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dpo
reviewed
Aug 18, 2025
Contributor
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@oihanc Il y a une erreur dans ton rapport qui l'empêche d'être généré... |
dpo
reviewed
Aug 19, 2025
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| return t, xt | ||
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| f = obj(model, x) | ||
| ∇f = grad(model, x) | ||
| ∇f_norm = norm(∇f) |
Contributor
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Pourquoi cette quantité est-il importante ?
| ∇f .= grad(model, x) | ||
| ∇f_norm = norm(∇f) | ||
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| if ∇f_norm <= tolerance |
Contributor
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Ce n'est pas un critère de stationnarité quand il y a des bornes...
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| ft = obj(nlp, xt) | ||
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| return t, xt |
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Pourquoi réimplémenter ??? La 1e implémentation doit fonctionner. C'est là le but de l'exercice.
| - `print_freq::Integer = 1`: print frequency | ||
| - `x_tol::T = eps(T)^(1/2)`: norm tolerance between two iterates xₖ | ||
| """ | ||
| function trust_region_descent(model::AbstractNLPModel{T,V}; radius::Float64 = Inf, callback = (args...) -> nothing, ϵa::T = eps(T)^(1/2), ϵr::T = eps(T)^(1/2), itmax::Integer = 10*model.meta.nvar, verbose::Integer = 0, print_freq::Integer = 1, x_tol::T = eps(T)^(1/2)) where {T, V} |
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Il ne faut pas réimplémenter. Le même solveur doit fonctionner.
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| g = rand(n) | ||
| A = rand(n, n) | ||
| H = 0.5 * (A + A') # H is symmetric positive semi-definite |
Contributor
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La quadratique est-elle non convexe ?
| - Recherche linéaire d'Armijo, pour des modèles quadratiques, il est préférable d'utiliser une recherche linéaire quadratique exacte (comme utilisé par la méthode du gradient conjugué) | ||
| - Conditions de KKT : ne les vérifie pas. Il faut vérifier la norme entre les itérés $x_k$. | ||
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| Pour ces raisons, l'utilisation d'une telle méthode n'est pas recommandée. |
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Mais la solution trouvée est-elle acceptable pour une méthode de région de confiance ?
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