How can I solve $inf_{x in X} x^T(Ax + b) + sum_{i=1}^n x_i log(x_i)$ in a closed form depending on...












0












$begingroup$


I'm trying to solve the minimization problem



$$inf_{x in X} x^T(Ax + b) + sum_{i=1}^n x_i log(x_i)$$



where $b in mathbb{R}^n$, $X subset mathbb{R}^n$ is a convex set. And $A$ is a symmetric and positive definite matrix. The optimality conditions give me something like



$$-b in 2Ax + vec{1} + L(x) + N_X(x)$$



where $L(x)$ represents the vector in $mathbb{R}^n$ obtained by taking the elementwise logarithm of $x$; i.e., $[L(x)]_i=log x_i$.



The problem is how do I take that cone out of the inclusion. If there wasn't that log term it would become a projection on the set $X$. Is there a trick to make this as a projection on $X$ of other vector or something?



I'd appreciate any help or any paper/book reference where there's some information about this kind of problem










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  • 1




    $begingroup$
    There is no closed form. You have to solve this numerically, even if $Xequivmathbb{R}^n$.
    $endgroup$
    – Michael Grant
    Mar 23 '15 at 19:38












  • $begingroup$
    I see. In this case do you know where can I find info on numerically solving this?
    $endgroup$
    – karlabos
    Mar 23 '15 at 22:46










  • $begingroup$
    I would recommend looking into CVX (incidentally, it is partially authored by the previous commenter): cvxr.com/cvx.
    $endgroup$
    – ChrKroer
    Mar 24 '15 at 18:49
















0












$begingroup$


I'm trying to solve the minimization problem



$$inf_{x in X} x^T(Ax + b) + sum_{i=1}^n x_i log(x_i)$$



where $b in mathbb{R}^n$, $X subset mathbb{R}^n$ is a convex set. And $A$ is a symmetric and positive definite matrix. The optimality conditions give me something like



$$-b in 2Ax + vec{1} + L(x) + N_X(x)$$



where $L(x)$ represents the vector in $mathbb{R}^n$ obtained by taking the elementwise logarithm of $x$; i.e., $[L(x)]_i=log x_i$.



The problem is how do I take that cone out of the inclusion. If there wasn't that log term it would become a projection on the set $X$. Is there a trick to make this as a projection on $X$ of other vector or something?



I'd appreciate any help or any paper/book reference where there's some information about this kind of problem










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    There is no closed form. You have to solve this numerically, even if $Xequivmathbb{R}^n$.
    $endgroup$
    – Michael Grant
    Mar 23 '15 at 19:38












  • $begingroup$
    I see. In this case do you know where can I find info on numerically solving this?
    $endgroup$
    – karlabos
    Mar 23 '15 at 22:46










  • $begingroup$
    I would recommend looking into CVX (incidentally, it is partially authored by the previous commenter): cvxr.com/cvx.
    $endgroup$
    – ChrKroer
    Mar 24 '15 at 18:49














0












0








0





$begingroup$


I'm trying to solve the minimization problem



$$inf_{x in X} x^T(Ax + b) + sum_{i=1}^n x_i log(x_i)$$



where $b in mathbb{R}^n$, $X subset mathbb{R}^n$ is a convex set. And $A$ is a symmetric and positive definite matrix. The optimality conditions give me something like



$$-b in 2Ax + vec{1} + L(x) + N_X(x)$$



where $L(x)$ represents the vector in $mathbb{R}^n$ obtained by taking the elementwise logarithm of $x$; i.e., $[L(x)]_i=log x_i$.



The problem is how do I take that cone out of the inclusion. If there wasn't that log term it would become a projection on the set $X$. Is there a trick to make this as a projection on $X$ of other vector or something?



I'd appreciate any help or any paper/book reference where there's some information about this kind of problem










share|cite|improve this question











$endgroup$




I'm trying to solve the minimization problem



$$inf_{x in X} x^T(Ax + b) + sum_{i=1}^n x_i log(x_i)$$



where $b in mathbb{R}^n$, $X subset mathbb{R}^n$ is a convex set. And $A$ is a symmetric and positive definite matrix. The optimality conditions give me something like



$$-b in 2Ax + vec{1} + L(x) + N_X(x)$$



where $L(x)$ represents the vector in $mathbb{R}^n$ obtained by taking the elementwise logarithm of $x$; i.e., $[L(x)]_i=log x_i$.



The problem is how do I take that cone out of the inclusion. If there wasn't that log term it would become a projection on the set $X$. Is there a trick to make this as a projection on $X$ of other vector or something?



I'd appreciate any help or any paper/book reference where there's some information about this kind of problem







optimization convex-analysis convex-optimization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 23 '15 at 19:40









Michael Grant

15.2k12045




15.2k12045










asked Mar 23 '15 at 18:58









karlaboskarlabos

401414




401414








  • 1




    $begingroup$
    There is no closed form. You have to solve this numerically, even if $Xequivmathbb{R}^n$.
    $endgroup$
    – Michael Grant
    Mar 23 '15 at 19:38












  • $begingroup$
    I see. In this case do you know where can I find info on numerically solving this?
    $endgroup$
    – karlabos
    Mar 23 '15 at 22:46










  • $begingroup$
    I would recommend looking into CVX (incidentally, it is partially authored by the previous commenter): cvxr.com/cvx.
    $endgroup$
    – ChrKroer
    Mar 24 '15 at 18:49














  • 1




    $begingroup$
    There is no closed form. You have to solve this numerically, even if $Xequivmathbb{R}^n$.
    $endgroup$
    – Michael Grant
    Mar 23 '15 at 19:38












  • $begingroup$
    I see. In this case do you know where can I find info on numerically solving this?
    $endgroup$
    – karlabos
    Mar 23 '15 at 22:46










  • $begingroup$
    I would recommend looking into CVX (incidentally, it is partially authored by the previous commenter): cvxr.com/cvx.
    $endgroup$
    – ChrKroer
    Mar 24 '15 at 18:49








1




1




$begingroup$
There is no closed form. You have to solve this numerically, even if $Xequivmathbb{R}^n$.
$endgroup$
– Michael Grant
Mar 23 '15 at 19:38






$begingroup$
There is no closed form. You have to solve this numerically, even if $Xequivmathbb{R}^n$.
$endgroup$
– Michael Grant
Mar 23 '15 at 19:38














$begingroup$
I see. In this case do you know where can I find info on numerically solving this?
$endgroup$
– karlabos
Mar 23 '15 at 22:46




$begingroup$
I see. In this case do you know where can I find info on numerically solving this?
$endgroup$
– karlabos
Mar 23 '15 at 22:46












$begingroup$
I would recommend looking into CVX (incidentally, it is partially authored by the previous commenter): cvxr.com/cvx.
$endgroup$
– ChrKroer
Mar 24 '15 at 18:49




$begingroup$
I would recommend looking into CVX (incidentally, it is partially authored by the previous commenter): cvxr.com/cvx.
$endgroup$
– ChrKroer
Mar 24 '15 at 18:49










1 Answer
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$begingroup$

Starting from your inclusion
$$-b in 2Ax + vec{1} + L(x) + N_X(x),$$
we can add $x$ on both sides and re-arrange to get:



$$x-2Ax-vec{1}-L(x)-b in (I+N_X)(x),$$
where $I$ denotes the identity operator $xmapsto x$.
We can now invert to get
$$x = P_Xbig(x-2Ax-vec{1}-L(x)-bbig).$$
This reformulation takes the normal cone operator out of
your problem and converts it to a fixed point problem
that you might try to approach with fixed-point theoretic
algorithms.






share|cite|improve this answer









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    0












    $begingroup$

    Starting from your inclusion
    $$-b in 2Ax + vec{1} + L(x) + N_X(x),$$
    we can add $x$ on both sides and re-arrange to get:



    $$x-2Ax-vec{1}-L(x)-b in (I+N_X)(x),$$
    where $I$ denotes the identity operator $xmapsto x$.
    We can now invert to get
    $$x = P_Xbig(x-2Ax-vec{1}-L(x)-bbig).$$
    This reformulation takes the normal cone operator out of
    your problem and converts it to a fixed point problem
    that you might try to approach with fixed-point theoretic
    algorithms.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Starting from your inclusion
      $$-b in 2Ax + vec{1} + L(x) + N_X(x),$$
      we can add $x$ on both sides and re-arrange to get:



      $$x-2Ax-vec{1}-L(x)-b in (I+N_X)(x),$$
      where $I$ denotes the identity operator $xmapsto x$.
      We can now invert to get
      $$x = P_Xbig(x-2Ax-vec{1}-L(x)-bbig).$$
      This reformulation takes the normal cone operator out of
      your problem and converts it to a fixed point problem
      that you might try to approach with fixed-point theoretic
      algorithms.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Starting from your inclusion
        $$-b in 2Ax + vec{1} + L(x) + N_X(x),$$
        we can add $x$ on both sides and re-arrange to get:



        $$x-2Ax-vec{1}-L(x)-b in (I+N_X)(x),$$
        where $I$ denotes the identity operator $xmapsto x$.
        We can now invert to get
        $$x = P_Xbig(x-2Ax-vec{1}-L(x)-bbig).$$
        This reformulation takes the normal cone operator out of
        your problem and converts it to a fixed point problem
        that you might try to approach with fixed-point theoretic
        algorithms.






        share|cite|improve this answer









        $endgroup$



        Starting from your inclusion
        $$-b in 2Ax + vec{1} + L(x) + N_X(x),$$
        we can add $x$ on both sides and re-arrange to get:



        $$x-2Ax-vec{1}-L(x)-b in (I+N_X)(x),$$
        where $I$ denotes the identity operator $xmapsto x$.
        We can now invert to get
        $$x = P_Xbig(x-2Ax-vec{1}-L(x)-bbig).$$
        This reformulation takes the normal cone operator out of
        your problem and converts it to a fixed point problem
        that you might try to approach with fixed-point theoretic
        algorithms.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 31 at 22:03









        max_zornmax_zorn

        3,44061429




        3,44061429






























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