How to convert constrainted optimization problem to an unconstrained one?












2












$begingroup$


Suppose I have an optimization problem



$$ mathbf{w}^* = max_{mathbf{w}} sum_{i=1}^{N} (y_i - mathbf{w}cdotmathbf{x}_i)^2 $$
$$text{subject to}$$
$$sum_{j=1}^m w_j^2 leq tau$$



I want to show this can be written as



$$ mathbf{w}^* = max_{mathbf{w}} sum_{i=1}^{N} (y_i - mathbf{w}cdotmathbf{x}_i)^2 + lambda sum_{j=1}^m w_j^2$$



What is the basic intuition behind this?



I have done basic Lagrangian multiplier problems like



$$max_{mathbf{x}} U(mathbf{x})$$
$$text{subject to}$$
$$mathbf{x}cdot mathbf{w} = tau$$



In those cases, I have written a Lagrangian like



$$L(mathbf{x},lambda) = U(mathbf{x}) + lambda (m - mathbf{x}cdot mathbf{w})$$



But there's two issues:




  1. A Lagrangian isn't itself something I minimize, correct? I am trying to rewrite my constrained problem as an unconstrained problem.

  2. There is an inequality, not an equality. I assume the Kuhn-Tucker conditions are relevant to this, but I am not sure how.






optimization lagrange-multiplier







share|cite|improve this question











$endgroup$












  • $begingroup$
    if an optimal solution to your original problem exists, what you want to show follows trivially from Lagrange duality (although determining $lambda$ is not trivial).
    $endgroup$
    – LinAlg
    Oct 19 '18 at 0:02










  • $begingroup$
    @LinAlg What information could I add to this post that would make it more answerable but that also wouldn't be like posting the original problem?
    $endgroup$
    – Stan Shunpike
    Oct 19 '18 at 1:31












  • $begingroup$
    I'm afraid I do not know what you are looking for. Lagrange duality means $min_x max_lambda L(x,lambda) = max_lambda min_x L(x,lambda)$, and you are rewriting the formulation on the left hand side to the one on the right hand side.
    $endgroup$
    – LinAlg
    Oct 19 '18 at 1:57










  • $begingroup$
    I've updated the question with a related one that will hopefully be more detailed.
    $endgroup$
    – Stan Shunpike
    Oct 19 '18 at 6:32
















2












$begingroup$


Suppose I have an optimization problem



$$ mathbf{w}^* = max_{mathbf{w}} sum_{i=1}^{N} (y_i - mathbf{w}cdotmathbf{x}_i)^2 $$
$$text{subject to}$$
$$sum_{j=1}^m w_j^2 leq tau$$



I want to show this can be written as



$$ mathbf{w}^* = max_{mathbf{w}} sum_{i=1}^{N} (y_i - mathbf{w}cdotmathbf{x}_i)^2 + lambda sum_{j=1}^m w_j^2$$



What is the basic intuition behind this?



I have done basic Lagrangian multiplier problems like



$$max_{mathbf{x}} U(mathbf{x})$$
$$text{subject to}$$
$$mathbf{x}cdot mathbf{w} = tau$$



In those cases, I have written a Lagrangian like



$$L(mathbf{x},lambda) = U(mathbf{x}) + lambda (m - mathbf{x}cdot mathbf{w})$$



But there's two issues:




  1. A Lagrangian isn't itself something I minimize, correct? I am trying to rewrite my constrained problem as an unconstrained problem.

  2. There is an inequality, not an equality. I assume the Kuhn-Tucker conditions are relevant to this, but I am not sure how.






optimization lagrange-multiplier







share|cite|improve this question











$endgroup$












  • $begingroup$
    if an optimal solution to your original problem exists, what you want to show follows trivially from Lagrange duality (although determining $lambda$ is not trivial).
    $endgroup$
    – LinAlg
    Oct 19 '18 at 0:02










  • $begingroup$
    @LinAlg What information could I add to this post that would make it more answerable but that also wouldn't be like posting the original problem?
    $endgroup$
    – Stan Shunpike
    Oct 19 '18 at 1:31












  • $begingroup$
    I'm afraid I do not know what you are looking for. Lagrange duality means $min_x max_lambda L(x,lambda) = max_lambda min_x L(x,lambda)$, and you are rewriting the formulation on the left hand side to the one on the right hand side.
    $endgroup$
    – LinAlg
    Oct 19 '18 at 1:57










  • $begingroup$
    I've updated the question with a related one that will hopefully be more detailed.
    $endgroup$
    – Stan Shunpike
    Oct 19 '18 at 6:32














2












2








2


1



$begingroup$


Suppose I have an optimization problem



$$ mathbf{w}^* = max_{mathbf{w}} sum_{i=1}^{N} (y_i - mathbf{w}cdotmathbf{x}_i)^2 $$
$$text{subject to}$$
$$sum_{j=1}^m w_j^2 leq tau$$



I want to show this can be written as



$$ mathbf{w}^* = max_{mathbf{w}} sum_{i=1}^{N} (y_i - mathbf{w}cdotmathbf{x}_i)^2 + lambda sum_{j=1}^m w_j^2$$



What is the basic intuition behind this?



I have done basic Lagrangian multiplier problems like



$$max_{mathbf{x}} U(mathbf{x})$$
$$text{subject to}$$
$$mathbf{x}cdot mathbf{w} = tau$$



In those cases, I have written a Lagrangian like



$$L(mathbf{x},lambda) = U(mathbf{x}) + lambda (m - mathbf{x}cdot mathbf{w})$$



But there's two issues:




  1. A Lagrangian isn't itself something I minimize, correct? I am trying to rewrite my constrained problem as an unconstrained problem.

  2. There is an inequality, not an equality. I assume the Kuhn-Tucker conditions are relevant to this, but I am not sure how.






optimization lagrange-multiplier







share|cite|improve this question











$endgroup$




Suppose I have an optimization problem



$$ mathbf{w}^* = max_{mathbf{w}} sum_{i=1}^{N} (y_i - mathbf{w}cdotmathbf{x}_i)^2 $$
$$text{subject to}$$
$$sum_{j=1}^m w_j^2 leq tau$$



I want to show this can be written as



$$ mathbf{w}^* = max_{mathbf{w}} sum_{i=1}^{N} (y_i - mathbf{w}cdotmathbf{x}_i)^2 + lambda sum_{j=1}^m w_j^2$$



What is the basic intuition behind this?



I have done basic Lagrangian multiplier problems like



$$max_{mathbf{x}} U(mathbf{x})$$
$$text{subject to}$$
$$mathbf{x}cdot mathbf{w} = tau$$



In those cases, I have written a Lagrangian like



$$L(mathbf{x},lambda) = U(mathbf{x}) + lambda (m - mathbf{x}cdot mathbf{w})$$



But there's two issues:




  1. A Lagrangian isn't itself something I minimize, correct? I am trying to rewrite my constrained problem as an unconstrained problem.

  2. There is an inequality, not an equality. I assume the Kuhn-Tucker conditions are relevant to this, but I am not sure how.






optimization lagrange-multiplier




optimization lagrange-multiplier






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 19 '18 at 6:31







Stan Shunpike

















asked Oct 18 '18 at 21:49









Stan ShunpikeStan Shunpike

1,81111438




1,81111438












  • $begingroup$
    if an optimal solution to your original problem exists, what you want to show follows trivially from Lagrange duality (although determining $lambda$ is not trivial).
    $endgroup$
    – LinAlg
    Oct 19 '18 at 0:02










  • $begingroup$
    @LinAlg What information could I add to this post that would make it more answerable but that also wouldn't be like posting the original problem?
    $endgroup$
    – Stan Shunpike
    Oct 19 '18 at 1:31












  • $begingroup$
    I'm afraid I do not know what you are looking for. Lagrange duality means $min_x max_lambda L(x,lambda) = max_lambda min_x L(x,lambda)$, and you are rewriting the formulation on the left hand side to the one on the right hand side.
    $endgroup$
    – LinAlg
    Oct 19 '18 at 1:57










  • $begingroup$
    I've updated the question with a related one that will hopefully be more detailed.
    $endgroup$
    – Stan Shunpike
    Oct 19 '18 at 6:32


















  • $begingroup$
    if an optimal solution to your original problem exists, what you want to show follows trivially from Lagrange duality (although determining $lambda$ is not trivial).
    $endgroup$
    – LinAlg
    Oct 19 '18 at 0:02










  • $begingroup$
    @LinAlg What information could I add to this post that would make it more answerable but that also wouldn't be like posting the original problem?
    $endgroup$
    – Stan Shunpike
    Oct 19 '18 at 1:31












  • $begingroup$
    I'm afraid I do not know what you are looking for. Lagrange duality means $min_x max_lambda L(x,lambda) = max_lambda min_x L(x,lambda)$, and you are rewriting the formulation on the left hand side to the one on the right hand side.
    $endgroup$
    – LinAlg
    Oct 19 '18 at 1:57










  • $begingroup$
    I've updated the question with a related one that will hopefully be more detailed.
    $endgroup$
    – Stan Shunpike
    Oct 19 '18 at 6:32
















$begingroup$
if an optimal solution to your original problem exists, what you want to show follows trivially from Lagrange duality (although determining $lambda$ is not trivial).
$endgroup$
– LinAlg
Oct 19 '18 at 0:02




$begingroup$
if an optimal solution to your original problem exists, what you want to show follows trivially from Lagrange duality (although determining $lambda$ is not trivial).
$endgroup$
– LinAlg
Oct 19 '18 at 0:02












$begingroup$
@LinAlg What information could I add to this post that would make it more answerable but that also wouldn't be like posting the original problem?
$endgroup$
– Stan Shunpike
Oct 19 '18 at 1:31






$begingroup$
@LinAlg What information could I add to this post that would make it more answerable but that also wouldn't be like posting the original problem?
$endgroup$
– Stan Shunpike
Oct 19 '18 at 1:31














$begingroup$
I'm afraid I do not know what you are looking for. Lagrange duality means $min_x max_lambda L(x,lambda) = max_lambda min_x L(x,lambda)$, and you are rewriting the formulation on the left hand side to the one on the right hand side.
$endgroup$
– LinAlg
Oct 19 '18 at 1:57




$begingroup$
I'm afraid I do not know what you are looking for. Lagrange duality means $min_x max_lambda L(x,lambda) = max_lambda min_x L(x,lambda)$, and you are rewriting the formulation on the left hand side to the one on the right hand side.
$endgroup$
– LinAlg
Oct 19 '18 at 1:57












$begingroup$
I've updated the question with a related one that will hopefully be more detailed.
$endgroup$
– Stan Shunpike
Oct 19 '18 at 6:32




$begingroup$
I've updated the question with a related one that will hopefully be more detailed.
$endgroup$
– Stan Shunpike
Oct 19 '18 at 6:32










1 Answer
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$begingroup$

Introducing the slack variable $epsilon$ you can write the unconstrained equivalent optimization problem with the Lagrangian



$$
L(mathbf{w},lambda,epsilon) = sum_{i=1}^{N} (y_i - mathbf{w}cdotmathbf{x}_i)^2 + lambda left(sum_{j=1}^m w_j^2-tau + epsilon^2right)
$$



NOTES



1) Solving the lagrangian we will have it's stationary points which should be qualified.



2) Analyzing the the values for $epsilon^*, lambda^*$ we can conclude the K-T conditions.






share|cite|improve this answer









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    1 Answer
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    1 Answer
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    $begingroup$

    Introducing the slack variable $epsilon$ you can write the unconstrained equivalent optimization problem with the Lagrangian



    $$
    L(mathbf{w},lambda,epsilon) = sum_{i=1}^{N} (y_i - mathbf{w}cdotmathbf{x}_i)^2 + lambda left(sum_{j=1}^m w_j^2-tau + epsilon^2right)
    $$



    NOTES



    1) Solving the lagrangian we will have it's stationary points which should be qualified.



    2) Analyzing the the values for $epsilon^*, lambda^*$ we can conclude the K-T conditions.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Introducing the slack variable $epsilon$ you can write the unconstrained equivalent optimization problem with the Lagrangian



      $$
      L(mathbf{w},lambda,epsilon) = sum_{i=1}^{N} (y_i - mathbf{w}cdotmathbf{x}_i)^2 + lambda left(sum_{j=1}^m w_j^2-tau + epsilon^2right)
      $$



      NOTES



      1) Solving the lagrangian we will have it's stationary points which should be qualified.



      2) Analyzing the the values for $epsilon^*, lambda^*$ we can conclude the K-T conditions.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Introducing the slack variable $epsilon$ you can write the unconstrained equivalent optimization problem with the Lagrangian



        $$
        L(mathbf{w},lambda,epsilon) = sum_{i=1}^{N} (y_i - mathbf{w}cdotmathbf{x}_i)^2 + lambda left(sum_{j=1}^m w_j^2-tau + epsilon^2right)
        $$



        NOTES



        1) Solving the lagrangian we will have it's stationary points which should be qualified.



        2) Analyzing the the values for $epsilon^*, lambda^*$ we can conclude the K-T conditions.






        share|cite|improve this answer









        $endgroup$



        Introducing the slack variable $epsilon$ you can write the unconstrained equivalent optimization problem with the Lagrangian



        $$
        L(mathbf{w},lambda,epsilon) = sum_{i=1}^{N} (y_i - mathbf{w}cdotmathbf{x}_i)^2 + lambda left(sum_{j=1}^m w_j^2-tau + epsilon^2right)
        $$



        NOTES



        1) Solving the lagrangian we will have it's stationary points which should be qualified.



        2) Analyzing the the values for $epsilon^*, lambda^*$ we can conclude the K-T conditions.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 17:04









        CesareoCesareo

        8,6393516




        8,6393516






























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