Is there a relation between Lagrange multipliers and the difficulty of solving a NLP?
$begingroup$
I heard a lot about the interpretation of Lagrange multipliers and the constraints qualification for the resolution of NLP but I still have a question:
Could we qualify the difficulty of solving a NLP based on the values of the Lagrange multipliers at the solutions?
In other words, could we say that a given NLP1 was more difficult to solve that NLP2 if for example max(Lagrange multipliers in 1)> max(Lagrange multipliers in 2)??
This question is motivated by a paragraph talking about sensitivity in the book numerical optimization of Nocedal.
optimization nonlinear-optimization
asked Dec 18 '18 at 14:39
yas areyas are
547
$endgroup$
add a comment |
$begingroup$
I heard a lot about the interpretation of Lagrange multipliers and the constraints qualification for the resolution of NLP but I still have a question:
Could we qualify the difficulty of solving a NLP based on the values of the Lagrange multipliers at the solutions?
In other words, could we say that a given NLP1 was more difficult to solve that NLP2 if for example max(Lagrange multipliers in 1)> max(Lagrange multipliers in 2)??
This question is motivated by a paragraph talking about sensitivity in the book numerical optimization of Nocedal.
optimization nonlinear-optimization
asked Dec 18 '18 at 14:39
yas areyas are
547
$endgroup$
$begingroup$
You can always rescale the constraints to make the Lagrange multipliers smaller; I would say the number of nonzero multipliers combined with the level of nonlinearity of the constraints is a better measure
$endgroup$
– LinAlg
Dec 18 '18 at 15:48
$begingroup$
could you please explain more with refering to the Hessian of the active constraints at the solution or the Jacobian of the active constraints at the solution? How could we estimate the nonlinearity of constraints? Thank you in advance.
$endgroup$
– yas are
Dec 18 '18 at 17:03
$begingroup$
I am afraid I can only give an intuitive explanation of nonlinearity: if you have $m$ constraints $f_i(x) leq 0$, you can combine them into a single constraint $max_i { f_i(x) } leq 0$, thus ending up with a problem with just one Lagrange multiplier.
$endgroup$
– LinAlg
Dec 18 '18 at 17:05
$begingroup$
This is funny because you stated the original problem I am trying to solve. In fact, I am trying to minimize the maximum of a set of functions. Since the objective function is not smooth I tried to transform it to a nonlinear programming. Thank you for your response.
$endgroup$
– yas are
Dec 18 '18 at 17:08
add a comment |
$begingroup$
I heard a lot about the interpretation of Lagrange multipliers and the constraints qualification for the resolution of NLP but I still have a question:
Could we qualify the difficulty of solving a NLP based on the values of the Lagrange multipliers at the solutions?
In other words, could we say that a given NLP1 was more difficult to solve that NLP2 if for example max(Lagrange multipliers in 1)> max(Lagrange multipliers in 2)??
This question is motivated by a paragraph talking about sensitivity in the book numerical optimization of Nocedal.
optimization nonlinear-optimization
asked Dec 18 '18 at 14:39
yas areyas are
547
$endgroup$
I heard a lot about the interpretation of Lagrange multipliers and the constraints qualification for the resolution of NLP but I still have a question:
Could we qualify the difficulty of solving a NLP based on the values of the Lagrange multipliers at the solutions?
In other words, could we say that a given NLP1 was more difficult to solve that NLP2 if for example max(Lagrange multipliers in 1)> max(Lagrange multipliers in 2)??
This question is motivated by a paragraph talking about sensitivity in the book numerical optimization of Nocedal.
optimization nonlinear-optimization
optimization nonlinear-optimization
asked Dec 18 '18 at 14:39
yas areyas are
547
asked Dec 18 '18 at 14:39
yas areyas are
547
asked Dec 18 '18 at 14:39
yas areyas are
547
asked Dec 18 '18 at 14:39
yas areyas are
547
asked Dec 18 '18 at 14:39
yas areyas are
547
547
$begingroup$
You can always rescale the constraints to make the Lagrange multipliers smaller; I would say the number of nonzero multipliers combined with the level of nonlinearity of the constraints is a better measure
$endgroup$
– LinAlg
Dec 18 '18 at 15:48
$begingroup$
could you please explain more with refering to the Hessian of the active constraints at the solution or the Jacobian of the active constraints at the solution? How could we estimate the nonlinearity of constraints? Thank you in advance.
$endgroup$
– yas are
Dec 18 '18 at 17:03
$begingroup$
I am afraid I can only give an intuitive explanation of nonlinearity: if you have $m$ constraints $f_i(x) leq 0$, you can combine them into a single constraint $max_i { f_i(x) } leq 0$, thus ending up with a problem with just one Lagrange multiplier.
$endgroup$
– LinAlg
Dec 18 '18 at 17:05
$begingroup$
This is funny because you stated the original problem I am trying to solve. In fact, I am trying to minimize the maximum of a set of functions. Since the objective function is not smooth I tried to transform it to a nonlinear programming. Thank you for your response.
$endgroup$
– yas are
Dec 18 '18 at 17:08
add a comment |
$begingroup$
You can always rescale the constraints to make the Lagrange multipliers smaller; I would say the number of nonzero multipliers combined with the level of nonlinearity of the constraints is a better measure
$endgroup$
– LinAlg
Dec 18 '18 at 15:48
$begingroup$
could you please explain more with refering to the Hessian of the active constraints at the solution or the Jacobian of the active constraints at the solution? How could we estimate the nonlinearity of constraints? Thank you in advance.
$endgroup$
– yas are
Dec 18 '18 at 17:03
$begingroup$
I am afraid I can only give an intuitive explanation of nonlinearity: if you have $m$ constraints $f_i(x) leq 0$, you can combine them into a single constraint $max_i { f_i(x) } leq 0$, thus ending up with a problem with just one Lagrange multiplier.
$endgroup$
– LinAlg
Dec 18 '18 at 17:05
$begingroup$
This is funny because you stated the original problem I am trying to solve. In fact, I am trying to minimize the maximum of a set of functions. Since the objective function is not smooth I tried to transform it to a nonlinear programming. Thank you for your response.
$endgroup$
– yas are
Dec 18 '18 at 17:08
$begingroup$
You can always rescale the constraints to make the Lagrange multipliers smaller; I would say the number of nonzero multipliers combined with the level of nonlinearity of the constraints is a better measure
$endgroup$
– LinAlg
Dec 18 '18 at 15:48
$begingroup$
You can always rescale the constraints to make the Lagrange multipliers smaller; I would say the number of nonzero multipliers combined with the level of nonlinearity of the constraints is a better measure
$endgroup$
– LinAlg
Dec 18 '18 at 15:48
$begingroup$
could you please explain more with refering to the Hessian of the active constraints at the solution or the Jacobian of the active constraints at the solution? How could we estimate the nonlinearity of constraints? Thank you in advance.
$endgroup$
– yas are
Dec 18 '18 at 17:03
$begingroup$
could you please explain more with refering to the Hessian of the active constraints at the solution or the Jacobian of the active constraints at the solution? How could we estimate the nonlinearity of constraints? Thank you in advance.
$endgroup$
– yas are
Dec 18 '18 at 17:03
$begingroup$
I am afraid I can only give an intuitive explanation of nonlinearity: if you have $m$ constraints $f_i(x) leq 0$, you can combine them into a single constraint $max_i { f_i(x) } leq 0$, thus ending up with a problem with just one Lagrange multiplier.
$endgroup$
– LinAlg
Dec 18 '18 at 17:05
$begingroup$
I am afraid I can only give an intuitive explanation of nonlinearity: if you have $m$ constraints $f_i(x) leq 0$, you can combine them into a single constraint $max_i { f_i(x) } leq 0$, thus ending up with a problem with just one Lagrange multiplier.
$endgroup$
– LinAlg
Dec 18 '18 at 17:05
$begingroup$
This is funny because you stated the original problem I am trying to solve. In fact, I am trying to minimize the maximum of a set of functions. Since the objective function is not smooth I tried to transform it to a nonlinear programming. Thank you for your response.
$endgroup$
– yas are
Dec 18 '18 at 17:08
$begingroup$
This is funny because you stated the original problem I am trying to solve. In fact, I am trying to minimize the maximum of a set of functions. Since the objective function is not smooth I tried to transform it to a nonlinear programming. Thank you for your response.
$endgroup$
– yas are
Dec 18 '18 at 17:08
add a comment |
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$begingroup$
You can always rescale the constraints to make the Lagrange multipliers smaller; I would say the number of nonzero multipliers combined with the level of nonlinearity of the constraints is a better measure
$endgroup$
– LinAlg
Dec 18 '18 at 15:48
$begingroup$
could you please explain more with refering to the Hessian of the active constraints at the solution or the Jacobian of the active constraints at the solution? How could we estimate the nonlinearity of constraints? Thank you in advance.
$endgroup$
– yas are
Dec 18 '18 at 17:03
$begingroup$
I am afraid I can only give an intuitive explanation of nonlinearity: if you have $m$ constraints $f_i(x) leq 0$, you can combine them into a single constraint $max_i { f_i(x) } leq 0$, thus ending up with a problem with just one Lagrange multiplier.
$endgroup$
– LinAlg
Dec 18 '18 at 17:05
$begingroup$
This is funny because you stated the original problem I am trying to solve. In fact, I am trying to minimize the maximum of a set of functions. Since the objective function is not smooth I tried to transform it to a nonlinear programming. Thank you for your response.
$endgroup$
– yas are
Dec 18 '18 at 17:08