Finding the Dual of an LP, where variables are not well isolated.











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Could someone guide me through how I could go about finding the dual of the following LP. Let $A subset Bbb R^{m x n}$, and $x = (x_1, ..., x_m)$.



$$text{max } z
\
text{st } z le x^TA_{.j} , forall j in {1, ldots, m}
\
x_1 + ldots + x_m = 1
\
x ge vec{0}
$$



, where $A_{.j}$ represents the j'th column of A.



My confusion is coming from the fact that in this case, $A$ is a constant matrix, but all the $x_i$'s are variables; as is z.
So I can't see how I can apply the standard primal-dual conversion.



Thanks!






optimization linear-programming duality-theorems







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    up vote
    0
    down vote

    favorite












    Could someone guide me through how I could go about finding the dual of the following LP. Let $A subset Bbb R^{m x n}$, and $x = (x_1, ..., x_m)$.



    $$text{max } z
    \
    text{st } z le x^TA_{.j} , forall j in {1, ldots, m}
    \
    x_1 + ldots + x_m = 1
    \
    x ge vec{0}
    $$



    , where $A_{.j}$ represents the j'th column of A.



    My confusion is coming from the fact that in this case, $A$ is a constant matrix, but all the $x_i$'s are variables; as is z.
    So I can't see how I can apply the standard primal-dual conversion.



    Thanks!






    optimization linear-programming duality-theorems







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Could someone guide me through how I could go about finding the dual of the following LP. Let $A subset Bbb R^{m x n}$, and $x = (x_1, ..., x_m)$.



      $$text{max } z
      \
      text{st } z le x^TA_{.j} , forall j in {1, ldots, m}
      \
      x_1 + ldots + x_m = 1
      \
      x ge vec{0}
      $$



      , where $A_{.j}$ represents the j'th column of A.



      My confusion is coming from the fact that in this case, $A$ is a constant matrix, but all the $x_i$'s are variables; as is z.
      So I can't see how I can apply the standard primal-dual conversion.



      Thanks!






      optimization linear-programming duality-theorems







      share|cite|improve this question













      Could someone guide me through how I could go about finding the dual of the following LP. Let $A subset Bbb R^{m x n}$, and $x = (x_1, ..., x_m)$.



      $$text{max } z
      \
      text{st } z le x^TA_{.j} , forall j in {1, ldots, m}
      \
      x_1 + ldots + x_m = 1
      \
      x ge vec{0}
      $$



      , where $A_{.j}$ represents the j'th column of A.



      My confusion is coming from the fact that in this case, $A$ is a constant matrix, but all the $x_i$'s are variables; as is z.
      So I can't see how I can apply the standard primal-dual conversion.



      Thanks!






      optimization linear-programming duality-theorems




      optimization linear-programming duality-theorems






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      asked 2 days ago









      Bojack Horseman

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          You could define $x'=begin{pmatrix}x \ zend{pmatrix}$, then the problem is:
          $$begin{align}
          max quad & begin{pmatrix}0 \ 0 \ vdots \ 0 \ 1end{pmatrix}^T x' \
          text{s.t.} quad & begin{pmatrix} A^T & -e \ e^T & 0 \ -e^T & 0end{pmatrix}x' leq begin{pmatrix} 0 \ 0 \ vdots \ 0 \ 1 \ -1end{pmatrix} \
          & x' geq 0.
          end{align}$$
          Deriving the dual is now simple.






          share|cite|improve this answer





















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            up vote
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            accepted










            You could define $x'=begin{pmatrix}x \ zend{pmatrix}$, then the problem is:
            $$begin{align}
            max quad & begin{pmatrix}0 \ 0 \ vdots \ 0 \ 1end{pmatrix}^T x' \
            text{s.t.} quad & begin{pmatrix} A^T & -e \ e^T & 0 \ -e^T & 0end{pmatrix}x' leq begin{pmatrix} 0 \ 0 \ vdots \ 0 \ 1 \ -1end{pmatrix} \
            & x' geq 0.
            end{align}$$
            Deriving the dual is now simple.






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              You could define $x'=begin{pmatrix}x \ zend{pmatrix}$, then the problem is:
              $$begin{align}
              max quad & begin{pmatrix}0 \ 0 \ vdots \ 0 \ 1end{pmatrix}^T x' \
              text{s.t.} quad & begin{pmatrix} A^T & -e \ e^T & 0 \ -e^T & 0end{pmatrix}x' leq begin{pmatrix} 0 \ 0 \ vdots \ 0 \ 1 \ -1end{pmatrix} \
              & x' geq 0.
              end{align}$$
              Deriving the dual is now simple.






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                You could define $x'=begin{pmatrix}x \ zend{pmatrix}$, then the problem is:
                $$begin{align}
                max quad & begin{pmatrix}0 \ 0 \ vdots \ 0 \ 1end{pmatrix}^T x' \
                text{s.t.} quad & begin{pmatrix} A^T & -e \ e^T & 0 \ -e^T & 0end{pmatrix}x' leq begin{pmatrix} 0 \ 0 \ vdots \ 0 \ 1 \ -1end{pmatrix} \
                & x' geq 0.
                end{align}$$
                Deriving the dual is now simple.






                share|cite|improve this answer












                You could define $x'=begin{pmatrix}x \ zend{pmatrix}$, then the problem is:
                $$begin{align}
                max quad & begin{pmatrix}0 \ 0 \ vdots \ 0 \ 1end{pmatrix}^T x' \
                text{s.t.} quad & begin{pmatrix} A^T & -e \ e^T & 0 \ -e^T & 0end{pmatrix}x' leq begin{pmatrix} 0 \ 0 \ vdots \ 0 \ 1 \ -1end{pmatrix} \
                & x' geq 0.
                end{align}$$
                Deriving the dual is now simple.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                LinAlg

                7,6141520




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