why is $sumlimits_{i=1}^{n}v_ix_i(1-x_i)=v^Tx+x^T diag(v)x?$ and its dual function












2












$begingroup$


I saw the solution of this question,but i have some problem



Q: min$_x c^T mathbf x$



$s.t. mathbf A mathbf x le mathbf b,mathbf x_i(1-mathbf x_i)=0,i=1,...,n,$ where $mathbf x =[x_1,...,x_n]^T$,find its dual function



Solution:



begin{align}
L(x, mu ,v) & = c^Tx+mu ^T(Ax-b)-sumlimits_{i=1}^{n}v_ix_i(1-x_i) \ & =c^Tx+mu ^T(Ax-b)-v^Tx+x^T diag(v)x \
end{align}



then minimizing over $x $ gives the dual function



$g(mu,v)=-b^Tmu - frac{1}{4} sumlimits_{i=1}^{n}(frac{c_i+a_i^T mu -v_i^2}{v_i})$,when $v ge 0$ ,otherwise,it is $-infty$



I want to ask



$1.$ why is $sumlimits_{i=1}^{n}v_ix_i(1-x_i)=v^Tx+x^T diag(v)x?$ it seems that $sumlimits_{i=1}^{n}v_i(1-x_i)=x^T diag(v)x, $ why?



$2.$Why is $ L(x, mu ,v)=g(mu,v)=-b^Tmu - frac{1}{4} sumlimits_{i=1}^{n}(frac{c_i+a_i^T mu -v_i^2}{v_i})$,when $v ge 0$






convex-analysis convex-optimization duality-theorems







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I saw the solution of this question,but i have some problem



    Q: min$_x c^T mathbf x$



    $s.t. mathbf A mathbf x le mathbf b,mathbf x_i(1-mathbf x_i)=0,i=1,...,n,$ where $mathbf x =[x_1,...,x_n]^T$,find its dual function



    Solution:



    begin{align}
    L(x, mu ,v) & = c^Tx+mu ^T(Ax-b)-sumlimits_{i=1}^{n}v_ix_i(1-x_i) \ & =c^Tx+mu ^T(Ax-b)-v^Tx+x^T diag(v)x \
    end{align}



    then minimizing over $x $ gives the dual function



    $g(mu,v)=-b^Tmu - frac{1}{4} sumlimits_{i=1}^{n}(frac{c_i+a_i^T mu -v_i^2}{v_i})$,when $v ge 0$ ,otherwise,it is $-infty$



    I want to ask



    $1.$ why is $sumlimits_{i=1}^{n}v_ix_i(1-x_i)=v^Tx+x^T diag(v)x?$ it seems that $sumlimits_{i=1}^{n}v_i(1-x_i)=x^T diag(v)x, $ why?



    $2.$Why is $ L(x, mu ,v)=g(mu,v)=-b^Tmu - frac{1}{4} sumlimits_{i=1}^{n}(frac{c_i+a_i^T mu -v_i^2}{v_i})$,when $v ge 0$






    convex-analysis convex-optimization duality-theorems







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I saw the solution of this question,but i have some problem



      Q: min$_x c^T mathbf x$



      $s.t. mathbf A mathbf x le mathbf b,mathbf x_i(1-mathbf x_i)=0,i=1,...,n,$ where $mathbf x =[x_1,...,x_n]^T$,find its dual function



      Solution:



      begin{align}
      L(x, mu ,v) & = c^Tx+mu ^T(Ax-b)-sumlimits_{i=1}^{n}v_ix_i(1-x_i) \ & =c^Tx+mu ^T(Ax-b)-v^Tx+x^T diag(v)x \
      end{align}



      then minimizing over $x $ gives the dual function



      $g(mu,v)=-b^Tmu - frac{1}{4} sumlimits_{i=1}^{n}(frac{c_i+a_i^T mu -v_i^2}{v_i})$,when $v ge 0$ ,otherwise,it is $-infty$



      I want to ask



      $1.$ why is $sumlimits_{i=1}^{n}v_ix_i(1-x_i)=v^Tx+x^T diag(v)x?$ it seems that $sumlimits_{i=1}^{n}v_i(1-x_i)=x^T diag(v)x, $ why?



      $2.$Why is $ L(x, mu ,v)=g(mu,v)=-b^Tmu - frac{1}{4} sumlimits_{i=1}^{n}(frac{c_i+a_i^T mu -v_i^2}{v_i})$,when $v ge 0$






      convex-analysis convex-optimization duality-theorems







      share|cite|improve this question











      $endgroup$




      I saw the solution of this question,but i have some problem



      Q: min$_x c^T mathbf x$



      $s.t. mathbf A mathbf x le mathbf b,mathbf x_i(1-mathbf x_i)=0,i=1,...,n,$ where $mathbf x =[x_1,...,x_n]^T$,find its dual function



      Solution:



      begin{align}
      L(x, mu ,v) & = c^Tx+mu ^T(Ax-b)-sumlimits_{i=1}^{n}v_ix_i(1-x_i) \ & =c^Tx+mu ^T(Ax-b)-v^Tx+x^T diag(v)x \
      end{align}



      then minimizing over $x $ gives the dual function



      $g(mu,v)=-b^Tmu - frac{1}{4} sumlimits_{i=1}^{n}(frac{c_i+a_i^T mu -v_i^2}{v_i})$,when $v ge 0$ ,otherwise,it is $-infty$



      I want to ask



      $1.$ why is $sumlimits_{i=1}^{n}v_ix_i(1-x_i)=v^Tx+x^T diag(v)x?$ it seems that $sumlimits_{i=1}^{n}v_i(1-x_i)=x^T diag(v)x, $ why?



      $2.$Why is $ L(x, mu ,v)=g(mu,v)=-b^Tmu - frac{1}{4} sumlimits_{i=1}^{n}(frac{c_i+a_i^T mu -v_i^2}{v_i})$,when $v ge 0$






      convex-analysis convex-optimization duality-theorems




      convex-analysis convex-optimization duality-theorems






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 15:26







      shineele

















      asked Dec 3 '18 at 13:18









      shineeleshineele

      377




      377






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          In terms of indices,



          $$v^Tx = sumlimits_i v_ix_i, text{ and } x^Tdiag(v)x = sumlimits_i x_iv_ix_i$$



          therefore their difference is $$v^Tx - x^Tdiag(v)x = sumlimits_i v_ix_i(1-x_i)$$



          I suspect you have an extra "$+x^Tv$" term in $L(x,mu,nu)$ that does not belong.



          Setting $L(x,mu,nu) = c^Tx + mu^T(Ax-b) - v^Tx + x^Tdiag(v)x$ we get the minimum by taking



          $$dL = c^T + mu^TA - v^T + 2v^T x = 0$$



          and solve that $2v^Tx = v^T - c^T - mu^TA$, in other words



          $$x_i = frac1{2v_i}(v_i-c_i-sum_j mu_jA_{j,i})$$



          Now plug this into $L(x,mu,nu)$ to get $g(mu,nu)$.




          ADDED: Ok, the reason you aren't getting the desired solution of $g(mu,nu)$ is that it has been copied down incorrectly. The correct solution is



          $$g(mu,nu) = begin{cases}-b^Tmu - (1/4)sum_{i=1}^n(c_i + a_i^Tmu - nu_i)^2/nu_i&nu geq 0\-infty&text{else}end{cases}$$



          Note the placement of the square outside of the parentheses. This problem is 5.13 from Boyd's textbook, and you may check that this is the correct solution by searching for a file from Stanford's course EE364a, Winter 2007-08 titled hw5sol.pdf.



          After making this change, you should check that the solution we arrived at above does give the correct answer.



          Lastly, it's a little easier to check the solution if you write the Lagrangian like this:
          $$L(x,mu,nu) = x^Ttext{diag}(nu)x + (c + A^Tmu - nu)^Tx - b^Tmu$$



          The algebra works like this: if $vX^2 + cX = -c^2/4v$ then the solution is $X = -c/2v$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            you are right!!
            $endgroup$
            – shineele
            Dec 3 '18 at 15:26










          • $begingroup$
            i still don't know how can i get $frac{1}{4} sumlimits_{i=1}^{n}(frac{c_i+a_i^T mu -v_i^2}{v_i})$,i have already plug $x_i$ into $L(x,mu,v)$,but there are just lots of $x_i$
            $endgroup$
            – shineele
            Dec 4 '18 at 15:02












          • $begingroup$
            @shineele I can't get it either, are you sure the formula for $g(mu,nu)$ is right? And did you double check the formula for $L(x,mu,nu)$? Where did this problem come from?
            $endgroup$
            – Ben
            Dec 4 '18 at 16:17










          • $begingroup$
            yes,both of them i am sure,this problem is from Convex Optimization 5.13
            $endgroup$
            – shineele
            Dec 4 '18 at 23:43










          • $begingroup$
            @shineele $g$ was wrong - see the new edit. A word of advice: include the source of the problem and where the solution is from when you ask such a question on here- you would get an answer much sooner this way!
            $endgroup$
            – Ben
            Dec 6 '18 at 3:25













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          active

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          active

          oldest

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          1












          $begingroup$

          In terms of indices,



          $$v^Tx = sumlimits_i v_ix_i, text{ and } x^Tdiag(v)x = sumlimits_i x_iv_ix_i$$



          therefore their difference is $$v^Tx - x^Tdiag(v)x = sumlimits_i v_ix_i(1-x_i)$$



          I suspect you have an extra "$+x^Tv$" term in $L(x,mu,nu)$ that does not belong.



          Setting $L(x,mu,nu) = c^Tx + mu^T(Ax-b) - v^Tx + x^Tdiag(v)x$ we get the minimum by taking



          $$dL = c^T + mu^TA - v^T + 2v^T x = 0$$



          and solve that $2v^Tx = v^T - c^T - mu^TA$, in other words



          $$x_i = frac1{2v_i}(v_i-c_i-sum_j mu_jA_{j,i})$$



          Now plug this into $L(x,mu,nu)$ to get $g(mu,nu)$.




          ADDED: Ok, the reason you aren't getting the desired solution of $g(mu,nu)$ is that it has been copied down incorrectly. The correct solution is



          $$g(mu,nu) = begin{cases}-b^Tmu - (1/4)sum_{i=1}^n(c_i + a_i^Tmu - nu_i)^2/nu_i&nu geq 0\-infty&text{else}end{cases}$$



          Note the placement of the square outside of the parentheses. This problem is 5.13 from Boyd's textbook, and you may check that this is the correct solution by searching for a file from Stanford's course EE364a, Winter 2007-08 titled hw5sol.pdf.



          After making this change, you should check that the solution we arrived at above does give the correct answer.



          Lastly, it's a little easier to check the solution if you write the Lagrangian like this:
          $$L(x,mu,nu) = x^Ttext{diag}(nu)x + (c + A^Tmu - nu)^Tx - b^Tmu$$



          The algebra works like this: if $vX^2 + cX = -c^2/4v$ then the solution is $X = -c/2v$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            you are right!!
            $endgroup$
            – shineele
            Dec 3 '18 at 15:26










          • $begingroup$
            i still don't know how can i get $frac{1}{4} sumlimits_{i=1}^{n}(frac{c_i+a_i^T mu -v_i^2}{v_i})$,i have already plug $x_i$ into $L(x,mu,v)$,but there are just lots of $x_i$
            $endgroup$
            – shineele
            Dec 4 '18 at 15:02












          • $begingroup$
            @shineele I can't get it either, are you sure the formula for $g(mu,nu)$ is right? And did you double check the formula for $L(x,mu,nu)$? Where did this problem come from?
            $endgroup$
            – Ben
            Dec 4 '18 at 16:17










          • $begingroup$
            yes,both of them i am sure,this problem is from Convex Optimization 5.13
            $endgroup$
            – shineele
            Dec 4 '18 at 23:43










          • $begingroup$
            @shineele $g$ was wrong - see the new edit. A word of advice: include the source of the problem and where the solution is from when you ask such a question on here- you would get an answer much sooner this way!
            $endgroup$
            – Ben
            Dec 6 '18 at 3:25


















          1












          $begingroup$

          In terms of indices,



          $$v^Tx = sumlimits_i v_ix_i, text{ and } x^Tdiag(v)x = sumlimits_i x_iv_ix_i$$



          therefore their difference is $$v^Tx - x^Tdiag(v)x = sumlimits_i v_ix_i(1-x_i)$$



          I suspect you have an extra "$+x^Tv$" term in $L(x,mu,nu)$ that does not belong.



          Setting $L(x,mu,nu) = c^Tx + mu^T(Ax-b) - v^Tx + x^Tdiag(v)x$ we get the minimum by taking



          $$dL = c^T + mu^TA - v^T + 2v^T x = 0$$



          and solve that $2v^Tx = v^T - c^T - mu^TA$, in other words



          $$x_i = frac1{2v_i}(v_i-c_i-sum_j mu_jA_{j,i})$$



          Now plug this into $L(x,mu,nu)$ to get $g(mu,nu)$.




          ADDED: Ok, the reason you aren't getting the desired solution of $g(mu,nu)$ is that it has been copied down incorrectly. The correct solution is



          $$g(mu,nu) = begin{cases}-b^Tmu - (1/4)sum_{i=1}^n(c_i + a_i^Tmu - nu_i)^2/nu_i&nu geq 0\-infty&text{else}end{cases}$$



          Note the placement of the square outside of the parentheses. This problem is 5.13 from Boyd's textbook, and you may check that this is the correct solution by searching for a file from Stanford's course EE364a, Winter 2007-08 titled hw5sol.pdf.



          After making this change, you should check that the solution we arrived at above does give the correct answer.



          Lastly, it's a little easier to check the solution if you write the Lagrangian like this:
          $$L(x,mu,nu) = x^Ttext{diag}(nu)x + (c + A^Tmu - nu)^Tx - b^Tmu$$



          The algebra works like this: if $vX^2 + cX = -c^2/4v$ then the solution is $X = -c/2v$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            you are right!!
            $endgroup$
            – shineele
            Dec 3 '18 at 15:26










          • $begingroup$
            i still don't know how can i get $frac{1}{4} sumlimits_{i=1}^{n}(frac{c_i+a_i^T mu -v_i^2}{v_i})$,i have already plug $x_i$ into $L(x,mu,v)$,but there are just lots of $x_i$
            $endgroup$
            – shineele
            Dec 4 '18 at 15:02












          • $begingroup$
            @shineele I can't get it either, are you sure the formula for $g(mu,nu)$ is right? And did you double check the formula for $L(x,mu,nu)$? Where did this problem come from?
            $endgroup$
            – Ben
            Dec 4 '18 at 16:17










          • $begingroup$
            yes,both of them i am sure,this problem is from Convex Optimization 5.13
            $endgroup$
            – shineele
            Dec 4 '18 at 23:43










          • $begingroup$
            @shineele $g$ was wrong - see the new edit. A word of advice: include the source of the problem and where the solution is from when you ask such a question on here- you would get an answer much sooner this way!
            $endgroup$
            – Ben
            Dec 6 '18 at 3:25
















          1












          1








          1





          $begingroup$

          In terms of indices,



          $$v^Tx = sumlimits_i v_ix_i, text{ and } x^Tdiag(v)x = sumlimits_i x_iv_ix_i$$



          therefore their difference is $$v^Tx - x^Tdiag(v)x = sumlimits_i v_ix_i(1-x_i)$$



          I suspect you have an extra "$+x^Tv$" term in $L(x,mu,nu)$ that does not belong.



          Setting $L(x,mu,nu) = c^Tx + mu^T(Ax-b) - v^Tx + x^Tdiag(v)x$ we get the minimum by taking



          $$dL = c^T + mu^TA - v^T + 2v^T x = 0$$



          and solve that $2v^Tx = v^T - c^T - mu^TA$, in other words



          $$x_i = frac1{2v_i}(v_i-c_i-sum_j mu_jA_{j,i})$$



          Now plug this into $L(x,mu,nu)$ to get $g(mu,nu)$.




          ADDED: Ok, the reason you aren't getting the desired solution of $g(mu,nu)$ is that it has been copied down incorrectly. The correct solution is



          $$g(mu,nu) = begin{cases}-b^Tmu - (1/4)sum_{i=1}^n(c_i + a_i^Tmu - nu_i)^2/nu_i&nu geq 0\-infty&text{else}end{cases}$$



          Note the placement of the square outside of the parentheses. This problem is 5.13 from Boyd's textbook, and you may check that this is the correct solution by searching for a file from Stanford's course EE364a, Winter 2007-08 titled hw5sol.pdf.



          After making this change, you should check that the solution we arrived at above does give the correct answer.



          Lastly, it's a little easier to check the solution if you write the Lagrangian like this:
          $$L(x,mu,nu) = x^Ttext{diag}(nu)x + (c + A^Tmu - nu)^Tx - b^Tmu$$



          The algebra works like this: if $vX^2 + cX = -c^2/4v$ then the solution is $X = -c/2v$






          share|cite|improve this answer











          $endgroup$



          In terms of indices,



          $$v^Tx = sumlimits_i v_ix_i, text{ and } x^Tdiag(v)x = sumlimits_i x_iv_ix_i$$



          therefore their difference is $$v^Tx - x^Tdiag(v)x = sumlimits_i v_ix_i(1-x_i)$$



          I suspect you have an extra "$+x^Tv$" term in $L(x,mu,nu)$ that does not belong.



          Setting $L(x,mu,nu) = c^Tx + mu^T(Ax-b) - v^Tx + x^Tdiag(v)x$ we get the minimum by taking



          $$dL = c^T + mu^TA - v^T + 2v^T x = 0$$



          and solve that $2v^Tx = v^T - c^T - mu^TA$, in other words



          $$x_i = frac1{2v_i}(v_i-c_i-sum_j mu_jA_{j,i})$$



          Now plug this into $L(x,mu,nu)$ to get $g(mu,nu)$.




          ADDED: Ok, the reason you aren't getting the desired solution of $g(mu,nu)$ is that it has been copied down incorrectly. The correct solution is



          $$g(mu,nu) = begin{cases}-b^Tmu - (1/4)sum_{i=1}^n(c_i + a_i^Tmu - nu_i)^2/nu_i&nu geq 0\-infty&text{else}end{cases}$$



          Note the placement of the square outside of the parentheses. This problem is 5.13 from Boyd's textbook, and you may check that this is the correct solution by searching for a file from Stanford's course EE364a, Winter 2007-08 titled hw5sol.pdf.



          After making this change, you should check that the solution we arrived at above does give the correct answer.



          Lastly, it's a little easier to check the solution if you write the Lagrangian like this:
          $$L(x,mu,nu) = x^Ttext{diag}(nu)x + (c + A^Tmu - nu)^Tx - b^Tmu$$



          The algebra works like this: if $vX^2 + cX = -c^2/4v$ then the solution is $X = -c/2v$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 6 '18 at 3:02

























          answered Dec 3 '18 at 14:12









          BenBen

          3,293616




          3,293616












          • $begingroup$
            you are right!!
            $endgroup$
            – shineele
            Dec 3 '18 at 15:26










          • $begingroup$
            i still don't know how can i get $frac{1}{4} sumlimits_{i=1}^{n}(frac{c_i+a_i^T mu -v_i^2}{v_i})$,i have already plug $x_i$ into $L(x,mu,v)$,but there are just lots of $x_i$
            $endgroup$
            – shineele
            Dec 4 '18 at 15:02












          • $begingroup$
            @shineele I can't get it either, are you sure the formula for $g(mu,nu)$ is right? And did you double check the formula for $L(x,mu,nu)$? Where did this problem come from?
            $endgroup$
            – Ben
            Dec 4 '18 at 16:17










          • $begingroup$
            yes,both of them i am sure,this problem is from Convex Optimization 5.13
            $endgroup$
            – shineele
            Dec 4 '18 at 23:43










          • $begingroup$
            @shineele $g$ was wrong - see the new edit. A word of advice: include the source of the problem and where the solution is from when you ask such a question on here- you would get an answer much sooner this way!
            $endgroup$
            – Ben
            Dec 6 '18 at 3:25




















          • $begingroup$
            you are right!!
            $endgroup$
            – shineele
            Dec 3 '18 at 15:26










          • $begingroup$
            i still don't know how can i get $frac{1}{4} sumlimits_{i=1}^{n}(frac{c_i+a_i^T mu -v_i^2}{v_i})$,i have already plug $x_i$ into $L(x,mu,v)$,but there are just lots of $x_i$
            $endgroup$
            – shineele
            Dec 4 '18 at 15:02












          • $begingroup$
            @shineele I can't get it either, are you sure the formula for $g(mu,nu)$ is right? And did you double check the formula for $L(x,mu,nu)$? Where did this problem come from?
            $endgroup$
            – Ben
            Dec 4 '18 at 16:17










          • $begingroup$
            yes,both of them i am sure,this problem is from Convex Optimization 5.13
            $endgroup$
            – shineele
            Dec 4 '18 at 23:43










          • $begingroup$
            @shineele $g$ was wrong - see the new edit. A word of advice: include the source of the problem and where the solution is from when you ask such a question on here- you would get an answer much sooner this way!
            $endgroup$
            – Ben
            Dec 6 '18 at 3:25


















          $begingroup$
          you are right!!
          $endgroup$
          – shineele
          Dec 3 '18 at 15:26




          $begingroup$
          you are right!!
          $endgroup$
          – shineele
          Dec 3 '18 at 15:26












          $begingroup$
          i still don't know how can i get $frac{1}{4} sumlimits_{i=1}^{n}(frac{c_i+a_i^T mu -v_i^2}{v_i})$,i have already plug $x_i$ into $L(x,mu,v)$,but there are just lots of $x_i$
          $endgroup$
          – shineele
          Dec 4 '18 at 15:02






          $begingroup$
          i still don't know how can i get $frac{1}{4} sumlimits_{i=1}^{n}(frac{c_i+a_i^T mu -v_i^2}{v_i})$,i have already plug $x_i$ into $L(x,mu,v)$,but there are just lots of $x_i$
          $endgroup$
          – shineele
          Dec 4 '18 at 15:02














          $begingroup$
          @shineele I can't get it either, are you sure the formula for $g(mu,nu)$ is right? And did you double check the formula for $L(x,mu,nu)$? Where did this problem come from?
          $endgroup$
          – Ben
          Dec 4 '18 at 16:17




          $begingroup$
          @shineele I can't get it either, are you sure the formula for $g(mu,nu)$ is right? And did you double check the formula for $L(x,mu,nu)$? Where did this problem come from?
          $endgroup$
          – Ben
          Dec 4 '18 at 16:17












          $begingroup$
          yes,both of them i am sure,this problem is from Convex Optimization 5.13
          $endgroup$
          – shineele
          Dec 4 '18 at 23:43




          $begingroup$
          yes,both of them i am sure,this problem is from Convex Optimization 5.13
          $endgroup$
          – shineele
          Dec 4 '18 at 23:43












          $begingroup$
          @shineele $g$ was wrong - see the new edit. A word of advice: include the source of the problem and where the solution is from when you ask such a question on here- you would get an answer much sooner this way!
          $endgroup$
          – Ben
          Dec 6 '18 at 3:25






          $begingroup$
          @shineele $g$ was wrong - see the new edit. A word of advice: include the source of the problem and where the solution is from when you ask such a question on here- you would get an answer much sooner this way!
          $endgroup$
          – Ben
          Dec 6 '18 at 3:25




















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