigning Employee 1 to Skill 0 costs 2 units. + Assigning Employee 2 to Skill 0 costs 1 unit, and assigning 0 units. Assigning Employee 3 to Skill 0 costs 2 units. + Assigning Employee 4 to Skill 0 costs 1 unit. + Assigning Employee 5 to Skill 0 costs 3 units. + Assigning Employee 6 to Skill 0 costs 2 units. Each unfulfilled position costs 100 units, which is significantly higher than preference costs. + +The goal is to assign employees to specific shifts and restaurants based on their skills and availability while determining the number of unfulfilled positions for each shift and restaurant. The total cost to minimize includes the cost of assigning employees to tasks they prefer and the penalty cost for unfulfilled positions. + +The total number of employees assigned to a shift at a restaurant, plus any unfulfilled positions, must equal the demand for that shift and skill. Employees can only be assigned to shifts they are available for and tasks they are qualified for. Each employee can be assigned to at most one shift across all restaurants and skills. + + Preference costs for assignments range from 1 to 3 units. The cost of an unfulfilled position is 100 units. Specific demands and constraints are as outlined above. skills."9 7Missing data about workers' availability for shifts and assignment costsmissing data
123Minimize the total operational costs, which include production setup costs, machine startup costs, inventory holding costs, and backlogging costs, while meeting customer demand and adhering to machine capacity constraints. The setup cost is 136.3292413778485 units per production run, and the startup cost is 72.14543590644504 units per machine startup. The inventory holding cost for Product 0 is 1.097996170356003 units per unit per period, and for Product 1, it is 0.9303010222262446 units per unit per period. The backlogging cost for Product 0 is 7.343834540408109 units per unit per period, and for Product 1, it is 6.589388559571906 units per unit per period. + +The production system must meet the following demand constraints: for Product 0, the demand is 17.38664149749027 units in Period 0, 18.25024949764659 units in Period 1, and 20.01141977261751 units in Period 2. For Product 1, the demand is 17.8541958922009 units in Period 0, 16.38473840141418 units in Period 1, and 15.49890198357548 units in Period 2. + +Machine 0 can handle up to 77.29936140903678 units of production in Period 0, adjusted for startup time, while Machine 1 can handle up to 74.94692092032173 units in the same period. Only one product can be produced on a machine in a given time period. If a machine was not producing in the previous period but is producing in the current period, a startup cost is incurred. Production quantities, inventory levels, and backlogs must be non-negative, and production and startup decisions are binary (0 or 1). + +The objective is to determine the optimal production schedule, machine startups, inventory levels, and backlogs to minimize total costs while meeting all operational constraints.525.01301077364Minimize the total operational costs, which include production setup costs, machine startup costs, inventory holding costs, and backlogging costs, while meeting customer demand and adhering to machine capacity constraints. The setup cost is 136.3292413778485 units per production run, and the startup cost is 72.14543590644504 units per machine startup. The inventory holding cost for Product 0 is 1.097996170356003 units per unit per period, and for Product 1, it is 0.9303010222262446 units per unit per period. The backlogging cost for Product 0 is 7.343834540408109 units per unit per period, and for Product 1, it is 6.589388559571906 units per unit per period. + +The production system must meet the following demand constraints: for Product 0, the demand is 17.38664149749027 units in Period 0, 18.25024949764659 units in Period 1, and 20.01141977261751 units in Period 2. For Product 1, the demand is 17.8541958922009 units in Period 0, 16.38473840141418 units in Period 1, and 15.49890198357548 units in Period 2. + +Machine 0 can handle up to 77.29936140903678 units of production in each period, while Machine 1 can handle up to 74.94692092032173 units in each period. Only one product can be produced on a machine in a given time period. If a machine was not producing in the previous period but is producing in the current period, a startup cost is incurred. Production quantities, inventory levels, and backlogs must be non-negative, and production and startup decisions are binary (0 or 1). + +The objective is to determine the optimal production schedule, machine startups, inventory levels, and backlogs to minimize total costs while meeting all operational constraints.525.01301077364The production system must meet the following demand constraints: for Product 0, the demand is 17.38664149749027 units in Period 0, 18.25024949764659 units in Period 1, and 20.01141977261751 units in Period 2. For Product 1, the demand is 17.8541958922009 units in Period 0, 16.38473840141418 units in Period 1, and 15.49890198357548 units in Period 2. + +Machine 0 can handle up to 77.29936140903678 units of production in Period 0, adjusted for startup time, each period, while Machine 1 can handle up to 74.94692092032173 units in the same each period. Only one product can be produced on a machine in a given time period. If a machine was not producing in the previous period but is producing in the current period, a startup cost is incurred. Production quantities, inventory levels, and backlogs must be non-negative, and production and startup decisions are binary (0 or 1). + +The objective is to determine the optimal production schedule, machine startups, inventory levels, and backlogs to minimize total costs while meeting all operational constraints.No differencesClarification needed for machine capacitiesambiguous
124Minimize the total operational cost, which includes the base cost of operating generators, the cost of power output, and the startup cost of generators. The base cost for operating a Type 0 generator is 79 units per period, Type 1 is 80 units, and Type 2 is 78 units. The cost of power output is 4 units per MW for Type 0, 3 units per MW for Type 1, and 4 units per MW for Type 2. The startup cost is 349 units for Type 0, 354 units for Type 1, and 354 units for Type 2. The total power output from all generators must meet or exceed the demand in each period: 176 MW in Period 0, 160 MW in Period 1, 178 MW in Period 2, 210 MW in Period 3, and 171 MW in Period 4. Each generator type has minimum and maximum output limits: Type 0 generators must produce at least 34 MW and no more than 130 MW, Type 1 generators must produce at least 34 MW and no more than 130 MW, and Type 2 generators must produce at least 35 MW and no more than 128 MW. The total maximum possible output from all generators must be at least 202.4 MW in Period 0, 184 MW in Period 1, 204.7 MW in Period 2, 241.5 MW in Period 3, and 196.65 MW in Period 4. The number of generators that can be operated is limited by the available units: up to 2 Type 0 generators, up to 5 Type 1 generators, and up to 1 Type 2 generator in each period. The number of generators started in a period cannot exceed the number of generators that were not operating in the previous period. The solution must determine the optimal number of generators to operate, the power output from each generator, and the number of generators to start in each period to minimize the total operational cost while meeting the power demand and adhering to all operational constraints.4193Minimize the total operational cost, which includes the base cost of operating generators, the cost of power output, and the startup cost of generators. The base cost for operating a Type 0 generator is 79 units per period, Type 1 is 80 units, and Type 2 is 78 units. The cost of power output is 4 units per MW for Type 0, 3 units per MW for Type 1, and 4 units per MW for Type 2. The startup cost is 349 units for Type 0, 354 units for Type 1, and 354 units for Type 2. The total power output from all generators must meet or exceed the demand in each period: 176 MW in Period 0, 160 MW in Period 1, 178 MW in Period 2, 210 MW in Period 3, and 171 MW in Period 4. Each generator type has minimum and maximum output limits: Type 0 generators must produce at least 34 MW and no more than 130 MW, Type 1 generators must produce at least 34 MW and no more than 130 MW, and Type 2 generators must produce at least 35 MW and no more than 128 MW. The total maximum possible output from all generators must be at least 202.4 MW in Period 0, 184 MW in Period 1, 204.7 MW in Period 2, 241.5 MW in Period 3, and 196.65 MW in Period 4. The number of generators that can be operated is limited by the available units: up to 2 Type 0 generators, up to 5 Type 1 generators, and up to 1 Type 2 generator in each period. The number of generators started in a period cannot exceed the number of generators that were not operating in the previous period. The solution must determine the optimal number of generators to operate, the power output from each generator, and the number of generators to start in each period to minimize the total operational cost while meeting the power demand and adhering to all operational constraints.4193No differencesNo differencesok
125You are tasked with planning the most efficient delivery route for a logistics company that serves seven cities: City 0, City 1, City 2, City 3, City 4, City 5, and City 6. The goal is to minimize the total transportation cost while ensuring that each city is visited exactly once and that the delivery route forms a complete tour without any subtours. The delivery vehicle must return to the starting city after visiting all others. + +The cost of traveling between each pair of cities varies. From City 0: to City 1 costs 277, to City 2 costs 304, to City 3 costs 280, to City 4 costs 321, to City 5 costs 285, and to City 6 costs 300. From City 1: to City 0 costs 280, to City 2 costs 317, to City 3 costs 301, to City 4 costs 306, to City 5 costs 312, and to City 6 costs 290. From City 2: to City 0 costs 304, to City 1 costs 318, to City 3 costs 283, to City 4 costs 289, to City 5 costs 301, and to City 6 costs 277. From City 3: to City 0 costs 304, to City 1 costs 282, to City 2 costs 290, to City 4 costs 307, to City 5 costs 291, and to City 6 costs 290. From City 4: to City 0 costs 322, to City 1 costs 291, to City 2 costs 277, to City 3 costs 300, to City 5 costs 318, and to City 6 costs 280. From City 5: to City 0 costs 308, to City 1 costs 297, to City 2 costs 288, to City 3 costs 314, to City 4 costs 290, and to City 6 costs 320. From City 6: to City 0 costs 316, to City 1 costs 313, to City 2 costs 301, to City 3 costs 324, to City 4 costs 280, and to City 5 costs 303. + +Each city must be visited exactly once. The delivery vehicle must arrive at each city from exactly one other city and depart to exactly one other city. The delivery route must form a single tour, meaning the vehicle cannot split into smaller loops or subtours. To enforce this, each city is assigned a position in the tour, with positions ranging from 1 to 7, ensuring a valid sequence. The positions must follow a logical sequence without gaps or repetitions. The route between any two cities is binary: it is either included (1) or excluded (0) in the final tour. + +The objective is to minimize the total transportation cost for the delivery route while satisfying all constraints. The solution must account for all the costs and constraints outlined above.1987You are tasked with planning the most efficient delivery route for a logistics company that serves seven cities: City 0, City 1, City 2, City 3, City 4, City 5, and City 6. The goal is to minimize the total transportation cost while ensuring that each city is visited exactly once and that the delivery route forms a complete tour without any subtours. The delivery vehicle must return to the starting city after visiting all others. + +The cost of traveling between each pair of cities varies. From City 0: to City 1 costs 277, to City 2 costs 304, to City 3 costs 280, to City 4 costs 321, to City 5 costs 285, and to City 6 costs 300. From City 1: to City 0 costs 280, to City 2 costs 317, to City 3 costs 301, to City 4 costs 306, to City 5 costs 312, and to City 6 costs 290. From City 2: to City 0 costs 304, to City 1 costs 318, to City 3 costs 283, to City 4 costs 289, to City 5 costs 301, and to City 6 costs 277. From City 3: to City 0 costs 304, to City 1 costs 282, to City 2 costs 290, to City 4 costs 307, to City 5 costs 291, and to City 6 costs 290. From City 4: to City 0 costs 322, to City 1 costs 291, to City 2 costs 277, to City 3 costs 300, to City 5 costs 318, and to City 6 costs 280. From City 5: to City 0 costs 308, to City 1 costs 297, to City 2 costs 288, to City 3 costs 314, to City 4 costs 290, and to City 6 costs 320. From City 6: to City 0 costs 316, to City 1 costs 313, to City 2 costs 301, to City 3 costs 324, to City 4 costs 280, and to City 5 costs 303. + +Each city must be visited exactly once. The delivery vehicle must arrive at each city from exactly one other city and depart to exactly one other city. The delivery route must form a single tour, meaning the vehicle cannot split into smaller loops or subtours. To enforce this, each city is assigned a position in the tour, with positions ranging from 1 to 7, ensuring a valid sequence. The positions must follow a logical sequence without gaps or repetitions. The route between any two cities is binary: it is either included (1) or excluded (0) in the final tour. + +The objective is to minimize the total transportation cost for the delivery route while satisfying all constraints. The solution must account for all the costs and constraints outlined above.1987No differencesNo differencesok
126You are tasked with optimizing the placement of emergency response centers across a network of five critical locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall coverage and effectiveness of the response system. Each location has unique characteristics, and the goal is to select exactly three locations to establish these centers. The effectiveness of the system is measured by the combined coverage scores between the selected centers, which are determined by predefined pairwise scores representing the strategic importance of connecting each pair of locations. + +#### Key Decisions: +1. **Selecting Locations**: You must choose exactly three out of the five locations to establish emergency response centers. Each location is represented by a binary decision: if a center is established at a location, it is selected (1); otherwise, it is not (0). +2. **Pairwise Coverage**: For each pair of selected locations, the coverage score is determined by the predefined pairwise importance values. These scores reflect the strategic benefit of having both locations in the network. + +#### Objective: +Maximize the total coverage score across all pairs of selected locations. The coverage scores for each pair are as follows: +- Node 0 and Node 1: 24 +- Node 0 and Node 2: 21 +- Node 0 and Node 3: 16 +- Node 0 and Node 4: 15 +- Node 1 and Node 0: 21 +- Node 1 and Node 2: 19 +- Node 1 and Node 3: 24 +- Node 1 and Node 4: 23 +- Node 2 and Node 0: 21 +- Node 2 and Node 1: 17 +- Node 2 and Node 3: 25 +- Node 2 and Node 4: 19 +- Node 3 and Node 0: 18 +- Node 3 and Node 1: 24 +- Node 3 and Node 2: 17 +- Node 3 and Node 4: 17 +- Node 4 and Node 0: 15 +- Node 4 and Node 1: 18 +- Node 4 and Node 2: 15 +- Node 4 and Node 3: 24 + +#### Constraints: +1. **Selection of Centers**: Exactly three locations must be selected to establish emergency response centers. This means the sum of selected locations must equal three. +2. **Pairwise Coverage Activation**: For any pair of locations, the coverage score is only counted if both locations in the pair are selected. This ensures that the coverage score is only applied to active pairs. +3. **Logical Relationships**: + - If a location is not selected, no coverage scores involving that location can be counted. + - If both locations in a pair are selected, the coverage score for that pair is fully counted. + - If only one location in a pair is selected, the coverage score for that pair is not counted. + +#### Numerical Parameters: +- The pairwise coverage scores are explicitly provided for all possible pairs of locations, as listed above. +- The total number of selected locations must be exactly three. + +#### Summary: +Your task is to select three out of the five locations to establish emergency response centers such that the total coverage score, based on the predefined pairwise importance values, is maximized. The selection must adhere to the constraints ensuring that only active pairs of selected locations contribute to the coverage score.130You are tasked with optimizing the placement of emergency response centers across a network of five critical locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall coverage and effectiveness of the response system. Each location has unique characteristics, and the goal is to select exactly three locations to establish these centers. The effectiveness of the system is measured by the combined coverage scores between the selected centers, which are determined by predefined pairwise scores representing the strategic importance of connecting each pair of locations. + +#### Key Decisions: +1. **Selecting Locations**: You must choose exactly three out of the five locations to establish emergency response centers. Each location is represented by a binary decision: if a center is established at a location, it is selected (1); otherwise, it is not (0). +2. **Pairwise Coverage**: For each pair of selected locations, the coverage score is determined by the predefined pairwise importance values. These scores reflect the strategic benefit of having both locations in the network. + +#### Objective: +Maximize the total coverage score across all pairs of selected locations. The coverage scores for each pair are as follows: +- Node 0 and Node 1: 24 +- Node 0 and Node 2: 21 +- Node 0 and Node 3: 16 +- Node 0 and Node 4: 15 +- Node 1 and Node 0: 21 +- Node 1 and Node 2: 19 +- Node 1 and Node 3: 24 +- Node 1 and Node 4: 23 +- Node 2 and Node 0: 21 +- Node 2 and Node 1: 17 +- Node 2 and Node 3: 25 +- Node 2 and Node 4: 19 +- Node 3 and Node 0: 18 +- Node 3 and Node 1: 24 +- Node 3 and Node 2: 17 +- Node 3 and Node 4: 17 +- Node 4 and Node 0: 15 +- Node 4 and Node 1: 18 +- Node 4 and Node 2: 15 +- Node 4 and Node 3: 24 + +#### Constraints: +1. **Selection of Centers**: Exactly three locations must be selected to establish emergency response centers. This means the sum of selected locations must equal three. +2. **Pairwise Coverage Activation**: For any pair of locations, the coverage score is only counted if both locations in the pair are selected. This ensures that the coverage score is only applied to active pairs. +3. **Logical Relationships**: + - If a location is not selected, no coverage scores involving that location can be counted. + - If both locations in a pair are selected, the coverage score for that pair is fully counted. + - If only one location in a pair is selected, the coverage score for that pair is not counted. + +#### Numerical Parameters: +- The pairwise coverage scores are explicitly provided for all possible pairs of locations, as listed above. +- The total number of selected locations must be exactly three. + +#### Summary: +Your task is to select three out of the five locations to establish emergency response centers such that the total coverage score, based on the predefined pairwise importance values, is maximized. The selection must adhere to the constraints ensuring that only active pairs of selected locations contribute to the coverage score.130No differencesNo differencesok
127The problem involves scheduling operations for 7 products, each requiring 2 operations to be performed in sequence. Each operation has a fixed processing time of 1 unit of time. The operations must be scheduled on shared machines, where each machine can handle only one operation at a time. The order of operations on shared machines is determined by binary decision variables, which decide whether one operation is performed before another. + +The goal is to minimize the total production time, known as the makespan, ensuring that all products are completed as quickly as possible. The makespan must be at least as long as the completion time of the last operation across all products. + +Constraints include precedence constraints, ensuring that the second operation of each product cannot start until the first operation is completed. Machine capacity constraints ensure that no two operations on the same machine overlap in time. These constraints are enforced using binary variables that determine the order of operations on shared machines. A large constant (100000) is used in the machine capacity constraints to ensure that the constraints are satisfied when the binary variables determine the order of operations. + +The start times of all operations must be non-negative, and the binary variables must take values of either 0 or 1. The objective is to minimize the makespan, ensuring that all operations are scheduled efficiently without violating the precedence and machine capacity constraints.4The problem involves scheduling operations for 7 products, each requiring 2 operations to be performed in sequence. Each operation has a fixed processing time of 1 unit of time. The operations must be scheduled on shared machines, where each machine can handle only one operation at a time. The order of operations on shared machines is determined by binary decision variables, which decide whether one operation is performed before another. + +The goal is to minimize the total production time, known as the makespan, ensuring that all products are completed as quickly as possible. The makespan must be at least as long as the completion time of the last operation across all products. + +Constraints include precedence constraints, ensuring that the second operation of each product cannot start until the first operation is completed. Machine capacity constraints ensure that no two operations on the same machine overlap in time. These constraints are enforced using binary variables that determine the order of operations on shared machines. A large constant (1000) is used in the machine capacity constraints to ensure that the constraints are satisfied when the binary variables determine the order of operations. + +The start times of all operations must be non-negative, and the binary variables must take values of either 0 or 1. The objective is to minimize the makespan, ensuring that all operations are scheduled efficiently without violating the precedence and machine capacity constraints. + +ADDITIONAL INPUT DATA : + +1) Sets and indices +- Jobs (products): J = {1, 2, 3, 4, 5, 6, 7} +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Machines: M = {1, 2, 3} + +2) Machine assignment for every operation (which machine executes the operation) +- Operation (1,1) -> Machine 1 +- Operation (1,2) -> Machine 2 +- Operation (2,1) -> Machine 1 +- Operation (2,2) -> Machine 3 +- Operation (3,1) -> Machine 2 +- Operation (3,2) -> Machine 3 +- Operation (4,1) -> Machine 1 +- Operation (4,2) -> Machine 2 +- Operation (5,1) -> Machine 2 +- Operation (5,2) -> Machine 3 +- Operation (6,1) -> Machine 1 +- Operation (6,2) -> Machine 3 +- Operation (7,1) -> Machine 2 +- Operation (7,2) -> Machine 1 + +3) Processing times +- All operations have processing time p_{j,k} = 1 (unit time) as in the original description.5The problem involves scheduling operations for 7 products, each requiring 2 operations to be performed in sequence. Each operation has a fixed processing time of 1 unit of time. The operations must be scheduled on shared machines, where each machine can handle only one operation at a time. The order of operations on shared machines is determined by binary decision variables, which decide whether one operation is performed before another. + +The goal is to minimize the total production time, known as the makespan, ensuring that all products are completed as quickly as possible. The makespan must be at least as long as the completion time of the last operation across all products. + +Constraints include precedence constraints, ensuring that the second operation of each product cannot start until the first operation is completed. Machine capacity constraints ensure that no two operations on the same machine overlap in time. These constraints are enforced using binary variables that determine the order of operations on shared machines. A large constant (100000) (1000) is used in the machine capacity constraints to ensure that the constraints are satisfied when the binary variables determine the order of operations. + +The start times of all operations must be non-negative, and the binary variables must take values of either 0 or 1. The objective is to minimize the makespan, ensuring that all operations are scheduled efficiently without violating the precedence and machine capacity constraints. + + ADDITIONAL INPUT DATA : + + 1) Sets and indices + - Jobs (products): J = {1, 2, 3, 4, 5, 6, 7} + - For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). + - Machines: M = {1, 2, 3} + + 2) Machine assignment for every operation (which machine executes the operation) + - Operation (1,1) -> Machine 1 + - Operation (1,2) -> Machine 2 + - Operation (2,1) -> Machine 1 + - Operation (2,2) -> Machine 3 + - Operation (3,1) -> Machine 2 + - Operation (3,2) -> Machine 3 + - Operation (4,1) -> Machine 1 + - Operation (4,2) -> Machine 2 + - Operation (5,1) -> Machine 2 + - Operation (5,2) -> Machine 3 + - Operation (6,1) -> Machine 1 + - Operation (6,2) -> Machine 3 + - Operation (7,1) -> Machine 2 + - Operation (7,2) -> Machine 1 + + 3) Processing times + - All operations have processing time p_{j,k} = 1 (unit time) as in the original description.4 5While the statement required to avoid overlap of operations on the same machines, it did not state how many machines are available or the assignment of operations to machines. missing data
128You are responsible for managing the distribution of resources across nine interconnected cities. Each city has a specific supply or demand for resources, and your goal is to ensure that all cities receive the resources they need while minimizing the total transportation cost. The transportation costs vary depending on the route taken between cities, and each route has a maximum capacity limit that cannot be exceeded. + +#### City Requirements: +- **City 0** has a surplus of 42 units of resources that need to be distributed to other cities. +- **City 1** requires an additional 3 units of resources to meet its demand. +- **City 2** has a surplus of 11 units of resources. +- **City 3** requires an additional 21 units of resources. +- **City 4** has a surplus of 4 units of resources. +- **City 5** has a surplus of 3 units of resources. +- **City 6** requires an additional 19 units of resources. +- **City 7** requires an additional 24 units of resources. +- **City 8** has a surplus of 7 units of resources. + +#### Transportation Costs: +The cost of transporting resources between cities varies. Below are the costs for each possible route: +- From **City 0** to other cities: + - To **City 1**: 8 units + - To **City 2**: 9 units + - To **City 3**: 8 units + - To **City 4**: 9 units + - To **City 5**: 8 units + - To **City 6**: 10 units + - To **City 7**: 7 units + - To **City 8**: 10 units + +- From **City 1** to other cities: + - To **City 0**: 6 units + - To **City 2**: 7 units + - To **City 3**: 6 units + - To **City 4**: 6 units + - To **City 5**: 6 units + - To **City 6**: 9 units + - To **City 7**: 10 units + - To **City 8**: 8 units + +- From **City 2** to other cities: + - To **City 0**: 9 units + - To **City 1**: 9 units + - To **City 3**: 8 units + - To **City 4**: 6 units + - To **City 5**: 8 units + - To **City 6**: 10 units + - To **City 7**: 7 units + - To **City 8**: 10 units + +- From **City 3** to other cities: + - To **City 0**: 8 units + - To **City 1**: 6 units + - To **City 2**: 10 units + - To **City 4**: 9 units + - To **City 5**: 10 units + - To **City 6**: 8 units + - To **City 7**: 10 units + - To **City 8**: 9 units + +- From **City 4** to other cities: + - To **City 0**: 6 units + - To **City 1**: 8 units + - To **City 2**: 9 units + - To **City 3**: 8 units + - To **City 5**: 10 units + - To **City 6**: 9 units + - To **City 7**: 7 units + - To **City 8**: 9 units + +- From **City 5** to other cities: + - To **City 0**: 10 units + - To **City 1**: 6 units + - To **City 2**: 7 units + - To **City 3**: 9 units + - To **City 4**: 8 units + - To **City 6**: 10 units + - To **City 7**: 7 units + - To **City 8**: 10 units + +- From **City 6** to other cities: + - To **City 0**: 10 units + - To **City 1**: 10 units + - To **City 2**: 10 units + - To **City 3**: 6 units + - To **City 4**: 9 units + - To **City 5**: 9 units + - To **City 7**: 7 units + - To **City 8**: 8 units + +- From **City 7** to other cities: + - To **City 0**: 6 units + - To **City 1**: 10 units + - To **City 2**: 9 units + - To **City 3**: 8 units + - To **City 4**: 8 units + - To **City 5**: 9 units + - To **City 6**: 8 units + - To **City 8**: 7 units + +- From **City 8** to other cities: + - To **City 0**: 10 units + - To **City 1**: 6 units + - To **City 2**: 8 units + - To **City 3**: 10 units + - To **City 4**: 7 units + - To **City 5**: 6 units + - To **City 6**: 9 units + - To **City 7**: 10 units + +#### Capacity Constraints: +Each route between cities has a maximum capacity limit, which cannot be exceeded: +- **City 0** to **City 1**: 40 units +- **City 0** to **City 2**: 56 units +- **City 0** to **City 3**: 41 units +- **City 0** to **City 4**: 49 units +- **City 0** to **City 5**: 50 units +- **City 0** to **City 6**: 36 units +- **City 0** to **City 7**: 59 units +- **City 0** to **City 8**: 55 units + +- **City 1** to **City 0**: 33 units +- **City 1** to **City 2**: 33 units +- **City 1** to **City 3**: 56 units +- **City 1** to **City 4**: 46 units +- **City 1** to **City 5**: 38 units +- **City 1** to **City 6**: 53 units +- **City 1** to **City 7**: 51 units +- **City 1** to **City 8**: 59 units + +- **City 2** to **City 0**: 59 units +- **City 2** to **City 1**: 52 units +- **City 2** to **City 3**: 41 units +- **City 2** to **City 4**: 38 units +- **City 2** to **City 5**: 36 units +- **City 2** to **City 6**: 45 units +- **City 2** to **City 7**: 32 units +- **City 2** to **City 8**: 41 units + +- **City 3** to **City 0**: 30 units +- **City 3** to **City 1**: 32 units +- **City 3** to **City 2**: 47 units +- **City 3** to **City 4**: 57 units +- **City 3** to **City 5**: 39 units +- **City 3** to **City 6**: 34 units +- **City 3** to **City 7**: 32 units +- **City 3** to **City 8**: 46 units + +- **City 4** to **City 0**: 55 units +- **City 4** to **City 1**: 40 units +- **City 4** to **City 2**: 55 units +- **City 4** to **City 3**: 41 units +- **City 4** to **City 5**: 59 units +- **City 4** to **City 6**: 51 units +- **City 4** to **City 7**: 35 units +- **City 4** to **City 8**: 53 units + +- **City 5** to **City 0**: 34 units +- **City 5** to **City 1**: 41 units +- **City 5** to **City 2**: 35 units +- **City 5** to **City 3**: 43 units +- **City 5** to **City 4**: 41 units +- **City 5** to **City 6**: 43 units +- **City 5** to **City 7**: 40 units +- **City 5** to **City 8**: 41 units + +- **City 6** to **City 0**: 34 units +- **City 6** to **City 1**: 32 units +- **City 6** to **City 2**: 47 units +- **City 6** to **City 3**: 43 units +- **City 6** to **City 4**: 46 units +- **City 6** to **City 5**: 34 units +- **City 6** to **City 7**: 60 units +- **City 6** to **City 8**: 59 units + +- **City 7** to **City 0**: 31 units +- **City 7** to **City 1**: 42 units +- **City 7** to **City 2**: 30 units +- **City 7** to **City 3**: 34 units +- **City 7** to **City 4**: 52 units +- **City 7** to **City 5**: 45 units +- **City 7** to **City 6**: 43 units +- **City 7** to **City 8**: 50 units + +- **City 8** to **City 0**: 49 units +- **City 8** to **City 1**: 31 units +- **City 8** to **City 2**: 53 units +- **City 8** to **City 3**: 46 units +- **City 8** to **City 4**: 42 units +- **City 8** to **City 5**: 36 units +- **City 8** to **City 6**: 30 units +- **City 8** to **City 7**: 41 units + +#### Objective: +Your task is to determine the optimal amount of resources to transport between cities to meet all demands while minimizing the total transportation cost. Ensure that the transportation does not exceed the capacity limits of any route and that all cities' resource requirements are satisfied. + +What is the minimum total cost to achieve this distribution?533You are responsible for managing the distribution of resources across nine interconnected cities. Each city has a specific supply or demand for resources, and your goal is to ensure that all cities receive the resources they need while minimizing the total transportation cost. The transportation costs vary depending on the route taken between cities, and each route has a maximum capacity limit that cannot be exceeded. + +#### City Requirements: +- **City 0** has a surplus of 42 units of resources that need to be distributed to other cities. +- **City 1** requires an additional 3 units of resources to meet its demand. +- **City 2** has a surplus of 11 units of resources. +- **City 3** requires an additional 21 units of resources. +- **City 4** has a surplus of 4 units of resources. +- **City 5** has a surplus of 3 units of resources. +- **City 6** requires an additional 19 units of resources. +- **City 7** requires an additional 24 units of resources. +- **City 8** has a surplus of 7 units of resources. + +#### Transportation Costs: +The cost of transporting resources between cities varies. Below are the costs for each possible route: +- From **City 0** to other cities: + - To **City 1**: 8 units + - To **City 2**: 9 units + - To **City 3**: 8 units + - To **City 4**: 9 units + - To **City 5**: 8 units + - To **City 6**: 10 units + - To **City 7**: 7 units + - To **City 8**: 10 units + +- From **City 1** to other cities: + - To **City 0**: 6 units + - To **City 2**: 7 units + - To **City 3**: 6 units + - To **City 4**: 6 units + - To **City 5**: 6 units + - To **City 6**: 9 units + - To **City 7**: 10 units + - To **City 8**: 8 units + +- From **City 2** to other cities: + - To **City 0**: 9 units + - To **City 1**: 9 units + - To **City 3**: 8 units + - To **City 4**: 6 units + - To **City 5**: 8 units + - To **City 6**: 10 units + - To **City 7**: 7 units + - To **City 8**: 10 units + +- From **City 3** to other cities: + - To **City 0**: 8 units + - To **City 1**: 6 units + - To **City 2**: 10 units + - To **City 4**: 9 units + - To **City 5**: 10 units + - To **City 6**: 8 units + - To **City 7**: 10 units + - To **City 8**: 9 units + +- From **City 4** to other cities: + - To **City 0**: 6 units + - To **City 1**: 8 units + - To **City 2**: 9 units + - To **City 3**: 8 units + - To **City 5**: 10 units + - To **City 6**: 9 units + - To **City 7**: 7 units + - To **City 8**: 9 units + +- From **City 5** to other cities: + - To **City 0**: 10 units + - To **City 1**: 6 units + - To **City 2**: 7 units + - To **City 3**: 9 units + - To **City 4**: 8 units + - To **City 6**: 10 units + - To **City 7**: 7 units + - To **City 8**: 10 units + +- From **City 6** to other cities: + - To **City 0**: 10 units + - To **City 1**: 10 units + - To **City 2**: 10 units + - To **City 3**: 6 units + - To **City 4**: 9 units + - To **City 5**: 9 units + - To **City 7**: 7 units + - To **City 8**: 8 units + +- From **City 7** to other cities: + - To **City 0**: 6 units + - To **City 1**: 10 units + - To **City 2**: 9 units + - To **City 3**: 8 units + - To **City 4**: 8 units + - To **City 5**: 9 units + - To **City 6**: 8 units + - To **City 8**: 7 units + +- From **City 8** to other cities: + - To **City 0**: 10 units + - To **City 1**: 6 units + - To **City 2**: 8 units + - To **City 3**: 10 units + - To **City 4**: 7 units + - To **City 5**: 6 units + - To **City 6**: 9 units + - To **City 7**: 10 units + +#### Capacity Constraints: +Each route between cities has a maximum capacity limit, which cannot be exceeded: +- **City 0** to **City 1**: 40 units +- **City 0** to **City 2**: 56 units +- **City 0** to **City 3**: 41 units +- **City 0** to **City 4**: 49 units +- **City 0** to **City 5**: 50 units +- **City 0** to **City 6**: 36 units +- **City 0** to **City 7**: 59 units +- **City 0** to **City 8**: 55 units + +- **City 1** to **City 0**: 33 units +- **City 1** to **City 2**: 33 units +- **City 1** to **City 3**: 56 units +- **City 1** to **City 4**: 46 units +- **City 1** to **City 5**: 38 units +- **City 1** to **City 6**: 53 units +- **City 1** to **City 7**: 51 units +- **City 1** to **City 8**: 59 units + +- **City 2** to **City 0**: 59 units +- **City 2** to **City 1**: 52 units +- **City 2** to **City 3**: 41 units +- **City 2** to **City 4**: 38 units +- **City 2** to **City 5**: 36 units +- **City 2** to **City 6**: 45 units +- **City 2** to **City 7**: 32 units +- **City 2** to **City 8**: 41 units + +- **City 3** to **City 0**: 30 units +- **City 3** to **City 1**: 32 units +- **City 3** to **City 2**: 47 units +- **City 3** to **City 4**: 57 units +- **City 3** to **City 5**: 39 units +- **City 3** to **City 6**: 34 units +- **City 3** to **City 7**: 32 units +- **City 3** to **City 8**: 46 units + +- **City 4** to **City 0**: 55 units +- **City 4** to **City 1**: 40 units +- **City 4** to **City 2**: 55 units +- **City 4** to **City 3**: 41 units +- **City 4** to **City 5**: 59 units +- **City 4** to **City 6**: 51 units +- **City 4** to **City 7**: 35 units +- **City 4** to **City 8**: 53 units + +- **City 5** to **City 0**: 34 units +- **City 5** to **City 1**: 41 units +- **City 5** to **City 2**: 35 units +- **City 5** to **City 3**: 43 units +- **City 5** to **City 4**: 41 units +- **City 5** to **City 6**: 43 units +- **City 5** to **City 7**: 40 units +- **City 5** to **City 8**: 41 units + +- **City 6** to **City 0**: 34 units +- **City 6** to **City 1**: 32 units +- **City 6** to **City 2**: 47 units +- **City 6** to **City 3**: 43 units +- **City 6** to **City 4**: 46 units +- **City 6** to **City 5**: 34 units +- **City 6** to **City 7**: 60 units +- **City 6** to **City 8**: 59 units + +- **City 7** to **City 0**: 31 units +- **City 7** to **City 1**: 42 units +- **City 7** to **City 2**: 30 units +- **City 7** to **City 3**: 34 units +- **City 7** to **City 4**: 52 units +- **City 7** to **City 5**: 45 units +- **City 7** to **City 6**: 43 units +- **City 7** to **City 8**: 50 units + +- **City 8** to **City 0**: 49 units +- **City 8** to **City 1**: 31 units +- **City 8** to **City 2**: 53 units +- **City 8** to **City 3**: 46 units +- **City 8** to **City 4**: 42 units +- **City 8** to **City 5**: 36 units +- **City 8** to **City 6**: 30 units +- **City 8** to **City 7**: 41 units + +#### Objective: +Your task is to determine the optimal amount of resources to transport between cities to meet all demands while minimizing the total transportation cost. Ensure that the transportation does not exceed the capacity limits of any route and that all cities' resource requirements are satisfied. + +What is the minimum total cost to achieve this distribution?533No differencesNo differencesok
129The manufacturing facility produces seven distinct products (Jobs 0 through 6) using four specialized machines (Machine 0, Machine 1, Machine 2, and Machine 3). Each product must pass through all four machines in a specific sequence, and the goal is to schedule production to minimize the total completion time (makespan), ensuring all products are completed as quickly as possible. The objective is to minimize the sum of the start time of the last product on Machine 3 and the weighted completion times of specific jobs in the final position. Each product must be assigned to exactly one position in the production sequence on each machine, and each position on each machine can only be assigned to one product. The start time of a product on a machine cannot begin until the product has completed processing on the previous machine, and the start time of a product in a later position on a machine cannot begin until the product in the previous position has completed processing on that machine. Processing times for each product on each machine are as follows: Job 0 takes 10 units on Machine 0, 5 units on Machine 1, 5 units on Machine 2, and 7 units on Machine 3; Job 1 takes 6 units on Machine 0, 5 units on Machine 1, 10 units on Machine 2, and 9 units on Machine 3; Job 2 takes 9 units on Machine 0, 8 units on Machine 1, 6 units on Machine 2, and 9 units on Machine 3; Job 3 takes 6 units on Machine 0, 5 units on Machine 1, 9 units on Machine 2, and 6 units on Machine 3; Job 4 takes 5 units on Machine 0, 8 units on Machine 1, 5 units on Machine 2, and 5 units on Machine 3; Job 5 takes 9 units on Machine 0, 10 units on Machine 1, 5 units on Machine 2, and 10 units on Machine 3; Job 6 takes 7 units on Machine 0, 6 units on Machine 1, 10 units on Machine 2, and 5 units on Machine 3. The objective function includes weighted coefficients for specific jobs in the final position, such as 7 for Job 0 in position 6, 9 for Job 1 in position 6, 9 for Job 2 in position 6, 6 for Job 3 in position 6, 5 for Job 4 in position 6, 10 for Job 5 in position 6, and 5 for Job 6 in position 6. The constraints ensure that each job is assigned to exactly one position on each machine, each position on each machine is assigned to exactly one job, and the start times respect the precedence relationships between machines and positions. The variables include binary assignment variables for each job and position on each machine, and continuous start time variables for each position on each machine. The bounds ensure that all start times are non-negative, and the binary variables are restricted to 0 or 1.72The manufacturing facility produces seven distinct products (Jobs 0 through 6) using four specialized machines (Machine 0, Machine 1, Machine 2, and Machine 3). Each product must pass through all four machines in a specific sequence, and the goal is to schedule production to minimize the total completion time (makespan), ensuring all products are completed as quickly as possible. The objective is to minimize the sum of the start time of the last product on Machine 3 and the weighted completion times of specific jobs in the final position. Each product must be assigned to exactly one position in the production sequence on each machine, and each position on each machine can only be assigned to one product. The start time of a product on a machine cannot begin until the product has completed processing on the previous machine, and the start time of a product in a later position on a machine cannot begin until the product in the previous position has completed processing on that machine. Processing times for each product on each machine are as follows: Job 0 takes 10 units on Machine 0, 5 units on Machine 1, 5 units on Machine 2, and 7 units on Machine 3; Job 1 takes 6 units on Machine 0, 5 units on Machine 1, 10 units on Machine 2, and 9 units on Machine 3; Job 2 takes 9 units on Machine 0, 8 units on Machine 1, 6 units on Machine 2, and 9 units on Machine 3; Job 3 takes 6 units on Machine 0, 5 units on Machine 1, 9 units on Machine 2, and 6 units on Machine 3; Job 4 takes 5 units on Machine 0, 8 units on Machine 1, 5 units on Machine 2, and 5 units on Machine 3; Job 5 takes 9 units on Machine 0, 10 units on Machine 1, 5 units on Machine 2, and 10 units on Machine 3; Job 6 takes 7 units on Machine 0, 6 units on Machine 1, 10 units on Machine 2, and 5 units on Machine 3. The objective function includes weighted coefficients for specific jobs in the final position, such as 7 for Job 0 in position 6, 9 for Job 1 in position 6, 9 for Job 2 in position 6, 6 for Job 3 in position 6, 5 for Job 4 in position 6, 10 for Job 5 in position 6, and 5 for Job 6 in position 6. The constraints ensure that each job is assigned to exactly one position on each machine, each position on each machine is assigned to exactly one job, and the start times respect the precedence relationships between machines and positions. The variables include binary assignment variables for each job and position on each machine, and continuous start time variables for each position on each machine. The bounds ensure that all start times are non-negative, and the binary variables are restricted to 0 or 1.72No differencesNo differencesok
130The goal is to minimize the total travel distance for a delivery route across six cities: City 0, City 1, City 2, City 3, City 4, and City 5. Each city must be visited exactly once, and the route must form a complete loop without any smaller loops (subtours). The distances between cities are as follows: from City 0 to City 1 (318 units), City 2 (278 units), City 3 (304 units), City 4 (287 units), and City 5 (277 units); from City 1 to City 0 (285 units), City 2 (277 units), City 3 (322 units), City 4 (277 units), and City 5 (306 units); from City 2 to City 0 (275 units), City 1 (302 units), City 3 (299 units), City 4 (277 units), and City 5 (291 units); from City 3 to City 0 (282 units), City 1 (292 units), City 2 (276 units), City 4 (283 units), and City 5 (304 units); from City 4 to City 0 (277 units), City 1 (301 units), City 2 (291 units), City 3 (306 units), and City 5 (325 units); from City 5 to City 0 (325 units), City 1 (319 units), City 2 (306 units), City 3 (279 units), and City 4 (275 units). + +Each city must have exactly one incoming and one outgoing route. To prevent subtours, each city is assigned a position number: City 0 is fixed at position 0, and the position numbers for the other cities must be between 1 and 5. If a route is taken from City i to City j, the position number of City j must be greater than the position number of City i. The decision to travel from one city to another is represented by binary variables, where 1 indicates the route is taken and 0 indicates it is not. The objective is to determine the optimal sequence of cities to visit, ensuring each city is visited exactly once, the route forms a complete loop without subtours, and the total distance traveled is minimized.1679The goal is to minimize the total travel distance for a delivery route across six cities: City 0, City 1, City 2, City 3, City 4, and City 5. Each city must be visited exactly once, and the route must form a complete loop without any smaller loops (subtours), and finally return to the origin city. The distances between cities are as follows: from City 0 to City 1 (318 units), City 2 (278 units), City 3 (304 units), City 4 (287 units), and City 5 (277 units); from City 1 to City 0 (285 units), City 2 (277 units), City 3 (322 units), City 4 (277 units), and City 5 (306 units); from City 2 to City 0 (275 units), City 1 (302 units), City 3 (299 units), City 4 (277 units), and City 5 (291 units); from City 3 to City 0 (282 units), City 1 (292 units), City 2 (276 units), City 4 (283 units), and City 5 (304 units); from City 4 to City 0 (277 units), City 1 (301 units), City 2 (291 units), City 3 (306 units), and City 5 (325 units); from City 5 to City 0 (325 units), City 1 (319 units), City 2 (306 units), City 3 (279 units), and City 4 (275 units). + +The task is to find the shortest possible route that visits each city exactly once, and return to the origin city. What is the minimal total distance travelled?1679The goal is to minimize the total travel distance for a delivery route across six cities: City 0, City 1, City 2, City 3, City 4, and City 5. Each city must be visited exactly once, and the route must form a complete loop without any smaller loops (subtours). (subtours), and finally return to the origin city. The distances between cities are as follows: from City 0 to City 1 (318 units), City 2 (278 units), City 3 (304 units), City 4 (287 units), and City 5 (277 units); from City 1 to City 0 (285 units), City 2 (277 units), City 3 (322 units), City 4 (277 units), and City 5 (306 units); from City 2 to City 0 (275 units), City 1 (302 units), City 3 (299 units), City 4 (277 units), and City 5 (291 units); from City 3 to City 0 (282 units), City 1 (292 units), City 2 (276 units), City 4 (283 units), and City 5 (304 units); from City 4 to City 0 (277 units), City 1 (301 units), City 2 (291 units), City 3 (306 units), and City 5 (325 units); from City 5 to City 0 (325 units), City 1 (319 units), City 2 (306 units), City 3 (279 units), and City 4 (275 units). + + Each city must have exactly one incoming and one outgoing route. To prevent subtours, each city is assigned a position number: City 0 is fixed at position 0, and the position numbers for the other cities must be between 1 and 5. If a route is taken from City i to City j, the position number of City j must be greater than the position number of City i. The decision to travel from one city to another is represented by binary variables, where 1 indicates the route is taken and 0 indicates it is not. The objective task is to determine find the optimal sequence of cities to visit, ensuring shortest possible route that visits each city is visited exactly once, the route forms a complete loop without subtours, and return to the origin city. What is the minimal total distance traveled is minimized. travelled?No differencesWe removed information that was not the description of the problem, but describing the variables/constraints of a possible model. Also, this additional information could lead to a wrong model (MTZ constraint on the return to the first node)ambiguous
131Cell Tower Deployment for Regional Coverage + +You are tasked with optimizing the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 33 distinct areas, each with a specific population size that needs to be served. Your goal is to determine which cell towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + +#### Key Decisions: +1. **Which cell towers to build**: There are 29 potential cell tower locations, each with a unique construction cost. You must decide whether to build each tower (a binary decision: build or not build). +2. **Which areas to cover**: Each of the 33 areas can be either covered or not covered by the deployed cell towers. Coverage for an area is only possible if at least one of the cell towers that can serve it is built. + +#### Objective: +Maximize the total population covered by the deployed cell towers. The population sizes for the 33 areas are as follows: +- Area 0: 3,522 people +- Area 1: 2,492 people +- Area 2: 2,865 people +- Area 3: 2,596 people +- Area 4: 2,194 people +- Area 5: 4,001 people +- Area 6: 3,027 people +- Area 7: 2,581 people +- Area 8: 4,848 people +- Area 9: 2,814 people +- Area 10: 3,543 people +- Area 11: 2,234 people +- Area 12: 2,960 people +- Area 13: 4,297 people +- Area 14: 2,792 people +- Area 15: 4,520 people +- Area 16: 2,685 people +- Area 17: 2,022 people +- Area 18: 4,858 people +- Area 19: 4,814 people +- Area 20: 3,432 people +- Area 21: 3,730 people +- Area 22: 2,526 people +- Area 23: 3,277 people +- Area 24: 4,363 people +- Area 25: 2,622 people +- Area 26: 3,740 people +- Area 27: 4,594 people +- Area 28: 3,193 people +- Area 29: 4,275 people +- Area 30: 4,407 people +- Area 31: 3,809 people +- Area 32: 2,163 people + +#### Constraints: +1. **Coverage Requirements**: For an area to be considered covered, at least one of the cell towers that can serve it must be built. The specific combinations of cell towers that can cover each area are provided in the data. For example: + - Area 0 can be covered if any of the following towers are built: Tower 3, Tower 5, Tower 7, Tower 10, Tower 11, Tower 13, Tower 15, Tower 17, Tower 19, Tower 22, Tower 23, Tower 26, Tower 27, or Tower 28. + - Similar constraints apply to all 33 areas, with each area having a unique set of towers that can provide coverage. + +2. **Budget Constraint**: The total cost of building the selected cell towers must not exceed the available budget of 3,059 units. The construction costs for the 29 towers are as follows: + - Tower 0: 303 units + - Tower 1: 378 units + - Tower 2: 327 units + - Tower 3: 444 units + - Tower 4: 391 units + - Tower 5: 357 units + - Tower 6: 444 units + - Tower 7: 244 units + - Tower 8: 433 units + - Tower 9: 407 units + - Tower 10: 305 units + - Tower 11: 296 units + - Tower 12: 315 units + - Tower 13: 302 units + - Tower 14: 495 units + - Tower 15: 240 units + - Tower 16: 379 units + - Tower 17: 211 units + - Tower 18: 388 units + - Tower 19: 220 units + - Tower 20: 408 units + - Tower 21: 430 units + - Tower 22: 327 units + - Tower 23: 470 units + - Tower 24: 381 units + - Tower 25: 476 units + - Tower 26: 215 units + - Tower 27: 239 units + - Tower 28: 372 units + +3. **Binary Decisions**: Each cell tower can either be built (1) or not built (0), and each area can either be covered (1) or not covered (0). + +#### Summary: +Your task is to select a combination of cell towers to build, ensuring that the total cost does not exceed 3,059 units, while maximizing the total population covered across the 33 areas. The coverage for each area depends on the specific towers built, and the population sizes and tower costs are provided in detail. The goal is to make strategic decisions about tower deployment to achieve the highest possible population coverage within the budget.111796Cell Tower Deployment for Regional Coverage + + You are tasked with optimizing the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 33 distinct areas, each with a specific population size that needs to be served. Your goal is to determine which cell towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + + #### Key Decisions: + 1. **Which cell towers to build**: There are 29 potential cell tower locations, each with a unique construction cost. You must decide whether to build each tower (a binary decision: build or not build). + 2. **Which areas to cover**: Each of the 33 areas can be either covered or not covered by the deployed cell towers. Coverage for an area is only possible if at least one of the cell towers that can serve it is built. + + #### Objective: + Maximize the total population covered by the deployed cell towers. The population sizes for the 33 areas are as follows: + - Area 0: 3,522 people +- Area 1: 2,492 people +- Area 2: 2,865 people +- Area 3: 2,596 people +- Area 4: 2,194 people +- Area 5: 4,001 people +- Area 6: 3,027 people +- Area 7: 2,581 people +- Area 8: 4,848 people +- Area 9: 2,814 people +- Area 10: 3,543 people +- Area 11: 2,234 people +- Area 12: 2,960 people +- Area 13: 4,297 people +- Area 14: 2,792 people +- Area 15: 4,520 people +- Area 16: 2,685 people +- Area 17: 2,022 people +- Area 18: 4,858 people +- Area 19: 4,814 people +- Area 20: 3,432 people +- Area 21: 3,730 people +- Area 22: 2,526 people +- Area 23: 3,277 people +- Area 24: 4,363 people +- Area 25: 2,622 people +- Area 26: 3,740 people +- Area 27: 4,594 people +- Area 28: 3,193 people +- Area 29: 4,275 people +- Area 30: 4,407 people +- Area 31: 3,809 people +- Area 32: 2,163 people + + #### Constraints: + 1. **Coverage Requirements**: For an area to be considered covered, at least one of the cell towers that can serve it must be built. The specific combinations of cell towers that can cover each area are: + - Area 0 can be covered if any of the following towers are built: Tower 3, Tower 5, Tower 7, Tower 10, Tower 11, Tower 13, Tower 15, Tower 17, Tower 19, Tower 22, Tower 23, Tower 26, Tower 27, Tower 28. + - Area 1 can be covered if any of the following towers are built: Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 7, Tower 8, Tower 13, Tower 16, Tower 17, Tower 18, Tower 23, Tower 24, Tower 27, Tower 28. + - Area 2 can be covered if any of the following towers are built: Tower 0, Tower 6, Tower 7, Tower 13, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 24, Tower 28. + - Area 3 can be covered if any of the following towers are built: Tower 0, Tower 4, Tower 5, Tower 6, Tower 8, Tower 10, Tower 13, Tower 18, Tower 20, Tower 22, Tower 24, Tower 25, Tower 27. + - Area 4 can be covered if any of the following towers are built: Tower 1, Tower 2, Tower 3, Tower 8, Tower 11, Tower 12, Tower 14, Tower 17, Tower 19, Tower 25. + - Area 5 can be covered if any of the following towers are built: Tower 2, Tower 6, Tower 9, Tower 11, Tower 12, Tower 17, Tower 18, Tower 19, Tower 20, Tower 27. + - Area 6 can be covered if any of the following towers are built: Tower 2, Tower 3, Tower 7, Tower 8, Tower 9, Tower 12, Tower 14, Tower 21, Tower 24, Tower 27. + - Area 7 can be covered if any of the following towers are built: Tower 2, Tower 5, Tower 6, Tower 7, Tower 8, Tower 11, Tower 16, Tower 17, Tower 18, Tower 20, Tower 21, Tower 25, Tower 26, Tower 27, Tower 28. + - Area 8 can be covered if any of the following towers are built: Tower 1, Tower 7, Tower 8, Tower 10, Tower 12, Tower 17, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24, Tower 26, Tower 28. + - Area 9 can be covered if any of the following towers are built: Tower 6, Tower 10, Tower 12, Tower 14, Tower 15, Tower 18, Tower 20, Tower 22, Tower 23, Tower 28. + - Area 10 can be covered if any of the following towers are built: Tower 4, Tower 7, Tower 8, Tower 12, Tower 13, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 28. + - Area 11 can be covered if any of the following towers are built: Tower 1, Tower 2, Tower 3, Tower 4, Tower 5, Tower 7, Tower 13, Tower 15, Tower 16, Tower 20, Tower 24, Tower 27. + - Area 12 can be covered if any of the following towers are built: Tower 0, Tower 3, Tower 8, Tower 10, Tower 12, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 21, Tower 23, Tower 24. + - Area 13 can be covered if any of the following towers are built: Tower 0, Tower 3, Tower 5, Tower 8, Tower 9, Tower 13, Tower 14, Tower 16, Tower 21, Tower 23, Tower 24, Tower 27. + - Area 14 can be covered if any of the following towers are built: Tower 0, Tower 3, Tower 4, Tower 5, Tower 6, Tower 10, Tower 11, Tower 14, Tower 15, Tower 16, Tower 17, Tower 19, Tower 20, Tower 27, Tower 28. + - Area 15 can be covered if any of the following towers are built: Tower 1, Tower 2, Tower 7, Tower 9, Tower 15, Tower 18, Tower 19, Tower 20, Tower 24, Tower 25, Tower 26, Tower 28. + - Area 16 can be covered if any of the following towers are built: Tower 4, Tower 5, Tower 6, Tower 8, Tower 9, Tower 13, Tower 15, Tower 16, Tower 17, Tower 19, Tower 21, Tower 23, Tower 24, Tower 27. + - Area 17 can be covered if any of the following towers are built: Tower 0, Tower 2, Tower 3, Tower 7, Tower 10, Tower 11, Tower 14, Tower 16, Tower 20, Tower 21, Tower 25, Tower 27, Tower 28. + - Area 18 can be covered if any of the following towers are built: Tower 0, Tower 1, Tower 2, Tower 7, Tower 8, Tower 10, Tower 16, Tower 20, Tower 21, Tower 22, Tower 23, Tower 25, Tower 26, Tower 28. + - Area 19 can be covered if any of the following towers are built: Tower 3, Tower 4, Tower 6, Tower 7, Tower 10, Tower 13, Tower 15, Tower 16, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 27, Tower 28. + - Area 20 can be covered if any of the following towers are built: Tower 1, Tower 3, Tower 10, Tower 11, Tower 12, Tower 13, Tower 14, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24, Tower 27. + - Area 21 can be covered if any of the following towers are built: Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 13, Tower 14, Tower 17, Tower 24, Tower 27. + - Area 22 can be covered if any of the following towers are built: Tower 0, Tower 1, Tower 2, Tower 3, Tower 14, Tower 17, Tower 20, Tower 24, Tower 25, Tower 27, Tower 28. + - Area 23 can be covered if any of the following towers are built: Tower 0, Tower 1, Tower 5, Tower 6, Tower 12, Tower 13, Tower 15, Tower 20, Tower 23, Tower 26, Tower 28. + - Area 24 can be covered if any of the following towers are built: Tower 4, Tower 6, Tower 9, Tower 13, Tower 14, Tower 15, Tower 17, Tower 21, Tower 22, Tower 25, Tower 27, Tower 28. + - Area 25 can be covered if any of the following towers are built: Tower 1, Tower 10, Tower 15, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 26, Tower 27, Tower 28. + - Area 26 can be covered if any of the following towers are built: Tower 1, Tower 2, Tower 3, Tower 5, Tower 7, Tower 12, Tower 16, Tower 18, Tower 19, Tower 21, Tower 23, Tower 26, Tower 27, Tower 28. + - Area 27 can be covered if any of the following towers are built: Tower 1, Tower 2, Tower 6, Tower 7, Tower 8, Tower 10, Tower 13, Tower 16, Tower 18, Tower 19, Tower 21, Tower 22, Tower 26, Tower 28. + - Area 28 can be covered if any of the following towers are built: Tower 0, Tower 2, Tower 3, Tower 4, Tower 9, Tower 10, Tower 12, Tower 14, Tower 18, Tower 20, Tower 21, Tower 23. + - Area 29 can be covered if any of the following towers are built: Tower 2, Tower 4, Tower 6, Tower 7, Tower 8, Tower 9, Tower 11, Tower 16, Tower 17, Tower 23. + - Area 30 can be covered if any of the following towers are built: Tower 0, Tower 9, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 24, Tower 26, Tower 27. + - Area 31 can be covered if any of the following towers are built: Tower 3, Tower 4, Tower 6, Tower 8, Tower 9, Tower 10, Tower 15, Tower 17, Tower 18, Tower 20, Tower 23, Tower 25, Tower 26, Tower 27, Tower 28. + - Area 32 can be covered if any of the following towers are built: Tower 0, Tower 1, Tower 2, Tower 4, Tower 5, Tower 8, Tower 10, Tower 13, Tower 20, Tower 22, Tower 26, Tower 27, Tower 28. + + 2. **Budget Constraint**: The total cost of building the selected cell towers must not exceed the available budget of 1,059 units. The construction costs for the 29 towers are as follows: + - Tower 0: 303 units + - Tower 1: 378 units + - Tower 2: 327 units + - Tower 3: 444 units + - Tower 4: 391 units + - Tower 5: 357 units + - Tower 6: 444 units + - Tower 7: 244 units + - Tower 8: 433 units + - Tower 9: 407 units + - Tower 10: 305 units + - Tower 11: 296 units + - Tower 12: 315 units + - Tower 13: 302 units + - Tower 14: 495 units + - Tower 15: 240 units + - Tower 16: 379 units + - Tower 17: 211 units + - Tower 18: 388 units + - Tower 19: 220 units + - Tower 20: 408 units + - Tower 21: 430 units + - Tower 22: 327 units + - Tower 23: 470 units + - Tower 24: 381 units + - Tower 25: 476 units + - Tower 26: 215 units + - Tower 27: 239 units + - Tower 28: 372 units + + 3. **Binary Decisions**: Each cell tower can either be built (1) or not built (0), and each area can either be covered (1) or not covered (0). + + #### Summary: + Your task is to select a combination of cell towers to build, ensuring that the total cost does not exceed 1,059 units, while maximizing the total population covered across the 33 areas. The coverage for each area depends on the specific towers built, and the population sizes and tower costs are provided in detail. The goal is to make strategic decisions about tower deployment to achieve the highest possible population coverage within the budget.111796Cell Tower Deployment for Regional Coverage + + You are tasked with optimizing the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 33 distinct areas, each with a specific population size that needs to be served. Your goal is to determine which cell towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + + #### Key Decisions: + 1. **Which cell towers to build**: There are 29 potential cell tower locations, each with a unique construction cost. You must decide whether to build each tower (a binary decision: build or not build). + 2. **Which areas to cover**: Each of the 33 areas can be either covered or not covered by the deployed cell towers. Coverage for an area is only possible if at least one of the cell towers that can serve it is built. + + #### Objective: + Maximize the total population covered by the deployed cell towers. The population sizes for the 33 areas are as follows: + - Area 0: 3,522 people +- Area 1: 2,492 people +- Area 2: 2,865 people +- Area 3: 2,596 people +- Area 31: 3,809 people +- Area 32: 2,163 people + + #### Constraints: + 1. **Coverage Requirements**: For an area to be considered covered, at least one of the cell towers that can serve it must be built. The specific combinations of cell towers that can cover each area are provided in the data. For example: are: + - Area 0 can be covered if any of the following towers are built: Tower 3, Tower 5, Tower 7, Tower 10, Tower 11, Tower 13, Tower 15, Tower 17, Tower 19, Tower 22, Tower 23, Tower 26, Tower 27, or Tower 28. + - Similar constraints apply to all 33 areas, with each area having a unique set Area 1 can be covered if any of the following towers that are built: Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 7, Tower 8, Tower 13, Tower 16, Tower 17, Tower 18, Tower 23, Tower 24, Tower 27, Tower 28. + - Area 2 can be covered if any of the following towers are built: Tower 0, Tower 6, Tower 7, Tower 13, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 24, Tower 28. + - Area 3 can be covered if any of the following towers are built: Tower 0, Tower 4, Tower 5, Tower 6, Tower 8, Tower 10, Tower 13, Tower 18, Tower 20, Tower 22, Tower 24, Tower 25, Tower 27. + - Area 4 can be covered if any of the following towers are built: Tower 1, Tower 2, Tower 3, Tower 8, Tower 11, Tower 12, Tower 14, Tower 17, Tower 19, Tower 25. + - Area 5 can provide coverage. be covered if any of the following towers are built: Tower 2, Tower 6, Tower 9, Tower 11, Tower 12, Tower 17, Tower 18, Tower 19, Tower 20, Tower 27. + - Area 6 can be covered if any of the following towers are built: Tower 2, Tower 3, Tower 7, Tower 8, Tower 9, Tower 12, Tower 14, Tower 21, Tower 24, Tower 27. + - Area 7 can be covered if any of the following towers are built: Tower 2, Tower 5, Tower 6, Tower 7, Tower 8, Tower 11, Tower 16, Tower 17, Tower 18, Tower 20, Tower 21, Tower 25, Tower 26, Tower 27, Tower 28. + - Area 8 can be covered if any of the following towers are built: Tower 1, Tower 7, Tower 8, Tower 10, Tower 12, Tower 17, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24, Tower 26, Tower 28. + - Area 9 can be covered if any of the following towers are built: Tower 6, Tower 10, Tower 12, Tower 14, Tower 15, Tower 18, Tower 20, Tower 22, Tower 23, Tower 28. + - Area 10 can be covered if any of the following towers are built: Tower 4, Tower 7, Tower 8, Tower 12, Tower 13, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 28. + - Area 11 can be covered if any of the following towers are built: Tower 1, Tower 2, Tower 3, Tower 4, Tower 5, Tower 7, Tower 13, Tower 15, Tower 16, Tower 20, Tower 24, Tower 27. + - Area 12 can be covered if any of the following towers are built: Tower 0, Tower 3, Tower 8, Tower 10, Tower 12, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 21, Tower 23, Tower 24. + - Area 13 can be covered if any of the following towers are built: Tower 0, Tower 3, Tower 5, Tower 8, Tower 9, Tower 13, Tower 14, Tower 16, Tower 21, Tower 23, Tower 24, Tower 27. + - Area 14 can be covered if any of the following towers are built: Tower 0, Tower 3, Tower 4, Tower 5, Tower 6, Tower 10, Tower 11, Tower 14, Tower 15, Tower 16, Tower 17, Tower 19, Tower 20, Tower 27, Tower 28. + - Area 15 can be covered if any of the following towers are built: Tower 1, Tower 2, Tower 7, Tower 9, Tower 15, Tower 18, Tower 19, Tower 20, Tower 24, Tower 25, Tower 26, Tower 28. + - Area 16 can be covered if any of the following towers are built: Tower 4, Tower 5, Tower 6, Tower 8, Tower 9, Tower 13, Tower 15, Tower 16, Tower 17, Tower 19, Tower 21, Tower 23, Tower 24, Tower 27. + - Area 17 can be covered if any of the following towers are built: Tower 0, Tower 2, Tower 3, Tower 7, Tower 10, Tower 11, Tower 14, Tower 16, Tower 20, Tower 21, Tower 25, Tower 27, Tower 28. + - Area 18 can be covered if any of the following towers are built: Tower 0, Tower 1, Tower 2, Tower 7, Tower 8, Tower 10, Tower 16, Tower 20, Tower 21, Tower 22, Tower 23, Tower 25, Tower 26, Tower 28. + - Area 19 can be covered if any of the following towers are built: Tower 3, Tower 4, Tower 6, Tower 7, Tower 10, Tower 13, Tower 15, Tower 16, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 27, Tower 28. + - Area 20 can be covered if any of the following towers are built: Tower 1, Tower 3, Tower 10, Tower 11, Tower 12, Tower 13, Tower 14, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24, Tower 27. + - Area 21 can be covered if any of the following towers are built: Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 13, Tower 14, Tower 17, Tower 24, Tower 27. + - Area 22 can be covered if any of the following towers are built: Tower 0, Tower 1, Tower 2, Tower 3, Tower 14, Tower 17, Tower 20, Tower 24, Tower 25, Tower 27, Tower 28. + - Area 23 can be covered if any of the following towers are built: Tower 0, Tower 1, Tower 5, Tower 6, Tower 12, Tower 13, Tower 15, Tower 20, Tower 23, Tower 26, Tower 28. + - Area 24 can be covered if any of the following towers are built: Tower 4, Tower 6, Tower 9, Tower 13, Tower 14, Tower 15, Tower 17, Tower 21, Tower 22, Tower 25, Tower 27, Tower 28. + - Area 25 can be covered if any of the following towers are built: Tower 1, Tower 10, Tower 15, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 26, Tower 27, Tower 28. + - Area 26 can be covered if any of the following towers are built: Tower 1, Tower 2, Tower 3, Tower 5, Tower 7, Tower 12, Tower 16, Tower 18, Tower 19, Tower 21, Tower 23, Tower 26, Tower 27, Tower 28. + - Area 27 can be covered if any of the following towers are built: Tower 1, Tower 2, Tower 6, Tower 7, Tower 8, Tower 10, Tower 13, Tower 16, Tower 18, Tower 19, Tower 21, Tower 22, Tower 26, Tower 28. + - Area 28 can be covered if any of the following towers are built: Tower 0, Tower 2, Tower 3, Tower 4, Tower 9, Tower 10, Tower 12, Tower 14, Tower 18, Tower 20, Tower 21, Tower 23. + - Area 29 can be covered if any of the following towers are built: Tower 2, Tower 4, Tower 6, Tower 7, Tower 8, Tower 9, Tower 11, Tower 16, Tower 17, Tower 23. + - Area 30 can be covered if any of the following towers are built: Tower 0, Tower 9, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 24, Tower 26, Tower 27. + - Area 31 can be covered if any of the following towers are built: Tower 3, Tower 4, Tower 6, Tower 8, Tower 9, Tower 10, Tower 15, Tower 17, Tower 18, Tower 20, Tower 23, Tower 25, Tower 26, Tower 27, Tower 28. + - Area 32 can be covered if any of the following towers are built: Tower 0, Tower 1, Tower 2, Tower 4, Tower 5, Tower 8, Tower 10, Tower 13, Tower 20, Tower 22, Tower 26, Tower 27, Tower 28. + + 2. **Budget Constraint**: The total cost of building the selected cell towers must not exceed the available budget of 3,059 1,059 units. The construction costs for the 29 towers are as follows: + - Tower 0: 303 units + - Tower 1: 378 units + - Tower 2: 327 units + - Tower 3: 444 units + - Tower 4: 391 units + - Tower 5: 357 units + - Tower 6: 444 units + - Tower 7: 244 units + - Tower 8: 433 units + - Tower 9: 407 units + - Tower 10: 305 units + - Tower 11: 296 units + - Tower 12: 315 units + - Tower 13: 302 units + - Tower 14: 495 units + - Tower 15: 240 units + - Tower 16: 379 units + - Tower 17: 211 units + - Tower 18: 388 units + - Tower 19: 220 units + - Tower 20: 408 units + - Tower 21: 430 units + - Tower 22: 327 units + - Tower 23: 470 units + - Tower 24: 381 units + - Tower 25: 476 units + - Tower 26: 215 units + - Tower 27: 239 units + - Tower 28: 372 units + + 3. **Binary Decisions**: Each cell tower can either be built (1) or not built (0), and each area can either be covered (1) or not covered (0). + + #### Summary: + Your task is to select a combination of cell towers to build, ensuring that the total cost does not exceed 3,059 1,059 units, while maximizing the total population covered across the 33 areas. The coverage for each area depends on the specific towers built, and the population sizes and tower costs are provided in detail. The goal is to make strategic decisions about tower deployment to achieve the highest possible population coverage within the budget.No differencesMissing data about tower coveragemissing data
132Imagine you are managing the placement of emergency response centers across five key locations in a region. Each location has the potential to host a center, but due to budget constraints, you can only establish **three centers** in total. The goal is to maximize the overall effectiveness of these centers by ensuring they are strategically placed to cover as much critical area as possible. + +Each pair of locations has a predefined **effectiveness score** that represents how well a center at one location can support or complement a center at another location. These scores are as follows: + +- **Location 0 and Location 1**: 18 +- **Location 0 and Location 2**: 20 +- **Location 0 and Location 3**: 22 +- **Location 0 and Location 4**: 19 +- **Location 1 and Location 0**: 20 +- **Location 1 and Location 2**: 15 +- **Location 1 and Location 3**: 22 +- **Location 1 and Location 4**: 16 +- **Location 2 and Location 0**: 23 +- **Location 2 and Location 1**: 23 +- **Location 2 and Location 3**: 17 +- **Location 2 and Location 4**: 22 +- **Location 3 and Location 0**: 20 +- **Location 3 and Location 1**: 19 +- **Location 3 and Location 2**: 23 +- **Location 3 and Location 4**: 20 +- **Location 4 and Location 0**: 24 +- **Location 4 and Location 1**: 16 +- **Location 4 and Location 2**: 24 +- **Location 4 and Location 3**: 24 + +Your task is to select **three locations** out of the five to establish the centers. The effectiveness of the selected centers is determined by the sum of the pairwise effectiveness scores between the chosen locations. For example, if you select Locations 0, 1, and 2, the total effectiveness would be the sum of the scores for pairs (0,1), (0,2), and (1,2). + +However, there are specific rules governing the selection: +1. A pair of locations can only contribute to the total effectiveness if **both locations in the pair are selected** as centers. +2. If only one location in a pair is selected, that pair does not contribute to the effectiveness score. +3. The total number of centers must be exactly three. + +Your objective is to determine which three locations to select to **maximize the total effectiveness** of the emergency response centers, ensuring all constraints are met.132Imagine you are managing the placement of emergency response centers across five key locations in a region. Each location has the potential to host a center, but due to budget constraints, you can only establish **three centers** in total. The goal is to maximize the overall effectiveness of these centers by ensuring they are strategically placed to cover as much critical area as possible. + +Each pair of locations has a predefined **effectiveness score** that represents how well a center at one location can support or complement a center at another location. These scores are as follows: + +- **Location 0 and Location 1**: 18 +- **Location 0 and Location 2**: 20 +- **Location 0 and Location 3**: 22 +- **Location 0 and Location 4**: 19 +- **Location 1 and Location 2**: 15 +- **Location 1 and Location 3**: 22 +- **Location 1 and Location 4**: 16 +- **Location 2 and Location 3**: 17 +- **Location 2 and Location 4**: 22 +- **Location 3 and Location 4**: 20 + +Your task is to select **three locations** out of the five to establish the centers. The effectiveness of the selected centers is determined by the sum of the pairwise effectiveness scores between the chosen locations. For example, if you select Locations 0, 1, and 2, the total effectiveness would be the sum of the scores for pairs (0,1), (0,2), and (1,2). + +However, there are specific rules governing the selection: +1. A pair of locations can only contribute to the total effectiveness if **both locations in the pair are selected** as centers. +2. If only one location in a pair is selected, that pair does not contribute to the effectiveness score. +3. The total number of centers must be exactly three. + +Your objective is to determine which three locations to select to **maximize the total effectiveness** of the emergency response centers, ensuring all constraints are met.62- **Location 0 and Location 2**: 20 +- **Location 0 and Location 3**: 22 +- **Location 0 and Location 4**: 19 + - **Location 1 and Location 0**: 20 +- **Location 1 and Location 2**: 15 +- **Location 1 and Location 3**: 22 +- **Location 1 and Location 4**: 16 + - **Location 2 and Location 0**: 23 + - **Location 2 and Location 1**: 23 +- **Location 2 and Location 3**: 17 +- **Location 2 and Location 4**: 22 + - **Location 3 and Location 0**: 20 + - **Location 3 and Location 1**: 19 + - **Location 3 and Location 2**: 23 +- **Location 3 and Location 4**: 20 + - **Location 4 and Location 0**: 24 + - **Location 4 and Location 1**: 16 + - **Location 4 and Location 2**: 24 + - **Location 4 and Location 3**: 24 + +Your task is to select **three locations** out of the five to establish the centers. The effectiveness of the selected centers is determined by the sum of the pairwise effectiveness scores between the chosen locations. For example, if you select Locations 0, 1, and 2, the total effectiveness would be the sum of the scores for pairs (0,1), (0,2), and (1,2).132 62Original statement had different dispersions between (0,1) and (1,0), which does not make sense in this context. We just kept one of themambiguous
133The goal is to determine the optimal allocation of resources for three products, Product A, Product B, and Product C, to effectively manage a logistics network. The decision variables include the quantities of each product, represented by \( x[0] \), \( x[1] \), and \( x[2] \), as well as auxiliary variables \( y_0[0] \) to \( y_0[3] \), \( y_1[0] \) to \( y_1[3] \), \( y_2[0] \) to \( y_2[3] \), and logistical thresholds \( t_0[0] \), \( t_1[0] \), and \( t_2[0] \). All variables are free to take any real value. + +The objective is to minimize the total cost, which is calculated as the sum of the following contributions: -1.109620416617511 times Product A, -0.6560359731414873 times Product B, and -0.7019449433705992 times Product C. This objective reflects the cost structure of the logistics network. + +Operational constraints must be satisfied, including three linear equality constraints that define exact relationships between the products. These constraints are: 0.973138297335475 times Product A plus 0.5753515721139748 times Product B minus 0.3345391015917421 times Product C equals -0.1751595502819031; 1.004855798779082 times Product A plus 1.155166374661675 times Product B minus 0.6908735867360416 times Product C equals -0.7437001018712311; and -2.110665713909855 times Product A plus 0.0465859108136975 times Product B plus 1.404349170472437 times Product C equals -2.786300004238841. + +Additionally, there are 14 linear constraints that ensure the allocation of resources adheres to specific conditions. These constraints involve combinations of the product quantities and the auxiliary variables \( y_0 \), \( y_1 \), \( y_2 \), and the logistical thresholds \( t_0 \), \( t_1 \), and \( t_2 \). For example, one constraint states that 0.0366410572471413 times Product A plus 0.0568820534300137 times Product B minus 0.000079998 times Product C plus \( y_0[0] \) equals 0.0535813009430022. Similar constraints are defined for the remaining auxiliary variables and thresholds. + +Furthermore, three second-order cone constraints ensure that the logistical thresholds satisfy specific quadratic relationships. These constraints are: the sum of the squares of \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), and \( y_0[3] \) minus the square of \( t_0[0] \) must be less than or equal to zero; the sum of the squares of \( y_1[0] \), \( y_1[1] \), \( y_1[2] \), and \( y_1[3] \) minus the square of \( t_1[0] \) must be less than or equal to zero; and the sum of the squares of \( y_2[0] \), \( y_2[1] \), \( y_2[2] \), and \( y_2[3] \) minus the square of \( t_2[0] \) must be less than or equal to zero. + +The task is to determine the optimal production and distribution levels for Products A, B, and C, along with the associated intermediate metrics and logistical thresholds, to minimize the total cost while satisfying all operational constraints.1.5708319915113720The goal is to determine the optimal allocation of resources for three products, Product A, Product B, and Product C, to effectively manage a logistics network. The decision variables include the quantities of each product, represented by \( x[0 \), \( x[1] \), and \( x[2] \), as well as auxiliary variables \( y_0[0] \) to \( y_0[3] \), \( y_1[0] \) to \( y_1[3] \), \( y_2[0] \) to \( y_2[3] \), and logistical thresholds \( t_0[0] \), \( t_1[0] \), and \( t_2[0] \). All variables are free to take any real value.] + + The objective is to minimize the total cost, which is calculated as the sum of the following contributions: -1.109620416617511 times Product A, -0.6560359731414873 times Product B, and -0.7019449433705992 times Product C. This objective reflects the cost structure of the logistics network. + + Operational constraints must be satisfied, including three linear equality constraints that define exact relationships between the products. These constraints are: 0.973138297335475 times Product A plus 0.5753515721139748 times Product B minus 0.3345391015917421 times Product C equals -0.1751595502819031; 1.004855798779082 times Product A plus 1.155166374661675 times Product B minus 0.6908735867360416 times Product C equals -0.7437001018712311; and -2.110665713909855 times Product A plus 0.0465859108136975 times Product B plus 1.404349170472437 times Product C equals -2.786300004238841. + + Additionally, there are 14 linear constraints that ensure the allocation of resources adheres to specific conditions. These constraints involve combinations of the product quantities and the auxiliary variables \( y_0 \), \( y_1 \), \( y_2 \), and the logistical thresholds \( t_0 \), \( t_1 \), and \( t_2 \). For example, one constraint states that 0.0366410572471413 times Product A plus 0.0568820534300137 times Product B minus 0.000079998 times Product C plus \( y_0[0 \) equals 0.0535813009430022. Similar constraints are defined for the remaining auxiliary variables and thresholds.] + + Furthermore, three second-order cone constraints ensure that the logistical thresholds satisfy specific quadratic relationships. These constraints are: the sum of the squares of \( y_0[0 \), \( y_0[1] \), \( y_0[2] \), and \( y_0[3] \) minus the square of \( t_0[0] \) must be less than or equal to zero; the sum of the squares of \( y_1[0] \), \( y_1[1] \), \( y_1[2] \), and \( y_1[3] \) minus the square of \( t_1[0] \) must be less than or equal to zero; and the sum of the squares of \( y_2[0] \), \( y_2[1] \), \( y_2[2] \), and \( y_2[3] \) minus the square of \( t_2[0] \) must be less than or equal to zero.] + + The task is to determine the optimal production and distribution levels for Products A, B, and C, along with the associated intermediate metrics and logistical thresholds, to minimize the total cost while satisfying all operational constraints.1.570831991511372Quadratic, out-of-scopenon-linear problem
134The problem involves minimizing the weighted sum of four decision variables, with weights -1.901009736372913, -0.7076764487261838, 0.083088283352758, and 0.4134020176281016, respectively. The objective is to minimize the expression -1.901009736372913 times the first variable minus 0.7076764487261838 times the second variable plus 0.083088283352758 times the third variable plus 0.4134020176281016 times the fourth variable. + +The problem includes five linear equality constraints. The first constraint requires 1.095257012774988 times the first variable plus 0.0405603467945054 times the second variable minus 0.4142813496378452 times the third variable plus 0.0769274791661522 times the fourth variable to equal -1.589822894954058. The second constraint requires 1.707274498669728 times the first variable minus 0.2477918751127619 times the second variable plus 1.014458021156339 times the third variable plus 0.0887060380647792 times the fourth variable to equal -1.412888255577238. The third constraint requires -0.1928047063342317 times the first variable plus 0.2495462876428885 times the second variable minus 1.347804555582835 times the third variable plus 0.5969103897857732 times the fourth variable to equal -0.7126523262327211. The fourth constraint requires 0.2539181754797121 times the first variable minus 0.4354680188894089 times the second variable minus 0.5625489084016247 times the third variable minus 0.2079253654505673 times the fourth variable to equal 0.0211399419876052. The fifth constraint requires 1.271683708095805 times the first variable plus 1.41227063919648 times the second variable plus 1.237138949722734 times the third variable plus 1.269794867507303 times the fourth variable to equal -3.001810041763422. + +Additionally, there are fifteen linear equality constraints involving intermediate variables. These constraints link the decision variables to intermediate variables \( y_0[0], y_0[1], y_1[0], y_1[1], y_2[0], y_2[1], y_3[0], y_3[1], y_4[0], y_4[1] \) and \( t_0[0], t_1[0], t_2[0], t_3[0], t_4[0] \). For example, the sixth constraint requires 0.0930293796109266 times the first variable plus 0.0314487985808588 times the second variable plus 0.0397727019597557 times the third variable plus 0.077024350952894 times the fourth variable plus \( y_0[0] \) to equal -0.0576022066753428. Similarly, the seventh constraint requires 0.3668170269095303 times the first variable plus 0.7382374711397578 times the second variable plus 1.389031691584004 times the third variable plus 0.4784897178942196 times the fourth variable plus \( t_0[0] \) to equal 0.5986251002407987. The remaining constraints follow a similar structure, connecting the decision variables to the intermediate variables with specific coefficients and right-hand side values. + +The problem also includes five second-order cone constraints, each involving the intermediate variables. These constraints require that the negative square of \( t_0[0] \) plus the sum of the squares of \( y_0[0] \) and \( y_0[1] \) be less than or equal to zero. Similarly, the negative square of \( t_1[0] \) plus the sum of the squares of \( y_1[0] \) and \( y_1[1] \) must be less than or equal to zero, and this pattern continues for the remaining second-order cone constraints. + +All variables, including the four decision variables and the intermediate variables, are free to take any real value. The goal is to determine the values of these variables that minimize the objective function while satisfying all the linear equality constraints and second-order cone constraints.3.193019492475020The problem involves minimizing the weighted sum of four decision variables, with weights -1.901009736372913, -0.7076764487261838, 0.083088283352758, and 0.4134020176281016, respectively. The objective is to minimize the expression -1.901009736372913 times the first variable minus 0.7076764487261838 times the second variable plus 0.083088283352758 times the third variable plus 0.4134020176281016 times the fourth variable. + + The problem includes five linear equality constraints. The first constraint requires 1.095257012774988 times the first variable plus 0.0405603467945054 times the second variable minus 0.4142813496378452 times the third variable plus 0.0769274791661522 times the fourth variable to equal -1.589822894954058. The second constraint requires 1.707274498669728 times the first variable minus 0.2477918751127619 times the second variable plus 1.014458021156339 times the third variable plus 0.0887060380647792 times the fourth variable to equal -1.412888255577238. The third constraint requires -0.1928047063342317 times the first variable plus 0.2495462876428885 times the second variable minus 1.347804555582835 times the third variable plus 0.5969103897857732 times the fourth variable to equal -0.7126523262327211. The fourth constraint requires 0.2539181754797121 times the first variable minus 0.4354680188894089 times the second variable minus 0.5625489084016247 times the third variable minus 0.2079253654505673 times the fourth variable to equal 0.0211399419876052. The fifth constraint requires 1.271683708095805 times the first variable plus 1.41227063919648 times the second variable plus 1.237138949722734 times the third variable plus 1.269794867507303 times the fourth variable to equal -3.001810041763422. + + Additionally, there are fifteen linear equality constraints involving intermediate variables. These constraints link the decision variables to intermediate variables \( y_0[0, y_0[1], y_1[0], y_1[1], y_2[0], y_2[1], y_3[0], y_3[1], y_4[0], y_4[1] \) and \( t_0[0], t_1[0], t_2[0], t_3[0], t_4[0] \). For example, the sixth constraint requires 0.0930293796109266 times the first variable plus 0.0314487985808588 times the second variable plus 0.0397727019597557 times the third variable plus 0.077024350952894 times the fourth variable plus \( y_0[0] \) to equal -0.0576022066753428. Similarly, the seventh constraint requires 0.3668170269095303 times the first variable plus 0.7382374711397578 times the second variable plus 1.389031691584004 times the third variable plus 0.4784897178942196 times the fourth variable plus \( t_0[0] \) to equal 0.5986251002407987. The remaining constraints follow a similar structure, connecting the decision variables to the intermediate variables with specific coefficients and right-hand side values.] + + The problem also includes five second-order cone constraints, each involving the intermediate variables. These constraints require that the negative square of \( t_0[0 \) plus the sum of the squares of \( y_0[0] \) and \( y_0[1] \) be less than or equal to zero. Similarly, the negative square of \( t_1[0] \) plus the sum of the squares of \( y_1[0] \) and \( y_1[1] \) must be less than or equal to zero, and this pattern continues for the remaining second-order cone constraints.] + + All variables, including the four decision variables and the intermediate variables, are free to take any real value. The goal is to determine the values of these variables that minimize the objective function while satisfying all the linear equality constraints and second-order cone constraints.3.19301949247502Quadratic, out-of-scopenon-linear problem
135The optimization problem involves minimizing the linear objective function with coefficients 0.4417937820038825 for variable \(x[0]\), 0.4540568683365753 for variable \(x[1]\), and 0.8269675169997344 for variable \(x[2]\). The problem is subject to a linear equality constraint: -0.8456455953416163 multiplied by \(x[0]\) plus 0.2274379327097405 multiplied by \(x[1]\) minus 0.77122733197339 multiplied by \(x[2]\) equals 0.0015019588399508. Additional linear constraints are imposed as follows: 0.0186179649018269 multiplied by \(x[0]\) plus 0.00822090536490999 multiplied by \(x[1]\) plus 0.0757498425936986 multiplied by \(x[2]\) plus \(y_0[0]\) equals 0.0928781722430411; -0.0864382752107571 multiplied by \(x[0]\) minus 0.0801306131882706 multiplied by \(x[1]\) plus 0.00147592929397403 multiplied by \(x[2]\) plus \(y_0[1]\) equals 0.0116010852586768; -0.0537777103532748 multiplied by \(x[0]\) minus 0.05178106148188 multiplied by \(x[1]\) plus 0.00775179400845416 multiplied by \(x[2]\) plus \(y_0[2]\) equals -0.0432037987392584; 0.0433708626106499 multiplied by \(x[0]\) minus 0.0555679005654094 multiplied by \(x[1]\) plus 0.1354122916870134 multiplied by \(x[2]\) plus \(y_0[3]\) equals -0.113448509613997; -0.13412 multiplied by \(x[0]\) minus 0.5390772678717964 multiplied by \(x[1]\) minus 1.224414665744034 multiplied by \(x[2]\) plus \(t_0[0]\) equals 2.324949537053539; 0.0894817797096893 multiplied by \(x[0]\) plus 0.1252203549610245 multiplied by \(x[1]\) plus 0.1674611083068293 multiplied by \(x[2]\) plus \(y_1[0]\) equals -0.1427178355564866; 0.0560619408765313 multiplied by \(x[0]\) plus 0.00020013926756253 multiplied by \(x[1]\) plus 0.00522276421792812 multiplied by \(x[2]\) plus \(y_1[1]\) equals -0.0793504157062673; 0.00954172270751238 multiplied by \(x[0]\) minus 0.0594530085708131 multiplied by \(x[1]\) minus 0.00960829189584381 multiplied by \(x[2]\) plus \(y_1[2]\) equals 0.098515604042304; -0.0150503497898215 multiplied by \(x[0]\) plus 0.1191163703340514 multiplied by \(x[1]\) plus 0.0214501936255565 multiplied by \(x[2]\) plus \(y_1[3]\) equals -0.0517634861476175; 0.0124789859358057 multiplied by \(x[0]\) plus 0.0900343661509949 multiplied by \(x[1]\) plus 1.223908566697448 multiplied by \(x[2]\) plus \(t_1[0]\) equals 0.7015191241991279. Two second-order cone constraints are imposed: the sum of the squares of \(y_0[0]\), \(y_0[1]\), \(y_0[2]\), and \(y_0[3]\) must not exceed the square of \(t_0[0]\), and the sum of the squares of \(y_1[0]\), \(y_1[1]\), \(y_1[2]\), and \(y_1[3]\) must not exceed the square of \(t_1[0]\). All decision variables, including \(x[0]\), \(x[1]\), \(x[2]\), \(y_0[0]\) through \(y_0[3]\), \(y_1[0]\) through \(y_1[3]\), \(t_0[0]\), and \(t_1[0]\), are free variables. The objective is to minimize the linear objective function while satisfying all linear and second-order cone constraints.-1.9814758603580050The optimization problem involves minimizing the linear objective function with coefficients 0.4417937820038825 for variable \(x[0\), 0.4540568683365753 for variable \(x[1]\), and 0.8269675169997344 for variable \(x[2]\). The problem is subject to a linear equality constraint: -0.8456455953416163 multiplied by \(x[0]\) plus 0.2274379327097405 multiplied by \(x[1]\) minus 0.77122733197339 multiplied by \(x[2]\) equals 0.0015019588399508. Additional linear constraints are imposed as follows: 0.0186179649018269 multiplied by \(x[0]\) plus 0.00822090536490999 multiplied by \(x[1]\) plus 0.0757498425936986 multiplied by \(x[2]\) plus \(y_0[0]\) equals 0.0928781722430411; -0.0864382752107571 multiplied by \(x[0]\) minus 0.0801306131882706 multiplied by \(x[1]\) plus 0.00147592929397403 multiplied by \(x[2]\) plus \(y_0[1]\) equals 0.0116010852586768; -0.0537777103532748 multiplied by \(x[0]\) minus 0.05178106148188 multiplied by \(x[1]\) plus 0.00775179400845416 multiplied by \(x[2]\) plus \(y_0[2]\) equals -0.0432037987392584; 0.0433708626106499 multiplied by \(x[0]\) minus 0.0555679005654094 multiplied by \(x[1]\) plus 0.1354122916870134 multiplied by \(x[2]\) plus \(y_0[3]\) equals -0.113448509613997; -0.13412 multiplied by \(x[0]\) minus 0.5390772678717964 multiplied by \(x[1]\) minus 1.224414665744034 multiplied by \(x[2]\) plus \(t_0[0]\) equals 2.324949537053539; 0.0894817797096893 multiplied by \(x[0]\) plus 0.1252203549610245 multiplied by \(x[1]\) plus 0.1674611083068293 multiplied by \(x[2]\) plus \(y_1[0]\) equals -0.1427178355564866; 0.0560619408765313 multiplied by \(x[0]\) plus 0.00020013926756253 multiplied by \(x[1]\) plus 0.00522276421792812 multiplied by \(x[2]\) plus \(y_1[1]\) equals -0.0793504157062673; 0.00954172270751238 multiplied by \(x[0]\) minus 0.0594530085708131 multiplied by \(x[1]\) minus 0.00960829189584381 multiplied by \(x[2]\) plus \(y_1[2]\) equals 0.098515604042304; -0.0150503497898215 multiplied by \(x[0]\) plus 0.1191163703340514 multiplied by \(x[1]\) plus 0.0214501936255565 multiplied by \(x[2]\) plus \(y_1[3]\) equals -0.0517634861476175; 0.0124789859358057 multiplied by \(x[0]\) plus 0.0900343661509949 multiplied by \(x[1]\) plus 1.223908566697448 multiplied by \(x[2]\) plus \(t_1[0]\) equals 0.7015191241991279. Two second-order cone constraints are imposed: the sum of the squares of \(y_0[0]\), \(y_0[1]\), \(y_0[2]\), and \(y_0[3]\) must not exceed the square of \(t_0[0]\), and the sum of the squares of \(y_1[0]\), \(y_1[1]\), \(y_1[2]\), and \(y_1[3]\) must not exceed the square of \(t_1[0]\). All decision variables, including \(x[0]\), \(x[1]\), \(x[2]\), \(y_0[0]\) through \(y_0[3]\), \(y_1[0]\) through \(y_1[3]\), \(t_0[0]\), and \(t_1[0]\), are free variables. The objective is to minimize the linear objective function while satisfying all linear and second-order cone constraints.]-1.981475860358005Quadratic, out-of-scopenon-linear problem
136Minimize 0.081229586973361 times x[0] minus 1.333096182188109 times x[1] plus 0.8259327119806184 times x[2] minus 1.123761222193983 times x[3] subject to five linear equality constraints: 0.566954322069251 times x[0] minus 0.7299810090585189 times x[1] plus 0.5965853226996864 times x[2] plus 0.7075700310276837 times x[3] equals -1.035685810880591; 0.1467098805498692 times x[0] plus 1.276004551518634 times x[1] plus 1.816969609021357 times x[2] minus 0.0616423385342787 times x[3] equals -3.620931266248864; 0.3340088790493272 times x[0] minus 0.7150468301081672 times x[1] minus 0.5567266298909594 times x[2] minus 0.8373889941281421 times x[3] equals -0.2488125442726408; -1.342305011059075 times x[0] plus 1.955549036585348 times x[1] plus 0.0602310986862688 times x[2] minus 0.4485566025786823 times x[3] equals 0.8148614925531417; and 0.662271009764529 times x[0] plus 1.235812465401366 times x[1] minus 0.6684584256202986 times x[2] plus 2.261775134584107 times x[3] equals 4.227927950879132. + +The model includes three second-order cone constraints defined using auxiliary variables y_0, y_1, and y_2, each with three components, and free variables t_0, t_1, and t_2. The auxiliary variables are defined by twelve linear equations involving the decision variables x and the auxiliary variables y and t. For instance, the first auxiliary variable y_0 is defined by three linear equations: -0.00807132776099688 times x[0] minus 0.00829904212968852 times x[1] plus 0.00729131816846323 times x[2] minus 0.0158474183769015 times x[3] plus y_0[0] equals 0.0740447156962421; 0.0398011692418615 times x[0] minus 0.1170083202748537 times x[1] plus 0.0605494567209828 times x[2] plus 0.0296817366716117 times x[3] plus y_0[1] equals -0.0221765972469103; and -0.00578839554637137 times x[0] minus 0.0410660878455097 times x[1] plus 0.021468433180218 times x[2] plus 0.03829764846326 times x[3] plus y_0[2] equals 0.0261113199527866. The variable t_0[0] is defined by -0.5485199622075393 times x[0] plus 2.161356714141078 times x[1] plus 0.4583402094085766 times x[2] plus 0.4120351464853843 times x[3] plus t_0[0] equals 1.070594461920216. Similar sets of equations define y_1 and y_2, and their associated t variables. + +The second-order cone constraints require that the square of t_0[0] must be greater than or equal to the sum of the squares of y_0[0], y_0[1], and y_0[2]; the square of t_1[0] must be greater than or equal to the sum of the squares of y_1[0], y_1[1], and y_1[2]; and the square of t_2[0] must be greater than or equal to the sum of the squares of y_2[0], y_2[1], and y_2[2]. All decision variables, including x, y_0, y_1, y_2, t_0, t_1, and t_2, are free to take any real value. The goal is to minimize the total cost while ensuring that all linear equality constraints and second-order cone constraints are satisfied.-3.6667437686164540Minimize 0.081229586973361 times x[0 minus 1.333096182188109 times x[1] plus 0.8259327119806184 times x[2] minus 1.123761222193983 times x[3] subject to five linear equality constraints: 0.566954322069251 times x[0] minus 0.7299810090585189 times x[1] plus 0.5965853226996864 times x[2] plus 0.7075700310276837 times x[3] equals -1.035685810880591; 0.1467098805498692 times x[0] plus 1.276004551518634 times x[1] plus 1.816969609021357 times x[2] minus 0.0616423385342787 times x[3] equals -3.620931266248864; 0.3340088790493272 times x[0] minus 0.7150468301081672 times x[1] minus 0.5567266298909594 times x[2] minus 0.8373889941281421 times x[3] equals -0.2488125442726408; -1.342305011059075 times x[0] plus 1.955549036585348 times x[1] plus 0.0602310986862688 times x[2] minus 0.4485566025786823 times x[3] equals 0.8148614925531417; and 0.662271009764529 times x[0] plus 1.235812465401366 times x[1] minus 0.6684584256202986 times x[2] plus 2.261775134584107 times x[3] equals 4.227927950879132.] + + The model includes three second-order cone constraints defined using auxiliary variables y_0, y_1, and y_2, each with three components, and free variables t_0, t_1, and t_2. The auxiliary variables are defined by twelve linear equations involving the decision variables x and the auxiliary variables y and t. For instance, the first auxiliary variable y_0 is defined by three linear equations: -0.00807132776099688 times x[0 minus 0.00829904212968852 times x[1] plus 0.00729131816846323 times x[2] minus 0.0158474183769015 times x[3] plus y_0[0] equals 0.0740447156962421; 0.0398011692418615 times x[0] minus 0.1170083202748537 times x[1] plus 0.0605494567209828 times x[2] plus 0.0296817366716117 times x[3] plus y_0[1] equals -0.0221765972469103; and -0.00578839554637137 times x[0] minus 0.0410660878455097 times x[1] plus 0.021468433180218 times x[2] plus 0.03829764846326 times x[3] plus y_0[2] equals 0.0261113199527866. The variable t_0[0] is defined by -0.5485199622075393 times x[0] plus 2.161356714141078 times x[1] plus 0.4583402094085766 times x[2] plus 0.4120351464853843 times x[3] plus t_0[0] equals 1.070594461920216. Similar sets of equations define y_1 and y_2, and their associated t variables.] + + The second-order cone constraints require that the square of t_0[0 must be greater than or equal to the sum of the squares of y_0[0], y_0[1], and y_0[2]; the square of t_1[0] must be greater than or equal to the sum of the squares of y_1[0], y_1[1], and y_1[2]; and the square of t_2[0] must be greater than or equal to the sum of the squares of y_2[0], y_2[1], and y_2[2]. All decision variables, including x, y_0, y_1, y_2, t_0, t_1, and t_2, are free to take any real value. The goal is to minimize the total cost while ensuring that all linear equality constraints and second-order cone constraints are satisfied.]-3.666743768616454Quadratic, out-of-scopenon-linear problem
137The problem involves minimizing the objective function, which is a linear combination of the decision variables \(x[0]\), \(x[1]\), \(x[2]\), and \(x[3]\) with coefficients -2.044113802210222, -0.591678008651819, -0.9834387040405338, and -0.4216128339992919, respectively. The variables \(x[0]\), \(x[1]\), \(x[2]\), \(x[3]\), \(y_0[0]\), \(y_0[1]\), \(y_1[0]\), and \(y_1[1]\) are free to take any real value. There are five linear equality constraints that relate \(x[0]\), \(x[1]\), \(x[2]\), and \(x[3]\) to specific constant values, with coefficients 1.583533632933326, -0.8648864187788813, 1.390091961404533, 1.822191081529753, -1.084701129083315, 0.8687853511014955, -1.061622663573521, 0.7234328322930561, 0.5805715615387739, -0.9882875862352942, 1.631714852918525, 0.1752665022129386, -0.2276163723137254, 0.3698287397442833, -1.107798916899263, 1.296533143808007, -1.295052034436542, 0.7550822815717648, 1.130836265772762, -1.018723280622327, 2.067786918679711, -1.455751318094515, -0.4504004313993055, 0.5779356751486215, and -1.462119472214241. + +Additionally, there are five constraints involving \(y_0[0]\), \(y_0[1]\), \(y_1[0]\), \(y_1[1]\), \(t_0[0]\), and \(t_1[0]\), with coefficients 0.0279053133377849, -0.00409039039473968, -0.0080383924189651, -0.2012607045179339, -0.1127331987985564, -0.1809080280498793, 0.094841149057402, -0.1684976328590128, 0.0145782756868836, -0.0421818911353208, 0.7645038993025399, -2.040190789758234, 0.225014606611158, 0.49714383406283, 0.3031138305570918, -0.0152851162771371, 0.198409707999221, 0.1277888002263441, 0.1653012900234249, 0.0294096537175579, 0.0764194503294455, 0.047258889505963, -0.0591645566302213, 0.0247500453597441, 0.066093810913026, -0.1089902399914438, 0.8938484212978033, 0.4688910009639788, -0.3004244670748821, and 1.093947030410214. + +The problem also includes two second-order cone constraints: the square of \(t_0[0]\) must be less than or equal to the sum of the squares of \(y_0[0]\) and \(y_0[1]\), and the square of \(t_1[0]\) must be less than or equal to the sum of the squares of \(y_1[0]\) and \(y_1[1]\). The goal is to determine the optimal values of all decision variables that minimize the objective function while satisfying all constraints. This is a Second-Order Cone Programming (SOCP) problem.2.1652315635553350The problem involves minimizing the objective function, which is a linear combination of the decision variables \(x[0\), \(x[1]\), \(x[2]\), and \(x[3]\) with coefficients -2.044113802210222, -0.591678008651819, -0.9834387040405338, and -0.4216128339992919, respectively. The variables \(x[0]\), \(x[1]\), \(x[2]\), \(x[3]\), \(y_0[0]\), \(y_0[1]\), \(y_1[0]\), and \(y_1[1]\) are free to take any real value. There are five linear equality constraints that relate \(x[0]\), \(x[1]\), \(x[2]\), and \(x[3]\) to specific constant values, with coefficients 1.583533632933326, -0.8648864187788813, 1.390091961404533, 1.822191081529753, -1.084701129083315, 0.8687853511014955, -1.061622663573521, 0.7234328322930561, 0.5805715615387739, -0.9882875862352942, 1.631714852918525, 0.1752665022129386, -0.2276163723137254, 0.3698287397442833, -1.107798916899263, 1.296533143808007, -1.295052034436542, 0.7550822815717648, 1.130836265772762, -1.018723280622327, 2.067786918679711, -1.455751318094515, -0.4504004313993055, 0.5779356751486215, and -1.462119472214241. ] + + Additionally, there are five constraints involving \(y_0[0\), \(y_0[1]\), \(y_1[0]\), \(y_1[1]\), \(t_0[0]\), and \(t_1[0]\), with coefficients 0.0279053133377849, -0.00409039039473968, -0.0080383924189651, -0.2012607045179339, -0.1127331987985564, -0.1809080280498793, 0.094841149057402, -0.1684976328590128, 0.0145782756868836, -0.0421818911353208, 0.7645038993025399, -2.040190789758234, 0.225014606611158, 0.49714383406283, 0.3031138305570918, -0.0152851162771371, 0.198409707999221, 0.1277888002263441, 0.1653012900234249, 0.0294096537175579, 0.0764194503294455, 0.047258889505963, -0.0591645566302213, 0.0247500453597441, 0.066093810913026, -0.1089902399914438, 0.8938484212978033, 0.4688910009639788, -0.3004244670748821, and 1.093947030410214. ] + + The problem also includes two second-order cone constraints: the square of \(t_0[0\) must be less than or equal to the sum of the squares of \(y_0[0]\) and \(y_0[1]\), and the square of \(t_1[0]\) must be less than or equal to the sum of the squares of \(y_1[0]\) and \(y_1[1]\). The goal is to determine the optimal values of all decision variables that minimize the objective function while satisfying all constraints. This is a Second-Order Cone Programming (SOCP) problem.]2.165231563555335Quadratic, out-of-scopenon-linear problem
138The problem involves minimizing a linear objective function with six decision variables \( x[0] \) through \( x[5] \), where each variable has a specific coefficient: \( x[0] \) with -2.058644777783222, \( x[1] \) with 0.2416010927247412, \( x[2] \) with -0.4181486412362984, \( x[3] \) with 0.5114234913219796, \( x[4] \) with 0.7386320402187463, and \( x[5] \) with -1.172672370815835. The problem includes five linear equality constraints that must be satisfied. The first constraint requires that the weighted sum of the variables, adjusted by factors -0.5367684107949979, 0.1034203788533778, 1.846625038230486, 0.9784353550415096, -1.878216393450231, and -0.7467874898339996, equals 3.5678313588737. The second constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients -0.5352309785306387, -0.8520650729788865, -0.8847016444802787, 1.600400673784742, -0.6170038459749025, and 0.5094850995322819 to 0.1979161022274378. The third constraint requires that the weighted sum of \( x[0] \) through \( x[5] \) with coefficients 0.7429110508496402, -0.7085502859117431, -1.428880331198872, -0.838487898882224, -0.4330191528056699, and -1.491505638645529 equals -4.293033952749457. The fourth constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients 1.389614629752773, 0.702810325846059, -1.416233193493607, 0.6259878762218291, -1.062718067935909, and 1.116816637099358 to 6.052880318149405. The fifth constraint requires that the weighted sum of \( x[0] \) through \( x[5] \) with coefficients 0.5035448845365915, -0.0513767067216146, -1.800678426974007, 1.198119996945881, -0.4513781697868756, and -0.1446651828337932 equals 0.3414915225001892. Additionally, the problem includes eight more linear equality constraints involving additional variables \( y_0[0] \), \( y_0[1] \), \( y_1[0] \), \( y_1[1] \), \( y_2[0] \), \( y_2[1] \), \( t_0[0] \), \( t_1[0] \), and \( t_2[0] \). These constraints link the decision variables \( x[0] \) through \( x[5] \) with the additional variables in specific linear combinations. For instance, the sixth constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients 0.0384343819842272, 0.0390068546116138, 0.1797102552048576, -0.00927856470872923, 0.1701908634063454, and -0.1248214943708578 plus \( y_0[0] \) to -0.00663951753988017. The seventh constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients -0.1362799231538006, -0.0831864520235293, 0.1112848630570577, 0.1060222181921134, 0.0402874170898317, and -0.0464976478913114 plus \( y_0[1] \) to 0.0242095284091262. The eighth constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients -0.0156129943224916, -0.896947647061286, -1.567254730229088, -0.7037621933777273, 0.180343362346215, and 1.006321968448518 plus \( t_0[0] \) to 0.5018287659872952. The ninth constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients -0.2332423383309324, -0.0330958699417269, -0.00964072153256004, 0.0585722335710037, 0.0217911804090265, and -0.051126178402744 plus \( y_1[0] \) to -0.0968796249195388. The tenth constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients 0.1151029171245048, 0.1503814324379974, -0.0781335582786794, 0.0402624396659884, -0.0358496162641436, and 0.211819753444535 plus \( y_1[1] \) to 0.0293392072741357. The eleventh constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients 0.5215036730393952, 0.9990730184126353, 1.195267064572967, 0.190014887816329, 1.638781801327033, and -1.411081833597206 plus \( t_1[0] \) to 0.2585473929490868. The twelfth constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients -0.0360929813745435, -0.00104760604299632, -0.1784046295918039, -0.0219795512892422, -0.1002032765225464, and 0.0553677513211222 plus \( y_2[0] \) to 0.0514139176871232. The thirteenth constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients -0.0830942200094662, -0.1767784677541842, 0.1439818445323525, -0.0133640836810608, 0.1360556372574424, and -0.1331263469306528 plus \( y_2[1] \) to -0.0667657097090114. The fourteenth constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients 0.3082403816753665, 0.1738249328101287, -0.3242604405259034, 0.1914616391261432, 0.8490602240068917, and -1.347767787369367 plus \( t_2[0] \) to 0.3484815971142778. The problem also includes three second-order cone constraints that ensure the sum of the squares of \( y_0[0] \) and \( y_0[1] \) minus the square of \( t_0[0] \) is less than or equal to zero, the sum of the squares of \( y_1[0] \) and \( y_1[1] \) minus the square of \( t_1[0] \) is less than or equal to zero, and the sum of the squares of \( y_2[0] \) and \( y_2[1] \) minus the square of \( t_2[0] \) is less than or equal to zero. All variables, including \( x[0] \) through \( x[5] \), \( y_0[0] \), \( y_0[1] \), \( y_1[0] \), \( y_1[1] \), \( y_2[0] \), \( y_2[1] \), \( t_0[0] \), \( t_1[0] \), and \( t_2[0] \), are free to take any real value. The goal is to minimize the objective function while satisfying all linear equality and second-order cone constraints.-6.8509831774247530The problem involves minimizing a linear objective function with six decision variables \( x[0 \) through \( x[5] \), where each variable has a specific coefficient: \( x[0] \) with -2.058644777783222, \( x[1] \) with 0.2416010927247412, \( x[2] \) with -0.4181486412362984, \( x[3] \) with 0.5114234913219796, \( x[4] \) with 0.7386320402187463, and \( x[5] \) with -1.172672370815835. The problem includes five linear equality constraints that must be satisfied. The first constraint requires that the weighted sum of the variables, adjusted by factors -0.5367684107949979, 0.1034203788533778, 1.846625038230486, 0.9784353550415096, -1.878216393450231, and -0.7467874898339996, equals 3.5678313588737. The second constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients -0.5352309785306387, -0.8520650729788865, -0.8847016444802787, 1.600400673784742, -0.6170038459749025, and 0.5094850995322819 to 0.1979161022274378. The third constraint requires that the weighted sum of \( x[0] \) through \( x[5] \) with coefficients 0.7429110508496402, -0.7085502859117431, -1.428880331198872, -0.838487898882224, -0.4330191528056699, and -1.491505638645529 equals -4.293033952749457. The fourth constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients 1.389614629752773, 0.702810325846059, -1.416233193493607, 0.6259878762218291, -1.062718067935909, and 1.116816637099358 to 6.052880318149405. The fifth constraint requires that the weighted sum of \( x[0] \) through \( x[5] \) with coefficients 0.5035448845365915, -0.0513767067216146, -1.800678426974007, 1.198119996945881, -0.4513781697868756, and -0.1446651828337932 equals 0.3414915225001892. Additionally, the problem includes eight more linear equality constraints involving additional variables \( y_0[0] \), \( y_0[1] \), \( y_1[0] \), \( y_1[1] \), \( y_2[0] \), \( y_2[1] \), \( t_0[0] \), \( t_1[0] \), and \( t_2[0] \). These constraints link the decision variables \( x[0] \) through \( x[5] \) with the additional variables in specific linear combinations. For instance, the sixth constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients 0.0384343819842272, 0.0390068546116138, 0.1797102552048576, -0.00927856470872923, 0.1701908634063454, and -0.1248214943708578 plus \( y_0[0] \) to -0.00663951753988017. The seventh constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients -0.1362799231538006, -0.0831864520235293, 0.1112848630570577, 0.1060222181921134, 0.0402874170898317, and -0.0464976478913114 plus \( y_0[1] \) to 0.0242095284091262. The eighth constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients -0.0156129943224916, -0.896947647061286, -1.567254730229088, -0.7037621933777273, 0.180343362346215, and 1.006321968448518 plus \( t_0[0] \) to 0.5018287659872952. The ninth constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients -0.2332423383309324, -0.0330958699417269, -0.00964072153256004, 0.0585722335710037, 0.0217911804090265, and -0.051126178402744 plus \( y_1[0] \) to -0.0968796249195388. The tenth constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients 0.1151029171245048, 0.1503814324379974, -0.0781335582786794, 0.0402624396659884, -0.0358496162641436, and 0.211819753444535 plus \( y_1[1] \) to 0.0293392072741357. The eleventh constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients 0.5215036730393952, 0.9990730184126353, 1.195267064572967, 0.190014887816329, 1.638781801327033, and -1.411081833597206 plus \( t_1[0] \) to 0.2585473929490868. The twelfth constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients -0.0360929813745435, -0.00104760604299632, -0.1784046295918039, -0.0219795512892422, -0.1002032765225464, and 0.0553677513211222 plus \( y_2[0] \) to 0.0514139176871232. The thirteenth constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients -0.0830942200094662, -0.1767784677541842, 0.1439818445323525, -0.0133640836810608, 0.1360556372574424, and -0.1331263469306528 plus \( y_2[1] \) to -0.0667657097090114. The fourteenth constraint equates a weighted sum of \( x[0] \) through \( x[5] \) with coefficients 0.3082403816753665, 0.1738249328101287, -0.3242604405259034, 0.1914616391261432, 0.8490602240068917, and -1.347767787369367 plus \( t_2[0] \) to 0.3484815971142778. The problem also includes three second-order cone constraints that ensure the sum of the squares of \( y_0[0] \) and \( y_0[1] \) minus the square of \( t_0[0] \) is less than or equal to zero, the sum of the squares of \( y_1[0] \) and \( y_1[1] \) minus the square of \( t_1[0] \) is less than or equal to zero, and the sum of the squares of \( y_2[0] \) and \( y_2[1] \) minus the square of \( t_2[0] \) is less than or equal to zero. All variables, including \( x[0] \) through \( x[5] \), \( y_0[0] \), \( y_0[1] \), \( y_1[0] \), \( y_1[1] \), \( y_2[0] \), \( y_2[1] \), \( t_0[0] \), \( t_1[0] \), and \( t_2[0] \), are free to take any real value. The goal is to minimize the objective function while satisfying all linear equality and second-order cone constraints.]-6.850983177424753Quadratic, out-of-scopenon-linear problem
139The problem is to minimize the linear objective function with coefficients -0.2234016903227589 for \(x[0]\), 0.0823239307040719 for \(x[1]\), -0.3139602147497034 for \(x[2]\), and -0.3588305741920246 for \(x[3]\). The variables \(x[0]\), \(x[1]\), \(x[2]\), \(x[3]\), \(y_0[0]\), \(y_0[1]\), and \(t_0[0]\) are free. The solution must satisfy six linear equality constraints. The first constraint requires the weighted sum of variables, with coefficients -2.379260712270257 for \(x[0]\), -0.0926191997384036 for \(x[1]\), 1.500019618890642 for \(x[2]\), and -2.240272675043434 for \(x[3]\), to equal -2.383475883203837. The second constraint requires the weighted sum, with coefficients 0.0880334820094267 for \(x[0]\), -1.902743286220365 for \(x[1]\), -1.061143421718847 for \(x[2]\), and -0.7298562539833904 for \(x[3]\), to equal -0.1651524265365722. The third constraint requires the weighted sum, with coefficients 0.879056298919617 for \(x[0]\), -0.0592926915590702 for \(x[1]\), 1.530225870090669 for \(x[2]\), and 0.7516434368490681 for \(x[3]\), to equal -1.840782058087354. The fourth constraint requires the weighted sum, with coefficients -0.7934367479385507 for \(x[0]\), 0.0232747502052466 for \(x[1]\), -0.6401263883470215 for \(x[2]\), and -0.1075051544628913 for \(x[3]\), to equal 0.5711825810091148. The fifth constraint requires the weighted sum of variables, with coefficients -0.0926233736251473 for \(x[0]\), 0.0599466213459983 for \(x[1]\), 0.2106990148269894 for \(x[2]\), and -0.139498666863404 for \(x[3]\), plus \(y_0[0]\), to equal 0.0103572858836726. The sixth constraint requires the weighted sum of variables, with coefficients -0.0912288940130291 for \(x[0]\), -0.1581472188037661 for \(x[1]\), -0.2376431900824324 for \(x[2]\), and -0.1641150951970651 for \(x[3]\), plus \(y_0[1]\), to equal 0.0072007946533593. The seventh constraint requires the weighted sum of variables, with coefficients -0.4718232097702756 for \(x[0]\), -1.786481176156424 for \(x[1]\), 5.276204984748865e-04 for \(x[2]\), and 0.2812610682665851 for \(x[3]\), plus \(t_0[0]\), to equal 0.3067145358130309. A quadratic constraint ensures that the sum of the squares of \(y_0[0]\) and \(y_0[1]\) is less than or equal to the square of \(t_0[0]\).0.467763458014580The problem is to minimize the linear objective function with coefficients -0.2234016903227589 for \(x[0\), 0.0823239307040719 for \(x[1]\), -0.3139602147497034 for \(x[2]\), and -0.3588305741920246 for \(x[3]\). The variables \(x[0]\), \(x[1]\), \(x[2]\), \(x[3]\), \(y_0[0]\), \(y_0[1]\), and \(t_0[0]\) are free. The solution must satisfy six linear equality constraints. The first constraint requires the weighted sum of variables, with coefficients -2.379260712270257 for \(x[0]\), -0.0926191997384036 for \(x[1]\), 1.500019618890642 for \(x[2]\), and -2.240272675043434 for \(x[3]\), to equal -2.383475883203837. The second constraint requires the weighted sum, with coefficients 0.0880334820094267 for \(x[0]\), -1.902743286220365 for \(x[1]\), -1.061143421718847 for \(x[2]\), and -0.7298562539833904 for \(x[3]\), to equal -0.1651524265365722. The third constraint requires the weighted sum, with coefficients 0.879056298919617 for \(x[0]\), -0.0592926915590702 for \(x[1]\), 1.530225870090669 for \(x[2]\), and 0.7516434368490681 for \(x[3]\), to equal -1.840782058087354. The fourth constraint requires the weighted sum, with coefficients -0.7934367479385507 for \(x[0]\), 0.0232747502052466 for \(x[1]\), -0.6401263883470215 for \(x[2]\), and -0.1075051544628913 for \(x[3]\), to equal 0.5711825810091148. The fifth constraint requires the weighted sum of variables, with coefficients -0.0926233736251473 for \(x[0]\), 0.0599466213459983 for \(x[1]\), 0.2106990148269894 for \(x[2]\), and -0.139498666863404 for \(x[3]\), plus \(y_0[0]\), to equal 0.0103572858836726. The sixth constraint requires the weighted sum of variables, with coefficients -0.0912288940130291 for \(x[0]\), -0.1581472188037661 for \(x[1]\), -0.2376431900824324 for \(x[2]\), and -0.1641150951970651 for \(x[3]\), plus \(y_0[1]\), to equal 0.0072007946533593. The seventh constraint requires the weighted sum of variables, with coefficients -0.4718232097702756 for \(x[0]\), -1.786481176156424 for \(x[1]\), 5.276204984748865e-04 for \(x[2]\), and 0.2812610682665851 for \(x[3]\), plus \(t_0[0]\), to equal 0.3067145358130309. A quadratic constraint ensures that the sum of the squares of \(y_0[0]\) and \(y_0[1]\) is less than or equal to the square of \(t_0[0]\).]0.46776345801458Quadratic, out-of-scopenon-linear problem
140Determine the optimal amount of resources \( x[0] \), \( x[1] \), and \( x[2] \) to allocate across three operational areas to minimize the total cost, calculated as 2.238472652501168 multiplied by \( x[0] \) plus 1.593692449804477 multiplied by \( x[1] \) plus 1.091018724438081 multiplied by \( x[2] \). The allocation must satisfy a linear resource balance constraint: 2.086630330627226 multiplied by \( x[0] \) plus 0.0977042297800921 multiplied by \( x[1] \) plus 0.3238066911742682 multiplied by \( x[2] \) equals 0.2728125978918034. + +The problem includes three quadratic cone constraints \( qc0 \), \( qc1 \), and \( qc2 \), which are part of a Second-Order Cone Programming (SOCP) problem. For the first operational area, the quadratic cone constraint \( qc0 \) requires that the sum of the squares of \( y_0[0] \) and \( y_0[1] \) must be less than or equal to the square of \( t_0[0] \). Here, \( y_0[0] \) is determined by 1.350013394321828 multiplied by \( x[0] \) minus 0.2433648689288818 multiplied by \( x[1] \) plus 1.170490297808538 multiplied by \( x[2] \), which equals -0.854030949971019. Similarly, \( y_0[1] \) is determined by -0.5827218400584065 multiplied by \( x[0] \) minus 1.214924181090006 multiplied by \( x[1] \) minus 0.7100967735067889 multiplied by \( x[2] \), which equals 0.1115410551073176. The variable \( t_0[0] \) is determined by -0.1042401171821044 multiplied by \( x[0] \) plus 0.2613612792965374 multiplied by \( x[1] \) plus 1.118900434050682 multiplied by \( x[2] \), which equals 0.6038954719472605. + +For the second operational area, the quadratic cone constraint \( qc1 \) requires that the sum of the squares of \( y_1[0] \) and \( y_1[1] \) must be less than or equal to the square of \( t_1[0] \). Here, \( y_1[0] \) is determined by 0.1242454238540142 multiplied by \( x[0] \) minus 0.2094940308550101 multiplied by \( x[1] \) plus 0.0218788315028139 multiplied by \( x[2] \), which equals -0.1826683487550716. Similarly, \( y_1[1] \) is determined by 0.0612039952350878 multiplied by \( x[0] \) plus 0.019467075730508 multiplied by \( x[1] \) minus 0.0279175737966843 multiplied by \( x[2] \), which equals -0.1133584554034866. The variable \( t_1[0] \) is determined by -0.1049189629762404 multiplied by \( x[0] \) plus 0.6171636592039963 multiplied by \( x[1] \) minus 0.385085431921569 multiplied by \( x[2] \), which equals 1.132753956602686. + +For the third operational area, the quadratic cone constraint \( qc2 \) requires that the sum of the squares of \( y_2[0] \) and \( y_2[1] \) must be less than or equal to the square of \( t_2[0] \). Here, \( y_2[0] \) is determined by -0.1504745044323135 multiplied by \( x[0] \) minus 0.1569086648551038 multiplied by \( x[1] \) minus 0.1058514836896046 multiplied by \( x[2] \), which equals 0.2363416370720583. Similarly, \( y_2[1] \) is determined by 0.0825995942902514 multiplied by \( x[0] \) minus 0.089460646459489 multiplied by \( x[1] \) minus 0.0359085847666962 multiplied by \( x[2] \), which equals 0.1827679172343429. The variable \( t_2[0] \) is determined by -0.2546431852316068 multiplied by \( x[0] \) minus 0.9584080837019103 multiplied by \( x[1] \) plus 0.3025850075547174 multiplied by \( x[2] \), which equals 1.774740955514529. + +All variables, including \( x[0] \), \( x[1] \), \( x[2] \), \( y_0[0] \), \( y_0[1] \), \( y_1[0] \), \( y_1[1] \), \( y_2[0] \), \( y_2[1] \), \( t_0[0] \), \( t_1[0] \), and \( t_2[0] \), are free to take any real value, as there are no bounds restricting their values. The quadratic cone constraints \( qc0 \), \( qc1 \), and \( qc2 \) are explicitly part of the problem constraints, and the variables \( y_0[0] \), \( y_0[1] \), \( y_1[0] \), \( y_1[1] \), \( y_2[0] \), \( y_2[1] \), \( t_0[0] \), \( t_1[0] \), and \( t_2[0] \) are auxiliary variables introduced to represent these quadratic cone constraints. The problem is a Second-Order Cone Programming (SOCP) problem, as indicated by the LP data header.-8.9494699077983580Determine the optimal amount of resources \( x[0 \), \( x[1] \), and \( x[2] \) to allocate across three operational areas to minimize the total cost, calculated as 2.238472652501168 multiplied by \( x[0] \) plus 1.593692449804477 multiplied by \( x[1] \) plus 1.091018724438081 multiplied by \( x[2] \). The allocation must satisfy a linear resource balance constraint: 2.086630330627226 multiplied by \( x[0] \) plus 0.0977042297800921 multiplied by \( x[1] \) plus 0.3238066911742682 multiplied by \( x[2] \) equals 0.2728125978918034.] + + The problem includes three quadratic cone constraints \( qc0 \), \( qc1 \), and \( qc2 \), which are part of a Second-Order Cone Programming (SOCP) problem. For the first operational area, the quadratic cone constraint \( qc0 \) requires that the sum of the squares of \( y_0[0 \) and \( y_0[1] \) must be less than or equal to the square of \( t_0[0] \). Here, \( y_0[0] \) is determined by 1.350013394321828 multiplied by \( x[0] \) minus 0.2433648689288818 multiplied by \( x[1] \) plus 1.170490297808538 multiplied by \( x[2] \), which equals -0.854030949971019. Similarly, \( y_0[1] \) is determined by -0.5827218400584065 multiplied by \( x[0] \) minus 1.214924181090006 multiplied by \( x[1] \) minus 0.7100967735067889 multiplied by \( x[2] \), which equals 0.1115410551073176. The variable \( t_0[0] \) is determined by -0.1042401171821044 multiplied by \( x[0] \) plus 0.2613612792965374 multiplied by \( x[1] \) plus 1.118900434050682 multiplied by \( x[2] \), which equals 0.6038954719472605.] + + For the second operational area, the quadratic cone constraint \( qc1 \) requires that the sum of the squares of \( y_1[0 \) and \( y_1[1] \) must be less than or equal to the square of \( t_1[0] \). Here, \( y_1[0] \) is determined by 0.1242454238540142 multiplied by \( x[0] \) minus 0.2094940308550101 multiplied by \( x[1] \) plus 0.0218788315028139 multiplied by \( x[2] \), which equals -0.1826683487550716. Similarly, \( y_1[1] \) is determined by 0.0612039952350878 multiplied by \( x[0] \) plus 0.019467075730508 multiplied by \( x[1] \) minus 0.0279175737966843 multiplied by \( x[2] \), which equals -0.1133584554034866. The variable \( t_1[0] \) is determined by -0.1049189629762404 multiplied by \( x[0] \) plus 0.6171636592039963 multiplied by \( x[1] \) minus 0.385085431921569 multiplied by \( x[2] \), which equals 1.132753956602686.] + + For the third operational area, the quadratic cone constraint \( qc2 \) requires that the sum of the squares of \( y_2[0 \) and \( y_2[1] \) must be less than or equal to the square of \( t_2[0] \). Here, \( y_2[0] \) is determined by -0.1504745044323135 multiplied by \( x[0] \) minus 0.1569086648551038 multiplied by \( x[1] \) minus 0.1058514836896046 multiplied by \( x[2] \), which equals 0.2363416370720583. Similarly, \( y_2[1] \) is determined by 0.0825995942902514 multiplied by \( x[0] \) minus 0.089460646459489 multiplied by \( x[1] \) minus 0.0359085847666962 multiplied by \( x[2] \), which equals 0.1827679172343429. The variable \( t_2[0] \) is determined by -0.2546431852316068 multiplied by \( x[0] \) minus 0.9584080837019103 multiplied by \( x[1] \) plus 0.3025850075547174 multiplied by \( x[2] \), which equals 1.774740955514529.] + + All variables, including \( x[0 \), \( x[1] \), \( x[2] \), \( y_0[0] \), \( y_0[1] \), \( y_1[0] \), \( y_1[1] \), \( y_2[0] \), \( y_2[1] \), \( t_0[0] \), \( t_1[0] \), and \( t_2[0] \), are free to take any real value, as there are no bounds restricting their values. The quadratic cone constraints \( qc0 \), \( qc1 \), and \( qc2 \) are explicitly part of the problem constraints, and the variables \( y_0[0] \), \( y_0[1] \), \( y_1[0] \), \( y_1[1] \), \( y_2[0] \), \( y_2[1] \), \( t_0[0] \), \( t_1[0] \), and \( t_2[0] \) are auxiliary variables introduced to represent these quadratic cone constraints. The problem is a Second-Order Cone Programming (SOCP) problem, as indicated by the LP data header.]-8.949469907798358Quadratic, out-of-scopenon-linear problem
141The problem involves minimizing a linear objective function with decision variables \( x[0] \), \( x[1] \), and \( x[2] \), along with auxiliary variables \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), and \( t_0[0] \). The objective function coefficients are 0.9775739758894315 for \( x[0] \), -1.622521430954005 for \( x[1] \), and -0.3823943916427769 for \( x[2] \). All variables \( x[0] \), \( x[1] \), \( x[2] \), \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), and \( t_0[0] \) are free. + +The constraints include four linear equality constraints. The first constraint ensures that 2.218395151719285 times \( x[0] \) minus 0.4934367321170816 times \( x[1] \) plus 2.037698386055625 times \( x[2] \) equals -7.639024318581815. The second constraint ensures that 0.1288997654276396 times \( x[0] \) plus 0.4134151782133442 times \( x[1] \) minus 1.016848910244227 times \( x[2] \) equals 0.8282605412220725. The third constraint ensures that -0.3888036525234204 times \( x[0] \) plus 1.582023744531529 times \( x[1] \) plus 0.4853205141377102 times \( x[2] \) equals 0.9861388989442696. The fourth constraint ensures that -0.8184959408216979 times \( x[0] \) plus 0.1508892622723634 times \( x[1] \) plus 0.4877918700202323 times \( x[2] \) equals 1.575829009785032. + +Additionally, there are four linear constraints defining \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), and \( t_0[0] \) in terms of \( x[0] \), \( x[1] \), and \( x[2] \). The first constraint ensures that 0.000682161 times \( x[0] \) minus 0.058594741163783 times \( x[1] \) minus 0.2214986665992643 times \( x[2] \) plus \( y_0[0] \) equals 0.0363203305835107. The second constraint ensures that -0.00108977088958764 times \( x[0] \) plus 0.0836225593865622 times \( x[1] \) minus 0.2893750072155089 times \( x[2] \) plus \( y_0[1] \) equals -0.1460000637840041. The third constraint ensures that -0.0231877971791914 times \( x[0] \) minus 0.0340439160737732 times \( x[1] \) minus 0.2441350484603667 times \( x[2] \) plus \( y_0[2] \) equals 0.0702001695223133. The fourth constraint ensures that 0.8200125377679498 times \( x[0] \) minus 0.2688442189065823 times \( x[1] \) plus 0.8884584718933239 times \( x[2] \) plus \( t_0[0] \) equals 0.0172532969418979. + +A second-order cone constraint ensures that the square of \( t_0[0] \) is less than or equal to the sum of the squares of \( y_0[0] \), \( y_0[1] \), and \( y_0[2] \), specifically expressed as \(-t_0[0]^2 + y_0[0]^2 + y_0[1]^2 + y_0[2]^2 \leq 0\). The goal is to determine the values of \( x[0] \), \( x[1] \), \( x[2] \), \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), and \( t_0[0] \) that minimize the objective function while satisfying all constraints. This is a Second-Order Cone Programming (SOCP) problem.-2.5488418807470690The problem involves minimizing a linear objective function with decision variables \( x[0 \), \( x[1] \), and \( x[2] \), along with auxiliary variables \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), and \( t_0[0] \). The objective function coefficients are 0.9775739758894315 for \( x[0] \), -1.622521430954005 for \( x[1] \), and -0.3823943916427769 for \( x[2] \). All variables \( x[0] \), \( x[1] \), \( x[2] \), \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), and \( t_0[0] \) are free.] + + The constraints include four linear equality constraints. The first constraint ensures that 2.218395151719285 times \( x[0 \) minus 0.4934367321170816 times \( x[1] \) plus 2.037698386055625 times \( x[2] \) equals -7.639024318581815. The second constraint ensures that 0.1288997654276396 times \( x[0] \) plus 0.4134151782133442 times \( x[1] \) minus 1.016848910244227 times \( x[2] \) equals 0.8282605412220725. The third constraint ensures that -0.3888036525234204 times \( x[0] \) plus 1.582023744531529 times \( x[1] \) plus 0.4853205141377102 times \( x[2] \) equals 0.9861388989442696. The fourth constraint ensures that -0.8184959408216979 times \( x[0] \) plus 0.1508892622723634 times \( x[1] \) plus 0.4877918700202323 times \( x[2] \) equals 1.575829009785032.] + + Additionally, there are four linear constraints defining \( y_0[0 \), \( y_0[1] \), \( y_0[2] \), and \( t_0[0] \) in terms of \( x[0] \), \( x[1] \), and \( x[2] \). The first constraint ensures that 0.000682161 times \( x[0] \) minus 0.058594741163783 times \( x[1] \) minus 0.2214986665992643 times \( x[2] \) plus \( y_0[0] \) equals 0.0363203305835107. The second constraint ensures that -0.00108977088958764 times \( x[0] \) plus 0.0836225593865622 times \( x[1] \) minus 0.2893750072155089 times \( x[2] \) plus \( y_0[1] \) equals -0.1460000637840041. The third constraint ensures that -0.0231877971791914 times \( x[0] \) minus 0.0340439160737732 times \( x[1] \) minus 0.2441350484603667 times \( x[2] \) plus \( y_0[2] \) equals 0.0702001695223133. The fourth constraint ensures that 0.8200125377679498 times \( x[0] \) minus 0.2688442189065823 times \( x[1] \) plus 0.8884584718933239 times \( x[2] \) plus \( t_0[0] \) equals 0.0172532969418979.] + + A second-order cone constraint ensures that the square of \( t_0[0 \) is less than or equal to the sum of the squares of \( y_0[0] \), \( y_0[1] \), and \( y_0[2] \), specifically expressed as \(-t_0[0]^2 + y_0[0]^2 + y_0[1]^2 + y_0[2]^2 \leq 0\). The goal is to determine the values of \( x[0] \), \( x[1] \), \( x[2] \), \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), and \( t_0[0] \) that minimize the objective function while satisfying all constraints. This is a Second-Order Cone Programming (SOCP) problem.]-2.548841880747069Quadratic, out-of-scopenon-linear problem
142The problem involves minimizing a linear objective function defined by the coefficients 0.0169907447059004 for x[0], -0.1354658035191434 for x[1], and 0.1439797319070909 for x[2]. The optimization is subject to four linear equality constraints that balance the relationships between the variables x[0], x[1], and x[2]. The first constraint requires that 0.8636590559284738 times x[0] plus 0.0966833012301337 times x[1] minus 0.3152291953090132 times x[2] equals 0.8843372418353203. The second constraint requires that -1.308791116694014 times x[0] plus 1.026765786674754 times x[1] minus 1.641892627931514 times x[2] equals -3.068584167132096. The third constraint requires that -2.022894561574939 times x[0] plus 1.078027292135305 times x[1] minus 0.6806946931130862 times x[2] equals -3.878545364964277. The fourth constraint requires that 0.5909183755526639 times x[0] minus 1.227754194746086 times x[1] plus 0.9978468809169225 times x[2] equals 2.373932955061617. + +The problem also includes twelve additional linear constraints that introduce auxiliary variables y_0, y_1, y_2, and t_0, t_1, t_2. These constraints balance the primary variables x[0], x[1], and x[2] with the auxiliary variables. For example, constraint R4 requires that -0.0830895372455731 times x[0] plus 0.140196500160931 times x[1] plus 0.0835731570777216 times x[2] plus y_0[0] equals -0.0542756929304628. Similarly, constraints R5 through R15 impose similar relationships involving y_0[1], y_0[2], y_1[0], y_1[1], y_1[2], y_2[0], y_2[1], y_2[2], t_0[0], t_1[0], and t_2[0]. + +The problem also includes three second-order cone constraints, qc0, qc1, and qc2, which ensure that the sum of the squares of the auxiliary variables y_0[0], y_0[1], and y_0[2] does not exceed the square of t_0[0], the sum of the squares of y_1[0], y_1[1], and y_1[2] does not exceed the square of t_1[0], and the sum of the squares of y_2[0], y_2[1], and y_2[2] does not exceed the square of t_2[0]. All variables, including x[0], x[1], x[2], y_0[0], y_0[1], y_0[2], y_1[0], y_1[1], y_1[2], y_2[0], y_2[1], y_2[2], t_0[0], t_1[0], and t_2[0], are free to take any real value. The goal is to determine the values of these variables that minimize the objective function while satisfying all constraints.0.2077973044930690The problem involves minimizing a linear objective function defined by the coefficients 0.0169907447059004 for x[0, -0.1354658035191434 for x[1], and 0.1439797319070909 for x[2]. The optimization is subject to four linear equality constraints that balance the relationships between the variables x[0], x[1], and x[2]. The first constraint requires that 0.8636590559284738 times x[0] plus 0.0966833012301337 times x[1] minus 0.3152291953090132 times x[2] equals 0.8843372418353203. The second constraint requires that -1.308791116694014 times x[0] plus 1.026765786674754 times x[1] minus 1.641892627931514 times x[2] equals -3.068584167132096. The third constraint requires that -2.022894561574939 times x[0] plus 1.078027292135305 times x[1] minus 0.6806946931130862 times x[2] equals -3.878545364964277. The fourth constraint requires that 0.5909183755526639 times x[0] minus 1.227754194746086 times x[1] plus 0.9978468809169225 times x[2] equals 2.373932955061617. ] + + The problem also includes twelve additional linear constraints that introduce auxiliary variables y_0, y_1, y_2, and t_0, t_1, t_2. These constraints balance the primary variables x[0, x[1], and x[2] with the auxiliary variables. For example, constraint R4 requires that -0.0830895372455731 times x[0] plus 0.140196500160931 times x[1] plus 0.0835731570777216 times x[2] plus y_0[0] equals -0.0542756929304628. Similarly, constraints R5 through R15 impose similar relationships involving y_0[1], y_0[2], y_1[0], y_1[1], y_1[2], y_2[0], y_2[1], y_2[2], t_0[0], t_1[0], and t_2[0]. ] + + The problem also includes three second-order cone constraints, qc0, qc1, and qc2, which ensure that the sum of the squares of the auxiliary variables y_0[0, y_0[1], and y_0[2] does not exceed the square of t_0[0], the sum of the squares of y_1[0], y_1[1], and y_1[2] does not exceed the square of t_1[0], and the sum of the squares of y_2[0], y_2[1], and y_2[2] does not exceed the square of t_2[0]. All variables, including x[0], x[1], x[2], y_0[0], y_0[1], y_0[2], y_1[0], y_1[1], y_1[2], y_2[0], y_2[1], y_2[2], t_0[0], t_1[0], and t_2[0], are free to take any real value. The goal is to determine the values of these variables that minimize the objective function while satisfying all constraints.]0.207797304493069Quadratic, out-of-scopenon-linear problem
143Minimize the total cost of resource allocation, where the cost coefficients for the variables x[0] to x[6] are -0.2747522145296182, -1.766479699359979, 0.7843967105070561, 0.1559911408895964, -0.891327664664444, -0.2108190112467947, and -0.4779719262070241, respectively. The allocation must satisfy five equality constraints: +1. The allocation for x[0] multiplied by 1.015905804200365, minus the allocation for x[1] multiplied by 1.075563368516513, plus the allocation for x[2] multiplied by 1.354422581651652, plus the allocation for x[3] multiplied by 1.231175719391763, minus the allocation for x[4] multiplied by 0.5695498149899738, minus the allocation for x[5] multiplied by 0.0845431272925385, minus the allocation for x[6] multiplied by 0.945839019408856, must equal -0.0350333551254621. +2. The allocation for x[0] multiplied by 0.1021906812588868, plus the allocation for x[1] multiplied by 1.017649603494175, plus the allocation for x[2] multiplied by 0.7113625469010652, minus the allocation for x[3] multiplied by 0.7845975403828425, minus the allocation for x[4] multiplied by 0.1330475619361555, minus the allocation for x[5] multiplied by 0.2027851245783147, minus the allocation for x[6] multiplied by 0.4907635167746233, must equal -0.0890746280206778. +3. The allocation for x[0] multiplied by -0.0691151611879505, minus the allocation for x[1] multiplied by 0.1726234142770679, plus the allocation for x[2] multiplied by 0.7490870928543292, minus the allocation for x[3] multiplied by 0.0017337760468129, minus the allocation for x[4] multiplied by 0.9675143456769632, minus the allocation for x[5] multiplied by 0.6700478590428591, minus the allocation for x[6] multiplied by 0.3470673767061095, must equal -2.198650485794045. +4. The allocation for x[0] multiplied by -0.1853251444505973, minus the allocation for x[1] multiplied by 1.849600638660094, minus the allocation for x[2] multiplied by 1.508370034295556, minus the allocation for x[3] multiplied by 0.9668525673953132, minus the allocation for x[4] multiplied by 1.21156196565678, minus the allocation for x[5] multiplied by 0.6645469549418966, plus the allocation for x[6] multiplied by 1.661638743910969, must equal -4.829481983044288. +5. The allocation for x[0] multiplied by 0.8891607518271019, plus the allocation for x[1] multiplied by 0.8384286881235276, minus the allocation for x[2] multiplied by 0.8035031028628979, minus the allocation for x[3] multiplied by 1.092054353478263, plus the allocation for x[4] multiplied by 0.3420691219898985, plus the allocation for x[5] multiplied by 0.1291789846691225, plus the allocation for x[6] multiplied by 2.02029578986399, must equal 0.8991262576526948. + +Additionally, there are four linear constraints involving auxiliary variables y_0[0], y_0[1], y_0[2], and t_0[0]: +1. The allocation for x[0] multiplied by 0.1370623147413594, minus the allocation for x[1] multiplied by 0.0955523959954012, minus the allocation for x[2] multiplied by 0.1070159267992108, minus the allocation for x[3] multiplied by 0.044283559484152, plus the allocation for x[4] multiplied by 0.2052751838377566, minus the allocation for x[5] multiplied by 0.0472147055355524, plus the allocation for x[6] multiplied by 0.0263779101242163, plus the auxiliary variable y_0[0], must equal -0.1372906598086168. +2. The allocation for x[0] multiplied by -0.0864033173525697, minus the allocation for x[1] multiplied by 0.0242389263268966, plus the allocation for x[2] multiplied by 0.0198400163659476, minus the allocation for x[3] multiplied by 0.0529154412212283, plus the allocation for x[4] multiplied by 0.0919575570765138, minus the allocation for x[5] multiplied by 0.0465630198887696, minus the allocation for x[6] multiplied by 0.0415047822548149, plus the auxiliary variable y_0[1], must equal -0.1090956146371356. +3. The allocation for x[0] multiplied by -0.0493566419623072, plus the allocation for x[1] multiplied by 0.0190877037065351, minus the allocation for x[2] multiplied by 0.0625676089375108, plus the allocation for x[3] multiplied by 0.0152602958492, minus the allocation for x[4] multiplied by 0.00439851178580794, plus the allocation for x[5] multiplied by 0.1593300015509853, plus the allocation for x[6] multiplied by 0.1099457671812097, plus the auxiliary variable y_0[2], must equal -0.018160978920798. +4. The allocation for x[0] multiplied by 2.17906321051382, plus the allocation for x[1] multiplied by 0.2404505380691451, minus the allocation for x[2] multiplied by 1.273156876389871, minus the allocation for x[3] multiplied by 1.521652877877997, minus the allocation for x[4] multiplied by 0.9431372726898773, minus the allocation for x[5] multiplied by 0.9371176801678375, minus the allocation for x[6] multiplied by 1.072428152750414, plus the auxiliary variable t_0[0], must equal 1.636356226721274. + +A second-order cone constraint requires that the negative square of the auxiliary variable t_0[0] plus the sum of the squares of the auxiliary variables y_0[0], y_0[1], and y_0[2] must be less than or equal to zero. All decision variables, including x[0] to x[6], y_0[0], y_0[1], y_0[2], and t_0[0], are free and unbounded. The goal is to minimize the total cost while satisfying all these constraints.-3.7400219903199680Minimize the total cost of resource allocation, where the cost coefficients for the variables x[0 to x[6] are -0.2747522145296182, -1.766479699359979, 0.7843967105070561, 0.1559911408895964, -0.891327664664444, -0.2108190112467947, and -0.4779719262070241, respectively. The allocation must satisfy five equality constraints: ] + 1. The allocation for x[0 multiplied by 1.015905804200365, minus the allocation for x[1] multiplied by 1.075563368516513, plus the allocation for x[2] multiplied by 1.354422581651652, plus the allocation for x[3] multiplied by 1.231175719391763, minus the allocation for x[4] multiplied by 0.5695498149899738, minus the allocation for x[5] multiplied by 0.0845431272925385, minus the allocation for x[6] multiplied by 0.945839019408856, must equal -0.0350333551254621. ] + 2. The allocation for x[0 multiplied by 0.1021906812588868, plus the allocation for x[1] multiplied by 1.017649603494175, plus the allocation for x[2] multiplied by 0.7113625469010652, minus the allocation for x[3] multiplied by 0.7845975403828425, minus the allocation for x[4] multiplied by 0.1330475619361555, minus the allocation for x[5] multiplied by 0.2027851245783147, minus the allocation for x[6] multiplied by 0.4907635167746233, must equal -0.0890746280206778. ] + 3. The allocation for x[0 multiplied by -0.0691151611879505, minus the allocation for x[1] multiplied by 0.1726234142770679, plus the allocation for x[2] multiplied by 0.7490870928543292, minus the allocation for x[3] multiplied by 0.0017337760468129, minus the allocation for x[4] multiplied by 0.9675143456769632, minus the allocation for x[5] multiplied by 0.6700478590428591, minus the allocation for x[6] multiplied by 0.3470673767061095, must equal -2.198650485794045. ] + 4. The allocation for x[0 multiplied by -0.1853251444505973, minus the allocation for x[1] multiplied by 1.849600638660094, minus the allocation for x[2] multiplied by 1.508370034295556, minus the allocation for x[3] multiplied by 0.9668525673953132, minus the allocation for x[4] multiplied by 1.21156196565678, minus the allocation for x[5] multiplied by 0.6645469549418966, plus the allocation for x[6] multiplied by 1.661638743910969, must equal -4.829481983044288. ] + 5. The allocation for x[0 multiplied by 0.8891607518271019, plus the allocation for x[1] multiplied by 0.8384286881235276, minus the allocation for x[2] multiplied by 0.8035031028628979, minus the allocation for x[3] multiplied by 1.092054353478263, plus the allocation for x[4] multiplied by 0.3420691219898985, plus the allocation for x[5] multiplied by 0.1291789846691225, plus the allocation for x[6] multiplied by 2.02029578986399, must equal 0.8991262576526948. ] + + Additionally, there are four linear constraints involving auxiliary variables y_0[0, y_0[1], y_0[2], and t_0[0]: ] + 1. The allocation for x[0 multiplied by 0.1370623147413594, minus the allocation for x[1] multiplied by 0.0955523959954012, minus the allocation for x[2] multiplied by 0.1070159267992108, minus the allocation for x[3] multiplied by 0.044283559484152, plus the allocation for x[4] multiplied by 0.2052751838377566, minus the allocation for x[5] multiplied by 0.0472147055355524, plus the allocation for x[6] multiplied by 0.0263779101242163, plus the auxiliary variable y_0[0], must equal -0.1372906598086168. ] + 2. The allocation for x[0 multiplied by -0.0864033173525697, minus the allocation for x[1] multiplied by 0.0242389263268966, plus the allocation for x[2] multiplied by 0.0198400163659476, minus the allocation for x[3] multiplied by 0.0529154412212283, plus the allocation for x[4] multiplied by 0.0919575570765138, minus the allocation for x[5] multiplied by 0.0465630198887696, minus the allocation for x[6] multiplied by 0.0415047822548149, plus the auxiliary variable y_0[1], must equal -0.1090956146371356. ] + 3. The allocation for x[0 multiplied by -0.0493566419623072, plus the allocation for x[1] multiplied by 0.0190877037065351, minus the allocation for x[2] multiplied by 0.0625676089375108, plus the allocation for x[3] multiplied by 0.0152602958492, minus the allocation for x[4] multiplied by 0.00439851178580794, plus the allocation for x[5] multiplied by 0.1593300015509853, plus the allocation for x[6] multiplied by 0.1099457671812097, plus the auxiliary variable y_0[2], must equal -0.018160978920798. ] + 4. The allocation for x[0 multiplied by 2.17906321051382, plus the allocation for x[1] multiplied by 0.2404505380691451, minus the allocation for x[2] multiplied by 1.273156876389871, minus the allocation for x[3] multiplied by 1.521652877877997, minus the allocation for x[4] multiplied by 0.9431372726898773, minus the allocation for x[5] multiplied by 0.9371176801678375, minus the allocation for x[6] multiplied by 1.072428152750414, plus the auxiliary variable t_0[0], must equal 1.636356226721274. ] + + A second-order cone constraint requires that the negative square of the auxiliary variable t_0[0 plus the sum of the squares of the auxiliary variables y_0[0], y_0[1], and y_0[2] must be less than or equal to zero. All decision variables, including x[0] to x[6], y_0[0], y_0[1], y_0[2], and t_0[0], are free and unbounded. The goal is to minimize the total cost while satisfying all these constraints.]-3.740021990319968Quadratic, out-of-scopenon-linear problem
144The production manager must determine the optimal levels of three components, referred to as Component A, Component B, and Component C, to minimize the objective function defined by 1.069640799533941 times the level of Component A minus 1.616200142169644 times the level of Component B minus 1.486492226609768 times the level of Component C. The production levels of Components A, B, and C, as well as the auxiliary variables y_0[0], y_0[1], y_0[2], y_0[3], and t_0[0], are free to vary, and the process is subject to several constraints. + +First, the production levels must satisfy three linear equality constraints. The first requires that -1.44151739228461 times the level of Component A plus 3.168186412960231 times the level of Component B minus 1.939318306738097 times the level of Component C equals -0.1920883547257966. The second requires that -2.518157292057484 times the level of Component A minus 0.045850744657515 times the level of Component B plus 0.8181912066930375 times the level of Component C equals 4.883115462374718. The third requires that -1.850469185521618 times the level of Component A plus 0.4312499690020329 times the level of Component B minus 0.7557135832325144 times the level of Component C equals 0.092955763950354. + +Additionally, the production process involves four auxiliary variables, y_0[0], y_0[1], y_0[2], and y_0[3], which must satisfy four resource allocation constraints. The first requires that -0.0609720704349455 times the level of Component A plus 0.0577626224097794 times the level of Component B plus 0.1279201721670446 times the level of Component C plus y_0[0] equals 0.1618518754288826. The second requires that -0.033065184399102 times the level of Component A plus 0.0448028105084557 times the level of Component B minus 0.1129590164777731 times the level of Component C plus y_0[1] equals -0.0270310934062265. The third requires that 0.084397043540958 times the level of Component A plus 0.0393016756676922 times the level of Component B plus 0.0346841557930454 times the level of Component C plus y_0[2] equals -0.0279180675961654. The fourth requires that 0.0184000253815845 times the level of Component A minus 0.00348695513114734 times the level of Component B minus 0.069624774030688 times the level of Component C plus y_0[3] equals 0.1874514710284287. + +A second-order cone constraint must also be satisfied, where the sum of the squares of y_0[0], y_0[1], y_0[2], and y_0[3] minus the square of t_0[0] must be less than or equal to zero. The variable t_0[0] is determined by the production levels of Components A, B, and C, such that -0.4776208627305155 times the level of Component A minus 0.0366363201101301 times the level of Component B minus 0.5186224190772866 times the level of Component C plus t_0[0] equals 0.309414690634093. + +The objective is to minimize the value of the objective function while satisfying all the above constraints and ensuring that the levels of Components A, B, and C, as well as the auxiliary variables y_0[0], y_0[1], y_0[2], y_0[3], and t_0[0], are free to vary within the given constraints. This problem is a Second-Order Cone Programming (SOCP) problem, characterized by the second-order cone constraint that ties the auxiliary variables together in a specific relationship.-7.7565049101077970The production manager must determine the optimal levels of three components, referred to as Component A, Component B, and Component C, to minimize the objective function defined by 1.069640799533941 times the level of Component A minus 1.616200142169644 times the level of Component B minus 1.486492226609768 times the level of Component C. The production levels of Components A, B, and C, as well as the auxiliary variables y_0[0, y_0[1], y_0[2], y_0[3], and t_0[0], are free to vary, and the process is subject to several constraints.] + + First, the production levels must satisfy three linear equality constraints. The first requires that -1.44151739228461 times the level of Component A plus 3.168186412960231 times the level of Component B minus 1.939318306738097 times the level of Component C equals -0.1920883547257966. The second requires that -2.518157292057484 times the level of Component A minus 0.045850744657515 times the level of Component B plus 0.8181912066930375 times the level of Component C equals 4.883115462374718. The third requires that -1.850469185521618 times the level of Component A plus 0.4312499690020329 times the level of Component B minus 0.7557135832325144 times the level of Component C equals 0.092955763950354. + + Additionally, the production process involves four auxiliary variables, y_0[0, y_0[1], y_0[2], and y_0[3], which must satisfy four resource allocation constraints. The first requires that -0.0609720704349455 times the level of Component A plus 0.0577626224097794 times the level of Component B plus 0.1279201721670446 times the level of Component C plus y_0[0] equals 0.1618518754288826. The second requires that -0.033065184399102 times the level of Component A plus 0.0448028105084557 times the level of Component B minus 0.1129590164777731 times the level of Component C plus y_0[1] equals -0.0270310934062265. The third requires that 0.084397043540958 times the level of Component A plus 0.0393016756676922 times the level of Component B plus 0.0346841557930454 times the level of Component C plus y_0[2] equals -0.0279180675961654. The fourth requires that 0.0184000253815845 times the level of Component A minus 0.00348695513114734 times the level of Component B minus 0.069624774030688 times the level of Component C plus y_0[3] equals 0.1874514710284287.] + + A second-order cone constraint must also be satisfied, where the sum of the squares of y_0[0, y_0[1], y_0[2], and y_0[3] minus the square of t_0[0] must be less than or equal to zero. The variable t_0[0] is determined by the production levels of Components A, B, and C, such that -0.4776208627305155 times the level of Component A minus 0.0366363201101301 times the level of Component B minus 0.5186224190772866 times the level of Component C plus t_0[0] equals 0.309414690634093.] + + The objective is to minimize the value of the objective function while satisfying all the above constraints and ensuring that the levels of Components A, B, and C, as well as the auxiliary variables y_0[0, y_0[1], y_0[2], y_0[3], and t_0[0], are free to vary within the given constraints. This problem is a Second-Order Cone Programming (SOCP) problem, characterized by the second-order cone constraint that ties the auxiliary variables together in a specific relationship.]-7.756504910107797Quadratic, out-of-scopenon-linear problem
145Minimize the objective function with cost coefficients -0.1919592146476727 for x[0], -1.473647303839492 for x[1], and 2.304735407761341 for x[2]. The problem is subject to three strict equality constraints: (1) -1.635895473616174 x[0] - 0.3973211001807447 x[1] + 0.9471007364101932 x[2] must equal -1.151224267166901, labeled as linear_eq[0]; (2) -0.5249535603075616 x[0] + 0.3668073807989349 x[1] - 0.7858336216136411 x[2] must equal 0.2111440590441619, labeled as linear_eq[1]; and (3) -1.91474276776438 x[0] - 0.0172618223950067 x[1] - 0.4534063203075578 x[2] must equal -0.0637133383926026, labeled as linear_eq[2]. + +Additionally, four auxiliary variables y_0[0], y_0[1], y_0[2], and y_0[3] are introduced, each subject to linear constraints: (1) -0.1884394252535835 x[0] + 0.0177104249784969 x[1] + 0.0986130842848287 x[2] + y_0[0] must equal 0.0043955531724021, labeled as R3; (2) -0.0708818331240574 x[0] - 0.0215819660876361 x[1] - 0.0616654799104094 x[2] + y_0[1] must equal 0.0879019749446116, labeled as R4; (3) -0.0907143884234421 x[0] + 0.0750697315409076 x[1] - 0.1169779589661307 x[2] + y_0[2] must equal 0.0357726532121181, labeled as R5; and (4) -0.1820309484664378 x[0] + 0.0261698752429371 x[1] + 0.0667294286775777 x[2] + y_0[3] must equal 0.0118088378651355, labeled as R6. + +A decision variable t_0[0] is introduced, subject to the constraint -0.2396501315769039 x[0] + 0.3255209569765619 x[1] + 0.2624492403208943 x[2] + t_0[0] must equal 1.588464205354944, labeled as R7. A second-order cone constraint labeled as qc0 is imposed, ensuring that the sum of the squares of y_0[0], y_0[1], y_0[2], and y_0[3] does not exceed the square of t_0[0], expressed as - t_0[0]^2 + y_0[0]^2 + y_0[1]^2 + y_0[2]^2 + y_0[3]^2 <= 0. All decision variables, including x[0], x[1], x[2], y_0[0], y_0[1], y_0[2], y_0[3], and t_0[0], are free to take any real value. The goal is to determine the optimal values for these variables to minimize the objective function while satisfying all constraints. This is a Second-Order Cone Programming (SOCP) problem.-0.5553452714909540Minimize the objective function with cost coefficients -0.1919592146476727 for x[0, -1.473647303839492 for x[1], and 2.304735407761341 for x[2]. The problem is subject to three strict equality constraints: (1) -1.635895473616174 x[0] - 0.3973211001807447 x[1] + 0.9471007364101932 x[2] must equal -1.151224267166901, labeled as linear_eq[0]; (2) -0.5249535603075616 x[0] + 0.3668073807989349 x[1] - 0.7858336216136411 x[2] must equal 0.2111440590441619, labeled as linear_eq[1]; and (3) -1.91474276776438 x[0] - 0.0172618223950067 x[1] - 0.4534063203075578 x[2] must equal -0.0637133383926026, labeled as linear_eq[2].] + + Additionally, four auxiliary variables y_0[0, y_0[1], y_0[2], and y_0[3] are introduced, each subject to linear constraints: (1) -0.1884394252535835 x[0] + 0.0177104249784969 x[1] + 0.0986130842848287 x[2] + y_0[0] must equal 0.0043955531724021, labeled as R3; (2) -0.0708818331240574 x[0] - 0.0215819660876361 x[1] - 0.0616654799104094 x[2] + y_0[1] must equal 0.0879019749446116, labeled as R4; (3) -0.0907143884234421 x[0] + 0.0750697315409076 x[1] - 0.1169779589661307 x[2] + y_0[2] must equal 0.0357726532121181, labeled as R5; and (4) -0.1820309484664378 x[0] + 0.0261698752429371 x[1] + 0.0667294286775777 x[2] + y_0[3] must equal 0.0118088378651355, labeled as R6.] + + A decision variable t_0[0 is introduced, subject to the constraint -0.2396501315769039 x[0] + 0.3255209569765619 x[1] + 0.2624492403208943 x[2] + t_0[0] must equal 1.588464205354944, labeled as R7. A second-order cone constraint labeled as qc0 is imposed, ensuring that the sum of the squares of y_0[0], y_0[1], y_0[2], and y_0[3] does not exceed the square of t_0[0], expressed as - t_0[0]^2 + y_0[0]^2 + y_0[1]^2 + y_0[2]^2 + y_0[3]^2 <= 0. All decision variables, including x[0], x[1], x[2], y_0[0], y_0[1], y_0[2], y_0[3], and t_0[0], are free to take any real value. The goal is to determine the optimal values for these variables to minimize the objective function while satisfying all constraints. This is a Second-Order Cone Programming (SOCP) problem.]-0.555345271490954Quadratic, out-of-scopenon-linear problem
146Minimize the objective function defined as 1.416733013504254 times the first decision variable minus 1.637403309183765 times the second decision variable plus 0.4022638456918747 times the third decision variable. The problem includes three linear equality constraints: the first equality constraint specifies that 0.1616012453467476 times the first variable minus 0.8306137706187448 times the second variable minus 0.1515288797410668 times the third variable equals -1.643419668997376. The second equality constraint specifies that 1.456169241005894 times the first variable plus 0.5395849002520692 times the second variable minus 0.8149223903970267 times the third variable equals 1.699940269626918. The third equality constraint specifies that 0.4277456985899902 times the first variable minus 0.5463295579766617 times the second variable plus 1.434710626115096 times the third variable equals -3.331337036084423. + +There are ten linear constraints involving auxiliary and throughput variables. The first three constraints involve the first set of auxiliary variables: the first constraint specifies that 0.0771566555713033 times the first variable plus 0.1116037485944799 times the second variable minus 0.0869350722337483 times the third variable plus the first auxiliary variable equals 0.0620112102329036. The second constraint specifies that 0.0616561304966069 times the first variable minus 0.0974104451697517 times the second variable plus 0.0731223386150885 times the third variable plus the second auxiliary variable equals -0.0587197478672789. The third constraint specifies that -0.0081168698754745 times the first variable plus 0.1006466302551332 times the second variable plus 0.0871949253632896 times the third variable plus the third auxiliary variable equals 0.1492287513423473. The fourth constraint involves the first throughput variable and specifies that 0.7878086620078162 times the first variable minus 1.237192350711952 times the second variable plus 0.0891701482981527 times the third variable plus the first throughput variable equals 0.0451513700840946. + +The next three constraints involve the second set of auxiliary variables: the fifth constraint specifies that -0.0877923363815518 times the first variable plus 0.0435833077366723 times the second variable plus 0.2441522313227069 times the third variable plus the fourth auxiliary variable equals -0.0455285314769021. The sixth constraint specifies that 0.1269499650822233 times the first variable minus 0.1249417310147306 times the second variable plus 0.0823756522034219 times the third variable plus the fifth auxiliary variable equals 0.1088319552421044. The seventh constraint specifies that -6.71937e-07 times the first variable plus 0.1218393680025435 times the second variable plus 0.1838094221276996 times the third variable plus the sixth auxiliary variable equals -0.0248992942856434. The eighth constraint involves the second throughput variable and specifies that 1.611447749097356 times the first variable minus 0.9783611539447722 times the second variable minus 0.3250051330062018 times the third variable plus the second throughput variable equals 0.8152009008108351. + +Two second-order cone constraints are included: the first specifies that the square of the first throughput variable subtracted from the sum of the squares of the first three auxiliary variables must be less than or equal to zero. The second specifies that the square of the second throughput variable subtracted from the sum of the squares of the second three auxiliary variables must be less than or equal to zero. All decision variables, auxiliary variables, and throughput variables are unrestricted in sign and can take any real value.-4.6516953177848110Minimize the objective function defined as 1.416733013504254 times the first decision variable minus 1.637403309183765 times the second decision variable plus 0.4022638456918747 times the third decision variable. The problem includes three linear equality constraints: the first equality constraint specifies that 0.1616012453467476 times the first variable minus 0.8306137706187448 times the second variable minus 0.1515288797410668 times the third variable equals -1.643419668997376. The second equality constraint specifies that 1.456169241005894 times the first variable plus 0.5395849002520692 times the second variable minus 0.8149223903970267 times the third variable equals 1.699940269626918. The third equality constraint specifies that 0.4277456985899902 times the first variable minus 0.5463295579766617 times the second variable plus 1.434710626115096 times the third variable equals -3.331337036084423. + + There are ten linear constraints involving auxiliary and throughput variables. The first three constraints involve the first set of auxiliary variables: the first constraint specifies that 0.0771566555713033 times the first variable plus 0.1116037485944799 times the second variable minus 0.0869350722337483 times the third variable plus the first auxiliary variable equals 0.0620112102329036. The second constraint specifies that 0.0616561304966069 times the first variable minus 0.0974104451697517 times the second variable plus 0.0731223386150885 times the third variable plus the second auxiliary variable equals -0.0587197478672789. The third constraint specifies that -0.0081168698754745 times the first variable plus 0.1006466302551332 times the second variable plus 0.0871949253632896 times the third variable plus the third auxiliary variable equals 0.1492287513423473. The fourth constraint involves the first throughput variable and specifies that 0.7878086620078162 times the first variable minus 1.237192350711952 times the second variable plus 0.0891701482981527 times the third variable plus the first throughput variable equals 0.0451513700840946. + + The next three constraints involve the second set of auxiliary variables: the fifth constraint specifies that -0.0877923363815518 times the first variable plus 0.0435833077366723 times the second variable plus 0.2441522313227069 times the third variable plus the fourth auxiliary variable equals -0.0455285314769021. The sixth constraint specifies that 0.1269499650822233 times the first variable minus 0.1249417310147306 times the second variable plus 0.0823756522034219 times the third variable plus the fifth auxiliary variable equals 0.1088319552421044. The seventh constraint specifies that -6.71937e-07 times the first variable plus 0.1218393680025435 times the second variable plus 0.1838094221276996 times the third variable plus the sixth auxiliary variable equals -0.0248992942856434. The eighth constraint involves the second throughput variable and specifies that 1.611447749097356 times the first variable minus 0.9783611539447722 times the second variable minus 0.3250051330062018 times the third variable plus the second throughput variable equals 0.8152009008108351. + + Two second-order cone constraints are included: the first specifies that the square of the first throughput variable subtracted from the sum of the squares of the first three auxiliary variables must be less than or equal to zero. The second specifies that the square of the second throughput variable subtracted from the sum of the squares of the second three auxiliary variables must be less than or equal to zero. All decision variables, auxiliary variables, and throughput variables are unrestricted in sign and can take any real value.-4.651695317784811Quadratic, out-of-scopenon-linear problem
147The goal is to minimize the total cost determined by the weighted sum of resources \( x[0] \), \( x[1] \), and \( x[2] \), where \( x[0] \) contributes 1.161237384198499 units of cost per unit allocated, \( x[1] \) reduces the cost by 0.166305801898613 units per unit allocated, and \( x[2] \) reduces the cost by 0.4010222263993871 units per unit allocated. The allocation must satisfy four strict equality conditions: the first requires -0.8171434027373111 units of \( x[0] \), 0.0148992990791839 units of \( x[1] \), and -1.624552817012578 units of \( x[2] \) to equal -2.266806032684915; the second requires 0.5642952588239045 units of \( x[0] \), -1.323990231764886 units of \( x[1] \), and 1.432095008297531 units of \( x[2] \) to equal 2.393666956177287; the third requires -0.5423126832826126 units of \( x[0] \), -0.6668037930363615 units of \( x[1] \), and -0.8987771739249211 units of \( x[2] \) to equal -1.048625110146119; and the fourth requires 0.398575737860184 units of \( x[0] \), -1.340131134270142 units of \( x[1] \), and 0.3595004047412258 units of \( x[2] \) to equal 0.8340459801039611. + +Additional constraints involve auxiliary variables \( y_0[0] \) to \( y_0[3] \), \( y_1[0] \) to \( y_1[3] \), \( y_2[0] \) to \( y_2[3] \), and \( t_0[0] \), \( t_1[0] \), \( t_2[0] \). For the first set, \( y_0[0] \) is determined by -0.0790956451487926 units of \( x[0] \), -0.00177045180427562 units of \( x[1] \), and -0.0135101531022542 units of \( x[2] \), equal to 4.3318049409104566e-04; \( y_0[1] \) is determined by -0.1888423336020157 units of \( x[0] \), 0.0468378089437413 units of \( x[1] \), and 0.0422915893690382 units of \( x[2] \), equal to -0.057473642380715; \( y_0[2] \) is determined by 0.1111684227208376 units of \( x[0] \), 0.1168866454616669 units of \( x[1] \), and 0.00994317726502118 units of \( x[2] \), equal to -0.0027025235150565; \( y_0[3] \) is determined by -0.1003054361114799 units of \( x[0] \), -0.1074280721090408 units of \( x[1] \), and 0.0347646765882297 units of \( x[2] \), equal to -0.0187719281760365; \( t_0[0] \) is determined by -0.0923314568919683 units of \( x[0] \), 0.3059364524963586 units of \( x[1] \), and -0.4891183949819865 units of \( x[2] \), equal to 1.559548175449872. + +For \( y_1[0] \), the sum is 0.0983461752470471, determined by 0.0260300607215097 units of \( x[0] \), 0.104222743854305 units of \( x[1] \), and 0.165095052940775 units of \( x[2] \); \( y_1[1] \) sums to 0.2590974156103631, determined by 3.9617645465962872e-04 units of \( x[0] \), -0.0842482365695222 units of \( x[1] \), and 0.2115742333860741 units of \( x[2] \); \( y_1[2] \) sums to 0.0656791263260323, determined by 0.0469551281711486 units of \( x[0] \), -0.0279863701043123 units of \( x[1] \), and -0.2267772085587545 units of \( x[2] \); \( y_1[3] \) sums to 0.0925284262906007, determined by -0.00769378398286824 units of \( x[0] \), -0.1620714406841336 units of \( x[1] \), and -0.137052404669175 units of \( x[2] \); \( t_1[0] \) sums to 0.9781052631063663, determined by -0.4553968316872111 units of \( x[0] \), -0.8477088153755726 units of \( x[1] \), and -0.4607927423478093 units of \( x[2] \). + +For \( y_2[0] \), the sum is -0.0316470399294818, determined by -0.0587561844433846 units of \( x[0] \), -0.0374159998289655 units of \( x[1] \), and -0.1432421286622242 units of \( x[2] \); \( y_2[1] \) sums to 0.1193203771742649, determined by -0.0483997505258602 units of \( x[0] \), 0.1896458148433305 units of \( x[1] \), and -0.0709856184240573 units of \( x[2] \); \( y_2[2] \) sums to 0.0249940394491055, determined by 0.0846693606546272 units of \( x[0] \), 0.0686759181099283 units of \( x[1] \), and -0.0216946264193857 units of \( x[2] \); \( y_2[3] \) sums to -0.0230055775535458, determined by 0.2218191833614268 units of \( x[0] \), 0.054166153139831 units of \( x[1] \), and 0.0802604337456637 units of \( x[2] \); \( t_2[0] \) sums to 0.0133770717575745, determined by -2.322162140550891 units of \( x[0] \), 0.6539919936190184 units of \( x[1] \), and -0.5823100315050422 units of \( x[2] \). + +The problem also includes second-order cone constraints: the first ensures that the sum of the squares of \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), and \( y_0[3] \) is less than or equal to the square of \( t_0[0] \); the second ensures that the sum of the squares of \( y_1[0] \), \( y_1[1] \), \( y_1[2] \), and \( y_1[3] \) is less than or equal to the square of \( t_1[0] \); the third ensures that the sum of the squares of \( y_2[0] \), \( y_2[1] \), \( y_2[2] \), and \( y_2[3] \) is less than or equal to the square of \( t_2[0] \). All decision variables, including \( x[0] \), \( x[1] \), \( x[2] \), \( y_0[0] \) to \( y_0[3] \), \( y_1[0] \) to \( y_1[3] \), \( y_2[0] \) to \( y_2[3] \), and \( t_0[0] \), \( t_1[0] \), \( t_2[0] \), are free and can take any real value. The task is to allocate resources \( x[0] \), \( x[1] \), and \( x[2] \) to minimize the total cost, while satisfying all equality constraints, resource allocation constraints, and second-order cone constraints. The specific numerical values and relationships provided must be strictly adhered to in order to achieve an optimal solution.-0.7682069972292490The goal is to minimize the total cost determined by the weighted sum of resources \( x[0 \), \( x[1] \), and \( x[2] \), where \( x[0] \) contributes 1.161237384198499 units of cost per unit allocated, \( x[1] \) reduces the cost by 0.166305801898613 units per unit allocated, and \( x[2] \) reduces the cost by 0.4010222263993871 units per unit allocated. The allocation must satisfy four strict equality conditions: the first requires -0.8171434027373111 units of \( x[0] \), 0.0148992990791839 units of \( x[1] \), and -1.624552817012578 units of \( x[2] \) to equal -2.266806032684915; the second requires 0.5642952588239045 units of \( x[0] \), -1.323990231764886 units of \( x[1] \), and 1.432095008297531 units of \( x[2] \) to equal 2.393666956177287; the third requires -0.5423126832826126 units of \( x[0] \), -0.6668037930363615 units of \( x[1] \), and -0.8987771739249211 units of \( x[2] \) to equal -1.048625110146119; and the fourth requires 0.398575737860184 units of \( x[0] \), -1.340131134270142 units of \( x[1] \), and 0.3595004047412258 units of \( x[2] \) to equal 0.8340459801039611.] + + Additional constraints involve auxiliary variables \( y_0[0 \) to \( y_0[3] \), \( y_1[0] \) to \( y_1[3] \), \( y_2[0] \) to \( y_2[3] \), and \( t_0[0] \), \( t_1[0] \), \( t_2[0] \). For the first set, \( y_0[0] \) is determined by -0.0790956451487926 units of \( x[0] \), -0.00177045180427562 units of \( x[1] \), and -0.0135101531022542 units of \( x[2] \), equal to 4.3318049409104566e-04; \( y_0[1] \) is determined by -0.1888423336020157 units of \( x[0] \), 0.0468378089437413 units of \( x[1] \), and 0.0422915893690382 units of \( x[2] \), equal to -0.057473642380715; \( y_0[2] \) is determined by 0.1111684227208376 units of \( x[0] \), 0.1168866454616669 units of \( x[1] \), and 0.00994317726502118 units of \( x[2] \), equal to -0.0027025235150565; \( y_0[3] \) is determined by -0.1003054361114799 units of \( x[0] \), -0.1074280721090408 units of \( x[1] \), and 0.0347646765882297 units of \( x[2] \), equal to -0.0187719281760365; \( t_0[0] \) is determined by -0.0923314568919683 units of \( x[0] \), 0.3059364524963586 units of \( x[1] \), and -0.4891183949819865 units of \( x[2] \), equal to 1.559548175449872.] + + For \( y_1[0 \), the sum is 0.0983461752470471, determined by 0.0260300607215097 units of \( x[0] \), 0.104222743854305 units of \( x[1] \), and 0.165095052940775 units of \( x[2] \); \( y_1[1] \) sums to 0.2590974156103631, determined by 3.9617645465962872e-04 units of \( x[0] \), -0.0842482365695222 units of \( x[1] \), and 0.2115742333860741 units of \( x[2] \); \( y_1[2] \) sums to 0.0656791263260323, determined by 0.0469551281711486 units of \( x[0] \), -0.0279863701043123 units of \( x[1] \), and -0.2267772085587545 units of \( x[2] \); \( y_1[3] \) sums to 0.0925284262906007, determined by -0.00769378398286824 units of \( x[0] \), -0.1620714406841336 units of \( x[1] \), and -0.137052404669175 units of \( x[2] \); \( t_1[0] \) sums to 0.9781052631063663, determined by -0.4553968316872111 units of \( x[0] \), -0.8477088153755726 units of \( x[1] \), and -0.4607927423478093 units of \( x[2] \).] + + For \( y_2[0 \), the sum is -0.0316470399294818, determined by -0.0587561844433846 units of \( x[0] \), -0.0374159998289655 units of \( x[1] \), and -0.1432421286622242 units of \( x[2] \); \( y_2[1] \) sums to 0.1193203771742649, determined by -0.0483997505258602 units of \( x[0] \), 0.1896458148433305 units of \( x[1] \), and -0.0709856184240573 units of \( x[2] \); \( y_2[2] \) sums to 0.0249940394491055, determined by 0.0846693606546272 units of \( x[0] \), 0.0686759181099283 units of \( x[1] \), and -0.0216946264193857 units of \( x[2] \); \( y_2[3] \) sums to -0.0230055775535458, determined by 0.2218191833614268 units of \( x[0] \), 0.054166153139831 units of \( x[1] \), and 0.0802604337456637 units of \( x[2] \); \( t_2[0] \) sums to 0.0133770717575745, determined by -2.322162140550891 units of \( x[0] \), 0.6539919936190184 units of \( x[1] \), and -0.5823100315050422 units of \( x[2] \).] + + The problem also includes second-order cone constraints: the first ensures that the sum of the squares of \( y_0[0 \), \( y_0[1] \), \( y_0[2] \), and \( y_0[3] \) is less than or equal to the square of \( t_0[0] \); the second ensures that the sum of the squares of \( y_1[0] \), \( y_1[1] \), \( y_1[2] \), and \( y_1[3] \) is less than or equal to the square of \( t_1[0] \); the third ensures that the sum of the squares of \( y_2[0] \), \( y_2[1] \), \( y_2[2] \), and \( y_2[3] \) is less than or equal to the square of \( t_2[0] \). All decision variables, including \( x[0] \), \( x[1] \), \( x[2] \), \( y_0[0] \) to \( y_0[3] \), \( y_1[0] \) to \( y_1[3] \), \( y_2[0] \) to \( y_2[3] \), and \( t_0[0] \), \( t_1[0] \), \( t_2[0] \), are free and can take any real value. The task is to allocate resources \( x[0] \), \( x[1] \), and \( x[2] \) to minimize the total cost, while satisfying all equality constraints, resource allocation constraints, and second-order cone constraints. The specific numerical values and relationships provided must be strictly adhered to in order to achieve an optimal solution.]-0.768206997229249Quadratic, out-of-scopenon-linear problem
148The problem is a Second-Order Cone Programming (SOCP) problem aimed at minimizing the expression: -0.0834997320390843 multiplied by x[0], minus 1.281157927338681 multiplied by x[1], minus 0.6997343607099248 multiplied by x[2], minus 1.124765322700406 multiplied by x[3], plus 1.075870893520574 multiplied by x[4], minus 0.873635307005547 multiplied by x[5]. The problem includes four linear equality constraints involving the decision variables x[0] through x[5]. The first constraint requires -0.5885607904893205 multiplied by x[0], minus 0.4065465515131446 multiplied by x[1], plus 0.3059688687379479 multiplied by x[2], plus 0.5752454203843212 multiplied by x[3], plus 0.8134273230222456 multiplied by x[4], minus 0.3137959036191741 multiplied by x[5] to equal 0.9326117395439714. The second constraint requires -1.925586727059667 multiplied by x[0], plus 1.52432983869767 multiplied by x[1], plus 0.7056564744559519 multiplied by x[2], plus 0.4429646642871883 multiplied by x[3], minus 1.191971501984802 multiplied by x[4], minus 1.423701744010086 multiplied by x[5] to equal -1.127568679343468. The third constraint specifies 0.3695976409523919 multiplied by x[0], minus 1.172799439640038 multiplied by x[1], plus 2.456480065266749 multiplied by x[2], minus 0.6028161578986598 multiplied by x[3], plus 0.1526235927792681 multiplied by x[4], minus 0.7698423890950179 multiplied by x[5] to equal -0.464660190344107. The fourth constraint mandates -1.43554468908569 multiplied by x[0], plus 1.029767278463664 multiplied by x[1], minus 0.22200928921289 multiplied by x[2], minus 1.475662550421019 multiplied by x[3], minus 0.4673380914761413 multiplied by x[4], plus 1.197928628533288 multiplied by x[5] to equal -1.088009446155389. + +There are ten additional linear constraints involving auxiliary variables y_0[0], y_0[1], y_0[2], y_0[3], y_1[0], y_1[1], y_1[2], y_1[3], t_0[0], and t_1[0]. These constraints define specific linear relationships between the decision variables x[0] through x[5] and the auxiliary variables. For example, one constraint requires 0.0893317393307157 multiplied by x[0], plus 0.0150647185847581 multiplied by x[1], plus 0.0261787249528073 multiplied by x[2], plus 0.0752497482880115 multiplied by x[3], plus 0.089362702614514 multiplied by x[4], minus 0.0903365860453957 multiplied by x[5], plus y_0[0] to equal 0.0348229641116467. Another constraint requires 0.0309053187877012 multiplied by x[0], plus 0.024045043885088 multiplied by x[1], plus 0.1070923676270683 multiplied by x[2], plus 0.2073521720181627 multiplied by x[3], minus 0.1770274961116007 multiplied by x[4], minus 0.0922005916406162 multiplied by x[5], plus y_0[1] to equal -0.1872964740542509. A third constraint requires -0.1262388132745864 multiplied by x[0], plus 0.0795068962664749 multiplied by x[1], plus 0.0187856757413385 multiplied by x[2], plus 0.1513494147153156 multiplied by x[3], minus 0.0163664274157314 multiplied by x[4], minus 0.00311286020002606 multiplied by x[5], plus y_0[2] to equal -0.0954444159908309. A fourth constraint requires -0.0757819715079799 multiplied by x[0], plus 0.0919093860274552 multiplied by x[1], minus 0.0364339288064356 multiplied by x[2], plus 0.083580945500982 multiplied by x[3], minus 0.1474689439069835 multiplied by x[4], plus 0.0301651918491998 multiplied by x[5], plus y_0[3] to equal 0.2579565194918507. A fifth constraint requires 0.7572820533712606 multiplied by x[0], minus 1.21352962362722 multiplied by x[1], plus 0.3803676681799136 multiplied by x[2], plus 0.4801395423918295 multiplied by x[3], minus 0.8927539207272388 multiplied by x[4], minus 1.447471364330223 multiplied by x[5], plus t_0[0] to equal 0.683762161476536. A sixth constraint requires -0.039176642120947 multiplied by x[0], minus 0.0594029213045902 multiplied by x[1], plus 0.0203899965676045 multiplied by x[2], minus 0.0006599505607145948 multiplied by x[3], plus 0.1993498474530564 multiplied by x[4], minus 0.1494561245854857 multiplied by x[5], plus y_1[0] to equal -0.0909307736334353. A seventh constraint requires -0.1551511441179779 multiplied by x[0], plus 0.0850834698172653 multiplied by x[1], minus 0.1602283190962608 multiplied by x[2], plus 0.0725598573219744 multiplied by x[3], minus 0.1055576967228311 multiplied by x[4], minus 0.1582958490297426 multiplied by x[5], plus y_1[1] to equal -0.1725026750262158. An eighth constraint requires -0.1933585329412353 multiplied by x[0], plus 0.0929980184151937 multiplied by x[1], plus 0.099592631874656 multiplied by x[2], plus 0.0797650799817848 multiplied by x[3], plus 0.0106313375995685 multiplied by x[4], minus 0.00903979011317048 multiplied by x[5], plus y_1[2] to equal -0.1167161263783248. A ninth constraint requires 0.0230142114746871 multiplied by x[0], plus 0.0796690661109758 multiplied by x[1], plus 0.0216327232273718 multiplied by x[2], minus 0.0337376899834936 multiplied by x[3], minus 0.1518320281674425 multiplied by x[4], plus 0.046720461629375 multiplied by x[5], plus y_1[3] to equal -0.1005698418344237. A tenth constraint requires -0.1501707210081249 multiplied by x[0], plus 0.7058311865593911 multiplied by x[1], plus 1.287754324454335 multiplied by x[2], plus 0.0373827245849037 multiplied by x[3], plus 0.1171725723905398 multiplied by x[4], plus 0.1069533502066662 multiplied by x[5], plus t_1[0] to equal 0.1223510811918817. + +Two quadratic cone constraints ensure that the squared sum of y_0[0], y_0[1], y_0[2], and y_0[3] does not exceed the square of t_0[0], and the squared sum of y_1[0], y_1[1], y_1[2], and y_1[3] does not exceed the square of t_1[0]. All variables, including the decision variables x[0] through x[5] and the auxiliary variables y_0[0], y_0[1], y_0[2], y_0[3], y_1[0], y_1[1], y_1[2], y_1[3], t_0[0], and t_1[0], are free to take any real value, with no bounds specified. The goal is to determine the values of these variables that minimize the objective function while satisfying all equality and quadratic cone constraints.-2.0438316416979130The problem is a Second-Order Cone Programming (SOCP) problem aimed at minimizing the expression: -0.0834997320390843 multiplied by x[0, minus 1.281157927338681 multiplied by x[1], minus 0.6997343607099248 multiplied by x[2], minus 1.124765322700406 multiplied by x[3], plus 1.075870893520574 multiplied by x[4], minus 0.873635307005547 multiplied by x[5]. The problem includes four linear equality constraints involving the decision variables x[0] through x[5]. The first constraint requires -0.5885607904893205 multiplied by x[0], minus 0.4065465515131446 multiplied by x[1], plus 0.3059688687379479 multiplied by x[2], plus 0.5752454203843212 multiplied by x[3], plus 0.8134273230222456 multiplied by x[4], minus 0.3137959036191741 multiplied by x[5] to equal 0.9326117395439714. The second constraint requires -1.925586727059667 multiplied by x[0], plus 1.52432983869767 multiplied by x[1], plus 0.7056564744559519 multiplied by x[2], plus 0.4429646642871883 multiplied by x[3], minus 1.191971501984802 multiplied by x[4], minus 1.423701744010086 multiplied by x[5] to equal -1.127568679343468. The third constraint specifies 0.3695976409523919 multiplied by x[0], minus 1.172799439640038 multiplied by x[1], plus 2.456480065266749 multiplied by x[2], minus 0.6028161578986598 multiplied by x[3], plus 0.1526235927792681 multiplied by x[4], minus 0.7698423890950179 multiplied by x[5] to equal -0.464660190344107. The fourth constraint mandates -1.43554468908569 multiplied by x[0], plus 1.029767278463664 multiplied by x[1], minus 0.22200928921289 multiplied by x[2], minus 1.475662550421019 multiplied by x[3], minus 0.4673380914761413 multiplied by x[4], plus 1.197928628533288 multiplied by x[5] to equal -1.088009446155389.] + + There are ten additional linear constraints involving auxiliary variables y_0[0, y_0[1], y_0[2], y_0[3], y_1[0], y_1[1], y_1[2], y_1[3], t_0[0], and t_1[0]. These constraints define specific linear relationships between the decision variables x[0] through x[5] and the auxiliary variables. For example, one constraint requires 0.0893317393307157 multiplied by x[0], plus 0.0150647185847581 multiplied by x[1], plus 0.0261787249528073 multiplied by x[2], plus 0.0752497482880115 multiplied by x[3], plus 0.089362702614514 multiplied by x[4], minus 0.0903365860453957 multiplied by x[5], plus y_0[0] to equal 0.0348229641116467. Another constraint requires 0.0309053187877012 multiplied by x[0], plus 0.024045043885088 multiplied by x[1], plus 0.1070923676270683 multiplied by x[2], plus 0.2073521720181627 multiplied by x[3], minus 0.1770274961116007 multiplied by x[4], minus 0.0922005916406162 multiplied by x[5], plus y_0[1] to equal -0.1872964740542509. A third constraint requires -0.1262388132745864 multiplied by x[0], plus 0.0795068962664749 multiplied by x[1], plus 0.0187856757413385 multiplied by x[2], plus 0.1513494147153156 multiplied by x[3], minus 0.0163664274157314 multiplied by x[4], minus 0.00311286020002606 multiplied by x[5], plus y_0[2] to equal -0.0954444159908309. A fourth constraint requires -0.0757819715079799 multiplied by x[0], plus 0.0919093860274552 multiplied by x[1], minus 0.0364339288064356 multiplied by x[2], plus 0.083580945500982 multiplied by x[3], minus 0.1474689439069835 multiplied by x[4], plus 0.0301651918491998 multiplied by x[5], plus y_0[3] to equal 0.2579565194918507. A fifth constraint requires 0.7572820533712606 multiplied by x[0], minus 1.21352962362722 multiplied by x[1], plus 0.3803676681799136 multiplied by x[2], plus 0.4801395423918295 multiplied by x[3], minus 0.8927539207272388 multiplied by x[4], minus 1.447471364330223 multiplied by x[5], plus t_0[0] to equal 0.683762161476536. A sixth constraint requires -0.039176642120947 multiplied by x[0], minus 0.0594029213045902 multiplied by x[1], plus 0.0203899965676045 multiplied by x[2], minus 0.0006599505607145948 multiplied by x[3], plus 0.1993498474530564 multiplied by x[4], minus 0.1494561245854857 multiplied by x[5], plus y_1[0] to equal -0.0909307736334353. A seventh constraint requires -0.1551511441179779 multiplied by x[0], plus 0.0850834698172653 multiplied by x[1], minus 0.1602283190962608 multiplied by x[2], plus 0.0725598573219744 multiplied by x[3], minus 0.1055576967228311 multiplied by x[4], minus 0.1582958490297426 multiplied by x[5], plus y_1[1] to equal -0.1725026750262158. An eighth constraint requires -0.1933585329412353 multiplied by x[0], plus 0.0929980184151937 multiplied by x[1], plus 0.099592631874656 multiplied by x[2], plus 0.0797650799817848 multiplied by x[3], plus 0.0106313375995685 multiplied by x[4], minus 0.00903979011317048 multiplied by x[5], plus y_1[2] to equal -0.1167161263783248. A ninth constraint requires 0.0230142114746871 multiplied by x[0], plus 0.0796690661109758 multiplied by x[1], plus 0.0216327232273718 multiplied by x[2], minus 0.0337376899834936 multiplied by x[3], minus 0.1518320281674425 multiplied by x[4], plus 0.046720461629375 multiplied by x[5], plus y_1[3] to equal -0.1005698418344237. A tenth constraint requires -0.1501707210081249 multiplied by x[0], plus 0.7058311865593911 multiplied by x[1], plus 1.287754324454335 multiplied by x[2], plus 0.0373827245849037 multiplied by x[3], plus 0.1171725723905398 multiplied by x[4], plus 0.1069533502066662 multiplied by x[5], plus t_1[0] to equal 0.1223510811918817.] + + Two quadratic cone constraints ensure that the squared sum of y_0[0, y_0[1], y_0[2], and y_0[3] does not exceed the square of t_0[0], and the squared sum of y_1[0], y_1[1], y_1[2], and y_1[3] does not exceed the square of t_1[0]. All variables, including the decision variables x[0] through x[5] and the auxiliary variables y_0[0], y_0[1], y_0[2], y_0[3], y_1[0], y_1[1], y_1[2], y_1[3], t_0[0], and t_1[0], are free to take any real value, with no bounds specified. The goal is to determine the values of these variables that minimize the objective function while satisfying all equality and quadratic cone constraints.]-2.043831641697913Quadratic, out-of-scopenon-linear problem
149The problem involves determining the optimal allocation of five types of resources, represented by variables \( x[0] \) through \( x[4] \), along with auxiliary variables \( y_0[0] \), \( y_0[1] \), and \( t_0[0] \), to minimize the total cost defined by the linear objective function: \( 1.126401441072199 x[0] - 0.3048391076081773 x[1] + 1.210295701208876 x[2] + 0.3629036707879342 x[3] - 1.541074052220742 x[4] \). + +The solution must satisfy four linear equality constraints: the first requires \( -0.9907409502145462 x[0] - 0.7264863472262246 x[1] + 0.2793136098129854 x[2] - 0.9349403902445268 x[3] + 1.444658005223368 x[4] = 1.116201276984661 \); the second requires \( 2.936722209688262 x[0] + 0.5035733953314866 x[1] + 0.3357143231898844 x[2] + 1.112758520098299 x[3] + 1.412597032089028 x[4] = -1.571971227040746 \); the third requires \( 0.7899614064673852 x[0] - 0.0724234240750259 x[1] - 0.7706602505688044 x[2] + 1.907488050972312 x[3] + 1.420058526737699 x[4] = -1.686571012140405 \); and the fourth requires \( 0.6068754572874664 x[0] + 0.7492819480236036 x[1] - 1.565830499338323 x[2] + 0.2718401004423009 x[3] + 0.5510027846961619 x[4] = 4.014442756917189 \). + +Additionally, three linear equality constraints, labeled R4, R5, and R6, tie the resource allocations to auxiliary variables \( y_0[0] \), \( y_0[1] \), and \( t_0[0] \): R4 requires \( -0.0124992573250491 x[0] - 0.1193568402945484 x[1] - 0.0200307964468845 x[2] + 0.0612863681420025 x[3] - 0.116537049877206 x[4] + y_0[0] = -0.0358590805559009 \); R5 requires \( 0.0484815211698107 x[0] - 0.0631246556642578 x[1] + 0.1244200448905836 x[2] + 0.0530487576814488 x[3] + 0.1361347655297342 x[4] + y_0[1] = 0.0271906214446268 \); and R6 requires \( -2.249909202694956 x[0] + 0.7776031115768471 x[1] - 0.2174454691465576 x[2] + 1.953508135119546 x[3] + 1.071251451084194 x[4] + t_0[0] = 0.9059772765697818 \). + +A second-order cone constraint, labeled \( qc0 \), ensures that the sum of the squares of \( y_0[0] \) and \( y_0[1] \) does not exceed the square of \( t_0[0] \), expressed as \( -t_0[0]^2 + y_0[0]^2 + y_0[1]^2 \leq 0 \). All decision variables, including \( x[0] \) through \( x[4] \), \( y_0[0] \), \( y_0[1] \), and \( t_0[0] \), are unrestricted in sign, meaning they can take any real value to satisfy the constraints and optimize the objective. This is a Second-Order Cone Programming (SOCP) problem that combines linear and quadratic constraints, requiring a balance between resource allocation and auxiliary variables to achieve the optimal solution while minimizing total cost.-5.0688282673918960The problem involves determining the optimal allocation of five types of resources, represented by variables \( x[0 \) through \( x[4] \), along with auxiliary variables \( y_0[0] \), \( y_0[1] \), and \( t_0[0] \), to minimize the total cost defined by the linear objective function: \( 1.126401441072199 x[0] - 0.3048391076081773 x[1] + 1.210295701208876 x[2] + 0.3629036707879342 x[3] - 1.541074052220742 x[4] \). ] + + The solution must satisfy four linear equality constraints: the first requires \( -0.9907409502145462 x[0 - 0.7264863472262246 x[1] + 0.2793136098129854 x[2] - 0.9349403902445268 x[3] + 1.444658005223368 x[4] = 1.116201276984661 \); the second requires \( 2.936722209688262 x[0] + 0.5035733953314866 x[1] + 0.3357143231898844 x[2] + 1.112758520098299 x[3] + 1.412597032089028 x[4] = -1.571971227040746 \); the third requires \( 0.7899614064673852 x[0] - 0.0724234240750259 x[1] - 0.7706602505688044 x[2] + 1.907488050972312 x[3] + 1.420058526737699 x[4] = -1.686571012140405 \); and the fourth requires \( 0.6068754572874664 x[0] + 0.7492819480236036 x[1] - 1.565830499338323 x[2] + 0.2718401004423009 x[3] + 0.5510027846961619 x[4] = 4.014442756917189 \).] + + Additionally, three linear equality constraints, labeled R4, R5, and R6, tie the resource allocations to auxiliary variables \( y_0[0 \), \( y_0[1] \), and \( t_0[0] \): R4 requires \( -0.0124992573250491 x[0] - 0.1193568402945484 x[1] - 0.0200307964468845 x[2] + 0.0612863681420025 x[3] - 0.116537049877206 x[4] + y_0[0] = -0.0358590805559009 \); R5 requires \( 0.0484815211698107 x[0] - 0.0631246556642578 x[1] + 0.1244200448905836 x[2] + 0.0530487576814488 x[3] + 0.1361347655297342 x[4] + y_0[1] = 0.0271906214446268 \); and R6 requires \( -2.249909202694956 x[0] + 0.7776031115768471 x[1] - 0.2174454691465576 x[2] + 1.953508135119546 x[3] + 1.071251451084194 x[4] + t_0[0] = 0.9059772765697818 \).] + + A second-order cone constraint, labeled \( qc0 \), ensures that the sum of the squares of \( y_0[0 \) and \( y_0[1] \) does not exceed the square of \( t_0[0] \), expressed as \( -t_0[0]^2 + y_0[0]^2 + y_0[1]^2 \leq 0 \). All decision variables, including \( x[0] \) through \( x[4] \), \( y_0[0] \), \( y_0[1] \), and \( t_0[0] \), are unrestricted in sign, meaning they can take any real value to satisfy the constraints and optimize the objective. This is a Second-Order Cone Programming (SOCP) problem that combines linear and quadratic constraints, requiring a balance between resource allocation and auxiliary variables to achieve the optimal solution while minimizing total cost.]-5.068828267391896Quadratic, out-of-scopenon-linear problem
150The production manager seeks to minimize the total production cost for four product lines, \( x[0] \), \( x[1] \), \( x[2] \), and \( x[3] \), with costs of \( -0.0211519215863677 \), \( -0.157914420869699 \), \( -1.276226636497168 \), and \( 1.887428393936258 \) per unit, respectively. Negative coefficients in the objective function represent costs to be minimized, not cost savings. Four linear equality constraints must be satisfied: the first requires a weighted sum of production levels using coefficients \( -0.6903544941297689 \), \( 0.184421264426265 \), \( 0.2271145373127337 \), and \( 0.3261334256865931 \) to equal \( -1.784682154876947 \). The second constraint uses coefficients \( 0.9527280569937944 \), \( -0.227563767653628 \), \( -0.1388448211471715 \), and \( -0.7259447957545259 \) to sum to \( 3.071741041299838 \). The third constraint uses coefficients \( -0.2904848690892891 \), \( 0.6067085131648451 \), \( 0.5360035950278403 \), and \( 1.175919727086906 \) to sum to \( -3.247120484557859 \). The fourth constraint uses coefficients \( -1.447328047167681 \), \( 1.078800425952911 \), \( -1.071986023688645 \), and \( 2.402600530074645 \) to sum to \( -9.240805229261266 \). + +Twelve auxiliary variables, \( y_0[0] \), \( y_0[1] \), \( y_1[0] \), \( y_1[1] \), \( y_2[0] \), \( y_2[1] \), \( y_3[0] \), \( y_3[1] \), \( t_0[0] \), \( t_1[0] \), \( t_2[0] \), and \( t_3[0] \), are introduced to model resource usage and operational safety. These variables are part of twelve additional linear constraints: R4 through R15. For example, R4 states that \( -0.00544743187557857 \) times \( x[0] \), minus \( 0.1023858515949546 \) times \( x[1] \), minus \( 0.2570937910987342 \) times \( x[2] \), minus \( 0.0303903824324441 \) times \( x[3] \), plus \( y_0[0] \), equals \( 0.0630402459981765 \). Similar relationships define R5 through R15. + +Four quadratic constraints ensure safe operational limits: the sum of the squares of \( y_0[0] \) and \( y_0[1] \) must be less than or equal to the square of \( t_0[0] \); the sum of the squares of \( y_1[0] \) and \( y_1[1] \) must be less than or equal to the square of \( t_1[0] \); the sum of the squares of \( y_2[0] \) and \( y_2[1] \) must be less than or equal to the square of \( t_2[0] \); and the sum of the squares of \( y_3[0] \) and \( y_3[1] \) must be less than or equal to the square of \( t_3[0] \). All decision variables, including production levels and auxiliary variables, are free to take any real value. The goal is to determine the optimal production levels for \( x[0] \), \( x[1] \), \( x[2] \), and \( x[3] \) that minimize total production cost while satisfying all constraints. This problem is a Second-Order Cone Programming (SOCP) problem, as indicated by the quadratic constraints and the structure of the model.-4.2987420204770840The production manager seeks to minimize the total production cost for four product lines, \( x[0 \), \( x[1] \), \( x[2] \), and \( x[3] \), with costs of \( -0.0211519215863677 \), \( -0.157914420869699 \), \( -1.276226636497168 \), and \( 1.887428393936258 \) per unit, respectively. Negative coefficients in the objective function represent costs to be minimized, not cost savings. Four linear equality constraints must be satisfied: the first requires a weighted sum of production levels using coefficients \( -0.6903544941297689 \), \( 0.184421264426265 \), \( 0.2271145373127337 \), and \( 0.3261334256865931 \) to equal \( -1.784682154876947 \). The second constraint uses coefficients \( 0.9527280569937944 \), \( -0.227563767653628 \), \( -0.1388448211471715 \), and \( -0.7259447957545259 \) to sum to \( 3.071741041299838 \). The third constraint uses coefficients \( -0.2904848690892891 \), \( 0.6067085131648451 \), \( 0.5360035950278403 \), and \( 1.175919727086906 \) to sum to \( -3.247120484557859 \). The fourth constraint uses coefficients \( -1.447328047167681 \), \( 1.078800425952911 \), \( -1.071986023688645 \), and \( 2.402600530074645 \) to sum to \( -9.240805229261266 \).] + + Twelve auxiliary variables, \( y_0[0 \), \( y_0[1] \), \( y_1[0] \), \( y_1[1] \), \( y_2[0] \), \( y_2[1] \), \( y_3[0] \), \( y_3[1] \), \( t_0[0] \), \( t_1[0] \), \( t_2[0] \), and \( t_3[0] \), are introduced to model resource usage and operational safety. These variables are part of twelve additional linear constraints: R4 through R15. For example, R4 states that \( -0.00544743187557857 \) times \( x[0] \), minus \( 0.1023858515949546 \) times \( x[1] \), minus \( 0.2570937910987342 \) times \( x[2] \), minus \( 0.0303903824324441 \) times \( x[3] \), plus \( y_0[0] \), equals \( 0.0630402459981765 \). Similar relationships define R5 through R15.] + + Four quadratic constraints ensure safe operational limits: the sum of the squares of \( y_0[0 \) and \( y_0[1] \) must be less than or equal to the square of \( t_0[0] \); the sum of the squares of \( y_1[0] \) and \( y_1[1] \) must be less than or equal to the square of \( t_1[0] \); the sum of the squares of \( y_2[0] \) and \( y_2[1] \) must be less than or equal to the square of \( t_2[0] \); and the sum of the squares of \( y_3[0] \) and \( y_3[1] \) must be less than or equal to the square of \( t_3[0] \). All decision variables, including production levels and auxiliary variables, are free to take any real value. The goal is to determine the optimal production levels for \( x[0] \), \( x[1] \), \( x[2] \), and \( x[3] \) that minimize total production cost while satisfying all constraints. This problem is a Second-Order Cone Programming (SOCP) problem, as indicated by the quadratic constraints and the structure of the model.]-4.298742020477084Quadratic, out-of-scopenon-linear problem
151The problem involves minimizing a linear objective function with coefficients 0.4662680322234939 for \( x[0] \), 0.1250396579327832 for \( x[1] \), and -0.1976509195604252 for \( x[2] \). The variables \( x[0] \), \( x[1] \), \( x[2] \), \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), and \( t_0[0] \) are free to take any real values. The problem includes three linear equality constraints: the first requires that -1.506294713918092 times \( x[0] \) minus 0.5786002519685364 times \( x[1] \) plus 1.651436537097151 times \( x[2] \) equals 1.525536343222344; the second requires that -2.426679243393074 times \( x[0] \) minus 0.4289126288561773 times \( x[1] \) plus 1.265936258705534 times \( x[2] \) equals 2.564935934767919; and the third requires that -0.8667404022651017 times \( x[0] \) minus 0.6788861516220543 times \( x[1] \) minus 0.0947089689368911 times \( x[2] \) equals 0.2370752919653094. Additionally, there are four linear constraints: the first requires that -0.0337374122584387 times \( x[0] \) plus 0.0926622976309022 times \( x[1] \) minus 0.0275747406547086 times \( x[2] \) plus \( y_0[0] \) equals 0.0767641469627959; the second requires that -0.0370852326700521 times \( x[0] \) minus 0.1174307145074716 times \( x[1] \) plus 0.2032292481351275 times \( x[2] \) plus \( y_0[1] \) equals 0.0370679052304449; the third requires that -0.0582508292397692 times \( x[0] \) plus 0.0721175797872196 times \( x[1] \) plus 0.1702423600392019 times \( x[2] \) plus \( y_0[2] \) equals 0.2075459354425953; and the fourth requires that -0.1277576901021839 times \( x[0] \) minus 1.799889451307227 times \( x[1] \) minus 1.238366353571482 times \( x[2] \) plus \( t_0[0] \) equals 0.9137569517560475. Finally, a second-order cone constraint requires that the sum of the squares of \( y_0[0] \), \( y_0[1] \), and \( y_0[2] \) minus the square of \( t_0[0] \) is less than or equal to zero. The constraint involving \( t_0[0] \) is part of the second-order cone constraint setup, and \( t_0[0] \) is explicitly recognized as a decision variable.-0.4374180719967340The problem involves minimizing a linear objective function with coefficients 0.4662680322234939 for \( x[0 \), 0.1250396579327832 for \( x[1] \), and -0.1976509195604252 for \( x[2] \). The variables \( x[0] \), \( x[1] \), \( x[2] \), \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), and \( t_0[0] \) are free to take any real values. The problem includes three linear equality constraints: the first requires that -1.506294713918092 times \( x[0] \) minus 0.5786002519685364 times \( x[1] \) plus 1.651436537097151 times \( x[2] \) equals 1.525536343222344; the second requires that -2.426679243393074 times \( x[0] \) minus 0.4289126288561773 times \( x[1] \) plus 1.265936258705534 times \( x[2] \) equals 2.564935934767919; and the third requires that -0.8667404022651017 times \( x[0] \) minus 0.6788861516220543 times \( x[1] \) minus 0.0947089689368911 times \( x[2] \) equals 0.2370752919653094. Additionally, there are four linear constraints: the first requires that -0.0337374122584387 times \( x[0] \) plus 0.0926622976309022 times \( x[1] \) minus 0.0275747406547086 times \( x[2] \) plus \( y_0[0] \) equals 0.0767641469627959; the second requires that -0.0370852326700521 times \( x[0] \) minus 0.1174307145074716 times \( x[1] \) plus 0.2032292481351275 times \( x[2] \) plus \( y_0[1] \) equals 0.0370679052304449; the third requires that -0.0582508292397692 times \( x[0] \) plus 0.0721175797872196 times \( x[1] \) plus 0.1702423600392019 times \( x[2] \) plus \( y_0[2] \) equals 0.2075459354425953; and the fourth requires that -0.1277576901021839 times \( x[0] \) minus 1.799889451307227 times \( x[1] \) minus 1.238366353571482 times \( x[2] \) plus \( t_0[0] \) equals 0.9137569517560475. Finally, a second-order cone constraint requires that the sum of the squares of \( y_0[0] \), \( y_0[1] \), and \( y_0[2] \) minus the square of \( t_0[0] \) is less than or equal to zero. The constraint involving \( t_0[0] \) is part of the second-order cone constraint setup, and \( t_0[0] \) is explicitly recognized as a decision variable.]-0.437418071996734Quadratic, out-of-scopenon-linear problem
152Minimize the total cost of allocating resources \( x[0] \) to \( x[4] \), where the cost coefficients are 0.190510737670758 for \( x[0] \), 1.129350959341043 for \( x[1] \), -1.45763004619859 for \( x[2] \), 0.4979171399142379 for \( x[3] \), and 0.217933194293673 for \( x[4] \). The allocation must satisfy four equality constraints. The first constraint requires 1.219567298556732 \( x[0] \) + 0.9289001747224751 \( x[1] \) - 1.336923760887845 \( x[2] \) - 1.115644167094657 \( x[3] \) - 0.4761759785063266 \( x[4] \) = -0.4344345340009604. The second constraint requires -1.662323208398086 \( x[0] \) + 0.8659229325025953 \( x[1] \) - 0.715611281293988 \( x[2] \) + 0.513394018474748 \( x[3] \) - 1.247456907892064 \( x[4] \) = -1.263487160460341. The third constraint requires -0.2849105246051509 \( x[0] \) - 1.235291728359102 \( x[1] \) + 1.089988844777487 \( x[2] \) - 1.059632720036963 \( x[3] \) + 0.2188818725225526 \( x[4] \) = 0.0598263158545554. The fourth constraint requires -1.037775353342995 \( x[0] \) - 1.808771173081519 \( x[1] \) + 1.006811579019581 \( x[2] \) + 0.2261443990014109 \( x[3] \) - 0.5001468099984979 \( x[4] \) = 0.6799624629319878. Additional linear constraints involve variables \( y_0[0] \), \( y_0[1] \), \( t_0[0] \), \( y_1[0] \), \( y_1[1] \), \( t_1[0] \), \( y_2[0] \), \( y_2[1] \), \( t_2[0] \), \( y_3[0] \), \( y_3[1] \), \( t_3[0] \), \( y_4[0] \), \( y_4[1] \), and \( t_4[0] \), which are free variables. These constraints are defined as follows: 0.1854712369852954 \( x[0] \) + 0.2126049926787192 \( x[1] \) - 0.0552426804013357 \( x[2] \) - 0.0300913423926646 \( x[3] \) + 0.0191442880734612 \( x[4] \) + \( y_0[0] \) = -0.00851123919328947; 0.033633911265618 \( x[0] \) + 0.0447803575912513 \( x[1] \) + 0.0906422549522432 \( x[2] \) - 0.140301925709981 \( x[3] \) + 0.0894655740323818 \( x[4] \) + \( y_0[1] \) = 0.1197073561594899; 1.032476848498754 \( x[0] \) + 0.5786229660982358 \( x[1] \) + 1.517117102281561 \( x[2] \) - 0.9892374377228246 \( x[3] \) - 0.7127788386839591 \( x[4] \) + \( t_0[0] \) = 0.0628164480170035; -0.0629876313760396 \( x[0] \) + 0.1186553496180699 \( x[1] \) + 0.0837866420373662 \( x[2] \) - 0.0175044400700457 \( x[3] \) + 0.0421466082438275 \( x[4] \) + \( y_1[0] \) = -0.0607801036942368; 0.0472913180476754 \( x[0] \) - 0.1672487320352799 \( x[1] \) + 0.00403030115525331 \( x[2] \) + 0.00774493287185739 \( x[3] \) - 0.0094688993206764 \( x[4] \) + \( y_1[1] \) = -0.0464456008702828; -0.2085979716989413 \( x[0] \) - 0.1092539540974228 \( x[1] \) - 2.174804559553748 \( x[2] \) - 1.69528753252808 \( x[3] \) - 1.525645556958908 \( x[4] \) + \( t_1[0] \) = 0.7445116249805878; 0.000912382 \( x[0] \) - 0.00863788041334999 \( x[1] \) - 0.0502948320610378 \( x[2] \) - 0.0667737933702229 \( x[3] \) + 0.2432277963883143 \( x[4] \) + \( y_2[0] \) = -0.139166601364713; 0.0232832507250414 \( x[0] \) + 0.2516736184489083 \( x[1] \) + 0.1386504617834828 \( x[2] \) - 0.0543682860611478 \( x[3] \) + 0.050704164298607 \( x[4] \) + \( y_2[1] \) = 0.0357160518360202; -0.5925926352567416 \( x[0] \) - 0.3891604825652387 \( x[1] \) + 0.0510217340581877 \( x[2] \) + 1.090606691890913 \( x[3] \) + 0.9938716757643966 \( x[4] \) + \( t_2[0] \) = 1.045297581678233; 0.1540622796961561 \( x[0] \) - 0.1883221064812006 \( x[1] \) + 0.1503453658091877 \( x[2] \) - 0.1474374678246578 \( x[3] \) + 0.0205212420510411 \( x[4] \) + \( y_3[0] \) = -0.00963411322153448; -0.0280861975467837 \( x[0] \) - 0.0105727074625737 \( x[1] \) + 0.0954257952878318 \( x[2] \) - 0.00589259567851985 \( x[3] \) + 0.145090882467532 \( x[4] \) + \( y_3[1] \) = -0.019437376136785; -1.252924325600617 \( x[0] \) + 1.244308803054627 \( x[1] \) - 2.21127388768641 \( x[2] \) - 1.885087766270206 \( x[3] \) - 1.006758968957397 \( x[4] \) + \( t_3[0] \) = 1.050824562362084; 0.1325971503040609 \( x[0] \) - 0.110249935172733 \( x[1] \) + 0.0558235975886938 \( x[2] \) + 0.0542813307961975 \( x[3] \) - 0.1156413261001238 \( x[4] \) + \( y_4[0] \) = 0.0325653324485877; -1.3730183823943952e-04 \( x[0] \) - 0.0268128227655337 \( x[1] \) + 0.0403141061350009 \( x[2] \) + 0.1864002689467194 \( x[3] \) - 0.0450317064978495 \( x[4] \) + \( y_4[1] \) = 0.0626883759594124; -1.090908275543176 \( x[0] \) + 0.3860841432731175 \( x[1] \) - 0.2953562653864454 \( x[2] \) - 0.2904011172481792 \( x[3] \) - 0.8273323625052638 \( x[4] \) + \( t_4[0] \) = 0.7353989816876624. Quadratic constraints require that the sum of the squares of the outcomes minus the square of the allocated resource must be less than or equal to zero for each project: \( y_0[0]^2 + y_0[1]^2 - t_0[0]^2 \leq 0 \), \( y_1[0]^2 + y_1[1]^2 - t_1[0]^2 \leq 0 \), \( y_2[0]^2 + y_2[1]^2 - t_2[0]^2 \leq 0 \), \( y_3[0]^2 + y_3[1]^2 - t_3[0]^2 \leq 0 \), and \( y_4[0]^2 + y_4[1]^2 - t_4[0]^2 \leq 0 \). All variables \( x[0] \) to \( x[4] \), \( y_0[0] \) to \( y_4[1] \), and \( t_0[0] \) to \( t_4[0] \) are free to take any real value.-0.3945968590238860Minimize the total cost of allocating resources \( x[0 \) to \( x[4] \), where the cost coefficients are 0.190510737670758 for \( x[0] \), 1.129350959341043 for \( x[1] \), -1.45763004619859 for \( x[2] \), 0.4979171399142379 for \( x[3] \), and 0.217933194293673 for \( x[4] \). The allocation must satisfy four equality constraints. The first constraint requires 1.219567298556732 \( x[0] \) + 0.9289001747224751 \( x[1] \) - 1.336923760887845 \( x[2] \) - 1.115644167094657 \( x[3] \) - 0.4761759785063266 \( x[4] \) = -0.4344345340009604. The second constraint requires -1.662323208398086 \( x[0] \) + 0.8659229325025953 \( x[1] \) - 0.715611281293988 \( x[2] \) + 0.513394018474748 \( x[3] \) - 1.247456907892064 \( x[4] \) = -1.263487160460341. The third constraint requires -0.2849105246051509 \( x[0] \) - 1.235291728359102 \( x[1] \) + 1.089988844777487 \( x[2] \) - 1.059632720036963 \( x[3] \) + 0.2188818725225526 \( x[4] \) = 0.0598263158545554. The fourth constraint requires -1.037775353342995 \( x[0] \) - 1.808771173081519 \( x[1] \) + 1.006811579019581 \( x[2] \) + 0.2261443990014109 \( x[3] \) - 0.5001468099984979 \( x[4] \) = 0.6799624629319878. Additional linear constraints involve variables \( y_0[0] \), \( y_0[1] \), \( t_0[0] \), \( y_1[0] \), \( y_1[1] \), \( t_1[0] \), \( y_2[0] \), \( y_2[1] \), \( t_2[0] \), \( y_3[0] \), \( y_3[1] \), \( t_3[0] \), \( y_4[0] \), \( y_4[1] \), and \( t_4[0] \), which are free variables. These constraints are defined as follows: 0.1854712369852954 \( x[0] \) + 0.2126049926787192 \( x[1] \) - 0.0552426804013357 \( x[2] \) - 0.0300913423926646 \( x[3] \) + 0.0191442880734612 \( x[4] \) + \( y_0[0] \) = -0.00851123919328947; 0.033633911265618 \( x[0] \) + 0.0447803575912513 \( x[1] \) + 0.0906422549522432 \( x[2] \) - 0.140301925709981 \( x[3] \) + 0.0894655740323818 \( x[4] \) + \( y_0[1] \) = 0.1197073561594899; 1.032476848498754 \( x[0] \) + 0.5786229660982358 \( x[1] \) + 1.517117102281561 \( x[2] \) - 0.9892374377228246 \( x[3] \) - 0.7127788386839591 \( x[4] \) + \( t_0[0] \) = 0.0628164480170035; -0.0629876313760396 \( x[0] \) + 0.1186553496180699 \( x[1] \) + 0.0837866420373662 \( x[2] \) - 0.0175044400700457 \( x[3] \) + 0.0421466082438275 \( x[4] \) + \( y_1[0] \) = -0.0607801036942368; 0.0472913180476754 \( x[0] \) - 0.1672487320352799 \( x[1] \) + 0.00403030115525331 \( x[2] \) + 0.00774493287185739 \( x[3] \) - 0.0094688993206764 \( x[4] \) + \( y_1[1] \) = -0.0464456008702828; -0.2085979716989413 \( x[0] \) - 0.1092539540974228 \( x[1] \) - 2.174804559553748 \( x[2] \) - 1.69528753252808 \( x[3] \) - 1.525645556958908 \( x[4] \) + \( t_1[0] \) = 0.7445116249805878; 0.000912382 \( x[0] \) - 0.00863788041334999 \( x[1] \) - 0.0502948320610378 \( x[2] \) - 0.0667737933702229 \( x[3] \) + 0.2432277963883143 \( x[4] \) + \( y_2[0] \) = -0.139166601364713; 0.0232832507250414 \( x[0] \) + 0.2516736184489083 \( x[1] \) + 0.1386504617834828 \( x[2] \) - 0.0543682860611478 \( x[3] \) + 0.050704164298607 \( x[4] \) + \( y_2[1] \) = 0.0357160518360202; -0.5925926352567416 \( x[0] \) - 0.3891604825652387 \( x[1] \) + 0.0510217340581877 \( x[2] \) + 1.090606691890913 \( x[3] \) + 0.9938716757643966 \( x[4] \) + \( t_2[0] \) = 1.045297581678233; 0.1540622796961561 \( x[0] \) - 0.1883221064812006 \( x[1] \) + 0.1503453658091877 \( x[2] \) - 0.1474374678246578 \( x[3] \) + 0.0205212420510411 \( x[4] \) + \( y_3[0] \) = -0.00963411322153448; -0.0280861975467837 \( x[0] \) - 0.0105727074625737 \( x[1] \) + 0.0954257952878318 \( x[2] \) - 0.00589259567851985 \( x[3] \) + 0.145090882467532 \( x[4] \) + \( y_3[1] \) = -0.019437376136785; -1.252924325600617 \( x[0] \) + 1.244308803054627 \( x[1] \) - 2.21127388768641 \( x[2] \) - 1.885087766270206 \( x[3] \) - 1.006758968957397 \( x[4] \) + \( t_3[0] \) = 1.050824562362084; 0.1325971503040609 \( x[0] \) - 0.110249935172733 \( x[1] \) + 0.0558235975886938 \( x[2] \) + 0.0542813307961975 \( x[3] \) - 0.1156413261001238 \( x[4] \) + \( y_4[0] \) = 0.0325653324485877; -1.3730183823943952e-04 \( x[0] \) - 0.0268128227655337 \( x[1] \) + 0.0403141061350009 \( x[2] \) + 0.1864002689467194 \( x[3] \) - 0.0450317064978495 \( x[4] \) + \( y_4[1] \) = 0.0626883759594124; -1.090908275543176 \( x[0] \) + 0.3860841432731175 \( x[1] \) - 0.2953562653864454 \( x[2] \) - 0.2904011172481792 \( x[3] \) - 0.8273323625052638 \( x[4] \) + \( t_4[0] \) = 0.7353989816876624. Quadratic constraints require that the sum of the squares of the outcomes minus the square of the allocated resource must be less than or equal to zero for each project: \( y_0[0]^2 + y_0[1]^2 - t_0[0]^2 \leq 0 \), \( y_1[0]^2 + y_1[1]^2 - t_1[0]^2 \leq 0 \), \( y_2[0]^2 + y_2[1]^2 - t_2[0]^2 \leq 0 \), \( y_3[0]^2 + y_3[1]^2 - t_3[0]^2 \leq 0 \), and \( y_4[0]^2 + y_4[1]^2 - t_4[0]^2 \leq 0 \). All variables \( x[0] \) to \( x[4] \), \( y_0[0] \) to \( y_4[1] \), and \( t_0[0] \) to \( t_4[0] \) are free to take any real value.]-0.394596859023886Quadratic, out-of-scopenon-linear problem
153The optimization problem involves minimizing a linear objective function influenced by decision variables \(x[0]\), \(x[1]\), and \(x[2]\), with coefficients \(-1.363345803670422\), \(1.361682302913205\), and \(-0.0230642010699799\), respectively. These variables are free to take any real value. The problem is subject to 18 linear equality constraints. The first three constraints enforce specific relationships: the sum of \(-1.712919234088719\) units from \(x[0]\), \(0.5094892843494099\) units from \(x[1]\), and \(-1.176977823298391\) units from \(x[2]\) must equal \(1.969673610649569\); the sum of \(0.370350858647903\) units from \(x[0]\), \(0.1859117162151037\) units from \(x[1]\), and \(0.4554018017289621\) units from \(x[2]\) must equal \(-0.42565571941171\); and the sum of \(1.568958943694931\) units from \(x[0]\), \(0.1377471346710665\) units from \(x[1]\), and \(-0.5120283579251602\) units from \(x[2]\) must equal \(-0.6850278359750025\). The remaining 15 constraints involve auxiliary variables \(y_i[j]\) and \(t_i[0]\), which are also free, and enforce linear relationships between these variables and the decision variables. For \(i=0\), the sum of \(0.0870156668345264\) units from \(x[0]\), \(0.0507128953480789\) units from \(x[1]\), \(0.2266461831421958\) units from \(x[2]\), and \(y_0[0]\) must equal \(-0.1485711357308674\). Similarly, for \(i=0\), the sum of \(-0.1414205820862144\) units from \(x[0]\), \(-0.3264725908092034\) units from \(x[1]\), \(-0.0734148848236748\) units from \(x[2]\), and \(y_0[1]\) must equal \(-0.0393674285854488\); and the sum of \(0.1205790377305938\) units from \(x[0]\), \(-0.0720533164152696\) units from \(x[1]\), \(1.0940570577090807e-04\) units from \(x[2]\), and \(y_0[2]\) must equal \(0.0713736217121076\). Additionally, for \(i=0\), the sum of \(1.069325952651591\) units from \(x[0]\), \(-1.138089180693524\) units from \(x[1]\), \(-0.2833695676847757\) units from \(x[2]\), and \(t_0[0]\) must equal \(0.5757538449770843\). These patterns continue for \(i=1\), \(i=2\), and \(i=3\), with similar linear relationships defined for \(y_i[j]\) and \(t_i[0]\). The problem also includes four quadratic constraints, which are second-order cone constraints. These constraints ensure that for each set of auxiliary variables \(y_i[0]\), \(y_i[1]\), and \(y_i[2]\), the squared sum of these variables does not exceed the square of \(t_i[0]\). The goal is to determine the values of the decision and auxiliary variables that minimize the objective function while satisfying all linear and quadratic constraints.1.420274769058680The optimization problem involves minimizing a linear objective function influenced by decision variables \(x[0\), \(x[1]\), and \(x[2]\), with coefficients \(-1.363345803670422\), \(1.361682302913205\), and \(-0.0230642010699799\), respectively. These variables are free to take any real value. The problem is subject to 18 linear equality constraints. The first three constraints enforce specific relationships: the sum of \(-1.712919234088719\) units from \(x[0]\), \(0.5094892843494099\) units from \(x[1]\), and \(-1.176977823298391\) units from \(x[2]\) must equal \(1.969673610649569\); the sum of \(0.370350858647903\) units from \(x[0]\), \(0.1859117162151037\) units from \(x[1]\), and \(0.4554018017289621\) units from \(x[2]\) must equal \(-0.42565571941171\); and the sum of \(1.568958943694931\) units from \(x[0]\), \(0.1377471346710665\) units from \(x[1]\), and \(-0.5120283579251602\) units from \(x[2]\) must equal \(-0.6850278359750025\). The remaining 15 constraints involve auxiliary variables \(y_i[j]\) and \(t_i[0]\), which are also free, and enforce linear relationships between these variables and the decision variables. For \(i=0\), the sum of \(0.0870156668345264\) units from \(x[0]\), \(0.0507128953480789\) units from \(x[1]\), \(0.2266461831421958\) units from \(x[2]\), and \(y_0[0]\) must equal \(-0.1485711357308674\). Similarly, for \(i=0\), the sum of \(-0.1414205820862144\) units from \(x[0]\), \(-0.3264725908092034\) units from \(x[1]\), \(-0.0734148848236748\) units from \(x[2]\), and \(y_0[1]\) must equal \(-0.0393674285854488\); and the sum of \(0.1205790377305938\) units from \(x[0]\), \(-0.0720533164152696\) units from \(x[1]\), \(1.0940570577090807e-04\) units from \(x[2]\), and \(y_0[2]\) must equal \(0.0713736217121076\). Additionally, for \(i=0\), the sum of \(1.069325952651591\) units from \(x[0]\), \(-1.138089180693524\) units from \(x[1]\), \(-0.2833695676847757\) units from \(x[2]\), and \(t_0[0]\) must equal \(0.5757538449770843\). These patterns continue for \(i=1\), \(i=2\), and \(i=3\), with similar linear relationships defined for \(y_i[j]\) and \(t_i[0]\). The problem also includes four quadratic constraints, which are second-order cone constraints. These constraints ensure that for each set of auxiliary variables \(y_i[0]\), \(y_i[1]\), and \(y_i[2]\), the squared sum of these variables does not exceed the square of \(t_i[0]\). The goal is to determine the values of the decision and auxiliary variables that minimize the objective function while satisfying all linear and quadratic constraints.]1.42027476905868Quadratic, out-of-scopenon-linear problem
154The problem involves optimizing the allocation of five free variables \( x[0] \) to \( x[4] \), along with auxiliary variables \( y_0[0] \) to \( y_2[3] \) and \( t_0[0] \) to \( t_2[0] \), which are also free. The objective is to minimize the expression \(-0.4183092460544717 x[0] - 0.2813052479554071 x[1] + 0.134543055273074 x[2] + 0.3293895938730942 x[3] - 0.5013729625677056 x[4]\). The problem is subject to five linear equality constraints. The first constraint requires \(-0.2042258848533399 x[0] + 0.2505367949866907 x[1] - 0.4309025998889995 x[2] + 1.309402520858531 x[3] + 2.586402096787606 x[4] = -3.156703022107761\). The second constraint imposes \(-0.9968810525037243 x[0] - 0.4546685233810487 x[1] + 0.4750317828663017 x[2] + 0.266787146193911 x[3] - 1.191880626574726 x[4] = -2.176991821226254\). The third constraint requires \(0.349956286844281 x[0] + 1.015584003075466 x[1] - 0.0836288458590485 x[2] - 0.0695677711676667 x[3] + 0.104824587707105 x[4] = 2.926695432150372\). The fourth constraint enforces \(0.7470800133528747 x[0] + 0.8253890222140957 x[1] + 0.0968330451370987 x[2] + 0.7049632173662657 x[3] + 0.7872792028260496 x[4] = 0.5732174371789283\). The fifth constraint requires \(-1.307269223716067 x[0] + 0.3813744238506016 x[1] - 0.4050739984159557 x[2] + 1.424197873139263 x[3] + 0.4653468565843648 x[4] = -3.193503084865396\). + +Additionally, the problem includes constraints linking the variables \( x[0] \) to \( x[4] \) with auxiliary variables \( y_0[0] \) to \( y_2[3] \). For example, the first linking constraint requires \(-0.3645044940419532 x[0] - 1.443832971588788 x[1] + 0.461589002970339 x[2] + 0.2606857473736294 x[3] + 1.721110703706202 x[4] + y_0[0] = -0.9152180704947402\). Similar constraints exist for \( y_0[1] \), \( y_0[2] \), \( y_0[3] \), \( y_1[0] \), \( y_1[1] \), \( y_1[2] \), \( y_1[3] \), \( y_2[0] \), \( y_2[1] \), \( y_2[2] \), and \( y_2[3] \), each with specific coefficients and target values. + +The problem also includes quadratic constraints involving the variables \( t_0[0] \), \( t_1[0] \), and \( t_2[0] \). The first quadratic constraint ensures that \(-t_0[0]^2 + y_0[0]^2 + y_0[1]^2 + y_0[2]^2 + y_0[3]^2 \leq 0\). The second quadratic constraint requires \(-t_1[0]^2 + y_1[0]^2 + y_1[1]^2 + y_1[2]^2 + y_1[3]^2 \leq 0\), and the third quadratic constraint enforces \(-t_2[0]^2 + y_2[0]^2 + y_2[1]^2 + y_2[2]^2 + y_2[3]^2 \leq 0\). These constraints ensure that the relationships between the auxiliary variables and the safety buffers are maintained. The goal is to minimize the objective function while satisfying all linear equality, linking, and quadratic constraints.-1.6505652629772340The problem involves optimizing the allocation of five free variables \( x[0 \) to \( x[4] \), along with auxiliary variables \( y_0[0] \) to \( y_2[3] \) and \( t_0[0] \) to \( t_2[0] \), which are also free. The objective is to minimize the expression \(-0.4183092460544717 x[0] - 0.2813052479554071 x[1] + 0.134543055273074 x[2] + 0.3293895938730942 x[3] - 0.5013729625677056 x[4]\). The problem is subject to five linear equality constraints. The first constraint requires \(-0.2042258848533399 x[0] + 0.2505367949866907 x[1] - 0.4309025998889995 x[2] + 1.309402520858531 x[3] + 2.586402096787606 x[4] = -3.156703022107761\). The second constraint imposes \(-0.9968810525037243 x[0] - 0.4546685233810487 x[1] + 0.4750317828663017 x[2] + 0.266787146193911 x[3] - 1.191880626574726 x[4] = -2.176991821226254\). The third constraint requires \(0.349956286844281 x[0] + 1.015584003075466 x[1] - 0.0836288458590485 x[2] - 0.0695677711676667 x[3] + 0.104824587707105 x[4] = 2.926695432150372\). The fourth constraint enforces \(0.7470800133528747 x[0] + 0.8253890222140957 x[1] + 0.0968330451370987 x[2] + 0.7049632173662657 x[3] + 0.7872792028260496 x[4] = 0.5732174371789283\). The fifth constraint requires \(-1.307269223716067 x[0] + 0.3813744238506016 x[1] - 0.4050739984159557 x[2] + 1.424197873139263 x[3] + 0.4653468565843648 x[4] = -3.193503084865396\).] + + Additionally, the problem includes constraints linking the variables \( x[0 \) to \( x[4] \) with auxiliary variables \( y_0[0] \) to \( y_2[3] \). For example, the first linking constraint requires \(-0.3645044940419532 x[0] - 1.443832971588788 x[1] + 0.461589002970339 x[2] + 0.2606857473736294 x[3] + 1.721110703706202 x[4] + y_0[0] = -0.9152180704947402\). Similar constraints exist for \( y_0[1] \), \( y_0[2] \), \( y_0[3] \), \( y_1[0] \), \( y_1[1] \), \( y_1[2] \), \( y_1[3] \), \( y_2[0] \), \( y_2[1] \), \( y_2[2] \), and \( y_2[3] \), each with specific coefficients and target values.] + + The problem also includes quadratic constraints involving the variables \( t_0[0 \), \( t_1[0] \), and \( t_2[0] \). The first quadratic constraint ensures that \(-t_0[0]^2 + y_0[0]^2 + y_0[1]^2 + y_0[2]^2 + y_0[3]^2 \leq 0\). The second quadratic constraint requires \(-t_1[0]^2 + y_1[0]^2 + y_1[1]^2 + y_1[2]^2 + y_1[3]^2 \leq 0\), and the third quadratic constraint enforces \(-t_2[0]^2 + y_2[0]^2 + y_2[1]^2 + y_2[2]^2 + y_2[3]^2 \leq 0\). These constraints ensure that the relationships between the auxiliary variables and the safety buffers are maintained. The goal is to minimize the objective function while satisfying all linear equality, linking, and quadratic constraints.]-1.650565262977234Quadratic, out-of-scopenon-linear problem
155The optimization problem involves minimizing the total cost defined by the weighted contributions of four decision variables: x[0] with a cost of 0.4568048651554198, x[1] with a cost of 0.6822815404386418, x[2] with a cost of 1.801831165946818, and x[3] with a cost of 1.056541158185233. The objective is to minimize the sum of these weighted variables. + +The problem is subject to several equality constraints. The first equality constraint is 1.222253726840453 times x[0] minus 0.9058589863483399 times x[1] plus 0.0693521718801037 times x[2] plus 2.178655656494156 times x[3] equals -0.4195106543118642. The second equality constraint is 0.1473169107230136 times x[0] plus 0.5172571290693887 times x[1] minus 1.293979451435283 times x[2] minus 1.054887597556351 times x[3] equals 0.3661434959821696. The third equality constraint is 0.4647687397733393 times x[0] plus 0.2341060054529592 times x[1] plus 1.778866018427452 times x[2] plus 0.8638766717259377 times x[3] equals -1.748015587968725. The fourth equality constraint is -0.0250593592360135 times x[0] minus 0.157487553997023 times x[1] minus 0.6980379871220622 times x[2] minus 0.1710905557322772 times x[3] equals 0.6138465007098338. The fifth equality constraint is -0.6647756257582154 times x[0] plus 0.5784421525498459 times x[1] plus 2.117378290823709 times x[2] minus 0.4463943452167778 times x[3] equals -1.25361991665147. + +Additionally, there are auxiliary equality constraints involving variables y_0 and y_1. The first auxiliary constraint is 0.0689239653107249 times x[0] plus 0.0678131299325625 times x[1] minus 0.2023493775147238 times x[2] minus 0.0739358112586078 times x[3] plus y_0[0] equals -0.093423190067388. The second auxiliary constraint is 0.0445323421724085 times x[0] minus 0.0374069933363333 times x[1] minus 0.0947746176074507 times x[2] plus 0.0625716234434831 times x[3] plus y_0[1] equals 0.0993152088054553. The third auxiliary constraint is -0.056297721725534 times x[0] plus 0.0263563878275638 times x[1] minus 0.1207830337883556 times x[2] minus 0.0345530874849539 times x[3] plus y_0[2] equals -0.0719891440967462. The fourth auxiliary constraint is -0.0389348469541953 times x[0] minus 0.0207466886867141 times x[1] minus 0.1633463289784982 times x[2] minus 0.0676994906553111 times x[3] plus y_0[3] equals 0.0610083958048695. + +The fifth auxiliary constraint is 0.5721177896133665 times x[0] minus 0.0673019079083626 times x[1] plus 0.0564335910547514 times x[2] minus 0.3663765912716627 times x[3] plus t_0[0] equals 1.173542716306169. The sixth auxiliary constraint is 0.0925151391451339 times x[0] minus 0.1031167146627894 times x[1] plus 0.000327681659276563 times x[2] minus 0.1038664121798131 times x[3] plus y_1[0] equals 0.00701092577147306. The seventh auxiliary constraint is -0.0311602517533045 times x[0] plus 0.0527217587580109 times x[1] plus 0.0357033538861844 times x[2] minus 0.0441292118120421 times x[3] plus y_1[1] equals -0.00804391349899621. The eighth auxiliary constraint is 0.088078411493411 times x[0] plus 0.0425752446328825 times x[1] plus 0.0742562500174835 times x[2] minus 0.0601203608650306 times x[3] plus y_1[2] equals 0.0417669069308694. The ninth auxiliary constraint is 0.0671054419934661 times x[0] minus 0.0455987092328547 times x[1] plus 0.00380146 times x[2] minus 0.0199997335030863 times x[3] plus y_1[3] equals 0.0438732344355552. The tenth auxiliary constraint is 0.6551923068010495 times x[0] plus 0.315985578120184 times x[1] minus 0.1539347685265308 times x[2] minus 0.9668187288441786 times x[3] plus t_1[0] equals 0.9921660803606838. + +The problem also includes quadratic inequality constraints to ensure operational feasibility. The first quadratic constraint is negative t_0[0] squared plus y_0[0] squared plus y_0[1] squared plus y_0[2] squared plus y_0[3] squared is less than or equal to zero. The second quadratic constraint is negative t_1[0] squared plus y_1[0] squared plus y_1[1] squared plus y_1[2] squared plus y_1[3] squared is less than or equal to zero. + +All decision variables x[0], x[1], x[2], x[3], as well as the auxiliary variables y_0 and y_1, are free to take any real value, meaning there are no explicit upper or lower bounds on these variables. The auxiliary variables t_0[0] and t_1[0] are also free to take any real value. The problem is a Second-Order Cone Programming (SOCP) problem, as indicated by the quadratic constraints.-2.2839944878866340The optimization problem involves minimizing the total cost defined by the weighted contributions of four decision variables: x[0 with a cost of 0.4568048651554198, x[1] with a cost of 0.6822815404386418, x[2] with a cost of 1.801831165946818, and x[3] with a cost of 1.056541158185233. The objective is to minimize the sum of these weighted variables.] + + The problem is subject to several equality constraints. The first equality constraint is 1.222253726840453 times x[0 minus 0.9058589863483399 times x[1] plus 0.0693521718801037 times x[2] plus 2.178655656494156 times x[3] equals -0.4195106543118642. The second equality constraint is 0.1473169107230136 times x[0] plus 0.5172571290693887 times x[1] minus 1.293979451435283 times x[2] minus 1.054887597556351 times x[3] equals 0.3661434959821696. The third equality constraint is 0.4647687397733393 times x[0] plus 0.2341060054529592 times x[1] plus 1.778866018427452 times x[2] plus 0.8638766717259377 times x[3] equals -1.748015587968725. The fourth equality constraint is -0.0250593592360135 times x[0] minus 0.157487553997023 times x[1] minus 0.6980379871220622 times x[2] minus 0.1710905557322772 times x[3] equals 0.6138465007098338. The fifth equality constraint is -0.6647756257582154 times x[0] plus 0.5784421525498459 times x[1] plus 2.117378290823709 times x[2] minus 0.4463943452167778 times x[3] equals -1.25361991665147.] + + Additionally, there are auxiliary equality constraints involving variables y_0 and y_1. The first auxiliary constraint is 0.0689239653107249 times x[0 plus 0.0678131299325625 times x[1] minus 0.2023493775147238 times x[2] minus 0.0739358112586078 times x[3] plus y_0[0] equals -0.093423190067388. The second auxiliary constraint is 0.0445323421724085 times x[0] minus 0.0374069933363333 times x[1] minus 0.0947746176074507 times x[2] plus 0.0625716234434831 times x[3] plus y_0[1] equals 0.0993152088054553. The third auxiliary constraint is -0.056297721725534 times x[0] plus 0.0263563878275638 times x[1] minus 0.1207830337883556 times x[2] minus 0.0345530874849539 times x[3] plus y_0[2] equals -0.0719891440967462. The fourth auxiliary constraint is -0.0389348469541953 times x[0] minus 0.0207466886867141 times x[1] minus 0.1633463289784982 times x[2] minus 0.0676994906553111 times x[3] plus y_0[3] equals 0.0610083958048695.] + + The fifth auxiliary constraint is 0.5721177896133665 times x[0 minus 0.0673019079083626 times x[1] plus 0.0564335910547514 times x[2] minus 0.3663765912716627 times x[3] plus t_0[0] equals 1.173542716306169. The sixth auxiliary constraint is 0.0925151391451339 times x[0] minus 0.1031167146627894 times x[1] plus 0.000327681659276563 times x[2] minus 0.1038664121798131 times x[3] plus y_1[0] equals 0.00701092577147306. The seventh auxiliary constraint is -0.0311602517533045 times x[0] plus 0.0527217587580109 times x[1] plus 0.0357033538861844 times x[2] minus 0.0441292118120421 times x[3] plus y_1[1] equals -0.00804391349899621. The eighth auxiliary constraint is 0.088078411493411 times x[0] plus 0.0425752446328825 times x[1] plus 0.0742562500174835 times x[2] minus 0.0601203608650306 times x[3] plus y_1[2] equals 0.0417669069308694. The ninth auxiliary constraint is 0.0671054419934661 times x[0] minus 0.0455987092328547 times x[1] plus 0.00380146 times x[2] minus 0.0199997335030863 times x[3] plus y_1[3] equals 0.0438732344355552. The tenth auxiliary constraint is 0.6551923068010495 times x[0] plus 0.315985578120184 times x[1] minus 0.1539347685265308 times x[2] minus 0.9668187288441786 times x[3] plus t_1[0] equals 0.9921660803606838.] + + The problem also includes quadratic inequality constraints to ensure operational feasibility. The first quadratic constraint is negative t_0[0 squared plus y_0[0] squared plus y_0[1] squared plus y_0[2] squared plus y_0[3] squared is less than or equal to zero. The second quadratic constraint is negative t_1[0] squared plus y_1[0] squared plus y_1[1] squared plus y_1[2] squared plus y_1[3] squared is less than or equal to zero.] + + All decision variables x[0, x[1], x[2], x[3], as well as the auxiliary variables y_0 and y_1, are free to take any real value, meaning there are no explicit upper or lower bounds on these variables. The auxiliary variables t_0[0] and t_1[0] are also free to take any real value. The problem is a Second-Order Cone Programming (SOCP) problem, as indicated by the quadratic constraints.]-2.283994487886634Quadratic, out-of-scopenon-linear problem
156The problem involves minimizing the total cost of allocating three resources, \(x_0\), \(x_1\), and \(x_2\), where allocating one unit of \(x_0\) reduces the cost by 0.4471510837992146 units, allocating one unit of \(x_1\) reduces the cost by 0.5249098983165018 units, and allocating one unit of \(x_2\) increases the cost by 1.523631136652647 units. The allocation must satisfy five linear equality constraints: +1. The combined allocation of \(x_0\), \(x_1\), and \(x_2\) must result in a net value of 0.7034913153072593, with \(x_0\) contributing -0.3702651403450675, \(x_1\) contributing 0.7297352123799089, and \(x_2\) contributing 0.418858390580152. +2. The allocation must balance out to -0.1118855888702875, with \(x_0\) contributing 0.4153630032109168, \(x_1\) contributing 1.22940502964378, and \(x_2\) contributing -0.369836178888419. +3. The allocation must balance out to -0.0870555137122539, with \(x_0\) contributing 0.6396510452716259, \(x_1\) contributing -0.9807875135502231, and \(x_2\) contributing 1.032537634661943. +4. The allocation must balance out to -1.503406234028836, with \(x_0\) contributing 1.154571984507194, \(x_1\) contributing -0.0797094266894998, and \(x_2\) contributing -1.259256932824717. +5. The allocation must balance out to 0.4968828204562829, with \(x_0\) contributing -0.7680742354583855, \(x_1\) contributing -1.463758817488339, and \(x_2\) contributing 0.7608599270643391. + +Additionally, four auxiliary variables \(y_0[0]\), \(y_0[1]\), \(y_0[2]\), and \(t_0[0]\) are introduced, which are determined by the following linear equations: +- \(y_0[0]\) is determined by \(0.149372637294763 \cdot x_0 - 0.0700461252415553 \cdot x_1 + 0.0637326829867655 \cdot x_2 = -0.1129095562448545\). +- \(y_0[1]\) is determined by \(-0.0658271548457527 \cdot x_0 + 0.0224831194568463 \cdot x_1 - 0.00639053916626198 \cdot x_2 = -0.0490454717367705\). +- \(y_0[2]\) is determined by \(-0.00559770209526484 \cdot x_0 - 0.1749117419573911 \cdot x_1 - 0.0549153174437265 \cdot x_2 = 0.0268845405956578\). +- \(t_0[0]\) is determined by \(0.2634865225900492 \cdot x_0 + 1.089839700108032 \cdot x_1 - 1.345710871710692 \cdot x_2 = 0.2701569757905559\). + +A second-order cone constraint is imposed, requiring that the sum of the squares of \(y_0[0]\), \(y_0[1]\), and \(y_0[2]\) should not exceed the square of \(t_0[0]\). All decision variables \(x_0\), \(x_1\), \(x_2\), \(y_0[0]\), \(y_0[1]\), \(y_0[2]\), and \(t_0[0]\) are free to take any real value, meaning there are no upper or lower bounds imposed on their values. The goal is to determine the optimal allocation of \(x_0\), \(x_1\), and \(x_2\) to minimize the total cost while satisfying all the above constraints, including the second-order cone constraint. This is a Second-Order Cone Programming (SOCP) problem.1.0303398836924220The problem involves minimizing the total cost of allocating three resources, \(x_0\), \(x_1\), and \(x_2\), where allocating one unit of \(x_0\) reduces the cost by 0.4471510837992146 units, allocating one unit of \(x_1\) reduces the cost by 0.5249098983165018 units, and allocating one unit of \(x_2\) increases the cost by 1.523631136652647 units. The allocation must satisfy five linear equality constraints: + 1. The combined allocation of \(x_0\), \(x_1\), and \(x_2\) must result in a net value of 0.7034913153072593, with \(x_0\) contributing -0.3702651403450675, \(x_1\) contributing 0.7297352123799089, and \(x_2\) contributing 0.418858390580152. + 2. The allocation must balance out to -0.1118855888702875, with \(x_0\) contributing 0.4153630032109168, \(x_1\) contributing 1.22940502964378, and \(x_2\) contributing -0.369836178888419. + 3. The allocation must balance out to -0.0870555137122539, with \(x_0\) contributing 0.6396510452716259, \(x_1\) contributing -0.9807875135502231, and \(x_2\) contributing 1.032537634661943. + 4. The allocation must balance out to -1.503406234028836, with \(x_0\) contributing 1.154571984507194, \(x_1\) contributing -0.0797094266894998, and \(x_2\) contributing -1.259256932824717. + 5. The allocation must balance out to 0.4968828204562829, with \(x_0\) contributing -0.7680742354583855, \(x_1\) contributing -1.463758817488339, and \(x_2\) contributing 0.7608599270643391. + + Additionally, four auxiliary variables \(y_0[0\), \(y_0[1]\), \(y_0[2]\), and \(t_0[0]\) are introduced, which are determined by the following linear equations:] + - \(y_0[0\) is determined by \(0.149372637294763 \cdot x_0 - 0.0700461252415553 \cdot x_1 + 0.0637326829867655 \cdot x_2 = -0.1129095562448545\).] + - \(y_0[1\) is determined by \(-0.0658271548457527 \cdot x_0 + 0.0224831194568463 \cdot x_1 - 0.00639053916626198 \cdot x_2 = -0.0490454717367705\).] + - \(y_0[2\) is determined by \(-0.00559770209526484 \cdot x_0 - 0.1749117419573911 \cdot x_1 - 0.0549153174437265 \cdot x_2 = 0.0268845405956578\).] + - \(t_0[0\) is determined by \(0.2634865225900492 \cdot x_0 + 1.089839700108032 \cdot x_1 - 1.345710871710692 \cdot x_2 = 0.2701569757905559\).] + + A second-order cone constraint is imposed, requiring that the sum of the squares of \(y_0[0\), \(y_0[1]\), and \(y_0[2]\) should not exceed the square of \(t_0[0]\). All decision variables \(x_0\), \(x_1\), \(x_2\), \(y_0[0]\), \(y_0[1]\), \(y_0[2]\), and \(t_0[0]\) are free to take any real value, meaning there are no upper or lower bounds imposed on their values. The goal is to determine the optimal allocation of \(x_0\), \(x_1\), and \(x_2\) to minimize the total cost while satisfying all the above constraints, including the second-order cone constraint. This is a Second-Order Cone Programming (SOCP) problem.]1.030339883692422Quadratic, out-of-scopenon-linear problem
157The optimization problem involves minimizing the total operational cost, calculated as -0.6407821570716368 times the resource allocation at Hub 0, plus 0.1282093451424064 times the allocation at Hub 1, plus 0.5610539844701511 times the allocation at Hub 2, minus 0.7387913261851949 times the allocation at Hub 3, minus 0.6386135273718804 times the allocation at Hub 4. The resource allocations at each hub must satisfy three operational balance equations. The first equation states that 0.8358787628285221 times the allocation at Hub 0, minus 0.3221455044008418 times the allocation at Hub 1, plus 0.1449025416792807 times the allocation at Hub 2, minus 1.143088864150698 times the allocation at Hub 3, minus 0.2961821681740678 times the allocation at Hub 4, equals -0.3533166906383274. The second equation requires that -0.00318416966765216 times the allocation at Hub 0, plus 0.0636663117252584 times the allocation at Hub 1, minus 0.6335115786242427 times the allocation at Hub 2, plus 0.7906235569218798 times the allocation at Hub 3, plus 1.611869546160813 times the allocation at Hub 4, equals 0.6687613216515473. The third equation specifies that 0.6413661217416131 times the allocation at Hub 0, plus 0.4316049771976506 times the allocation at Hub 1, minus 2.705808657534711 times the allocation at Hub 2, minus 0.4442117847616459 times the allocation at Hub 3, plus 0.1130949341290518 times the allocation at Hub 4, equals -3.522255392214968. + +Each hub has additional constraints linking resource allocations to secondary operational metrics. For Hub 0, 1.486108997408437 times the allocation at Hub 0, plus 1.031757962743657 times the allocation at Hub 1, minus 0.4518720885240323 times the allocation at Hub 2, minus 0.3480442946563601 times the allocation at Hub 3, plus 1.580022153897245 times the allocation at Hub 4, plus the auxiliary variable y_0[0], equals -0.2475605897441492. Similarly, for Hub 0, -0.4066096523615669 times the allocation at Hub 0, minus 1.175211752170571 times the allocation at Hub 1, minus 0.7069411728647763 times the allocation at Hub 2, plus 0.836056412101241 times the allocation at Hub 3, minus 0.8926238938498942 times the allocation at Hub 4, plus the auxiliary variable y_0[1], equals 0.3074449602324441. Additionally, for Hub 0, -0.0569892486285832 times the allocation at Hub 0, plus 0.1233562548906215 times the allocation at Hub 1, minus 0.7134682305447539 times the allocation at Hub 2, minus 0.6038275844685674 times the allocation at Hub 3, minus 1.619511303009083 times the allocation at Hub 4, plus the auxiliary variable t_0[0], equals 0.2204862656053741. + +Similar constraints are specified for Hubs 1 to 4. For Hub 1, -1.294423006398278 times the allocation at Hub 0, plus 1.346172260047954 times the allocation at Hub 1, minus 0.358451283614951 times the allocation at Hub 2, minus 1.928440987292954 times the allocation at Hub 3, plus 1.126221395948665 times the allocation at Hub 4, plus y_1[0], equals 1.376491310478152. For Hub 1, 0.5609955781941032 times the allocation at Hub 0, minus 0.0458140884642076 times the allocation at Hub 1, minus 0.0344324048732548 times the allocation at Hub 2, plus 0.4252922167535212 times the allocation at Hub 3, minus 1.28973345438307 times the allocation at Hub 4, plus y_1[1], equals 0.051955328245377. For Hub 1, 0.0627617087678124 times the allocation at Hub 0, plus 1.313586666171607 times the allocation at Hub 1, minus 0.0750373702925165 times the allocation at Hub 2, plus 0.8637943589659705 times the allocation at Hub 3, minus 2.128672133715813 times the allocation at Hub 4, plus t_1[0], equals 0.4369998177620138. + +For Hub 2, 0.00108296658143404 times the allocation at Hub 0, plus 0.1150269209288099 times the allocation at Hub 1, minus 0.058430562821969 times the allocation at Hub 2, plus 0.1809260750957712 times the allocation at Hub 3, minus 0.107839460895758 times the allocation at Hub 4, plus y_2[0], equals -0.0466779941627568. For Hub 2, 0.1638929872378048 times the allocation at Hub 0, minus 0.0440335287917339 times the allocation at Hub 1, plus 0.1119306383643067 times the allocation at Hub 2, plus 0.0829079267212388 times the allocation at Hub 3, minus 0.0762160399972494 times the allocation at Hub 4, plus y_2[1], equals 0.1065935090322246. For Hub 2, 1.78119670849211 times the allocation at Hub 0, minus 0.9909783216060062 times the allocation at Hub 1, minus 1.08396186916195 times the allocation at Hub 2, minus 0.2137547994124323 times the allocation at Hub 3, minus 0.2410135384227864 times the allocation at Hub 4, plus t_2[0], equals 0.5794902323601145. + +For Hub 3, -0.0641439226811574 times the allocation at Hub 0, minus 0.0718053951291886 times the allocation at Hub 1, plus 0.0269862978161679 times the allocation at Hub 2, plus 0.1102601358034707 times the allocation at Hub 3, plus 7.27392e-05 times the allocation at Hub 4, plus y_3[0], equals 0.00315977947277087. For Hub 3, -0.0632288388926973 times the allocation at Hub 0, minus 0.1241779187569768 times the allocation at Hub 1, plus 0.0064629944919384 times the allocation at Hub 2, plus 0.0116896390416125 times the allocation at Hub 3, plus 0.079988047761199 times the allocation at Hub 4, plus y_3[1], equals 0.0925241950659326. For Hub 3, 0.4429906926189995 times the allocation at Hub 0, plus 0.0991542311544501 times the allocation at Hub 1, minus 0.3240726565893556 times the allocation at Hub 2, minus 0.6543525499518488 times the allocation at Hub 3, minus 0.318896820227852 times the allocation at Hub 4, plus t_3[0], equals 0.1889924405903281. + +For Hub 4, 0.0341804938689715 times the allocation at Hub 0, minus 0.1200246209041026 times the allocation at Hub 1, plus 0.0477968354488662 times the allocation at Hub 2, plus 0.0486460651624679 times the allocation at Hub 3, plus 0.0259780092785045 times the allocation at Hub 4, plus y_4[0], equals 0.1858842417400162. For Hub 4, -0.0427025044864235 times the allocation at Hub 0, minus 0.1161557760668262 times the allocation at Hub 1, minus 0.1281191768228226 times the allocation at Hub 2, plus 0.1296447474115312 times the allocation at Hub 3, minus 0.16357870291776 times the allocation at Hub 4, plus y_4[1], equals 0.0601306829794548. For Hub 4, 0.2140065823499792 times the allocation at Hub 0, minus 0.2328142561302691 times the allocation at Hub 1, minus 0.848196376440762 times the allocation at Hub 2, minus 0.5073797164375169 times the allocation at Hub 3, minus 2.928182945709474 times the allocation at Hub 4, plus t_4[0], equals 1.310467204578853. + +To ensure operational feasibility, each hub must satisfy a second-order cone constraint. For Hub 0, the sum of the squares of y_0[0] and y_0[1] must be less than or equal to the square of t_0[0]. Similarly, for Hub 1, the sum of the squares of y_1[0] and y_1[1] must be less than or equal to the square of t_1[0]. For Hub 2, the sum of the squares of y_2[0] and y_2[1] must be less than or equal to the square of t_2[0]. For Hub 3, the sum of the squares of y_3[0] and y_3[1] must be less than or equal to the square of t_3[0]. For Hub 4, the sum of the squares of y_4[0] and y_4[1] must be less than or equal to the square of t_4[0]. + +All variables, including resource allocations and auxiliary variables, are free to take any real value, allowing for flexible resource allocation to minimize costs while satisfying all operational constraints. The problem is a Second-Order Cone Programming (SOCP) problem, as it involves quadratic constraints represented by the second-order cone constraints. The equality constraints are linear, while the additional constraints (R3 to R17) link resource allocations to secondary operational metrics. The auxiliary variables (y_0[0], y_0[1], t_0[0], etc.) are part of the second-order cone constraints, and all variables (x[0] to x[4], y_0[0], y_0[1], y_1[0], y_1[1], etc., and t_0[0], t_1[0], t_2[0], etc.) are free variables.-0.179653919116790The optimization problem involves minimizing the total operational cost, calculated as -0.6407821570716368 times the resource allocation at Hub 0, plus 0.1282093451424064 times the allocation at Hub 1, plus 0.5610539844701511 times the allocation at Hub 2, minus 0.7387913261851949 times the allocation at Hub 3, minus 0.6386135273718804 times the allocation at Hub 4. The resource allocations at each hub must satisfy three operational balance equations. The first equation states that 0.8358787628285221 times the allocation at Hub 0, minus 0.3221455044008418 times the allocation at Hub 1, plus 0.1449025416792807 times the allocation at Hub 2, minus 1.143088864150698 times the allocation at Hub 3, minus 0.2961821681740678 times the allocation at Hub 4, equals -0.3533166906383274. The second equation requires that -0.00318416966765216 times the allocation at Hub 0, plus 0.0636663117252584 times the allocation at Hub 1, minus 0.6335115786242427 times the allocation at Hub 2, plus 0.7906235569218798 times the allocation at Hub 3, plus 1.611869546160813 times the allocation at Hub 4, equals 0.6687613216515473. The third equation specifies that 0.6413661217416131 times the allocation at Hub 0, plus 0.4316049771976506 times the allocation at Hub 1, minus 2.705808657534711 times the allocation at Hub 2, minus 0.4442117847616459 times the allocation at Hub 3, plus 0.1130949341290518 times the allocation at Hub 4, equals -3.522255392214968. + + Each hub has additional constraints linking resource allocations to secondary operational metrics. For Hub 0, 1.486108997408437 times the allocation at Hub 0, plus 1.031757962743657 times the allocation at Hub 1, minus 0.4518720885240323 times the allocation at Hub 2, minus 0.3480442946563601 times the allocation at Hub 3, plus 1.580022153897245 times the allocation at Hub 4, plus the auxiliary variable y_0[0, equals -0.2475605897441492. Similarly, for Hub 0, -0.4066096523615669 times the allocation at Hub 0, minus 1.175211752170571 times the allocation at Hub 1, minus 0.7069411728647763 times the allocation at Hub 2, plus 0.836056412101241 times the allocation at Hub 3, minus 0.8926238938498942 times the allocation at Hub 4, plus the auxiliary variable y_0[1], equals 0.3074449602324441. Additionally, for Hub 0, -0.0569892486285832 times the allocation at Hub 0, plus 0.1233562548906215 times the allocation at Hub 1, minus 0.7134682305447539 times the allocation at Hub 2, minus 0.6038275844685674 times the allocation at Hub 3, minus 1.619511303009083 times the allocation at Hub 4, plus the auxiliary variable t_0[0], equals 0.2204862656053741.] + + Similar constraints are specified for Hubs 1 to 4. For Hub 1, -1.294423006398278 times the allocation at Hub 0, plus 1.346172260047954 times the allocation at Hub 1, minus 0.358451283614951 times the allocation at Hub 2, minus 1.928440987292954 times the allocation at Hub 3, plus 1.126221395948665 times the allocation at Hub 4, plus y_1[0, equals 1.376491310478152. For Hub 1, 0.5609955781941032 times the allocation at Hub 0, minus 0.0458140884642076 times the allocation at Hub 1, minus 0.0344324048732548 times the allocation at Hub 2, plus 0.4252922167535212 times the allocation at Hub 3, minus 1.28973345438307 times the allocation at Hub 4, plus y_1[1], equals 0.051955328245377. For Hub 1, 0.0627617087678124 times the allocation at Hub 0, plus 1.313586666171607 times the allocation at Hub 1, minus 0.0750373702925165 times the allocation at Hub 2, plus 0.8637943589659705 times the allocation at Hub 3, minus 2.128672133715813 times the allocation at Hub 4, plus t_1[0], equals 0.4369998177620138.] + + For Hub 2, 0.00108296658143404 times the allocation at Hub 0, plus 0.1150269209288099 times the allocation at Hub 1, minus 0.058430562821969 times the allocation at Hub 2, plus 0.1809260750957712 times the allocation at Hub 3, minus 0.107839460895758 times the allocation at Hub 4, plus y_2[0, equals -0.0466779941627568. For Hub 2, 0.1638929872378048 times the allocation at Hub 0, minus 0.0440335287917339 times the allocation at Hub 1, plus 0.1119306383643067 times the allocation at Hub 2, plus 0.0829079267212388 times the allocation at Hub 3, minus 0.0762160399972494 times the allocation at Hub 4, plus y_2[1], equals 0.1065935090322246. For Hub 2, 1.78119670849211 times the allocation at Hub 0, minus 0.9909783216060062 times the allocation at Hub 1, minus 1.08396186916195 times the allocation at Hub 2, minus 0.2137547994124323 times the allocation at Hub 3, minus 0.2410135384227864 times the allocation at Hub 4, plus t_2[0], equals 0.5794902323601145.] + + For Hub 3, -0.0641439226811574 times the allocation at Hub 0, minus 0.0718053951291886 times the allocation at Hub 1, plus 0.0269862978161679 times the allocation at Hub 2, plus 0.1102601358034707 times the allocation at Hub 3, plus 7.27392e-05 times the allocation at Hub 4, plus y_3[0, equals 0.00315977947277087. For Hub 3, -0.0632288388926973 times the allocation at Hub 0, minus 0.1241779187569768 times the allocation at Hub 1, plus 0.0064629944919384 times the allocation at Hub 2, plus 0.0116896390416125 times the allocation at Hub 3, plus 0.079988047761199 times the allocation at Hub 4, plus y_3[1], equals 0.0925241950659326. For Hub 3, 0.4429906926189995 times the allocation at Hub 0, plus 0.0991542311544501 times the allocation at Hub 1, minus 0.3240726565893556 times the allocation at Hub 2, minus 0.6543525499518488 times the allocation at Hub 3, minus 0.318896820227852 times the allocation at Hub 4, plus t_3[0], equals 0.1889924405903281.] + + For Hub 4, 0.0341804938689715 times the allocation at Hub 0, minus 0.1200246209041026 times the allocation at Hub 1, plus 0.0477968354488662 times the allocation at Hub 2, plus 0.0486460651624679 times the allocation at Hub 3, plus 0.0259780092785045 times the allocation at Hub 4, plus y_4[0, equals 0.1858842417400162. For Hub 4, -0.0427025044864235 times the allocation at Hub 0, minus 0.1161557760668262 times the allocation at Hub 1, minus 0.1281191768228226 times the allocation at Hub 2, plus 0.1296447474115312 times the allocation at Hub 3, minus 0.16357870291776 times the allocation at Hub 4, plus y_4[1], equals 0.0601306829794548. For Hub 4, 0.2140065823499792 times the allocation at Hub 0, minus 0.2328142561302691 times the allocation at Hub 1, minus 0.848196376440762 times the allocation at Hub 2, minus 0.5073797164375169 times the allocation at Hub 3, minus 2.928182945709474 times the allocation at Hub 4, plus t_4[0], equals 1.310467204578853.] + + To ensure operational feasibility, each hub must satisfy a second-order cone constraint. For Hub 0, the sum of the squares of y_0[0 and y_0[1] must be less than or equal to the square of t_0[0]. Similarly, for Hub 1, the sum of the squares of y_1[0] and y_1[1] must be less than or equal to the square of t_1[0]. For Hub 2, the sum of the squares of y_2[0] and y_2[1] must be less than or equal to the square of t_2[0]. For Hub 3, the sum of the squares of y_3[0] and y_3[1] must be less than or equal to the square of t_3[0]. For Hub 4, the sum of the squares of y_4[0] and y_4[1] must be less than or equal to the square of t_4[0].] + + All variables, including resource allocations and auxiliary variables, are free to take any real value, allowing for flexible resource allocation to minimize costs while satisfying all operational constraints. The problem is a Second-Order Cone Programming (SOCP) problem, as it involves quadratic constraints represented by the second-order cone constraints. The equality constraints are linear, while the additional constraints (R3 to R17) link resource allocations to secondary operational metrics. The auxiliary variables (y_0[0, y_0[1], t_0[0], etc.) are part of the second-order cone constraints, and all variables (x[0] to x[4], y_0[0], y_0[1], y_1[0], y_1[1], etc., and t_0[0], t_1[0], t_2[0], etc.) are free variables.]-0.17965391911679Quadratic, out-of-scopenon-linear problem
158The problem involves minimizing the total cost calculated as 0.408544260832344 times the value of x[0] minus 1.177163209326489 times the value of x[1] minus 0.3033727028272864 times the value of x[2]. This minimization is subject to several constraints: the first linear constraint requires that 0.5462940531635887 times x[0] minus 1.241668902664911 times x[1] minus 0.5998249994903344 times x[2] equals 0.3835911096798181. The second linear constraint requires that negative 0.030455142845263 times x[0] minus 0.7427336810498593 times x[1] plus 0.3957667498639689 times x[2] equals negative 0.6803383944729795. Another constraint, R2, requires that negative 0.0677712816971205 times x[0] plus 0.170365390174673 times x[1] minus 0.00413868631847959 times x[2] plus y_0[0] equals 0.00802717043497164. Constraint R3 requires that 0.0215916329150542 times x[0] minus 0.0479685166742133 times x[1] plus 0.0677847235104595 times x[2] plus y_0[1] equals negative 0.1511858316417728. Constraint R4 requires that negative 0.219725429611857 times x[0] minus 1.453471058496226 times x[1] minus 0.5544082541070903 times x[2] plus t_0[0] equals 1.182764044680309. Additionally, the square of t_0[0] must be less than or equal to the sum of the squares of y_0[0] and y_0[1], represented by the constraint negative t_0[0] squared plus y_0[0] squared plus y_0[1] squared is less than or equal to zero. All variables x[0], x[1], x[2], y_0[0], y_0[1], and t_0[0] are free to take any real value. The problem is a Second-Order Cone Programming (SOCP) problem due to the presence of the quadratic constraint.0.0266909529063380The problem involves minimizing the total cost calculated as 0.408544260832344 times the value of x[0 minus 1.177163209326489 times the value of x[1] minus 0.3033727028272864 times the value of x[2]. This minimization is subject to several constraints: the first linear constraint requires that 0.5462940531635887 times x[0] minus 1.241668902664911 times x[1] minus 0.5998249994903344 times x[2] equals 0.3835911096798181. The second linear constraint requires that negative 0.030455142845263 times x[0] minus 0.7427336810498593 times x[1] plus 0.3957667498639689 times x[2] equals negative 0.6803383944729795. Another constraint, R2, requires that negative 0.0677712816971205 times x[0] plus 0.170365390174673 times x[1] minus 0.00413868631847959 times x[2] plus y_0[0] equals 0.00802717043497164. Constraint R3 requires that 0.0215916329150542 times x[0] minus 0.0479685166742133 times x[1] plus 0.0677847235104595 times x[2] plus y_0[1] equals negative 0.1511858316417728. Constraint R4 requires that negative 0.219725429611857 times x[0] minus 1.453471058496226 times x[1] minus 0.5544082541070903 times x[2] plus t_0[0] equals 1.182764044680309. Additionally, the square of t_0[0] must be less than or equal to the sum of the squares of y_0[0] and y_0[1], represented by the constraint negative t_0[0] squared plus y_0[0] squared plus y_0[1] squared is less than or equal to zero. All variables x[0], x[1], x[2], y_0[0], y_0[1], and t_0[0] are free to take any real value. The problem is a Second-Order Cone Programming (SOCP) problem due to the presence of the quadratic constraint.]0.026690952906338Quadratic, out-of-scopenon-linear problem
159The problem involves minimizing a linear objective function defined as 0.7117963359087778 times the value of variable x[0], plus 1.495363634959802 times the value of variable x[1], plus 0.0634582749300659 times the value of variable x[2]. The variables x[0], x[1], and x[2] are free to take any real value. The optimization is subject to two linear equality constraints. The first constraint requires that -0.0258717563775294 times x[0], plus 0.4400877796975226 times x[1], plus 1.085219312984192 times x[2], equals -0.399404323103251. The second constraint requires that -1.063740820798877 times x[0], plus 0.0262817424591634 times x[1], plus 0.3002430009821057 times x[2], equals -0.4003348752264555. + +Additionally, the problem includes two second-order cone constraints. The first cone constraint involves variables y_0[0], y_0[1], y_0[2], and t_0[0], and requires that the square root of the sum of the squares of y_0[0], y_0[1], and y_0[2] is less than or equal to t_0[0]. This constraint is associated with the following linear relationships: 0.0709198094546279 times x[0], minus 0.0210367192622051 times x[1], plus 0.0895063725378235 times x[2], plus y_0[0], equals -0.1178974392901265; 0.0489330070748356 times x[0], minus 0.0910855567896424 times x[1], minus 0.0299901411888931 times x[2], plus y_0[1], equals 0.1528438418108943; -0.0102367576685978 times x[0], minus 0.0809953969784582 times x[1], minus 0.0326792481068488 times x[2], plus y_0[2], equals 0.1304009358682119; and -0.7205007161713685 times x[0], plus 0.8402710079185606 times x[1], plus 1.174037649876418 times x[2], plus t_0[0], equals 1.142365596215822. + +The second cone constraint involves variables y_1[0], y_1[1], y_1[2], and t_1[0], and requires that the square root of the sum of the squares of y_1[0], y_1[1], and y_1[2] is less than or equal to t_1[0]. This constraint is associated with the following linear relationships: -0.1053047361476771 times x[0], minus 0.1098916681441547 times x[1], plus 0.0725443868339502 times x[2], plus y_1[0], equals -0.0464104086014158; -0.1846597649031371 times x[0], minus 0.0474185137216481 times x[1], minus 0.0085470813934405 times x[2], plus y_1[1], equals -0.0958578667235037; -0.1177843202999298 times x[0], minus 0.0387875095823661 times x[1], minus 0.0741126306954255 times x[2], plus y_1[2], equals -0.013119694402389; and 1.090558517373246 times x[0], minus 0.0977574216352732 times x[1], plus 0.0671448858489764 times x[2], plus t_1[0], equals 0.6764507139985713. + +All variables, including x[0], x[1], x[2], y_0[0], y_0[1], y_0[2], y_1[0], y_1[1], y_1[2], t_0[0], and t_1[0], are free to take any real value. The goal is to determine the values of these variables that minimize the objective function while satisfying all the specified constraints.-1.3735384528365860The problem involves minimizing a linear objective function defined as 0.7117963359087778 times the value of variable x[0, plus 1.495363634959802 times the value of variable x[1], plus 0.0634582749300659 times the value of variable x[2]. The variables x[0], x[1], and x[2] are free to take any real value. The optimization is subject to two linear equality constraints. The first constraint requires that -0.0258717563775294 times x[0], plus 0.4400877796975226 times x[1], plus 1.085219312984192 times x[2], equals -0.399404323103251. The second constraint requires that -1.063740820798877 times x[0], plus 0.0262817424591634 times x[1], plus 0.3002430009821057 times x[2], equals -0.4003348752264555.] + + Additionally, the problem includes two second-order cone constraints. The first cone constraint involves variables y_0[0, y_0[1], y_0[2], and t_0[0], and requires that the square root of the sum of the squares of y_0[0], y_0[1], and y_0[2] is less than or equal to t_0[0]. This constraint is associated with the following linear relationships: 0.0709198094546279 times x[0], minus 0.0210367192622051 times x[1], plus 0.0895063725378235 times x[2], plus y_0[0], equals -0.1178974392901265; 0.0489330070748356 times x[0], minus 0.0910855567896424 times x[1], minus 0.0299901411888931 times x[2], plus y_0[1], equals 0.1528438418108943; -0.0102367576685978 times x[0], minus 0.0809953969784582 times x[1], minus 0.0326792481068488 times x[2], plus y_0[2], equals 0.1304009358682119; and -0.7205007161713685 times x[0], plus 0.8402710079185606 times x[1], plus 1.174037649876418 times x[2], plus t_0[0], equals 1.142365596215822.] + + The second cone constraint involves variables y_1[0, y_1[1], y_1[2], and t_1[0], and requires that the square root of the sum of the squares of y_1[0], y_1[1], and y_1[2] is less than or equal to t_1[0]. This constraint is associated with the following linear relationships: -0.1053047361476771 times x[0], minus 0.1098916681441547 times x[1], plus 0.0725443868339502 times x[2], plus y_1[0], equals -0.0464104086014158; -0.1846597649031371 times x[0], minus 0.0474185137216481 times x[1], minus 0.0085470813934405 times x[2], plus y_1[1], equals -0.0958578667235037; -0.1177843202999298 times x[0], minus 0.0387875095823661 times x[1], minus 0.0741126306954255 times x[2], plus y_1[2], equals -0.013119694402389; and 1.090558517373246 times x[0], minus 0.0977574216352732 times x[1], plus 0.0671448858489764 times x[2], plus t_1[0], equals 0.6764507139985713.] + + All variables, including x[0, x[1], x[2], y_0[0], y_0[1], y_0[2], y_1[0], y_1[1], y_1[2], t_0[0], and t_1[0], are free to take any real value. The goal is to determine the values of these variables that minimize the objective function while satisfying all the specified constraints.]-1.373538452836586Quadratic, out-of-scopenon-linear problem
160Minimize the linear objective function involving five variables with coefficients: -1.27457045729125 for x[0], 0.3829428410019269 for x[1], 0.8634766321930054 for x[2], -0.8495951803940098 for x[3], and 0.7818351274236842 for x[4]. + +The solution must satisfy five linear equality constraints: +1. A weighted combination of x[0] (0.3779915609931544), x[1] (-0.2098956852913138), x[2] (-0.2045186307690221), x[3] (1.70241687084974), and x[4] (-0.0830495385416335) must equal -0.8872046273965938. +2. A weighted combination of x[0] (-0.1980432307652125), x[1] (0.5734762682253645), x[2] (0.8168321608033806), x[3] (0.4534299593541723), and x[4] (-1.551957669500629) must equal -1.682642259460053. +3. A weighted combination of x[0] (2.09940339313846), x[1] (-0.5382869592818297), x[2] (-0.4711126728138521), x[3] (2.41705884114686), and x[4] (0.3807643962655075) must equal -1.178472853778845. +4. A weighted combination of x[0] (-0.0319043926204142), x[1] (1.274635913461541), x[2] (0.2494503935760785), x[3] (-1.178243710464141), and x[4] (-0.658162908023939) must equal 1.097123236331887. +5. A weighted combination of x[0] (-2.38990451126146), x[1] (-0.4143841871813654), x[2] (-0.1278674033858785), x[3] (-0.0913945158661918), and x[4] (-0.6463312417465074) must equal 0.1278954511759081. + +Additionally, four auxiliary variables y_0[0], y_0[1], y_0[2], and t_0[0] are introduced as part of a second-order cone constraint. The solution must satisfy four linear equality constraints involving these variables: +1. A weighted combination of x[0] (0.0811031804701355), x[1] (0.0614084577317061), x[2] (-0.0221560152059943), x[3] (0.2071308897950201), and x[4] (0.1014268263641295), plus y_0[0], must equal 0.0612422178599395. +2. A weighted combination of x[0] (-0.0913274075374076), x[1] (0.0084232463574275), x[2] (0.0231115788245813), x[3] (-0.0528298653734004), and x[4] (-0.1547249726938889), plus y_0[1], must equal 0.0531856861000841. +3. A weighted combination of x[0] (0.0682088948475448), x[1] (-0.1651917752267487), x[2] (-0.0413346954944568), x[3] (-0.0532349804461065), and x[4] (0.0783732630578606), plus y_0[2], must equal -0.0703701003068877. +4. A weighted combination of x[0] (0.7179187345461103), x[1] (-0.5038326161213537), x[2] (0.9210968531714094), x[3] (-0.4759925591110667), and x[4] (-1.776260360500366), plus t_0[0], must equal 0.6058017967567707. + +The solution must also satisfy a second-order cone constraint: the square of t_0[0] must be less than or equal to the sum of the squares of y_0[0], y_0[1], and y_0[2], expressed as \(-t_0[0]^2 + y_0[0]^2 + y_0[1]^2 + y_0[2]^2 \leq 0\). + +All variables, including x[0] to x[4], y_0[0], y_0[1], y_0[2], and t_0[0], are free to take any real value. The goal is to determine the values of these variables that minimize the objective function while satisfying all the equality and cone constraints.-0.51483576230749010Minimize the linear objective function involving five variables with coefficients: -1.27457045729125 for x[0, 0.3829428410019269 for x[1], 0.8634766321930054 for x[2], -0.8495951803940098 for x[3], and 0.7818351274236842 for x[4]. ] + + The solution must satisfy five linear equality constraints: + 1. A weighted combination of x[0 (0.3779915609931544), x[1] (-0.2098956852913138), x[2] (-0.2045186307690221), x[3] (1.70241687084974), and x[4] (-0.0830495385416335) must equal -0.8872046273965938. ] + 2. A weighted combination of x[0 (-0.1980432307652125), x[1] (0.5734762682253645), x[2] (0.8168321608033806), x[3] (0.4534299593541723), and x[4] (-1.551957669500629) must equal -1.682642259460053. ] + 3. A weighted combination of x[0 (2.09940339313846), x[1] (-0.5382869592818297), x[2] (-0.4711126728138521), x[3] (2.41705884114686), and x[4] (0.3807643962655075) must equal -1.178472853778845. ] + 4. A weighted combination of x[0 (-0.0319043926204142), x[1] (1.274635913461541), x[2] (0.2494503935760785), x[3] (-1.178243710464141), and x[4] (-0.658162908023939) must equal 1.097123236331887. ] + 5. A weighted combination of x[0 (-2.38990451126146), x[1] (-0.4143841871813654), x[2] (-0.1278674033858785), x[3] (-0.0913945158661918), and x[4] (-0.6463312417465074) must equal 0.1278954511759081. ] + + Additionally, four auxiliary variables y_0[0, y_0[1], y_0[2], and t_0[0] are introduced as part of a second-order cone constraint. The solution must satisfy four linear equality constraints involving these variables: ] + 1. A weighted combination of x[0 (0.0811031804701355), x[1] (0.0614084577317061), x[2] (-0.0221560152059943), x[3] (0.2071308897950201), and x[4] (0.1014268263641295), plus y_0[0], must equal 0.0612422178599395. ] + 2. A weighted combination of x[0 (-0.0913274075374076), x[1] (0.0084232463574275), x[2] (0.0231115788245813), x[3] (-0.0528298653734004), and x[4] (-0.1547249726938889), plus y_0[1], must equal 0.0531856861000841. ] + 3. A weighted combination of x[0 (0.0682088948475448), x[1] (-0.1651917752267487), x[2] (-0.0413346954944568), x[3] (-0.0532349804461065), and x[4] (0.0783732630578606), plus y_0[2], must equal -0.0703701003068877. ] + 4. A weighted combination of x[0 (0.7179187345461103), x[1] (-0.5038326161213537), x[2] (0.9210968531714094), x[3] (-0.4759925591110667), and x[4] (-1.776260360500366), plus t_0[0], must equal 0.6058017967567707. ] + + The solution must also satisfy a second-order cone constraint: the square of t_0[0 must be less than or equal to the sum of the squares of y_0[0], y_0[1], and y_0[2], expressed as \(-t_0[0]^2 + y_0[0]^2 + y_0[1]^2 + y_0[2]^2 \leq 0\). ] + + All variables, including x[0 to x[4], y_0[0], y_0[1], y_0[2], and t_0[0], are free to take any real value. The goal is to determine the values of these variables that minimize the objective function while satisfying all the equality and cone constraints.]-0.5148357623074901Quadratic, out-of-scopenon-linear problem
161The goal is to minimize the total cost, which is calculated as 0.786272624432491 times x[0] plus 0.8625634188757304 times x[1] minus 1.12398756204435 times x[2]. The decision variables include x[0], x[1], x[2], y_0[0], y_0[1], y_0[2], y_1[0], y_1[1], y_1[2], y_2[0], y_2[1], y_2[2], y_3[0], y_3[1], y_3[2], t_0[0], t_1[0], t_2[0], and t_3[0]. All variables are free to take any real value. + +The problem is subject to linear equality constraints. The first constraint requires that 1.265089926033166 times x[0] plus 0.8265511590295959 times x[1] plus 0.6425143729335491 times x[2] equals -1.340261867227508. The second constraint requires that -1.038046026005065 times x[0] minus 0.978681730406505 times x[1] minus 0.3158917890099441 times x[2] equals 0.9895473898354382. The third constraint requires that 0.3708381346357463 times x[0] plus 0.2148963219535531 times x[1] minus 1.83990668416334 times x[2] equals 3.062677436779988. The fourth constraint requires that -1.872418427024533 times x[0] minus 1.159027962632082 times x[1] plus 0.3313042271069708 times x[2] equals -0.2396557035046597. + +Additional constraints ensure that specific combinations of the decision variables and auxiliary variables satisfy certain conditions. For the first set of auxiliary variables, 0.0845901604928491 times x[0] plus 0.00653062634511069 times x[1] minus 0.0396187361282293 times x[2] plus y_0[0] equals 0.1028911134186171, -0.1655645187396123 times x[0] minus 0.00481490476108966 times x[1] plus 0.1091868022965404 times x[2] plus y_0[1] equals -0.0791773986380616, and 0.0657601002643706 times x[0] plus 0.0770866340133718 times x[1] plus 0.0717168532804696 times x[2] plus y_0[2] equals 0.1545300907589844. For the second set of auxiliary variables, 0.3298234267560927 times x[0] plus 1.022831383811937 times x[1] minus 0.00493650012051474 times x[2] plus t_0[0] equals 0.9587050025563594, 0.8976617894842607 times x[0] minus 1.960565868013685 times x[1] plus 0.2084251295126738 times x[2] plus y_1[0] equals -0.1849711881061493, -0.7094023471201548 times x[0] plus 0.6353877939866961 times x[1] minus 0.0876866180562744 times x[2] plus y_1[1] equals -0.4919818599100178, and 2.698160782648244 times x[0] minus 0.4574765835824846 times x[1] plus 0.6985665960561629 times x[2] plus y_1[2] equals 0.2614076505075524. + +For the third set of auxiliary variables, -0.7226507325429118 times x[0] plus 0.0568702304829931 times x[1] plus 1.471507498629236 times x[2] plus t_1[0] equals 0.320143628846921, 0.0533180689091473 times x[0] minus 0.1728737890908431 times x[1] plus 0.0319070165322495 times x[2] plus y_2[0] equals 0.0152872162677227, -0.0512090278756547 times x[0] minus 0.0557654636443004 times x[1] minus 0.2980053053415462 times x[2] plus y_2[1] equals -0.0964105256925016, and 0.176881384588485 times x[0] plus 0.0368327469045029 times x[1] plus 0.0368571461194376 times x[2] plus y_2[2] equals 0.0672644472047825. For the fourth set of auxiliary variables, 0.7106618613482838 times x[0] minus 0.7138000011448565 times x[1] plus 0.1976996075797842 times x[2] plus t_2[0] equals 1.260005801822901, 0.0443424626310529 times x[0] plus 0.1363607191306093 times x[1] minus 0.0668054006192922 times x[2] plus y_3[0] equals 0.0932454917759326, 0.0536583684212656 times x[0] minus 0.0491976587190737 times x[1] plus 0.06269906389831 times x[2] plus y_3[1] equals -0.00327479473248982, and -0.1482437619827962 times x[0] minus 0.1780634692898283 times x[1] plus 0.1091477105610945 times x[2] plus y_3[2] equals 0.0218802336021696. + +The problem also includes Second-Order Cone Constraints. The first Second-Order Cone Constraint ensures that the negative square of t_0[0] plus the squares of y_0[0], y_0[1], and y_0[2] is less than or equal to zero. The second Second-Order Cone Constraint ensures that the negative square of t_1[0] plus the squares of y_1[0], y_1[1], and y_1[2] is less than or equal to zero. The third Second-Order Cone Constraint ensures that the negative square of t_2[0] plus the squares of y_2[0], y_2[1], and y_2[2] is less than or equal to zero. The fourth Second-Order Cone Constraint ensures that the negative square of t_3[0] plus the squares of y_3[0], y_3[1], and y_3[2] is less than or equal to zero. These constraints are critical for ensuring the problem's feasibility and optimality.1.458029591903660The goal is to minimize the total cost, which is calculated as 0.786272624432491 times x[0 plus 0.8625634188757304 times x[1] minus 1.12398756204435 times x[2]. The decision variables include x[0], x[1], x[2], y_0[0], y_0[1], y_0[2], y_1[0], y_1[1], y_1[2], y_2[0], y_2[1], y_2[2], y_3[0], y_3[1], y_3[2], t_0[0], t_1[0], t_2[0], and t_3[0]. All variables are free to take any real value.] + + The problem is subject to linear equality constraints. The first constraint requires that 1.265089926033166 times x[0 plus 0.8265511590295959 times x[1] plus 0.6425143729335491 times x[2] equals -1.340261867227508. The second constraint requires that -1.038046026005065 times x[0] minus 0.978681730406505 times x[1] minus 0.3158917890099441 times x[2] equals 0.9895473898354382. The third constraint requires that 0.3708381346357463 times x[0] plus 0.2148963219535531 times x[1] minus 1.83990668416334 times x[2] equals 3.062677436779988. The fourth constraint requires that -1.872418427024533 times x[0] minus 1.159027962632082 times x[1] plus 0.3313042271069708 times x[2] equals -0.2396557035046597.] + + Additional constraints ensure that specific combinations of the decision variables and auxiliary variables satisfy certain conditions. For the first set of auxiliary variables, 0.0845901604928491 times x[0 plus 0.00653062634511069 times x[1] minus 0.0396187361282293 times x[2] plus y_0[0] equals 0.1028911134186171, -0.1655645187396123 times x[0] minus 0.00481490476108966 times x[1] plus 0.1091868022965404 times x[2] plus y_0[1] equals -0.0791773986380616, and 0.0657601002643706 times x[0] plus 0.0770866340133718 times x[1] plus 0.0717168532804696 times x[2] plus y_0[2] equals 0.1545300907589844. For the second set of auxiliary variables, 0.3298234267560927 times x[0] plus 1.022831383811937 times x[1] minus 0.00493650012051474 times x[2] plus t_0[0] equals 0.9587050025563594, 0.8976617894842607 times x[0] minus 1.960565868013685 times x[1] plus 0.2084251295126738 times x[2] plus y_1[0] equals -0.1849711881061493, -0.7094023471201548 times x[0] plus 0.6353877939866961 times x[1] minus 0.0876866180562744 times x[2] plus y_1[1] equals -0.4919818599100178, and 2.698160782648244 times x[0] minus 0.4574765835824846 times x[1] plus 0.6985665960561629 times x[2] plus y_1[2] equals 0.2614076505075524.] + + For the third set of auxiliary variables, -0.7226507325429118 times x[0 plus 0.0568702304829931 times x[1] plus 1.471507498629236 times x[2] plus t_1[0] equals 0.320143628846921, 0.0533180689091473 times x[0] minus 0.1728737890908431 times x[1] plus 0.0319070165322495 times x[2] plus y_2[0] equals 0.0152872162677227, -0.0512090278756547 times x[0] minus 0.0557654636443004 times x[1] minus 0.2980053053415462 times x[2] plus y_2[1] equals -0.0964105256925016, and 0.176881384588485 times x[0] plus 0.0368327469045029 times x[1] plus 0.0368571461194376 times x[2] plus y_2[2] equals 0.0672644472047825. For the fourth set of auxiliary variables, 0.7106618613482838 times x[0] minus 0.7138000011448565 times x[1] plus 0.1976996075797842 times x[2] plus t_2[0] equals 1.260005801822901, 0.0443424626310529 times x[0] plus 0.1363607191306093 times x[1] minus 0.0668054006192922 times x[2] plus y_3[0] equals 0.0932454917759326, 0.0536583684212656 times x[0] minus 0.0491976587190737 times x[1] plus 0.06269906389831 times x[2] plus y_3[1] equals -0.00327479473248982, and -0.1482437619827962 times x[0] minus 0.1780634692898283 times x[1] plus 0.1091477105610945 times x[2] plus y_3[2] equals 0.0218802336021696.] + + The problem also includes Second-Order Cone Constraints. The first Second-Order Cone Constraint ensures that the negative square of t_0[0 plus the squares of y_0[0], y_0[1], and y_0[2] is less than or equal to zero. The second Second-Order Cone Constraint ensures that the negative square of t_1[0] plus the squares of y_1[0], y_1[1], and y_1[2] is less than or equal to zero. The third Second-Order Cone Constraint ensures that the negative square of t_2[0] plus the squares of y_2[0], y_2[1], and y_2[2] is less than or equal to zero. The fourth Second-Order Cone Constraint ensures that the negative square of t_3[0] plus the squares of y_3[0], y_3[1], and y_3[2] is less than or equal to zero. These constraints are critical for ensuring the problem's feasibility and optimality.]1.45802959190366Quadratic, out-of-scopenon-linear problem
162Minimize the total cost of allocating six resources, where the cost coefficients for each resource are -0.2230299351960377, 0.2260048024316069, -0.1766835611048999, 0.5573216224422132, -1.208376221550068, and 0.5847966392616851 units per unit allocated, respectively. The allocation must satisfy sixteen linear equality constraints and three second-order cone constraints. The first five linear equality constraints are as follows: the first constraint requires that 0.0864123382415046 times the first resource minus 0.2999520944563616 times the second resource minus 0.4373153026838008 times the third resource plus 0.2236844833244651 times the fourth resource minus 0.1726687049023944 times the fifth resource plus 1.131231237141819 times the sixth resource equals -0.2555719673916053. The second constraint requires that 0.6475024542623146 times the first resource minus 0.2411773085202067 times the second resource minus 1.152365276965448 times the third resource plus 0.0681294271686064 times the fourth resource minus 0.460006367650037 times the fifth resource minus 0.8626734903353966 times the sixth resource equals -1.611099466848817. The third constraint requires that -0.1892421564042141 times the first resource plus 0.4442286186625635 times the second resource plus 0.0351084932857791 times the third resource minus 1.65945657674057 times the fourth resource plus 1.813083837788898 times the fifth resource minus 0.1731007007836295 times the sixth resource equals -2.302900632383548. The fourth constraint requires that 0.6105836251433618 times the first resource plus 1.224758698265837 times the second resource minus 0.6569021430924753 times the third resource minus 1.723125872166611 times the fourth resource plus 0.6658852466385591 times the fifth resource plus 0.8417973521416725 times the sixth resource equals -3.826954722671106. The fifth constraint requires that -0.1167867054929794 times the first resource plus 0.1666505226611245 times the second resource minus 0.0580663537752885 times the third resource minus 0.1929108225442672 times the fourth resource minus 1.118313643238442 times the fifth resource plus 1.582899027912414 times the sixth resource equals 0.5701541914439772. + +The remaining eleven linear equality constraints involve auxiliary variables introduced for the second-order cone constraints. The sixth constraint requires that -0.0231503875559003 times the first resource plus 0.00454721771555977 times the second resource minus 0.0401563754472746 times the third resource plus 0.0319081283767895 times the fourth resource minus 0.0445254387459282 times the fifth resource plus 0.0666388070919248 times the sixth resource plus y_0[0] equals 0.1220045375876136. The seventh constraint requires that -0.0850790157464754 times the first resource minus 0.0476072870612039 times the second resource minus 0.068828152035275 times the third resource minus 0.0919526883283748 times the fourth resource minus 0.0484614346821255 times the fifth resource minus 0.157778975935517 times the sixth resource plus y_0[1] equals 0.00586701808623117. The eighth constraint requires that 0.0578622514811724 times the first resource minus 0.1018459374054135 times the second resource plus 0.028959008397348 times the third resource plus 0.0124914167995535 times the fourth resource minus 0.1217493644708651 times the fifth resource minus 0.0455371326070688 times the sixth resource plus y_0[2] equals 0.00103506581258387. The ninth constraint requires that 0.5438484424289765 times the first resource minus 0.4058548451270487 times the second resource plus 0.00464438168201298 times the third resource plus 0.6829984219379786 times the fourth resource minus 0.813089431925804 times the fifth resource plus 0.1465879382567549 times the sixth resource plus t_0[0] equals 0.8921404899384406. + +The tenth constraint requires that -0.1565490941785506 times the first resource minus 0.0151974010871207 times the second resource plus 0.0330083659733866 times the third resource plus 0.0409835559769911 times the fourth resource minus 0.025926735354538 times the fifth resource minus 0.0791575923837095 times the sixth resource plus y_1[0] equals 0.0252004385932512. The eleventh constraint requires that -0.00679732652950372 times the first resource minus 0.0159930899677971 times the second resource plus 0.00044788212331377231 times the third resource minus 0.00870755949376334 times the fourth resource plus 0.0106774891090615 times the fifth resource minus 0.0819505944729122 times the sixth resource plus y_1[1] equals -0.2461745595888191. The twelfth constraint requires that -0.0767410168100431 times the first resource minus 0.1162981298613443 times the second resource plus 0.2516807060221921 times the third resource minus 0.00442617885794681 times the fourth resource minus 0.0897268862621352 times the fifth resource minus 0.1001649216336424 times the sixth resource plus y_1[2] equals 0.0676538189988438. The thirteenth constraint requires that 2.12708121691458 times the first resource plus 0.5354675680200557 times the second resource plus 0.1244147628194144 times the third resource plus 1.32540724304661 times the fourth resource minus 0.6888780286074161 times the fifth resource plus 1.1055685780941 times the sixth resource plus t_1[0] equals 0.8216483573755342. + +The fourteenth constraint requires that 0.1045779404782078 times the first resource plus 0.117211531954123 times the second resource minus 0.093303260093519 times the third resource plus 0.114920382796502 times the fourth resource plus 0.0892956784939136 times the fifth resource minus 0.1270063126632456 times the sixth resource plus y_2[0] equals 0.1487112654209727. The fifteenth constraint requires that 0.1031929746661822 times the first resource plus 0.0366834994452099 times the second resource minus 0.0811775022010661 times the third resource minus 0.0941975746436261 times the fourth resource plus 0.034402649559877 times the fifth resource minus 0.2382614709448476 times the sixth resource plus y_2[1] equals -0.00259845523628418. The sixteenth constraint requires that -0.0649606742324766 times the first resource plus 0.1500400272048616 times the second resource plus 0.0157455125935616 times the third resource minus 0.0173529909921789 times the fourth resource minus 0.045730709242314 times the fifth resource minus 0.068092234748515 times the sixth resource plus y_2[2] equals -0.1982729375753503. + +The three second-order cone constraints are as follows: the first cone constraint requires that the sum of the squares of y_0[0], y_0[1], and y_0[2] is at most the square of t_0[0]. The second cone constraint requires that the sum of the squares of y_1[0], y_1[1], and y_1[2] is at most the square of t_1[0]. The third cone constraint requires that the sum of the squares of y_2[0], y_2[1], and y_2[2] is at most the square of t_2[0]. All decision variables, including the resource allocations x[0], x[1], x[2], x[3], x[4], and x[5], and the auxiliary variables y_0[0], y_0[1], y_0[2], y_1[0], y_1[1], y_1[2], y_2[0], y_2[1], y_2[2], t_0[0], t_1[0], and t_2[0], are free to take any real value. The goal is to determine the optimal allocation of the six resources that minimizes the total cost while satisfying all linear equality and second-order cone constraints.0.99294876483910210Minimize the total cost of allocating six resources, where the cost coefficients for each resource are -0.2230299351960377, 0.2260048024316069, -0.1766835611048999, 0.5573216224422132, -1.208376221550068, and 0.5847966392616851 units per unit allocated, respectively. The allocation must satisfy sixteen linear equality constraints and three second-order cone constraints. The first five linear equality constraints are as follows: the first constraint requires that 0.0864123382415046 times the first resource minus 0.2999520944563616 times the second resource minus 0.4373153026838008 times the third resource plus 0.2236844833244651 times the fourth resource minus 0.1726687049023944 times the fifth resource plus 1.131231237141819 times the sixth resource equals -0.2555719673916053. The second constraint requires that 0.6475024542623146 times the first resource minus 0.2411773085202067 times the second resource minus 1.152365276965448 times the third resource plus 0.0681294271686064 times the fourth resource minus 0.460006367650037 times the fifth resource minus 0.8626734903353966 times the sixth resource equals -1.611099466848817. The third constraint requires that -0.1892421564042141 times the first resource plus 0.4442286186625635 times the second resource plus 0.0351084932857791 times the third resource minus 1.65945657674057 times the fourth resource plus 1.813083837788898 times the fifth resource minus 0.1731007007836295 times the sixth resource equals -2.302900632383548. The fourth constraint requires that 0.6105836251433618 times the first resource plus 1.224758698265837 times the second resource minus 0.6569021430924753 times the third resource minus 1.723125872166611 times the fourth resource plus 0.6658852466385591 times the fifth resource plus 0.8417973521416725 times the sixth resource equals -3.826954722671106. The fifth constraint requires that -0.1167867054929794 times the first resource plus 0.1666505226611245 times the second resource minus 0.0580663537752885 times the third resource minus 0.1929108225442672 times the fourth resource minus 1.118313643238442 times the fifth resource plus 1.582899027912414 times the sixth resource equals 0.5701541914439772. + + The remaining eleven linear equality constraints involve auxiliary variables introduced for the second-order cone constraints. The sixth constraint requires that -0.0231503875559003 times the first resource plus 0.00454721771555977 times the second resource minus 0.0401563754472746 times the third resource plus 0.0319081283767895 times the fourth resource minus 0.0445254387459282 times the fifth resource plus 0.0666388070919248 times the sixth resource plus y_0[0 equals 0.1220045375876136. The seventh constraint requires that -0.0850790157464754 times the first resource minus 0.0476072870612039 times the second resource minus 0.068828152035275 times the third resource minus 0.0919526883283748 times the fourth resource minus 0.0484614346821255 times the fifth resource minus 0.157778975935517 times the sixth resource plus y_0[1] equals 0.00586701808623117. The eighth constraint requires that 0.0578622514811724 times the first resource minus 0.1018459374054135 times the second resource plus 0.028959008397348 times the third resource plus 0.0124914167995535 times the fourth resource minus 0.1217493644708651 times the fifth resource minus 0.0455371326070688 times the sixth resource plus y_0[2] equals 0.00103506581258387. The ninth constraint requires that 0.5438484424289765 times the first resource minus 0.4058548451270487 times the second resource plus 0.00464438168201298 times the third resource plus 0.6829984219379786 times the fourth resource minus 0.813089431925804 times the fifth resource plus 0.1465879382567549 times the sixth resource plus t_0[0] equals 0.8921404899384406.] + + The tenth constraint requires that -0.1565490941785506 times the first resource minus 0.0151974010871207 times the second resource plus 0.0330083659733866 times the third resource plus 0.0409835559769911 times the fourth resource minus 0.025926735354538 times the fifth resource minus 0.0791575923837095 times the sixth resource plus y_1[0 equals 0.0252004385932512. The eleventh constraint requires that -0.00679732652950372 times the first resource minus 0.0159930899677971 times the second resource plus 0.00044788212331377231 times the third resource minus 0.00870755949376334 times the fourth resource plus 0.0106774891090615 times the fifth resource minus 0.0819505944729122 times the sixth resource plus y_1[1] equals -0.2461745595888191. The twelfth constraint requires that -0.0767410168100431 times the first resource minus 0.1162981298613443 times the second resource plus 0.2516807060221921 times the third resource minus 0.00442617885794681 times the fourth resource minus 0.0897268862621352 times the fifth resource minus 0.1001649216336424 times the sixth resource plus y_1[2] equals 0.0676538189988438. The thirteenth constraint requires that 2.12708121691458 times the first resource plus 0.5354675680200557 times the second resource plus 0.1244147628194144 times the third resource plus 1.32540724304661 times the fourth resource minus 0.6888780286074161 times the fifth resource plus 1.1055685780941 times the sixth resource plus t_1[0] equals 0.8216483573755342.] + + The fourteenth constraint requires that 0.1045779404782078 times the first resource plus 0.117211531954123 times the second resource minus 0.093303260093519 times the third resource plus 0.114920382796502 times the fourth resource plus 0.0892956784939136 times the fifth resource minus 0.1270063126632456 times the sixth resource plus y_2[0 equals 0.1487112654209727. The fifteenth constraint requires that 0.1031929746661822 times the first resource plus 0.0366834994452099 times the second resource minus 0.0811775022010661 times the third resource minus 0.0941975746436261 times the fourth resource plus 0.034402649559877 times the fifth resource minus 0.2382614709448476 times the sixth resource plus y_2[1] equals -0.00259845523628418. The sixteenth constraint requires that -0.0649606742324766 times the first resource plus 0.1500400272048616 times the second resource plus 0.0157455125935616 times the third resource minus 0.0173529909921789 times the fourth resource minus 0.045730709242314 times the fifth resource minus 0.068092234748515 times the sixth resource plus y_2[2] equals -0.1982729375753503.] + + The three second-order cone constraints are as follows: the first cone constraint requires that the sum of the squares of y_0[0, y_0[1], and y_0[2] is at most the square of t_0[0]. The second cone constraint requires that the sum of the squares of y_1[0], y_1[1], and y_1[2] is at most the square of t_1[0]. The third cone constraint requires that the sum of the squares of y_2[0], y_2[1], and y_2[2] is at most the square of t_2[0]. All decision variables, including the resource allocations x[0], x[1], x[2], x[3], x[4], and x[5], and the auxiliary variables y_0[0], y_0[1], y_0[2], y_1[0], y_1[1], y_1[2], y_2[0], y_2[1], y_2[2], t_0[0], t_1[0], and t_2[0], are free to take any real value. The goal is to determine the optimal allocation of the six resources that minimizes the total cost while satisfying all linear equality and second-order cone constraints.]0.9929487648391021Quadratic, out-of-scopenon-linear problem
163The optimization problem involves minimizing a linear objective function with coefficients 1.878721051112547 for variable x[0], -1.28487812168959 for x[1], and 0.4011175494566619 for x[2]. The problem is subject to five linear equality constraints: the first requires -0.1975342265737635 times x[0], -0.2719873754935637 times x[1], and -1.783486559633287 times x[2] to sum to 0.7278205554841768; the second requires -1.768263279220658 times x[0], 0.0736070884409486 times x[1], and -0.0509245482723127 times x[2] to sum to -2.868156724043526; the third requires 0.7748053077719607 times x[0], 3.021004813587345 times x[1], and 0.1666219482636987 times x[2] to sum to 2.766284453816483; the fourth requires 0.5458915584128072 times x[0], -1.22059114128179 times x[1], and 1.494593953041657 times x[2] to sum to -0.7394828371831723; and the fifth requires 2.433843249484103 times x[0], 0.841951398178463 times x[1], and 0.282600820817885 times x[2] to sum to 4.301152000602892. Additionally, there are twelve auxiliary constraints involving variables y_0[0], y_0[1], y_1[0], y_1[1], y_2[0], y_2[1], y_3[0], y_3[1], t_0[0], t_1[0], t_2[0], and t_3[0], which are all free to vary. These constraints include: 0.091171079971759 times x[0], 0.0086023 times x[1], 0.0626180219067843 times x[2], and y_0[0] summing to 0.0142231362041137; 8.9989367704573157e-04 times x[0], 0.0708190370286829 times x[1], -0.1447953767846306 times x[2], and y_0[1] summing to 0.0212301017034705; -0.8573356126193631 times x[0], 0.4752467573146003 times x[1], 0.4816251155416254 times x[2], and t_0[0] summing to 0.4620815487077579; 0.1204662490077911 times x[0], -0.010719970766377 times x[1], 0.0320226968522765 times x[2], and y_1[0] summing to 0.0954368100893973; -0.0364566909694576 times x[0], -0.0272692243126929 times x[1], -0.0454010664159346 times x[2], and y_1[1] summing to -0.1175842378552672; -0.1011799433173105 times x[0], -1.207915192355112 times x[1], 0.4912270735806125 times x[2], and t_1[0] summing to 0.6938004510085989; -0.00734887111481677 times x[0], -0.00717377976753872 times x[1], -1.672448115208352 times x[2], and y_2[0] summing to -0.1718293002638799; 0.4244121952583523 times x[0], -1.705679041306142 times x[1], 0.7925934720092012 times x[2], and y_2[1] summing to 1.440028055824932; -0.7883828402480292 times x[0], -0.5798703112590434 times x[1], 0.7119997733584963 times x[2], and t_2[0] summing to 0.4478948452160413; 0.1052425837766692 times x[0], 0.0226225196363593 times x[1], 0.1733283965993159 times x[2], and y_3[0] summing to 0.0432589010425869; 0.0134537413244718 times x[0], -0.00900079302918625 times x[1], -0.2071251065640836 times x[2], and y_3[1] summing to -0.0100284400478248; and 0.6853622167005485 times x[0], -0.7295354399357989 times x[1], 0.2093882195931892 times x[2], and t_3[0] summing to 0.8991800948338391. The problem also includes four second-order cone constraints: the first requires the square of t_0[0] to be less than or equal to the sum of the squares of y_0[0] and y_0[1]; the second requires the square of t_1[0] to be less than or equal to the sum of the squares of y_1[0] and y_1[1]; the third requires the square of t_2[0] to be less than or equal to the sum of the squares of y_2[0] and y_2[1]; and the fourth requires the square of t_3[0] to be less than or equal to the sum of the squares of y_3[0] and y_3[1]. All variables are free to vary, and the goal is to determine the optimal values of x[0], x[1], x[2], and the auxiliary variables to minimize the objective function while satisfying all constraints.2.1789982327638350The optimization problem involves minimizing a linear objective function with coefficients 1.878721051112547 for variable x[0, -1.28487812168959 for x[1], and 0.4011175494566619 for x[2]. The problem is subject to five linear equality constraints: the first requires -0.1975342265737635 times x[0], -0.2719873754935637 times x[1], and -1.783486559633287 times x[2] to sum to 0.7278205554841768; the second requires -1.768263279220658 times x[0], 0.0736070884409486 times x[1], and -0.0509245482723127 times x[2] to sum to -2.868156724043526; the third requires 0.7748053077719607 times x[0], 3.021004813587345 times x[1], and 0.1666219482636987 times x[2] to sum to 2.766284453816483; the fourth requires 0.5458915584128072 times x[0], -1.22059114128179 times x[1], and 1.494593953041657 times x[2] to sum to -0.7394828371831723; and the fifth requires 2.433843249484103 times x[0], 0.841951398178463 times x[1], and 0.282600820817885 times x[2] to sum to 4.301152000602892. Additionally, there are twelve auxiliary constraints involving variables y_0[0], y_0[1], y_1[0], y_1[1], y_2[0], y_2[1], y_3[0], y_3[1], t_0[0], t_1[0], t_2[0], and t_3[0], which are all free to vary. These constraints include: 0.091171079971759 times x[0], 0.0086023 times x[1], 0.0626180219067843 times x[2], and y_0[0] summing to 0.0142231362041137; 8.9989367704573157e-04 times x[0], 0.0708190370286829 times x[1], -0.1447953767846306 times x[2], and y_0[1] summing to 0.0212301017034705; -0.8573356126193631 times x[0], 0.4752467573146003 times x[1], 0.4816251155416254 times x[2], and t_0[0] summing to 0.4620815487077579; 0.1204662490077911 times x[0], -0.010719970766377 times x[1], 0.0320226968522765 times x[2], and y_1[0] summing to 0.0954368100893973; -0.0364566909694576 times x[0], -0.0272692243126929 times x[1], -0.0454010664159346 times x[2], and y_1[1] summing to -0.1175842378552672; -0.1011799433173105 times x[0], -1.207915192355112 times x[1], 0.4912270735806125 times x[2], and t_1[0] summing to 0.6938004510085989; -0.00734887111481677 times x[0], -0.00717377976753872 times x[1], -1.672448115208352 times x[2], and y_2[0] summing to -0.1718293002638799; 0.4244121952583523 times x[0], -1.705679041306142 times x[1], 0.7925934720092012 times x[2], and y_2[1] summing to 1.440028055824932; -0.7883828402480292 times x[0], -0.5798703112590434 times x[1], 0.7119997733584963 times x[2], and t_2[0] summing to 0.4478948452160413; 0.1052425837766692 times x[0], 0.0226225196363593 times x[1], 0.1733283965993159 times x[2], and y_3[0] summing to 0.0432589010425869; 0.0134537413244718 times x[0], -0.00900079302918625 times x[1], -0.2071251065640836 times x[2], and y_3[1] summing to -0.0100284400478248; and 0.6853622167005485 times x[0], -0.7295354399357989 times x[1], 0.2093882195931892 times x[2], and t_3[0] summing to 0.8991800948338391. The problem also includes four second-order cone constraints: the first requires the square of t_0[0] to be less than or equal to the sum of the squares of y_0[0] and y_0[1]; the second requires the square of t_1[0] to be less than or equal to the sum of the squares of y_1[0] and y_1[1]; the third requires the square of t_2[0] to be less than or equal to the sum of the squares of y_2[0] and y_2[1]; and the fourth requires the square of t_3[0] to be less than or equal to the sum of the squares of y_3[0] and y_3[1]. All variables are free to vary, and the goal is to determine the optimal values of x[0], x[1], x[2], and the auxiliary variables to minimize the objective function while satisfying all constraints.]2.178998232763835Quadratic, out-of-scopenon-linear problem
164Minimize the linear combination of decision variables \( x[0] \), \( x[1] \), \( x[2] \), and \( x[3] \) with coefficients -0.4846985478797502, -0.5148337459861181, -0.103222308633488, and -0.4982619511067216, respectively. The variables \( x[0] \), \( x[1] \), \( x[2] \), and \( x[3] \) must satisfy five linear equality constraints: a combination weighted by 0.6711266895702664, 0.3859509788606931, 0.4567515243481577, and 0.9480688150913272 must equal -0.722783786716152; a combination weighted by 0.8243709699260372, -1.258592224290495, 0.6114769465783959, and 0.1988705802173026 must equal 0.4645396774435609; a combination weighted by -1.502871694222167, 0.7239127905595304, -0.2373110244288602, and -0.1260210332851125 must equal 0.3023538191780689; a combination weighted by -0.1738503900411032, 1.795035445633136, 0.4912858705710282, and 1.011593878467344 must equal -0.4494297548628751; and a combination weighted by 0.4863065784013345, 1.917253921273967, 0.0113609495408681, and 1.178604249156841 must equal -1.701240670398688. The resource allocation variables \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), and \( y_0[3] \) must satisfy specific linear combinations with \( x[0] \), \( x[1] \), \( x[2] \), and \( x[3] \) to meet exact values of 0.0827964599101699, 0.00176755725217533, -0.1846363653081483, and 0.0643720488743014, respectively. Similarly, the resource allocation variables \( y_1[0] \), \( y_1[1] \), \( y_1[2] \), and \( y_1[3] \) must satisfy specific linear combinations with \( x[0] \), \( x[1] \), \( x[2] \), and \( x[3] \) to meet exact values of 0.0789339456076434, 0.0889865820408794, 0.1048309970353658, and 0.0769986096204651, respectively. The variables \( t_0[0] \) and \( t_1[0] \) must satisfy specific linear combinations with \( x[0] \), \( x[1] \), \( x[2] \), and \( x[3] \) to meet exact values of 0.6054001385216589 and 0.9698797396720825, respectively. The system must also satisfy two second-order cone constraints: the sum of the squares of \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), and \( y_0[3] \) minus the square of \( t_0[0] \) must be less than or equal to zero, and the sum of the squares of \( y_1[0] \), \( y_1[1] \), \( y_1[2] \), and \( y_1[3] \) minus the square of \( t_1[0] \) must be less than or equal to zero. All decision variables \( x[0] \), \( x[1] \), \( x[2] \), \( x[3] \), \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), \( y_0[3] \), \( y_1[0] \), \( y_1[1] \), \( y_1[2] \), \( y_1[3] \), \( t_0[0] \), and \( t_1[0] \) are free to take any real value.0.6541956704126570Minimize the linear combination of decision variables \( x[0 \), \( x[1] \), \( x[2] \), and \( x[3] \) with coefficients -0.4846985478797502, -0.5148337459861181, -0.103222308633488, and -0.4982619511067216, respectively. The variables \( x[0] \), \( x[1] \), \( x[2] \), and \( x[3] \) must satisfy five linear equality constraints: a combination weighted by 0.6711266895702664, 0.3859509788606931, 0.4567515243481577, and 0.9480688150913272 must equal -0.722783786716152; a combination weighted by 0.8243709699260372, -1.258592224290495, 0.6114769465783959, and 0.1988705802173026 must equal 0.4645396774435609; a combination weighted by -1.502871694222167, 0.7239127905595304, -0.2373110244288602, and -0.1260210332851125 must equal 0.3023538191780689; a combination weighted by -0.1738503900411032, 1.795035445633136, 0.4912858705710282, and 1.011593878467344 must equal -0.4494297548628751; and a combination weighted by 0.4863065784013345, 1.917253921273967, 0.0113609495408681, and 1.178604249156841 must equal -1.701240670398688. The resource allocation variables \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), and \( y_0[3] \) must satisfy specific linear combinations with \( x[0] \), \( x[1] \), \( x[2] \), and \( x[3] \) to meet exact values of 0.0827964599101699, 0.00176755725217533, -0.1846363653081483, and 0.0643720488743014, respectively. Similarly, the resource allocation variables \( y_1[0] \), \( y_1[1] \), \( y_1[2] \), and \( y_1[3] \) must satisfy specific linear combinations with \( x[0] \), \( x[1] \), \( x[2] \), and \( x[3] \) to meet exact values of 0.0789339456076434, 0.0889865820408794, 0.1048309970353658, and 0.0769986096204651, respectively. The variables \( t_0[0] \) and \( t_1[0] \) must satisfy specific linear combinations with \( x[0] \), \( x[1] \), \( x[2] \), and \( x[3] \) to meet exact values of 0.6054001385216589 and 0.9698797396720825, respectively. The system must also satisfy two second-order cone constraints: the sum of the squares of \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), and \( y_0[3] \) minus the square of \( t_0[0] \) must be less than or equal to zero, and the sum of the squares of \( y_1[0] \), \( y_1[1] \), \( y_1[2] \), and \( y_1[3] \) minus the square of \( t_1[0] \) must be less than or equal to zero. All decision variables \( x[0] \), \( x[1] \), \( x[2] \), \( x[3] \), \( y_0[0] \), \( y_0[1] \), \( y_0[2] \), \( y_0[3] \), \( y_1[0] \), \( y_1[1] \), \( y_1[2] \), \( y_1[3] \), \( t_0[0] \), and \( t_1[0] \) are free to take any real value.]0.654195670412657Quadratic, out-of-scopenon-linear problem
165The problem is to minimize the total cost, represented by the objective function: -0.1903710639101614 multiplied by x[0], plus 0.2471205466943747 multiplied by x[1], plus 2.348193401972641 multiplied by x[2], plus 0.9152449416117181 multiplied by x[3], minus 1.742613003351053 multiplied by x[4]. The problem is subject to five linear equality constraints. The first constraint requires that -0.2030886237352828 multiplied by x[0], minus 0.1763724551444181 multiplied by x[1], minus 0.5941927118633189 multiplied by x[2], plus 0.830361694227971 multiplied by x[3], minus 0.5802412833432681 multiplied by x[4] equals 1.780877672468478. The second constraint requires that -0.6639499053878003 multiplied by x[0], plus 1.188142824766808 multiplied by x[1], minus 1.177300154867763 multiplied by x[2], minus 0.3969063969716068 multiplied by x[3], plus 0.2889021746134166 multiplied by x[4] equals -0.3331997792148634. The third constraint requires that -0.8719514624527444 multiplied by x[0], plus 0.1808126068216141 multiplied by x[1], minus 0.9665774590076043 multiplied by x[2], plus 1.100084951306936 multiplied by x[3], minus 0.9042747703694114 multiplied by x[4] equals 2.800183976297944. The fourth constraint requires that -0.2015270532241274 multiplied by x[0], minus 1.293118253679058 multiplied by x[1], minus 0.4545577007105097 multiplied by x[2], minus 0.1304722040029908 multiplied by x[3], plus 0.6064299665361618 multiplied by x[4] equals 0.3418400860293037. The fifth constraint requires that -0.6152966616157548 multiplied by x[0], minus 2.528352204302152 multiplied by x[1], plus 1.007133413111276 multiplied by x[2], plus 0.4961854953912131 multiplied by x[3], minus 0.8422906933926506 multiplied by x[4] equals 2.844445804939649. + +Additionally, the problem includes four constraints involving additional variables y_0[0], y_0[1], y_0[2], and t_0[0]. The first constraint, labeled R5, requires that 0.8380405126438697 multiplied by x[0], plus 1.725727253419972 multiplied by x[1], plus 0.2150240770195237 multiplied by x[2], plus 0.9187896715161638 multiplied by x[3], minus 0.336923104799861 multiplied by x[4], plus y_0[0] equals -0.457902185585681. The second constraint, labeled R6, requires that -1.68549282142258 multiplied by x[0], minus 0.2316401110005971 multiplied by x[1], plus 2.523836638490076 multiplied by x[2], minus 0.2575764188061656 multiplied by x[3], plus 0.3440917268091695 multiplied by x[4], plus y_0[1] equals 0.1406890072115732. The third constraint, labeled R7, requires that 1.252648113065757 multiplied by x[0], minus 0.6790006571802352 multiplied by x[1], plus 0.7626153811807369 multiplied by x[2], minus 0.2640373876912787 multiplied by x[3], minus 1.630978171354607 multiplied by x[4], plus y_0[2] equals 0.0216487444612001. The fourth constraint, labeled R8, requires that -0.2911339846463995 multiplied by x[0], minus 0.2752818731665084 multiplied by x[1], minus 0.284762721024827 multiplied by x[2], plus 0.842789585430482 multiplied by x[3], plus 1.448230318030449 multiplied by x[4], plus t_0[0] equals 0.9299709419894615. + +The problem also includes a second-order cone constraint, labeled qc0, which requires that the square of t_0[0] is less than or equal to the sum of the squares of y_0[0], y_0[1], and y_0[2], expressed as -t_0[0]^2 + y_0[0]^2 + y_0[1]^2 + y_0[2]^2 ≤ 0. All decision variables, including x[0], x[1], x[2], x[3], x[4], y_0[0], y_0[1], y_0[2], and t_0[0], are free to take any real value. The task is to determine the optimal values for these variables to minimize the total cost while satisfying all constraints.1.6515636613254360The problem is to minimize the total cost, represented by the objective function: -0.1903710639101614 multiplied by x[0, plus 0.2471205466943747 multiplied by x[1], plus 2.348193401972641 multiplied by x[2], plus 0.9152449416117181 multiplied by x[3], minus 1.742613003351053 multiplied by x[4]. The problem is subject to five linear equality constraints. The first constraint requires that -0.2030886237352828 multiplied by x[0], minus 0.1763724551444181 multiplied by x[1], minus 0.5941927118633189 multiplied by x[2], plus 0.830361694227971 multiplied by x[3], minus 0.5802412833432681 multiplied by x[4] equals 1.780877672468478. The second constraint requires that -0.6639499053878003 multiplied by x[0], plus 1.188142824766808 multiplied by x[1], minus 1.177300154867763 multiplied by x[2], minus 0.3969063969716068 multiplied by x[3], plus 0.2889021746134166 multiplied by x[4] equals -0.3331997792148634. The third constraint requires that -0.8719514624527444 multiplied by x[0], plus 0.1808126068216141 multiplied by x[1], minus 0.9665774590076043 multiplied by x[2], plus 1.100084951306936 multiplied by x[3], minus 0.9042747703694114 multiplied by x[4] equals 2.800183976297944. The fourth constraint requires that -0.2015270532241274 multiplied by x[0], minus 1.293118253679058 multiplied by x[1], minus 0.4545577007105097 multiplied by x[2], minus 0.1304722040029908 multiplied by x[3], plus 0.6064299665361618 multiplied by x[4] equals 0.3418400860293037. The fifth constraint requires that -0.6152966616157548 multiplied by x[0], minus 2.528352204302152 multiplied by x[1], plus 1.007133413111276 multiplied by x[2], plus 0.4961854953912131 multiplied by x[3], minus 0.8422906933926506 multiplied by x[4] equals 2.844445804939649.] + + Additionally, the problem includes four constraints involving additional variables y_0[0, y_0[1], y_0[2], and t_0[0]. The first constraint, labeled R5, requires that 0.8380405126438697 multiplied by x[0], plus 1.725727253419972 multiplied by x[1], plus 0.2150240770195237 multiplied by x[2], plus 0.9187896715161638 multiplied by x[3], minus 0.336923104799861 multiplied by x[4], plus y_0[0] equals -0.457902185585681. The second constraint, labeled R6, requires that -1.68549282142258 multiplied by x[0], minus 0.2316401110005971 multiplied by x[1], plus 2.523836638490076 multiplied by x[2], minus 0.2575764188061656 multiplied by x[3], plus 0.3440917268091695 multiplied by x[4], plus y_0[1] equals 0.1406890072115732. The third constraint, labeled R7, requires that 1.252648113065757 multiplied by x[0], minus 0.6790006571802352 multiplied by x[1], plus 0.7626153811807369 multiplied by x[2], minus 0.2640373876912787 multiplied by x[3], minus 1.630978171354607 multiplied by x[4], plus y_0[2] equals 0.0216487444612001. The fourth constraint, labeled R8, requires that -0.2911339846463995 multiplied by x[0], minus 0.2752818731665084 multiplied by x[1], minus 0.284762721024827 multiplied by x[2], plus 0.842789585430482 multiplied by x[3], plus 1.448230318030449 multiplied by x[4], plus t_0[0] equals 0.9299709419894615.] + + The problem also includes a second-order cone constraint, labeled qc0, which requires that the square of t_0[0 is less than or equal to the sum of the squares of y_0[0], y_0[1], and y_0[2], expressed as -t_0[0]^2 + y_0[0]^2 + y_0[1]^2 + y_0[2]^2 ≤ 0. All decision variables, including x[0], x[1], x[2], x[3], x[4], y_0[0], y_0[1], y_0[2], and t_0[0], are free to take any real value. The task is to determine the optimal values for these variables to minimize the total cost while satisfying all constraints.]1.651563661325436Quadratic, out-of-scopenon-linear problem
166Determine the optimal levels of five resources, labeled x[0] through x[4], along with auxiliary variables y_i[0], y_i[1], and t_i[0] for i = 0, 1, 2, 3, to minimize the total operational cost. The cost coefficients for the resources are as follows: x[0] has a coefficient of -0.1583313894422966, x[1] has a coefficient of 0.3001605310606917, x[2] has a coefficient of -0.1036603523790828, x[3] has a coefficient of -0.248725637286869, and x[4] has a coefficient of -0.6446466512774527. The optimization must satisfy 18 constraints, including 3 linear equality constraints, 11 additional constraints involving auxiliary variables, and 4 second-order cone constraints. The first linear equality constraint requires 1.340455763918918 units of x[0], 0.6122011761088476 units of x[1], 0.0241146255663964 units of x[2], 0.3597528706289873 units of x[3], and 1.140437589082963 units of x[4] to equal -2.525763513585038. The second linear equality constraint requires -0.3516433576310653 units of x[0], 0.0965558577316191 units of x[1], 0.3529378836507989 units of x[2], -0.0202375554076704 units of x[3], and -0.1677595357452861 units of x[4] to equal 0.2235505634382546. The third linear equality constraint requires 0.3075615720609267 units of x[0], 0.8085248764397751 units of x[1], 0.8880007171842481 units of x[2], 0.9386933269367855 units of x[3], and 1.224241726116549 units of x[4] to equal -3.117140788520132. The additional constraints involving auxiliary variables are as follows: 0.2455202814808017 units of x[0], 0.0472232513833277 units of x[1], 0.047517849850401 units of x[2], -0.00744390387231037 units of x[3], and 0.1666805170718211 units of x[4] plus y_0[0] must equal 0.1079335532146737; 0.13246246186774 units of x[0], 0.00659412390607974 units of x[1], 0.1114369110680747 units of x[2], -0.2665760991911567 units of x[3], and -0.18331535073365 units of x[4] plus y_0[1] must equal 0.1309744569827054; 1.420088014213417 units of x[0], 0.0293620814227688 units of x[1], -0.2441905598608127 units of x[2], -0.5623611776952713 units of x[3], and 1.040148376816393 units of x[4] plus t_0[0] must equal 0.1035292994181456; -0.10415918739318 units of x[0], 0.0857458 units of x[1], -0.110689170404098 units of x[2], -0.0521857879268957 units of x[3], and -0.1552718932805042 units of x[4] plus y_1[0] must equal -0.1542023398547745; 0.1066277951798869 units of x[0], -0.2407462132103433 units of x[1], 0.0250995224738677 units of x[2], 0.0697439542403621 units of x[3], and 0.0913997005017029 units of x[4] plus y_1[1] must equal -0.1125960948765685; -1.307066723082076 units of x[0], -0.1389121435606677 units of x[1], 0.3296112909294153 units of x[2], -0.3486230654088014 units of x[3], and 0.991038664821373 units of x[4] plus t_1[0] must equal 1.974360972589342; 0.0338157458141306 units of x[0], -0.0691173193642718 units of x[1], 0.06921235157118 units of x[2], 0.1912305441179409 units of x[3], and 0.0307745904197725 units of x[4] plus y_2[0] must equal 0.1826196734978178; -0.0850017860817879 units of x[0], 0.032183697895793 units of x[1], 0.0975233971012199 units of x[2], 0.1552972331129627 units of x[3], and 0.0274993364499619 units of x[4] plus y_2[1] must equal -0.00804524201171049; 1.435354965811914 units of x[0], 0.0916142411098695 units of x[1], 0.6979687102235776 units of x[2], -0.603509686532014 units of x[3], and 0.4772240882858254 units of x[4] plus t_2[0] must equal 1.174433212798432; -0.8060483631976895 units of x[0], 2.079831334431379 units of x[1], -0.0295281669371715 units of x[2], -1.61454817201267 units of x[3], and 0.5192117784670098 units of x[4] plus y_3[0] must equal 0.2657818441153715; -0.9441542385781047 units of x[0], 2.176756986944978 units of x[1], 0.6386189181592981 units of x[2], 1.295414594438755 units of x[3], and -0.6941305016601239 units of x[4] plus y_3[1] must equal -0.9771957594358285; 0.1874916310113519 units of x[0], -2.40692481111106 units of x[1], 1.487697722815744 units of x[2], 1.078504407492902 units of x[3], and 0.1412227274745621 units of x[4] plus t_3[0] must equal 0.0280865701316309. Additionally, four second-order cone constraints must be satisfied: the squared value of y_0[0] plus the squared value of y_0[1] must not exceed the squared value of t_0[0]; the squared value of y_1[0] plus the squared value of y_1[1] must not exceed the squared value of t_1[0]; the squared value of y_2[0] plus the squared value of y_2[1] must not exceed the squared value of t_2[0]; the squared value of y_3[0] plus the squared value of y_3[1] must not exceed the squared value of t_3[0]. All variables, including x[0] through x[4], y_i[0], y_i[1], and t_i[0], are free to take any real value.0.90260872858278410Determine the optimal levels of five resources, labeled x[0 through x[4], along with auxiliary variables y_i[0], y_i[1], and t_i[0] for i = 0, 1, 2, 3, to minimize the total operational cost. The cost coefficients for the resources are as follows: x[0] has a coefficient of -0.1583313894422966, x[1] has a coefficient of 0.3001605310606917, x[2] has a coefficient of -0.1036603523790828, x[3] has a coefficient of -0.248725637286869, and x[4] has a coefficient of -0.6446466512774527. The optimization must satisfy 18 constraints, including 3 linear equality constraints, 11 additional constraints involving auxiliary variables, and 4 second-order cone constraints. The first linear equality constraint requires 1.340455763918918 units of x[0], 0.6122011761088476 units of x[1], 0.0241146255663964 units of x[2], 0.3597528706289873 units of x[3], and 1.140437589082963 units of x[4] to equal -2.525763513585038. The second linear equality constraint requires -0.3516433576310653 units of x[0], 0.0965558577316191 units of x[1], 0.3529378836507989 units of x[2], -0.0202375554076704 units of x[3], and -0.1677595357452861 units of x[4] to equal 0.2235505634382546. The third linear equality constraint requires 0.3075615720609267 units of x[0], 0.8085248764397751 units of x[1], 0.8880007171842481 units of x[2], 0.9386933269367855 units of x[3], and 1.224241726116549 units of x[4] to equal -3.117140788520132. The additional constraints involving auxiliary variables are as follows: 0.2455202814808017 units of x[0], 0.0472232513833277 units of x[1], 0.047517849850401 units of x[2], -0.00744390387231037 units of x[3], and 0.1666805170718211 units of x[4] plus y_0[0] must equal 0.1079335532146737; 0.13246246186774 units of x[0], 0.00659412390607974 units of x[1], 0.1114369110680747 units of x[2], -0.2665760991911567 units of x[3], and -0.18331535073365 units of x[4] plus y_0[1] must equal 0.1309744569827054; 1.420088014213417 units of x[0], 0.0293620814227688 units of x[1], -0.2441905598608127 units of x[2], -0.5623611776952713 units of x[3], and 1.040148376816393 units of x[4] plus t_0[0] must equal 0.1035292994181456; -0.10415918739318 units of x[0], 0.0857458 units of x[1], -0.110689170404098 units of x[2], -0.0521857879268957 units of x[3], and -0.1552718932805042 units of x[4] plus y_1[0] must equal -0.1542023398547745; 0.1066277951798869 units of x[0], -0.2407462132103433 units of x[1], 0.0250995224738677 units of x[2], 0.0697439542403621 units of x[3], and 0.0913997005017029 units of x[4] plus y_1[1] must equal -0.1125960948765685; -1.307066723082076 units of x[0], -0.1389121435606677 units of x[1], 0.3296112909294153 units of x[2], -0.3486230654088014 units of x[3], and 0.991038664821373 units of x[4] plus t_1[0] must equal 1.974360972589342; 0.0338157458141306 units of x[0], -0.0691173193642718 units of x[1], 0.06921235157118 units of x[2], 0.1912305441179409 units of x[3], and 0.0307745904197725 units of x[4] plus y_2[0] must equal 0.1826196734978178; -0.0850017860817879 units of x[0], 0.032183697895793 units of x[1], 0.0975233971012199 units of x[2], 0.1552972331129627 units of x[3], and 0.0274993364499619 units of x[4] plus y_2[1] must equal -0.00804524201171049; 1.435354965811914 units of x[0], 0.0916142411098695 units of x[1], 0.6979687102235776 units of x[2], -0.603509686532014 units of x[3], and 0.4772240882858254 units of x[4] plus t_2[0] must equal 1.174433212798432; -0.8060483631976895 units of x[0], 2.079831334431379 units of x[1], -0.0295281669371715 units of x[2], -1.61454817201267 units of x[3], and 0.5192117784670098 units of x[4] plus y_3[0] must equal 0.2657818441153715; -0.9441542385781047 units of x[0], 2.176756986944978 units of x[1], 0.6386189181592981 units of x[2], 1.295414594438755 units of x[3], and -0.6941305016601239 units of x[4] plus y_3[1] must equal -0.9771957594358285; 0.1874916310113519 units of x[0], -2.40692481111106 units of x[1], 1.487697722815744 units of x[2], 1.078504407492902 units of x[3], and 0.1412227274745621 units of x[4] plus t_3[0] must equal 0.0280865701316309. Additionally, four second-order cone constraints must be satisfied: the squared value of y_0[0] plus the squared value of y_0[1] must not exceed the squared value of t_0[0]; the squared value of y_1[0] plus the squared value of y_1[1] must not exceed the squared value of t_1[0]; the squared value of y_2[0] plus the squared value of y_2[1] must not exceed the squared value of t_2[0]; the squared value of y_3[0] plus the squared value of y_3[1] must not exceed the squared value of t_3[0]. All variables, including x[0] through x[4], y_i[0], y_i[1], and t_i[0], are free to take any real value.]0.9026087285827841Quadratic, out-of-scopenon-linear problem
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+ + + \ No newline at end of file diff --git a/optimind/data/hints.csv b/optimind/data/hints.csv new file mode 100644 index 0000000..8000898 --- /dev/null +++ b/optimind/data/hints.csv @@ -0,0 +1,26 @@ +problem_class,error analysis +Assignment Problem,"[(""When calculating profit, it is easy to confuse the per-unit profit with the total profit."", ""Whenever defining profit, be careful to judge if it should be per unit or total. You may have to multiply the per-unit amount with the total amount of units to get the final profit. Always align profit definition with the scale of the decision (per-unit vs total) based on the problem description"")]" +Blending Problem,"[ (""When solving blending problems with multiple types of products/materials and lower/upper bounds for quantities, it is easy to make mistakes by confusing whether these bounds should apply to each type separately or to the total aggregate amount across all types."", ""When you encounter blending problems with multiple types of products/materials and lower/upper bounds for quantities, think carefully whether they should apply to each type separately or to the total aggregate quantity across all types.""), (""When solving blending/production problems with multiple types of production sources (e.g., energy sources) with maximum quantities for each type, it is easy to make mistakes by assuming a source has to deliver its maximum quantity when used, instead of allowing partial production up to the capacity limit"", ""When solving blending/production problems with multiple types of production sources (e.g., energy sources) with maximum quantities for each type, please define variables y_i for the quantity that is provided from each type of source i, such that 0 <= y_i <= maxQuantity_i. Use these variables both for any costs that depend on the production amounts in the objective, as well as to make sure enough quantities have been produced to satisfy the demand.""), (""When solving blending/production problems with multiple types of production sources (e.g., energy sources) and installation costs for each type, it is easy to make mistakes by considering the installation cost to be dependent on the produced quantity of each type, instead of a fixed cost for using the type/source"", ""When solving blending/production problems with multiple types of production sources (e.g., energy sources) and installation costs for each type, please define binary variables x_i representing whether the type/source i is used and multiply this with the installation cost of i in the objective in order to capture the installation cost for each type/source as fixed cost for using that type/source, and not dependent on the amount produced by the source, unless explictly stated otherwise.""), (""When solving allocation problems with multiple considerations (e.g., costs, preferences), it is easy to confuse the different aspects of the problem and optimize for the wrong objective"", ""When solving allocation problems with multiple considerations (e.g., costs, preferences), think carefully which aspect must be in the objective and which in the constraints. For example, if the problem asks to maximize the total preferences under budget constraints, the preferences must be in the objective, and the costs must be considered in the budget constraint.""), (""When solving blending problems with batch percentages and cost bounds, it is easy to make mistakes by mixing units."", ""When you encountered blending problems with batch normalization, please always decide on one normalization—either set total mix = 1 (fractions) or = W (mass)—and then convert every percentage constraint and price to the same base (percent×W for bounds, price/W if needed).""), (""When solving problems with tiered purchase costs, it is easy to make mistakes by incorrectly allowing higher-tier costs."", ""When you encountered a problem with tiered purchase costs, use a binary variable such as U_i to denote that tier i can be used. To guarantee purchases only from allowed tiers, apply constraint such as P_i <= num_i * U_i for every tier i, where P_i is the amount purchase at tier i and num_i is the number of units that can be purchased at this tier, and for every tier i except the last use constraints such as U_{i+1} * num_i <= P_i to disallow the next tier i+1 unless the current tier i is fully used."") ]" +Capacitated Facility Location Problem,"[ ( ""When solving capacitated facility location or network-flow–style models with node supplies/demands, it is easy to make mistakes by flipping the sign of supply versus demand in the flow-conservation constraints."", ""When you encounter a model that stores the net supply on a vector B (so B[i] is positive for supply surplus and negative for net demand), please enforce flow conservation as outflow from i − inflow into i <= B[i], i.e., ∑_j x[i,j] − ∑_j x[j,i] <= B[i]; when B stores the net demand (so B[i] is negative for supply surplus and positive for net demand), please enforce inflow into i − outflow from i >= B[i], i.e., ∑_j x[j,i] − ∑_j x[i,j] >= B[i]."" ) ]" +Capacitated Lot-sizing Problem (CLSP),"[ (""When solving capacitated lot-sizing or production planning problems that refer to countable “units” of items, it is easy to make mistakes by allowing fractional production quantities (continuous variables) instead of enforcing integrality."", ""When you encounter a production planning problem that mentions “unit” of items, “per unit” costs, “how many units,” or related notions, please consider that it is only possible to produce integral quantities of the items. The appropriate decision variables should be restricted to be integral to reflect this.""), (""When solving lot-sizing problems that allow backlog, it is easy to make mistakes by mis-handling backlog terms in the inventory balance (e.g., wrong sign or missing B_{t-1} or B_t)."", ""When you encounter lot-sizing with backlog, please take backlog into account in the inventory balance constraint at period t as follows: x_t + I_{t-1} = d_t + B_{t-1} - B_t + I_t, where B_{t-1} is the backlog from the previous time, B_t is the backlog at the end of the present time, I_{t-1} is the inventory from the previous time, I_t is the inventory at the end of the present time, and x_t is the production level of the current time. Notice that the backlog from the previous time should be added to the demand, and the present backlog should be subtracted.""), (""When solving multi-product, multi-period production planning with a shared time/machine capacity, it is easy to make mistakes by relying only on per-product limits and missing the aggregate shared-capacity coupling, which can permit total capacity to be exceeded."", ""When you work on multi-period and production planning models, please focus on correctly linking time periods and constraining shared resources. Treat initial conditions (e.g., starting inventory at the beginning of the horizon) as fixed parameters, not decision variables, and ensure your primary balance constraint starts with the very first decision period to correctly connect its decisions to these fixed starting values. Additionally, look for phrases like “if all lines are devoted to one product,” which signal a shared capacity (like total machine-days); this must be modeled with a single, aggregate constraint that sums the consumption of that resource across all competing activities, rather than just using separate, independent limits."")]" +Contract Allocation,"[ (""When solving budget-constrained allocation with branch/category preferences and supplier prices/capacities, it is easy to make mistakes by multiplying preference-weighted book counts by unit prices in the objective, which effectively maximizes money spent and favors expensive suppliers rather than maximizing preference-weighted quantities of books delivered."", ""When you encounter budget-constrained allocation with branch/category preferences and supplier prices/capacities, please define x_{b,s,c} as integer book counts, maximize ∑{b,c} w{b,c} (∑{s} x{b,s,c}) (preference-weighted quantities), and enforce budgets ∑{s,c} p{s,c} x_{b,s,c} ≤ budget_b and supplier-category capacities ∑{b} x{b,s,c} ≤ supply_{s,c}; do not include prices in the objective."") ]" +Diet Problem,"[ (""When solving diet optimization problems where servings are countable items (whole servings), it is easy to make mistakes by modeling the serving decision variables as continuous (or binary) rather than integer, which permits fractional (or only 0–1) servings and leads to unrealistic or incorrect costs."", ""When you encounter a diet problem with countable servings, please define servings as integer variables (don’t allow fractional values for servings) unless the problem statement explicitly states otherwise."") ]" +Discrete Lot-sizing and Scheduling Problem,"[ (""When solving production-planning / lot-sizing problems where items are countable units, it is easy to make mistakes by allowing fractional production/inventory/sales variables instead of using integrality."", ""When you encounter a production planning problem that mentions “unit” of items, “per unit” costs, “how many units”, or related notions, please restrict the appropriate decision variables to be integral to reflect indivisible units.""), (""When solving multi-period lot-sizing models with backlog, it is easy to make mistakes by omitting backlog terms in the inventory balance or using the wrong signs."", ""When you encounter a model where backlog is allowed, please take it into account in the period-t balance as: x_t + I_{t-1} = d_t + B_{t-1} - B_t + I_t, where B_{t-1} is the backlog at the end of period t-1, B_t is the backlog at the end of period t, I_{t-1} is the inventory at the end of period t-1, I_t is the inventory at the end of period t, and x_t is the production level of period t. Note that B_{t-1} must be added to the demand of period t and B_t must be subtracted.""), (""When solving multi-period lot-sizing/inventory models with specified initial inventories, it is easy to make mistakes by mis-indexing the inventory state (e.g., reusing I[0] as both the fixed initial inventory and the inventory at the end of the first period), which forces incorrect carryover and mischarges holding cost."", ""If there is given initial inventory, make sure to use a different name for the initial inventory versus the inventory at the end of the first period.""), (""When solving multi-period lot-sizing problems, if there are periods that are unavailable for production, it is easy to make mistakes by relying on implicit restrictions (e.g., zero capacity in a needed period) instead of explicitly forbidding production. Also, it is easy to make mistakes in the treatment of the inventory at the end of the first period, t=0."", ""When you encounter multi-period production–inventory problems, if some periods are unavailable for production, please add explicit no-production constraints for unavailable periods (x_{p,t}=0), where x_{p,t} is the production amount for product p on period t. If there is no given initial inventory and backlogs are not allowed, then the inventory at the end of period t=0 is given by I_{p,0}=x_{p,0}-s_{p,0} where I_{p,0} is the inventory at the end of period 0, x_{p,0} the amount produced in period 0, and s_{p,0} the sales in period 0; for t>0, we need to account for the inventory from the previous period I_{p, t-1} as follows I_{p,t}=I_{p,t-1}+x_{p,t}-s_{p,t}). Please ensure the objective charges holding costs only on end-of-period inventories."") ]" +Facility Location Problem (FLP),"[(""When solving linear facility-sizing or resource-allocation models, it is easy to make mistakes by misinterpreting the objective’s units or confusing costs/benefits for different aspects of the problem in the objective."", ""When solving linear facility-sizing or resource-allocation models, think carefully what are the correct units in the objective, for example if we are allowed to allocate partial amounts (e.g., there is a total available area and we can allocate in some of it), then we may pay partial cost, but also may have partial benefits, i.e., only the proportion of benefits that correspond to this allocation fraction are obtained."")]" +Factory Planning Problem,"[ (""When solving production planning problems with countable 'units', it is easy to make mistakes by modeling decision variables as fractional rather than integer, which permits unrealistic partial units."", ""When you encounter a production planning problem that mentions 'unit' of items, 'per unit' costs, 'how many units', or related notions, please restrict the appropriate decision variables to be integral to reflect this.""), (""When solving multi-period lot-sizing problems, if there are periods that are unavailable for production, it is easy to make mistakes by relying on implicit restrictions (e.g., zero capacity in a needed period) instead of explicitly forbidding production. Also, it is easy to make mistakes in the treatment of the inventory at the end of the first period, t=0."", ""When you encounter multi-period production–inventory problems, if some periods are unavailable for production, please add explicit no-production constraints for unavailable periods (x_{p,t}=0), where x_{p,t} is the production amount for product p on period t. If there is no given initial inventory and backlogs are not allowed, then the inventory at the end of period t=0 is given by I_{p,0}=x_{p,0}-s_{p,0} where I_{p,0} is the inventory at the end of period 0, x_{p,0} the amount produced in period 0, and s_{p,0} the sales in period 0; for t>0, we need to account for the inventory from the previous period I_{p, t-1} as follows I_{p,t}=I_{p,t-1}+x_{p,t}-s_{p,t}). Please ensure the objective charges holding costs only on end-of-period inventories."") ]" +Job Shop,"[ (""When solving job-shop scheduling with precedence chains and no-overlap constraints, it is easy to make mistakes by creating bilinear terms and by introducing two directional binaries per unordered operation pair without exclusivity, which can permit overlapping on the same machine."", ""When you encounter job-shop scheduling with no-overlap constraints or shared machine, introduce a single binary variable for each unordered pair of operations that cannot overlap to indicate which of them will start first (e.g. y_{(j1,o1),(j2,o2)} to indicate whether operation o1 of job j1 should start before operation o2 of job j2), and impose this relation on the start times of the operations via two disjunctive inequalities with a big-M (e.g, s_{j1,o1} + p_{j1,o1} <= s_{j2,o2} + M*(1 - y_{(j1,o1),(j2,o2)}) and s_{j2,o2} + p_{j2,o2} <= s_{j1,o1} + M*y_{(j1,o1),(j2,o2)}, where s_{j,o} denotes the start time of operation o of job j and p_{j,o} its processing time). Never multiply a binary variable by the makespan variable. If the objective is to minimize the makespan Cmax, enforce Cmax >= start_time + processing_time for every operation.""), (""When solving scheduling problems, it is easy to make mistakes by fabricating instance details (e.g., assuming exactly two operations per job or inventing fixed machine assignments)."", ""When you encounter a scheduling problem that does not explicitly give the assignment of the operations to machines, do not assume or fabricate machines or machine assignments, and impose the no-overlap constraints using the 2 disjunctive big-M inequalities.""), (""Setting the big-M too small can lead to errors."", ""Set the big-M to be equal to the sum of the processing times of all operations plus 100.""), ]" +Knapsack,"[(""It easy to make mistakes by using placeholders like '...' and not listing the full values and costs arrays with all items in the code. That makes the script incomplete."", ""In the final Gurobi/Python code, expand every parameter fully. Do not use placeholders or ellipses."")]" +Lot-Sizing Problem,"[(""When solving production planning problems where items are counted in whole units (e.g., the statement uses ‘unit’, ‘per unit’, or ‘how many units’), it is easy to make mistakes by allowing fractional production for inherently discrete items."", ""When you encounter production planning problems that mention ‘unit’ of items or similar notions, please restrict the relevant decision variables to be integral so that only whole units can be produced, unless the problem explicitly allows fractional production.""), (""When solving multi-period lot-sizing/inventory models with specified initial inventories, it is easy to make mistakes by mis-indexing the inventory state (e.g., reusing I[0] as both the fixed initial inventory and the inventory at the end of the first period), which forces incorrect carryover and mischarges holding cost."", ""If there is given initial inventory, make sure to use a different name for the initial inventory versus the inventory at the end of the first period.""), (""When solving multi-period lot-sizing, if there are periods that are unavailable for production, it is easy to make mistakes by relying on implicit restrictions (e.g., zero capacity in a needed period) instead of explicitly forbidding production. Also, it is easy to make mistakes in the treatment of the inventory at the end of the first period, t=0."", ""When you encounter multi-period production–inventory problems, if some periods are unavailable for production, please add explicit no-production constraints for unavailable periods (x_{p,t}=0), where x_{p,t} is the production amount for product p on period t. If there is no given initial inventory and backlogs are not allowed, then the inventory at the end of period t=0 is given by I_{p,0}=x_{p,0}-s_{p,0} where I_{p,0} is the inventory at the end of period 0, x_{p,0} the amount produced in period 0, and s_{p,0} the sales in period 0; for t>0, we need to account for the inventory from the previous period I_{p, t-1} as follows I_{p,t}=I_{p,t-1}+x_{p,t}-s_{p,t}). Please ensure the objective charges holding costs only on end-of-period inventories.""), (""When solving multi-period lot-sizing problems with recurring per-period costs (e.g., maintenance costs), it is easy to make mistakes by modeling these costs as a one-time charge rather than accruing it each period the asset remains in service."", ""When you encounter multi-period lot-sizing problems with recurring costs, please model the recurring costs correctly by accruing them for all periods the asset remains in use, and not only one time when it is first installed, unless explicitly stated otherwise."" ), (""When solving multi-period inventory models that may allow unmet demand (lost sales) or backorders, it is easy to make mistakes by writing the inventory-balance equation with the wrong sign/placement of the shortage term, effectively treating shortage as extra demand."", ""When you encounter multi-period inventory problems that allow unmet demand or backorders, please define the unmet-demand/backlog variable precisely and write a consistent flow: for example, for lost sales use I_t = I_{t-1} + x_t - D_t + L_t (L_t >= 0) where I_t is the inventory of period t, x_t the quantity produced in period t, D_t the demand in period t, and L_t the lost sales/unmet demand in period t; for backorders introduce B_t for the backorders in period t and use I_t = I_{t-1} + x_t - D_t - B_{t-1} + B_t. Also be careful to set the initial inventory correctly if an initial inventory quantity is given (do not confuse it with the inventory I_0 at the end of the first period 0), and apply capacity constraints to peak on-hand inventory (I_{t-1}+x_t) and end inventory (I_t) for each period when applicable."")]" +Minimum Cost Flow Problem,"[ (""When solving min-cost flow models with node supplies/demands stored in a vector, it is easy to make mistakes by flipping the sign in the flow-balance constraints (treating supply as demand or vice versa)."", ""When you model min-cost flow with node imbalances recorded in a vector B, choose one convention and apply it consistently: if B[i]>0 denotes net supply (so B[i] is positive for supply surplus and negative for net demand), enforce flow conservation as outflow from i − inflow into i <= B[i], i.e., ∑_j x[i,j] − ∑_j x[j,i] <= B[i]; if B[i]>0 denotes net demand (so B[i] is negative for supply surplus and positive for net demand), then enforce inflow into i − outflow from i >= B[i], i.e., ∑_j x[j,i] − ∑_j x[i,j] >= B[i].""), (""When solving min-cost flow problems that actually include fixed costs for using an arc, it is easy to make mistakes by modeling the fixed cost as a per-unit cost rather than a one-time activation cost."", ""When you encounter a flow with per-arc fixed costs, please introduce a binary variable y[i,j] corresponding to whether the arc is used and link it to the flow (e.g., x[i,j] ≤ M·y[i,j]); include the fixed cost term c_fixed[i,j]·y[i,j] (and any per-unit cost term c_var[i,j]·x[i,j]) in the objective.""), (""When solving general min-cost flow on arbitrary networks, it is easy to make mistakes by allowing only to send flow from a node with net supply directly to a node with net demand."", ""When you face a network flow on a general graph, you do not need to directly send the flow from a node with net supply to a node with net demand, you can allow the use of intermediate nodes unless explicitly forbidden."") ]" +Network Optimization,"[ (""When solving network optimization problems with node supplies/demands, it is easy to make mistakes by flipping the sign in the flow-balance constraints (treating supply as demand or vice versa)."", ""When you encounter network optimization with node imbalances recorded in a vector B, please choose one convention and apply it consistently: if B[i]>0 denotes net supply (so B[i] is positive for supply surplus and negative for net demand), enforce flow conservation as outflow from i − inflow into i <= B[i], i.e., ∑_j x[i,j] − ∑_j x[j,i] <= B[i]; if B[i]>0 denotes net demand (so B[i] is negative for supply surplus and positive for net demand), then enforce inflow into i − outflow from i >= B[i], i.e., ∑_j x[j,i] − ∑_j x[i,j] >= B[i].""), (""When solving network optimization problems that include fixed costs for using an arc, it is easy to make mistakes by modeling a fixed cost as a per-unit cost rather than a one-time activation cost."", ""When you encounter a network optimization problem with per-arc fixed costs, please introduce a binary variable y[i,j] corresponding to whether the arc is used and link it to the flow (e.g., x[i,j] ≤ M·y[i,j]); include the fixed cost term c_fixed[i,j]·y[i,j] (and any per-unit cost term c_var[i,j]·x[i,j]) in the objective."") ]" +Operations Optimization,"[ (""When solving multi-period inventory models that may allow unmet demand (lost sales) or backorders, it is easy to make mistakes by writing the inventory-balance equation with the wrong sign/placement of the shortage term or by omitting the backlog."", ""When you encounter multi-period inventory problems that allow unmet demand or backorders, please define the unmet-demand/backlog variable precisely and write a consistent flow: for example, for lost sales use I_t = I_{t-1} + x_t - D_t + L_t (L_t >= 0) where I_t is the inventory of period t, x_t the quantity produced in period t, D_t the demand in period t, and L_t the lost sales/unmet demand in period t; for backorders introduce B_t for the backorders in period t and use I_t = I_{t-1} + x_t - D_t - B_{t-1} + B_t. Also be careful to set the initial inventory correctly if an initial inventory quantity is given (do not confuse it with the inventory I_0 at the end of the first period 0), and apply capacity constraints to peak on-hand inventory (I_{t-1}+x_t) and end inventory (I_t) for each period when applicable.""), (""When solving multi-period lot-sizing/inventory models with specified initial inventories, it is easy to make mistakes by mis-indexing the inventory state (e.g., reusing I[0] as both the fixed initial inventory and the inventory at the end of the first period), which forces incorrect carryover and mischarges holding cost."", ""If there is given initial inventory, make sure to use a different name for the initial inventory versus the inventory at the end of the first period.""), (""When solving multi-period lot-sizing problems with recurring per-period costs (e.g., maintenance costs), it is easy to make mistakes by modeling these costs as a one-time charge rather than accruing it each period the asset remains in service."", ""When you encounter multi-period lot-sizing problems with recurring costs, please model the recurring costs correctly by accruing them for all periods the asset remains in use, and not only one time when it is first installed, unless explicitly stated otherwise.""), (""When solving problems with multiple metrics that are combined in a single objective, it is easy to make mistakes by using wrong coefficients for each metric in the objective."", ""When solving problems with multiple metrics that are combined in a single objective, be careful to use any coefficient values that are provided in the problem description in order to define correctly the objective function and make sure everything is in the right units."")" +Production Planning Problem,"[ (""When solving production planning problems where items are counted in whole units (e.g., the statement uses ‘unit’, ‘per unit’, or ‘how many units’), it is easy to make mistakes by allowing fractional production for inherently discrete items."", ""When you encounter production planning problems that mention ‘unit’ of items or similar notions, please restrict the relevant decision variables to be integral so that only whole units can be produced, unless the problem explicitly allows fractional production.""), (""When solving multi-period inventory problems with optional shortages or backorders, it is easy to make mistakes by not taking backlog into account correctly in the balance equation (e.g., using the wrong signs), which distorts feasibility and costs."", ""When you encounter multi-period inventory problems that allow backlog, please take backlog into account explicitly at period t as follows: x_t + I_{t-1} = d_t + B_{t-1} - B_t + I_t, where B_{t-1} is the backlog from the previous time, B_t is the backlog at the end of the present time, I_{t-1} is the inventory from the previous time, I_t is the inventory at the end of the present time, and x_t is the production level of the current time. Notice that the backlog from the previous time should be added to the demand, and the present backlog should be subtracted.""), (""When solving multi-period lot-sizing/inventory models with specified initial inventories, it is easy to make mistakes by mis-indexing the inventory state (e.g., reusing I[0] as both the fixed initial inventory and the inventory at the end of the first period), which forces incorrect carryover and mischarges holding cost."", ""If there is given initial inventory, make sure to use a different name for the initial inventory versus the inventory at the end of the first period."")]" +Profit Maximization Problem,"[(""When solving blending/production problems with multiple types of production sources (e.g., energy sources) with maximum quantities for each type, it is easy to make mistakes by assuming a source has to deliver its maximum quantity when used, instead of allowing partial production up to the capacity limit"", ""When solving blending/production problems with multiple types of production sources (e.g., energy sources) with maximum quantities for each type, please define variables y_i for the quantity that is provided from each type of source i, such that 0 <= y_i <= maxQuantity_i. Use these variables both for any costs that depend on the production amounts in the objective, as well as to make sure enough quantities have been produced to satisfy the demand.""), (""When solving blending/production problems with multiple types of production sources (e.g., energy sources) and installation costs for each type, it is easy to make mistakes by considering the installation cost to be dependent on the produced quantity of each type, instead of a fixed cost for using the type/source"", ""When solving blending/production problems with multiple types of production sources (e.g., energy sources) and installation costs for each type, please define binary variables x_i representing whether the type/source i is used and multiply this with the installation cost of i in the objective in order to capture the installation cost for each type/source as fixed cost for using that type/source, and not dependent on the amount produced by the source, unless explictly stated otherwise.""), (""When solving linear facility-sizing or resource-allocation models, it is easy to make mistakes by misinterpreting the objective’s units or confusing costs/benefits for different aspects of the problem in the objective."", ""When solving linear facility-sizing or resource-allocation models, think carefully what are the correct units in the objective, for example if we are allowed to allocate partial amounts (e.g., there is a total available area and we can allocate in some of it), then we may pay partial cost, but also may have partial benefits, i.e., only the proportion of benefits that correspond to this allocation fraction are obtained.""), (""When solving problems with multiple demands/orders, it is easy to make mistakes by returning the optimal objective for a single demand, instead of the total objective for all demands."", ""When solving problems with multiple demands/orders, make sure that the objective captures the total objective for all demands and not only for a single demand.""), (""When solving problems with multiple metrics that are combined in a single objective, it is easy to make mistakes by using wrong coefficients for each metric in the objective."", ""When solving problems with multiple metrics that are combined in a single objective, be careful to use any coefficient values that are provided in the problem description in order to define correctly the objective function and make sure everything is in the right units."")]" +SupplyChain,"[ (""When solving supply chain network-flow or transportation models with node supplies/demands, it is easy to make mistakes by flipping the sign convention in the flow-conservation equations."", ""When you encounter supply chain network-flow problems with node imbalances recorded in a vector B, please choose one convention and apply it consistently: if B[i]>0 denotes net supply (so B[i] is positive for supply surplus and negative for net demand), enforce flow conservation as outflow from i − inflow into i <= B[i], i.e., ∑_j x[i,j] − ∑_j x[j,i] <= B[i]; if B[i]>0 denotes net demand (so B[i] is negative for supply surplus and positive for net demand), then enforce inflow into i − outflow from i >= B[i], i.e., ∑_j x[j,i] − ∑_j x[i,j] >= B[i].""), (""When solving multi-period supply chain lot-sizing/inventory models with specified initial inventory, it is easy to make mistakes by mis-indexing the inventory state (e.g., using I[0] as both the fixed initial inventory and the inventory at the end of the first period), which forces incorrect carryover and distorts costs."", ""If there is given initial inventory, make sure to use a different name for the initial inventory versus the inventory at the end of the first period."") ]" +Transportation Problem,"[ (""When solving transportation problems with node supplies/demands, it is easy to make mistakes by flipping the sign convention in the flow-conservation equations."", ""When you encounter transportation/network-flow models with supplies/demands on the nodes, enforce flow conservation as demand_i + outflow_from_i <= supply_i + inflow_into_i. Note that if you have more total supply than total demand you must use inequality instead of equality.""), (""When solving transportation problems that include a fixed cost to open/use a route (edge), it is easy to make mistakes by treating that cost as per-unit and omitting the open/close decision."", ""When solving transportation problems that include a fixed cost to open/use a route (edge), please introduce a binary variable y[i,j] to represent using edge (i,j) and include the fixed charge c[i,j]·y[i,j] in the objective; link the flow x[i,j] on the edge to the corresponding binary variable with a capacity big-M constraint x[i,j] ≤ M·y[i,j] so the fixed costs are paid only when the edge carries flow.""), (""It is easy to make mistakes by allowing only to send flow from a node with net supply directly to a node with net demand."", ""You do not need to directly send the flow from a node with net supply to a node with net demand, you can allow the use of intermediate nodes unless explicitly forbidden.""), (""It is easy to make mistakes by trying to solve the problem by brute force"", ""Do not try to solve the problem by enumerating all possible solutions or by using any kind of reasoning alone. Instead, focus on formulating the optimization model and the python code to solve it"") ]" +TravelingSalesman,"[ (""When solving TSP, it is easy to make mistakes by not enforcing the MTZ subtour elimination constraints correctly."", ""When solving TSP using MTZ, fix one city’s position (e.g., u[0]=0) and apply u[i] − u[j] + n·x[i,j] ≤ n−1 for the *remaining* cities i, j≠0 with i≠j to ensure no subtours are created; do *not* apply this constraint u[i] − u[j] + n·x[i,j] ≤ n−1 for the city whose position you fixed (e.g., city 0)""), (""When solving TSPs that use MTZ order variables, it is easy to make mistakes by declaring the position/order variables u[i] as continuous, allowing fractional positions that do not correspond to a valid city ordering."", ""When you implement MTZ order variables, please declare the position variable of node i, denote for example by u[i], as integer with tight bounds (e.g., u[0]=0 and 1≤u[i]≤n−1 for i>0) so the solver enforces a discrete tour ordering.""), (""When solving TSP formulations, it is easy to make mistakes by allowing self-loops (x[i,i]) and counting them in the degree constraints, enabling zero-length “tours” that satisfy in/out-degree but are invalid."", ""When you set up the TSP arc set, please exclude self-loops by defining x[i,j] only for i≠j (or fix x[i,i]=0) and enforce degree constraints as ∑_{j: j≠i} x[i,j]=1 and ∑_{i: i≠j} x[i,j]=1 for each city.""), (""If in a TSP problem not all cities/nodes need to be visited or if there is a given visit order preference/constraint between the nodes, you have to be careful."", ""If in a TSP problem not all cities/nodes need to be visited or if there is a given visit order preference/constraint between the nodes, please use binary selection x_i and assignment y_{i,k} with ∑_k y{i,k}=x_i and ∑_i y{i,k}≤1, define pos_i=∑_k k·y{i,k}, enforce any stated precedences (e.g., pos_A+1≤pos_B, …), and avoid hard-coding the number of stops or irrelevant capacity limits.""), (""When solving TSP or routing variants that include per-visit service/dwell times, it is easy to make mistakes by minimizing only travel time and omitting node service times from the objective, thereby understating total time/cost."", ""When you handle TSP/routing with service times, please include both arc travel times and node service/dwell times in the objective and ensure the tour returns to the start if required.""), (""It is easy to make mistakes by trying to solve the problem by brute force"", ""Do not try to solve the problem by enumerating all possible solutions or by using any kind of reasoning alone. Instead, focus on formulating the optimization model and the python code to solve it"")]" +Unit Commitment Problem,"[ (""When solving unit-commitment models, it is easy to make mistakes by forcing all generators online at period 0, by charging incorrect startup costs at period 0, and by making mistakes with shutdowns."", ""When you encounter unit-commitment models, please handle initial commitments carefully. In particular, don’t automatically force all of the generators to startup at period 0, only those that are needed, and think carefully what startup costs should be charged in the initial period (don't forget to charge startup costs, unless it is stated that some generators can start for free). Also, check if shutdowns are allowed, and if there are any constraints or costs associated with them.""), (""When solving unit-commitment models with both variable operating cost based on generated production (e.g., yuan/kWh) and fixed/no-load or maintenance cost that do not depend on the generated production quantity (e.g., yuan/hour), it is easy to make mistakes by mixing units and tying the wrong costs to the wrong variables, or by failing to link on/off decisions to producible output."", ""When you encounter unit-commitment costs with different units, please tie fixed costs that do not depend on production quantities (e.g., yuan/hour) to binary on/off commitment variables, tie variable/operating costs that depend on production quantities (yuan/kWh) to continuous generation variables, and link the binary commitment to feasible production via capacity (big-M) constraints - keep costs consistent with their units, separate fixed and variable decisions, and use linking constraints so the model captures both commitment and output correctly."") ]" +Flow Shop Problem,"[(""When solving flow shop problems, it is easy to make mistakes by confusing it with the job shop problem and assigning the same operation across jobs to different machines, instead of ensuring that the i-th operation of each job is scheduled on the i-th machine."", ""When solving flow shop problems, please ensure that the i-th operation of all jobs is scheduled on the i-th machine, add precedence constraints for the operations of each job, as well as precedence constraints for the jobs on each machine so that there are no overlaps on the same machine."")] " +Farm planning problem,"[(""When solving a farm planning problem, with crops that require multiple units of land, it is easy to make mistakes in the objective by capturing only the per-unit profit, instead of the total profit over the entire crop area."", ""When solving a farm planning problem, think carefully when defining the objective if it should be per unit or total. If a crop has a fixed required area, multiply the per-unit profit by that area; if a plot is fully planted, multiply the per-unit profit by the plot’s area. Always align profit definition with the scale of the decision (per-unit vs whole plot) based on the problem description."")]" +Capacitated Vehicle Routing Problem with Time Windows (CVRPTW),"[(""When you encounter Vehicle Routing Problems and a depot node is not provided, it is easy to make mistakes by taking one of the other nodes to be the depot, instead of introducing a separate node 0 with zero travel costs to the other nodes."", ""When you encounter Vehicle Routing Problems, if a depot node is not provided, introduce a new node 0 to be the depot with 0 travel costs to other nodes.""), (""When you encounter Vehicle Routing Problems, it is easy to make mistakes by allowing travel from a node to itself."", ""When you encounter Vehicle Routing Problems, do not allow going from one node to itself.""), (""When solving Vehicle Routing Problems, it is easy to make mistakes by imposing a constraint on the arrival time at the depot at the end of the route that causes infeasibility"", ""When you encounter Vehicle Routing Problems with Time Windows, unless it is explicitly stated, do not impose the time propagation constraint from any node back to the depot. For example, if 0 is the depot and t[k,0] is the arrival time at the depot for vehicle k and x[k,i,j] is the binary variable denoting if vehicle k goes from node i to j, do not add the time propagation constraint t[k,j] >= t[k,i] + travel[i][j] - M*(1 - x[k,i,j]) for j=0."")]" +Facility Dispersion Problem,"[(""It is easy to ignore values on both directions of the chosen nodes/items."", ""If the problem defines values or distances for both directions of a pair of nodes (e.g., defines a value for the pair (0,1) and also for (1,0)), then you should add up the values in both directions to obtain the total contribution of selecting the nodes in the pair."")] " diff --git a/optimind/data/optimind_cleaned_classified_industryor.csv b/optimind/data/optimind_cleaned_classified_industryor.csv new file mode 100644 index 0000000..125e4ca --- /dev/null +++ b/optimind/data/optimind_cleaned_classified_industryor.csv @@ -0,0 +1,982 @@ +problem_class,question,answer +"['Production Planning Problem', 'Capacitated Lot-sizing Problem (CLSP)']","A factory produces two types of food, I and II, and currently has 50 skilled workers. It is known that one skilled worker can produce $10 \ \mathrm{kg} / \ \mathrm{h}$ of food I or $6 \ \mathrm{kg} / \ \mathrm{h}$ of food II. According to contract bookings, the weekly demand for these two foods will rise sharply, as shown in Table 1-11. Therefore, the factory has decided to train 50 new workers by the end of the 8th week. It is known that a worker works $40 \ \mathrm{h}$ per week, and a skilled worker can train up to three new workers in two weeks (during the training period, both the skilled worker and the trainees do not participate in production). The weekly wage of a skilled worker is 360 yuan, the weekly wage of a trainee during the training period is 120 yuan, and after training, the wage is 240 yuan per week, with the same production efficiency as skilled workers. During the transition period of training, many skilled workers are willing to work overtime, and the factory has decided to arrange some workers to work $60 \ \mathrm{h}$ per week, with a weekly wage of 540 yuan. If the booked food cannot be delivered on time, the compensation fee for each week of delay per $ \ \mathrm{kg}$ is 0.5 yuan for food I and 0.6 yuan for food II. Under these conditions, how should the factory make comprehensive arrangements to minimize the total cost? + +Table 1-11 + +| Week | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | +|------|---|---|---|---|---|---|---|---| +| I | 10000 | 10000 | 12000 | 12000 | 16000 | 16000 | 20000 | 20000 | +| II | 6000 | 7200 | 8400 | 10800 | 10800 | 12000 | 12000 | 12000 |",219816.0 +"['Capacitated Lot-sizing Problem (CLSP)', 'Lot-Sizing Problem', 'Production Planning Problem', 'Discrete Lot-sizing and Scheduling Problem', 'Operations Optimization']","Each year $t=1,\dots ,n$ two production lines deliver $a_1=10$ and $a_2=15$ new fighter jets (25 total). $n=10$. Decide how many of that year's 25 aircraft, $x_t$, enter combat immediately and how many, $y_t=25-x_t$, become training platforms. A training jet produces five newly qualified pilots who are available at the start of the next year; every combat jet must be matched with one trained pilot to be operational, and training jets can be reassigned to combat in later years. Starting with no aircraft or pilots, choose integer sequences $\{x_t,y_t\}_{t=1}^n$ to maximise the cumulative number of operational combat jet-years $\sum_{t=1}^{n} x_t$, subject to annual pilot-availability and fleet-balance constraints.",1350.0 +"['Capacitated Lot-sizing Problem (CLSP)', 'Production Planning Problem']","A company specializing in foldable tables needs to create an optimal production and human resources plan for a six-month period (January to June) to maximize its total net profit. The plan must detail monthly in-house production levels, outsourcing quantities, and workforce management (hiring/firing). + +**Initial Conditions (at the start of January):** +- Initial Workforce: 1,000 employees +- Initial Inventory: 15,000 units + +**Revenue and Cost Structure:** +- **Sales Price:** 300 Yuan per unit sold. +- **Raw Material Cost:** 90 Yuan per unit, applicable *only* to units produced in-house. +- **Outsourcing Cost:** 200 Yuan per unit for finished tables acquired from a third-party supplier. This is an all-inclusive cost. +- **Inventory Holding Cost:** 15 Yuan per unit for any inventory held at the end of a month. +- **Backorder Cost:** 35 Yuan per unit for any unfulfilled demand (stockout) carried over to the next month. + +**Labor and Production Parameters:** +- **Labor Requirement:** Each in-house unit requires 5 labor hours to produce. +- **Regular Labor:** Each worker provides 160 regular working hours per month (8 hours/day * 20 days/month). The company pays a regular wage of 30 Yuan/hour for these 160 hours, regardless of full utilization. +- **Overtime Labor:** Workers can perform overtime. Total overtime hours per month for the entire workforce cannot exceed 20 hours per worker. The overtime wage is 40 Yuan/hour. +- **Workforce Management:** The company can hire or fire workers each month. The cost to hire a new worker is 5,000 Yuan, and the cost to fire a worker is 8,000 Yuan. + +**Demand and Fulfillment Logic:** +- Unfulfilled demand from one month is back-ordered and must be met in subsequent months. +- The company fulfills orders (both current demand and backorders) using available inventory from the previous month, current in-house production, and outsourced units. + +**Terminal Condition (at the end of June):** +- The ending inventory must be at least 10,000 units. +- All backorders must be cleared (i.e., ending backorders must be zero). + +**Forecasted Demand:** +| Month | January | February | March | April | May | June | +|:---:|:---:|:---:|:---:|:---:|:---:|:---:| +| Demand Forecast | 20,000 | 40,000 | 42,000 | 35,000 | 19,000 | 18,500 | + +Based on this information, formulate the optimal six-month operational plan.",10349920.0 +"['Farm Planning', 'Profit Maximization Problem']","A farmer needs to decide how many cows, sheep, and chickens to raise in order to achieve maximum profit. The farmer can sell cows, sheep, and chickens for $500, $200, and $8 each, respectively. The feed costs for each cow, sheep, and chicken are $100, $80, and $5, respectively. The profit is the difference between the selling price and the feed cost. Each cow, sheep, and chicken produces 10, 5, and 3 units of manure per day, respectively. Due to the limited time the farm staff has for cleaning the farm each day, they can handle up to 800 units of manure. Additionally, because of the limited farm size, the farmer can raise at most 50 chickens. Furthermore, the farmer must have at least 10 cows to meet customer demand. The farmer must also raise at least 20 sheep. Finally, the total number of animals cannot exceed 100.",30400.0 +['Diet Problem'],"Mary is planning her dinner tonight. Every 100 grams of okra contains 3.2 grams of fiber, every 100 grams of carrots contains 2.7 grams of fiber, every 100 grams of celery contains 1.6 grams of fiber, and every 100 grams of cabbage contains 2 grams of fiber. How many grams of each type of food should Mary buy to maximize her fiber intake? + +She is considering choosing one among salmon, beef, and pork as a protein source. For the chosen protein she must take at least one gram of it. + +She also considers choosing at least two kinds of vegetables among okra, carrots, celery, and cabbage. For each of the selected vegetables, she must take at least one gram. + +The price of salmon is $4 per 100 grams, beef is $3.6 per 100 grams, pork is $1.8 per 100 grams. The price of okra is $2.6 per 100 grams, carrots are $1.2 per 100 grams, celery is $1.6 per 100 grams, and cabbage is $2.3 per 100 grams. Mary has a budget of $15 for this meal. + +The total food intake should be 600 grams.",18.95657143 +['Capacitated Lot-sizing Problem (CLSP)'],"The contract reservations for the next year for products I, II, and III of a certain factory in each quarter are shown in Table 1-10. + +Table 1-10 +| Product | 1 | 2 | 3 | 4 | +|---------|------|------|------|------| +| I | 1500 | 1000 | 2000 | 1200 | +| II | 1500 | 1500 | 1200 | 1500 | +| III | 1000 | 2000 | 1500 | 2500 | + +At the beginning of the first quarter, there is no inventory for these three products, and it is required to have 150 units in stock for each product by the end of the fourth quarter. It is known that the factory has 15,000 production hours per quarter, and each unit of products I, II, and III requires 2, 4, and 3 hours respectively. Due to a change in equipment, product I cannot be produced in the second quarter. It is stipulated that if the products cannot be delivered on time, a compensation of 20 yuan per unit per quarter delay is required for products I and II, while for product III, the compensation is 10 yuan. Additionally, for products produced but not delivered in the current quarter, the inventory cost is 5 yuan per unit per quarter. How should the factory schedule production to minimize the total cost of compensation and inventory?",10755.0 +['Transportation Problem'],"An Italian transportation company needs to move some empty containers from its 6 warehouses (located in Verona, Perugia, Rome, Pescara, Taranto, and Lamezia) to major national ports (Genoa, Venice, Ancona, Naples, Bari). The container inventory at the warehouses is as follows: + +| | Empty Containers | +|:---:|:---:| +| Verona | 10 | +| Perugia | 12 | +| Rome | 20 | +| Pescara | 24 | +| Taranto | 18 | +| Lamezia | 40 | + +The demand at the ports is as follows: + +| | Container Demand | +|:---:|:---:| +| Genoa | 20 | +| Venice | 15 | +| Ancona | 25 | +| Naples | 33 | +| Bari | 21 | + +The transport is carried out by a fleet of trucks. The cost to transport each container is proportional to the distance traveled by the trucks, with a rate of 30 euros per kilometer. Each truck can carry up to 2 containers. The distances are as follows: + +| | Genoa | Venice | Ancona | Naples | Bari | +|:---:|:---:|:---:|:---:|:---:|:---:| +| Verona | $290 \mathrm{~km}$ | $115 \mathrm{~km}$ | $355 \mathrm{~km}$ | $715 \mathrm{~km}$ | $810 \mathrm{~km}$ | +| Perugia | $380 \mathrm{~km}$ | $340 \mathrm{~km}$ | $165 \mathrm{~km}$ | $380 \mathrm{~km}$ | $610 \mathrm{~km}$ | +| Rome | $505 \mathrm{~km}$ | $530 \mathrm{~km}$ | $285 \mathrm{~km}$ | $220 \mathrm{~km}$ | $450 \mathrm{~km}$ | +| Pescara | $655 \mathrm{~km}$ | $450 \mathrm{~km}$ | $155 \mathrm{~km}$ | $240 \mathrm{~km}$ | $315 \mathrm{~km}$ | +| Taranto | $1010 \mathrm{~km}$ | $840 \mathrm{~km}$ | $550 \mathrm{~km}$ | $305 \mathrm{~km}$ | $95 \mathrm{~km}$ | +| Lamezia | $1072 \mathrm{~km}$ | $1097 \mathrm{~km}$ | $747 \mathrm{~km}$ | $372 \mathrm{~km}$ | $333 \mathrm{~km}$ | + +Write a mathematical program to find the minimum cost transportation policy and solve it.",904590.0 +['Assignment Problem'],"Now, we need to determine 4 out of 5 workers to complete one of the four tasks respectively. Due to each worker's different technical specialties, the time required for them to complete each task varies. The hours required by each worker to complete each task are shown in Table 5-2. + +Table 5-2 +| Worker | $A$ | $B$ | $C$ | $D$ | +|--------|-----|-----|-----|-----| +| I | 9 | 4 | 3 | 7 | +| II | 4 | 6 | 5 | 6 | +| III | 5 | 4 | 7 | 5 | +| IV | 7 | 5 | 2 | 3 | +| V | 10 | 6 | 7 | 4 | + +Try to find a job assignment plan that minimizes the total working hours.",14.0 +"['Profit Maximization Problem', 'Production Planning Problem']","Haus Toys can manufacture and sell toy trucks, toy airplanes, toy boats, and toy trains. The profit for each truck sold is $5, each airplane $10, each boat $8, and each train $7. How many types of toys should Haus Toys manufacture to maximize profits? + +There are 890 units of wood available. Each truck requires 12 units, each airplane 20 units, each boat 15 units, and each train 10 units. + +There are 500 units of steel available. Each airplane requires 3 units, each boat 5 units, each train 4 units, and each truck 6 units. + +If Haus Toys manufactures trucks, they will not manufacture trains. + +However, if they manufacture boats, they will also manufacture airplanes. + +The number of toy boats manufactured cannot exceed the number of toy trains manufactured.",623.0 +['Set Cover'],"A convenience supermarket is planning to open several chain stores in a newly built residential area in the northwest suburb of the city. For shopping convenience, the distance from any residential area to one of the chain stores should not exceed $800 \mathrm{~m}$. Table 5-1 shows the new residential areas and the residential areas within a radius of $800 \mathrm{~m}$ from each of them. Question: What is the minimum number of chain stores the supermarket needs to build among the mentioned residential areas, and in which residential areas should they be built? + +| Area Code | Residential Areas within $800 \mathrm{~m}$ Radius | +|-----------|---------------------------------------------------| +| A | A, C, E, G, H, I | +| B | B, H, I | +| C | A, C, G, H, I | +| D | D, J | +| E | A, E, G | +| F | F, J, K | +| G | A, C, E, G | +| H | A, B, C, H, I | +| I | A, B, C, H, I | +| J | D, F, J, K, L | +| K | F, J, K, L | +| L | J, K, L |",3.0 +"['Production Planning Problem', 'Operations Optimization']","A company produces two types of small motorcycles, where type A is entirely manufactured by the company, and type B is assembled from imported parts. The production, assembly, and inspection time required for each unit of these two products are shown in Table 3.2. + +Table 3.2 + +| Type | Process | | | Selling Price
(Yuan/unit) | +| :---: | :---: | :---: | :---: | :---: | +| | Manufacturing | Assembly | Inspection | | +| Type A (hours/unit) | 20 | 5 | 3 | 650 | +| Type B (hours/unit) | 0 | 7 | 6 | 725 | +| Max production capacity per week (hours) | 120 | 80 | 40 | | +| Production cost per hour (Yuan) | 12 | 8 | 10 | | + +If the company's operational goals and targets are as follows: + +$p_{1}$ : The total profit per week should be at least 3000 yuan; + +$p_{2}$ : At least 5 units of type A motorcycles should be produced per week; + +$p_{3}$ : Minimize the idle time of each process as much as possible. The weight coefficients of the three processes are their hourly costs, and overtime is not allowed. + +Try to establish a model for this problem.",272.0 +['Facility Location Problem'],"Red Star Plastics Factory produces six distinct types of plastic containers. Each container type is characterized by a specific volume, market demand, and unit variable production cost, as detailed in Table 5-11. + +**Table 5-11: Container Data** +| Container Type (Code) | 1 | 2 | 3 | 4 | 5 | 6 | +| :------------------------------ | :--- | :--- | :--- | :--- | :--- | :---- | +| Volume ($\text{cm}^3$) | 1500 | 2500 | 4000 | 6000 | 9000 | 12000 | +| Market Demand (units) | 500 | 550 | 700 | 900 | 400 | 300 | +| Unit Variable Production Cost (Yuan/unit) | 5 | 8 | 10 | 12 | 16 | 18 | + +The production of any container type necessitates the use of its dedicated specialized equipment. If the decision is made to **activate** the production equipment for a particular container type (i.e., if the production quantity of that type is greater than zero), a fixed setup cost of 1200 Yuan is incurred for that specific equipment. + +Should the production quantity of a certain container type be insufficient to meet its direct demand, the factory has the option to utilize other container types with **larger or equal volume** as substitutes to fulfill this unmet demand. For instance, type 2 containers (volume 2500 $\text{cm}^3$) can be used to satisfy the demand for type 1 containers (requiring a volume of 1500 $\text{cm}^3$), but type 1 containers cannot be used for type 2 demand. In this problem, the container type codes are pre-sorted in ascending order of their volumes. + +**Question:** +How should the factory organize its production? The objective is to develop a production plan that minimizes the total cost—comprising the sum of variable production costs for all containers produced and the fixed costs for all activated equipment—while ensuring that the demand for all container types is fully met.",43200.0 +['Profit Maximization Problem'],"Tom and Jerry just bought a farm in Sunshine Valley, and they are considering using it to plant corn, wheat, soybeans, and sorghum. The profit per acre for planting corn is $1500, the profit per acre for planting wheat is $1200, the profit per acre for planting soybeans is $1800, and the profit per acre for planting sorghum is $1600. To maximize their profit, how many acres of land should they allocate to each crop? Tom and Jerry’s farm has a total area of 100 acres. + +The land area used for planting corn must be at least twice the land area used for planting wheat. + +The land area used for planting soybeans must be at least half the land area used for planting sorghum. + +The land area used for planting wheat must be three times the land area used for planting sorghum.",180000.0 +"['Knapsack', 'Diet Problem']","Mary is planning tonight's dinner. She wants to choose a combination of protein and vegetables to maximize her protein intake for the meal. Her protein options are chicken, salmon, and tofu, which can be bought in any quantity. + +- Chicken: 23g protein, $3.00 cost, per 100g. +- Salmon: 20g protein, $5.00 cost, per 100g. +- Tofu: 8g protein, $1.50 cost, per 100g. + +She also wants to choose from a list of five vegetables, sold in 100g packs. She must select at least three different types of vegetables. + +- Broccoli (100g pack): 2.8g protein, $1.20 cost. +- Carrots (100g pack): 0.9g protein, $0.80 cost. +- Spinach (100g pack): 2.9g protein, $1.50 cost. +- Bell Pepper (100g pack): 1.0g protein, $1.00 cost. +- Mushrooms (100g pack): 3.1g protein, $2.00 cost. + +Mary has two main constraints: +1. Her total budget is $20. +2. The total weight of all food must not exceed 800 grams. + +How should Mary choose her ingredients to get the maximum possible amount of protein?",123.8 +['Lot-Sizing Problem'],"A certain factory needs to use a special tool over $n$ planning stages. At stage $j$, $r_j$ specialized tools are needed. At the end of this stage, all tools used within this stage must be sent for repair before they can be reused. There are two repair methods: one is slow repair, which is cheaper (costs $b$ per tool) but takes longer ($p$ stages to return, e.g. if a tool goes to repair after stage 1, it will return at stage 1+p); the other is fast repair, which costs $c$ per tool $(c > b)$ and is faster, requiring only $q$ stages to return $(q < p)$. If the repaired tools cannot meet the needs, new ones must be purchased, with a cost of $a$ per new tool $(a > c)$. This special tool will no longer be used after $n$ stages. Determine an optimal plan for purchasing and repairing the tools to minimize the cost spent on tools during the planning period.\n\nn = 10 # number of stages\nr = [3, 5, 2, 4, 6, 5, 4, 3, 2, 1] # tool requirements per stage, indexing starts at 1\na = 10 # cost of buying a new tool\nb = 1 # cost of slow repair\nc = 3 # cost of fast repair\np = 3 # slow repair duration\nq = 1 # fast repair duration",134.0 +"['Lot-Sizing Problem', 'Capacitated Lot-sizing Problem (CLSP)']","A store plans to formulate the purchasing and sales plan for a certain product for the first quarter of next year. It is known that the warehouse capacity of the store can store up to 500 units of the product, and there are 200 units in stock at the end of this year. The store purchases goods once at the beginning of each month. The purchasing and selling prices of the product in each month are shown in Table 1.3. + +Table 1.3 + +| Month | 1 | 2 | 3 | +| :---: | :---: | :---: | :---: | +| Purchasing Price (Yuan) | 8 | 6 | 9 | +| Selling Price (Yuan) | 9 | 8 | 10 | + +Now, determine how many units should be purchased and sold each month to maximize the total profit, and express this problem as a linear programming model.",4100.0 +['Production Planning Problem'],"A textile factory produces two types of fabrics: one for clothing and the other for curtains. The factory operates two shifts, with a weekly production time set at 110 hours. Both types of fabrics are produced at a rate of 1000 meters per hour. Assuming that up to 70,000 meters of curtain fabric can be sold per week, with a profit of 2.5 yuan per meter, and up to 45,000 meters of clothing fabric can be sold per week, with a profit of 1.5 yuan per meter, the factory has the following objectives in formulating its production plan: + +$p_{1}$ : The weekly production time must fully utilize 110 hours; + +$p_{2}$ : Overtime should not exceed 10 hours per week; + +$p_{3}$ : At least 70,000 meters of curtain fabric and 45,000 meters of clothing fabric must be sold per week; + +$p_{4}$ : Minimize overtime as much as possible. + +Formulate a model for this problem.",5.0 +['Production Planning Problem'],"A furniture store can choose to order chairs from three different manufacturers: A, B, and C. The cost of ordering each chair from manufacturer A is $50, from manufacturer B is $45, and from manufacturer C is $40. The store needs to minimize the total cost of the order. + +Additionally, each order from manufacturer A will include 15 chairs, while each order from manufacturers B and C will include 10 chairs. The number of orders must be an integer. The store needs to order at least 100 chairs. + +Each order from manufacturer A will include 15 chairs, while each order from manufacturers B and C will include 10 chairs. The store needs to order at most 500 chairs. + +If the store decides to order chairs from manufacturer A, it must also order at least 10 chairs from manufacturer B. + +Furthermore, if the store decides to order chairs from manufacturer B, it must also order chairs from manufacturer C.",4000.0 +"['Production Planning Problem', 'Profit Maximization Problem']","Bright Future Toys wants to build and sell robots, model cars, building blocks, and dolls. The profit for each robot sold is $15, for each model car sold is $8, for each set of building blocks sold is $12, and for each doll sold is $5. How many types of toys should Bright Future Toys manufacture to maximize profit? +There are 1200 units of plastic available. Each robot requires 30 units of plastic, each model car requires 10 units of plastic, each set of building blocks requires 20 units of plastic, and each doll requires 15 units of plastic. + +There are 800 units of electronic components available. Each robot requires 8 units of electronic components, each model car requires 5 units of electronic components, each set of building blocks requires 3 units of electronic components, and each doll requires 2 units of electronic components. + +If Bright Future Toys manufactures robots, they will not manufacture dolls. + +However, if they manufacture model cars, they will also manufacture building blocks. + +The number of dolls manufactured cannot exceed the number of model cars manufactured.",956.0 +"['Lot-Sizing Problem', 'Operations Optimization']","A restaurant needs to order dining tables from three different suppliers, A, B, and C. The cost of ordering each dining table from Supplier A is $120, from Supplier B is $110, and from Supplier C is $100. The restaurant needs to minimize the total cost of the order. + +Additionally, each order from Supplier A will include 20 tables, while each order from Suppliers B and C will include 15 tables. The number of orders must be an integer. The restaurant needs to order at least 150 tables. + +Each order from Supplier A will include 20 tables, and each order from Suppliers B and C will include 15 tables. The restaurant needs to order no more than 600 tables. + +If the restaurant decides to order tables from Supplier A, it must also order at least 30 tables from Supplier B. + +Additionally, if the restaurant decides to order tables from Supplier B, it must also order tables from Supplier C.",15000.0 +"['Production Planning Problem', 'Capacitated Lot-sizing Problem (CLSP)']","A company plans to produce 3 types of products $A_{1}, A_{2}, A_{3}$. It can produce for 22 days in a month. The following table gives the maximum demand (unit $=100 \mathrm{~kg}$), price ($\$ / 100 \mathrm{Kg}$), production cost (per 100Kg product), and production quota (the maximum number of 100kg units that can be produced in one day if all production lines are devoted to this product). + +| Product | $A_{1}$ | $A_{2}$ | $A_{3}$ | +| :---: | :---: | :---: | :---: | +| Maximum Demand | 5300 | 4500 | 5400 | +| Selling Price | $124$ | $109$ | $115$ | +| Production Cost | $73.30$ | $52.90$ | $65.40$ | +| Production Quota | 500 | 450 | 550 | + +The fixed activation cost of the production line is as follows: + +| Product | $A_{1}$ | $A_{2}$ | $A_{3}$ | +| :---: | :---: | :---: | :---: | +| Activation Cost | $170000$ | $150000$ | $100000$ | + +Minimum production batch: + +$$ +\begin{array}{c|ccc} +Product & A_{1} & A_{2} & A_{3} \\ +\hline +Minimum Batch & 20 & 20 & 16 +\end{array} +$$ + +Please formulate an operations research model to determine a production plan that maximizes total revenue while accommodating fixed activation costs and minimum production batch constraints.",270290.0 +"['Profit Maximization Problem', 'Production Planning Problem']","Hongdou Clothing Factory uses three special equipment to produce shirts, short-sleeved shirts, and casual clothes respectively. It is known that the labor, material usage, selling price, and variable cost of each of the above products are as shown in Table 5-10. + +Table 5-10 + +| Product Name | Labor per unit | Material per unit | Selling Price | Variable Cost | +|--------------|----------------|------------------|---------------|---------------| +| Shirt | 3 | 4 | 120 | 60 | +| Short-sleeve | 2 | 3 | 80 | 40 | +| Casual Cloth | 6 | 6 | 180 | 80 | + +It is known that the available labor per week is 1500 units, the available material is 1600 units, and the weekly fixed costs for the three special equipment for producing shirts, short-sleeved shirts, and casual clothes are 2000, 1500, and 1000 respectively. Design a weekly production plan for the factory to maximize its profit.",24000.0 +['Transportation Problem'],"A manufacturing company needs to transport 1800 units of product from the warehouse to three different sales points. The company has four transportation options to choose from: truck, van, motorcycle, and electric vehicle. Since the van and electric vehicle both consume a lot of energy, the company wants to choose only one of these two options. Each trip with a truck generates 100 units of pollution, a van generates 50 units of pollution, a motorcycle generates 10 units of pollution, and an electric vehicle generates 0 units of pollution. The total pollution generated from all trips cannot exceed 2000 units. At least 10 trips must use a truck. Trucks, vans, motorcycles, and electric vehicles can transport 100 units, 80 units, 40 units, and 60 units of product per trip, respectively. The company needs to ensure that the total amount of transported product is at least 1800 units. Return the minimized pollution in units while meeting all constraints.",1000.0 +['PortfolioOptimization'],"An investor plans to invest 100,000 yuan, with two investment options to choose from. The first investment guarantees a return of 0.7 yuan for every 1 yuan invested after one year. The second investment guarantees a return of 2 yuan for every 1 yuan invested after two years, but the investment time must be in multiples of two years. In order to maximize the investor's earnings by the end of the third year, how should the investments be made? Formulate this as a linear programming problem.",510000.0 +['Set Multi-Cover'],"The number of salespeople required at a 24-hour convenience store in different time periods is as follows: 2:00-6:00 - 10 people, 6:00-10:00 - 15 people, 10:00-14:00 - 25 people, 14:00-18:00 - 20 people, 18:00-22:00 - 18 people, 22:00-2:00 - 12 people. Salespeople start their shifts at 2:00, 6:00, 10:00, 14:00, 18:00, and 22:00, working continuously for 8 hours. Determine the minimum number of salespeople needed to meet the requirements.",53.0 +"['Factory Planning Problem', 'Production Planning Problem']","A factory produces three types of products: I, II, and III. Each product needs to go through two processing procedures, A and B. The factory has two pieces of equipment that can complete process A, denoted as A1 and A2; it has three pieces of equipment that complete process B, denoted as B1, B2, and B3. Product I can be processed on any equipment for A and B; Product II can be processed on any A equipment but only on B1 for process B; Product III can only be processed on A2 and B2. Given the unit processing time on various machines, raw material costs, product sale prices, effective machine hours, and the costs of operating the machines at full capacity as shown in Table 1-4, the task is to arrange the optimal production plan to maximize the factory's profit. + +Table 1-4 +| Equipment | Product I | Product II | Product III | Effective Machine Hours | Operating Costs at Full Capacity (Yuan) | +|------------|-----------|------------|-------------|--------------------------|------------------------------------------| +| A1 | 5 | 10 | | 6000 | 300 | +| A2 | 7 | 9 | 12 | 10000 | 321 | +| B1 | 6 | 8 | | 4000 | 250 | +| B2 | 4 | | 11 | 7000 | 783 | +| B3 | 7 | | | 4000 | 200 | +| Raw Material Cost (Yuan/Unit) | 0.25 | 0.35 | 0.50 | | | +| Unit Price (Yuan/Unit) | 1.25 | 2.00 | 2.80 | | |",1146.4142 +['Profit Maximization Problem'],"Someone has a fund of 300,000 yuan and has the following investment projects in the next three years: +(1) Investment can be made at the beginning of each year within three years, with an annual profit of 20% of the investment amount, and the principal and interest can be used for investment in the following year; +(2) Investment is only allowed at the beginning of the first year, and it can be recovered at the end of the second year, with the total principal and interest amounting to 150% of the investment amount, but the investment limit is no more than 150,000 yuan; +(3) Investment is allowed at the beginning of the second year within three years, and it can be recovered at the end of the third year, with the total principal and interest amounting to 160% of the investment amount, and the investment limit is 200,000 yuan; +(4) Investment is allowed at the beginning of the third year within three years, and it can be recovered in one year with a profit of 40%, and the investment limit is 100,000 yuan. +Chapter One: Linear Programming and Simplex Method +Try to determine an investment plan for this person that maximizes the principal and interest at the end of the third year.",580000.0 +['Assignment Problem'],"Jieli Company needs to recruit three types of professionals to work in the two regional branches located in Donghai City and Nanjiang City. The demand for different professionals in these regional branches is shown in Table 4-3. After assessing the situation of the applicants, the company has categorized them into 6 types. Table 4-4 lists the specialties each type of person can handle, the specialty they prefer, and the city they prefer to work in. The company's personnel arrangement considers the following three priorities: +$p_1$: All three types of professionals needed are fully met; +$p_2$: 4000 recruited personnel meet their preferred specialty; +$p_3$: 4000 recruited personnel meet their preferred city. +Formulate a plan to minimize the total number of people that need to move from one city to another to meet these priorities. Return the minimized objective value. + +Table 4-3 +| Branch Location | Specialty | Demand | +|-----------------|-----------|--------| +| Donghai City | 1 | 1000 | +| Donghai City | 2 | 2000 | +| Donghai City | 3 | 1500 | +| Nanjiang City | 1 | 2000 | +| Nanjiang City | 2 | 1000 | +| Nanjiang City | 3 | 1000 | + +Table 4-4 + +| Type | Number of People | Suitable Specialty | Preferred Specialty | Preferred City | +|------|------------------|--------------------|---------------------|----------------| +| 1 | 1500 | 1,2 | 1 | Donghai | +| 2 | 1500 | 2,3 | 2 | Donghai | +| 3 | 1500 | 1,3 | 1 | Nanjiang | +| 4 | 1500 | 1,3 | 3 | Nanjiang | +| 5 | 1500 | 2,3 | 3 | Donghai | +| 6 | 1500 | 3 | 3 | Nanjiang |",2000.0 +['Diet Problem'],"Suppose a certain animal needs at least $700 \mathrm{~g}$ of protein, $30 \mathrm{~g}$ of minerals, and $100 \mathrm{mg}$ of vitamins daily. There are 5 types of feed available, and the nutritional content and price per kilogram of each type of feed are shown in Table 1-5: +Try to formulate a linear programming model that meets the animal's growth needs while minimizing the cost of selecting the feed. +Table 1-6 +| Feed | Protein (g) | Minerals (g) | Vitamins (mg) | Price (¥/kg) | Feed | Protein (g) | Minerals (g) | Vitamins (mg) | Price (¥/kg) | +|------|-------------|--------------|---------------|--------------|------|-------------|--------------|---------------|--------------| +| 1 | 3 | 1 | 0.5 | 0.2 | 4 | 6 | 2 | 2 | 0.3 | +| 2 | 2 | 0.5 | 1 | 0.7 | 5 | 18 | 0.5 | 0.8 | 0.8 | +| 3 | 1 | 0.2 | 0.2 | 0.4 | | | | | |",32.43589744 +"['Factory Planning Problem', 'Production Planning Problem', 'Profit Maximization Problem']","A factory produces three types of products: I, II, and III. Each product must undergo two processing stages, A and B. The factory has two types of equipment to complete stage A (A1, A2) and three types of equipment to complete stage B (B1, B2, B3). + +The production rules are as follows: +- Product I can be processed on any type of A equipment (A1 or A2) and any type of B equipment (B1, B2, or B3). +- Product II can be processed on any type of A equipment (A1 or A2), but for stage B, it can only be processed on B1 equipment. +- Product III can only be processed on A2 equipment for stage A and B2 equipment for stage B. + +The detailed data for processing time per piece, costs, sales price, and machine availability is provided in the table below. The objective is to determine the optimal production plan to maximize the factory's total profit. + +Data Table +| Equipment | Product I | Product II | Product III | Effective Machine Hours | Full - load Equipment Cost (Yuan) | Processing Cost per Machine Hour (Yuan/hour) | +| :--- | :--- | :--- | :--- | :--- | :--- | :--- | +| A1 | 5 | 10 | - | 6000 | 300 | 0.05 | +| A2 | 7 | 9 | 12 | 10000 | 321 | 0.03 | +| B1 | 6 | 8 | - | 4000 | 250 | 0.06 | +| B2 | 4 | - | 11 | 7000 | 783 | 0.11 | +| B3 | 7 | - | - | 4000 | 200 | 0.05 | +| Raw Material Cost (Yuan/piece) | 0.25 | 0.35 | 0.5 | - | - | - | +| Unit Price (Yuan/piece) | 1.25 | 2 | 2.8 | - | - | - |",1190.38 +['Production Planning Problem'],"A product consists of three components produced by four workshops, each with a limited number of production hours. Table 1.4 below provides the production rates of the three components. The objective is to determine the number of hours each workshop should allocate to each component to maximize the number of completed products. Formulate this problem. + +Table 1.4 + +| Workshop | Production Capacity (hours) | Production Rate (units/hour) | | | +| :------: | :-------------------------: | :--------------------------: | - | - | +| | | Component 1 | Component 2 | Component 3 | +| A | 100 | 10 | 15 | 5 | +| B | 150 | 15 | 10 | 5 | +| C | 80 | 20 | 5 | 10 | +| D | 200 | 10 | 15 | 20 |",2924.0 +['Knapsack'],"A wealthy noble passed away, leaving the following inheritance: + +- A painting by Caillebotte: $25000 +- A bust of Diocletian: $5000 +- A Yuan dynasty Chinese vase: $20000 +- A 911 Porsche: $40000 +- Three diamonds: each $12000 +- A Louis XV sofa: $3000 +- Two very precious Jack Russell racing dogs: each $3000 (will stipulates they must not be separated) +- A sculpture from 200 AD: $10000 +- A sailing boat: $15000 +- A Harley Davidson motorcycle: $10000 +- A piece of furniture once belonging to Cavour: $13000, + +which must be shared between two sons. How to formulate a mathematical program and solve it to minimize the difference in value between the two parts?",1000.0 +['Bin Packing'],"The current problem faced by the company is how to use the fewest number of containers to pack the currently needed goods for transportation, while considering the weight of the goods, specific packaging requirements, and inventory limitations. Professional modeling and analysis are needed for a batch of goods’ transportation strategy to ensure maximum utilization of the limited container space. + +The company currently has a batch to be transported, with each container able to hold a maximum of 60 tons of goods and each container used must load at least 18 tons of goods. The goods to be loaded include five types: A, B, C, D, and E, with quantities of 120, 90, 300, 90, and 120 respectively. The weights are 0.5 tons for A, 1 ton for B, 0.4 tons for C, 0.6 tons for D, and 0.65 tons for E. Additionally, to meet specific usage requirements, every time A goods are loaded, at least 1 unit of C must also be loaded, but loading C alone does not require simultaneously loading A; and considering the demand limitation for D goods, each container must load at least 12 units of D. + +Establish an operations research model so that the company can use the fewest number of containers to pack this batch of goods.",7.0 +['Flow Shop Scheduling'],"A fabric dyeing plant has 3 dyeing vats. Each batch of fabric must be dyed in sequence in each vat: first, the second, and third vats. The plant must color five batches of fabric of different sizes. The time required in hours to dye batch $i$ in vat $j$ is given in the following matrix: + +$$ +\left(\begin{array}{ccc} +3 & 1 & 1 \\ +2 & 1.5 & 1 \\ +3 & 1.2 & 1.3 \\ +2 & 2 & 2 \\ +2.1 & 2 & 3 +\end{array}\right) +$$ + +Schedule the dyeing operations in the vats to minimize the completion time of the last batch.",14.1 +['Capacitated Vehicle Routing Problem with Time Windows (CVRPTW)'],"The Vehicle Routing Problem (VRP) was first proposed by Dantzig and Ramser in 1959. It is a classic combinatorial optimization problem. The basic VRP can be described as follows: in a certain area, there is a number of customers and a distribution center or depot. Customers are generally located at different positions, and each has a specific demand for goods. The distribution center needs to dispatch a fleet of vehicles and design appropriate delivery routes to fulfill the demands of all customers. The objective of VRP is to optimize a certain benefit metric while satisfying all customer demands. The benefit metric is usually presented as an objective function, which varies according to the company's requirements. Common objective functions include minimizing the total distance traveled by vehicles, minimizing the total delivery time, or minimizing the number of vehicles used. In addition to satisfying customer demands, VRP often needs to consider various other constraints, leading to several variants. For example, if the vehicle's load cannot exceed its maximum capacity, the problem becomes the Capacitated Vehicle Routing Problem (CVRP). If each customer's delivery must be made within a specific time frame, the problem becomes the Vehicle Routing Problem with Time Windows (VRPTW). + +The Vehicle Routing Problem with Time Windows (VRPTW) is a classic variant of the VRP. There are many real-world applications of VRPTW, as customer locations often have service time windows. For instance, some logistics centers need to stock parcels during off-peak hours, and large supermarkets need to replenish goods outside of business hours. Real-time delivery services like food delivery also require strict delivery time windows. Time windows can be categorized as hard or soft. A Hard Time Window (HTW) means that a vehicle must arrive at the delivery point within or before the time window; late arrivals are not permitted. If a vehicle arrives early, it must wait until the time window opens to begin service. This is common in scenarios like supermarket restocking and logistics center inbound operations. A Soft Time Window (STW) means that a vehicle is not strictly required to arrive within the time window, but it is encouraged to do so. A penalty is incurred for early or late arrivals. This is applicable in scenarios such as meal delivery, school bus services, and industrial deliveries. + +The Vehicle Routing Problem with Hard Time Windows (VRPHTW) can be described as follows: within a region, there is a set of customer locations and a central depot. Vehicles must start from the depot and return to the depot, following continuous paths. Each customer must be served by exactly one vehicle, and vehicles have a limited capacity. Each customer has a specific service time window, and service is only accepted within this window. A vehicle can arrive at a customer location early and wait for the time window to open, or it can arrive within the time window to provide service. Service can only begin within the time window, and the service duration is known. The distribution center must arrange an optimal delivery plan to both complete the delivery tasks and minimize travel costs. Because VRPHTW does not allow for delays, it, like the VRP, primarily emphasizes the minimization of travel costs along the routes. + + Now we consider a major enterprise logistics provider, 'Global Logistics', is responsible for providing precise material delivery services for multiple high-end office buildings and shops in a city's central business district (CBD). Due to traffic control in the CBD and the specific receiving requirements of the customers, the delivery task is highly challenging. + +**Specific Requirements:** + +1. **Delivery Task**: There are 20 customers requiring delivery service on the day, and the demands of all customers must be met. +2. **Vehicle Constraints**: The company can use at most 5 trucks, and the capacity of each truck is 200 units. +3. **Capacity Constraint**: The total demand of all customers on a single route must not exceed the truck's maximum capacity (200 units). +4. **Time Window Constraint**: Each customer has a strict 'hard time window.' Service must begin within this specified time window. Early arrivals must wait, and late arrivals are not permitted. +5. **Service Time**: Due to the complex handover procedures at customer sites, a fixed service time of 90 minutes is required for unloading, handover, and paperwork at each customer location. +6. **Optimization Objective**: While satisfying all constraints, the company's objective is to **minimize the total distance traveled by all vehicles** to reduce operational costs. + +**Data Details:** + +* **Central Depot (Depot 0)**: + * Coordinates: (40, 50) + * Operating Time Window: [0, 1236] (minutes) +* **Customer Locations (Customers 1-20)**: The coordinates, demand, service time window, and service duration for each customer are shown in the table below. + +| Customer ID | Coordinates (X, Y) | Demand (units) | Time Window (minutes) | Service Duration (minutes) | +| :--- | :--- | :--- |:--- | :--- | +| 1 | (45, 68) | 10 | [912, 967] | 90 | +| 2 | (45, 70) | 30 | [825, 870] | 90 | +| 3 | (42, 66) | 10 | [65, 146] | 90 | +| 4 | (42, 68) | 10 | [727, 782] | 90 | +| 5 | (42, 65) | 10 | [15, 67] | 90 | +| 6 | (40, 69) | 20 | [621, 702] | 90 | +| 7 | (40, 66) | 20 | [170, 225] | 90 | +| 8 | (38, 68) | 20 | [255, 324] | 90 | +| 9 | (38, 70) | 10 | [534, 605] | 90 | +| 10 | (35, 66) | 10 | [357, 410] | 90 | +| 11 | (35, 69) | 10 | [448, 505] | 90 | +| 12 | (25, 85) | 20 | [652, 721] | 90 | +| 13 | (22, 75) | 30 | [30, 92] | 90 | +| 14 | (22, 85) | 10 | [567, 620] | 90 | +| 15 | (20, 80) | 40 | [384, 429] | 90 | +| 16 | (20, 85) | 40 | [475, 528] | 90 | +| 17 | (18, 75) | 20 | [99, 148] | 90 | +| 18 | (15, 75) | 20 | [179, 254] | 90 | +| 19 | (15, 80) | 10 | [278, 345] | 90 | +| 20 | (30, 50) | 10 | [10, 73] | 90 | + +Now, please provide an operations research model for this VRPHTW.",175.37 +['Production Planning Problem'],"A factory produces two types of microcomputers, A and B. Each type of microcomputer requires the same two production processes. The processing time, profit from sales, and the maximum weekly processing capacity for each type are shown in Table 3.1. + +Table 3.1 + +| Process | Model | | Maximum Weekly Processing Capacity | +| :---: | :---: | :---: | :---: | +| | $\\mathrm{A}$ | $\\mathrm{B}$ | | +| I (hours / unit) | 4 | 6 | 150 | +| II (hours / unit) | 3 | 2 | 70 | +| Profit ($ per unit) | 300 | 450 | | + +The expected values for the factory's operational goals are as follows: + +$p_{1}$: The total weekly profit must not be less than $10,000. + +$p_{2}$: Due to contractual requirements, at least 10 units of Model A and at least 15 units of Model B must be produced per week. + +$p_{3}$: The weekly production time for Process I should be exactly 150 hours, and the production time for Process II should be fully utilized, with potential overtime if necessary. + +Try to establish the mathematical programming model for this problem in oder to maximize total profit.",11250.0 +['Flow Shop Scheduling'],"There are three different products to be processed on three machine tools. Each product must first be processed on machine 1, then sequentially on machines 2 and 3. The order of processing the three products on each machine should remain the same. Assuming $t_{ij}$ represents the time to process the $i$-th product on the $j$-th machine, how should the schedule be arranged to minimize the total processing cycle for the three products? The timetable is as follows: +| Product | Machine 1 | Machine 2 | Machine 3 | +|---------|-----------|-----------|-----------| +| Product 1 | 2 | 3 | 1 | +| Product 2 | 4 | 2 | 3 | +| Product 3 | 3 | 5 | 2 |",14.0 +"[""Transportation, Airline Industry, Resource Allocation"", ""Knapsack""]","A company plans to transport goods between the city and the suburb and needs to choose the most environmentally friendly transportation method. The company can choose from the following three methods: motorcycle, small truck, and large truck. Each motorcycle trip produces 40 units of pollution, each small truck trip produces 70 units of pollution, and each large truck trip produces 100 units of pollution. The company's goal is to minimize total pollution. + +The company can only choose two out of these three transportation methods. + +Due to certain road restrictions, the number of motorcycle trips cannot exceed 8. + +Each motorcycle trip can transport 10 units of products, each small truck trip can transport 20 units of products, and each large truck trip can transport 50 units of products. The company needs to transport at least 300 units of products. + +The total number of trips must be less than or equal to 20.",600.0 +"['Production Planning Problem', 'Operations Optimization']","The independent country of Carelland mainly exports four commodities: steel, engines, electronic components, and plastic. Carelland's Minister of Finance (i.e., Minister of Economy) wants to maximize exports and minimize imports. The unit prices of steel, engines, electronics, and plastic on the world market are, in local currency (Klunz), 500, 1500, 300, 1200 respectively. Producing 1 unit of steel requires 0.02 units of engines, 0.01 units of plastic, 250 Klunz of other imported goods, and 6 person-months of labor. Producing 1 unit of engines requires 0.8 units of steel, 0.15 units of electronic components, 0.11 units of plastic, 300 Klunz of imported goods, and 1 person-year. One unit of electronics requires: 0.01 units of steel, 0.01 units of engines, 0.05 units of plastic, 50 Klunz of imported goods, and 6 person-months of labor. One unit of plastic requires: 0.03 units of engines, 0.2 units of steel, 0.05 units of electronic components, 300 Klunz of imported goods, and 2 person-years. Engine production is limited to 650000 units, and plastic production is limited to 60000 units. The total available labor force per year is 830000 person-months. Write a mathematical program to maximize domestic GDP and solve the problem.",36288567.0 +['Profit Maximization Problem'],"A person has a fund of 500,000 yuan and the following investment projects available in the next three years: + +(1) Investment can be made at the beginning of each year within three years, and the annual profit is 20% of the investment amount. + +(2) Investment is only allowed at the beginning of the first year, and can be recovered at the end of the second year, with the total principal and interest being 150% of the investment amount. However, this type of investment is limited to no more than 120,000 yuan. + +(3) Investment at the beginning of the second year, recoverable at the end of the second year, with the total principal and interest being 160% of the investment amount. This type of investment is limited to 150,000 yuan. + +(4) Investment is allowed at the beginning of the third year, recoverable in one year, with a profit of 40%, and the investment limit is 100,000 yuan. + +Determine an investment plan for the person that maximizes the total principal and interest by the end of the third year.",964640.0 +"['Production Planning Problem', 'Operations Optimization']","Two steel furnaces at a steel plant each use two methods of steelmaking simultaneously. The first method takes $a=2$ hours per furnace and costs $m=50$ in fuel expenses; the second method takes $b=3$ hours per furnace and costs $n=70$ in fuel expenses. Assuming each furnace produces $k=10$ tons of steel regardless of the method used, and that at least $d=30$ tons of steel must be produced within $c=12$ hours, how should these two methods be allocated to minimize fuel expenses? Formulate this problem as a linear programming model.",150.0 +['Transportation Problem'],"A production base needs to extract raw materials from warehouses A and B every day for production. The required raw materials are: at least 240 pieces of raw material A, at least 80 kg of raw material B, and at least 120 tons of raw material C. It is known that: Each truck from warehouse A can transport back to the production base 4 pieces of raw material A, 2 kg of raw material B, 6 tons of raw material C, with a freight cost of 200 yuan per truck; each truck from warehouse B can transport back to the production base 7 pieces of raw material A, 2 kg of raw material B, 2 tons of raw material C per day, with a freight cost of 160 yuan per truck. Question: In order to meet production needs, how many trucks should be dispatched daily from warehouse A and warehouse B to minimize the total freight cost?",6800.0 +['Capacitated Facility Location Problem'],"Given that there are $m=2$ production points for a certain type of material, where the output at the $i$-th point $(i=1,2)$ is $a_i$, $a_1 = 100$, and $a_2 = 150$. This material is to be shipped to $n=2$ demand points, where the demand at the $j$-th point $(j=1, 2)$ is $b_j$, $b_1 = 80$, and $b_2 = 120$. It is known that $\sum_i a_i \geqslant \sum_j b_j$. It is also known that when shipping from production points to demand points, it must pass through one of the $p=2$ intermediate marshaling stations. If the $k$-th $(k=1, 2)$ intermediate marshaling station is used, a fixed cost $f_k$ is incurred regardless of the transshipment volume, where $f_1 = 10$ and $f_2 = 15$. The $k$-th intermediate marshaling station has a maximum transshipment capacity limitation $q_k$, where $q_1 = 100$ and $q_2 = 100$. Let $c_{i k}$ and $c'_{k j}$ denote the unit transportation cost from $i$ to $k$ and from $k$ to $j$, respectively, where $c_{11}=2$, $c_{12}=3$, $c_{21}=4$, $c_{22}=1$, $c'_{11}=3$, $c'_{12}=2$, $c'_{21}=1$, and $c'_{22}=4$. Try to determine a transportation plan for this material that minimizes the total cost.",685.0 +"['Production Planning Problem', 'Profit Maximization Problem']","A factory produces three types of products, A, B, and C. Each unit of product A requires 1 hour for technical preparation, 10 hours of direct labor, and 3 kg of materials. Each unit of product B requires 2 hours for technical preparation, 4 hours of labor, and 2 kg of materials. Each unit of product C requires 1 hour for technical preparation, 5 hours of labor, and 1 kg of materials. The available technical preparation time is 100 hours, labor time is 700 hours, and materials are 400 kg. The company offers larger discounts for bulk purchases, as detailed in Table 1-22. Determine the company's production plan to maximize profit. +Table 1-22 +| Product A | | Product B | | Product C | | +|:---------------|:---------:|:---------------|:---------:|:---------------|:---------:| +| Sales Volume (pieces) | Profit (yuan) | Sales Volume (pieces) | Profit (yuan) | Sales Volume (pieces) | Profit (yuan) | +| 0 ~ 40 | 10 | 0 ~ 50 | 6 | 0 ~ 100 | 5 | +| 40 ~ 100 | 9 | 50 ~ 100 | 4 | Above 100 | 4 | +| 100 ~ 150 | 8 | Above 100 | 3 | | | +| Above 150 | 7 | | | | |",712.0 +['Assignment Problem'],"A university computer lab hires 4 undergraduates (designated 1, 2, 3, and 4) and 2 graduate students (designated 5 and 6) for duty answering questions. The maximum duty hours from Monday to Friday and the hourly wage for each person are shown in Table 5-9. + +Table 5-9 +Student ID | Wage (CNY/h) | Monday | Tuesday | Wednesday | Thursday | Friday +1 | 10.0 | 6 | 0 | 6 | 0 | 7 +2 | 10.0 | 0 | 6 | 0 | 6 | 7 +3 | 9.9 | 4 | 8 | 4 | 0 | 5 +4 | 9.8 | 5 | 5 | 6 | 0 | 4 +5 | 10.8 | 4 | 0 | 4 | 8 | 0 +6 | 11.3 | 5 | 6 | 0 | 6 | 3 + +The lab operates from 8:00 AM to 10:00 PM, and there must be one and only one student on duty during open hours. It is also required that each undergraduate must work at least 8 hours per week, and each graduate student must work at least 7 hours per week. Additionally, each student can work no more than 2 shifts per week, and no more than 3 students can be scheduled for duty each day. + +Based on these conditions, establish a mathematical model to determine the work schedule that satisfies all requirements.",717.9 +['Farm Planning'],"A certain farm has 100 hectares of land and 15,000 yuan in funds for production development. The labor force situation on the farm is 3,500 person-days in autumn and winter, and 4,000 person-days in spring and summer. If the labor force itself is not fully utilized, they can work externally, earning 2.1 yuan/person-day in spring and summer and 1.8 yuan/person-day in autumn and winter. + +The farm cultivates three types of crops: soybeans, corn, and wheat, and also raises dairy cows and chickens. Crop cultivation requires no specialized investment, but raising animals involves an investment of 400 yuan per dairy cow and 3 yuan per chicken. Raising dairy cows requires allocating 1.5 hectares of land per cow to grow feed, and involves 100 person-days in autumn and winter, and 50 person-days in spring and summer per cow. The annual net income is 400 yuan per dairy cow. Raising chickens does not use land, requires 0.6 person-days in autumn and winter, and 0.3 person-days in spring and summer per chicken. Annual net income is 2 yuan per chicken. The current chicken coop can accommodate up to 3,000 chickens, and the cow barn can accommodate up to 32 dairy cows. The labor and income requirements for the three types of crops per year are shown in Table 1-9. + +Table 1-9 +| Item | Soybean | Corn | Wheat | +|----------------|---------|------|-------| +| Person-days (Autumn/Winter) | 20 | 35 | 10 | +| Person-days (Spring/Summer) | 50 | 75 | 40 | +| Annual Net Income (Yuan/hectare) | 175 | 300 | 120 | + +Determine the farm's operating plan to maximize annual net income. Please note that workers can only work externally for full days, fractions are not allowed. It is not possible to change the crop and animal raising plans from season to season.",20241.8 +['Production Planning Problem'],"A factory produces two models of microcomputers, A and B. Each model requires the same two processes. The processing time, sales profit, and the factory’s maximum weekly processing capacity for each model are shown in Table 3.1. + +Table 3.1 + +| Process | Model | | Maximum Weekly Processing Capacity | +| :---: | :---: | :---: | :---: | +| | $A$ | $B$ | | +| I (hours/unit) | 4 | 6 | 150 | +| II (hours/unit) | 3 | 2 | 70 | +| Profit (yuan/unit) | 300 | 450 | | + +Given the factory's business goals: + +$p_{1}$: The total weekly profit should not be less than 10,000 yuan; + +$p_{2}$: Due to contract requirements, at least 10 units of model A and at least 15 units of model B must be produced each week; + +$p_{3}$: The processing time for Process I should be exactly 150 hours per week, and the processing time for Process II should ideally be fully utilized, with potential for appropriate overtime; + +$p_{4}$: If products are produced during overtime in Process II, the profit per unit is reduced by 20 yuan for model A and 25 yuan for model B, and the maximum overtime for Process II is 30 hours per week. Formulate the mathematical model for this problem.",11250.0 +['Lot-Sizing Problem'],"A factory must rent warehouse space to cover storage needs over the next four months. The required storage areas are: +Month 1: 1500 m² +Month 2: 1000 m² +Month 3: 2000 m² +Month 4: 1200 m² + +Warehouse space can be rented via contracts of fixed duration. A contract of length k months (k ? {1, 2, 3, 4}) may start at the beginning of any month t provided it ends no later than Month 4 (i.e., t + k ? 1 ? 4). A contract starting in month t covers months t through t + k ? 1. The rental fee is charged per square meter per month and depends on the contract length as follows: +1-month contract: 22 yuan per m² per month +2-month contract: 21 yuan per m² per month +3-month contract: 20 yuan per m² per month +4-month contract: 19 yuan per m² per month + +Additional rules and assumptions: + +You may sign any number of contracts. + +Rented area is divisible (you may rent any nonnegative real number of m²). + +Supply is unlimited at the listed rates. + +In each month, the total active rented area must be at least the required area for that month. + +You pay for the entire area specified in each contract for every month it is active, even if some capacity is unused. + +Your task is to choose the start times, durations, and areas of contracts to minimize the total rental cost over the four-month horizon while satisfying the monthly area requirements.",113000.0 +"['Lot-Sizing Problem', 'SupplyChain']","A store has formulated a purchase and sales plan for a certain product from July to December. It is known that the warehouse capacity must not exceed 500 units, with 200 units in stock at the end of June. Thereafter, purchases are made at the beginning of each month. Assume the purchase and selling prices of this product for each month are shown in Table 1-21. How much should be purchased and sold each month to maximize the total revenue? + +Table 1-21 +| Month | 7 | 8 | 9 | 10 | 11 | 12 | +|-------|----|----|----|----|----|----| +| Buy | 28 | 24 | 25 | 27 | 23 | 23 | +| Sell | 29 | 24 | 26 | 28 | 22 | 25 |",9100.0 +"['Military Personnel Deployment Problem', 'Assignment Problem']","The number of nurses required in each time period over 24 hours at a certain hospital is as follows: 2:00-6:00 - 10 people, 6:00-10:00 - 15 people, 10:00-14:00 - 25 people, 14:00-18:00 - 20 people, 18:00-22:00 - 18 people, 22:00-2:00 - 12 people. Nurses start shifts in 6 batches at 2:00, 6:00, 10:00, 14:00, 18:00, and 22:00 and work continuously for 8 hours. Please determine: If the hospital can hire contract nurses with the same working hours as regular nurses, and if the pay for regular nurses is 10 yuan/hour and for contract nurses is 15 yuan/hour, should the hospital hire contract nurses and if so, how many?",4240.0 +['Set Multi-Cover'],"For a certain 24-hour bus service, the number of drivers and crew members required during different time periods each day is shown in Table 1-2: +Table 1-2 +\begin{tabular}{|c|c|c||c|c|c|} +\hline Shift & Time & Required number & Shift & Time & Required number \\ +\hline 1 & $6: 00 \sim 10: 00$ & 60 & 4 & $18 ; 00 \sim 22 ; 00$ & 50 \\ +\hline 2 & $10 ; 00 \sim 14 ; 00$ & 70 & 5 & $22 ; 00 \sim 2 ; 00$ & 20 \\ +\hline 3 & $14 ; 00 \sim 18 ; 00$ & 60 & 6 & $2: 00 \sim 6 ; 00$ & 30 \\ +\hline +\end{tabular} + +Assuming that drivers and crew members start their shifts at the beginning of each time period and work continuously for 8 hours, determine the minimum number of drivers and crew members needed for this bus route. Formulate the linear programming model for this problem.",150.0 +['Knapsack'],"The Zhang family has 6 children: Harry, Hermione, Ron, Fred, George, and Ginny. The cost of taking Harry is $1200, Hermione is $1650, Ron is $750, Fred is $800, George is $800, and Ginny is $1500. Which children should the couple take to minimize the total cost of taking the children? They can take up to four children on the upcoming trip. + +Ginny is the youngest, so the Zhang family will definitely take her. + +If the couple takes Harry, they will not take Fred because Harry does not get along with him. + +If the couple takes Harry, they will not take George because Harry does not get along with him. + +If they take George, they must also take Fred. + +If they take George, they must also take Hermione. + +Even though it will cost them a lot of money, the Zhang family has decided to take at least three children.",3050.0 +['Production Planning Problem'],"Given that a certain factory plans to produce three types of products, I, II, and III, each product needs to be processed on equipment $A, B, C$ as shown in Table 2-3: + +Table 2-3 +| Equipment Code | I | II | III | Effective Monthly Equipment Hours | +|----------------|----|----|-----|----------------------------------| +| A | 8 | 2 | 10 | 300 | +| B | 10 | 5 | 8 | 400 | +| C | 2 | 13 | 10 | 420 | +| Unit Product Profit (per thousand yuan) | 3 | 2 | 2.9 | | + +How can the equipment capacity be fully utilized to maximize production profit? The quantity of each product must be an integer.",134.5 +['Set Multi-Cover'],"A master's student in Operations Research at a certain university is required to select two courses in mathematics, two in operations research, and two in computer science from a total of seven courses: Calculus, Operations Research, Data Structures, Management Statistics, Computer Simulation, Computer Programming, and Forecasting. Some courses belong to only one category: Calculus falls under Mathematics, Computer Programming under Computer Science. However, some courses fall under multiple categories: Operations Research can be considered both Operations Research and Mathematics, Data Structures both Computer Science and Mathematics, Management Statistics both Mathematics and Operations Research, Computer Simulation both Computer Science and Operations Research, and Forecasting both Operations Research and Mathematics. Courses that fall under multiple categories can fulfill the requirement of both categories simultaneously. Additionally, some courses have prerequisites: Computer Simulation or Data Structures requires Computer Programming first, Management Statistics requires Calculus first, and Forecasting requires Management Statistics first. The question is: What is the minimum number of courses a master's student must take, and which specific courses, to meet the above requirements?",4.0 +['Lot-Sizing Problem'],"A trading company specializes in the wholesale business of certain grains. The company currently has a warehouse with a capacity of 5000 dan. On January 1, the company has 1000 dan of grain in stock and 20,000 yuan in funds. The estimated grain prices for the first quarter are shown in Table 1-8. + +Table 1-8 +| Month | Purchase Price (yuan/dan) | Selling Price (yuan/dan) | +|-------|---------------------------|--------------------------| +| 1 | 2.85 | 3.10 | +| 2 | 3.05 | 3.25 | +| 3 | 2.90 | 2.95 | + +The purchased grains will be delivered in the same month but can only be sold in the next month, and payment is required upon delivery. The company hopes to have an inventory of 2000 dan at the end of the quarter. What purchasing and selling strategy should be adopted to maximize the total profit over the three months?",-700.0 +['Cutting Stock Problem'],"Assuming a paper mill receives three orders for rolls of paper, with length and width requirements as shown in Table 1.2. + +Table 1.2 + +| Order Number | Width (meters) | Length (meters) | +| :---: | :---: | :---: | +| 1 | 0.5 | 1000 | +| 2 | 0.7 | 3000 | +| 3 | 0.9 | 2000 | + +The mill produces rolls of paper with standard widths of 1 meter and 2 meters. Assuming the length of the rolls is unlimited and can be spliced to reach the required length, how should the rolls be cut to minimize the area of waste?",600.0 +"['Farm Planning', 'Profit Maximization Problem']","Vicky and David have just bought a farm in the Yarra Valley, and they are considering using it to grow apples, pears, oranges, and lemons. The profit for growing one acre of apples is $2000, for one acre of pears is $1800, for one acre of oranges is $2200, and for one acre of lemons is $3000. To achieve maximum profit, how many acres of land should they use to grow each type of fruit? Vicky and David have just bought a farm in the Yarra Valley with a total area of 120 acres. + +The land used to grow apples should be at least twice the land used to grow pears. + +The land used to grow apples should be at least three times the land used to grow lemons. + +The land used to grow oranges must be twice the land used to grow lemons if lemons are grown. If no lemons are grown, then we do not have this constraint. + +Vicky and David are unwilling to grow more than two types of fruit.",264000.0 +"['Blending Problem', 'Profit Maximization Problem']","A candy factory uses raw materials A, B, and C to process three different brands of candies, A, B, and C. It is known that the content of A, B, and C in each brand of candy, the cost of raw materials, the monthly limit of each raw material, and the unit processing fee and selling price of the three brands of candies are shown in Table 1-7. + +Table 1-7 + +| Item | A | B | C | Raw Material Cost (Yuan/kg) | Monthly Limit (kg) | +|:----------------|:---------------|:---------------|:---------------|:-----------------------------|:-------------------| +| A | ? 60% | ? 15% | | 2.00 | 2000 | +| B | | | | 1.50 | 2500 | +| C | ? 20% | ? 60% | ? 50% | 1.00 | 1200 | +| Processing Fee (Yuan/kg) | 0.50 | 0.40 | 0.30 | | | +| Selling Price (Yuan/kg) | 3.40 | 2.85 | 2.25 | | | + +How many kilograms of each of the three brands of candies should the factory produce each month to maximize the profit?",6160.0 +['TravelingSalesman'],"A traveling salesman must visit 7 customers at 7 different locations, with the (symmetric) distance matrix as follows: + +| | 1 | 2 | 3 | 4 | 5 | 6 | 7 | +| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | +| 1 | - | 86 | 49 | 57 | 31 | 69 | 50 | +| 2 | | - | 68 | 79 | 93 | 24 | 5 | +| 3 | | | - | 16 | 7 | 72 | 67 | +| 4 | | | | - | 90 | 69 | 1 | +| 5 | | | | | - | 86 | 59 | +| 6 | | | | | | - | 81 | + +Formulate a mathematical program to determine the visiting order starting and ending at location 1 to minimize the travel distance.",153.0 +['Capacitated Facility Location Problem'],"A product can be processed on any one of the four devices: A, B, C, or D. The preparation completion costs when each device is enabled, the unit production cost for the product, and the maximum processing capacity of each device are shown in Table 5-7. If 2000 units of the product need to be produced, how can the total cost be minimized? Try to establish a mathematical model. + +Table 5-7 +| Device | Prep Completion Cost (Yuan) | Unit Production Cost (Yuan/Unit) | Maximum Processing Capacity (Units) | +|--------|------------------------------|----------------------------------|------------------------------------| +| A | 1000 | 20 | 900 | +| B | 920 | 24 | 1000 | +| C | 800 | 16 | 1200 | +| D | 700 | 28 | 1600 |",37000.0 +['Knapsack'],"The Zhang family is deciding to invest in several different restaurants. The annual revenue of Restaurant A is $15,000, Restaurant B is $40,000, Restaurant C is $30,000, and Restaurant D is $50,000. They need to decide whether to purchase each restaurant, with each restaurant being able to be purchased only once. Help them decide which restaurants to buy to maximize their annual income. +The cost of Restaurant A is 1.6 million, Restaurant B is 2.5 million, Restaurant C is 1.8 million, and Restaurant D is 3 million. The Zhang family's investment budget is 6 million. + +If they purchase Restaurant D, then they cannot purchase Restaurant A.",90000.0 +['Transportation Problem'],"A farmer needs to transport 1000 units of fresh produce from the farm to a nearby market. The farmer has three transportation options: a horse, a bicycle, and a handcart. Since both the bicycle and handcart are very physically demanding, the farmer wants to choose only one of these two transportation methods. The horse generates 80 units of pollution per trip, the bicycle generates 0 units of pollution, and the handcart generates 0 units of pollution. The total amount of pollution generated by all trips must not exceed 1000 units. At least 8 trips must be made using the horse. The horse, bicycle, and handcart can carry 55 units, 30 units, and 40 units of produce per trip respectively. The farmer needs to ensure that the total amount of transported produce is at least 1000 units while minimizing the total amount of pollution. What is the minimum amount of pollution that the farmer can achieve?",640.0 +"['Knapsack', 'Team Formulation Problem']","A company needs to decide whether to hire some of the five candidates to join their R&D team. The salary requirements for candidates F, G, H, I, and J are $12,000, $15,000, $18,000, $5,000, and $10,000 respectively. The company wants to minimize the total amount paid to candidates without exceeding the budget. + +The company's budget is $40,000 and they wish to hire a maximum of 4 new employees. + +The skill levels of the candidates are as follows: +Candidate F: Level 2 +Candidate G: Level 3 +Candidate H: Level 4 +Candidate I: Level 1 +Candidate J: Level 2 + +The company needs to ensure that the total skill level of the hired employees is at least 8. + +The project management experience years of each candidate are as follows: +Candidate F: 1 year +Candidate G: 2 years +Candidate H: 2 years +Candidate I: 5 years +Candidate J: 4 years + +They hope the total project management experience of the team is at least 8 years. + +Due to the similar technical background of candidates G and J, the company can choose at most one of them.",38000.0 +['Production Planning Problem'],"A company produces two types of products: microwave ovens and water heaters, which are manufactured in both workshops A and B. It is known that apart from the purchased parts, the production of one microwave oven requires 2 hours of processing in workshop A and 1 hour of assembly in workshop B. The production of one water heater requires 1 hour of processing in workshop A and 3 hours of assembly in workshop B. After production, both products need inspection, sales, and other procedures. The inspection and sales cost for each microwave oven is 30 yuan, and for each water heater is 50 yuan. Workshop A has 250 hours of available production time per month, with each hour costing 80 yuan; workshop B has 150 hours of available production time per month, with each hour costing 20 yuan. It is estimated that an average of 80 microwave ovens and 50 water heaters can be sold per month next year. Based on these actual conditions, the company has established the following monthly plan constraints: + +1. Inspection and sales costs should not exceed 5500 yuan per month; +2. At least 80 microwave ovens should be sold per month; +3. The production hours of both workshops A and B should be fully utilized, and overtime for workshop A and B are allowed. +4. Overtime in workshop A should not exceed 20 hours; we do not have upper limit on workshop B's overtime. +5. At least 50 water heaters should be sold per month. + +Try to determine the monthly production plan for the company.",30500.0 +"['Production Planning Problem', 'Profit Maximization Problem']","A toy company manufactures three types of tabletop golf toys, each requiring different manufacturing techniques. The high-end type requires 17 hours of manufacturing labor, 8 hours of inspection, and yields a profit of 300 yuan per unit. The mid-range type requires 10 hours of labor, 4 hours of inspection, and yields a profit of 200 yuan per unit. The low-end type requires 2 hours of labor, 2 hours of inspection, and yields a profit of 100 yuan per unit. Available labor hours are 1000, and available inspection hours are 500. Additionally, market forecasts indicate a demand of no more than 50 units for the high-end type, no more than 80 units for the mid-range type, and no more than 150 units for the low-end type. Determine the production plan for the company to maximize profit.",25000.0 +"['Lot-Sizing Problem', 'Capacitated Lot-sizing Problem (CLSP)', 'Production Planning Problem']","The market demand for products I and II is as follows: Product I requires 10,000 units per month from January to April, 30,000 units per month from May to September, and 100,000 units per month from October to December. Product II requires 15,000 units per month from March to September and 50,000 units per month during other months. The cost of producing these two products at a certain factory is as follows: Product I costs 5 yuan per unit to produce from January to May, and 4.50 yuan per unit from June to December; Product II costs 8 yuan per unit to produce from January to May, and 7 yuan per unit from June to December. The factory's combined production capacity for both products should not exceed 120,000 units per month. Product I has a volume of 0.2 cubic meters per unit, Product II has a volume of 0.4 cubic meters per unit, and the factory's warehouse capacity is 15,000 cubic meters. If the factory's warehouse space is insufficient, external warehouse space can be rented. Using the factory’s own warehouse costs 1 yuan per cubic meter per month, while renting an external warehouse increases this cost to 1.5 yuan per cubic meter per month. Given that the initial inventory of both products at the beginning of July is zero, how should production be scheduled from July to December to minimize the total production and inventory costs while meeting market demand?",3160500.0 +['Transportation Problem'],"There are two coal yards A and B, each receiving no less than 80 tons and 100 tons of coal per month, respectively. They are responsible for supplying coal to three residential areas, which need 55 tons, 75 tons, and 50 tons of coal per month, respectively. Coal yard A is located 10 kilometers, 5 kilometers, and 6 kilometers from these three residential areas. Coal yard B is located 4 kilometers, 8 kilometers, and 15 kilometers from these three residential areas. How should these two coal yards distribute coal to the three residential areas to minimize the ton-kilometers of transportation?",1030.0 +['Cutting Stock Problem'],"A steel reinforcement workshop produces a batch of steel bars (with the same diameter), consisting of 90 pieces of 3 meters in length and 60 pieces of 4 meters in length. It is known that each piece of raw steel bar used is 10 meters in length. How can the raw material be cut most efficiently? Establish a linear programming model for this problem.",53.0 +['TravelingSalesman'],"The famous Traveling Salesman Problem (TSP) in operations research can be described as follows: A traveling salesman departs from a certain city, and must visit each city exactly once before returning to the original starting city. The distances between the cities are provided in the table below (the entry at row i and column j represents the cost of going from city i to city j) +| City | 1 | 2 | 3 | 4 | +| ---- | ------ | ------ | ------ | ------ | +| 1 | 0 | 10 | 20 | 12 | +| 2 | 10 | 0 | 5 | 10 | +| 3 | 20 | 5 | 0 | 8 | +| 4 | 15 | 12 | 8 | 0 | + +What route should the salesman choose to travel in order to minimize the total distance? Try to formulate an integer programming model for this problem.",35.0 +['Assignment Problem'],"Consider assigning $n=2$ factories to $n$ locations. The transportation volume between factory $i$ and factory $j$ is $d_{ij}$, and the unit transportation cost from location $p$ to location $q$ is $c_{pq}$. The specific values are shown in the following table: Table 1.1 + +| | Transportation volume to Location 1 | Transportation volume to Location 2 | Transportation cost to Location 1 | Transportation cost to Location 2 | +| :----: | :---------------------------------: | :---------------------------------: | :-------------------------------: | :-------------------------------: | +| Factory 1 | 10 | 20 | 5 | 8 | +| Factory 2 | 30 | 40 | 6 | 7 | + +In order to minimize the total transportation cost, formulate this problem as an integer model.",330.0 +['Knapsack'],"The Li family plans to invest their retirement fund in commercial real estate. The annual income from Property 1 is $12,500, Property 2 is $35,000, Property 3 is $23,000, and Property 4 is $100,000. The decision to be made is whether to buy each property or not, rather than how many to buy, as there is only one of each property available. Help them decide which properties to purchase to maximize their annual income. + +The cost of Property 1 is $1.5 million, Property 2 is $2.1 million, Property 3 is $2.3 million, and Property 4 is $4.2 million. The Li family's budget is $7 million. + +If they purchase Property 4, they cannot purchase Property 3.",135000.0 +['Knapsack'],"The Li family has 5 children: Alice, Bob, Charlie, Diana, and Ella. The cost to take Alice is $1000, Bob is $900, Charlie is $600, Diana is $500, and Ella is $700. Which children should the couple take to minimize the total cost of taking the children? + +They can take up to 3 children on the upcoming trip. + +Bob is the youngest, so the Li family will definitely take him. + +If the couple takes Alice, they will not take Diana because Alice does not get along with her. + +If the couple takes Bob, they will not take Charlie because Bob does not get along with him. + +If they take Charlie, they must also take Diana. + +If they take Diana, they must also take Ella. + +Despite the cost, the Li family has decided to take at least two children.",1600.0 +['Operations Optimization'],"A project includes the following 7 activities, with their durations (in days) as follows: $A(4), B(3), C(5), D(2), E(10), F(10), G(1)$. The precedence relationships are also given as: $A \rightarrow G, D ; E, G \rightarrow F; D, F \rightarrow C ; F \rightarrow B$. The cost of work per day is 1000 Euros; additionally, a special machine must be rented from the start of activity $A$ to the end of activity $B$, costing 5000 Euros per day. Formulate this as a linear programming problem to minimize cost and complete all activities.",115000.0 +"['Production Planning Problem', 'Profit Maximization Problem']","There are $\mathrm{A}$ and $\mathrm{B}$ two products, both requiring two successive chemical reaction processes. Each unit of product $\mathrm{A}$ needs 2 hours for the first process and 3 hours for the second process. Each unit of product $\mathrm{B}$ needs 3 hours for the first process and 4 hours for the second process. Available time for the first process is 16 hours, and available time for the second process is 24 hours. + +For each unit of product $\mathrm{B}$ produced, 2 units of by-product $\mathrm{C}$ are generated simultaneously, requiring no additional cost. By-product $\mathrm{C}$ can be sold up to 5 units, and the rest must be disposed of at a cost of 2 yuan per unit. + +Each unit of product $\mathrm{A}$ sold yields a profit of 4 yuan, each unit of product $\mathrm{B}$ yields a profit of 10 yuan, and each unit of by-product $\mathrm{C}$ sold yields a profit of 3 yuan. + +In order to maximize total profit, establish the linear programming model for this problem.",57.0 +"['Lot-Sizing Problem', 'Production Planning Problem']","A timber storage and transport company has a large warehouse for storing and transporting timber for sale. Due to seasonal price fluctuations, the company purchases timber at the beginning of each quarter, with part of it being sold within the quarter and part being stored for future sales. It is known that the maximum storage capacity of the company's warehouse is 200,000 m³, and the storage cost is $(a+b u)$ yuan/m³, where $a=70$, $b=100$, and $u$ is the storage time (in quarters). The purchase and sale prices for each quarter and the estimated maximum sales volumes are shown in Table 1-18. + +Table 1-18 +| Quarter | Purchase Price (10,000 yuan/10,000 m²) | Sale Price (10,000 yuan/10,000 m²) | Estimated Maximum Sales Volume (10,000 m³) | +|---------|----------------------------------------|------------------------------------|---------------------------------------------| +| Winter | 410 | 425 | 100 | +| Spring | 430 | 440 | 140 | +| Summer | 460 | 465 | 200 | +| Autumn | 450 | 455 | 160 | + +Since timber is not suitable for long-term storage, all inventory should be sold by the end of autumn. Try to establish a linear programming model for this problem to maximize the company's annual profit. Return your answer in the unit of 10000 yuan.",4700.0 +['Capacitated Facility Location Problem'],"There are 10 different parts, and they can all be processed on machine \( A \), machine \( B \), or machine \( C \). The unit processing costs are shown in Table 5-6. Additionally, as long as any part is processed on the aforementioned machines, a one-time setup cost will be incurred regardless of whether one or multiple types of parts are processed, with the respective costs being \( d_A = 100 \), \( d_B = 135 \), and \( d_C = 200 \) yuan. If the requirements are: + +1. One piece of each of the aforementioned 10 types of parts needs to be processed; +2. If the 1st part is processed on machine \( A \), then the 2nd part must be processed on machine \( B \) or \( C \); conversely, if the 1st part is processed on machine \( B \) or \( C \), then the 2nd part must be processed on machine \( A \); +3. Parts 3, 4, and 5 must be processed on machines A, B, and C respectively; +4. The number of parts processed on machine \( C \) should not exceed 3 types. + +Try to establish an integer programming mathematical model for this problem with the objective of minimizing the total cost. + +Table 5-6 +| Machine/Part | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | +|--------------|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----| +| A | $10$ | $20$ | $30$ | $40$ | $50$ | $60$ | $70$ | $80$ | $90$ | $100$ | +| B | $15$ | $25$ | $35$ | $45$ | $55$ | $65$ | $75$ | $85$ | $95$ | $105$ | +| C | $20$ | $30$ | $40$ | $50$ | $60$ | $70$ | $80$ | $90$ | $100$ | $110$ |",1005.0 +['Operations Optimization'],"A shoe store employs 5 full-time sales clerks and 4 part-time sales clerks. Their working hours and wage conditions are shown in Table 3.3. + +Table 3.3 + +| | Monthly Working Hours | Sales Volume (Pairs/Hour) | Wage (Yuan/Hour) | Overtime Pay (Yuan/Hour) | +| :---: | :---: | :---: | :---: | :---: | +| Full-time | 160 | 5 | 1 | 1.5 | +| Part-time | 80 | 2 | 0.6 | 0.7 | + +Each pair of shoes sold earns a profit of 0.3 yuan. The store has set the following goals: + +$p_{1}$: Achieve monthly sales of 5500 pairs; + +$p_{2}$: Ensure full employment of all sales clerks; + +$p_{3}$: Minimize overtime hours. + +Try to establish a model for this problem.",172.0 +"['Production Planning Problem', 'Profit Maximization Problem']","A furniture factory needs to decide how many tables, chairs, and bookshelves to produce in order to maximize its profit. The factory can sell each table for $200, each chair for $50, and each bookshelf for $150. The manufacturing costs for each table, chair, and bookshelf are $120, $20, and $90 respectively. The profit is the difference between the selling price and the manufacturing cost. Each table, chair, and bookshelf occupy 5, 2, and 3 square meters of warehouse space respectively. Due to limited warehouse space, the total space cannot exceed 500 square meters. In addition, due to market demand, the factory needs to produce at least 10 tables and 20 bookshelves. Finally, the total number of items produced by the factory cannot exceed 200.",9800.0 +['Operations Optimization'],"A company requires skilled workers and laborers for three tasks. The first task can be completed by one skilled worker alone, or by a group of one skilled worker and two laborers. The second task can be done by one skilled worker or one laborer alone. The third task can be completed by a group of five laborers, or by one skilled worker leading three laborers. The weekly wages for skilled workers and laborers are 100 yuan and 80 yuan respectively. They work 48 hours per week, but their actual effective working hours are 42 hours and 36 hours respectively. To complete these tasks, the company needs a total effective working time of 8400 hours for the first task, 10800 hours for the second task, and 18000 hours for the third task per week. The number of workers that can be recruited is limited to a maximum of 400 skilled workers and 800 laborers. Establish a mathematical model to determine how many skilled workers and laborers should be hired in order to minimize the total wage expenditure.",84000.0 +"['Assignment Problem', 'Bin Packing']","On Danzig Street, vehicles can park on both sides of the street. Mr. Edmonds, who lives at No. 1, is organizing a party with about 30 participants, and they will arrive in 15 cars. The length of the i-th car is ?_i, in meters, as follows: + +| i | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | +|----|----|-----|----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----| +| ?_i | 4 | 4.5 | 5 | 4.1 | 2.4 | 5.2 | 3.7 | 3.5 | 3.2 | 4.5 | 2.3 | 3.3 | 3.8 | 4.6 | 3 | + +In order to avoid disturbing the neighbors, Mr. Edmonds wants to arrange parking on both sides of the street so that the total length of the street occupied by his friends' vehicles is minimized. Please provide a mathematical programming formulation and solve this problem. +How does the program change if the cars on one side of the street cannot occupy more than 30 meters?",28.6 +['Knapsack'],"Changjiang Comprehensive Shopping Mall has 5000 m² of space for lease and plans to attract the following 5 types of stores as tenants. The table below shows the area occupied by each type of store for one shop, the minimum and maximum number of shops for each type within the mall, and the expected annual profit (in ten thousand yuan) per store for different numbers of stores. Each store pays 20% of its annual profit as rent to the mall. Question: How many of each type of store should the mall lease to maximize total rental income? + +Table 5-12 + +| Code | Store Type | Area per Shop / m² | Min | Max | 1 Store | 2 Stores | 3 Stores | +|------|------------|--------------------|-----|-----|---------|----------|----------| +| 1 | Jewelry | 250 | 1 | 3 | 9 | 8 | 7 | +| 2 | Shoes & Hats | 350 | 1 | 2 | 10 | 9 | - | +| 3 | General Merchandise | 800 | 1 | 3 | 27 | 21 | 20 | +| 4 | Bookstore | 400 | 0 | 2 | 16 | 10 | - | +| 5 | Catering | 500 | 1 | 3 | 17 | 15 | 12 |",28.0 +['Set Multi-Cover'],"A certain restaurant operates around the clock, and the number of waiters needed in 24 hours is shown in Table 1.1. + +Table 1.1 + +| Time | Minimum Number of Waiters Needed | Time | Minimum Number of Waiters Needed | +|:-----------:|:-------------------------------:|:-----------:|:-------------------------------:| +| $2 \sim 6$ | 4 | $14 \sim 18$| 7 | +| $6 \sim 10$ | 8 | $18 \sim 22$| 12 | +| $10 \sim 14$| 10 | $22 \sim 2$ | 4 | + +Each waiter works continuously for 8 hours a day. The goal is to find the minimum number of waiters that meet the above conditions and represent this problem as a linear programming model.",26.0 +['Knapsack'],"A company hopes to recruit new employees for its team. The salary requirements for candidates A, B, C, D, and E are $8100, $20000, $21000, $3000, and $8000 respectively. They need to decide whether to hire each candidate. The team wants to minimize the total amount paid to the candidates. + +They hope to hire a maximum of 3 new employees. + +The team has a limited budget of $35,000. They need to ensure that the total payment to the selected candidates does not exceed the budget. + +The qualifications of the five candidates are as follows: +Candidate A: Bachelor's degree; +Candidate B: Master's degree; +Candidate C: Doctoral degree; +Candidate D: No degree; +Candidate E: No degree. +They will select at least one candidate with a Master's or Doctoral degree. + +The work experience of the five candidates is as follows: +Candidate A: 3 years of work experience; +Candidate B: 10 years of work experience; +Candidate C: 4 years of work experience; +Candidate D: 3 years of work experience; +Candidate E: 7 years of work experience. +They hope the total work experience of the selected candidates is no less than 12 years. + +Due to the equivalent professional skills of candidates A and E, the company will choose at most one from the two. + +They will hire at least 2 new employees.",23000.0 +"['Production Planning Problem', 'Profit Maximization Problem']","A company is producing two products (X and Y). The resources required for the production of X and Y are divided into two parts: machine time for automated processing and craftsman time for manual finishing. The table below shows the number of minutes required for each product: + +| Item | Machine Time (minutes) | Craftsman Time (minutes) | +| :---: | :---: | :---: | +| X | 13 | 20 | +| Y | 19 | 29 | + +The company has 40 hours of machine time available in the next working week, but only 35 hours of craftsman time. The cost of machine time is £10 per hour, and the cost of craftsman time is £2 per hour. Idle time for machines and craftsmen incurs no cost. For each product produced (all products produced will be sold), the revenue for product X is £20, and the revenue for product Y is £30. Products can only be produced in whole units. The company has a specific contract that requires 10 units of product X to be produced for a customer each week. Formulate a model for this problem.",1861.466667 +"['Profit Maximization Problem', 'Production Planning Problem']","Healthy Pet Foods Company produces two types of dog food: Meaties and Yummies. Each pack of Meaties contains 2 pounds of grains and 3 pounds of meat; each pack of Yummies contains 3 pounds of grains and 1.5 pounds of meat. The company believes it can sell any quantity of dog food that it can produce. Meaties sell for $2.80 per pack, and Yummies sell for $2.00 per pack. The company's production is subject to several constraints. First, a maximum of 400,000 pounds of grains can be purchased each month at a price of $0.20 per pound of grains. A maximum of 300,000 pounds of meat can be purchased each month at a price of $0.50 per pound of meat. Additionally, a special machine is required to produce Meaties, with a monthly capacity of 90,000 packs. The variable costs for mixing and packaging dog food are $0.25 per pack (Meaties) and $0.20 per pack (Yummies). Detailed information is provided in Table B-1. + +**Table B-1 Healthy Pet Foods Data** + +| | Meaties | Yummies | +|--------------------|--------------|------------| +| Price per pack | $2.80 | $2.00 | +| Raw materials | | | +| - Grains | 2.0 lbs | 3.0 lbs | +| - Meat | 3.0 lbs | 1.5 lbs | +| Variable cost | $0.25/pack | $0.20/pack | +| Resources | | | +| Meaties capacity | 90,000 packs/month | | +| Monthly available grains | 400,000 lbs | | +| Monthly available meat | 300,000 lbs | | + +Assume you are the manager of the dog food department at Healthy Pet Foods Company. Your salary is based on the department's profit, so you will try to maximize profit. How should you operate the department to maximize both the profit and your salary?",77500.0 +['Multi-Commodity Transportation Problem'],"A transportation company has two types of trucks, Type A and Type B. Type A trucks have 20 cubic meters of refrigerated capacity and 40 cubic meters of non-refrigerated capacity. In contrast, Type B trucks have the same total capacity, but the capacities for refrigerated and non-refrigerated cargo are equal. A grocer needs to rent trucks to transport 3000 cubic meters of refrigerated cargo and 4000 cubic meters of non-refrigerated cargo. The rental cost per kilometer for Type A trucks is £30, while the rental cost per kilometer for Type B trucks is £40. How many of each type of truck should the grocer rent to minimize the total cost? + +Try to formulate a model for this problem.",4170.0 +"['Production Planning Problem', 'Lot-Sizing Problem']","A company uses two machines (Machine 1 and Machine 2) to produce two types of products (liquid fertilizer and solid fertilizer). To produce one unit of liquid fertilizer, it takes 50 minutes on Machine 1 and 30 minutes on Machine 2. To produce one unit of solid fertilizer, it takes 24 minutes on Machine 1 and 33 minutes on Machine 2. Fertilizers must be produced in whole units, and fractional amounts are not allowed. At the beginning of the week, there are 30 units of liquid fertilizer and 90 units of solid fertilizer in inventory. The available processing time for Machine 1 this week is expected to be 40 hours, and for Machine 2 it is expected to be 35 hours. The demand for liquid fertilizer this week is estimated at 75 units, and for solid fertilizer at 95 units. The company's policy is to maximize the total number of units of liquid fertilizer and solid fertilizer in inventory at the end of the week. + +Formulate a model for this problem.",1.0 +"['Production Planning Problem', 'Profit Maximization Problem']","A company produces product A and product B. Each unit of product A sold generates a profit of £30, while each unit of product B sold generates a profit of £10. The company can allocate a maximum of 40 hours per week for production. Producing one unit of product A requires 6 hours, while producing one unit of product B requires 3 hours, and products can only be produced in whole units. Market demand requires that the quantity of product B produced must be at least three times the quantity of product A. The storage space occupied by product A is four times that of product B. The storage space's capacity is such that it can store 4 units of product A when only product A is stored. + +Formulate a model for this problem.",140.0 +['Revenue Management Problem'],"A store wants to clear out 200 shirts and 100 pairs of pants from last season. They decide to introduce two promotional packages, A and B. Package A includes one shirt and two pairs of pants, priced at £30. Package B includes three shirts and one pair of pants, priced at £50. The store does not want to sell fewer than 20 A packages and 10 B packages. How many of each package do they need to sell to maximize the revenue from the promotion? + +Try to establish a model for this problem.",3600.0 +"['Profit Maximization Problem', 'Production Planning Problem']","A company produces two products (A and B), with a profit of £3 and £5 per unit sold, respectively. Each product must be assembled on a specific machine, requiring 12 minutes of assembly time per unit for product A and 25 minutes per unit for product B. The company's estimated effective machine working time per week is only 30 hours (due to maintenance or malfunctions). Technical constraints mean that for every five units of product A produced, at least two units of product B must be produced. + +Try to formulate a model for this problem.",408.0 +"['Transportation, Airline Industry, Resource Allocation', 'Transportation Problem']","A school is preparing a trip for 400 students. The transportation company has 10 buses with 50 seats each and 8 minibuses with 40 seats each, but only 9 drivers are available. The rental cost for a bus is £800, and the rental cost for a minibus is £600. Calculate how many of each type of bus should be used to achieve the lowest cost. + +Try to formulate a model for this problem.",6200.0 +['Production Planning Problem'],"A dairy processing plant uses milk to produce two dairy products, \( A_{1} \) and \( A_{2} \). One barrel of milk can be processed into 3 kg of \( A_{1} \) in 12 hours on Type A equipment or into 4 kg of \( A_{2} \) in 8 hours on Type B equipment. According to market demand, all produced \( A_{1} \) and \( A_{2} \) can be sold. The profit is 24 yuan per kilogram of \( A_{1} \) and 16 yuan per kilogram of \( A_{2} \). The processing plant can get a daily supply of 50 barrels of milk, with a total of 480 hours of labor time available from regular workers each day. The Type A equipment can process up to 100 kg of \( A_{1} \) per day, while the processing capacity of Type B equipment is not limited. Formulate a production plan for the plant to maximize daily profit.",3360.0 +['Blending Problem'],"A company blends two types of crude oil (A and B) to produce two types of gasoline (Type I and Type II). The minimum proportion of crude oil A in gasoline Types I and II is 50% and 60%, respectively. The selling prices are 4800 yuan/t and 5600 yuan/t, respectively. The company has current inventories of 500 t of crude oil A and 1000 t of crude oil B, and they can purchase up to 1500 t of crude oil A from the market. The market price for crude oil A is: 10,000 yuan/t for purchases up to 500 t; 8,000 yuan/t for the portion exceeding 500 t but not exceeding 1000 t; 6,000 yuan/t for the portion exceeding 1000 t. How should the company plan its purchasing and processing of crude oil? Return the maximized profit in yuan.",5000000.0 +"['Capacitated Lot-sizing Problem (CLSP)', 'Lot-Sizing Problem', 'Production Planning Problem']","A beverage factory produces a kind of beverage to meet market demand. According to market forecasts, the sales department of the factory has determined the demand for the beverage for the next 4 weeks. The planning department, based on the actual situation of the factory, has provided the production capacity and production cost for the next 4 weeks, as shown in Table 1. When there is a surplus of beverages after meeting the demand each week, a storage cost of 0.2 thousand yuan per week per thousand boxes of beverages needs to be paid. How should the production plan be arranged to minimize the total cost (the sum of production cost and storage cost) over the four weeks while meeting the weekly market demand? + +Table 1 Beverage Production and Demand Data: + +\begin{tabular}{c|c|c|c} +\hline +Week & Demand/1000 boxes & Production Capacity/1000 boxes & Cost per 1000 boxes/1000 yuan \\ +\hline +1 & 15 & 30 & 5.0 \\ +\hline +2 & 25 & 40 & 5.1 \\ +\hline +3 & 35 & 45 & 5.4 \\ +\hline +4 & 25 & 20 & 5.5 \\ +\hline +Total & 100 & 135 & \\ +\hline +\end{tabular}",528.0 +['Cutting Stock Problem'],"A steel pipe retailer sources raw steel pipes from a steel pipe factory, cuts the pipes according to customer requirements, and sells them. The raw steel pipes obtained from the factory are all 1850 mm in length. A customer now needs 15 pieces of 290 mm, 28 pieces of 315 mm, 21 pieces of 350 mm, and 30 pieces of 455 mm steel pipes. To simplify the production process, it is required that no more than 4 types of cutting patterns are used. The most frequently used cutting pattern incurs an additional cost of 1/10 of the value of a raw steel pipe, the second most frequent incurs an additional cost of 2/10, and so on. Moreover, the number of cuts for each pattern cannot be too many (a single raw steel pipe can produce up to 5 products). Additionally, to minimize waste, the leftover material for each cutting pattern should not exceed 100 mm. How should the material be cut to minimize total cost, and what is the total cost in this case?",21.5 +['Blending Problem'],"A company mixes four types of liquid raw materials with different sulfur contents (denoted as A, B, C, and D, respectively) to produce two products (denoted as \( \mathrm{A} \) and \( \mathrm{B} \)). According to the production process requirements, raw materials A, B, and D must first be mixed in a mixing tank, and then the mixed liquid is further mixed with raw material C to produce \( \mathrm{A} \) and \( \mathrm{B} \). The sulfur contents of raw materials A, B, C, and D are \( 3\%, 1\%, 2\%, 1\% \) respectively, and their purchase prices are 6, 16, 10, 15 (thousand yuan per ton) respectively. The sulfur content of products \( \mathrm{A} \) and \( \mathrm{B} \) must not exceed \( 2.5\% \) and \( 1.5\% \) respectively, and their selling prices are 9, 15 (thousand yuan per ton) respectively. According to market information, there is no limit to the supply of raw materials A, B, and C, but the supply of raw material D is limited to a maximum of 50 tons. The market demand for products \( \mathrm{A} \) and \( \mathrm{B} \) is 100 tons and 200 tons respectively. How should the production be arranged to maximize the total profit?",450.0 +"['Production Planning Problem', 'Profit Maximization Problem']","A company uses steel and aluminum as raw materials to produce two products (A and B). A single unit of product A requires 6 kg of steel, 8 kg of aluminum, 11 hours of labor, and yields a profit of 5000 yuan (excluding worker overtime pay). A single unit of product B requires 12 kg of steel, 20 kg of aluminum, 24 hours of labor, and yields a profit of 11000 yuan (excluding worker overtime pay). Products can only be produced in whole units. The company currently has 200 kg of steel, 300 kg of aluminum, and 300 hours of labor available. If workers need to work overtime, the overtime pay is 100 yuan per hour. Please develop a production plan to maximize the company's overall profit taking into account worker overtime.",165900.0 +['Knapsack'],"An electronic system is composed of 3 types of components. The system operates normally if all three components function properly. By installing one or more spare parts for any of the components, the reliability of the components can be improved. The system's operational reliability is the product of the reliabilities of each component, and the reliability of each component is a function of the number of spare parts installed. The first half of the table below shows the function relationship between the number of spare parts and the reliability of a specific component. The prices and weights of the 3 types of components are shown in rows 8 to 9 of the table. Given that the total budget for all spare parts is limited to 150 yuan, and the weight limit is 20 kg, how should spare parts be installed to maximize the system's operational reliability? + +\begin{table}[h] +\centering +\begin{tabular}{|c|c|c|c|} +\hline +\textbf{Component Number} & \textbf{1} & \textbf{2} & \textbf{3} \\ \hline +\textbf{Number of Spares} & & & \\ \hline +0 & 0.5 & 0.6 & 0.7 \\ \hline +1 & 0.6 & 0.75 & 0.9 \\ \hline +2 & 0.7 & 0.95 & 1.0 \\ \hline +3 & 0.8 & 1.0 & 1.0 \\ \hline +4 & 0.9 & 1.0 & 1.0 \\ \hline +5 & 1.0 & 1.0 & 1.0 \\ \hline +\textbf{Unit Price (yuan)} & 20 & 30 & 40 \\ \hline +\textbf{Unit Weight (kg)} & 2 & 4 & 6 \\ \hline +\end{tabular} +\caption{Spare Component Data Table} +\end{table}",0.6075 +['Network Optimization'],"In network communication services, bandwidth plays an important role. Below is a bandwidth communication table between several communication nodes, showing the bandwidth between any two nodes. If two nodes cannot be directly connected, the corresponding bandwidth is $0$. It is required to establish a link between node $A$ and node $E$ that must pass through service node $C$ (without loops). The bandwidth of this link is defined as the minimum bandwidth value on the link. Please propose a reasonable link arrangement to maximize the bandwidth of this link and find out the maximum bandwidth. + +\begin{table}[h] + \centering + \begin{tabular}{|c|c|c|c|c|c|} + \hline + & A & B & C & D & E \\ + \hline + A & 0 & 90 & 85 & 0 & 65 \\ + \hline + B & 95 & 0 & 70 & 65 & 34 \\ + \hline + C & 60 & 0 & 0 & 88 & 80 \\ + \hline + D & 67 & 30 & 25 & 0 & 84 \\ + \hline + E & 0 & 51 & 0 & 56 & 0 \\ + \hline + \end{tabular} +\end{table}",84.0 diff --git a/optimind/data/optimind_cleaned_classified_mamo_complex.csv b/optimind/data/optimind_cleaned_classified_mamo_complex.csv new file mode 100644 index 0000000..893c894 --- /dev/null +++ b/optimind/data/optimind_cleaned_classified_mamo_complex.csv @@ -0,0 +1,3640 @@ +problem_class,question,answer +"[""Diet Problem""]","Imagine you're a dietitian, tasked with planning a nutritious and balanced meal for an athlete. The athlete needs to maintain a high protein, moderate carb diet, with enough calories to sustain their intense training regimen. The goal is to meet these dietary requirements at the lowest cost possible. You have five different food items to choose from: Steak, Rice, Chicken, Iceberg Lettuce, and Almonds. Each of these foods contains varying amounts of protein, carbohydrates, and calories, and each comes with a specific cost. + +Here's a detailed breakdown of the nutritional value of each food item and its cost: + +- Steak: Offers a whopping 20 grams of protein, 12 grams of carbohydrates, and 167 calories for $7. +- Rice: Provides 6 grams of protein, an impressive 29 grams of carbohydrates, and a hefty 275 calories for $8. +- Chicken: Packs 17 grams of protein, 26 grams of carbohydrates, and 112 calories for only $2. +- Iceberg Lettuce: Contains only 1 gram of protein, 8 grams of carbohydrates, and 148 calories for $8. +- Almonds: Provide 10 grams of protein, 2 grams of carbohydrates, and a high 286 calories for just $2. + +The athlete's daily dietary requirements are as follows: at least 83 grams of protein, 128 grams of carbohydrates, and 1686 calories. + +Your challenge is to select a combination of these foods that meets the athletes' nutritional needs, while keeping the costs as low as possible. So, what is the lowest possible cost to meet these dietary requirements? You could only assign integral portions of each type of food.",[18.0] +['Diet Problem'],"Imagine you are on a mission to take control of your diet, meeting all your nutritional needs without breaking the bank. As a savvy shopper, you've narrowed down your food options to five different items, each providing different amounts of nutrients at varying costs. + +Here is a summary of what each food option offers and their respective costs: + +- Chicken: It provides 18 grams of protein, 6 grams of carbohydrates, and 142 calories at a cost of $4. +- Rice: This food item gives you 12 grams of protein, 28 grams of carbohydrates, and 88 calories for $8. +- Broccoli: For $7, it offers 13 grams of protein, 19 grams of carbohydrates, and 137 calories. +- Tofu: At a cost of $10, you get 5 grams of protein, 16 grams of carbohydrates, and 141 calories. +- Beans: This item provides an impressive 10 grams of protein, 20 grams of carbohydrates, and 150 calories for only $1. + +Your daily nutritional targets are to consume at least 56 grams of protein, 194 grams of carbohydrates, and 2012 calories. The key challenge here is to meet these dietary requirements in the most cost-effective way possible, using any combination of the food items listed above. + +Now, the question is, what is the minimal cost to meet your daily nutritional requirements using these food items? You could only assign integral portions of each type of food.",[14.0] +"[""Network Optimization""]","In the heart of a bustling metropolis, an expansive network of waterways and canals forms the lifeline for its residents, connecting 9 critical distribution centers that manage the flow of water from the city's reservoirs to its farthest suburbs. This intricate system, designed to ensure the smooth and efficient delivery of water, consists of channels with varying capacities, reflecting the maximum volumes of water (in thousands of gallons per hour) that can traverse these connections without causing overflow or damage. + +Here's a detailed look at the capacities of each channel in this vital network: + +- From Distribution Center 0 (Source), the water can be directed to Centers 1 through 8 with capacities of 9, 7, 18, 0, 0, 7, 0, and 2 thousand gallons per hour, respectively. +- Center 1 has the capability to forward water to other centers, including sending 15, 11, 5, 20, 15, 18, 9, and 15 thousand gallons per hour to Centers 0 through 8 (excluding itself). +- Similarly, the capacities from Center 2 to others are marked by flows of 3, 8, 0, 10, 16, 18, 8, 0, and 20 thousand gallons per hour to Centers 0 through 8, respectively. +- The flow continues from Center 3 with capacities of 13, 3, 7, 0, 20, 13, 16, 6, and 20 thousand gallons per hour to the respective centers. +- From Center 4, water can be distributed with capacities of 7, 20, 2, 4, 0, 9, 3, 14, and 2 thousand gallons per hour. +- Center 5 allows for a distribution flow of 17, 10, 5, 1, 4, 0, 4, 3, and 20 thousand gallons per hour towards the respective centers. +- The network from Center 6 can support flows of 0, 20, 10, 10, 15, 5, 0, 1, and 10 thousand gallons per hour. +- From Center 7, water can be directed with capacities of 10, 3, 9, 4, 3, 13, 18, 0, and 20 thousand gallons per hour. +- Finally, Center 8 can send back or redistribute water with capacities of 2, 12, 10, 5, 15, 14, 0, 10, and 0 thousand gallons per hour to the respective centers. + +In this network, a capacity of ""0"" indicates the absence of a direct channel between two centers, necessitating the redirection of flow through other centers to ensure connectivity. + +Given this complex and dynamic system, the city's water management team faces the challenging task of determining the maximum volume of water that can be efficiently distributed from Distribution Center 0 (the source) to all other centers, culminating at the final node (Center 8) which represents the city's outskirts. This calculation is crucial for optimizing the water supply and meeting the growing needs of the city's population. + +What is the maximum amount of water (in thousands of gallons per hour) that can be distributed from the source to the outskirts without exceeding the specified capacities of the channels in the network?",[43.0] +['Capacitated Facility Location Problem'],"The nationwide retail chain, MarketHub, is in the process of restructuring its distribution network to efficiently cater to the demands of its five retail stores spread across different regions. The strategic decision involves selecting from eight potential distribution centers, each with its unique operational costs and capacities. The goal is to ensure that each retail store's demand is met with the lowest possible total cost, a comprehensive figure that includes the opening costs of the distribution centers and the transportation costs of the goods. + +Given Data: +- Number of Potential Distribution Centers: 8 +- Number of Retail Stores Needing Supply: 5 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 +- Center 6: $120,000 +- Center 7: $182,000 +- Center 8: $186,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $3 to Store 1, $3 to Store 2, $5 to Store 3, $4 to Store 4, $3 to Store 5 +- From Center 2: $5 to Store 1, $2 to Store 2, $4 to Store 3, $2 to Store 4, $4 to Store 5 +- From Center 3: $5 to Store 1, $1 to Store 2, $4 to Store 3, $2 to Store 4, $5 to Store 5 +- From Center 4: $4 to Store 1, $1 to Store 2, $1 to Store 3, $3 to Store 4, $3 to Store 5 +- From Center 5: $2 to Store 1, $4 to Store 2, $4 to Store 3, $3 to Store 4, $4 to Store 5 +- From Center 6: $4 to Store 1, $1 to Store 2, $3 to Store 3, $5 to Store 4, $3 to Store 5 +- From Center 7: $5 to Store 1, $1 to Store 2, $2 to Store 3, $4 to Store 4, $1 to Store 5 +- From Center 8: $4 to Store 1, $2 to Store 2, $2 to Store 3, $1 to Store 4, $2 to Store 5 + +Demand of Each Retail Store (in units): +- Store 1: 964 units +- Store 2: 739 units +- Store 3: 491 units +- Store 4: 766 units +- Store 5: 854 units + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1308 units +- Center 2: 1575 units +- Center 3: 834 units +- Center 4: 1005 units +- Center 5: 1904 units +- Center 6: 1825 units +- Center 7: 1821 units +- Center 8: 1365 units + +MarketHub aims to optimize the balance between maintaining an efficient supply chain and minimizing costs. The challenge is to select the right combination of distribution centers to open and manage the flow of goods to the stores in a way that meets all the demand while keeping both opening and transportation costs as low as possible. + +Question: +What is the optimal total cost for MarketHub to open distribution centers and transport goods to fully satisfy the demands of all five retail stores, including both the opening costs of the centers and the transportation costs of supplying the stores?",[394198.0] +['TravelingSalesman'],"Imagine a scenario involving a salesperson who needs to visit five distinct cities to promote their products. These cities are uniquely numbered from 1 to 5. The objective for the salesperson is to undertake a journey starting from any one of these cities and cover all others once before returning to the starting point, all while incurring the least travel cost possible. The travel costs vary depending on the route between each city pair. + +Here's a breakdown of the travel costs between the cities: +- The cost to travel from City 1 to City 2 is 73 units, to City 3 is 39 units, to City 4 is 79 units, and to City 5 is 38 units. +- From City 2, it costs 73 units to reach City 1, 30 units to go to City 3, 46 units to visit City 4, and 29 units to get to City 5. +- Traveling from City 3, the costs are 39 units to City 1, 30 units to City 2, 88 units to City 4, and 57 units to City 5. +- From City 4, it costs 79 units to go to City 1, 46 units to City 2, 88 units to City 3, and 59 units to City 5. +- Lastly, from City 5, the travel costs are 38 units to City 1, 29 units to City 2, 57 units to City 3, and 59 units to City 4. + +Given these details, what is the minimum total travel cost for the salesperson to visit each city exactly once and return to the starting city?",[212.0] +['TravelingSalesman'],"Consider a scenario where a tour guide is planning a bus tour across five cities, named E, F, G, H, and I. The tour must start and end in the same city, and each city should be visited exactly once. The objective is to minimize the total cost of the tour, which could be influenced by factors such as distance, tolls, and fuel expenses. + +Here are the travel costs between the cities: + +From City E, it costs 37 units to travel to F, 72 units to G, 66 units to H, and 33 units to I. +Travelling from City F, the costs are 37 units to E, 26 units to G, 56 units to H, and a whopping 100 units to I. +If the journey starts from City G, it takes 72 units to reach E, 26 units to F, 49 units to H, and 32 units to I. +From City H, it costs 66 units to get to E, 56 units to F, 49 units to G, and a mere 13 units to I. +Lastly, from City I, the travel costs are 33 units to E, 100 units to F, 32 units to G, and 13 units to H. + +Keeping these costs in mind, what is the minimum total travel cost for the tour guide to take the bus to each city exactly once and return to the starting city?",[158.0] +['Diet Problem'],"Imagine you're a dietitian, tasked with crafting a healthy and balanced meal plan for a client on a budget. The tricky part is that you must choose from a list of nine food items, each with their own cost and nutritional values for protein, carbohydrates, and calories. + +Here's the rundown of what each food item provides and its cost: + +- Chicken Breast: Delivers 17 grams of protein, 28 grams of carbohydrates, and 136 calories for $3. +- Tofu: Yields 15 grams of protein, 11 grams of carbohydrates, and 198 calories for $2. +- Brown Rice: Offers just 1 gram of protein, 11 grams of carbohydrates, and 93 calories for $2. +- Salmon: Gives you 11 grams of protein, 3 grams of carbohydrates, and a hefty 269 calories for $6. +- Greek Yogurt: Provides 15 grams of protein, 27 grams of carbohydrates, and 121 calories for $6. +- Steak: Comes with 12 grams of protein, 3 grams of carbohydrates, and 80 calories for a steeper price of $10. +- Avocado: Provides 4 grams of protein, 8 grams of carbohydrates, and a whopping 300 calories for $7. +- Eggs: Offer 16 grams of protein, 25 grams of carbohydrates, and 99 calories for $5. +- Quinoa: Comes with 11 grams of protein, 20 grams of carbohydrates, and 87 calories for just $2. + +Your goal is to make sure your client gets at least 72 grams of protein, 158 grams of carbohydrates, and 1896 calories from the combination of foods you choose. But you also need to keep costs to a minimum. What is the least amount of money you would need to spend to meet all of these nutritional targets from these nine food options? You could only assign integral portions of each type of food.",[24.0] +"[""TravelingSalesman""]","Imagine a salesperson tasked with visiting five distinct cities to expand their business network. These cities are uniquely identified as City 1, City 2, City 3, City 4, and City 5. The salesperson is free to choose any of these cities as the starting point. However, they must travel to each city exactly once, concluding their journey at the starting city. The primary objective is to minimize the total travel cost associated with the entire trip, which includes expenses such as transportation costs between the cities. + +Here is a detailed breakdown of the travel costs between the cities: +- The cost to travel from City 1 to City 2 is 87 units, to City 3 is 41 units, to City 4 is 55 units, and to City 5 is 57 units. +- From City 2, it costs 87 units to reach City 1, 80 units to travel to City 3, 33 units to City 4, and 40 units to City 5. +- Traveling from City 3, the costs are 41 units to City 1, 80 units to City 2, 90 units to City 4, and 42 units to City 5. +- From City 4, the journey costs 55 units to City 1, 33 units to City 2, 90 units to City 3, and 53 units to City 5. +- Lastly, from City 5, it costs 57 units to travel to City 1, 40 units to City 2, 42 units to City 3, and 53 units to City 4. + +Given this cost matrix, what is the minimum total travel cost for the salesperson to visit each city exactly once and return to the starting city?",[211.0] +"[""Profit Maximization Problem""]","Star Oil Company is considering five different investment opportunities. The cash outflows and net present values (NPVs) (in millions of dollars) for each opportunity are given in Table 7. Specifically, Investment 1 has a time 0 cash outflow of $11 million and a time 1 cash outflow of $3 million, with an NPV of $13 million. Investment 2 requires a time 0 outflow of $53 million and a time 1 outflow of $6 million, yielding an NPV of $16 million. Investment 3 has smaller outflows of $5 million at both time 0 and time 1 and also an NPV of $16 million. Investment 4 asks for a time 0 outflow of $5 million and a minimal $1 million at time 1, with an NPV of $14 million. Lastly, Investment 5 requires a $29 million outflow at time 0 and a significant $34 million at time 1, resulting in the highest NPV of $39 million. Star Oil has $40 million available for investment at time 0, and it estimates that $20 million will be available for investment at time 1. Star Oil may purchase any fraction of each investment, in which case, the cash outflows and NPV are adjusted proportionally. For example, if Star Oil purchases one-fifth of investment 3, then a cash outflow of (1/5)*5=1 million dollars or $1 million would be required at time 0, and a cash outflow of (1/5)*5=1 million dollars or $1 million would be required at time 1. The one-fifth share of investment 3 would yield an NPV of (1/5)*16=3.2 million dollars or $3.2 million. Star Oil wants to maximize the NPV that can be obtained by investing in investments 1-5. The goal is to formulate an LP that will help achieve this objective. Assume that any funds left over at time 0 cannot be used at time 1. Find the maximum NPV in million dollars.",[57.4490172] +['Capacitated Facility Location Problem'],"In the bustling marketplace of today, GlobalSupply Inc. is at a critical juncture in strategizing the logistics of its supply chain. The company is faced with the challenge of efficiently distributing products to eight different retail stores. Each store has its unique demand for products, and GlobalSupply Inc. must decide on the best locations out of six possible distribution centers to fulfill these needs. The goal is to minimize the overall cost, which includes both the opening expenses of these distribution centers and the transportation costs associated with moving goods to the retail stores. + +Given Data: +- Number of Potential Distribution Centers: 6 +- Number of Retail Stores Needing Supply: 8 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 +- Center 6: $120,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $2 to Store 1, $3 to Store 2, and so on, up to $5 to Store 8 +- From Center 2: $2 to Store 1, $4 to Store 2, and so on, up to $2 to Store 8 +- From Center 3: $5 to Store 1, $4 to Store 2, and so on, up to $4 to Store 8 +- From Center 4: $4 to Store 1, $3 to Store 2, and so on, up to $3 to Store 8 +- From Center 5: $5 to Store 1, $1 to Store 2, and so on, up to $2 to Store 8 +- From Center 6: $1 to Store 1, $2 to Store 2, and so on, up to $4 to Store 8 + +Demand of Each Retail Store (in units): +- Store 1: 908 +- Store 2: 434 +- Store 3: 605 +- Store 4: 480 +- Store 5: 961 +- Store 6: 787 +- Store 7: 401 +- Store 8: 789 + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1365 +- Center 2: 1929 +- Center 3: 1502 +- Center 4: 1201 +- Center 5: 1529 +- Center 6: 961 + +The challenge for GlobalSupply Inc. is to find the most cost-effective way to open the necessary distribution centers and plan the logistics so that each retail store's demand is met. The company must carefully consider the opening costs of these centers, the capacities they offer, and the varying transportation costs to each retail store to devise a plan that minimizes total expenses. + +Question: +What is the optimal total cost for GlobalSupply Inc. to open distribution centers and transport goods to meet all retail stores' demands, including both the opening costs of the centers and the transportation costs of supplying the stores?",[587428.0] +['Capacitated Facility Location Problem'],"BestWay Logistics is in the process of determining the best strategy to distribute products from its warehouses to four retail stores. The company has six potential distribution centers to choose from, each with its own costs associated with opening the center and transporting goods to the stores. Each center also has a specified maximum capacity it can handle. The aim is to satisfy the demand at each store at the lowest possible total cost, while taking into account the cost of opening the distribution centers and the transportation costs of the goods. + +The details of the scenario are as follows: + +Number of Potential Distribution Centers: 6 +Number of Retail Stores Needing Supply: 4 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: 151,000 +- Center 2: 192,000 +- Center 3: 114,000 +- Center 4: 171,000 +- Center 5: 160,000 +- Center 6: 120,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $2 to Store 1, $3 to Store 2, $3 to Store 3, $3 to Store 4 +- From Center 2: $5 to Store 1, $4 to Store 2, $3 to Store 3, $5 to Store 4 +- From Center 3: $2 to Store 1, $4 to Store 2, $2 to Store 3, $4 to Store 4 +- From Center 4: $5 to Store 1, $1 to Store 2, $4 to Store 3, $2 to Store 4 +- From Center 5: $5 to Store 1, $4 to Store 2, $1 to Store 3, $1 to Store 4 +- From Center 6: $3 to Store 1, $3 to Store 2, $2 to Store 3, $4 to Store 4 + +Demand of Each Retail Store (in units): +- Store 1: 589 +- Store 2: 962 +- Store 3: 966 +- Store 4: 643 + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1631 +- Center 2: 1954 +- Center 3: 1446 +- Center 4: 820 +- Center 5: 1640 +- Center 6: 966 + +Question: +What is the optimal total cost for BestWay Logistics to open the necessary distribution centers and transport goods to meet the demand of all retail stores? This cost should include both the opening costs of the centers and the transportation costs of supplying the stores. Please provide only the optimal value.",[316282.0] +['Diet Problem'],"Imagine you're a college student living off-campus with a tight budget, but you're committed to maintaining a healthy diet. You have identified a set of nine different food items available in your local supermarket. Each of these items provides specific amounts of protein, carbohydrates, and calories and has its own price. + +Here is the breakdown of what each food item provides along with its cost: + +1. Chicken (Cost: $10): Gives you 7 grams of protein, 17 grams of carbohydrates, and 95 calories. +2. Tofu (Cost: $2): Provides 12 grams of protein, 12 grams of carbohydrates, and 233 calories. +3. Brown Rice (Cost: $7): Contains 14 grams of protein, 27 grams of carbohydrates, and 134 calories. +4. Fish (Cost: $8): Offers 9 grams of protein, 6 grams of carbohydrates, and 109 calories. +5. Lentils (Cost: $4): Deliver 11 grams of protein, 19 grams of carbohydrates, and 202 calories. +6. Quinoa (Cost: $9): Packs 10 grams of protein, 23 grams of carbohydrates, and 148 calories. +7. Eggs (Cost: $2): Provides 10 grams of protein, 6 grams of carbohydrates, and 184 calories. +8. Beef (Cost: $9): Contains 9 grams of protein, 16 grams of carbohydrates, and 136 calories. +9. Beans (Cost: $9): Offers 8 grams of protein, 25 grams of carbohydrates, and 229 calories. + +Your daily dietary requirements are that you need at least 71 grams of protein, 107 grams of carbohydrates, and 1623 calories. Your challenge is to meet these nutritional targets by choosing from these nine food items and spending as little money as possible. How much is the minimum cost you need to spend to meet your nutritional needs? Remember, your response should only provide the optimal value. You could only assign integral portions of each type of food.",[18.0] +['Diet Problem'],"Imagine you are a fitness enthusiast who takes your daily nutritional needs very seriously. Your dietitian has provided you with specific benchmarks to achieve every day: 91 grams of protein, 136 grams of carbohydrates, and 1563 calories. + +You have a variety of food items available to you, each with its own cost and nutritional content. Let's take a look at what each of these food items offers: + +- Chicken breast: For $7, it provides 12g of protein, 5g of carbs, and 231 calories. +- Tofu: Also $7, it delivers 10g of protein, 1g of carbs, and 210 calories. +- Spinach: This cost-effective option at just $1 provides 11g of protein, 30g of carbs, and 295 calories. +- Beef steak: Available at $9, it contributes 8g of protein, 26g of carbs, and 211 calories. +- Eggs: For $6, eggs add 16g of protein, 9g of carbs, and 232 calories to your daily intake. +- Brown rice: This $7 meal component provides 4g of protein, 25g of carbs, and 146 calories. +- Bananas: At just $2, they contribute 4g of protein, 16g of carbs, and 255 calories. +- Quinoa: For $6, quinoa provides 4g of protein, 20g of carbs, and 205 calories. + +Now here is your challenge: Using these food options, how can you meet your daily dietary requirements while keeping the cost as low as possible? What is the minimum cost you need to spend to meet your daily nutritional needs? Keep in mind that your response should be a single answer question, asking for only the optimal value, under the scenario of food selection. You could only assign integral portions of each type of food.",[9.0] +['Diet Problem'],"Imagine you're meticulously planning your weekly meal prep and want to ensure that your diet is nutritionally balanced while also being cost-effective. You have a choice between five different food items, each with its own nutritional profile and cost. Your challenge is to choose the right combination of these foods to meet your nutritional needs without overspending. + +Here's what each food item offers along with its cost: + +- Food_1: Provides 10 grams of protein, 22 grams of carbohydrates, and 80 calories for $2. +- Food_2: Offers 3 grams of protein, 4 grams of carbohydrates, and 189 calories for $8. +- Food_3: Contains 17 grams of protein, 10 grams of carbohydrates, and 153 calories for $4. +- Food_4: Delivers 6 grams of protein, 13 grams of carbohydrates, and 232 calories for $1. +- Food_5: Packs 20 grams of protein, 5 grams of carbohydrates, and 204 calories for $8. + +Your dietary goals are to consume at least 74 grams of protein, 102 grams of carbohydrates, and 1564 calories in total. What is the minimum cost required to meet these nutritional targets using any combination of the available food items? You could only assign integral portions of each type of food.",[13.0] +['Diet Problem'],"Imagine you're a dietitian crafting a meal plan for someone with specific nutritional goals. The challenge is not just to meet these nutritional needs but to do so in the most cost-effective way possible. The individual requires a precise amount of protein, carbohydrates, and calories each day, but the budget is tight, making it crucial to minimize the cost of the meal plan. + +You have a list of eight food items to choose from, each with distinct nutritional content and associated cost: + +- Food_1: A modest source of protein and carbs with 1 gram of protein, 4 grams of carbohydrates, and 93 calories at a cost of $6. +- Food_2: Rich in protein with 18 grams of protein, 2 grams of carbohydrates, and 219 calories, priced at $4. +- Food_3: A balanced choice providing 19 grams of protein, 26 grams of carbs, and 155 calories for $4. +- Food_4: High in protein with 19 grams of protein, 10 grams of carbs, and 105 calories, costing $3. +- Food_5: Carb-heavy with 3 grams of protein, 24 grams of carbohydrates, and 255 calories at just $2. +- Food_6: A good mix with 8 grams of protein, 26 grams of carbs, and 182 calories for $7. +- Food_7: Impressive protein and carb content with 15 grams of protein, 27 grams of carbohydrates, and 173 calories, all for $3. +- Food_8: Carb-rich with 3 grams of protein, 30 grams of carbohydrates, and 191 calories, also at $3. + +The daily dietary requirements for the individual are: + +- Protein: At least 98 grams +- Carbohydrates: At least 113 grams +- Calories: At least 1768 + +With these food options and nutritional needs in mind, what is the minimum cost required to meet the daily protein, carbohydrate, and calorie requirements? You could only assign integral portions of each type of food.",[24.0] +['Diet Problem'],"Imagine you are on a mission to ensure you're eating healthily but also on a tight budget. You have five different food items to choose from: Chicken, Rice, Beans, Broccoli, and Fruit. Each of these foods has a specific amount of protein, carbohydrates, and calories, and each comes with its own cost. + +Here's an overview of what each food provides, along with its cost: + +- Chicken: Gives you 12 grams of protein, 2 grams of carbohydrates, and 186 calories for $6. +- Rice: Provides 12 grams of protein, 24 grams of carbohydrates, and 250 calories for $6. +- Beans: Offers 19 grams of protein, 18 grams of carbohydrates, and 134 calories for $6. +- Broccoli: Supplies 19 grams of protein, 7 grams of carbohydrates, and 84 calories for $7. +- Fruit: Gives you 4 grams of protein, a high 27 grams of carbohydrates, and 104 calories for $9. + +Your goal is to make sure you get at least 97 grams of protein, 163 grams of carbohydrates, and 2217 calories from the combination of these foods that you choose. Remember, you’re trying to figure out the least expensive way to meet these nutritional goals with these five food options. + +So, what is the minimal cost you need to spend to satisfy all these dietary requirements? You could only assign integral portions of each type of food.",[54.0] +['Transportation Problem'],"Imagine you're the logistics coordinator in charge of distributing medical supplies among four different hospitals during an unexpected health crisis. Each hospital starts with a specific number of supplies but has varying needs based on the urgency and scale of cases they are handling. + +Here's what each hospital has and what they need: + +- Hospital 1 has 480 units of supplies but needs only 61 units. +- Hospital 2 has 492 units but needs 313 units. +- Hospital 3 has 74 units but needs 93 units. +- Hospital 4 has 317 units but needs 199 units. + +Due to the urgent situation, you can transfer supplies between hospitals, but each transfer incurs a cost, which varies depending on the hospitals involved. Here’s a breakdown of the cost to transfer supplies from one hospital to another: + +- Moving supplies from/to Hospital 1: To Hospital 2 costs 8, to Hospital 3 costs 42, to Hospital 4 costs 27. +- Moving supplies from/to Hospital 2: To Hospital 1 costs 17, to Hospital 3 costs 13, to Hospital 4 costs 30. +- Moving supplies from/to Hospital 3: To Hospital 1 costs 4, to Hospital 2 costs 14, to Hospital 4 costs 16. +- Moving supplies from/to Hospital 4: To Hospital 1 costs 33, to Hospital 2 costs 9, to Hospital 3 costs 42. + +Your objective is to ensure each hospital has the supplies it needs while minimizing the total cost of transfers. What would be the minimum cost to achieve this?",[247.0] +['TravelingSalesman'],"Imagine a delivery driver who is tasked with delivering packages to six different locations, each uniquely identified as Location 1, 2, 3, 4, 5, and 6. The driver can kick off his route from any location, but he must visit each one exactly once before returning to the starting point. His goal is to find the route that minimizes the total delivery cost. This cost could be based on a combination of factors, such as distance, fuel consumption, or traffic congestion. + +Here's a summary of the delivery costs between the locations: +From Location 1, it costs 48 units to get to Location 2, 23 units to Location 3, 70 units to Location 4, 14 units to Location 5, and 20 units to Location 6. +If the driver starts from Location 2, it would cost 48 units to reach Location 1, 79 units to reach Location 3, 37 units to reach Location 4, 63 units to reach Location 5, and 96 units to reach Location 6. +From Location 3, the costs are 23 units to reach Location 1, 79 units to reach Location 2, 40 units to reach Location 4, 66 units to reach Location 5, and 68 units to reach Location 6. +Starting at Location 4, it would cost the driver 70 units to get to Location 1, 37 units to get to Location 2, 40 units to get to Location 3, 70 units to get to Location 5, and 28 units to get to Location 6. +From Location 5, the costs are 14 units to get to Location 1, 63 units to get to Location 2, 66 units to get to Location 3, 70 units to get to Location 4, and 73 units to get to Location 6. +And finally, from Location 6, it would cost 20 units to drive to Location 1, 96 units to drive to Location 2, 68 units to drive to Location 3, 28 units to drive to Location 4, and 73 units to drive to Location 5. + +The question is, what is the minimum total delivery cost for the driver to deliver packages to each location exactly once and return to the starting location?",[233.0] +['TravelingSalesman'],"Imagine a scenario where a salesperson is tasked with visiting four distinct cities to explore new market opportunities. These cities are uniquely identified as City 1, City 2, City 3, and City 4. The task for the salesperson is to plan a route that starts from any of these cities, traverses each city exactly once, and concludes by returning to the starting city. The overarching aim is to achieve this while incurring the minimum possible travel cost. + +Here’s a detailed breakdown of the travel costs between the cities: +- The cost to travel from City 1 to City 2 is 17 units, to City 3 is 25 units, and to City 4 is 31 units. +- From City 2, it costs 17 units to reach City 1, 93 units to get to City 3, and 45 units to City 4. +- Traveling from City 3, the expenses are 25 units to City 1, 93 units to City 2, and 95 units to City 4. +- Lastly, from City 4, the journey costs 31 units to City 1, 45 units to City 2, and 95 units to City 3. + +What is the minimum total travel cost for the salesperson to visit each city exactly once and return to the starting city?",[182.0] +['TravelingSalesman'],"In the scenario of the Travelling Salesman Problem (TSP), we consider four distinct cities labeled as City 1, City 2, City 3, and City 4. A salesperson must visit each of these cities starting from any one of them, travel to each of the other cities exactly once, and then return to the starting city. The primary objective for the salesperson is to minimize the total travel cost during this circuit. + +Here's a detailed look at the travel costs between each pair of cities: +- The cost to travel from City 1 to City 2 is 76 units, to City 3 is 17 units, and to City 4 is 24 units. +- From City 2, the travel costs are 76 units to City 1, 84 units to City 3, and just 11 units to City 4. +- For journeys from City 3, it costs 17 units to reach City 1, a steep 84 units to get to City 2, and 90 units to go to City 4. +- Lastly, from City 4, the costs involved are 24 units to City 1, 11 units to City 2, and 90 units to City 3. + +Given these conditions, what is the minimum total travel cost for the salesperson to complete their route of visiting each city exactly once and returning to the starting point?",[136.0] +['Network Optimization'],"Welcome to a challenge set in the bustling realm of digital communication within a fictional city, where data flows through a network of 9 strategic relay stations, connecting a central data source to a primary data sink. Each relay station serves as a node in this intricate web, tasked with handling the transmission of information packets. The objective is to optimize this network for the highest possible data throughput from the source to the sink, ensuring the city's communication infrastructure operates at peak efficiency. + +Here's a detailed layout of the network's capacity, illustrating the maximum number of data packets per hour that can be transmitted from one station to another: + +- From Station 0 (Source): Has the capacity to send data to Station 1 (14 packets), Station 2 (20 packets), Station 3 (7 packets), Station 4 (9 packets), Station 5 (1 packet), Station 6 (4 packets), Station 7 (8 packets), and Station 8 (19 packets). +- From Station 1: Can transmit data to Station 0 (3 packets), Station 2 (15 packets), Station 3 (12 packets), Station 4 (17 packets), Station 6 (15 packets), Station 7 (3 packets), and Station 8 (3 packets). +- From Station 2: Can send data to Station 0 (11 packets), Station 1 (8 packets), Station 3 (1 packet), Station 4 (19 packets), Station 5 (16 packets), Station 6 (5 packets), Station 7 (1 packet), and Station 8 (2 packets). +- From Station 3: Has the ability to route data to Station 0 (13 packets), Station 1 (13 packets), Station 2 (2 packets), Station 4 (16 packets), Station 5 (20 packets), Station 6 (16 packets), Station 7 (17 packets), and Station 8 (3 packets). +- From Station 4: Can forward data to Station 0 (6 packets), Station 1 (15 packets), Station 2 (7 packets), Station 3 (3 packets), Station 5 (18 packets), Station 6 (1 packet), Station 7 (3 packets), and Station 8 (9 packets). +- From Station 5: Capable of sending data to Station 0 (15 packets), Station 1 (20 packets), Station 2 (8 packets), Station 3 (4 packets), Station 4 (6 packets), Station 6 (3 packets), Station 7 (1 packet), and Station 8 (4 packets). +- From Station 6: Can transmit data to Station 0 (5 packets), Station 1 (15 packets), Station 2 (19 packets), Station 3 (4 packets), Station 4 (15 packets), Station 5 (12 packets), Station 7 (3 packets), and Station 8 (8 packets). +- From Station 7: Has the capacity to route data to Station 0 (7 packets), Station 1 (14 packets), Station 2 (13 packets), Station 3 (12 packets), Station 4 (9 packets), Station 5 (10 packets), Station 6 (3 packets), and Station 8 (4 packets). +- From Station 8 (Destination): Can receive data back from Station 0 (3 packets), Station 1 (12 packets), Station 2 (6 packets), Station 3 (6 packets), Station 4 (1 packet), Station 5 (11 packets), Station 6 (6 packets), and Station 7 (1 packet). + +In this scenario, a ""0"" capacity means there is no direct data transmission link between those two particular stations. + +Given this network configuration, your task is to calculate the maximum number of data packets that can be optimally transmitted from Station 0 (the source) to Station 8 (the destination) per hour, without exceeding the specified transmission capacities between stations. What is the maximum data throughput the city's communication network can achieve in packets per hour?",[52.0] +"[""Network Optimization""]","Imagine a vibrant metropolitan area planning to upgrade its electrical grid to ensure a stable power supply across six crucial distribution centers, from the primary power plant all the way to the main distribution hub. This network of centers is intricately connected by high-capacity cables, each capable of transmitting a certain amount of electricity measured in megawatts (MW). + +Here's the layout of the electrical transmission capabilities between each center, detailing the maximum power flow in megawatts: + +- From the Power Plant (Station 0): Can transmit electricity to Station 1 (4 MW), Station 2 (12 MW), Station 3 (19 MW), Station 4 (13 MW), and Station 5 (11 MW). +- From Station 1: Can transmit to the Power Plant (20 MW), Station 2 (13 MW), Station 3 (9 MW), and Station 4 (10 MW). +- From Station 2: Capable of sending back to the Power Plant (14 MW), to Station 1 (8 MW), Station 3 (5 MW), Station 4 (13 MW), and Station 5 (11 MW). +- From Station 3: Has connections back to the Power Plant (5 MW), to Station 1 (15 MW), Station 2 (18 MW), Station 4 (6 MW), and Station 5 (7 MW). +- From Station 4: Can send power back to the Power Plant (8 MW), to Station 1 (17 MW), Station 2 (20 MW), Station 3 (14 MW), and Station 5 (9 MW). + +In this proposed network, a capacity of ""0"" means there's no direct electrical connection between those two centers. + +The challenge here is to compute the maximum amount of electricity that can be efficiently transmitted from the Power Plant (Station 0) to the Main Distribution Hub (Station 5), ensuring that the transmission through each cable does not exceed its carrying capacity. + +What is the maximum amount of electricity in megawatts (MW) that can be transmitted from the source to the destination?",[38.0] +['Network Optimization'],"Welcome to the challenge of optimizing the network of a futuristic transportation system that interconnects 7 strategic hubs across a thriving metropolis. These hubs, ranging from the central dispatch center to the main distribution hub, are connected by a network of high-speed transit lanes, each with a distinct capacity limitation for vehicles per hour. + +Here's a breakdown of the transit system's capabilities, indicating the maximum number of vehicles that can move from one hub to another within an hour: + +- From Hub 0 (Central Dispatch): Vehicles can move to Hub 1 (15 vehicles), Hub 2 (2 vehicles), Hub 3 (16 vehicles), Hub 4 (3 vehicles), Hub 5 (5 vehicles), and Hub 6 (3 vehicles). +- From Hub 1: Vehicles can move to Hub 0 (7 vehicles), Hub 4 (19 vehicles), Hub 5 (15 vehicles), and Hub 6 (16 vehicles). +- From Hub 2: Vehicles can move to Hub 0 (11 vehicles), Hub 1 (14 vehicles), Hub 3 (14 vehicles), Hub 4 (8 vehicles), Hub 5 (11 vehicles), and Hub 6 (6 vehicles). +- From Hub 3: Vehicles can move to Hub 0 (17 vehicles), Hub 1 (11 vehicles), Hub 2 (11 vehicles), Hub 4 (11 vehicles), Hub 5 (16 vehicles), and Hub 6 (19 vehicles). +- From Hub 4: Vehicles can move to Hub 0 (10 vehicles), Hub 1 (4 vehicles), Hub 2 (17 vehicles), Hub 3 (20 vehicles), Hub 5 (12 vehicles), and Hub 6 (17 vehicles). +- From Hub 5: Vehicles can move to Hub 0 (3 vehicles), Hub 2 (4 vehicles), Hub 3 (5 vehicles), Hub 4 (20 vehicles), and Hub 6 (4 vehicles). +- From Hub 6 (Main Distribution Hub): Vehicles can move back to Hub 0 (15 vehicles), Hub 1 (2 vehicles), Hub 2 (15 vehicles), Hub 3 (18 vehicles), Hub 4 (17 vehicles), and Hub 5 (7 vehicles). + +In this advanced transit network, a capacity of ""0"" indicates that there is no direct transit lane connecting two specific hubs. + +The objective is to determine the maximum capacity of vehicles that can be dispatched from Hub 0 (the Central Dispatch Center) to Hub 6 (the Main Distribution Hub) per hour, while ensuring that the flow through each transit lane does not exceed its specified capacity limit. + +What is the maximum number of vehicles that can be optimally dispatched from the Central Dispatch Center to the Main Distribution Hub within an hour?",[44.0] +"[""Capacitated Facility Location Problem""]","In the bustling landscape of retail and distribution, EfficientDistro Inc. is at a pivotal point of re-evaluating its logistics strategy to ensure the timely and cost-effective delivery of goods to its consumer base. The company has narrowed down its choice to four potential distribution centers, each with its own set of operational costs and capacities. These centers are tasked with the responsibility of distributing products to four retail stores, spread across different locations, each with its unique demand. The challenge lies in selecting the right mix of distribution centers to open and determining the most efficient distribution routes to minimize the overall operational costs, which include both the opening costs of the distribution centers and the transportation costs of goods to the retail stores. + +Given Data: +- Number of Potential Distribution Centers: 4 +- Number of Retail Stores Needing Supply: 4 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $5 to Store 1, $5 to Store 2, $2 to Store 3, $3 to Store 4 +- From Center 2: $3 to Store 1, $3 to Store 2, $5 to Store 3, $4 to Store 4 +- From Center 3: $3 to Store 1, $5 to Store 2, $2 to Store 3, $4 to Store 4 +- From Center 4: $2 to Store 1, $4 to Store 2, $5 to Store 3, $1 to Store 4 + +Demand of Each Retail Store (in units): +- Store 1: 859 +- Store 2: 713 +- Store 3: 421 +- Store 4: 652 + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1,547 +- Center 2: 1,656 +- Center 3: 1,274 +- Center 4: 1,882 + +EfficientDistro Inc. now faces the complex task of deciding which distribution centers to open and how to allocate shipments to meet the demand at each retail store in the most cost-effective manner. The solution must take into account the varying opening costs of the distribution centers, their capacities, the transportation costs involved in shipping goods to the retail stores, and the specific demand at each store location. + +Question: +What is the optimal total cost for EfficientDistro Inc. to open the necessary distribution centers and transport goods to meet all retail stores' demands, including both the opening costs of the centers and the transportation costs of supplying the stores?",[273940.0] +"[""Capacitated Facility Location Problem""]","In the bustling world of retail, SupplyChain Innovations is at a crucial crossroads. With a diverse network of five retail stores, each with its own unique demand, the company is exploring options to establish a network of distribution centers that can efficiently fulfill these demands at the lowest possible cost. The challenge lies in selecting from seven potential distribution centers, each with its own set of opening costs and capacities, alongside varying transportation costs to the stores. The ultimate goal is to devise a solution that minimizes the total expense involved, which includes the opening costs of the distribution centers and the transportation costs of delivering goods to the retail outlets. + +Given Data: +- Number of Potential Distribution Centers: 7 +- Number of Retail Stores Needing Supply: 5 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 +- Center 6: $120,000 +- Center 7: $182,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $3 to Store 1, $3 to Store 2, $5 to Store 3, $4 to Store 4, $3 to Store 5 +- From Center 2: $5 to Store 1, $2 to Store 2, $4 to Store 3, $2 to Store 4, $4 to Store 5 +- From Center 3: $5 to Store 1, $1 to Store 2, $4 to Store 3, $2 to Store 4, $5 to Store 5 +- From Center 4: $4 to Store 1, $1 to Store 2, $1 to Store 3, $3 to Store 4, $3 to Store 5 +- From Center 5: $2 to Store 1, $4 to Store 2, $4 to Store 3, $3 to Store 4, $4 to Store 5 +- From Center 6: $4 to Store 1, $1 to Store 2, $3 to Store 3, $5 to Store 4, $3 to Store 5 +- From Center 7: $5 to Store 1, $1 to Store 2, $2 to Store 3, $4 to Store 4, $1 to Store 5 + +Demand of Each Retail Store (in units): +- Store 1: 715 +- Store 2: 413 +- Store 3: 641 +- Store 4: 745 +- Store 5: 964 + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1697 +- Center 2: 891 +- Center 3: 1755 +- Center 4: 1308 +- Center 5: 1575 +- Center 6: 834 +- Center 7: 1005 + +SupplyChain Innovations seeks to establish a distribution framework that ensures each retail store's demand is met without exceeding the capacities of the chosen distribution centers. The varying transportation costs from each center present a complex puzzle that requires an optimal combination of distribution centers to be activated and the best logistical paths to be chosen to fulfill the store demands efficiently. + +Question: +What is the optimal total cost for SupplyChain Innovations to open the necessary distribution centers and transport goods to meet all the retail stores' demands, considering both the opening costs of the centers and the transportation costs of supplying the stores?",[393863.0] +"[""Network Optimization""]","There are four nodes connected by directed edges. Node 1, which could represent the starting point, is connected to two other nodes: There is an edge from node 1 to node 2 with a capacity of 8. There is an edge from node 1 to node 3 with a capacity of 7. Node 2 is situated at the top of the graph and acts as an intermediate node, with three edges connected to it: An edge leading from node 2 to node 3 with a capacity of 2. An edge leading from node 2 to node 4 with a capacity of 4. Node 3 is another intermediate node and has an edge from node 3 to node 4 with a substantial capacity of 7. Node 4, which could represent the target or terminal, does not have any edges leading to other nodes, implying that it could be the end point. Find the corresponding maximum flow of the graph. Provide your answer rounded to the nearest integer.",[11.0] +"['Production Planning Problem', 'Profit Maximization Problem']","Rylon Corporation manufactures Brute and Chanelle perfumes. The raw material needed to manufacture each type of perfume can be purchased for $3 per pound. Processing 1 lb of raw material requires 1 hour of laboratory time. Each pound of processed raw material yields 3 oz of Regular Brute Perfume and 4 oz of Regular Chanelle Perfume. Regular Brute can be sold for $7/oz and Regular Chanelle for $6/oz. Rylon also has the option of further processing Regular Brute and Regular Chanelle to produce Luxury Brute, sold at $18/oz, and Luxury Chanelle, sold at $14/oz. Each ounce of Regular Brute processed further requires an additional 3 hours of laboratory time and $4 processing cost and yields 1 oz of Luxury Brute. Each ounce of Regular Chanelle processed further requires an additional 2 hours of laboratory time and $4 processing cost and yields 1 oz of Luxury Chanelle. Each year, Rylon has 6,000 hours of laboratory time available and can purchase up to 4,000 lb of raw material. Formulate an LP that can be used to determine how Rylon can maximize profits. Assume that the cost of the laboratory hours is a fixed cost. Find the maximal profit.",[172666.67] +['Diet Problem'],"Consider the following table of food and corresponding nutritional values: Bread contains 4 grams of protein, 7 grams of carbohydrates, has a calorie count of 130, and costs 3 dollar. Milk contains 6 grams of protein, 10 grams of carbohydrates, has a calorie count of 120, and costs 4 dollars. Fish is high in protein with 20 grams, contains no carbohydrates, has a calorie count of 150, and is the most expensive at 8 dollars. Potato contains 1 gram of protein, is high in carbohydrates with 30 grams, has the lowest calorie count at 70, and is the cheapest at 2 dollars. The ideal intake for an adult is at least 40 grams of protein, 50 grams of carbohydrates, and 450 calories per day. The problem is to find the least costly way to achieve those amounts of nutrition by using the four types of food shown in the table. ",[22] +"[""TravelingSalesman""]","Imagine a salesperson who is required to visit four distinct cities to manage business meetings. These cities are labeled as 1, 2, 3, and 4. The salesperson has the flexibility to start their route from any of these cities but must meet a critical condition: they must visit each of the other cities exactly once before returning to their starting city. The primary goal is to minimize the total travel cost incurred during the trip, considering factors such as fuel, time, or transport fares. + +Here’s a breakdown of the travel costs between the cities: +- The cost to travel from City 1 to City 2 is 61 units, to City 3 is 17 units, and to City 4 is 81 units. +- From City 2, the costs are 61 units to City 1, 40 units to City 3, and 52 units to City 4. +- Moving from City 3, the journey costs 17 units to City 1, 40 units to City 2, and 23 units to City 4. +- Lastly, from City 4, it costs 81 units to travel to City 1, 52 units to City 2, and 23 units to City 3. + +The question now is, what is the minimum total travel cost for the salesperson to visit each city exactly once and then return to the starting city?",[153.0] +"[""Network Optimization""]","Welcome to the enchanting world of Rivertown, a bustling town known for its complex and sophisticated water supply network. This network consists of 9 crucial pumping stations, each interconnected by a series of underground pipelines designed to ensure the smooth flow of water from the town's primary water source to its final destination, ensuring that every resident of Rivertown has access to clean water. + +The capacities of these pipelines, indicating the maximum volume of water (in thousands of gallons per hour) that can flow from one station to another, are meticulously planned as follows: + +- From Station 0 (Source): Water is pumped to Station 1 (5 gallons), Station 2 (2 gallons), Station 3 (7 gallons), Station 4 (17 gallons), Station 5 (4 gallons), Station 6 (1 gallon), Station 7 (2 gallons), and Station 8 (13 gallons). +- From Station 1: Water flows to Station 0 (12 gallons), Station 2 (15 gallons), Station 3 (18 gallons), Station 4 (8 gallons), Station 5 (9 gallons), Station 6 (8 gallons), Station 7 (1 gallon), and Station 8 (15 gallons). +- From Station 2: Water is directed to Station 0 (17 gallons), Station 1 (16 gallons), Station 3 (20 gallons), Station 4 (17 gallons), Station 5 (13 gallons), Station 6 (2 gallons), Station 7 (19 gallons), and Station 8 (6 gallons). +- From Station 3: Water can be sent to Station 0 (5 gallons), Station 1 (9 gallons), Station 2 (14 gallons), Station 4 (10 gallons), Station 5 (18 gallons), Station 6 (20 gallons), Station 7 (9 gallons), and Station 8 (18 gallons). +- From Station 4: Water is supplied to Station 0 (12 gallons), Station 1 (14 gallons), Station 2 (7 gallons), Station 3 (6 gallons), Station 5 (6 gallons), Station 6 (17 gallons), Station 7 (4 gallons), and Station 8 (9 gallons). +- From Station 5: Water flows to Station 0 (12 gallons), Station 1 (11 gallons), Station 2 (10 gallons), Station 3 (1 gallon), Station 4 (13 gallons), Station 6 (14 gallons), Station 7 (12 gallons), and Station 8 (14 gallons). +- From Station 6: Water is channeled to Station 0 (13 gallons), Station 1 (7 gallons), Station 2 (3 gallons), Station 3 (9 gallons), Station 4 (1 gallon), Station 5 (14 gallons), Station 7 (7 gallons), and Station 8 (16 gallons). +- From Station 7: Water is directed from here to Station 0 (18 gallons), Station 1 (15 gallons), Station 2 (8 gallons), Station 3 (0 gallons), Station 4 (12 gallons), Station 5 (13 gallons), Station 6 (5 gallons), and Station 8 (17 gallons). +- From Station 8 (Destination): Water can flow back to Station 0 (3 gallons), Station 1 (11 gallons), Station 2 (15 gallons), Station 3 (14 gallons), Station 4 (0 gallons), Station 5 (4 gallons), Station 6 (18 gallons), and Station 7 (16 gallons). + +In this scenario, a capacity of ""0"" signifies that there is no direct pipeline connection between those two stations. + +Your mission, should you choose to accept it, is to calculate the maximum volume of water that can be distributed from Station 0 (the source) to Station 8 (the destination) per hour, ensuring no pipeline exceeds its capacity limits. What is the maximum amount of water, in thousands of gallons per hour, that can flow from the source to the destination?",[51.0] +"[""Network Optimization""]","Welcome to the heart of an intricate transportation network, designed to efficiently distribute a critical resource across 8 bustling hubs of a futuristic city. Each hub, from a massive distribution center to the ultimate delivery point, is interconnected through a series of high-capacity conduits, each with a unique capability to handle the flow of resources. + +Here's the detailed layout of this advanced network: + +- From Hub 0 (Source): Can dispatch resources to Hub 1 (19 units), Hub 2 (5 units), Hub 3 (7 units), Hub 5 (17 units), and Hub 6 (19 units), with a special direct line to Hub 7 (9 units). +- From Hub 1: Channels resources to Hub 0 (6 units), Hub 2 (15 units), Hub 3 (15 units), Hub 4 (16 units), Hub 5 (8 units), and Hub 6 (5 units). +- From Hub 2: Feeds into Hub 0 (5 units), Hub 1 (6 units), Hub 3 (3 units), Hub 4 (10 units), Hub 5 (7 units), Hub 6 (6 units), and directly to Hub 7 (3 units). +- From Hub 3: Transfers resources back to Hub 0 (8 units), to Hub 1 (2 units), Hub 2 (6 units), Hub 4 (5 units), Hub 5 (17 units), and has a pathway to Hub 7 (1 unit). +- From Hub 4: Sends its resources to Hub 0 (17 units), Hub 1 (4 units), Hub 2 (10 units), Hub 3 (10 units), Hub 5 (6 units), Hub 6 (15 units), and to Hub 7 (14 units) directly. +- From Hub 5: Moves resources to Hub 0 (11 units), Hub 1 (11 units), Hub 2 (9 units), Hub 3 (7 units), Hub 4 (7 units), Hub 6 (16 units), and directly to Hub 7 (19 units). +- From Hub 6: Directs resources back to Hub 0 (13 units), to Hub 1 (19 units), Hub 2 (14 units), Hub 3 (14 units), Hub 4 (16 units), Hub 5 (1 unit), and to Hub 7 (16 units). +- From Hub 7 (Destination): Receives backflow to Hub 0 (0 units), from Hub 1 (14 units), Hub 3 (17 units), Hub 4 (1 unit), Hub 5 (7 units), and Hub 6 (9 units). + +In this scenario, a capacity of ""0"" indicates an absence of a direct conduit between two specific hubs. + +The challenge ahead is to determine the maximum amount of resources that can be efficiently transported from Hub 0 (the source) to Hub 7 (the destination) within this network, ensuring the flow through each conduit does not exceed its stated capacity. + +What is the maximum unit of resources that can be delivered from the source to the destination?",[62.0] +['Diet Problem'],"Imagine you're on a mission to revamp your diet and want to ensure you're meeting your nutritional needs without putting a hole in your wallet. You've narrowed down your food options to four items: Chicken, Tofu, Rice, and Avocado. Each of these foods contributes different amounts of protein, carbs, and calories, and they each come at a different cost. + +Here's a detailed breakdown of what each food offers and the cost involved: + +- Chicken: Provides you 2 grams of protein, 4 grams of carbohydrates, and 218 calories at the price of $3. +- Tofu: Offers a generous 18 grams of protein, 17 grams of carbohydrates, and 290 calories for $6. +- Rice: Supplies you with 2 grams of protein, a hefty 28 grams of carbohydrates, and 298 calories for just $1. +- Avocado: Packs in 18 grams of protein, 27 grams of carbohydrates, and 158 calories for $10. + +Your dietary goal is to consume at least 86 grams of protein, 120 grams of carbohydrates, and 2435 calories from your meals. Your challenge is to figure out the least expensive method to meet these nutritional targets using only these four food choices. What is the lowest cost to meet your nutritional requirements? You could only assign integral portions of each type of food.",[31.0] +['Capacitated Facility Location Problem'],"GlobalSupply Inc. is at a strategic crossroads, looking to streamline its operations by identifying the most cost-effective way to distribute its products to six retail stores from a pool of five potential distribution centers. Each distribution center has not only a unique opening cost but also a specific supply capacity, which limits the amount of product it can dispatch. Moreover, the transport cost for shipping a unit of product from these centers to the retail stores varies, adding another layer of complexity to the decision-making process. The ultimate goal for GlobalSupply Inc. is to ensure that the demand at each retail store is met while keeping the total costs—comprising both the opening costs of the distribution centers and the transportation costs of the products—as low as possible. + +Given Data: +- Number of Potential Distribution Centers: 5 +- Number of Retail Stores Needing Supply: 6 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $5 to Store 1, $2 to Store 2, $3 to Store 3, $3 to Store 4, $3 to Store 5, $5 to Store 6 +- From Center 2: $4 to Store 1, $3 to Store 2, $5 to Store 3, $2 to Store 4, $4 to Store 5, $2 to Store 6 +- From Center 3: $4 to Store 1, $5 to Store 2, $1 to Store 3, $4 to Store 4, $2 to Store 5, $5 to Store 6 +- From Center 4: $4 to Store 1, $1 to Store 2, $1 to Store 3, $3 to Store 4, $3 to Store 5, $2 to Store 6 +- From Center 5: $4 to Store 1, $4 to Store 2, $3 to Store 3, $4 to Store 4, $4 to Store 5, $1 to Store 6 + +Demand of Each Retail Store (in units): +- Store 1: 530 +- Store 2: 884 +- Store 3: 420 +- Store 4: 566 +- Store 5: 673 +- Store 6: 787 + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1400 +- Center 2: 1115 +- Center 3: 813 +- Center 4: 1041 +- Center 5: 1576 + +Question: +What is the optimal total cost for GlobalSupply Inc. to open the necessary distribution centers and transport goods in order to satisfy the demand of all six retail stores, including both the opening costs of the distribution centers and the transportation costs of supplying the stores?",[490293.0] +"[""Blending Problem""]","The Healthy Pet Food Company manufactures two types of dog food: Meaties and Yummies. Each package of Meaties contains 2 pounds of cereal and 3 pounds of meat; each package of Yummies contains 3 pounds of cereal and 1.5 pounds of meat. Healthy believes it can sell as much of each dog food as it can make. Meaties sell for $2.80 per package and Yummies sell for $2.00 per package. Healthy’s production is limited in several ways. First, Healthy can buy only up to 400,000 pounds of cereal each month at $0.20 per pound. It can buy only up to 300,000 pounds of meat per month at $0.50 per pound. In addition, a special piece of machinery is required to make Meaties, and this machine has a capacity of 90,000 packages per month. The variable cost of blending and packing the dog food is $0.25 per package for Meaties and $0.20 per package for Yummies. For Meaties: The sales price per package is $2.80. Each package requires 2.0 pounds of cereal and 3.0 pounds of meat. The variable cost for blending and packing is $0.25 per package. The production capacity for Meaties is 90,000 packages per month. For Yummies: The sales price per package is $2.00. Each package requires 3.0 pounds of cereal and 1.5 pounds of meat. The variable cost for blending and packing is $0.20 per package.There are 400,000 pounds of cereal available per month. There are 300,000 pounds of meat available per month. Suppose you are the manager of the Dog Food Division of the Healthy Pet Food Company. Your salary is based on division profit, so you try to maximize the profit.",[77500.0] +"[""TravelingSalesman""]","Consider a delivery company that operates in six cities, labelled from 1 to 6. The company needs to deliver packages to all cities, starting from any one of them, but following these constraints: each city should be visited exactly once, and after visiting all the cities, the delivery van should return to the city where it started. We need to minimize the total cost of this delivery route, taking into account factors such as fuel, road taxes, and driver fees. +Here are the associated costs between the cities: +The cost to deliver from City 1 to City 2 is 43 units, to City 3 is 39 units, to City 4 is 97 units, to City 5 is 84 units, and to City 6 is 67 units. +From City 2, the costs are 43 units to City 1, 18 units to City 3, 92 units to City 4, 27 units to City 5, and 51 units to City 6. +For deliveries from City 3, it costs 39 units to reach City 1, 18 units to reach City 2, 84 units to reach City 4, 60 units to reach City 5, and 15 units to reach City 6. +From City 4, the costs are 97 units to City 1, 92 units to City 2, 84 units to City 3, 42 units to City 5, and 32 units to City 6. +Delivering from City 5, the costs are 84 units to City 1, 27 units to City 2, 60 units to City 3, 42 units to City 4, and 18 units to City 6. +Lastly, from City 6, it costs 67 units to reach City 1, 51 units to reach City 2, 15 units to reach City 3, 32 units to reach City 4, and 18 units to reach City 5. +What is the minimum total delivery cost for the company to deliver to each city exactly once and return to the starting city?",[192.0] +['TravelingSalesman'],"Imagine a scenario where a sales representative needs to travel to six distinct cities to promote their company's products. The cities, identified as City 1, City 2, City 3, City 4, City 5, and City 6, are all part of the salesperson's designated territory. The objective is for the salesperson to visit each city exactly once, starting and ending at any one of the cities, while minimizing the total travel cost incurred during the trip. The costs associated with traveling between the cities are as follows: +- From City 1, the travel costs are 65 units to City 2, 41 units to City 3, 14 units to City 4, 16 units to City 5, and 35 units to City 6. +- From City 2, the costs are 65 units to City 1, 76 units to City 3, 52 units to City 4, 89 units to City 5, and 45 units to City 6. +- From City 3, the travel costs are 41 units to City 1, 76 units to City 2, 72 units to City 4, 38 units to City 5, and 62 units to City 6. +- From City 4, the costs to travel are 14 units to City 1, 52 units to City 2, 72 units to City 3, 22 units to City 5, and 24 units to City 6. +- From City 5, the costs are 16 units to City 1, 89 units to City 2, 38 units to City 3, 22 units to City 4, and 89 units to City 6. +- Lastly, from City 6, the travel costs are 35 units to City 1, 45 units to City 2, 62 units to City 3, 24 units to City 4, and 89 units to City 5. + +Given these costs, what is the minimum total travel cost for the salesperson to visit each city exactly once and return to the starting city?",[213.0] +['Capacitated Lot-sizing Problem (CLSP)'],"Basel Tool and Die Company (BTD) makes large industrial pipe wrenches in one of its factories. The marketing department estimates demand for this product during the next 6 months to be: In January, the demand is 430, in February, the demand is 430, in March, the demand is 380, in April, the demand is 450, in May, the demand is 520, in June, the demand is 440. With the current labor force, BTD believes it can make approximately 420 pipe wrenches per month at a cost of $40 per wrench using regular-time production. An additional 80 wrenches per month can be made using overtime production at a cost per wrench of $45. Wrenches can be made in advance and held in inventory for later shipment at a cost of $3 per month per wrench. The monthly demand for wrenches must be satisfied every month. At the end of December (beginning of January) BTD has 10 wrenches in inventory. BTD wants to plan its production, including overtime, and inventory for the next 6 months so as to maximize profit. Assuming the revenue for these wrenches is fixed, the production manager can maximize profit by minimizing the total costs incurred in producing and delivering the wrenches. Find the minimal total cost in dollar. Provide your answer rounded to nearest integar.",[106380.0] +"[""Network Optimization""]","In the bustling metropolis of FlowCity, the local government has embarked on an ambitious project to revamp its old water distribution network, which connects 9 critical reservoirs across the city. This network, vital for the city's water supply, is designed to maximize the efficiency of water flow from the primary reservoir to a central distribution hub. + +Here's a detailed look at the current capacities of the pipelines (measured in millions of gallons per day) that link these reservoirs: + +- From Reservoir 0 (Source): Water can flow to Reservoir 1 (6), Reservoir 2 (16), Reservoir 3 (12), Reservoir 4 (16), Reservoir 5 (3), Reservoir 6 (9), Reservoir 7 (3), and Reservoir 8 (13). +- From Reservoir 1: Water can be distributed to Reservoir 0 (2), Reservoir 3 (20), Reservoir 4 (13), Reservoir 5 (7), Reservoir 6 (20), Reservoir 7 (11), and Reservoir 8 (13). +- From Reservoir 2: Water can flow to Reservoir 0 (4), Reservoir 1 (15), Reservoir 3 (14), Reservoir 4 (13), Reservoir 5 (5), Reservoir 6 (12), Reservoir 7 (14), and Reservoir 8 (16). +- From Reservoir 3: Water can be distributed to Reservoir 0 (1), Reservoir 2 (2), Reservoir 4 (8), Reservoir 5 (7), Reservoir 6 (11), Reservoir 7 (14), and Reservoir 8 (16). +- From Reservoir 4: Water can flow to Reservoir 0 (4), Reservoir 1 (19), Reservoir 2 (7), Reservoir 3 (14), Reservoir 5 (1), Reservoir 6 (20), Reservoir 7 (14), and Reservoir 8 (16). +- From Reservoir 5: Water can be distributed to Reservoir 2 (14), Reservoir 3 (4), Reservoir 4 (17), Reservoir 6 (2), and Reservoir 7 (20), and Reservoir 8 (9). +- From Reservoir 6: Water can flow to Reservoir 0 (12), Reservoir 1 (19), Reservoir 2 (3), Reservoir 3 (3), Reservoir 7 (13), and Reservoir 8 (3). +- From Reservoir 7: Water can be distributed to Reservoir 0 (8), Reservoir 1 (19), Reservoir 2 (16), Reservoir 3 (4), Reservoir 4 (6), Reservoir 5 (13), Reservoir 6 (9), and Reservoir 8 (15). +- From Reservoir 8 (Destination): Water can flow back to Reservoir 0 (11), Reservoir 1 (6), Reservoir 3 (6), Reservoir 4 (10), Reservoir 5 (9), and Reservoir 6 (3), and Reservoir 7 (1). + +In this complex network, a pipeline capacity of ""0"" indicates that there is no direct connection between those two reservoirs. + +The city's goal is to ensure that the water supply can meet the growing demands by calculating the maximum amount of water that can be distributed from Reservoir 0 (the primary source) to Reservoir 8 (the main distribution hub) each day, without exceeding the capacities of the pipelines. + +What is the optimal value of water (in millions of gallons per day) that can be maximally distributed from the source to the destination? +",[78.0] +"[""Capacitated Facility Location Problem""]","In the bustling world of retail and distribution, a major company, GlobalSupply, is facing a strategic challenge. It needs to efficiently supply its products to eight different retail stores, each with its own demand for goods. To achieve this, GlobalSupply is considering opening up to eight different distribution centers, but each center comes with its own set of opening costs and has a specific capacity limit. Additionally, the cost of transporting goods from these distribution centers to the retail stores varies, adding another layer of complexity to the decision-making process. GlobalSupply's primary objective is to fulfill the demand at each retail store while minimizing the total cost, which includes both the opening costs of the distribution centers and the transportation costs of the goods. + +Given Data: +- Number of Potential Distribution Centers: 8 +- Number of Retail Stores Needing Supply: 8 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 +- Center 6: $120,000 +- Center 7: $182,000 +- Center 8: $186,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1 to Stores: $3 to Store 1, $3 to Store 2, $5 to Store 3, $4 to Store 4, $3 to Store 5, $5 to Store 6, $2 to Store 7, $4 to Store 8 +- From Center 2 to Stores: $2 to Store 1, $4 to Store 2, $5 to Store 3, $1 to Store 4, $4 to Store 5, $2 to Store 6, $5 to Store 7, $4 to Store 8 +- From Center 3 to Stores: $1 to Store 1, $1 to Store 2, $3 to Store 3, $3 to Store 4, $2 to Store 5, $4 to Store 6, $4 to Store 7, $3 to Store 8 +- From Center 4 to Stores: $4 to Store 1, $4 to Store 2, $1 to Store 3, $3 to Store 4, $5 to Store 5, $3 to Store 6, $5 to Store 7, $1 to Store 8 +- From Center 5 to Stores: $2 to Store 1, $4 to Store 2, $1 to Store 3, $4 to Store 4, $2 to Store 5, $2 to Store 6, $1 to Store 7, $2 to Store 8 +- From Center 6 to Stores: $5 to Store 1, $2 to Store 2, $4 to Store 3, $4 to Store 4, $4 to Store 5, $4 to Store 6, $5 to Store 7, $3 to Store 8 +- From Center 7 to Stores: $1 to Store 1, $4 to Store 2, $2 to Store 3, $4 to Store 4, $2 to Store 5, $2 to Store 6, $4 to Store 7, $5 to Store 8 +- From Center 8 to Stores: $2 to Store 1, $2 to Store 2, $4 to Store 3, $2 to Store 4, $2 to Store 5, $4 to Store 6, $4 to Store 7, $1 to Store 8 + +Demand of Each Retail Store (in units): +- Store 1: 695 units +- Store 2: 737 units +- Store 3: 452 units +- Store 4: 616 units +- Store 5: 587 units +- Store 6: 779 units +- Store 7: 892 units +- Store 8: 440 units + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1980 units +- Center 2: 814 units +- Center 3: 864 units +- Center 4: 1320 units +- Center 5: 1952 units +- Center 6: 1447 units +- Center 7: 1886 units +- Center 8: 1962 units + +Question: +Considering these details, what is the optimal total cost for GlobalSupply to open distribution centers and transport goods to meet all retail stores' demands? This cost includes both the opening costs of the centers and the transportation costs of supplying the stores.",[443177.0] +['Capacitated Facility Location Problem'],"LogiSphere Inc. is at a crossroads in determining the most cost-efficient strategy to supply nine retail stores with goods from a selection of five potential distribution centers. Each center has a unique opening cost and a maximum supply capacity. The challenge is further complicated by the variable transportation costs from these centers to the retail stores. LogiSphere's goal is to fulfill the demand at each retail store while minimizing the total cost, which encompasses the opening costs of the distribution centers and the transportation costs of the goods. + +**Given Data:** + +- **Number of Potential Distribution Centers:** 5 +- **Number of Retail Stores Needing Supply:** 9 + +**Opening Costs for Each Distribution Center (in dollars):** +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 + +**Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars):** +- From Center 1 to Stores: $5 to Store 1, $2 to Store 2, $3 to Store 3, $3 to Store 4, $3 to Store 5, $5 to Store 6, $4 to Store 7, $3 to Store 8, $5 to Store 9 +- From Center 2 to Stores: $2 to Store 1, $4 to Store 2, $2 to Store 3, $4 to Store 4, $5 to Store 5, $1 to Store 6, $4 to Store 7, $2 to Store 8, $5 to Store 9 +- From Center 3 to Stores: $4 to Store 1, $1 to Store 2, $1 to Store 3, $3 to Store 4, $3 to Store 5, $2 to Store 6, $4 to Store 7, $4 to Store 8, $3 to Store 9 +- From Center 4 to Stores: $4 to Store 1, $4 to Store 2, $1 to Store 3, $3 to Store 4, $5 to Store 5, $3 to Store 6, $5 to Store 7, $1 to Store 8, $2 to Store 9 +- From Center 5 to Stores: $4 to Store 1, $1 to Store 2, $4 to Store 3, $2 to Store 4, $2 to Store 5, $1 to Store 6, $2 to Store 7, $5 to Store 8, $2 to Store 9 + +**Demand of Each Retail Store (in units):** +- Store 1: 739 units +- Store 2: 491 units +- Store 3: 766 units +- Store 4: 854 units +- Store 5: 827 units +- Store 6: 908 units +- Store 7: 434 units +- Store 8: 605 units +- Store 9: 480 units + +**Supply Capacity of Each Distribution Center (in units):** +- Center 1: 1825 units +- Center 2: 1821 units +- Center 3: 1365 units +- Center 4: 1929 units +- Center 5: 1502 units + +**Question:** + +What is the optimal total cost for LogiSphere Inc. to open distribution centers and transport goods in order to meet all the retail stores' demands? This cost includes both the opening costs of the centers and the transportation costs of supplying the stores.",[608437.0] +['Diet Problem'],"Imagine you're a nutrition enthusiast, planning your daily meals while aiming to meet all your dietary needs without breaking the bank. You're considering five different types of food to include in your meal plan: Chicken, Rice, Apples, Cheese, and Tofu. Each of these foods contributes different amounts of protein, carbohydrates, and calories, and comes with its own price. + +Here's the nutritional content and cost for each food: + +- Chicken: Provides 14 grams of protein, 28 grams of carbohydrates, and 184 calories for $5. +- Rice: Offers 10 grams of protein, 27 grams of carbohydrates, and 298 calories for just $1. +- Apples: Come with 4 grams of protein, 13 grams of carbohydrates, and 162 calories for $5. +- Cheese: Is a protein powerhouse with 19 grams, contains only 5 grams of carbohydrates and provides 81 calories for $9. +- Tofu: Balances 10 grams of protein, 8 grams of carbohydrates, and 147 calories for $9. + +Your goal is to consume at least 97 grams of protein, 140 grams of carbohydrates, and 2304 calories from your daily meals. You aim to achieve all these nutritional targets with the cheapest possible combination of these five food options. + +What is the minimum cost you need to spend to meet all your dietary requirements? Please note, your response should be a single value, indicating the the optimal and minimal cost. You could only assign integral portions of each type of food.",[10.0] +"[""Diet Problem""]","Imagine you're trying to put together a meal plan that ticks all the boxes: it meets all your nutritional needs and it doesn't break the bank. You're considering six different food items: Chicken, Rice, Eggs, Broccoli, Lentils, and Apples. Each of these foods provides different amounts of protein, carbohydrates, and calories, and they all come at different prices. + +Here's a rundown of what each food offers and how much it costs: + +- Chicken: It gives you 18 grams of protein, 5 grams of carbohydrates, and 202 calories for just $1. +- Rice: With 14 grams of protein, 13 grams of carbs, and 234 calories, Rice is a bit pricier at $7. +- Eggs: Like Chicken, Eggs also offer 18 grams of protein, along with 4 grams of carbs and 220 calories for $4. +- Broccoli: A great source of protein at 7 grams, with 15 grams of carbs and 247 calories, this vegetable will cost you $3. +- Lentils: They provide 15 grams of protein, 17 grams of carbs, and 88 calories for only $1. +- Apples: They offer 8 grams of protein, 13 grams of carbohydrates, and 77 calories for $2. + +You know that your meal plan needs to give you at least 70 grams of protein, 187 grams of carbohydrates, and 2181 calories. Your challenge is to determine the least expensive way to get all the nutrients you need from this group of six food items. + +So, what is the minimum total cost you would need to spend to meet all your dietary requirements for the day? You could only assign integral portions of each type of food.",[16.0] +"[""Diet Problem""]","You're in charge of crafting a meal plan for a week, aiming to meet certain nutritional requirements while keeping your grocery bill to a minimum. There are seven food options you're considering: Chicken, Tuna, Eggs, Beef, Rice, Apples, and Bread. Each food item provides distinct amounts of protein, carbohydrates, and calories, and has a different price. + +Here are the details for each food: + +- Chicken: Provides 18 grams of protein, 23 grams of carbohydrates, and 217 calories for $3. +- Tuna: Offers 12 grams of protein, 30 grams of carbohydrates, and 182 calories for $10. +- Eggs: Contains 7 grams of protein, 16 grams of carbohydrates, and 79 calories for $10. +- Beef: Delivers 17 grams of protein, 26 grams of carbohydrates, and 295 calories for $7. +- Rice: Supplies 3 grams of protein, 30 grams of carbohydrates, and 264 calories for just $4. +- Apples: Gives 4 grams of protein, 6 grams of carbohydrates, and 160 calories for $6. +- Bread: Carries 4 grams of protein, 23 grams of carbohydrates, and 141 calories for $5. + +Your nutritional targets for the week are as follows: 59 grams of protein, 160 grams of carbohydrates, and 1809 calories. + +Your challenge is to find the most economical way of meeting these nutritional requirements with the food items available. What is the least possible amount you can spend to meet your nutritional goals for the week? You could only assign integral portions of each type of food.",[26.0] +['Transportation Problem'],"Imagine you're responsible for managing the distribution of medical supplies across five different regions during a critical healthcare crisis. Each region starts with a certain amount of supplies but has different requirements based on the healthcare demands. + +Here is the current stock and the needed amounts in each region: + +- Region 1 has 410 units of supplies but needs 274 units. +- Region 2 has 266 units but needs 331 units. +- Region 3 has 132 units but needs 232 units. +- Region 4 has only 6 units but urgently needs 359 units. +- Region 5 has 485 units but needs just 74 units. + +To ensure each region has the medical supplies it needs, you can transfer supplies between regions. However, transferring supplies incurs costs which vary between regions. Below is the cost matrix that shows the cost of transferring supplies from one region to another: + +- From/to Region 1: To Region 2 costs 24, to Region 3 costs 3, to Region 4 costs 27, to Region 5 costs 31. +- From/to Region 2: To Region 1 costs 24, to Region 3 costs 11, to Region 4 costs 7, to Region 5 costs 37. +- From/to Region 3: To Region 1 costs 30, to Region 2 costs 1, to Region 4 costs 38, to Region 5 costs 22. +- From/to Region 4: To Region 1 costs 36, to Region 2 costs 18, to Region 3 costs 17, to Region 5 costs 35. +- From/to Region 5: To Region 1 costs 41, to Region 2 costs 35, to Region 3 costs 33, to Region 4 costs 33. + +Your objective is to ensure every region has the medical supplies it requires at the lowest possible transportation cost. What would be the minimum cost to ensure all regions have the necessary medical supplies?",[13079.0] +"[""TravelingSalesman""]","Consider a scenario where a logistics company needs to organize a delivery route across five warehouses. These warehouses are located in different cities labeled as A, B, C, D, and E. A delivery truck has to start from any one of these warehouses, deliver goods at each of the remaining four exactly once, and finally return to the starting warehouse. The challenge here is to plan the route in a way that minimizes the total travel cost. The cost could be based on various factors like distance, fuel usage, or time spent. + +Here are the details of the travel costs between the warehouses: + +The cost to travel from Warehouse A to B is 15 units, to C is 88 units, to D is 54 units, and to E is 92 units. + +From Warehouse B, the cost to reach A is 15 units, to C is 29 units, to D is 98 units, and to E is 33 units. + +For Warehouse C, the cost to get to A is 88 units, to B is 29 units, to D is 96 units, and to E is 96 units. + +Moving from Warehouse D, it costs 54 units to reach A, 98 units to get to B, 96 units to C, and 63 units to E. + +Lastly, from Warehouse E, the cost to go to A is 92 units, to B is 33 units, to C is 96 units, and to D is 63 units. + +How can the logistics company plan the route so that the total travel cost is minimized for the delivery truck to visit each warehouse exactly once and return to the starting warehouse?",[257.0] +['TravelingSalesman'],"Consider a scenario where a delivery company needs to deliver packages to four different cities: Alpha, Bravo, Charlie, and Delta. The company can start the delivery from any city, but they must ensure that each city is visited exactly once and then return to the city they started from. The objective is to minimize the total delivery cost, which could include gas expenses, labour hours, and vehicle maintenance costs. +Here is an overview of the delivery costs between the cities: +The cost to deliver from Alpha to Bravo is 41 units, to Charlie is 55 units, and to Delta is 22 units. +From Bravo, it costs 41 units to reach Alpha, 47 units to get to Charlie, and 47 units to Delta. +If the company starts from Charlie, the costs are 55 units to Alpha, 47 units to Bravo, and 35 units to Delta. +And if they start at Delta, it costs 22 units to reach Alpha, 47 units to Bravo, and 35 units to Charlie. +Based on these costs, what is the minimum total delivery cost for the company to deliver packages to each city exactly once and return to the starting point?",[145.0] +['TravelingSalesman'],"In a scenario designed to test the efficiency of travel strategies, a salesperson is tasked with visiting four distinct cities to conduct business. The cities, referred to simply as City 1, City 2, City 3, and City 4, are connected by varying travel costs that represent either the distance, time, or expense associated with moving between these locations. The objective for the salesperson is to minimize the total cost of traveling, which includes visiting each city exactly once and then returning to the starting city. + +Here's a detailed look at the travel costs between each city: +- From City 1 to City 2 the travel cost is 26 units, to City 3 is 49 units, and to City 4 is 75 units. +- From City 2, it costs 26 units to return to City 1, 19 units to proceed to City 3, and 52 units to go to City 4. +- Traveling from City 3, the costs are 49 units to City 1, 19 units back to City 2, and 49 units to City 4. +- Finally, from City 4, the travel costs are 75 units to City 1, 52 units to City 2, and 49 units to City 3. + +Given these costs, what is the minimum total travel cost for the salesperson to complete their circuit of visiting each city exactly once and returning to the starting city?",[169.0] +"[""Network Optimization""]","Imagine an intricate and bustling transportation network connecting six major cities in a region. Goods are transported from City 0 (the source) to City 5 (the destination) through numerous highways, each with varying capacities to handle the flow of goods. + +Here is a rundown of the maximum amount of goods (in thousands of tons) that can be transferred between cities per day: + +- From City 0 (Source): Can send goods to City 1 (6 tons), City 2 (8 tons), City 3 (1 ton), City 4 (9 tons), and City 5 (8 tons). +- From City 1: Can send goods to City 0 (16 tons), City 2 (6 tons), City 4 (12 tons), and City 5 (16 tons). It has no connection to City 3. +- From City 2: Can send goods to City 0 (7 tons), City 1 (16 tons), City 3 (20 tons), City 4 (17 tons), and City 5 (9 tons). +- From City 3: Can send goods to City 0 (3 tons), City 1 (11 tons), City 2 (14 tons), and City 4 (7 tons). It has no connection to City 5. +- From City 4: Can send goods to City 0 (6 tons), City 1 (19 tons), City 2 (19 tons), City 3 (3 tons), and City 5 (20 tons). +- From City 5 (Destination): Can send goods back to City 0 (20 tons), City 1 (5 tons), City 2 (16 tons), City 3 (12 tons), and City 4 (2 tons). + +In this scenario, a capacity of ""0"" means there is no direct highway connection between those two cities. + +The question at hand is to calculate the maximum amount of goods that can be transported from City 0 (the source) to City 5 (the destination) per day while ensuring that the flow of goods on each highway doesn't exceed its capacity. + +So what's the maximum amount of goods that can be transported from the source to the destination per day in thousands of tons?",[32.0] +['Network Optimization'],"In the heart of a bustling metropolis, there lies an intricate network of digital information pipelines connecting 8 crucial data centers. These centers, from the main server hub to the final relay station, are vital for processing and distributing vast amounts of data across the city. The connections between these data centers have specific bandwidth capacities that dictate the maximum flow of data (measured in terabytes per hour) that can be transmitted from one center to another. + +Here's a breakdown of the network's layout and its capacities: + +- From Data Center 0 (Source): Data can be sent to Center 1 (6 TB/h), Center 2 (16 TB/h), Center 3 (5 TB/h), Center 4 (6 TB/h), Center 5 (11 TB/h), Center 6 (6 TB/h), and Center 7 (4 TB/h). +- From Data Center 1: Data can be transmitted to Center 2 (8 TB/h), Center 3 (10 TB/h), Center 4 (6 TB/h), Center 5 (3 TB/h), Center 6 (6 TB/h), and Center 7 (14 TB/h). +- From Data Center 2: Can send data to Center 1 (6 TB/h), Center 3 (20 TB/h), Center 4 (7 TB/h), Center 5 (16 TB/h), Center 6 (13 TB/h), and Center 7 (14 TB/h). +- From Data Center 3: Has capacities to Center 1 (11 TB/h), Center 2 (19 TB/h), Center 4 (18 TB/h), Center 5 (5 TB/h), and Center 7 (8 TB/h). +- From Data Center 4: Can route data to Center 1 (1 TB/h), Center 2 (14 TB/h), Center 3 (18 TB/h), Center 5 (12 TB/h), Center 6 (4 TB/h), and Center 7 (7 TB/h). +- From Data Center 5: Can dispatch data to Center 1 (16 TB/h), Center 2 (3 TB/h), Center 3 (20 TB/h), Center 4 (7 TB/h), Center 6 (15 TB/h), and Center 7 (6 TB/h). +- From Data Center 6: Capable of sending to Center 1 (11 TB/h), Center 2 (4 TB/h), Center 3 (19 TB/h), Center 4 (7 TB/h), Center 5 (6 TB/h), and Center 7 (17 TB/h). +- From Data Center 7 (Destination): Can receive return data flow from Center 1 (5 TB/h), Center 2 (8 TB/h), Center 3 (17 TB/h), Center 4 (2 TB/h), Center 5 (5 TB/h), and Center 6 (4 TB/h). + +In this network, a capacity of ""0"" signifies an absence of direct data transmission capability between any two given data centers. + +The challenge is to determine the maximum amount of data that can be efficiently transmitted from Data Center 0 (the main server hub) to Data Center 7 (the final relay station) per hour, without exceeding the bandwidth capacities of the connections between the centers. What is the optimal value of data flow (in terabytes per hour) that can be achieved from the source to the destination?",[54.0] +"[""Network Optimization""]","Imagine a network of highways that connect six major cities in a country. These highways are designed to facilitate the transportation of goods from a central warehouse (City 0) to a main retail hub (City 5). Each highway has a specific transportation capacity, indicating the maximum number of trucks it can accommodate per hour. + +Here is the transportation capacity of each highway connecting the cities: + +- From City 0 (Warehouse): Can send goods to City 1 (14 trucks), City 2 (15 trucks), City 3 (13 trucks), and City 5 (4 trucks). +- From City 1: Can send goods to City 0 (10 trucks), City 2 (7 trucks), City 3 (18 trucks), City 4 (9 trucks), and City 5 (5 trucks). +- From City 2: Can send goods to City 0 (11 trucks), City 3 (7 trucks), City 4 (13 trucks), and City 5 (6 trucks). +- From City 3: Can send goods to City 0 (7 trucks), City 1 (11 trucks), City 2 (18 trucks), City 4 (17 trucks), and City 5 (2 trucks). +- From City 4: Can send goods to City 0 (19 trucks), City 1 (8 trucks), City 2 (15 trucks), City 3 (15 trucks), and City 5 (5 trucks). +- From City 5 (Retail Hub): Can send goods back to City 0 (10 trucks), City 1 (19 trucks), City 2 (17 trucks), City 3 (2 trucks), and City 4 (3 trucks). + +A capacity of ""0"" indicates that there is no direct highway link between those two cities. + +Now, consider a situation where you're tasked with finding the maximum number of trucks that can be dispatched from the warehouse in City 0 to the retail hub in City 5 per hour, without exceeding the transportation capacity of any highway. + +What is the maximum number of trucks that can be dispatched from the warehouse to the retail hub per hour, while maintaining the transportation capacity of each highway?",[22.0] +['Capacitated Facility Location Problem'],"GlobalSupply Inc. is at a crossroads in revamping its logistics operations to efficiently cater to the needs of five retail stores spread across different locations. The company can choose from nine potential distribution centers, each with its unique setup cost and supply capacity. Moreover, the transportation costs per unit from these distribution centers to the retail stores vary, making the decision complex. The ultimate goal for GlobalSupply Inc. is to fulfill the demand at each retail store while minimizing the total cost, which includes the opening costs of the distribution centers and the transportation costs of the goods to the stores. + +Given Data: + +- Number of Potential Distribution Centers: 9 +- Number of Retail Stores Needing Supply: 5 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 +- Center 6: $120,000 +- Center 7: $182,000 +- Center 8: $186,000 +- Center 9: $174,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $3 to Store 1, $5 to Store 2, $4 to Store 3, $3 to Store 4, $5 to Store 5 +- From Center 2: $2 to Store 1, $4 to Store 2, $2 to Store 3, $4 to Store 4, $5 to Store 5 +- From Center 3: $1 to Store 1, $4 to Store 2, $2 to Store 3, $5 to Store 4, $4 to Store 5 +- From Center 4: $1 to Store 1, $1 to Store 2, $3 to Store 3, $3 to Store 4, $2 to Store 5 +- From Center 5: $4 to Store 1, $4 to Store 2, $3 to Store 3, $4 to Store 4, $4 to Store 5 +- From Center 6: $1 to Store 1, $3 to Store 2, $5 to Store 3, $3 to Store 4, $5 to Store 5 +- From Center 7: $1 to Store 1, $2 to Store 2, $4 to Store 3, $1 to Store 4, $4 to Store 5 +- From Center 8: $2 to Store 1, $2 to Store 2, $1 to Store 3, $2 to Store 4, $5 to Store 5 +- From Center 9: $2 to Store 1, $4 to Store 2, $4 to Store 3, $4 to Store 4, $4 to Store 5 + +Demand of Each Retail Store (in units): +- Store 1: 908 units +- Store 2: 434 units +- Store 3: 605 units +- Store 4: 480 units +- Store 5: 961 units + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1825 units +- Center 2: 1821 units +- Center 3: 1365 units +- Center 4: 1929 units +- Center 5: 1502 units +- Center 6: 1201 units +- Center 7: 1529 units +- Center 8: 961 units +- Center 9: 1001 units + +Question: +What is the optimal total cost for GlobalSupply Inc. to open distribution centers and transport goods to meet all retail stores' demands, including both the opening costs of the centers and the transportation costs of supplying the stores?",[329872.0] +['Profit Maximization Problem'],"A company produces and sells two different products. The demand for each product is unlimited, but the company is constrained by cash availability and machine capacity. Each unit of the first and second product requires 3 and 4 machine hours, respectively. There are 20,000 machine hours available in the current production period. The production costs are $3 and $2 per unit of the first and second product, respectively. The selling prices of the first and second product are $6 and $5.40 per unit, respectively. The available cash is $4,000; furthermore, 45% of the sales revenues from the first product and 30% of the sales revenues from the second product will be made available to finance operations during the current period. Suppose that the company could increase its available machine hours by 2,000, after spending $400 for certain repairs. Formulate a mixed integer linear programming problem and find the maximum net income subject to the cash availability and machine capacity limitations. ",[21599.0] +"[""Production Planning Problem""]","A company produces two kinds of products. A product of the first type requires 1/4 hours of assembly labor, 1/8 hours of testing, and $1.2 worth of raw materials. A product of the second type requires 1/3 hours of assembly, 1/3 hours of testing, and $0.9 worth of raw materials. Given the current personnel of the company, there can be at most 90 hours of assembly labor and 80 hours of testing each day. Suppose that up to 50 hours of overtime assembly labor can be scheduled, at a cost of $7 per hour. Products of the first and second type have a market value of $9 and $8 respectively. Try to maximize the daily profit of the company. ",[4018.0] +['Production Planning Problem'],"Semicond is a small electronics company that manufactures tape recorders and radios. The per-unit labor costs for tape recorders and radios are $50 and $35 respectively, and the raw material costs are $30 for tape recorders and $40 for radios. The selling prices are $100 for a tape recorder and $90 for a radio, as outlined in Table 10. On December 1, 2002, Semicond has enough raw material to manufacture 100 tape recorders and 100 radios. On the same date, the company’s balance sheet shows cash at $10,000, accounts receivable at $3,000, and inventory outstanding valued at $7,000, with a bank loan liability of $10,000, resulting in a current ratio of 2 ($20,000 in total assets over $10,000 in liabilities), as shown in Table 11. Semicond must decide how many tape recorders and radios to produce during December, considering that demand is high enough to sell all units produced. Although all sales in December will be on credit, with payment not received until February 1, 2003, Semicond will collect $2,000 in accounts receivable in December. Furthermore, Semicond has to pay off $1,000 of its loan and a monthly rent of $1,000 in December. On January 1, 2003, Semicond will receive raw materials worth $2,000, to be paid for on February 1, 2003. Management has mandated a minimum cash balance of $4,000 for January 1, 2003. Additionally, to satisfy bank requirements, the current ratio on January 1 must be at least 2. Find the maximal contribution to profit from December’s production in $.",[2500.0] +"[""Capacitated Facility Location Problem""]","In an intricate dance of supply chain optimization, GlobalSupply Inc. is at a crossroads, tasked with the pivotal decision of efficiently distributing its products to seven distinct retail stores. This challenge is compounded by the need to select the most cost-effective combination of distribution centers from a pool of nine potential sites. Each potential distribution center is not only geographically unique but also comes with its own set of opening costs and maximum supply capacities. Additionally, the transportation cost per unit of product varies significantly from one distribution center to another, adding layers of complexity to the decision-making process. + +Given Data: +- Number of Potential Distribution Centers: 9 +- Number of Retail Stores in Need of Supply: 7 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 +- Center 6: $120,000 +- Center 7: $182,000 +- Center 8: $186,000 +- Center 9: $174,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): + +From Center 1: $3 to Store 1, $5 to Store 2, $4 to Store 3, $7 to Store 4, $6 to Store 5, $5 to Store 6, $8 to Store 7 + +From Center 2: $2 to Store 1, $4 to Store 2, $3 to Store 3, $6 to Store 4, $5 to Store 5, $4 to Store 6, $7 to Store 7 + +From Center 3: $4 to Store 1, $6 to Store 2, $5 to Store 3, $8 to Store 4, $7 to Store 5, $6 to Store 6, $9 to Store 7 + +From Center 4: $3 to Store 1, $5 to Store 2, $4 to Store 3, $7 to Store 4, $6 to Store 5, $5 to Store 6, $8 to Store 7 + +From Center 5: $2 to Store 1, $4 to Store 2, $3 to Store 3, $6 to Store 4, $5 to Store 5, $4 to Store 6, $7 to Store 7 + +From Center 6: $3 to Store 1, $5 to Store 2, $4 to Store 3, $7 to Store 4, $6 to Store 5, $5 to Store 6, $8 to Store 7 + +From Center 7: $2 to Store 1, $4 to Store 2, $3 to Store 3, $6 to Store 4, $5 to Store 5, $4 to Store 6, $7 to Store 7 + +From Center 8: $4 to Store 1, $6 to Store 2, $5 to Store 3, $8 to Store 4, $7 to Store 5, $6 to Store 6, $9 to Store 7 + +From Center 9: $3 to Store 1, $5 to Store 2, $4 to Store 3, $7 to Store 4, $6 to Store 5, $5 to Store 6, $8 to Store 7 + +Demand of Each Retail Store (in units): +- Store 1: 695 +- Store 2: 737 +- Store 3: 452 +- Store 4: 616 +- Store 5: 587 +- Store 6: 779 +- Store 7: 892 + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1864 +- Center 2: 1980 +- Center 3: 814 +- Center 4: 864 +- Center 5: 1320 +- Center 6: 1952 +- Center 7: 1447 +- Center 8: 1886 +- Center 9: 1962 + +The ultimate goal for GlobalSupply Inc. is to fulfill the demand at each retail store while minimizing the total cost. This total cost encompasses both the opening costs of selecting certain distribution centers and the ongoing transportation costs associated with moving goods from these centers to the retail stores. + +Question: +What is the optimal total cost for GlobalSupply Inc. to open distribution centers and transport goods to meet all retail stores' demands, thereby minimizing the combination of opening and transportation costs?",[456123.0] +['Diet Problem'],"Imagine you are going on a fitness journey and have set specific nutritional goals for your daily intake. You aim to consume 74 grams of protein, 166 grams of carbohydrates, and 2210 calories each day. To achieve this, you've found four food items that you can incorporate into your meals: Chicken, Rice, Beans, and Avocados. Each of these foods has its own cost and provides a specific amount of protein, carbohydrates, and calories. + +Here's what each food offers: + +- Chicken: Costs $4 and provides 14 grams of protein, 7 grams of carbohydrates, and 208 calories. +- Rice: Costs $4 and provides 5 grams of protein, 13 grams of carbohydrates, and 140 calories. +- Beans: Costs $2 and are quite nutritious, providing 9 grams of protein, 26 grams of carbohydrates, and 280 calories. +- Avocado: Costs $4 and offers 10 grams of protein, 14 grams of carbohydrates, and 199 calories. + +Your challenge is to figure out the least expensive way to satisfy your protein, carbohydrate, and calorie goals using only these four foods. So the question is, what is the minimum cost required to meet your daily nutritional needs? You can only buy the food in integer servings. Please, your response should be a single numerical answer representing the optimal cost value.",[18.0] +['Capacitated Facility Location Problem'],"SupplyTek is a retail supply company that needs to optimize its distribution network. The company has four retail stores that require supply from a choice of seven possible distribution centers. Each center comes with its own opening cost and a limit on how much it can supply. The cost of transporting goods from these centers to the retail stores varies. The objective is to meet the demand at each retail store at the minimum possible total cost, which includes the opening costs of the distribution centers and the transportation costs of the goods. + +Given Data: +- Number of Potential Distribution Centers: 7 +- Number of Retail Stores Needing Supply: 4 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: 151,000 +- Center 2: 192,000 +- Center 3: 114,000 +- Center 4: 171,000 +- Center 5: 160,000 +- Center 6: 120,000 +- Center 7: 182,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $3 to Store 1, $3 to Store 2, $5 to Store 3, $4 to Store 4 +- From Center 2: $3 to Store 1, $5 to Store 2, $2 to Store 3, $4 to Store 4 +- From Center 3: $2 to Store 1, $4 to Store 2, $5 to Store 3, $1 to Store 4 +- From Center 4: $4 to Store 1, $2 to Store 2, $5 to Store 3, $4 to Store 4 +- From Center 5: $1 to Store 1, $1 to Store 2, $3 to Store 3, $3 to Store 4 +- From Center 6: $2 to Store 1, $4 to Store 2, $4 to Store 3, $3 to Store 4 +- From Center 7: $4 to Store 1, $4 to Store 2, $1 to Store 3, $3 to Store 4 + +Demand of Each Retail Store (in units): +- Store 1: 884 +- Store 2: 420 +- Store 3: 566 +- Store 4: 673 + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1187 +- Center 2: 1400 +- Center 3: 1115 +- Center 4: 813 +- Center 5: 1041 +- Center 6: 1576 +- Center 7: 1364 + +Question: +What is the minimum total cost for SupplyTek to open distribution centers and transport goods to meet all retail stores' demands? The total cost should include both the opening costs of the centers and the transportation costs of supplying the stores.",[240385.0] +"[""Diet Problem""]","Imagine you are tasked with planning a nutritious and cost-effective meal plan for the day, ensuring that you meet specific nutritional targets. You have a selection of six different food items, each with unique nutritional values and costs. Your goal is to meet the daily dietary requirements for protein, carbohydrates, and calories without overspending. + +Here's a breakdown of the food items available: + +- Food_1: Costs $4, provides 17 grams of protein, 8 grams of carbohydrates, and 237 calories. +- Food_2: Costs $2, provides 4 grams of protein, 24 grams of carbohydrates, and 213 calories. +- Food_3: Costs $6, provides 7 grams of protein, 27 grams of carbohydrates, and 133 calories. +- Food_4: Costs $2, provides 14 grams of protein, 16 grams of carbohydrates, and 118 calories. +- Food_5: Costs $6, provides 13 grams of protein, 1 gram of carbohydrates, and 136 calories. +- Food_6: Costs $8, provides 1 gram of protein, 13 grams of carbohydrates, and 225 calories. + +The dietary requirements you need to meet are at least 76 grams of protein, 173 grams of carbohydrates, and 1751 calories for the day. + +What is the minimum cost required to meet these nutritional needs using any combination of these food items? You could only assign integral portions of each type of food.",[22.0] +"[""Transportation Problem""]","Imagine that you are responsible for distributing a supply of vaccines among four different cities to tackle a healthcare crisis. Each city starts with a certain number of vaccines but needs a specific quantity to ensure that all its residents can be vaccinated. + +Here's what each city currently has and what they require: + +- City 1 has 483 vaccines but needs 346 vaccines. +- City 2 has 265 vaccines but needs 221 vaccines. +- City 3 has 130 vaccines but needs 217 vaccines. +- City 4 has 361 vaccines but needs 293 vaccines. + +In order to meet these healthcare demands, you can transport vaccines from one city to another. However, it costs money to do so. The cost varies depending on which cities you're moving vaccines between. Here's a list showing how much it costs to move vaccines from one city to any other: + +- Moving vaccines from/to City 1: To City 2 costs 2, to City 3 costs 19, to City 4 costs 13. +- Moving vaccines from/to City 2: To City 1 costs 34, to City 3 costs 10, to City 4 costs 48. +- Moving vaccines from/to City 3: To City 1 costs 46, to City 2 costs 22, to City 4 costs 46. +- Moving vaccines from/to City 4: To City 1 costs 48, to City 2 costs 15, to City 3 costs 23. + +Your goal is to ensure every city has the vaccines it needs at the lowest possible cost. What would be the minimum cost to ensure all cities have enough vaccines?",[956.0] +['Capacitated Facility Location Problem'],"In the bustling market of Logistico, a strategic dilemma looms over the horizon for the newly-established retail giant, SupplyChainMax. With an expansive network of five retail stores to cater to, the company stands at the crossroads of selecting the most cost-effective distribution strategy from a pool of eight potential distribution centers. Each center not only comes with its unique price tag for opening but also a specific capacity limit and variable transportation costs to each store. The crux of SupplyChainMax's challenge lies in opening the right blend of distribution centers that would minimally cost them to meet the retail stores' demands, ensuring a smooth flow of goods at the lowest possible overall cost. + +Given Data: +- Number of Potential Distribution Centers: 8 +- Number of Retail Stores to be Supplied: 5 + +Opening Costs for Each Distribution Center (in dollars): +- Distribution Center 1: $151,000 +- Distribution Center 2: $192,000 +- Distribution Center 3: $114,000 +- Distribution Center 4: $171,000 +- Distribution Center 5: $160,000 +- Distribution Center 6: $120,000 +- Distribution Center 7: $182,000 +- Distribution Center 8: $186,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Distribution Center 1: $3 to Store 1, $3 to Store 2, $5 to Store 3, $4 to Store 4, $3 to Store 5 +- From Distribution Center 2: $5 to Store 1, $2 to Store 2, $4 to Store 3, $2 to Store 4, $4 to Store 5 +- From Distribution Center 3: $5 to Store 1, $1 to Store 2, $4 to Store 3, $2 to Store 4, $5 to Store 5 +- From Distribution Center 4: $4 to Store 1, $1 to Store 2, $1 to Store 3, $3 to Store 4, $3 to Store 5 +- From Distribution Center 5: $2 to Store 1, $4 to Store 2, $4 to Store 3, $3 to Store 4, $4 to Store 5 +- From Distribution Center 6: $4 to Store 1, $1 to Store 2, $3 to Store 3, $5 to Store 4, $3 to Store 5 +- From Distribution Center 7: $5 to Store 1, $1 to Store 2, $2 to Store 3, $4 to Store 4, $1 to Store 5 +- From Distribution Center 8: $4 to Store 1, $2 to Store 2, $2 to Store 3, $1 to Store 4, $2 to Store 5 + +Demand of Each Retail Store (in units): +- Store 1: 964 units +- Store 2: 739 units +- Store 3: 491 units +- Store 4: 766 units +- Store 5: 854 units + +Supply Capacity of Each Distribution Center (in units): +- Distribution Center 1: 1,308 units +- Distribution Center 2: 1,575 units +- Distribution Center 3: 834 units +- Distribution Center 4: 1,005 units +- Distribution Center 5: 1,904 units +- Distribution Center 6: 1,825 units +- Distribution Center 7: 1,821 units +- Distribution Center 8: 1,365 units + +Question: +Considering all the given data about opening costs, transportation costs per unit to stores, demand of each retail store, and supply capacity of each distribution center, what is the optimal total cost for SupplyChainMax to establish distribution centers and transport goods to fully meet the demands of all five retail stores? This optimal total cost includes both the opening costs of the selected distribution centers and the transportation costs of supplying the stores.",[394198.0] +"[""Transportation Problem""]","Imagine you're coordinating a crucial resource distribution across six different regions to meet specific needs. Each region has a certain stock of resources, but the demands vary, requiring a strategic reallocation to ensure every region's needs are adequately met. + +Here's the current and required resource allocation for each region: + +- Region 1 has 319 units but needs 243 units. +- Region 2 has 281 units but needs 243 units. +- Region 3 has 247 units but needs 124 units. +- Region 4 has 237 units but needs 202 units. +- Region 5 has 29 units but needs 84 units. +- Region 6 has 453 units but needs 28 units. + +Transporting resources between these regions incurs varying costs. The cost matrix detailing the expense of moving resources from one region to another is as follows: + +- From Region 1 to other regions: To Region 2 costs 11, to Region 3 costs 30, to Region 4 costs 12, to Region 5 costs 48, to Region 6 costs 23. +- From Region 2 to other regions: To Region 1 costs 1, to Region 3 costs 27, to Region 4 costs 9, to Region 5 costs 26, to Region 6 costs 13. +- From Region 3 to other regions: To Region 1 costs 29, to Region 2 costs 6, to Region 4 costs 27, to Region 5 costs 13, to Region 6 costs 44. +- From Region 4 to other regions: To Region 1 costs 6, to Region 2 costs 50, to Region 3 costs 12, to Region 5 costs 19, to Region 6 costs 32. +- From Region 5 to other regions: To Region 1 costs 26, to Region 2 costs 13, to Region 3 costs 29, to Region 4 costs 2, to Region 6 costs 24. +- From Region 6 to other regions: To Region 1 costs 9, to Region 2 costs 26, to Region 3 costs 2, to Region 4 costs 37, to Region 5 costs 22. + +Your task is to calculate the minimum cost required to redistribute the resources so that all regions have exactly what they need. What is the optimal value for the minimum cost of achieving this balance?",[715.0] +"[""Network Optimization""]","Welcome to the fictional city of Aquaville, where the city's administration is working on a groundbreaking project to maximize the efficiency of its new water distribution network. This network is designed to ensure that every corner of the city has access to water, especially in times of need. The system is made up of 9 pivotal stations - including the main water source and the central distribution hub - connected by a series of pipelines with varying capacities. + +Here's a closer look at the capacities (in thousands of gallons per hour) that each pipeline can handle, showcasing the maximum flow from one station to another across the city: + +- From Station 0 (Source): Water can be directed to Station 1 (2 gallons), Station 2 (8 gallons), Station 3 (17 gallons), Station 4 (16 gallons), Station 5 (20 gallons), Station 6 (15 gallons), Station 7 (12 gallons), and Station 8 (9 gallons). +- From Station 1: Water can flow to Station 0 (4 gallons), Station 2 (11 gallons), Station 3 (7 gallons), Station 4 (3 gallons), Station 5 (19 gallons), Station 6 (10 gallons), Station 7 (19 gallons), and Station 8 (8 gallons). +- From Station 2: Water can move to Station 0 (14 gallons), Station 1 (7 gallons), Station 3 (7 gallons), Station 4 (13 gallons), Station 5 (19 gallons), Station 6 (18 gallons), Station 7 (20 gallons), and Station 8 (5 gallons). +- From Station 3: Water can be sent to Station 0 (17 gallons), Station 1 (17 gallons), Station 2 (7 gallons), Station 4 (11 gallons), Station 5 (7 gallons), Station 6 (17 gallons), Station 7 (0 gallons), and Station 8 (10 gallons). +- From Station 4: Water can be distributed to Station 0 (20 gallons), Station 1 (10 gallons), Station 2 (2 gallons), Station 3 (18 gallons), Station 5 (15 gallons), Station 6 (8 gallons), Station 7 (6 gallons), and Station 8 (18 gallons). +- From Station 5: Water can be channeled to Station 0 (4 gallons), Station 1 (9 gallons), Station 2 (12 gallons), Station 3 (1 gallon), Station 4 (17 gallons), Station 6 (0 gallons), Station 7 (13 gallons), and Station 8 (18 gallons). +- From Station 6: Water can flow to Station 0 (12 gallons), Station 1 (15 gallons), Station 2 (12 gallons), Station 3 (19 gallons), Station 4 (18 gallons), Station 5 (19 gallons), Station 7 (6 gallons), and Station 8 (17 gallons). +- From Station 7: Water can be directed to Station 0 (14 gallons), Station 1 (1 gallon), Station 2 (17 gallons), Station 3 (20 gallons), Station 4 (11 gallons), Station 5 (20 gallons), Station 6 (6 gallons), and Station 8 (4 gallons). +- From Station 8 (Destination): Water can be sent back to Station 0 (19 gallons), Station 1 (3 gallons), Station 2 (19 gallons), Station 3 (12 gallons), Station 4 (8 gallons), Station 5 (18 gallons), Station 6 (17 gallons), and Station 7 (13 gallons). + +In this intricate network, a capacity of ""0"" implies that there is no direct connection for water flow between those two specific stations. + +The critical mission is to determine the maximum quantity of water that can be efficiently distributed from Station 0, the primary source, to Station 8, the central distribution hub, per hour. The flow through each pipeline must not exceed its maximum specified capacity. + +What is the optimal amount of water (in thousands of gallons per hour) that can be transported from the source to the distribution hub (equivalently, the net outflow from the source or net amount that goes to the distribution hub), ensuring the network operates at its maximum potential?",[89.0] +['Capacitated Facility Location Problem'],"GlobalSupply Inc. is at a crucial juncture in streamlining its logistics to efficiently service six retail outlets spread across different locations. The company has narrowed down its options to four possible distribution centers, each with unique characteristics, including varying operational costs and different capacities for handling supply. Moreover, the cost to transport goods from these centers to each retail outlet is not uniform, adding layers of complexity to the decision-making process. The overarching goal for GlobalSupply Inc. is to ensure that the demands of all retail stores are met at the lowest possible total cost, incorporating both the opening expenses for these distribution centers and the costs associated with transporting goods to the stores. + +Given Data: +- Number of Potential Distribution Centers: 4 +- Number of Retail Stores to be Supplied: 6 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $5 to Store 1, $5 to Store 2, $2 to Store 3, $3 to Store 4, $3 to Store 5, $3 to Store 6 +- From Center 2: $5 to Store 1, $4 to Store 2, $3 to Store 3, $5 to Store 4, $2 to Store 5, $4 to Store 6 +- From Center 3: $2 to Store 1, $4 to Store 2, $5 to Store 3, $1 to Store 4, $4 to Store 5, $2 to Store 6 +- From Center 4: $5 to Store 1, $4 to Store 2, $1 to Store 3, $1 to Store 4, $3 to Store 5, $3 to Store 6 + +Demand of Each Retail Store (in units): +- Store 1: 910 units +- Store 2: 875 units +- Store 3: 589 units +- Store 4: 962 units +- Store 5: 966 units +- Store 6: 643 units + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1631 units +- Center 2: 1954 units +- Center 3: 1446 units +- Center 4: 820 units + +Question: +What is the optimal total cost for GlobalSupply Inc. to open the necessary distribution centers and transport goods in such a way that all retail stores' demands are fulfilled? This total should encompass both the operational opening costs and the logistical costs of supply to the outlets.",[469200.0] +"[""Capacitated Facility Location Problem""]","SupplyChain Solutions Inc. is facing a complex logistical challenge in optimizing its distribution network. With five retail stores requiring steady supplies and six possible distribution centers to choose from, the task at hand is to find the most cost-effective way to keep the stores well-stocked. Each distribution center has its own fixed opening cost and a maximum supply capacity, while the cost to transport goods from these centers to the stores varies. The goal is straightforward yet daunting: to fulfill the demand of each retail store while minimizing the total operational cost, which includes both the opening of distribution centers and the transportation of goods. + +Given Data: +- Number of Potential Distribution Centers: 6 +- Number of Retail Stores Needing Supply: 5 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 +- Center 6: $120,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $2 to Store 1, $3 to Store 2, $3 to Store 3, $3 to Store 4, $5 to Store 5 +- From Center 2: $4 to Store 1, $3 to Store 2, $5 to Store 3, $2 to Store 4, $4 to Store 5 +- From Center 3: $2 to Store 1, $4 to Store 2, $5 to Store 3, $1 to Store 4, $4 to Store 5 +- From Center 4: $2 to Store 1, $5 to Store 2, $4 to Store 3, $1 to Store 4, $1 to Store 5 +- From Center 5: $3 to Store 1, $3 to Store 2, $2 to Store 3, $4 to Store 4, $4 to Store 5 +- From Center 6: $3 to Store 1, $4 to Store 2, $4 to Store 3, $1 to Store 4, $3 to Store 5 + +Demand of Each Retail Store (in units): +- Store 1: 884 +- Store 2: 420 +- Store 3: 566 +- Store 4: 673 +- Store 5: 787 + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1400 +- Center 2: 1115 +- Center 3: 813 +- Center 4: 1041 +- Center 5: 1576 +- Center 6: 1364 + +Question: +What is the optimal total cost for SupplyChain Solutions Inc. to open distribution centers and transport goods to meet all retail stores' demands, including both the opening costs of the centers and the transportation costs of supplying the stores?",[392760.0] +"[""Network Optimization""]","Welcome to the scenario of an intricate transportation network connecting 8 bustling cities, designed to efficiently distribute essential supplies during a crisis. Each city is a node in this network, ranging from the primary distribution center (the source) to a central hub designated as the destination. The connections between these cities are roads, each with a specific carrying capacity that denotes the maximum amount of supplies (in tons) that can traverse these roads per day. + +Here's the layout of the road capacities between cities: + +- From City 0 (Source): Supplies can be sent to City 1 (11 tons), City 2 (1 ton), City 3 (6 tons), City 4 (4 tons), City 5 (5 tons), City 6 (7 tons), and City 7 (1 ton). +- From City 1: Supplies can be routed to City 0 (3 tons), City 2 (15 tons), City 3 (15 tons), City 4 (16 tons), City 5 (9 tons), City 6 (14 tons), and City 7 (12 tons). +- From City 2: Can dispatch supplies to City 0 (19 tons), City 1 (9 tons), City 3 (5 tons), City 4 (9 tons), City 5 (11 tons), City 6 (16 tons), and City 7 (3 tons). +- From City 3: Can send supplies to City 0 (15 tons), City 1 (1 ton), City 2 (14 tons), City 4 (0 tons, indicating no direct road), City 5 (4 tons), City 6 (1 ton), and City 7 (15 tons). +- From City 4: Can transport supplies to City 0 (8 tons), City 1 (16 tons), City 2 (7 tons), City 3 (3 tons), City 5 (16 tons), City 6 (10 tons), and City 7 (11 tons). +- From City 5: Supplies can be moved to City 0 (7 tons), City 1 (8 tons), City 2 (1 ton), City 3 (11 tons), City 4 (19 tons), City 6 (3 tons), and City 7 (3 tons). +- From City 6: Can route supplies to City 0 (2 tons), City 1 (19 tons), City 2 (2 tons), City 3 (6 tons), City 4 (12 tons), City 5 (20 tons), and City 7 (6 tons). +- From City 7 (Destination): Supplies can be sent back to City 0 (15 tons), City 1 (17 tons), City 2 (4 tons), City 3 (7 tons), City 4 (8 tons), City 5 (2 tons), and City 6 (16 tons). + +In this system, a capacity of ""0"" signifies no direct roadway connection between the two cities. + +The challenge is to determine the maximum amount of supplies that can be delivered from City 0 (the source) to City 7 (the destination) per day, without exceeding the road capacities. + +What is the maximum amount of supplies, in tons, that can be transported from the source to the destination in one day?",[35.0] +['Diet Problem'],"Let's imagine you are a student on a budget, trying to create a meal plan to meet your nutritional needs while keeping your finances in check. You have five different food items to consider: Lentils, Rice, Chicken, Beef, and Tofu. Each of these foods provides a varying amount of protein, carbs, and calories, and they also come with different price tags. + +Here is the nutritional content and cost of each food: + +- Lentils: For 7 dollars, Lentils provide you with 13 grams of protein, 21 grams of carbohydrates, and 284 calories. +- Rice: Providing 2 grams of protein, a substantial 30 grams of carbohydrates, and 55 calories, Rice comes at 6 dollars. +- Chicken: At the same cost as Rice, Chicken offers 7 grams of protein, 13 grams of carbohydrates, and a hefty 276 calories. +- Beef: Priced at 10 dollars, Beef gives you 18 grams of protein, 8 grams of carbohydrates, and 204 calories. +- Tofu: The cheapest option at 2 dollars, Tofu still provides a good amount of nutrients with 12 grams of protein, 28 grams of carbohydrates, and 160 calories. + +In your meal planning, you aim to get at least 79 grams of protein, 170 grams of carbohydrates, and 1699 calories from your selection of these foods. The challenge is to meet these nutritional targets in the most cost-effective way possible. What is the least amount of money you have to spend to meet your dietary needs with these five food options? + +Remark: Please note that your response should be a single answer question, asking for only the optimal value. This problem is set under the scenario of food selection. You could only assign integral portions of each type of food.",[22.0] +['Network Optimization'],"Imagine we're managing a telecommunications network that spans 6 key points, from a primary data center (Point 0) to a major user hub (Point 5). These points are interconnected by data lines, each with a specific capacity for data transfer. + +Here are the capacities of each data line, indicating the maximum amount of data (in Gigabytes per second) that can flow from one point to another: + +- From Point 0 (Data Center): Can send data to Point 1 (14 GB/s), Point 2 (9 GB/s), Point 3 (8 GB/s), Point 4 (7 GB/s), and Point 5 (8 GB/s). +- From Point 1: Can send data to Point 0 (4 GB/s), Point 2 (9 GB/s), Point 3 (10 GB/s), and Point 5 (12 GB/s). +- From Point 2: Can send data to Point 0 (4 GB/s), Point 1 (12 GB/s), Point 3 (13 GB/s), Point 4 (20 GB/s), and Point 5 (16 GB/s). +- From Point 3: Can send data to Point 0 (10 GB/s), Point 1 (8 GB/s), Point 2 (12 GB/s), and Point 5 (18 GB/s). +- From Point 4: Can send data to Point 0 (3 GB/s), Point 1 (13 GB/s), Point 2 (11 GB/s), Point 3 (20 GB/s), and Point 5 (2 GB/s). +- From Point 5 (User Hub): Can send data back to Point 0 (17 GB/s), Point 1 (4 GB/s), Point 2 (8 GB/s), Point 3 (2 GB/s), and Point 4 (12 GB/s). + +In this network, a capacity of ""0"" means there is no direct data line between those two points. + +The objective is to find out the maximum amount of data that can be transferred from Point 0 (Data Center) to Point 5 (User Hub) per second, while making sure that the data transfer through each line does not exceed its capacity. What is the maximum amount of data that can be sent from the data center to the user hub per second in Gigabytes?",[46.0] +"[""Diet Problem""]","Imagine you're planning your meals for the day and you want to ensure you're meeting all your nutritional needs without overspending. You have five food options: Chicken, Pasta, Rice, Beans, and Cheese. Each of these foods provides different amounts of protein, carbohydrates, and calories, along with their respective costs. + +Here's a breakdown of what each food offers and how much it costs: + +- Chicken: Provides 5 grams of protein, 13 grams of carbohydrates, and 121 calories for $7. +- Pasta: Offers 3 grams of protein, 23 grams of carbohydrates, and 274 calories for $10. +- Rice: Contains 11 grams of protein, 27 grams of carbohydrates, and 251 calories for $7. +- Beans: Supplies 10 grams of protein, 27 grams of carbohydrates, and 259 calories for $9. +- Cheese: Gives you 4 grams of protein, 18 grams of carbohydrates, and 199 calories for $7. + +Your goal is to get at least 60 grams of protein, 146 grams of carbohydrates, and 1727 calories from any combination of these foods. What is the cheapest cost to meet your nutritional needs? You could only assign integral portions of each type of food.",[49.0] +['Diet Problem'],"Imagine you're responsible for preparing the meals for a professional athlete who has specific dietary needs. You have six different food items available to purchase and cook: Chicken, Rice, Tuna, Beef, Corn, and Whey Protein. Each food provides various quantities of protein, carbohydrates, and calories, and they each have their own cost. + +Here's a quick rundown of what each food item provides and its cost: + +- Chicken: Offers 13 grams of protein, 27 grams of carbohydrates, and 81 calories for $1. +- Rice: Provides 1 gram of protein, 1 gram of carbohydrate, and a whopping 137 calories for $9. +- Tuna: Delivers 17 grams of protein, 2 grams of carbohydrates, and 53 calories for $2. +- Beef: Contains 15 grams of protein, 18 grams of carbohydrates, and 126 calories for $9. +- Corn: Comes with 2 grams of protein, 19 grams of carbohydrates, and 117 calories for $4. +- Whey Protein: Provides a hefty 15 grams of protein, 11 grams of carbohydrates, and 277 calories for $6. + +The athlete's daily dietary targets are at least 58 grams of protein, 148 grams of carbohydrates, and 2454 calories. Your mission is to determine the lowest-cost way to satisfy all these nutritional targets utilizing these six food options. So, what is the least amount of money you'll have to spend to meet the athlete's dietary requirements? You could only assign integral portions of each type of food.",[31.0] +['Diet Problem'],"Imagine you're creating a meal plan for yourself and want to ensure you're getting the right amount of nutrients, but you also want to keep your budget in check. You're considering four different food items: Steak, Eggs, Pasta, and Chicken. Each of these food items has a different nutritional content and cost. + +Here's a brief overview of what each food item provides and how much it costs: + +- Steak: Provides you with 20 grams of protein, 2 grams of carbohydrates, and 68 calories for $4. +- Eggs: Offer 7 grams of protein, 3 grams of carbohydrates, and 143 calories for just $1. +- Pasta: Gives you 1 gram of protein, a large 26 grams of carbohydrates, and 52 calories for $9. +- Chicken: Packs 10 grams of protein, 1 gram of carbohydrate, and 69 calories for $4. + +You need to ensure that you consume at least 71 grams of protein, 141 grams of carbohydrates, and 1839 calories from the food combination you choose. The challenge is to determine the least expensive way to meet these nutritional requirements using these four food options. What is the least amount of money you can spend to meet your nutritional needs? Note that your response should only include the optimal cost value.",[47.0] +"[""Transportation Problem""]","Imagine you are tasked with managing the distribution of medical supplies among six clinics in a large healthcare system. Each clinic currently holds a certain quantity of supplies but requires a specific amount to adequately meet the healthcare needs for the upcoming season. + +Here's the status of each of these clinics: + +- Clinic 1 has 467 units of supplies but needs 261 units. +- Clinic 2 has 121 units but needs 209 units. +- Clinic 3 has 109 units but needs 135 units. +- Clinic 4 has 131 units but needs 146 units. +- Clinic 5 has 500 units but needs 488 units. +- Clinic 6 has 10 units but needs 30 units. + +To ensure each clinic meets its demand, you can transfer supplies between them. However, each transfer incurs a cost which varies depending on the clinics involved. Below is the matrix showing the transportation costs (in dollars per unit) between each pair of clinics: + +- Moving supplies from/to Clinic 1: To Clinic 2 costs $43, to Clinic 3 costs $8, to Clinic 4 costs $15, to Clinic 5 costs $37, to Clinic 6 costs $9. +- Moving supplies from/to Clinic 2: To Clinic 1 costs $9, to Clinic 3 costs $50, to Clinic 4 costs $37, to Clinic 5 costs $50, to Clinic 6 costs $29. +- Moving supplies from/to Clinic 3: To Clinic 1 costs $3, to Clinic 2 costs $37, to Clinic 4 costs $7, to Clinic 5 costs $18, to Clinic 6 costs $27. +- Moving supplies from/to Clinic 4: To Clinic 1 costs $15, to Clinic 2 costs $3, to Clinic 3 costs $3, to Clinic 5 costs $2, to Clinic 6 costs $48. +- Moving supplies from/to Clinic 5: To Clinic 1 costs $34, to Clinic 2 costs $18, to Clinic 3 costs $44, to Clinic 4 costs $2, to Clinic 6 costs $27. +- Moving supplies from/to Clinic 6: To Clinic 1 costs $36, to Clinic 2 costs $49, to Clinic 3 costs $29, to Clinic 4 costs $12, to Clinic 5 costs $33. + +Your objective is to ensure that each clinic has the amount of supplies it needs while minimizing the total transportation costs. What would be the minimum cost to achieve this balance?",[2041.0] +['TravelingSalesman'],"Imagine a scenario where a salesperson needs to visit five distinct cities to meet clients and finalize deals. These cities are labeled as City 1, City 2, City 3, City 4, and City 5. The salesperson can initiate their journey from any city, but the challenge is to visit each other city exactly once and then return to the starting city, all while attempting to minimize the total travel cost involved. + +Here's a summary of the travel costs between the cities: +- From City 1, the travel costs are: to City 2 is 40 units, to City 3 is 40 units, to City 4 is 56 units, and to City 5 is 93 units. +- From City 2, the travel costs are: to City 1 is 40 units, to City 3 is 52 units, to City 4 is 71 units, and to City 5 is 90 units. +- From City 3, the travel costs are: to City 1 is 40 units, to City 2 is 52 units, to City 4 is 57 units, and to City 5 is 14 units. +- From City 4, the travel costs are: to City 1 is 56 units, to City 2 is 71 units, to City 3 is 57 units, and to City 5 is 13 units. +- From City 5, the travel costs are: to City 1 is 93 units, to City 2 is 90 units, to City 3 is 14 units, and to City 4 is 13 units. + +Now, what is the minimum total travel cost for the salesperson to visit each city exactly once and return to their starting city?",[175.0] +['TravelingSalesman'],"Imagine a logistics manager tasked with planning a route for a courier who needs to deliver packages to six different cities. The cities are labeled from 1 to 6. The courier has to start from any one of these cities, visit each of the other cities exactly once, and then return to the starting city. The main objective for the logistics manager is to design the route in such a way that the total travel cost is minimized. + +Here is a breakdown of the travel costs between each pair of cities: +- The cost to travel from City 1 to City 2 is 23 units, to City 3 is 50 units, to City 4 is 92 units, to City 5 is 56 units, and to City 6 is 91 units. +- From City 2, the cost is 23 units to City 1, 38 units to City 3, 25 units to City 4, 25 units to City 5, and 69 units to City 6. +- From City 3, the travel costs are 50 units to City 1, 38 units to City 2, 66 units to City 4, 51 units to City 5, and 76 units to City 6. +- Traveling from City 4, it costs 92 units to City 1, 25 units to City 2, 66 units to City 3, 97 units to City 5, and 37 units to City 6. +- From City 5, the costs are 56 units to City 1, 25 units to City 2, 51 units to City 3, 97 units to City 4, and 30 units to City 6. +- Lastly, from City 6, it costs 91 units to go to City 1, 69 units to City 2, 76 units to City 3, 37 units to City 4, and 30 units to City 5. + +What is the minimum total travel cost for the courier to visit each city exactly once and return to the starting city?",[216.0] +['Network Optimization'],"Embark on a journey through a complex network of digital data highways, designed to facilitate the swift and secure transmission of information across 9 strategic relay stations. This network, vital for the operation of a futuristic city, relies on its capacity to handle vast amounts of data, ensuring that communication flows from its source to the intended destination without a hitch. + +Here's a detailed blueprint of the data flow capacities (in terabytes per hour) from one relay station to another within this intricate system: + +- From Station 0 (Source): Has the capability to dispatch data to Station 1 (15 TB), Station 2 (1 TB), Station 3 (9 TB), Station 4 (18 TB), Station 5 (12 TB), Station 6 (18 TB), Station 7 (6 TB), and Station 8 (17 TB). +- From Station 1: Can transmit data to Station 0 (14 TB), Station 2 (13 TB), Station 3 (11 TB), Station 4 (14 TB), Station 5 (10 TB), Station 6 (15 TB), Station 7 (4 TB), and Station 8 (15 TB). +- From Station 2: Can route data to Station 0 (3 TB), Station 3 (2 TB), Station 4 (14 TB), Station 5 (16 TB), Station 6 (7 TB), Station 7 (12 TB), and Station 8 (4 TB). +- From Station 3: Can forward data to Station 0 (5 TB), Station 1 (20 TB), Station 4 (9 TB), Station 5 (1 TB), Station 6 (12 TB), Station 7 (1 TB), and Station 8 (20 TB). +- From Station 4: Can push data to Station 0 (11 TB), Station 1 (18 TB), Station 2 (12 TB), Station 3 (3 TB), Station 5 (4 TB), Station 6 (16 TB), Station 7 (15 TB), and Station 8 (2 TB). +- From Station 5: Can channel data to Station 0 (13 TB), Station 1 (8 TB), Station 2 (8 TB), Station 3 (6 TB), Station 4 (4 TB), Station 6 (8 TB), Station 7 (10 TB), and Station 8 (11 TB). +- From Station 6: Can send data to Station 0 (12 TB), Station 1 (1 TB), Station 2 (3 TB), Station 3 (4 TB), Station 4 (11 TB), Station 5 (8 TB), Station 7 (10 TB), and Station 8 (19 TB). +- From Station 7: Can relay data to Station 0 (18 TB), Station 1 (3 TB), Station 2 (9 TB), Station 3 (5 TB), Station 4 (9 TB), Station 5 (17 TB), Station 6 (10 TB), and Station 8 (13 TB). +- From Station 8 (Destination): Data can be reflected back to Station 0 (9 TB), Station 1 (8 TB), Station 2 (12 TB), Station 3 (2 TB), Station 4 (5 TB), Station 5 (2 TB), Station 6 (16 TB), and Station 7 (5 TB). + +In this virtual scenario, a capacity of ""0"" indicates that there is no direct data link between the specified stations, requiring the data to be rerouted through other nodes in the network. + +Your mission is to calculate the maximum amount of data (in terabytes per hour) that can be transmitted from Station 0 (the source) to Station 8 (the destination), ensuring that the traffic flowing through each data link does not exceed its maximum capacity. What is the optimal amount of data that can be relayed from the source to the destination per hour?",[96.0] +"[""Network Optimization""]","Welcome to the virtual challenge of constructing a futuristic energy distribution network spanning across 8 pivotal nodes, designed to optimize the flow of a newly discovered renewable energy source from its origin to a central processing facility. This network, conceptualized within a high-tech cityscape, involves complex routes and capacities that dictate the maximum energy flow possible between nodes, each representing energy substations or relay points. + +Here’s a closer look at the energy flow capacities (in megawatts) between these substations: + +- From Substation 0 (Source): Energy can be directed to Substation 1 (19 MW), Substation 2 (4 MW), Substation 3 (1 MW), Substation 4 (9 MW), Substation 5 (13 MW), Substation 6 (8 MW), and Substation 7 (20 MW). +- From Substation 1: Can relay energy to Substation 0 (10 MW), Substation 2 (12 MW), Substation 3 (12 MW), Substation 4 (8 MW), Substation 5 (12 MW), Substation 6 (7 MW), and Substation 7 (7 MW). +- From Substation 2: Can transfer energy to Substation 0 (13 MW), Substation 1 (11 MW), Substation 3 (8 MW), Substation 4 (4 MW), Substation 5 (1 MW), Substation 6 (2 MW), and Substation 7 (3 MW). +- From Substation 3: Can distribute energy to Substation 0 (4 MW), Substation 1 (12 MW), Substation 2 (13 MW), Substation 4 (18 MW), Substation 5 (6 MW), Substation 6 (19 MW), and Substation 7 (8 MW). +- From Substation 4: Can send energy to Substation 0 (9 MW), Substation 1 (16 MW), Substation 2 (2 MW), Substation 5 (14 MW), Substation 6 (18 MW), and Substation 7 (7 MW). +- From Substation 5: Can dispatch energy to Substation 0 (10 MW), Substation 1 (11 MW), Substation 2 (7 MW), Substation 3 (11 MW), Substation 4 (16 MW), Substation 6 (16 MW), and Substation 7 (5 MW). +- From Substation 6: Capable of forwarding energy to Substation 0 (14 MW), Substation 1 (11 MW), Substation 2 (9 MW), Substation 3 (11 MW), Substation 4 (3 MW), Substation 5 (17 MW), and Substation 7 (19 MW). +- From Substation 7 (Destination): Can receive energy and has the potential to send it back to Substation 0 (12 MW), Substation 1 (20 MW), Substation 2 (5 MW), Substation 3 (11 MW), Substation 4 (10 MW), Substation 5 (4 MW), and Substation 6 (6 MW). + +In this setup, a capacity of ""0"" signifies the absence of a direct energy transfer link between the two respective substations. + +The ultimate objective is to compute the maximum energy that can be transmitted from Substation 0 (the source) to Substation 7 (the destination) per time unit, with the condition that the energy flow through each link does not surpass its maximum capacity. What is the optimal value of energy (in megawatts) that can be efficiently relayed from the source to the destination per time unit?",[69.0] +"[""Network Optimization""]","Welcome to a complex and engaging scenario involving a futuristic transportation network connecting 8 critical hubs across a sprawling metropolis. This network, designed with cutting-edge technology, facilitates the seamless flow of resources - be it energy, data, or materials - from the primary distribution center to the ultimate receiving facility, ensuring the city's heartbeat never skips a beat. + +Here's a detailed layout of the transportation capacities between these hubs, measured in units of resources per hour that can be transmitted: + +- From Hub 0 (Source): Can dispatch resources to Hub 1 (15 units), Hub 2 (19 units), Hub 3 (4 units), Hub 4 (4 units), Hub 5 (3 units), Hub 6 (20 units), and Hub 7 (17 units). +- From Hub 1: Can transmit to Hub 0 (8 units), Hub 2 (6 units), Hub 3 (5 units), Hub 4 (20 units), Hub 5 (3 units), Hub 6 (15 units), and Hub 7 (1 unit). +- From Hub 2: Can route resources to Hub 0 (11 units), Hub 1 (13 units), Hub 3 (16 units), Hub 4 (15 units), Hub 5 (7 units), Hub 6 (10 units), and Hub 7 (13 units). +- From Hub 3: Can send to Hub 0 (17 units), Hub 1 (20 units), Hub 2 (6 units), Hub 4 (11 units), Hub 5 (3 units), Hub 6 (2 units). +- From Hub 4: Can distribute to Hub 0 (10 units), Hub 1 (15 units), Hub 2 (5 units), Hub 5 (9 units), Hub 6 (9 units), and Hub 7 (15 units). +- From Hub 5: Can forward to Hub 0 (9 units), Hub 1 (17 units), Hub 2 (5 units), Hub 3 (1 unit), Hub 4 (9 units), Hub 6 (20 units), and Hub 7 (2 units). +- From Hub 6: Can transfer to Hub 0 (4 units), Hub 1 (16 units), Hub 2 (10 units), Hub 3 (2 units), Hub 4 (1 unit), Hub 5 (6 units), and Hub 7 (4 units). +- Lastly, Hub 7 (Destination): Can receive back from Hubs 0 through 6 various capacities, with the specifics already detailed above, marking the end of our resource flow chain. + +In this futuristic city, a ""0"" capacity means there is no established direct transmission link between those particular hubs. + +The critical task at hand is to determine the maximum amount of resources that can be transported from Hub 0 (the source) to Hub 7 (the destination) per hour without overloading any part of this intricate network. What is the maximum resource flow, in units per hour, that can be achieved from the source to the destination?",[52.0] +['Capacitated Facility Location Problem'],"SupplyLink, a major supplier, is looking to optimize its distribution network to ensure the efficient supply of goods to its five retail stores from a selection of five potential distribution centers. Each distribution center incurs an initial opening cost and has a limit on how much it can supply. The transportation cost of goods from these centers to the retail stores also varies. The goal is to meet the demand at each retail store at minimal total cost, which includes the opening costs of distribution centers and the transportation cost of the goods. + +Given Data: +- Number of Potential Distribution Centers: 5 +- Number of Retail Stores Needing Supply: 5 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: 151,000 +- Center 2: 192,000 +- Center 3: 114,000 +- Center 4: 171,000 +- Center 5: 160,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $5 to Store 1, $2 to Store 2, $3 to Store 3, $3 to Store 4, $3 to Store 5 +- From Center 2: $5 to Store 1, $4 to Store 2, $3 to Store 3, $5 to Store 4, $2 to Store 5 +- From Center 3: $4 to Store 1, $2 to Store 2, $4 to Store 3, $5 to Store 4, $1 to Store 5 +- From Center 4: $4 to Store 1, $2 to Store 2, $5 to Store 3, $4 to Store 4, $1 to Store 5 +- From Center 5: $1 to Store 1, $3 to Store 2, $3 to Store 3, $2 to Store 4, $4 to Store 5 + +Demand of Each Retail Store (in units): +- Store 1: 589 +- Store 2: 962 +- Store 3: 966 +- Store 4: 643 +- Store 5: 904 + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1954 +- Center 2: 1446 +- Center 3: 820 +- Center 4: 1640 +- Center 5: 966 + +Question: +What is the minimum total cost that SupplyLink will incur to open distribution centers and transport goods in order to satisfy all retail stores' demands? This total cost includes both the opening costs of the centers and the transportation costs of supplying the stores.",[446011.0] +['Capacitated Facility Location Problem'],"Welcome to the strategic challenge faced by DistribuMax, a company at a crossroads in optimizing its distribution network to efficiently supply nine unique retail stores. Each store has a specific demand that must be met to ensure customer satisfaction and operational success. DistribuMax can select from nine potential distribution centers, each with its own set of opening costs and capabilities. Furthermore, the cost of transporting goods from these centers to the retail stores varies, adding a layer of complexity to the decision-making process. The goal is to meet the demands of each retail store at the lowest possible total cost, which includes both the opening costs of the distribution centers and the transportation costs of the goods. + +Given Data: +- Number of Potential Distribution Centers: 9 +- Number of Retail Stores Needing Supply: 9 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 +- Center 6: $120,000 +- Center 7: $182,000 +- Center 8: $186,000 +- Center 9: $174,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +Costs vary from $1 to $5 per unit, with each distribution center offering a different rate to each store, reflecting the logistical complexity and geographical challenges of the distribution network: +- From Center 1: $2 to Store 1, $3 to Store 2, $4 to Store 3, $1 to Store 4, $5 to Store 5, $2 to Store 6, $3 to Store 7, $4 to Store 8, and $1 to Store 9. +- From Center 2: $5 to Store 1, $2 to Store 2, $3 to Store 3, $4 to Store 4, $1 to Store 5, $3 to Store 6, $5 to Store 7, $2 to Store 8, and $4 to Store 9. +- From Center 3: $3 to Store 1, $4 to Store 2, $2 to Store 3, $5 to Store 4, $3 to Store 5, $1 to Store 6, $2 to Store 7, $3 to Store 8, and $5 to Store 9. +- From Center 4: $1 to Store 1, $5 to Store 2, $4 to Store 3, $2 to Store 4, $4 to Store 5, $5 to Store 6, $3 to Store 7, $1 to Store 8, and $2 to Store 9. +- From Center 5: $4 to Store 1, $1 to Store 2, $5 to Store 3, $3 to Store 4, $2 to Store 5, $4 to Store 6, $1 to Store 7, $5 to Store 8, and $3 to Store 9. +- From Center 6: $2 to Store 1, $3 to Store 2, $1 to Store 3, $4 to Store 4, $5 to Store 5, $2 to Store 6, $4 to Store 7, $3 to Store 8, and $1 to Store 9. +- From Center 7: $3 to Store 1, $2 to Store 2, $3 to Store 3, $1 to Store 4, $4 to Store 5, $3 to Store 6, $5 to Store 7, $2 to Store 8, and $4 to Store 9. +- From Center 8: $5 to Store 1, $4 to Store 2, $2 to Store 3, $3 to Store 4, $1 to Store 5, $4 to Store 6, $2 to Store 7, $4 to Store 8, and $3 to Store 9. +- From Center 9: $4 to Store 1, $3 to Store 2, $4 to Store 3, $2 to Store 4, $3 to Store 5, $1 to Store 6, $3 to Store 7, $5 to Store 8, and $2 to Store 9. + +Demand of Each Retail Store (in units): +- Store 1: 871 units +- Store 2: 462 units +- Store 3: 538 units +- Store 4: 898 units +- Store 5: 992 units +- Store 6: 791 units +- Store 7: 818 units +- Store 8: 688 units +- Store 9: 778 units + +Supply Capacity of Each Distribution Center (in units): +- Capacities range from 840 units to 1856 units, with each center having a unique capacity to provide for the stores. + +DistribuMax seeks to identify the optimal combination of distribution centers to open and the distribution of supplies to stores that will minimize the total operational cost. This includes meticulously planning which distribution centers to activate and how to distribute their inventory to meet the demands of the retail stores in the most cost-efficient manner. + +Question: +What is the optimal total cost for DistribuMax to open distribution centers and transport goods to meet the demands of all nine retail stores, considering both the opening costs of the centers and the transportation costs of supplying the stores?",[749272.0] +"[""Set Multi-Cover""]","A hospital wants to make a weekly night shift schedule for its nurses.The demand for the night shift on day j is dj , for j = 1, ..., 7. d1 = 5, d2 = 4, d3 = 7, d4 = 3, d5 = 8, d6 = 4, d7 = 3. Every nurse works 5 days in a row. We want to minimize the total number of nurses used while meeting all demand. Ignore the integrality constraints for now (i.e., we allow “half” nurse if necessary). Provide your answer rounded to the nearest integer.",[8.0] +"[""Transportation Problem""]","The China Railroad Ministry is in the process of planning relocations of freight cars among 5 regions of the country to get ready for the fall harvest. The following presents the costs of moving a car between each pair of regions. The costs are as follows: moving a car from region 1 to region 2 costs 10 units, to region 3 costs 12 units, to region 4 costs 17 units, and to region 5 costs 34 units. From region 2 to region 1 costs 10 units, to region 3 costs 18 units, to region 4 costs 8 units, and to region 5 costs 46 units. From region 3, moving a car to region 1 costs 12 units, to region 2 costs 18 units, to region 4 costs 9 units, and to region 5 costs 27 units. From region 4, the cost to move a car to region 1 is 17 units, to region 2 is 8 units, to region 3 is 9 units, and to region 5 is 20 units. Lastly, from region 5 to region 1 the cost is 34 units, to region 2 is 46 units, to region 3 is 27 units, and to region 4 is 20 units. The following displays the current number of cars and the number needed for harvest shipping in each region. Region 1 currently has 115 cars but needs 300. Region 2 has 385 cars available but requires 400. Region 3 has 410 cars against a need of 800. Region 4 is currently equipped with 480 cars, yet the need is only for 200. Finally, region 5 has 610 cars available, with a need for 300 cars. Minimize the cost to move the cars such as the need is met. ",[12170.0] +['Diet Problem'],"You've decided to take control of your health and focus on your nutritional intake. After researching and consulting with a dietitian, you've determined that your daily requirements are 78 grams of protein, 140 grams of carbs, and 1537 calories. Now, the next challenge is to meet these daily requirements in the most cost-effective way possible. + +Your local supermarket offers a variety of food items, each with its own cost and nutritional content. Here's what you have to choose from: + +- Chicken Breast: For $4, you'll get 14 grams of protein, 4 grams of carbohydrates, and 275 calories. +- Brown Rice: A $10 serving gives you 11 grams of protein, 17 grams of carbohydrates, and 151 calories. +- Tofu: For $6, you can get 20 grams of protein, 12 carbs, and 155 calories. +- Spinach: At only $1, you get 6 grams of protein, 20 carbs, and 106 calories. This is the cheapest option but also the least calorie-dense. +- Almonds: For $8, you get 9 grams of protein, 11 grams of carbohydrates, and 279 calories. +- Salmon: Finally, for $6, you get a whopping 20 grams of protein, 19 grams of carbohydrates, and 93 calories. + +With these food options available, your task is to figure out how to meet your daily nutritional requirements while keeping your costs to a minimum. So what is the least amount of money you need to spend to meet your daily nutritional needs? You could only assign integral portions of each type of food.",[15.0] +['Network Optimization'],"There are two special nodes marked as S (likely the start) and T (likely the target or terminal). The other nodes are numbered from 2 to 7. Edges connect these nodes and each edge is labeled with a number indicating its weight. Node S is connected to nodes 2 and 3 with edge weights of 4 and 5, respectively. Node 2 is connected to node S with a weight of 5, to node 4 with a weight of 3, and to node 3 with a weight of 2. Node 3 is connected to node S with a weight of 4, to node 2 with a weight of 2, to node 5 with a weight of 1, and to node 6 with a weight of 7. Node 4 is connected to node 2 with a weight of 3 and to node 6 with a weight of 2. Node 5 is connected to node 3 with a weight of 1 and to node T with a weight of 5. Node 6 is connected to nodes 3, 4, and 7 with edge weights of 7, 2, and 3, respectively. Node 7 is connected to nodes 6 and T with edge weights of 3 and 1, respectively. Node T is connected to nodes 5 and 7 with edge weights of 5 and 1, respectively. Considering the weight as distance, find the shortest distance from S to T.Provide your answer rounded to the nearest meter.",[11.0] +['TravelingSalesman'],"Imagine a salesperson planning a trip across four cities, labeled City 1, City 2, City 3, and City 4. This salesperson must devise a travel route that allows them to visit each of these cities exactly once and return to their starting point, all while aiming to keep their travel costs as low as possible. + +Here's a breakdown of the travel costs between each pair of cities: +- The travel cost from City 1 to City 2 is 94 units, to City 3 is 34 units, and to City 4 is 87 units. +- From City 2, it costs 94 units to get back to City 1, 62 units to go to City 3, and 87 units to reach City 4. +- Traveling from City 3, the costs are 34 units back to City 1, 62 units to City 2, and 97 units to City 4. +- Lastly, from City 4, it costs 87 units to travel back to City 1, 87 units to City 2, and 97 units to City 3. + +The challenge for the salesperson is to find the route that minimizes the total travel cost. What is the minimum total travel cost for the salesperson to visit each city exactly once and return to the starting city?",[270.0] +"[""TravelingSalesman""]","Consider a scenario where a delivery driver has a list of five different locations: Shop1, Shop2, Shop3, Shop4, and Shop5. The driver's task is to deliver packages to each shop and then return to the starting point. The catch is, the driver can visit each shop only once. The cost associated with traveling from one location to another varies, and might include factors such as fuel costs, traffic considerations, and road tolls. The driver's goal is to minimize the total travel cost. + +Here's a quick summary of the travel costs between the shops: + +The cost to travel from Shop1 to Shop2 is 13 units, to Shop3 is 87 units, to Shop4 is 76 units, and to Shop5 is 50 units. +From Shop2, it costs 13 units to reach Shop1, 15 units to get to Shop3, 65 units to Shop4, and 85 units to Shop5. +Traveling from Shop3, the costs are 87 units to Shop1, 15 units to Shop2, 45 units to Shop4, and 81 units to Shop5. +From Shop4, it costs 76 units to go to Shop1, 65 units to Shop2, 45 units to Shop3, and 80 units to Shop5. +Lastly, from Shop5, it takes 50 units to reach Shop1, 85 units to Shop2, 81 units to Shop3, and 80 units to Shop4. + +What is the optimal travel cost for the driver to deliver packages to each shop exactly once and return to the starting point, while minimizing the total travel cost?",[203.0] +"[""Transportation Problem""]","Imagine you are tasked with coordinating the distribution of emergency food supplies across six different regions following a natural disaster. Each region started with a certain amount of supplies but now has specific needs to adequately support their populations until further aid can arrive. + +Here are the current food supply quantities and the required amounts for each region: + +- Region 1 has 268 units of food but needs 165 units. +- Region 2 has 222 units of food but needs 246 units. +- Region 3 has 357 units of food but needs 262 units. +- Region 4 has 128 units of food but needs 286 units. +- Region 5 has 465 units of food but needs 439 units. +- Region 6 has 48 units of food but needs 83 units. + +To transport food supplies between the regions, there are associated costs which vary depending on the route taken. Here is the cost matrix showing the transportation costs per unit of food between each pair of regions: + +- From Region 1 to other regions: To Region 2 costs 49, to Region 3 costs 11, to Region 4 costs 23, to Region 5 costs 2, to Region 6 costs 39. +- From Region 2 to other regions: To Region 1 costs 33, to Region 3 costs 26, to Region 4 costs 2, to Region 5 costs 2, to Region 6 costs 14. +- From Region 3 to other regions: To Region 1 costs 12, to Region 2 costs 31, to Region 4 costs 36, to Region 5 costs 30, to Region 6 costs 34. +- From Region 4 to other regions: To Region 1 costs 18, to Region 2 costs 50, to Region 3 costs 12, to Region 5 costs 19, to Region 6 costs 20. +- From Region 5 to other regions: To Region 1 costs 3, to Region 2 costs 13, to Region 3 costs 20, to Region 4 costs 29, to Region 6 costs 1. +- From Region 6 to other regions: To Region 1 costs 2, to Region 2 costs 9, to Region 3 costs 50, to Region 4 costs 36, to Region 5 costs 39. + +Your objective is to ensure each region has the food supplies it needs at the lowest possible transportation cost. What is the minimum cost required to achieve this?",[3609.0] +"[""Profit Maximization Problem""]","A company produces two kinds of products. A product of the first type requires 1/4 hours of assembly labor, 1/8 hours of testing, and $1.2 worth of raw materials. A product of the second type requires 1/3 hours of assembly, 1/3 hours of testing, and $0.9 worth of raw materials. Given the current personnel of the company, there can be at most 90 hours of assembly labor and 80 hours of testing each day. Suppose that up to 50 hours of overtime assembly labor can be scheduled, at a cost of $9 per hour. Products of the first and second type have a market value of $9 and $8 respectively. Try to maximize the daily profit of the company. ",[3918.0] +"[""Set Multi-Cover""]","A hospital wants to make a weekly night shift schedule for its nurses.The demand for the night shift on day j is dj , for j = 1, ..., 7 we have d1 = 2, d2 = 4, d3 = 4, d4 = 3, d5 = 1, d6 = 2, d7 = 3. Every nurse works 5 days in a row. We want to minimize the total number of nurses used while meeting all demands. Ignore the integrality constraints for now (i.e., we allow “fraction” of nurse if necessary). Provide your answer rounded to the nearest integer.",[4.0] +"[""Capacitated Facility Location Problem""]","In the bustling landscape of retail and distribution, LogicFlow Inc. is at a pivotal juncture aiming to streamline its supply chain for maximum efficiency and minimal cost. The company needs to determine the best strategy for distributing products to its eight retail stores, which are scattered across different locations. The challenge lies in selecting from eight potential distribution centers, each with its unique opening cost and varying transportation costs per unit to the stores. The goal is clear: to fulfill the demand of every retail store while keeping the overall costs as low as possible. + +The specifics of this optimization scenario are as follows: + +- There are 8 potential Distribution Centers that LogicFlow Inc. is considering. +- The company operates 8 Retail Stores requiring supplies. + +The Opening Costs for Each Distribution Center (in dollars) are listed below: +- Distribution Center 1: $151,000 +- Distribution Center 2: $192,000 +- Distribution Center 3: $114,000 +- Distribution Center 4: $171,000 +- Distribution Center 5: $160,000 +- Distribution Center 6: $120,000 +- Distribution Center 7: $182,000 +- Distribution Center 8: $186,000 + +Transportation Costs Per Unit from Each Distribution Center to Each Retail Store (in dollars) are detailed as follows: +- From Distribution Center 1 to Stores 1–8: 3, 4, 5, 6, 5, 4, 3, 4 +- From Distribution Center 2 to Stores 1–8: 2, 3, 4, 5, 4, 3, 2, 4 +- From Distribution Center 3 to Stores 1–8: 4, 3, 2, 3, 4, 5, 6, 3 +- From Distribution Center 4 to Stores 1–8: 5, 4, 3, 2, 3, 4, 5, 2 +- From Distribution Center 5 to Stores 1–8: 6, 5, 4, 3, 2, 3, 4, 3 +- From Distribution Center 6 to Stores 1–8: 4, 4, 4, 4, 4, 4, 4, 4 +- From Distribution Center 7 to Stores 1–8: 3, 2, 3, 4, 5, 6, 2, 3 +- From Distribution Center 8 to Stores 1–8: 4, 3, 4, 3, 4, 5, 4, 2 + +The Demand of Each Retail Store (in units) is: +Store 1: 695 +Store 2: 520 +Store 3: 610 +Store 4: 430 +Store 5: 575 +Store 6: 680 +Store 7: 490 +Store 8: 440 + +The Supply Capacity of Each Distribution Center (in units) is as follows: +Distribution Center 1: 1,980 +Distribution Center 2: 1,750 +Distribution Center 3: 2,100 +Distribution Center 4: 1,600 +Distribution Center 5: 1,450 +Distribution Center 6: 1,900 +Distribution Center 7: 1,850 +Distribution Center 8: 1,962 + +Given these particulars, LogicFlow Inc. must make a calculated decision on which distribution centers to open and the most cost-efficient way to transport goods to meet the demands of all eight retail stores. The decision matrix includes opening costs for each center, the transportation cost per unit to each store, and the capacity to meet the store demands. + +Question: +What is the optimal total cost for LogicFlow Inc. to open the necessary distribution centers and transport goods in order to satisfy the demands of all eight retail stores? This total cost must include both the opening costs of the centers and the transportation costs of supplying the stores.",[398965.0] +"[""Transportation Problem""]","Imagine you're coordinating the distribution of medical supplies to four different clinics to prepare for an upcoming health drive. Each clinic starts with a certain stock of supplies, but each has a specific requirement to ensure they are adequately prepared. + +Here’s the initial stock and the required quantities for each clinic: +- Clinic 1 has 330 units but needs 476 units. +- Clinic 2 has 294 units but needs 92 units. +- Clinic 3 has 474 units but needs 385 units. +- Clinic 4 has 86 units but needs 203 units. + +To transport medical supplies from one clinic to another, there are associated costs which vary depending on the route between clinics. Here are the costs for transferring supplies between clinics: +- From Clinic 1: To Clinic 2 costs 2, to Clinic 3 costs 12, to Clinic 4 costs 19. +- From Clinic 2: To Clinic 1 costs 28, to Clinic 3 costs 9, to Clinic 4 costs 10. +- From Clinic 3: To Clinic 1 costs 23, to Clinic 2 costs 46, to Clinic 4 costs 7. +- From Clinic 4: To Clinic 1 costs 19, to Clinic 2 costs 16, to Clinic 3 costs 23. + +The challenge is to fulfill the needs of all clinics by redistributing the supplies among them at the lowest possible transportation cost. What is the minimum cost required to ensure all clinics have the necessary supplies?",[4813.0] +"[""Network Optimization""]","Imagine a bustling city where seven primary communication stations are in operation. These stations, ranging from a central source (Station 0) to the main endpoint (Station 6), are interconnected by a network of telecommunication lines with varying transfer capacities. + +Here are the capacities of each connection, representing the maximum amount of data (in terabytes) that can be transferred from one station to another: + +- From Station 0 (Source): Can send data to Station 1 (17TB), Station 2 (3TB), Station 3 (20TB), Station 4 (11TB), Station 5 (11TB), and Station 6 (3TB). +- From Station 1: Can send data to Station 0 (11TB), Station 2 (12TB), Station 3 (18TB), Station 4 (5TB), Station 5 (6TB), and Station 6 (4TB). +- From Station 2: Can send data to Station 0 (7TB), Station 1 (2TB), Station 3 (15TB), Station 4 (13TB), Station 5 (20TB), and Station 6 (18TB). +- From Station 3: Can send data to Station 0 (7TB), Station 1 (4TB), Station 2 (2TB), Station 4 (2TB), Station 5 (15TB), and Station 6 (3TB). +- From Station 4: Can send data to Station 0 (19TB), Station 1 (12TB), Station 2 (16TB), Station 3 (4TB), Station 5 (19TB), and Station 6 (4TB). +- From Station 5: Can send data to Station 0 (13TB), Station 1 (20TB), Station 2 (10TB), Station 3 (14TB), Station 4 (15TB), and Station 6 (1TB). +- From Station 6 (Endpoint): Can send data back to Station 0 (3TB), Station 1 (3TB), Station 2 (6TB), Station 3 (20TB), Station 4 (9TB), and Station 5 (10TB). + +In this scenario, a capacity of ""0"" indicates there is no direct telecommunication link between the two respective stations. + +Your challenge is to determine the maximum amount of data that can be transferred from Station 0 (the source) to Station 6 (the endpoint) per hour, while ensuring the data flow along each telecommunication line does not exceed its set capacity. What is the maximum amount of data, in terabytes, that can be transferred from the source to the endpoint within one hour?",[33.0] +"[""TravelingSalesman""]","In a scenario involving a traveling salesperson, there are six cities labeled 1 through 6. The salesperson needs to visit each city exactly once, starting and ending at the same city, with the objective to minimize the total travel cost incurred during the trip. The costs of traveling between each pair of cities are provided in a cost matrix and the specific travel costs between cities are as follows: + +- The cost from city 1 to city 2 is 86, to city 3 is 81, to city 4 is 64, to city 5 is 65, and to city 6 is 24. +- From city 2, the travel costs are 86 to city 1, 44 to city 3, 80 to city 4, 91 to city 5, and 23 to city 6. +- Traveling from city 3, the costs are 81 to city 1, 44 to city 2, 15 to city 4, 25 to city 5, and 89 to city 6. +- From city 4, it costs 64 to travel to city 1, 80 to city 2, 15 to city 3, 89 to city 5, and 41 to city 6. +- From city 5, the travel costs are 65 to city 1, 91 to city 2, 25 to city 3, 89 to city 4, and 29 to city 6. +- Lastly, from city 6, the travel costs are 24 to city 1, 23 to city 2, 89 to city 3, 41 to city 4, and 29 to city 5. + +Considering these travel costs, what is the minimum total travel cost for the salesperson to visit each city exactly once and return to the starting city?",[232.0] +"[""Network Optimization""]","There are two special nodes marked as S (likely the start) and T (likely the target or terminal). The other nodes are numbered from 2 to 7. Edges connect these nodes and each edge is labeled with a number indicating its weight. Node S is connected to nodes 2 and 3 with edge weights of 5 and 4, respectively. Node 2 is connected to node S with a weight of 5, to node 4 with a weight of 3, and to node 3 with a weight of 2. Node 3 is connected to node S with a weight of 4, to node 2 with a weight of 2, to node 5 with a weight of 1, and to node 6 with a weight of 7. Node 4 is connected to node 2 with a weight of 3 and to node 6 with a weight of 2. Node 5 is connected to node 3 with a weight of 1 and to node T with a weight of 5. Node 6 is connected to nodes 3, 4, and 7 with edge weights of 7, 2, and 3, respectively. Node 7 is connected to nodes 6 and T with edge weights of 3 and 1, respectively. Node T is connected to nodes 5 and 7 with edge weights of 5 and 1, respectively. Considering the weight as distance, find the shortest distance from S to T.Provide your answer rounded to the nearest meter.",[10.0] +"[""TravelingSalesman""]","Imagine a salesperson tasked with visiting four distinct cities—let's call them A, B, C, and D—to promote their products. The salesperson can commence their journey from any of these cities, with the requirement that they visit each city exactly once before returning to the starting point. The primary objective is to minimize the total travel cost incurred during this journey. + +Here's a breakdown of the travel costs between the cities: +- To travel from City A to B, it costs 61 units. +- To journey from City A to C, the cost is 65 units. +- The travel cost from City A to D is 63 units. +- If the salesperson begins from City B, it costs 61 units to reach A, 94 units to get to C, and 31 units to D. +- From City C, the costs are 65 units to A, 94 units to B, and 19 units to D. +- Finally, from City D, the costs are 63 units to A, 31 units to B, and 19 units to C. + +Now, what is the minimum total travel cost for the salesperson to visit each city exactly once and return to the starting city?",[176.0] +"[""Diet Problem""]","Imagine you're a dietitian tasked with planning a daily meal for a client with specific dietary goals. Your challenge is to select from four available food options to meet the nutritional requirements at the minimal cost. Each food item has its distinct protein, carbohydrate, and caloric content along with a price tag. + +Here’s a quick overview of the food items and their respective nutritional values and costs: + +- Food_1: Contains 2 grams of protein, 9 grams of carbohydrates, and 165 calories, all for $3. +- Food_2: Provides 11 grams of protein, 29 grams of carbohydrates, and 230 calories for $3. +- Food_3: Offers 19 grams of protein, 16 grams of carbohydrates, and 68 calories for $6. +- Food_4: Delivers 17 grams of protein, 9 grams of carbohydrates, and 258 calories for $9. + +The nutritional targets set for your client are to consume at least 88 grams of protein, 107 grams of carbohydrates, and 1944 calories in a day. Your goal is to determine the least expensive way to meet these dietary requirements using the available food options. What is the cheapest cost to achieve these nutritional goals? You could only assign integral portions of each type of food. You could only assign integral portions of each type of food.",[27.0] +"[""Diet Problem""]","Imagine you're trying to balance your nutritional intake and budget for the week. You have an array of eight different food items to choose from: Chicken, Rice, Eggs, Beans, Spinach, Corn, Beef, and Broccoli. These foods each offer different levels of protein, carbohydrates, and calories and come with various price tags. + +Here's the nutritional breakdown and cost of each food item: + +- Chicken: Provides 5 grams of protein, 15 grams of carbohydrates, and 162 calories at a cost of $8. +- Rice: Supplies 8 grams of protein, 17 grams of carbohydrates, and 282 calories for $4. +- Eggs: Deliver a solid 18 grams of protein, 20 grams of carbohydrates, and 234 calories for just $1. +- Beans: Offer 16 grams of protein, 15 grams of carbohydrates, and 85 calories for $4. +- Spinach: Adds 3 grams of protein, 28 grams of carbohydrates, and 194 calories for $7. +- Corn: Gives 3 grams of protein, 19 grams of carbohydrates, and 153 calories for a modest $2. +- Beef: Brings in 6 grams of protein, 24 grams of carbohydrates, and 167 calories for $9. +- Broccoli: Offers 17 grams of protein, 9 grams of carbohydrates, and 192 calories for just $3. + +Your dietary goal for the week is to consume at least 68 grams of protein, 189 grams of carbohydrates, and 1840 calories. What is the least amount of money you need to spend to meet these nutritional requirements by choosing from these eight food items? You could only assign integral portions of each type of food.",[10.0] +['Transportation Problem'],"Welcome to the world of international logistics! You are a resource manager at a global company, responsible for the allocation of a valuable product among your six international warehouses. + +Here's what each warehouse currently has and what they need: + +- Warehouse 1 currently has 497 units but needs 161 units. +- Warehouse 2 currently has 39 units but needs 257 units. +- Warehouse 3 currently has 416 units but needs 252 units. +- Warehouse 4 currently has 71 units but needs 251 units. +- Warehouse 5 currently has 445 units but needs 372 units. +- Warehouse 6 currently has 471 units but needs 370 units. + +You need to orchestrate the movement of these resources to ensure every warehouse meets its required quantity. However, transportation between warehouses carries costs, and these costs vary depending on the pair of warehouses involved. Here's the cost matrix that shows the cost of moving a unit from one warehouse to another: + +- Moving product from/to Warehouse 1: To Warehouse 2 costs 5, to Warehouse 3 costs 7, to Warehouse 4 costs 5, to Warehouse 5 costs 27, and to Warehouse 6 costs 49. +- Moving product from/to Warehouse 2: To Warehouse 1 costs 31, to Warehouse 3 costs 14, to Warehouse 4 costs 31, to Warehouse 5 costs 3, and to Warehouse 6 costs 4. +- Moving product from/to Warehouse 3: To Warehouse 1 costs 9, to Warehouse 2 costs 13, to Warehouse 4 costs 23, to Warehouse 5 costs 38, and to Warehouse 6 costs 19. +- Moving product from/to Warehouse 4: To Warehouse 1 costs 47, to Warehouse 2 costs 34, to Warehouse 3 costs 7, to Warehouse 5 costs 34, and to Warehouse 6 costs 35. +- Moving product from/to Warehouse 5: To Warehouse 1 costs 21, to Warehouse 2 costs 11, to Warehouse 3 costs 6, to Warehouse 4 costs 7, and to Warehouse 6 costs 50. +- Moving product from/to Warehouse 6: To Warehouse 1 costs 28, to Warehouse 2 costs 41, to Warehouse 3 costs 34, to Warehouse 4 costs 17, to Warehouse 5 costs 32. + +Your objective is to relocate the products such that all warehouse needs are met, but with the lowest possible total transportation cost. What is the minimum cost to ensure all warehouses have the quantities they need?",[2114.0] +"[""Transportation Problem""]","Imagine you are managing resources for five different cities, each with its own specific needs and supplies. The cities are facing a scarcity of a critical resource, and your task is to ensure that all cities meet their requirements by redistributing the available resources most efficiently. + +Initially, the cities have the following amounts of resources: +- City 1 has 221 units. +- City 2 has 98 units. +- City 3 has 337 units. +- City 4 has 425 units. +- City 5 has 250 units. + +The requirements for each city to adequately meet its needs are: +- City 1 needs 128 units. +- City 2 needs 317 units. +- City 3 needs 394 units. +- City 4 needs 108 units. +- City 5 needs 101 units. + +To transfer resources between these cities, there are associated costs, which differ depending on the route taken. Here’s the cost matrix for transporting resources between each pair of cities: +- From/to City 1: To City 2 costs 20, to City 3 costs 39, to City 4 costs 4, to City 5 costs 45. +- From/to City 2: To City 1 costs 1, to City 3 costs 8, to City 4 costs 23, to City 5 costs 44. +- From/to City 3: To City 1 costs 5, to City 2 costs 46, to City 4 costs 36, to City 5 costs 15. +- From/to City 4: To City 1 costs 3, to City 2 costs 5, to City 3 costs 10, to City 5 costs 27. +- From/to City 5: To City 1 costs 37, to City 2 costs 15, to City 3 costs 7, to City 4 costs 45. + +Your mission is to redistribute the resources so that each city meets its required amount at the lowest possible transportation cost. What would be the minimum cost to ensure all cities have the resources they need?",[1494.0] +['TravelingSalesman'],"In a scenario that mirrors the challenges faced by a global logistics company, imagine that there is a need to send a single shipment to six different regional distribution centers. These centers are located in various cities, identified by the numbers 1 to 6. The shipment must be delivered to each center exactly once before returning to the city of origin, and the objective is to minimize the total cost of travel between these cities. + +The costs associated with travel between the centers are encapsulated in a cost matrix, which represents the diverse expenses of logistics, such as fuel, labor, tolls, and fees associated with each route. Each pair of cities has a unique cost associated with the travel between them, with no possibility of avoiding a cost when moving from a city to itself. + +Here is a breakdown of some specific travel costs between the cities: +- From City 1, the travel costs are: to City 2 is 83 units, to City 3 is 30 units, to City 4 is 38 units, to City 5 is 87 units, to City 6 is 66 units, +- From City 2, the travel costs are: to City 1 is 83 units, to City 3 is 93 units, to City 4 is 13 units, to City 5 is 42 units, to City 6 is 32 units, +- From City 3, the travel costs are: to City 1 is 30 units, to City 2 is 93 units, to City 4 is 85 units, to City 5 is 17 units, to City 6 is 41 units, +- From City 4, the travel costs are: to City 1 is 38 units, to City 2 is 13 units, to City 3 is 85 units, to City 5 is 23 units, to City 6 is 18 units, +- From City 5, the travel costs are: to City 1 is 87 units, to City 2 is 42 units, to City 3 is 17 units, to City 4 is 23 units, to City 6 is 59 units, +- From City 6, the travel costs are: to City 1 is 66 units, to City 2 is 32 units, to City 3 is 41 units, to City 4 is 18 units, to City 5 is 59 units + +Given the complexity of the routes and the varying costs, the logistical challenge is to determine the sequence of deliveries that will result in the lowest total cost for transporting the shipment to all six regional distribution centers and then returning it to the starting city. + +What is the minimum total travel cost for completing this delivery circuit, visiting each city exactly once and then returning to the city of origin?",[177.0] +['TravelingSalesman'],"Imagine a business consultant who needs to visit six unique cities to evaluate their company branches. These cities are labeled as 1 through 6. The consultant can begin their travels from any of these cities, but must adhere to specific rules: they are required to visit each city exactly once and then return to their starting city. The primary objective is to minimize the total travel expense, which could include factors like distance, time, or monetary costs. + +Here are the costs of traveling between the cities: +- From City 1, the costs are 63 to City 2, 62 to City 3, 49 to City 4, 58 to City 5, and 38 to City 6. +- From City 2, the travel costs are 63 to City 1, 93 to City 3, 21 to City 4, 66 to City 5, and 42 to City 6. +- Travelling from City 3 involves costs of 62 to City 1, 93 to City 2, 26 to City 4, 99 to City 5, and 96 to City 6. +- From City 4, the costs are 49 to City 1, 21 to City 2, 26 to City 3, 73 to City 5, and 80 to City 6. +- From City 5, it costs 58 to travel to City 1, 66 to City 2, 99 to City 3, 73 to City 4, and 76 to City 6. +- Lastly, from City 6, it costs 38 to go to City 1, 42 to City 2, 96 to City 3, 80 to City 4, and 76 to City 5. + +What is the minimum total travel cost for the consultant to visit each city exactly once and return to the starting city?",[284.0] +"[""Network Optimization""]","Let's consider a complex transportation network comprised of 6 major cargo hubs, stretching from a main dispatch center to a primary delivery destination. These hubs are interconnected by roads, each with its own weight limit for cargo transportation. + +Here's how the weight limits of each road are distributed, reflecting the maximum amount of cargo (in tons) that can be transported from one hub to another: + +- From Hub 0 (Dispatch Center): Can send cargo to Hub 1 (7 tons), Hub 2 (8 tons), Hub 3 (1 ton), Hub 4 (2 tons), and Hub 5 (4 tons). +- From Hub 1: Can send cargo to Hub 0 (15 tons), Hub 2 (16 tons), Hub 4 (9 tons), and Hub 5 (13 tons). +- From Hub 2: Can send cargo to Hub 0 (9 tons), Hub 1 (1 ton), Hub 3 (9 tons), Hub 4 (13 tons), and Hub 5 (13 tons). +- From Hub 3: Can send cargo to Hub 0 (10 tons), Hub 1 (11 tons), Hub 2 (19 tons), Hub 4 (1 ton), and Hub 5 (12 tons). +- From Hub 4: Can send cargo to Hub 0 (15 tons), Hub 1 (16 tons), Hub 2 (7 tons), Hub 3 (11 tons), and Hub 5 (19 tons). +- From Hub 5 (Destination): Can send cargo back to Hub 0 (10 tons), Hub 1 (12 tons), Hub 2 (12 tons), Hub 3 (12 tons), and Hub 4 (20 tons). + +In this model, a capacity of ""0"" indicates there is no direct road connection between those two hubs. + +The challenge is to determine the maximum amount of cargo that can be transported from Hub 0 (the dispatch center) to Hub 5 (the destination) per day, in a manner that the cargo on each road does not surpass its listed weight limit. What is the maximum amount of cargo that can be delivered from the dispatch center to the destination in tons per day? Please note that the answer should provide only the optimal value under the maximum flow scenario, and all the data in the problem details should be included in the created question.",[22.0] +"[""Network Optimization""]","Imagine a complex transportation network across 9 key hub cities that are tasked with delivering critical supplies from a central distribution center to a remote area in need. Each city (or hub) in this network can send and receive supplies through established routes, each with a specific maximum capacity. The goal is to maximize the flow of supplies from the source hub to the destination hub, ensuring that the delivery system operates at its peak efficiency without overloading any route. + +Here's a detailed look at the capacities, in tons of supplies, that each route can handle between the cities: + +- From City 0 (Source): Can send supplies to City 2 (20 tons), City 3 (10 tons), City 5 (2 tons), City 6 (11 tons), City 7 (5 tons), and City 8 (12 tons). +- City 1: Can receive and then send supplies to City 0 (7 tons), City 2 (5 tons), City 3 (9 tons), City 4 (9 tons), City 5 (16 tons), City 6 (1 ton), City 7 (17 tons), and City 8 (7 tons). +- City 2: Can manage supplies from City 0 (3 tons), and then distribute to City 1 (17 tons), City 3 (1 ton), City 4 (5 tons), City 5 (18 tons), City 6 (13 tons), City 7 (12 tons), and City 8 (14 tons). +- City 3: Starts with supplies from City 0 (9 tons), forwarding to City 1 (10 tons), City 2 (2 tons), City 4 (10 tons), City 5 (11 tons), City 6 (11 tons), City 7 (5 tons), and City 8 (10 tons). +- City 4: Receives from City 0 (7 tons) and redistributes to City 1 (5 tons), City 2 (1 ton), City 3 (19 tons), City 5 (19 tons), City 6 (12 tons), City 7 (14 tons), and City 8 (15 tons). +- City 5: Gathers supplies from City 0 (16 tons) and can send out to City 1 (4 tons), City 2 (6 tons), City 3 (15 tons), City 4 (19 tons), City 6 (0 tons), and City 8 (20 tons). +- City 6: Channels supplies from City 0 (3 tons) towards City 1 (0 tons), City 2 (17 tons), City 3 (16 tons), City 4 (19 tons), City 5 (8 tons), City 7 (8 tons), and City 8 (9 tons). +- City 7: Starts with resources from City 0 (14 tons), sending to City 1 (11 tons), City 2 (13 tons), City 3 (17 tons), City 4 (10 tons), City 5 (13 tons), City 6 (20 tons), and City 8 (16 tons). + +In this scenario, a ""0"" indicates that there is no established route for supplies to move directly between those two cities. + +The challenge lies in determining the maximum tonnage of supplies that can be transported from City 0 (the source) to City 8 (the destination) without exceeding the capacity of any individual route in the network. What is the optimal value of supplies, in tons, that can be moved through this network?",[60.0] +"[""Network Optimization""]","Imagine a futuristic transportation network that connects 9 major cities across a vast region, designed to maximize the efficiency of cargo transport from the central hub to a key distribution center. Each city in this network is connected by high-capacity routes, with the ability to transport a significant volume of goods measured in thousands of tons. + +Here's a detailed layout of the transportation capacities between each pair of cities: + +- From City 0 (Central Hub): Cargo can be sent to City 1 (1 ton), City 2 (16 tons), City 3 (8 tons), City 4 (18 tons), City 5 (4 tons), City 6 (0 tons, indicating no direct route), City 7 (8 tons), and City 8 (5 tons). +- From City 1: Cargo can be dispatched to City 0 (15 tons), City 2 (11 tons), City 3 (8 tons), City 4 (13 tons), City 5 (2 tons), City 6 (1 ton), City 7 (7 tons), and City 8 (6 tons). +- From City 2: Shipments can go to City 0 (15 tons), City 1 (4 tons), City 3 (3 tons), City 4 (1 ton), City 5 (13 tons), City 6 (13 tons), City 7 (11 tons), and City 8 (9 tons). +- From City 3: Goods can be transported to City 0 (4 tons), City 1 (16 tons), City 2 (8 tons), City 4 (12 tons), City 5 (18 tons), City 6 (19 tons), City 7 (8 tons), and City 8 (11 tons). +- From City 4: Cargo movement is possible to City 0 (4 tons), City 1 (19 tons), City 2 (3 tons), City 3 (2 tons), City 5 (6 tons), City 6 (1 ton), City 7 (13 tons), and City 8 (14 tons). +- From City 5: Shipments can be directed to City 0 (6 tons), City 1 (19 tons), City 2 (15 tons), City 3 (10 tons), City 4 (16 tons), City 6 (17 tons), City 7 (15 tons), and City 8 (17 tons). +- From City 6: Cargo can be routed to City 0 (13 tons), City 1 (7 tons), City 2 (13 tons), City 3 (4 tons), City 4 (14 tons), City 5 (3 tons), City 7 (8 tons), and City 8 (8 tons). +- From City 7: Goods can be moved to City 0 (10 tons), City 1 (6 tons), City 2 (19 tons), City 3 (1 ton), City 4 (12 tons), City 5 (6 tons), City 6 (9 tons), and City 8 (14 tons). +- From City 8 (Key Distribution Center): This city receives and dispatches goods but primarily serves as the endpoint for cargo coming from the Central Hub. + +A capacity of ""0"" in this context indicates that there is no established route for direct cargo transport between those two cities. + +The challenge is to determine the maximum amount of cargo, in thousands of tons, that can be transported from the Central Hub (City 0) to the Key Distribution Center (City 8), ensuring that the flow of goods through each route does not surpass its highest capacity. What is the optimal value of cargo that can be moved from the Central Hub to the Key Distribution Center?",[60.0] +"[""Capacitated Facility Location Problem""]","Global Supply Network Inc. is at a pivotal moment in redesigning its distribution strategy to ensure efficient delivery of products to six retail stores. The decision revolves around selecting the most cost-effective combination of distribution centers from nine possible locations. Each distribution center has its unique opening cost and a maximum capacity it can supply. Furthermore, the cost of transporting goods from these centers to the retail stores varies significantly. The challenge lies in fulfilling the demand at each retail store while minimizing the overall cost, including the expenses of opening distribution centers and the transportation cost of goods. + +Given Data: +- Number of Potential Distribution Centers: 9 +- Number of Retail Stores Needing Supply: 6 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: 151,000 +- Center 2: 192,000 +- Center 3: 114,000 +- Center 4: 171,000 +- Center 5: 160,000 +- Center 6: 120,000 +- Center 7: 182,000 +- Center 8: 186,000 +- Center 9: 174,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $3 to Store 1, $5 to Store 2, $4 to Store 3, $3 to Store 4, $5 to Store 5, $2 to Store 6 +- From Center 2: $4 to Store 1, $2 to Store 2, $4 to Store 3, $5 to Store 4, $1 to Store 5, $4 to Store 6 +- From Center 3: $2 to Store 1, $5 to Store 2, $4 to Store 3, $1 to Store 4, $1 to Store 5, $3 to Store 6 +- From Center 4: $3 to Store 1, $2 to Store 2, $4 to Store 3, $4 to Store 4, $3 to Store 5, $4 to Store 6 +- From Center 5: $4 to Store 1, $1 to Store 2, $3 to Store 3, $5 to Store 4, $3 to Store 5, $5 to Store 6 +- From Center 6: $1 to Store 1, $2 to Store 2, $4 to Store 3, $1 to Store 4, $4 to Store 5, $2 to Store 6 +- From Center 7: $2 to Store 1, $1 to Store 2, $2 to Store 3, $5 to Store 4, $2 to Store 5, $4 to Store 6 +- From Center 8: $4 to Store 1, $4 to Store 2, $4 to Store 3, $5 to Store 4, $3 to Store 5, $1 to Store 6 +- From Center 9: $4 to Store 1, $2 to Store 2, $4 to Store 3, $2 to Store 4, $2 to Store 5, $4 to Store 6 + +Demand of Each Retail Store (in units): +- Store 1: 876 units +- Store 2: 801 units +- Store 3: 955 units +- Store 4: 561 units +- Store 5: 601 units +- Store 6: 669 units + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1615 units +- Center 2: 1255 units +- Center 3: 1816 units +- Center 4: 1095 units +- Center 5: 1519 units +- Center 6: 1137 units +- Center 7: 1678 units +- Center 8: 1876 units +- Center 9: 1591 units + +Question: +What is the optimal total cost for Global Supply Network Inc. to open the necessary distribution centers and transport goods to fully meet the demands of all six retail stores? This cost includes the opening costs of the chosen distribution centers and the transportation costs of supplying the stores.",[394338.0] +['Capacitated Facility Location Problem'],"In the bustling marketplace of today, the fictional company GlobalSupply Inc. stands at a crucial juncture. It seeks to efficiently bridge its supply chain, connecting six potential distribution centers with nine eager retail stores scattered across the region. The objective is clear yet challenging: to minimize the total cost involved in opening distribution centers and transporting goods to fulfill the demands of the retail stores. This total cost encapsulates both the opening expenses of the distribution centers and the variable transportation costs per unit of goods delivered to the stores. + +Here are the specifics: + +- There are six Distribution Centers (DCs) under consideration for opening. +- Nine Retail Stores are in need of regular supplies. + +The Opening Costs for each Distribution Center (in dollars) are as follows: +- DC 1: $151,000 +- DC 2: $192,000 +- DC 3: $114,000 +- DC 4: $171,000 +- DC 5: $160,000 +- DC 6: $120,000 + +Transportation Cost Per Unit from each Distribution Center to each Retail Store (in dollars): +- From DC 1 to Stores: $2 to Store 1, $3 to Store 2, $4 to Store 3, $5 to Store 4, $6 to Store 5, $2 to Store 6, $3 to Store 7, $4 to Store 8, and $2 to Store 9 +- From DC 2 to Stores: $4 to Store 1, $2 to Store 2, $3 to Store 3, $5 to Store 4, $4 to Store 5, $3 to Store 6, $4 to Store 7, $2 to Store 8, and $4 to Store 9 +- From DC 3 to Stores: $1 to Store 1, $1 to Store 2, $2 to Store 3, $3 to Store 4, $4 to Store 5, $2 to Store 6, $1 to Store 7, $2 to Store 8, and $4 to Store 9 +- From DC 4 to Stores: $4 to Store 1, $1 to Store 2, $2 to Store 3, $4 to Store 4, $3 to Store 5, $4 to Store 6, $2 to Store 7, $1 to Store 8, and $4 to Store 9 +- From DC 5 to Stores: $1 to Store 1, $4 to Store 2, $5 to Store 3, $2 to Store 4, $3 to Store 5, $1 to Store 6, $2 to Store 7, $4 to Store 8, and $4 to Store 9 +- From DC 6 to Stores: $4 to Store 1, $4 to Store 2, $4 to Store 3, $4 to Store 4, $4 to Store 5, $4 to Store 6, $4 to Store 7, $4 to Store 8, and $4 to Store 9 + +The Demand of each Retail Store (in units) ranges from 401 units for Store 1, 522 units for Store 2, 634 units for Store 3, 478 units for Store 4, 558 units for Store 5, 712 units for Store 6, 955 units for Store 7, 812 units for Store 8 and 603 units for Store 9, with varying demands for the others in between. + +The Supply Capacity of each Distribution Center (in units) is also diverse, with DC 1 capable of supplying 1795 units, DC2 capable of supplying 1540 units, DC3 capable of supplying 1435 units, DC4 capable of supplying 1325 units, and DC5 capable of supplying 1325 units and DC6 capable of supplying 1095 units. + +GlobalSupply Inc. now faces the daunting task of determining the optimal combination of distribution centers to open and the distribution of supplies to the retail stores in a way that the total opening and transportation costs are minimized. + +**Question:** +What is the optimal total cost for GlobalSupply Inc. to both open the necessary distribution centers and transport goods to meet the demands of all nine retail stores, considering the given capacities and costs?",[567010.0] +['AircraftLanding'],"An air traffic controller needs to control the landing times of n aircrafts: Flights must land in the order 1, ..., n. Flight j must land in time interval [aj , bj ]. n = 10. [a1, b1] = [40, 67], [a2, b2] = [11, 87], [a3, b3] = [35,94], [a4, b4] = [38,96], [a5, b5] = [74,105], [a6, b6] = [87,137], [a7, b7] = [62,121], [a8, b8] = [75,143], [a9, b9] = [94,172], [a10, b10] = [128,151]. The objective is to maximize the minimum separation time, which is the interval between two landings. Find the maximal minimum separation. Provide your answer rounded to the nearest time.",[12.0] +['PortfolioOptimization'],"Before one World Cup championship, there are 5 types of securities available for sale. The price of each share of the security is fixed and the payoff of it will be contingent on the outcome of the outcome of the world cup. The information of the 5 securities are shown in Table 1. Here, for example, Security 1’s payoff is 1 if either Argentina, Brazil, or England wins the world cup, and the payoff is 0 if Germany or Spain wins. The Share Limit is the maximum number of shares one can purchase, and Price is the current purchasing price per share of each security. There are five countries, Argentina, Brazil, England, Spain, Germany. Security 1 has a price of 0.75 and a share limit of 10. The payoff for this security is $1 in Argentina, $1 in Brazil, $1 in England, 0 in Germany, and 0 in Spain. Security 2 is priced at 0.35 with a share limit of 5. The payoff is 0 for Argentina, Brazil, or England, but it's $1 for Germany and Spain. Security 3 has a price of 0.40 and a share limit of 10. Similar to Security 2, the payoff is $1 for Argentina, $1 for Spain, and $1 for England, and 0 payoff for Germany and Brazil. Security 4, with a price of 0.75 and a share limit of 10, has payoff of $1 across all countries except for Spain. Security 5 is priced at 0.65, has a share limit of 5, and has payoff of $1 in Brazil, Germany and Spain, 0 in Argentina and England. Find the maximum worst-case revenue. ",[1.0] +['TravelingSalesman'],"Consider a delivery company that needs to deliver packages to five different cities, named E, F, G, H, and I. The delivery truck can start its route from any of these cities, but needs to visit each city exactly once and then return to the starting point. The challenge is to find the route that minimizes the total delivery cost, which could be influenced by factors such as fuel, distance, or time. +Here is a breakdown of the delivery costs between the cities: +The cost to deliver from City E to F is 31 units, to G is 19 units, to H is 71 units, and to I is 74 units. +From City F, it costs 31 units to reach E, 39 units to get to G, 77 units to H, and 84 units to I. +When delivering from City G, the costs are 19 units to E, 39 units to F, 65 units to H, and 78 units to I. +From City H, it costs 71 units to go to E, 77 units to F, 65 units to G, and 38 units to I. +Finally, from City I, it costs 74 units to reach E, 84 units to F, 78 units to G, and 38 units to H. +What is the minimum total delivery cost for the truck to visit each city exactly once and return to the starting city?",[237.0] +"[""Capacitated Facility Location Problem""]","In the scenario presented, SupplyChain Solutions is faced with a complex decision in establishing its distribution network to efficiently serve six retail stores from a selection of seven distribution centers. Each distribution center has its own specific opening cost and a maximum supply capacity. Furthermore, the cost of transporting goods from these centers to the retail stores varies significantly. The goal is to fulfill the demand at each retail store while minimizing the total cost incurred, which encompasses both the opening costs of the distribution centers and the transportation costs of the goods. + +Given Data: +- Number of Potential Distribution Centers: 7 +- Number of Retail Stores Needing Supply: 6 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 +- Center 6: $120,000 +- Center 7: $182,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $3 to Store 1, $3 to Store 2, $5 to Store 3, $4 to Store 4, $3 to Store 5, $5 to Store 6 +- From Center 2: $2 to Store 1, $4 to Store 2, $2 to Store 3, $4 to Store 4, $5 to Store 5, $1 to Store 6 +- From Center 3: $4 to Store 1, $2 to Store 2, $5 to Store 3, $4 to Store 4, $1 to Store 5, $1 to Store 6 +- From Center 4: $3 to Store 1, $3 to Store 2, $2 to Store 3, $4 to Store 4, $4 to Store 5, $3 to Store 6 +- From Center 5: $4 to Store 1, $4 to Store 2, $1 to Store 3, $3 to Store 4, $5 to Store 5, $3 to Store 6 +- From Center 6: $5 to Store 1, $1 to Store 2, $2 to Store 3, $4 to Store 4, $1 to Store 5, $4 to Store 6 +- From Center 7: $2 to Store 1, $2 to Store 2, $1 to Store 3, $2 to Store 4, $5 to Store 5, $2 to Store 6 + +Demand of Each Retail Store (in units): +- Store 1: 739 +- Store 2: 491 +- Store 3: 766 +- Store 4: 854 +- Store 5: 827 +- Store 6: 908 + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1575 +- Center 2: 834 +- Center 3: 1005 +- Center 4: 1904 +- Center 5: 1825 +- Center 6: 1821 +- Center 7: 1365 + +Question: +What is the optimal total cost for SupplyChain Solutions to select distribution centers and transport goods to meet the demands of all retail stores, considering both the opening costs of the centers and the transportation costs of supplying the stores?",[402947.0] +"[""Network Optimization""]","Welcome to the grand challenge of designing the optimal network for the newest, state-of-the-art high-speed data transfer system connecting 9 strategic data centers across the country. From the primary data hub (Station 0) to the primary backup center (Station 8), the network is a complex system of fiber optic cables, each with a distinct data capacity dictating the maximum amount of data (in terabits per second, Tbps) that can be transmitted from one center to another. + +Here's a detailed breakdown of the capacities between each pair of data centers: + +From Data Center 0 (Source): 1 → 6 Tbps, 2 → 20, 3 → 15, 4 → 9, 5 → 20, 6 → 17, 7 → 16, 8 → 20. + +From Data Center 1 (all capacities ≤ 18 Tbps): 0 → 12, 2 → 8, 3 → 18, 4 → 14, 5 → 10, 6 → 15, 7 → 7, 8 → 18. + +From Data Center 2 (peak 19 Tbps to DC 4): 0 → 11, 1 → 13, 3 → 17, 4 → 19, 5 → 14, 6 → 18, 7 → 9, 8 → 17. + +From Data Center 3: 0 → 14, 1 → 16, 2 → 13, 4 → 11, 5 → 19, 6 → 6, 7 → 15, 8 → 20. + +From Data Center 4: 0 → 8, 1 → 12, 2 → 18, 3 → 10, 5 → 16, 6 → 19, 7 → 13, 8 → 15. + +From Data Center 5: 0 → 18, 1 → 9, 2 → 15, 3 → 19, 4 → 12, 6 → 17, 7 → 14, 8 → 17. + +From Data Center 6: 0 → 10, 1 → 14, 2 → 16, 3 → 12, 4 → 18, 5 → 13, 7 → 19, 8 → 18. + +From Data Center 7: 0 → 11, 1 → 8, 2 → 14, 3 → 20, 4 → 12, 5 → 15, 6 → 9, 8 → 16. + +Data Center 8 (Destination): receives the flows listed above but has no outbound links—all (8, j) capacities are 0. +In this network, a capacity of ""0"" indicates that there is no direct data transmission link between those two centers. + +The ultimate goal is to maximize the data flow from Data Center 0, the heart of our network, to Data Center 8, ensuring the highest level of efficiency and reliability in data backup and disaster recovery operations. + +**What is the maximum amount of data (in Tbps) that can be transmitted from the source to the destination per second, without exceeding the specified capacities of the data transmission links?**",[123.0] +['Transportation Problem'],"Imagine you're overseeing the distribution of medical supplies across five different regions to ensure everyone has access to the resources they need. Each region currently has a certain quantity of supplies, but the demand varies from region to region. + +Here's the breakdown of the current supplies and the required quantities for each region: + +- Region 1 currently has 133 units but needs 146 units. +- Region 2 currently has 189 units but needs 270 units. +- Region 3 currently has 446 units but needs 313 units. +- Region 4 currently has 346 units but needs 102 units. +- Region 5 currently has 184 units but needs 412 units. + +You have the ability to redistribute supplies from one region to another, but there are transportation costs associated with this. + +Here's a cost matrix showing how much it costs to transport supplies between each pair of regions: + +- Transporting supplies from/to Region 1: To Region 2 costs 24, to Region 3 costs 32, to Region 4 costs 44, to Region 5 costs 46. +- Transporting supplies from/to Region 2: To Region 1 costs 41, to Region 3 costs 6, to Region 4 costs 11, to Region 5 costs 49. +- Transporting supplies from/to Region 3: To Region 1 costs 40, to Region 2 costs 10, to Region 4 costs 40, to Region 5 costs 45. +- Transporting supplies from/to Region 4: To Region 1 costs 22, to Region 2 costs 22, to Region 3 costs 34, to Region 5 costs 1. +- Transporting supplies from/to Region 5: To Region 1 costs 46, to Region 2 costs 14, to Region 3 costs 35, to Region 4 costs 36. + +Your objective is to ensure that every region receives the necessary supplies while minimizing transportation costs. What would be the minimum cost to meet the demand for medical supplies in all regions?",[1324.0] +"[""Network Optimization""]","Imagine a city that has a complex electrical network with 8 power stations (including source and the main receiving stations). The stations are connected with transmission lines, each capable of handling a certain amount of electricity. + +Here's the capacity of each transmission line, showing the maximum amount of electricity (in megawatts) that can flow from one station to another: + +- From Station 0 (Source): Can send electricity to Station 1 (9 MW), Station 2 (6 MW), Station 3 (1 MW), Station 4 (8 MW), Station 5 (14 MW), Station 6 (1 MW) and Station 7 (17 MW). +- From Station 1: Can send electricity to Station 0 (1 MW), Station 2 (6 MW), Station 3 (14 MW), Station 4 (16 MW), Station 5 (8 MW), Station 6 (19 MW) and Station 7 (15 MW). +- From Station 2: Can send electricity to Station 0 (4 MW), Station 1 (7 MW), Station 3 (9 MW), Station 4 (1 MW), Station 5 (2 MW), Station 6 (9 MW) and Station 7 (4 MW). +- From Station 3: Can send electricity to Station 0 (11 MW), Station 1 (5 MW), Station 2 (14 MW), Station 4 (17 MW), Station 5 (6 MW), Station 6 (20 MW) and Station 7 (13 MW). +- From Station 4: Can send electricity to Station 0 (8 MW), Station 1 (5 MW), Station 2 (20 MW), Station 3 (0 MW), Station 5 (16 MW), Station 6 (5 MW) and Station 7 (5 MW). +- From Station 5: Can send electricity to Station 0 (15 MW), Station 1 (18 MW), Station 2 (9 MW), Station 3 (1 MW), Station 4 (19 MW), Station 6 (16 MW) and Station 7 (0 MW). +- From Station 6: Can send electricity to Station 0 (20 MW), Station 1 (15 MW), Station 2 (1 MW), Station 3 (0 MW), Station 4 (11 MW), Station 5 (6 MW) and Station 7 (19 MW). +- From Station 7 (Destination): Can send electricity back to Station 0 (4 MW), Station 1 (18 MW), Station 2 (10 MW), Station 3 (2 MW), Station 4 (1 MW), Station 5 (20 MW) and Station 6 (14 MW). + +In this network, a capacity of ""0"" means there is no direct transmission line between those two stations. + +The challenge is to calculate the maximum amount of electricity that can be transmitted from Station 0 (the source) to Station 7 (the main receiving station), ensuring that the flow through each transmission line does not exceed its capacity. What is the maximum amount of electricity that can be transmitted from the source to the destination in megawatts?",[56.0] +"[""Transportation Problem"",""Minimum Cost Flow Problem""]","The China Railroad Ministry is in the process of planning relocations of freight cars among 5 regions of the country to get ready for the fall harvest. Moving a car from locations 1 to 2 costs 12 units. Moving a car from location 1 to 3 costs 11 units. Moving a car from location 1 to 4 costs 13 units. Moving a car from location 1 to 5 costs 28 units. Moving a car from location 2 to 3 costs 18 units. Moving a car from location 2 to 4 costs 8 units. Moving a car from location 2 to 5 costs 46 units. Moving a car from location 3 to 4 costs 9 units. Moving a car from location 3 to 5 costs 27 units. Moving a car from location 4 to 5 costs 20 units. At location 1, there are currently 120 cars present, but 150 cars are needed. At location 2, there are currently 330 cars present, but 200 cars are needed. At location 3, there are currently 400 cars present, but 600 cars are needed. At location 4, there are currently 400 cars present, but 200 cars are needed. At location 5, there are currently 600 cars present, but 400 cars are needed. Write down a linear optimization to compute the least costly way to move the cars such us the need is met. ",[2160.0] +"[""Diet Problem""]","Let's say you're on a mission to create the perfect meal plan for yourself. You're focused on getting the right balance of nutrients without digging too deep into your pockets. You have six different food options to consider: Chicken, Oats, Tofu, Rice, Beef, and Salmon. Each of these foods provides a certain amount of protein, carbohydrates, and calories, and they all come with their own distinct prices. + +Here's a detailed breakdown of the nutritional content and cost of each food item: + +- Chicken: Delivers 14 grams of protein, a single gram of carbohydrates, and 164 calories at a cost of $6. +- Oats: Gives you 11 grams of protein, 6 grams of carbohydrates, and 210 calories for just $2. +- Tofu: Offers 8 grams of protein, 12 grams of carbohydrates, and 98 calories at a cost of $9. +- Rice: Provides 1 gram of protein, a generous 17 grams of carbohydrates, and 92 calories for $3. +- Beef: Comes packed with 16 grams of protein, 11 grams of carbohydrates, and 211 calories, priced at $7. +- Salmon: Brings a hefty 19 grams of protein, 13 grams of carbohydrates, and 211 calories but costs $9. + +You want to make sure your meal plan meets the following nutritional targets: at least 70 grams of protein, 117 grams of carbohydrates, and 1837 calories. Considering these six foods, what is the least amount of money you need to spend to meet these dietary needs? +Remember, your response should only contain the optimal value of the cost to meet the requirements. You could only assign integral portions of each type of food.",[29.0] +"[""TravelingSalesman""]","Imagine a salesperson embarking on a journey to visit five different cities—let's call them City A, City B, City C, City D, and City E—to promote their products. The objective of this journey is to minimize the total travel cost incurred. The salesperson can start from any city but must visit each city exactly once before returning to the starting point. + +Here's an overview of the travel costs between these cities: +- To travel from City A to City B, it costs 95 units. +- From City A to City C, the cost is 36 units. +- Traveling from City A to City D incurs a cost of 93 units. +- To reach City A from City E, the cost is 16 units. +- From City B, it costs 95 units to get to City A, 95 units to City C, 29 units to City D, and 96 units to City E. +- Moving from City C, it costs 36 units to City A, 95 units to City B, 72 units to City D, and 38 units to City E. +- City D is connected to City A with a cost of 93 units, to City B with 29 units, to City C with 72 units, and to City E with 77 units. +- Lastly, from City E, it costs 16 units to reach City A, 96 units to City B, 38 units to City C, and 77 units to City D. + +Given these costs, what is the minimum total travel cost for the salesperson to visit each city exactly once and return to the starting city?",[249.0] +['Capacitated Facility Location Problem'],"In the bustling world of retail, GlobalSupply Inc. is at a crossroads, needing to efficiently distribute its products to eight retail stores scattered across the region. To achieve this, GlobalSupply Inc. must select from six potential distribution centers, each with its unique set of opening costs and capacities. The challenge lies not only in selecting which distribution centers to open but also in determining how to distribute their products in a way that minimizes the total cost, including both the opening of distribution centers and the transportation of goods to the stores. + +Given Data: + +Number of Potential Distribution Centers: 6 + +Number of Retail Stores Needing Supply: 8 + +Opening Costs for Each Distribution Center (in dollars): + +Center 1: $151,000 +Center 2: $192,000 +Center 3: $114,000 +Center 4: $171,000 +Center 5: $160,000 +Center 6: $120,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): + +From Center 1: $2 to Store 1, $3 to Store 2, $4 to Store 3, $2 to Store 4, $3 to Store 5, $4 to Store 6, $4 to Store 7, $5 to Store 8 +From Center 2: $2 to Store 1, $4 to Store 2, $3 to Store 3, $2 to Store 4, $4 to Store 5, $3 to Store 6, $2 to Store 7, $2 to Store 8 +From Center 3: $5 to Store 1, $4 to Store 2, $3 to Store 3, $5 to Store 4, $4 to Store 5, $3 to Store 6, $4 to Store 7, $4 to Store 8 +From Center 4: $4 to Store 1, $3 to Store 2, $2 to Store 3, $4 to Store 4, $3 to Store 5, $2 to Store 6, $3 to Store 7, $3 to Store 8 +From Center 5: $5 to Store 1, $1 to Store 2, $2 to Store 3, $3 to Store 4, $4 to Store 5, $2 to Store 6, $3 to Store 7, $2 to Store 8 +From Center 6: $1 to Store 1, $2 to Store 2, $2 to Store 3, $3 to Store 4, $1 to Store 5, $3 to Store 6, $4 to Store 7, $4 to Store 8 + +Demand of Each Retail Store (in units): + +Store 1: 908 units +Store 2: 434 units +Store 3: 605 units +Store 4: 480 units +Store 5: 961 units +Store 6: 787 units +Store 7: 401 units +Store 8: 789 units + +Supply Capacity of Each Distribution Center (in units): + +Center 1: 1365 units +Center 2: 1929 units +Center 3: 1502 units +Center 4: 1201 units +Center 5: 1529 units +Center 6: 961 units + +The optimal solution must navigate the intricate balance between opening costs, transportation costs, supply capacities, and the demands of the retail stores. + +Question: +What is the optimal total cost for GlobalSupply Inc. to open distribution centers and transport goods to meet all the retail stores' demands, considering the cost of opening the centers and the transportation costs of supplying the stores?",[58895.0] +['Capacitated Facility Location Problem'],"In the strategic planning of WareCo's distribution network, the company is evaluating how to efficiently supply seven retail stores from seven available distribution centers. Each distribution center has a unique opening cost and capacity that limits the volume of goods it can distribute. Additionally, the cost to transport goods from these centers to each retail store varies, making the task of minimizing the overall operational cost a complex puzzle. The goal is to determine the optimal setup that satisfies the demand at each retail store while incurring the lowest possible total cost, including both the opening of selected distribution centers and the transportation of goods. + +Given Data: + +- Number of Potential Distribution Centers: 7 +- Number of Retail Stores to be Supplied: 7 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 +- Center 6: $120,000 +- Center 7: $182,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $3 to Store 1, $3 to Store 2, $5 to Store 3, $4 to Store 4, $3 to Store 5, $5 to Store 6, $2 to Store 7 +- From Center 2: $4 to Store 1, $2 to Store 2, $4 to Store 3, $5 to Store 4, $1 to Store 5, $4 to Store 6, $2 to Store 7 +- From Center 3: $5 to Store 1, $4 to Store 2, $1 to Store 3, $1 to Store 4, $3 to Store 5, $3 to Store 6, $2 to Store 7 +- From Center 4: $4 to Store 1, $4 to Store 2, $3 to Store 3, $4 to Store 4, $4 to Store 5, $1 to Store 6, $3 to Store 7 +- From Center 5: $5 to Store 1, $3 to Store 2, $5 to Store 3, $1 to Store 4, $2 to Store 5, $4 to Store 6, $1 to Store 7 +- From Center 6: $4 to Store 1, $2 to Store 2, $2 to Store 3, $1 to Store 4, $2 to Store 5, $5 to Store 6, $2 to Store 7 +- From Center 7: $4 to Store 1, $4 to Store 2, $4 to Store 3, $4 to Store 4, $5 to Store 5, $3 to Store 6, $1 to Store 7 + +Demand of Each Retail Store (in units): +- Store 1: 961 +- Store 2: 787 +- Store 3: 401 +- Store 4: 789 +- Store 5: 965 +- Store 6: 505 +- Store 7: 876 + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1502 +- Center 2: 1201 +- Center 3: 1529 +- Center 4: 961 +- Center 5: 1001 +- Center 6: 1795 +- Center 7: 1069 + +Question: +What is the optimal total cost for WareCo to meet the demand at all seven retail stores, considering both the opening costs of the distribution centers and the transportation costs of supplying the stores?",[554968.0] +['Diet Problem'],"You're on a mission to carefully plan your weekly meal prep, aiming to meet your specific dietary needs while keeping your budget in check. After doing some research, you've found seven different food items that you're considering incorporating into your meals. Each food item has its own nutritional content and cost, which you need to factor into your planning. + +Here are the details of each food option: + +- Food_1: For $6, it offers 6 grams of protein, 3 grams of carbohydrates, and 73 calories. +- Food_2: Costs $2 and provides 16 grams of protein, 3 grams of carbs, and 181 calories. +- Food_3: Priced at $10, this item brings 12 grams of protein, 17 grams of carbs, and 231 calories to the table. +- Food_4: Another $6 option, giving you 9 grams of protein, 28 grams of carbs, and 160 calories. +- Food_5: This one costs $4 and offers 12 grams of protein, 17 grams of carbohydrates, and 209 calories. +- Food_6: The cheapest at $1, with 12 grams of protein, 14 grams of carbs, and 184 calories. +- Food_7: Costs $9, providing 13 grams of protein, 20 grams of carbs, and 118 calories. + +Your nutritional goals for the week are quite specific: you aim to consume at least 59 grams of protein, 179 grams of carbohydrates, and 2,414 calories. Your challenge is to select from these seven food options in such a way that you meet your dietary requirements at the lowest possible cost. + +What is the cheapest cost to meet your nutritional needs? You could only assign integral portions of each type of food.",[14.0] +"[""Capacitated Facility Location Problem""]","In the bustling market landscape of SupplyChainVille, the logistics company LogiTech has embarked on a mission to efficiently distribute its products to nine retail stores scattered across the town. The challenge at hand is to select from eight potential distribution centers, each with its unique operational cost and a maximum supply capacity. The objective is clear: to minimize the total operational cost, which includes both the opening costs of the distribution centers and the transportation costs of shipping goods to the retail stores. + +Given Data: + +- Number of Potential Distribution Centers: 8 +- Number of Retail Stores to be Supplied: 9 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 +- Center 6: $120,000 +- Center 7: $182,000 +- Center 8: $186,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1 to Stores 1–9: $2, $5, $4, $2, $3, $5, $4, $3, $5 +- From Center 2 to Stores 1–9: $1, $3, $5, $2, $4, $1, $3, $5, $2 +- From Center 3 to Stores 1–9: $1, $4, $2, $3, $1, $4, $2, $3, $4 +- From Center 4 to Stores 1–9: $2, $1, $5, $3, $4, $2, $5, $1, $3 +- From Center 5 to Stores 1–9: $3, $2, $4, $1, $5, $3, $2, $4, $1 +- From Center 6 to Stores 1–9: $1, $5, $3, $2, $4, $1, $5, $3, $2 +- From Center 7 to Stores 1–9: $2, $3, $5, $4, $2, $3, $5, $4, $2 +- From Center 8 to Stores 1–9: $1, $4, $2, $5, $3, $1, $4, $2, $5 + +Demand of Each Retail Store (in units): +- Store 1: 779 units +- Store 2: 892 units +- Store 3: 440 units +- Store 4: 556 units +- Store 5: 414 units +- Store 6: 464 units +- Store 7: 920 units +- Store 8: 743 units +- Store 9: 528 units + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1447 units +- Center 2: 1886 units +- Center 3: 1962 units +- Center 4: 1392 units +- Center 5: 1191 units +- Center 6: 1218 units +- Center 7: 1088 units +- Center 8: 1178 units + +The conundrum LogiTech must solve revolves around which distribution centers to activate and how to route the distribution of products such that all retail stores' demands are met at the lowest possible total cost. This puzzle encapsulates not just the hefty opening costs of the distribution centers but also the intricate web of transportation costs associated with moving goods to each retail store. + +Question: +What is the optimal total cost for LogiTech to open distribution centers and transport goods to meet all retail stores' demands, considering both the opening costs of the centers and the transportation costs of supplying the stores?",[555325.0] +['Diet Problem'],"Imagine you are a dietitian and you have been tasked with creating a meal plan for a bodybuilder. You have six food items to choose from: Steak, Tofu, Chicken, Broccoli, Rice, and Spinach. Each food provides certain amounts of protein, carbohydrates, and calories, and each has its own cost. You can buy a partial serving of a food. + +Here's the nutritional value and cost of each food: + +- Steak: It gives you 14 grams of protein, 23 grams of carbohydrates, and 63 calories for $4. +- Tofu: It offers 2 grams of protein, 13 grams of carbohydrates, and 162 calories for $6. +- Chicken: It packs a punch with 17 grams of protein, 13 grams of carbohydrates, and gives you 260 calories for $6. +- Broccoli: It provides 3 grams of protein, a mere 1 gram of carbohydrates, and 55 calories for $8. +- Rice: It gives a hearty 15 grams of protein, 23 grams of carbohydrates, and 231 calories for $8. +- Spinach: It provides 2 grams of protein, 8 grams of carbohydrates, and a huge 297 calories for just $5. + +Your goal is to ensure that the bodybuilder gets at least 83 grams of protein, 192 grams of carbohydrates, and 2089 calories from whatever combination of these foods you choose. The challenge is to keep the cost as low as possible while meeting these nutritional targets. + +What is the minimum cost to meet these nutritional requirements with the available food options? You could only assign integral portions of each type of food.",[57] +"[""Diet Problem""]","Imagine you're a student living on a tight budget, but you're committed to maintaining a healthy, balanced diet. You've got five potential food items to select from: Steak, Apples, Eggs, Chicken, and Tofu. Each of these offers specific amounts of protein, carbohydrates, and calories, and of course, they each come at a certain cost. + +Here's a brief overview of what each food offers and its cost: + +- Steak: Provides you with 1 gram of protein, 3 grams of carbohydrates, and 210 calories for $6. +- Apples: They give you 7 grams of protein, 22 grams of carbohydrates, and 96 calories for just $1. +- Eggs: Deliver 6 grams of protein, 18 grams of carbohydrates, and 189 calories for only $1. +- Chicken: It's a protein powerhouse with 20 grams of protein, but only 6 grams of carbohydrates, and it provides you 132 calories for $3. +- Tofu: Brings in 8 grams of protein, 4 grams of carbohydrates, and 75 calories for $1. + +Your dietary goals are to consume at least 87 grams of protein, 135 grams of carbohydrates, and 1926 calories in a day. Your task is to figure out what combination of these foods can meet all those nutritional requirements for the least amount of money. So, what is the least amount you would have to spend to meet your nutritional needs? You could only assign integral portions of each type of food.",[14.0] +['Diet Problem'],"Imagine you're on a mission to plan your meals to meet specific dietary needs for the day, but you also want to keep your spending within limits. You have four types of food items on your list: Chicken, Rice, Broccoli, and Eggs. Each of these foods has its own nutritional value in terms of protein, carbohydrates, and calories, and also comes at a different price. + +Here's a quick overview of what each food item offers and how much they cost: + +- Chicken: Provides 17 grams of protein, 6 grams of carbohydrates, and 94 calories for $7. +- Rice: Offers 20 grams of protein, 13 grams of carbohydrates, and 74 calories for $10. +- Broccoli: Comes with 3 grams of protein, a high 25 grams of carbohydrates, and 73 calories for $9. +- Eggs: Are a budget-friendly option providing 18 grams of protein, 20 grams of carbohydrates, and 119 calories for only $2. + +You need to ensure that your daily intake should consist of at least 84 grams of protein, 117 grams of carbohydrates, and 1554 calories. Your task is to figure out the most cost-effective way to meet these nutritional targets with the available food options. So, what is the minimal cost you need to incur to satisfy your dietary requirements? Remember, your response should indicate the optimal cost value, under the scenario of food selection. You could only assign integral portions of each type of food.",[28.0] +"[""Transportation Problem""]","Imagine you are the director of a non-profit organization tasked with providing food supplies to six regions suffering from a famine. Each region has a certain amount of food already, but they require more to sustain their population through the hardship. + +Here are the current quantities of food (in tons) available and the required quantities for each region: + +- Region 1 has 42 tons but needs 74 tons. +- Region 2 has 32 tons but needs 476 tons. +- Region 3 has 398 tons but only needs 2 tons. +- Region 4 has 224 tons but needs 235 tons. +- Region 5 has 210 tons but needs 221 tons. +- Region 6 has 209 tons but only needs 72 tons. + +You have the ability to transfer food supplies from one region to another. However, the cost of transportation varies depending on which regions you are transferring food between. Below is a list detailing the cost of moving food from one region to any other: + +- To move food from/to Region 1: To Region 2 costs 16, to Region 3 costs 48, to Region 4 costs 42, to Region 5 costs 50, to Region 6 costs 8. +- To move food from/to Region 2: To Region 1 costs 27, to Region 3 costs 23, to Region 4 costs 37, to Region 5 costs 39, to Region 6 costs 29. +- To move food from/to Region 3: To Region 1 costs 49, to Region 2 costs 39, to Region 4 costs 33, to Region 5 costs 50, to Region 6 costs 6. +- To move food from/to Region 4: To Region 1 costs 23, to Region 2 costs 49, to Region 3 costs 46, to Region 5 costs 50, to Region 6 costs 6. +- To move food from/to Region 5: To Region 1 costs 45, to Region 2 costs 47, to Region 3 costs 48, to Region 4 costs 26, to Region 6 costs 39. +- To move food from/to Region 6: To Region 1 costs 33, to Region 2 costs 11, to Region 3 costs 9, to Region 4 costs 4, to Region 5 costs 12. + +Your mission is to ensure every region receives the food it needs while keeping the transportation cost as low as possible. What would be the minimum cost to make sure all regions have enough food?",[8090.0] +"[""Transportation Problem""]","Imagine you are in charge of managing food supplies for six different cities in a region. Each city has a certain amount of food resources, but they need a specific amount to make sure they're well-prepared for an unexpected food crisis. + +Here's what each city has and what they need: + +- City 1 has 255 tons of food, but the requirement is 251 tons. +- City 2 has 250 tons of food, but needs 200 tons. +- City 3 has 314 tons of food, but the demand here is 366 tons. +- City 4 has only 8 tons of food, but it needs 114 tons to be safe. +- City 5 has an abundance of 385 tons but only needs 59 tons. +- City 6 has 440 tons but needs 287 tons. + +You have been given the responsibility to transport food from one city to another, but it incurs transportation costs. The cost varies depending on which cities you're moving food between. Here's the cost matrix that shows how much it costs to move food from one city to any other: + +- Relocating supplies from/to City 1: To City 2 costs 26, to City 3 costs 39, to City 4 costs 13, to City 5 costs 37, to City 6 costs 28. +- Relocating from/to City 2: To City 1 costs 36, to City 3 costs 48, to City 4 costs 5, to City 5 costs 22, to City 6 costs 14. +- Relocating from/to City 3: To City 1 costs 38, to City 2 costs 36, to City 4 costs 12, to City 5 costs 45, to City 6 costs 36. +- Relocating from/to City 4: To City 1 costs 32, to City 2 costs 18, to City 3 costs 47, to City 5 costs 7, to City 6 costs 9. +- Relocating from/to City 5: To City 1 costs 18, to City 2 costs 6, to City 3 costs 12, to City 4 costs 20, to City 6 costs 42. +- Relocating from/to City 6: To City 1 costs 4, to City 2 costs 28, to City 3 costs 42, to City 4 costs 43, to City 5 costs 7. + +Your objective is to ensure every city has the food resources it needs at the lowest possible transportation cost. Can you determine what would be the minimum cost to ensure all cities have enough food resources?",[1490.0] +['TravelingSalesman'],"In a scenario involving a salesperson who needs to visit five different cities to conduct business, each city is uniquely numbered from 1 to 5. The salesperson's objective is to minimize the total travel expenses, which could be influenced by factors such as distance, fuel costs, or transportation fees. The salesperson can start their journey from any of these cities but must ensure they visit each city exactly once before returning to their starting point. + +The travel costs between the cities are as follows: +- From City 1, the travel costs are 58 units to City 2, 15 units to City 3, 75 units to City 4, and 91 units to City 5. +- From City 2, it costs 58 units to City 1, 54 units to City 3, 85 units to City 4, and 11 units to City 5. +- Traveling from City 3, the expenses are 15 units to City 1, 54 units to City 2, 28 units to City 4, and 61 units to City 5. +- From City 4, the costs are 75 units to City 1, 85 units to City 2, 28 units to City 3, and 47 units to City 5. +- Lastly, from City 5, it costs 91 units to City 1, 11 units to City 2, 61 units to City 3, and 47 units to City 4. + +Given this setup, what is the minimum total travel cost for the salesperson to visit each city exactly once and then return to the starting city?",[159.0] +"[""TravelingSalesman""]","Imagine a delivery driver who needs to deliver packages to five different cities. These cities are represented as City 1, City 2, City 3, City 4, and City 5. The driver can start their route from any city, but they must deliver a package to each city exactly once before returning to the starting city. The objective is to find the route that minimizes the total travel cost. The cost could be based on factors such as distance, fuel consumption, or time spent. + +Here are the details of the travel costs between the cities: + +The cost to travel from City 1 to City 2 is 82 units, to City 3 is 76 units, to City 4 is 92 units, and to City 5 is 59 units. +From City 2, it costs 82 units to return to City 1, 98 units to travel to City 3, 58 units to City 4, and 82 units to City 5. +When leaving from City 3, it costs 76 units to reach City 1, 98 units to get to City 2, 26 units to travel to City 4 and 74 units to travel to City 5. +From City 4, the costs are 92 units to City 1, 58 units to City 2, 26 units to City 3 and 96 units to City 5. +Lastly, from City 5, it costs 59 units to go to City 1, 82 units to travel to City 2, 74 units to City 3, and 96 units to reach City 4. + +Given these costs, what is the minimum total cost for the delivery driver to deliver packages to each city exactly once and return to the starting city?",[299.0] +"[""Network Optimization""]","In an expansive digital communication network designed to connect 7 critical data centers across a region, data flows from a central processing unit to a primary data storage facility. This network is meticulously architected with fiber optic cables of varying capacities to ensure data integrity and speed. The network infrastructure is as follows: + +- From Data Center 0 (Source): Data can flow to Data Center 1 (18 terabits per second - Tbps), Data Center 2 (2 Tbps), Data Center 3 (10 Tbps), Data Center 4 (17 Tbps), and Data Center 6 (12 Tbps). There is no direct connection to Data Center 5. +- From Data Center 1: Data can flow to Data Center 0 (14 Tbps), Data Center 2 (3 Tbps), Data Center 3 (1 Tbps), Data Center 4 (19 Tbps), Data Center 5 (18 Tbps), and Data Center 6 (2 Tbps). +- From Data Center 2: Data can flow to Data Center 0 (17 Tbps), Data Center 1 (5 Tbps), Data Center 3 (15 Tbps), Data Center 4 (15 Tbps), Data Center 5 (19 Tbps), but not directly to Data Center 6. +- From Data Center 3: Data can flow to Data Center 0 (14 Tbps), Data Center 1 (17 Tbps), Data Center 4 (19 Tbps), Data Center 5 (9 Tbps), and Data Center 6 (16 Tbps). There's no direct connection to Data Center 2. +- From Data Center 4: Data can flow to Data Center 0 (2 Tbps), Data Center 1 (13 Tbps), Data Center 2 (15 Tbps), Data Center 3 (12 Tbps), Data Center 5 (20 Tbps), and Data Center 6 (5 Tbps). +- From Data Center 5: Data can flow to Data Center 0 (9 Tbps), Data Center 1 (16 Tbps), Data Center 2 (4 Tbps), Data Center 3 (15 Tbps), Data Center 4 (15 Tbps), and Data Center 6 (11 Tbps). +- From Data Center 6 (Destination): Data can flow back to Data Center 0 (8 Tbps), Data Center 1 (10 Tbps), Data Center 2 (19 Tbps), Data Center 3 (19 Tbps), Data Center 4 (17 Tbps), and Data Center 5 (16 Tbps). + +In this setup, a capacity of ""0"" indicates the absence of a direct fiber optic link between the two data centers. + +The goal is to maximize the flow of data from the central processing unit (Data Center 0) to the primary data storage facility (Data Center 6), adhering strictly to the transfer capacity limitations of the fiber optic cables between each data center. + +What is the maximum amount of data, in terabits per second, that can be transferred from the source to the destination without exceeding any of the network's capacity limitations?",[46.0] +"[""Network Optimization""]","Imagine a country with six major cities, each city being a node in a complex railway network, spanning from City 0 (the source) to City 5 (the destination). These cities are connected by multiple railway tracks, each capable of handling a specific number of trains per day. + +Here's a detailed description of the railway tracks and their capacities, which indicates the maximum number of trains that can travel between two cities in a day: + +- From City 0: Up to 17 trains can travel to City 1, 14 trains to City 2, 13 trains to City 3, 11 trains to City 4, and 3 trains to City 5. +- From City 1: 1 train can travel back to City 0, 18 trains to City 2, 14 trains to City 3, 12 trains to City 4, and 16 trains to City 5. +- From City 2: 5 trains can travel back to City 0, 1 train to City 1, 7 trains to City 3, 2 trains to City 4, and 5 trains to City 5. +- From City 3: 10 trains can travel back to City 0, 13 trains to City 1, 19 trains to City 2, 4 trains to City 4, and 16 trains to City 5. +- From City 4: 3 trains can travel back to City 0, no trains to City 1, 13 trains to City 2, no trains to City 3, and 14 trains to City 5. +- From City 5 (Destination): 15 trains can travel back to City 0, 16 trains to City 1, 4 trains to City 2, 1 train to City 3, and 16 trains to City 4. + +In this scenario, a capacity of ""0"" indicates that there is no direct railway track between those two cities. + +Given this information, your task is to determine the maximum number of trains that can be dispatched from City 0 (the source) to City 5 (the destination) in a single day, without exceeding the capacity of any individual track. What is the maximum number of trains that can travel from the source to the destination in a day?",[54.0] +"[""Network Optimization""]","Welcome to the challenge of optimizing the power grid of a futuristic city, interconnected through 7 strategic substations - from the main power plant to the central hub that distributes electricity to various sectors of the city. Each substation is connected by power lines that have a maximum capacity, measured in megawatts (MW), for transmitting power. + +Here's a breakdown of the power transmission capacities between substations, detailing the maximum amount of electricity that can flow through the network: + +- From Substation 0 (Main Power Plant): Power can be transmitted to Substation 1 (16 MW), Substation 2 (6 MW), Substation 3 (16 MW), Substation 4 (2 MW), Substation 5 (17 MW), and Substation 6 (15 MW). +- From Substation 1: Power can be transmitted to Substation 0 (7 MW), Substation 2 (13 MW), Substation 3 (3 MW), Substation 5 (8 MW), and Substation 6 (17 MW). +- From Substation 2: Power can be transmitted to Substation 0 (3 MW), Substation 1 (17 MW), Substation 3 (18 MW), Substation 4 (11 MW), Substation 5 (18 MW), and Substation 6 (11 MW). +- From Substation 3: Power can be transmitted to Substation 0 (9 MW), Substation 2 (15 MW), Substation 4 (3 MW), Substation 5 (8 MW), and Substation 6 (10 MW). +- From Substation 4: Power can be transmitted to Substation 0 (3 MW), Substation 1 (6 MW), Substation 2 (1 MW), Substation 3 (16 MW), Substation 5 (13 MW), and Substation 6 (10 MW). +- From Substation 5: Power can be transmitted to Substation 0 (15 MW), Substation 1 (20 MW), Substation 2 (2 MW), Substation 3 (6 MW), Substation 4 (18 MW), and Substation 6 (9 MW). +- From Substation 6 (Central Hub): Power can be transmitted back to Substation 0 (4 MW), Substation 1 (6 MW), Substation 2 (8 MW), Substation 3 (13 MW), Substation 4 (16 MW), and Substation 5 (10 MW). + +A ""0"" capacity means there is no direct power line connection between those two substations. + +The city's goal is to ensure that the maximum amount of electricity is efficiently transmitted from the Main Power Plant (Substation 0) to the Central Hub (Substation 6), without overloading any of the power lines beyond their capacity. + +What is the maximum amount of electricity, in megawatts (MW), that can be transmitted from the Main Power Plant to the Central Hub while adhering to these constraints?",[72.0] +"[""Capacitated Facility Location Problem""]","In a challenging and complex market, SwiftLogistics is at a crossroads to design an efficient distribution network to fulfill the needs of eight retail stores scattered across various locations. The company has the opportunity to choose from six potential distribution centers, each with its unique opening costs and capacities. The primary goal is to ensure all retail stores are adequately supplied while minimizing the total operational costs, which include both the opening of distribution centers and the transportation of goods. + +Given Data: +- Number of Potential Distribution Centers: 6 +- Number of Retail Stores Needing Supply: 8 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 +- Center 6: $120,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $2 to Store 1, $3 to Store 2, $3 to Store 3, $3 to Store 4, $5 to Store 5, $4 to Store 6, $3 to Store 7, $5 to Store 8 +- From Center 2: $2 to Store 1, $4 to Store 2, $2 to Store 3, $4 to Store 4, $5 to Store 5, $1 to Store 6, $4 to Store 7, $2 to Store 8 +- From Center 3: $5 to Store 1, $4 to Store 2, $1 to Store 3, $1 to Store 4, $3 to Store 5, $3 to Store 6, $2 to Store 7, $4 to Store 8 +- From Center 4: $4 to Store 1, $3 to Store 2, $4 to Store 3, $4 to Store 4, $1 to Store 5, $3 to Store 6, $5 to Store 7, $3 to Store 8 +- From Center 5: $5 to Store 1, $1 to Store 2, $2 to Store 3, $4 to Store 4, $1 to Store 5, $4 to Store 6, $2 to Store 7, $2 to Store 8 +- From Center 6: $1 to Store 1, $2 to Store 2, $5 to Store 3, $2 to Store 4, $4 to Store 5, $4 to Store 6, $4 to Store 7, $4 to Store 8 + +Demand of Each Retail Store (in units): +- Store 1: 908 units +- Store 2: 434 units +- Store 3: 605 units +- Store 4: 480 units +- Store 5: 961 units +- Store 6: 787 units +- Store 7: 401 units +- Store 8: 789 units + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1365 units +- Center 2: 1929 units +- Center 3: 1502 units +- Center 4: 1201 units +- Center 5: 1529 units +- Center 6: 961 units + +SwiftLogistics now faces the intricate task of determining which distribution centers to open and how to distribute goods to meet the demands of all eight retail stores efficiently. The strategy must ensure that the total cost, including opening and transportation expenses, is minimized. + +Question: +What is the optimal total cost for SwiftLogistics to open the necessary distribution centers and transport goods to satisfy the demands of all eight retail stores, keeping in mind the goal of minimizing the combined opening and transportation costs?",[587428.0] +['Capacitated Facility Location Problem'],"LogistiCorp is at a pivotal juncture, faced with the challenge of efficiently supplying nine retail stores from a pool of seven possible distribution centers. Each center presents a unique opening cost and has a specific capacity limit. Moreover, transporting goods from these centers to the retail stores incurs varying costs. The aim is to fulfill the demand at each retail store while minimizing the overall cost, which encompasses the opening expenses of the distribution centers and the transportation costs of the goods. + +Given Data: +- Number of Possible Distribution Centers: 7 +- Number of Retail Stores to be Supplied: 9 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 +- Center 6: $120,000 +- Center 7: $182,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1 to Stores: $3 to Store 1, $3 to Store 2, $5 to Store 3, $4 to Store 4, $3 to Store 5, $5 to Store 6, $2 to Store 7, $4 to Store 8, $2 to Store 9 +- From Center 2 to Stores: $4 to Store 1, $5 to Store 2, $1 to Store 3, $4 to Store 4, $2 to Store 5, $5 to Store 6, $4 to Store 7, $1 to Store 8, $1 to Store 9 +- From Center 3 to Stores: $3 to Store 1, $3 to Store 2, $2 to Store 3, $4 to Store 4, $4 to Store 5, $3 to Store 6, $4 to Store 7, $4 to Store 8, $1 to Store 9 +- From Center 4 to Stores: $3 to Store 1, $5 to Store 2, $3 to Store 3, $5 to Store 4, $1 to Store 5, $2 to Store 6, $4 to Store 7, $1 to Store 8, $4 to Store 9 +- From Center 5 to Stores: $2 to Store 1, $2 to Store 2, $1 to Store 3, $2 to Store 4, $5 to Store 5, $2 to Store 6, $4 to Store 7, $4 to Store 8, $4 to Store 9 +- From Center 6 to Stores: $4 to Store 1, $5 to Store 2, $3 to Store 3, $1 to Store 4, $4 to Store 5, $2 to Store 6, $4 to Store 7, $2 to Store 8, $2 to Store 9 +- From Center 7 to Stores: $4 to Store 1, $5 to Store 2, $2 to Store 3, $2 to Store 4, $4 to Store 5, $2 to Store 6, $2 to Store 7, $4 to Store 8, $4 to Store 9 + +Demand of Each Retail Store (in units): +- Store 1: 695 units +- Store 2: 737 units +- Store 3: 452 units +- Store 4: 616 units +- Store 5: 587 units +- Store 6: 779 units +- Store 7: 892 units +- Store 8: 440 units +- Store 9: 556 units + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 814 units +- Center 2: 864 units +- Center 3: 1,320 units +- Center 4: 1,952 units +- Center 5: 1,447 units +- Center 6: 1,886 units +- Center 7: 1,962 units + +Question: +What is the optimal total cost for LogistiCorp to open distribution centers and transport goods to meet all retail stores' demands, including both the opening costs of the centers and the transportation costs of supplying the stores?",[485771.0] +['Profit Maximization Problem'],"One unit of fruit salad A requires 1/4 of a mango, 1/8 of a pineapple, and 3 strawberries, with a net profit of 10 RMB. Making one unit of fruit salad B requires 1/2 a mango, 1/4 of a pineapple, and 1 strawberry, with a net profit of 20 RMB. The stock/resources row shows that there are 25 mangos, 10 pineapples, and 120 strawberries available for making these fruit salads. Suppose all fruits need to be processed and completely used to make the fruit salads A and B. All fruits must satisfy the constraint of having at most 5 units left or 5 additional units used. Find the maximal total profits of the fruit store. Note: Since we want to produce “complete fruit salads”, the variables x1 and x2 should actually be modeled as integer variables: x1, x2 ∈ Z.",[1200.0] +['Diet Problem'],"International Wool Company operates a large farm on which sheep are raised. The farm manager determined that for the sheep to grow in the desired fashion, they need at least minimum amounts of four nutrients (the nutrients are nontoxic so the sheep can consume more than the minimum without harm). The manager is considering three different grains to feed the sheep. The nutrients listed are A, B, C, and D. For Grain 1: Nutrient A is present in the amount of 20 units per pound. Nutrient B is present in the amount of 10 units per pound. Nutrient C is present in the amount of 50 units per pound. Nutrient D is present in the amount of 6 units per pound. The cost of Grain 1 is 41 cents per pound. For Grain 2: Nutrient A is present in the amount of 30 units per pound. Nutrient B is present in the amount of 10 units per pound. Nutrient C is present in the amount of 30 units per pound. Nutrient D is present in the amount of 2.5 units per pound. The cost of Grain 2 is 36 cents per pound. For Grain 3: Nutrient A is present in the amount of 70 units per pound. Nutrient B is present in the amount of 0 units per pound. Nutrient C is present in the amount of 0 units per pound. Nutrient D is present in the amount of 10 units per pound. The cost of Grain 3 is 96 cents per pound. The minimum daily requirement for each nutrient is also listed: Nutrient A requires 110 units daily. Nutrient B requires 18 units daily. Nutrient C requires 90 units daily. Nutrient D requires 14 units daily. The manager believes that as long as a sheep receives the minimum daily amount of each nutrient, it will be healthy and produce a standard amount of wool. The manager wants to raise the sheep at minimum cost. Find the minimum cost in cents.",[178] +"[""Capacitated Facility Location Problem""]","SupplyCo is presented with a strategic conundrum to optimize its distribution network. They need to supply their four retail outlets from potential distribution centers. Each center has its own operating cost and supply capacity. Transportation costs from each center to the respective retail outlets also need to be factored in. SupplyCo's objective is to satisfy the demand at each retail outlet at the minimum possible total cost. + +Here are the case specifics: + +- Number of Potential Distribution Centers: 4 +- Number of Retail Outlets: 4 + +The opening costs for each center are as follows (in dollars): + +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 + +The transportation cost per unit from each distribution center to the outlets (in dollars) is as follows: + +- From Center 1: $5 to Outlet 1, $5 to Outlet 2, $2 to Outlet 3, $3 to Outlet 4 +- From Center 2: $3 to Outlet 1, $3 to Outlet 2, $5 to Outlet 3, $4 to Outlet 4 +- From Center 3: $3 to Outlet 1, $5 to Outlet 2, $2 to Outlet 3, $4 to Outlet 4 +- From Center 4: $2 to Outlet 1, $4 to Outlet 2, $5 to Outlet 3, $1 to Outlet 4 + +The demand at each retail outlet (in units) is: + +- Outlet 1: 859 units +- Outlet 2: 713 units +- Outlet 3: 421 units +- Outlet 4: 652 units + +The supply capacity of each distribution center (in units) is: + +- Center 1: 1547 units +- Center 2: 1656 units +- Center 3: 1274 units +- Center 4: 1882 units + +Question: +What is the optimal total cost for SupplyCo to open the distribution centers and transport goods to meet all retail outlets' demands, considering both the opening costs of the centers and the transportation costs for the goods?",[273940.0] +"[""Diet Problem""]","Imagine you're a nutrition-conscious shopper who is trying to meet specific dietary requirements without breaking the bank. You have eight food items to choose from: Eggs, Grains, Berries, Nuts, Salmon, Vegetables, Rice, and Meat. Each of these foods has its own nutritional content and cost. + +Let's break down the nutritional content and cost of each food item: + +- Eggs: For $4, you get 2 grams of protein, 4 grams of carbohydrates, and 282 calories. +- Grains: For $3, you get 7 grams of protein, 9 grams of carbohydrates, and 104 calories. +- Berries: For $2, you get 6 grams of protein, 18 grams of carbohydrates, and 71 calories. +- Nuts: For $4, you get 16 grams of protein, 3 grams of carbohydrates, and 116 calories. +- Salmon: For $9, you get 20 grams of protein, 11 grams of carbohydrates, and 175 calories. +- Vegetables: For $3, you get 6 grams of protein, 27 grams of carbohydrates, and 132 calories. +- Rice: For $6, you get 6 grams of protein, 30 grams of carbohydrates, and 251 calories. +- Meat: For $6, you get 5 grams of protein, 1 gram of carbohydrates, and 74 calories. + +Your goal is to get at least 84 grams of protein, 195 grams of carbohydrates, and 1941 calories within a day from a combination of these food items. The challenge here is to figure out the least expensive way to meet these nutritional targets with the given food options. So, what is the minimum cost you need to spend to meet your daily nutritional requirements? Keep in mind, the answer should be the optimal value under the scenario of food selection. You could only assign integral portions of each type of food.",[40.0] +"[""TravelingSalesman""]","In this scenario, a salesperson is tasked with visiting six distinct cities to conduct business meetings. The cities are sequentially numbered from 1 to 6. The objective for the salesperson is to devise a route that allows them to visit each city exactly once, starting and ending at the same city, while ensuring the total travel cost is minimized. + +The cost of traveling between the cities is as follows: +- From City 1, the travel costs are 49 units to City 2, 34 units to City 3, 18 units to City 4, 62 units to City 5, and 14 units to City 6. +- From City 2, the costs are 49 units back to City 1, 35 units to City 3, 24 units to City 4, 57 units to City 5, and 17 units to City 6. +- From City 3, travel involves 34 units to City 1, 35 units to City 2, 20 units to City 4, 34 units to City 5, and 58 units to City 6. +- From City 4, the costs are 18 units to City 1, 24 units to City 2, 20 units to City 3, 47 units to City 5, and 45 units to City 6. +- From City 5, it costs 62 units to reach City 1, 57 units to City 2, 34 units to City 3, 47 units to City 4, and 71 units to City 6. +- From City 6, the costs are 14 units to City 1, 17 units to City 2, 58 units to City 3, 45 units to City 4, and 71 units to City 5. + +What is the minimum total travel cost for the salesperson to visit each of the six cities exactly once and return to the starting city?",[160.0] +['Capacitated Facility Location Problem'],"In the quest to streamline its distribution network, MarketFlow Inc. is faced with a critical decision regarding the allocation of resources to efficiently supply six retail stores. The company has narrowed down its options to four potential distribution centers, each with distinct opening costs and capabilities. The challenge lies in selecting the right mix of distribution centers and determining the optimal transportation routes to ensure that the demands of all retail stores are met at the lowest possible total cost. This cost encompasses both the opening expenses of the distribution centers and the costs associated with transporting goods to the stores. + +Given Data: +- Number of Potential Distribution Centers: 4 +- Number of Retail Stores Needing Supply: 6 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $5 to Store 1, $5 to Store 2, $2 to Store 3, $3 to Store 4, $3 to Store 5, $3 to Store 6 +- From Center 2: $5 to Store 1, $4 to Store 2, $3 to Store 3, $5 to Store 4, $2 to Store 5, $4 to Store 6 +- From Center 3: $2 to Store 1, $4 to Store 2, $5 to Store 3, $1 to Store 4, $4 to Store 5, $2 to Store 6 +- From Center 4: $5 to Store 1, $4 to Store 2, $1 to Store 3, $1 to Store 4, $3 to Store 5, $3 to Store 6 + +Demand of Each Retail Store (in units): +- Store 1: 910 +- Store 2: 875 +- Store 3: 589 +- Store 4: 962 +- Store 5: 966 +- Store 6: 643 + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1631 +- Center 2: 1954 +- Center 3: 1446 +- Center 4: 820 + +MarketFlow Inc.'s objective is to fulfill the demand at each of its six retail stores efficiently while minimizing the total expenditure incurred from opening distribution centers and transporting goods. This scenario requires a strategic allocation of resources that entails choosing which distribution centers to open and determining the volume of goods to be transported from these centers to each retail store, all while staying within the supply constraints of the centers. + +Question: +What is the optimal total cost for MarketFlow Inc. to open the necessary distribution centers and transport goods to all six retail stores, thereby meeting their demands while ensuring the cost is minimized, including both the opening costs of the centers and the transportation costs of supplying the stores?",[469200.0] +"[""AircraftLanding""]","An air traffic controller needs to control the landing times of n aircrafts: Flights must land in the order 1, ..., n. Flight j must land in time interval [aj , bj ]. n = 10. [a1, b1] = [0, 60], [a2, b2] = [10, 70], [a3, b3] = [20,80], [a4, b4] = [30, 90], [a5, b5] = [40, 100], [a6, b6] = [50, 110], [a7, b7] = [60, 120], [a8, b8] = [70, 130], [a9, b9] = [80, 140], [a10, b10] = [90, 150]. The objective is to maximize the minimum separation time, which is the interval between two landings. Find the maximal minimum separation. ",[16.66666667] +"[""Network Optimization""]","In the heart of a bustling metropolis, there exists a complex network of tunnels designed to transport electricity from a central power plant to various substations across the city. This network is composed of 8 crucial nodes, each representing a key junction point in the electrical distribution system. The nodes range from the primary power plant, where electricity is generated, to a final distribution center, from which electricity is routed to meet the city's demands. + +The capacity of each tunnel, measured in thousands of kilowatts per hour (kWh), dictates the maximum amount of electricity that can flow between any two nodes. The connections and their capacities are as follows: + +- From the Power Plant (Node 0): Electricity can be sent to Node 1 (2 kWh), Node 2 (1 kWh), Node 3 (1 kWh), Node 4 (8 kWh), Node 5 (17 kWh), Node 6 (6 kWh), and Node 7 (7 kWh). +- From Node 1: Electricity can flow to the Power Plant (4 kWh), Node 2 (14 kWh), Node 3 (14 kWh), Node 4 (13 kWh), Node 5 (19 kWh), Node 6 (4 kWh), and Node 7 (19 kWh). +- From Node 2: Electricity can be distributed to the Power Plant (6 kWh), Node 1 (16 kWh), Node 3 (3 kWh), Node 4 (10 kWh), Node 5 (7 kWh), Node 6 (18 kWh), and Node 7 (10 kWh). +- From Node 3: Electricity can travel to the Power Plant (19 kWh), Node 1 (4 kWh), Node 2 (15 kWh), Node 4 (6 kWh), Node 5 (1 kWh), Node 6 (17 kWh), and Node 7 (9 kWh). +- From Node 4: Electricity can be dispatched to the Power Plant (12 kWh), Node 1 (15 kWh), Node 2 (3 kWh), Node 3 (14 kWh), Node 5 (18 kWh), Node 6 (20 kWh), and Node 7 (14 kWh). +- From Node 5: Electricity can move to the Power Plant (20 kWh), Node 1 (5 kWh), Node 2 (19 kWh), Node 3 (9 kWh), Node 4 (20 kWh), Node 6 (10 kWh), and Node 7 (6 kWh). +- From Node 6: Electricity can be sent to the Power Plant (9 kWh), Node 1 (14 kWh), Node 2 (12 kWh), Node 3 (16 kWh), Node 4 (19 kWh), Node 5 (3 kWh), and Node 7 (10 kWh). +- From Node 7 (Final Distribution Center): Electricity can return to the Power Plant (20 kWh), Node 1 (16 kWh), Node 2 (6 kWh), Node 3 (16 kWh), Node 4 (8 kWh), Node 5 (16 kWh), and Node 6 (4 kWh). + +In this electrical network, a capacity of ""0"" indicates no direct tunnel connection for electricity flow between those two nodes. + +The challenge is to compute the maximum amount of electricity that can be efficiently distributed from the Power Plant (Node 0) to the Final Distribution Center (Node 7) per hour, ensuring that the flow through each tunnel does not exceed its maximum capacity. What is the optimal value of electricity (in thousands of kilowatts per hour) that can be transported from the source to the sink in this network?",[42.0] +"[""Network Optimization""]","Imagine a country's interstate highway system containing 6 major cities. These cities are connected by highways, each highway has a specific capacity for the number of vehicles that can travel on it per hour. + +Here's how the capacity of each highway is laid out, showing the maximum number of vehicles (in thousands) that can travel from one city to another per hour: + +- From City 0 (Source): Can send vehicles to City 1 (9,000 vehicles), City 2 (13,000 vehicles), City 4 (20,000 vehicles), and City 5 (16,000 vehicles). +- From City 1: Can send vehicles to City 0 (6,000 vehicles), City 2 (19,000 vehicles), City 3 (11,000 vehicles), City 4 (14,000 vehicles), and City 5 (15,000 vehicles). +- From City 2: Can send vehicles to City 0 (11,000 vehicles), City 1 (6,000 vehicles), City 3 (8,000 vehicles), City 4 (3,000 vehicles), and City 5 (10,000 vehicles). +- From City 3: Can send vehicles to City 0 (16,000 vehicles), City 1 (2,000 vehicles), City 2 (17,000 vehicles), City 4 (1,000 vehicles), and City 5 (17,000 vehicles). +- From City 4: Can send vehicles to City 0 (4,000 vehicles), City 1 (12,000 vehicles), City 2 (5,000 vehicles), City 3 (16,000 vehicles), and City 5 (15,000 vehicles). +- From City 5 (Destination): Can send vehicles back to City 0 (7,000 vehicles), City 1 (19,000 vehicles), City 2 (18,000 vehicles), City 3 (6,000 vehicles), and City 4 (8,000 vehicles). + +In this setup, a capacity of ""0"" means there is no direct highway connection between those two cities. + +The task is to calculate the maximum number of vehicules that can be distributed from City 0 (the source) to City 5 (the destination) per hour, ensuring that the flow through each highway does not exceed its specified capacity. What is the maximum number of vehicles that can be sent from the source to the destination in thousands per hour?",[58.0] +"[""Capacitated Facility Location Problem""]","In the bustling world of retail, SupplyChain Solutions is at a crossroads, deciding how best to structure its distribution network to efficiently supply nine retail stores scattered across the region. The company has narrowed down its choices to five potential distribution centers, each with its unique costs for opening and varying capacities for fulfilling store demands. Additionally, the transportation costs for shipping goods from these centers to the stores differ, making the decision all the more complex. The challenge lies in opening the right combination of distribution centers and planning the supply routes to minimize the total cost, which includes both the opening expenses of the distribution centers and the transportation costs of delivering goods to the stores. + +Given Data: +- Number of Potential Distribution Centers: 5 +- Number of Retail Stores to be Supplied: 9 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $5 to Store 1, $2 to Store 2, $3 to Store 3, $3 to Store 4, $3 to Store 5, $5 to Store 6, $4 to Store 7, $3 to Store 8, $5 to Store 9 +- From Center 2: $2 to Store 1, $4 to Store 2, $2 to Store 3, $4 to Store 4, $5 to Store 5, $1 to Store 6, $4 to Store 7, $2 to Store 8, $5 to Store 9 +- From Center 3: $4 to Store 1, $1 to Store 2, $1 to Store 3, $3 to Store 4, $3 to Store 5, $2 to Store 6, $4 to Store 7, $4 to Store 8, $3 to Store 9 +- From Center 4: $4 to Store 1, $4 to Store 2, $1 to Store 3, $3 to Store 4, $5 to Store 5, $3 to Store 6, $5 to Store 7, $1 to Store 8, $2 to Store 9 +- From Center 5: $4 to Store 1, $1 to Store 2, $4 to Store 3, $2 to Store 4, $2 to Store 5, $1 to Store 6, $2 to Store 7, $5 to Store 8, $2 to Store 9 + +Demand of Each Retail Store (in units): +- Store 1: 739 +- Store 2: 491 +- Store 3: 766 +- Store 4: 854 +- Store 5: 827 +- Store 6: 908 +- Store 7: 434 +- Store 8: 605 +- Store 9: 480 + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1,825 +- Center 2: 1,821 +- Center 3: 1,365 +- Center 4: 1,929 +- Center 5: 1,502 + +Question: +What is the optimal total cost for SupplyChain Solutions to open distribution centers and transport goods to meet all nine retail stores' demands, including both the opening costs of the centers and the transportation costs of supplying the stores?",[608437.0] +"[""Network Optimization""]","Imagine a futuristic transportation network connecting 9 major hubs across a sprawling metropolis, each designed to swiftly move resources from the main distribution center to a critical operations base. This network is meticulously engineered with roads of varying capacities, ensuring the smooth flow of essential supplies. + +The city's network is as follows, with each road's capacity measured in thousands of units of resources that can be transported per hour: + +- From Hub 0 (Source): Roads lead to Hub 1 (4 units), Hub 2 (7 units), Hub 3 (15 units), Hub 4 (4 units), Hub 5 (7 units), Hub 6 (2 units), Hub 7 (10 units), and Hub 8 (12 units). +- From Hub 1: Roads lead to Hub 2 (17 units), Hub 3 (2 units), Hub 4 (5 units), Hub 5 (1 unit), Hub 6 (10 units), Hub 7 (14 units), and Hub 8 (5 units). +- From Hub 2: Roads lead to Hub 1 (7 units), Hub 3 (17 units), Hub 4 (9 units), Hub 5 (13 units), Hub 6 (10 units), Hub 7 (7 units), and Hub 8 (5 units). +- From Hub 3: Roads lead to Hub 1 (16 units), Hub 2 (4 units), Hub 4 (11 units), Hub 5 (9 units), Hub 6 (12 units), Hub 7 (18 units), and Hub 8 (12 units). +- From Hub 4: Roads lead to Hub 1 (3 units), Hub 2 (0 units), Hub 3 (18 units), Hub 5 (5 units), Hub 6 (2 units), and Hub 8 (19 units). +- From Hub 5: Roads lead to Hub 1 (15 units), Hub 2 (7 units), Hub 3 (20 units), Hub 4 (20 units), Hub 6 (20 units), Hub 7 (12 units), and Hub 8 (9 units). +- From Hub 6: Roads lead to Hub 1 (8 units), Hub 2 (14 units), Hub 3 (13 units), Hub 4 (1 unit), Hub 5 (17 units), Hub 7 (10 units), and Hub 8 (19 units). +- From Hub 7: Roads lead to Hub 1 (5 units), Hub 2 (5 units), Hub 3 (16 units), Hub 4 (5 units), Hub 5 (7 units), Hub 6 (3 units), and Hub 8 (19 units). +- From Hub 8 (Destination): Roads lead back to Hub 1 (19 units), Hub 2 (18 units), Hub 3 (0 units), Hub 4 (4 units), Hub 5 (7 units), Hub 6 (1 unit), and Hub 7 (20 units). + +A capacity of ""0"" indicates that there is no direct road connecting those two hubs. + +This complex system has been designed with an aim to maximize the resources transported from Hub 0, the primary distribution center, to Hub 8, the critical operations base, without exceeding the capacity of the roads connecting these hubs. + +The challenge: What is the maximum amount of resources, in thousands of units, that can be transported from the primary distribution center (Hub 0) to the critical operations base (Hub 8) per hour?",[61.0] +['TravelingSalesman'],"Consider a delivery driver who must deliver parcels to six different locations named A through F. The driver can begin the delivery route from any location, but they must visit each location exactly once before returning to the starting point. The objective for the driver is to find the route that minimizes the total delivery cost, which could be based on distance, time, or fuel consumption. + +Here are the specific delivery costs between the locations: +It costs 77 units to deliver from Location A to B, 89 units to C, 19 units to D, 31 units to E, and 62 units to F. +From Location B, the delivery cost to A is 77 units, to C is 36 units, to D is 81 units, to E is 71 units, and to F is 93 units. +For Location C, it costs 89 units to deliver to A, 36 units to B, 32 units to D, 48 units to E, and 36 units to F. +From Location D, the delivery costs are 19 units to A, 81 units to B, 32 units to C, 98 units to E, and 82 units to F. +For deliveries from Location E, the costs are 31 units to A, 71 units to B, 48 units to C, 98 units to D, and 66 units to F. +Finally, from Location F, it costs 62 units to deliver to A, 93 units to B, 36 units to C, 82 units to D, and 66 units to E. + +Given this information, what is the minimum total delivery cost for the driver to deliver parcels to each location exactly once and return to the starting location?",[269.0] +"[""Network Optimization""]","There are four nodes connected by directed edges. Node 1, which could represent the starting point, is connected to two other nodes: There is an edge from node 1 to node 2 with a capacity of 8. There is an edge from node 1 to node 3 with a capacity of 7. Node 2 is situated at the top of the graph and acts as an intermediate node, with three edges connected to it: An edge leading from node 2 to node 3 with a capacity of 2. An edge leading from node 2 to node 4 with a capacity of 4. Node 3 is another intermediate node and has two edges: An edge from node 3 to node 4 with a substantial capacity of 12. Node 4, which could represent the target or terminal, does not have any edges leading to other nodes, implying that it could be the end point. Find the corresponding maximum flow of the graph. ",[13.0] +"[""Set Multi-Cover""]","A post office requires different numbers of full-time employees on different days of the week. The number of full-time employees required on each day is given in Table 4. Monday requires 17 employees,Tuesday requires13, Wednesday requires 15, Thursday requires 19, Friday requires 14, Saturday requires 16, Sunday requires 11. Union rules state that each full-time employee must work five consecutive days and then receive two days off. For example, an employee who works Monday to Friday must be off on Saturday and Sunday. The post office wants to meet its daily requirements using only fulltime employees. Formulate an LP that the post office can use to minimize the number of full-time employees who must be hired. Find the minimal number.",[23.0] +['Diet Problem'],"Imagine you are in charge of planning a week's worth of meals for a small group and need to ensure everyone gets the right amount of nutrients without overspending. You have four different food items to choose from, each with its own nutritional content and cost. You can buy a partial serving of a food. + +Here's what each food item offers: +- Food_1: Provides 19 grams of protein, 9 grams of carbohydrates, and 83 calories for $7. +- Food_2: Offers 4 grams of protein, 16 grams of carbohydrates, and 166 calories for $7. +- Food_3: Contains 3 grams of protein, 11 grams of carbohydrates, and 71 calories for $10. +- Food_4: Delivers 8 grams of protein, 7 grams of carbohydrates, and 56 calories for $2. + +Your goal is to meet the following nutritional requirements for the group: +- At least 84 grams of protein, +- At least 132 grams of carbohydrates, +- At least 1990 calories. + +Your challenge is to determine the most cost-effective way to purchase these food items to meet or exceed the nutritional requirements. What is the minimal cost to meet these dietary needs? You could only assign integral portions of each type of food.",[72] +"[""Diet Problem""]","Suppose you are a student on a budget, trying to balance the need for good nutrition with your financial constraints. You have identified four different food items that you could include in your meals: Eggs, Rice, Chicken, and Broccoli. Each of these items provides different amounts of protein, carbohydrates, and calories, and also has a different cost. + +Here's a rundown of what each food item offers and its cost: + +- Eggs: For a cost of $6, you get 4 grams of protein, 13 grams of carbohydrates, and 238 calories. +- Rice: For $10, you can get 1 gram of protein, 13 grams of carbohydrates, and 85 calories. +- Chicken: A true bargain at only $1, it offers 10 grams of protein, 29 grams of carbohydrates, and 146 calories. +- Broccoli: Costs $7 and provides 8 grams of protein, 26 grams of carbohydrates, and 139 calories. + +Your daily dietary goal is to consume at least 93 grams of protein, 165 grams of carbohydrates, and 1553 calories. Your ultimate goal is to work out the least expensive way to meet all these nutritional targets with the four food options you have. What is the cheapest cost to meet your nutrition requirements? You could only assign integral portions of each type of food.",[11.0] +['Diet Problem'],"Imagine yourself as a dedicated fitness enthusiast on a strict diet and you're trying to plan your meals for an entire day. You aim to meet specific nutritional goals without burning a hole in your pocket. You have five different food items to choose from: Chicken, Pasta, Apples, Tuna, and Rice. Each food item contributes different amounts of proteins, carbohydrates, and calories, and each one has a different price. + +Here is an overview of the nutritional values and cost of each food: + +- Chicken: Provides 8 grams of protein, 8 grams of carbohydrates, and 118 calories for $6. +- Pasta: Delivers a hefty 10 grams of protein, a substantial 30 grams of carbohydrates, and 131 calories for $10. +- Apples: Offer a modest 3 grams of protein, 10 grams of carbohydrates, and a surprising 151 calories for just $1. +- Tuna: Packs a stunning 16 grams of protein, 29 grams of carbohydrates, and a whopping 235 calories for $9. +- Rice: Supplies 4 grams of protein, a solid 24 grams of carbohydrates, and 101 calories for a mere $4. + +Your dietary goal for the day is to consume at least 93 grams of protein, 181 grams of carbohydrates, and 1620 calories. Your challenge is to figure out the most cost-effective way to achieve these nutritional targets using a combination of these five food options. What is the minimum cost required to meet your daily nutritional requirements? Please note that your response should provide only the optimal value in the context of food selection. You could only assign integral portions of each type of food.",[31.0] +"[""Transportation Problem""]","Imagine you are the head of a massive logistics company responsible for distributing medical supplies among seven regions during a global health crisis. Each region currently has a certain amount of supplies, but they each need different quantities to adequately meet their healthcare demands. + +Here's a breakdown of the current supply and demand in each region: + +- Region 1 currently has 318 units of supplies but needs 309 units. +- Region 2 has 349 units but needs merely 9 units. +- Region 3 has 129 units but needs a substantial 279 units. +- Region 4 has 381 units but only needs 71 units. +- Region 5 has 430 units but requires 365 units. +- Region 6, quite undersupplied, has 52 units but needs 233 units. +- Region 7 has 124 units but the demand is very high at 483 units. + +The transportation of these supplies from one region to another comes at a cost. This cost varies depending on the regions involved. Here's a matrix depicting the cost to transport supplies from one region to any other: + +- Transporting from/to Region 1: To Region 2 costs 2, to Region 3 costs 26, to Region 4 costs 19, to Region 5 costs 16, to Region 6 costs 26, to Region 7 costs 36. +- Transporting from/to Region 2: To Region 1 costs 27, to Region 3 costs 26, to Region 4 costs 4, to Region 5 costs 15, to Region 6 costs 43, to Region 7 costs 25. +- Transporting from/to Region 3: To Region 1 costs 17, to Region 2 costs 45, to Region 4 costs 18, to Region 5 costs 36, to Region 6 costs 37, to Region 7 costs 39. +- Transporting from/to Region 4: To Region 1 costs 19, to Region 2 costs 16, to Region 3 costs 14, to Region 5 costs 5, to Region 6 costs 35, to Region 7 costs 9. +- Transporting from/to Region 5: To Region 1 costs 44, to Region 2 costs 47, to Region 3 costs 4, to Region 4 costs 3, to Region 6 costs 2, to Region 7 costs 4. +- Transporting from/to Region 6: To Region 1 costs 30, to Region 2 costs 12, to Region 3 costs 28, to Region 4 costs 10, to Region 5 costs 16, to Region 7 costs 25. +- Transporting from/to Region 7: To Region 1 costs 19, to Region 2 costs 16, to Region 3 costs 10, to Region 4 costs 2, to Region 5 costs 50, to Region 6 costs 4. + +Your mission is to devise a plan that ensures each region gets the medical supplies it needs, while keeping transportation costs as low as possible. Considering these constraints, what is the minimum cost to ensure all regions receive the necessary supplies?",[6783.0] +['TravelingSalesman'],"Consider a courier company that needs to deliver packages to five distinct cities, denoted as E, F, G, H, and I. The courier can start from any city, but they must visit each city only once and then return to the starting point. The aim is to find a route that would minimize the total delivery cost. The cost might include factors like distance, fuel expenses, or traffic conditions. Here's an outline of the delivery cost between these cities: +The cost to deliver from City E to F is 50 units, to G is 48 units, to H is 99 units, and to I is 91 units. +From City F, it costs 50 units to deliver to E, 57 units to deliver to G, 84 units to H, and 72 units to I. +For City G, the delivery costs are 48 units to E, 57 units to F, 46 units to H, and 86 units to I. +If the package starts from City H, it costs 99 units to deliver to E, 84 units to F, 46 units to G, and 29 units to I. +Lastly, from City I, it costs 91 units to deliver to E, 72 units to F, 86 units to G, and 29 units to H. +What is the least total delivery cost for the courier to visit each city exactly once and then return to the starting point?",[245.0] +['TravelingSalesman'],"Imagine a scenario in which a business consultant needs to visit four distinct cities to evaluate the operations of regional offices. These cities are labeled as City 1, City 2, City 3, and City 4. The consultant can begin their route from any of these cities. The key requirement is that each city must be visited exactly once before returning to the starting city. The main objective is to minimize the total cost of travel, which includes expenses such as transportation, lodging, and time spent traveling. + +Here’s a quick summary of the travel costs between the cities: +- The cost to travel from City 1 to City 2 is 29 units, to City 3 is 97 units, and to City 4 is 72 units. +- From City 2, the journey costs 29 units to reach City 1, 69 units to City 3, and 52 units to City 4. +- Traveling from City 3, the expenses are 97 units to City 1, 69 units to City 2, and 91 units to City 4. +- Lastly, from City 4, it costs 72 units to go to City 1, 52 units to City 2, and 91 units to City 3. + +What is the minimum total travel cost for the consultant to visit each city exactly once and then return to the starting city?",[261.0] +"[""TravelingSalesman""]","Imagine a courier company that needs to deliver packages to five different cities, labeled as E, F, G, H, and I. The company requires that their courier starts from any city, delivers packages to each city exactly once, and finally returns to the starting city. The aim is to devise a route that minimizes the total delivery cost, which could be influenced by factors like distance, traffic, and fuel expenses. +Let's break down the delivery costs between the cities: +To deliver from City E to F, it costs 64 units, to G costs 78 units, to H costs 23 units, and to I costs 15 units. +From City F, it costs 64 units to reach E, 15 units to deliver to G, 44 units to H, and 93 units to I. +If the courier starts from City G, the costs are 78 units to E, 15 units to F, 53 units to H, and 57 units to I. +From City H, it costs 23 units to deliver to E, 44 units to F, 53 units to G, and 56 units to I. +Finally, from City I, it costs 15 units to reach E, 93 units to F, 57 units to G, and 56 units to H. +What is the minimum total delivery cost for the courier to deliver packages to each city exactly once and return to the starting city?",[154.0] +"[""Network Optimization""]","In a vast network of canals interconnecting 9 strategic reservoirs, engineers have meticulously mapped out the maximum capacities for water transfer (in millions of gallons per day) between each reservoir to facilitate efficient water distribution across a sprawling agricultural region. These reservoirs, from the source (Reservoir 0) to the final distribution point (Reservoir 8), are crucial in ensuring that water reaches the farthest corners of the region, especially during peak farming season. + +Here's a detailed breakdown of the water transfer capacities between reservoirs: + +- From Reservoir 0: Water can be directed to Reservoir 1 (6 million gallons), Reservoir 2 (1 million gallons), Reservoir 3 (14 million gallons), Reservoir 4 (20 million gallons), Reservoir 5 (6 million gallons), Reservoir 6 (13 million gallons), Reservoir 7 (13 million gallons), and Reservoir 8 (15 million gallons). +- From Reservoir 1: Water can flow to Reservoir 0 (2 million gallons), Reservoir 2 (5 million gallons), Reservoir 3 (5 million gallons), Reservoir 4 (13 million gallons), Reservoir 5 (5 million gallons), Reservoir 6 (12 million gallons), Reservoir 7 (9 million gallons), and Reservoir 8 (15 million gallons). +- From Reservoir 2: Capacities are to Reservoir 0 (15 million gallons), Reservoir 3 (8 million gallons), Reservoir 4 (16 million gallons), Reservoir 5 (9 million gallons), Reservoir 6 (19 million gallons), Reservoir 7 (15 million gallons), and Reservoir 8 (6 million gallons). +- From Reservoir 3: Water can be transferred to Reservoir 0 (13 million gallons), Reservoir 1 (8 million gallons), Reservoir 4 (14 million gallons), Reservoir 5 (12 million gallons), Reservoir 6 (9 million gallons), Reservoir 7 (1 million gallons), and Reservoir 8 (2 million gallons). +- From Reservoir 4: Channels lead to Reservoir 0 (4 million gallons), Reservoir 1 (14 million gallons), Reservoir 2 (19 million gallons), Reservoir 3 (16 million gallons), Reservoir 5 (14 million gallons), Reservoir 6 (11 million gallons), Reservoir 7 (11 million gallons), and Reservoir 8 (2 million gallons). +- From Reservoir 5: Water is distributed to Reservoir 0 (19 million gallons), Reservoir 1 (18 million gallons), Reservoir 2 (5 million gallons), Reservoir 3 (20 million gallons), Reservoir 4 (9 million gallons), Reservoir 6 (11 million gallons), Reservoir 7 (5 million gallons), and Reservoir 8 (7 million gallons). +- From Reservoir 6: The flow is to Reservoir 0 (11 million gallons), Reservoir 1 (14 million gallons), Reservoir 2 (20 million gallons), Reservoir 3 (17 million gallons), Reservoir 4 (11 million gallons), Reservoir 5 (13 million gallons), Reservoir 7 (11 million gallons), and Reservoir 8 (16 million gallons). +- From Reservoir 7: It can send to Reservoir 0 (6 million gallons), Reservoir 1 (2 million gallons), Reservoir 2 (17 million gallons), Reservoir 4 (10 million gallons), Reservoir 5 (13 million gallons), Reservoir 6 (17 million gallons), and Reservoir 8 (9 million gallons). +- From Reservoir 8: Water can flow back to Reservoir 0 (8 million gallons), Reservoir 1 (9 million gallons), Reservoir 2 (7 million gallons), Reservoir 3 (18 million gallons), Reservoir 4 (12 million gallons), Reservoir 5 (13 million gallons), Reservoir 6 (5 million gallons), and Reservoir 7 (9 million gallons). + +A ""0"" indicates that there is no direct water transfer capability between those two reservoirs. + +The challenge is to optimize this water distribution system to ascertain the maximum possible water transfer from Reservoir 0 (the source) to Reservoir 8 (the final distribution point) per day. What is the optimal value of water (in millions of gallons) that can be transferred from the source to the final distribution point per day?",[72.0] +"[""Network Optimization""]","Imagine you're overseeing a logistics network that connects 7 major cities - from a central distribution hub (City 0) to a final destination (City 6). These cities are linked by routes, each with a specific capacity for goods transportation. + +Here's how the capacity of each route is laid out, showing the maximum amount of goods (in thousands of kilograms) that can be transported from one city to another: + +- From City 0 (Distribution Hub): Can send goods to City 1 (20 tons), City 2 (19 tons), City 3 (18 tons), City 5 (11 tons), and City 6 (2 tons). +- From City 1: Can send goods to City 0 (18 tons), City 2 (15 tons), City 3 (10 tons), and City 5 (9 tons). +- From City 2: Can send goods to City 0 (18 tons), City 1 (4 tons), City 3 (11 tons), City 4 (15 tons), and City 6 (13 tons). +- From City 3: Can send goods to City 0 (17 tons), City 1 (3 tons), City 2 (15 tons), City 4 (2 tons), City 5 (4 tons), and City 6 (4 tons). +- From City 4: Can send goods to City 0 (2 tons), City 1 (11 tons), City 2 (17 tons), City 3 (17 tons), City 5 (2 tons), and City 6 (17 tons). +- From City 5: Can send goods to City 0 (12 tons), City 1 (4 tons), City 2 (16 tons), City 3 (13 tons), City 4 (5 tons), and City 6 (3 tons). +- From City 6 (Destination): Can send goods back to City 0 (7 tons), City 1 (9 tons), City 2 (17 tons), City 3 (18 tons), City 4 (16 tons), and City 5 (8 tons). + +In this system, a capacity of ""0"" implies that there is no direct route between those two cities. + +The objective is to calculate the maximum amount of goods that can be transported from City 0 (the distribution hub) to City 6 (the destination) per day, ensuring that the flow on each route does not exceed its capacity. So the question is, what is the maximum amount of goods that can be transported from the distribution hub to the final destination in thousands of kilograms per day?",[39.0] +"[""Network Optimization""]","Let's consider a scenario where we have a set of interconnected computer servers intended to distribute data packets. The servers are set up in a network of 6 nodes, ranging from Node 0, the source, to Node 5, the destination. The connections between servers have a certain data transfer capacity, representing the maximum number of data packets that can be sent per unit time. + +Here's how the capacities of each connection, represented in thousands of data packets per unit time, are distributed: + +- From Node 0 (Source): Can send data to Node 1 (11 packets), Node 2 (8 packets), Node 3 (11 packets), Node 4 (20 packets), and Node 5 (16 packets). +- From Node 1: Can send data to Node 0 (15 packets), Node 2 (11 packets), Node 3 (16 packets), and Node 5 (20 packets). +- From Node 2: Can send data to Node 0 (7 packets), Node 1 (4 packets), Node 3 (4 packets), and Node 4 (11 packets). +- From Node 3: Can send data to Node 0 (4 packets), Node 1 (8 packets), Node 2 (14 packets), and Node 4 (19 packets). +- From Node 4: Can send data to Node 0 (10 packets), Node 2 (6 packets), Node 3 (19 packets), and Node 5 (13 packets). +- From Node 5 (Destination): Can send data back to Node 0 (10 packets), Node 1 (15 packets), Node 2 (20 packets), Node 3 (19 packets), and Node 4 (10 packets). + +In this network, a capacity of ""0"" indicates there is no direct connection between the two nodes. + +The task is to calculate the maximum number of data packets that can be sent from Node 0 (the source) to Node 5 (the destination) without exceeding the capacity of any individual connection. In other words, what is the maximum flow possible in this network? What is the maximum data that can be sent from the source to the destination in thousands of packets per unit time?",[49.0] +['Capacitated Facility Location Problem'],"In a quest to revamp its supply chain, MegaMart is at the crossroads of deciding the best way to distribute its products to eight retail stores scattered across the region. The company has identified nine potential locations for distribution centers, each with its own set of opening costs and capabilities. The challenge lies in selecting which distribution centers to open and determining the most cost-effective way to transport goods, ensuring that the demands of all retail stores are met without overshooting the budget. + +Given Data: +- Number of Potential Distribution Centers: 9 +- Number of Retail Stores Needing Supply: 8 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 +- Center 6: $120,000 +- Center 7: $182,000 +- Center 8: $186,000 +- Center 9: $174,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- Starting from Center 1: Costs are $3 to Store 1, $7 to Store 2, $4 to Store 3, $6 to Store 4, $5 to Store 5, $8 to Store 6, $9 to Store 7, and $2 to Store 8. +- Starting from Center 2: Costs are $5 to Store 1, $3 to Store 2, $6 to Store 3, $7 to Store 4, $4 to Store 5, $5 to Store 6, $10 to Store 7, and $3 to Store 8. +- Starting from Center 3: Costs are $6 to Store 1, $8 to Store 2, $5 to Store 3, $6 to Store 4, $7 to Store 5, $4 to Store 6, $8 to Store 7, and $5 to Store 8. +- Starting from Center 4: Costs are $4 to Store 1, $6 to Store 2, $7 to Store 3, $5 to Store 4, $6 to Store 5, $7 to Store 6, $9 to Store 7, and $4 to Store 8. +- Starting from Center 5: Costs are $7 to Store 1, $5 to Store 2, $6 to Store 3, $8 to Store 4, $5 to Store 5, $6 to Store 6, $7 to Store 7, and $3 to Store 8. +- Starting from Center 6: Costs are $8 to Store 1, $4 to Store 2, $5 to Store 3, $6 to Store 4, $7 to Store 5, $5 to Store 6, $6 to Store 7, and $4 to Store 8. +- Starting from Center 7: Costs are $5 to Store 1, $7 to Store 2, $4 to Store 3, $7 to Store 4, $6 to Store 5, $8 to Store 6, $5 to Store 7, and $6 to Store 8. +- Starting from Center 8: Costs are $6 to Store 1, $5 to Store 2, $8 to Store 3, $4 to Store 4, $7 to Store 5, $6 to Store 6, $9 to Store 7, and $5 to Store 8. +- Starting from Center 9: Costs are $7 to Store 1, $6 to Store 2, $5 to Store 3, $7 to Store 4, $4 to Store 5, $7 to Store 6, $8 to Store 7, and $6 to Store 8. + +Demand of Each Retail Store (in units): +- The demands are 414 units for Store 1, 920 units for Store 2, 530 units for Store 3, 780 units for Store 4, 625 units for Store 5, 475 units for Store 6, 880 units for Store 7, 610 units for Store 8 + +Supply Capacity of Each Distribution Center (in units): +- Capacities are 1,030 units for Center 1, $1250 for Center 2, $980 for Center 3, $1962 for Center 4, $1480 for Center 5, $1120 for Center 6, $1310 for Center 7, $1675 for Center 8 and $1420 for Center 9. + +The objective of MegaMart is clear: to unfurl a distribution strategy that minimizes the total cost, including both the opening costs of selected distribution centers and the transportation costs of goods to the retail stores. The company must navigate through the maze of potential costs to find the most economical route that ensures all retail stores' demands are satisfied without excess. + +Question: +What is the optimal total cost for MegaMart to open the necessary distribution centers and transport goods to meet all retail stores' demands, considering both the opening and transportation costs? + +This question seeks a singular answer, the minimum possible total cost, encapsulating the strategic decision-making challenge MegaMart faces.",[590186.0] +"[""Capacitated Facility Location Problem""]","In the bustling world of retail, a company named RetailOpt is faced with the challenge of efficiently supplying nine of its stores from a selection of five possible distribution centers. This task involves not only selecting which distribution centers to open but also determining how to distribute goods in a way that minimizes the total cost, including both the opening costs of the distribution centers and the costs of transportation from these centers to the stores. + +Given Data: +- Number of Potential Distribution Centers: 5 +- Number of Retail Stores to be Supplied: 9 + +The Opening Costs for Each Distribution Center (in dollars) are as follows: +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 + +The Transportation Cost Per Unit from Each Distribution Center to Each Retail Store (in dollars) varies considerably, presenting a complex matrix of options: +- From Center 1: $5 to Store 1, $2 to Store 2, $9 to Store 3, $1 to Store 4, $7 to Store 5, $4 to Store 6, $3 to Store 7, $8 to Store 8, and $5 to Store 9. +- From Center 2: $2 to Store 1, $4 to Store 2, $6 to Store 3, $5 to Store 4, $8 to Store 5, $3 to Store 6, $9 to Store 7, $7 to Store 8, and $5 to Store 9. +- From Center 3: $9 to Store 1, $1 to Store 2, $3 to Store 3, $4 to Store 4, $2 to Store 5, $7 to Store 6, $6 to Store 7, $5 to Store 8, and $8 to Store 9. +- From Center 4: $3 to Store 1, $5 to Store 2, $2 to Store 3, $8 to Store 4, $6 to Store 5, $9 to Store 6, $1 to Store 7, $4 to Store 8, and $7 to Store 9. +- From Center 5: $4 to Store 1, $3 to Store 2, $7 to Store 3, $2 to Store 4, $1 to Store 5, $5 to Store 6, $8 to Store 7, $6 to Store 8, and $9 to Store 9. + +Each Retail Store has its own demand for units, totaling: +- Store 1 needs 739 units; Store 2 needs 491 units; Store 3 needs 832 units; Store 4 needs 567 units; Store 5 needs 913 units; Store 6 needs 642 units; Store 7 needs 755 units; Store 8 needs 981 units; Store 9 needs 480 units. + +Meanwhile, the Supply Capacity of Each Distribution Center (in units) determines how much each center can distribute: +- Center 1: 1825 units +- Center 2: 1821 units +- Center 3: 1365 units +- Center 4: 1929 units +- Center 5: 1502 units + +This scenario presents RetailOpt with a complex optimization puzzle: How to meet the demands of all nine retail stores in the most cost-effective manner. The decision involves selecting which distribution centers to open and determining the distribution routes to minimize the total cost incurred from opening costs and transportation costs. + +Question: +What is the optimal total cost for RetailOpt to open the necessary distribution centers and transport goods to meet all nine retail stores' demands, considering both the opening costs of the centers and the transportation costs of supplying the stores?",[612877.0] +"[""Transportation Problem""]","The China Railroad Ministry is in the process of planning relocations of freight cars among 5 regions of the country to get ready for the fall harvest. Moving a car from locations 1 to 2 costs 15 units. Moving a car from location 1 to 3 costs 18 units. Moving a car from location 1 to 4 costs 11 units. Moving a car from location 1 to 5 costs 28 units. Moving a car from location 2 to 3 costs 18 units. Moving a car from location 2 to 4 costs 8 units. Moving a car from location 2 to 5 costs 49 units. Moving a car from location 3 to 4 costs 9 units. Moving a car from location 3 to 5 costs 27 units. Moving a car from location 4 to 5 costs 20 units. At location 1, there are currently 120 cars present, but 150 cars are needed. At location 2, there are currently 330 cars present, but 200 cars are needed. At location 3, there are currently 400 cars present, but 600 cars are needed. At location 4, there are currently 400 cars present, but 200 cars are needed. At location 5, there are currently 600 cars present, but 400 cars are needed. Write down a linear optimization to compute the least costly way to move the cars such us the need is met.",[2250.0] +['Blending Problem'],"Solar Oil Company is a gasoline refiner and wholesaler. It sells two products to gas stations: regular and premium gasoline. It makes these two final products by blending together four raw gasolines and some chemical additives (the amount and cost of the additives per barrel are assumed to be independent of the mixture). Each gasoline has an octane rating that reflects its energy content. The four types of raw gasoline have octane ratings and daily availability as follows: raw gasoline 1 with 86 octane is available up to 20,000 barrels per day at a cost of $17.00 per barrel, raw gasoline 2 with 88 octane can be purchased up to 15,000 barrels per day at $18.00 per barrel, raw gasoline 3 with 92 octane is available up to 15,000 barrels daily at $20.50 per barrel, and raw gasoline 4 with 96 octane has a daily availability of 10,000 barrels at $23.00 per barrel. The required minimum octane for each final gasoline product is 89 for regular, which sells at $19.50 per barrel with a maximum daily demand of 35,000 barrels, and 93 for premium gasoline, priced at $22.00 per barrel with a demand of up to 23,000 barrels per day. Solar Oil can sell all the gas it produces up to that amount. The blending of gasoline is approximately a linear operation in terms of volume and octane. If x barrels of 80 octane gasoline are blended with y barrels of 90 octane gasoline, this produces x + y barrels of gasoline with an octane of (80x + 90y)/(x+y). There is no significant volume gain or loss, and octane of the mixture is a weighted average of the octanes of the inputs. Find the maximal profit of the company.",[42141.0] +"[""Diet Problem""]","Imagine you're planning your meals for the week and you want to keep everything balanced: enough nutrients for a healthy lifestyle but also manageable costs. You have six food options to consider: Chicken, Rice, Vegetables, Bread, Almonds, and Salmon. Each of these foods has its own nutritional content in terms of proteins, carbs, and calories, and also its own cost. + +Here's a summary of what each food offers and its cost: + +- Chicken: Offers 16 grams of protein, 8 grams of carbohydrates, and 50 calories for $2. +- Rice: Provides 11 grams of protein, 24 grams of carbohydrates, and 90 calories for $3. +- Vegetables: Gives you 1 gram of protein, 5 grams of carbohydrates, and a hefty 164 calories for $3. +- Bread: Packs in 6 grams of protein, 26 grams of carbohydrates, and 165 calories for $2. +- Almonds: Offer 2 grams of protein, 11 grams of carbohydrates, and 51 calories for $4. +- Salmon: Packs in a big punch with 19 grams of protein, 20 grams of carbohydrates, and provides 238 calories for $10. + +You aim to hit at least 58 grams of protein, 150 grams of carbohydrates, and 2392 calories from whichever combination of these foods you choose. Your challenge is to determine the most affordable way to meet these nutritional targets with the six food options at your disposal. What is the minimum cost to satisfy all your nutritional needs for the week? Remember, your answer should indicate only the optimal cost value. You could only assign integral portions of each type of food.",[30.0] +['Capacitated Facility Location Problem'],"LogiChain Solutions is in the process of refining its logistics network to efficiently supply goods to nine retail stores, which are scattered across different locations. The company is considering the feasibility of opening up to eight different distribution centers. Each distribution center has a unique set of operational costs and capacities, as well as varying transportation costs for shipping goods to the retail stores. The challenge lies in selecting the most cost-effective combination of distribution centers to open and determining the optimal shipping routes to fulfill the demands of all retail stores while keeping the total operational and transportation costs as low as possible. + +Given Data: +- Number of Potential Distribution Centers: 8 +- Number of Retail Stores Needing Supply: 9 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 +- Center 6: $120,000 +- Center 7: $182,000 +- Center 8: $186,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1 to Stores: $3 to Store 1, $3 to Store 2, $5 to Store 3, $4 to Store 4, $3 to Store 5, $5 to Store 6, $2 to Store 7, $4 to Store 8, $2 to Store 9 +- From Center 2 to Stores: $4 to Store 1, $5 to Store 2, $1 to Store 3, $4 to Store 4, $2 to Store 5, $5 to Store 6, $4 to Store 7, $1 to Store 8, $1 to Store 9 +- From Center 3 to Stores: $3 to Store 1, $3 to Store 2, $2 to Store 3, $4 to Store 4, $4 to Store 5, $3 to Store 6, $4 to Store 7, $4 to Store 8, $1 to Store 9 +- From Center 4 to Stores: $3 to Store 1, $5 to Store 2, $3 to Store 3, $5 to Store 4, $1 to Store 5, $2 to Store 6, $4 to Store 7, $1 to Store 8, $4 to Store 9 +- From Center 5 to Stores: $2 to Store 1, $2 to Store 2, $1 to Store 3, $2 to Store 4, $5 to Store 5, $2 to Store 6, $4 to Store 7, $4 to Store 8, $4 to Store 9 +- From Center 6 to Stores: $4 to Store 1, $5 to Store 2, $3 to Store 3, $1 to Store 4, $4 to Store 5, $2 to Store 6, $4 to Store 7, $2 to Store 8, $2 to Store 9 +- From Center 7 to Stores: $4 to Store 1, $5 to Store 2, $2 to Store 3, $2 to Store 4, $4 to Store 5, $2 to Store 6, $2 to Store 7, $4 to Store 8, $4 to Store 9 +- From Center 8 to Stores: $1 to Store 1, $5 to Store 2, $5 to Store 3, $2 to Store 4, $5 to Store 5, $2 to Store 6, $1 to Store 7, $4 to Store 8, $4 to Store 9 + +Demand of Each Retail Store (in units): +- Store 1: 779 units +- Store 2: 892 units +- Store 3: 440 units +- Store 4: 556 units +- Store 5: 414 units +- Store 6: 464 units +- Store 7: 920 units +- Store 8: 743 units +- Store 9: 528 units + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1447 units +- Center 2: 1886 units +- Center 3: 1962 units +- Center 4: 1392 units +- Center 5: 1191 units +- Center 6: 1218 units +- Center 7: 1088 units +- Center 8: 1178 units + +Question: +What is the optimal total cost for LogiChain Solutions to open the necessary distribution centers and transport goods to meet all the demands of the nine retail stores? This cost includes both the opening costs of the centers and the transportation costs of supplying the stores.",[556908.0] +['Diet Problem'],"Imagine you're on a mission to optimize your daily meal plan. You're aiming to fulfill your daily nutritional intake while keeping your expenses as low as possible. In front of you, you have a selection of nine different food items: Beef, Pasta, Broccoli, Chicken, Apple, Pork, Lentils, Eggs, and Cheese. Each of these foods has its own cost, protein, carbohydrate, and calorie content. + +Let's take a look at the nutritional details and price of each food item: + +- Beef: For a cost of $5, serves you with 18 grams of protein, 13 grams of carbohydrates, and 233 calories. +- Pasta: Comes at $6, but it provides 2 grams of protein, 25 grams of carbohydrates, and 251 calories. +- Broccoli: Costs $5, and gives 7 grams of protein, 30 grams of carbohydrates, and 191 calories. +- Chicken: Offers 20 grams of protein, 16 grams of carbohydrates, and 231 calories at a cost of $4. +- Apple: At $1, provides 5 grams of protein, 6 grams of carbohydrates, and 111 calories. +- Pork: Priced at $6, gives you 13 grams of protein, 9 grams of carbohydrates, and 275 calories. +- Lentils: Costing $1, it provides 15 grams of protein, 18 grams of carbohydrates, and 99 calories. +- Eggs: For $2, they serve 20 grams of protein, 18 grams of carbohydrates, and 210 calories. +- Cheese: At $7, it provides 9 grams of protein, 14 grams of carbohydrates, and 74 calories. + +Your daily dietary requirements are to consume at least 68 grams of protein, 151 grams of carbohydrates, and 1900 calories. The question before you is: How can you meet these nutritional targets in the most cost-effective way using the food options provided? In other words, what is the least amount of money you need to spend in order to fulfill your daily nutritional requirements? You could only assign integral portions of each type of food.",[18.0] +['TravelingSalesman'],"Imagine a scenario where a salesperson has to visit five different cities, labeled as City 1, City 2, City 3, City 4, and City 5. The objective for the salesperson is to minimize the travel costs associated with visiting each city exactly once and then returning to the starting city. The salesperson can begin their journey from any of the cities. + +Here are the travel costs between the cities: +- Traveling from City 1 to City 2 costs 46 units, to City 3 costs 63 units, to City 4 costs 54 units, and to City 5 costs 45 units. +- From City 2, it costs 46 units to reach City 1, 48 units to get to City 3, 50 units to reach City 4, and 51 units to travel to City 5. +- From City 3, the travel costs are 63 units to City 1, 48 units to City 2, 31 units to City 4, and 64 units to City 5. +- From City 4, the expenses are 54 units to City 1, 50 units to City 2, 31 units to City 3, and a steep 94 units to get to City 5. +- Lastly, traveling from City 5 involves costs of 45 units to City 1, 51 units to City 2, 64 units to City 3, and 94 units to City 4. + +What is the minimum total travel cost for the salesperson to visit each city exactly once and return to the starting point?",[229.0] +['Transportation Problem'],"In a region divided into four districts, each district has a certain amount of food supplies available but requires a specific amount to adequately prepare for an upcoming season. Here's the breakdown of the current food supplies and the needed quantities: + +- District 1 currently has 412 units of food but only needs 74 units. +- District 2 has 92 units but needs a substantial increase to 334 units. +- District 3 has 265 units, yet it requires only 61 units. +- District 4 holds 342 units of food and needs 352 units. + +To redistribute these food supplies, there are costs associated with transporting food from one district to another. The transportation costs between the districts are as follows: + +- From District 1: To District 2 costs 20 units, to District 3 costs 43 units, to District 4 costs 1 unit. +- From District 2: To District 1 costs 42 units, to District 3 costs 32 units, to District 4 costs 13 units. +- From District 3: To District 1 costs 27 units, to District 2 costs 5 units, to District 4 costs 10 units. +- From District 4: To District 1 costs 30 units, to District 2 costs 22 units, to District 3 costs 9 units. + +The goal is to ensure each district has the amount of food it needs, and to achieve this at the lowest possible transportation cost. What is the minimum cost required to achieve the optimal distribution of food supplies across all districts?",[1600.0] +"[""Diet Problem""]","You have been assigned as the personal dietitian for a professional athlete, who is intent on maintaining a strict diet while also minimizing their grocery bill. You have six different food items to utilize: Eggs, Chicken, Tofu, Salmon, Brown Rice, and Spinach. Each of these food items provides specific amounts of protein, carbohydrates, and calories, and each comes at a different cost. + +Here's a quick breakdown of what each food offers and its cost: + +- Eggs: Provides 1 gram of protein, 13 grams of carbohydrates, and 211 calories for $10. +- Chicken: Offers 4 grams of protein, 10 grams of carbohydrates, and 256 calories for $9. +- Tofu: Delivers a whopping 13 grams of protein, 9 grams of carbohydrates, and 273 calories for just $2. +- Salmon: Gives 8 grams of protein, 3 grams of carbohydrates, and 54 calories for $10. +- Brown Rice: Provides 7 grams of protein, 13 grams of carbohydrates, and 93 calories for $8. +- Spinach: Packs in 8 grams of protein, 16 grams of carbohydrates, and 52 calories for only $1. + +The athlete's nutritional goals are at least 60 grams of protein, 122 grams of carbohydrates, and 2028 calories daily. + +Your challenge is to design the most cost-effective daily meal plan that meets all these nutritional targets using these six food options. What is the least amount of money the athlete would have to spend to meet their nutritional requirements? You could only assign integral portions of each type of food.",[18.0] +"[""Transportation Problem""]","The China Railroad Ministry is in the process of planning relocations of freight cars among 5 regions of the country to get ready for the fall harvest. The following presents the costs of moving a car between each pair of regions. The costs are as follows: moving a car from region 1 to region 2 costs 10 units, to region 3 costs 12 units, to region 4 costs 17 units, and to region 5 costs 34 units. From region 2 to region 1 costs 10 units, to region 3 costs 18 units, to region 4 costs 8 units, and to region 5 costs 46 units. From region 3, moving a car to region 1 costs 12 units, to region 2 costs 18 units, to region 4 costs 9 units, and to region 5 costs 27 units. From region 4, the cost to move a car to region 1 is 17 units, to region 2 is 8 units, to region 3 is 9 units, and to region 5 is 20 units. Lastly, from region 5 to region 1 the cost is 34 units, to region 2 is 46 units, to region 3 is 27 units, and to region 4 is 20 units. The following displays the current number of cars and the number needed for harvest shipping in each region. Region 1 currently has 115 cars but needs 200. Region 2 has 385 cars available but requires 500. Region 3 has 410 cars against a need of 800. Region 4 is currently equipped with 480 cars, yet the need is only for 200. Finally, region 5 has 610 cars available, with a need for 300 cars. Minimize the cost to move the cars such as the need is met. ",[11370.0] +"[""Minimum Cost Flow Problem""]","Imagine you are tasked with managing the distribution of medical supplies among seven regional hospitals to ensure each has enough supplies to handle an impending health crisis. Each hospital starts with a certain stock of supplies but requires a different amount to be fully prepared. + +Here’s the current stock and the required supplies for each hospital: + +- Hospital 1 has 160 units but needs 19 units. +- Hospital 2 has 353 units but needs 138 units. +- Hospital 3 has 171 units but needs 110 units. +- Hospital 4 has 193 units but needs 446 units. +- Hospital 5 has 102 units but needs 341 units. +- Hospital 6 has 429 units but needs 428 units. +- Hospital 7 has 284 units but needs 47 units. + +To aid in your task, you can transport supplies between hospitals, but there are costs associated with each transfer depending on the routes taken. Here is the cost matrix showing the transfer costs between each pair of hospitals: + +- From/To Hospital 1: To 2 costs 12, to 3 costs 47, to 4 costs 39, to 5 costs 4, to 6 costs 21, to 7 costs 31. +- From/To Hospital 2: To 1 costs 14, to 3 costs 39, to 4 costs 12, to 5 costs 31, to 6 costs 29, to 7 costs 15. +- From/To Hospital 3: To 1 costs 15, to 2 costs 6, to 4 costs 26, to 5 costs 12, to 6 costs 5, to 7 costs 3. +- From/To Hospital 4: To 1 costs 39, to 2 costs 47, to 3 costs 42, to 5 costs 33, to 6 costs 48, to 7 costs 2. +- From/To Hospital 5: To 1 costs 35, to 2 costs 25, to 3 costs 46, to 4 costs 12, to 6 costs 25, to 7 costs 45. +- From/To Hospital 6: To 1 costs 39, to 2 costs 28, to 3 costs 28, to 4 costs 11, to 5 costs 16, to 7 costs 13. +- From/To Hospital 7: To 1 costs 42, to 2 costs 43, to 3 costs 49, to 4 costs 41, to 5 costs 41, to 6 costs 18. + +Your goal is to ensure that every hospital has the supplies it requires at the lowest possible transportation cost. What would be the minimum cost to ensure all hospitals are adequately supplied?",[6218.0] +"[""TravelingSalesman""]","In a scenario involving a sales representative tasked with visiting four distinct cities to promote their product, we identify these cities as A, B, C, and D. The representative has the flexibility to commence their travels from any city but is required to make a stop at each city exactly once before returning to their starting point. The primary objective for the sales representative is to minimize their overall travel expenses, which might encompass aspects like distance, time, or monetary costs. + +Here’s a breakdown of the travel costs between each city: +- From City A, the cost is 19 units to City B, 45 units to City C, and 30 units to City D. +- Traveling from City B, it costs 19 units to reach City A, 89 units to City C, and 46 units to City D. +- From City C, the journey costs 45 units to City A, 89 units to City B, and merely 11 units to City D. +- Lastly, from City D, the travel expenses are 30 units to City A, 46 units to City B, and 11 units to City C. + +Given this setup, what is the minimum total travel cost for the salesperson to visit each city exactly once and return to the starting city?",[121.0] +['Transportation Problem'],"Suppose you are the chief logistic officer of a large food distribution company and you have the responsibility to ensure that food resources are adequately distributed to four district warehouses. Each warehouse currently holds a certain quantity of food but there are specific requirements that must be met to cater to the demand in each district. + +Here is the present quantity of food stocks and the required quantity for each warehouse: + +- Warehouse 1 has 500 tons of food but needs only 17 tons. +- Warehouse 2 has 497 tons of food but requires 65 tons. +- Warehouse 3 is stocked with 354 tons but needs 26 tons. +- Warehouse 4 currently holds 223 tons but needs a more substantial 332 tons of food. + +You have the capacity to move food from one warehouse to another, but each movement comes with a transportation cost. The cost varies depending on which warehouses you are moving the food between. The following list outlines the cost of moving food from one warehouse to any other: + +- Moving food from/to Warehouse 1: To Warehouse 2 costs 17, to Warehouse 3 costs 3, to Warehouse 4 costs 5. +- Moving food from/to Warehouse 2: To Warehouse 1 costs 39, to Warehouse 3 costs 18, to Warehouse 4 costs 50. +- Moving food from/to Warehouse 3: To Warehouse 1 costs 8, to Warehouse 2 costs 45, to Warehouse 4 costs 15. +- Moving food from/to Warehouse 4: To Warehouse 1 costs 11, to Warehouse 2 costs 4, to Warehouse 3 costs 29. + +Your task is to ensure each warehouse has the required amount of food while keeping the transportation cost as minimal as possible. What would be the lowest cost to ensure all warehouses are adequately stocked with food? Remember, your response should only contain the optimal value and the scenario should be based on relocation.",[545.0] +['TravelingSalesman'],"Consider a scenario where a courier delivery person needs to deliver packages to five different locations, labeled as E, F, G, H, and I. The courier can start from any location, but they need to deliver the package at each location exactly once and then return to their starting point. The task is to find the optimal route that minimizes total delivery cost. In this case, the delivery cost could represent distance, time, or fuel consumption. Here's a summary of the delivery costs between different locations: + +From Location E, it costs 65 units to reach F, 10 units to reach G, 25 units to reach H, and 82 units to reach I. +From Location F, it costs 65 units to deliver to E, 98 units to G, 42 units to H, and 26 units to I. +Moving from Location G, the costs are 10 units to E, 98 units to F, 75 units to H, and 72 units to I. +From Location H, it costs 25 units to go to E, 42 units to F, 75 units to G, and 46 units to I. +Finally, from Location I, it costs 82 units to deliver to E, 26 units to F, 72 units to G, and 46 units to H. + +Considering these costs, what is the minimum total delivery cost for the courier to deliver a package at each location exactly once and return to the starting location? The optimal value is asked for in this problem.",[175.0] +['Capacitated Facility Location Problem'],"In the bustling world of retail, LogiChain, a renowned logistics company, is at a crossroads in determining the most efficient way to distribute products to its network of eight retail stores. The challenge lies in selecting from four strategic distribution centers, each with its distinct opening costs and supply capacities. The goal is to ensure that each store's demand is met with the least total cost, incorporating both the opening of distribution centers and the transportation costs of delivering goods. + +Here are the specifics of the situation: + +- LogiChain can open any of the four distribution centers, each with a different opening cost. +- There are eight retail stores in need of supplies, each with its unique demand. +- The transportation cost per unit of product varies not only from one distribution center to another but also from one store to another. + +The details are as follows: + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $5 to Store 1, $5 to Store 2, $2 to Store 3, $3 to Store 4, $3 to Store 5, $3 to Store 6, $5 to Store 7, $4 to Store 8 +- From Center 2: $3 to Store 1, $5 to Store 2, $2 to Store 3, $4 to Store 4, $2 to Store 5, $4 to Store 6, $5 to Store 7, $1 to Store 8 +- From Center 3: $4 to Store 1, $2 to Store 2, $5 to Store 3, $4 to Store 4, $1 to Store 5, $1 to Store 6, $3 to Store 7, $3 to Store 8 +- From Center 4: $2 to Store 1, $4 to Store 2, $4 to Store 3, $3 to Store 4, $4 to Store 5, $4 to Store 6, $1 to Store 7, $3 to Store 8 + +Demand of Each Retail Store (in units): +- Store 1: 884 units +- Store 2: 420 units +- Store 3: 566 units +- Store 4: 673 units +- Store 5: 787 units +- Store 6: 1000 units +- Store 7: 715 units +- Store 8: 413 units + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1041 units +- Center 2: 1576 units +- Center 3: 1364 units +- Center 4: 1697 units + +Given these variables, LogiChain must make a calculated decision on which distribution centers to open and how to allocate shipments to meet the demands of the eight stores in the most cost-effective manner. + +Question: +What is the optimal total cost for LogiChain to open distribution centers and transport goods to satisfy the demand of all eight retail stores, taking both the opening costs and transportation costs into account?",[637573.0] +"[""Diet Problem""]","Imagine you're meticulously planning your weekly meals to ensure you're hitting all your nutritional goals while sticking to a budget. You're considering five diverse food items, each with unique nutritional benefits and costs. + +Here's a breakdown of what each food offers and its cost: + +- Food_1: Delivers 3 grams of protein, 16 grams of carbohydrates, and 96 calories for $9. +- Food_2: Provides 17 grams of protein, 24 grams of carbohydrates, and 182 calories for $9. +- Food_3: Offers 16 grams of protein, 27 grams of carbohydrates, and 114 calories for just $2. +- Food_4: Contains 8 grams of protein, 16 grams of carbohydrates, and 208 calories for $9. +- Food_5: Supplies 6 grams of protein, 6 grams of carbohydrates, and 236 calories for $5. + +Your dietary targets for the week are to consume at least 100 grams of protein, 180 grams of carbohydrates, and 1796 calories. Your challenge is to determine the most cost-effective combination of these food items to meet or exceed these nutritional requirements. What is the cheapest cost to achieve your dietary goals? You could only assign integral portions of each type of food.",[32.0] +['Diet Problem'],"Picture this: you're trying to plan your meals for the day and you're aiming to get the right amount of nutrients, but also want to keep your expenses low. You have four different food options to choose from: Chicken, Cheese, Rice, and Beans. Each of these foods has a certain amount of protein, carbs, and calories, and each comes with its own price. + +Here's a brief overview of what each food provides and its cost: + +- Chicken: Gives you 16 grams of protein, 22 grams of carbohydrates, and 292 calories for $5. +- Cheese: Offers 10 grams of protein, 4 grams of carbohydrates, and 82 calories for $9. +- Rice: Provides 12 grams of protein, 21 grams of carbohydrates, and 207 calories for just $2. +- Beans: Packs 13 grams of protein, 14 grams of carbohydrates, and 81 calories for a single dollar. + +You need to ensure you consume at least 94 grams of protein, 191 grams of carbohydrates, and 2271 calories from the combination of these foods you choose. Your goal is to determine the most cost-effective way to reach these nutritional targets with these four food options. What is the minimum cost to meet your nutritional requirements?",[22.0] +['Diet Problem'],"Imagine you're preparing your meal plan for the week, with the goal of meeting specific dietary requirements without breaking the bank. You have six different types of food to choose from: Chicken, Rice, Lentils, Apples, Eggs, and Broccoli. Each of these foods offers specific amounts of protein, carbohydrates, and calories, and each comes with its own cost. + +Here's a breakdown of what each food item provides and its cost: + +- Chicken: Offers 20 grams of protein, 9 grams of carbohydrates, and 89 calories for $6. +- Rice: Provides 2 grams of protein, 15 grams of carbohydrates, and 276 calories for $8. +- Lentils: Contains 10 grams of protein, 30 grams of carbohydrates, and 255 calories for $4. +- Apples: Deliver 5 grams of protein, 19 grams of carbohydrates, and 258 calories for just $1. +- Eggs: Give you 17 grams of protein, 12 grams of carbohydrates, and 212 calories for $3. +- Broccoli: Brings 6 grams of protein, 18 grams of carbohydrates, and 246 calories for $5. + +Your goal is to ensure you get at least 90 grams of protein, 162 grams of carbohydrates, and 1926 calories from your combined food choices. You're trying to figure out the minimum cost to meet these nutritional goals with the food options provided. + +So, what is the least amount of money you need to spend to meet your dietary requirements? Keep in mind that you should only provide the optimal value, and no additional responses are necessary.",[17.0] +['Transportation Problem'],"Imagine you are responsible for managing the distribution of medical supplies across seven different healthcare facilities in a region to ensure each has what they need for an impending health crisis. Each facility begins with a certain number of supplies but has a specific required amount to ensure they are well-prepared. + +Here is a rundown of the current and required supplies at each facility: +- Facility 1 has 229 units but needs 158 units. +- Facility 2 has 345 units but needs 14 units. +- Facility 3 has 231 units but needs 326 units. +- Facility 4 has 164 units but needs 432 units. +- Facility 5 has 354 units but needs 207 units. +- Facility 6 has 437 units but needs 215 units. +- Facility 7 has 265 units but needs 233 units. + +To transfer supplies from one facility to another, there are associated costs, which vary depending on the facilities involved. Here is the cost matrix for transferring supplies between facilities: +- From/To Facility 1: To Facility 2 costs 26, to Facility 3 costs 8, to Facility 4 costs 7, to Facility 5 costs 34, to Facility 6 costs 35, to Facility 7 costs 34. +- From/To Facility 2: To Facility 1 costs 41, to Facility 3 costs 27, to Facility 4 costs 31, to Facility 5 costs 3, to Facility 6 costs 33, to Facility 7 costs 36. +- From/To Facility 3: To Facility 1 costs 20, to Facility 2 costs 49, to Facility 4 costs 48, to Facility 5 costs 21, to Facility 6 costs 38, to Facility 7 costs 42. +- From/To Facility 4: To Facility 1 costs 10, to Facility 2 costs 35, to Facility 3 costs 26, to Facility 5 costs 21, to Facility 6 costs 33, to Facility 7 costs 31. +- From/To Facility 5: To Facility 1 costs 21, to Facility 2 costs 30, to Facility 3 costs 24, to Facility 4 costs 25, to Facility 6 costs 5, to Facility 7 costs 8. +- From/To Facility 6: To Facility 1 costs 27, to Facility 2 costs 34, to Facility 3 costs 34, to Facility 4 costs 26, to Facility 5 costs 27, to Facility 7 costs 43. +- From/To Facility 7: To Facility 1 costs 38, to Facility 2 costs 20, to Facility 3 costs 11, to Facility 4 costs 43, to Facility 5 costs 49, to Facility 6 costs 9. + +Your objective is to redistribute the supplies so that every facility has exactly what it needs while minimizing the total transportation costs. What would be the minimum cost to ensure all facilities have the required medical supplies?",[7084.0] +"[""TravelingSalesman""]","Imagine a logistics manager tasked with planning a delivery route for a truck that needs to visit four different cities to distribute goods. The cities are identified numerically as 1, 2, 3, and 4. The truck can start its journey from any of these cities but must travel to each city exactly once and then return to the starting point. The objective is to arrange this route in such a way that the total travel cost is minimized. The costs associated with traveling between the cities are as follows: +- The cost to travel from City 1 to City 2 is 52 units, to City 3 is 89 units, and to City 4 is 11 units. +- From City 2, it costs 52 units to reach City 1, 14 units to get to City 3, and 13 units to City 4. +- Traveling from City 3, the costs are 89 units to City 1, 14 units to City 2, and 87 units to City 4. +- Lastly, from City 4, it costs 11 units to go to City 1, 13 units to City 2, and 87 units to City 3. + +What is the minimum total travel cost for the truck to visit each city exactly once and return to the starting city?",[127.0] +['TravelingSalesman'],"Imagine a scenario where a salesperson must visit four distinct cities to conduct business meetings. Let's label these cities as City 1, City 2, City 3, and City 4. The salesperson can start their trip from any city, but they must visit each city exactly once and then return to the starting point. The main objective is to minimize the total cost of traveling between these cities. + +Here’s a detailed breakdown of the travel costs: +- From City 1, the travel cost is 27 units to City 2, 48 units to City 3, and 76 units to City 4. +- From City 2, it costs 27 units to return to City 1, 82 units to reach City 3, and 88 units to go to City 4. +- From City 3, traveling back to City 1 costs 48 units, moving to City 2 costs 82 units, and it is 97 units to journey to City 4. +- Lastly, from City 4, it costs 76 units to go back to City 1, 88 units to travel to City 2, and 97 units to move to City 3. + +What is the minimum total travel cost for the salesperson to visit each of these cities exactly once and return to the starting city?",[260.0] +"[""TravelingSalesman""]","Consider a courier company that delivers packages to four distinct locations, labeled as E, F, G, and H. The courier can begin their delivery route from any of these locations, but they need to ensure that each location is visited exactly once before returning to the starting point. Their objective is to minimize the total delivery cost, which might be based on factors like distance, fuel consumption, or delivery time. +Here is a brief rundown of the delivery costs between the locations: +The cost to deliver a package from Location E to F is 62 units, to G is 51 units, and to H is 50 units. +From Location F, it costs 62 units to deliver to E, 42 units to G, and 69 units to H. +For delivery from Location G, the costs are 51 units to E, 42 units to F, and 11 units to H. +Lastly, from Location H, the delivery costs are 50 units to E, 69 units to F, and 11 units to G. +What is the minimum total delivery cost for the courier to visit each location exactly once and return to the starting point?",[165.0] +"[""Network Optimization""]","Welcome to the challenge of optimizing the digital traffic flow in CyberCity, a bustling metropolis of the future. In this city, data packets travel through a sophisticated network of 8 digital relay stations, from the primary source node to a key destination node, facilitating seamless communication across the city. + +The network is structured as follows, with each connection representing the maximum data capacity (measured in terabits per second, Tbps) that can flow from one relay station to another: + +- From Relay Station 0 (Source): Data can flow to Station 1 (3 Tbps), Station 2 (3 Tbps), Station 3 (5 Tbps), Station 4 (13 Tbps), Station 5 (7 Tbps), Station 6 (14 Tbps), and Station 7 (17 Tbps). +- From Relay Station 1: Data can flow to Station 0 (5 Tbps), Station 3 (2 Tbps), Station 4 (18 Tbps), Station 5 (20 Tbps), Station 6 (1 Tbps), and Station 7 (17 Tbps). +- From Relay Station 2: Data can flow to Station 0 (2 Tbps), Station 1 (18 Tbps), Station 3 (10 Tbps), Station 4 (14 Tbps), Station 5 (15 Tbps), Station 6 (8 Tbps), and Station 7 (18 Tbps). +- From Relay Station 3: Data can flow to Station 0 (20 Tbps), Station 1 (7 Tbps), Station 2 (11 Tbps), Station 4 (13 Tbps), Station 5 (2 Tbps), Station 6 (8 Tbps), and Station 7 (12 Tbps). +- From Relay Station 4: Data can flow to Station 0 (20 Tbps), Station 1 (20 Tbps), Station 2 (12 Tbps), Station 3 (7 Tbps), Station 5 (11 Tbps), Station 6 (2 Tbps), and Station 7 (14 Tbps). +- From Relay Station 5: Data can flow to Station 0 (5 Tbps), Station 1 (16 Tbps), Station 2 (15 Tbps), Station 3 (19 Tbps), Station 4 (11 Tbps), Station 6 (1 Tbps), and Station 7 (8 Tbps). +- From Relay Station 6: Data can flow to Station 0 (10 Tbps), Station 1 (3 Tbps), Station 2 (14 Tbps), Station 3 (18 Tbps), Station 4 (13 Tbps), Station 5 (6 Tbps), and Station 7 (7 Tbps). +- From Relay Station 7 (Destination): Data can loop back to Station 0 (9 Tbps), Station 1 (3 Tbps), Station 2 (14 Tbps), Station 3 (13 Tbps), Station 4 (6 Tbps), Station 5 (5 Tbps), and Station 6 (9 Tbps). + +In this system, a capacity of ""0"" means there is no direct data link between those two stations. The objective of this problem is to find the maximum amount of data that can be transmitted from Station 0 (the source) to Station 7 (the destination) per second, while ensuring the data flow through each connection doesn't exceed its maximum capacity. + +What is the maximum data flow, in terabits per second, that can be achieved from the source to the destination in this network?",[62.0] +"[""Minimum Cost Flow Problem""]","Imagine a futuristic transportation network connecting 9 bustling sectors of a newly established colony on Mars, where each sector specializes in different aspects of Martian life, from agriculture to tech development. The network is designed to transport vital resources, such as water, food, and technology, from the colony's central supply hub (Sector 0) to its farthest outpost (Sector 8). The sectors are interconnected through a series of high-speed transport tubes, each with a specific capacity for carrying resources per Martian day. + +The capacity for resource flow from one sector to another is as follows: + +- From Sector 0 (Central Supply Hub): Can send resources to Sector 1 (7 units), Sector 2 (4 units), Sector 3 (17 units), Sector 4 (4 units), Sector 5 (18 units), Sector 6 (17 units), Sector 7 (5 units), and Sector 8 (14 units). +- From Sector 1: Can send resources to Sector 0 (2 units), Sector 2 (11 units), Sector 3 (10 units), Sector 4 (1 unit), Sector 5 (19 units), Sector 6 (6 units), Sector 7 (16 units), and Sector 8 (13 units). +- From Sector 2: Can send resources to Sector 0 (8 units), Sector 1 (15 units), Sector 3 (5 units), Sector 4 (1 unit), Sector 5 (5 units), Sector 6 (11 units), Sector 7 (17 units), and Sector 8 (18 units). +- From Sector 3: Can send resources to Sector 0 (2 units), Sector 1 (5 units), Sector 4 (6 units), Sector 5 (17 units), Sector 6 (1 unit), Sector 7 (16 units), and Sector 8 (14 units). +- From Sector 4: Can send resources to Sector 0 (15 units), Sector 1 (6 units), Sector 2 (20 units), Sector 3 (15 units), Sector 5 (7 units), Sector 6 (13 units), Sector 7 (6 units), and Sector 8 (19 units). +- From Sector 5: Can send resources to Sector 0 (15 units), Sector 1 (2 units), Sector 2 (8 units), Sector 3 (2 units), Sector 4 (3 units), Sector 6 (2 units), Sector 7 (9 units), and Sector 8 (2 units). +- From Sector 6: Can send resources to Sector 0 (10 units), Sector 1 (12 units), Sector 2 (11 units), Sector 3 (10 units), Sector 4 (19 units), Sector 5 (9 units), Sector 7 (17 units), and Sector 8 (1 unit). +- From Sector 7: Can send resources to Sector 0 (0 units), Sector 1 (20 units), Sector 2 (15 units), Sector 3 (19 units), Sector 4 (20 units), Sector 5 (18 units), Sector 6 (14 units), and Sector 8 (8 units). +- From Sector 8 (Farthest Outpost): Can send resources back to Sector 0 (16 units), Sector 1 (13 units), Sector 2 (3 units), Sector 3 (3 units), Sector 4 (20 units), Sector 5 (19 units), Sector 6 (2 units), and Sector 7 (7 units). + +In this Martian colony, a capacity of ""0"" means there is no direct transport tube connection between those two sectors. + +Your task is to calculate the maximum amount of resources that can be transported from the Central Supply Hub (Sector 0) to the Farthest Outpost (Sector 8) per Martian day, ensuring that the flow through each transport tube does not exceed its specified capacity. What is the maximum amount of resources that can be sent from the Central Supply Hub to the Farthest Outpost in units per Martian day?",[86.0] +"[""Network Optimization""]","Welcome to the challenge of optimizing the logistics network of a futuristic city, connected by an advanced transportation system comprising 8 crucial nodes. This city's system is designed to maximize the efficiency of goods transport from the central distribution hub to a primary delivery station, passing through various checkpoints and distribution centers along the way. + +Here's a detailed view of the transportation capacities (in thousands of units) that each route can handle: + +- From Node 0 (Source): Can dispatch goods to Node 1 (3 units), Node 2 (7 units), Node 3 (2 units), Node 4 (18 units), Node 5 (18 units), Node 6 (14 units), and Node 7 (10 units). +- From Node 1: Can send goods to Node 0 (1 unit), Node 2 (10 units), Node 3 (16 units), Node 4 (14 units), Node 6 (6 units), and Node 7 (10 units). +- From Node 2: Can transport goods to Node 0 (4 units), Node 3 (7 units), Node 4 (2 units), Node 5 (2 units), Node 6 (6 units), and Node 7 (10 units). +- From Node 3: Can move goods to Node 0 (2 units), Node 1 (3 units), Node 2 (7 units), Node 4 (20 units), Node 5 (8 units), Node 6 (6 units), and Node 7 (10 units). +- From Node 4: Can dispatch goods to Node 0 (1 unit), Node 1 (20 units), Node 2 (4 units), Node 3 (8 units), Node 5 (13 units), Node 6 (8 units), and Node 7 (1 unit). +- From Node 5: Can send goods to Node 0 (16 units), Node 1 (9 units), Node 2 (17 units), Node 3 (17 units), Node 4 (11 units), Node 6 (7 units), and Node 7 (8 units). +- From Node 6: Can transport goods to Node 0 (3 units), Node 1 (7 units), Node 2 (14 units), Node 3 (1 unit), Node 4 (20 units), Node 5 (5 units), and Node 7 (13 units). + +A capacity of ""0"" indicates there is no direct route available between those two nodes. + +The goal is to determine the maximum amount of goods (in thousands of units) that can be transported from the central distribution hub (Node 0) to the primary delivery station (Node 7) without exceeding the transport capacity of any route. + +What is the optimal value of goods (in thousands of units) that can be transported from the source to the destination?",[62.0] +"[""Capacitated Facility Location Problem""]","In the bustling world of retail, SupplyChain Solutions is at a crossroads, facing the challenge of efficiently distributing goods to nine of its bustling retail stores. To do this, they have the option of opening up to six different distribution centers, each with its own set of costs and capacities. The goal is crystal clear: to ensure that all retail stores are adequately supplied while keeping the total cost, including both the opening of distribution centers and the transportation of goods, at an absolute minimum. + +Given Data: +- Number of Potential Distribution Centers: 6 +- Number of Retail Stores Needing Supply: 9 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: 151,000 +- Center 2: 192,000 +- Center 3: 114,000 +- Center 4: 171,000 +- Center 5: 160,000 +- Center 6: 120,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $2 to Store 1, $3 to Store 2, $4 to Store 3, $5 to Store 4, $6 to Store 5, $2 to Store 6, $3 to Store 7, $4 to Store 8, and $2 to Store 9. +- From Center 2: $4 to Store 1, $2 to Store 2, $3 to Store 3, $5 to Store 4, $4 to Store 5, $3 to Store 6, $4 to Store 7, $2 to Store 8, and $4 to Store 9. +- From Center 3: $3 to Store 1, $4 to Store 2, $2 to Store 3, $3 to Store 4, $5 to Store 5, $4 to Store 6, $5 to Store 7, $3 to Store 8, and $5 to Store 9. +- From Center 4: $5 to Store 1, $3 to Store 2, $4 to Store 3, $2 to Store 4, $3 to Store 5, $5 to Store 6, $2 to Store 7, $4 to Store 8, and $3 to Store 9. +- From Center 5: $2 to Store 1, $5 to Store 2, $3 to Store 3, $4 to Store 4, $2 to Store 5, $3 to Store 6, $4 to Store 7, $5 to Store 8, and $3 to Store 9. +- From Center 6: $4 to Store 1, $3 to Store 2, $5 to Store 3, $3 to Store 4, $4 to Store 5, $2 to Store 6, $3 to Store 7, $4 to Store 8, and $2 to Store 9. + +Demand of Each Retail Store (in units): +- Store 1: 401 +- Store 2: 789 +- Store 3: 965 +- Store 4: 505 +- Store 5: 876 +- Store 6: 801 +- Store 7: 955 +- Store 8: 561 +- Store 9: 601 + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1795 +- Center 2: 1069 +- Center 3: 1615 +- Center 4: 1255 +- Center 5: 1816 +- Center 6: 1095 + +SupplyChain Solutions now faces the daunting task of identifying which distribution centers should be operational and how goods should be routed to the stores to ensure every store's demand is met. The solution must be cost-effective, taking into consideration the initial opening costs of the distribution centers and the variable costs of transporting goods to the stores. + +Question: +What is the optimal total cost for SupplyChain Solutions to open the necessary distribution centers and transport goods to satisfy the demands of all nine retail stores? This total cost should include both the opening costs of the distribution centers and the transportation costs of supplying the stores.",[610975.0] +['Capacitated Facility Location Problem'],"In the bustling world of retail, SupplyChain Solutions is faced with a critical challenge of efficiently supplying seven diverse retail stores. To achieve this goal, the company is considering opening up to six potential distribution centers, each with its unique operating costs and capacities. However, the logistical conundrum doesn't end there, as transporting goods from these centers to the stores incurs varying costs. The ultimate aim for SupplyChain Solutions is to ensure that each retail store's demand is met, while simultaneously minimizing the total operational and transportation expenses. Let's delve into the specifics of this optimization problem. + +Given Data: +- Number of Potential Distribution Centers: 6 +- Number of Retail Stores to be Supplied: 7 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 +- Center 6: $120,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $2 to Store 1, $3 to Store 2, $3 to Store 3, $3 to Store 4, $5 to Store 5, $4 to Store 6, $3 to Store 7 +- From Center 2: $5 to Store 1, $2 to Store 2, $4 to Store 3, $2 to Store 4, $4 to Store 5, $5 to Store 6, $1 to Store 7 +- From Center 3: $4 to Store 1, $2 to Store 2, $5 to Store 3, $4 to Store 4, $1 to Store 5, $1 to Store 6, $3 to Store 7 +- From Center 4: $3 to Store 1, $2 to Store 2, $4 to Store 3, $4 to Store 4, $3 to Store 5, $4 to Store 6, $4 to Store 7 +- From Center 5: $1 to Store 1, $3 to Store 2, $5 to Store 3, $3 to Store 4, $5 to Store 5, $1 to Store 6, $2 to Store 7 +- From Center 6: $4 to Store 1, $1 to Store 2, $4 to Store 3, $2 to Store 4, $2 to Store 5, $1 to Store 6, $2 to Store 7 + +Demand of Each Retail Store (in units): +- Store 1: 964 units +- Store 2: 739 units +- Store 3: 491 units +- Store 4: 766 units +- Store 5: 854 units +- Store 6: 827 units +- Store 7: 908 units + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1575 units +- Center 2: 834 units +- Center 3: 1005 units +- Center 4: 1904 units +- Center 5: 1825 units +- Center 6: 1821 units + +The puzzle that SupplyChain Solutions needs to solve involves choosing which distribution centers to open and determining the most cost-efficient way to supply each retail store from these centers. The goal is to cover the demand at each store without exceeding the supply capacity of the chosen distribution centers, all while keeping an eye on minimizing the sum of opening and transportation costs. + +Question: +What is the optimal total cost for SupplyChain Solutions to open the necessary distribution centers and transport goods to meet all seven retail stores' demands? This cost should include both the opening costs of the distribution centers and the transportation costs of supplying the stores.",[461962.0] +['Diet Problem'],"Consider the following table of food and corresponding nutritional values: Bread contains 4 grams of protein, 7 grams of carbohydrates, has a calorie count of 130, and costs 3 dollar. Milk contains 6 grams of protein, 10 grams of carbohydrates, has a calorie count of 120, and costs 4 dollars. Fish is high in protein with 20 grams, contains no carbohydrates, has a calorie count of 150, and is the most expensive at 8 dollars. Potato contains 1 gram of protein, is high in carbohydrates with 30 grams, has the lowest calorie count at 70, and is the cheapest at 2 dollars. The ideal intake for an adult is at least 30 grams of protein, 40 grams of carbohydrates, and 400 calories per day. The problem is to find the least costly way to achieve those amounts of nutrition by using the four types of food shown in the table.",[17] +"[""Blending Problem""]","Chip Green is the head groundskeeper at Birdie Valley Golf Club. For the mix of grass for the golf course, Chip has decided that the best fertilizer would be a 10-8-12 mixture. Fertilizer is defined by three values—a, b, and c—where a is the percentage of nitrogen (Ni), b is the percentage of phosphorus (Ph), and c is the percentage of potash (Po) in the fertilizer. The remaining material is inert matter. Chip can buy a 10-8-12 mix of fertilizer for $21.75 per 100 pounds, but there are other fertilizers on the market at a variety of prices. The chemical content and prices per 100 pounds for these fertilizers are as follows: fertilizer 1 has a 10-8-12 mix at $21.75, fertilizer 2 has an 8-11-15 mix at $23.50, fertilizer 3 has a 12-7-12 mix at $22.00, fertilizer 4 has a 10-10-10 mix at $19.50, and fertilizer 5 has a 15-10-6 mix at $18.50. Chip would like to determine whether or not he could buy several fertilizers and mix them together to obtain a 10-8-12 mixture at a lower cost than $21.75 per 100 pounds. Recognizing that it might be impossible to obtain an exact 10-8-12 mix from the fertilizers, Chip is willing to accept chemical percentages of at least the target amounts, but no more than 0.5% above them (so the nitrogen level should be between 10% and 10.5%, phosphorus between 8% and 8.5%, and potash between 12% and 12.5%). Find the minimal cost per pound of fertilizer that meets these conditions.",[0.2160625] +"[""Diet Problem""]","My diet requires that all the food I eat come from one of the four 'basic food groups' (chocolate cake, ice cream, soda, and cheesecake). At present, the following four foods are available for consumption: brownies, chocolate ice cream, cola, and pineapple cheesecake. The nutritional content per unit of each food is as follows: each brownie provides 400 calories, 3 oz of chocolate, 2 oz of sugar, and 2 oz of fat; each scoop of chocolate ice cream provides 200 calories, 2 oz of chocolate, 2 oz of sugar, and 4 oz of fat; each bottle of cola provides 150 calories, 0 oz of chocolate, 4 oz of sugar, and 1 oz of fat; and each piece of pineapple cheesecake provides 500 calories, 0 oz of chocolate, 4 oz of sugar, and 5 oz of fat. The costs are 50¢ for each brownie, 20¢ for each scoop of chocolate ice cream, 30¢ for each bottle of cola, and 80¢ for each piece of pineapple cheesecake. Each day, I must ingest at least 500 calories, 6 oz of chocolate, 10 oz of sugar, and 8 oz of fat. Formulate a linear programming model that can be used to satisfy my daily nutritional requirements at minimum cost. Find the minimal cost in dollars. You could only assign integral portions of each type of food.",[0.9] +['Network Optimization'],"Embark on a journey through a futuristic transportation network, connecting 8 bustling metropolises - from a cutting-edge logistics hub to a far-reaching distribution center. This network is not ordinary; it's a complex web of superhighways, each capable of supporting a vast amount of traffic, represented as the capacity for vehicles (in thousands) that can travel per hour. + +Here's the breakdown of this ultra-modern network's capacities, detailing the maximum flow of traffic that each superhighway can handle from one city to another: + +- From City 0 (Source): The hub can dispatch vehicles to City 1 (18,000), City 2 (5,000), City 3 (6,000), City 4 (8,000), City 5 (10,000), and City 6 (16,000). +- From City 1: Vehicles can flow to City 0 (18,000), City 2 (1,000), City 3 (15,000), City 4 (20,000), City 5 (13,000), City 6 (1,000), and City 7 (2,000). +- From City 2: Traffic can move to City 0 (18,000), City 1 (16,000), City 3 (19,000), City 4 (15,000), City 5 (4,000), City 6 (9,000), and City 7 (10,000). +- From City 3: Vehicles can head to City 0 (20,000), City 1 (2,000), City 2 (18,000), City 4 (19,000), City 5 (17,000), City 6 (15,000), and City 7 (19,000). +- From City 4: Traffic can be directed to City 0 (14,000), City 1 (13,000), City 2 (14,000), City 3 (1,000), City 5 (8,000), City 6 (16,000), and City 7 (20,000). +- From City 5: Vehicles can travel to City 0 (0), City 1 (17,000), City 2 (18,000), City 3 (10,000), City 4 (17,000), City 6 (16,000), and City 7 (6,000). +- From City 6: Traffic flow can include movements to City 0 (5,000), City 1 (14,000), City 2 (19,000), City 3 (6,000), City 4 (0), City 5 (19,000), and City 7 (12,000). +- From City 7 (Destination): Vehicles can return to City 0 (7,000), City 1 (16,000), City 2 (17,000), City 3 (13,000), City 4 (6,000), City 5 (9,000), and City 6 (5,000). + +In this network, a capacity of ""0"" symbolizes that there is no direct superhighway linking those two cities. + +Your challenge is to compute the maximum number of vehicles (in thousands) that can be transported from City 0 (the source) to City 7 (the destination) per hour, without exceeding the superhighways' capacities. What is the maximum amount of traffic that can be efficiently moved from the source to the destination in thousands of vehicles per hour?",[63.0] +['TravelingSalesman'],"Consider four cities: E, F, G, and H. A delivery driver is tasked with delivering packages to each of these cities. The driver can start their route from any one of these cities. However, the driver must ensure that they visit each city exactly once and then return back to the city they started from. The ultimate goal is to minimize the total travel cost, which might include expenses such as fuel, toll fees, or time spent on the road. + +Here's a detailed breakdown of the travel costs between these cities: + +To travel from City E to F costs 17 units, to G costs 52 units, and to H costs 79 units. +From City F, the cost is 17 units to go to E, 15 units to reach G, and 71 units to go to H. +If the driver starts from City G, it costs 52 units to reach E, 15 units to go to F, and 54 units to reach H. +Lastly, if the journey begins from City H, it costs 79 units to go to E, 71 units to go to F, and 54 units to reach G. + +Now the question is: What is the minimum total travel cost for the driver to deliver packages in each city exactly once and then return to the starting city?",[165.0] +"[""Network Optimization""]","Embark on a journey through a bustling transport network in the heart of a bustling metropolis, featuring 9 critical nodes that form the backbone of this city's logistic framework. From the primary dispatch center to the ultimate distribution hub, each node plays a pivotal role in ensuring goods and resources flow seamlessly across the city. + +Here's a detailed glimpse into the transport capacities (in thousands of units) between these nodes, ensuring that every node is a cog in the machine that keeps the city running: + +- From Node 0 (The Source): Goods can be dispatched to Node 1 (13 units), Node 2 (19 units), Node 3 (17 units), Node 4 (7 units), Node 5 (7 units), Node 6 (11 units), Node 7 (7 units), and Node 8 (7 units). +- From Node 1: Goods can move on to Node 0 (9 units), Node 2 (19 units), Node 3 (20 units), Node 4 (10 units), Node 5 (19 units), Node 6 (20 units), Node 7 (19 units), and Node 8 (3 units). +- From Node 2: Goods can be transported to Node 0 (0 units), Node 1 (16 units), Node 3 (0 units), Node 4 (10 units), Node 5 (2 units), Node 6 (14 units), Node 7 (9 units), and Node 8 (9 units). +- From Node 3: Goods can be routed to Node 0 (6 units), Node 1 (0 units), Node 2 (5 units), Node 4 (7 units), Node 5 (18 units), Node 6 (15 units), Node 7 (20 units), and Node 8 (7 units). +- From Node 4: Goods can be forwarded to Node 0 (18 units), Node 1 (19 units), Node 2 (12 units), Node 3 (0 units), Node 5 (16 units), Node 6 (20 units), Node 7 (10 units), and Node 8 (4 units). +- From Node 5: Goods can be sent to Node 0 (3 units), Node 1 (4 units), Node 2 (8 units), Node 3 (9 units), Node 4 (20 units), Node 6 (7 units), Node 7 (8 units), and Node 8 (1 unit). +- From Node 6: Goods can be transported to Node 0 (3 units), Node 1 (14 units), Node 2 (14 units), Node 3 (16 units), Node 4 (1 unit), Node 5 (12 units), Node 7 (10 units), and Node 8 (10 units). +- From Node 7: Goods can move on to Node 0 (9 units), Node 1 (20 units), Node 2 (19 units), Node 3 (4 units), Node 4 (9 units), Node 5 (16 units), Node 6 (1 unit), and Node 8 (6 units). +- From Node 8 (The Destination): Goods from Node 0 (7 units), Node 1 (15 units), Node 2 (14 units), Node 3 (18 units), Node 4 (18 units), Node 5 (11 units), Node 6 (18 units) can be finally distributed. + +A capacity of ""0"" signifies the absence of a direct route between those particular nodes, indicating the need for strategic planning to navigate goods through this intricate network. + +Your mission is to calculate the maximum volume of goods (in thousands of units) that can be transported from Node 0, the primary dispatch center, to Node 8, the ultimate distribution hub, per unit time. This calculation must ensure that the flow of goods through each connection does not exceed its transportation capacity. What is the maximum volume of goods that can be moved from the source to the destination in thousands of units?",[47.0] +"[""Capacitated Facility Location Problem""]","Distribution Dynamics Inc. has a strategic challenge to efficiently supply its seven retail stores from a choice of five potential distribution centers. This operation has to manage not only the differing opening costs of each distribution center but also the variability in their capacity to meet demands and the transportation costs involved in shipping goods to each retail store. The company aims to fulfill the demand at each retail location while minimizing the total cost involved, which encompasses the opening costs of the distribution centers and the transportation costs of the goods. + +Given Data: +- Number of Potential Distribution Centers: 5 +- Number of Retail Stores Needing Supply: 7 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 +- Center 5: $160,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $5 to Store 1, $2 to Store 2, $3 to Store 3, $3 to Store 4, $3 to Store 5, $5 to Store 6, $4 to Store 7 +- From Center 2: $3 to Store 1, $5 to Store 2, $2 to Store 3, $4 to Store 4, $2 to Store 5, $4 to Store 6, $5 to Store 7 +- From Center 3: $1 to Store 1, $4 to Store 2, $2 to Store 3, $5 to Store 4, $4 to Store 5, $1 to Store 6, $1 to Store 7 +- From Center 4: $3 to Store 1, $3 to Store 2, $2 to Store 3, $4 to Store 4, $4 to Store 5, $3 to Store 6, $4 to Store 7 +- From Center 5: $4 to Store 1, $1 to Store 2, $3 to Store 3, $5 to Store 4, $3 to Store 5, $5 to Store 6, $1 to Store 7 + +Demand of Each Retail Store (in units): +- Store 1: 566 units +- Store 2: 673 units +- Store 3: 787 units +- Store 4: 1000 units +- Store 5: 715 units +- Store 6: 413 units +- Store 7: 641 units + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1576 units +- Center 2: 1364 units +- Center 3: 1697 units +- Center 4: 891 units +- Center 5: 1755 units + +Question: +What is the optimal total cost for Distribution Dynamics Inc. to open the necessary distribution centers and transport goods in order to meet all retail stores' demands at the lowest possible cost, taking into account both the opening costs of the centers and the transportation costs of supplying the stores?",[434081.0] +"[""Network Optimization""]","Visualize a sophisticated network of digital information pathways connecting 9 pivotal data centers across a vast metropolitan area. These centers, numbered from 0 to 8, form the backbone of the city's communication infrastructure, with the primary data hub at Center 0 acting as the source of all transmissions. The objective is to ensure the maximum flow of information reaches Center 8, the main processing facility, thus optimizing the network's overall efficiency. + +The capacities of the digital pathways, measured in terabytes per hour, dictate how much information can flow between each pair of data centers: + +From Data Center 0 (Source): Can transmit information to Center 1 (11 TB), Center 2 (11 TB), Center 3 (14 TB), Center 4 (6 TB), Center 5 (11 TB), Center 6 (4 TB), Center 7 (15 TB), and Center 8 (9 TB). + +From Data Center 1: Can transmit information to Center 2 (7 TB), Center 3 (12 TB), Center 4 (5 TB), Center 5 (3 TB), Center 6 (20 TB), Center 7 (8 TB), and Center 8 (10 TB), with no direct connection back to Center 0. + +From Data Center 2: Can transmit information to Center 1 (18 TB), Center 3 (6 TB), Center 4 (4 TB), Center 5 (18 TB), Center 6 (9 TB), Center 7 (2 TB), and Center 8 (14 TB). + +From Data Center 3: Can transmit information to Center 1 (13 TB), Center 2 (7 TB), Center 4 (11 TB), Center 5 (5 TB), Center 6 (1 TB), Center 7 (20 TB), and Center 8 (8 TB). + +From Data Center 4: Can transmit information to Center 1 (2 TB), Center 2 (10 TB), Center 3 (9 TB), Center 5 (7 TB), Center 6 (12 TB), Center 7 (4 TB), and Center 8 (6 TB). + +From Data Center 5: Can transmit information to Center 1 (3 TB), Center 2 (4 TB), Center 3 (6 TB), Center 4 (10 TB), Center 6 (15 TB), Center 7 (9 TB), and Center 8 (11 TB). + +From Data Center 6: Can transmit information to Center 1 (5 TB), Center 2 (8 TB), Center 3 (7 TB), Center 4 (6 TB), Center 5 (4 TB), Center 7 (3 TB), and Center 8 (12 TB). + +From Data Center 7: Can transmit information to Center 1 (14 TB), Center 2 (9 TB), Center 3 (11 TB), Center 4 (7 TB), Center 5 (20 TB), Center 6 (10 TB), and Center 8 (16 TB). + +From Data Center 8: Can transmit information back to Center 1 (4 TB), Center 2 (3 TB), Center 3 (6 TB), Center 4 (2 TB), Center 5 (5 TB), Center 6 (1 TB), and Center 7 (7 TB). + +This intricate network is devoid of direct connections in certain instances, symbolizing the absence of direct communication lines between specific data centers. + +The paramount task at hand is to calculate the maximum possible transmission of information, from the source at Data Center 0 to the main processing facility at Data Center 8, per hour. This calculation must ensure that the flow through each digital pathway does not exceed its maximum capacity. + +What is the maximum amount of information, in terabytes per hour, that can be transmitted from the source to the main processing facility within this network?",[81.0] +"[""Diet Problem""]","You're in charge of planning your meals for the week with the aim of balancing your nutritional needs while keeping your expenses within a budget. You have four main food items to choose from: Chicken, Broccoli, Beef, and Rice. Each food provides specific amounts of protein, carbohydrates, and calories, and they each have their own cost. + +Here's the nutritional value and cost of each food item: + +- Chicken: Provides you with 10 grams of protein, 17 grams of carbohydrates, and 159 calories at a cost of $5. +- Broccoli: You will receive 4 grams of protein, 3 grams of carbohydrates, and 126 calories for $5. +- Beef: This high-protein food provides 13 grams of protein, 8 grams of carbohydrates, and 230 calories for $8. +- Rice: This carb-rich food offers 15 grams of protein, a substantial 25 grams of carbohydrates, and 112 calories for just $3. + +Your nutritional target for the week is to consume at least 66 grams of protein, 101 grams of carbohydrates, and 1701 calories from your combination of these food items. The challenge is to figure out the most cost-effective way to meet these nutritional goals with these four food options. What is the minimal cost required to meet your nutritional requirements? Remember, the answer should be a single optimal value. You could only assign integral portions of each type of food.",[47.0] +"[""TravelingSalesman""]","Consider a scenario where a delivery company needs to deliver packages to five different cities. These cities are labeled as E, F, G, H and I. The delivery van can start its route from any of these cities but must follow one rule: it must visit each city exactly once and then return to the starting city. The aim is to find the route that minimizes the total travel cost, which could be a function of fuel costs, labor costs, or road tolls. + +Here's a rundown of the travel costs between these cities: + +From City E (1), the cost to reach City F (2) is 35 units, to G (3) is 90 units, to H (4) is 23 units, and to I (5) is 62 units. + +From City F (2), the cost to City E (1) is 35 units, to G (3) is 53 units, to H (4) is 58 units, and to I (5) is 46 units. + +If we start from City G (3), the cost to reach City E (1) is 90 units, to F (2) is 53 units, to H (4) is 28 units, and to I (5) is 91 units. + +From City H (4), the cost to City E (1) is 23 units, to F (2) is 58 units, to G (3) is 28 units, and to I (5) is 77 units. + +Lastly, from City I (5), it costs 62 units to reach City E (1), 46 units to get to F (2), 91 units to reach G (3), and 77 units to reach H (4). + +What is the minimum total travel cost for the delivery van to visit each city exactly once and return to the starting city?",[212.0] +['Capacitated Facility Location Problem'],"GlobalSupply Inc. is embarking on a strategic overhaul of its distribution network to efficiently serve six retail stores scattered across different regions. The company has pinpointed nine potential locations for establishing distribution centers, each with its unique operational costs and capacities. The challenge lies in selecting the right combination of distribution centers to open and determining the best way to distribute goods to the retail stores in order to fulfill their demands. The overarching goal is to minimize the total cost involved, which includes the initial opening costs of the distribution centers and the ongoing transportation costs of delivering goods to the retail stores. + +Given Data: +- Number of Prospective Distribution Centers: 9 +- Number of Retail Stores to be Supplied: 6 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: 151,000 +- Center 2: 192,000 +- Center 3: 114,000 +- Center 4: 171,000 +- Center 5: 160,000 +- Center 6: 120,000 +- Center 7: 182,000 +- Center 8: 186,000 +- Center 9: 174,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1 to Stores: $3 to Store 1, $5 to Store 2, $4 to Store 3, $3 to Store 4, $5 to Store 5, $2 to Store 6 +- From Center 2 to Stores: $4 to Store 1, $2 to Store 2, $4 to Store 3, $5 to Store 4, $1 to Store 5, $4 to Store 6 +- From Center 3 to Stores: $2 to Store 1, $5 to Store 2, $4 to Store 3, $1 to Store 4, $1 to Store 5, $3 to Store 6 +- From Center 4 to Stores: $3 to Store 1, $2 to Store 2, $4 to Store 3, $4 to Store 4, $3 to Store 5, $4 to Store 6 +- From Center 5 to Stores: $4 to Store 1, $1 to Store 2, $3 to Store 3, $5 to Store 4, $3 to Store 5, $5 to Store 6 +- From Center 6 to Stores: $1 to Store 1, $2 to Store 2, $4 to Store 3, $1 to Store 4, $4 to Store 5, $2 to Store 6 +- From Center 7 to Stores: $2 to Store 1, $1 to Store 2, $2 to Store 3, $5 to Store 4, $2 to Store 5, $4 to Store 6 +- From Center 8 to Stores: $4 to Store 1, $4 to Store 2, $4 to Store 3, $5 to Store 4, $3 to Store 5, $1 to Store 6 +- From Center 9 to Stores: $4 to Store 1, $2 to Store 2, $4 to Store 3, $2 to Store 4, $2 to Store 5, $4 to Store 6 + +Demand of Each Retail Store (in units): +- Store 1: 876 units +- Store 2: 801 units +- Store 3: 955 units +- Store 4: 561 units +- Store 5: 601 units +- Store 6: 669 units + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1615 units +- Center 2: 1255 units +- Center 3: 1816 units +- Center 4: 1095 units +- Center 5: 1519 units +- Center 6: 1137 units +- Center 7: 1678 units +- Center 8: 1876 units +- Center 9: 1591 units + +Question: +Given the challenge of selecting the optimal number and combination of distribution centers to open, along with configuring the most cost-effective way to supply the six retail stores to meet their demands, what is the minimal possible total cost for GlobalSupply Inc.? This total cost should include both the opening costs of the distribution centers and the transportation costs for delivering goods to the stores.",[394338.0] +['Transportation Problem'],"You are managing the supplies of seven warehouses located in different regions. Each warehouse starts with a certain amount of supplies but needs a specific amount to fulfill the demands. Your task is to devise a plan to move supplies from one warehouse to another to meet the demand, considering the movement costs. + +Here are the current and required quantities of supplies in each warehouse: + +- Warehouse 1 has 57 units but needs 499 units. +- Warehouse 2 has 180 units but needs 173 units. +- Warehouse 3 has 288 units but needs 293 units. +- Warehouse 4 has 255 units but needs 158 units. +- Warehouse 5 has 477 units but needs 15 units. +- Warehouse 6 has 222 units but needs 78 units. +- Warehouse 7 has 314 units but needs 278 units. + +Movement of supplies between warehouses incurs transportation costs. The cost varies depending on which warehouses you are moving supplies between. Here's a list showing how much it costs to move supplies from one warehouse to another: + +- Moving supplies from/to Warehouse 1: To Warehouse 2 costs 36, to Warehouse 3 costs 36, to Warehouse 4 costs 48, to Warehouse 5 costs 10, to Warehouse 6 costs 20, to Warehouse 7 costs 48. +- Moving supplies from/to Warehouse 2: To Warehouse 1 costs 28, to Warehouse 3 costs 16, to Warehouse 4 costs 18, to Warehouse 5 costs 36, to Warehouse 6 costs 42, to Warehouse 7 costs 12. +- Moving supplies from/to Warehouse 3: To Warehouse 1 costs 6, to Warehouse 2 costs 40, to Warehouse 4 costs 49, to Warehouse 5 costs 31, to Warehouse 6 costs 1, to Warehouse 7 costs 46. +- Moving supplies from/to Warehouse 4: To Warehouse 1 costs 14, to Warehouse 2 costs 22, to Warehouse 3 costs 4, to Warehouse 5 costs 9, to Warehouse 6 costs 12, to Warehouse 7 costs 1. +- Moving supplies from/to Warehouse 5: To Warehouse 1 costs 50, to Warehouse 2 costs 44, to Warehouse 3 costs 7, to Warehouse 4 costs 48, to Warehouse 6 costs 7, to Warehouse 7 costs 25. +- Moving supplies from/to Warehouse 6: To Warehouse 1 costs 26, to Warehouse 2 costs 9, to Warehouse 3 costs 33, to Warehouse 4 costs 38, to Warehouse 5 costs 8, to Warehouse 7 costs 30. +- Moving supplies from/to Warehouse 7: To Warehouse 1 costs 8, to Warehouse 2 costs 9, to Warehouse 3 costs 2, to Warehouse 4 costs 13, to Warehouse 5 costs 8, to Warehouse 6 costs 49. + +Given the above scenario, your goal is to ensure that every warehouse has the supplies it needs at the lowest possible cost. What is the minimum cost needed to relocate the supplies and meet the demand in each warehouse? +",[5213.0] +"[""Facility Location Problem (FLP)""]","The Chinese University of Hong Kong, Shenzhen decides to build a circular fountain on the campus. The school wants the fountain to be round and as large as possible but it must be restricted in a polygonal construction field, which is given by the following points: (0, 1), (0, 6), (4, 10), (8, 10), (11, 7), (11, 4), (7, 0), and (1, 0), the unit is m. Give a linear optimization formulation and find the maximal area. Keep your answer in four significant digit number.",[66.39] +['Set Cover'],"A graph with 10 vertices and a series of edges connecting them. Each vertex is labeled with a lowercase letter from 'a' to 'j'. The vertices labeled 'a', 'c', 'd', 'f', 'g', and 'j' are colored in red, while 'b', 'e', 'h', and 'i' are colored in blue. The red vertices seem to form a set that covers all the edges of the graph, indicating a solution to the vertex cover problem. The vertex cover problem is a classical optimization problem where the goal is to find the smallest set of vertices such that every edge in the graph is incident to at least one vertex in the set. The graph itself has a pentagonal structure on the outer edges, with an additional five vertices inside the pentagon that are fully interconnected, forming a complete graph K5 (every vertex inside the pentagon is connected to every other vertex inside). The graph is not planar as there are edges that cross each other. In terms of vertex connections: Vertex 'a' connects to vertices 'f', 'e', and 'b'. Vertex 'b' connect to vertices 'a', 'g', and 'c'. Vertex 'c' connects to vertices 'b', 'h', and 'd'. Vertex 'd' connects to vertices 'c', 'i', and 'e'. Vertex 'e' connects to vertices 'd', 'j', and 'a'. The vertices inside the pentagon ('f', 'g', 'h', 'i', 'j') are all interconnected. The red-colored vertices ('a', 'c', 'd', 'f', 'g', 'j') suggest a possible vertex cover, indicating they are a proposed solution to ensure that all edges of the graph are covered. Find the minimum vertex cover. Provide your answer rounded to nearest integer.",[7.0] +['Diet Problem'],"Imagine you're a college student aiming to balance your diet and budget. You have identified nine different food items from your local grocery store that you can include in your menu: Chicken, Rice, Apples, Steak, Lentils, Fish, Tofu, Cheese, and Bread. Each of these foods provides varying amounts of protein, carbohydrates, and calories, and each comes with its own price. + +Here is the detailed nutritional content and cost for each food item: + +- Chicken: Gives you 15 grams of protein, 18 grams of carbohydrates, and 300 calories for $4. +- Rice: Offers 1 gram of protein, 25 grams of carbohydrates, and 267 calories for $2. +- Apples: Provide 1 gram of protein, 21 grams of carbohydrates, and 266 calories for $5. +- Steak: Contains 6 grams of protein, 3 grams of carbohydrates, and 119 calories for a higher cost of $10. +- Lentils: These give 3 grams of protein, 7 grams of carbohydrates, and 166 calories for just $2. +- Fish: Delivers 17 grams of protein, 13 grams of carbohydrates, and 129 calories for $8. +- Tofu: Offers a substantial 18 grams of protein, 27 grams of carbohydrates, and 216 calories for $10. +- Cheese: Gives 12 grams of protein, 17 grams of carbohydrates, and 76 calories for $9. +- Bread: Provides 2 grams of protein, a massive 30 grams of carbohydrates, and 258 calories for $4. + +Your daily dietary goal is to consume at least 90 grams of protein, 105 grams of carbohydrates, and 1805 calories. Your challenge is to figure out how to meet these nutritional requirements from the food options mentioned above while spending the least amount of money. So, what is the least amount of money you need to spend to meet your daily dietary requirements? Please note that the response should be a single answer, asking for only the optimal value. You could only assign integral portions of each type of food.",[26.0] +['Diet Problem'],"Imagine you are a dietitian, working diligently to provide a balanced diet plan for a client while keeping the cost to a minimum. The client's daily dietary requirement is 88 grams of protein, 144 grams of protein, and 1900 calories. To achieve this, you have a variety of six food items to choose from. Each food item comes with its specific nutritional values and price. + +Here's a detailed breakdown of each food item: + +- Food 1: A high-protein food that offers 15 grams of protein, 10 grams of carbohydrates, and 163 calories, costing $10. +- Food 2: A balanced food offering 16 grams of protein, 25 grams of carbohydrates, and 108 calories, costing just $4. +- Food 3: A calorie-dense food providing 3 grams of protein, 8 grams of carbohydrates, and a whopping 198 calories, costing just $1. +- Food 4: A well-rounded food item that gives 19 grams of protein, 20 grams of carbohydrates, and 141 calories, priced at $7. +- Food 5: A modest food item that offers 11 grams of protein, 8 grams of carbohydrates, and 105 calories, costing $6. +- Food 6: Another calorie-dense food that delivers 10 grams of protein, 9 grams of carbohydrates, and 199 calories, costing $8. + +Your task is to design a daily meal plan that meets all the client's requirements using the given food items while keeping the cost as low as possible. What is the minimum cost you can achieve while meeting the client's nutritional requirements? You could only assign integral portions of each type of food.",[24.0] +"[""Diet Problem""]","Imagine you are planning your meals for the week, and you need to make sure you're receiving the right amount of nutrients without breaking the bank. You're considering six different food items, namely Pasta, Avocado, Chicken, Tofu, Spinach, and Cheese. Each of these foods offers a certain amount of protein, carbohydrates, and calories, and each comes with its own cost. + +Here's what each food offers and its cost: + +- Pasta: Gives you 15 grams of protein, 30 grams of carbohydrates, and 154 calories for $3. +- Avocado: Offers 20 grams of protein, 16 grams of carbohydrates, and 111 calories for $8. +- Chicken: Packs 15 grams of protein, 8 grams of carbohydrates, and 211 calories for $7. +- Tofu: Delivers 19 grams of protein, 5 grams of carbohydrates, and 87 calories for only $2. +- Spinach: Provides 19 grams of protein, 28 grams of carbohydrates, and 98 calories for $9. +- Cheese: Contributes 12 grams of protein, 17 grams of carbohydrates, and 231 calories for $10. + +Your goal is to ensure that you get at least 73 grams of protein, 111 grams of carbohydrates, and 2196 calories from whatever combination of these foods you choose. The challenge is to figure out the most cost-effective way to hit all these nutritional targets with these six food options. What is the minimum cost to meet your nutritional needs? +Note: your response should be a single answer question, asking for only the optimal value. The scenario revolves around the selection of food. You could only assign integral portions of each type of food.",[44.0] +"[""Transportation Problem""]","Imagine you are coordinating a logistics effort to redistribute essential supplies among seven regional distribution centers. Each center starts with a specific quantity of supplies but has different needs to ensure smooth operations across the regions. + +Here's the initial setup: + +- Distribution Center 1 has 97 units but needs 119 units. +- Distribution Center 2 has 458 units but needs 275 units. +- Distribution Center 3 has 473 units but needs only 36 units. +- Distribution Center 4 has 380 units but needs 459 units. +- Distribution Center 5 has 34 units but needs 17 units. +- Distribution Center 6 has 399 units but needs 128 units. +- Distribution Center 7 has 163 units but needs 425 units. + +To move supplies between these centers, there is a cost associated with each transfer, which varies depending on the route. Here's the cost matrix showing the cost to transfer supplies from one center to another: + +- From/To Center 1: To 2 costs 41, to 3 costs 42, to 4 costs 31, to 5 costs 39, to 6 costs 2, to 7 costs 18. +- From/To Center 2: To 1 costs 48, to 3 costs 9, to 4 costs 22, to 5 costs 47, to 6 costs 29, to 7 costs 39. +- From/To Center 3: To 1 costs 36, to 2 costs 22, to 4 costs 29, to 5 costs 45, to 6 costs 33, to 7 costs 45. +- From/To Center 4: To 1 costs 19, to 2 costs 5, to 3 costs 10, to 5 costs 3, to 6 costs 42, to 7 costs 20. +- From/To Center 5: To 1 costs 13, to 2 costs 10, to 3 costs 21, to 4 costs 7, to 6 costs 35, to 7 costs 11. +- From/To Center 6: To 1 costs 5, to 2 costs 35, to 3 costs 29, to 4 costs 16, to 5 costs 40, to 7 costs 9. +- From/To Center 7: To 1 costs 7, to 2 costs 24, to 3 costs 26, to 4 costs 38, to 5 costs 33, to 6 costs 13. + +Your task is to ensure each distribution center has the supplies it needs while minimizing the total cost of redistribution. What would be the minimum cost to achieve this?",[4172.0] +['TravelingSalesman'],"Imagine a scenario where a salesperson needs to visit five different cities, identified as Cities 1 through 5, to present their products. The salesperson has the flexibility to choose any city as a starting point; however, they are required to visit each city exactly once and then return to the starting city. The primary goal is to complete this circuit while incurring the lowest possible travel cost. + +Here’s a summary of the travel costs between the cities: +- The cost to travel from City 1 to City 2 is 28 units, to City 3 is 89 units, to City 4 is 59 units, and to City 5 is 41 units. +- From City 2, it costs 28 units to reach City 1, 85 units to get to City 3, 12 units to City 4, and 68 units to City 5. +- Traveling from City 3, the costs are 89 units to City 1, 85 units to City 2, 22 units to City 4, and 70 units to City 5. +- From City 4, the journey costs 59 units to City 1, 12 units to City 2, 22 units to City 3, and 48 units to City 5. +- Lastly, from City 5, it costs 41 units to travel to City 1, 68 units to City 2, 70 units to City 3, and 48 units to City 4. + +What is the minimum total travel cost for the salesperson to visit each of the five cities exactly once and return to the starting city?",[173.0] +"[""TravelingSalesman""]","Imagine a salesperson embarking on a journey to visit five distinct cities—let's name them A, B, C, D, and E—to promote their products. The salesperson has the flexibility to commence their expedition from any of these cities. However, there's a pivotal condition: they must travel to each city exactly once and then return to their initial departure point. The primary objective is to minimize the overall travel cost incurred during this excursion, which encompasses factors such as distance, time, or expenses. + +Here's a comprehensive breakdown of the travel costs between the cities: +- Traveling from City A incurs a cost of 44 units to reach B, 19 units to reach C, 92 units to reach D, and 71 units to reach E. +- City B entails a cost of 44 units to travel to A, 22 units to C, 84 units to D, and 63 units to E. +- For City C, the costs are 19 units to A, 22 units to B, 47 units to D, and 89 units to E. +- Moving on to City D, it involves a cost of 92 units to reach A, 84 units to B, 47 units to C, and 78 units to E. +- Finally, from City E, the costs are 71 units to A, 63 units to B, 89 units to C, and 78 units to D. + +Given these details, the pivotal question arises: What is the minimum total travel cost for the salesperson to visit each city exactly once and return to the starting city?",[251.0] +['TravelingSalesman'],"Imagine a salesperson tasked with visiting six unique cities to promote their products. These cities are conveniently numbered from 1 to 6, and the salesperson can start their journey from any of these cities. The essential requirement is that each city must be visited exactly once, with the journey ending back at the starting city. The primary objective is to minimize the total travel costs, which include expenses like fuel, tolls, and time spent on the road. + +Here is a breakdown of the travel costs between the cities: +- From City 1, the costs to travel to the other cities are as follows: City 2 costs 21 units, City 3 costs 43 units, City 4 costs 25 units, City 5 costs 38 units, and City 6 costs 91 units. +- From City 2, it costs 21 units to City 1, 32 units to City 3, 25 units to City 4, 16 units to City 5, and 14 units to City 6. +- Traveling from City 3, the costs are 43 units to City 1, 32 units to City 2, 96 units to City 4, 71 units to City 5, and 98 units to City 6. +- From City 4, it costs 25 units to reach City 1, 25 units to City 2, 96 units to City 3, 53 units to City 5, and 22 units to City 6. +- From City 5, the travel costs are 38 units to City 1, 16 units to City 2, 71 units to City 3, 53 units to City 4, and 100 units to City 6. +- Lastly, from City 6, it costs 91 units to travel to City 1, 14 units to City 2, 98 units to City 3, 22 units to City 4, and 100 units to City 5. + +What is the minimum total travel cost for the salesperson to visit each city exactly once and return to the starting city?",[191.0] +"[""Network Optimization""]","Consider a complex transportation network connecting 9 cities, where goods are moved from a central warehouse (City 0) to a major marketplace (City 8). Trucks of different capacities are assigned to each route between the cities. + +Here's the maximum capacity of goods (in tons) each truck can move from one city to another: + +- From City 0 (Warehouse): Trucks can move goods to City 1 (15 tons), City 2 (12 tons), City 3 (9 tons), City 4 (7 tons), City 5 (12 tons), City 6 (5 tons), City 7 (20 tons), and City 8 (10 tons). +- From City 1: Trucks can move goods to City 0 (6 tons), City 2 (8 tons), City 4 (19 tons), City 5 (14 tons), City 6 (17 tons), City 7 (9 tons), and City 8 (16 tons). +- From City 2: Trucks can move goods to City 0 (19 tons), City 1 (11 tons), City 4 (13 tons), City 5 (14 tons), City 6 (17 tons), City 7 (10 tons), and City 8 (13 tons). +- From City 3: Trucks can move goods to City 0 (15 tons), City 1 (4 tons), City 2 (5 tons), City 4 (2 tons), City 5 (18 tons), City 6 (6 tons), City 7 (17 tons), and City 8 (19 tons). +- From City 4: Trucks can move goods to City 0 (2 tons), City 1 (8 tons), City 2 (2 tons), City 3 (4 tons), City 5 (6 tons), City 6 (14 tons), City 7 (20 tons), and City 8 (18 tons). +- From City 5: Trucks can move goods to City 0 (14 tons), City 1 (9 tons), City 2 (18 tons), City 3 (7 tons), City 4 (4 tons), City 6 (11 tons), City 7 (11 tons), and City 8 (13 tons). +- From City 6: Trucks can move goods to City 0 (12 tons), City 1 (14 tons), City 2 (12 tons), City 3 (4 tons), City 4 (18 tons), City 5 (6 tons), City 7 (5 tons), and City 8 (4 tons). +- From City 7: Trucks can move goods to City 0 (20 tons), City 1 (1 ton), City 2 (6 tons), City 3 (17 tons), City 4 (4 tons), City 5 (2 tons), City 6 (11 tons), and City 8 (9 tons). +- From City 8 (Marketplace): Trucks can move goods back to City 0 (16 tons), City 1 (6 tons), City 2 (14 tons), City 3 (0 tons), City 4 (17 tons), City 5 (17 tons), City 6 (6 tons), and City 7 (9 tons). + +In this setup, a capacity of ""0"" means there is no truck allocated to move goods between those two cities. + +The challenge is to calculate the maximum amount of goods that can be transported from City 0 (the Warehouse) to City 8 (the Marketplace) without exceeding the capacity of any truck on its route. What is the maximum quantity of goods that can be transported from the warehouse to the marketplace in tons?",[90.0] +"[""Network Optimization""]","Imagine we're planning the logistics for a new high-speed data network connecting 8 major data centers across the country. This network is designed to optimize the flow of information, functioning much like a water distribution system, but for data. Each data center is connected to others through high-capacity data transmission lines, each with a specific capacity for data flow measured in terabits per second (Tbps). + +Here's the setup of this futuristic data network, detailing the maximum data flow capacity from one center to another: + +- From Data Center 0 (Source): Can send data to Data Center 1 (5 Tbps), Data Center 2 (16 Tbps), Data Center 3 (11 Tbps), Data Center 4 (2 Tbps), Data Center 5 (10 Tbps), Data Center 6 (17 Tbps), and Data Center 7 (2 Tbps). +- From Data Center 1: Can send data to Data Center 0 (8 Tbps), Data Center 2 (14 Tbps), Data Center 3 (14 Tbps), Data Center 4 (2 Tbps), Data Center 5 (7 Tbps), Data Center 6 (11 Tbps), and Data Center 7 (1 Tbps). +- From Data Center 2: Can send data to Data Center 0 (17 Tbps), Data Center 1 (20 Tbps), Data Center 3 (16 Tbps), Data Center 4 (13 Tbps), Data Center 6 (6 Tbps), and Data Center 7 (1 Tbps). +- From Data Center 3: Can send data to Data Center 0 (3 Tbps), Data Center 1 (16 Tbps), Data Center 2 (10 Tbps), Data Center 4 (19 Tbps), Data Center 5 (12 Tbps), Data Center 6 (3 Tbps), and Data Center 7 (15 Tbps). +- From Data Center 4: Can send data to Data Center 0 (15 Tbps), Data Center 1 (12 Tbps), Data Center 2 (19 Tbps), Data Center 3 (18 Tbps), Data Center 5 (14 Tbps), Data Center 6 (12 Tbps), and Data Center 7 (15 Tbps). +- From Data Center 5: Can send data to Data Center 0 (2 Tbps), Data Center 1 (20 Tbps), Data Center 2 (20 Tbps), Data Center 3 (17 Tbps), Data Center 4 (4 Tbps), Data Center 6 (10 Tbps), and Data Center 7 (17 Tbps). +- From Data Center 6: Can send data to Data Center 0 (15 Tbps), Data Center 1 (13 Tbps), Data Center 2 (13 Tbps), Data Center 3 (5 Tbps), Data Center 4 (10 Tbps), Data Center 5 (15 Tbps), and Data Center 7 (19 Tbps). +- From Data Center 7 (Destination): Can send data back to Data Center 0 (10 Tbps), Data Center 1 (3 Tbps), Data Center 2 (14 Tbps), Data Center 3 (11 Tbps), Data Center 4 (5 Tbps), Data Center 5 (20 Tbps), and Data Center 6 (4 Tbps). + +In this network, a capacity of ""0"" indicates that there is no direct data line connecting those two centers. + +The objective is to determine the maximum amount of data (in Tbps) that can be transmitted from Data Center 0 (the source) to Data Center 7 (the destination), ensuring that the flow through each data line does not exceed its specified capacity. What is the maximum amount of data that can be sent from the source to the destination in terabits per second?",[63.0] +"[""Network Optimization""]","Consider a complex railway network connecting 7 major cities - from the central hub to a key end destination. These cities are interconnected by railway lines, each having a specific capacity for passenger flow. + +The capacity of each railway line is as follows, indicating the maximum number of passengers that can travel from one city to another per day: + +- From City 0 (Central Hub): Can dispatch passengers to City 1 (18,000 passengers), City 2 (11,000 passengers), City 3 (15,000 passengers), City 4 (2,000 passengers), City 5 (4,000 passengers), and City 6 (20,000 passengers). +- From City 1: Can dispatch passengers to City 0 (4,000 passengers), City 2 (6,000 passengers), City 3 (5,000 passengers), City 4 (19,000 passengers), City 5 (15,000 passengers). +- From City 2: Can dispatch passengers to City 0 (15,000 passengers), City 1 (10,000 passengers), City 3 (7,000 passengers), City 5 (7,000 passengers), and City 6 (5,000 passengers). +- From City 3: Can dispatch passengers to City 0 (14,000 passengers), City 1 (5,000 passengers), City 2 (3,000 passengers), City 4 (10,000 passengers), City 5 (15,000 passengers), and City 6 (19,000 passengers). +- From City 4: Can dispatch passengers to City 0 (2,000 passengers), City 1 (11,000 passengers), City 2 (1,000 passengers), City 3 (10,000 passengers), City 5 (14,000 passengers), and City 6 (6,000 passengers). +- From City 5: Can dispatch passengers to City 0 (10,000 passengers), City 1 (14,000 passengers), City 2 (6,000 passengers), City 3 (5,000 passengers), City 4 (3,000 passengers), City 6 (18,000 passengers). +- From City 6 (End Destination): Can receive passengers from City 0 (8,000 passengers), City 1 (1,000 passengers), City 2 (11,000 passengers), City 3 (8,000 passengers), City 4 (6,000 passengers), and City 5 (20,000 passengers). + +In this system, a capacity of ""0"" signifies there is no direct railway line between those two cities. + +Given this situation, we need to determine the maximum number of passengers that can be transported from City 0 (the central hub) to City 6 (the end destination) per day, without exceeding the transportation capacity of any railway line. What is the maximum number of passengers that can be dispatched from the central hub to the end destination per day?",[70000.0] +['Capacitated Facility Location Problem'],"In the bustling world of retail, a company named SupplyChain Masters is faced with the challenge of efficiently supplying its nine retail stores with products from a network of seven potential distribution centers. The company's goal is to ensure that each store's demand is met, while simultaneously minimizing the total cost involved. This total cost encompasses the opening expenses of the distribution centers and the transportation costs for delivering the goods to the stores. + +Let's dive into the specifics: + +- The company is contemplating between seven distribution centers to serve its needs. +- There are nine retail stores scattered in different locations, each with its unique demand for products. + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: 151,000 +- Center 2: 192,000 +- Center 3: 114,000 +- Center 4: 171,000 +- Center 5: 160,000 +- Center 6: 120,000 +- Center 7: 182,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $2 to Store 1, $3 to Store 2, $4 to Store 3, $5 to Store 4, $3 to Store 5, $2 to Store 6, $3 to Store 7, $4 to Store 8, and $2 to Store 9. +- From Center 2: $3 to Store 1, $2 to Store 2, $5 to Store 3, $4 to Store 4, $3 to Store 5, $2 to Store 6, $4 to Store 7, $3 to Store 8, and $4 to Store 9. +- From Center 3: $4 to Store 1, $3 to Store 2, $2 to Store 3, $3 to Store 4, $5 to Store 5, $4 to Store 6, $5 to Store 7, $3 to Store 8, and $5 to Store 9. +- From Center 4: $5 to Store 1, $4 to Store 2, $3 to Store 3, $2 to Store 4, $4 to Store 5, $5 to Store 6, $2 to Store 7, $4 to Store 8, and $3 to Store 9. +- From Center 5: $2 to Store 1, $5 to Store 2, $4 to Store 3, $3 to Store 4, $2 to Store 5, $3 to Store 6, $4 to Store 7, $5 to Store 8, and $3 to Store 9. +- From Center 6: $3 to Store 1, $4 to Store 2, $5 to Store 3, $3 to Store 4, $4 to Store 5, $2 to Store 6, $3 to Store 7, $4 to Store 8, and $2 to Store 9. +- From Center 7: $1 to Store 1, $2 to Store 2, $3 to Store 3, $4 to Store 4, $5 to Store 5, $3 to Store 6, $2 to Store 7, $1 to Store 8, and $2 to Store 9. + +Demand of Each Retail Store (in units): +- Store 1: 440 +- Store 2: 712 +- Store 3: 860 +- Store 4: 504 +- Store 5: 892 +- Store 6: 738 +- Store 7: 651 +- Store 8: 580 +- Store 9: 790 + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1800 +- Center 2: 1300 +- Center 3: 1600 +- Center 4: 1200 +- Center 5: 1700 +- Center 6: 814 +- Center 7: 1962 + +SupplyChain Masters' challenge is to calculate the optimal setup of distribution centers that should be opened and determine the distribution of supplies to the retail stores in a way that all stores' demands are met at the lowest possible total cost. + +Question: +What is the optimal total cost for SupplyChain Masters to open the necessary distribution centers and transport goods to all retail stores, considering both the opening costs and the transportation costs?",[610975.0] +['Capacitated Facility Location Problem'],"In the bustling world of retail, SupplyChain Solutions is faced with the challenge of efficiently distributing products to eight eager retail stores. With four potential distribution centers to choose from, each with its own unique costs and capacities, the company aims to devise a strategic plan to minimize overall expenses. This plan must account for the initial cost of opening distribution centers and the ongoing expenses of transporting goods to stores. The objective is clear: to fulfill the demand of each retail store at the lowest possible total cost. + +Given Data: +- Number of Potential Distribution Centers: 4 +- Number of Retail Stores Needing Supply: 8 + +Opening Costs for Each Distribution Center (in dollars): +- Center 1: $151,000 +- Center 2: $192,000 +- Center 3: $114,000 +- Center 4: $171,000 + +Transportation Cost Per Unit from Each Distribution Center to Retail Stores (in dollars): +- From Center 1: $5 to Store 1, $5 to Store 2, $2 to Store 3, $3 to Store 4, $3 to Store 5, $3 to Store 6, $5 to Store 7, $4 to Store 8 +- From Center 2: $3 to Store 1, $5 to Store 2, $2 to Store 3, $4 to Store 4, $2 to Store 5, $4 to Store 6, $5 to Store 7, $1 to Store 8 +- From Center 3: $4 to Store 1, $2 to Store 2, $5 to Store 3, $4 to Store 4, $1 to Store 5, $1 to Store 6, $3 to Store 7, $3 to Store 8 +- From Center 4: $2 to Store 1, $4 to Store 2, $4 to Store 3, $3 to Store 4, $4 to Store 5, $4 to Store 6, $1 to Store 7, $3 to Store 8 + +Demand of Each Retail Store (in units): +- Store 1: 884 units +- Store 2: 420 units +- Store 3: 566 units +- Store 4: 673 units +- Store 5: 787 units +- Store 6: 1000 units +- Store 7: 715 units +- Store 8: 413 units + +Supply Capacity of Each Distribution Center (in units): +- Center 1: 1041 units +- Center 2: 1576 units +- Center 3: 1364 units +- Center 4: 1697 units + +Question: +What is the optimal total cost for SupplyChain Solutions to open distribution centers and transport goods to meet all retail stores' demands? This cost includes both the opening costs of the centers and the transportation costs of supplying the stores.",[637573.0] +"[""Network Optimization""]","There are four nodes connected by directed edges. Node 1, which could represent the starting point, is connected to two other nodes: There is an edge from node 1 to node 2 with a capacity of 5. There is an edge from node 1 to node 3 with a capacity of 7. Node 2 is situated at the top of the graph and acts as an intermediate node, with three edges connected to it: An edge leading from node 2 to node 3 with a capacity of 2. An edge leading from node 2 to node 4 with a capacity of 4. Node 3 is another intermediate node and has two edges: An edge from node 3 to node 4 with a substantial capacity of 12. Node 4, which could represent the target or terminal, does not have any edges leading to other nodes, implying that it could be the end point. Find the corresponding maximum flow of the graph. Provide your answer rounded to the nearest integar.",[12.0] +"[""Lot-Sizing Problem""]","Sailco Corporation must determine how many sailboats should be produced during each of the next four quarters (one quarter three months). The demand during each of the next four quarters is as follows: first quarter, 40 sailboats; second quarter, 60 sailboats; third quarter, 75 sailboats; fourth quarter, 25 sailboats. Sailco must meet demands on time. At the beginning of the first quarter, Sailco has an inventory of 10 sailboats. At the beginning of each quarter, Sailco must decide how many sailboats should be produced during that quarter. For simplicity, we assume that sailboats manufactured during a quarter can be used to meet demand for that quarter. During each quarter, Sailco can produce up to 40 sailboats with regular-time labor at a total cost of $400 per sailboat. By having employees work overtime during a quarter, Sailco can produce additional sailboats with overtime labor at a total cost of $450 per sailboat. At the end of each quarter (after production has occurred and the current quarter’s demand has been satisfied), a carrying or holding cost of $20 per sailboat is incurred. Use linear programming to determine a production schedule to minimize the sum of production and inventory costs during the next four quarters. Find the minimal sum.",[78450.0] diff --git a/optimind/data/optimind_cleaned_classified_optmath.csv b/optimind/data/optimind_cleaned_classified_optmath.csv new file mode 100644 index 0000000..13a853c --- /dev/null +++ b/optimind/data/optimind_cleaned_classified_optmath.csv @@ -0,0 +1,3938 @@ +problem_class,question,answer +['Job Shop'],"The manufacturing facility produces custom components for two jobs, Job 0 and Job 1, each consisting of a sequence of operations that must be performed in a specific order. The goal is to schedule these operations to minimize the total completion time (makespan) while satisfying all operational constraints. Job 0 has five operations with processing times: Operation 0 takes 4 units, Operation 1 takes 1 unit, Operation 2 takes 6 units, Operation 3 takes 6 units, and Operation 4 takes 8 units. Job 1 has four operations with processing times: Operation 0 takes 9 units, Operation 1 takes 1 unit, Operation 2 takes 4 units, and Operation 3 takes 2 units. + +Precedence constraints ensure that operations within each job are performed in sequence with specific gaps. For Job 0, Operation 1 must start at least 4 units after Operation 0 starts, Operation 2 must start at least 1 unit after Operation 1 starts, Operation 3 must start at least 6 units after Operation 2 starts, and Operation 4 must start at least 6 units after Operation 3 starts. For Job 1, Operation 1 must start at least 9 units after Operation 0 starts, Operation 2 must start at least 1 unit after Operation 1 starts, and Operation 3 must start at least 4 units after Operation 2 starts. + +Machine capacity constraints ensure that operations assigned to the same machine do not overlap. In particular, there are 3 machines (M0, M1, M2). Operation 1 of Job 0, Operation 3 of Job 0, and Operation 2 of Job 1 are on machine M0. Operation 0of Job 0, Operation 2 of Job 0, and Operation 3 of Job 1 are on machine M1. Finally, Operation 4 of Job 0, Operation 0 of Job 1, and Operation 1 of Job 1 are on machine M2. Binary variables determine the order of operations on shared machines. Constraints apply to all operation pairs on shared machines, ensuring no overlap and maintaining required time gaps (one operation must finish before another can start). The large constant of 100,000 is used in these constraints to enforce the sequencing logic by ensuring that the constraints are only active when the binary variable is set to 1. + +The makespan must be at least as large as the completion time of every operation. Specifically, the makespan must be at least 4 units after Operation 0 of Job 0 starts, 1 unit after Operation 1 of Job 0 starts, 6 units after Operation 2 of Job 0 starts, 6 units after Operation 3 of Job 0 starts, 8 units after Operation 4 of Job 0 starts, 9 units after Operation 0 of Job 1 starts, 1 unit after Operation 1 of Job 1 starts, 4 units after Operation 2 of Job 1 starts, and 2 units after Operation 3 of Job 1 starts. + +The objective is to determine the start times for all operations and the sequence of operations on shared machines to minimize the makespan while respecting all precedence constraints, machine capacity constraints, and the makespan definition. Binary variables are used to enforce the correct sequencing of operations on shared machines, with a large constant of 100,000 ensuring the constraints are properly applied. The start times for all operations and the makespan must be non-negative. The binary variables are explicitly defined for all relevant operation pairs on shared machines, and the start times and makespan are continuous variables.",25.0 +['AircraftLanding'],"The problem involves scheduling the landing times of three aircraft to minimize the total penalty costs associated with deviations from their target landing times while ensuring safe separation between aircraft and adhering to their respective time windows. Each aircraft has an earliest and latest possible landing time, and penalty costs for landing earlier or later than their target landing time. Aircraft 0 has a target landing time of 231 minutes, an earliest landing time of 201 minutes, a latest landing time of 261 minutes, an early penalty cost of \$37 per minute, and a late penalty cost of \$73 per minute. Aircraft 1 has a target landing time of 264 minutes, an earliest landing time of 234 minutes, a latest landing time of 294 minutes, an early penalty cost of \$99 per minute, and a late penalty cost of \$84 per minute. Aircraft 2 has a target landing time of 59 minutes, an earliest landing time of 29 minutes, a latest landing time of 89 minutes, an early penalty cost of \$90 per minute, and a late penalty cost of \$56 per minute. + +The separation requirements are enforced using binary decision variables that determine the order in which the aircraft land. If Aircraft 0 lands before Aircraft 1, there must be at least 11 minutes of separation; if Aircraft 1 lands before Aircraft 0, the separation must be at least 27 minutes. If Aircraft 0 lands before Aircraft 2, there must be at least 9 minutes of separation; if Aircraft 2 lands before Aircraft 0, the separation must be at least 232 minutes. If Aircraft 1 lands before Aircraft 2, there must be at least 9 minutes of separation; if Aircraft 2 lands before Aircraft 1, the separation must be at least 265 minutes. + +The landing time windows must be respected: Aircraft 0 must land between 201 and 261 minutes, Aircraft 1 between 234 and 294 minutes, and Aircraft 2 between 29 and 89 minutes. The penalties for early and late landings are calculated based on the difference between the actual landing time and the target landing time, multiplied by the respective penalty costs. The objective is to determine the optimal landing times for all three aircraft to minimize the total penalty costs while ensuring all separation and time window constraints are satisfied. The total penalty cost is the sum of the early and late penalties for all aircraft. The order in which the aircraft land affects the separation requirements and landing times, so the schedule must balance these costs to achieve the lowest total penalty. Binary decision variables are used to determine the order of aircraft landings, ensuring that the separation constraints are enforced correctly. The constraints also ensure that the landing times are within the specified windows and that the penalties for early and late landings are calculated accurately based on the deviations from the target landing times.",7410.0 +['TravelingSalesman'],"A logistics company must deliver packages to four cities: City 0, City 1, City 2, and City 3, minimizing the total delivery cost while ensuring each city is visited exactly once and the route forms a continuous loop without subtours. The delivery costs between cities are as follows: City 0 to City 1 costs 616 units, City 0 to City 2 costs 567 units, City 0 to City 3 costs 143 units, City 1 to City 0 costs 388 units, City 1 to City 2 costs 947 units, City 1 to City 3 costs 628 units, City 2 to City 0 costs 495 units, City 2 to City 1 costs 650 units, City 2 to City 3 costs 604 units, City 3 to City 0 costs 77 units, City 3 to City 1 costs 630 units, and City 3 to City 2 costs 23 units. The company uses binary decision variables to indicate whether a route is used and assigns position variables to each city to prevent subtours. The position variable for City 0 is fixed at 0, while the position variables for City 1, City 2, and City 3 must be integers between 0 and 3. The objective is to minimize the total delivery cost, which is the sum of the costs of all selected routes, including the term 0 times the position variable for City 0. Constraints ensure each city is visited and departed exactly once. Subtour elimination is enforced by ensuring that if a route from City i to City j is used, the position of City j must be greater than the position of City i, except for City 0, which is fixed as the starting point. These constraints are implemented through specific relationships between the route variables and the position variables, guaranteeing the delivery route forms a single loop without smaller loops. The constraints also explicitly bound the position variables for City 1, City 2, and City 3 to be between 0 and 3, and ensure that the position variables are correctly related to the route variables to eliminate subtours.",1204.0 +['Job Shop'],"The problem involves scheduling operations for three jobs (Job 0, Job 1, and Job 2) on shared machines to minimize the total completion time (makespan). Each job consists of a sequence of operations with specific processing times: Job 0 has two operations (Operation 0 takes 9 units of time, Operation 1 takes 8 units), Job 1 has three operations (Operation 0 takes 8 units, Operation 1 takes 8 units, Operation 2 takes 4 units), and Job 2 has four operations (Operation 0 takes 1 unit, Operation 1 takes 6 units, Operation 2 takes 4 units, Operation 3 takes 2 units). Operations within a job must follow a strict order, meaning each operation cannot start until the previous one in the same job has finished. For example, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 has completed after 9 units of time, Operation 1 of Job 1 cannot start until Operation 0 of Job 1 has completed after 8 units of time, and Operation 2 of Job 1 cannot start until Operation 1 of Job 1 has completed after 8 units of time. Similarly, Operation 1 of Job 2 cannot start until Operation 0 of Job 2 has completed after 1 unit of time, Operation 2 of Job 2 cannot start until Operation 1 of Job 2 has completed after 6 units of time, and Operation 3 of Job 2 cannot start until Operation 2 of Job 2 has completed after 4 units of time. + +Certain operations from different jobs share the same machine and cannot overlap in time: Operation 1 of Job 0 and Operation 1 of Job 2 are assigned to the same machine, so one must finish before the other starts. Similarly, Operation 0 of Job 0 and Operation 1 of Job 1 share a machine, and their schedules must not overlap. This also applies to the following additional pairs of operations: Operation 0 of Job 0 and Operation 0 of Job 2, Operation 1 of Job 1 and Operation 0 of Job 2, Operation 0 of Job 1 and Operation 2 of Job 2, Operation 0 of Job 1 and Operation 3 of Job 2, Operation 2 of Job 1 and Operation 2 of Job 2, Operation 2 of Job 1 and Operation 3 of Job 2, and Operation 2 of Job 2 and Operation 3 of Job 2. + +To enforce these constraints, binary variables are used to determine the order of operations on shared machines. A large constant (1000) is used in the constraints to ensure that operations on the same machine do not overlap. The binary variables are restricted to values of 0 or 1, and the start times of operations are continuous and non-negative. + +The makespan must be at least as large as the completion time of every operation. For example, it must be at least 9 units of time (completion of Operation 0 of Job 0), 8 units (completion of Operation 1 of Job 0), 8 units (completion of Operation 0 of Job 1), 8 units (completion of Operation 1 of Job 1), 4 units (completion of Operation 2 of Job 1), 1 unit (completion of Operation 0 of Job 2), 6 units (completion of Operation 1 of Job 2), 4 units (completion of Operation 2 of Job 2), and 2 units (completion of Operation 3 of Job 2). + +The objective is to determine the start times for each operation on each machine such that all precedence and machine capacity constraints are satisfied, and the total completion time (makespan) is minimized. This involves deciding the order of operations on shared machines and ensuring no overlaps, while respecting the processing times and sequence constraints of each job.",22.0 +"['SupplyChain', 'Network Optimization']","Imagine you are managing a supply chain network for a company that operates across three key locations: **Node 0**, **Node 1**, and **Node 2**. Each location has specific supply and demand requirements, and your goal is to optimize the flow of goods between these nodes while minimizing the total cost of establishing connections and shipping products. + +#### Supply and Demand: +- **Node 0** has a net supply of **236 units**, meaning it can provide this amount to other nodes. +- **Node 1** has a net demand of **29 units**, requiring this amount to be fulfilled from other nodes. +- **Node 2** has a net demand of **207 units**, which must also be met through shipments from other nodes. + +#### Shipping Costs and Capacities: +To move goods between nodes, you must decide whether to establish a connection (which incurs a fixed cost) and how much to ship (which incurs a variable cost per unit). The costs and capacities for each possible connection are as follows: + +1. **Node 0 to Node 1**: + - Fixed cost to establish the connection: **$3,532** + - Variable shipping cost per unit: **$45** + - Maximum shipping capacity: **773 units** + +2. **Node 0 to Node 2**: + - Fixed cost: **$3,895** + - Variable shipping cost: **$49** + - Maximum capacity: **973 units** + +3. **Node 1 to Node 0**: + - Fixed cost: **$3,485** + - Variable shipping cost: **$11** + - Maximum capacity: **785 units** + +4. **Node 1 to Node 2**: + - Fixed cost: **$1,845** + - Variable shipping cost: **$40** + - Maximum capacity: **787 units** + +5. **Node 2 to Node 0**: + - Fixed cost: **$1,291** + - Variable shipping cost: **$13** + - Maximum capacity: **602 units** + +6. **Node 2 to Node 1**: + - Fixed cost: **$1,818** + - Variable shipping cost: **$47** + - Maximum capacity: **766 units** + +#### Key Decisions: +- For each connection between nodes, you must decide whether to establish it (a binary decision: yes or no). +- For each established connection, you must determine the amount of goods to ship, ensuring it does not exceed the connection's capacity. + +#### Objective: +Your goal is to minimize the total cost, which includes: +1. The fixed costs of establishing the necessary connections. +2. The variable shipping costs based on the amount of goods transported. + +#### Constraints: +1. **Capacity Constraints**: The amount of goods shipped between any two nodes cannot exceed the maximum capacity of that connection. For example, if you decide to ship goods from **Node 0 to Node 1**, the shipment must be **773 units or less**. +2. **Flow Balance Constraints**: The total inflow and outflow of goods at each node must match its supply or demand. For example: + - At **Node 0**, the total inflow from **Node 1** and **Node 2** minus the outflow to **Node 1** and **Node 2** must equal **236 units**. + - At **Node 1**, the total inflow from **Node 0** and **Node 2** minus the outflow to **Node 0** and **Node 2** must equal **-29 units** (indicating a net demand). + - At **Node 2**, the total inflow from **Node 0** and **Node 1** minus the outflow to **Node 0** and **Node 1** must equal **-207 units** (indicating a net demand). + +#### Summary: +Your task is to determine the most cost-effective way to establish connections and ship goods between **Node 0**, **Node 1**, and **Node 2** to meet the supply and demand requirements while minimizing the total cost. This involves balancing fixed and variable costs, adhering to capacity limits, and ensuring the flow of goods aligns with the needs of each node.",18640.0 +['AircraftLanding'],"The problem involves scheduling the landing times of four aircraft (aircraft_0, aircraft_1, aircraft_2, aircraft_3) at an airport to minimize the total penalty costs associated with deviations from their earliest possible landing times while adhering to separation and timing constraints. Each aircraft has an earliest and latest possible landing time, and penalty costs for landing early or late. +Aircraft_0 has an earliest time of 0 minutes, latest time of 37 minutes, early penalty of \$61 per minute, and late penalty of \$82 per minute. +Aircraft_1 has an earliest time of 7 minutes, latest time of 67 minutes, early penalty of \$35 per minute, and late penalty of \$34 per minute. +Aircraft_2 has an earliest time of 225 minutes, latest time of 285 minutes, early penalty of \$74 per minute, and late penalty of \$41 per minute. +Aircraft_3 has an earliest time of 119 minutes, latest time of 179 minutes, early penalty of \$93 per minute, and late penalty of \$56 per minute. + +The landing order between any two aircraft must be explicitly determined using binary decision variables, ensuring one lands before the other. Separation constraints are enforced based on these landing orders, with two different separation times for each unordered pair {i,j} depending on which aircraft lands first. The required separation times (in minutes) are: + +• Pair (0,1): + – If aircraft_0 lands before aircraft_1, separation = 30 + – If aircraft_1 lands before aircraft_0, separation = 2 + +• Pair (0,2): + – If aircraft_0 lands before aircraft_2, separation = 30 + – If aircraft_2 lands before aircraft_0, separation = 2 + +• Pair (0,3): + – If aircraft_0 lands before aircraft_3, separation = 30 + – If aircraft_3 lands before aircraft_0, separation = 2 + +• Pair (1,2): + – If aircraft_1 lands before aircraft_2, separation = 20 + – If aircraft_2 lands before aircraft_1, separation = 3 + +• Pair (1,3): + – If aircraft_1 lands before aircraft_3, separation = 25 + – If aircraft_3 lands before aircraft_1, separation = 4 + +• Pair (2,3): + – If aircraft_2 lands before aircraft_3, separation = 15 + – If aircraft_3 lands before aircraft_2, separation = 2 + +These constraints ensure that the separation between consecutive landings is maintained according to the specific coefficients provided above. + +Each aircraft should land within its specified earliest and latest possible landing times. Early deviations are calculated as the difference between the earliest time and the actual landing time if the aircraft lands early, and late deviations are calculated as the difference between the actual landing time and the earliest time if the aircraft lands late. The total penalty cost is the sum of the early and late penalties for all aircraft. The objective is to determine the optimal landing times for all four aircraft to minimize the total penalty costs while ensuring all separation and timing constraints are satisfied.",738.0 +['AircraftLanding'],"The problem involves scheduling the landings of six aircraft to minimize the total penalty costs associated with deviations from their target landing times. Each aircraft has a target landing time, an earliest possible landing time, and a latest possible landing time. + +• Target landing times (minutes): +– aircraft_0: 244 +– aircraft_1: 192 +– aircraft_2: 293 +– aircraft_3: 28 +– aircraft_4: 150 +– aircraft_5: 287 + +Penalties are incurred for landing early or late, with specific penalty rates for each aircraft: aircraft_0 has an early penalty rate of 72 and a late penalty rate of 59, aircraft_1 has an early penalty rate of 81 and a late penalty rate of 40, aircraft_2 has an early penalty rate of 46 and a late penalty rate of 34, aircraft_3 has an early penalty rate of 78 and a late penalty rate of 24, aircraft_4 has an early penalty rate of 40 and a late penalty rate of 80, and aircraft_5 has an early penalty rate of 60 and a late penalty rate of 14. + +The landing times must respect minimum separation times between consecutive landings, which vary depending on the pair of aircraft. The required minimum separation times s_ij (in minutes, where i lands before j) are: + +j → 0 1 2 3 4 5 +i ↓ +0 0 29 24 31 26 25 +1 29 0 35 29 28 29 +2 24 35 0 29 26 27 +3 31 29 29 0 29 24 +4 26 28 26 29 0 25 +5 25 29 27 24 25 0 + +The separation constraints are enforced using binary decision variables that determine the order in which the aircraft land. The objective is to minimize the sum of early and late landing penalties, weighted by the respective penalty rates for each aircraft. The landing times must fall within the specified time windows for each aircraft: aircraft_0 must land between 221 and 281, aircraft_1 between 179 and 239, aircraft_2 between 268 and 300, aircraft_3 between 0 and 36, aircraft_4 between 131 and 191, and aircraft_5 between 265 and 300. The order of landings must ensure that the separation constraints are satisfied, with the specific separation times shown above for each pair of aircraft. The binary decision variables ensure that the separation constraints are met by determining the sequence of landings. + +The problem requires determining the exact landing times for each aircraft within their respective time windows, ensuring that the separation requirements are met, and minimizing the total penalty costs. The binary variables are used to enforce the separation constraints, ensuring that if one aircraft lands before another, the required separation time is maintained. The constraints also ensure that the landing times fall within the specified time windows and that the penalties for early or late landings are correctly calculated based on the deviation from the target landing times. The binary variables are explicitly defined as binary (0 or 1) to determine the sequence of landings and enforce the separation constraints.",1078.0 +['Job Shop'],"The manufacturing facility produces custom orders for three jobs: Job 0, Job 1, and Job 2. Each job consists of a sequence of operations that must be performed in a specific order, with each operation requiring a specific duration. The facility has limited machines, and some operations from different jobs must share the same machine. The goal is to schedule all operations to minimize the total completion time (makespan). + +Job 0 has two operations: Operation 0 and Operation 1. Operation 1 cannot start until Operation 0 has finished, with at least 8 hours between their start times. The makespan must be at least 8 hours after the start of Operation 0 and at least 5 hours after the start of Operation 1. + +Job 1 has five operations: Operation 0, Operation 1, Operation 2, Operation 3, and Operation 4. Operation 1 cannot start until Operation 0 has finished, with at least 7 hours between their start times. Operation 2 cannot start until Operation 1 has finished, with at least 9 hours between their start times. Operation 3 cannot start until Operation 2 has finished, with at least 5 hours between their start times. Operation 4 cannot start until Operation 3 has finished, with at least 6 hours between their start times. The makespan must be at least 7 hours after the start of Operation 0, 9 hours after the start of Operation 1, 5 hours after the start of Operation 2, 6 hours after the start of Operation 3, and 2 hours after the start of Operation 4. + +Job 2 has five operations: Operation 0, Operation 1, Operation 2, Operation 3, and Operation 4. Operation 1 cannot start until Operation 0 has finished, with at least 9 hours between their start times. Operation 2 cannot start until Operation 1 has finished, with at least 5 hours between their start times. Operation 3 cannot start until Operation 2 has finished, with at least 8 hours between their start times. Operation 4 cannot start until Operation 3 has finished, with at least 3 hours between their start times. The makespan must be at least 9 hours after the start of Operation 0, 5 hours after the start of Operation 1, 8 hours after the start of Operation 2, 3 hours after the start of Operation 3, and 1 hour after the start of Operation 4. + +Certain operations from different jobs share machines and must be scheduled without overlap. The following pairs of operations share machines: Operation 0 of Job 1 and Operation 2 of Job 1; Operation 1 of Job 0 and Operation 3 of Job 2; Operation 0 of Job 0 and Operation 1 of Job 2; Operation 0 of Job 0 and Operation 2 of Job 2; Operation 1 of Job 2 and Operation 2 of Job 2; Operation 4 of Job 1 and Operation 0 of Job 2. The scheduling system must determine the order of these operations to avoid conflicts, using binary variables to enforce the order. A large constant (100000) is used in the constraints to ensure that the order of operations sharing machines is respected. + +The objective is to determine the start times for all operations such that all precedence constraints within each job are satisfied, no two operations sharing the same machine overlap in time, and the total completion time (makespan) is minimized. The makespan must be at least as large as the completion time of every operation. All start times must be non-negative. The binary variables are used to enforce the order of operations sharing machines, and the large constant ensures that the constraints are properly applied. The start times and the makespan are general variables, and the objective is explicitly to minimize the makespan.",29.0 +['TravelingSalesman'],"Determine the optimal sequence of six cities (City 0, City 1, City 2, City 3, City 4, and City 5) to minimize the total travel distance while ensuring each city is visited exactly once and return to the origin city. The distances between the cities are as follows: From City 0 to City 1 (722 units), City 2 (758 units), City 3 (688 units), City 4 (897 units), and City 5 (100 units); from City 1 to City 0 (746 units), City 2 (76 units), City 3 (49 units), City 4 (86 units), and City 5 (385 units); from City 2 to City 0 (479 units), City 1 (154 units), City 3 (11 units), City 4 (374 units), and City 5 (350 units); from City 3 to City 0 (645 units), City 1 (842 units), City 2 (117 units), City 4 (619 units), and City 5 (535 units); from City 4 to City 0 (278 units), City 1 (829 units), City 2 (499 units), City 3 (268 units), and City 5 (973 units); from City 5 to City 0 (951 units), City 1 (144 units), City 2 (810 units), City 3 (943 units), and City 4 (989 units). The task is to find the shortest possible route that visits each city exactly once, and returns to the origin city. What is the minimal total distance travelled?",1062.0 +['Job Shop'],"A manufacturing facility produces custom orders for two jobs, Job 0 and Job 1. Each job consists of a sequence of operations that must be performed in a specific order, and some operations require the use of shared machinery. The goal is to schedule these operations to minimize the total production time, or makespan, ensuring all orders are completed as quickly as possible. + +Job 0 has three operations: Operation 0, Operation 1, and Operation 2. Operation 1 must start at least 8 units of time after Operation 0 completes, and Operation 2 must start at least 1 unit of time after Operation 1 completes. Job 1 has six operations: Operation 0, Operation 1, Operation 2, Operation 3, Operation 4, and Operation 5. Operation 1 must start at least 5 units of time after Operation 0 completes, Operation 2 must start at least 5 units of time after Operation 1 completes, Operation 3 must start at least 7 units of time after Operation 2 completes, Operation 4 must start at least 4 units of time after Operation 3 completes, and Operation 5 must start at least 2 units of time after Operation 4 completes. + +Some operations from both jobs require the use of the same machine, and only one operation can be processed on a machine at any given time. The following pairs of operations share the same machine: Operation 3 and Operation 4 of Job 1, Operation 0 and Operation 1 of Job 0, Operation 0 of Job 0 and Operation 2 of Job 1, and Operation 1 of Job 0 and Operation 2 of Job 1. For each pair of operations sharing a machine, a decision must be made about which operation is processed first. This decision is represented by a binary variable, where a value of 1 indicates that the first operation in the pair is processed before the second, and a value of 0 indicates the opposite. + +The primary objective is to minimize the total production time, or makespan, which is defined as the time at which the last operation across all jobs is completed. The makespan must be at least as large as the completion time of every operation in both jobs. Specifically, the makespan must be at least 8 units of time after Operation 0 of Job 0 completes, at least 1 unit of time after Operation 1 of Job 0 completes, at least 4 units of time after Operation 2 of Job 0 completes, at least 5 units of time after Operation 0 of Job 1 completes, at least 5 units of time after Operation 1 of Job 1 completes, at least 7 units of time after Operation 2 of Job 1 completes, at least 4 units of time after Operation 3 of Job 1 completes, at least 2 units of time after Operation 4 of Job 1 completes, and at least 9 units of time after Operation 5 of Job 1 completes. + +All operation start times must be non-negative, meaning no operation can start before time zero. For each pair of operations sharing a machine, the following must hold: if Operation A is scheduled before Operation B, then Operation B cannot start until Operation A has been completed, and conversely, if Operation B is scheduled before Operation A, then Operation A cannot start until Operation B has been completed. These constraints are enforced using a large constant (100,000) to ensure that the scheduling logic is maintained. The specific values derived from this constant, such as 99996, 99992, and 99999, are used in the constraints to enforce the sequencing logic. + +The task is to determine the optimal schedule for all operations in both jobs, ensuring that all precedence and machine capacity constraints are satisfied, and that the total production time (makespan) is minimized. This involves deciding the order of operations on shared machines and ensuring that all operations are completed as quickly as possible.",32.0 +['AircraftLanding'],"The problem involves scheduling the landing times of three aircraft (Aircraft 0, Aircraft 1, and Aircraft 2) to minimize the total penalty costs associated with landing early or late. Each aircraft has a target landing time, and the actual landing time must fall within a specified earliest and latest possible landing time window. Aircraft 0 must land between 24 and 84 minutes, Aircraft 1 between 180 and 240 minutes, and Aircraft 2 between 215 and 275 minutes. The penalties for landing early are \$21, \$16, and \$49 per minute for Aircraft 0, 1, and 2, respectively. The penalties for landing late are \$57, \$34, and \$45 per minute for Aircraft 0, 1, and 2, respectively. The target landing times are defined such that Aircraft 0 should land at 54 minutes, Aircraft 1 should land at 210 minutes, and Aircraft 2 should land at 245 minutes. + +To ensure safe landings, separation times between aircraft must be maintained based on the order in which they land. Binary variables determine the order of landings between any two aircraft. If Aircraft 0 lands before Aircraft 1, there must be at least 5 minutes between their landings; if Aircraft 1 lands before Aircraft 0, the separation must be at least 96 minutes. If Aircraft 0 lands before Aircraft 2, the separation must be at least 4 minutes; if Aircraft 2 lands before Aircraft 0, the separation must be at least 131 minutes. If Aircraft 1 lands before Aircraft 2, the separation must be at least 5 minutes; if Aircraft 2 lands before Aircraft 1, the separation must be at least 25 minutes. + +The objective is to determine the landing times for each aircraft that minimize the total penalty costs, which are calculated as the sum of the early and late penalties for all aircraft. The early penalty for each aircraft is the number of minutes it lands before its target time multiplied by its early penalty rate, and the late penalty is the number of minutes it lands after its target time multiplied by its late penalty rate. The solution must ensure that all aircraft land within their respective time windows and that the separation times between landings are respected based on the order in which the aircraft land. The decision variables include the actual landing times for each aircraft and binary variables that determine the order of landings between any two aircraft. The goal is to find the optimal landing schedule that minimizes the total penalty costs while satisfying all constraints.",0.0 +['Job Shop'],"In a manufacturing facility, you are tasked with scheduling the production of two distinct jobs, Job 0 and Job 1, across a set of machines. Each job consists of a sequence of operations that must be performed in a specific order, and each operation requires a certain amount of time to complete. The goal is to minimize the total production time, known as the makespan, while ensuring that all operations are scheduled without overlapping on the same machine and that the sequence of operations within each job is respected. + +Job 0 consists of two operations: Operation 0 takes 5 units of time to complete, and Operation 1 can only start after Operation 0 has finished. The makespan must be at least 5 units of time after the start of Operation 0 and at least 2 units of time after the start of Operation 1. + +Job 1 consists of five operations: Operation 0 takes 9 units of time to complete, Operation 1 takes 9 units of time and can only start after Operation 0 has finished, Operation 2 takes 7 units of time and can only start after Operation 1 has finished, Operation 3 takes 2 units of time and can only start after Operation 2 has finished, and Operation 4 takes 8 units of time and can only start after Operation 3 has finished. The makespan must be at least 9 units of time after the start of Operation 0, at least 9 units of time after the start of Operation 1, at least 7 units of time after the start of Operation 2, at least 2 units of time after the start of Operation 3, and at least 8 units of time after the start of Operation 4. + +Certain operations must be performed on the same machine, and their scheduling must avoid overlaps. + +The primary objective is to minimize the makespan, which is the total time taken to complete all operations across both jobs. The makespan must be at least as long as the completion time of the last operation in either job. All operations must start at or after time 0, and the order of operations within each job must be strictly followed. The scheduling of operations on shared machines must respect the no-overlap constraints described above. Binary variables are used to enforce the no-overlap constraints on the shared machines, ensuring that the operations do not overlap and the sequence constraints are satisfied. + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 2 + - Operation 1 of Job 1: Machine 2 + - Operation 2 of Job 1: Machine 2 + - Operation 3 of Job 1: Machine 1 + - Operation 4 of Job 1: Machine 1",37.0 +['AircraftLanding'],"There are two aircraft, **aircraft_0** and **aircraft_1**, that need to be scheduled for landing. The goal is to minimize the total penalty costs associated with deviations from their target landing times. **aircraft_0** must land between 0 and 30 units of time, while **aircraft_1** must land between 128 and 188 units of time. The target landing time for **aircraft_0** is 0 units, and for **aircraft_1**, it is 158 units. Penalties are incurred for deviations from the target times: **aircraft_0** incurs a penalty of 42 units per unit time for landing early and 32 units per unit time for landing late, while **aircraft_1** incurs a penalty of 62 units per unit time for landing early and 55 units per unit time for landing late. + +To ensure safety, a minimum separation time must be maintained between the landings of the two aircraft. If **aircraft_0** lands before **aircraft_1**, there must be at least 3 units of separation time. If **aircraft_1** lands before **aircraft_0**, there must be at least 98 units of separation time. The order of landings is determined by binary variables **AircraftOrder[aircraft_0,aircraft_1]** and **AircraftOrder[aircraft_1,aircraft_0]**, which are mutually exclusive and must sum to 1. The landing times must respect the separation requirements based on the landing order, and each aircraft must land within its specified time window. + +The early and late deviations for each aircraft are calculated based on the difference between the actual landing time and the target time. For **aircraft_0**, the early deviation is non-negative, and the late deviation is non-negative. For **aircraft_1**, the early deviation is calculated as the difference between the target time and the actual landing time, and the late deviation is calculated as the difference between the actual landing time and the target time. The objective is to minimize the total penalty costs, which are calculated as the sum of the penalties for early and late deviations for both aircraft.",0.0 +['Job Shop'],"The manufacturing facility produces custom parts for two jobs, Job 0 and Job 1, each with a sequence of operations that must follow specific precedence constraints. Job 0 has two operations: Operation 0 with duration 2, and Operation 1 with duration 4. Job 1 has five operations: Operation 0 with duration 4, Operation 1 with duration 2, Operation 2 with duration 1, Operation 3 with duration 3, and Operation 4 with duration 7. For each job, all of its operations need to be processed in order, and one operation can only start once the previous operation has been completed. In particular, for Job 0, Operation 1 must start at least 2 hours after Operation 0 begins. For Job 1, Operation 1 must start at least 4 hours after Operation 0 begins, Operation 2 must start at least 2 hours after Operation 1 begins, Operation 3 must start at least 1 hour after Operation 2 begins, and Operation 4 must start at least 3 hours after Operation 3 begins. + +There are 3 machines (M0, M1, M2). Operations from different jobs share machines: Operation 0 of Job 0, Operation 0 of Job 1, Operation 1 of Job 0, and Operation 1 of Job 1 share machine M0. Operation 2 of Job 1 and Operation 4 of Job 1 share machine M1. Operation 3 of Job 1 is on machine M2. To prevent overlapping, binary variables determine the order of operations on shared machines. If Operation A is scheduled before Operation B on a shared machine, the binary variable is 1; otherwise, it is 0. + +The objective is to minimize the makespan, the total completion time for all operations. The makespan must be at least as large as the completion time of the last operation in either job. Constraints ensure that operations within the same job follow their precedence requirements and that operations sharing a machine do not overlap. + +The goal is to determine the optimal schedule for all operations across both jobs, ensuring all constraints are met while minimizing the makespan.",17.0 +['Job Shop'],"The problem involves scheduling four jobs (Job 0, Job 1, Job 2, and Job 3) on a set of machines to minimize the total completion time (makespan). Each job consists of a sequence of operations, and each operation must be processed on a specific machine with a given processing time. Job 0 has two operations: Operation 0 takes 2 units of time, and Operation 1 takes 3 units of time. Job 1 has five operations: Operation 0 takes 8 units of time, Operation 1 takes 7 units of time, Operation 2 takes 4 units of time, Operation 3 takes 3 units of time, and Operation 4 takes 2 units of time. Job 2 has five operations: Operation 0 takes 8 units of time, Operation 1 takes 3 units of time, Operation 2 takes 7 units of time, Operation 3 takes 8 units of time, and Operation 4 takes 4 units of time. Job 3 has seven operations: Operation 0 takes 6 units of time, Operation 1 takes 4 units of time, Operation 2 takes 7 units of time, Operation 3 takes 6 units of time, Operation 4 takes 2 units of time, Operation 5 takes 8 units of time, and Operation 6 takes 9 units of time. + +Each operation in a job must start only after the previous operation in the same job has completed. For example, in Job 1, Operation 1 cannot start until Operation 0 has finished, and in Job 3, Operation 6 cannot start until Operation 5 has finished. No two operations can be processed on the same machine at the same time. For example, if Operation 3 of Job 1 and Operation 4 of Job 2 are assigned to the same machine, they cannot overlap, and the start time of one must be at least 4 units after the other, depending on the order in which they are scheduled. Similarly, Operation 1 of Job 1 and Operation 2 of Job 3 cannot overlap on the same machine, and the start time of one must be at least 7 units after the other. + +The total completion time (makespan) must be at least as large as the completion time of every operation. For example, the makespan must be at least 2 units to account for the completion of Operation 0 in Job 0, and it must also be at least 9 units to account for the completion of Operation 6 in Job 3. The objective is to minimize the makespan, which involves determining the start times for each operation on each machine while respecting the precedence and machine capacity constraints. All start times must be non-negative, and the order of operations on each machine is determined by binary decision variables, which decide whether one operation starts before another on the same machine. A large constant (1000) is used in the machine capacity constraints to ensure that the constraints are only active when the binary variable is 1 or 0. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3} +- For each job j in J the operations are indexed from 0 to (n_j - 1). We denote operation (j,k) where k is the operation index in job j. +- Machines: M = {0, 1, 2} + +2) Machine assignment for every operation +- Job 0: + - (0,0) -> Machine 0 + - (0,1) -> Machine 1 + +- Job 1: + - (1,0) -> Machine 0 + - (1,1) -> Machine 2 + - (1,2) -> Machine 1 + - (1,3) -> Machine 0 + - (1,4) -> Machine 2 + +- Job 2: + - (2,0) -> Machine 2 + - (2,1) -> Machine 1 + - (2,2) -> Machine 0 + - (2,3) -> Machine 2 + - (2,4) -> Machine 1 + +- Job 3: + - (3,0) -> Machine 1 + - (3,1) -> Machine 0 + - (3,2) -> Machine 2 + - (3,3) -> Machine 1 + - (3,4) -> Machine 0 + - (3,5) -> Machine 2 + - (3,6) -> Machine 1 + +3) Processing times (duration) for every operation (repeated from above for clarity) +- Job 0: p[(0,0)] = 2, p[(0,1)] = 3 +- Job 1: p[(1,0)] = 8, p[(1,1)] = 7, p[(1,2)] = 4, p[(1,3)] = 3, p[(1,4)] = 2 +- Job 2: p[(2,0)] = 8, p[(2,1)] = 3, p[(2,2)] = 7, p[(2,3)] = 8, p[(2,4)] = 4 +- Job 3: p[(3,0)] = 6, p[(3,1)] = 4, p[(3,2)] = 7, p[(3,3)] = 6, p[(3,4)] = 2, p[(3,5)] = 8, p[(3,6)] = 9",46.0 +['AircraftLanding'],"The problem involves scheduling the landing times of six aircraft at a busy airport to minimize the total penalty costs associated with deviations from their target landing times while ensuring safe separation between landings. Each aircraft has a target landing time, earliest and latest possible landing times, and penalty costs for landing early or late. The landing times must fall within the specified bounds for each aircraft: Aircraft 0 between 8 and 68 minutes, Aircraft 1 between 46 and 106 minutes, Aircraft 2 between 90 and 150 minutes, Aircraft 3 between 16 and 76 minutes, Aircraft 4 between 0 and 57 minutes, and Aircraft 5 between 39 and 99 minutes. The target landing times are 38 for Aircraft 0, 76 for Aircraft 1, 120 for Aircraft 2, 46 for Aircraft 3, 28 for Aircraft 4, and 69 for Aircraft 5. The penalties are calculated based on the deviation from the target landing times, with specific costs per minute for early or late landings: 55 for early landing of Aircraft 0, 94 for Aircraft 1, 61 for Aircraft 2, 64 for Aircraft 3, 74 for Aircraft 4, and 84 for Aircraft 5; and 38 for late landing of Aircraft 0, 81 for Aircraft 1, 71 for Aircraft 2, 80 for Aircraft 3, 53 for Aircraft 4, and 84 for Aircraft 5. The binary decision variables ensure that for each pair of aircraft, one lands before the other, and the separation times are enforced based on the landing order. The separation constraints include specific times such as 22 minutes between Aircraft 0 and 1, 22 minutes between Aircraft 0 and 2, 5 minutes between Aircraft 0 and 3, 3 minutes between Aircraft 0 and 4, 3 minutes between Aircraft 0 and 5, 4 minutes between Aircraft 1 and 2, 3 minutes between Aircraft 1 and 3, 1 minute between Aircraft 1 and 4, 5 minutes between Aircraft 1 and 5, 2 minutes between Aircraft 2 and 3, 1 minute between Aircraft 2 and 4, 5 minutes between Aircraft 2 and 5, 5 minutes between Aircraft 3 and 4, 3 minutes between Aircraft 3 and 5, and 5 minutes between Aircraft 4 and 5. The goal is to determine the actual landing times for all six aircraft such that the total penalty cost is minimized, and all separation and landing time constraints are satisfied. ",0.0 +['AircraftLanding'],"The problem involves scheduling the landing times of three aircraft (aircraft_0, aircraft_1, and aircraft_2) to minimize the total penalty costs associated with deviations from their target landing times. Each aircraft has specific time windows within which it must land: aircraft_0 must land between 19 and 79 minutes, aircraft_1 between 58 and 118 minutes, and aircraft_2 between 205 and 265 minutes. The target times are 49 for aircraft_0, 88 for aircraft_1, and 235 for aircraft_2. The penalties for landing early or late are as follows: aircraft_0 incurs a penalty of 58 units per minute early and 57 units per minute late, aircraft_1 incurs 25 units per minute early and 61 units per minute late, and aircraft_2 incurs 37 units per minute early and 70 units per minute late. + +The landing sequence between any two aircraft is determined by binary variables, where a value of 1 indicates that one aircraft lands before the other. The constraints ensure that the landing times respect the sequence and separation requirements. For example, if aircraft_0 lands before aircraft_1, there must be at least 4 minutes of separation, but if aircraft_1 lands before aircraft_0, the required separation is 21 minutes. Similarly, if aircraft_0 lands before aircraft_2, the separation must be at least 12 minutes, but if aircraft_2 lands before aircraft_0, the separation must be at least 126 minutes. For aircraft_1 and aircraft_2, the separation is at least 13 minutes if aircraft_1 lands first, and at least 87 minutes if aircraft_2 lands first. + +The objective is to minimize the total penalty costs, which include the penalties for early and late landings for all three aircraft. The constraints ensure that each aircraft lands within its specified time window, the landing sequence respects the separation requirements, and the binary variables enforce the correct sequence between aircraft pairs. The penalties for early or late landings are calculated based on the deviation from the target landing times.",0.0 +['AircraftLanding'],"The problem involves scheduling the landing times of five aircraft to minimize the total penalty costs associated with deviations from their target landing times, while ensuring that each aircraft lands within its specified time window and that the required separation times between consecutive landings are maintained. Each aircraft has a target landing time: Aircraft 0's target is 200, aircraft 1's target is 65, aircraft 2's target is 283, aircraft 3's target is 20, aircraft 4's target is 99. Each aircraft also has an earliest possible landing time, and a latest possible landing time. Aircraft 0 must land between 178 and 238, aircraft 1 between 35 and 95, aircraft 2 between 266 and 300, aircraft 3 between 0 and 40, and aircraft 4 between 69 and 129. Penalties are incurred for landing early or late, with specific costs per unit time for each aircraft: aircraft 0 has an early penalty of 62 and a late penalty of 82, aircraft 1 has an early penalty of 10 and a late penalty of 65, aircraft 2 has an early penalty of 89 and a late penalty of 72, aircraft 3 has an early penalty of 81 and a late penalty of 69, and aircraft 4 has an early penalty of 50 and a late penalty of 42. The separation requirements between aircraft depend on the order in which they land, and these requirements are enforced using binary decision variables that determine the landing sequence. The separation times are as follows: between aircraft 0 and 1, the required separation is 203 if 0 lands before 1 and 3 if 1 lands before 0; between aircraft 0 and 2, the required separation is 2 if 0 lands before 2 and 28 if 2 lands before 0; between aircraft 0 and 3, the required separation is 4 if 0 lands before 3 and 238 if 3 lands before 0; between aircraft 0 and 4, the required separation is 5 if 0 lands before 4 and 169 if 4 lands before 0; between aircraft 1 and 2, the required separation is 3 if 1 lands before 2 and 171 if 2 lands before 1; between aircraft 1 and 3, the required separation is 2 if 1 lands before 3 and 95 if 3 lands before 1; between aircraft 1 and 4, the required separation is 4 if 1 lands before 4 and 26 if 4 lands before 1; between aircraft 2 and 3, the required separation is 5 if 2 lands before 3 and 300 if 3 lands before 2; between aircraft 2 and 4, the required separation is 2 if 2 lands before 4 and 231 if 4 lands before 2; between aircraft 3 and 4, the required separation is 5 if 3 lands before 4 and 29 if 4 lands before 3. The objective is to minimize the sum of the penalties for early and late landings across all aircraft. The constraints include ensuring that each aircraft lands within its time window, enforcing the separation times based on the landing order, and calculating the penalties for deviations from the target landing times. The binary decision variables ensure that only one of the two possible orders between any pair of aircraft is chosen, and the separation constraints are applied accordingly. The landing times must also satisfy bounds that are specific to each aircraft. The penalties are calculated based on the difference between the actual landing time and the target landing time, with separate variables for early and late deviations. The goal is to find the optimal landing times and sequence that minimize the total penalty costs while satisfying all constraints.",9410.0 +['Job Shop'],"There are six jobs, each with a specific sequence of operations that must be performed on shared machines to complete their tasks. The goal is to schedule these operations to minimize the total time (makespan) required to complete all jobs. Each job has a predefined sequence of operations with specific processing times. Job 0 has two operations: the first takes 1 unit of time, and the second takes 6 units. Job 1 has two operations: the first takes 4 units, and the second takes 2 units. Job 2 has three operations: the first takes 9 units, the second takes 2 units, and the third takes 2 units. Job 3 has four operations: the first takes 5 units, the second takes 7 units, the third takes 6 units, and the fourth takes 3 units. Job 4 has four operations: the first takes 2 units, the second takes 9 units, the third takes 8 units, and the fourth takes 7 units. Job 5 has two operations: the first takes 4 units, and the second takes 8 units. + +Operations from different jobs may share the same machine, and the order in which they are processed must be determined. For example, the second operation of Job 0 and the fourth operation of Job 3 share a machine. If Job 0’s second operation starts first, it must finish before Job 3’s fourth operation begins, or vice versa. Similarly, the second operation of Job 1 and the second operation of Job 2 share a machine. If Job 1’s second operation starts first, it must finish before Job 2’s second operation begins, or vice versa. + +Each job’s operations must be performed in a specific order. For Job 0, the second operation cannot start until the first operation is completed. For Job 1, the second operation cannot start until the first operation is completed. For Job 2, the second operation cannot start until the first operation is completed, and the third operation cannot start until the second operation is completed. For Job 3, the second operation cannot start until the first operation is completed, the third operation cannot start until the second operation is completed, and the fourth operation cannot start until the third operation is completed. For Job 4, the second operation cannot start until the first operation is completed, the third operation cannot start until the second operation is completed, and the fourth operation cannot start until the third operation is completed. For Job 5, the second operation cannot start until the first operation is completed. + +The objective is to minimize the total time (makespan) required to complete all operations for all jobs. The makespan is determined by the latest completion time among all operations. The constraints include precedence constraints, which ensure that each operation starts only after its preceding operation in the same job’s sequence is completed. For example, Job 0’s second operation must start at least 1 unit of time after the first operation starts. Job 1’s second operation must start at least 4 units of time after the first operation starts. Job 2’s second operation must start at least 9 units of time after the first operation starts, and the third operation must start at least 2 units of time after the second operation starts. Job 3’s second operation must start at least 5 units of time after the first operation starts, the third operation must start at least 7 units of time after the second operation starts, and the fourth operation must start at least 6 units of time after the third operation starts. Job 4’s second operation must start at least 2 units of time after the first operation starts, the third operation must start at least 9 units of time after the second operation starts, and the fourth operation must start at least 8 units of time after the third operation starts. Job 5’s second operation must start at least 4 units of time after the first operation starts. + +Machine capacity constraints ensure that operations sharing the same machine do not overlap. For example, if Job 0’s second operation starts first on a shared machine, it must finish before Job 3’s fourth operation begins, or vice versa. A large constant (100,000) is used to enforce the order of operations on shared machines. Non-negativity constraints ensure that all operation start times are non-negative. Binary decision variables determine the order of operations on shared machines. + +The processing times for each operation are fixed and provided. The precedence constraints specify the minimum time gaps between operations. The machine capacity constraints ensure that operations on shared machines do not overlap. The decisions to be made include determining the start time for each operation for all jobs, deciding the order of operations on shared machines using binary variables, and ensuring that all constraints are satisfied while minimizing the total makespan. The objective is to minimize the total time (makespan) required to complete all operations for all jobs, ensuring that all precedence and machine capacity constraints are met.",26.0 +['Job Shop'],"The manufacturing facility produces custom parts for two jobs, Job 0 and Job 1, each consisting of operations performed on machines. The goal is to minimize the total completion time (makespan) by scheduling operations efficiently while respecting constraints. +Job 0 has three operations: Operation 0 takes 2 hours, Operation 1 takes 8 hours, and Operation 2 takes 5 hours. Job 1 has two operations: Operation 0 takes 1 hour, and Operation 1 takes 7 hours. +The makespan must be at least as large as the completion time of every operation: Operation 0 of Job 0 takes 2 hours, Operation 1 of Job 0 takes 8 hours, Operation 2 of Job 0 takes 5 hours, Operation 0 of Job 1 takes 1 hour, and Operation 1 of Job 1 takes 7 hours. Each machine can only process one operation at a time, and the operations of a job must be processes sequentially. For example, operation 1 of job 0 can only be processed after the completion of operation 0 of job 0. +The objective is to determine the start times for all operations and their processing order on shared machines to minimize the makespan while satisfying all constraints. + +Additional Information: + +There are 3 machines: M1, M2, M3 +Assignment of operations to machines: +Job 0: +Operation 0 → Machine M1 +Operation 1 → Machine M2 +Operation 2 → Machine M3 + +Job 1: +Operation 0 → Machine M2 +Operation 1 → Machine M1",15.0 +['Job Shop'],"The problem involves scheduling operations for three jobs (Job 0, Job 1, and Job 2) on a set of machines in a manufacturing facility to minimize the total completion time (makespan). Each job consists of a sequence of operations with specific processing times. Job 0 has 5 operations with processing times of 7, 9, 8, 3, and 9 hours. Job 1 has 6 operations with processing times of 1, 3, 4, 1, 2, and 9 hours. Job 2 has 7 operations with processing times of 9, 3, 1, 3, 3, 3, and 5 hours. Each operation must start only after the previous operation in the same job has been completed. For example, Operation 1 of Job 0 cannot start until Operation 0 has been completed (7 hours), and Operation 1 of Job 1 cannot start until Operation 0 has been completed (1 hour). Operations from different jobs that require the same machine must be scheduled in a specific order to avoid overlapping. Binary decision variables determine the order of operations on shared machines. For instance, if Operation 0 of Job 0 is scheduled before Operation 0 of Job 1, the binary variable is set to 1; otherwise, it is set to 0. A large constant (1000) is used in the constraints to ensure they are only active when necessary. The makespan must be at least as long as the completion time of the last operation across all jobs. The start times for all operations must be non-negative, and the objective is to minimize the makespan while satisfying all precedence and machine capacity constraints. Specifically, the constraints include precedence constraints within each job, machine capacity constraints to prevent overlapping operations on the same machine, and makespan constraints to ensure the makespan is at least as large as the completion time of every operation. The binary variables are used to enforce the order of operations on shared machines, and the large constant ensures the constraints are only active when necessary. The start times for all operations must be non-negative, and the objective is to minimize the makespan while satisfying all constraints. + +ADDITIONAL INPUT DATA (explicit, needed to implement Gurobi model): + +1) Sets and indices +- Jobs: J = {0, 1, 2} +- Operations per job (indexed by k starting at 0): + - Job 0: K0 = {0, 1, 2, 3, 4} + - Job 1: K1 = {0, 1, 2, 3, 4, 5} + - Job 2: K2 = {0, 1, 2, 3, 4, 5, 6} + +2) Machines (explicit list) +- Machines: M = {M0, M1, M2, M3} + +3) Machine assignment for every operation (explicit mapping) +- Job 0 operations: + - (0,0) -> M0 + - (0,1) -> M1 + - (0,2) -> M2 + - (0,3) -> M0 + - (0,4) -> M3 +- Job 1 operations: + - (1,0) -> M1 + - (1,1) -> M2 + - (1,2) -> M0 + - (1,3) -> M3 + - (1,4) -> M1 + - (1,5) -> M2 +- Job 2 operations: + - (2,0) -> M2 + - (2,1) -> M0 + - (2,2) -> M1 + - (2,3) -> M3 + - (2,4) -> M2 + - (2,5) -> M1 + - (2,6) -> M3 + +4) Processing times (repeated here explicitly per operation) +- Job 0 durations: + - p(0,0) = 7 + - p(0,1) = 9 + - p(0,2) = 8 + - p(0,3) = 3 + - p(0,4) = 9 +- Job 1 durations: + - p(1,0) = 1 + - p(1,1) = 3 + - p(1,2) = 4 + - p(1,3) = 1 + - p(1,4) = 2 + - p(1,5) = 9 +- Job 2 durations: + - p(2,0) = 9 + - p(2,1) = 3 + - p(2,2) = 1 + - p(2,3) = 3 + - p(2,4) = 3 + - p(2,5) = 3 + - p(2,6) = 5",41.0 +['Job Shop'],"The problem involves scheduling four jobs (Job 0, Job 1, Job 2, and Job 3) across multiple machines to minimize the total completion time (makespan). Each job consists of a sequence of operations that must be performed in a specific order, and each operation is assigned to a particular machine. The processing times for each operation are as follows: Job 0 has Operation 0 (6 units), Operation 1 (8 units), and Operation 2 (9 units); Job 1 has Operation 0 (3 units), Operation 1 (8 units), Operation 2 (5 units), Operation 3 (5 units), and Operation 4 (2 units); Job 2 has Operation 0 (8 units), Operation 1 (1 unit), and Operation 2 (6 units); Job 3 has Operation 0 (1 unit), Operation 1 (9 units), Operation 2 (4 units), and Operation 3 (5 units). + +Each machine can process only one operation at a time. If two operations are assigned to the same machine, they cannot overlap in time. The order of operations on the same machine is determined by binary decision variables. For example, if Operation 0 of Job 0 and Operation 3 of Job 1 are on the same machine, the binary variable determines whether Operation 0 of Job 0 is processed before Operation 3 of Job 1 or vice versa. Similar constraints apply to other operation pairs on the same machine, such as Operation 2 of Job 1 and Operation 4 of Job 1, Operation 2 of Job 0 and Operation 0 of Job 2, Operation 2 of Job 0 and Operation 3 of Job 3, Operation 0 of Job 2 and Operation 3 of Job 3, Operation 0 of Job 1 and Operation 1 of Job 2, Operation 0 of Job 1 and Operation 2 of Job 3, and Operation 1 of Job 2 and Operation 2 of Job 3. + +The objective is to minimize the makespan, which must be at least as large as the completion time of every operation. For example, the completion time of Operation 0 of Job 0 is at least 6 units, so the makespan must be at least 6 units. Similarly, the completion time of Operation 1 of Job 0 is at least 8 units, so the makespan must be at least 8 units. This applies to all operations, with the final operation of Job 3 requiring at least 5 units. + +Constraints include precedence constraints within jobs, ensuring that each operation starts only after the previous operation in the same job has completed. For example, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 has completed, which takes at least 6 units. Machine capacity constraints ensure that operations assigned to the same machine do not overlap, with binary variables determining the order of operations on the same machine. The makespan must be at least as large as the completion time of every operation, and all start times for operations must be non-negative. Binary variables must be either 0 or 1, determining the order of operations on the same machine. + +The goal is to find the optimal schedule that meets all these constraints while minimizing the total time required to complete all jobs. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3} +- For each job j in J, operations are indexed sequentially starting at 0. We denote operation (j,k) for job j and operation index k. +- Machines: M = {0, 1, 2} # machines are indexed 0..2 + +2) Machine assignment for every operation (explicit mapping) +- Job 0: Operation (0,0) -> Machine 0 +- Job 0: Operation (0,1) -> Machine 1 +- Job 0: Operation (0,2) -> Machine 2 + +- Job 1: Operation (1,0) -> Machine 0 +- Job 1: Operation (1,1) -> Machine 1 +- Job 1: Operation (1,2) -> Machine 2 +- Job 1: Operation (1,3) -> Machine 1 +- Job 1: Operation (1,4) -> Machine 2 + +- Job 2: Operation (2,0) -> Machine 2 +- Job 2: Operation (2,1) -> Machine 0 +- Job 2: Operation (2,2) -> Machine 1 + +- Job 3: Operation (3,0) -> Machine 0 +- Job 3: Operation (3,1) -> Machine 2 +- Job 3: Operation (3,2) -> Machine 1 +- Job 3: Operation (3,3) -> Machine 2 + +3) Processing durations (explicit list matching the original description) +- Job 0: durations = [6, 8, 9] +- Job 1: durations = [3, 8, 5, 5, 2] +- Job 2: durations = [8, 1, 6] +- Job 3: durations = [1, 9, 4, 5]",38.0 +"['Multicommodity Capacitated Network Design', 'Network Optimization']","Imagine you are managing a supply chain network for a company that operates across six key locations (nodes). Each location has specific supply and demand requirements, and your goal is to optimize the flow of goods between these locations while minimizing the total cost, which includes both fixed costs for establishing connections and variable costs for shipping goods. + +#### Supply and Demand: +- **Location 0** has a demand of 116 units. +- **Location 1** has a demand of 134 units. +- **Location 2** has no net supply or demand (balance is 0). +- **Location 3** has a demand of 141 units. +- **Location 4** has a demand of 8 units. +- **Location 5** has a demand of 101 units. + +#### Costs and Capacities: +To establish a connection between any two locations, there is a fixed cost. Additionally, shipping goods between locations incurs a variable cost per unit. Each connection also has a maximum capacity, limiting the amount of goods that can be shipped. + +Here are the specific costs and capacities for each possible connection: + +- **Fixed Costs (for establishing connections):** + - From Location 0: To Location 1 costs 1972, to Location 2 costs 2030, to Location 3 costs 2048, to Location 4 costs 1951, to Location 5 costs 2043. + - From Location 1: To Location 0 costs 1993, to Location 2 costs 2029, to Location 3 costs 1975, to Location 4 costs 1967, to Location 5 costs 2026. + - From Location 2: To Location 0 costs 2042, to Location 1 costs 1973, to Location 3 costs 2001, to Location 4 costs 2034, to Location 5 costs 2035. + - From Location 3: To Location 0 costs 2007, to Location 1 costs 1968, to Location 2 costs 2004, to Location 4 costs 1996, to Location 5 costs 1969. + - From Location 4: To Location 0 costs 1973, to Location 1 costs 1955, to Location 2 costs 2038, to Location 3 costs 1953, to Location 5 costs 2000. + - From Location 5: To Location 0 costs 1999, to Location 1 costs 2031, to Location 2 costs 2011, to Location 3 costs 1965, to Location 4 costs 1984. + +- **Variable Costs (per unit shipped):** + - From Location 0: To Location 1 costs 21, to Location 2 costs 19, to Location 3 costs 20, to Location 4 costs 21, to Location 5 costs 21. + - From Location 1: To Location 0 costs 21, to Location 2 costs 19, to Location 3 costs 21, to Location 4 costs 19, to Location 5 costs 20. + - From Location 2: To Location 0 costs 20, to Location 1 costs 20, to Location 3 costs 21, to Location 4 costs 21, to Location 5 costs 20. + - From Location 3: To Location 0 costs 19, to Location 1 costs 19, to Location 2 costs 20, to Location 4 costs 19, to Location 5 costs 20. + - From Location 4: To Location 0 costs 20, to Location 1 costs 19, to Location 2 costs 20, to Location 3 costs 20, to Location 5 costs 19. + - From Location 5: To Location 0 costs 21, to Location 1 costs 20, to Location 2 costs 20, to Location 3 costs 19, to Location 4 costs 21. + +- **Maximum Capacities (units that can be shipped):** + - From Location 0: To Location 1 can handle up to 137 units, to Location 2 up to 154 units, to Location 3 up to 189 units, to Location 4 up to 188 units, to Location 5 up to 247 units. + - From Location 1: To Location 0 up to 163 units, to Location 2 up to 197 units, to Location 3 up to 242 units, to Location 4 up to 191 units, to Location 5 up to 135 units. + - From Location 2: To Location 0 up to 226 units, to Location 1 up to 155 units, to Location 3 up to 182 units, to Location 4 up to 189 units, to Location 5 up to 160 units. + - From Location 3: To Location 0 up to 179 units, to Location 1 up to 232 units, to Location 2 up to 210 units, to Location 4 up to 146 units, to Location 5 up to 241 units. + - From Location 4: To Location 0 up to 238 units, to Location 1 up to 164 units, to Location 2 up to 238 units, to Location 3 up to 218 units, to Location 5 up to 195 units. + - From Location 5: To Location 0 up to 237 units, to Location 1 up to 191 units, to Location 2 up to 165 units, to Location 3 up to 133 units, to Location 4 up to 191 units. + +#### Objective: +Your task is to determine which connections to establish and how much to ship between locations to meet all demand requirements while minimizing the total cost. The total cost includes both the fixed costs of establishing connections and the variable costs of shipping goods. + +#### Constraints: +1. **Capacity Constraints:** The amount of goods shipped between any two locations cannot exceed the maximum capacity of that connection. +2. **Flow Balance:** The total amount of goods entering a location minus the total amount leaving must equal the net supply or demand at that location. +3. **Binary Decisions:** For each connection, you must decide whether to establish it (yes or no), which incurs the fixed cost. + +By carefully analyzing these factors, you aim to find the most cost-effective way to manage the supply chain network, ensuring all locations receive the goods they need without exceeding capacity limits or incurring unnecessary costs.",13033.0 +"['Facility Location Problem', 'Set Cover']","You are tasked with optimizing the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 33 distinct areas, each with a specific population size that needs to be covered by cell tower signals. The goal is to determine which cell towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + +For each of the 29 available cell tower locations (tower_0 to tower_28), you must decide whether to build the tower (a binary decision: build or not build). For each of the 33 regions (region_0 to region_32), you must decide whether the region will be covered by at least one active cell tower (a binary decision: cover or not cover). + +The objective is to maximize the total population covered by the deployed cell towers. The population sizes for the regions are as follows: Region 0: 4,901; Region 1: 2,727; Region 2: 2,994; Region 3: 2,984; Region 4: 3,693; Region 5: 2,615; Region 6: 3,354; Region 7: 4,965; Region 8: 4,223; Region 9: 4,738; Region 10: 2,345; Region 11: 3,915; Region 12: 4,068; Region 13: 2,613; Region 14: 3,140; Region 15: 4,894; Region 16: 3,503; Region 17: 3,733; Region 18: 4,275; Region 19: 3,795; Region 20: 3,803; Region 21: 3,497; Region 22: 2,403; Region 23: 3,441; Region 24: 3,116; Region 25: 2,091; Region 26: 4,494; Region 27: 3,593; Region 28: 2,090; Region 29: 3,397; Region 30: 3,973; Region 31: 2,362; Region 32: 2,068. + +Each region can only be considered covered if at least one of the cell towers that can serve it is built. - Region 0 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 7, Tower 9, Tower 16, Tower 17, Tower 20, Tower 22, Tower 24, Tower 25, Tower 26, Tower 27. + - Region 1 can be covered by towers Tower 0, Tower 1, Tower 15, Tower 19, Tower 22, Tower 23, Tower 25, Tower 26, Tower 27, Tower 28. + - Region 2 can be covered by towers Tower 1, Tower 3, Tower 5, Tower 6, Tower 9, Tower 10, Tower 15, Tower 16, Tower 18, Tower 19, Tower 20, Tower 21, Tower 23, Tower 25. + - Region 3 can be covered by towers Tower 2, Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 11, Tower 13, Tower 16, Tower 17, Tower 18. + - Region 4 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 4, Tower 7, Tower 8, Tower 9, Tower 13, Tower 15, Tower 18, Tower 19, Tower 25, Tower 26, Tower 27. + - Region 5 can be covered by towers Tower 1, Tower 2, Tower 5, Tower 6, Tower 11, Tower 13, Tower 16, Tower 18, Tower 20, Tower 21, Tower 23, Tower 25, Tower 26. + - Region 6 can be covered by towers Tower 1, Tower 2, Tower 5, Tower 6, Tower 7, Tower 10, Tower 11, Tower 14, Tower 25, Tower 26, Tower 27. + - Region 7 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 9, Tower 12, Tower 15, Tower 16, Tower 19, Tower 21, Tower 22, Tower 26, Tower 27. + - Region 8 can be covered by towers Tower 1, Tower 4, Tower 6, Tower 9, Tower 12, Tower 14, Tower 20, Tower 21, Tower 22, Tower 24, Tower 25, Tower 26. + - Region 9 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 6, Tower 7, Tower 9, Tower 10, Tower 13, Tower 15, Tower 16, Tower 19, Tower 20, Tower 22, Tower 27. + - Region 10 can be covered by towers Tower 3, Tower 5, Tower 7, Tower 11, Tower 13, Tower 15, Tower 18, Tower 19, Tower 20, Tower 21, Tower 27. + - Region 11 can be covered by towers Tower 0, Tower 1, Tower 4, Tower 9, Tower 12, Tower 13, Tower 17, Tower 20, Tower 21, Tower 22, Tower 24, Tower 26, Tower 27. + - Region 12 can be covered by towers Tower 3, Tower 4, Tower 6, Tower 8, Tower 10, Tower 13, Tower 15, Tower 16, Tower 19, Tower 20, Tower 23, Tower 24, Tower 26, Tower 28. + - Region 13 can be covered by towers Tower 0, Tower 3, Tower 8, Tower 9, Tower 11, Tower 14, Tower 17, Tower 20, Tower 21, Tower 22, Tower 24, Tower 25, Tower 26, Tower 27. + - Region 14 can be covered by towers Tower 0, Tower 2, Tower 4, Tower 12, Tower 13, Tower 15, Tower 17, Tower 18, Tower 22, Tower 24, Tower 25. + - Region 15 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 7, Tower 12, Tower 13, Tower 16, Tower 19, Tower 21, Tower 24, Tower 25, Tower 26, Tower 28. + - Region 16 can be covered by towers Tower 1, Tower 2, Tower 9, Tower 11, Tower 18, Tower 19, Tower 22, Tower 23, Tower 24, Tower 25, Tower 26, Tower 27, Tower 28. + - Region 17 can be covered by towers Tower 5, Tower 7, Tower 10, Tower 11, Tower 15, Tower 16, Tower 17, Tower 19, Tower 20, Tower 25. + - Region 18 can be covered by towers Tower 6, Tower 7, Tower 9, Tower 10, Tower 17, Tower 19, Tower 20, Tower 21, Tower 22, Tower 24, Tower 25, Tower 27. + - Region 19 can be covered by towers Tower 0, Tower 2, Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 12, Tower 15, Tower 21, Tower 22, Tower 23, Tower 25, Tower 27. + - Region 20 can be covered by towers Tower 1, Tower 2, Tower 5, Tower 6, Tower 10, Tower 12, Tower 14, Tower 17, Tower 19, Tower 25, Tower 26. + - Region 21 can be covered by towers Tower 3, Tower 7, Tower 9, Tower 10, Tower 12, Tower 14, Tower 16, Tower 20, Tower 22, Tower 24, Tower 25. + - Region 22 can be covered by towers Tower 0, Tower 2, Tower 3, Tower 7, Tower 14, Tower 16, Tower 17, Tower 22, Tower 26, Tower 28. + - Region 23 can be covered by towers Tower 0, Tower 3, Tower 5, Tower 6, Tower 7, Tower 8, Tower 9, Tower 13, Tower 14, Tower 16, Tower 17, Tower 20, Tower 23, Tower 25. + - Region 24 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 6, Tower 8, Tower 9, Tower 10, Tower 14, Tower 20, Tower 22, Tower 25, Tower 26. + - Region 25 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 6, Tower 7, Tower 9, Tower 13, Tower 14, Tower 18, Tower 20, Tower 21, Tower 22, Tower 23, Tower 26, Tower 27. + - Region 26 can be covered by towers Tower 0, Tower 2, Tower 4, Tower 5, Tower 7, Tower 10, Tower 16, Tower 17, Tower 18, Tower 22, Tower 23, Tower 25, Tower 26. + - Region 27 can be covered by towers Tower 0, Tower 6, Tower 8, Tower 11, Tower 12, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 23, Tower 24, Tower 26, Tower 28. + - Region 28 can be covered by towers Tower 0, Tower 2, Tower 4, Tower 5, Tower 6, Tower 12, Tower 15, Tower 16, Tower 18, Tower 19, Tower 21, Tower 24. + - Region 29 can be covered by towers Tower 1, Tower 4, Tower 7, Tower 10, Tower 12, Tower 13, Tower 15, Tower 18, Tower 27, Tower 28. + - Region 30 can be covered by towers Tower 1, Tower 2, Tower 4, Tower 8, Tower 12, Tower 13, Tower 17, Tower 19, Tower 21, Tower 22, Tower 23, Tower 25, Tower 27, Tower 28. + - Region 31 can be covered by towers Tower 0, Tower 6, Tower 8, Tower 9, Tower 12, Tower 16, Tower 17, Tower 18, Tower 22, Tower 27, Tower 28. + - Region 32 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 5, Tower 6, Tower 7, Tower 9, Tower 12, Tower 15, Tower 17, Tower 23, Tower 26. +The total cost of building the selected cell towers must not exceed the available budget of 2,996 units. The costs for building each tower are as follows: Tower 0: 227; Tower 1: 355; Tower 2: 287; Tower 3: 487; Tower 4: 497; Tower 5: 412; Tower 6: 495; Tower 7: 294; Tower 8: 212; Tower 9: 290; Tower 10: 427; Tower 11: 399; Tower 12: 361; Tower 13: 253; Tower 14: 350; Tower 15: 314; Tower 16: 234; Tower 17: 203; Tower 18: 257; Tower 19: 248; Tower 20: 406; Tower 21: 264; Tower 22: 277; Tower 23: 419; Tower 24: 426; Tower 25: 297; Tower 26: 474; Tower 27: 333; Tower 28: 491. + +Your task is to select which cell towers to build and which regions to cover, ensuring that the total population covered is maximized while staying within the budget. Each region can only be covered if at least one of its associated towers is built, and the total cost of building the selected towers must not exceed 2,996 units. The solution must balance the trade-off between coverage and cost to achieve the best possible outcome.",113803.0 +['Minimum Cost Flow Problem'],"You are tasked with managing the distribution of resources across a network of nine cities. Each city has a specific supply or demand for resources, and your goal is to ensure that all cities meet their requirements while minimizing the total transportation cost. The transportation costs vary depending on the route taken between cities, and each route has a maximum capacity limit for resource shipments. + +#### City Requirements: +- **City 0** has a net demand of **41 units**. +- **City 1** has a net demand of **23 units**. +- **City 2** has a net supply of **15 units** (indicated by a negative demand of -15). +- **City 3** has a net demand of **2 units**. +- **City 4** has a net supply of **13 units** (indicated by a negative demand of -13). +- **City 5** has a net supply of **4 units** (indicated by a negative demand of -4). +- **City 6** has a net supply of **7 units** (indicated by a negative demand of -7). +- **City 7** has a net supply of **2 units** (indicated by a negative demand of -2). +- **City 8** has a net supply of **25 units** (indicated by a negative demand of -25). + +#### Transportation Costs and Capacities: +The cost of shipping resources between cities varies, and each route has a maximum capacity. Below are the costs and capacities for each possible route: + +- **From City 0**: + - To City 1: Cost = 7, Capacity = 35 + - To City 2: Cost = 9, Capacity = 60 + - To City 3: Cost = 8, Capacity = 48 + - To City 4: Cost = 8, Capacity = 58 + - To City 5: Cost = 10, Capacity = 39 + - To City 6: Cost = 8, Capacity = 60 + - To City 7: Cost = 6, Capacity = 54 + - To City 8: Cost = 9, Capacity = 31 + +- **From City 1**: + - To City 0: Cost = 10, Capacity = 39 + - To City 2: Cost = 8, Capacity = 31 + - To City 3: Cost = 9, Capacity = 56 + - To City 4: Cost = 8, Capacity = 60 + - To City 5: Cost = 9, Capacity = 37 + - To City 6: Cost = 10, Capacity = 54 + - To City 7: Cost = 9, Capacity = 39 + - To City 8: Cost = 7, Capacity = 50 + +- **From City 2**: + - To City 0: Cost = 10, Capacity = 31 + - To City 1: Cost = 6, Capacity = 35 + - To City 3: Cost = 7, Capacity = 50 + - To City 4: Cost = 7, Capacity = 53 + - To City 5: Cost = 9, Capacity = 36 + - To City 6: Cost = 10, Capacity = 48 + - To City 7: Cost = 8, Capacity = 46 + - To City 8: Cost = 10, Capacity = 60 + +- **From City 3**: + - To City 0: Cost = 7, Capacity = 50 + - To City 1: Cost = 6, Capacity = 38 + - To City 2: Cost = 7, Capacity = 39 + - To City 4: Cost = 10, Capacity = 47 + - To City 5: Cost = 10, Capacity = 37 + - To City 6: Cost = 6, Capacity = 54 + - To City 7: Cost = 7, Capacity = 44 + - To City 8: Cost = 10, Capacity = 55 + +- **From City 4**: + - To City 0: Cost = 9, Capacity = 31 + - To City 1: Cost = 8, Capacity = 31 + - To City 2: Cost = 7, Capacity = 30 + - To City 3: Cost = 7, Capacity = 40 + - To City 5: Cost = 6, Capacity = 60 + - To City 6: Cost = 6, Capacity = 46 + - To City 7: Cost = 7, Capacity = 31 + - To City 8: Cost = 9, Capacity = 49 + +- **From City 5**: + - To City 0: Cost = 7, Capacity = 34 + - To City 1: Cost = 7, Capacity = 30 + - To City 2: Cost = 6, Capacity = 51 + - To City 3: Cost = 8, Capacity = 32 + - To City 4: Cost = 9, Capacity = 33 + - To City 6: Cost = 7, Capacity = 47 + - To City 7: Cost = 7, Capacity = 42 + - To City 8: Cost = 8, Capacity = 43 + +- **From City 6**: + - To City 0: Cost = 8, Capacity = 55 + - To City 1: Cost = 7, Capacity = 48 + - To City 2: Cost = 7, Capacity = 35 + - To City 3: Cost = 7, Capacity = 31 + - To City 4: Cost = 8, Capacity = 43 + - To City 5: Cost = 8, Capacity = 56 + - To City 7: Cost = 6, Capacity = 41 + - To City 8: Cost = 6, Capacity = 55 + +- **From City 7**: + - To City 0: Cost = 7, Capacity = 31 + - To City 1: Cost = 6, Capacity = 31 + - To City 2: Cost = 9, Capacity = 38 + - To City 3: Cost = 8, Capacity = 34 + - To City 4: Cost = 7, Capacity = 31 + - To City 5: Cost = 8, Capacity = 40 + - To City 6: Cost = 9, Capacity = 39 + - To City 8: Cost = 6, Capacity = 42 + +- **From City 8**: + - To City 0: Cost = 8, Capacity = 48 + - To City 1: Cost = 8, Capacity = 60 + - To City 2: Cost = 8, Capacity = 36 + - To City 3: Cost = 10, Capacity = 47 + - To City 4: Cost = 8, Capacity = 32 + - To City 5: Cost = 7, Capacity = 36 + - To City 6: Cost = 6, Capacity = 57 + - To City 7: Cost = 10, Capacity = 35 + +#### Objective: +Your task is to determine the optimal amount of resources to ship between cities to meet all demands and supplies while minimizing the total transportation cost. The shipments must respect the capacity limits of each route, and the total flow of resources into and out of each city must balance according to their respective requirements. + +#### Constraints: +1. **Flow Balance**: For each city, the total resources shipped into the city plus its supply must equal the total resources shipped out of the city plus its demand. +2. **Capacity Limits**: The amount of resources shipped on any route cannot exceed the maximum capacity for that route. +3. **Non-Negativity**: All shipments must be non-negative. + +By carefully planning the shipments, you aim to achieve the lowest possible total cost while ensuring all cities receive the resources they need.",493.0 +"['Minimum Cost Flow Problem', 'Transportation Problem']","Imagine you are managing the distribution of goods across a network of eight cities. Each city has a specific supply or demand for goods, and your goal is to ensure that all cities receive the goods they need while minimizing the total transportation costs. The transportation costs vary depending on the route taken between cities, and each route has a maximum capacity for the amount of goods that can be shipped. + +#### **Cities and Their Requirements:** +- **City 0** has a net demand of **86 units** of goods. +- **City 1** has a net demand of **19 units** of goods. +- **City 2** has a net supply of **23 units** of goods (meaning it can send out 23 units). +- **City 3** has a net supply of **37 units** of goods. +- **City 4** has a net supply of **24 units** of goods. +- **City 5** has a balanced supply and demand, meaning it neither supplies nor demands goods. +- **City 6** has a net demand of **4 units** of goods. +- **City 7** has a net supply of **25 units** of goods. + +#### **Transportation Costs and Capacities:** +The cost of shipping goods between cities varies, and each route has a maximum capacity. Below are the costs and capacities for each possible route: + +- **From City 0:** + - To City 1: Cost = **9**, Capacity = **44** + - To City 2: Cost = **6**, Capacity = **47** + - To City 3: Cost = **8**, Capacity = **60** + - To City 4: Cost = **7**, Capacity = **40** + - To City 5: Cost = **7**, Capacity = **39** + - To City 6: Cost = **10**, Capacity = **40** + - To City 7: Cost = **7**, Capacity = **52** + +- **From City 1:** + - To City 0: Cost = **6**, Capacity = **32** + - To City 2: Cost = **7**, Capacity = **59** + - To City 3: Cost = **6**, Capacity = **58** + - To City 4: Cost = **8**, Capacity = **58** + - To City 5: Cost = **7**, Capacity = **51** + - To City 6: Cost = **6**, Capacity = **31** + - To City 7: Cost = **6**, Capacity = **34** + +- **From City 2:** + - To City 0: Cost = **10**, Capacity = **59** + - To City 1: Cost = **10**, Capacity = **49** + - To City 3: Cost = **9**, Capacity = **51** + - To City 4: Cost = **9**, Capacity = **39** + - To City 5: Cost = **10**, Capacity = **34** + - To City 6: Cost = **6**, Capacity = **47** + - To City 7: Cost = **6**, Capacity = **55** + +- **From City 3:** + - To City 0: Cost = **9**, Capacity = **60** + - To City 1: Cost = **7**, Capacity = **34** + - To City 2: Cost = **7**, Capacity = **30** + - To City 4: Cost = **7**, Capacity = **32** + - To City 5: Cost = **9**, Capacity = **43** + - To City 6: Cost = **7**, Capacity = **30** + - To City 7: Cost = **10**, Capacity = **43** + +- **From City 4:** + - To City 0: Cost = **9**, Capacity = **57** + - To City 1: Cost = **9**, Capacity = **38** + - To City 2: Cost = **6**, Capacity = **54** + - To City 3: Cost = **10**, Capacity = **46** + - To City 5: Cost = **6**, Capacity = **30** + - To City 6: Cost = **9**, Capacity = **48** + - To City 7: Cost = **7**, Capacity = **43** + +- **From City 5:** + - To City 0: Cost = **9**, Capacity = **44** + - To City 1: Cost = **6**, Capacity = **35** + - To City 2: Cost = **8**, Capacity = **32** + - To City 3: Cost = **7**, Capacity = **58** + - To City 4: Cost = **8**, Capacity = **56** + - To City 6: Cost = **6**, Capacity = **48** + - To City 7: Cost = **8**, Capacity = **49** + +- **From City 6:** + - To City 0: Cost = **9**, Capacity = **52** + - To City 1: Cost = **10**, Capacity = **48** + - To City 2: Cost = **7**, Capacity = **33** + - To City 3: Cost = **8**, Capacity = **39** + - To City 4: Cost = **10**, Capacity = **36** + - To City 5: Cost = **10**, Capacity = **47** + - To City 7: Cost = **9**, Capacity = **49** + +- **From City 7:** + - To City 0: Cost = **6**, Capacity = **38** + - To City 1: Cost = **8**, Capacity = **44** + - To City 2: Cost = **10**, Capacity = **33** + - To City 3: Cost = **6**, Capacity = **30** + - To City 4: Cost = **10**, Capacity = **60** + - To City 5: Cost = **7**, Capacity = **39** + - To City 6: Cost = **8**, Capacity = **55** + +#### **Objective:** +Your task is to determine the optimal amount of goods to ship between each pair of cities to meet all demands and supplies while minimizing the total transportation cost. The total cost is calculated by summing the cost of shipping goods across all routes, weighted by the amount shipped on each route. + +#### **Constraints:** +1. **Flow Balance:** For each city, the total amount of goods shipped into the city plus its own supply must equal the total amount shipped out of the city plus its demand. +2. **Capacity Limits:** The amount of goods shipped on any route cannot exceed the maximum capacity of that route. +3. **Non-Negativity:** The amount of goods shipped on any route must be zero or positive. + +By carefully planning the shipments, you aim to ensure that all cities receive the goods they need while keeping the total transportation costs as low as possible.",875.0 +"['Team Formulation Problem', 'Assignment Problem']","Assign seven staff members (Person 0 to Person 6) to three projects (Project 0, Project 1, and Project 2) such that each staff member is assigned to exactly one project. Each project has specific skill requirements: Project 0 requires a skill level of 8 for Skill 0 and 8 for Skill 1, Project 1 requires a skill level of 7 for Skill 0 and 9 for Skill 1, and Project 2 requires a skill level of 9 for Skill 0 and 7 for Skill 1. Each staff member has specific skill levels: Person 0 has Skill 0 level 5 and Skill 1 level 7, Person 1 has Skill 0 level 6 and Skill 1 level 7, Person 2 has Skill 0 level 5 and Skill 1 level 7, Person 3 has Skill 0 level 6 and Skill 1 level 7, Person 4 has Skill 0 level 5 and Skill 1 level 6, Person 5 has Skill 0 level 5 and Skill 1 level 6, and Person 6 has Skill 0 level 5 and Skill 1 level 7. The attained skill level for each project is the sum of the skill levels of the assigned staff members. The skill shortage for each project is the difference between the required skill level and the attained skill level. The goal is to minimize the maximum skill shortage across all projects and skills. Constraints include ensuring that each staff member is assigned to exactly one project, and that the skill shortages for Project 0, Skill 0 must be less than or equal to -8, for Project 0, Skill 1 must be less than or equal to -8, for Project 1, Skill 0 must be less than or equal to -7, for Project 1, Skill 1 must be less than or equal to -9, for Project 2, Skill 0 must be less than or equal to -9, and for Project 2, Skill 1 must be less than or equal to -7.",0.0 +['Job Shop'],"The problem involves scheduling six jobs, labeled Job 0 through Job 5, in a manufacturing facility. Each job consists of two operations that must be performed in a specific sequence. The goal is to minimize the total completion time, known as the makespan, by determining the start times for all operations and the order of operations on shared machines. Each operation has a processing time of 1 unit of time. Operations within the same job must follow a strict sequence, meaning the second operation cannot start until the first operation has completed. Certain operations are assigned to the same machine, and these operations cannot overlap in time. To enforce this, binary decision variables are used to determine the order of operations on shared machines. For example, if Operation 1 of Job 1 is scheduled before Operation 1 of Job 3 on the same machine, the start time of Operation 1 of Job 3 must be at least 1 unit of time after the completion of Operation 1 of Job 1. The makespan is defined as the maximum completion time across all operations, and it must be at least as large as the completion time of every operation. The constraints include precedence constraints within jobs, ensuring that each operation starts only after the previous operation in the same job has completed; machine capacity constraints, ensuring that operations assigned to the same machine do not overlap; and non-negativity constraints, ensuring that all start times are non-negative. The objective is to minimize the makespan, ensuring that all jobs are completed as quickly as possible. The binary decision variables, such as those determining the order of Operation 1 of Job 1 and Operation 1 of Job 3, are critical in enforcing the machine capacity constraints. The numerical parameters include a large constant (1000) used in the constraints involving binary variables to ensure the correct sequencing of operations on shared machines. The start times and the makespan are general variables that can take on any real value within the bounds defined by the constraints. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3, 4, 5} +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Machines: M = {A, B, C} + +2) Machine assignment for every operation (explicit) +- Operation (0,1) -> Machine A +- Operation (0,2) -> Machine B +- Operation (1,1) -> Machine A +- Operation (1,2) -> Machine C +- Operation (2,1) -> Machine B +- Operation (2,2) -> Machine C +- Operation (3,1) -> Machine A +- Operation (3,2) -> Machine B +- Operation (4,1) -> Machine C +- Operation (4,2) -> Machine B +- Operation (5,1) -> Machine A +- Operation (5,2) -> Machine C",5.0 +['TravelingSalesman'],"You are tasked with planning the most efficient route for a delivery service that must visit seven cities: City 0, City 1, City 2, City 3, City 4, City 5, and City 6. The goal is to minimize the total travel cost while ensuring that each city is visited exactly once and that the delivery route forms a complete loop, starting and ending at the same city. The travel costs between each pair of cities are as follows: From City 0 to City 1 costs 285, to City 2 costs 275, to City 3 costs 303, to City 4 costs 282, to City 5 costs 308, and to City 6 costs 290. From City 1 to City 0 costs 293, to City 2 costs 291, to City 3 costs 319, to City 4 costs 295, to City 5 costs 294, and to City 6 costs 324. From City 2 to City 0 costs 309, to City 1 costs 325, to City 3 costs 313, to City 4 costs 292, to City 5 costs 275, and to City 6 costs 320. From City 3 to City 0 costs 307, to City 1 costs 286, to City 2 costs 305, to City 4 costs 297, to City 5 costs 299, and to City 6 costs 278. From City 4 to City 0 costs 320, to City 1 costs 313, to City 2 costs 278, to City 3 costs 300, to City 5 costs 318, and to City 6 costs 325. From City 5 to City 0 costs 281, to City 1 costs 291, to City 2 costs 305, to City 3 costs 285, to City 4 costs 298, and to City 6 costs 292. From City 6 to City 0 costs 323, to City 1 costs 296, to City 2 costs 307, to City 3 costs 295, to City 4 costs 293, and to City 5 costs 301. The task is to find the shortest possible route that visits each city exactly once, and return to the origin city. What is the minimal total distance travelled?",1987.0 +['Job Shop'],"There are six jobs, each consisting of two operations that must be performed in a specific order. Each operation takes 1 unit of time to complete. Operations are assigned to shared machines, and each machine can handle only one operation at a time. The goal is to schedule the operations to minimize the total completion time, known as the makespan, which is the latest completion time of all operations. + +For each job, the second operation cannot start until the first operation is completed. If two operations are assigned to the same machine, they must be scheduled in a specific order, determined by binary decision variables. These variables indicate whether one operation starts before or after the other on the same machine. A large constant, 1000, is used to enforce the sequencing logic in the constraints. + +The start times of all operations must be non-negative, and the makespan must be at least as large as the completion time of every operation. The objective is to determine the start times for all operations and the order in which they are processed on shared machines, ensuring that all precedence constraints are satisfied, overlaps on shared machines are avoided, and the makespan is minimized. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {1, 2, 3, 4, 5, 6} +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Machines: M = {1, 2, 3} + +2) Machine assignment for every operation (which machine executes the operation) +- Operation (1,1) -> Machine 1 +- Operation (1,2) -> Machine 2 +- Operation (2,1) -> Machine 1 +- Operation (2,2) -> Machine 3 +- Operation (3,1) -> Machine 2 +- Operation (3,2) -> Machine 3 +- Operation (4,1) -> Machine 1 +- Operation (4,2) -> Machine 2 +- Operation (5,1) -> Machine 2 +- Operation (5,2) -> Machine 3 +- Operation (6,1) -> Machine 1 +- Operation (6,2) -> Machine 3 + +3) Operation durations +- All operations have processing time p[j,k] = 1 for every job j and operation k. + +Notes about index conventions and mapping +- Each operation (j,k) will be treated as a single task in the model. For pairwise sequencing constraints we consider only pairs of operations that share the same machine.",5.0 +['Job Shop'],"There are five jobs, each consisting of two operations that must be performed in a specific order. The first operation of each job must finish before the second operation of the same job can start. All operations are assigned to shared machines, and no two operations on the same machine can overlap in time. The goal is to schedule the operations such that the total completion time, or makespan, is minimized. + +Each operation has a processing time of 1 unit. For each job, the start time of the second operation must be at least 1 unit after the start time of the first operation. For every pair of operations assigned to the same machine, one operation must finish before the other can start, enforced by binary decision variables that determine the order of operations on the machine. A large constant (1000) is used in the constraints to ensure that operations do not overlap on the same machine. The makespan must be at least as large as the completion time of every operation. + +The decision variables include the start time of each operation and the binary variables that determine the order of operations on shared machines. The objective is to minimize the makespan while satisfying all precedence and machine constraints. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {1, 2, 3, 4, 5} +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Machines: M = {1, 2} + +2) Machine assignment for every operation (which machine executes the operation) +- Operation (1,1) -> Machine 1 +- Operation (1,2) -> Machine 2 +- Operation (2,1) -> Machine 2 +- Operation (2,2) -> Machine 1 +- Operation (3,1) -> Machine 1 +- Operation (3,2) -> Machine 2 +- Operation (4,1) -> Machine 2 +- Operation (4,2) -> Machine 1 +- Operation (5,1) -> Machine 1 +- Operation (5,2) -> Machine 2 + +3) Processing times +- All operations have processing time p_{j,k} = 1 for all jobs j and operations k.",5.0 +['Capacitated Facility Location Problem'],"You are tasked with managing the logistics of a supply chain network involving seven key facilities (nodes) that produce and distribute goods. Each facility has specific supply and demand requirements, and the goal is to minimize the total cost of establishing connections between facilities and transporting goods while ensuring all demands are met. + +Facility 0 has a net demand of 106 units, Facility 1 has a net demand of 144 units, Facility 2 has a supply of 250 units, Facility 3 has a supply of 250 units, Facility 4 has a net demand of 202 units, Facility 5 has a net demand of 40 units, and Facility 6 has a net demand of 8 units. + +Establishing a connection between any two facilities incurs a fixed cost, and transporting goods between them incurs a variable cost per unit. Each connection has a maximum capacity for goods that can be transported. The fixed costs, variable costs, and maximum capacities for each possible connection are as follows: + +- Fixed Costs (for establishing connections): + - Facility 0 to Facility 1: 1985 units + - Facility 0 to Facility 2: 1968 units + - Facility 0 to Facility 3: 2043 units + - Facility 0 to Facility 4: 2035 units + - Facility 0 to Facility 5: 2043 units + - Facility 0 to Facility 6: 1987 units + - Facility 1 to Facility 0: 2035 units + - Facility 1 to Facility 2: 2044 units + - Facility 1 to Facility 3: 1985 units + - Facility 1 to Facility 4: 2047 units + - Facility 1 to Facility 5: 2003 units + - Facility 1 to Facility 6: 2005 units + - Facility 2 to Facility 0: 2000 units + - Facility 2 to Facility 1: 1991 units + - Facility 2 to Facility 3: 2005 units + - Facility 2 to Facility 4: 1969 units + - Facility 2 to Facility 5: 2010 units + - Facility 2 to Facility 6: 1965 units + - Facility 3 to Facility 0: 1962 units + - Facility 3 to Facility 1: 2024 units + - Facility 3 to Facility 2: 2025 units + - Facility 3 to Facility 4: 2043 units + - Facility 3 to Facility 5: 2034 units + - Facility 3 to Facility 6: 2017 units + - Facility 4 to Facility 0: 1974 units + - Facility 4 to Facility 1: 2012 units + - Facility 4 to Facility 2: 2009 units + - Facility 4 to Facility 3: 1993 units + - Facility 4 to Facility 5: 1991 units + - Facility 4 to Facility 6: 2040 units + - Facility 5 to Facility 0: 2044 units + - Facility 5 to Facility 1: 2022 units + - Facility 5 to Facility 2: 2025 units + - Facility 5 to Facility 3: 1955 units + - Facility 5 to Facility 4: 2001 units + - Facility 5 to Facility 6: 2029 units + - Facility 6 to Facility 0: 1968 units + - Facility 6 to Facility 1: 1956 units + - Facility 6 to Facility 2: 1986 units + - Facility 6 to Facility 3: 2030 units + - Facility 6 to Facility 4: 2001 units + - Facility 6 to Facility 5: 1961 units + +- Variable Costs (per unit transported): + - Facility 0 to Facility 1: 19 units + - Facility 0 to Facility 2: 19 units + - Facility 0 to Facility 3: 19 units + - Facility 0 to Facility 4: 19 units + - Facility 0 to Facility 5: 21 units + - Facility 0 to Facility 6: 21 units + - Facility 1 to Facility 0: 21 units + - Facility 1 to Facility 2: 19 units + - Facility 1 to Facility 3: 20 units + - Facility 1 to Facility 4: 20 units + - Facility 1 to Facility 5: 21 units + - Facility 1 to Facility 6: 21 units + - Facility 2 to Facility 0: 21 units + - Facility 2 to Facility 1: 19 units + - Facility 2 to Facility 3: 21 units + - Facility 2 to Facility 4: 21 units + - Facility 2 to Facility 5: 21 units + - Facility 2 to Facility 6: 20 units + - Facility 3 to Facility 0: 21 units + - Facility 3 to Facility 1: 19 units + - Facility 3 to Facility 2: 21 units + - Facility 3 to Facility 4: 19 units + - Facility 3 to Facility 5: 19 units + - Facility 3 to Facility 6: 20 units + - Facility 4 to Facility 0: 21 units + - Facility 4 to Facility 1: 21 units + - Facility 4 to Facility 2: 21 units + - Facility 4 to Facility 3: 20 units + - Facility 4 to Facility 5: 20 units + - Facility 4 to Facility 6: 19 units + - Facility 5 to Facility 0: 21 units + - Facility 5 to Facility 1: 19 units + - Facility 5 to Facility 2: 20 units + - Facility 5 to Facility 3: 20 units + - Facility 5 to Facility 4: 20 units + - Facility 5 to Facility 6: 21 units + - Facility 6 to Facility 0: 19 units + - Facility 6 to Facility 1: 21 units + - Facility 6 to Facility 2: 19 units + - Facility 6 to Facility 3: 20 units + - Facility 6 to Facility 4: 21 units + - Facility 6 to Facility 5: 21 units + +- Maximum Capacities (for goods transported): + - Facility 0 to Facility 1: 153 units + - Facility 0 to Facility 2: 202 units + - Facility 0 to Facility 3: 185 units + - Facility 0 to Facility 4: 134 units + - Facility 0 to Facility 5: 248 units + - Facility 0 to Facility 6: 170 units + - Facility 1 to Facility 0: 214 units + - Facility 1 to Facility 2: 184 units + - Facility 1 to Facility 3: 193 units + - Facility 1 to Facility 4: 181 units + - Facility 1 to Facility 5: 153 units + - Facility 1 to Facility 6: 139 units + - Facility 2 to Facility 0: 217 units + - Facility 2 to Facility 1: 238 units + - Facility 2 to Facility 3: 231 units + - Facility 2 to Facility 4: 130 units + - Facility 2 to Facility 5: 203 units + - Facility 2 to Facility 6: 232 units + - Facility 3 to Facility 0: 147 units + - Facility 3 to Facility 1: 222 units + - Facility 3 to Facility 2: 129 units + - Facility 3 to Facility 4: 206 units + - Facility 3 to Facility 5: 202 units + - Facility 3 to Facility 6: 134 units + - Facility 4 to Facility 0: 183 units + - Facility 4 to Facility 1: 210 units + - Facility 4 to Facility 2: 238 units + - Facility 4 to Facility 3: 193 units + - Facility 4 to Facility 5: 157 units + - Facility 4 to Facility 6: 137 units + - Facility 5 to Facility 0: 134 units + - Facility 5 to Facility 1: 128 units + - Facility 5 to Facility 2: 241 units + - Facility 5 to Facility 3: 143 units + - Facility 5 to Facility 4: 198 units + - Facility 5 to Facility 6: 213 units + - Facility 6 to Facility 0: 213 units + - Facility 6 to Facility 1: 206 units + - Facility 6 to Facility 2: 184 units + - Facility 6 to Facility 3: 231 units + - Facility 6 to Facility 4: 207 units + - Facility 6 to Facility 5: 165 units + +Decide whether to establish a connection between each pair of facilities, where establishing a connection incurs a fixed cost. Determine the amount of goods to transport between connected facilities, ensuring that the total flow does not exceed the connection's capacity. The total incoming goods minus the total outgoing goods must equal the facility's net demand for each facility. The decision to establish a connection is binary. The goal is to minimize the total cost, which includes both the fixed costs of establishing connections and the variable costs of transporting goods.",19805.0 +['TravelingSalesman'],"A logistics company must deliver packages to six cities: City 0, City 1, City 2, City 3, City 4, and City 5. Each city must be visited exactly once, and each city must be departed exactly once, ensuring the route forms a continuous loop, and finally return to the origin city. The goal is to minimize the total delivery cost, which is the sum of the costs of traveling between the cities. The costs between cities are as follows: + +- From City 0: To City 1 costs 321, to City 2 costs 294, to City 3 costs 323, to City 4 costs 286, and to City 5 costs 325. +- From City 1: To City 0 costs 290, to City 2 costs 319, to City 3 costs 321, to City 4 costs 280, and to City 5 costs 324. +- From City 2: To City 0 costs 304, to City 1 costs 317, to City 3 costs 322, to City 4 costs 313, and to City 5 costs 300. +- From City 3: To City 0 costs 312, to City 1 costs 324, to City 2 costs 280, to City 4 costs 318, and to City 5 costs 284. +- From City 4: To City 0 costs 289, to City 1 costs 279, to City 2 costs 278, to City 3 costs 301, and to City 5 costs 324. +- From City 5: To City 0 costs 325, to City 1 costs 309, to City 2 costs 304, to City 3 costs 305, and to City 4 costs 310. +The task is to find the shortest possible route that visits each city exactly once, and return to the origin city. What is the minimal total distance travelled?",1766.0 +['Set Cover'],"""Maximize the total population covered by deploying cell towers across 35 distinct areas, ensuring that the total cost of building the towers does not exceed the available budget of 697 units. The population of each area is as follows: Area 0 has 3,870 people, Area 1 has 4,020 people, Area 2 has 2,440 people, Area 3 has 4,263 people, Area 4 has 2,727 people, Area 5 has 3,650 people, Area 6 has 3,155 people, Area 7 has 4,805 people, Area 8 has 2,213 people, Area 9 has 3,635 people, Area 10 has 3,262 people, Area 11 has 4,015 people, Area 12 has 4,991 people, Area 13 has 2,657 people, Area 14 has 2,150 people, Area 15 has 3,655 people, Area 16 has 2,141 people, Area 17 has 2,108 people, Area 18 has 4,200 people, Area 19 has 3,137 people, Area 20 has 4,255 people, Area 21 has 3,452 people, Area 22 has 2,155 people, Area 23 has 3,142 people, Area 24 has 4,909 people, Area 25 has 2,865 people, Area 26 has 3,204 people, Area 27 has 4,244 people, Area 28 has 4,975 people, Area 29 has 2,131 people, Area 30 has 4,710 people, Area 31 has 4,721 people, Area 32 has 4,061 people, Area 33 has 3,162 people, and Area 34 has 3,377 people. + +For each of the 27 potential cell tower locations, decide whether to build the tower, where building a tower incurs a specific cost: Tower 0 costs 233 units, Tower 1 costs 216 units, Tower 2 costs 366 units, Tower 3 costs 264 units, Tower 4 costs 413 units, Tower 5 costs 332 units, Tower 6 costs 414 units, Tower 7 costs 233 units, Tower 8 costs 432 units, Tower 9 costs 461 units, Tower 10 costs 465 units, Tower 11 costs 248 units, Tower 12 costs 392 units, Tower 13 costs 244 units, Tower 14 costs 416 units, Tower 15 costs 399 units, Tower 16 costs 448 units, Tower 17 costs 212 units, Tower 18 costs 216 units, Tower 19 costs 224 units, Tower 20 costs 489 units, Tower 21 costs 288 units, Tower 22 costs 256 units, Tower 23 costs 308 units, Tower 24 costs 325 units, Tower 25 costs 359 units, and Tower 26 costs 339 units. + +Each area must be covered by at least one of the built towers + - Area 0 can be covered if any of the towers 20, 3, 0, 23, 8, 7, 21, 4, 25, 17, 2, or 13 are built. + - Area 1 can be covered if any of the towers 1, 0, 2, 6, 7, 16, 19, 25, 17, 23, 13, or 22 are built. + - Area 2 can be covered if any of the towers 14, 18, 8, 0, 5, 13, 10, 24, 4, 6, 20, or 3 are built. + - Area 3 can be covered if any of the towers 2, 12, 3, 11, 23, 19, 8, 1, 14, 17, 24, or 25 are built. + - Area 4 can be covered if any of the towers 2, 17, 9, 20, 19, 11, 18, 6, 26, 1, 7, or 24 are built. + - Area 5 can be covered if any of the towers 2, 7, 3, 12, 8, 14, 20, 11, 5, 19, 17, or 6 are built. + - Area 6 can be covered if any of the towers 21, 8, 22, 26, 20, 2, 19, 5, 17, 7, 24, or 14 are built. + - Area 7 can be covered if any of the towers 12, 8, 20, 22, 17, 7, 10, 1, 21, 19, 24, or 26 are built. + - Area 8 can be covered if any of the towers 8, 2, 6, 18, 22, 10, 24, 15, 12, 14, 4, or 26 are built. + - Area 9 can be covered if any of the towers 4, 7, 23, 17, 24, 8, 18, 13, 20, 12, 11, or 25 are built. + - Area 10 can be covered if any of the towers 4, 16, 15, 2, 1, 3, 26, 5, 13, 23, 12, or 25 are built. + - Area 11 can be covered if any of the towers 19, 14, 16, 8, 17, 0, 3, 22, 23, 10, 20, or 9 are built. + - Area 12 can be covered if any of the towers 13, 5, 14, 0, 8, 16, 25, 21, 3, 9, 19, or 6 are built. + - Area 13 can be covered if any of the towers 4, 11, 24, 5, 17, 16, 0, 19, 10, 15, 20, or 3 are built. + - Area 14 can be covered if any of the towers 11, 25, 9, 7, 1, 23, 18, 2, 19, 15, 20, or 4 are built. + - Area 15 can be covered if any of the towers 4, 21, 15, 17, 5, 8, 16, 19, 13, 6, 23, or 9 are built. + - Area 16 can be covered if any of the towers 12, 21, 20, 11, 14, 16, 22, 3, 7, 18, 2, or 10 are built. + - Area 17 can be covered if any of the towers 0, 18, 17, 7, 25, 23, 26, 2, 1, 21, 19, or 22 are built. + - Area 18 can be covered if any of the towers 10, 2, 16, 7, 8, 21, 15, 6, 17, 4, 20, or 23 are built. + - Area 19 can be covered if any of the towers 25, 15, 13, 6, 3, 22, 24, 11, 20, 18, 14, or 1 are built. + - Area 20 can be covered if any of the towers 21, 20, 25, 3, 1, 12, 10, 23, 7, 6, 17, or 14 are built. + - Area 21 can be covered if any of the towers 4, 13, 5, 8, 14, 7, 2, 22, 17, 3, 1, or 0 are built. + - Area 22 can be covered if any of the towers 2, 24, 7, 5, 13, 15, 21, 6, 12, 1, 23, or 18 are built. + - Area 23 can be covered if any of the towers 0, 12, 8, 14, 9, 13, 17, 15, 4, 6, 22, or 20 are built. + - Area 24 can be covered if any of the towers 1, 18, 23, 17, 26, 10, 22, 20, 25, 15, 16, or 5 are built. + - Area 25 can be covered if any of the towers 1, 16, 2, 5, 24, 19, 22, 7, 12, 3, 21, or 26 are built. + - Area 26 can be covered if any of the towers 19, 2, 13, 21, 18, 22, 16, 10, 8, 6, 26, or 7 are built. + - Area 27 can be covered if any of the towers 8, 12, 4, 21, 20, 9, 14, 10, 2, 0, 22, or 3 are built. + - Area 28 can be covered if any of the towers 2, 17, 6, 16, 8, 4, 11, 26, 7, 20, 9, or 5 are built. + - Area 29 can be covered if any of the towers 14, 17, 22, 9, 19, 20, 16, 0, 25, 23, 3, or 4 are built. + - Area 30 can be covered if any of the towers 8, 3, 25, 23, 17, 4, 26, 9, 6, 10, 18, or 20 are built. + - Area 31 can be covered if any of the towers 16, 15, 8, 1, 2, 20, 13, 24, 23, 0, 10, or 4 are built. + - Area 32 can be covered if any of the towers 20, 8, 5, 23, 14, 17, 13, 21, 0, 3, 2, or 4 are built. + - Area 33 can be covered if any of the towers 17, 1, 11, 18, 26, 4, 13, 21, 25, 9, 24, or 23 are built. + - Area 34 can be covered if any of the towers 11, 6, 21, 7, 24, 3, 26, 17, 13, 4, 23, or 5 are built.. + The decisions to build towers and cover areas are binary, meaning you cannot partially build a tower or partially cover an area. The goal is to maximize the total population covered while ensuring the total cost of building the selected towers does not exceed the budget.""",117736.0 +['Job Shop'],"The problem involves scheduling 7 jobs, each consisting of 2 operations, on a set of machines to minimize the total production time, or makespan. Each operation has a processing time of 1 unit of time. Operations within the same job must follow a specific sequence: operation 0 must be completed before operation 1 can begin. Some operations from different jobs are assigned to the same machine, and the machine can only process one operation at a time. The order in which operations are scheduled on a shared machine is determined by binary decision variables, which take a value of 1 if one operation is scheduled before another and 0 otherwise. A large constant of 100 is used to enforce that operations on the same machine do not overlap in time. The makespan must be at least as large as the completion time of every operation, ensuring it accounts for the longest-running job. Start times for all operations must be non-negative, and the binary variables must be either 0 or 1. The objective is to determine the start times for all operations and the order in which operations are processed on shared machines to minimize the makespan while respecting the sequence of operations within each job and ensuring no overlaps on shared machines. + + +Additional information: +- There are 4 machines, numbered from 0 to 3 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 3 + - Operation 1 of Job 1: Machine 3 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 3 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 3 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 0 + - Operation 0 of Job 5: Machine 0 + - Operation 1 of Job 5: Machine 3 + - Operation 0 of Job 6: Machine 0 + - Operation 1 of Job 6: Machine 1",5.0 +['Minimum Cost Flow Problem'],"The problem involves managing the distribution of a critical resource across seven interconnected cities: City 0, City 1, City 2, City 3, City 4, City 5, and City 6. The goal is to minimize the total cost of transporting resources between cities while ensuring that all demands are met and all operational constraints are satisfied. The cost of transporting resources from City 0 to City 1 is 2 units, from City 0 to City 2 is 1 unit, from City 0 to City 3 is 1 unit, from City 0 to City 4 is 3 units, from City 0 to City 5 is 3 units, and from City 0 to City 6 is 3 units. The cost of transporting resources from City 1 to City 0 is 2 units, from City 1 to City 2 is 1 unit, from City 1 to City 3 is 1 unit, from City 1 to City 4 is 1 unit, from City 1 to City 5 is 1 unit, and from City 1 to City 6 is 3 units. The cost of transporting resources from City 2 to City 0 is 2 units, from City 2 to City 1 is 2 units, from City 2 to City 3 is 3 units, from City 2 to City 4 is 2 units, from City 2 to City 5 is 1 unit, and from City 2 to City 6 is 3 units. The cost of transporting resources from City 3 to City 0 is 2 units, from City 3 to City 1 is 2 units, from City 3 to City 2 is 3 units, from City 3 to City 4 is 2 units, from City 3 to City 5 is 3 units, and from City 3 to City 6 is 1 unit. The cost of transporting resources from City 4 to City 0 is 1 unit, from City 4 to City 1 is 3 units, from City 4 to City 2 is 2 units, from City 4 to City 3 is 2 units, from City 4 to City 5 is 3 units, and from City 4 to City 6 is 2 units. The cost of transporting resources from City 5 to City 0 is 3 units, from City 5 to City 1 is 1 unit, from City 5 to City 2 is 1 unit, from City 5 to City 3 is 1 unit, from City 5 to City 4 is 2 units, and from City 5 to City 6 is 1 unit. The cost of transporting resources from City 6 to City 0 is 3 units, from City 6 to City 1 is 2 units, from City 6 to City 2 is 2 units, from City 6 to City 3 is 2 units, from City 6 to City 4 is 3 units, and from City 6 to City 5 is 1 unit. + +Each city must maintain a balance between incoming and outgoing resources. City 0 must ensure that the total resources it sends out are balanced by the resources it receives from other cities. City 1 has a net demand of 1 unit, City 2 has a net surplus of 2 units, City 3 has a net demand of 2 units, City 4 has a net surplus of 2 units, City 5 has a net demand of 1 unit, and City 6 has no net demand or surplus. The route from City 0 to City 1 can handle up to 7 units of resources, from City 0 to City 2 up to 8 units, from City 0 to City 3 up to 9 units, from City 0 to City 4 up to 9 units, from City 0 to City 5 up to 9 units, and from City 0 to City 6 up to 8 units. The route from City 1 to City 0 can handle up to 8 units, from City 1 to City 2 up to 8 units, from City 1 to City 3 up to 8 units, from City 1 to City 4 up to 8 units, from City 1 to City 5 up to 7 units, and from City 1 to City 6 up to 8 units. The route from City 2 to City 0 can handle up to 8 units, from City 2 to City 1 up to 8 units, from City 2 to City 3 up to 8 units, from City 2 to City 4 up to 8 units, from City 2 to City 5 up to 7 units, and from City 2 to City 6 up to 9 units. The route from City 3 to City 0 can handle up to 8 units, from City 3 to City 1 up to 9 units, from City 3 to City 2 up to 8 units, from City 3 to City 4 up to 9 units, from City 3 to City 5 up to 9 units, and from City 3 to City 6 up to 8 units. The route from City 4 to City 0 can handle up to 7 units, from City 4 to City 1 up to 7 units, from City 4 to City 2 up to 9 units, from City 4 to City 3 up to 7 units, from City 4 to City 5 up to 8 units, and from City 4 to City 6 up to 7 units. The route from City 5 to City 0 can handle up to 9 units, from City 5 to City 1 up to 8 units, from City 5 to City 2 up to 7 units, from City 5 to City 3 up to 9 units, from City 5 to City 4 up to 9 units, and from City 5 to City 6 up to 9 units. The route from City 6 to City 0 can handle up to 7 units, from City 6 to City 1 up to 9 units, from City 6 to City 2 up to 9 units, from City 6 to City 3 up to 9 units, from City 6 to City 4 up to 9 units, and from City 6 to City 5 up to 7 units. + +Each city has a maximum capacity for handling incoming resources. City 0 can receive up to 16 units, City 1 up to 15 units, City 2 up to 12 units, City 3 up to 15 units, City 4 up to 15 units, City 5 up to 15 units, and City 6 up to 17 units. The amount of resources transported between any two cities cannot be negative. The objective is to determine the optimal amount of resources to transport between cities to minimize the total transportation cost while ensuring that all cities meet their resource demands or surpluses, no transportation route exceeds its capacity, and no city exceeds its capacity for receiving resources.",7.0 +['Capacitated Lot-sizing Problem (CLSP)'],"The manufacturing facility produces three products over a six-month period, aiming to minimize total costs, including production, setup, and inventory holding costs, while meeting customer demand and adhering to resource constraints. Product 0 has cumulative demands of 75, 141, 241, 315, 386, and 469 units over the six months. Product 1 has cumulative demands of 65, 151, 228, 286, 338, and 390 units. Product 2 has cumulative demands of 57, 143, 193, 263, 363, and 428 units. Production costs per unit for Product 0 are $42, $47, $41, $40, $43, and $47 across the six months. For Product 1, they are $43, $50, $41, $50, $49, and $46. For Product 2, they are $42, $49, $40, $47, $49, and $45. Setup costs for Product 0 are $1,282, $1,493, $1,006, $1,286, $1,305, and $1,360. For Product 1, they are $1,402, $1,498, $1,296, $1,366, $1,376, and $1,172. For Product 2, they are $1,362, $1,094, $1,283, $1,279, $1,430, and $1,146. Inventory holding costs per unit for Product 0 are $4, $5, $4, $5, $4, and $4. For Product 1, they are $5, $5, $5, $4, $4, and $4. For Product 2, they are $5, $5, $4, $5, $5, and $4. The facility has monthly resource capacities of 1,173, 1,102, 1,018, 1,077, 1,190, and 1,132 units. Each unit of Product 0 consumes 1.778 resource units, Product 1 consumes 1.502, and Product 2 consumes 1.556. If a product is produced in a month, a setup cost is incurred, and production quantities are limited by setup decisions: Product 0 can produce up to 469, 394, 328, 228, 154, and 83 units; Product 1 up to 390, 325, 239, 162, 104, and 52 units; and Product 2 up to 428, 371, 285, 235, 165, and 65 units. Inventory at the end of each month must satisfy demand, and no initial or final inventory is allowed. The objective is to determine optimal production quantities, setup decisions, and inventory levels to minimize total costs while meeting demand and resource constraints.",67501.0 +"['Job Shop', 'Flow Shop Scheduling']","There are five clients, labeled Client 0 to Client 4, each requiring two operations to be performed in sequence. Each operation takes exactly 1 unit of time to complete. The operations for each client must follow a strict sequence: the second operation cannot start until the first operation is completed, with a minimum gap of 1 unit of time between them. Operations from different clients may need to be performed on the same machine, and no two operations can overlap on the same machine. To enforce this, binary decision variables determine the order of operations on shared machines. The goal is to minimize the total production time, known as the makespan, which is the time when the last operation across all clients is completed. The makespan must be at least as long as the completion time of every operation. Start times for all operations are continuous variables, and the binary variables ensure that operations on the same machine do not overlap. A large number (1000) is used in the constraints to enforce the scheduling logic, ensuring that one operation finishes before another starts on the same machine. The problem involves determining the optimal start times for all operations and the sequence of operations on shared machines to minimize the total production time while satisfying all constraints. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3, 4} (these correspond to Client 0..Client 4) +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Machines: M = {0, 1} (two machines, indexed 0 and 1) + +2) Machine assignment for every operation (explicit): +- Operation (0,1) -> Machine 0 +- Operation (0,2) -> Machine 1 +- Operation (1,1) -> Machine 0 +- Operation (1,2) -> Machine 1 +- Operation (2,1) -> Machine 0 +- Operation (2,2) -> Machine 1 +- Operation (3,1) -> Machine 0 +- Operation (3,2) -> Machine 1 +- Operation (4,1) -> Machine 0 +- Operation (4,2) -> Machine 1 + +3) Operation durations (time units): +- All operations have duration p_{j,k} = 1 for every job j and operation k",7.0 +['Minimum Cost Flow Problem'],"You are responsible for managing the distribution of resources across a network of eight cities. Each city has a specific supply of resources and a demand that must be met. The goal is to minimize the total transportation cost while ensuring that all cities receive the resources they need, without exceeding the capacity of the transportation links or the storage capacity of the cities. + +City 0 has a net demand of 2 units, City 1 has a balanced supply and demand, City 2 has a net supply of 1 unit, City 3 has a net demand of 2 units, City 4 has a net supply of 2 units, City 5 has a balanced supply and demand, City 6 has a net demand of 1 unit, and City 7 has a net supply of 2 units. + +The cost of transporting resources from City 0 to City 1 is 1 unit, to City 2 is 1 unit, to City 3 is 1 unit, to City 4 is 1 unit, to City 5 is 3 units, to City 6 is 3 units, and to City 7 is 3 units. From City 1 to City 0 is 1 unit, to City 2 is 2 units, to City 3 is 2 units, to City 4 is 2 units, to City 5 is 3 units, to City 6 is 2 units, and to City 7 is 1 unit. From City 2 to City 0 is 2 units, to City 1 is 3 units, to City 3 is 3 units, to City 4 is 2 units, to City 5 is 2 units, to City 6 is 3 units, and to City 7 is 3 units. From City 3 to City 0 is 2 units, to City 1 is 3 units, to City 2 is 1 unit, to City 4 is 3 units, to City 5 is 1 unit, to City 6 is 3 units, and to City 7 is 3 units. From City 4 to City 0 is 3 units, to City 1 is 1 unit, to City 2 is 3 units, to City 3 is 1 unit, to City 5 is 1 unit, to City 6 is 2 units, and to City 7 is 1 unit. From City 5 to City 0 is 1 unit, to City 1 is 1 unit, to City 2 is 3 units, to City 3 is 2 units, to City 4 is 1 unit, to City 6 is 3 units, and to City 7 is 3 units. From City 6 to City 0 is 2 units, to City 1 is 3 units, to City 2 is 3 units, to City 3 is 2 units, to City 4 is 3 units, to City 5 is 1 unit, and to City 7 is 1 unit. From City 7 to City 0 is 2 units, to City 1 is 1 unit, to City 2 is 2 units, to City 3 is 2 units, to City 4 is 1 unit, to City 5 is 1 unit, and to City 6 is 1 unit. + +The maximum capacity for transporting resources from City 0 to City 1 is 7 units, to City 2 is 9 units, to City 3 is 7 units, to City 4 is 8 units, to City 5 is 9 units, to City 6 is 7 units, and to City 7 is 7 units. From City 1 to City 0 is 8 units, to City 2 is 8 units, to City 3 is 8 units, to City 4 is 7 units, to City 5 is 9 units, to City 6 is 8 units, and to City 7 is 9 units. From City 2 to City 0 is 7 units, to City 1 is 9 units, to City 3 is 9 units, to City 4 is 7 units, to City 5 is 9 units, to City 6 is 8 units, and to City 7 is 9 units. From City 3 to City 0 is 8 units, to City 1 is 8 units, to City 2 is 9 units, to City 4 is 7 units, to City 5 is 9 units, to City 6 is 7 units, and to City 7 is 8 units. From City 4 to City 0 is 7 units, to City 1 is 7 units, to City 2 is 7 units, to City 3 is 9 units, to City 5 is 8 units, to City 6 is 7 units, and to City 7 is 7 units. From City 5 to City 0 is 9 units, to City 1 is 9 units, to City 2 is 9 units, to City 3 is 9 units, to City 4 is 7 units, to City 6 is 7 units, and to City 7 is 7 units. From City 6 to City 0 is 8 units, to City 1 is 9 units, to City 2 is 7 units, to City 3 is 9 units, to City 4 is 9 units, to City 5 is 7 units, and to City 7 is 9 units. From City 7 to City 0 is 9 units, to City 1 is 7 units, to City 2 is 8 units, to City 3 is 8 units, to City 4 is 8 units, to City 5 is 9 units, and to City 6 is 8 units. + +The maximum storage capacity for City 0 is 16 units, City 1 is 16 units, City 2 is 18 units, City 3 is 12 units, City 4 is 14 units, City 5 is 16 units, City 6 is 12 units, and City 7 is 14 units. + +Your task is to determine the optimal amount of resources to transport between cities to minimize the total transportation cost while ensuring that each city's net demand or supply is satisfied, the transportation links are not overloaded, and the storage capacities of the cities are not exceeded.",7.0 +['AircraftLanding'],"Minimize the total penalty costs for landing five aircraft, where penalties are incurred for landing earlier or later than the target times. The target landing times are 89, 63, 64, 97, and 85 for aircraft 0, 1, 2, 3, and 4, respectively. Each aircraft must land within its specified time window: aircraft 0 between 60 and 100, aircraft 1 between 60 and 93, aircraft 2 between 60 and 94, aircraft 3 between 67 and 100, and aircraft 4 between 60 and 100. The penalty costs per unit time for landing early are 19, 10, 14, 12, and 19 for aircraft 0, 1, 2, 3, and 4, respectively. The penalty costs per unit time for landing late are 14, 13, 15, 15, and 14 for aircraft 0, 1, 2, 3, and 4, respectively. + +To ensure safety, a minimum separation time must be maintained between consecutive landings. The required separation times between aircraft pairs are as follows: between aircraft 0 and 1, 7 units if aircraft 0 lands first or 8 units if aircraft 1 lands first; between aircraft 0 and 2, 6 units if aircraft 0 lands first or 7 units if aircraft 2 lands first; between aircraft 0 and 3, 5 units if aircraft 0 lands first or 8 units if aircraft 3 lands first; between aircraft 0 and 4, 5 units if aircraft 0 lands first or 7 units if aircraft 4 lands first; between aircraft 1 and 2, 5 units if aircraft 1 lands first or 6 units if aircraft 2 lands first; between aircraft 1 and 3, 5 units if aircraft 1 lands first or 7 units if aircraft 3 lands first; between aircraft 1 and 4, 6 units if aircraft 1 lands first or 5 units if aircraft 4 lands first; between aircraft 2 and 3, 5 units if aircraft 2 lands first or 6 units if aircraft 3 lands first; between aircraft 2 and 4, 5 units if aircraft 2 lands first or 7 units if aircraft 4 lands first; between aircraft 3 and 4, 6 units if aircraft 3 lands first or 6 units if aircraft 4 lands first. + +The order in which aircraft land must be determined, and for each pair of aircraft, one must land before the other. The separation time must be enforced based on the landing order. The actual landing times for each aircraft must be determined, and the penalties for landing early or late must be calculated based on the deviations from the target landing times. The objective is to minimize the total penalty costs while ensuring that all constraints are satisfied.",87.0 +['Facility Dispersion Problem'],"Facility Location for Maximum Dispersion + +You are tasked with selecting a set of facilities from a network of five potential locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall dispersion between the selected facilities. Dispersion is measured by the weighted distances between pairs of facilities, where higher weights indicate greater desirability for separation. The goal is to ensure that the selected facilities are as spread out as possible, given the constraints. + +#### Key Decisions: +1. **Facility Selection**: Decide which facilities to open. Each facility can either be selected (1) or not selected (0). Specifically, you must select exactly 2 facilities out of the 5 available. +2. **Pairwise Dispersion**: Determine whether a pair of facilities contributes to the dispersion score. A pair contributes only if both facilities in the pair are selected. + +#### Objective: +Maximize the total dispersion score, which is the sum of the weighted distances between all pairs of selected facilities. The weights for each pair are as follows: +- Node 0 and Node 1: 22 +- Node 0 and Node 2: 19 +- Node 0 and Node 3: 24 +- Node 0 and Node 4: 22 +- Node 1 and Node 0: 19 +- Node 1 and Node 2: 23 +- Node 1 and Node 3: 16 +- Node 1 and Node 4: 25 +- Node 2 and Node 0: 19 +- Node 2 and Node 1: 16 +- Node 2 and Node 3: 19 +- Node 2 and Node 4: 21 +- Node 3 and Node 0: 25 +- Node 3 and Node 1: 19 +- Node 3 and Node 2: 21 +- Node 3 and Node 4: 17 +- Node 4 and Node 0: 19 +- Node 4 and Node 1: 22 +- Node 4 and Node 2: 19 +- Node 4 and Node 3: 21 + +#### Constraints: +1. **Facility Selection Limit**: Exactly 2 facilities must be selected from the 5 available. This means the sum of the selection decisions for all facilities must equal 2. +2. **Pairwise Contribution Rules**: + - A pair of facilities can only contribute to the dispersion score if both facilities in the pair are selected. + - If either facility in a pair is not selected, the pair cannot contribute to the dispersion score. + - If both facilities in a pair are selected, the pair must contribute to the dispersion score. + +#### Numerical Parameters: +- The weights for each pair of facilities are explicitly provided in the objective section above. +- The total number of facilities to select is fixed at 2. + +#### Summary: +Your task is to select 2 facilities out of 5 such that the total dispersion score, calculated as the sum of the weighted distances between all selected pairs, is maximized. The selection must adhere to the constraints that ensure only valid pairs contribute to the dispersion score and that exactly 2 facilities are chosen.",49.0 +['Job Shop'],"The problem involves scheduling production tasks for five jobs (Job 0 through Job 4) in a manufacturing facility. Each job consists of a sequence of 2 operations that must be performed in sequential order, with each operation taking exactly 1 unit of time to complete. The facility has a limited number of machines, and certain operations from different jobs may require the same machine. The goal is to minimize the total production time, known as the makespan, while ensuring that all operational constraints are satisfied. + +Each operation within a job must start only after the previous operation in the same job is completed. For example, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 is finished. Operations that require the same machine cannot overlap in time. To enforce this, binary decision variables are used to determine the order of operations on each machine. For instance, if Operation 0 of Job 0 is scheduled before Operation 0 of Job 1 on Machine 1, the binary variable \( X_{(0,0),(1,0)} \) is set to 1; otherwise, it is set to 0. A large constant (100) is used in the constraints to ensure that the sequencing is properly enforced. + +The makespan must be at least as long as the completion time of the last operation across all jobs. For example, if Operation 1 of Job 4 finishes at time 10, the makespan must be at least 10. All operation start times must be non-negative, and the binary decision variables can only take values of 0 or 1. + +The objective is to determine the start times for all operations and the sequence of operations on shared machines such that the total production time is minimized. This must be done while ensuring that all precedence constraints, machine capacity constraints, and non-negativity constraints are satisfied. The processing time for each operation is fixed at 1 unit of time, and the binary decision variables are used to enforce the order of operations on shared machines. + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 1 + - Operation 1 of Job 0: Machine 0 + - Operation 0 of Job 1: Machine 2 + - Operation 1 of Job 1: Machine 0 + - Operation 0 of Job 2: Machine 1 + - Operation 1 of Job 2: Machine 0 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 1",4.0 +['Minimum Cost Flow Problem'],"Managing Resource Distribution Across Cities + +Imagine you are responsible for managing the distribution of a critical resource (e.g., medical supplies, food, or energy) across seven interconnected cities. Each city has a specific supply of the resource, a demand that must be met, and limited capacity to store or process incoming resources. Your goal is to minimize the total cost of transporting the resource between cities while ensuring that all demands are satisfied and no capacity limits are exceeded. + +#### City Details: +- **City 0** has a net demand of 2 units (it requires 2 more units than it currently has). +- **City 1** has a net demand of 2 units. +- **City 2** has a net supply of 1 unit (it has 1 more unit than it needs). +- **City 3** has a net supply of 2 units. +- **City 4** is balanced, with no net demand or supply. +- **City 5** has a net supply of 1 unit. +- **City 6** is balanced, with no net demand or supply. + +#### Transportation Costs: +The cost of transporting resources between cities varies depending on the route. Below are the costs per unit transported: +- From **City 0**: + - To City 1: 1 unit cost + - To City 2: 2 units cost + - To City 3: 3 units cost + - To City 4: 2 units cost + - To City 5: 3 units cost + - To City 6: 3 units cost +- From **City 1**: + - To City 0: 2 units cost + - To City 2: 3 units cost + - To City 3: 2 units cost + - To City 4: 2 units cost + - To City 5: 3 units cost + - To City 6: 3 units cost +- From **City 2**: + - To City 0: 2 units cost + - To City 1: 1 unit cost + - To City 3: 3 units cost + - To City 4: 2 units cost + - To City 5: 3 units cost + - To City 6: 2 units cost +- From **City 3**: + - To City 0: 2 units cost + - To City 1: 2 units cost + - To City 2: 2 units cost + - To City 4: 3 units cost + - To City 5: 1 unit cost + - To City 6: 1 unit cost +- From **City 4**: + - To City 0: 1 unit cost + - To City 1: 1 unit cost + - To City 2: 1 unit cost + - To City 3: 1 unit cost + - To City 5: 1 unit cost + - To City 6: 1 unit cost +- From **City 5**: + - To City 0: 3 units cost + - To City 1: 2 units cost + - To City 2: 3 units cost + - To City 3: 2 units cost + - To City 4: 3 units cost + - To City 6: 3 units cost +- From **City 6**: + - To City 0: 3 units cost + - To City 1: 3 units cost + - To City 2: 2 units cost + - To City 3: 3 units cost + - To City 4: 3 units cost + - To City 5: 3 units cost + +#### Transportation Constraints: +Each route between cities has a maximum capacity for resource transportation: +- **City 0**: + - To City 1: Maximum 8 units + - To City 2: Maximum 7 units + - To City 3: Maximum 7 units + - To City 4: Maximum 8 units + - To City 5: Maximum 7 units + - To City 6: Maximum 8 units +- **City 1**: + - To City 0: Maximum 8 units + - To City 2: Maximum 9 units + - To City 3: Maximum 8 units + - To City 4: Maximum 7 units + - To City 5: Maximum 7 units + - To City 6: Maximum 8 units +- **City 2**: + - To City 0: Maximum 9 units + - To City 1: Maximum 9 units + - To City 3: Maximum 7 units + - To City 4: Maximum 9 units + - To City 5: Maximum 9 units + - To City 6: Maximum 9 units +- **City 3**: + - To City 0: Maximum 9 units + - To City 1: Maximum 7 units + - To City 2: Maximum 7 units + - To City 4: Maximum 9 units + - To City 5: Maximum 8 units + - To City 6: Maximum 9 units +- **City 4**: + - To City 0: Maximum 8 units + - To City 1: Maximum 9 units + - To City 2: Maximum 9 units + - To City 3: Maximum 9 units + - To City 5: Maximum 8 units + - To City 6: Maximum 8 units +- **City 5**: + - To City 0: Maximum 8 units + - To City 1: Maximum 7 units + - To City 2: Maximum 9 units + - To City 3: Maximum 8 units + - To City 4: Maximum 8 units + - To City 6: Maximum 8 units +- **City 6**: + - To City 0: Maximum 9 units + - To City 1: Maximum 8 units + - To City 2: Maximum 8 units + - To City 3: Maximum 7 units + - To City 4: Maximum 9 units + - To City 5: Maximum 8 units + +#### City Capacity Constraints: +Each city has a maximum capacity for incoming resources: +- **City 0**: Maximum 18 units +- **City 1**: Maximum 13 units +- **City 2**: Maximum 14 units +- **City 3**: Maximum 16 units +- **City 4**: Maximum 15 units +- **City 5**: Maximum 18 units +- **City 6**: Maximum 14 units + +#### Objective: +Your task is to determine the optimal amount of resources to transport between cities to meet all demands while minimizing the total transportation cost. Ensure that: +1. All net demands and supplies are balanced. +2. No transportation route exceeds its capacity. +3. No city exceeds its capacity for incoming resources. +4. All transportation amounts are non-negative. + +By solving this problem, you will ensure efficient resource distribution across the cities at the lowest possible cost.",7.0 +['Assignment Problem'],"You are responsible for assigning 8 employees to 7 projects to minimize the total cost while meeting all project requirements and respecting employee availability and capacity limits. Each employee has a fixed total number of hours they must contribute, and each project requires a specific number of hours to be completed. The decision variables are binary, indicating whether an employee is assigned to a project or not. + +Employee 0 must contribute exactly 1 hour in total, Employee 1 must contribute exactly 18 hours, Employee 2 must contribute exactly 15 hours, Employee 3 must contribute exactly 19 hours, Employee 4 must contribute exactly 18 hours, Employee 5 must contribute exactly 17 hours, Employee 6 must contribute exactly 15 hours, and Employee 7 must contribute exactly 19 hours. + +Project 0 requires exactly 15 hours, Project 1 requires exactly 19 hours, Project 2 requires exactly 16 hours, Project 3 requires exactly 20 hours, Project 4 requires exactly 17 hours, Project 5 requires exactly 16 hours, and Project 6 requires exactly 19 hours. + +Each employee has a maximum number of hours they can contribute to each project: Employee 0 can contribute up to 8 hours to Project 0, 9 hours to Project 1, 5 hours to Project 2, 10 hours to Project 3, 7 hours to Project 4, 10 hours to Project 5, and 10 hours to Project 6. Employee 1 can contribute up to 8 hours to Project 0, 10 hours to Project 1, 6 hours to Project 2, 6 hours to Project 3, 6 hours to Project 4, 8 hours to Project 5, and 5 hours to Project 6. Employee 2 can contribute up to 5 hours to Project 0, 7 hours to Project 1, 9 hours to Project 2, 7 hours to Project 3, 8 hours to Project 4, 9 hours to Project 5, and 7 hours to Project 6. Employee 3 can contribute up to 5 hours to Project 0, 6 hours to Project 1, 7 hours to Project 2, 7 hours to Project 3, 6 hours to Project 4, 8 hours to Project 5, and 9 hours to Project 6. Employee 4 can contribute up to 7 hours to Project 0, 10 hours to Project 1, 7 hours to Project 2, 6 hours to Project 3, 9 hours to Project 4, 5 hours to Project 5, and 8 hours to Project 6. Employee 5 can contribute up to 6 hours to Project 0, 10 hours to Project 1, 5 hours to Project 2, 9 hours to Project 3, 9 hours to Project 4, 6 hours to Project 5, and 10 hours to Project 6. Employee 6 can contribute up to 7 hours to Project 0, 8 hours to Project 1, 10 hours to Project 2, 5 hours to Project 3, 10 hours to Project 4, 10 hours to Project 5, and 6 hours to Project 6. Employee 7 can contribute up to 10 hours to Project 0, 7 hours to Project 1, 6 hours to Project 2, 9 hours to Project 3, 10 hours to Project 4, 5 hours to Project 5, and 7 hours to Project 6. + +The cost of assigning Employee 0 to Project 0 is 29, to Project 1 is 30, to Project 2 is 32, to Project 3 is 34, to Project 4 is 26, to Project 5 is 35, and to Project 6 is 27. The cost of assigning Employee 1 to Project 0 is 32, to Project 1 is 25, to Project 2 is 35, to Project 3 is 25, to Project 4 is 28, to Project 5 is 26, and to Project 6 is 33. The cost of assigning Employee 2 to Project 0 is 31, to Project 1 is 29, to Project 2 is 35, to Project 3 is 29, to Project 4 is 32, to Project 5 is 25, and to Project 6 is 27. The cost of assigning Employee 3 to Project 0 is 30, to Project 1 is 29, to Project 2 is 28, to Project 3 is 35, to Project 4 is 29, to Project 5 is 27, and to Project 6 is 28. The cost of assigning Employee 4 to Project 0 is 33, to Project 1 is 28, to Project 2 is 26, to Project 3 is 31, to Project 4 is 26, to Project 5 is 29, and to Project 6 is 32. The cost of assigning Employee 5 to Project 0 is 35, to Project 1 is 34, to Project 2 is 29, to Project 3 is 32, to Project 4 is 33, to Project 5 is 31, and to Project 6 is 28. The cost of assigning Employee 6 to Project 0 is 27, to Project 1 is 30, to Project 2 is 27, to Project 3 is 30, to Project 4 is 32, to Project 5 is 32, and to Project 6 is 26. The cost of assigning Employee 7 to Project 0 is 30, to Project 1 is 27, to Project 2 is 31, to Project 3 is 32, to Project 4 is 34, to Project 5 is 32, and to Project 6 is 26. + +The objective is to assign employees to projects such that all project requirements are met, no employee exceeds their total available hours, no employee exceeds their maximum contribution limit for any specific project, and the total cost of all assignments is minimized.",3320.0 +['Facility Location Problem'],"You are tasked with optimizing the placement of facilities across a network of five locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall benefit derived from their interactions. Each location has the potential to host a facility, but due to budget constraints, you can only select **exactly 3 locations** to establish facilities. The goal is to maximize the total benefit generated by the interactions between the selected facilities. + +#### Key Details: +1. **Facility Selection**: You must choose exactly 3 out of the 5 locations to establish facilities. Each location can either be selected (1) or not selected (0). +2. **Interaction Benefits**: The benefit of interaction between any two facilities depends on the specific pair of locations. For example: + - If facilities are placed at Node 0 and Node 1, the interaction benefit is **16**. + - If facilities are placed at Node 0 and Node 2, the interaction benefit is **20**. + - Similarly, the interaction benefits for all other pairs are provided in the table below. + +3. **Interaction Rules**: + - An interaction benefit is only counted if **both facilities in the pair are selected**. For example, if Node 0 and Node 1 are both selected, the interaction benefit of 16 is added to the total. + - If only one facility in a pair is selected, no interaction benefit is counted for that pair. + +4. **Interaction Benefit Table**: + | Pair of Nodes | Interaction Benefit | + |---------------|---------------------| + | Node 0 & Node 1 | 16 | + | Node 0 & Node 2 | 20 | + | Node 0 & Node 3 | 18 | + | Node 0 & Node 4 | 22 | + | Node 1 & Node 0 | 25 | + | Node 1 & Node 2 | 24 | + | Node 1 & Node 3 | 17 | + | Node 1 & Node 4 | 16 | + | Node 2 & Node 0 | 19 | + | Node 2 & Node 1 | 22 | + | Node 2 & Node 3 | 18 | + | Node 2 & Node 4 | 23 | + | Node 3 & Node 0 | 21 | + | Node 3 & Node 1 | 21 | + | Node 3 & Node 2 | 25 | + | Node 3 & Node 4 | 17 | + | Node 4 & Node 0 | 20 | + | Node 4 & Node 1 | 19 | + | Node 4 & Node 2 | 22 | + | Node 4 & Node 3 | 16 | + +5. **Constraints**: + - **Facility Selection Constraint**: Exactly 3 facilities must be selected out of the 5 locations. + - **Interaction Constraints**: For each pair of locations, the interaction benefit is only counted if both facilities in the pair are selected. This is enforced by ensuring that: + - If either facility in a pair is not selected, the interaction benefit for that pair is not counted. + - If both facilities in a pair are selected, the interaction benefit is fully counted. + +6. **Objective**: + - Maximize the total interaction benefit across all selected facility pairs. + +#### Example: +Suppose you select facilities at Node 0, Node 1, and Node 2. The total interaction benefit would be calculated as follows: +- Node 0 & Node 1: 16 +- Node 0 & Node 2: 20 +- Node 1 & Node 2: 24 +Total Benefit = 16 + 20 + 24 = **60** + +Your task is to determine the optimal selection of 3 facilities that maximizes the total interaction benefit, considering all the constraints and interaction rules provided.",127.0 +['Flow Shop Scheduling'],"There are 5 jobs labeled Job 0 to Job 4, each requiring processing on 4 machines labeled Machine 0 to Machine 3. Each job has a specific processing time on each machine: Job 0 takes 5 hours on Machine 0, 9 hours on Machine 1, 7 hours on Machine 2, and 7 hours on Machine 3; Job 1 takes 10 hours on Machine 0, 8 hours on Machine 1, 8 hours on Machine 2, and 9 hours on Machine 3; Job 2 takes 6 hours on Machine 0, 8 hours on Machine 1, 9 hours on Machine 2, and 7 hours on Machine 3; Job 3 takes 7 hours on Machine 0, 7 hours on Machine 1, 8 hours on Machine 2, and 8 hours on Machine 3; Job 4 takes 9 hours on Machine 0, 5 hours on Machine 1, 5 hours on Machine 2, and 9 hours on Machine 3. Each job must be assigned to exactly one position in the schedule, and each position must be assigned exactly one job. A job cannot start on a machine until it has completed processing on the previous machine in the sequence. Additionally, a job in a later position in the schedule cannot start on a machine until the job in the previous position has completed processing on that machine. The objective is to minimize the total completion time, which includes the start time of the last job on the last machine (Machine 3) and the processing time of the last job on Machine 3. The start times for each job on each machine must be non-negative, and all constraints must be satisfied to avoid delays and resource conflicts.",62.0 +['Unit Commitment Problem'],"Minimize the total operational cost of a power generation system over six periods, consisting of Type 0 and Type 1 generators. The cost includes a fixed base cost of 80 units per generator per period, a power generation cost of 4 units per MW produced, and startup costs of 354 units for Type 0 generators and 347 units for Type 1 generators. The system must meet the following demand in each period: 201 MW in period 0, 191 MW in period 1, 182 MW in period 2, 190 MW in period 3, 194 MW in period 4, and 168 MW in period 5. Each Type 0 generator can produce between 34 MW and 127 MW, and no more than 4 Type 0 generators can operate in any period. Each Type 1 generator can produce between 35 MW and 129 MW, and no more than 1 Type 1 generator can operate in any period. The total maximum power output of all generators in each period must be at least 15% higher than the demand for that period, ensuring a reserve margin. For Type 0 generators, the number of operational generators in any period cannot exceed the number from the previous period plus the number started in the current period, and the same applies to Type 1 generators. The number of generators and startup decisions must be integers. The goal is to determine the optimal number of generators to operate, their power output levels, and startup decisions to minimize total costs while satisfying all constraints.",6165.0 +['Minimum Cost Flow Problem'],"You are tasked with managing the distribution of resources across a network of eight cities. Each city has a specific supply of resources and a demand that must be met. The goal is to minimize the total cost of transporting resources between cities while ensuring that all demands are satisfied and all constraints are respected. + +#### City Details: +- **City 0** has a net supply of 4 units. +- **City 1** has a net demand of 1 unit. +- **City 2** has a net demand of 2 units. +- **City 3** has a net supply of 1 unit. +- **City 4** has a net demand of 1 unit. +- **City 5** has a net demand of 1 unit. +- **City 6** and **City 7** are balanced, with no net supply or demand. + +#### Transportation Costs: +The cost of transporting resources between cities varies. Here are the costs for each possible route: +- From **City 0** to **City 1**: 3 units +- From **City 0** to **City 2**: 3 units +- From **City 0** to **City 3**: 3 units +- From **City 0** to **City 4**: 3 units +- From **City 0** to **City 5**: 1 unit +- From **City 0** to **City 6**: 3 units +- From **City 0** to **City 7**: 2 units +- From **City 1** to **City 0**: 3 units +- From **City 1** to **City 2**: 2 units +- From **City 1** to **City 3**: 2 units +- From **City 1** to **City 4**: 3 units +- From **City 1** to **City 5**: 2 units +- From **City 1** to **City 6**: 2 units +- From **City 1** to **City 7**: 1 unit +- From **City 2** to **City 0**: 1 unit +- From **City 2** to **City 1**: 1 unit +- From **City 2** to **City 3**: 3 units +- From **City 2** to **City 4**: 1 unit +- From **City 2** to **City 5**: 2 units +- From **City 2** to **City 6**: 2 units +- From **City 2** to **City 7**: 3 units +- From **City 3** to **City 0**: 1 unit +- From **City 3** to **City 1**: 3 units +- From **City 3** to **City 2**: 1 unit +- From **City 3** to **City 4**: 1 unit +- From **City 3** to **City 5**: 1 unit +- From **City 3** to **City 6**: 2 units +- From **City 3** to **City 7**: 3 units +- From **City 4** to **City 0**: 3 units +- From **City 4** to **City 1**: 2 units +- From **City 4** to **City 2**: 1 unit +- From **City 4** to **City 3**: 2 units +- From **City 4** to **City 5**: 3 units +- From **City 4** to **City 6**: 2 units +- From **City 4** to **City 7**: 3 units +- From **City 5** to **City 0**: 1 unit +- From **City 5** to **City 1**: 2 units +- From **City 5** to **City 2**: 3 units +- From **City 5** to **City 3**: 2 units +- From **City 5** to **City 4**: 1 unit +- From **City 5** to **City 6**: 3 units +- From **City 5** to **City 7**: 3 units +- From **City 6** to **City 0**: 3 units +- From **City 6** to **City 1**: 1 unit +- From **City 6** to **City 2**: 1 unit +- From **City 6** to **City 3**: 3 units +- From **City 6** to **City 4**: 2 units +- From **City 6** to **City 5**: 1 unit +- From **City 6** to **City 7**: 3 units +- From **City 7** to **City 0**: 3 units +- From **City 7** to **City 1**: 3 units +- From **City 7** to **City 2**: 1 unit +- From **City 7** to **City 3**: 2 units +- From **City 7** to **City 4**: 2 units +- From **City 7** to **City 5**: 1 unit +- From **City 7** to **City 6**: 3 units + +#### Constraints: +1. **Flow Balance**: The total resources entering a city must equal the total resources leaving that city, adjusted for its supply or demand. +2. **Link Capacity**: The amount of resources transported between any two cities cannot exceed the capacity of the link: + - **City 0** to **City 1**: 9 units + - **City 0** to **City 2**: 7 units + - **City 0** to **City 3**: 9 units + - **City 0** to **City 4**: 7 units + - **City 0** to **City 5**: 8 units + - **City 0** to **City 6**: 9 units + - **City 0** to **City 7**: 9 units + - **City 1** to **City 0**: 9 units + - **City 1** to **City 2**: 7 units + - **City 1** to **City 3**: 9 units + - **City 1** to **City 4**: 7 units + - **City 1** to **City 5**: 9 units + - **City 1** to **City 6**: 8 units + - **City 1** to **City 7**: 7 units + - **City 2** to **City 0**: 8 units + - **City 2** to **City 1**: 7 units + - **City 2** to **City 3**: 8 units + - **City 2** to **City 4**: 9 units + - **City 2** to **City 5**: 9 units + - **City 2** to **City 6**: 7 units + - **City 2** to **City 7**: 8 units + - **City 3** to **City 0**: 7 units + - **City 3** to **City 1**: 7 units + - **City 3** to **City 2**: 9 units + - **City 3** to **City 4**: 9 units + - **City 3** to **City 5**: 9 units + - **City 3** to **City 6**: 7 units + - **City 3** to **City 7**: 9 units + - **City 4** to **City 0**: 8 units + - **City 4** to **City 1**: 8 units + - **City 4** to **City 2**: 8 units + - **City 4** to **City 3**: 8 units + - **City 4** to **City 5**: 8 units + - **City 4** to **City 6**: 8 units + - **City 4** to **City 7**: 9 units + - **City 5** to **City 0**: 7 units + - **City 5** to **City 1**: 9 units + - **City 5** to **City 2**: 8 units + - **City 5** to **City 3**: 7 units + - **City 5** to **City 4**: 7 units + - **City 5** to **City 6**: 9 units + - **City 5** to **City 7**: 7 units + - **City 6** to **City 0**: 9 units + - **City 6** to **City 1**: 8 units + - **City 6** to **City 2**: 8 units + - **City 6** to **City 3**: 7 units + - **City 6** to **City 4**: 7 units + - **City 6** to **City 5**: 8 units + - **City 6** to **City 7**: 8 units + - **City 7** to **City 0**: 9 units + - **City 7** to **City 1**: 9 units + - **City 7** to **City 2**: 7 units + - **City 7** to **City 3**: 8 units + - **City 7** to **City 4**: 8 units + - **City 7** to **City 5**: 8 units + - **City 7** to **City 6**: 8 units + +3. **City Capacity**: The total resources entering a city cannot exceed its capacity: + - **City 0**: 19 units + - **City 1**: 17 units + - **City 2**: 15 units + - **City 3**: 16 units + - **City 4**: 14 units + - **City 5**: 15 units + - **City 6**: 14 units + - **City 7**: 13 units + +#### Objective: +Your objective is to determine the optimal amount of resources to transport between each pair of cities to minimize the total transportation cost while ensuring that all demands are met and all constraints are satisfied. + +This problem requires careful planning to balance the supply and demand across the network, respecting the capacity limits of both the transportation links and the cities themselves. The solution will provide the most cost-effective way to distribute resources across the cities.",10.0 +"['Minimum Cost Flow Problem', 'Network Optimization']","The problem involves determining the most cost-effective way to transport medical supplies across a network of 10 hospitals (node_0 to node_9) while minimizing total transportation costs. Each hospital is interconnected, and the cost of transporting supplies between any two hospitals varies. The decision variables are binary, representing whether a route between two hospitals is used (1) or not (0). The objective is to minimize the total cost of transporting supplies, calculated as the sum of the costs of all activated routes. + +The constraints ensure that the supply balance is maintained across all hospitals. Specifically, node_0 must have a net supply of +1, indicating it is the source of supplies, while node_9 must have a net supply of -1, indicating it is the destination. All intermediate hospitals (node_1 to node_8) must have a net supply of 0, meaning they neither gain nor lose supplies overall. The flow balance constraints ensure that the total flow into each hospital equals the total flow out of it, with the exception of node_0 and node_9, which have specific net supply requirements. All decision variables are binary, meaning routes are either fully used or not used at all. The goal is to find the set of routes that satisfies these constraints while minimizing the total transportation cost. + +Cost matrix c_(i,j) for i,j ∈ {0,…,9} + 0 1 2 3 4 5 6 7 8 9 + ------------------------------------------ +0 | 0 12 23 31 22 44 37 55 28 60 +1 | 12 0 9 17 35 27 31 48 19 58 +2 | 23 9 0 8 14 26 22 33 12 45 +3 | 31 17 8 0 11 16 29 38 21 40 +4 | 22 35 14 11 0 13 18 27 17 32 +5 | 44 27 26 16 13 0 7 21 29 26 +6 | 37 31 22 29 18 7 0 10 16 24 +7 | 55 48 33 38 27 21 10 0 14 18 +8 | 28 19 12 21 17 29 16 14 0 20 +9 | 60 58 45 40 32 26 24 18 20 0",48.0 +['Job Shop'],"There are 7 jobs, each with 2 operations that must be performed in sequence. Each operation has a processing time of 1 unit of time. The facility has multiple machines, and some operations from different jobs must be performed on the same machine. Operations sharing the same machine cannot overlap in time. The order of operations on shared machines is determined by binary decision variables, which can only take values of 0 or 1. The goal is to minimize the total completion time, known as the makespan, which is the latest completion time among all operations. + +Each operation in a job must start only after the previous operation in the same job is completed. Operations that share the same machine must be scheduled such that one starts only after the other is completed, as determined by the binary variables. The makespan must be at least as large as the completion time of every operation. All operation start times must be non-negative. The binary variables enforce the order of operations on shared machines, ensuring no overlap. The processing time for all operations is 1 unit of time. The makespan is determined by the latest completion time among all operations, and the objective is to minimize this makespan. + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 1 + - Operation 1 of Job 0: Machine 0 + - Operation 0 of Job 1: Machine 2 + - Operation 1 of Job 1: Machine 0 + - Operation 0 of Job 2: Machine 1 + - Operation 1 of Job 2: Machine 0 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 1 + - Operation 0 of Job 5: Machine 1 + - Operation 1 of Job 5: Machine 0 + - Operation 0 of Job 6: Machine 0 + - Operation 1 of Job 6: Machine 2",5.0 +['AircraftLanding'],"Minimize the total penalty costs associated with landing five aircraft either earlier or later than their target times, while ensuring safe separation between consecutive landings. Each aircraft has a target landing time, earliest and latest possible landing times, and penalties for landing early or late. The penalties are \$15, \$10, \$17, \$15, and \$15 per minute early, and \$11, \$16, \$17, \$15, and \$12 per minute late for Aircraft 0, 1, 2, 3, and 4, respectively. The target landing times are 71, 95, 73, 80, and 78 minutes, with earliest landing times of 60, 65, 60, 60, and 60 minutes, and latest landing times of 100 minutes for all aircraft. + +To ensure safety, specific separation times must be maintained between consecutive landings: there must be at least 7 minutes between Aircraft 0 and Aircraft 1, 5 minutes between Aircraft 0 and Aircraft 2, 5 minutes between Aircraft 0 and Aircraft 3, 6 minutes between Aircraft 0 and Aircraft 4, 5 minutes between Aircraft 1 and Aircraft 2, 8 minutes between Aircraft 1 and Aircraft 3, 6 minutes between Aircraft 1 and Aircraft 4, 7 minutes between Aircraft 2 and Aircraft 3, 7 minutes between Aircraft 2 and Aircraft 4, and 7 minutes between Aircraft 3 and Aircraft 4. + +The landing time for each aircraft must be between its earliest and latest possible landing times. The separation times between consecutive landings must be enforced based on the order in which the aircraft land. The penalties for landing early or late are applied based on the deviation from the target landing times. The objective is to determine the actual landing times for all five aircraft such that the total penalty cost is minimized, each aircraft lands within its specified earliest and latest landing times, and the required separation times between consecutive landings are strictly maintained. Binary decision variables are used to determine the order of landings between aircraft pairs, ensuring proper separation based on the landing order. Constraints enforce the relationship between landing times and the binary variables, ensuring proper separation based on the landing order. The binary variables are restricted to 0 or 1.",153.0 +['Capacitated Lot-sizing Problem (CLSP)'],"You are managing a manufacturing facility that produces three types of products (Product 0, Product 1, and Product 2) using two machines (Machine 0 and Machine 1) over three time periods (Period 0, Period 1, and Period 2). Your goal is to minimize the total operational costs, which include production setup costs, machine startup costs, inventory holding costs, and backlogging costs, while meeting customer demand for each product in each period. + +For each product, machine, and time period, decide whether to produce the product and how much to produce. For each machine and time period, decide whether to start up the machine to begin production. For each product and time period, decide how much inventory to carry over to the next period or how much demand to backlog if production falls short. + +Minimize the total cost, which includes production setup costs of 145.4242073380383 per production decision, machine startup costs of 79.18604813454679 per startup decision, inventory holding costs of 0.9008900951620358 for Product 0, 1.068925395295676 for Product 1, and 1.066244796883431 for Product 2 per unit per period, and backlogging costs of 6.734178682720012 for Product 0, 7.442225721133133 for Product 1, and 7.104627209109436 for Product 2 per unit per period. + +For each product and time period, the total production across machines, plus any inventory carried over from the previous period, minus any backlog from the previous period, must equal the demand for that period plus the current inventory minus the current backlog. The demands for each product and period are as follows: Product 0: Period 0 = 20.33410585875594, Period 1 = 22.57961652322203, Period 2 = 15.89714240515548; Product 1: Period 0 = 24.05647618194615, Period 1 = 20.36522893983323, Period 2 = 17.47466293074195; Product 2: Period 0 = 16.36553930950429, Period 1 = 22.7831030020558, Period 2 = 19.7805033393852. + +For each machine and time period, the production quantity multiplied by -71.79935969683524 for Machine 0 or -73.09691843304661 for Machine 1, plus the startup time multiplied by 6.342821301312698 for Machine 0 or 6.171760083826499 for Machine 1, plus the amount produced must be less than or equal to zero. Each machine can only produce one product at a time in any given period, meaning the sum of production decisions across all products for a given machine and period must be less than or equal to 1. + +If a machine is producing a product in a period but was not producing it in the previous period, a startup decision must be made. Production quantities, inventory levels, and backlogs must be non-negative. Production and startup decisions are binary (either 0 or 1).",919.3047225564642 +['Job Shop'],"There are 5 jobs, labeled Job 0 to Job 4, each consisting of 2 operations. Each operation must be performed on a specific machine, and the order of operations within a job is fixed. For example, Operation 1 of Job 0 must be completed before Operation 2 of Job 0 can begin. The processing time for each operation is 1 unit of time. The facility has a limited number of machines, and some operations from different jobs must be performed on the same machine. To avoid conflicts, no two operations can be scheduled on the same machine at the same time. This requires determining the order in which operations are processed on each machine. For example, Operations from Job 1 and Job 2 both require Machine 1, and you must decide whether to process Job 1's operation before Job 2's, or vice versa. Similarly, Operations from Job 4 and Job 0 must be sequenced on Machine 2. Binary variables are used to indicate whether one operation is scheduled before another on the same machine, ensuring that the scheduling constraints are satisfied. The primary objective is to minimize the makespan, which is the total time required to complete all jobs, determined by the latest completion time among all operations. For example, if the last operation finishes at time 10, the makespan is 10. Constraints include precedence constraints, ensuring each operation within a job starts only after the previous operation in the same job has finished; machine capacity constraints, ensuring operations sharing the same machine do not overlap in time; makespan definition, ensuring the makespan is at least as large as the completion time of every operation; non-negativity constraints, ensuring all operation start times are non-negative; and binary variables, determining the order of operations on shared machines. The processing time for each operation is 1 unit of time, and binary variables are used with a large constant (1000) to enforce sequencing constraints. For example, if a binary variable is 1, then one operation must finish before another starts on the same machine. The task is to determine the start times for all operations across all jobs, ensuring operations within the same job are performed in the correct sequence, operations sharing the same machine do not overlap, and the total completion time is minimized. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3, 4} +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Machines: M = {1, 2, 3} + +2) Processing times (in time units) +- All operations have processing time p_{j,k} = 1 for every job j and operation k. + +3) Machine assignment for every operation (explicit mapping) +- Operation (0,1) -> Machine 2 +- Operation (0,2) -> Machine 2 +- Operation (1,1) -> Machine 1 +- Operation (1,2) -> Machine 3 +- Operation (2,1) -> Machine 1 +- Operation (2,2) -> Machine 3 +- Operation (3,1) -> Machine 3 +- Operation (3,2) -> Machine 1 +- Operation (4,1) -> Machine 2 +- Operation (4,2) -> Machine 1",4.0 +"['Facility Location Problem (FLP)', 'Set Cover']","Cell Tower Coverage Planning + +You are tasked with planning the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 35 distinct areas, each with a specific population size that needs to be served. Your goal is to determine which cell towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + +#### Key Decisions: +1. **Build or Not Build**: For each of the 26 potential cell tower locations, you must decide whether to build the tower (a binary decision: build or not build). +2. **Coverage Assignment**: For each of the 35 areas, you must determine whether it will be covered by the deployed towers (a binary decision: covered or not covered). + +#### Objective: +Maximize the total population covered by the deployed cell towers. The population sizes for the 35 areas are as follows: +- Area 0: 3,215 people +- Area 1: 2,395 people +- Area 2: 4,466 people +- Area 3: 2,520 people +- Area 4: 2,192 people +- Area 5: 4,625 people +- Area 6: 4,796 people +- Area 7: 2,973 people +- Area 8: 2,034 people +- Area 9: 2,754 people +- Area 10: 2,143 people +- Area 11: 2,473 people +- Area 12: 3,099 people +- Area 13: 4,136 people +- Area 14: 2,154 people +- Area 15: 3,411 people +- Area 16: 3,230 people +- Area 17: 2,155 people +- Area 18: 4,907 people +- Area 19: 2,665 people +- Area 20: 4,708 people +- Area 21: 4,494 people +- Area 22: 2,381 people +- Area 23: 3,077 people +- Area 24: 2,699 people +- Area 25: 3,320 people +- Area 26: 4,931 people +- Area 27: 4,645 people +- Area 28: 2,346 people +- Area 29: 4,393 people +- Area 30: 4,051 people +- Area 31: 4,507 people +- Area 32: 4,610 people +- Area 33: 3,412 people +- Area 34: 3,059 people + +#### Constraints: +1. **Coverage Requirements**: + - Each area can only be considered covered if at least one of the cell towers assigned to it is built. For example: + - Area 0 can be covered if any of the towers 1, 5, 10, 11, 13, 15, 18, 19, 20, 22, 23, or 24 are built. + - Area 1 can be covered if any of the towers 3, 4, 6, 11, 12, 14, 15, 17, 18, 19, 21, 23, or 24 are built. + - Area 2 can be covered if any of the towers 23, 14, 0, 10, 15, 6, 21, 11, 20, 25, 9, or 17 are built. + - Area 3 can be covered if any of the towers 17, 11, 19, 2, 7, 18, 6, 16, 8, 1, 0, or 23 are built. + - Area 4 can be covered if any of the towers 12, 11, 19, 9, 6, 23, 3, 24, 15, 8, 22, or 20 are built. + - Area 5 can be covered if any of the towers 24, 5, 13, 17, 22, 21, 15, 19, 1, 4, 0, or 6 are built. + - Area 6 can be covered if any of the towers 3, 5, 19, 24, 0, 6, 20, 15, 18, 17, 25, or 2 are built. + - Area 7 can be covered if any of the towers 1, 16, 5, 7, 20, 11, 12, 19, 8, 14, 21, or 15 are built. + - Area 8 can be covered if any of the towers 7, 20, 15, 14, 4, 10, 1, 0, 22, 3, 11, or 9 are built. + - Area 9 can be covered if any of the towers 19, 5, 8, 13, 0, 6, 4, 22, 20, 12, 24, or 2 are built. + - Area 10 can be covered if any of the towers 2, 1, 3, 24, 6, 10, 14, 17, 11, 7, 23, or 8 are built. + - Area 11 can be covered if any of the towers 24, 6, 15, 18, 16, 9, 11, 17, 0, 2, 10, or 12 are built. + - Area 12 can be covered if any of the towers 18, 11, 1, 10, 8, 16, 9, 13, 19, 7, 20, or 6 are built. + - Area 13 can be covered if any of the towers 7, 21, 6, 10, 2, 16, 9, 4, 15, 0, 17, or 18 are built. + - Area 14 can be covered if any of the towers 11, 13, 16, 10, 1, 24, 4, 8, 0, 25, 7, or 19 are built. + - Area 15 can be covered if any of the towers 7, 11, 1, 24, 15, 14, 21, 12, 9, 0, 16, or 2 are built. + - Area 16 can be covered if any of the towers 7, 10, 9, 0, 13, 15, 12, 2, 1, 25, 20, or 6 are built. + - Area 17 can be covered if any of the towers 24, 8, 21, 2, 0, 18, 25, 4, 23, 7, 10, or 5 are built. + - Area 18 can be covered if any of the towers 11, 19, 14, 0, 18, 24, 5, 23, 2, 10, 7, or 8 are built. + - Area 19 can be covered if any of the towers 4, 20, 25, 17, 14, 12, 8, 7, 10, 22, 21, or 23 are built. + - Area 20 can be covered if any of the towers 21, 19, 25, 20, 4, 6, 10, 3, 2, 14, 8, or 15 are built. + - Area 21 can be covered if any of the towers 16, 25, 4, 11, 3, 18, 5, 19, 9, 2, 22, or 1 are built. + - Area 22 can be covered if any of the towers 23, 18, 20, 12, 21, 17, 10, 25, 2, 9, 4, or 6 are built. + - Area 23 can be covered if any of the towers 14, 25, 20, 18, 11, 23, 7, 13, 2, 4, 19, or 12 are built. + - Area 24 can be covered if any of the towers 0, 5, 16, 24, 7, 23, 14, 6, 25, 10, 19, or 22 are built. + - Area 25 can be covered if any of the towers 5, 20, 3, 23, 10, 2, 6, 19, 13, 25, 9, or 15 are built. + - Area 26 can be covered if any of the towers 18, 2, 5, 22, 3, 12, 6, 15, 9, 10, 21, or 1 are built. + - Area 27 can be covered if any of the towers 13, 10, 23, 7, 9, 3, 18, 17, 21, 25, 2, or 14 are built. + - Area 28 can be covered if any of the towers 11, 17, 0, 10, 22, 21, 2, 3, 14, 25, 9, or 6 are built. + - Area 29 can be covered if any of the towers 1, 16, 4, 14, 3, 19, 2, 18, 9, 7, 0, or 13 are built. + - Area 30 can be covered if any of the towers 2, 6, 14, 25, 15, 21, 13, 22, 23, 12, 11, or 9 are built. + - Area 31 can be covered if any of the towers 8, 6, 11, 23, 16, 1, 13, 18, 14, 2, 20, or 19 are built. + - Area 32 can be covered if any of the towers 19, 25, 16, 24, 1, 4, 22, 11, 13, 15, 21, or 12 are built. + - Area 33 can be covered if any of the towers 11, 2, 9, 3, 4, 21, 25, 5, 12, 6, 19, or 7 are built. + - Area 34 can be covered if any of the towers 10, 19, 0, 6, 18, 21, 12, 25, 14, 13, 11, or 20 are built. + +1. **Budget Constraint**: + - The total cost of building the selected towers must not exceed the available budget of 894 units. The costs for building each tower are as follows: + - Tower 0: 411 units + - Tower 1: 294 units + - Tower 2: 376 units + - Tower 3: 447 units + - Tower 4: 414 units + - Tower 5: 498 units + - Tower 6: 437 units + - Tower 7: 272 units + - Tower 8: 381 units + - Tower 9: 323 units + - Tower 10: 356 units + - Tower 11: 200 units + - Tower 12: 297 units + - Tower 13: 428 units + - Tower 14: 427 units + - Tower 15: 484 units + - Tower 16: 494 units + - Tower 17: 436 units + - Tower 18: 218 units + - Tower 19: 428 units + - Tower 20: 399 units + - Tower 21: 484 units + - Tower 22: 367 units + - Tower 23: 204 units + - Tower 24: 289 units + - Tower 25: 283 units + +3. **Binary Decisions**: + - Each tower can either be built (1) or not built (0). + - Each area can either be covered (1) or not covered (0). + +#### Summary: +Your task is to select which towers to build and which areas to cover, ensuring that the total population covered is maximized while staying within the budget. The coverage of each area depends on the specific combination of towers built, and the total cost of building the selected towers must not exceed 894 units.",116311.0 +"['Minimum Cost Flow Problem', 'Transportation Problem']","You are responsible for managing the distribution of resources across a network of nine cities. Each city has a specific supply or demand for resources, and your goal is to minimize the total transportation cost while ensuring that all cities receive the resources they need. The transportation costs vary depending on the route taken between cities, and there are capacity limits on how much can be transported between any two cities. + +City 0 has a net demand of 63 units of resources. City 1 has a net demand of 2 units of resources. City 2 has a net demand of 5 units of resources. City 3 has a net demand of 16 units of resources. City 4 has a net supply of 13 units of resources. City 5 has a net supply of 13 units of resources. City 6 has a net supply of 26 units of resources. City 7 has a net supply of 19 units of resources. City 8 has a net supply of 15 units of resources. + +The cost of transporting resources from City 0 to City 1 is 6, with a capacity of 48. From City 0 to City 2, the cost is 7, with a capacity of 37. From City 0 to City 3, the cost is 10, with a capacity of 57. From City 0 to City 4, the cost is 9, with a capacity of 41. From City 0 to City 5, the cost is 9, with a capacity of 40. From City 0 to City 6, the cost is 10, with a capacity of 48. From City 0 to City 7, the cost is 9, with a capacity of 38. From City 0 to City 8, the cost is 8, with a capacity of 44. + +The cost of transporting resources from City 1 to City 0 is 10, with a capacity of 44. From City 1 to City 2, the cost is 8, with a capacity of 58. From City 1 to City 3, the cost is 9, with a capacity of 57. From City 1 to City 4, the cost is 6, with a capacity of 34. From City 1 to City 5, the cost is 8, with a capacity of 49. From City 1 to City 6, the cost is 8, with a capacity of 57. From City 1 to City 7, the cost is 8, with a capacity of 46. From City 1 to City 8, the cost is 6, with a capacity of 59. + +The cost of transporting resources from City 2 to City 0 is 8, with a capacity of 31. From City 2 to City 1, the cost is 8, with a capacity of 43. From City 2 to City 3, the cost is 7, with a capacity of 43. From City 2 to City 4, the cost is 6, with a capacity of 46. From City 2 to City 5, the cost is 6, with a capacity of 45. From City 2 to City 6, the cost is 8, with a capacity of 41. From City 2 to City 7, the cost is 8, with a capacity of 59. From City 2 to City 8, the cost is 10, with a capacity of 40. + +The cost of transporting resources from City 3 to City 0 is 9, with a capacity of 44. From City 3 to City 1, the cost is 10, with a capacity of 46. From City 3 to City 2, the cost is 10, with a capacity of 32. From City 3 to City 4, the cost is 7, with a capacity of 56. From City 3 to City 5, the cost is 9, with a capacity of 42. From City 3 to City 6, the cost is 6, with a capacity of 52. From City 3 to City 7, the cost is 8, with a capacity of 51. From City 3 to City 8, the cost is 7, with a capacity of 39. + +The cost of transporting resources from City 4 to City 0 is 9, with a capacity of 49. From City 4 to City 1, the cost is 10, with a capacity of 31. From City 4 to City 2, the cost is 7, with a capacity of 47. From City 4 to City 3, the cost is 7, with a capacity of 60. From City 4 to City 5, the cost is 10, with a capacity of 46. From City 4 to City 6, the cost is 7, with a capacity of 32. From City 4 to City 7, the cost is 8, with a capacity of 42. From City 4 to City 8, the cost is 7, with a capacity of 45. + +The cost of transporting resources from City 5 to City 0 is 8, with a capacity of 31. From City 5 to City 1, the cost is 8, with a capacity of 51. From City 5 to City 2, the cost is 10, with a capacity of 53. From City 5 to City 3, the cost is 7, with a capacity of 47. From City 5 to City 4, the cost is 8, with a capacity of 36. From City 5 to City 6, the cost is 6, with a capacity of 30. From City 5 to City 7, the cost is 9, with a capacity of 57. From City 5 to City 8, the cost is 9, with a capacity of 40. + +The cost of transporting resources from City 6 to City 0 is 10, with a capacity of 57. From City 6 to City 1, the cost is 8, with a capacity of 36. From City 6 to City 2, the cost is 10, with a capacity of 56. From City 6 to City 3, the cost is 9, with a capacity of 51. From City 6 to City 4, the cost is 7, with a capacity of 48. From City 6 to City 5, the cost is 7, with a capacity of 56. From City 6 to City 7, the cost is 7, with a capacity of 60. From City 6 to City 8, the cost is 6, with a capacity of 56. + +The cost of transporting resources from City 7 to City 0 is 10, with a capacity of 37. From City 7 to City 1, the cost is 10, with a capacity of 36. From City 7 to City 2, the cost is 8, with a capacity of 48. From City 7 to City 3, the cost is 7, with a capacity of 60. From City 7 to City 4, the cost is 9, with a capacity of 38. From City 7 to City 5, the cost is 7, with a capacity of 43. From City 7 to City 6, the cost is 9, with a capacity of 60. From City 7 to City 8, the cost is 9, with a capacity of 53. + +The cost of transporting resources from City 8 to City 0 is 7, with a capacity of 54. From City 8 to City 1, the cost is 8, with a capacity of 42. From City 8 to City 2, the cost is 8, with a capacity of 36. From City 8 to City 3, the cost is 10, with a capacity of 46. From City 8 to City 4, the cost is 8, with a capacity of 58. From City 8 to City 5, the cost is 7, with a capacity of 54. From City 8 to City 6, the cost is 10, with a capacity of 56. From City 8 to City 7, the cost is 10, with a capacity of 34. + +Your task is to determine the optimal amount of resources to transport between each pair of cities to meet all demands and supplies while minimizing the total transportation cost. The transportation must respect the capacity limits on each route, and the amount of resources transported cannot be negative. For each city, the total amount of resources sent out plus its supply must equal the total amount received plus its demand.",714.0 +['Job Shop'],"Job Shop Scheduling in a Manufacturing Facility + +You are managing a manufacturing facility that produces custom parts for various clients. The facility has multiple machines, each capable of performing specific operations. There are seven jobs (Job 0 to Job 6) that need to be processed, and each job consists of 2 operations that must be performed in a specific order. All operations take 1 time unit. + +#### Objective: +Your goal is to minimize the **makespan**, which is the total time required to complete all jobs. This ensures that all jobs are finished as quickly as possible, allowing the facility to meet client deadlines and optimize resource utilization. + +#### Decisions to Be Made: +1. **Start Times**: Determine the start time for each operation of every job. +2. **Operation Order**: Decide the order in which operations are processed on each machine, ensuring no overlaps. + +#### Constraints: +1. **Precedence Constraints**: + - Each operation in a job must start only after the previous operation in the same job has been completed. For example: + - Operation 1 of Job 0 must start at least 1 time unit after Operation 0 of Job 0 has finished. + - Operation 1 of Job 1 must start at least 1 time unit after Operation 0 of Job 1 has finished. + - This pattern applies to all jobs and their respective operations. + +2. **Machine Capacity Constraints**: + - Certain additional pairs of operations also cannot overlap in time. These additional pairs are: + - Operation 1 of Job 3 and Operation 1 of Job 4. + - Operation 1 of Job 3 and Operation 1 of Job 6. + - Operation 0 of Job 4 and Operation 1 of Job 4. + - Operation 0 of Job 4 and Operation 1 of Job 6. + - Operation 1 of Job 4 and Operation 1 of Job 6. + - Operation 1 of Job 0 and Operation 0 of Job 1. + - Operation 1 of Job 0 and Operation 0 of Job 2. + - Operation 0 of Job 1 and Operation 0 of Job 2. + - Operation 0 of Job 0 and Operation 1 of Job 1. + - Operation 0 of Job 0 and Operation 1 of Job 2. + - Operation 0 of Job 0 and Operation 0 of Job 3. + - Operation 0 of Job 0 and Operation 1 of Job 5. + - Operation 0 of Job 0 and Operation 0 of Job 6. + - Operation 1 of Job 1 and Operation 1 of Job 2. + - Operation 1 of Job 1 and Operation 0 of Job 3. + - Operation 1 of Job 1 and Operation 1 of Job 5. + - Operation 1 of Job 1 and Operation 0 of Job 6. + - Operation 1 of Job 2 and Operation 0 of Job 3. + - Operation 1 of Job 2 and Operation 1 of Job 5. + - Operation 1 of Job 2 and Operation 0 of Job 6. + - Operation 0 of Job 3 and Operation 1 of Job 5. + - Operation 0 of Job 3 and Operation 0 of Job 6. + - Operation 1 of Job 5 and Operation 0 of Job 6. + +3. **Makespan Definition**: + - The makespan must be at least as large as the completion time of every operation + +4. **Non-Negativity**: + - All start times must be non-negative, meaning no operation can start before time zero. + +5. **Binary Variables**: + - Binary variables are used to determine the order of operations on shared machines. For example: + - A binary variable decides whether Operation 1 of Job 3 is processed before or after Operation 0 of Job 4 on the same machine. + - Similar binary variables are used for all relevant pairs of operations sharing the same machine. + +#### Numerical Parameters: +- The large constant \( N = 100 \) is used in the machine capacity constraints to ensure that the binary variables correctly enforce the order of operations. + +#### Summary: +Your task is to schedule the start times of all operations across all jobs and machines in such a way that the total completion time (makespan) is minimized. You must ensure that all precedence and machine capacity constraints are satisfied, and that the makespan reflects the latest completion time of any operation. The use of binary variables is critical to determining the correct order of operations on shared machines.",6.0 +['Job Shop'],"The problem involves scheduling operations for five jobs (Job 0 through Job 4) in a manufacturing facility. Each job consists of two operations that must be performed in sequence, with the second operation starting only after the first operation is completed. All operations have a processing time of 1 unit. Operations from different jobs may share the same machine, and each machine can handle only one operation at a time. The goal is to minimize the total time required to complete all jobs, known as the makespan. + +The start time of each operation must be determined, ensuring that the sequence of operations within each job is respected and that no two operations on the same machine overlap. Binary variables are used to determine the order of operations on shared machines. For example, if Operation 0 of Job 0 and Operation 0 of Job 2 are on the same machine, a binary variable indicates whether Operation 0 of Job 0 precedes Operation 0 of Job 2 or vice versa. A large constant (100) is used in the constraints to enforce the correct ordering of operations on shared machines. + +The makespan must be at least as large as the completion time of every operation. For example, if Operation 0 of Job 0 finishes at time 2, the makespan must be at least 2. The constraints include precedence constraints within jobs, ensuring that the start time of the second operation is at least 1 unit after the completion of the first operation. Machine capacity constraints ensure that two operations assigned to the same machine do not overlap in time. The solution will provide a schedule that ensures all jobs are completed as quickly as possible, respecting the constraints of machine availability and operation precedence. + + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 0 + - Operation 0 of Job 1: Machine 1 + - Operation 1 of Job 1: Machine 1 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 0 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 0 + - Operation 0 of Job 4: Machine 1 + - Operation 1 of Job 4: Machine 1",5.0 +['Flow Shop Scheduling'],"There are 5 jobs labeled as Job 0, Job 1, Job 2, Job 3, and Job 4, and 4 machines labeled as Machine 0, Machine 1, Machine 2, and Machine 3. Each job must be assigned to one of 5 schedule positions (Position 0 to Position 4) on each machine. The goal is to minimize the total completion time of all jobs, which includes the start time of the job in the last schedule position (Position 4) on the last machine (Machine 3) and the processing time of that job. Each job has specific processing times on each machine: Job 0 takes 9 units on Machine 0, 7 units on Machine 1, 5 units on Machine 2, and 6 units on Machine 3; Job 1 takes 7 units on Machine 0, 10 units on Machine 1, 10 units on Machine 2, and 8 units on Machine 3; Job 2 takes 6 units on Machine 0, 9 units on Machine 1, 9 units on Machine 2, and 8 units on Machine 3; Job 3 takes 10 units on Machine 0, 10 units on Machine 1, 7 units on Machine 2, and 8 units on Machine 3; Job 4 takes 6 units on Machine 0, 10 units on Machine 1, 5 units on Machine 2, and 10 units on Machine 3. Each schedule position on each machine must be assigned exactly one job, and each job must be assigned to exactly one schedule position across all machines. The start time of a job on the next machine must be at least equal to its completion time on the current machine. Additionally, the start time of a job in the next schedule position must be at least equal to the completion time of the job in the current position on the same machine. The processing times for the jobs in the last schedule position (Position 4) are: Job 0 takes 6 units, Job 1 takes 8 units, Job 2 takes 8 units, Job 3 takes 8 units, and Job 4 takes 10 units. The start time of the job in Position 4 on Machine 3 is a critical factor in the objective function. The decision variables include assigning each job to a specific schedule position on each machine and determining the start times of each job on each machine to ensure all constraints are satisfied. The objective is to minimize the total completion time of all jobs while adhering to the constraints of the manufacturing process.",67.0 +['Job Shop'],"There are 7 jobs, each consisting of 2 operations that must be processed in sequence. Each operation takes exactly 1 unit of time to complete. The operations are assigned to specific machines, and no two operations on the same machine can overlap in time. For each pair of operations assigned to the same machine, a binary decision variable determines the order in which they are processed. If one operation is scheduled before the other, the corresponding binary variable is set to 1; otherwise, it is set to 0. A large constant value of 100 is used to enforce the order of operations on shared machines. The start time of each operation must be non-negative, and the start time of the second operation in each job must be at least 1 unit of time after the start time of the first operation in the same job. The total completion time, or makespan, is defined as the maximum completion time across all operations. The objective is to minimize the makespan by determining the optimal start times for all operations and the order of operations on shared machines, while ensuring that all constraints are satisfied. + + + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 2 + - Operation 1 of Job 1: Machine 2 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 0 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 0 + - Operation 0 of Job 5: Machine 0 + - Operation 1 of Job 5: Machine 1 + - Operation 0 of Job 6: Machine 0 + - Operation 1 of Job 6: Machine 1",5.0 +['Facility Dispersion Problem'],"Select two facilities from five potential locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the minimum distance between any two selected facilities. Each location can either be selected (1) or not selected (0). For every pair of locations, determine whether both are selected as facilities, represented by a binary decision variable. The distances between the locations are scaled by a factor of 1,000,000 and are as follows: Node 0 and Node 1: 19 units, Node 0 and Node 2: 21 units, Node 0 and Node 3: 31 units, Node 0 and Node 4: 35 units, Node 1 and Node 0: 29 units, Node 1 and Node 2: 13 units, Node 1 and Node 3: 46 units, Node 1 and Node 4: 37 units, Node 2 and Node 0: 50 units, Node 2 and Node 1: 22 units, Node 2 and Node 3: 49 units, Node 2 and Node 4: 33 units, Node 3 and Node 0: 18 units, Node 3 and Node 1: 20 units, Node 3 and Node 2: 46 units, Node 3 and Node 4: 33 units, Node 4 and Node 0: 18 units, Node 4 and Node 1: 26 units, Node 4 and Node 2: 39 units, Node 4 and Node 3: 43 units. The total number of selected facilities must equal two. For every pair of locations, the selection of both facilities in the pair is only possible if both individual facilities are selected, enforced through constraints linking the pairwise selection variables to the individual facility selection variables. All decision variables are binary, ensuring facilities are either fully selected or not selected at all. The objective is to maximize the minimum distance between any two selected facilities, ensuring geographical dispersion while adhering to the constraints on pairwise selection and total facility count.",46.0 +['Job Shop'],"The problem involves scheduling operations for six jobs (Job 0 through Job 5) in a manufacturing facility, where each job consists of two operations that must be performed in sequence. Each operation has a processing time of 1 unit of time. Operations from different jobs may share the same machine, and a machine can only process one operation at a time. The goal is to minimize the total time to complete all jobs, known as the makespan. + +For each job, the second operation cannot start until the first operation is completed, ensuring a sequence within each job. Binary variables are used to determine the order of operations on shared machines. If a binary variable is set to 1, it indicates that one operation is scheduled before another on the same machine; if set to 0, the order is reversed. A large constant (1000) is used in the constraints to enforce this sequencing logic. + +The makespan must be at least as large as the completion time of every operation across all jobs. All start times for operations must be non-negative, and the binary variables must be either 0 or 1. The objective is to determine the start times for all operations and the sequence of operations on shared machines to minimize the makespan, ensuring efficient completion of all jobs. + +Constraints include ensuring that the start time of the second operation in each job is at least 1 unit of time after the start time of the first operation. For operations sharing the same machine, one operation must finish before the other starts, enforced by the binary variables and the large constant. The makespan must be at least as large as the completion time of every operation, and all start times must be non-negative. The binary variables must be either 0 or 1 to determine the correct sequence of operations on shared machines. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3, 4, 5} +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Operation identifiers: we'll flatten operations into a set O with elements o = (j,k) and also give each a unique integer id o_id = 0..11 for modeling convenience. +- Machines: M = {A, B, C} + +2) Machine assignment for every operation (explicit) +- Operation (0,1) -> Machine A +- Operation (0,2) -> Machine B +- Operation (1,1) -> Machine A +- Operation (1,2) -> Machine C +- Operation (2,1) -> Machine B +- Operation (2,2) -> Machine C +- Operation (3,1) -> Machine A +- Operation (3,2) -> Machine B +- Operation (4,1) -> Machine C +- Operation (4,2) -> Machine A +- Operation (5,1) -> Machine B +- Operation (5,2) -> Machine C + +3) Processing durations (time units) +All operations have processing time p(o) = 1",4.0 +['Job Shop'],"In a manufacturing facility, six distinct jobs need to be processed on shared machines. Each job consists of two operations that must be performed in a specific order, with the second operation starting only after the first operation is completed. Each operation has a processing time of 1 unit. The goal is to minimize the total completion time, known as the makespan, by scheduling operations on the machines efficiently. + +Each operation is assigned to a specific machine, and some operations from different jobs share the same machine, creating potential conflicts. To resolve these conflicts, the scheduler must determine the order in which operations are processed on each machine. This is achieved using binary variables, where a value of 1 indicates that one operation is scheduled before another, and 0 indicates the opposite. A large constant of 1000 is used in the constraints to ensure that only one of the two possible orders is active at any time, preventing overlapping operations on the same machine. + +The start time of each operation must be non-negative, and the makespan must be at least as large as the completion time of every operation. The precedence constraints ensure that for each job, the start time of the second operation is at least 1 unit of time after the start time of the first operation. The machine capacity constraints ensure that for every pair of operations assigned to the same machine, one operation must finish before the other starts, determined by the binary variables. + +The objective is to minimize the makespan by scheduling the operations in a way that respects the precedence constraints within each job, avoids overlapping operations on the same machine, and ensures all start times are non-negative. The solution must also accurately reflect the completion time of the last operation, ensuring the makespan is minimized. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {1, 2, 3, 4, 5, 6} +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Machines: M = {1, 2, 3} + +2) Machine assignment for every operation (which machine executes the operation) +- Operation (1,1) -> Machine 1 +- Operation (1,2) -> Machine 2 +- Operation (2,1) -> Machine 1 +- Operation (2,2) -> Machine 3 +- Operation (3,1) -> Machine 2 +- Operation (3,2) -> Machine 3 +- Operation (4,1) -> Machine 1 +- Operation (4,2) -> Machine 2 +- Operation (5,1) -> Machine 2 +- Operation (5,2) -> Machine 3 +- Operation (6,1) -> Machine 3 +- Operation (6,2) -> Machine 1 + +3) Processing times (duration) for each operation +- All operations have processing time p_{j,k} = 1",4.0 +['TravelingSalesman'],"The goal is to minimize the total travel cost for a delivery vehicle visiting six cities: City 0, City 1, City 2, City 3, City 4, and City 5. The vehicle must visit each city exactly once and return to the starting point, ensuring the route forms a single tour without subtours. The decision variables include binary choices for whether the vehicle travels directly from one city to another and general integer variables representing the sequence in which cities are visited. The travel costs between cities are as follows: from City 0 to City 1 costs 281, to City 2 costs 300, to City 3 costs 307, to City 4 costs 313, and to City 5 costs 308; from City 1 to City 0 costs 314, to City 2 costs 313, to City 3 costs 324, to City 4 costs 321, and to City 5 costs 323; from City 2 to City 0 costs 317, to City 1 costs 277, to City 3 costs 297, to City 4 costs 299, and to City 5 costs 308; from City 3 to City 0 costs 287, to City 1 costs 304, to City 2 costs 291, to City 4 costs 299, and to City 5 costs 320; from City 4 to City 0 costs 312, to City 1 costs 314, to City 2 costs 292, to City 3 costs 322, and to City 5 costs 275; from City 5 to City 0 costs 320, to City 1 costs 282, to City 2 costs 280, to City 3 costs 313, and to City 4 costs 321. The constraints include ensuring the vehicle arrives at and departs from each city exactly once, assigning sequence numbers to cities to prevent subtours, and ensuring sequence numbers are unique and range from 0 to 5. Subtour elimination is enforced by linking route decisions with sequence numbers, ensuring logical visit order and preventing invalid loops. The solution must respect all constraints to form a valid tour while minimizing total travel costs.",1741.0 +['Capacitated Lot-sizing Problem (CLSP)'],"Minimize the total operational costs, which include production setup costs of 145.2319299071881 per production run, machine startup costs of 65.79360965435782 per startup, inventory holding costs of 0.8652264753415143 for Product 0, 0.9341662877492469 for Product 1, and 1.07339072521956 for Product 2 per unit, and backlogging penalties of 7.607410261300208 for Product 0, 6.920344391340704 for Product 1, and 6.497781941350707 for Product 2 per unit. The decisions include whether to produce each product on each machine in each period, whether to start up a machine for a specific product in a given period, the exact amount of each product to produce on each machine in each period, how much of each product to keep in inventory at the end of each period, and how much to backlog if demand cannot be met. + +The constraints include meeting customer demand in each period, where the total production of each product across both machines, plus any inventory carried over from the previous period, minus any backlog from the previous period, must equal the demand for that period plus the current inventory minus the current backlog. The demands are 21.53845915883999 for Product 0 in Period 0, 22.99877365370049 in Period 1, and 20.41750946364395 in Period 2; 24.34292430654904 for Product 1 in Period 0, 18.69161579278047 in Period 1, and 16.96075239636408 in Period 2; and 15.44135005337596 for Product 2 in Period 0, 24.75403844533352 in Period 1, and 17.09921582696376 in Period 2. + +The total production quantity for each product on each machine in each period, plus any startup time required, must not exceed the machine's capacity. The startup time for Machine 0 is 6.900949355420833, and for Machine 1, it is 7.255877688483005. The capacity of Machine 0 is 73.62985121550206, and for Machine 1, it is 69.3191618930891. Each machine can only be used for one product at a time in each period, meaning the sum of production decisions for all products on a single machine in a given period cannot exceed 1. + +If a machine is used to produce a product in a period but was not used for that product in the previous period, a startup cost is incurred. For the first period, a startup is automatically required if production occurs. Production quantities, inventory levels, and backlogs must be non-negative, and production and startup decisions are binary (0 or 1).",856.1453747955272 +['TravelingSalesman'],"A logistics company must deliver packages to seven cities: City 0, City 1, City 2, City 3, City 4, City 5, and City 6. The goal is to determine the most efficient route that minimizes the total delivery cost while ensuring each city is visited exactly once and the route forms a single continuous loop. The delivery costs between cities are as follows: + +- From City 0: To City 1 costs 277, to City 2 costs 301, to City 3 costs 299, to City 4 costs 301, to City 5 costs 280, and to City 6 costs 282. +- From City 1: To City 0 costs 311, to City 2 costs 291, to City 3 costs 291, to City 4 costs 310, to City 5 costs 292, and to City 6 costs 313. +- From City 2: To City 0 costs 315, to City 1 costs 310, to City 3 costs 285, to City 4 costs 320, to City 5 costs 278, and to City 6 costs 311. +- From City 3: To City 0 costs 290, to City 1 costs 289, to City 2 costs 313, to City 4 costs 289, to City 5 costs 320, and to City 6 costs 305. +- From City 4: To City 0 costs 277, to City 1 costs 313, to City 2 costs 325, to City 3 costs 300, to City 5 costs 300, and to City 6 costs 281. +- From City 5: To City 0 costs 281, to City 1 costs 301, to City 2 costs 311, to City 3 costs 292, to City 4 costs 299, and to City 6 costs 305. +- From City 6: To City 0 costs 280, to City 1 costs 291, to City 2 costs 311, to City 3 costs 298, to City 4 costs 275, and to City 5 costs 298. + +The problem requires that each city is visited exactly once and that each city is the starting point for exactly one trip. To ensure the route forms a single continuous loop without smaller disconnected loops, the company uses constraints involving the position of each city in the tour. Each city is assigned a position in the delivery sequence, ranging from 1 to 7, and these positions must satisfy constraints that prevent subtours. For example, if City 1 is visited before City 2, the position of City 1 must be less than that of City 2. The objective is to minimize the total delivery cost, which is the sum of the costs of all trips taken, while satisfying all constraints.",1988.0 +['Set Cover'],"Cell Tower Deployment for Regional Coverage + + You are tasked with deploying cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 34 distinct areas, each with a specific population size that needs to be covered by the cell towers. Your goal is to determine which towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + + #### Key Decisions: + 1. **Tower Deployment**: Decide whether to build each of the 30 available cell towers. Each tower has a unique construction cost, and building it will enable coverage for certain areas. + 2. **Area Coverage**: Determine which of the 34 areas will be covered by the deployed towers. An area is considered covered if at least one of the towers that can serve it is built. + + #### Objective: + Maximize the total population covered by the deployed cell towers. The population sizes for the 34 areas are as follows: +- Area 0: 3,328 people +- Area 1: 2,066 people +- Area 2: 2,261 people +- Area 3: 4,527 people +- Area 4: 4,205 people +- Area 5: 4,927 people +- Area 6: 2,941 people +- Area 7: 4,277 people +- Area 8: 3,276 people +- Area 9: 4,326 people +- Area 10: 2,852 people +- Area 11: 3,957 people +- Area 12: 2,068 people +- Area 13: 4,724 people +- Area 14: 3,226 people +- Area 15: 3,575 people +- Area 16: 2,922 people +- Area 17: 3,850 people +- Area 18: 4,898 people +- Area 19: 2,288 people +- Area 20: 4,726 people +- Area 21: 2,513 people +- Area 22: 2,336 people +- Area 23: 4,669 people +- Area 24: 4,626 people +- Area 25: 4,585 people +- Area 26: 4,367 people +- Area 27: 4,981 people +- Area 28: 4,049 people +- Area 29: 4,948 people +- Area 30: 2,182 people +- Area 31: 2,421 people +- Area 32: 3,061 people +- Area 33: 4,310 people + + #### Constraints: + 1. **Coverage Requirements**: Each area must be covered by at least one of the towers that can serve it. For example: + - Area 0 can be covered by Towers 0, 3, 4, 7, 9, 16, 17, 20, 22, 24, 25, 26, or 27. + - Area 1 can be covered by Towers 0, 1, 15, 19, 22, 23, 25, 26, 27, or 29. + - Full coverage relationships for all areas: + - Area 0 can be covered by Towers Tower 0, Tower 3, Tower 4, Tower 7, Tower 9, Tower 16, Tower 17, Tower 20, Tower 22, Tower 24, Tower 25, Tower 26, Tower 27. + - Area 1 can be covered by Towers Tower 0, Tower 1, Tower 15, Tower 19, Tower 22, Tower 23, Tower 25, Tower 26, Tower 27, Tower 29. + - Area 2 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 7, Tower 8, Tower 13, Tower 17, Tower 18, Tower 21, Tower 23, Tower 25, Tower 28, Tower 29. + - Area 3 can be covered by Towers Tower 0, Tower 6, Tower 7, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 24, Tower 29. + - Area 4 can be covered by Towers Tower 0, Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 10, Tower 13, Tower 14, Tower 18, Tower 22, Tower 25, Tower 26. + - Area 5 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 8, Tower 11, Tower 12, Tower 14, Tower 17, Tower 19, Tower 25, Tower 26, Tower 29. + - Area 6 can be covered by Towers Tower 1, Tower 2, Tower 6, Tower 7, Tower 9, Tower 11, Tower 17, Tower 18, Tower 19, Tower 20, Tower 26, Tower 27, Tower 29. + - Area 7 can be covered by Towers Tower 3, Tower 5, Tower 7, Tower 8, Tower 11, Tower 12, Tower 14, Tower 20, Tower 27, Tower 29. + - Area 8 can be covered by Towers Tower 2, Tower 5, Tower 6, Tower 8, Tower 11, Tower 17, Tower 19, Tower 20, Tower 21, Tower 22, Tower 23, Tower 27. + - Area 9 can be covered by Towers Tower 1, Tower 5, Tower 7, Tower 8, Tower 10, Tower 12, Tower 14, Tower 17, Tower 19, Tower 20, Tower 21, Tower 22, Tower 24, Tower 26, Tower 29. + - Area 10 can be covered by Towers Tower 2, Tower 4, Tower 6, Tower 10, Tower 12, Tower 14, Tower 15, Tower 18, Tower 20, Tower 22, Tower 23, Tower 28. + - Area 11 can be covered by Towers Tower 4, Tower 7, Tower 8, Tower 11, Tower 12, Tower 13, Tower 17, Tower 18, Tower 22, Tower 23, Tower 26, Tower 27. + - Area 12 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 4, Tower 5, Tower 13, Tower 15, Tower 16, Tower 20, Tower 24, Tower 29. + - Area 13 can be covered by Towers Tower 0, Tower 2, Tower 3, Tower 8, Tower 10, Tower 12, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 23, Tower 28. + - Area 14 can be covered by Towers Tower 0, Tower 3, Tower 5, Tower 8, Tower 9, Tower 13, Tower 14, Tower 16, Tower 22, Tower 23, Tower 25, Tower 28. + - Area 15 can be covered by Towers Tower 0, Tower 3, Tower 4, Tower 5, Tower 6, Tower 10, Tower 11, Tower 15, Tower 16, Tower 17, Tower 19, Tower 20, Tower 24, Tower 28, Tower 29. + - Area 16 can be covered by Towers Tower 1, Tower 2, Tower 7, Tower 9, Tower 15, Tower 18, Tower 20, Tower 21, Tower 25, Tower 26, Tower 27, Tower 28. + - Area 17 can be covered by Towers Tower 4, Tower 5, Tower 6, Tower 8, Tower 13, Tower 15, Tower 16, Tower 17, Tower 18, Tower 19, Tower 21, Tower 24, Tower 28, Tower 29. + - Area 18 can be covered by Towers Tower 0, Tower 2, Tower 3, Tower 7, Tower 9, Tower 10, Tower 11, Tower 12, Tower 14, Tower 16, Tower 19, Tower 20, Tower 21, Tower 24, Tower 26. + - Area 19 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 7, Tower 8, Tower 10, Tower 16, Tower 20, Tower 22, Tower 23, Tower 24, Tower 26, Tower 27, Tower 29. + - Area 20 can be covered by Towers Tower 3, Tower 4, Tower 6, Tower 7, Tower 13, Tower 15, Tower 17, Tower 18, Tower 19, Tower 22, Tower 23, Tower 24, Tower 27, Tower 28, Tower 29. + - Area 21 can be covered by Towers Tower 1, Tower 3, Tower 10, Tower 12, Tower 13, Tower 14, Tower 20, Tower 21, Tower 23, Tower 25, Tower 26, Tower 29. + - Area 22 can be covered by Towers Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 13, Tower 14, Tower 17, Tower 25, Tower 28. + - Area 23 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 14, Tower 17, Tower 20, Tower 25, Tower 26, Tower 27, Tower 28. + - Area 24 can be covered by Towers Tower 0, Tower 1, Tower 5, Tower 6, Tower 12, Tower 13, Tower 15, Tower 21, Tower 24, Tower 27, Tower 29. + - Area 25 can be covered by Towers Tower 4, Tower 6, Tower 9, Tower 13, Tower 14, Tower 15, Tower 17, Tower 21, Tower 22, Tower 25, Tower 28, Tower 29. + - Area 26 can be covered by Towers Tower 1, Tower 6, Tower 10, Tower 15, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 26, Tower 27, Tower 28. + - Area 27 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 5, Tower 7, Tower 12, Tower 16, Tower 19, Tower 21, Tower 23, Tower 26, Tower 27, Tower 28, Tower 29. + - Area 28 can be covered by Towers Tower 1, Tower 2, Tower 6, Tower 7, Tower 8, Tower 10, Tower 13, Tower 16, Tower 18, Tower 19, Tower 21, Tower 22, Tower 27, Tower 28. + - Area 29 can be covered by Towers Tower 0, Tower 2, Tower 3, Tower 4, Tower 7, Tower 8, Tower 9, Tower 10, Tower 12, Tower 14, Tower 20, Tower 21, Tower 22, Tower 28, Tower 29. + - Area 30 can be covered by Towers Tower 2, Tower 4, Tower 6, Tower 7, Tower 8, Tower 9, Tower 11, Tower 16, Tower 17, Tower 24. + - Area 31 can be covered by Towers Tower 0, Tower 9, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 25, Tower 26, Tower 27. + - Area 32 can be covered by Towers Tower 3, Tower 4, Tower 6, Tower 8, Tower 9, Tower 10, Tower 17, Tower 18, Tower 20, Tower 23, Tower 25, Tower 26, Tower 27, Tower 28, Tower 29. + - Area 33 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 4, Tower 5, Tower 8, Tower 10, Tower 13, Tower 15, Tower 20, Tower 22, Tower 26, Tower 27, Tower 28. + + 2. **Budget Constraint**: The total cost of building the selected towers must not exceed the available budget of 675 units. The construction costs for the towers are: + - Tower 0: 248 units + - Tower 1: 226 units + - Tower 2: 454 units + - Tower 3: 466 units + - Tower 4: 241 units + - Tower 5: 201 units + - Tower 6: 415 units + - Tower 7: 486 units + - Tower 8: 485 units + - Tower 9: 492 units + - Tower 10: 249 units + - Tower 11: 486 units + - Tower 12: 417 units + - Tower 13: 327 units + - Tower 14: 257 units + - Tower 15: 486 units + - Tower 16: 438 units + - Tower 17: 266 units + - Tower 18: 473 units + - Tower 19: 268 units + - Tower 20: 293 units + - Tower 21: 379 units + - Tower 22: 476 units + - Tower 23: 482 units + - Tower 24: 367 units + - Tower 25: 239 units + - Tower 26: 368 units + - Tower 27: 492 units + - Tower 28: 454 units + - Tower 29: 319 units + + 3. **Binary Decisions**: Each tower can either be built (1) or not built (0), and each area can either be covered (1) or not covered (0). + + #### Summary: + Select which towers to build and which areas to cover to maximize the total population served, ensuring coverage requirements are met and the total construction cost does not exceed the budget. The solution must account for the specific coverage capabilities of each tower and the population sizes of the areas.",107698.0 +"['Capacitated Lot-sizing Problem (CLSP)', 'Production Planning Problem', 'Operations Optimization']","Minimize the total operational costs, which include production setup costs, machine startup costs, inventory holding costs, and backlogging costs, while meeting customer demand for each product in each period. The decisions involve determining whether to produce each product on each machine in each period, deciding whether to start up each machine for each product in each period, determining the exact quantity of each product to produce on each machine in each period, and deciding how much inventory to carry over or how much demand to backlog for each product in each period. The total cost includes production setup costs of 149.1695690337089 per production run, machine startup costs of 69.24269269143792 per machine startup, inventory holding costs of 1.080318154052025 for Product 0 and 1.143069947055348 for Product 1, and backlogging costs of 6.798895023015501 for Product 0 and 6.372651501311341 for Product 1. The constraints ensure that the total production of each product across all machines, plus any inventory carried over from the previous period, minus any backlog from the previous period, equals the demand for that product in the current period, plus any inventory carried forward, minus any backlogged demand. The demands are 22.92965094851524 units for Product 0 in Period 0, 19.03396834281949 units in Period 1, and 18.90167422155575 units in Period 2, and 19.80185049682801 units for Product 1 in Period 0, 24.22810173621122 units in Period 1, and 16.84556169689551 units in Period 2. The production quantity on each machine, plus the startup time, cannot exceed the machine's capacity, which is 76.7045484622153 for Machine 0 and 69.3695364951939 for Machine 1. Each machine can only produce one product at a time in any given period. If a machine is used for production in a period but was not used in the previous period, a startup cost is incurred. Production quantities, inventory levels, and backlogs must be non-negative, and production and startup decisions must be binary.",564.4326342045792 +[],, +['Facility Dispersion Problem'],"You are tasked with optimizing the placement of facilities across a network of five locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall dispersion between the selected facilities. Dispersion is measured by the weighted distances between pairs of facilities, where the weights represent the importance or benefit of separating specific pairs. Your goal is to select exactly three facilities out of the five possible locations while ensuring that the total dispersion benefit is maximized. + +#### Key Decisions: +- **Facility Selection:** Decide which three out of the five locations will host the facilities. Each location can either be selected (1) or not selected (0). +- **Pairwise Dispersion Activation:** Determine whether the dispersion benefit between a pair of selected facilities is activated. This is represented by a binary decision (1 if both facilities in the pair are selected, 0 otherwise). + +#### Objective: +Maximize the total dispersion benefit across all pairs of selected facilities. The dispersion benefit between each pair of locations is as follows: +- Node 0 and Node 1: 17 units +- Node 0 and Node 2: 18 units +- Node 0 and Node 3: 21 units +- Node 0 and Node 4: 22 units +- Node 1 and Node 0: 22 units +- Node 1 and Node 2: 15 units +- Node 1 and Node 3: 21 units +- Node 1 and Node 4: 21 units +- Node 2 and Node 0: 16 units +- Node 2 and Node 1: 21 units +- Node 2 and Node 3: 23 units +- Node 2 and Node 4: 17 units +- Node 3 and Node 0: 21 units +- Node 3 and Node 1: 22 units +- Node 3 and Node 2: 19 units +- Node 3 and Node 4: 24 units +- Node 4 and Node 0: 24 units +- Node 4 and Node 1: 16 units +- Node 4 and Node 2: 15 units +- Node 4 and Node 3: 23 units + +#### Constraints: +1. **Facility Selection Limit:** Exactly three facilities must be selected out of the five locations. This means the sum of selected facilities must equal three. +2. **Pairwise Dispersion Activation Rules:** + - The dispersion benefit between two locations can only be activated if both locations are selected. + - If either location in a pair is not selected, the dispersion benefit for that pair cannot be activated. + - If both locations in a pair are selected, the dispersion benefit must be activated. + +#### Numerical Parameters: +- The dispersion benefits between each pair of locations are explicitly provided in the objective description above. +- The total number of facilities to be selected is fixed at three. + +#### Practical Implications: +This problem ensures that the selected facilities are spread out in a way that maximizes their overall separation benefit, which could be critical in scenarios such as minimizing risk, optimizing coverage, or ensuring redundancy. The constraints ensure that the solution is both feasible and aligned with the operational requirements of selecting exactly three facilities. + +By solving this problem, you will determine the optimal set of three facilities and the corresponding pairwise dispersion benefits that maximize the total separation value across the network.",135.0 +['Lot-Sizing Problem'],"Inventory Management with Backlogging in a Retail Supply Chain + +You are managing the inventory and ordering process for a retail supply chain over a 19-week period. Your goal is to minimize the total costs associated with ordering, holding inventory, and backlogging unmet demand, while ensuring that customer demand is met as efficiently as possible. The costs and constraints are as follows: + +#### **Cost Components:** +1. **Ordering Costs**: Each time an order is placed in a given week, a fixed cost is incurred. These costs vary by week and are as follows: + - Week 1: \$930, Week 2: \$881, Week 3: \$925, Week 4: \$808, Week 5: \$711, Week 6: \$785, Week 7: \$997, Week 8: \$820, Week 9: \$843, Week 10: \$631, Week 11: \$726, Week 12: \$1,071, Week 13: \$741, Week 14: \$1,157, Week 15: \$916, Week 16: \$791, Week 17: \$1,055, Week 18: \$932, Week 19: \$794. + +2. **Variable Ordering Costs**: In addition to the fixed costs, there is a variable cost per unit ordered in each week. These costs are: + - Week 1: \$40, Week 2: \$26, Week 3: \$38, Week 4: \$20, Week 5: \$25, Week 6: \$22, Week 7: \$40, Week 8: \$39, Week 9: \$25, Week 10: \$26, Week 11: \$34, Week 12: \$25, Week 13: \$26, Week 14: \$27, Week 15: \$27, Week 16: \$35, Week 17: \$23, Week 18: \$26, Week 19: \$36. + +3. **Inventory Holding Costs**: Holding inventory at the end of each week incurs a cost per unit. These costs are: + - Week 1: \$19, Week 2: \$16, Week 3: \$24, Week 4: \$23, Week 5: \$24, Week 6: \$23, Week 7: \$20, Week 8: \$21, Week 9: \$21, Week 10: \$25, Week 11: \$15, Week 12: \$20, Week 13: \$24, Week 14: \$16, Week 15: \$16, Week 16: \$16, Week 17: \$19, Week 18: \$15, Week 19: \$20. + +4. **Backlogging Costs**: If demand cannot be met in a given week, the unmet demand is backlogged to the next week, incurring a cost per backlogged unit. These costs are: + - Week 1: \$55, Week 2: \$54, Week 3: \$71, Week 4: \$58, Week 5: \$51, Week 6: \$55, Week 7: \$45, Week 8: \$52, Week 9: \$60, Week 10: \$52, Week 11: \$52, Week 12: \$52, Week 13: \$76, Week 14: \$50, Week 15: \$47, Week 16: \$66, Week 17: \$70, Week 18: \$48, Week 19: \$69. + +#### **Demand Constraints:** +- The demand for each week must be met either by the inventory on hand, the units ordered in that week, or by backlogging to the next week. The demand for each week is as follows: + - Week 1: 495 units, Week 2: 316 units, Week 3: 375 units, Week 4: 467 units, Week 5: 418 units, Week 6: 398 units, Week 7: 533 units, Week 8: 525 units, Week 9: 361 units, Week 10: 452 units, Week 11: 470 units, Week 12: 378 units, Week 13: 302 units, Week 14: 463 units, Week 15: 438 units, Week 16: 383 units, Week 17: 377 units, Week 18: 429 units, Week 19: 324 units. + +#### **Inventory and Backlogging Rules:** +- The inventory at the start of Week 1 and the end of Week 19 must be zero. +- The backlogged demand at the start of Week 1 and the end of Week 19 must also be zero. +- The amount ordered in any week cannot exceed the total demand over the 19-week period (7,904 units) if an order is placed in that week. + +#### **Objective:** +Your task is to determine the optimal ordering schedule, inventory levels, and backlogged amounts for each week to minimize the total cost, which includes fixed ordering costs, variable ordering costs, inventory holding costs, and backlogging costs. + +#### **Key Decisions:** +1. **Order Placement**: Decide whether to place an order in each week (a binary decision: yes or no). +2. **Order Quantity**: Determine the number of units to order in each week. +3. **Inventory Levels**: Decide how much inventory to hold at the end of each week. +4. **Backlogged Demand**: Determine the amount of unmet demand to backlog to the next week. + +By carefully balancing these decisions, you aim to minimize the total cost while ensuring that all customer demand is met either immediately or through backlogging.",252401.0 +['Assignment Problem'],"The problem involves assigning employees to shifts across two restaurants to minimize costs associated with unfulfilled positions and employee assignments. There are two restaurants, each with two shifts per day, and eight employees available for assignment. Each shift at each restaurant has specific staffing demands for a particular skill, and employees have varying availability and skill sets. + +Restaurant 0, Shift 0, Skill 0 requires 1 employee. Restaurant 0, Shift 0, Skill 1 requires 1 employee. Restaurant 0, Shift 1, Skill 0 requires 1 employee. Restaurant 1, Shift 0, Skill 0 requires 2 employees. Restaurant 1, Shift 1, Skill 1 requires 1 employee. + +Employees can only be assigned to shifts they are available for and to roles that match their skills. Employee 0 is available for Shift 0 but not for Shift 1. Employee 1 is available for Shift 0 but not for Shift 1. Employee 2 is available for Shifts 0 and 1. Employee 3 is available for Shift 1 but not for Shift 0. Employee 4 is available for Shifts 0 and 1. Employee 5 is available for Shift 1 but not for Shift 0. Employees 6 and 7 are available for Shifts 0 and 1. + +Each employee is qualified for specific skills. For instance, Employees 0, 6, 7 are not qualified for Skill 0 but qualified for Skill 1, while Employees 3 and 5 are not qualified for Skill 1 but qualified for Skill 0; Employees 1, 2, 4 are qualified by both Skill 0 and 1. Each employee can be assigned to at most one shift across all restaurants. Assigning employees to shifts incurs specific costs: assigning Employee 0 to Skill 1 costs 1 unit, assigning Employee 1 to Skill 0 costs 2 units and to Skill 1 costs 1 unit, assigning Employee 2 to Skill 0 costs 3 units and to Skill 1 costs 2 units, assigning Employee 3 to Skill 0 costs 2 units, assigning Employee 4 to Skill 0 costs 2 units and to Skill 1 costs 4 units, assigning Employee 5 to Skill 0 costs 3 units, assigning Employee 6 to Skill 1 costs 2 units, and assigning Employee 7 to Skill 1 costs 1 unit. If a restaurant cannot meet the demand for a specific skill in a shift, unfulfilled positions incur a penalty cost of 100 units per position. The goal is to minimize the total cost, which includes the penalty costs for unfulfilled positions and the costs of assigning employees to shifts. Employees can only be assigned to shifts they are available for and to roles that match their skills.",11.0 +['Flow Shop Scheduling'],"Minimize the total completion time of six jobs processed on three machines, where each job must be assigned to exactly one of six schedule positions and each position must have exactly one job. The processing times for each job on each machine are as follows: Job 0 takes 5, 5, and 7 hours on Machines 0, 1, and 2 respectively; Job 1 takes 5, 9, and 10 hours; Job 2 takes 6, 7, and 10 hours; Job 3 takes 8, 7, and 9 hours; Job 4 takes 9, 5, and 9 hours; and Job 5 takes 8, 8, and 7 hours. Each job must start on Machine 1 only after completing on Machine 0, and on Machine 2 only after completing on Machine 1. Additionally, the start time of a job in a later position must be at least the start time of the job in the previous position plus its processing time on the same machine. The objective is to minimize the total completion time, which includes the start time of the last job on the last machine plus its processing time. All start times must be non-negative, and job assignments to positions are binary (either assigned or not). The objective is to minimize the weighted combination of the completion times of the jobs (on the last machine). The weights for Jobs 0 to 5 are, respectively, 5, 6, 10, 7, 8, and 8. The constraints include ensuring one job per schedule position, one schedule position per job, machine precedence, and job precedence.",1746.0 +['Facility Dispersion Problem'],"Facility Dispersion in a Network + +You are tasked with selecting a set of facilities from a network of five locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall dispersion between the selected facilities. Dispersion is measured by the pairwise distances between the facilities, and your goal is to ensure that the selected facilities are as spread out as possible. + +#### Key Decisions: +1. **Facility Selection**: You must choose exactly 2 facilities out of the 5 available locations. Each location can either be selected (1) or not selected (0). +2. **Pairwise Dispersion Activation**: For every pair of facilities, a binary decision determines whether both facilities in the pair are selected. This ensures that dispersion is only calculated for pairs where both facilities are chosen. + +#### Objective: +Maximize the total dispersion across all selected facility pairs. The dispersion values between the facilities are as follows: +- Node 0 and Node 1: 24 units +- Node 0 and Node 2: 25 units +- Node 0 and Node 3: 18 units +- Node 0 and Node 4: 24 units +- Node 1 and Node 2: 17 units +- Node 1 and Node 3: 22 units +- Node 1 and Node 4: 24 units +- Node 2 and Node 3: 25 units +- Node 2 and Node 4: 20 units +- Node 3 and Node 4: 17 units + +#### Constraints: +1. **Facility Selection Limit**: Exactly 2 facilities must be selected from the 5 available locations. +2. **Pairwise Activation Rules**: + - For any pair of facilities, the pairwise dispersion can only be counted if both facilities in the pair are selected. + - If either facility in a pair is not selected, the pairwise dispersion for that pair cannot be counted. + - If both facilities in a pair are selected, the pairwise dispersion must be counted. + +#### Numerical Parameters: +- Total facilities to select: 2 +- Dispersion values between facilities are explicitly provided in the objective section above. + +#### Goal: +Determine which 2 facilities to select and which pairwise dispersions to activate to maximize the total dispersion across the network while adhering to the constraints.",25.0 +['Unit Commitment Problem'],"Minimize the total operational cost, which includes the base cost for active generators, the per MW cost for power output, and the startup cost for generators. The base cost is 79 units per generator for Type 0, 78 units for Type 1, and 78 units for Type 2. The per MW cost is 4 units for Type 0, 3 units for Type 1, and 4 units for Type 2. The startup cost is 355 units for Type 0, 352 units for Type 1, and 346 units for Type 2. The total power output from all generators must meet or exceed the demand in each period, with demands of 173 MW in period 0, 197 MW in period 1, 227 MW in period 2, 201 MW in period 3, and 196 MW in period 4. Each active generator must produce at least its minimum output and no more than its maximum output, with minimum outputs of 35 MW for all types and maximum outputs of 127 MW for Type 0, 129 MW for Type 1, and 128 MW for Type 2. The total maximum output capacity of all active generators must be at least 15% higher than the demand in each period. The number of active generators cannot exceed the available units for each type, with a maximum of 3 generators for Type 0, 1 generator for Type 1, and 5 generators for Type 2. Each generator of each type that is active in a period but was not active in the previous period needs to pay the appropriate startup cost. Whenever a generator is active, it needs to pay its base cost. The number of active generators and startups must be non-negative integers, and the power output must be non-negative.",5233.0 +['Capacitated Lot-sizing Problem (CLSP)'],"Minimize the total operational costs, which include setup costs, startup costs, inventory holding costs, and backlogging costs, while meeting the demand for each item in each period. The setup cost is **136.4441474007699** per production run, and the startup cost is **72.08169936581939** per startup. The holding cost for Item 0 is **1.045681545748948** per unit, for Item 1 is **1.076278851472706** per unit, and for Item 2 is **1.141130030903227** per unit. The backlogging cost for Item 0 is **7.516688931886524** per unit, for Item 1 is **6.810219227144024** per unit, and for Item 2 is **7.989326755092092** per unit. + +The production system involves three items (Item 0, Item 1, and Item 2), two machines (Machine 0 and Machine 1), and three time periods (Period 0, Period 1, and Period 2). For each item, machine, and time period, decide whether to produce the item (binary decision) and how much to produce (continuous decision). Determine whether to start up a machine for a specific item in a given period (binary decision). Decide how much inventory to carry over or how much demand to backlog for each item in each period (continuous decisions). + +The total production, inventory, and backlog must balance the demand for each item in each period. For example, in Period 0, the demand for Item 0 is **15.13481645049243**, for Item 1 is **15.96509719647714**, and for Item 2 is **15.05799468831989**. In Period 1, the demand for Item 0 is **23.22348134905647**, for Item 1 is **24.89682877145792**, and for Item 2 is **20.28924491987942**. In Period 2, the demand for Item 0 is **19.81696900208589**, for Item 1 is **22.4702270586731**, and for Item 2 is **18.30123580755789**. + +The production quantity and startup time must not exceed the machine's capacity. Machine 0 has a capacity of **74.37564803880434**, and starting it up requires **6.598818345230923** units of time. Machine 1 has a capacity of **61.80119129480688**, and starting it up requires **6.132700881798776** units of time. Each machine can produce only one item at a time in any given period. For example, in Period 0, Machine 0 can produce either Item 0, Item 1, or Item 2, but not more than one. Similarly, Machine 1 can produce only one item at a time in any period. + +If a machine is used to produce an item in a period after being idle in the previous period, a startup cost is incurred. For example, if Machine 0 is used to produce Item 0 in Period 1 but was not used in Period 0, a startup cost of **72.08169936581939** is incurred. Production quantities, inventory levels, and backlogs must be non-negative. Production and startup decisions are binary (either 0 or 1). + +The goal is to determine the optimal production, startup, inventory, and backlog decisions for each item, machine, and time period to minimize the total cost while satisfying all constraints.",889.1372720402694 +['Job Shop'],"In a manufacturing facility, six distinct jobs labeled Job 0 to Job 5 need to be processed on a set of machines. Each job consists of two operations that must be performed in a specific order, with each operation assigned to a specific machine. The goal is to schedule these operations to minimize the total completion time, known as the makespan, ensuring all jobs are completed as quickly as possible. + +Each operation has a processing time of 1 unit. For each job, the second operation cannot start until the first operation is completed. For example, Operation 1 of Job 0 cannot start until Operation 0 of Job 0 is finished, and this applies to all jobs. + +If two operations are assigned to the same machine, they cannot overlap in time. The order in which they are processed is determined by a binary decision variable. For example, if Operation 1 of Job 0 and Operation 1 of Job 4 are assigned to the same machine, one must be scheduled before the other. The binary variable X_(0,1)_(4,1) determines the order: if X_(0,1)_(4,1) is 1, then Operation 1 of Job 0 is scheduled before Operation 1 of Job 4; if it is 0, then Operation 1 of Job 4 is scheduled before Operation 1 of Job 0. This logic applies to all pairs of operations on shared machines. + +The total completion time, C_max, must be at least as large as the completion time of every operation. For example, C_max must be greater than or equal to the completion time of Operation 0 of Job 0, which is S_(0,0) plus 1 unit. This applies to all operations across all jobs. + +The start time of every operation must be non-negative. For example, S_(0,0) must be greater than or equal to 0, and this applies to all operations. + +The binary variables X_(j,i)_(k,l) are used to enforce the order of operations on shared machines and can only take values of 0 or 1. + +The objective is to minimize the makespan, C_max, which represents the total time required to complete all jobs. This ensures that the manufacturing process is as efficient as possible, reducing idle time and maximizing throughput. All processing times are 1 unit, and the makespan must be at least 1 unit greater than the completion time of every operation. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3, 4, 5} +- For each job j in J, there are 2 operations indexed i = 0..1. We denote operation (j,i). +- Machines: M = {M0, M1, M2} + +2) Machine assignment for every operation (explicit listing) +- Operation (0,0) -> Machine M0 +- Operation (0,1) -> Machine M1 +- Operation (1,0) -> Machine M0 +- Operation (1,1) -> Machine M2 +- Operation (2,0) -> Machine M1 +- Operation (2,1) -> Machine M2 +- Operation (3,0) -> Machine M2 +- Operation (3,1) -> Machine M0 +- Operation (4,0) -> Machine M1 +- Operation (4,1) -> Machine M0 +- Operation (5,0) -> Machine M2 +- Operation (5,1) -> Machine M1 + +3) Operation durations +- All operations have processing time p_(j,i) = 1 for every job j and operation i.",4.0 +[],, +['Unit Commitment Problem'],"Determine the number of generators of each type (Type 0 and Type 1) to operate, the power output for each generator type, and the number of startups in each of the five consecutive time periods to minimize the total cost of operation. The total cost includes a base cost of 80 units per Type 0 generator and 79 units per Type 1 generator for each period they are active, a power output cost of 3 units per MW for Type 0 generators and 4 units per MW for Type 1 generators, and a startup cost of 352 units for Type 0 generators and 346 units for Type 1 generators. The total power output from all generators must meet or exceed the demand in each period: 217 MW in period 0, 209 MW in period 1, 194 MW in period 2, 182 MW in period 3, and 164 MW in period 4. Each Type 0 generator must produce at least 34 MW and at most 129 MW when active, and each Type 1 generator must produce at least 34 MW and at most 127 MW when active. The combined maximum output of all active generators must meet or exceed specific capacity requirements: 249.55 MW in period 0, 240.35 MW in period 1, 223.1 MW in period 2, 209.3 MW in period 3, and 188.6 MW in period 4. No more than 1 Type 0 generator and 2 Type 1 generators can be active in any period. The number of startups for each generator type is constrained by the number of generators available and the number of generators active in the previous period. The variables representing the number of generators and startups are general integer variables.",4712.0 +['Job Shop'],"A manufacturing facility produces custom parts for seven jobs, labeled Job 0 through Job 6. Each job consists of two operations that must be performed in sequence, with the second operation starting only after the first is completed. Operations are assigned to specific machines, and some operations share the same machine, meaning they cannot be processed simultaneously. The goal is to schedule all operations to minimize the total completion time, known as the makespan. + +Each operation has a processing time of 1 unit. For example, Operation 0 of Job 0 takes 1 unit of time, and Operation 1 of Job 0 also takes 1 unit of time. This applies to all operations across all jobs. Operations sharing the same machine must be scheduled such that one finishes before the other starts. Binary decision variables determine the order of operations on shared machines. For instance, if Operation 1 of Job 0 and Operation 0 of Job 1 share a machine, a binary variable decides whether Operation 1 of Job 0 finishes before Operation 0 of Job 1 starts, or vice versa. + +The makespan must be at least as large as the completion time of every operation. For example, if Operation 1 of Job 0 finishes at time 2, the makespan must be at least 2. A large constant, 100, is used in the constraints to ensure that operations on the same machine do not overlap. The objective is to minimize the makespan, ensuring the facility operates as efficiently as possible while respecting all precedence and machine capacity constraints. + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 0 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 1 + - Operation 1 of Job 1: Machine 2 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 0 + - Operation 0 of Job 3: Machine 0 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 1 + - Operation 1 of Job 4: Machine 0 + - Operation 0 of Job 5: Machine 2 + - Operation 1 of Job 5: Machine 1 + - Operation 0 of Job 6: Machine 0 + - Operation 1 of Job 6: Machine 1",5.0 +['Flow Shop Scheduling'],"There are 5 jobs labeled Job 0 to Job 4 and 4 machines labeled Machine 0 to Machine 3. Each job must be processed on all machines in sequence, with specific processing times for each job on each machine. The goal is to assign each job to a unique schedule position (Position 0 to Position 4) and determine the start times for each job on each machine to minimize the total completion time. Each schedule position must be assigned to exactly one job, and each job must be assigned to exactly one schedule position. A job cannot start on a machine until it has finished processing on the previous machine, and a job in a later schedule position cannot start on a machine until the job in the previous position has finished processing on that machine. The objective function includes coefficients for the schedule variables in the last position: 6 for Job 0, 5 for Job 1, 10 for Job 2, 10 for Job 3, and 8 for Job 4, as well as the start time of the last job on the last machine. The processing times for each job on each machine are as follows: Job 0 takes 9, 9, 7, and 6 hours on Machines 0 to 3, respectively; Job 1 takes 6, 6, 7, and 5 hours; Job 2 takes 9, 10, 10, and 10 hours; Job 3 takes 6, 10, 6, and 10 hours; and Job 4 takes 10, 7, 6, and 8 hours. The JobSchedule variables are binary, indicating whether a job is assigned to a specific schedule position, and the StartTime variables represent the start times of jobs on machines, which must be non-negative.",65.0 +['Facility Dispersion Problem'],"You are tasked with selecting exactly 2 out of 5 nodes (node_0, node_1, node_2, node_3, node_4) to host facilities, with the goal of maximizing the minimum distance between any two selected facilities. The decision variables are binary: each node can either be selected (1) or not selected (0), and for every pair of nodes, a binary variable indicates whether both nodes in the pair are selected. The distances between nodes are derived from the LP constraints, where the distance between node_0 and node_1 is 18 units (1,000,018 - 1,000,000), node_0 and node_2 is 36 units (1,000,036 - 1,000,000), node_0 and node_3 is 28 units (1,000,028 - 1,000,000), node_0 and node_4 is 20 units (1,000,020 - 1,000,000), node_1 and node_0 is 25 units (1,000,025 - 1,000,000), node_1 and node_2 is 37 units (1,000,037 - 1,000,000), node_1 and node_3 is 36 units (1,000,036 - 1,000,000), node_1 and node_4 is 17 units (1,000,017 - 1,000,000), node_2 and node_0 is 42 units (1,000,042 - 1,000,000), node_2 and node_1 is 42 units (1,000,042 - 1,000,000), node_2 and node_3 is 23 units (1,000,023 - 1,000,000), node_2 and node_4 is 46 units (1,000,046 - 1,000,000), node_3 and node_0 is 31 units (1,000,031 - 1,000,000), node_3 and node_1 is 36 units (1,000,036 - 1,000,000), node_3 and node_2 is 11 units (1,000,011 - 1,000,000), node_3 and node_4 is 50 units (1,000,050 - 1,000,000), node_4 and node_0 is 10 units (1,000,010 - 1,000,000), node_4 and node_1 is 29 units (1,000,029 - 1,000,000), node_4 and node_2 is 24 units (1,000,024 - 1,000,000), and node_4 and node_3 is 42 units (1,000,042 - 1,000,000). The constraints ensure that exactly 2 nodes are selected, and for each pair of nodes, the minimum distance between any two selected facilities must be at least as large as the distance between those two nodes if both nodes in the pair are selected. If either node in the pair is not selected, the constraint is relaxed using a large constant (1,000,000). Additionally, the selection of a pair of nodes is only valid if both nodes in the pair are selected, enforced by ensuring that if a pair is selected, both nodes in the pair must be selected, and if both nodes in a pair are selected, the pair must be selected. The solution must ensure that the facilities are optimally dispersed across the network.",42.0 +['Facility Location Problem (FLP)'],"Cell Tower Coverage Planning + +You are tasked with planning the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 30 distinct areas, each with a specific population size that needs to be served. Your goal is to determine which cell towers to build and which areas to cover, ensuring that the total cost of building the towers does not exceed the available budget. + +#### Key Decisions: +1. **Build or Not Build a Tower**: For each of the 25 potential cell tower locations, you must decide whether to build the tower (a binary decision: build or not build). +2. **Coverage of Areas**: For each of the 30 areas, you must determine whether it will be covered by at least one of the built towers (a binary decision: covered or not covered). + +#### Objective: +Maximize the total population covered by the deployed cell towers. The population sizes for the 30 areas are as follows: +- Area 0: 2,882 people +- Area 1: 4,926 people +- Area 2: 3,930 people +- Area 3: 3,468 people +- Area 4: 3,415 people +- Area 5: 2,697 people +- Area 6: 2,746 people +- Area 7: 3,817 people +- Area 8: 4,806 people +- Area 9: 2,889 people +- Area 10: 2,276 people +- Area 11: 2,087 people +- Area 12: 4,452 people +- Area 13: 3,015 people +- Area 14: 2,305 people +- Area 15: 3,196 people +- Area 16: 4,854 people +- Area 17: 4,466 people +- Area 18: 2,154 people +- Area 19: 4,661 people +- Area 20: 2,587 people +- Area 21: 4,013 people +- Area 22: 3,519 people +- Area 23: 4,837 people +- Area 24: 2,465 people +- Area 25: 3,807 people +- Area 26: 4,498 people +- Area 27: 3,423 people +- Area 28: 2,832 people +- Area 29: 3,064 people + +#### Constraints: +1. **Coverage Requirements**: + - Each area must be covered by at least one of the built towers. For example: + - Area 0 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 7, Tower 9, Tower 16, Tower 17, Tower 20, Tower 22, Tower 24. + - Area 1 can be covered by towers Tower 0, Tower 1, Tower 15, Tower 19, Tower 22, Tower 23. + - Area 2 can be covered by towers Tower 1, Tower 4, Tower 8, Tower 11, Tower 12, Tower 13, Tower 16, Tower 18, Tower 20, Tower 22, Tower 23, Tower 24. + - Area 3 can be covered by towers Tower 1, Tower 3, Tower 4, Tower 7, Tower 8, Tower 10, Tower 14, Tower 16, Tower 19, Tower 20, Tower 22, Tower 24. + - Area 4 can be covered by towers Tower 0, Tower 3, Tower 6, Tower 10, Tower 15, Tower 16, Tower 17, Tower 18, Tower 20, Tower 21, Tower 23. + - Area 5 can be covered by towers Tower 3, Tower 5, Tower 7, Tower 10, Tower 11, Tower 12, Tower 13, Tower 14, Tower 16, Tower 17, Tower 18, Tower 19, Tower 21, Tower 22, Tower 24. + - Area 6 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 6, Tower 8, Tower 12, Tower 13, Tower 14, Tower 15, Tower 17, Tower 20, Tower 23, Tower 24. + - Area 7 can be covered by towers Tower 2, Tower 3, Tower 4, Tower 5, Tower 9, Tower 12, Tower 14, Tower 16, Tower 17, Tower 20. + - Area 8 can be covered by towers Tower 0, Tower 2, Tower 7, Tower 12, Tower 13, Tower 14, Tower 17, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24. + - Area 9 can be covered by towers Tower 0, Tower 2, Tower 3, Tower 5, Tower 6, Tower 7, Tower 8, Tower 9, Tower 10, Tower 12, Tower 17, Tower 18, Tower 19, Tower 21, Tower 24. + - Area 10 can be covered by towers Tower 1, Tower 2, Tower 6, Tower 7, Tower 12, Tower 14, Tower 16, Tower 17, Tower 19, Tower 23, Tower 24. + - Area 11 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 7, Tower 12, Tower 13, Tower 15, Tower 16, Tower 17, Tower 20, Tower 22, Tower 23. + - Area 12 can be covered by towers Tower 2, Tower 4, Tower 5, Tower 8, Tower 10, Tower 11, Tower 13, Tower 14, Tower 17, Tower 19, Tower 22, Tower 23, Tower 24. + - Area 13 can be covered by towers Tower 4, Tower 5, Tower 6, Tower 8, Tower 9, Tower 11, Tower 12, Tower 15, Tower 17, Tower 19, Tower 20, Tower 22, Tower 23. + - Area 14 can be covered by towers Tower 4, Tower 8, Tower 10, Tower 11, Tower 12, Tower 18, Tower 19, Tower 20, Tower 23, Tower 24. + - Area 15 can be covered by towers Tower 0, Tower 2, Tower 3, Tower 4, Tower 6, Tower 7, Tower 8, Tower 9, Tower 11, Tower 13, Tower 14, Tower 15, Tower 18, Tower 22. + - Area 16 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 4, Tower 6, Tower 12, Tower 14, Tower 17, Tower 20, Tower 21, Tower 22. + - Area 17 can be covered by towers Tower 2, Tower 3, Tower 4, Tower 7, Tower 12, Tower 17, Tower 18, Tower 19, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24. + - Area 18 can be covered by towers Tower 1, Tower 3, Tower 4, Tower 5, Tower 7, Tower 8, Tower 9, Tower 11, Tower 12, Tower 15, Tower 18, Tower 19, Tower 20, Tower 22, Tower 23. + - Area 19 can be covered by towers Tower 1, Tower 3, Tower 8, Tower 9, Tower 11, Tower 14, Tower 16, Tower 18, Tower 19, Tower 23. + - Area 20 can be covered by towers Tower 1, Tower 3, Tower 4, Tower 5, Tower 6, Tower 7, Tower 11, Tower 12, Tower 13, Tower 18, Tower 20, Tower 21. + - Area 21 can be covered by towers Tower 0, Tower 3, Tower 4, Tower 6, Tower 8, Tower 10, Tower 12, Tower 16, Tower 18, Tower 21, Tower 24. + - Area 22 can be covered by towers Tower 1, Tower 4, Tower 5, Tower 7, Tower 8, Tower 10, Tower 12, Tower 13, Tower 16, Tower 17, Tower 19, Tower 22, Tower 24. + - Area 23 can be covered by towers Tower 0, Tower 1, Tower 3, Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 12, Tower 14, Tower 15, Tower 17, Tower 21. + - Area 24 can be covered by towers Tower 1, Tower 2, Tower 3, Tower 5, Tower 6, Tower 7, Tower 8, Tower 9, Tower 10, Tower 13, Tower 14, Tower 16, Tower 17, Tower 18, Tower 21. + - Area 25 can be covered by towers Tower 1, Tower 2, Tower 6, Tower 7, Tower 9, Tower 10, Tower 11, Tower 12, Tower 13, Tower 14, Tower 17, Tower 18, Tower 19, Tower 24. + - Area 26 can be covered by towers Tower 1, Tower 2, Tower 4, Tower 6, Tower 9, Tower 12, Tower 15, Tower 16, Tower 18, Tower 19, Tower 20, Tower 22. + - Area 27 can be covered by towers Tower 0, Tower 1, Tower 4, Tower 8, Tower 9, Tower 10, Tower 13, Tower 15, Tower 17, Tower 19, Tower 23, Tower 24. + - Area 28 can be covered by towers Tower 0, Tower 1, Tower 3, Tower 5, Tower 11, Tower 12, Tower 13, Tower 16, Tower 18, Tower 19, Tower 20, Tower 21, Tower 22, Tower 24. + - Area 29 can be covered by towers Tower 0, Tower 1, Tower 2, Tower 8, Tower 10, Tower 11, Tower 12, Tower 14, Tower 15, Tower 16, Tower 17, Tower 18, Tower 20, Tower 23. + +2. **Budget Constraint**: + - The total cost of building the selected towers must not exceed the available budget of 2,501 units. The costs for building each tower are as follows: + - Tower 0: 453 units + - Tower 1: 208 units + - Tower 2: 223 units + - Tower 3: 336 units + - Tower 4: 473 units + - Tower 5: 274 units + - Tower 6: 257 units + - Tower 7: 436 units + - Tower 8: 440 units + - Tower 9: 242 units + - Tower 10: 272 units + - Tower 11: 277 units + - Tower 12: 232 units + - Tower 13: 292 units + - Tower 14: 369 units + - Tower 15: 318 units + - Tower 16: 475 units + - Tower 17: 238 units + - Tower 18: 440 units + - Tower 19: 496 units + - Tower 20: 275 units + - Tower 21: 240 units + - Tower 22: 291 units + - Tower 23: 320 units + - Tower 24: 462 units + +3. **Binary Decisions**: + - Each tower can either be built (1) or not built (0). + - Each area can either be covered (1) or not covered (0). + +#### Summary: +Your task is to select which towers to build and which areas to cover, ensuring that the total population covered is maximized while staying within the budget. The coverage of each area depends on the specific combination of towers built, and the total cost of building the selected towers must not exceed 2,501 units.",104087.0 +"['Transportation Problem', 'Minimum Cost Flow Problem']","Imagine you are managing the distribution of critical medical supplies across eight cities in a region. Each city has a specific supply or demand for these supplies, and your goal is to ensure that all cities receive the necessary resources while minimizing the total transportation cost. The transportation network connects these cities, and each route has a specific cost per unit of supply shipped, as well as a maximum capacity limit. + +#### Cities and Their Supply/Demand: +- **City 0** has a net demand of **34 units** (it needs 34 more units than it currently has). +- **City 1** has a surplus of **11 units** (it can supply 11 units to other cities). +- **City 2** has a surplus of **22 units**. +- **City 3** has a surplus of **10 units**. +- **City 4** has a surplus of **2 units**. +- **City 5** has a surplus of **9 units**. +- **City 6** has a surplus of **6 units**. +- **City 7** has a net demand of **26 units**. + +#### Transportation Costs and Capacities: +The cost of shipping one unit of supply between cities varies depending on the route. Below are the costs and maximum capacities for each route: + +- **From City 0**: + - To City 1: Cost = **9**, Capacity = **59** + - To City 2: Cost = **8**, Capacity = **59** + - To City 3: Cost = **8**, Capacity = **59** + - To City 4: Cost = **9**, Capacity = **43** + - To City 5: Cost = **10**, Capacity = **56** + - To City 6: Cost = **10**, Capacity = **40** + - To City 7: Cost = **9**, Capacity = **37** + +- **From City 1**: + - To City 0: Cost = **10**, Capacity = **36** + - To City 2: Cost = **9**, Capacity = **36** + - To City 3: Cost = **6**, Capacity = **40** + - To City 4: Cost = **9**, Capacity = **39** + - To City 5: Cost = **6**, Capacity = **57** + - To City 6: Cost = **6**, Capacity = **44** + - To City 7: Cost = **8**, Capacity = **56** + +- **From City 2**: + - To City 0: Cost = **6**, Capacity = **52** + - To City 1: Cost = **6**, Capacity = **40** + - To City 3: Cost = **6**, Capacity = **39** + - To City 4: Cost = **7**, Capacity = **31** + - To City 5: Cost = **9**, Capacity = **38** + - To City 6: Cost = **6**, Capacity = **44** + - To City 7: Cost = **10**, Capacity = **58** + +- **From City 3**: + - To City 0: Cost = **10**, Capacity = **30** + - To City 1: Cost = **10**, Capacity = **49** + - To City 2: Cost = **6**, Capacity = **52** + - To City 4: Cost = **10**, Capacity = **47** + - To City 5: Cost = **9**, Capacity = **47** + - To City 6: Cost = **6**, Capacity = **30** + - To City 7: Cost = **8**, Capacity = **34** + +- **From City 4**: + - To City 0: Cost = **6**, Capacity = **39** + - To City 1: Cost = **8**, Capacity = **45** + - To City 2: Cost = **9**, Capacity = **55** + - To City 3: Cost = **8**, Capacity = **36** + - To City 5: Cost = **6**, Capacity = **48** + - To City 6: Cost = **6**, Capacity = **35** + - To City 7: Cost = **10**, Capacity = **54** + +- **From City 5**: + - To City 0: Cost = **9**, Capacity = **43** + - To City 1: Cost = **7**, Capacity = **58** + - To City 2: Cost = **9**, Capacity = **49** + - To City 3: Cost = **6**, Capacity = **33** + - To City 4: Cost = **9**, Capacity = **59** + - To City 6: Cost = **10**, Capacity = **30** + - To City 7: Cost = **7**, Capacity = **59** + +- **From City 6**: + - To City 0: Cost = **10**, Capacity = **33** + - To City 1: Cost = **7**, Capacity = **37** + - To City 2: Cost = **6**, Capacity = **50** + - To City 3: Cost = **7**, Capacity = **34** + - To City 4: Cost = **9**, Capacity = **30** + - To City 5: Cost = **6**, Capacity = **56** + - To City 7: Cost = **10**, Capacity = **44** + +- **From City 7**: + - To City 0: Cost = **7**, Capacity = **46** + - To City 1: Cost = **7**, Capacity = **58** + - To City 2: Cost = **9**, Capacity = **60** + - To City 3: Cost = **7**, Capacity = **34** + - To City 4: Cost = **8**, Capacity = **39** + - To City 5: Cost = **6**, Capacity = **57** + - To City 6: Cost = **9**, Capacity = **42** + +#### Objective: +Your task is to determine the optimal amount of supplies to ship between cities to meet all demands and surpluses while minimizing the total transportation cost. The shipments must respect the capacity limits of each route, and no negative shipments are allowed. + +#### Constraints: +1. **Flow Balance**: For each city, the total amount of supplies shipped into the city must equal the total amount shipped out, adjusted for the city's supply or demand. +2. **Capacity Limits**: The amount of supplies shipped on any route cannot exceed the maximum capacity for that route. +3. **Non-Negativity**: All shipments must be non-negative. + +By solving this problem, you will ensure that all cities receive the necessary medical supplies at the lowest possible transportation cost.",443.0 +['AircraftLanding'],"You are managing the landing schedule for six aircraft at a busy airport. Each aircraft has a target landing time, but due to operational constraints, the actual landing time may vary. Your goal is to minimize the total penalty costs associated with landing aircraft either earlier or later than their target times, while ensuring safe separation between consecutive landings. + +Aircraft 0 has a target landing time of 86 units, can land as early as 60 units or as late as 100 units, and incurs a penalty of 12 units per time unit for landing early and 20 units per time unit for landing late. Aircraft 1 has a target landing time of 60 units, can land as early as 60 units or as late as 90 units, and incurs a penalty of 13 units per time unit for landing early and 16 units per time unit for landing late. Aircraft 2 has a target landing time of 71 units, can land as early as 60 units or as late as 100 units, and incurs a penalty of 16 units per time unit for landing early and 15 units per time unit for landing late. Aircraft 3 has a target landing time of 97 units, can land as early as 67 units or as late as 100 units, and incurs a penalty of 14 units per time unit for landing early and 13 units per time unit for landing late. Aircraft 4 has a target landing time of 89 units, can land as early as 60 units or as late as 100 units, and incurs a penalty of 15 units per time unit for landing early and 18 units per time unit for landing late. Aircraft 5 has a target landing time of 74 units, can land as early as 60 units or as late as 100 units, and incurs a penalty of 14 units per time unit for landing early and 19 units per time unit for landing late. + +To ensure safety, there must be a minimum separation time between the landings of any two aircraft. If Aircraft 0 lands before Aircraft 1, there must be at least 7 units of separation. If Aircraft 1 lands before Aircraft 0, the separation must be at least 5 units. If Aircraft 0 lands before Aircraft 2, there must be at least 7 units of separation. If Aircraft 2 lands before Aircraft 0, the separation must be at least 5 units. If Aircraft 0 lands before Aircraft 3, there must be at least 5 units of separation. If Aircraft 3 lands before Aircraft 0, the separation must be at least 8 units. If Aircraft 0 lands before Aircraft 4, there must be at least 5 units of separation. If Aircraft 4 lands before Aircraft 0, the separation must be at least 7 units. If Aircraft 0 lands before Aircraft 5, there must be at least 5 units of separation. If Aircraft 5 lands before Aircraft 0, the separation must be at least 7 units. If Aircraft 1 lands before Aircraft 2, there must be at least 7 units of separation. If Aircraft 2 lands before Aircraft 1, the separation must be at least 8 units. If Aircraft 1 lands before Aircraft 3, there must be at least 5 units of separation. If Aircraft 3 lands before Aircraft 1, the separation must be at least 8 units. If Aircraft 1 lands before Aircraft 4, there must be at least 8 units of separation. If Aircraft 4 lands before Aircraft 1, the separation must be at least 7 units. If Aircraft 1 lands before Aircraft 5, there must be at least 7 units of separation. If Aircraft 5 lands before Aircraft 1, the separation must be at least 5 units. If Aircraft 2 lands before Aircraft 3, there must be at least 8 units of separation. If Aircraft 3 lands before Aircraft 2, the separation must be at least 8 units. If Aircraft 2 lands before Aircraft 4, there must be at least 8 units of separation. If Aircraft 4 lands before Aircraft 2, the separation must be at least 6 units. If Aircraft 2 lands before Aircraft 5, there must be at least 6 units of separation. If Aircraft 5 lands before Aircraft 2, the separation must be at least 8 units. If Aircraft 3 lands before Aircraft 4, there must be at least 6 units of separation. If Aircraft 4 lands before Aircraft 3, the separation must be at least 8 units. If Aircraft 3 lands before Aircraft 5, there must be at least 5 units of separation. If Aircraft 5 lands before Aircraft 3, the separation must be at least 7 units. If Aircraft 4 lands before Aircraft 5, there must be at least 6 units of separation. If Aircraft 5 lands before Aircraft 4, the separation must be at least 7 units. + +Your task is to determine the actual landing times for all six aircraft such that each aircraft lands within its allowed time window, the separation requirements between consecutive landings are satisfied, and the total penalty cost, which is the sum of the penalties for landing early or late for all aircraft, is minimized. For each pair of aircraft, one must land before the other, and the separation time must be respected. The landing time for each aircraft must fall within its specified earliest and latest landing times. The penalties for landing early or late are calculated based on the difference between the actual landing time and the target landing time.",72.0 +['Job Shop'],"In a manufacturing facility, six jobs labeled Job 0 through Job 5 must be scheduled, each consisting of two operations. The goal is to minimize the total production time (makespan), ensuring all jobs are completed efficiently. Each operation has a processing time of 1 unit of time. Operations within the same job must follow a sequence: the second operation cannot start until the first operation is completed. For example, Operation (0,1) of Job 0 cannot start until Operation (0,0) of Job 0 is finished, and this applies to all jobs. + +Certain operations share the same machine and cannot overlap in time. The machine assignments are listed explicitly below so a solver can be implemented directly. When two operations share the same machine, their order is decided by a binary variable which enforces a non-overlap constraint using a large constant N. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3, 4, 5} +- For each job j in J, there are 2 operations indexed k = 0..1. We denote operation (j,k) where k=0 is the first operation and k=1 is the second operation. +- Machines: M = {M0, M1, M2} + +2) Machine assignment for every operation (which machine executes the operation) +- Operation (0,0) -> M0 +- Operation (0,1) -> M1 +- Operation (1,0) -> M0 +- Operation (1,1) -> M2 +- Operation (2,0) -> M1 +- Operation (2,1) -> M2 +- Operation (3,0) -> M0 +- Operation (3,1) -> M1 +- Operation (4,0) -> M1 +- Operation (4,1) -> M2 +- Operation (5,0) -> M0 +- Operation (5,1) -> M2 + +3) Processing times (duration) for every operation (in time units) +- All operations have processing time p(j,k) = 1 for all j in J and k in {0,1}.",5.0 +"['Set Cover', 'Facility Location Problem']","Cell Tower Coverage Planning + +You are tasked with planning the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 32 distinct areas, each with a specific population size that needs to be covered by cell tower signals. Your goal is to determine which cell towers to build and which areas to cover, ensuring that the total cost of building the towers does not exceed the available budget. + +#### Key Details: +1. **Population Coverage**: Each area has a specific population size that can be covered if a cell tower is built in a location that serves that area. +2. **Cell Tower Options**: There are 29 potential cell tower locations, each with a unique construction cost. The costs range from 210 units for Tower 1 to 498 units for Tower 6. Building a tower at any of these locations will enable coverage for specific areas, depending on the tower's range and location. + +3. **Coverage Requirements**: For an area to be considered covered, at least one of the cell towers that can serve it must be built. Here is the list of which towers can cover which areas: + - Area 0 can be covered if any of the towers 20, 3, 0, 23, 8, 7, 25, 4, 27, 17, 2, or 13 are built. + - Area 1 can be covered if any of the towers 1, 0, 2, 6, 7, 16, 19, 27, 17, 25, 20, or 13 are built. + - Area 2 can be covered if any of the towers 7, 14, 18, 8, 0, 5, 22, 13, 10, 25, 4, or 6 are built. + - Area 3 can be covered if any of the towers 24, 10, 3, 2, 12, 26, 11, 22, 19, 8, 1, or 14 are built. + - Area 4 can be covered if any of the towers 17, 3, 12, 2, 28, 9, 20, 19, 11, 18, 6, or 25 are built. + - Area 5 can be covered if any of the towers 1, 21, 7, 24, 9, 2, 26, 3, 12, 8, 14, or 11 are built. + - Area 6 can be covered if any of the towers 5, 11, 27, 6, 21, 8, 22, 24, 20, 2, 28, or 17 are built. + - Area 7 can be covered if any of the towers 23, 7, 5, 14, 12, 8, 20, 17, 27, 10, 1, or 22 are built. + - Area 8 can be covered if any of the towers 26, 1, 25, 10, 12, 8, 2, 6, 18, 28, 21, or 15 are built. + - Area 9 can be covered if any of the towers 12, 20, 14, 4, 8, 25, 7, 17, 21, 24, 18, or 13 are built. + - Area 10 can be covered if any of the towers 28, 18, 12, 11, 7, 4, 16, 15, 2, 1, 3, or 23 are built. + - Area 11 can be covered if any of the towers 20, 5, 25, 21, 13, 19, 2, 12, 26, 23, 14, or 16 are built. + - Area 12 can be covered if any of the towers 8, 17, 0, 21, 23, 3, 25, 27, 28, 10, 24, or 9 are built. + - Area 13 can be covered if any of the towers 13, 5, 14, 0, 23, 24, 8, 16, 27, 21, 3, or 9 are built. + - Area 14 can be covered if any of the towers 26, 20, 16, 19, 6, 4, 11, 5, 17, 28, 0, or 10 are built. + - Area 15 can be covered if any of the towers 15, 0, 3, 11, 9, 7, 1, 23, 18, 2, 19, or 28 are built. + - Area 16 can be covered if any of the towers 26, 2, 24, 17, 28, 4, 23, 21, 15, 25, 5, or 8 are built. + - Area 17 can be covered if any of the towers 16, 27, 19, 13, 6, 17, 22, 24, 9, 12, 11, or 14 are built. + - Area 18 can be covered if any of the towers 28, 16, 14, 3, 7, 24, 2, 10, 0, 18, 17, or 23 are built. + - Area 19 can be covered if any of the towers 18, 7, 0, 2, 22, 20, 1, 27, 25, 24, 10, or 23 are built. + - Area 20 can be covered if any of the towers 16, 7, 8, 21, 15, 6, 17, 4, 18, 20, 24, or 27 are built. + - Area 21 can be covered if any of the towers 25, 15, 28, 13, 6, 3, 23, 21, 26, 11, 20, or 18 are built. + - Area 22 can be covered if any of the towers 14, 27, 23, 1, 21, 20, 26, 3, 25, 12, 10, or 24 are built. + - Area 23 can be covered if any of the towers 7, 6, 27, 17, 14, 4, 13, 5, 8, 24, 28, or 2 are built. + - Area 24 can be covered if any of the towers 14, 25, 17, 3, 1, 20, 26, 0, 2, 7, 5, or 13 are built. + - Area 25 can be covered if any of the towers 15, 28, 6, 12, 1, 5, 25, 0, 22, 8, 14, or 9 are built. + - Area 26 can be covered if any of the towers 13, 22, 23, 25, 17, 21, 27, 15, 4, 6, 9, or 19 are built. + - Area 27 can be covered if any of the towers 1, 18, 23, 17, 28, 26, 10, 24, 21, 27, 15, or 16 are built. + - Area 28 can be covered if any of the towers 27, 16, 5, 1, 28, 2, 26, 23, 19, 21, 7, or 12 are built. + - Area 29 can be covered if any of the towers 3, 18, 7, 27, 19, 1, 24, 2, 13, 25, 22, or 16 are built. + - Area 30 can be covered if any of the towers 10, 8, 6, 21, 22, 28, 7, 27, 12, 4, 9, or 14 are built. + - Area 31 can be covered if any of the towers 10, 24, 2, 0, 14, 19, 18, 3, 26, 17, 6, or 16 are built. +4. **Budget Constraint**: The total cost of building the selected cell towers must not exceed 3,226 units. This budget must be carefully allocated to maximize population coverage while staying within the financial limit. + +5. **Decision Variables**: + - **Build Decisions**: For each of the 29 cell tower locations, you must decide whether to build the tower (yes or no). + - **Coverage Decisions**: For each of the 32 areas, you must determine whether it will be covered (yes or no), based on the towers built. + +#### Objective: +Your primary objective is to maximize the total population covered by the cell towers. This means selecting a combination of towers to build that ensures the largest possible number of people across the 32 areas have access to cell service, while staying within the budget. + +#### Constraints: +1. **Coverage Logic**: An area can only be considered covered if at least one of the cell towers capable of serving it is built. For example, Area 1 can only be covered if at least one of Towers 0, 2, 3, 4, 5, 7, 10, 11, 15, 16, 17, 20, 23, or 28 is built. + +2. **Budget Limit**: The total cost of building the selected towers must not exceed 3,226 units. This requires careful selection of towers to balance coverage and cost. + +3. **Binary Decisions**: Both the decision to build a tower and the decision to cover an area are binary (yes or no). You cannot partially build a tower or partially cover an area. + +#### Numerical Parameters: +- **Population Sizes**: The populations of the 32 areas are as follows: + - Area 0: 2,285 + - Area 1: 3,023 + - Area 2: 3,557 + - Area 3: 3,938 + - Area 4: 3,689 + - Area 5: 2,426 + - Area 6: 4,697 + - Area 7: 3,214 + - Area 8: 2,580 + - Area 9: 2,010 + - Area 10: 3,180 + - Area 11: 4,083 + - Area 12: 2,787 + - Area 13: 3,095 + - Area 14: 2,898 + - Area 15: 4,374 + - Area 16: 4,970 + - Area 17: 3,689 + - Area 18: 4,612 + - Area 19: 3,232 + - Area 20: 4,177 + - Area 21: 2,211 + - Area 22: 3,123 + - Area 23: 4,194 + - Area 24: 3,774 + - Area 25: 3,391 + - Area 26: 3,959 + - Area 27: 3,010 + - Area 28: 3,213 + - Area 29: 3,106 + - Area 30: 2,652 + - Area 31: 3,441 + +- **Tower Costs**: The construction costs for the 29 towers are as follows: + - Tower 0: 323 units + - Tower 1: 210 units + - Tower 2: 360 units + - Tower 3: 480 units + - Tower 4: 375 units + - Tower 5: 491 units + - Tower 6: 498 units + - Tower 7: 267 units + - Tower 8: 413 units + - Tower 9: 417 units + - Tower 10: 291 units + - Tower 11: 388 units + - Tower 12: 284 units + - Tower 13: 463 units + - Tower 14: 309 units + - Tower 15: 382 units + - Tower 16: 476 units + - Tower 17: 325 units + - Tower 18: 428 units + - Tower 19: 265 units + - Tower 20: 323 units + - Tower 21: 282 units + - Tower 22: 441 units + - Tower 23: 467 units + - Tower 24: 491 units + - Tower 25: 232 units + - Tower 26: 396 units + - Tower 27: 370 units + - Tower 28: 309 units + +#### Summary: +Your task is to select which cell towers to build and which areas to cover, ensuring that the total population covered is maximized while staying within the budget of 3,226 units. The decisions must respect the coverage requirements for each area and the binary nature of the build and coverage decisions.",108590.0 +['Team Formulation Problem'],"Assign six staff members to four projects such that each staff member is assigned to exactly one project. Each staff member has specific skill levels in three skills, and each project has required skill levels for each skill. The skill level attained for each project is determined by the highest skill level among the staff members assigned to that project. The skill shortage for each project is the difference between the required skill level and the attained skill level. The goal is to minimize the maximum skill shortage across all projects and skills. Staff members and their skill levels are as follows: Person 0 has Skill 0 (Level 5), Skill 1 (Level 5), Skill 2 (Level 5); Person 1 has Skill 0 (Level 6), Skill 1 (Level 5), Skill 2 (Level 7); Person 2 has Skill 0 (Level 6), Skill 1 (Level 5), Skill 2 (Level 7); Person 3 has Skill 0 (Level 5), Skill 1 (Level 5), Skill 2 (Level 6); Person 4 has Skill 0 (Level 7), Skill 1 (Level 5), Skill 2 (Level 6); Person 5 has Skill 0 (Level 6), Skill 1 (Level 7), Skill 2 (Level 5). Project skill requirements are as follows: Project 0 requires Skill 0 (Level 9), Skill 1 (Level 9), Skill 2 (Level 7); Project 1 requires Skill 0 (Level 7), Skill 1 (Level 7), Skill 2 (Level 8); Project 2 requires Skill 0 (Level 9), Skill 1 (Level 9), Skill 2 (Level 7); Project 3 requires Skill 0 (Level 7), Skill 1 (Level 8), Skill 2 (Level 9). The objective is to minimize the maximum skill shortage across all projects and skills.",0.0 +['Facility Dispersion Problem'],"Facility Dispersion in a Network of Locations + +Imagine you are tasked with selecting a subset of facilities to open across a network of five locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall dispersion between the selected facilities. Dispersion is measured by the pairwise distances between the facilities, and your goal is to ensure that the selected facilities are as spread out as possible to optimize coverage and minimize redundancy. + +#### Key Details: +1. **Facility Selection**: You must select exactly **2 facilities** to open from the five available locations. Each location can either be chosen (1) or not chosen (0) to host a facility. +2. **Dispersion Values**: The pairwise distances (dispersion values) between the locations are as follows: + - Node 0 and Node 1: 20 units + - Node 0 and Node 2: 20 units + - Node 0 and Node 3: 18 units + - Node 0 and Node 4: 18 units + - Node 1 and Node 0: 22 units + - Node 1 and Node 2: 21 units + - Node 1 and Node 3: 16 units + - Node 1 and Node 4: 17 units + - Node 2 and Node 0: 16 units + - Node 2 and Node 1: 23 units + - Node 2 and Node 3: 15 units + - Node 2 and Node 4: 24 units + - Node 3 and Node 0: 23 units + - Node 3 and Node 1: 16 units + - Node 3 and Node 2: 23 units + - Node 3 and Node 4: 20 units + - Node 4 and Node 0: 19 units + - Node 4 and Node 1: 21 units + - Node 4 and Node 2: 15 units + - Node 4 and Node 3: 25 units + +3. **Objective**: Maximize the total dispersion between the selected facilities. This means you want to choose the two facilities such that the sum of their pairwise distances is as large as possible. + +4. **Constraints**: + - **Selection Constraint**: Exactly two facilities must be selected from the five locations. + - **Pairwise Activation**: For any two locations, the pairwise dispersion value is only counted if both locations are selected. + - **Logical Relationships**: If a location is not selected, no pairwise dispersion involving that location can be included in the total. + +#### Decision to Make: +- Which two facilities to open from the five available locations to maximize the total pairwise dispersion. + +#### Numerical Parameters: +- All pairwise dispersion values are provided above. +- The total number of facilities to select is fixed at 2. + +This problem requires balancing the selection of facilities to ensure they are as dispersed as possible, while adhering to the constraint of selecting exactly two locations. The solution will provide the optimal pair of facilities to open, maximizing the overall dispersion across the network.",45.0 +['Facility Dispersion Problem'],"Select a set of facilities from five potential locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the minimum distance between any two selected facilities. The decision variables include binary choices for selecting each location and binary variables for selecting pairs of locations. The objective is to maximize the minimum distance (MinDistance) between any two selected facilities. Exactly two facilities must be selected. For each pair of locations, if both are selected, the distance between them must be at least MinDistance. If either or both locations in a pair are not selected, the distance constraint is relaxed using a large constant (1,000,000). The pairwise selection variables are symmetric, meaning selecting Node i and Node j is the same as selecting Node j and Node i. Additionally, the pairwise selection variables for the same node (e.g., Node 0 and Node 0) are set to 0. The distances between locations are as follows: Node 0 and Node 1 (41 units), Node 0 and Node 2 (31 units), Node 0 and Node 3 (43 units), Node 0 and Node 4 (10 units), Node 1 and Node 2 (34 units), Node 1 and Node 3 (46 units), Node 1 and Node 4 (12 units), Node 2 and Node 3 (41 units), Node 2 and Node 4 (24 units), and Node 3 and Node 4 (23 units). All decision variables are binary, and the constraints ensure that pairwise selections are consistent with individual selections.",46.0 +"['Transportation Problem', 'Minimum Cost Flow Problem']","Managing Resource Distribution Across Cities + +You are responsible for managing the distribution of a critical resource (e.g., medical supplies, food, or energy) across eight interconnected cities. Each city has a specific supply or demand for the resource, and your goal is to ensure that all cities receive the required amount while minimizing the total transportation cost. The transportation costs vary depending on the routes between cities, and each route has a maximum capacity limit for resource shipments. + +#### City Requirements: +- **City 0** has a surplus of 51 units of the resource. +- **City 1** requires an additional 2 units to meet its demand. +- **City 2** requires an additional 7 units. +- **City 3** requires an additional 11 units. +- **City 4** requires an additional 9 units. +- **City 5** requires an additional 25 units. +- **City 6** has a surplus of 19 units. +- **City 7** requires an additional 16 units. + +#### Transportation Costs and Capacities: +The cost of transporting one unit of the resource between cities varies. Below are the costs and maximum capacities for each route: + +- **From City 0**: + - To City 1: Cost = 10, Capacity = 42 + - To City 2: Cost = 10, Capacity = 47 + - To City 3: Cost = 10, Capacity = 31 + - To City 4: Cost = 9, Capacity = 54 + - To City 5: Cost = 8, Capacity = 41 + - To City 6: Cost = 9, Capacity = 55 + - To City 7: Cost = 10, Capacity = 52 + +- **From City 1**: + - To City 0: Cost = 9, Capacity = 40 + - To City 2: Cost = 7, Capacity = 57 + - To City 3: Cost = 10, Capacity = 56 + - To City 4: Cost = 10, Capacity = 57 + - To City 5: Cost = 10, Capacity = 44 + - To City 6: Cost = 9, Capacity = 60 + - To City 7: Cost = 7, Capacity = 50 + +- **From City 2**: + - To City 0: Cost = 7, Capacity = 45 + - To City 1: Cost = 7, Capacity = 43 + - To City 3: Cost = 8, Capacity = 50 + - To City 4: Cost = 9, Capacity = 36 + - To City 5: Cost = 10, Capacity = 42 + - To City 6: Cost = 6, Capacity = 58 + - To City 7: Cost = 10, Capacity = 33 + +- **From City 3**: + - To City 0: Cost = 9, Capacity = 57 + - To City 1: Cost = 9, Capacity = 42 + - To City 2: Cost = 6, Capacity = 30 + - To City 4: Cost = 9, Capacity = 55 + - To City 5: Cost = 10, Capacity = 60 + - To City 6: Cost = 10, Capacity = 41 + - To City 7: Cost = 8, Capacity = 40 + +- **From City 4**: + - To City 0: Cost = 10, Capacity = 42 + - To City 1: Cost = 6, Capacity = 42 + - To City 2: Cost = 9, Capacity = 33 + - To City 3: Cost = 8, Capacity = 52 + - To City 5: Cost = 10, Capacity = 33 + - To City 6: Cost = 7, Capacity = 36 + - To City 7: Cost = 8, Capacity = 31 + +- **From City 5**: + - To City 0: Cost = 7, Capacity = 48 + - To City 1: Cost = 8, Capacity = 42 + - To City 2: Cost = 9, Capacity = 49 + - To City 3: Cost = 6, Capacity = 34 + - To City 4: Cost = 9, Capacity = 48 + - To City 6: Cost = 7, Capacity = 58 + - To City 7: Cost = 7, Capacity = 38 + +- **From City 6**: + - To City 0: Cost = 7, Capacity = 41 + - To City 1: Cost = 7, Capacity = 31 + - To City 2: Cost = 8, Capacity = 31 + - To City 3: Cost = 7, Capacity = 34 + - To City 4: Cost = 6, Capacity = 42 + - To City 5: Cost = 7, Capacity = 48 + - To City 7: Cost = 9, Capacity = 49 + +- **From City 7**: + - To City 0: Cost = 9, Capacity = 49 + - To City 1: Cost = 6, Capacity = 31 + - To City 2: Cost = 6, Capacity = 34 + - To City 3: Cost = 9, Capacity = 50 + - To City 4: Cost = 10, Capacity = 59 + - To City 5: Cost = 10, Capacity = 49 + - To City 6: Cost = 10, Capacity = 60 + +#### Objective: +Your task is to determine the optimal amount of resources to transport between cities to meet all demands while minimizing the total transportation cost. Ensure that: +1. The total amount of resources shipped into each city equals the total amount shipped out, adjusted for the city's supply or demand. +2. No shipment exceeds the capacity of its respective route. +3. All shipments are non-negative. + +By solving this problem, you will ensure efficient resource distribution across the cities at the lowest possible cost.",584.0 +['Assignment Problem'],"The problem involves assigning employees to shifts across two restaurants to meet specific skill demands while minimizing costs. There are seven employees, each with specific availability and skill capabilities. Each shift at each restaurant requires one employee with Skill 0. Assigning an employee to a shift incurs a cost: some assignments cost 1 unit, while others cost 3 units. If a shift cannot be filled, a penalty of 100 units is incurred per unfulfilled position. + +Employees are only available for certain shifts, as specified by the availability constraints. For example, Employee 0 is available for Shift 0 but not for Shift 1 at Restaurant 1, and Employee 1 is available for Shift 0 but not for Shift 1 at Restaurant 1. Employees can only be assigned to roles that match their skills, and each employee can be assigned to at most one shift across all restaurants and roles. The total number of employees assigned to a shift plus any unfulfilled positions must equal the demand for that shift. + +The objective is to minimize the total cost, which includes the cost of assigning employees to roles based on their preferences and the penalty cost for any unfulfilled positions. The preference costs for assignments range from 1 to 3 units, and the penalty cost for unfulfilled positions is 100 units each. The demand for each shift is one employee with the required skill.",4.0 +['TravelingSalesman'],"Minimize the total travel distance for a delivery route across six cities, labeled City 0 through City 5, starting and ending at City 0. Each city must be visited exactly once, and the route must form a single loop without subtours, and finally returning to the origin city. The distances between cities are as follows: +- From City 0: To City 1: 301 km, To City 2: 313 km, To City 3: 298 km, To City 4: 280 km, To City 5: 300 km. +- From City 1: To City 0: 309 km, To City 2: 317 km, To City 3: 317 km, To City 4: 279 km, To City 5: 321 km. +- From City 2: To City 0: 297 km, To City 1: 290 km, To City 3: 302 km, To City 4: 278 km, To City 5: 323 km. +- From City 3: To City 0: 292 km, To City 1: 305 km, To City 2: 321 km, To City 4: 292 km, To City 5: 321 km. +- From City 4: To City 0: 299 km, To City 1: 295 km, To City 2: 282 km, To City 3: 284 km, To City 5: 300 km. +- From City 5: To City 0: 318 km, To City 1: 317 km, To City 2: 304 km, To City 3: 278 km, To City 4: 284 km. + +The task is to find the shortest possible route that visits each city exactly once, and return to the origin city. What is the minimal total distance travelled?",1741.0 +['Facility Location Problem (FLP)'],"""You are tasked with optimizing the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 35 distinct areas, each with a specific population size that needs to be covered by the cell towers. The goal is to determine which towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + +#### Key Decisions: +1. **Build or Not Build**: For each of the 29 potential tower locations, you must decide whether to build the tower (a binary decision: build or not build). +2. **Coverage**: For each of the 35 regions, you must determine whether it will be covered by at least one of the built towers (a binary decision: covered or not covered). + +#### Objective: +Maximize the total population covered by the deployed cell towers. The population sizes for each region are as follows: +- Region 0: 2,328 people +- Region 1: 2,364 people +- Region 2: 2,101 people +- Region 3: 4,451 people +- Region 4: 4,175 people +- Region 5: 3,684 people +- Region 6: 3,871 people +- Region 7: 4,018 people +- Region 8: 2,922 people +- Region 9: 3,610 people +- Region 10: 4,982 people +- Region 11: 3,296 people +- Region 12: 3,211 people +- Region 13: 3,875 people +- Region 14: 4,545 people +- Region 15: 4,382 people +- Region 16: 3,873 people +- Region 17: 3,481 people +- Region 18: 2,656 people +- Region 19: 3,621 people +- Region 20: 4,167 people +- Region 21: 2,027 people +- Region 22: 4,889 people +- Region 23: 3,684 people +- Region 24: 3,987 people +- Region 25: 2,597 people +- Region 26: 4,578 people +- Region 27: 2,159 people +- Region 28: 2,635 people +- Region 29: 2,699 people +- Region 30: 3,364 people +- Region 31: 3,164 people +- Region 32: 4,867 people +- Region 33: 3,938 people +- Region 34: 3,341 people + +#### Constraints: +1. **Coverage Requirement**: For a region to be considered covered, at least one of the towers that can serve it must be built. The specific towers that can cover each region are listed in the problem data. +- Region 0 can be covered by Towers Tower 0, Tower 3, Tower 4, Tower 7, Tower 9, Tower 16, Tower 17, Tower 20, Tower 22, Tower 24, Tower 25, Tower 26, Tower 27. + - Region 1 can be covered by Towers Tower 0, Tower 1, Tower 15, Tower 19, Tower 22, Tower 23, Tower 25, Tower 26, Tower 27, Tower 29. + - Region 2 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 7, Tower 8, Tower 13, Tower 16, Tower 17, Tower 18, Tower 23, Tower 24, Tower 27, Tower 28. + - Region 3 can be covered by Towers Tower 0, Tower 6, Tower 7, Tower 13, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 24, Tower 28. + - Region 4 can be covered by Towers Tower 0, Tower 4, Tower 5, Tower 6, Tower 8, Tower 10, Tower 13, Tower 18, Tower 20, Tower 22, Tower 24, Tower 25, Tower 27. + - Region 5 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 8, Tower 11, Tower 12, Tower 14, Tower 17, Tower 19, Tower 25. + - Region 6 can be covered by Towers Tower 2, Tower 6, Tower 9, Tower 11, Tower 12, Tower 17, Tower 18, Tower 19, Tower 20, Tower 27. + - Region 7 can be covered by Towers Tower 2, Tower 3, Tower 7, Tower 8, Tower 9, Tower 12, Tower 14, Tower 21, Tower 24, Tower 27. + - Region 8 can be covered by Towers Tower 2, Tower 5, Tower 6, Tower 7, Tower 8, Tower 11, Tower 16, Tower 17, Tower 18, Tower 20, Tower 21, Tower 25, Tower 26, Tower 27, Tower 28. + - Region 9 can be covered by Towers Tower 1, Tower 7, Tower 8, Tower 10, Tower 12, Tower 17, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24, Tower 26, Tower 28. + - Region 10 can be covered by Towers Tower 6, Tower 10, Tower 12, Tower 14, Tower 15, Tower 18, Tower 20, Tower 22, Tower 23, Tower 28. + - Region 11 can be covered by Towers Tower 4, Tower 7, Tower 8, Tower 12, Tower 13, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 28. + - Region 12 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 4, Tower 5, Tower 7, Tower 13, Tower 15, Tower 16, Tower 20, Tower 24, Tower 27. + - Region 13 can be covered by Towers Tower 0, Tower 3, Tower 8, Tower 10, Tower 12, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 21, Tower 23, Tower 24. + - Region 14 can be covered by Towers Tower 0, Tower 3, Tower 5, Tower 8, Tower 9, Tower 13, Tower 14, Tower 16, Tower 21, Tower 23, Tower 24, Tower 27. + - Region 15 can be covered by Towers Tower 0, Tower 3, Tower 4, Tower 5, Tower 6, Tower 10, Tower 11, Tower 14, Tower 15, Tower 16, Tower 17, Tower 19, Tower 20, Tower 27, Tower 28. + - Region 16 can be covered by Towers Tower 1, Tower 2, Tower 7, Tower 9, Tower 15, Tower 18, Tower 19, Tower 20, Tower 24, Tower 25, Tower 26, Tower 28. + - Region 17 can be covered by Towers Tower 4, Tower 5, Tower 6, Tower 8, Tower 9, Tower 13, Tower 15, Tower 16, Tower 17, Tower 19, Tower 21, Tower 23, Tower 24, Tower 27. + - Region 18 can be covered by Towers Tower 0, Tower 2, Tower 3, Tower 7, Tower 10, Tower 11, Tower 14, Tower 16, Tower 20, Tower 21, Tower 25, Tower 27, Tower 28. + - Region 19 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 7, Tower 8, Tower 10, Tower 16, Tower 20, Tower 21, Tower 22, Tower 23, Tower 25, Tower 26, Tower 28. + - Region 20 can be covered by Towers Tower 3, Tower 4, Tower 6, Tower 7, Tower 10, Tower 13, Tower 15, Tower 16, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 27, Tower 28. + - Region 21 can be covered by Towers Tower 1, Tower 3, Tower 10, Tower 11, Tower 12, Tower 13, Tower 14, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24, Tower 27. + - Region 22 can be covered by Towers Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 13, Tower 14, Tower 17, Tower 24, Tower 27. + - Region 23 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 14, Tower 17, Tower 20, Tower 24, Tower 25, Tower 27, Tower 28. + - Region 24 can be covered by Towers Tower 0, Tower 1, Tower 5, Tower 6, Tower 12, Tower 13, Tower 15, Tower 20, Tower 23, Tower 26, Tower 28. + - Region 25 can be covered by Towers Tower 4, Tower 6, Tower 9, Tower 13, Tower 14, Tower 15, Tower 17, Tower 21, Tower 22, Tower 25, Tower 27, Tower 28. + - Region 26 can be covered by Towers Tower 1, Tower 10, Tower 15, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 26, Tower 27, Tower 28. + - Region 27 can be covered by Towers Tower 1, Tower 2, Tower 3, Tower 5, Tower 7, Tower 12, Tower 16, Tower 18, Tower 19, Tower 21, Tower 23, Tower 26, Tower 27, Tower 28. + - Region 28 can be covered by Towers Tower 1, Tower 2, Tower 6, Tower 7, Tower 8, Tower 10, Tower 13, Tower 16, Tower 18, Tower 19, Tower 21, Tower 22, Tower 26, Tower 28. + - Region 29 can be covered by Towers Tower 0, Tower 2, Tower 3, Tower 4, Tower 9, Tower 10, Tower 12, Tower 14, Tower 18, Tower 20, Tower 21, Tower 23. + - Region 30 can be covered by Towers Tower 2, Tower 4, Tower 6, Tower 7, Tower 8, Tower 9, Tower 11, Tower 16, Tower 17, Tower 23. + - Region 31 can be covered by Towers Tower 0, Tower 9, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 24, Tower 26, Tower 27. + - Region 32 can be covered by Towers Tower 3, Tower 4, Tower 6, Tower 8, Tower 9, Tower 10, Tower 15, Tower 17, Tower 18, Tower 20, Tower 23, Tower 25, Tower 26, Tower 27, Tower 28. + - Region 33 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 4, Tower 5, Tower 8, Tower 10, Tower 13, Tower 20, Tower 22, Tower 26, Tower 27, Tower 28. + - Region 34 can be covered by Towers Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 11, Tower 13, Tower 14, Tower 17, Tower 20, Tower 21, Tower 22, Tower 24, Tower 25, Tower 27. + +2. **Budget Constraint**: The total cost of building the selected towers must not exceed the available budget of 2,968 units. The cost of building each tower is as follows: + - Tower 0: 476 units + - Tower 1: 388 units + - Tower 2: 444 units + - Tower 3: 230 units + - Tower 4: 449 units + - Tower 5: 363 units + - Tower 6: 458 units + - Tower 7: 400 units + - Tower 8: 208 units + - Tower 9: 373 units + - Tower 10: 454 units + - Tower 11: 348 units + - Tower 12: 465 units + - Tower 13: 445 units + - Tower 14: 378 units + - Tower 15: 288 units + - Tower 16: 335 units + - Tower 17: 361 units + - Tower 18: 254 units + - Tower 19: 362 units + - Tower 20: 290 units + - Tower 21: 287 units + - Tower 22: 291 units + - Tower 23: 213 units + - Tower 24: 237 units + - Tower 25: 232 units + - Tower 26: 356 units + - Tower 27: 221 units + - Tower 28: 289 units + +#### Summary: +Your task is to select which towers to build and which regions to cover, ensuring that the total population covered is maximized while staying within the budget. The coverage of each region depends on the specific towers built, and the total cost of building the selected towers must not exceed 2,968 units. The goal is to make strategic decisions that balance coverage and cost to achieve the best possible outcome for the region.""",123542.0 +['Discrete Lot-sizing and Scheduling Problem'],"Minimize the total production costs, which include setup costs, startup costs, inventory holding costs, and backlogging costs, while meeting customer demand for each product in each period. The setup cost is **121.0782541217011** per production run, and the startup cost is **69.65555018358044** per startup. The inventory holding cost for Product 0 is **1.196682091019819** per unit per period, and for Product 1, it is **1.146459407730801** per unit per period. The backlogging cost for Product 0 is **7.384397449671641** per unit per period, and for Product 1, it is **6.58642780982885** per unit per period. + +The production facility manufactures two types of products (Product 0 and Product 1) using two machines (Machine 0 and Machine 1) over four time periods (Period 0 to Period 3). The demand for Product 0 in Period 0 is **22.28694038228271**, in Period 1 is **15.40553617195959**, in Period 2 is **16.19622794603683**, and in Period 3 is **23.77133167016131**. The demand for Product 1 in Period 0 is **16.51020937307772**, in Period 1 is **15.41395287058592**, in Period 2 is **23.98409764196543**, and in Period 3 is **18.84683167137874**. + +The total production quantity for each product on each machine in each period, plus any startup time required, must not exceed the machine's capacity. Each machine has a limited capacity: Machine 0 has a capacity of **63.4339018881536**, and Machine 1 has a capacity of **67.51600987446864**. The startup time for Machine 0 is **7.917022863645049**, and for Machine 1, it is **7.682853596177669**. Each machine can produce only one product at a time in any given period. A machine can only be started up for a product if it was not producing that product in the previous period. + +The total production, inventory, and backlog must balance customer demand for each product in each period. Production quantities, inventory, and backlog must be non-negative. Production and startup decisions are binary (either 0 or 1). The goal is to determine the optimal production schedule, inventory levels, and backlog amounts to minimize the total cost while meeting all constraints and customer demand.",763.2675651474576 +['Job Shop'],"Minimize the total completion time (makespan) required to process all jobs in a manufacturing facility. There are 7 jobs, each consisting of 2 operations that must be performed in sequence. Each operation has a processing time of 1 unit. Operations from different jobs may share the same machine, but a machine can only handle one operation at a time. To avoid overlaps, the order of operations on shared machines must be determined. + +The start time of each operation must be non-negative, and the start time of an operation in a job must be at least 1 unit after the completion of the previous operation in the same job. For operations sharing a machine, one operation must finish before the other starts. This is enforced using binary decision variables that determine the order of operations on shared machines. A large constant (100) is used in the constraints to ensure logical consistency when determining the order of operations. + +The makespan must be at least as large as the completion time of every operation across all jobs. The objective is to minimize the makespan while ensuring that all precedence and machine capacity constraints are satisfied. The decision variables include the start times of all operations, the binary variables determining the order of operations on shared machines, and the makespan itself. + + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 1 + - Operation 1 of Job 0: Machine 2 + - Operation 0 of Job 1: Machine 0 + - Operation 1 of Job 1: Machine 1 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 0 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 1 + - Operation 1 of Job 4: Machine 1 + - Operation 0 of Job 5: Machine 1 + - Operation 1 of Job 5: Machine 0 + - Operation 0 of Job 6: Machine 0 + - Operation 1 of Job 6: Machine 2",6.0 +['Job Shop'],"There are five jobs, labeled Job 0 through Job 4, each consisting of two operations that must be processed in sequence. Each operation takes exactly 1 unit of time to complete and is assigned to a specific machine. The goal is to schedule these operations to minimize the total completion time, known as the makespan, which is the maximum completion time across all operations. + +For each job, the second operation cannot start until the first operation is completed. This ensures that the operations within a job follow a strict sequence. Additionally, each machine can process only one operation at a time. If two operations are assigned to the same machine, they cannot overlap. A binary decision variable determines the order in which these operations are scheduled on the machine. If the variable is 1, the first operation is scheduled before the second; if it is 0, the order is reversed. + +The constraints ensure that the start time of the second operation in a job is at least 1 unit of time after the start time of the first operation. For operations assigned to the same machine, the start time of one operation must be at least 1 unit of time after the completion of the other operation. A large number (100) is used in the constraints to enforce this order without overlapping. + +The objective is to determine the start times for all operations and the order in which operations are processed on shared machines, such that the total makespan is minimized. This involves ensuring that all precedence constraints within jobs are satisfied, preventing overlapping operations on the same machine, and minimizing the latest completion time across all operations. + + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 1 + - Operation 1 of Job 0: Machine 2 + - Operation 0 of Job 1: Machine 0 + - Operation 1 of Job 1: Machine 1 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 0 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 1 + - Operation 1 of Job 4: Machine 1",5.0 +['Facility Dispersion Problem'],"Facility Dispersion in a Network + +Imagine you are tasked with selecting a set of facilities to maximize their dispersion across a network of five locations (nodes). The goal is to ensure that the selected facilities are as spread out as possible, which is measured by the pairwise distances between them. You must select exactly two facilities out of the five available locations. + +Each pair of locations has a specific distance value associated with it, representing how far apart they are. These distances are as follows: + +- Distance between Node 0 and Node 1: 24 units +- Distance between Node 0 and Node 2: 22 units +- Distance between Node 0 and Node 3: 23 units +- Distance between Node 0 and Node 4: 21 units +- Distance between Node 1 and Node 0: 20 units +- Distance between Node 1 and Node 2: 19 units +- Distance between Node 1 and Node 3: 19 units +- Distance between Node 1 and Node 4: 25 units +- Distance between Node 2 and Node 0: 17 units +- Distance between Node 2 and Node 1: 20 units +- Distance between Node 2 and Node 3: 18 units +- Distance between Node 2 and Node 4: 20 units +- Distance between Node 3 and Node 0: 25 units +- Distance between Node 3 and Node 1: 18 units +- Distance between Node 3 and Node 2: 21 units +- Distance between Node 3 and Node 4: 24 units +- Distance between Node 4 and Node 0: 21 units +- Distance between Node 4 and Node 1: 18 units +- Distance between Node 4 and Node 2: 15 units +- Distance between Node 4 and Node 3: 23 units + +Your objective is to maximize the total dispersion, which is the sum of the distances between all pairs of selected facilities. However, there are specific rules governing the selection process: + +1. **Selection Constraint**: Exactly two facilities must be selected from the five available locations. +2. **Pairwise Activation**: A distance between two locations is only counted if both locations are selected as facilities. +3. **Logical Consistency**: If a pair of locations is selected, the distance between them must be fully accounted for in the total dispersion. Conversely, if either location in a pair is not selected, the distance between them must not contribute to the total dispersion. + +Your task is to determine which two locations to select as facilities to maximize the total dispersion while adhering to the constraints above. The solution must ensure that the selection of facilities is logically consistent and that the total dispersion is calculated accurately based on the pairwise distances provided.",48.0 +['Job Shop'],"There are 5 jobs, each consisting of 2 operations that must be performed in sequence. Each operation has a processing time of 1 unit of time. Operations are assigned to machines, and each machine can process only one operation at a time. The goal is to minimize the total completion time (makespan) of all jobs by determining the start times of each operation and the order in which operations are processed on the machines. + +For each job, the second operation cannot start until the first operation has been completed. For example, Operation 1 of Job 0 must start at least 1 unit of time after Operation 0 of Job 0 starts. This applies to all jobs. Operations assigned to the same machine cannot overlap in time. For every pair of operations on the same machine, one must be processed before the other. If Operation A is processed before Operation B, then Operation B must start at least 1 unit of time after Operation A completes. Conversely, if Operation B is processed before Operation A, then Operation A must start at least 1 unit of time after Operation B completes. These constraints apply to all relevant pairs of operations. + +The makespan must be at least as large as the completion time of every operation. For example, the makespan must be at least 1 unit of time after the start of Operation 0 of Job 0, and this applies to all operations across all jobs. All start times must be non-negative, and the order variables, which determine the sequence of operations on the machines, must be binary (0 or 1). A large constant (100) is used in the constraints to ensure the correct order of operations is enforced. The objective is to minimize the makespan, ensuring all operations are scheduled efficiently without violating the constraints. + + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 2 + - Operation 1 of Job 1: Machine 2 + - Operation 0 of Job 2: Machine 2 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 1 + - Operation 1 of Job 3: Machine 0 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 0",5.0 +['Facility Dispersion Problem'],"Facility Dispersion in a Network of Locations + +You are tasked with selecting a subset of facilities to open across a network of five locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall dispersion between the selected facilities. Dispersion is measured by the pairwise distances between the facilities, with higher distances indicating better dispersion. The goal is to ensure that the selected facilities are as spread out as possible within the network. + +#### Key Decisions: +1. **Facility Selection**: You must decide which facilities to open. Specifically, you need to select exactly 2 out of the 5 available locations to open facilities. Each location can either be selected (1) or not selected (0). +2. **Pairwise Dispersion Activation**: For each pair of selected facilities, you must determine whether their dispersion contributes to the overall objective. This is represented by a binary decision variable that activates only if both facilities in the pair are selected. + +#### Objective: +Maximize the total dispersion across all pairs of selected facilities. The dispersion values between each pair of locations are as follows: +- Node 0 and Node 1: 18 units +- Node 0 and Node 2: 24 units +- Node 0 and Node 3: 21 units +- Node 0 and Node 4: 24 units +- Node 1 and Node 0: 17 units +- Node 1 and Node 2: 21 units +- Node 1 and Node 3: 21 units +- Node 1 and Node 4: 17 units +- Node 2 and Node 0: 19 units +- Node 2 and Node 1: 15 units +- Node 2 and Node 3: 19 units +- Node 2 and Node 4: 22 units +- Node 3 and Node 0: 15 units +- Node 3 and Node 1: 17 units +- Node 3 and Node 2: 22 units +- Node 3 and Node 4: 16 units +- Node 4 and Node 0: 25 units +- Node 4 and Node 1: 24 units +- Node 4 and Node 2: 16 units +- Node 4 and Node 3: 23 units + +The objective is to maximize the sum of these dispersion values for all pairs of selected facilities. + +#### Constraints: +1. **Facility Selection Limit**: Exactly 2 facilities must be selected out of the 5 available locations. This ensures that the solution is neither too sparse nor too dense. +2. **Pairwise Dispersion Activation Rules**: + - For any pair of locations, their dispersion can only contribute to the objective if both facilities in the pair are selected. + - If either facility in a pair is not selected, their dispersion cannot be counted. + - If both facilities in a pair are selected, their dispersion must be counted. + +#### Numerical Parameters: +- The number of facilities to select: 2. +- The dispersion values between each pair of locations are explicitly provided above. + +#### Summary: +Your task is to select 2 out of 5 locations to open facilities such that the total dispersion between the selected facilities is maximized. The dispersion values between each pair of locations are given, and the solution must adhere to the constraints on facility selection and pairwise dispersion activation. The goal is to ensure that the selected facilities are as spread out as possible within the network.",49.0 +"['Capacitated Lot-sizing Problem (CLSP)', 'Production Planning Problem']","You are managing a factory that produces three types of products: **product_0**, **product_1**, and **product_2**, over a five-period planning horizon. The factory operates with three types of machines: **grinders**, **drills**, and **borers**, each with specific capacities and maintenance schedules. Your goal is to maximize the factory's profit by optimizing production, inventory, and sales decisions while adhering to operational constraints. + +The profit per unit sold is **$249** for **product_0**, **$222** for **product_1**, and **$249** for **product_2**. The holding cost for each unit in inventory is **$15** per period. + +In the first period, the production of each product must equal the sum of its sales and inventory for that period. For subsequent periods, the inventory from the previous period plus the current production must equal the sum of current sales and inventory. + +Each unit of **product_0** requires **2 hours** on grinders, **2 hours** on drills, and **1 hour** on borers. Each unit of **product_1** requires **1 hour** on grinders, **2 hours** on drills, and **1 hour** on borers. Each unit of **product_2** requires **1 hour** on grinders, **2 hours** on drills, and **2 hours** on borers. + +The total available grinding capacity is **480 hours** per period, except in period 3, where it is **480 hours**. The total available drilling capacity is **320 hours** per period, except in period 1, where it is reduced to **160 hours**. The total available boring capacity is **160 hours** per period, except in period 0, where it is **0 hours**. + +The maximum inventory capacity for each product is **80 units** per period. The maximum number of units that can be sold for each product in each period is as follows: **product_0**: **63** (period 0), **68** (period 1), **42** (period 2), **58** (period 3), **54** (period 4); **product_1**: **53** (period 0), **62** (period 1), **67** (period 2), **46** (period 3), **56** (period 4); **product_2**: **64** (period 0), **67** (period 1), **65** (period 2), **65** (period 3), **58** (period 4). + +At the end of the planning horizon (period 4), the inventory for each product must be exactly **20 units**. All production, inventory, and sales quantities must be non-negative. Machine downtime for maintenance is accounted for in the capacity constraints. + +Your task is to determine the optimal production, inventory, and sales plan that maximizes profit while satisfying all operational constraints.",104148.0 +['Facility Location Problem'],"Cell Tower Coverage Planning + +You are tasked with planning the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 32 distinct areas, each with a specific population size that needs to be served. Your goal is to determine which cell towers to build and which areas to cover, ensuring that the total cost of building the towers does not exceed the available budget. + +#### Key Decisions: +1. **Build or Not Build**: For each of the 30 potential cell tower locations, you must decide whether to build the tower (yes or no). Building a tower incurs a specific cost, which varies by location. +2. **Coverage Assignment**: For each of the 32 areas, you must determine whether it will be covered by at least one of the built towers (yes or no). An area is considered covered if at least one of the towers that can serve it is built. + +#### Objective: +Maximize the total population covered by the deployed cell towers. The population sizes for the 32 areas are as follows: +- Area 0: 4,657 +- Area 1: 4,663 +- Area 2: 3,139 +- Area 3: 2,579 +- Area 4: 3,171 +- Area 5: 2,796 +- Area 6: 3,502 +- Area 7: 4,951 +- Area 8: 4,211 +- Area 9: 2,885 +- Area 10: 4,948 +- Area 11: 2,195 +- Area 12: 4,002 +- Area 13: 3,856 +- Area 14: 2,663 +- Area 15: 3,197 +- Area 16: 4,465 +- Area 17: 4,648 +- Area 18: 3,796 +- Area 19: 4,714 +- Area 20: 4,952 +- Area 21: 3,088 +- Area 22: 4,029 +- Area 23: 3,379 +- Area 24: 2,002 +- Area 25: 3,840 +- Area 26: 4,037 +- Area 27: 4,246 +- Area 28: 4,852 +- Area 29: 4,423 +- Area 30: 2,630 +- Area 31: 4,951 + +#### Constraints: +1. **Coverage Requirements**: Each area must be covered by at least one of the built towers. The specific towers that can cover each area are listed below: + - Area 0 can be covered by Towers 2, 3, 4, 5, 8, 9, 10, 13, 14, 15, 20, 21, 22, 23, 26, or 28. + - Area 1 can be covered by Towers 1, 3, 5, 6, 9, 14, 15, 17, 18, 20, 21, 25, or 27. + - Area 2 can be covered if any of the towers 20, 3, 0, 23, 8, 7, 24, 4, 28, 17, 2, or 18 are built. + - Area 3 can be covered if any of the towers 13, 1, 0, 2, 6, 7, 16, 19, 27, 17, 25, or 20 are built. + - Area 4 can be covered if any of the towers 13, 7, 14, 18, 8, 0, 5, 22, 29, 10, 25, or 4 are built. + - Area 5 can be covered if any of the towers 6, 24, 10, 3, 2, 12, 26, 11, 22, 19, 8, or 1 are built. + - Area 6 can be covered if any of the towers 23, 14, 17, 3, 12, 2, 27, 9, 20, 19, 11, or 18 are built. + - Area 7 can be covered if any of the towers 6, 22, 2, 1, 21, 7, 9, 27, 24, 3, 12, or 8 are built. + - Area 8 can be covered if any of the towers 14, 20, 26, 11, 5, 27, 24, 6, 21, 8, 2, or 25 are built. + - Area 9 can be covered if any of the towers 17, 23, 7, 5, 14, 12, 8, 20, 29, 27, 10, or 1 are built. + - Area 10 can be covered if any of the towers 7, 26, 1, 25, 10, 12, 8, 2, 6, 18, 28, or 21 are built. + - Area 11 can be covered if any of the towers 20, 15, 12, 29, 14, 4, 8, 24, 7, 17, 26, or 23 are built. + - Area 12 can be covered if any of the towers 23, 18, 13, 28, 12, 11, 7, 4, 16, 15, 2, or 1 are built. + - Area 13 can be covered if any of the towers 27, 3, 4, 20, 5, 21, 13, 19, 2, 12, 26, or 14 are built. + - Area 14 can be covered if any of the towers 16, 8, 17, 0, 21, 23, 3, 25, 27, 28, 10, or 24 are built. + - Area 15 can be covered if any of the towers 9, 13, 5, 14, 0, 23, 24, 8, 16, 27, 21, or 3 are built. + - Area 16 can be covered if any of the towers 27, 20, 9, 26, 28, 16, 19, 6, 4, 11, 5, or 17 are built. + - Area 17 can be covered if any of the towers 24, 16, 0, 19, 10, 15, 27, 3, 11, 9, 7, or 1 are built. + - Area 18 can be covered if any of the towers 7, 28, 18, 2, 26, 23, 15, 25, 17, 4, 20, or 24 are built. + - Area 19 can be covered if any of the towers 17, 5, 8, 16, 19, 13, 6, 29, 23, 9, 12, or 11 are built. + - Area 20 can be covered if any of the towers 14, 28, 16, 29, 3, 7, 24, 2, 10, 0, 18, or 17 are built. + - Area 21 can be covered if any of the towers 7, 18, 29, 0, 2, 22, 20, 1, 27, 25, 24, or 10 are built. + - Area 22 can be covered if any of the towers 2, 16, 7, 8, 21, 15, 6, 17, 4, 18, 20, or 24 are built. + - Area 23 can be covered if any of the towers 7, 25, 15, 28, 13, 6, 3, 23, 21, 26, 11, or 20 are built. + - Area 24 can be covered if any of the towers 13, 14, 27, 23, 1, 21, 20, 26, 3, 25, 12, or 10 are built. + - Area 25 can be covered if any of the towers 25, 27, 3, 7, 6, 29, 17, 14, 4, 13, 5, or 8 are built. + - Area 26 can be covered if any of the towers 14, 7, 27, 2, 29, 17, 3, 1, 20, 24, 0, or 26 are built. + - Area 27 can be covered if any of the towers 29, 24, 27, 7, 5, 13, 15, 23, 6, 12, 1, or 25 are built. + - Area 28 can be covered if any of the towers 12, 0, 29, 8, 25, 14, 9, 13, 17, 15, 4, or 6 are built. + - Area 29 can be covered if any of the towers 9, 6, 1, 18, 23, 17, 27, 10, 25, 21, 26, or 15 are built. + - Area 30 can be covered if any of the towers 16, 27, 29, 5, 1, 28, 2, 26, 23, 19, 21, or 7 are built. + - Area 31 can be covered if any of the towers 12, 3, 18, 7, 27, 19, 1, 24, 2, 13, 25, or 22 are built. +2. **Budget Constraint**: The total cost of building the selected towers must not exceed the available budget of 1,204 units. The costs for building each tower are as follows: + - Tower 0: 455 + - Tower 1: 288 + - Tower 2: 208 + - Tower 3: 441 + - Tower 4: 428 + - Tower 5: 424 + - Tower 6: 285 + - Tower 7: 414 + - Tower 8: 459 + - Tower 9: 488 + - Tower 10: 362 + - Tower 11: 233 + - Tower 12: 341 + - Tower 13: 251 + - Tower 14: 422 + - Tower 15: 261 + - Tower 16: 437 + - Tower 17: 384 + - Tower 18: 442 + - Tower 19: 310 + - Tower 20: 218 + - Tower 21: 491 + - Tower 22: 412 + - Tower 23: 243 + - Tower 24: 311 + - Tower 25: 247 + - Tower 26: 352 + - Tower 27: 318 + - Tower 28: 380 + - Tower 29: 375 + +3. **Binary Decisions**: Both the decision to build a tower and the decision to cover an area are binary (yes or no). You cannot partially build a tower or partially cover an area. + +#### Summary: +Your task is to select which cell towers to build and which areas to cover, ensuring that: +- Each area is served by at least one built tower. +- The total cost of building the towers does not exceed the budget of 1,204 units. +- The total population covered by the selected towers is maximized. + +This problem requires careful balancing of costs and coverage to achieve the best possible outcome within the given constraints.",121467.0 +['Job Shop'],"In a manufacturing facility, you are tasked with scheduling a set of jobs across multiple machines to minimize the total completion time, known as the makespan. Each job consists of a sequence of operations that must be performed in a specific order, and each operation is assigned to a particular machine. The goal is to determine the start times for each operation on its assigned machine while respecting the following constraints: + +Each job has a predefined sequence of operations. For example, Job 0 has two operations: Operation 0 must be completed before Operation 1 can begin. Similarly, Job 1 has Operation 0 and Operation 1, with Operation 0 preceding Operation 1. This pattern applies to all jobs, ensuring that no operation starts before its predecessor in the same job has finished. Specifically, for Job 0, Operation 1 cannot start until Operation 0 has been completed, and for Job 1, Operation 1 cannot start until Operation 0 has been completed. This logic applies to all jobs, ensuring the correct sequence of operations. + +Each machine can process only one operation at a time. To prevent overlapping, the start times of operations assigned to the same machine must be carefully scheduled. For example, if Operation 0 of Job 0 and Operation 0 of Job 1 are assigned to the same machine, they cannot overlap. A binary decision variable determines the order in which these operations are processed. If Operation 0 of Job 0 is scheduled first, Operation 0 of Job 1 must start only after Operation 0 of Job 0 has finished, and vice versa. Similar constraints apply to all pairs of operations assigned to the same machine, ensuring no two operations overlap. A large constant (1000) is used in the constraints to enforce the order of operations on shared machines. + +The makespan is defined as the total time required to complete all jobs. It must be at least as large as the completion time of the last operation across all jobs. For example, if Operation 1 of Job 6 finishes at time 100, the makespan must be at least 100. This ensures that the makespan accurately reflects the total time needed to complete all operations. + +The start time of every operation must be non-negative, meaning no operation can start before time zero. Binary variables are used to determine the order of operations on shared machines. These variables can only take values of 0 or 1, representing the scheduling decisions. The processing time for each operation is implicitly defined by the constraints. For example, the precedence constraints specify that Operation 1 of Job 0 cannot start until Operation 0 of Job 0 has been completed, with a required gap of at least 1 time unit. The binary variables determine the order of operations on shared machines. For instance, if the binary variable for Operation 0 of Job 0 and Operation 0 of Job 1 is 1, Operation 0 of Job 0 is scheduled before Operation 0 of Job 1 on the same machine. The makespan must be at least as large as the completion time of every operation. For example, if Operation 0 of Job 0 finishes at time 10, the makespan must be at least 10. + +The objective is to minimize the makespan, ensuring that all jobs are completed in the shortest possible time while respecting the precedence and machine capacity constraints. This problem requires determining the optimal start times for all operations and the sequence of operations on shared machines to achieve the minimum total completion time. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {0, 1, 2, 3} +- For each job j in J, there are 2 operations indexed k = 0..1. We denote operation (j,k). +- Machines: M = {0, 1} + +2) Machine assignment for every operation (which machine executes the operation) +- Operation (0,0) -> Machine 0 +- Operation (0,1) -> Machine 1 +- Operation (1,0) -> Machine 0 +- Operation (1,1) -> Machine 1 +- Operation (2,0) -> Machine 1 +- Operation (2,1) -> Machine 0 +- Operation (3,0) -> Machine 1 +- Operation (3,1) -> Machine 0 + +3) Processing times (duration) for every operation (in time units) +- p(0,0) = 3 +- p(0,1) = 2 +- p(1,0) = 2 +- p(1,1) = 4 +- p(2,0) = 4 +- p(2,1) = 3 +- p(3,0) = 1 +- p(3,1) = 2",11.0 +['Job Shop'],"There are 7 jobs, labeled Job 0 through Job 6, each consisting of 2 operations that must be performed in sequence. Each operation takes 1 unit of time to complete. The facility has a limited number of machines, and some operations from different jobs must be performed on the same machine. To avoid conflicts, no two operations can be scheduled on the same machine at the same time. A binary decision variable determines the order in which these operations are scheduled. The makespan, defined as the total time required to complete all jobs, must be at least as large as the completion time of the last operation across all jobs. For every pair of operations that share the same machine, a binary variable determines the order in which they are scheduled. The objective is to minimize the makespan, ensuring that all jobs are completed in the shortest possible time. The start time of the second operation in each job must be at least 1 unit of time after the start time of the first operation. For every pair of operations assigned to the same machine, the schedule must ensure that they do not overlap. The makespan must be at least as large as the completion time of every operation. All start times must be non-negative. The binary variables that determine the order of operations on the same machine must be either 0 or 1. A large constant, 100, is used in the constraints to enforce the order of operations on shared machines. The processing time for every operation is 1 unit of time. The objective is to minimize the total time required to complete all jobs, ensuring that all operations are scheduled efficiently without violating any constraints. + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 1 + - Operation 1 of Job 1: Machine 0 + - Operation 0 of Job 2: Machine 0 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 0 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 0 + - Operation 0 of Job 5: Machine 0 + - Operation 1 of Job 5: Machine 1 + - Operation 0 of Job 6: Machine 2 + - Operation 1 of Job 6: Machine 1",5.0 +['Minimum Cost Flow Problem'],"The problem involves managing the distribution of a critical resource across seven interconnected cities to minimize the total transportation cost while meeting the demand or surplus of each city and respecting capacity constraints. + +• Net demand / surplus of every city +– City 0 requires 3 additional units +– City 1 requires 1 additional unit +– City 2 has a surplus of 2 units +– City 3 is balanced (0) +– City 4 has a surplus of 2 units +– City 5 has a surplus of 1 unit +– City 6 requires 1 additional unit + +• Maximum amount a city is allowed to receive +– City 0: 18 units +– City 1: 18 units +– City 2: 12 units +– City 3: 12 units +– City 4: 14 units +– City 5: 18 units +– City 6: 18 units + +• Transportation links (all links are bidirectional; each direction has the same +cost and capacity). +For every ordered pair (i, j) listed below, the first number is the cost per +unit shipped, the second number is the maximum capacity of that link. + +(0, 1): cost 1, capacity 9  (1, 0): cost 1, capacity 9 +(0, 2): cost 2, capacity 8  (2, 0): cost 2, capacity 8 +(0, 3): cost 2, capacity 10 (3, 0): cost 2, capacity 10 +(0, 4): cost 3, capacity 10 (4, 0): cost 3, capacity 10 +(0, 5): cost 3, capacity 12 (5, 0): cost 3, capacity 12 +(0, 6): cost 2, capacity 10 (6, 0): cost 2, capacity 10 + +(1, 2): cost 1, capacity 9  (2, 1): cost 1, capacity 9 +(1, 3): cost 2, capacity 9  (3, 1): cost 2, capacity 9 +(1, 4): cost 3, capacity 10 (4, 1): cost 3, capacity 10 +(1, 5): cost 2, capacity 11 (5, 1): cost 2, capacity 11 +(1, 6): cost 2, capacity 10 (6, 1): cost 2, capacity 10 + +(2, 3): cost 1, capacity 8  (3, 2): cost 1, capacity 8 +(2, 4): cost 1, capacity 7  (4, 2): cost 1, capacity 7 +(2, 5): cost 2, capacity 8  (5, 2): cost 2, capacity 8 +(2, 6): cost 3, capacity 9  (6, 2): cost 3, capacity 9 + +(3, 4): cost 2, capacity 8  (4, 3): cost 2, capacity 8 +(3, 5): cost 1, capacity 9  (5, 3): cost 1, capacity 9 +(3, 6): cost 2, capacity 9  (6, 3): cost 2, capacity 9 + +(4, 5): cost 1, capacity 7  (5, 4): cost 1, capacity 7 +(4, 6): cost 1, capacity 8  (6, 4): cost 1, capacity 8 + +(5, 6): cost 2, capacity 12 (6, 5): cost 2, capacity 12 + +The objective is to determine the optimal quantities to transport over every link so that all city demands are satisfied, no link or city capacity is violated, and the total transportation cost is minimized. All shipment amounts must be non-negative.",10.0 +"['Minimum Cost Flow Problem', 'Transportation Problem', 'Network Optimization']","Imagine you are managing the distribution of resources across nine interconnected cities. Each city has a specific supply or demand for resources, and your task is to ensure that the flow of resources between cities meets these requirements while minimizing the total transportation cost. The transportation costs vary depending on the route taken between cities, and each route has a maximum capacity limit that cannot be exceeded. + +#### City Requirements: +- **City 0** has a surplus of 69 units of resources. +- **City 1** has a deficit of 3 units. +- **City 2** has a surplus of 8 units. +- **City 3** has a deficit of 10 units. +- **City 4** has a deficit of 28 units. +- **City 5** has a deficit of 15 units. +- **City 6** has a surplus of 8 units. +- **City 7** has a deficit of 20 units. +- **City 8** has a deficit of 9 units. + +#### Transportation Costs: +The cost of transporting resources between cities varies as follows: +- From **City 0**: To City 1 costs 7, City 2 costs 6, City 3 costs 6, City 4 costs 10, City 5 costs 6, City 6 costs 8, City 7 costs 7, and City 8 costs 6. +- From **City 1**: To City 0 costs 10, City 2 costs 7, City 3 costs 9, City 4 costs 10, City 5 costs 10, City 6 costs 9, City 7 costs 8, and City 8 costs 6. +- From **City 2**: To City 0 costs 6, City 1 costs 6, City 3 costs 7, City 4 costs 8, City 5 costs 6, City 6 costs 7, City 7 costs 7, and City 8 costs 6. +- From **City 3**: To City 0 costs 8, City 1 costs 9, City 2 costs 6, City 4 costs 8, City 5 costs 8, City 6 costs 8, City 7 costs 10, and City 8 costs 10. +- From **City 4**: To City 0 costs 8, City 1 costs 7, City 2 costs 8, City 3 costs 7, City 5 costs 8, City 6 costs 7, City 7 costs 10, and City 8 costs 10. +- From **City 5**: To City 0 costs 10, City 1 costs 9, City 2 costs 6, City 3 costs 8, City 4 costs 10, City 6 costs 9, City 7 costs 8, and City 8 costs 9. +- From **City 6**: To City 0 costs 9, City 1 costs 7, City 2 costs 10, City 3 costs 7, City 4 costs 10, City 5 costs 10, City 7 costs 7, and City 8 costs 9. +- From **City 7**: To City 0 costs 10, City 1 costs 6, City 2 costs 6, City 3 costs 10, City 4 costs 9, City 5 costs 9, City 6 costs 10, and City 8 costs 10. +- From **City 8**: To City 0 costs 9, City 1 costs 10, City 2 costs 10, City 3 costs 10, City 4 costs 7, City 5 costs 9, City 6 costs 6, and City 7 costs 8. + +#### Capacity Constraints: +Each route between cities has a maximum capacity limit: +- **City 0**: To City 1 (59), City 2 (51), City 3 (49), City 4 (36), City 5 (38), City 6 (37), City 7 (55), City 8 (33). +- **City 1**: To City 0 (59), City 2 (36), City 3 (43), City 4 (38), City 5 (35), City 6 (41), City 7 (45), City 8 (47). +- **City 2**: To City 0 (47), City 1 (50), City 3 (58), City 4 (48), City 5 (53), City 6 (55), City 7 (50), City 8 (60). +- **City 3**: To City 0 (32), City 1 (34), City 2 (54), City 4 (31), City 5 (55), City 6 (45), City 7 (34), City 8 (49). +- **City 4**: To City 0 (59), City 1 (50), City 2 (57), City 3 (47), City 5 (33), City 6 (59), City 7 (56), City 8 (58). +- **City 5**: To City 0 (38), City 1 (34), City 2 (37), City 3 (40), City 4 (35), City 6 (46), City 7 (36), City 8 (59). +- **City 6**: To City 0 (51), City 1 (46), City 2 (54), City 3 (48), City 4 (35), City 5 (34), City 7 (39), City 8 (45). +- **City 7**: To City 0 (33), City 1 (55), City 2 (54), City 3 (35), City 4 (44), City 5 (56), City 6 (52), City 8 (35). +- **City 8**: To City 0 (59), City 1 (43), City 2 (56), City 3 (53), City 4 (53), City 5 (55), City 6 (38), City 7 (56). + +#### Objective: +Your goal is to determine the optimal flow of resources between cities to meet all supply and demand requirements while minimizing the total transportation cost. Ensure that the flow on each route does not exceed its capacity and that all resource flows are non-negative. + +This problem requires balancing the flow of resources across the network of cities, respecting the capacity limits of each route, and achieving the lowest possible transportation cost.",629.0 +['Unit Commitment Problem'],"Minimize the total operational cost, which includes the base cost of operating generators, the variable cost based on power output, and the startup cost for generators. The base cost is 80 units per generator per period for Type 0 and Type 2 generators, and 78 units for Type 1 generators. The variable cost is 4 units per MW for Type 0 and Type 1 generators, and 3 units per MW for Type 2 generators. The startup cost is 353 units for Type 0 generators, 349 units for Type 1 generators, and 351 units for Type 2 generators. + +The total power output from all generators must meet or exceed the demand in each period, which is 163 MW in Period 0, 180 MW in Period 1, 208 MW in Period 2, 183 MW in Period 3, and 227 MW in Period 4. Each generator type has a minimum and maximum power output per generator: Type 0 generators must produce at least 35 MW and at most 126 MW, Type 1 generators must produce at least 35 MW and at most 130 MW, and Type 2 generators must produce at least 35 MW and at most 125 MW. + +The total maximum possible output from all generators must be at least 15% higher than the demand in each period. For example, in Period 0, the total maximum output must be at least 187.45 MW. The number of generators operating in each period cannot exceed the available capacity: up to 5 Type 0 generators, 2 Type 1 generators, and 1 Type 2 generator can be operated in any period. + +The number of generators started in a period is limited by the change in the number of generators operating from the previous period to the current period. For example, in Period 1, the number of Type 0 generators started cannot exceed the difference between the number of Type 0 generators operating in Period 1 and Period 0. This rule applies similarly for Type 1 and Type 2 generators across all periods. + +The goal is to determine the optimal number of generators to operate, the power output for each generator type, and the number of generators to start in each period to minimize the total operational cost while meeting all constraints.",5136.0 +['Flow Shop Scheduling'],"A manufacturing facility processes six jobs (Job 0 through Job 5) on four machines (Machine 0 through Machine 3). Each job must be processed on all four machines in sequence, and the facility must determine the optimal schedule to minimize the total production time. Each job has a unique processing time on each machine, as follows: + +- **Job 0**: 5 hours on Machine 0, 8 hours on Machine 1, 8 hours on Machine 2, and 5 hours on Machine 3. +- **Job 1**: 7 hours on Machine 0, 6 hours on Machine 1, 7 hours on Machine 2, and 5 hours on Machine 3. +- **Job 2**: 10 hours on Machine 0, 10 hours on Machine 1, 8 hours on Machine 2, and 7 hours on Machine 3. +- **Job 3**: 7 hours on Machine 0, 10 hours on Machine 1, 5 hours on Machine 2, and 8 hours on Machine 3. +- **Job 4**: 9 hours on Machine 0, 9 hours on Machine 1, 5 hours on Machine 2, and 6 hours on Machine 3. +- **Job 5**: 6 hours on Machine 0, 9 hours on Machine 1, 6 hours on Machine 2, and 6 hours on Machine 3. + +The facility must adhere to the following constraints: + +1. **One Job per Schedule Position**: Each of the six schedule positions (Position 0 through Position 5) must be assigned exactly one job. No two positions can have the same job. + +2. **One Schedule Position per Job**: Each job must be assigned to exactly one schedule position. No job can be assigned to multiple positions. + +3. **Machine Precedence**: The processing of a job on Machine 1 cannot start until it has completed processing on Machine 0. Similarly, processing on Machine 2 cannot begin until the job has finished on Machine 1, and processing on Machine 3 cannot start until the job has completed on Machine 2. + +4. **Job Precedence**: The processing of a job in a later schedule position cannot start on any machine until the job in the previous position has completed its processing on that machine. For example, if Job 0 is in Position 0 and Job 1 is in Position 1, Job 1 cannot start on Machine 0 until Job 0 has finished its processing on Machine 0. + +The objective is to determine the optimal schedule for processing the six jobs across the four machines to minimize the total production time, ensuring that all constraints are satisfied. The total production time is defined as the time when the last job completes its processing on Machine 3, plus any additional processing time required for the last job in the schedule.",68.0 +['AircraftLanding'],"The problem involves scheduling the landing times of five aircraft to minimize the total penalty costs associated with deviations from their target landing times while ensuring safe separation between landings and adhering to time windows. Each aircraft has a target landing time, earliest and latest possible landing times, and penalty costs for landing early or late. The aircraft details are as follows: Aircraft 0 has a target landing time of 83 minutes, earliest landing time of 60 minutes, latest landing time of 100 minutes, early penalty cost of \$16 per minute, and late penalty cost of \$16 per minute. Aircraft 1 has a target landing time of 79 minutes, earliest landing time of 60 minutes, latest landing time of 100 minutes, early penalty cost of \$16 per minute, and late penalty cost of \$10 per minute. Aircraft 2 has a target landing time of 75 minutes, earliest landing time of 60 minutes, latest landing time of 100 minutes, early penalty cost of \$16 per minute, and late penalty cost of \$13 per minute. Aircraft 3 has a target landing time of 95 minutes, earliest landing time of 65 minutes, latest landing time of 100 minutes, early penalty cost of \$19 per minute, and late penalty cost of \$19 per minute. Aircraft 4 has a target landing time of 98 minutes, earliest landing time of 68 minutes, latest landing time of 100 minutes, early penalty cost of \$14 per minute, and late penalty cost of \$15 per minute. + +To ensure safety, the following minimum separation times must be maintained between landings: 5 minutes between Aircraft 0 and Aircraft 1, 5 minutes between Aircraft 0 and Aircraft 2, 5 minutes between Aircraft 0 and Aircraft 3, 5 minutes between Aircraft 0 and Aircraft 4, 7 minutes between Aircraft 1 and Aircraft 2, 7 minutes between Aircraft 1 and Aircraft 3, 5 minutes between Aircraft 1 and Aircraft 4, 6 minutes between Aircraft 2 and Aircraft 3, 7 minutes between Aircraft 2 and Aircraft 4, and 5 minutes between Aircraft 3 and Aircraft 4. + +The landing times of each aircraft must fall within their specified earliest and latest possible landing times. The separation constraints must be respected, meaning that the landing times of any two aircraft must maintain the required separation based on their order of landing. The penalty for landing early is calculated as the number of minutes early multiplied by the early penalty cost, and the penalty for landing late is calculated as the number of minutes late multiplied by the late penalty cost. The objective is to determine the landing times for all five aircraft such that the total penalty cost is minimized, and all separation requirements and landing time windows are strictly adhered to. This involves deciding the exact landing time for each aircraft and the order in which they land, ensuring that the schedule is both safe and cost-effective.",94.0 +['Assignment Problem'],"The problem involves assigning employees to shifts across two restaurants to meet skill demands while minimizing costs. There are 2 restaurants, 6 shifts per restaurant, and 6 employees. Each employee has specific skills and availability for shifts. Employees have preference costs for certain skills, and unfulfilled positions incur a penalty cost of 100 units. The demand for skills varies by restaurant, shift, and skill type. For example, Restaurant 0, Shift 0 requires 2 employees with Skill 0, and Restaurant 1, Shift 0 requires 1 employee with Skill 0. Employees can only be assigned to shifts they are available for and to roles they are skilled in. Each employee can be assigned to at most one shift across all restaurants and skills. The objective is to minimize the total cost, which includes preference costs for employee assignments and penalty costs for unfulfilled positions. Constraints ensure that the total number of employees assigned to a specific skill in a shift, plus any unfulfilled positions, equals the demand for that skill. Employee availability and skill requirements are respected, and each employee is limited to one assignment. The preference costs for assignments are as follows: Employee 0 to Skill 0 costs 1 unit, Employee 1 to Skill 0 costs 2 units, and Employee 3 to Skill 0 costs 3 units. The penalty cost for unfulfilled positions is 100 units per position. The goal is to assign employees to shifts and roles in a way that meets all demands while minimizing the total cost.",6.0 +"['Transportation Problem', 'Minimum Cost Flow Problem', 'Network Optimization']","You are responsible for managing the distribution of emergency medical supplies across eight cities in a region. Each city has a specific supply of medical resources and a demand that must be met to ensure adequate healthcare coverage. The goal is to minimize the total transportation cost while ensuring that all cities receive the necessary supplies and that no distribution routes exceed their capacity. + +### City Supply and Demand: +- **City 0** has a net demand of 1 unit of medical supplies. +- **City 1** has a balanced supply and demand (net demand of 0 units). +- **City 2** has a net supply of 1 unit of medical supplies. +- **City 3** has a net demand of 2 units of medical supplies. +- **City 4** has a balanced supply and demand (net demand of 0 units). +- **City 5** has a balanced supply and demand (net demand of 0 units). +- **City 6** has a net supply of 2 units of medical supplies. +- **City 7** has a balanced supply and demand (net demand of 0 units). + +### Transportation Costs: +The cost of transporting medical supplies between cities varies depending on the route. Below are the transportation costs per unit of supplies: + +- **From City 0**: To City 1 costs 3, to City 2 costs 2, to City 3 costs 2, to City 4 costs 2, to City 5 costs 3, to City 6 costs 3, and to City 7 costs 1. +- **From City 1**: To City 0 costs 1, to City 2 costs 2, to City 3 costs 3, to City 4 costs 1, to City 5 costs 2, to City 6 costs 1, and to City 7 costs 2. +- **From City 2**: To City 0 costs 2, to City 1 costs 2, to City 3 costs 3, to City 4 costs 3, to City 5 costs 2, to City 6 costs 1, and to City 7 costs 2. +- **From City 3**: To City 0 costs 1, to City 1 costs 2, to City 2 costs 1, to City 4 costs 3, to City 5 costs 3, to City 6 costs 2, and to City 7 costs 3. +- **From City 4**: To City 0 costs 3, to City 1 costs 2, to City 2 costs 1, to City 3 costs 1, to City 5 costs 3, to City 6 costs 2, and to City 7 costs 2. +- **From City 5**: To City 0 costs 1, to City 1 costs 2, to City 2 costs 1, to City 3 costs 2, to City 4 costs 1, to City 6 costs 2, and to City 7 costs 1. +- **From City 6**: To City 0 costs 2, to City 1 costs 3, to City 2 costs 1, to City 3 costs 1, to City 4 costs 1, to City 5 costs 1, and to City 7 costs 1. +- **From City 7**: To City 0 costs 1, to City 1 costs 1, to City 2 costs 3, to City 3 costs 1, to City 4 costs 2, to City 5 costs 3, and to City 6 costs 2. + +### Route Capacity Constraints: +Each route between cities has a maximum capacity for transporting medical supplies: + +- **From City 0**: To City 1 (7 units), to City 2 (7 units), to City 3 (7 units), to City 4 (7 units), to City 5 (8 units), to City 6 (8 units), and to City 7 (8 units). +- **From City 1**: To City 0 (8 units), to City 2 (7 units), to City 3 (8 units), to City 4 (8 units), to City 5 (7 units), to City 6 (7 units), and to City 7 (9 units). +- **From City 2**: To City 0 (8 units), to City 1 (7 units), to City 3 (7 units), to City 4 (7 units), to City 5 (7 units), to City 6 (9 units), and to City 7 (7 units). +- **From City 3**: To City 0 (7 units), to City 1 (7 units), to City 2 (9 units), to City 4 (8 units), to City 5 (7 units), to City 6 (7 units), and to City 7 (9 units). +- **From City 4**: To City 0 (9 units), to City 1 (7 units), to City 2 (8 units), to City 3 (9 units), to City 5 (7 units), to City 6 (7 units), and to City 7 (7 units). +- **From City 5**: To City 0 (7 units), to City 1 (8 units), to City 2 (9 units), to City 3 (9 units), to City 4 (8 units), to City 6 (9 units), and to City 7 (8 units). +- **From City 6**: To City 0 (9 units), to City 1 (8 units), to City 2 (7 units), to City 3 (8 units), to City 4 (8 units), to City 5 (7 units), and to City 7 (8 units). +- **From City 7**: To City 0 (9 units), to City 1 (8 units), to City 2 (7 units), to City 3 (9 units), to City 4 (9 units), to City 5 (8 units), and to City 6 (8 units). + +### City Capacity Constraints: +Each city has a maximum capacity for receiving medical supplies: + +- **City 0**: Can receive up to 19 units. +- **City 1**: Can receive up to 15 units. +- **City 2**: Can receive up to 15 units. +- **City 3**: Can receive up to 14 units. +- **City 4**: Can receive up to 15 units. +- **City 5**: Can receive up to 15 units. +- **City 6**: Can receive up to 14 units. +- **City 7**: Can receive up to 16 units. + +### Objective: +Your task is to determine the optimal distribution of medical supplies between the cities to minimize the total transportation cost while ensuring that all cities meet their supply and demand requirements, no route exceeds its capacity, and no city exceeds its receiving capacity.",4.0 +['Unit Commitment Problem'],"Imagine you are managing a power generation system responsible for supplying electricity to a region over five distinct periods. The system consists of three types of power generators, each with unique operational characteristics and costs. Your goal is to determine the optimal number of generators to operate, the amount of power each generator should produce, and when to start up additional generators to meet the region's electricity demand at the lowest possible cost. + +#### Key Decisions: +1. **Number of Generators to Operate**: For each generator type (Type 0, Type 1, Type 2) and each period (Period 0 to Period 4), decide how many generators should be active. +2. **Power Output**: Determine the amount of power (in MW) each active generator should produce during each period. +3. **Generator Startups**: Decide whether to start up additional generators in each period, incurring a startup cost. + +#### Objective: +Minimize the total cost of operating the power generation system, which includes: +- **Base Costs**: Each active generator incurs a fixed cost of 79 units per period, regardless of its power output. +- **Per-MW Costs**: Each generator also incurs a variable cost based on its power output: 3 units per MW for Type 0, 4 units per MW for Type 1, and 4 units per MW for Type 2. +- **Startup Costs**: Starting up a generator incurs a one-time cost of 355 units for Type 0 and 349 units for Types 1 and 2. + +#### Constraints: +1. **Demand Satisfaction**: The total power output across all generators must meet or exceed the region's electricity demand in each period. The demands are: + - Period 0: 167 MW + - Period 1: 212 MW + - Period 2: 224 MW + - Period 3: 221 MW + - Period 4: 189 MW + +2. **Generator Capacity Limits**: + - Each generator has a minimum and maximum power output: + - Type 0: Minimum 35 MW, Maximum 128 MW + - Type 1: Minimum 35 MW, Maximum 129 MW + - Type 2: Minimum 35 MW, Maximum 125 MW + - The power output of each generator must be within these limits based on whether it is active. + +3. **Reserve Margin**: To ensure reliability, the total maximum possible power output across all generators must be at least 15% higher than the demand in each period. This translates to: + - Period 0: 192.05 MW + - Period 1: 243.8 MW + - Period 2: 257.6 MW + - Period 3: 254.15 MW + - Period 4: 217.35 MW + +4. **Generator Availability**: + - The number of generators that can be active is limited: + - Type 0: Maximum 1 generator per period + - Type 1: Maximum 4 generators per period + - Type 2: Maximum 3 generators per period + +5. **Startup Logic**: + - If a generator is started in a period, it must have been inactive in the previous period. + - For Type 2 generators, up to 2 generators can be started in Period 0 without requiring a startup cost. + +#### Numerical Parameters: +- **Base Costs**: 79 units per active generator per period. +- **Per-MW Costs**: 3 units for Type 0, 4 units for Types 1 and 2. +- **Startup Costs**: 355 units for Type 0, 349 units for Types 1 and 2. +- **Demands**: 167 MW (Period 0), 212 MW (Period 1), 224 MW (Period 2), 221 MW (Period 3), 189 MW (Period 4). +- **Capacity Limits**: + - Type 0: 35 MW (min), 128 MW (max) + - Type 1: 35 MW (min), 129 MW (max) + - Type 2: 35 MW (min), 125 MW (max) +- **Reserve Margins**: 192.05 MW (Period 0), 243.8 MW (Period 1), 257.6 MW (Period 2), 254.15 MW (Period 3), 217.35 MW (Period 4). + +Your task is to determine the optimal combination of active generators, their power outputs, and startup decisions to meet the region's electricity demand at the lowest total cost while adhering to all operational constraints.",4904.0 +"['Capacitated Lot-sizing Problem (CLSP)', 'Discrete Lot-sizing and Scheduling Problem']","Minimize the total cost, which includes production costs of 143.7514410203047 units per production decision, startup costs of 65.60228476581376 units per machine startup, inventory holding costs of 1.164609943740513 units for Product 0, 1.188814519462053 units for Product 1, and 0.8889238919759239 units for Product 2 per period, and backlogging costs of 6.772870896644705 units for Product 0, 7.291249122946056 units for Product 1, and 6.513011399630116 units for Product 2 per period. The production decisions are binary, indicating whether a product is produced on a specific machine in a given period, and the startup decisions are binary, indicating whether a machine is started up for a specific product in a given period. The production quantities must be non-negative, and the inventory and backlog levels must balance to meet the demand for each product in each period. The demand for Product 0 is 16.86823685495585 units in Period 0, 21.66307881187974 units in Period 1, and 22.77330720589323 units in Period 2. The demand for Product 1 is 23.97615061004427 units in Period 0, 19.60905873481086 units in Period 1, and 23.31252847872402 units in Period 2. The demand for Product 2 is 23.6195969696504 units in Period 0, 18.33370330296734 units in Period 1, and 22.95855759036257 units in Period 2. Each machine has a limited capacity, and the total production on a machine in a given period cannot exceed this capacity. For Machine 0, the capacity constraints are -71.6204295699619 times the production decision plus 6.477399095211152 times the startup decision plus the production quantity must be less than or equal to zero for each product and period. For Machine 1, the capacity constraints are -71.47086166745689 times the production decision plus 6.845777110793686 times the startup decision plus the production quantity must be less than or equal to zero for each product and period. Each machine can only produce one product at a time in a given period, and the production and startup decisions must follow logical constraints to ensure that a machine is only started up if it was not already producing the same product in the previous period. The production and startup decisions are binary, and the production quantities, inventory, and backlog levels must be non-negative.",914.6636376994904 +['Job Shop'],"In a manufacturing facility, the goal is to schedule the production of seven jobs, each consisting of two operations that must be performed in sequence, to minimize the total production time (makespan). Each operation has a processing time of 1 unit of time. Operations from different jobs may share machines, requiring careful scheduling to avoid overlaps. Binary variables determine the order of operations on shared machines, ensuring that one operation finishes before another starts. A large constant (100) is used to enforce this order in the constraints. + +The constraints include precedence constraints within each job, ensuring that the second operation cannot start until the first operation is completed. Machine capacity constraints prevent overlapping operations on shared machines, with specific pairs of operations requiring explicit ordering. The pairs of operations that cannot be overlapped are: Operation 1 of Job 2 and Operation 1 of Job 4, and Operation 0 of Job 2 and Operation 0 of Job 6. The makespan is defined as the maximum completion time across all operations, ensuring it is at least as long as the last operation's finish time. All start times for operations must be non-negative, and binary variables are restricted to 0 or 1. + +The objective is to minimize the makespan, ensuring all operations are completed efficiently while respecting machine usage and operation sequencing constraints.",3.0 +['Job Shop'],"""The problem involves scheduling operations for six products, each requiring two operations to be performed in sequence. Each operation is assigned to a specific machine, and the goal is to minimize the total production time (makespan). Each operation has a processing time of 1 unit. The start time of each operation must be determined, and the order of operations on shared machines must be decided to avoid overlaps. Precedence constraints ensure that the second operation of a product cannot start until the first operation is completed. Machine capacity constraints ensure that operations assigned to the same machine do not overlap in time. The makespan must be at least as long as the completion time of the last operation across all products. Binary decision variables determine the order of operations on shared machines, and a large number (100) is used in the constraints to enforce this order without overlapping. The objective is to minimize the makespan by determining the start times of all operations and the order of operations on shared machines, ensuring all products are completed as quickly as possible. + + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Product 0: Machine 2 + - Operation 1 of Product 0: Machine 1 + - Operation 0 of Product 1: Machine 2 + - Operation 1 of Product 1: Machine 2 + - Operation 0 of Product 2: Machine 2 + - Operation 1 of Product 2: Machine 1 + - Operation 0 of Product 3: Machine 1 + - Operation 1 of Product 3: Machine 0 + - Operation 0 of Product 4: Machine 2 + - Operation 1 of Product 4: Machine 0 + - Operation 0 of Product 5: Machine 0 + - Operation 1 of Product 5: Machine 1""",5.0 +['Job Shop'],"There are 7 jobs, each consisting of 2 operations that must be performed in sequence. Each operation takes exactly 1 unit of time to complete. The operations are assigned to specific machines, and each machine can process only one operation at a time. The goal is to schedule the operations on the machines to minimize the total completion time, known as the makespan, while satisfying all constraints. + +For each job, the second operation cannot start until the first operation is completed. If two operations are assigned to the same machine, they cannot overlap in time. Binary decision variables are used to determine the order of operations on shared machines. If one operation is scheduled before another on the same machine, the corresponding binary variable is set to 1; otherwise, it is set to 0. A large constant (1000) is used in the constraints to ensure that the binary variables correctly enforce the order of operations on shared machines. + +The start time of each operation must be non-negative, and the makespan must be at least as large as the completion time of every operation. The objective is to determine the start times for all operations on their respective machines such that all precedence and machine capacity constraints are satisfied, and the total completion time is minimized. + +ADDITIONAL INPUT DATA: + +1) Sets and indices +- Jobs: J = {1, 2, 3, 4, 5, 6, 7} +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Machines: M = {A, B, C, D} + +2) Operation durations (all durations are in time units) +- Duration of every operation (j,k) is p_{j,k} = 1 for all j in J and k = 1,2. + +3) Machine assignment for every operation (explicit mapping) +- Operation (1,1) -> Machine A +- Operation (1,2) -> Machine B +- Operation (2,1) -> Machine A +- Operation (2,2) -> Machine B +- Operation (3,1) -> Machine C +- Operation (3,2) -> Machine D +- Operation (4,1) -> Machine B +- Operation (4,2) -> Machine C +- Operation (5,1) -> Machine D +- Operation (5,2) -> Machine A +- Operation (6,1) -> Machine B +- Operation (6,2) -> Machine D +- Operation (7,1) -> Machine C +- Operation (7,2) -> Machine A",4.0 +"['Capacitated Lot-sizing Problem (CLSP)', 'Lot-Sizing Problem', 'Production Planning Problem']","Determine the production amounts, inventory levels, and setup decisions for two products, Product A and Product B, over a seven-period planning horizon to minimize total costs, including production, setup, and inventory holding costs, while meeting demand and adhering to resource constraints. Production costs vary by period: producing one unit of Product A costs 42 units in Period 0, 44 in Period 1, 40 in Period 2, 43 in Period 3, 45 in Period 4, 45 in Period 5, and 42 in Period 6. Producing one unit of Product B costs 46 units in Period 0, 44 in Period 1, 50 in Period 2, 40 in Period 3, 41 in Period 4, 41 in Period 5, and 43 in Period 6. Setup costs are fixed: setting up production for Product A costs 1032 units in Period 0, 1318 in Period 1, 1002 in Period 2, 1073 in Period 3, 1387 in Period 4, 1491 in Period 5, and 1063 in Period 6. For Product B, setup costs are 1260 units in Period 0, 1418 in Period 1, 1455 in Period 2, 1266 in Period 3, 1095 in Period 4, 1431 in Period 5, and 1420 in Period 6. Inventory holding costs are constant: holding one unit of Product A costs 5 units in all periods except Periods 5 and 6, where it costs 4 units. Holding one unit of Product B costs 5 units in Periods 0, 2, 4, 5, and 6, and 4 units in Periods 1 and 3. + +Demand must be satisfied in each period: cumulative demand for Product A is 67 units in Period 0, 160 in Period 1, 213 in Period 2, 309 in Period 3, 376 in Period 4, 464 in Period 5, and 564 in Period 6. For Product B, cumulative demand is 100 units in Period 0, 199 in Period 1, 250 in Period 2, 332 in Period 3, 394 in Period 4, 450 in Period 5, and 516 in Period 6. Resource capacity limits production: in Period 0, the combined production of Product A and Product B cannot exceed 1138 units of resource capacity, with Product A consuming 1.547487551721541 units and Product B consuming 1.91801512951683 units per unit produced. Capacity limits are 1019 units in Period 1, 1166 in Period 2, 1163 in Period 3, 1042 in Period 4, 1098 in Period 5, and 1166 in Period 6. + +Production of a product in a period can only occur if the setup for that product is activated in that period. For Product A, production in Period 0 cannot exceed 564 units if the setup is activated, 497 in Period 1, 404 in Period 2, 351 in Period 3, 255 in Period 4, 188 in Period 5, and 100 in Period 6. For Product B, production in Period 0 cannot exceed 516 units if the setup is activated, 416 in Period 1, 317 in Period 2, 266 in Period 3, 184 in Period 4, 122 in Period 5, and 66 in Period 6. + +Initial inventory for both products is zero at the start of Period 0, and all inventory must be depleted by the end of Period 6. Production amounts and inventory levels must be non-negative. The objective is to minimize the total cost, which includes production costs, setup costs, and inventory holding costs, while ensuring all constraints are satisfied.",56408.0 +['Lot-Sizing Problem'],"Inventory Management for a Retail Supply Chain + +You are responsible for managing the inventory and ordering process for a retail supply chain over a 21-week period. The goal is to minimize the total costs associated with ordering and holding inventory while ensuring that customer demand is met each week. The costs include ordering costs, unit purchase costs, and inventory holding costs. Additionally, there are specific constraints on inventory levels and ordering decisions. + +#### Key Decisions: +1. **Ordering Decisions**: Each week, you must decide whether to place an order (a binary decision) and, if so, how much to order. Placing an order incurs a fixed cost, and the amount ordered cannot exceed a certain limit based on the total demand over the 21-week period. +2. **Inventory Levels**: You must manage the ending inventory each week, ensuring that it meets the flow balance constraints (i.e., the inventory carried over from the previous week plus the amount ordered equals the ending inventory plus the demand for that week). + +#### Objective: +Minimize the total cost, which includes: +- **Ordering Costs**: Fixed costs for placing an order in a given week. These costs vary by week and are as follows: 459 in week 1, 555 in week 2, 547 in week 3, 530 in week 4, 436 in week 5, 401 in week 6, 490 in week 7, 462 in week 8, 479 in week 9, 563 in week 10, 456 in week 11, 442 in week 12, 530 in week 13, 427 in week 14, 512 in week 15, 574 in week 16, 430 in week 17, 424 in week 18, 484 in week 19, 507 in week 20, and 446 in week 21. +- **Unit Purchase Costs**: Variable costs for each unit ordered. These costs also vary by week and are as follows: 49 in week 1, 48 in week 2, 55 in week 3, 44 in week 4, 51 in week 5, 48 in week 6, 53 in week 7, 43 in week 8, 42 in week 9, 52 in week 10, 45 in week 11, 42 in week 12, 43 in week 13, 42 in week 14, 59 in week 15, 57 in week 16, 57 in week 17, 44 in week 18, 57 in week 19, 60 in week 20, and 43 in week 21. +- **Holding Costs**: Costs for holding inventory at the end of each week. These costs are as follows: 23 in week 1, 28 in week 2, 28 in week 3, 25 in week 4, 30 in week 5, 25 in week 6, 28 in week 7, 22 in week 8, 30 in week 9, 30 in week 10, 23 in week 11, 28 in week 12, 27 in week 13, 22 in week 14, 22 in week 15, 23 in week 16, 20 in week 17, 28 in week 18, 30 in week 19, 30 in week 20, and 24 in week 21. + +#### Constraints: +1. **Flow Balance**: The ending inventory for each week must satisfy the following relationship: + - Ending inventory from the previous week + amount ordered = ending inventory for the current week + demand for the current week. + - The demand for each week is as follows: 150 in week 1, 132 in week 2, 135 in week 3, 178 in week 4, 133 in week 5, 173 in week 6, 161 in week 7, 195 in week 8, 110 in week 9, 118 in week 10, 131 in week 11, 198 in week 12, 118 in week 13, 120 in week 14, 129 in week 15, 148 in week 16, 189 in week 17, 178 in week 18, 145 in week 19, 161 in week 20, and 155 in week 21. + +2. **Ordering Limits**: The amount ordered in any week cannot exceed the total demand over the 21-week period (3157 units) if an order is placed. If no order is placed, the amount ordered must be zero. + +3. **Starting and Ending Inventory**: The starting inventory at the beginning of week 1 is zero, and the ending inventory at the end of week 21 must also be zero. + +#### Goal: +Your task is to determine the optimal ordering and inventory management plan that minimizes the total cost while meeting all constraints. This includes deciding when to place orders, how much to order each week, and how much inventory to carry over each week to meet demand without exceeding the budget or violating any constraints.",165425.0 +"['Minimum Cost Flow Problem', 'Network Optimization']","You are tasked with managing the distribution of resources across eight interconnected cities to ensure each city meets its specific demand while minimizing the total transportation cost. Each city has a unique supply of resources, a specific demand, and a limited capacity to handle incoming shipments. Additionally, the transportation links between cities have varying costs and capacity constraints. + +City 0 has a net outflow of 5 units, meaning it is a supply node, and it can handle up to 13 units of incoming shipments. City 1 has a net inflow of 2 units, meaning it is a demand node, and it can accommodate up to 14 units of incoming shipments. City 2 is balanced, with no net demand or surplus, and can handle up to 17 units of incoming shipments. City 3 is also balanced, with no net demand or surplus, and can handle up to 12 units of incoming shipments. City 4 has a net outflow of 1 unit, meaning it is a supply node, and it can handle up to 16 units of incoming shipments. City 5 has a net inflow of 2 units, meaning it is a demand node, and it can accommodate up to 19 units of incoming shipments. City 6 is balanced, with no net demand or surplus, and can handle up to 16 units of incoming shipments. City 7 has a net inflow of 2 units, meaning it is a demand node, and it can accommodate up to 14 units of incoming shipments. + +The cost of transporting resources between cities varies, and each transportation link has a maximum capacity. From City 0: To City 1: Cost = 2, Capacity = 8; To City 2: Cost = 2, Capacity = 7; To City 3: Cost = 2, Capacity = 7; To City 4: Cost = 2, Capacity = 9; To City 5: Cost = 2, Capacity = 8; To City 6: Cost = 1, Capacity = 9; To City 7: Cost = 2, Capacity = 7. From City 1: To City 0: Cost = 1, Capacity = 8; To City 2: Cost = 1, Capacity = 8; To City 3: Cost = 3, Capacity = 7; To City 4: Cost = 2, Capacity = 8; To City 5: Cost = 1, Capacity = 7; To City 6: Cost = 3, Capacity = 9; To City 7: Cost = 1, Capacity = 9. From City 2: To City 0: Cost = 1, Capacity = 8; To City 1: Cost = 2, Capacity = 9; To City 3: Cost = 3, Capacity = 7; To City 4: Cost = 2, Capacity = 9; To City 5: Cost = 2, Capacity = 8; To City 6: Cost = 3, Capacity = 8; To City 7: Cost = 1, Capacity = 8. From City 3: To City 0: Cost = 3, Capacity = 7; To City 1: Cost = 3, Capacity = 7; To City 2: Cost = 3, Capacity = 7; To City 4: Cost = 2, Capacity = 7; To City 5: Cost = 3, Capacity = 7; To City 6: Cost = 2, Capacity = 9; To City 7: Cost = 3, Capacity = 9. From City 4: To City 0: Cost = 2, Capacity = 9; To City 1: Cost = 3, Capacity = 8; To City 2: Cost = 3, Capacity = 8; To City 3: Cost = 1, Capacity = 8; To City 5: Cost = 1, Capacity = 7; To City 6: Cost = 3, Capacity = 8; To City 7: Cost = 2, Capacity = 9. From City 5: To City 0: Cost = 1, Capacity = 7; To City 1: Cost = 3, Capacity = 7; To City 2: Cost = 3, Capacity = 7; To City 3: Cost = 3, Capacity = 9; To City 4: Cost = 2, Capacity = 8; To City 6: Cost = 2, Capacity = 9; To City 7: Cost = 3, Capacity = 8. From City 6: To City 0: Cost = 2, Capacity = 8; To City 1: Cost = 2, Capacity = 7; To City 2: Cost = 3, Capacity = 7; To City 3: Cost = 2, Capacity = 7; To City 4: Cost = 1, Capacity = 9; To City 5: Cost = 1, Capacity = 8; To City 7: Cost = 3, Capacity = 7. From City 7: To City 0: Cost = 2, Capacity = 8; To City 1: Cost = 2, Capacity = 8; To City 2: Cost = 2, Capacity = 7; To City 3: Cost = 3, Capacity = 9; To City 4: Cost = 1, Capacity = 7; To City 5: Cost = 2, Capacity = 8; To City 6: Cost = 3, Capacity = 7. + +Your goal is to determine the optimal amount of resources to transport between cities to meet all demands while minimizing the total transportation cost. Ensure that each city's net resource flow matches its demand or surplus, no transportation link exceeds its capacity, and no city exceeds its capacity to handle incoming shipments.",11.0 +['Job Shop'],"The problem involves scheduling operations for five jobs (Job 0 through Job 4) in a manufacturing facility, where each job consists of two operations that must be performed in sequence. Each operation takes exactly 1 unit of time to complete. The facility has a limited number of machines, and multiple operations from different jobs may need to be assigned to the same machine. A machine can only handle one operation at a time, so the order of operations on shared machines must be determined to avoid overlaps. + +The goal is to minimize the total completion time, known as the makespan, which is the time at which the last operation across all jobs finishes. To achieve this, the start times of all operations must be scheduled such that: (1) for each job, the second operation starts only after the first operation is completed; (2) if two operations are assigned to the same machine, one must finish before the other starts; and (3) the makespan is at least as large as the completion time of every operation. + +The order of operations on shared machines is determined using binary decision variables, which indicate whether one operation is scheduled before another. A large constant (100) is used in the constraints to enforce the sequencing of operations on shared machines. The start times of operations and the makespan are continuous variables, while the binary variables are restricted to values of 0 or 1. The objective is to minimize the makespan, ensuring all jobs are completed as quickly as possible. + + + +Additional information: +- There are 3 machines, numbered from 0 to 2 +- The assignment or operations to machines is the following: + - Operation 0 of Job 0: Machine 2 + - Operation 1 of Job 0: Machine 1 + - Operation 0 of Job 1: Machine 1 + - Operation 1 of Job 1: Machine 0 + - Operation 0 of Job 2: Machine 0 + - Operation 1 of Job 2: Machine 1 + - Operation 0 of Job 3: Machine 0 + - Operation 1 of Job 3: Machine 2 + - Operation 0 of Job 4: Machine 2 + - Operation 1 of Job 4: Machine 0",4.0 +['Assignment Problem'],"""There are 2 restaurants, 7 employees, and 2 shifts per day. Each shift requires employees with specific skills, and each employee has a set of 2 skills they are qualified to perform. Employees have preferences for certain skills, and assigning them to tasks they prefer incurs lower costs. The cost of leaving a position unfulfilled is significantly higher, so the goal is to minimize both the preference costs and the number of unfulfilled positions. + +Restaurant 0, Shift 0, Skill 0 requires 1 employee. Restaurant 0, Shift 1, Skill 0 requires 1 employee. Restaurant 1, Shift 0, Skill 0 requires 2 employees. Restaurant 1, Shift 1, Skill 0 requires 1 employee. + +Employees have varying availability for shifts. Employee 0 is available for Shift 0 but not for Shift 1. Employee 1 is available for Shift 0 but not for Shift 1. Employee 2 is available for Shifts 0 and 1. Employee 3 is available for Shift 1 but not for Shift 0. Employee 4 is available for Shifts 0 and 1. Employee 5 is available for Shift 1 but not for Shift 0. Employee 6 is available for Shifts 0 and 1. Each employee is qualified for specific skills. For instance, Employee 0 is not qualified for Skill 0 but qualified for Skill 1, while Employees 3 and 5 are not qualified for Skill 1 but qualified for Skill 0; Employees 1, 2, 4 and 6 are qualified by both Skill 0 and 1. +Assigning employees to tasks they prefer incurs lower costs. In particular, assigning Employee 1 to Skill 0 costs 2 units. +Assigning Employee 2 to Skill 0 costs 0 units. Assigning Employee 3 to Skill 0 costs 2 units. +Assigning Employee 4 to Skill 0 costs 1 unit. +Assigning Employee 5 to Skill 0 costs 3 units. +Assigning Employee 6 to Skill 0 costs 2 units. Each unfulfilled position costs 100 units, which is significantly higher than preference costs. + +The goal is to assign employees to specific shifts and restaurants based on their skills and availability while determining the number of unfulfilled positions for each shift and restaurant. The total cost to minimize includes the cost of assigning employees to tasks they prefer and the penalty cost for unfulfilled positions. + +The total number of employees assigned to a shift at a restaurant, plus any unfulfilled positions, must equal the demand for that shift and skill. Employees can only be assigned to shifts they are available for and tasks they are qualified for. Each employee can be assigned to at most one shift across all restaurants and skills.""",7.0 +"['Discrete Lot-sizing and Scheduling Problem', 'Capacitated Lot-sizing Problem (CLSP)']","Minimize the total operational costs, which include production setup costs, machine startup costs, inventory holding costs, and backlogging costs, while meeting customer demand and adhering to machine capacity constraints. The setup cost is 136.3292413778485 units per production run, and the startup cost is 72.14543590644504 units per machine startup. The inventory holding cost for Product 0 is 1.097996170356003 units per unit per period, and for Product 1, it is 0.9303010222262446 units per unit per period. The backlogging cost for Product 0 is 7.343834540408109 units per unit per period, and for Product 1, it is 6.589388559571906 units per unit per period. + +The production system must meet the following demand constraints: for Product 0, the demand is 17.38664149749027 units in Period 0, 18.25024949764659 units in Period 1, and 20.01141977261751 units in Period 2. For Product 1, the demand is 17.8541958922009 units in Period 0, 16.38473840141418 units in Period 1, and 15.49890198357548 units in Period 2. + +Machine 0 can handle up to 77.29936140903678 units of production in each period, while Machine 1 can handle up to 74.94692092032173 units in each period. Only one product can be produced on a machine in a given time period. If a machine was not producing in the previous period but is producing in the current period, a startup cost is incurred. Production quantities, inventory levels, and backlogs must be non-negative, and production and startup decisions are binary (0 or 1). + +The objective is to determine the optimal production schedule, machine startups, inventory levels, and backlogs to minimize total costs while meeting all operational constraints.",525.01301077364 +['Unit Commitment Problem'],"Minimize the total operational cost, which includes the base cost of operating generators, the cost of power output, and the startup cost of generators. The base cost for operating a Type 0 generator is 79 units per period, Type 1 is 80 units, and Type 2 is 78 units. The cost of power output is 4 units per MW for Type 0, 3 units per MW for Type 1, and 4 units per MW for Type 2. The startup cost is 349 units for Type 0, 354 units for Type 1, and 354 units for Type 2. The total power output from all generators must meet or exceed the demand in each period: 176 MW in Period 0, 160 MW in Period 1, 178 MW in Period 2, 210 MW in Period 3, and 171 MW in Period 4. Each generator type has minimum and maximum output limits: Type 0 generators must produce at least 34 MW and no more than 130 MW, Type 1 generators must produce at least 34 MW and no more than 130 MW, and Type 2 generators must produce at least 35 MW and no more than 128 MW. The total maximum possible output from all generators must be at least 202.4 MW in Period 0, 184 MW in Period 1, 204.7 MW in Period 2, 241.5 MW in Period 3, and 196.65 MW in Period 4. The number of generators that can be operated is limited by the available units: up to 2 Type 0 generators, up to 5 Type 1 generators, and up to 1 Type 2 generator in each period. The number of generators started in a period cannot exceed the number of generators that were not operating in the previous period. The solution must determine the optimal number of generators to operate, the power output from each generator, and the number of generators to start in each period to minimize the total operational cost while meeting the power demand and adhering to all operational constraints.",4193.0 +['TravelingSalesman'],"You are tasked with planning the most efficient delivery route for a logistics company that serves seven cities: City 0, City 1, City 2, City 3, City 4, City 5, and City 6. The goal is to minimize the total transportation cost while ensuring that each city is visited exactly once and that the delivery route forms a complete tour without any subtours. The delivery vehicle must return to the starting city after visiting all others. + +The cost of traveling between each pair of cities varies. From City 0: to City 1 costs 277, to City 2 costs 304, to City 3 costs 280, to City 4 costs 321, to City 5 costs 285, and to City 6 costs 300. From City 1: to City 0 costs 280, to City 2 costs 317, to City 3 costs 301, to City 4 costs 306, to City 5 costs 312, and to City 6 costs 290. From City 2: to City 0 costs 304, to City 1 costs 318, to City 3 costs 283, to City 4 costs 289, to City 5 costs 301, and to City 6 costs 277. From City 3: to City 0 costs 304, to City 1 costs 282, to City 2 costs 290, to City 4 costs 307, to City 5 costs 291, and to City 6 costs 290. From City 4: to City 0 costs 322, to City 1 costs 291, to City 2 costs 277, to City 3 costs 300, to City 5 costs 318, and to City 6 costs 280. From City 5: to City 0 costs 308, to City 1 costs 297, to City 2 costs 288, to City 3 costs 314, to City 4 costs 290, and to City 6 costs 320. From City 6: to City 0 costs 316, to City 1 costs 313, to City 2 costs 301, to City 3 costs 324, to City 4 costs 280, and to City 5 costs 303. + +Each city must be visited exactly once. The delivery vehicle must arrive at each city from exactly one other city and depart to exactly one other city. The delivery route must form a single tour, meaning the vehicle cannot split into smaller loops or subtours. To enforce this, each city is assigned a position in the tour, with positions ranging from 1 to 7, ensuring a valid sequence. The positions must follow a logical sequence without gaps or repetitions. The route between any two cities is binary: it is either included (1) or excluded (0) in the final tour. + +The objective is to minimize the total transportation cost for the delivery route while satisfying all constraints. The solution must account for all the costs and constraints outlined above.",1987.0 +['Facility Dispersion Problem'],"You are tasked with optimizing the placement of emergency response centers across a network of five critical locations (Node 0, Node 1, Node 2, Node 3, and Node 4) to maximize the overall coverage and effectiveness of the response system. Each location has unique characteristics, and the goal is to select exactly three locations to establish these centers. The effectiveness of the system is measured by the combined coverage scores between the selected centers, which are determined by predefined pairwise scores representing the strategic importance of connecting each pair of locations. + +#### Key Decisions: +1. **Selecting Locations**: You must choose exactly three out of the five locations to establish emergency response centers. Each location is represented by a binary decision: if a center is established at a location, it is selected (1); otherwise, it is not (0). +2. **Pairwise Coverage**: For each pair of selected locations, the coverage score is determined by the predefined pairwise importance values. These scores reflect the strategic benefit of having both locations in the network. + +#### Objective: +Maximize the total coverage score across all pairs of selected locations. The coverage scores for each pair are as follows: +- Node 0 and Node 1: 24 +- Node 0 and Node 2: 21 +- Node 0 and Node 3: 16 +- Node 0 and Node 4: 15 +- Node 1 and Node 0: 21 +- Node 1 and Node 2: 19 +- Node 1 and Node 3: 24 +- Node 1 and Node 4: 23 +- Node 2 and Node 0: 21 +- Node 2 and Node 1: 17 +- Node 2 and Node 3: 25 +- Node 2 and Node 4: 19 +- Node 3 and Node 0: 18 +- Node 3 and Node 1: 24 +- Node 3 and Node 2: 17 +- Node 3 and Node 4: 17 +- Node 4 and Node 0: 15 +- Node 4 and Node 1: 18 +- Node 4 and Node 2: 15 +- Node 4 and Node 3: 24 + +#### Constraints: +1. **Selection of Centers**: Exactly three locations must be selected to establish emergency response centers. This means the sum of selected locations must equal three. +2. **Pairwise Coverage Activation**: For any pair of locations, the coverage score is only counted if both locations in the pair are selected. This ensures that the coverage score is only applied to active pairs. +3. **Logical Relationships**: + - If a location is not selected, no coverage scores involving that location can be counted. + - If both locations in a pair are selected, the coverage score for that pair is fully counted. + - If only one location in a pair is selected, the coverage score for that pair is not counted. + +#### Numerical Parameters: +- The pairwise coverage scores are explicitly provided for all possible pairs of locations, as listed above. +- The total number of selected locations must be exactly three. + +#### Summary: +Your task is to select three out of the five locations to establish emergency response centers such that the total coverage score, based on the predefined pairwise importance values, is maximized. The selection must adhere to the constraints ensuring that only active pairs of selected locations contribute to the coverage score.",130.0 +['Job Shop'],"The problem involves scheduling operations for 7 products, each requiring 2 operations to be performed in sequence. Each operation has a fixed processing time of 1 unit of time. The operations must be scheduled on shared machines, where each machine can handle only one operation at a time. The order of operations on shared machines is determined by binary decision variables, which decide whether one operation is performed before another. + +The goal is to minimize the total production time, known as the makespan, ensuring that all products are completed as quickly as possible. The makespan must be at least as long as the completion time of the last operation across all products. + +Constraints include precedence constraints, ensuring that the second operation of each product cannot start until the first operation is completed. Machine capacity constraints ensure that no two operations on the same machine overlap in time. These constraints are enforced using binary variables that determine the order of operations on shared machines. A large constant (1000) is used in the machine capacity constraints to ensure that the constraints are satisfied when the binary variables determine the order of operations. + +The start times of all operations must be non-negative, and the binary variables must take values of either 0 or 1. The objective is to minimize the makespan, ensuring that all operations are scheduled efficiently without violating the precedence and machine capacity constraints. + +ADDITIONAL INPUT DATA : + +1) Sets and indices +- Jobs (products): J = {1, 2, 3, 4, 5, 6, 7} +- For each job j in J, there are 2 operations indexed k = 1..2. We denote operation (j,k). +- Machines: M = {1, 2, 3} + +2) Machine assignment for every operation (which machine executes the operation) +- Operation (1,1) -> Machine 1 +- Operation (1,2) -> Machine 2 +- Operation (2,1) -> Machine 1 +- Operation (2,2) -> Machine 3 +- Operation (3,1) -> Machine 2 +- Operation (3,2) -> Machine 3 +- Operation (4,1) -> Machine 1 +- Operation (4,2) -> Machine 2 +- Operation (5,1) -> Machine 2 +- Operation (5,2) -> Machine 3 +- Operation (6,1) -> Machine 1 +- Operation (6,2) -> Machine 3 +- Operation (7,1) -> Machine 2 +- Operation (7,2) -> Machine 1 + +3) Processing times +- All operations have processing time p_{j,k} = 1 (unit time) as in the original description.",5.0 +"['Minimum Cost Flow Problem', 'Transportation Problem']","You are responsible for managing the distribution of resources across nine interconnected cities. Each city has a specific supply or demand for resources, and your goal is to ensure that all cities receive the resources they need while minimizing the total transportation cost. The transportation costs vary depending on the route taken between cities, and each route has a maximum capacity limit that cannot be exceeded. + +#### City Requirements: +- **City 0** has a surplus of 42 units of resources that need to be distributed to other cities. +- **City 1** requires an additional 3 units of resources to meet its demand. +- **City 2** has a surplus of 11 units of resources. +- **City 3** requires an additional 21 units of resources. +- **City 4** has a surplus of 4 units of resources. +- **City 5** has a surplus of 3 units of resources. +- **City 6** requires an additional 19 units of resources. +- **City 7** requires an additional 24 units of resources. +- **City 8** has a surplus of 7 units of resources. + +#### Transportation Costs: +The cost of transporting resources between cities varies. Below are the costs for each possible route: +- From **City 0** to other cities: + - To **City 1**: 8 units + - To **City 2**: 9 units + - To **City 3**: 8 units + - To **City 4**: 9 units + - To **City 5**: 8 units + - To **City 6**: 10 units + - To **City 7**: 7 units + - To **City 8**: 10 units + +- From **City 1** to other cities: + - To **City 0**: 6 units + - To **City 2**: 7 units + - To **City 3**: 6 units + - To **City 4**: 6 units + - To **City 5**: 6 units + - To **City 6**: 9 units + - To **City 7**: 10 units + - To **City 8**: 8 units + +- From **City 2** to other cities: + - To **City 0**: 9 units + - To **City 1**: 9 units + - To **City 3**: 8 units + - To **City 4**: 6 units + - To **City 5**: 8 units + - To **City 6**: 10 units + - To **City 7**: 7 units + - To **City 8**: 10 units + +- From **City 3** to other cities: + - To **City 0**: 8 units + - To **City 1**: 6 units + - To **City 2**: 10 units + - To **City 4**: 9 units + - To **City 5**: 10 units + - To **City 6**: 8 units + - To **City 7**: 10 units + - To **City 8**: 9 units + +- From **City 4** to other cities: + - To **City 0**: 6 units + - To **City 1**: 8 units + - To **City 2**: 9 units + - To **City 3**: 8 units + - To **City 5**: 10 units + - To **City 6**: 9 units + - To **City 7**: 7 units + - To **City 8**: 9 units + +- From **City 5** to other cities: + - To **City 0**: 10 units + - To **City 1**: 6 units + - To **City 2**: 7 units + - To **City 3**: 9 units + - To **City 4**: 8 units + - To **City 6**: 10 units + - To **City 7**: 7 units + - To **City 8**: 10 units + +- From **City 6** to other cities: + - To **City 0**: 10 units + - To **City 1**: 10 units + - To **City 2**: 10 units + - To **City 3**: 6 units + - To **City 4**: 9 units + - To **City 5**: 9 units + - To **City 7**: 7 units + - To **City 8**: 8 units + +- From **City 7** to other cities: + - To **City 0**: 6 units + - To **City 1**: 10 units + - To **City 2**: 9 units + - To **City 3**: 8 units + - To **City 4**: 8 units + - To **City 5**: 9 units + - To **City 6**: 8 units + - To **City 8**: 7 units + +- From **City 8** to other cities: + - To **City 0**: 10 units + - To **City 1**: 6 units + - To **City 2**: 8 units + - To **City 3**: 10 units + - To **City 4**: 7 units + - To **City 5**: 6 units + - To **City 6**: 9 units + - To **City 7**: 10 units + +#### Capacity Constraints: +Each route between cities has a maximum capacity limit, which cannot be exceeded: +- **City 0** to **City 1**: 40 units +- **City 0** to **City 2**: 56 units +- **City 0** to **City 3**: 41 units +- **City 0** to **City 4**: 49 units +- **City 0** to **City 5**: 50 units +- **City 0** to **City 6**: 36 units +- **City 0** to **City 7**: 59 units +- **City 0** to **City 8**: 55 units + +- **City 1** to **City 0**: 33 units +- **City 1** to **City 2**: 33 units +- **City 1** to **City 3**: 56 units +- **City 1** to **City 4**: 46 units +- **City 1** to **City 5**: 38 units +- **City 1** to **City 6**: 53 units +- **City 1** to **City 7**: 51 units +- **City 1** to **City 8**: 59 units + +- **City 2** to **City 0**: 59 units +- **City 2** to **City 1**: 52 units +- **City 2** to **City 3**: 41 units +- **City 2** to **City 4**: 38 units +- **City 2** to **City 5**: 36 units +- **City 2** to **City 6**: 45 units +- **City 2** to **City 7**: 32 units +- **City 2** to **City 8**: 41 units + +- **City 3** to **City 0**: 30 units +- **City 3** to **City 1**: 32 units +- **City 3** to **City 2**: 47 units +- **City 3** to **City 4**: 57 units +- **City 3** to **City 5**: 39 units +- **City 3** to **City 6**: 34 units +- **City 3** to **City 7**: 32 units +- **City 3** to **City 8**: 46 units + +- **City 4** to **City 0**: 55 units +- **City 4** to **City 1**: 40 units +- **City 4** to **City 2**: 55 units +- **City 4** to **City 3**: 41 units +- **City 4** to **City 5**: 59 units +- **City 4** to **City 6**: 51 units +- **City 4** to **City 7**: 35 units +- **City 4** to **City 8**: 53 units + +- **City 5** to **City 0**: 34 units +- **City 5** to **City 1**: 41 units +- **City 5** to **City 2**: 35 units +- **City 5** to **City 3**: 43 units +- **City 5** to **City 4**: 41 units +- **City 5** to **City 6**: 43 units +- **City 5** to **City 7**: 40 units +- **City 5** to **City 8**: 41 units + +- **City 6** to **City 0**: 34 units +- **City 6** to **City 1**: 32 units +- **City 6** to **City 2**: 47 units +- **City 6** to **City 3**: 43 units +- **City 6** to **City 4**: 46 units +- **City 6** to **City 5**: 34 units +- **City 6** to **City 7**: 60 units +- **City 6** to **City 8**: 59 units + +- **City 7** to **City 0**: 31 units +- **City 7** to **City 1**: 42 units +- **City 7** to **City 2**: 30 units +- **City 7** to **City 3**: 34 units +- **City 7** to **City 4**: 52 units +- **City 7** to **City 5**: 45 units +- **City 7** to **City 6**: 43 units +- **City 7** to **City 8**: 50 units + +- **City 8** to **City 0**: 49 units +- **City 8** to **City 1**: 31 units +- **City 8** to **City 2**: 53 units +- **City 8** to **City 3**: 46 units +- **City 8** to **City 4**: 42 units +- **City 8** to **City 5**: 36 units +- **City 8** to **City 6**: 30 units +- **City 8** to **City 7**: 41 units + +#### Objective: +Your task is to determine the optimal amount of resources to transport between cities to meet all demands while minimizing the total transportation cost. Ensure that the transportation does not exceed the capacity limits of any route and that all cities' resource requirements are satisfied. + +What is the minimum total cost to achieve this distribution?",533.0 +['Flow Shop Scheduling'],"The manufacturing facility produces seven distinct products (Jobs 0 through 6) using four specialized machines (Machine 0, Machine 1, Machine 2, and Machine 3). Each product must pass through all four machines in a specific sequence, and the goal is to schedule production to minimize the total completion time (makespan), ensuring all products are completed as quickly as possible. The objective is to minimize the sum of the start time of the last product on Machine 3 and the weighted completion times of specific jobs in the final position. Each product must be assigned to exactly one position in the production sequence on each machine, and each position on each machine can only be assigned to one product. The start time of a product on a machine cannot begin until the product has completed processing on the previous machine, and the start time of a product in a later position on a machine cannot begin until the product in the previous position has completed processing on that machine. Processing times for each product on each machine are as follows: Job 0 takes 10 units on Machine 0, 5 units on Machine 1, 5 units on Machine 2, and 7 units on Machine 3; Job 1 takes 6 units on Machine 0, 5 units on Machine 1, 10 units on Machine 2, and 9 units on Machine 3; Job 2 takes 9 units on Machine 0, 8 units on Machine 1, 6 units on Machine 2, and 9 units on Machine 3; Job 3 takes 6 units on Machine 0, 5 units on Machine 1, 9 units on Machine 2, and 6 units on Machine 3; Job 4 takes 5 units on Machine 0, 8 units on Machine 1, 5 units on Machine 2, and 5 units on Machine 3; Job 5 takes 9 units on Machine 0, 10 units on Machine 1, 5 units on Machine 2, and 10 units on Machine 3; Job 6 takes 7 units on Machine 0, 6 units on Machine 1, 10 units on Machine 2, and 5 units on Machine 3. The objective function includes weighted coefficients for specific jobs in the final position, such as 7 for Job 0 in position 6, 9 for Job 1 in position 6, 9 for Job 2 in position 6, 6 for Job 3 in position 6, 5 for Job 4 in position 6, 10 for Job 5 in position 6, and 5 for Job 6 in position 6. The constraints ensure that each job is assigned to exactly one position on each machine, each position on each machine is assigned to exactly one job, and the start times respect the precedence relationships between machines and positions. The variables include binary assignment variables for each job and position on each machine, and continuous start time variables for each position on each machine. The bounds ensure that all start times are non-negative, and the binary variables are restricted to 0 or 1.",72.0 +['TravelingSalesman'],"The goal is to minimize the total travel distance for a delivery route across six cities: City 0, City 1, City 2, City 3, City 4, and City 5. Each city must be visited exactly once, and the route must form a complete loop without any smaller loops (subtours), and finally return to the origin city. The distances between cities are as follows: from City 0 to City 1 (318 units), City 2 (278 units), City 3 (304 units), City 4 (287 units), and City 5 (277 units); from City 1 to City 0 (285 units), City 2 (277 units), City 3 (322 units), City 4 (277 units), and City 5 (306 units); from City 2 to City 0 (275 units), City 1 (302 units), City 3 (299 units), City 4 (277 units), and City 5 (291 units); from City 3 to City 0 (282 units), City 1 (292 units), City 2 (276 units), City 4 (283 units), and City 5 (304 units); from City 4 to City 0 (277 units), City 1 (301 units), City 2 (291 units), City 3 (306 units), and City 5 (325 units); from City 5 to City 0 (325 units), City 1 (319 units), City 2 (306 units), City 3 (279 units), and City 4 (275 units). + +The task is to find the shortest possible route that visits each city exactly once, and return to the origin city. What is the minimal total distance travelled?",1679.0 +['Facility Location Problem'],"Cell Tower Deployment for Regional Coverage + + You are tasked with optimizing the deployment of cell towers across a region to maximize population coverage while adhering to a strict budget constraint. The region is divided into 33 distinct areas, each with a specific population size that needs to be served. Your goal is to determine which cell towers to build and which areas to cover to ensure the maximum number of people have access to cellular service. + + #### Key Decisions: + 1. **Which cell towers to build**: There are 29 potential cell tower locations, each with a unique construction cost. You must decide whether to build each tower (a binary decision: build or not build). + 2. **Which areas to cover**: Each of the 33 areas can be either covered or not covered by the deployed cell towers. Coverage for an area is only possible if at least one of the cell towers that can serve it is built. + + #### Objective: + Maximize the total population covered by the deployed cell towers. The population sizes for the 33 areas are as follows: + - Area 0: 3,522 people +- Area 1: 2,492 people +- Area 2: 2,865 people +- Area 3: 2,596 people +- Area 4: 2,194 people +- Area 5: 4,001 people +- Area 6: 3,027 people +- Area 7: 2,581 people +- Area 8: 4,848 people +- Area 9: 2,814 people +- Area 10: 3,543 people +- Area 11: 2,234 people +- Area 12: 2,960 people +- Area 13: 4,297 people +- Area 14: 2,792 people +- Area 15: 4,520 people +- Area 16: 2,685 people +- Area 17: 2,022 people +- Area 18: 4,858 people +- Area 19: 4,814 people +- Area 20: 3,432 people +- Area 21: 3,730 people +- Area 22: 2,526 people +- Area 23: 3,277 people +- Area 24: 4,363 people +- Area 25: 2,622 people +- Area 26: 3,740 people +- Area 27: 4,594 people +- Area 28: 3,193 people +- Area 29: 4,275 people +- Area 30: 4,407 people +- Area 31: 3,809 people +- Area 32: 2,163 people + + #### Constraints: + 1. **Coverage Requirements**: For an area to be considered covered, at least one of the cell towers that can serve it must be built. The specific combinations of cell towers that can cover each area are: + - Area 0 can be covered if any of the following towers are built: Tower 3, Tower 5, Tower 7, Tower 10, Tower 11, Tower 13, Tower 15, Tower 17, Tower 19, Tower 22, Tower 23, Tower 26, Tower 27, Tower 28. + - Area 1 can be covered if any of the following towers are built: Tower 0, Tower 1, Tower 2, Tower 3, Tower 4, Tower 7, Tower 8, Tower 13, Tower 16, Tower 17, Tower 18, Tower 23, Tower 24, Tower 27, Tower 28. + - Area 2 can be covered if any of the following towers are built: Tower 0, Tower 6, Tower 7, Tower 13, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 24, Tower 28. + - Area 3 can be covered if any of the following towers are built: Tower 0, Tower 4, Tower 5, Tower 6, Tower 8, Tower 10, Tower 13, Tower 18, Tower 20, Tower 22, Tower 24, Tower 25, Tower 27. + - Area 4 can be covered if any of the following towers are built: Tower 1, Tower 2, Tower 3, Tower 8, Tower 11, Tower 12, Tower 14, Tower 17, Tower 19, Tower 25. + - Area 5 can be covered if any of the following towers are built: Tower 2, Tower 6, Tower 9, Tower 11, Tower 12, Tower 17, Tower 18, Tower 19, Tower 20, Tower 27. + - Area 6 can be covered if any of the following towers are built: Tower 2, Tower 3, Tower 7, Tower 8, Tower 9, Tower 12, Tower 14, Tower 21, Tower 24, Tower 27. + - Area 7 can be covered if any of the following towers are built: Tower 2, Tower 5, Tower 6, Tower 7, Tower 8, Tower 11, Tower 16, Tower 17, Tower 18, Tower 20, Tower 21, Tower 25, Tower 26, Tower 27, Tower 28. + - Area 8 can be covered if any of the following towers are built: Tower 1, Tower 7, Tower 8, Tower 10, Tower 12, Tower 17, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24, Tower 26, Tower 28. + - Area 9 can be covered if any of the following towers are built: Tower 6, Tower 10, Tower 12, Tower 14, Tower 15, Tower 18, Tower 20, Tower 22, Tower 23, Tower 28. + - Area 10 can be covered if any of the following towers are built: Tower 4, Tower 7, Tower 8, Tower 12, Tower 13, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 28. + - Area 11 can be covered if any of the following towers are built: Tower 1, Tower 2, Tower 3, Tower 4, Tower 5, Tower 7, Tower 13, Tower 15, Tower 16, Tower 20, Tower 24, Tower 27. + - Area 12 can be covered if any of the following towers are built: Tower 0, Tower 3, Tower 8, Tower 10, Tower 12, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 21, Tower 23, Tower 24. + - Area 13 can be covered if any of the following towers are built: Tower 0, Tower 3, Tower 5, Tower 8, Tower 9, Tower 13, Tower 14, Tower 16, Tower 21, Tower 23, Tower 24, Tower 27. + - Area 14 can be covered if any of the following towers are built: Tower 0, Tower 3, Tower 4, Tower 5, Tower 6, Tower 10, Tower 11, Tower 14, Tower 15, Tower 16, Tower 17, Tower 19, Tower 20, Tower 27, Tower 28. + - Area 15 can be covered if any of the following towers are built: Tower 1, Tower 2, Tower 7, Tower 9, Tower 15, Tower 18, Tower 19, Tower 20, Tower 24, Tower 25, Tower 26, Tower 28. + - Area 16 can be covered if any of the following towers are built: Tower 4, Tower 5, Tower 6, Tower 8, Tower 9, Tower 13, Tower 15, Tower 16, Tower 17, Tower 19, Tower 21, Tower 23, Tower 24, Tower 27. + - Area 17 can be covered if any of the following towers are built: Tower 0, Tower 2, Tower 3, Tower 7, Tower 10, Tower 11, Tower 14, Tower 16, Tower 20, Tower 21, Tower 25, Tower 27, Tower 28. + - Area 18 can be covered if any of the following towers are built: Tower 0, Tower 1, Tower 2, Tower 7, Tower 8, Tower 10, Tower 16, Tower 20, Tower 21, Tower 22, Tower 23, Tower 25, Tower 26, Tower 28. + - Area 19 can be covered if any of the following towers are built: Tower 3, Tower 4, Tower 6, Tower 7, Tower 10, Tower 13, Tower 15, Tower 16, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 27, Tower 28. + - Area 20 can be covered if any of the following towers are built: Tower 1, Tower 3, Tower 10, Tower 11, Tower 12, Tower 13, Tower 14, Tower 20, Tower 21, Tower 22, Tower 23, Tower 24, Tower 27. + - Area 21 can be covered if any of the following towers are built: Tower 4, Tower 5, Tower 6, Tower 7, Tower 8, Tower 13, Tower 14, Tower 17, Tower 24, Tower 27. + - Area 22 can be covered if any of the following towers are built: Tower 0, Tower 1, Tower 2, Tower 3, Tower 14, Tower 17, Tower 20, Tower 24, Tower 25, Tower 27, Tower 28. + - Area 23 can be covered if any of the following towers are built: Tower 0, Tower 1, Tower 5, Tower 6, Tower 12, Tower 13, Tower 15, Tower 20, Tower 23, Tower 26, Tower 28. + - Area 24 can be covered if any of the following towers are built: Tower 4, Tower 6, Tower 9, Tower 13, Tower 14, Tower 15, Tower 17, Tower 21, Tower 22, Tower 25, Tower 27, Tower 28. + - Area 25 can be covered if any of the following towers are built: Tower 1, Tower 10, Tower 15, Tower 17, Tower 18, Tower 21, Tower 23, Tower 24, Tower 26, Tower 27, Tower 28. + - Area 26 can be covered if any of the following towers are built: Tower 1, Tower 2, Tower 3, Tower 5, Tower 7, Tower 12, Tower 16, Tower 18, Tower 19, Tower 21, Tower 23, Tower 26, Tower 27, Tower 28. + - Area 27 can be covered if any of the following towers are built: Tower 1, Tower 2, Tower 6, Tower 7, Tower 8, Tower 10, Tower 13, Tower 16, Tower 18, Tower 19, Tower 21, Tower 22, Tower 26, Tower 28. + - Area 28 can be covered if any of the following towers are built: Tower 0, Tower 2, Tower 3, Tower 4, Tower 9, Tower 10, Tower 12, Tower 14, Tower 18, Tower 20, Tower 21, Tower 23. + - Area 29 can be covered if any of the following towers are built: Tower 2, Tower 4, Tower 6, Tower 7, Tower 8, Tower 9, Tower 11, Tower 16, Tower 17, Tower 23. + - Area 30 can be covered if any of the following towers are built: Tower 0, Tower 9, Tower 14, Tower 16, Tower 17, Tower 19, Tower 20, Tower 22, Tower 24, Tower 26, Tower 27. + - Area 31 can be covered if any of the following towers are built: Tower 3, Tower 4, Tower 6, Tower 8, Tower 9, Tower 10, Tower 15, Tower 17, Tower 18, Tower 20, Tower 23, Tower 25, Tower 26, Tower 27, Tower 28. + - Area 32 can be covered if any of the following towers are built: Tower 0, Tower 1, Tower 2, Tower 4, Tower 5, Tower 8, Tower 10, Tower 13, Tower 20, Tower 22, Tower 26, Tower 27, Tower 28. + + 2. **Budget Constraint**: The total cost of building the selected cell towers must not exceed the available budget of 1,059 units. The construction costs for the 29 towers are as follows: + - Tower 0: 303 units + - Tower 1: 378 units + - Tower 2: 327 units + - Tower 3: 444 units + - Tower 4: 391 units + - Tower 5: 357 units + - Tower 6: 444 units + - Tower 7: 244 units + - Tower 8: 433 units + - Tower 9: 407 units + - Tower 10: 305 units + - Tower 11: 296 units + - Tower 12: 315 units + - Tower 13: 302 units + - Tower 14: 495 units + - Tower 15: 240 units + - Tower 16: 379 units + - Tower 17: 211 units + - Tower 18: 388 units + - Tower 19: 220 units + - Tower 20: 408 units + - Tower 21: 430 units + - Tower 22: 327 units + - Tower 23: 470 units + - Tower 24: 381 units + - Tower 25: 476 units + - Tower 26: 215 units + - Tower 27: 239 units + - Tower 28: 372 units + + 3. **Binary Decisions**: Each cell tower can either be built (1) or not built (0), and each area can either be covered (1) or not covered (0). + + #### Summary: + Your task is to select a combination of cell towers to build, ensuring that the total cost does not exceed 1,059 units, while maximizing the total population covered across the 33 areas. The coverage for each area depends on the specific towers built, and the population sizes and tower costs are provided in detail. The goal is to make strategic decisions about tower deployment to achieve the highest possible population coverage within the budget.",111796.0 +['Facility Dispersion Problem'],"Imagine you are managing the placement of emergency response centers across five key locations in a region. Each location has the potential to host a center, but due to budget constraints, you can only establish **three centers** in total. The goal is to maximize the overall effectiveness of these centers by ensuring they are strategically placed to cover as much critical area as possible. + +Each pair of locations has a predefined **effectiveness score** that represents how well a center at one location can support or complement a center at another location. These scores are as follows: + +- **Location 0 and Location 1**: 18 +- **Location 0 and Location 2**: 20 +- **Location 0 and Location 3**: 22 +- **Location 0 and Location 4**: 19 +- **Location 1 and Location 2**: 15 +- **Location 1 and Location 3**: 22 +- **Location 1 and Location 4**: 16 +- **Location 2 and Location 3**: 17 +- **Location 2 and Location 4**: 22 +- **Location 3 and Location 4**: 20 + +Your task is to select **three locations** out of the five to establish the centers. The effectiveness of the selected centers is determined by the sum of the pairwise effectiveness scores between the chosen locations. For example, if you select Locations 0, 1, and 2, the total effectiveness would be the sum of the scores for pairs (0,1), (0,2), and (1,2). + +However, there are specific rules governing the selection: +1. A pair of locations can only contribute to the total effectiveness if **both locations in the pair are selected** as centers. +2. If only one location in a pair is selected, that pair does not contribute to the effectiveness score. +3. The total number of centers must be exactly three. + +Your objective is to determine which three locations to select to **maximize the total effectiveness** of the emergency response centers, ensuring all constraints are met.",62.0 diff --git a/optimind/eval.py b/optimind/eval.py new file mode 100644 index 0000000..085992f --- /dev/null +++ b/optimind/eval.py @@ -0,0 +1,911 @@ +#!/usr/bin/env python3 + +import asyncio, json, logging, os, re, sys, time, random +import pandas as pd, numpy as np +from functools import partial +from concurrent.futures import ThreadPoolExecutor +from json_repair import repair_json +import nest_asyncio, pdb +from datetime import datetime +from collections import Counter +import argparse, os, logging, warnings, subprocess, math, random, ast, re +from pathlib import Path + +from typing import Any, Dict, List, Optional, Tuple, Callable, Awaitable, NamedTuple +import math +import copy + +from eval_utils.utils_general import ( + setup_model_and_env, + as_list, + get_hint_instructions, + load_ea_pairs, + build_error_analysis_str, +) +from eval_utils.build_prompt import ( + build_prompt_optmath, + build_prompt_sirl, + build_prompt_sirl_system, +) + +# ────────────────── ARGUMENT PARSER ───────────────────────────────── +parser = argparse.ArgumentParser(description="Run multi-turn evaluation with hybrid backend (SGLang or openai).") +# data name +parser.add_argument("--model_path", type=str, default="Qwen/Qwen3-8B", help="Path to the model.") +parser.add_argument("--data", type=str, required=True, help="Path to a .csv, .jsonl, or .parquet file to evaluate.") +parser.add_argument("--max-turns", type=int, default=5, help="maximum number of turns") +parser.add_argument("--num-majority", type=int, default=1, help="Number of parallel samples per turn for majority vote. 1 = no voting.") + + +# Use cached model only +parser.add_argument("--offline_only", action="store_true", + help="Force fully-offline mode (error if files are not in local cache).") +parser.add_argument("--hf_cache_dir", type=str, default=None, + help="Path to HF cache (default: ~/.cache/huggingface). If set, we only read from here.") +parser.add_argument("--hf_revision", type=str, default=None, + help="Specific revision/commit to load from local cache (optional, default fetching the newest snapshot from local cache).") + +# backend params +parser.add_argument("--backend_str", type=str, default="sglang", help="Backend to use: hybrid, sglang, openai.") +parser.add_argument("--tp_size", type=int, default=1, help="Tensor parallel size for SGLang.") +parser.add_argument("--dp_size", type=int, default=1, help="Data parallel size for SGLang.") +parser.add_argument("--suffix", type=str, default="", help="Suffix for the output directory.") +parser.add_argument("--openapi_api_key_name", type=str, default="OPENROUTER_API_KEY", help="Environment variable name for OpenAI API key.") +parser.add_argument("--openai_base_url", type=str, default="https://openrouter.ai/api/v1", help="Base URL for OpenAI SDK when backend_str=openai.") +parser.add_argument("--openai_model_name", type=str, default="x-ai/grok-4.1-fast", help="OpenAI SDK model name (e.g. 'x-ai/grok-4.1-fast') when backend_str=openai.") +parser.add_argument("--seed", default=None, type=int, help="Random seed for reproducibility.") +parser.add_argument("--batch_size_sglang", type=int, default=256, help="Batch size for sglang.") + +# sampling params +parser.add_argument("--temp", type=float, default=0.6, help="Temperature for sampling.") +parser.add_argument("--top-p", type=float, default=0.95, help="Top-p for sampling.") + +# gpt oss params +parser.add_argument("--gpt-oss", action="store_true", help="use gpt oss based model") +parser.add_argument("--reasoning-effort", type=str, default="medium", help="gpt oss reasoning effort") + +# prompt params +parser.add_argument("--user_prompt_type", default="default", type=str, choices=["default", "optmath", "sirl"], + help="Type of prompt to use (default, optmath, sirl.).") +parser.add_argument("--tool_prompt_type", default="default", type=str, choices=["default", "add_question", "add_hint", "add_initial_prompt"], + help="Type of prompt to use (default, optmath, sirl.).") +parser.add_argument("--system_prompt_type", default="default", type=str, choices=["default", "simple"]) +parser.add_argument("--force", action="store_true", help="Force re-evaluation even if results exist.") +parser.add_argument("--debug", action="store_true", help="Debug the code.") + + +# error analysis params +parser.add_argument("--apply_error_analysis", action="store_true", help="Use error analysis in the prompt") +parser.add_argument("--max_edit", type=int, default=5, + help="Max edit distance for fuzzy matching 'problem_class' to error-analysis types.") +parser.add_argument("--error_analysis_file", type=str, help="CSV with error analysis & hints.") +parser.add_argument("--apply_hint_instructions", action="store_true", help="Add instructions for using hints and general hints") + +# early stopping in multi-turn correction +parser.add_argument("--no_early_stop", action="store_true", help="Do not early terminate if the model outputs a correct answer before max turns.") + +# majority voting +parser.add_argument("--majority-rtol", type=float, default=1e-6, help="Relative tolerance for majority voting.") +parser.add_argument("--majority-atol", type=float, default=1e-4, help="Absolute tolerance for majority voting.") +parser.add_argument("--majority_vote_final", action="store_true", help="Use majority vote to select the final answer among all turns.") + +# Utils +parser.add_argument("--precision", type=float, default=1e-6, help="Precision for comparing float answers.") +parser.add_argument("--output_dir", type=str, default="eval_results", help="Output directory to save results.") + +# parse arguments +args = parser.parse_args() + +# ────────────────── NEST ASYNCIO ──────────────────────────────────── +nest_asyncio.apply() +asyncio.set_event_loop(asyncio.new_event_loop()) + +# ───────────────────────── CONFIG ──────────────────────────────────── +cfg_model = setup_model_and_env(args) +MODEL_PATH = cfg_model["MODEL_PATH"] +MODEL_NAME = cfg_model["MODEL_NAME"] +MODEL_LOAD_PATH = cfg_model["MODEL_LOAD_PATH"] + +DATA_PATH = args.data +MAX_TURNS = args.max_turns +NUM_MAJORITY = args.num_majority +MAJORITY_FINAL = args.majority_vote_final +MAX_NEW_TOKENS = 8_000 if not args.gpt_oss else 24_000 # qwen is subject to context limit +TEMP, TOP_P = args.temp, args.top_p +BATCH_SIZE_SGLANG = args.batch_size_sglang # prompts per GPU pass +THREADS_POOL = 60 # parallel api calls +MAX_TOOL_PAR = 60 # concurrent CodeExecutionTool +TP_SIZE = args.tp_size +DP_SIZE = args.dp_size + +# gpt oss is not subject to this token limit, so only for qwen +MAX_PROMPT_TOKEN = 24_000 # a few tokens for assistant message +PRECISION = args.precision + +if args.backend_str not in ["hybrid", "sglang", "openai"]: + BACKEND_SCHEDULE = ["sglang"] * MAX_TURNS + BACKEND_STR = "sglang" if all(b == "sglang" for b in BACKEND_SCHEDULE) else "hybrid" +else: + BACKEND_SCHEDULE = ["sglang"] + ["openai"] * (MAX_TURNS - 1) if args.backend_str == "hybrid" else \ + [args.backend_str] * MAX_TURNS + BACKEND_STR = args.backend_str + +TOOLS = [{ + "type": "function", + "function": { + "name": "code_execution", + "description": "Execute Gurobi code and return stdout / stderr", + "parameters": {"type": "object", + "properties": {"code": {"type": "string", "description": "The gurobipy code to execute"}}, + "required": ["code"]}, + "strict": False, + }, +}] +CODE_TOOL_IDX = 0 + +# ───────────────── CodeExecutionTool ───────────────────────────────── +path = os.path.dirname(os.path.dirname(os.path.abspath(__file__))) +sys.path.append(path) +print("Load parent path", path) + +from recipe.tools.code_execution import CodeExecutionTool +from rewards.rm_objective_gurobipy_full import compute_score +from rewards.utils import compute_score_history_output +from rewards.utils import extract_python_code_block as extract_python_code_block_helper + +from recipe.tools.schemas import OpenAIFunctionToolSchema + +exec_tool = CodeExecutionTool({"max_assistant_turns": MAX_TURNS, "tool_format": "interaction"}, + OpenAIFunctionToolSchema.model_validate(TOOLS[CODE_TOOL_IDX])) +tool_sem = asyncio.Semaphore(MAX_TOOL_PAR) + + +####################### TOOL PROMPT and SYSTEM PROMPT ########################## +CODE_USER_STR = """ +Running the given gurobipy code block result in the following standard output and error. +Please analyze the standard output and error and provide your thoughts on the correctness of the code and the output, and if it is not correct, provide a corrected version of the code. + +{tool_output} + +# Note: +- The Code must include:```python +import gurobipy as gp +from gurobipy import GRB +``` +- Make sure the model variable is named `model`. +- Avoid using "<" and ">" in Gurobi constraints; instead, use "<=" or ">=". +- Carefully determine whether the variable is an integer or continuous. +""" + +CODE_SYSTEM_PROMPT = """You are an expert in optimization and mixed integer programming. You are given an optimization problem and you need to solve it using gurobipy. +Reasoning step by step before generating the gurobipy code. The user will execute the code and return the standard output and error to you, and you should use the user response to refine your previously generated code. +When you are satisfied with the code, you can put your final answer in the format of `#### `.""" + +SIMPLE_SYSTEM_PROMPT = """You are an expert in optimization and mixed integer programming. You are given an optimization problem and you need to solve it using gurobipy. +Reasoning step by step before generating the gurobipy code. """ + + + +# ─────────────────── Build prompt with error analysis ─────────────────────────────────────────────── +def build_first_turn_user_prompt_list(question: str, error_pairs: List = []) -> str: + lines = [] + lines.append("You are an expert in optimization and mixed integer programming. You are given an optimization problem and you need to solve it using gurobipy.\n") + + lines.append(question.strip()) + + local_hint, global_hint = "", "" + if len(error_pairs) > 0: + local_hint = build_error_analysis_str(error_pairs) + lines.append(local_hint) + + if args.apply_hint_instructions: + global_hint = get_hint_instructions() + lines.append(global_hint) + + combined_hint = "" if not (local_hint or global_hint) else "\n".join([local_hint, global_hint]) + + lines.append( + "Reason step by step before generating the gurobipy code.\n" + "When you respond, first think carefully.\n" + "After thinking, output the math modeling of the problem.\n" + "Finally output a ```python ...``` code block that solves the problem.\n" + "The code must include:\n" + "import gurobipy as gp\n" + "from gurobipy import GRB\n" + ) + + print(combined_hint) + return "".join(lines), combined_hint + +def extract_python_code_block(response: str) -> str | None: + return extract_python_code_block_helper(response, return_none_if_failed=True) + +# helper function to get more tool prompt if needed +def get_more_tool_prompt(item): + if args.tool_prompt_type == "default": + return "" + + if args.tool_prompt_type == "add_question": + more_tool_prompt = f"""Recap: the optimization problem that you need to solve using gurobipy is +{item['initial_question']}""" + elif args.tool_prompt_type == "add_hint": + more_tool_prompt = f"""Recap: {item['initial_hint'].lstrip()}""" + elif args.tool_prompt_type == "add_initial_prompt": + more_tool_prompt = f"""Recap: the optimization problem that you need to solve using gurobipy is +{item['initial_question']} +{item['initial_hint'].lstrip()} +""" + + return more_tool_prompt + +async def run_tool_as_user(item, call_arg, more_tool_prompt=""): + if call_arg is None or not call_arg: + if len(item["metrics"]) == 0 or args.no_early_stop: + # in these cases, we may ask the model to generate new code; + item["history"].append({"role": "user", "content": "No python code block found. Skipping code execution. If you believe the last provided code is correct, you can stop. Otherwise, please provide a corrected code block."}) + + # We default the metric to the previous round if it exists. + # However, if valid initial code is extracted, we have no previous solution to default to, hence we set all metrics to 0.0 + if len(item["metrics"]) == 0: + item["metrics"].append({"score": 0.0, "no_execution_error": 0.0, + "have_objective": 0.0, "objective": 0.0, "accuracy": 0.0}) + else: + item["metrics"].append(copy.deepcopy(item["metrics"][-1])) + return + + async with tool_sem: + out, _, m = await exec_tool.execute(item["iid"], call_arg, precision=PRECISION) + + + user_output = CODE_USER_STR.format(tool_output=out) + if more_tool_prompt: + user_output += "\n" + more_tool_prompt + + item["history"].append({"role": "user", "content": user_output}) + + item["metrics"].append(m) + item["found_code"] = True + +### find a way to convert the assistant history into a tool call history by splitting the text and the code and use that for tool call + +# ─────────────────── Tokenize prompts ──────────────────────────────── +from transformers import AutoTokenizer + +# if global variable tokenizer is not set, load the tokenizer +if 'tokenizer' not in globals(): + tokenizer = AutoTokenizer.from_pretrained( + MODEL_LOAD_PATH, + trust_remote_code=True, + local_files_only=args.offline_only, # <-- offline only if requested + ) +else: + tokenizer = globals()['tokenizer'] + + +if args.gpt_oss: + from eval_utils.harmony_utils import harmony_parse_calls + +def to_prompt_no_tools(msgs, add_generation_prompt=True): + if args.backend_str in ["azure", "openai"]: + # TOOLS are encoded in the openai query api, not by the string template + # concatenate all messages directly, assuming no tokenizer + segs_str = "\n".join(f"{m['role'].capitalize()}: {m['content']}\n" for m in msgs) + if add_generation_prompt: + segs_str += "Assistant: " + return segs_str + + + segs_str = tokenizer.apply_chat_template(msgs, add_generation_prompt=add_generation_prompt, tokenize=False) + + if args.gpt_oss and args.reasoning_effort != "medium": + # search for "Reasoning: (low | mid | high)", replace the first + segs_str = re.sub(r"(Reasoning:\s*)(low|medium|high)", + lambda m: m.group(1) + args.reasoning_effort, segs_str, count=1, flags=re.IGNORECASE) + + # cap the prompt window for qwen models + if not args.gpt_oss: + segs_tokens = tokenizer.encode(segs_str) + if len(segs_tokens) > MAX_PROMPT_TOKEN: + print(f"Warning: the prompt exceeds the maximum token length, truncating to fit (Current length {len(segs_tokens)} | Length limit {MAX_PROMPT_TOKEN}).") + segs_str = tokenizer.decode(segs_tokens[-MAX_PROMPT_TOKEN:]) + + return segs_str + +def majority_vote_select(answers: List[Optional[float]], rtol: float = 1e-6, atol: float = 1e-4) -> Tuple[int, Optional[float], int]: + clusters: List[Tuple[float, List[int]]] = [] + for idx, val in enumerate(answers): + if val is None: + continue + placed = False + for rep, idxs in clusters: + if math.isclose(float(val), float(rep), rel_tol=rtol, abs_tol=atol): + idxs.append(idx); placed = True; break + if not placed: + clusters.append((float(val), [idx])) + if not clusters: + return 0, None, 0 + clusters.sort(key=lambda x: (-len(x[1]), min(x[1]))) + rep_val, idxs = clusters[0] + selected_j = min(idxs) + return selected_j, rep_val, len(idxs) + +##### General +class GenOutput(NamedTuple): + text: str + token_ids: Optional[List[int]] = None # only for sglang + +# type alias: async function (histories, pool, loop) -> list[GenOutput] +BackendGenFn = Callable[ + [List[List[Dict[str, Any]]], ThreadPoolExecutor, asyncio.AbstractEventLoop], + Awaitable[List[GenOutput]], +] + +async def generic_gen( + todo_items: List[Dict[str, Any]], + backend_label: str, + generate_batch: BackendGenFn, + pool: ThreadPoolExecutor, + loop: asyncio.AbstractEventLoop, +): + """ + Shared generation logic for all backends. + + - `generate_batch(histories)` returns List[GenOutput] with same length. + - `histories` is a list of chat histories: List[{"role", "content"}]. + """ + print(f"Generating with {backend_label}...") + + # ---------- 1) one or many samples per item ---------- + if MAJORITY_FINAL or NUM_MAJORITY == 1: + histories = [it["history"] for it in todo_items] + outs = await generate_batch(histories, pool, loop) + else: + histories = [it["history"] for it in todo_items for _ in range(NUM_MAJORITY)] + repeated_outs = await generate_batch(histories, pool, loop) + + outs = [] + for idx, it in enumerate(todo_items): + outs_idx = repeated_outs[idx * NUM_MAJORITY:(idx + 1) * NUM_MAJORITY] + + repeated_objectives_idx = [] + for o in outs_idx: + res = compute_score("nl4ilp", o.text, it["ground_truth"], extra_info=None, precision=PRECISION) + obj = res["objective"] if res["have_objective"] else None + repeated_objectives_idx.append(obj) + + if repeated_objectives_idx: + chosen_j, majority_objective, majority_count = majority_vote_select( + repeated_objectives_idx, rtol=args.majority_rtol, atol=args.majority_atol, + ) + chosen = outs_idx[chosen_j] if majority_objective is not None else outs_idx[0] + else: + chosen = outs_idx[0] + + outs.append(chosen) + + # ---------- 2) common postprocessing: history + code + calls_args ---------- + for it, o in zip(todo_items, outs): + txt = o.text + token_ids = o.token_ids + + if args.gpt_oss and token_ids is not None: + # only for gpt-oss models (sglang path) + reasoning_content, final_content = harmony_parse_calls(tokenizer, token_ids, txt, keep_tools=False) + code = extract_python_code_block(txt) + + if reasoning_content and final_content: + it["history"].append({"role": "assistant", "content": final_content, "thinking": reasoning_content}) + elif reasoning_content or final_content: + content = reasoning_content if reasoning_content else final_content + it["history"].append({"role": "assistant", "content": content}) + else: + it["history"].append({"role": "assistant", "content": txt}) + else: + code = extract_python_code_block(txt) + it["history"].append({"role": "assistant", "content": txt}) + + if not code: + print(f"No code block found in {backend_label} response, skipping item.") + it["calls_args"] = [] + else: + # we store the full reply as "code" so the tool can parse the python block inside + it["calls_args"] = [{"code": txt}] + + print(f"{backend_label} generation done.") + + +# ─────────────────── S G L A N G ───────────────────────────────────── +if "sglang" in BACKEND_SCHEDULE: + import sglang as sgl + + async def sglang_generate_batch(histories, pool, loop) -> List[GenOutput]: + prompts = [to_prompt_no_tools(h) for h in histories] + outs = await engine.async_generate(prompts, sglang_sampling_params) + + rets: List[GenOutput] = [] + for o in outs: + text = o["text"] + token_ids = o.get("output_ids") if args.gpt_oss else None + rets.append(GenOutput(text=text, token_ids=token_ids)) + return rets + + +# ──────────────────── O P E N A I (remote backend) ───────── +if "openai" in BACKEND_SCHEDULE: + from openai import OpenAI + import dotenv + dotenv.load_dotenv(os.path.join(os.path.dirname(__file__), "eval_utils/.env")) + + OPENAI_BASE_URL = args.openai_base_url # e.g. "https://openrouter.ai/api/v1" + OPENAI_MODEL = args.openai_model_name + OPENAI_API_KEY = os.environ.get(args.openapi_api_key_name) + + if not OPENAI_API_KEY: + raise RuntimeError( + f"{args.openapi_api_key_name} environment variable is not set. " + f"Export it before running when backend_str=openai." + ) + + openai_client = OpenAI( + base_url=OPENAI_BASE_URL, + api_key=OPENAI_API_KEY, + ) + + def _openai_block(prompt, client, model_name): + if not isinstance(prompt, list): + prompt = [{"role": "user", "content": prompt}] + + m = None + for _ in range(5): + try: + r = client.chat.completions.create( + model=model_name, + messages=prompt, + # temperature=TEMP, + # top_p=TOP_P, + ) + response = r.choices[0].message + m = response.content + except Exception as e: + print(f"OpenAI call error: {e}, retrying...") + import pdb; pdb.set_trace() + time.sleep(random.uniform(2, 5)) + # import pdb; pdb.set_trace() + return m if m else "" + + async def openai_generate_batch(histories, pool, loop) -> List[GenOutput]: + replies = await asyncio.gather( + *[loop.run_in_executor(pool, partial(_openai_block, h, client=openai_client, + model_name=OPENAI_MODEL)) for h in histories]) + return [GenOutput(text=reply) for reply in replies] + + +# ──────────────────── hybrid dispatch ─────────────────────────────── +async def dispatch_backend(todo_items, backend, pool, loop): + if backend == "sglang": + await generic_gen(todo_items, backend_label="SGLANG", + generate_batch=sglang_generate_batch, pool=pool, loop=loop) + elif backend == "openai": + await generic_gen(todo_items, backend_label=f"OpenAI {OPENAI_MODEL} at {OPENAI_BASE_URL}", + generate_batch=openai_generate_batch, pool=pool, loop=loop) + else: + raise ValueError(backend) + + +# ──────────────────── Solve batch ──────────────────────────────────── +from eval_utils.build_prompt import build_prompt_optmath, build_prompt_sirl, build_prompt_sirl_system + +def get_assistant_messages(history): + assistants = [d for d in history if isinstance(d, dict) and d.get("role") == "assistant"] + sol_assistant = "\n".join([x for a in assistants for x in [a.get("thinking", ""), a.get("content", "")] if x]) + return sol_assistant + +def get_all_messages(history): + sol_str = "\n".join([x for d in history for x in [d.get("thinking", ""), d.get("content", ""), "-"*30] if x]) + return sol_str + + + +async def final_compute_score(items, pool, loop): + sol_extracted = [None] * len(items) + + recs = [] + for it, sol_ex in zip(items, sol_extracted): + if it["history"][0]["role"] == "system": + pstr = to_prompt_no_tools(it["history"][:2], add_generation_prompt=False) + fstr = to_prompt_no_tools([h for i, h in enumerate(it["history"]) if i < 2 or h["role"] == "assistant"], add_generation_prompt=False) + else: + pstr = to_prompt_no_tools(it["history"][:1], add_generation_prompt=False) + fstr = to_prompt_no_tools([h for i, h in enumerate(it["history"]) if i == 0 or h["role"] == "assistant"], add_generation_prompt=False) + pl = len(tokenizer.encode(pstr, add_special_tokens=False)) + sl = len(tokenizer.encode(fstr[len(pstr):], add_special_tokens=False)) + + res = compute_score_history_output("nl4ilp", it["history"], it["ground_truth"], extra_info=None, precision=PRECISION) + + it.update(objective=res["objective"], score=res["accuracy"], + no_execution_error=res["no_execution_error"], + have_objective=res["have_objective"], + prompt_len=pl, solution_len=sl) + + ##### save record + for t in range(MAX_TURNS): + it[f"accuracy_{t}"] = (it["metrics"][t]["accuracy"] if t < len(it["metrics"]) else it.get(f"accuracy_{t-1}", 0.0)) + it[f"objective_{t}"] = (it["metrics"][t]["objective"] if t < len(it["metrics"]) else it.get(f"objective_{t-1}", 0.0)) + it[f"no_execution_error_{t}"] = (it["metrics"][t]["no_execution_error"] if t < len(it["metrics"]) else it.get(f"no_execution_error_{t-1}", True)) + it[f"have_objective_{t}"] = (it["metrics"][t]["have_objective"] if t < len(it["metrics"]) else it.get(f"have_objective_{t-1}", False)) + + if MAX_TURNS > 0 and it[f"accuracy_{MAX_TURNS-1}"] != res["accuracy"]: + save_idx = 0 + os.makedirs("tmp_compute_score", exist_ok=True) + while os.path.exists(f"tmp_compute_score/item_{save_idx}.json"): + save_idx += 1 + with open(f"tmp_compute_score/item_{save_idx}.json", "w") as fp: + json.dump({"history": it["history"], "ground_truth": it["ground_truth"], + f"accuracy_{MAX_TURNS-1}": it[f"accuracy_{MAX_TURNS-1}"], "accuracy": res["accuracy"]}, fp) + with open(f"tmp_compute_score/history_{save_idx}.txt", "w") as f: + f.write(get_all_messages) + + recs.append({"id": it["iid"], + "objective": it["objective"], + "ground_truth": it["ground_truth"], + "score": it["score"], + "no_execution_error": it["no_execution_error"], + "have_objective": it["have_objective"], + "accuracy_per_turn": {f"accuracy_{t}": it[f"accuracy_{t}"] for t in range(MAX_TURNS)}, + "objective_per_turn": {f"objective_{t}": it[f"objective_{t}"] for t in range(MAX_TURNS)}, + "no_execution_error_per_turn": {f"no_execution_error_{t}": it[f"no_execution_error_{t}"] for t in range(MAX_TURNS)}, + "have_objective_per_turn": {f"have_objective_{t}": it[f"have_objective_{t}"] for t in range(MAX_TURNS)}, + "dialogue": it["history"], + "prompt_len": it["prompt_len"], + "solution_len": it["solution_len"]}) + + return recs + + +async def build_items(rows, majority_final=MAJORITY_FINAL, num_majority=NUM_MAJORITY): + items = [] + ##################### load error analysis prompt ##################### + # (Optional) early file existence check + if args.apply_error_analysis and args.error_analysis_file and not Path(args.error_analysis_file).exists(): + raise FileNotFoundError(f"error_analysis_file not found: {args.error_analysis_file}") + + if args.apply_error_analysis: + # read error analysis file + ea = pd.read_csv(Path(args.error_analysis_file)) + # define once (removed duplicate) + class_to_pairs = load_ea_pairs(ea, args) + + ##################### load system (optional) and user prompts, build items ##################### + for ir, r in enumerate(rows): + question = getattr(r, "question", None) + answer = getattr(r, "answer", None) + answer = ast.literal_eval(answer) if isinstance(answer, str) else answer + classes = as_list(getattr(r, "problem_class", [])) + combined_hint = "" + + if args.apply_error_analysis and classes: + error_pairs: List[Tuple[str, str]] = [] + for c in classes: + error_analyses = class_to_pairs[c] if c in class_to_pairs else [] + error_pairs.extend(error_analyses) + + # row-level dedup (type corrected) + seen = set() + dedup: List[Tuple[str, str]] = [] + for a in error_pairs: + key = a + if key not in seen: + seen.add(key) + dedup.append(key) + + if args.user_prompt_type == "optmath": + user_content = build_prompt_optmath(question) + "\n" + build_error_analysis_str(dedup) + elif args.user_prompt_type == "sirl": + user_content = build_prompt_sirl(question) + "\n" + build_error_analysis_str(dedup) + else: + try: + user_content, combined_hint = build_first_turn_user_prompt_list(question, dedup) + except Exception as e: + import pdb; pdb.set_trace() + else: + if args.user_prompt_type == "optmath": + user_content = build_prompt_optmath(question) + if args.apply_hint_instructions: + user_content += "\n" + get_hint_instructions() + elif args.user_prompt_type == "sirl": + user_content = build_prompt_sirl(question) + if args.apply_hint_instructions: + user_content += "\n" + get_hint_instructions() + else: + try: + user_content, combined_hint = build_first_turn_user_prompt_list(question) + except Exception as e: + import pdb; pdb.set_trace() + + if args.user_prompt_type == "sirl": + # If build_prompt_sirl_system has parameters, pass them here instead. + system_content = build_prompt_sirl_system() + else: + system_content = CODE_SYSTEM_PROMPT if args.system_prompt_type == "default" else SIMPLE_SYSTEM_PROMPT + + if args.no_early_stop and args.user_prompt_type != "sirl": + hist = [{"role": "user", "content": user_content}] + else: + hist = [ + {"role": "system", "content": system_content}, + {"role": "user", "content": user_content}, + ] + + if majority_final and num_majority > 1: + for i_majority in range(num_majority): + iid = await exec_tool.create(instance_id=ir * num_majority + i_majority, ground_truth=answer) + it = {"iid": iid, "initial_question": question, + "ground_truth": answer, "ground_truth_code": getattr(r, "code", None), + "history": copy.deepcopy(hist), "metrics": [], "done": False, "found_code": False} + if args.tool_prompt_type != "default": + it.update({"initial_hint": combined_hint}) + items.append(it) + else: + iid = await exec_tool.create(instance_id=ir, ground_truth=answer) + it = {"iid": iid, "initial_question": question, + "ground_truth": answer, "history": hist, "metrics": [], "done": False, "found_code": False} + if args.tool_prompt_type != "default": + it.update({"initial_hint": combined_hint}) + items.append(it) + + return items + + +async def solve_batch_no_tools(rows): + items = await build_items(rows) + + ##################### main loop ##################### + pool = ThreadPoolExecutor(max_workers=THREADS_POOL) + loop = asyncio.get_running_loop() + + # if majority_final: repeat the items and then aggregate at the end + + print(f"Data: {DATA_PATH}") + print(f"Model: {MODEL_PATH}") + print(f"Seed: {args.seed}") + print(f"OUT_DIR: {OUT_DIR}") + + print(f"Error analysis file: {args.error_analysis_file}") + for turn in range(MAX_TURNS): + print("Start turn", turn + 1) + backend = BACKEND_SCHEDULE[min(turn, len(BACKEND_SCHEDULE)-1)] + todo = [it for it in items if not it["done"]] + if not todo: + break + + await dispatch_backend(todo, backend, pool, loop) + + print("Finished llm generation for turn", turn + 1) + + calls_args_list = [it["calls_args"][0] if "calls_args" in it and len(it["calls_args"]) > 0 else None for it in todo] + tool_tasks = [run_tool_as_user(it, calls_args, + more_tool_prompt=get_more_tool_prompt(it) if args.tool_prompt_type != "default" else "") + for it, calls_args in zip(todo, calls_args_list)] + if tool_tasks: + await asyncio.gather(*tool_tasks) + + accuracies_turn = [it["metrics"][-1]["accuracy"] if ("metrics" in it and len(it["metrics"]) > 0) else 0.0 for it in items] + print("Finished tool calls for turn", turn + 1, f"accuracy: {np.mean(accuracies_turn)} +- {np.std(accuracies_turn)})") + + for it in todo: + # Note: calls_args is empty if no code is extracted + if args.no_early_stop or not it["found_code"]: + it["done"] = False # we never done until max turns | or if no code has been found yet, we continue + else: + # stop if cannot find any code; condition in some previous turn has found some code + it["done"] = len(it["metrics"]) > 0 and "calls_args" in it and len(it["calls_args"]) == 0 + + pool.shutdown(wait=True) + await asyncio.gather(*(exec_tool.release(it["iid"]) for it in items)) + + if MAJORITY_FINAL and NUM_MAJORITY > 1: + # aggregate the results for each group of NUM_MAJORITY items using majority_vote_select + aggregated_items = [] + for i in range(0, len(items), NUM_MAJORITY): + group = items[i:i+NUM_MAJORITY] + # aggregate base on the objectives + group_objectives = [ + it["metrics"][-1]["objective"] + for it in group if ("metrics" in it and len(it["metrics"]) > 0 and it["metrics"][-1]["have_objective"]) + ] + if group_objectives: + chosen_j, majority_objective, majority_count = majority_vote_select(group_objectives, rtol=args.majority_rtol, atol=args.majority_atol) + chosen = group[chosen_j] if majority_objective is not None else group[0] + else: + majority_objective = None + chosen = group[0] + + # build the aggregated item + aggregated_item = { + "iid": chosen["iid"] // NUM_MAJORITY, + "ground_truth": chosen["ground_truth"], + "ground_truth_code": chosen.get("ground_truth_code", None), + "history": chosen["history"], + "metrics": chosen["metrics"], + "done": chosen["done"], + } + aggregated_items.append(aggregated_item) + # print("Aggregated item from iids:", [it["iid"] for it in group], "to iid:", aggregated_item["iid"], "with objective:", majority_objective) + + items = aggregated_items + + # final compute score + pool = ThreadPoolExecutor(max_workers=THREADS_POOL) + loop = asyncio.get_running_loop() + recs = await final_compute_score(items, pool, loop) + pool.shutdown(wait=True) + + return recs + + +# ────────────────────── DRIVER ─────────────────────────────────────── +async def main(): + ext = os.path.splitext(DATA_PATH)[1].lower() + if ext == ".csv": + df = pd.read_csv(DATA_PATH) + elif ext in (".jsonl", ".json"): + # jsonl: one JSON object per line + with open(DATA_PATH, "r", encoding="utf-8") as f: + recs = [json.loads(line) for line in f if line.strip()] + df = pd.DataFrame(recs) + else: + # fallback keeps previous behavior for .parquet + df = pd.read_parquet(DATA_PATH) + + # drop rows where question is nan or answer is nan or question not in df or answer not in df + df = df.dropna(subset=["question", "answer"]) + df = df.reset_index(drop=True) + # only load first 10 examples for debugging | remember to set back! + print(f"Loaded {len(df)} valid records from {DATA_PATH}") + + if os.path.exists(OUT_PATH): + os.remove(OUT_PATH) + + all_recs = [] + for i in range(0, len(df), BATCH_SIZE_SGLANG): + recs = await solve_batch_no_tools(df.iloc[i:i+BATCH_SIZE_SGLANG].itertuples(index=False)) + # import pdb; pdb.set_trace() + with open(OUT_PATH, "a") as fh: + for idx, rec in enumerate(recs): + rec["id"] = idx + i # fix the index + fh.write(json.dumps(rec, indent=4) + "\n") + all_recs.extend(recs) + # import pdb; pdb.set_trace() + + # also write to .json + with open(OUT_PATH.replace(".jsonl", ".json"), "w") as f: + json.dump(all_recs, f, indent=4) + + print("Results write to:", OUT_PATH, "and ", OUT_PATH.replace(".jsonl", ".json")) + + # quick stats + scores = [rec["score"] for rec in all_recs] + no_execution_errors = [rec["no_execution_error"] for rec in all_recs] + have_objectives = [rec["have_objective"] for rec in all_recs] + plens = [rec["prompt_len"] for rec in all_recs] + slens = [rec["solution_len"] for rec in all_recs] + print(f"Avg prompt: {np.mean(plens):.1f} tok | Avg response: {np.mean(slens):.1f} tok") + print(f"Avg accuracy: {100*np.mean(scores):.2f}% (n={len(scores)})") + print(f"Avg no execution error: {100*np.mean(no_execution_errors):.2f}% (n={len(no_execution_errors)})") + print(f"Avg have objective: {100*np.mean(have_objectives):.2f}% (n={len(have_objectives)})") + for t in range(MAX_TURNS): + ts = [rec["accuracy_per_turn"][f"accuracy_{t}"] for rec in all_recs] + print(f"Turn {t}: {100*np.mean(ts):.2f}%") + + # save the above stats in stats.json + stats = { + "num_items": len(all_recs), + "avg_prompt_len": np.mean(plens), + "avg_solution_len": np.mean(slens), + "avg_accuracy": np.mean(scores), + "avg_no_execution_error": np.mean(no_execution_errors), + "avg_have_objective": np.mean(have_objectives), + "accuracy_per_turn": {f"turn_{t}": np.mean([rec["accuracy_per_turn"][f"accuracy_{t}"] for rec in all_recs]) for t in range(MAX_TURNS)}, + "no_execution_error_per_turn": {f"turn_{t}": np.mean([rec["no_execution_error_per_turn"][f"no_execution_error_{t}"] for rec in all_recs]) for t in range(MAX_TURNS)}, + "have_objective_per_turn": {f"turn_{t}": np.mean([rec["have_objective_per_turn"][f"have_objective_{t}"] for rec in all_recs]) for t in range(MAX_TURNS)}, + } + + with open(os.path.join(OUT_DIR, "stats.json"), "w") as f: + json.dump(stats, f, indent=4) + + with open(os.path.join(OUT_DIR, "backend_schedule.json"), "w") as f: + json.dump(BACKEND_SCHEDULE, f) + + +if __name__ == "__main__": + # if error analysis assert error_analysis_file is given + if args.apply_error_analysis and not args.error_analysis_file: + raise ValueError("error_analysis_file must be provided if apply_error_analysis is set.") + + if not args.apply_error_analysis and not args.apply_hint_instructions: + assert args.tool_prompt_type in ("default", "add_question"), "If not applying hint instructions or error analysis, tool_prompt_type must be 'default' or 'add_question'." + + + SYSTEM_PROMPT_TYPE_STR = "" if args.system_prompt_type == "default" else f"_{args.system_prompt_type}" + USER_PROMPT_TYPE_STR = "" if args.user_prompt_type == "default" else f"_{args.user_prompt_type}" + TOOL_PROMPT_TYPE_STR = "" if args.tool_prompt_type == "default" else f"_{args.tool_prompt_type}" + + MAX_TURNS_STR = "" if args.max_turns == 5 else f"-turns-{args.max_turns}" + NUM_MAJORITY_STR = "" if args.num_majority == 1 else f"-majority-{args.num_majority}" + if MAJORITY_FINAL and NUM_MAJORITY > 1: + NUM_MAJORITY_STR += "-final" + + DEFAULT_TEMP, DEFAULT_TOP_P = 0.6, 0.95 + TEMP_STR = "" if TEMP == DEFAULT_TEMP else f"-temp-{TEMP}" + TOP_P_STR = "" if TOP_P == DEFAULT_TOP_P else f"-topp-{TOP_P}" + HINT_STR = "" if args.apply_error_analysis else "-no-hint" + if args.apply_hint_instructions: + HINT_STR += "-with-hint-instructions" + + DEBUG_STR = "-debug" if args.debug else "" + EARLY_STOP_STR = "" + if args.max_turns > 1: + EARLY_STOP_STR = "-no-early-stop" if args.no_early_stop else "" + + if not args.error_analysis_file and args.apply_error_analysis: + raise ValueError("Please provide --error_analysis_file to indicate the error analysis file used.") + + ERROR_ANALYSIS_STR = "" if (not args.apply_error_analysis or not args.error_analysis_file or "training_error_analysis_0911" in args.error_analysis_file) else f"-{os.path.basename(args.error_analysis_file).replace('.csv','').replace('.txt','')}" + + if BACKEND_STR == "openai": + model_name_str = f"openai-{OPENAI_MODEL}{MAX_TURNS_STR}{NUM_MAJORITY_STR}{TEMP_STR}{TOP_P_STR}{SYSTEM_PROMPT_TYPE_STR}{USER_PROMPT_TYPE_STR}{TOOL_PROMPT_TYPE_STR}{ERROR_ANALYSIS_STR}{EARLY_STOP_STR}{HINT_STR}{DEBUG_STR}" + else: + GPT_REASONING_STR = f"-reasoning-{args.reasoning_effort}" if args.gpt_oss else "" + model_name_str = f"{MODEL_NAME}{GPT_REASONING_STR}{MAX_TURNS_STR}{NUM_MAJORITY_STR}{TEMP_STR}{TOP_P_STR}{SYSTEM_PROMPT_TYPE_STR}{USER_PROMPT_TYPE_STR}{TOOL_PROMPT_TYPE_STR}{ERROR_ANALYSIS_STR}{EARLY_STOP_STR}{HINT_STR}{DEBUG_STR}" + + backend_str = BACKEND_STR + (f"_{args.suffix}" if args.suffix else f"_{args.seed}" if args.seed else "_" + datetime.now().strftime("%Y%m%d_%H%M%S")) + OUT_DIR = os.path.join(args.output_dir, os.path.basename(DATA_PATH).replace(".parquet", "").replace(".csv", "").replace(".json", "").replace(".jsonl", ""), model_name_str, backend_str) + + print("Output dir", OUT_DIR) + + os.makedirs(OUT_DIR, exist_ok=True) + OUT_PATH = os.path.join(OUT_DIR, "multi_turn_results.jsonl") + + + if not args.force and os.path.exists(OUT_PATH): + print(f"Output file {OUT_PATH} already exists, exit.") + exit(0) + + if args.gpt_oss: + from openai_harmony import ( + HarmonyEncodingName, + load_harmony_encoding, + ) + + encoding = load_harmony_encoding(HarmonyEncodingName.HARMONY_GPT_OSS) + + if "sglang" in BACKEND_SCHEDULE: + engine = sgl.Engine( + model_path=MODEL_LOAD_PATH, + tp_size=TP_SIZE, + dp_size=DP_SIZE, + dtype="bfloat16", + kv_cache_dtype="auto", + random_seed=args.seed, + max_running_requests=64, + ) + if args.gpt_oss: + stop_token_ids = encoding.stop_tokens_for_assistant_actions() + stop_tokens = list(set([tokenizer.decode([tid]) for tid in stop_token_ids] + [tokenizer.pad_token])) + else: + stop_tokens = [tokenizer.eos_token, tokenizer.pad_token] + print(">>> SGLANG stop tokens:", stop_tokens) + sglang_sampling_params = {"temperature": TEMP, "top_p": TOP_P, "max_new_tokens": MAX_NEW_TOKENS, "stop": stop_tokens} + + + print("="*45) + print("BACKEND_SCHEDULE:", " | ".join([f"{b.upper()}" for b in BACKEND_SCHEDULE])) + print("="*45) + + try: + asyncio.run(main()) + finally: + if "sglang" in BACKEND_SCHEDULE: + engine.shutdown() + + print("Done!") diff --git a/optimind/eval_utils/__init__.py b/optimind/eval_utils/__init__.py new file mode 100644 index 0000000..e69de29 diff --git a/optimind/eval_utils/build_prompt.py b/optimind/eval_utils/build_prompt.py new file mode 100644 index 0000000..732dd7c --- /dev/null +++ b/optimind/eval_utils/build_prompt.py @@ -0,0 +1,101 @@ +import random + +def build_prompt_sirl_system(solver="gurobipy"): + return f"""You are a helpful Assistant with expertise in mathmetical modeling and the {solver} solver. When the User provides an OR question, you will analyze it, build a detailed mathematical model, and provide the {solver} code to solve it. + + Your response should follow these steps: + 1. +Carefully analyze the problem to identify decision variables, objective, and constraints. + + 2. Develop a complete mathematical model, explicitly defining: + * Sets + * Parameters + * Decision Variables (and their types) + * Objective Function + * Constraints + 3. Provide the corresponding {solver} Python code to implement the model. + + The output must be in Markdown format, with each step enclosed in the specified tags. + +The output must be in Markdown format, with each step enclosed in the specified tags. +""" + +def build_prompt_sirl(question, solver="gurobipy"): + return f"""Solve the following mathematical modeling problem: + +{question} + +Think step by step.""" + + +instruction_list = [ + "Below is an operations research question. Build a mathematical model and corresponding Python code using `{solver}` to solve it.", + "Analyze the given operations research problem, construct a mathematical model, and write Python code with {solver} to find the optimal solution.", + "Solve this optimization challenge by first writing out the complete mathematical model, then translating it into Python code using the {solver} package. Include comments explaining your implementation.", + "This is an operations research problem. Follow this structured approach: Begin with understanding the problem → Identify the key variables → Analyze the constraints → Develop the mathematical model → Solve it programmatically using {solver} in Python. Provide clear reasoning and explanations at each stage.", + "Build an optimization model for this problem and solve it using {solver}. Your solution should include both the mathematical formulation and working Python code.", + """Take the following steps to solve this optimization problem: +- Start with careful problem analysis +- Determine appropriate variable types +- Develop mathematical model with proper constraints +- Implement solution using {solver} +Provide clear explanations for your modeling decisions.""", + """Let's solve this optimization challenge methodically: +1. First share your initial analysis of the problem +2. Then identify and classify all variables needed +3. Develop the complete mathematical formulation +4. Show how to implement in {solver} +Explain your reasoning throughout the process.""", + "Using {solver}, write Python code to solve the following optimization problem. Start by developing the mathematical model before implementing the solution.", + "We are presented with an operations research problem. Solve it step-by-step: 1) Identify and list the decision variables 2) Clearly define the objective 3) Outline all relevant constraints 4) Formulate the mathematical model 5) Implement and solve it using {solver} in Python, providing explanations at every step.", + "Create a complete solution that includes: 1) Mathematical formulation 2) Python code using {solver} 3) Results interpretation. Ensure all variables and constraints are properly defined.", + "Presented below is an operations research problem. Approach this systematically: What are we trying to optimize? What are the main constraints? After analyzing these elements, construct a mathematical model and solve it using {solver} in Python.", + "Here is an operations research problem to solve. Tackle this problem carefully: Start by determining the goal we want to optimize, then pinpoint the constraints. Build the mathematical model and implement it using {solver} in Python, explaining each decision along the way.", + "Please solve this optimization problem by first creating a mathematical formulation, then implement the solution using {solver} in Python.", + """Let's solve this step-by-step: +1) Analyze the problem requirements carefully +2) Identify decision variables and their domains +3) Write out the complete mathematical model +4) Implement solution in {solver} +Remember to explain your thought process.""", + "Please provide a step-by-step solution that includes the mathematical formulation and corresponding Python implementation using {solver}. Ensure proper documentation of your approach." +] + + +#### different solver import languages +solver_import_dict = { + "gurobipy": "import gurobipy as gp\nfrom gurobipy import Model, GRB, quicksum, LinExpr, QuadExpr, Var, Constr", +} + +solver_extra_info_dict = {} + + +input_format = """# Question: {question} + +# Note: +- The Code must include:```python +{solver_import} +``` +{solver_extra_info} +- Make sure the model variable is named `model`. +- Avoid using "<" and ">" in {solver} constraints; instead, use "<=" or ">=". +- Carefully determine whether the variable is an integer or continuous. +""" + +def get_solver_import(solver: str) -> str: + solver = "gurobipy" if solver not in solver_import_dict else solver + return solver_import_dict.get(solver, "") + +def get_solver_extra_info(solver: str) -> str: + solver = "gurobipy" if solver not in solver_extra_info_dict else solver + extra_info = solver_extra_info_dict.get(solver, "") + return "- " + extra_info if extra_info != "" else "" + +def build_prompt_optmath(question: str, instruction: str = "", solver="gurobipy") -> str: + if instruction == "": + instruction = random.choice(instruction_list).strip().format(solver=solver) + + solver_import = get_solver_import(solver) + solver_extra_info = get_solver_extra_info(solver) + input_format_str = input_format.format(question=question.strip(),solver_import=solver_import, solver=solver, solver_extra_info=solver_extra_info) + return f"{instruction}\n{input_format_str}\n{question.strip()}\nThink step by step." diff --git a/optimind/eval_utils/harmony_utils.py b/optimind/eval_utils/harmony_utils.py new file mode 100644 index 0000000..815e01e --- /dev/null +++ b/optimind/eval_utils/harmony_utils.py @@ -0,0 +1,705 @@ +# A majority of this file is adapted from the vLLM project, which is licensed under the Apache 2.0 license. +# SPDX-License-Identifier: Apache-2.0 +# SPDX-FileCopyrightText: Copyright contributors to the vLLM project +# +import datetime +import re +import os +import json +from collections.abc import Iterable, Sequence +from typing import Literal, Optional, Union + +from openai.types.responses import (ResponseFunctionToolCall, + ResponseOutputItem, ResponseOutputMessage, + ResponseOutputText, ResponseReasoningItem) +from openai.types.responses.response_function_web_search import ( + ActionFind, ActionOpenPage, ActionSearch, ResponseFunctionWebSearch) +from openai.types.responses.response_reasoning_item import ( + Content as ResponseReasoningTextContent) +from openai.types.responses.tool import Tool +from openai_harmony import (Author, Conversation, DeveloperContent, + HarmonyEncodingName, Message, ReasoningEffort, + Role, StreamableParser, SystemContent, TextContent, + ToolDescription, load_harmony_encoding, + RenderConversationConfig) +from json_repair import repair_json +from uuid import uuid4 as random_uuid + +REASONING_EFFORT = { + "high": ReasoningEffort.HIGH, + "medium": ReasoningEffort.MEDIUM, + "low": ReasoningEffort.LOW, +} + +_harmony_encoding = None + + +def get_encoding(): + global _harmony_encoding + if _harmony_encoding is None: + _harmony_encoding = load_harmony_encoding( + HarmonyEncodingName.HARMONY_GPT_OSS) + return _harmony_encoding + + +def get_system_message( + model_identity: Optional[str] = None, + reasoning_effort: Optional[Literal["high", "medium", "low"]] = None, + start_date: Optional[str] = None, + browser_description: Optional[str] = None, + python_description: Optional[str] = None, +) -> Message: + sys_msg_content = SystemContent.new() + if model_identity is not None: + sys_msg_content = sys_msg_content.with_model_identity(model_identity) + if reasoning_effort is not None: + sys_msg_content = sys_msg_content.with_reasoning_effort( + REASONING_EFFORT[reasoning_effort]) + if start_date is None: + # NOTE(woosuk): This brings non-determinism in vLLM. Be careful. + start_date = datetime.datetime.now().strftime("%Y-%m-%d") + sys_msg_content = sys_msg_content.with_conversation_start_date(start_date) + if browser_description is not None: + sys_msg_content = sys_msg_content.with_tools(browser_description) + if python_description is not None: + sys_msg_content = sys_msg_content.with_tools(python_description) + sys_msg = Message.from_role_and_content(Role.SYSTEM, sys_msg_content) + return sys_msg + + +def get_developer_message(instructions: Optional[str] = None, + tools: Optional[list[Tool]] = None) -> Message: + dev_msg_content = DeveloperContent.new() + if instructions is not None: + dev_msg_content = dev_msg_content.with_instructions(instructions) + if tools is not None: + function_tools = [] + for tool in tools: + if tool.type in ("web_search_preview", "code_interpreter"): + # These are built-in tools that are added to the system message. + pass + elif tool.type == "function": + function_tools.append(tool) + else: + raise ValueError(f"tool type {tool.type} not supported") + if function_tools: + function_tool_descriptions = [ + ToolDescription.new( + name=tool.name, + description=tool.description, + parameters=tool.parameters, + ) for tool in function_tools + ] + dev_msg_content = dev_msg_content.with_function_tools( + function_tool_descriptions) + dev_msg = Message.from_role_and_content(Role.DEVELOPER, dev_msg_content) + return dev_msg + + +def get_user_message(content: str) -> Message: + return Message.from_role_and_content(Role.USER, content) + + +def parse_chat_input(chat_msg) -> Message: + role = chat_msg["role"] + content = chat_msg["content"] + if isinstance(content, str): + contents = [TextContent(text=content)] + else: + # TODO: Support refusal. + contents = [TextContent(text=c["text"]) for c in content] + msg = Message.from_role_and_contents(role, contents) + return msg + + +def render_for_completion(messages: list[Message]) -> list[int]: + conversation = Conversation.from_messages(messages) + token_ids = get_encoding().render_conversation_for_completion( + conversation, Role.ASSISTANT) + return token_ids + + +def parse_output_message(message: Message) -> list[ResponseOutputItem]: + """ + Parse a Harmony message into a list of output response items. + """ + if message.author.role != "assistant": + # This is a message from a tool to the assistant (e.g., search result). + # Don't include it in the final output for now. This aligns with + # OpenAI's behavior on models like o4-mini. + return [] + + output_items: list[ResponseOutputItem] = [] + recipient = message.recipient + content_type = message.content_type + + if recipient is not None and recipient.startswith("browser."): + if len(message.content) != 1: + raise ValueError("Invalid number of contents in browser message") + content = message.content[0] + try: + browser_call = json.loads(content.text) + except: + for content in message.content: + reasoning_item = ResponseReasoningItem( + id=f"rs_{random_uuid()}", + summary=[], + type="reasoning", + content=[ + ResponseReasoningTextContent( + text=content.text, type="reasoning_text" + ) + ], + status=None, + ) + output_items.append(reasoning_item) + return output_items + + if recipient == "browser.search": + action = ActionSearch( + query=f"cursor:{browser_call.get('query', '')}", type="search") + elif recipient == "browser.open": + action = ActionOpenPage( + url=f"cursor:{browser_call.get('url', '')}", type="open_page") + elif recipient == "browser.find": + action = ActionFind(pattern=browser_call["pattern"], + url=f"cursor:{browser_call.get('url', '')}", + type="find") + else: + raise ValueError(f"Unknown browser action: {recipient}") + web_search_item = ResponseFunctionWebSearch( + id=f"ws_{random_uuid()}", + action=action, + status="completed", + type="web_search_call", + ) + output_items.append(web_search_item) + elif message.channel == "analysis": + for content in message.content: + reasoning_item = ResponseReasoningItem( + id=f"rs_{random_uuid()}", + summary=[], + type="reasoning", + content=[ + ResponseReasoningTextContent(text=content.text, + type="reasoning_text") + ], + status=None, + ) + output_items.append(reasoning_item) + elif message.channel == "commentary": + if recipient is None: + # add to reasoning + for content in message.content: + reasoning_item = ResponseReasoningItem( + id=f"rs_{random_uuid()}", + summary=[], + type="reasoning", + content=[ + ResponseReasoningTextContent(text=content.text, + type="reasoning_text") + ], + status=None, + ) + output_items.append(reasoning_item) + elif recipient.startswith("functions"): # functions. + function_name = recipient.split(".")[-1] if "." in recipient else recipient.replace("functions", "").strip() + if function_name == "" and ("code" in content_type or "python" in content_type): + function_name = "code" + + for content in message.content: + arguments = content.text + local_function_name = function_name + if not function_name: + try: + parsed_args = json.loads(arguments) + except: + try: + parsed_args = json.loads(repair_json(arguments)) + except: + parsed_args = {} + + if isinstance(parsed_args, dict) and "name" in parsed_args and "arguments" in parsed_args: + local_function_name = parsed_args["name"] + arguments = parsed_args["arguments"] + if not isinstance(arguments, str): + arguments = json.dumps(arguments) + + random_id = random_uuid() + response_item = ResponseFunctionToolCall( + arguments=arguments, + call_id=f"call_{random_id}", + type="function_call", + name=local_function_name, + id=f"ft_{random_id}", + ) + output_items.append(response_item) + elif recipient.startswith("python") or message.recipient.startswith("browser") \ + or recipient.startswith("container"): + print("Recipient is", recipient, "!!!") + for content in message.content: + reasoning_item = ResponseReasoningItem( + id=f"rs_{random_uuid()}", + summary=[], + type="reasoning", + content=[ + ResponseReasoningTextContent(text=content.text, + type="reasoning_text") + ], + status=None, + ) + output_items.append(reasoning_item) + else: + # Unknown / missing recipient: don't crash, just treat as reasoning + for content in message.content: + reasoning_item = ResponseReasoningItem( + id=f"rs_{random_uuid()}", + summary=[], + type="reasoning", + content=[ + ResponseReasoningTextContent( + text=content.text, type="reasoning_text" + ) + ], + status=None, + ) + output_items.append(reasoning_item) + # raise ValueError(f"Unknown recipient: {recipient}") + elif message.channel == "final": + contents = [] + for content in message.content: + output_text = ResponseOutputText( + text=content.text, + annotations=[], # TODO + type="output_text", + logprobs=None, # TODO + ) + contents.append(output_text) + text_item = ResponseOutputMessage( + id=f"msg_{random_uuid()}", + content=contents, + role=message.author.role, + # if the parser still has messages (ie if the generator got cut + # abruptly), this should be incomplete + status="incomplete", + type="message", + ) + output_items.append(text_item) + else: + raise ValueError(f"Unknown channel: {message.channel}") + return output_items + + +def parse_remaining_state( + parser: StreamableParser) -> list[ResponseOutputItem]: + if not parser.current_content: + return [] + if parser.current_role != Role.ASSISTANT: + return [] + current_recipient = parser.current_recipient + if (current_recipient is not None + and current_recipient.startswith("browser.")): + return [] + + if parser.current_channel == "analysis": + reasoning_item = ResponseReasoningItem( + id=f"rs_{random_uuid()}", + summary=[], + type="reasoning", + content=[ + ResponseReasoningTextContent(text=parser.current_content, + type="reasoning_text") + ], + status=None, + ) + return [reasoning_item] + elif parser.current_channel == "final": + output_text = ResponseOutputText( + text=parser.current_content, + annotations=[], # TODO + type="output_text", + logprobs=None, # TODO + ) + text_item = ResponseOutputMessage( + id=f"msg_{random_uuid()}", + content=[output_text], + role="assistant", + status="completed", + type="message", + ) + return [text_item] + return [] + + +def get_stop_tokens_for_assistant_actions() -> list[int]: + return get_encoding().stop_tokens_for_assistant_actions() + + +def get_streamable_parser_for_assistant() -> StreamableParser: + return StreamableParser(get_encoding(), role=Role.ASSISTANT) + + +def parse_output_into_messages(token_ids: Iterable[int]) -> StreamableParser: + parser = get_streamable_parser_for_assistant() + for token_id in token_ids: + parser.process(token_id) + return parser + + +def parse_chat_output(token_ids: Sequence[int]) -> tuple[str | None, str | None, bool]: + parser = parse_output_into_messages(token_ids) + output_msgs = parser.messages + is_tool_call = False # TODO: update this when tool call is supported + + if len(output_msgs) == 0: + # The generation has stopped during reasoning. + reasoning = parser.current_content + final_content = None + elif len(output_msgs) == 1: + # The generation has stopped during final message. + reasoning = output_msgs[0].content[0].text + final_content = parser.current_content + else: + reasoning_msg = output_msgs[:-1] + final_msg = output_msgs[-1] + reasoning = "\n".join([msg.content[0].text for msg in reasoning_msg]) + final_content = final_msg.content[0].text + return reasoning, final_content, is_tool_call + + + +def sanitize_harmony_headers_keep_tools(token_txt: str) -> str: + """ + Normalize legacy / non-Harmony headers into something the Harmony parser + understands, while PRESERVING tool calls. + """ + # 0) Normalize all <|return|> (end of inference) to <|end|> (end of assistant turn) + token_txt = token_txt.replace("<|return|>", "<|end|>") + # Strip legacy / unsupported control tokens <|call|><|end|> -> <|end|> (strip empty calls) + token_txt = re.sub(r"<\|call\|>\s*<\|end\|>", "<|end|>", token_txt) + + # 1) Legacy "final" header -> assistant final + token_txt = token_txt.replace( + "<|start|>final<|message|>", + "<|start|>assistant<|channel|>final<|message|>", + ) + + # <|start|>assistant<|channel|>analysis to=functions. -> <|start|>assistant<|channel|>commentary to=functions. + token_txt = re.sub( + r"<\|start\|>assistant<\|channel\|>analysis\s+to=(?Pfunctions[^\s<]+)", + r"<|start|>assistant<|channel|>commentary to=\g", + token_txt, + ) + + # <|start|>assistant<|channel|>commentary to=functions. -> <|start|>assistant<|channel|>commentary to=functions. + token_txt = re.sub( + r"<\|start\|>assistant\s*<\|channel\|>commentary\s*to=functions\s+(\w+)(\s*<\|message\|>)", + r"<|start|>assistant<|channel|>commentary to=functions.\1\2", + token_txt, + ) + + # 3) Assistant headers that completely lack a channel: + token_txt = re.sub( + r"<\|start\|>assistant\s*<\|message\|>", + "<|start|>assistant<|channel|>analysis<|message|>", + token_txt, + ) + + # 3b) Ensure <|channel|>analysis / final are followed by <|message|> + token_txt = re.sub( + r"(<\|channel\|>\s*(analysis|final))(?!\s*<\|message\|>)", + r"\1<|message|>", + token_txt, + ) + + # 3c) Commentary: insert <|message|> only if NOT followed by <|message|>, 'to=' or <|constrain|> + # This avoids breaking legacy "commentary to=..." or "commentary <|constrain|>..." + token_txt = re.sub( + r"(<\|channel\|>\s*commentary)(?!\s*(<\|message\||to=|<\|constrain\|>))", r"\1<|message|>", + token_txt, + ) + + # 3d) Commentary: allow optional "to=..." and "<|constrain|>...", then ensure there's a <|message|> + def _insert_message_if_missing(m: re.Match) -> str: + header = m.group(1) + end = m.end(1) + # Look at what actually follows in the original string + if token_txt[end:].lstrip().startswith("<|message|>"): + return header # already correct + return header + "<|message|>" + + token_txt = re.sub( + r'(<\|channel\|>\s*commentary(?:\s+to=[^\s<]+)?(?:\s*<\|constrain\|>[^\s<]+)?)', + _insert_message_if_missing, + token_txt, + flags=re.DOTALL, + ) + + # 4) normalize all <|return|> to <|end|>, and remove everything after the last <|end|> + last_end_idx = token_txt.rfind("<|end|>") + if last_end_idx != -1: + token_txt = token_txt[: last_end_idx + len("<|end|>")] + else: + # If there's no <|end|>, ensure each assistant message is properly closed + token_txt = token_txt + "<|end|>" + + return token_txt + + +def sanitize_start_end_message(prefix_txt: str) -> str: + msg = prefix_txt.lstrip() + msg = msg.replace("<|return|>", "<|end|>") + if not msg.endswith("<|end|>"): + msg += "<|end|>" + + # Heuristic: looks like a message header fragment + if msg.startswith("<|channel|>"): + msg = "<|start|>assistant" + msg + return msg + # Optionally: support other incomplete headers here + return msg + + +def extract_tool_args(raw: str) -> str: + if raw is None: + return "" + + s = raw.strip() + + # 1) Split after first <|message|> if it exists + marker = "<|message|>" + idx = s.find(marker) + if idx != -1: + end_idx = idx + len(marker) + s = s[end_idx:].strip() + else: + # 2) Otherwise, start from first '{' if present + brace_idx = s.find("{") + if brace_idx != -1: + s = s[brace_idx:].strip() + + # 3) Trim after the last closing brace if there is one + last_brace_idx = s.rfind("}") + if last_brace_idx != -1: + s = s[: last_brace_idx + 1].strip() + + # 4) Balance braces: if we have more '{' than '}', append missing '}' + open_braces = s.count("{") + close_braces = s.count("}") + if open_braces > close_braces: + s = s + ("}" * (open_braces - close_braces)) + + return s, open_braces + +def save_arguments(arguments, prefix="tmp_args"): + subdir = os.path.join("tmp_harmony", prefix) + os.makedirs(subdir, exist_ok=True) + save_idx = 0 + while os.path.exists(os.path.join(subdir, f"{save_idx}.json")): + save_idx += 1 + with open(os.path.join(subdir, f"{save_idx}.json"), "w") as f: + json.dump(arguments, f, indent=2) + print(f"Saved arguments to {subdir}/{save_idx}.json for debugging.") + + +def save_txt_examples(save_txt_examples, reasoning_content=None, final_content=None, tool_calls=None, prefix="tmp"): + save_idx = 0 + subdir = os.path.join("tmp_harmony", prefix) + os.makedirs(subdir, exist_ok=True) + while os.path.exists(os.path.join(subdir, f"{save_idx}.txt")): + save_idx += 1 + with open(os.path.join(subdir, f"{save_idx}.txt"), "w") as f: + f.write(save_txt_examples) + if reasoning_content: + with open(os.path.join(subdir, f"{save_idx}_reasoning.txt"), "w") as f: + f.write(reasoning_content) + if final_content: + with open(os.path.join(subdir, f"{save_idx}_final.txt"), "w") as f: + f.write(final_content) + if tool_calls: + with open(os.path.join(subdir, f"{save_idx}_tools.json"), "w") as f: + json.dump(tool_calls, f, indent=2) + print(f"Saved to {subdir}/{save_idx}.txt for debugging.") + + + +def harmony_parse_calls(tokenizer, token_ids, txt, keep_tools: bool = True): + # 1. Keep track of everything, but also split on first <|start|> + token_ids_input = token_ids + token_txt_input = tokenizer.decode(token_ids_input) + txt_input = txt + + start_id = tokenizer.encode("<|start|>")[0] + + prefix_txt = "" + if start_id in token_ids_input: + start_pos = token_ids_input.index(start_id) + prefix_ids = token_ids_input[:start_pos] # what you used to throw away + token_ids = token_ids_input[start_pos:] # the structured part + if prefix_ids: + prefix_txt = tokenizer.decode(prefix_ids) + else: + # No structured header at all, treat everything as prefix + prefix_txt = tokenizer.decode(token_ids_input) + token_ids = [] # nothing left for Harmony parser + + # 2. Decode once for the structured part + token_txt = tokenizer.decode(token_ids) if token_ids else "" + + # --- try to treat prefix as a Harmony message --- + synthetic_prefix = sanitize_start_end_message(prefix_txt) if prefix_txt else "" + token_txt = sanitize_start_end_message(token_txt) if token_txt else "" + + if synthetic_prefix: + # Prepend synthetic message to the structured text + # If wrap prefix failed: we do not trust the prefix and hence we would throw it away (instead of prepending it to reasoning_content) + token_txt = synthetic_prefix + token_txt + # prefix_txt = "" + + # 3. Patch corner cases + token_txt = sanitize_harmony_headers_keep_tools(token_txt) + + # 4. Re-encode after sanitization (if there is anything left) + token_ids = tokenizer.encode(token_txt) if token_txt else [] + + try: + reasoning_content = "" + final_content = "" + tool_calls = [] + + if token_ids: + # Parse into Harmony messages + parser = parse_output_into_messages(token_ids) + messages = parser.messages + + # Use the new helpers to turn messages + remaining state into ResponseOutputItems + output_items: list[ResponseOutputItem] = [] + for msg in messages: + output_items.extend(parse_output_message(msg)) + + # Capture any partially streamed content + output_items.extend(parse_remaining_state(parser)) + + # Accumulate: + # - raw reasoning chunks (reasoning items + non-final messages + fn-call args if keep_tools=False) + # - assistant message texts (for picking the last as final) + # - tool_calls (if keep_tools=True) + reasoning_chunks: list[str] = [] + assistant_msg_texts: list[str] = [] + tool_calls: list[dict] = [] + + for item in output_items: + item_type = getattr(item, "type", None) + + # --- Reasoning items --- + if item_type == "reasoning": + for c in getattr(item, "content", []): + if getattr(c, "type", None) == "reasoning_text": + text = getattr(c, "text", "") or "" + if text: + reasoning_chunks.append(text) + + # --- Assistant messages (collect all texts first) --- + elif item_type == "message": + role = getattr(item, "role", None) + if role == "assistant": + texts: list[str] = [] + for c in getattr(item, "content", []): + if getattr(c, "type", None) == "output_text": + t = getattr(c, "text", "") or "" + if t: + texts.append(t) + if texts: + assistant_msg_texts.append("\n".join(texts)) + + # --- Tool calls --- + elif item_type == "function_call": + fn_name = getattr(item, "name", None) + arguments = getattr(item, "arguments", "") or "" + + is_code = fn_name in ["code", "python", "code_execution"] or "code" in fn_name.lower() or "python" in fn_name.lower() + + # try strip arguments, some times to=functions code<|message|> xxx mess up parsing, parsed to code<|message|> xxx as the content + arguments, open_braces = extract_tool_args(arguments) + + parsed_args = {} + if open_braces == 0: + if is_code: + parsed_args = {"code": arguments} + else: + print("Empty function name!!!") + save_txt_examples(token_txt_input, prefix="tmp_args_no_code") + parsed_args = {"text": arguments} + else: + try: + parsed_args = json.loads(arguments) + except: + try: + parsed_args = json.loads(repair_json(arguments)) + print("argument error (basic) but ok with json reapir!") + save_txt_examples(token_txt_input, prefix="tmp_args_json_error_ok_repair") + save_arguments(arguments, prefix="tmp_args_json_error_ok_repair") + except: + print("argument error (json repair)!!!") + save_txt_examples(token_txt_input, prefix="tmp_args_json_error_after_repair") + save_arguments(arguments, prefix="tmp_args_json_error_after_repair") + if keep_tools: + tool_calls.append({"name": fn_name, "arguments": arguments, + "role": "assistant", "channel": "commentary"}) + else: + # Sometimes even if we tell it to output ```python ... ```, it still tries to do a python tool call. + # Let's handle that case for keep_tools = False by mapping it back to ```python ... ``` syntax. + # Otherwise, fold tool-call args into reasoning if we're not keeping tools + if parsed_args: + found_code = False + # edge case: no name + if fn_name in ["code", "python", "code_execution", ""] or "code" in fn_name.lower() or "python" in fn_name.lower(): + for key in ["code", "python", "python_code"]: + if isinstance(parsed_args, dict) and key in parsed_args: + code_snippet = parsed_args[key] + assistant_msg_texts.append(f"```python\n{code_snippet}\n```") + found_code = True + break + + if not found_code: + if isinstance(parsed_args, dict): + for key in parsed_args: + if key == "text": + assistant_msg_texts.append(parsed_args[key]) + else: + assistant_msg_texts.append(f"{key}: {parsed_args[key]}") + else: + assistant_msg_texts.append(f"{parsed_args}") + else: + raise ValueError(f"Unknown output item type: {item_type}") + + # Now split assistant messages into reasoning vs final + last_final_text: str | None = None + if assistant_msg_texts: + # All but the last go into reasoning + last_final_text = assistant_msg_texts[-1] + reasoning_chunks.extend(assistant_msg_texts[:-1]) + + reasoning_content = "\n".join(reasoning_chunks) + final_content = last_final_text or "" + + # Debug: check if we lost a lot of content + if len(reasoning_content) + len(final_content) + len(str(tool_calls)) + 200 < len(token_txt_input): + diff_token = len(token_txt_input) - (len(reasoning_content) + len(final_content) - len(str(tool_calls))) + print(f"Warning: parsed content is much ({diff_token} tokens) shorter than original token_txt_input.") + save_txt_examples(token_txt_input, reasoning_content, final_content, tool_calls, prefix="tmp") + + if keep_tools: + return reasoning_content, final_content, tool_calls + else: + return reasoning_content, final_content + + except Exception as e: + print(f"Error **parse_output_into_messages**: {str(e)[:200]}") + save_txt_examples(token_txt_input, prefix="error") + # On error, at least return the original txt as "final" + prefix as reasoning + if keep_tools: + return "", txt, [] + else: + return "", txt +############################################################# diff --git a/optimind/eval_utils/utils_general.py b/optimind/eval_utils/utils_general.py new file mode 100644 index 0000000..a9e06ba --- /dev/null +++ b/optimind/eval_utils/utils_general.py @@ -0,0 +1,325 @@ +import os +import json +from pathlib import Path +from typing import List, Dict, Any, Optional, Tuple +import math +import ast +import pandas as pd + +import sys + +# file's parent's parent directory +path = os.path.dirname(os.path.dirname(os.path.dirname(os.path.abspath(__file__)))) +sys.path.append(path) +from rewards.utils import extract_python_code_block as extract_python_code_block_helper + + +# ─────────────────── Helper: load indented JSON objects ─────────────────────────────── +def load_indented_json_objects(file_path: str) -> List[Dict[str, Any]]: + objects: List[Dict[str, Any]] = [] + buffer: List[str] = [] + with open(file_path, "r") as f: + for line in f: + if not line.strip(): + continue + buffer.append(line) + try: + obj = json.loads("".join(buffer)) + objects.append(obj) + buffer = [] # reset for next object + except json.JSONDecodeError: + # incomplete object, keep accumulating + continue + return objects + + +# ─────────────── Helper: resolve local snapshot from HF cache ─────────────── +def resolve_model_path( + model_path: str, + cache_dir: Optional[str] = None, + revision: Optional[str] = None, + local_files_only: bool = False, +) -> str: + """ + If model_path is a local dir, return it. + Otherwise, return a *local snapshot path* for our private repos + (using snapshot_download). For public HF models, return the + original model_path so that transformers/vLLM can handle it. + """ + p = Path(model_path).expanduser() + if p.exists(): + return str(p) + + + # If we are offline and don't have a specific revision, try to find the + # most recent snapshot in the Hugging Face cache structure. + if local_files_only and not revision: + try: + if cache_dir: + base_cache = Path(cache_dir) + else: + base_cache = Path(os.getenv("HF_HOME", "~/.cache/huggingface/hub")).expanduser() + + repo_dir_name = f"models--{model_path.replace('/', '--')}" + snapshot_parent = base_cache / repo_dir_name / "snapshots" + + if snapshot_parent.exists(): + snapshots = [x for x in snapshot_parent.iterdir() if x.is_dir()] + + if snapshots: + # Sort by modification time, newest first + snapshots.sort(key=lambda x: x.stat().st_mtime, reverse=True) + # Set revision to the most recent folder name (the hash) + revision = snapshots[0].name + print(f"[_resolve_model_path] Offline mode: Auto-detected latest snapshot revision: {revision}") + except Exception as e: + # We do not want this logic to crash the script; strictly fallback to standard behavior + print(f"[_resolve_model_path] Warning: Attempted to find cached snapshot but failed: {e}") + + from huggingface_hub import snapshot_download + + try: + return snapshot_download( + repo_id=model_path, + cache_dir=cache_dir, + revision=revision, + local_files_only=local_files_only, + ) + except Exception as e: + if local_files_only: + raise FileNotFoundError( + f"Offline mode: model '{model_path}' not found in local cache. " + f"Run once without --offline_only to download. Original error:\n{e}" + ) + raise + + return model_path + + +def setup_hf_env(args) -> None: + """Configure HF cache / offline flags from CLI args.""" + if args.hf_cache_dir: + os.environ["HF_HOME"] = os.path.expanduser(args.hf_cache_dir) + + if args.offline_only: + os.environ["HF_HUB_OFFLINE"] = "1" + os.environ["TRANSFORMERS_OFFLINE"] = "1" + else: + os.environ.pop("HF_HUB_OFFLINE", None) + os.environ.pop("TRANSFORMERS_OFFLINE", None) + + +def setup_model_and_env(args): + """ + Compute derived global config that depends on args and environment. + Returns a dict with MODEL_PATH, MODEL_NAME, MODEL_LOAD_PATH, etc. + """ + setup_hf_env(args) + + model_path = args.model_path.rstrip("/") + + if model_path.startswith("/mnt/ddn"): + pieces = model_path.split("/") + if "actor_hf" in model_path: + name_parts = pieces[-3:-1] + else: + name_parts = pieces[-2:] + model_name = "-".join([p.lower() for p in name_parts]) + else: + model_name = os.path.basename(model_path).lower() + + if "lmsys" in model_path: + model_name = "lmsys-" + model_name + + if args.backend_str == "azure": + model_load_path = model_path + else: + model_load_path = resolve_model_path( + model_path, + cache_dir=os.path.expanduser(args.hf_cache_dir) if args.hf_cache_dir else None, + revision=args.hf_revision, + local_files_only=args.offline_only, + ) + print(f"[HF load] Using path: {model_load_path} (offline={args.offline_only})") + + return { + "MODEL_PATH": model_path, + "MODEL_NAME": model_name, + "MODEL_LOAD_PATH": model_load_path, + } + + +################ utils for error analysis ################ + +def as_list(obj) -> List[str]: + """Parse a list-like field that may be a string repr, list, or NaN.""" + if obj is None or (isinstance(obj, float) and math.isnan(obj)): + return [] + if isinstance(obj, list): + return [str(x) for x in obj] + s = str(obj).strip() + if not s: + return [] + try: + val = ast.literal_eval(s) + if isinstance(val, list): + return [str(x) for x in val] + return [s] + except Exception: + if "," in s: + return [t.strip() for t in s.split(",") if t.strip()] + return [s] + +def parse_list_of_pairs(obj: Any) -> List[Tuple[str, str]]: + """ + Expect a list of (mistake, hint) pairs. Robust to strings via literal_eval. + """ + if obj is None or (isinstance(obj, float) and math.isnan(obj)): + return [] + if isinstance(obj, list): + out: List[Tuple[str, str]] = [] + for x in obj: + if isinstance(x, (list, tuple)) and len(x) >= 2: + out.append((str(x[0]), str(x[1]))) + else: + out.append((str(x), "")) + return out + s = str(obj).strip() + if not s: + return [] + try: + val = ast.literal_eval(s) + return parse_list_of_pairs(val) + except Exception: + pairs = manual_parse_list_of_pairs(s) + return pairs + +def manual_parse_list_of_pairs(s): + pairs: List[Tuple[str, str]] = [] + current_pair = [] + for line in s.splitlines(): + line = line.strip().rstrip(",") + if line: + if line.startswith("(") and line.endswith(")"): + # eval this line + try: + tup = ast.literal_eval(line) + if isinstance(tup, (list, tuple)) and len(tup) >= 2: + pairs.append((str(tup[0]), str(tup[1]))) + except: + pairs.append((line, "")) + elif line.startswith("("): + # print("Found start") + current_pair = [line.lstrip("(")] + elif line.endswith(")") and current_pair: + # print("Found end", line) + current_pair.append(line.rstrip(")")) + if len(current_pair) >= 2: + pairs.append((current_pair[0], current_pair[1])) + current_pair = [] + pairs = [(a.strip().strip('"').strip("'"), b.strip().strip('"').strip("'")) for a, b in pairs] + + return pairs + +def load_ea_pairs(ea, args): + # define once (removed duplicate) + class_to_pairs: Dict[str, List[Tuple[str, str]]] = {} + + if "problem_class" not in ea.columns or "error analysis" not in ea.columns: + raise ValueError("error_analysis_file must have columns 'problem_class' and 'error analysis'.") + + for _, row in ea.iterrows(): + classes = as_list(row["problem_class"]) + + # format error analysis in case it's just a single tuple instead of a list + error_analysis_str = row["error analysis"] + if not isinstance(error_analysis_str, str): + # cannot find a valid error analysis for this + continue + + error_analysis_str = error_analysis_str.strip() + if not error_analysis_str.startswith("["): + error_analysis_str = "[" + error_analysis_str + if not error_analysis_str.endswith("]"): + error_analysis_str = error_analysis_str + "]" + + pairs_list = parse_list_of_pairs(error_analysis_str) + if not classes or not pairs_list: + continue + for c in classes: + class_to_pairs.setdefault(c, []).extend(pairs_list) + + # ----- move dedup BEFORE building all_hints_block ----- + for c, pairs in list(class_to_pairs.items()): + seen = set() + dedup: List[Tuple[str, str]] = [] # <— type corrected + for a, b in pairs: + key = (a.strip(), b.strip()) + if key not in seen: + seen.add(key) + dedup.append(key) + class_to_pairs[c] = dedup + return class_to_pairs + +# hint instruction +def get_hint_instructions(): + lines = [] + lines.append( + "Instructions for applying error-analysis hints:\n" + "- Review the provided hints and identify which ones are applicable to this problem.\n" + "- Please apply ALL applicable hints.\n" + "- Before applying any hint or writing constraints, check the sign and direction of every variable and coefficient for consistency (e.g., profit = revenue - cost; capacities as ≤; flow conservation as =).\n" + "\n" + "General modeling checklist (follow rigorously):\n" + "- Units: Use correct units everywhere and ensure the objective’s units match the goal (e.g., dollars for cost/profit, distance for TSP, time for scheduling). Do not mix units (e.g., minutes with hours, dollars with 1000 dollars) without converting.\n" + ) + return "".join(lines) + +def _format_error_pair(error_pair) -> str: + """ + Normalize one error-analysis pair into a single string. + Accepts tuples/lists (mistake, hint) or a bare string. + Tries to recover if one side is empty by literal_eval of the other. + """ + # Case 1: pair-like (tuple/list) + if isinstance(error_pair, (list, tuple)): + # make sure we're working with a tuple, then pad with empty strings + a, b = (tuple(error_pair) + ("", "",))[:2] # <-- FIX: tuple(...) + ("","") + # If one side is blank, try to parse the other as a tuple-like string + if isinstance(a, str) and not a.strip(): + try: + parsed = ast.literal_eval(b) + if isinstance(parsed, (list, tuple)) and len(parsed) >= 2: + a, b = parsed[0], parsed[1] + except Exception: + pass + if isinstance(b, str) and not b.strip(): + try: + parsed = ast.literal_eval(a) + if isinstance(parsed, (list, tuple)) and len(parsed) >= 2: + a, b = parsed[0], parsed[1] + except Exception: + pass + + a = (a if isinstance(a, str) else str(a)).strip() + b = (b if isinstance(b, str) else str(b)).strip() + + if a and b: + return f"{a}, {b}" + return a or b or "" + + # Case 2: bare string (or anything else) + s = str(error_pair).strip() + return s + + +def build_error_analysis_str(error_pairs): + lines = [] + if len(error_pairs) > 0: + lines.append("\nBelow are hints for avoiding common mistakes often seen for this problem type. Avoid them if applicable.\n") + for k, error_pair in enumerate(error_pairs): + body = _format_error_pair(error_pair) + error_analysis_str = f"Error analysis #{k}: {body}" + lines.append(error_analysis_str + "\n") + return "".join(lines) + diff --git a/optimind/recipe/tools/__init__.py b/optimind/recipe/tools/__init__.py new file mode 100644 index 0000000..e69de29 diff --git a/optimind/recipe/tools/base_tool.py b/optimind/recipe/tools/base_tool.py new file mode 100644 index 0000000..890f5d9 --- /dev/null +++ b/optimind/recipe/tools/base_tool.py @@ -0,0 +1,92 @@ +# Copyright 2023-2024 SGLang Team +# Copyright 2025 ModelBest Inc. and/or its affiliates +# +# Licensed under the Apache License, Version 2.0 (the "License"); +# you may not use this file except in compliance with the License. +# You may obtain a copy of the License at +# +# http://www.apache.org/licenses/LICENSE-2.0 +# +# Unless required by applicable law or agreed to in writing, software +# distributed under the License is distributed on an "AS IS" BASIS, +# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. +# See the License for the specific language governing permissions and +# limitations under the License. +import json +from typing import Any, Optional, Tuple +from uuid import uuid4 + +from recipe.tools.rollout_traces import rollout_trace_op + +from recipe.tools.schemas import OpenAIFunctionToolSchema + + +class BaseTool: + """Base class for tools. + + A tool should support the following methods: + + - `to_openai_function_tool_schema`: return the tool schema in OpenAI format. + - `create`: create a tool instance for a trajectory. + - `execute`: execute the tool. + - `calc_reward`: calculate the reward respect to tool state. + - `release`: release the tool instance. + """ + + def __init__(self, config: dict, tool_schema: OpenAIFunctionToolSchema): + self.config = config + self.tool_schema = tool_schema or self.get_openai_tool_schema() + assert self.tool_schema is not None, "Tool schema is not set!" + self.name = self.tool_schema.function.name + # print(json.dumps(self.tool_schema.model_dump(exclude_unset=True, exclude_none=True), indent=2)) + + def get_openai_tool_schema(self) -> OpenAIFunctionToolSchema: + return self.tool_schema + + async def create(self, instance_id: Optional[str] = None, **kwargs) -> str: + """Create a tool instance. + + Args: + instance_id: The instance id of the tool. + + Returns: + The instance id of the tool. + """ + if instance_id is None: + return str(uuid4()) + else: + return instance_id + + @rollout_trace_op + async def execute(self, instance_id: str, parameters: dict[str, Any], **kwargs) -> Tuple[str, float, dict]: + """Execute the tool. + + Args: + instance_id: The instance id of the tool. + parameters: The json string of the parameters of the tool. + + Returns: tool_response, tool_reward_score, tool_metrics + tool_response: The response str of the tool. + tool_reward_score: The step reward score of the tool. + tool_metrics: The metrics of the tool. + """ + return "Updated the tool state.", 0.0, {} + + async def calc_reward(self, instance_id: str, **kwargs) -> float: + """Calculate the reward of the tool. + + Args: + instance_id: The instance id of the tool. + + Returns: + The reward of the tool. + """ + return 0.0 + + async def release(self, instance_id: str, **kwargs) -> None: + """Release the tool instance. + + Args: + instance_id: The instance id of the tool. + """ + pass \ No newline at end of file diff --git a/optimind/recipe/tools/code_execution.py b/optimind/recipe/tools/code_execution.py new file mode 100644 index 0000000..8577bff --- /dev/null +++ b/optimind/recipe/tools/code_execution.py @@ -0,0 +1,165 @@ +import re +import subprocess +import sys +import tempfile +import uuid + +import logging +import os +from typing import Any, Optional, Tuple +from uuid import uuid4 + + +from recipe.tools.base_tool import BaseTool +from recipe.tools.schemas import OpenAIFunctionToolSchema +from recipe.tools.rollout_traces import rollout_trace_op + +# sys import parent's parent directory utils +current_dir = os.path.dirname(os.path.abspath(__file__)) +parent_dir = os.path.dirname(os.path.dirname(current_dir)) +parent_parent_dir = os.path.dirname(parent_dir) +sys.path.append(parent_dir) +from rewards.utils import extract_python_code_block, run_full_code, compute_acc + +logger = logging.getLogger(__name__) +logger.setLevel(os.getenv("VERL_LOGGING_LEVEL", "WARN")) +######################################################################################## + +class CodeExecutionTool(BaseTool): + """A demo tool for executing the code and return the standard output and error + + - `to_openai_function_tool_schema`: return the tool schema in OpenAI format. + - `create`: create a tool instance for a trajectory. + - `execute`: execute the tool. + - `calc_reward`: calculate the reward respect to tool state. + - `release`: release the tool instance. + """ + + def __init__(self, config: dict, tool_schema: OpenAIFunctionToolSchema): + """ + _tool_schema = OpenAIFunctionToolSchema.model_validate({ + "type": "function", + "function": { + "name": "code_execution", + "description": "A tool for executing the {solver} code and return the standard output and error", + "parameters": { + "type": "object", + "properties": { + "code": { + "type": "string", + "description": "The {solver} code to be executed", + }, + }, + "required": ["code"], + }, + } + }) + """ + super().__init__(config, tool_schema) + self.max_assistant_turns = config.get("max_assistant_turns", 3) + self.tool_format = config.get("tool_format", "interaction") + + if self.tool_format == "tool": + self.tool_str = """Based on the output and error message (if any), please think step by step to determine if the code is correct. +Based on your thoughts, please invoke a `code_execute` function call with an (improved) {solver} code if you have identified mistakes in the previous code. +If the previous code is correct, you do not need to invoke the `code_execute` tool again. + +For each function call, Do not directly return the code in the ```python ``` block, as it will not be detected by the system. +Instead, you should return a json object with function name and arguments by making the appropriate tool call. +""" + elif self.tool_format == "minimal": + self.tool_str = "" + else: + self.tool_str = """Based on the output and error message (if any), please think step by step to determine if the code is correct. +Based on your thoughts, please provide an (improved) {solver} code in the ```python\n``` code block if you have identified mistakes in the previous code. +If the previous code is correct, you do not need to provide a new code block again. +Your performance will be evaluated based on the correctness of the final ```python\n``` code block. +""" + + self._instance_dict = {} + + def get_openai_tool_schema(self) -> OpenAIFunctionToolSchema: + return self.tool_schema + + async def create(self, instance_id: Optional[str] = None, ground_truth: Optional[str] = None, **kwargs) -> str: + if instance_id is None: + instance_id = str(uuid4()) + self._instance_dict[instance_id] = { + "response": "", + "ground_truth": ground_truth, + "reward": 0.0, + "round": 0, + } + return instance_id + + @rollout_trace_op + async def execute(self, instance_id: str, parameters: dict[str, Any], **kwargs) -> Tuple[str, float, dict]: + code_response = parameters.get("code", "") + if not isinstance(code_response, str): + code_response = str(code_response) + + self._instance_dict[instance_id]["response"] = code_response + + ## update one step + self._instance_dict[instance_id]["round"] += 1 + cur_round = self._instance_dict[instance_id]["round"] + solver = kwargs.get("solver", "gurobipy") + results = await self.calc_reward(instance_id, precision=kwargs.get("precision", None), + solver=solver) + reward, output, error = results["score"], results["output"], results["traceback_str"] + + # penalty for non improved answer submission + # tool_reward = 0.0 if reward > self._instance_dict[instance_id]["reward"] else -0.05 + tool_reward = self._instance_dict[instance_id]["reward"] - reward + # update the reward + self._instance_dict[instance_id]["reward"] = reward + + metrics = {"score": reward, "no_execution_error": results["no_execution_error"], + "have_objective": results["have_objective"], + "objective": 0.0 if results["objective"] is None else float(results["objective"]), + "accuracy": results["accuracy"] if "accuracy" in results else 0.0} + + return_str = f"""The standard output and error message (if any) for your generated code are as follows: +``` +[STDOUT] {output} +[STDERR] {error} +``` +""" + if cur_round < self.max_assistant_turns: + # add the suffix to ask the model to generate the next code, if not the last round + return_str += self.tool_str.format(solver=solver) + + return return_str, tool_reward, metrics + + async def calc_reward(self, instance_id: str, **kwargs) -> float: + # find the code block in the content + response = self._instance_dict[instance_id]["response"] + ground_truth = self._instance_dict[instance_id]["ground_truth"] + + code_block = extract_python_code_block(response) + solver = kwargs.get("solver", "gurobipy") + solution_dict = run_full_code(code_block, solver=solver) + objective = solution_dict.get("objective_value", None) + + if isinstance(ground_truth, list) or isinstance(ground_truth, tuple): + # if any is correct then correct + score = float(any([compute_acc(objective, gt, precision=kwargs.get("precision", None)) for gt in ground_truth])) + else: + score = float(compute_acc(objective, ground_truth, precision=kwargs.get("precision", None))) + + no_execution_error = float("failed" not in solution_dict["output"].lower() and solution_dict["traceback_str"].strip() == "" ) + have_objective = float(objective is not None) + + return { + "score": score, + "accuracy": score, + "no_execution_error": no_execution_error, + "have_objective": have_objective, + "objective": objective, + "output": solution_dict["output"], + "traceback_str": solution_dict["traceback_str"] + } + + async def release(self, instance_id: str, **kwargs) -> None: + del self._instance_dict[instance_id] + \ No newline at end of file diff --git a/optimind/recipe/tools/rollout_traces.py b/optimind/recipe/tools/rollout_traces.py new file mode 100644 index 0000000..1bf0e6a --- /dev/null +++ b/optimind/recipe/tools/rollout_traces.py @@ -0,0 +1,192 @@ +# Copyright 2025 Bytedance Ltd. and/or its affiliates +# +# Licensed under the Apache License, Version 2.0 (the "License"); +# you may not use this file except in compliance with the License. +# You may obtain a copy of the License at +# +# http://www.apache.org/licenses/LICENSE-2.0 +# +# Unless required by applicable law or agreed to in writing, software +# distributed under the License is distributed on an "AS IS" BASIS, +# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. +# See the License for the specific language governing permissions and +# limitations under the License. + +import asyncio +import contextlib +import functools +import inspect +from typing import Optional + + +class RolloutTraceConfig: + _instance: Optional["RolloutTraceConfig"] = None + backend: Optional[str] = None + client: Optional[object] = None + token2text: bool = False + + def __new__(cls, *args, **kwargs): + if cls._instance is None: + cls._instance = super().__new__(cls) + return cls._instance + + @classmethod + def get_instance(cls) -> "RolloutTraceConfig": + if cls._instance is None: + cls._instance = cls() + return cls._instance + + @classmethod + def init(cls, project_name: str, experiment_name: str, backend: str, token2text: bool = False): + config = cls.get_instance() + config.backend = backend + config.token2text = token2text + config.project_name = project_name + config.experiment_name = experiment_name + if backend == "weave": + import weave + + config.client = weave.init(project_name) + elif backend == "mlflow": + import mlflow + + config.client = mlflow + else: + config.client = None + + @classmethod + def get_backend(cls) -> Optional[str]: + return cls.get_instance().backend + + @classmethod + def get_client(cls) -> Optional[object]: + return cls.get_instance().client + + @classmethod + def enable_token2text(cls) -> Optional[bool]: + return cls.get_instance().token2text + + @classmethod + def reset(cls): + cls._instance = None + + +@contextlib.contextmanager +def rollout_trace_attr(sample_index=None, step=None, rollout_n=None): + """A context manager to add attributes to a trace for the configured backend.""" + backend = RolloutTraceConfig.get_backend() + attributes = {} + if sample_index is not None: + attributes["sample_index"] = sample_index + if step is not None: + attributes["step"] = step + if rollout_n is not None: + attributes["rollout_n"] = rollout_n + attributes["experiment_name"] = RolloutTraceConfig.get_instance().experiment_name + + if not attributes or backend is None: + yield + return + + if backend == "weave": + import weave + + with weave.attributes(attributes): + yield + # TODO implement mlfow trace + # elif backend == "mlflow": + # import mlflow + # # This assumes a run is already active. + # # We are setting tags for the current active run. + # try: + # mlflow.set_tags(attributes) + # except Exception: + # # Silently fail if there is no active run. + # pass + # yield + else: + yield + + +def rollout_trace_op(func): + @functools.wraps(func) + async def async_wrapper(self, *args, **kwargs): + backend = RolloutTraceConfig.get_backend() + enable_token2text = RolloutTraceConfig.enable_token2text() + if backend is None: + return await func(self, *args, **kwargs) + + sig = inspect.signature(func) + bound_args = sig.bind(self, *args, **kwargs) + bound_args.apply_defaults() + inputs = dict(bound_args.arguments) + del inputs["self"] + + async def add_token2text(self, result): + if hasattr(result, "prompt_ids") and hasattr(self, "tokenizer") and hasattr(self.tokenizer, "decode"): + _result = [result] + loop = asyncio.get_running_loop() + if hasattr(result, "prompt_ids"): + prompt_text = await loop.run_in_executor(None, self.tokenizer.decode, result.prompt_ids) + _result.append(prompt_text) + + if hasattr(result, "response_ids"): + response_text = await loop.run_in_executor(None, self.tokenizer.decode, result.response_ids) + _result.append(response_text) + return _result + return result + + if backend == "weave": + tracer = RolloutTraceConfig.get_client() + from weave.trace.context import call_context + + cur_attributes = {**call_context.call_attributes.get()} + call = tracer.create_call(op=func.__qualname__, inputs=inputs, attributes=cur_attributes) + try: + result = await func(self, *args, **kwargs) + + if enable_token2text: + _result = await add_token2text(self, result) + tracer.finish_call(call, output=_result) + else: + tracer.finish_call(call, output=result) + + return result + + except Exception as e: + tracer.finish_call(call, exception=e) + raise e + # TODO implement other backends such as mlflow + else: + return await func(self, *args, **kwargs) + + @functools.wraps(func) + def wrapper(self, *args, **kwargs): + backend = RolloutTraceConfig.get_backend() + if backend is None: + return func(self, *args, **kwargs) + + sig = inspect.signature(func) + bound_args = sig.bind(self, *args, **kwargs) + bound_args.apply_defaults() + inputs = dict(bound_args.arguments) + del inputs["self"] + + if backend == "weave": + tracer = RolloutTraceConfig.get_client() + from weave.trace.context import call_context + + cur_attributes = {**call_context.call_attributes.get()} + call = tracer.create_call(op=func.__qualname__, inputs=inputs, attributes=cur_attributes) + try: + result = func(self, *args, **kwargs) + tracer.finish_call(call, output=result) + return result + except Exception as e: + tracer.finish_call(call, exception=e) + raise e + # TODO implement other backends such as mlflow + else: + return func(self, *args, **kwargs) + + return async_wrapper if inspect.iscoroutinefunction(func) else wrapper \ No newline at end of file diff --git a/optimind/recipe/tools/schemas.py b/optimind/recipe/tools/schemas.py new file mode 100644 index 0000000..221b765 --- /dev/null +++ b/optimind/recipe/tools/schemas.py @@ -0,0 +1,89 @@ +# Copyright 2023-2024 SGLang Team +# Copyright 2025 ModelBest Inc. and/or its affiliates +# +# Licensed under the Apache License, Version 2.0 (the "License"); +# you may not use this file except in compliance with the License. +# You may obtain a copy of the License at +# +# http://www.apache.org/licenses/LICENSE-2.0 +# +# Unless required by applicable law or agreed to in writing, software +# distributed under the License is distributed on an "AS IS" BASIS, +# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. +# See the License for the specific language governing permissions and +# limitations under the License. +import json +from typing import Any, Literal + +from pydantic import BaseModel + + +class OpenAIFunctionPropertySchema(BaseModel): + """The schema of a parameter in OpenAI format.""" + + type: str + description: str | None = None + enum: list[str] | None = None + + +class OpenAIFunctionParametersSchema(BaseModel): + """The schema of parameters in OpenAI format.""" + + type: str + properties: dict[str, OpenAIFunctionPropertySchema] + required: list[str] + + +class OpenAIFunctionSchema(BaseModel): + """The schema of a function in OpenAI format.""" + + name: str + description: str + parameters: OpenAIFunctionParametersSchema + strict: bool = False + + +class OpenAIFunctionToolSchema(BaseModel): + """The schema of a tool in OpenAI format.""" + + type: str + function: OpenAIFunctionSchema + + +class OpenAIFunctionParsedSchema(BaseModel): + """The parsed schema of a tool in OpenAI format.""" + + name: str + arguments: str # JSON string + + +class OpenAIFunctionCallSchema(BaseModel): + """The parsed schema of a tool in OpenAI format.""" + + name: str + arguments: dict[str, Any] + + @staticmethod + def from_openai_function_parsed_schema( + parsed_schema: OpenAIFunctionParsedSchema, + ) -> tuple["OpenAIFunctionCallSchema", bool]: + has_decode_error = False + try: + arguments = json.loads(parsed_schema.arguments) + except json.JSONDecodeError: + arguments = {} + has_decode_error = True + # If the arguments is not a dict, it means the arguments is not a valid JSON string + if not isinstance(arguments, dict): + arguments = {} + has_decode_error = True + + return OpenAIFunctionCallSchema(name=parsed_schema.name, arguments=arguments), has_decode_error + + +class OpenAIFunctionToolCall(BaseModel): + """The tool call in OpenAI format.""" + + id: str + type: Literal["function"] = "function" + function: OpenAIFunctionCallSchema \ No newline at end of file diff --git a/optimind/rewards/__init__.py b/optimind/rewards/__init__.py new file mode 100644 index 0000000..e69de29 diff --git a/optimind/rewards/rm_objective_gurobipy_full.py b/optimind/rewards/rm_objective_gurobipy_full.py new file mode 100644 index 0000000..efb2bff --- /dev/null +++ b/optimind/rewards/rm_objective_gurobipy_full.py @@ -0,0 +1,64 @@ +import torch +import os +import re +import copy +import time +import subprocess +import random +import sys +import importlib +import traceback +import types +import uuid +import multiprocessing as mp +from io import StringIO +import contextlib +import tempfile + +# import the current file's directory's utils +current_dir = os.path.dirname(os.path.abspath(__file__)) +sys.path.append(current_dir) # add parent directory to path + +from utils import * + +################## helper functions end ################## +def compute_score(data_source, solution_str, ground_truth, extra_info=None, precision=None): + """ + label: the expected objective + """ + + code_block = extract_python_code_block(solution_str) + solution_dict = run_full_code(code_block, solver="gurobipy") + objective = solution_dict.get("objective_value", None) + + if isinstance(ground_truth, list) or isinstance(ground_truth, tuple): + # if any is correct then correct + accuracy = score = float(any([compute_acc(objective, gt, precision=precision) for gt in ground_truth])) + else: + accuracy = score = float(compute_acc(objective, ground_truth, precision=precision)) + + no_execution_error = float("failed" not in solution_dict["output"].lower() and solution_dict["traceback_str"].strip() == "") + have_objective = float(objective is not None) + + return { + "score": score, + "accuracy": accuracy, + "objective": objective if objective is not None else 0.0, + "no_execution_error": no_execution_error, + "have_objective": have_objective, + } + +if __name__ == '__main__': + import os + import argparse + import pdb + import pandas as pd + + solution_str = """ +analysisWe need to formulate: decision variables = pounds of each grain (continuous, non-negative). Minimize cost = 41 x1 + 36 x2 + 96 x3. Subject to nutrient constraints: 20x1+30x2+70x3 >= 110 (A); 10x1+10x2+0x3 >= 18 (B); 50x1+30x2+0x3 >= 90 (C); 6x1+2.5x2+10x3 >= 14 (D). Non-negativity.\n\nSolve linear program. Compute optimal solution manually? Let's try solve quickly.\n\nWe can attempt to find feasible combination with minimal cost.\n\nWe could use python after code runs. But we need to produce code; after execution we will get solution. So we will write code accordingly.\n\nWe also need to output minimum cost in cents.\n\nWe will implement, solve, retrieve objective value.\n\nReturn result.\n\nWe'll wait for user to execute.assistantfinal**Mathematical formulation**\n\nLet \n\n* \\(x_1\\) = pounds of Grain\u202f1 to use (continuous, \\(x_1 \\ge 0\\)) \n* \\(x_2\\) = pounds of Grain\u202f2 to use (continuous, \\(x_2 \\ge 0\\)) \n* \\(x_3\\) = pounds of Grain\u202f3 to use (continuous, \\(x_3 \\ge 0\\))\n\n**Objective (minimum cost in cents)** \n\n\\[\n\\min \\; 41x_1 + 36x_2 + 96x_3\n\\]\n\n**Nutrient constraints**\n\n\\[\n\\begin{aligned}\n20x_1 + 30x_2 + 70x_3 &\\ge 110 \\qquad\\text{(Nutrient\u202fA)}\\\\\n10x_1 + 10x_2 + 0x_3 &\\ge 18 \\qquad\\text{(Nutrient\u202fB)}\\\\\n50x_1 + 30x_2 + 0x_3 &\\ge 90 \\qquad\\text{(Nutrient\u202fC)}\\\\\n 6x_1 + 2.5x_2 + 10x_3 &\\ge 14 \\qquad\\text{(Nutrient\u202fD)}\n\\end{aligned}\n\\]\n\n\\(x_1, x_2, x_3 \\ge 0\\)\n\n---\n\n**Gurobi implementation**\n\n```python\nimport gurobipy as gp\nfrom gurobipy import GRB\n\ndef solve_wool_feeding():\n # create model\n model = gp.Model(\"wool_feeding\")\n\n # decision variables: pounds of each grain (continuous, non\u2011negative)\n x1 = model.addVar(name=\"Grain1\", lb=0.0, vtype=GRB.CONTINUOUS)\n x2 = model.addVar(name=\"Grain2\", lb=0.0, vtype=GRB.CONTINUOUS)\n x3 = model.addVar(name=\"Grain3\", lb=0.0, vtype=GRB.CONTINUOUS)\n\n # objective: minimize total cost (cents)\n model.setObjective(41 * x1 + 36 * x2 + 96 * x3, GRB.MINIMIZE)\n\n # nutrient constraints\n model.addConstr(20 * x1 + 30 * x2 + 70 * x3 >= 110, name=\"Nutrient_A\")\n model.addConstr(10 * x1 + 10 * x2 >= 18, name=\"Nutrient_B\")\n model.addConstr(50 * x1 + 30 * x2 >= 90, name=\"Nutrient_C\")\n model.addConstr( 6 * x1 + 2.5 * x2 + 10 * x3 >= 14, name=\"Nutrient_D\")\n\n # solve\n model.optimize()\n\n if model.status == GRB.OPTIMAL:\n cost = model.objVal\n print(f\"Optimal cost: {cost:.2f} cents\")\n print(f\"Grain 1: {x1.X:.4f} lb\")\n print(f\"Grain 2: {x2.X:.4f} lb\")\n print(f\"Grain 3: {x3.X:.4f} lb\")\n return cost\n else:\n print(\"No optimal solution found.\")\n return None\n\nif __name__ == \"__main__\":\n solve_wool_feeding()\n```\n\nRunning the script will display the minimum cost (in cents) together with the optimal amounts of each grain. The variable `model` is used as required, all variables are continuous, and constraints are expressed with `>=` (implemented via `>=` in Gurobi, which internally translates to the appropriate linear form)." +""" + + ground_truth = 2.0 + + score = compute_score("math", solution_str, ground_truth) + print("score:", score) diff --git a/optimind/rewards/utils.py b/optimind/rewards/utils.py new file mode 100644 index 0000000..c6892c8 --- /dev/null +++ b/optimind/rewards/utils.py @@ -0,0 +1,265 @@ +import re +import json + + +def extract_python_code_block( + response: str, allow_overlapping: bool = True, return_status: bool = False, + return_none_if_failed: bool = False, +) -> str | None: + """ + Return the last non-empty Python code block in *response*, + scanning from the end to the beginning. + + Non-empty = contains at least one character other than whitespace or a dot. + Supports three block syntaxes: + 1) ```python ... ``` but when matching this we do not want to match ```python ... ```python + (the closing ``` should not be immediately followed by python) + 2) ... + 3) ... + + If no closed fence is found, falls back to the tail after the last ```python. + """ + + def non_empty(block: str) -> bool: + return bool(re.search(r"[^\s\.]", block)) + + def get_output(result, success: bool): + if return_status: + return result, success + else: + return result + + # --- patterns --- + # Standard (non-overlapping) + # NOTE: closing ``` must NOT be immediately followed by "python" + rx_fence = re.compile(r"```python\s*(.*?)\s*```(?!python)", re.DOTALL) + rx_py = re.compile(r"(.*?)", re.DOTALL) + rx_code = re.compile(r"(.*?)", re.DOTALL) + + if not allow_overlapping: + # Reverse order by consuming all matches, then iterate backwards. + for block in reversed([m.group(1) for m in rx_fence.finditer(response)]): + if non_empty(block): + return get_output(block.strip(), True) + + for block in reversed([m.group(1) for m in rx_py.finditer(response)]): + if non_empty(block): + return get_output(block.strip(), True) + + for block in reversed([m.group(1) for m in rx_code.finditer(response)]): + if non_empty(block): + return get_output(block.strip(), True) + else: + # Overlapping via lookahead (captures the inner group only). + # Same constraint: closing ``` must NOT be immediately followed by "python" + rx_fence_ov = re.compile(r"(?s)(?=(```python\s*(.*?)\s*```(?!python)))") + rx_py_ov = re.compile(r"(?s)(?=((.*?)))") + rx_code_ov = re.compile(r"(?s)(?=((.*?)))") + + for block in reversed([m.group(2) for m in rx_fence_ov.finditer(response)]): + if non_empty(block): + return get_output(block.strip(), True) + for block in reversed([m.group(2) for m in rx_py_ov.finditer(response)]): + if non_empty(block): + return get_output(block.strip(), True) + for block in reversed([m.group(2) for m in rx_code_ov.finditer(response)]): + if non_empty(block): + return get_output(block.strip(), True) + + if return_none_if_failed: + return get_output(None, False) + + # Fallback: unterminated ```python at the end — return tail if non-empty + start = response.rfind("```python") + if start != -1: + tail = response[start + len("```python"):].strip() + if non_empty(tail): + return get_output(tail, False) + + # Nothing suitable found — return stripped whole response (original fallback) + # This is also for cases where the response is a code (extracted from a code block before), so we just return it. + return get_output(response.strip(), False) + + +def safe_eval(value): + if isinstance(value, (int, float)): + return round(value, 5) + + # First attempt: direct safe eval + try: + return round(eval(value), 5) + except: + # Second attempt: strip non-numeric/operator chars + try: + cleaned = re.sub(r"[^0-9eE\.\+\-\*/\(\)]", "", value) + return round(eval(cleaned), 5) + except: + return None + +def compute_acc(pred, true, precision=None): + if precision is None: precision = 1e-6 + + if pred is None or true is None: + return float(pred is None and true is None) + + try: + if precision > 1e-3: + acc = abs(pred - true) < precision + else: + acc = abs(pred - true) / (abs(true) + 1) < precision + except: + acc = 0 + return acc + + +def run_full_code(code, solver="gurobipy", time_limit=10, tol=1e-7, return_code=False): + IMPORTS = { + "gurobipy": "import os\nimport gurobipy as gp\nfrom gurobipy import Model, GRB, quicksum, LinExpr, QuadExpr, Var, Constr", + } + + ADD_SCRIPT_TOL = { + "gurobipy": '{m}.setParam("IntFeasTol", {tol})', + } # Only support gurobipy right now + + # Use {m} placeholder; we'll format with detected model var name + ADD_SCRIPT_FMT = { + "gurobipy": 'if {m}.status == GRB.OPTIMAL:\n print("Just print the best solution:", {m}.objVal)\nelse:\n print("No Best Solution")', + } + + if solver not in IMPORTS: + raise ValueError(f"Unsupported solver: {solver}") + + if "```python" in code: + maybe = extract_python_code_block(code) + if maybe: + code = maybe + + code_lines = code.strip().splitlines() + + # --- Detect solve/optimize line by solver (keep the last match) --- + model_name = None + solve_index = None + solve_indent = 0 + + pat = re.compile(r'([A-Za-z_]\w*)\s*\.\s*optimize\s*\(') + # Now: use the *last* such call in the code. + for i, line in enumerate(code_lines): + m = pat.search(line) + if m: + model_name = m.group(1) + solve_index = i + solve_indent = len(line) - len(line.lstrip()) + + if solve_index is None or model_name is None: + ret = { + "output": f"Failed: No detectable optimize call for solver '{solver}'.", + "sol_output": f"Failed: No detectable optimize call for solver '{solver}'.", + "objective_value": None, + "traceback_str": "", + } + if return_code: + ret["code"] = code + return ret + + # Optional tol injection + insert_idx_tol = solve_index + if tol is not None and solver in ADD_SCRIPT_TOL and ADD_SCRIPT_TOL[solver]: + tol_line = ADD_SCRIPT_TOL[solver].format(m=model_name, tol=tol) + aligned_tol_line = (" " * solve_indent) + tol_line + code_lines.insert(insert_idx_tol, aligned_tol_line) + solve_index += 1 # Adjust since we inserted a line + + insert_idx = solve_index + 1 + + # Align inserted script to the same indentation as the solve/optimize call + template = ADD_SCRIPT_FMT[solver] + add_script = template.format(m=model_name) if "{m}" in template else template + aligned_add_script = [(" " * solve_indent) + l for l in add_script.splitlines()] + new_code_lines = code_lines[:insert_idx] + aligned_add_script + code_lines[insert_idx:] + + program = f"{IMPORTS[solver]}\n\n" + "\n".join(new_code_lines) + + import tempfile, uuid, subprocess, sys + try: + import os # may already be in IMPORTS; safe to import again + except Exception: + pass + + with tempfile.TemporaryDirectory() as workdir: + script_path = os.path.join(workdir, f"prog_{uuid.uuid4().hex}.py") + with open(script_path, "w") as f: + f.write(program) + + try: + res = subprocess.run( + [sys.executable, script_path], + cwd=workdir, + stdout=subprocess.PIPE, + stderr=subprocess.PIPE, + text=True, + timeout=time_limit, + ) + output = res.stdout + traceback_str = res.stderr + except subprocess.TimeoutExpired: + err = "Failed: Execution timed out" + return {"output": err, "sol_output": err, + "objective_value": None, "traceback_str": err} + except Exception as e: + err = f"Failed: {str(e)[:200]}" + return {"output": err, "sol_output": err, + "objective_value": None, "traceback_str": err} + + # Extract numeric objective from the standardized print + match = re.search(r"Just print the best solution:\s*([-+]?\d*\.?\d+(?:[eE][-+]?\d+)?)", output) + objective = float(match.group(1)) if match else None + + ret = { + "output": output, + "sol_output": output, + "objective_value": objective, + "traceback_str": traceback_str + } + if return_code: + ret["code"] = program + return ret + + +def compute_score_history_output(data_source, history_list, ground_truth, extra_info=None, precision=None): + user_idxs = [i for i, turn in enumerate(history_list) if turn['role'] == 'user'] + objective = None + for idx in reversed(user_idxs): + output_str = history_list[idx]['content'] + potential_query = history_list[idx].get("query", "") + if "No python code block found. Skipping code execution." in output_str or potential_query: + continue + else: + match = re.search(r"Just print the best solution:\s*([-+]?\d*\.?\d+(?:[eE][-+]?\d+)?)", output_str) + if match: + objective = float(match.group(1)) + break + + if isinstance(ground_truth, list) or isinstance(ground_truth, tuple): + # if any is correct then correct + accuracy = score = float(any([compute_acc(objective, gt, precision=precision) for gt in ground_truth])) + else: + accuracy = score = float(compute_acc(objective, ground_truth, precision=precision)) + + no_execution_error = float(objective is not None) + have_objective = float(objective is not None) + + return { + "score": score, + "accuracy": accuracy, + "objective": objective if objective is not None else 0.0, + "no_execution_error": no_execution_error, # this isnt accurate + "have_objective": have_objective, + } + +# if __name__ == '__main__': +# s = "```python\nimport gurobipy as gp\nfrom gurobipy import GRB\n\n# ------------------------------------------------------------------\n# Data (0\u2011based indices)\n# ------------------------------------------------------------------\nopen_cost = [\n 151000, 192000, 114000, 171000, 160000,\n 120000, 182000, 186000, 174000\n]\n\n# Shipping costs per unit: 9 centers \u00d7 5 stores\n# Store1 Store2 Store3 Store4 Store5\nship_cost = [\n [3, 5, 4, 3, 5], # Center 1\n [2, 4, 2, 4, 5], # Center 2\n [1, 4, 2, 5, 4], # Center 3\n [1, 1, 3, 3, 2], # Center 4\n [4, 4, 3, 4, 4], # Center 5\n [1, 3, 5, 3, 5], # Center 6\n [1, 2, 4, 1, 4], # Center 7\n [2, 2, 1, 2, 5], # Center 8\n [2, 4, 4, 4, 4], # Center 9\n]\n\ndemand = [908, 434, 605, 480, 961]\n\ncapacity = [1825, 1821, 1365, 1929, 1502,\n 1201, 1529, 961, 1001]\n\n# ------------------------------------------------------------------\n# Model\n# ------------------------------------------------------------------\nmodel = gp.Model(\"GlobalSupply\")\n\ny = model.addVars(len(open_cost), vtype=GRB.BINARY, name=\"open\")\nx = model.addVars(len(demand), len(open_cost), vtype=GRB.CONTINUOUS, name=\"ship\")\n\n# Demand constraints\nfor i in range(len(demand)):\n model.addConstr(\n gp.quicksum(x[i, j] for j in range(len(open_cost))) == demand[i],\n name=f\"demand_{i}\"\n )\n\n# Capacity constraints\nfor j in range(len(open_cost)):\n model.addConstr(\n gp.quicksum(x[i, j] for i in range(len(demand))) <= capacity[j] * y[j],\n name=f\"capacity_{j}\"\n )\n\n# Objective\nobj = gp.quicksum(open_cost[j] * y[j] for j in range(len(open_cost))) + \\\n gp.quicksum(ship_cost[j][i] * x[i, j]\n for i in range(len(demand))\n for j in range(len(open_cost)))\nmodel.setObjective(obj, GRB.MINIMIZE)\n\n# ------------------------------------------------------------------\n# Solve\n# ------------------------------------------------------------------\n# In this execution environment the method is named `solve`\nif hasattr(model, \"solve\"):\n model.solve()\nelse:\n model.optimize()\n\n# ------------------------------------------------------------------\n# Output\n# ------------------------------------------------------------------\nif model.status == GRB.OPTIMAL:\n total_cost = model.objVal\n print(f\"\\nOptimal total cost: ${total_cost:,.0f}\")\n\n print(\"\\nCenters to open:\")\n for j in range(len(open_cost)):\n if y[j].X > 0.5:\n print(f\" Center {j+1}: open (${open_cost[j]:,.0f})\")\n\n print(\"\\nShipping plan (units):\")\n for i in range(len(demand)):\n plan = []\n for j in range(len(open_cost)):\n if x[i, j].X > 1e-6:\n plan.append(f\"C{j+1}:{x[i, j].X:.1f}\")\n print(f\"Store {i+1:>2}: \" + \" -> \".join(plan))\nelse:\n print(\"No optimal solution found.\")\n```" +# code_block = extract_python_code_block(s) +# print("Extracted code block:") +# print(code_block) +# ret = run_full_code(code_block) +# print(ret) diff --git a/optimind/submit_eval.sh b/optimind/submit_eval.sh new file mode 100644 index 0000000..cb7bebe --- /dev/null +++ b/optimind/submit_eval.sh @@ -0,0 +1,48 @@ +#!/bin/bash +industryor="data/optimind_cleaned_classified_industryor.csv" +mamo_complex="data/optimind_cleaned_classified_mamo_complex.csv" +optmath="data/optimind_cleaned_classified_optmath.csv" + +hint="data/hints.csv" + +qwen3_32b="Qwen/Qwen3-32B" + +user_prompt_type=${1:-default} +system_prompt_type=${2:-default} + + +MAX_TURNS=5 + +TP_SIZE=4 +DP_SIZE=2 + +TEMP=0.6 +TOP_P=0.95 + +for num_majority in 1 8; do + for seed in 1 2 3 4 5 6 7 8 9 10; do + echo "============================== Starting evaluations ==============================" + for data_name in $industryor $mamo_complex $optmath ; do + echo "============================== Dataset: $data_name ==============================" + for model_path in $qwen3_32b ; do + echo "======================================= S G L A N G (NO HINTS, NO EARLY STOP) ==========================================" + echo "Running evaluation for dataset: $data_name" + echo "Model path: $model_path" + CUDA_VISIBLE_DEVICES=0,1,2,3,4,5,6,7 python3 eval.py --data "$data_name" --backend_str sglang --model_path "$model_path" --dp_size $DP_SIZE --tp_size $TP_SIZE --seed $seed \ + --user_prompt_type $user_prompt_type --system_prompt_type $system_prompt_type \ + --num-majority $num_majority --temp $TEMP --top-p $TOP_P --debug --max-turns $MAX_TURNS --no_early_stop + echo "================================================================================================" + + echo "======================================= S G L A N G (W/ HINTS + HINT INSTRUCTION, NO EARLY STOP) ==========================================" + echo "Running evaluation for dataset: $data_name" + echo "Model path: $model_path" + CUDA_VISIBLE_DEVICES=0,1,2,3,4,5,6,7 python3 eval.py --data "$data_name" --backend_str sglang --model_path "$model_path" --dp_size $DP_SIZE --tp_size $TP_SIZE --seed $seed \ + --user_prompt_type $user_prompt_type --system_prompt_type $system_prompt_type \ + --num-majority $num_majority --temp $TEMP --top-p $TOP_P \ + --error_analysis_file $hint --apply_error_analysis --apply_hint_instructions --debug --max-turns $MAX_TURNS --no_early_stop + echo "================================================================================================" + done + done + done +done + diff --git a/optimind/submit_eval_gptoss.sh b/optimind/submit_eval_gptoss.sh new file mode 100644 index 0000000..f959881 --- /dev/null +++ b/optimind/submit_eval_gptoss.sh @@ -0,0 +1,53 @@ +#!/bin/bash +industryor="data/optimind_cleaned_classified_industryor.csv" +mamo_complex="data/optimind_cleaned_classified_mamo_complex.csv" +optmath="data/optimind_cleaned_classified_optmath.csv" + +hint="data/hints.csv" + +unsloth_gpt_path="unsloth/gpt-oss-20b-BF16" +optimind_sft="microsoft/OptiMind-SFT" + +user_prompt_type=${1:-default} +system_prompt_type=${2:-default} + + +MAX_TURNS=5 + +TP_SIZE=4 +DP_SIZE=2 + +TEMP=0.9 +TOP_P=1 + +REASONING="medium" +HF_HOME="~/.cache/huggingface/hub" + +for num_majority in 1 8; do + for seed in 1 2 3 4 5 6 7 8 9 10; do + echo "============================== Starting evaluations ==============================" + for data_name in $industryor $mamo_complex $optmath ; do + echo "============================== Dataset: $data_name ==============================" + for model_path in $unsloth_gpt_path $optimind_sft ; do + echo "======================================= S G L A N G (NO HINTS, NO EARLY STOP) ==========================================" + echo "Running evaluation for dataset: $data_name" + echo "Model path: $model_path" + CUDA_VISIBLE_DEVICES=0,1,2,3,4,5,6,7 python3 eval.py --data "$data_name" --backend_str sglang --model_path "$model_path" --dp_size $DP_SIZE --tp_size $TP_SIZE --seed $seed \ + --user_prompt_type $user_prompt_type --system_prompt_type $system_prompt_type --gpt-oss --reasoning-effort $REASONING \ + --num-majority $num_majority --temp $TEMP --top-p $TOP_P --debug --max-turns $MAX_TURNS --no_early_stop + echo "================================================================================================" + + echo "======================================= S G L A N G (W/ HINTS + HINT INSTRUCTION, NO EARLY STOP) ==========================================" + echo "Running evaluation for dataset: $data_name" + echo "Model path: $model_path" + CUDA_VISIBLE_DEVICES=0,1,2,3,4,5,6,7 python3 eval.py --data "$data_name" --backend_str sglang --model_path "$model_path" --dp_size $DP_SIZE --tp_size $TP_SIZE --seed $seed \ + --user_prompt_type $user_prompt_type --system_prompt_type $system_prompt_type --gpt-oss --reasoning-effort $REASONING \ + --num-majority $num_majority --temp $TEMP --top-p $TOP_P \ + --error_analysis_file $hint --apply_error_analysis --apply_hint_instructions --debug --max-turns $MAX_TURNS --no_early_stop + echo "================================================================================================" + done + done + done +done + +