Duality gap in non-convex optimization: Do KKT conditions+constraint qualification imply strong duality? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Help me organize these concepts — KKT conditions and dual problemRecovering the solution of optimization problem from the dual problemStrong duality for nonconvex quadratic program (with multiple constraints)KKT conditions for a convex optimization problem with a L1-penalty and box constraintsDuality theory and nonlinear optimizationWhen do constraint qualifications imply strong duality?Convex optimization and strong dualityIs KKT conditions necessary and sufficient for any convex problems?KKT conditions for general conic optimization problemDuality gap in polynomial optimization problem Lasserre relaxation

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Duality gap in non-convex optimization: Do KKT conditions+constraint qualification imply strong duality?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Help me organize these concepts — KKT conditions and dual problemRecovering the solution of optimization problem from the dual problemStrong duality for nonconvex quadratic program (with multiple constraints)KKT conditions for a convex optimization problem with a L1-penalty and box constraintsDuality theory and nonlinear optimizationWhen do constraint qualifications imply strong duality?Convex optimization and strong dualityIs KKT conditions necessary and sufficient for any convex problems?KKT conditions for general conic optimization problemDuality gap in polynomial optimization problem Lasserre relaxation










0












$begingroup$


Consider the non-convex optimization problem:
$$undersetxin mathbbR^nmin ~f(x) \
mboxs.t.~h_i(x)=0 ~~~mboxfor~~~i=1,ldots,p \
~~~~~~ g_j(x)leq0 ~~~mboxfor~~~j=1,ldots,m$$

where $f(x),h_i(x)$ and $g_j(x)$ are continuously differential functions, but not necessarily convex. Let's say $barx$ is a global minimizer of the problem, and $barx$ is a KKT point (i.e., the KKT conditions hold at $barx$) with some constraint qualification (e.g: Mangasarian-Fromovitz constraint qualification holds at $barx$). What can be said about the duality gap (assuming that the dual solution exists and is attainable)? Does strong duality hold?










share|cite|improve this question









$endgroup$
















    0












    $begingroup$


    Consider the non-convex optimization problem:
    $$undersetxin mathbbR^nmin ~f(x) \
    mboxs.t.~h_i(x)=0 ~~~mboxfor~~~i=1,ldots,p \
    ~~~~~~ g_j(x)leq0 ~~~mboxfor~~~j=1,ldots,m$$

    where $f(x),h_i(x)$ and $g_j(x)$ are continuously differential functions, but not necessarily convex. Let's say $barx$ is a global minimizer of the problem, and $barx$ is a KKT point (i.e., the KKT conditions hold at $barx$) with some constraint qualification (e.g: Mangasarian-Fromovitz constraint qualification holds at $barx$). What can be said about the duality gap (assuming that the dual solution exists and is attainable)? Does strong duality hold?










    share|cite|improve this question









    $endgroup$














      0












      0








      0





      $begingroup$


      Consider the non-convex optimization problem:
      $$undersetxin mathbbR^nmin ~f(x) \
      mboxs.t.~h_i(x)=0 ~~~mboxfor~~~i=1,ldots,p \
      ~~~~~~ g_j(x)leq0 ~~~mboxfor~~~j=1,ldots,m$$

      where $f(x),h_i(x)$ and $g_j(x)$ are continuously differential functions, but not necessarily convex. Let's say $barx$ is a global minimizer of the problem, and $barx$ is a KKT point (i.e., the KKT conditions hold at $barx$) with some constraint qualification (e.g: Mangasarian-Fromovitz constraint qualification holds at $barx$). What can be said about the duality gap (assuming that the dual solution exists and is attainable)? Does strong duality hold?










      share|cite|improve this question









      $endgroup$




      Consider the non-convex optimization problem:
      $$undersetxin mathbbR^nmin ~f(x) \
      mboxs.t.~h_i(x)=0 ~~~mboxfor~~~i=1,ldots,p \
      ~~~~~~ g_j(x)leq0 ~~~mboxfor~~~j=1,ldots,m$$

      where $f(x),h_i(x)$ and $g_j(x)$ are continuously differential functions, but not necessarily convex. Let's say $barx$ is a global minimizer of the problem, and $barx$ is a KKT point (i.e., the KKT conditions hold at $barx$) with some constraint qualification (e.g: Mangasarian-Fromovitz constraint qualification holds at $barx$). What can be said about the duality gap (assuming that the dual solution exists and is attainable)? Does strong duality hold?







      optimization nonlinear-optimization duality-theorems non-convex-optimization






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      share|cite|improve this question











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      asked Apr 2 at 17:19









      Mateus de FreitasMateus de Freitas

      84




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          1 Answer
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          0












          $begingroup$

          Presume $f(x),h_i(x)$ and $g_j(x)$ are continuously differential function, then



          A constraint qualification is necessary to guarantee that local minimum implies KKT point. I.e., constraint qualification is required to make KKT conditions necessary for minimum.



          If one or more of $f(x),h_i(x)$ and $g_j(x)$ are non-convex, then (in general) strong duality does not necessarily hold. That's what makes non-convex optimization difficult.



          See https://en.wikipedia.org/wiki/Strong_duality






          share|cite|improve this answer









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            1 Answer
            1






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            active

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            0












            $begingroup$

            Presume $f(x),h_i(x)$ and $g_j(x)$ are continuously differential function, then



            A constraint qualification is necessary to guarantee that local minimum implies KKT point. I.e., constraint qualification is required to make KKT conditions necessary for minimum.



            If one or more of $f(x),h_i(x)$ and $g_j(x)$ are non-convex, then (in general) strong duality does not necessarily hold. That's what makes non-convex optimization difficult.



            See https://en.wikipedia.org/wiki/Strong_duality






            share|cite|improve this answer









            $endgroup$

















              0












              $begingroup$

              Presume $f(x),h_i(x)$ and $g_j(x)$ are continuously differential function, then



              A constraint qualification is necessary to guarantee that local minimum implies KKT point. I.e., constraint qualification is required to make KKT conditions necessary for minimum.



              If one or more of $f(x),h_i(x)$ and $g_j(x)$ are non-convex, then (in general) strong duality does not necessarily hold. That's what makes non-convex optimization difficult.



              See https://en.wikipedia.org/wiki/Strong_duality






              share|cite|improve this answer









              $endgroup$















                0












                0








                0





                $begingroup$

                Presume $f(x),h_i(x)$ and $g_j(x)$ are continuously differential function, then



                A constraint qualification is necessary to guarantee that local minimum implies KKT point. I.e., constraint qualification is required to make KKT conditions necessary for minimum.



                If one or more of $f(x),h_i(x)$ and $g_j(x)$ are non-convex, then (in general) strong duality does not necessarily hold. That's what makes non-convex optimization difficult.



                See https://en.wikipedia.org/wiki/Strong_duality






                share|cite|improve this answer









                $endgroup$



                Presume $f(x),h_i(x)$ and $g_j(x)$ are continuously differential function, then



                A constraint qualification is necessary to guarantee that local minimum implies KKT point. I.e., constraint qualification is required to make KKT conditions necessary for minimum.



                If one or more of $f(x),h_i(x)$ and $g_j(x)$ are non-convex, then (in general) strong duality does not necessarily hold. That's what makes non-convex optimization difficult.



                See https://en.wikipedia.org/wiki/Strong_duality







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Apr 3 at 0:36









                Mark L. StoneMark L. Stone

                1,99558




                1,99558



























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