Standard name for the computation of Lagrange multipliers iteratively by fixing other multipliers?











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Dear Optimization Experts,



Background:



I have a convex optimization problem on hand that can be shown in general form as given below
begin{equation}
begin{aligned}
& underset{x in mathbb{R}^N}{text{minimize}}
& & f(x) \
& text{subject to}
& & g_i(x) - alpha_i leq 0 forall i = 1,cdots,K ; ,
end{aligned}
end{equation}

where both functions $g_i: mathbb{R}^N rightarrowmathbb{R}$ and $f: mathbb{R}^N rightarrowmathbb{R}$ are convex, and $alpha_i in mathbb{R}$ is given.



The Lagrangian is:
begin{align}
Lleft(x, left{lambda_iright}right)
&= f(x) + sum limits_{i=1}^{K} lambda_i left(g_i(x) - alpha_i right) ; .
end{align}



Question:



If $K=1$ then I can obtain the closed-form solution $x$ (and analytical solution of the Lagrange multiplier $lambda_1$) by following the KKT conditions.



Now, the question arises when $K > 1$ then I can't obtain the closed-form solution, but I can compute $x$ analytically which is dependent on all the $lambda_i$. So, to compute the Lagrange multiplier say $lambda_i$, I resort to iterative solution where I fix other Lagrange multipliers ($lambda_j forall j = 1,cdots,K$ except $i$). Then repeat the above process for other Lagrange multipliers iteratively.




  • do you have any standard name for such scheme to compute Lagrange multipliers cyclically?

  • If not, can I say that this cyclic/iterative scheme is nothing but Coordinate Descent (or like)?


Thank you so much for your time in advance.










share|cite|improve this question




















  • 1




    Depending on how you select $lambda$, this may be coordinate descent in the dual problem.
    – LinAlg
    Nov 18 at 13:53










  • thank you LinAlg!
    – user550103
    Nov 18 at 15:51















up vote
0
down vote

favorite
2












Dear Optimization Experts,



Background:



I have a convex optimization problem on hand that can be shown in general form as given below
begin{equation}
begin{aligned}
& underset{x in mathbb{R}^N}{text{minimize}}
& & f(x) \
& text{subject to}
& & g_i(x) - alpha_i leq 0 forall i = 1,cdots,K ; ,
end{aligned}
end{equation}

where both functions $g_i: mathbb{R}^N rightarrowmathbb{R}$ and $f: mathbb{R}^N rightarrowmathbb{R}$ are convex, and $alpha_i in mathbb{R}$ is given.



The Lagrangian is:
begin{align}
Lleft(x, left{lambda_iright}right)
&= f(x) + sum limits_{i=1}^{K} lambda_i left(g_i(x) - alpha_i right) ; .
end{align}



Question:



If $K=1$ then I can obtain the closed-form solution $x$ (and analytical solution of the Lagrange multiplier $lambda_1$) by following the KKT conditions.



Now, the question arises when $K > 1$ then I can't obtain the closed-form solution, but I can compute $x$ analytically which is dependent on all the $lambda_i$. So, to compute the Lagrange multiplier say $lambda_i$, I resort to iterative solution where I fix other Lagrange multipliers ($lambda_j forall j = 1,cdots,K$ except $i$). Then repeat the above process for other Lagrange multipliers iteratively.




  • do you have any standard name for such scheme to compute Lagrange multipliers cyclically?

  • If not, can I say that this cyclic/iterative scheme is nothing but Coordinate Descent (or like)?


Thank you so much for your time in advance.










share|cite|improve this question




















  • 1




    Depending on how you select $lambda$, this may be coordinate descent in the dual problem.
    – LinAlg
    Nov 18 at 13:53










  • thank you LinAlg!
    – user550103
    Nov 18 at 15:51













up vote
0
down vote

favorite
2









up vote
0
down vote

favorite
2






2





Dear Optimization Experts,



Background:



I have a convex optimization problem on hand that can be shown in general form as given below
begin{equation}
begin{aligned}
& underset{x in mathbb{R}^N}{text{minimize}}
& & f(x) \
& text{subject to}
& & g_i(x) - alpha_i leq 0 forall i = 1,cdots,K ; ,
end{aligned}
end{equation}

where both functions $g_i: mathbb{R}^N rightarrowmathbb{R}$ and $f: mathbb{R}^N rightarrowmathbb{R}$ are convex, and $alpha_i in mathbb{R}$ is given.



The Lagrangian is:
begin{align}
Lleft(x, left{lambda_iright}right)
&= f(x) + sum limits_{i=1}^{K} lambda_i left(g_i(x) - alpha_i right) ; .
end{align}



Question:



If $K=1$ then I can obtain the closed-form solution $x$ (and analytical solution of the Lagrange multiplier $lambda_1$) by following the KKT conditions.



Now, the question arises when $K > 1$ then I can't obtain the closed-form solution, but I can compute $x$ analytically which is dependent on all the $lambda_i$. So, to compute the Lagrange multiplier say $lambda_i$, I resort to iterative solution where I fix other Lagrange multipliers ($lambda_j forall j = 1,cdots,K$ except $i$). Then repeat the above process for other Lagrange multipliers iteratively.




  • do you have any standard name for such scheme to compute Lagrange multipliers cyclically?

  • If not, can I say that this cyclic/iterative scheme is nothing but Coordinate Descent (or like)?


Thank you so much for your time in advance.










share|cite|improve this question















Dear Optimization Experts,



Background:



I have a convex optimization problem on hand that can be shown in general form as given below
begin{equation}
begin{aligned}
& underset{x in mathbb{R}^N}{text{minimize}}
& & f(x) \
& text{subject to}
& & g_i(x) - alpha_i leq 0 forall i = 1,cdots,K ; ,
end{aligned}
end{equation}

where both functions $g_i: mathbb{R}^N rightarrowmathbb{R}$ and $f: mathbb{R}^N rightarrowmathbb{R}$ are convex, and $alpha_i in mathbb{R}$ is given.



The Lagrangian is:
begin{align}
Lleft(x, left{lambda_iright}right)
&= f(x) + sum limits_{i=1}^{K} lambda_i left(g_i(x) - alpha_i right) ; .
end{align}



Question:



If $K=1$ then I can obtain the closed-form solution $x$ (and analytical solution of the Lagrange multiplier $lambda_1$) by following the KKT conditions.



Now, the question arises when $K > 1$ then I can't obtain the closed-form solution, but I can compute $x$ analytically which is dependent on all the $lambda_i$. So, to compute the Lagrange multiplier say $lambda_i$, I resort to iterative solution where I fix other Lagrange multipliers ($lambda_j forall j = 1,cdots,K$ except $i$). Then repeat the above process for other Lagrange multipliers iteratively.




  • do you have any standard name for such scheme to compute Lagrange multipliers cyclically?

  • If not, can I say that this cyclic/iterative scheme is nothing but Coordinate Descent (or like)?


Thank you so much for your time in advance.







optimization convex-optimization numerical-optimization






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share|cite|improve this question













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edited Nov 18 at 8:27

























asked Nov 18 at 8:12









user550103

8911315




8911315








  • 1




    Depending on how you select $lambda$, this may be coordinate descent in the dual problem.
    – LinAlg
    Nov 18 at 13:53










  • thank you LinAlg!
    – user550103
    Nov 18 at 15:51














  • 1




    Depending on how you select $lambda$, this may be coordinate descent in the dual problem.
    – LinAlg
    Nov 18 at 13:53










  • thank you LinAlg!
    – user550103
    Nov 18 at 15:51








1




1




Depending on how you select $lambda$, this may be coordinate descent in the dual problem.
– LinAlg
Nov 18 at 13:53




Depending on how you select $lambda$, this may be coordinate descent in the dual problem.
– LinAlg
Nov 18 at 13:53












thank you LinAlg!
– user550103
Nov 18 at 15:51




thank you LinAlg!
– user550103
Nov 18 at 15:51















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