Solve robust minimax optimization problem in two subsequent steps?












0












$begingroup$


I want to solve a robust optimization problem (worst-case optimization) of the form



$$
min_{x} max_q f(x,q) tag{1}
$$



with $x in mathbb{R}^n$ and $q subset mathbb{R}^m$ where $q_i in [underline{q}_i, overline{q}_i]$ and $underline{q}_i < overline{q}_i ,forall i dots m$. The variables in $x$ are the desicion variables, the $q$ are uncertain parameters (box-uncertainty).



Assume now we can analytically solve the following problem



$$
M(q) = min_x f(x, q) tag{2}
$$



and i.e. derive the minimum $M(q)$ of $f$ over $x$ as a function of the parameters $q$. Further assume I can then solve the following optimization problem



$$
max_{q} M(q). tag{3}
$$



Question: Is the solution to $(3)$, computed with the solution of $(2)$ also a solution of $(1)$? I.e., can I solve a problem like $(1)$ by first finding a parameter dependent minimizer for $x$ and then a parameter combination $q$ that is a maximizer for this minimum?






optimization convex-optimization linear-programming nonlinear-optimization







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I want to solve a robust optimization problem (worst-case optimization) of the form



    $$
    min_{x} max_q f(x,q) tag{1}
    $$



    with $x in mathbb{R}^n$ and $q subset mathbb{R}^m$ where $q_i in [underline{q}_i, overline{q}_i]$ and $underline{q}_i < overline{q}_i ,forall i dots m$. The variables in $x$ are the desicion variables, the $q$ are uncertain parameters (box-uncertainty).



    Assume now we can analytically solve the following problem



    $$
    M(q) = min_x f(x, q) tag{2}
    $$



    and i.e. derive the minimum $M(q)$ of $f$ over $x$ as a function of the parameters $q$. Further assume I can then solve the following optimization problem



    $$
    max_{q} M(q). tag{3}
    $$



    Question: Is the solution to $(3)$, computed with the solution of $(2)$ also a solution of $(1)$? I.e., can I solve a problem like $(1)$ by first finding a parameter dependent minimizer for $x$ and then a parameter combination $q$ that is a maximizer for this minimum?






    optimization convex-optimization linear-programming nonlinear-optimization







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I want to solve a robust optimization problem (worst-case optimization) of the form



      $$
      min_{x} max_q f(x,q) tag{1}
      $$



      with $x in mathbb{R}^n$ and $q subset mathbb{R}^m$ where $q_i in [underline{q}_i, overline{q}_i]$ and $underline{q}_i < overline{q}_i ,forall i dots m$. The variables in $x$ are the desicion variables, the $q$ are uncertain parameters (box-uncertainty).



      Assume now we can analytically solve the following problem



      $$
      M(q) = min_x f(x, q) tag{2}
      $$



      and i.e. derive the minimum $M(q)$ of $f$ over $x$ as a function of the parameters $q$. Further assume I can then solve the following optimization problem



      $$
      max_{q} M(q). tag{3}
      $$



      Question: Is the solution to $(3)$, computed with the solution of $(2)$ also a solution of $(1)$? I.e., can I solve a problem like $(1)$ by first finding a parameter dependent minimizer for $x$ and then a parameter combination $q$ that is a maximizer for this minimum?






      optimization convex-optimization linear-programming nonlinear-optimization







      share|cite|improve this question









      $endgroup$




      I want to solve a robust optimization problem (worst-case optimization) of the form



      $$
      min_{x} max_q f(x,q) tag{1}
      $$



      with $x in mathbb{R}^n$ and $q subset mathbb{R}^m$ where $q_i in [underline{q}_i, overline{q}_i]$ and $underline{q}_i < overline{q}_i ,forall i dots m$. The variables in $x$ are the desicion variables, the $q$ are uncertain parameters (box-uncertainty).



      Assume now we can analytically solve the following problem



      $$
      M(q) = min_x f(x, q) tag{2}
      $$



      and i.e. derive the minimum $M(q)$ of $f$ over $x$ as a function of the parameters $q$. Further assume I can then solve the following optimization problem



      $$
      max_{q} M(q). tag{3}
      $$



      Question: Is the solution to $(3)$, computed with the solution of $(2)$ also a solution of $(1)$? I.e., can I solve a problem like $(1)$ by first finding a parameter dependent minimizer for $x$ and then a parameter combination $q$ that is a maximizer for this minimum?






      optimization convex-optimization linear-programming nonlinear-optimization




      optimization convex-optimization linear-programming nonlinear-optimization






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 31 '18 at 13:07









      SampleTimeSampleTime

      55139




      55139






















          1 Answer
          1






          active

          oldest

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          1












          $begingroup$

          According to the minimax theorem, if $f$ is continuous function which is concave in $q$ and convex in $x$ (roughly speaking), then
          $$
          min_xmax_q f(x,q) =max_qmin_x f(x,q)=max_q M(q)
          $$
          holds. Hence in this case, your method can solve the problem. However, in general cases, we can say at most
          $$
          min_xmax_q f(x,q) gemax_qmin_x f(x,q),
          $$           and equality may not hold. In this case, your method does not give the solution.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            note that the minimax theorem only applies if the domain of either $x$ or $q$ is compact (which holds for $q$ here), if both domains are convex, and if the function is lower semicontinuous & quasiconvex in $x$ and upper semicontinuous & quasiconcave in $q$
            $endgroup$
            – LinAlg
            Dec 31 '18 at 14:10












          • $begingroup$
            @LinAlg Thanks. One more question, what if $f$ is both concave and convex (i.e. linear) in $q$?
            $endgroup$
            – SampleTime
            Dec 31 '18 at 14:18






          • 1




            $begingroup$
            @SampleTime then it is concave in $q$ and satisfies that part of the conditions.
            $endgroup$
            – LinAlg
            Dec 31 '18 at 14:19











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          According to the minimax theorem, if $f$ is continuous function which is concave in $q$ and convex in $x$ (roughly speaking), then
          $$
          min_xmax_q f(x,q) =max_qmin_x f(x,q)=max_q M(q)
          $$
          holds. Hence in this case, your method can solve the problem. However, in general cases, we can say at most
          $$
          min_xmax_q f(x,q) gemax_qmin_x f(x,q),
          $$           and equality may not hold. In this case, your method does not give the solution.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            note that the minimax theorem only applies if the domain of either $x$ or $q$ is compact (which holds for $q$ here), if both domains are convex, and if the function is lower semicontinuous & quasiconvex in $x$ and upper semicontinuous & quasiconcave in $q$
            $endgroup$
            – LinAlg
            Dec 31 '18 at 14:10












          • $begingroup$
            @LinAlg Thanks. One more question, what if $f$ is both concave and convex (i.e. linear) in $q$?
            $endgroup$
            – SampleTime
            Dec 31 '18 at 14:18






          • 1




            $begingroup$
            @SampleTime then it is concave in $q$ and satisfies that part of the conditions.
            $endgroup$
            – LinAlg
            Dec 31 '18 at 14:19
















          1












          $begingroup$

          According to the minimax theorem, if $f$ is continuous function which is concave in $q$ and convex in $x$ (roughly speaking), then
          $$
          min_xmax_q f(x,q) =max_qmin_x f(x,q)=max_q M(q)
          $$
          holds. Hence in this case, your method can solve the problem. However, in general cases, we can say at most
          $$
          min_xmax_q f(x,q) gemax_qmin_x f(x,q),
          $$           and equality may not hold. In this case, your method does not give the solution.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            note that the minimax theorem only applies if the domain of either $x$ or $q$ is compact (which holds for $q$ here), if both domains are convex, and if the function is lower semicontinuous & quasiconvex in $x$ and upper semicontinuous & quasiconcave in $q$
            $endgroup$
            – LinAlg
            Dec 31 '18 at 14:10












          • $begingroup$
            @LinAlg Thanks. One more question, what if $f$ is both concave and convex (i.e. linear) in $q$?
            $endgroup$
            – SampleTime
            Dec 31 '18 at 14:18






          • 1




            $begingroup$
            @SampleTime then it is concave in $q$ and satisfies that part of the conditions.
            $endgroup$
            – LinAlg
            Dec 31 '18 at 14:19














          1












          1








          1





          $begingroup$

          According to the minimax theorem, if $f$ is continuous function which is concave in $q$ and convex in $x$ (roughly speaking), then
          $$
          min_xmax_q f(x,q) =max_qmin_x f(x,q)=max_q M(q)
          $$
          holds. Hence in this case, your method can solve the problem. However, in general cases, we can say at most
          $$
          min_xmax_q f(x,q) gemax_qmin_x f(x,q),
          $$           and equality may not hold. In this case, your method does not give the solution.






          share|cite|improve this answer









          $endgroup$



          According to the minimax theorem, if $f$ is continuous function which is concave in $q$ and convex in $x$ (roughly speaking), then
          $$
          min_xmax_q f(x,q) =max_qmin_x f(x,q)=max_q M(q)
          $$
          holds. Hence in this case, your method can solve the problem. However, in general cases, we can say at most
          $$
          min_xmax_q f(x,q) gemax_qmin_x f(x,q),
          $$           and equality may not hold. In this case, your method does not give the solution.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 31 '18 at 13:22









          SongSong

          18.6k21651




          18.6k21651








          • 1




            $begingroup$
            note that the minimax theorem only applies if the domain of either $x$ or $q$ is compact (which holds for $q$ here), if both domains are convex, and if the function is lower semicontinuous & quasiconvex in $x$ and upper semicontinuous & quasiconcave in $q$
            $endgroup$
            – LinAlg
            Dec 31 '18 at 14:10












          • $begingroup$
            @LinAlg Thanks. One more question, what if $f$ is both concave and convex (i.e. linear) in $q$?
            $endgroup$
            – SampleTime
            Dec 31 '18 at 14:18






          • 1




            $begingroup$
            @SampleTime then it is concave in $q$ and satisfies that part of the conditions.
            $endgroup$
            – LinAlg
            Dec 31 '18 at 14:19














          • 1




            $begingroup$
            note that the minimax theorem only applies if the domain of either $x$ or $q$ is compact (which holds for $q$ here), if both domains are convex, and if the function is lower semicontinuous & quasiconvex in $x$ and upper semicontinuous & quasiconcave in $q$
            $endgroup$
            – LinAlg
            Dec 31 '18 at 14:10












          • $begingroup$
            @LinAlg Thanks. One more question, what if $f$ is both concave and convex (i.e. linear) in $q$?
            $endgroup$
            – SampleTime
            Dec 31 '18 at 14:18






          • 1




            $begingroup$
            @SampleTime then it is concave in $q$ and satisfies that part of the conditions.
            $endgroup$
            – LinAlg
            Dec 31 '18 at 14:19








          1




          1




          $begingroup$
          note that the minimax theorem only applies if the domain of either $x$ or $q$ is compact (which holds for $q$ here), if both domains are convex, and if the function is lower semicontinuous & quasiconvex in $x$ and upper semicontinuous & quasiconcave in $q$
          $endgroup$
          – LinAlg
          Dec 31 '18 at 14:10






          $begingroup$
          note that the minimax theorem only applies if the domain of either $x$ or $q$ is compact (which holds for $q$ here), if both domains are convex, and if the function is lower semicontinuous & quasiconvex in $x$ and upper semicontinuous & quasiconcave in $q$
          $endgroup$
          – LinAlg
          Dec 31 '18 at 14:10














          $begingroup$
          @LinAlg Thanks. One more question, what if $f$ is both concave and convex (i.e. linear) in $q$?
          $endgroup$
          – SampleTime
          Dec 31 '18 at 14:18




          $begingroup$
          @LinAlg Thanks. One more question, what if $f$ is both concave and convex (i.e. linear) in $q$?
          $endgroup$
          – SampleTime
          Dec 31 '18 at 14:18




          1




          1




          $begingroup$
          @SampleTime then it is concave in $q$ and satisfies that part of the conditions.
          $endgroup$
          – LinAlg
          Dec 31 '18 at 14:19




          $begingroup$
          @SampleTime then it is concave in $q$ and satisfies that part of the conditions.
          $endgroup$
          – LinAlg
          Dec 31 '18 at 14:19


















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