Geometric programming: Why are the constraints defined to be less than/equal to 1?Has the expressibility of 'non-integrality testing' as extension to MILP been studied before?What are common abbreviations in Operations Research?

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Geometric programming: Why are the constraints defined to be less than/equal to 1?


Has the expressibility of 'non-integrality testing' as extension to MILP been studied before?What are common abbreviations in Operations Research?













4












$begingroup$


In a simple convex optimisation problem, the standard form is given by beginalignmin_bf x&qquad f(bf x)\texts.t.&qquad g_i(bf x)le 0,quad i=1,cdots,m\&qquad h_j(bf x)=0,quad j=1,cdots,pendalign with $bf xinBbb R^n$ being the vector of variables to be optimised.



By changing the RHS of the constraints, we obtain a geometric optimisation problem, whose standard form is given by beginalignmin_bf x&qquad f(bf x)\texts.t.&qquad g_i(bf x)le 1,quad i=1,cdots,m\&qquad h_j(bf x)=1,quad j=1,cdots,pendalign with $bf xinBbb R^n$ being the vector of variables to be optimised. This has the additional restrictions that $f,g_i$ are posynomials and $h_j$ are simple monomials.




Questions



  1. Why is it so common to have 1 as the RHS of the constraints? That is, is there anything significant/geometrically convenient about it or is it by convention? For example, in a simple LP, the constraint is of the form $Abf xle bf b$ and not usually written as $Abf xbf b^-1le 1$.


  2. Suppose that we replace 1 by the more general set of constraints begincasesg_i(bf x)le a_i,quad i=1,cdots,m\h_j(bf x)=b_j,quad j=1,cdots,p.endcases What is the terminology for this generalised problem?











share|improve this question











$endgroup$


















    4












    $begingroup$


    In a simple convex optimisation problem, the standard form is given by beginalignmin_bf x&qquad f(bf x)\texts.t.&qquad g_i(bf x)le 0,quad i=1,cdots,m\&qquad h_j(bf x)=0,quad j=1,cdots,pendalign with $bf xinBbb R^n$ being the vector of variables to be optimised.



    By changing the RHS of the constraints, we obtain a geometric optimisation problem, whose standard form is given by beginalignmin_bf x&qquad f(bf x)\texts.t.&qquad g_i(bf x)le 1,quad i=1,cdots,m\&qquad h_j(bf x)=1,quad j=1,cdots,pendalign with $bf xinBbb R^n$ being the vector of variables to be optimised. This has the additional restrictions that $f,g_i$ are posynomials and $h_j$ are simple monomials.




    Questions



    1. Why is it so common to have 1 as the RHS of the constraints? That is, is there anything significant/geometrically convenient about it or is it by convention? For example, in a simple LP, the constraint is of the form $Abf xle bf b$ and not usually written as $Abf xbf b^-1le 1$.


    2. Suppose that we replace 1 by the more general set of constraints begincasesg_i(bf x)le a_i,quad i=1,cdots,m\h_j(bf x)=b_j,quad j=1,cdots,p.endcases What is the terminology for this generalised problem?











    share|improve this question











    $endgroup$
















      4












      4








      4





      $begingroup$


      In a simple convex optimisation problem, the standard form is given by beginalignmin_bf x&qquad f(bf x)\texts.t.&qquad g_i(bf x)le 0,quad i=1,cdots,m\&qquad h_j(bf x)=0,quad j=1,cdots,pendalign with $bf xinBbb R^n$ being the vector of variables to be optimised.



      By changing the RHS of the constraints, we obtain a geometric optimisation problem, whose standard form is given by beginalignmin_bf x&qquad f(bf x)\texts.t.&qquad g_i(bf x)le 1,quad i=1,cdots,m\&qquad h_j(bf x)=1,quad j=1,cdots,pendalign with $bf xinBbb R^n$ being the vector of variables to be optimised. This has the additional restrictions that $f,g_i$ are posynomials and $h_j$ are simple monomials.




      Questions



      1. Why is it so common to have 1 as the RHS of the constraints? That is, is there anything significant/geometrically convenient about it or is it by convention? For example, in a simple LP, the constraint is of the form $Abf xle bf b$ and not usually written as $Abf xbf b^-1le 1$.


      2. Suppose that we replace 1 by the more general set of constraints begincasesg_i(bf x)le a_i,quad i=1,cdots,m\h_j(bf x)=b_j,quad j=1,cdots,p.endcases What is the terminology for this generalised problem?











      share|improve this question











      $endgroup$




      In a simple convex optimisation problem, the standard form is given by beginalignmin_bf x&qquad f(bf x)\texts.t.&qquad g_i(bf x)le 0,quad i=1,cdots,m\&qquad h_j(bf x)=0,quad j=1,cdots,pendalign with $bf xinBbb R^n$ being the vector of variables to be optimised.



      By changing the RHS of the constraints, we obtain a geometric optimisation problem, whose standard form is given by beginalignmin_bf x&qquad f(bf x)\texts.t.&qquad g_i(bf x)le 1,quad i=1,cdots,m\&qquad h_j(bf x)=1,quad j=1,cdots,pendalign with $bf xinBbb R^n$ being the vector of variables to be optimised. This has the additional restrictions that $f,g_i$ are posynomials and $h_j$ are simple monomials.




      Questions



      1. Why is it so common to have 1 as the RHS of the constraints? That is, is there anything significant/geometrically convenient about it or is it by convention? For example, in a simple LP, the constraint is of the form $Abf xle bf b$ and not usually written as $Abf xbf b^-1le 1$.


      2. Suppose that we replace 1 by the more general set of constraints begincasesg_i(bf x)le a_i,quad i=1,cdots,m\h_j(bf x)=b_j,quad j=1,cdots,p.endcases What is the terminology for this generalised problem?








      terminology geometric-programming






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 10 hours ago







      TheSimpliFire

















      asked 10 hours ago









      TheSimpliFireTheSimpliFire

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          2 Answers
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          active

          oldest

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          6












          $begingroup$

          Log(1) = 0 could explain that. I.e., take log of both sides.



          Edit: I have now looked and found this in "A tutorial on geometric programming", Stephen Boyd · Seung-Jean Kim · Lieven Vandenberghe · Arash Hassibi. Look on p. 73 equation (6), and the paragraph which precedes it.



          As for the RHS not being 1, that is still basically a Geometric Program, which can be converted to standard form by dividing both the LHS and RHS by the RHS. That is addressed in the first paragraph of section 2.3 on p. 71 of the link.






          share|improve this answer











          $endgroup$














          • $begingroup$
            it is down to 19 minutes. Do you remember last time it took 20 minutes for me to write up my answer... anyway it is small but PROGRESS...
            $endgroup$
            – Oguz Toragay
            9 hours ago






          • 2




            $begingroup$
            @Oguz Toragay Sometimes someone else posts basically the same answer as mine while I am composing. When that happens, I just delete my post.
            $endgroup$
            – Mark L. Stone
            9 hours ago










          • $begingroup$
            I will do the same then. Thank you for letting me know what happens when two people post exactly the same answer.
            $endgroup$
            – Oguz Toragay
            9 hours ago


















          2












          $begingroup$

          Some of the reasons are convention (and the reason $1$ is the convention is that after substituting $x_i = e^y_i$ and taking logs of both sides, we get a convex program where all right-hand sides are $0$). But the primary practical reason to put the constraints in a standard form - that is, $g_i(mathbf x) le 1$ - is that then the dual of a geometric program is slightly more convenient to state.



          I'm going to be lazy and link to my own notes on the geometric programming dual. The dual geometric program is described on the first page. The description is, uh, kind of long. The example at the bottom of page $3$ might be more helpful.



          There is a dual variable $delta_i$ for each term $C_i x_1^alpha_i1 x_2^alpha_i2 dotsm x_m^alpha_im$ that appears in the geometric program, wherever it appears: in the objective function, or in one of the constraints. The key point to highlight for your question is that, provided your geometric program is in standard form, the dual objective function has a $left(fracC_idelta_iright)^delta_i$ factor.



          If we're working with a geometric program that's not in standard form, and the $i^textth$ term comes from a constraint with right-hand side $B$, then this factor should look like $left(fracC_i/Bdelta_iright)^delta_i$ instead. Effectively, we are forced to divide through by $B$ before we can take the dual.




          I would still consider programs that have the more general constraints of the form $textposynomial le B$ (for $B>0$) to be geometric programs. They're not in standard form, but depending on what you're about to do with them, you may not care.



          In fact, we could even allow the right-hand side to be any monomial with a positive coefficient. For example, the constraint $$x_1^2 + x_2^2 le x_3^2$$ is not any harder to deal with. The reason, of course, is that we can divide through by such a monomial to get a posynomial constraint $$x_1^2 x_3^-2 + x_2^2 x_3^-2 le 1$$ in standard form. I would still call programs with such constraints geometric programs, though somewhat less neatly stated ones.






          share|improve this answer








          New contributor



          Misha Lavrov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          $endgroup$

















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            2 Answers
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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

            votes









            6












            $begingroup$

            Log(1) = 0 could explain that. I.e., take log of both sides.



            Edit: I have now looked and found this in "A tutorial on geometric programming", Stephen Boyd · Seung-Jean Kim · Lieven Vandenberghe · Arash Hassibi. Look on p. 73 equation (6), and the paragraph which precedes it.



            As for the RHS not being 1, that is still basically a Geometric Program, which can be converted to standard form by dividing both the LHS and RHS by the RHS. That is addressed in the first paragraph of section 2.3 on p. 71 of the link.






            share|improve this answer











            $endgroup$














            • $begingroup$
              it is down to 19 minutes. Do you remember last time it took 20 minutes for me to write up my answer... anyway it is small but PROGRESS...
              $endgroup$
              – Oguz Toragay
              9 hours ago






            • 2




              $begingroup$
              @Oguz Toragay Sometimes someone else posts basically the same answer as mine while I am composing. When that happens, I just delete my post.
              $endgroup$
              – Mark L. Stone
              9 hours ago










            • $begingroup$
              I will do the same then. Thank you for letting me know what happens when two people post exactly the same answer.
              $endgroup$
              – Oguz Toragay
              9 hours ago















            6












            $begingroup$

            Log(1) = 0 could explain that. I.e., take log of both sides.



            Edit: I have now looked and found this in "A tutorial on geometric programming", Stephen Boyd · Seung-Jean Kim · Lieven Vandenberghe · Arash Hassibi. Look on p. 73 equation (6), and the paragraph which precedes it.



            As for the RHS not being 1, that is still basically a Geometric Program, which can be converted to standard form by dividing both the LHS and RHS by the RHS. That is addressed in the first paragraph of section 2.3 on p. 71 of the link.






            share|improve this answer











            $endgroup$














            • $begingroup$
              it is down to 19 minutes. Do you remember last time it took 20 minutes for me to write up my answer... anyway it is small but PROGRESS...
              $endgroup$
              – Oguz Toragay
              9 hours ago






            • 2




              $begingroup$
              @Oguz Toragay Sometimes someone else posts basically the same answer as mine while I am composing. When that happens, I just delete my post.
              $endgroup$
              – Mark L. Stone
              9 hours ago










            • $begingroup$
              I will do the same then. Thank you for letting me know what happens when two people post exactly the same answer.
              $endgroup$
              – Oguz Toragay
              9 hours ago













            6












            6








            6





            $begingroup$

            Log(1) = 0 could explain that. I.e., take log of both sides.



            Edit: I have now looked and found this in "A tutorial on geometric programming", Stephen Boyd · Seung-Jean Kim · Lieven Vandenberghe · Arash Hassibi. Look on p. 73 equation (6), and the paragraph which precedes it.



            As for the RHS not being 1, that is still basically a Geometric Program, which can be converted to standard form by dividing both the LHS and RHS by the RHS. That is addressed in the first paragraph of section 2.3 on p. 71 of the link.






            share|improve this answer











            $endgroup$



            Log(1) = 0 could explain that. I.e., take log of both sides.



            Edit: I have now looked and found this in "A tutorial on geometric programming", Stephen Boyd · Seung-Jean Kim · Lieven Vandenberghe · Arash Hassibi. Look on p. 73 equation (6), and the paragraph which precedes it.



            As for the RHS not being 1, that is still basically a Geometric Program, which can be converted to standard form by dividing both the LHS and RHS by the RHS. That is addressed in the first paragraph of section 2.3 on p. 71 of the link.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 9 hours ago

























            answered 10 hours ago









            Mark L. StoneMark L. Stone

            3,1977 silver badges27 bronze badges




            3,1977 silver badges27 bronze badges














            • $begingroup$
              it is down to 19 minutes. Do you remember last time it took 20 minutes for me to write up my answer... anyway it is small but PROGRESS...
              $endgroup$
              – Oguz Toragay
              9 hours ago






            • 2




              $begingroup$
              @Oguz Toragay Sometimes someone else posts basically the same answer as mine while I am composing. When that happens, I just delete my post.
              $endgroup$
              – Mark L. Stone
              9 hours ago










            • $begingroup$
              I will do the same then. Thank you for letting me know what happens when two people post exactly the same answer.
              $endgroup$
              – Oguz Toragay
              9 hours ago
















            • $begingroup$
              it is down to 19 minutes. Do you remember last time it took 20 minutes for me to write up my answer... anyway it is small but PROGRESS...
              $endgroup$
              – Oguz Toragay
              9 hours ago






            • 2




              $begingroup$
              @Oguz Toragay Sometimes someone else posts basically the same answer as mine while I am composing. When that happens, I just delete my post.
              $endgroup$
              – Mark L. Stone
              9 hours ago










            • $begingroup$
              I will do the same then. Thank you for letting me know what happens when two people post exactly the same answer.
              $endgroup$
              – Oguz Toragay
              9 hours ago















            $begingroup$
            it is down to 19 minutes. Do you remember last time it took 20 minutes for me to write up my answer... anyway it is small but PROGRESS...
            $endgroup$
            – Oguz Toragay
            9 hours ago




            $begingroup$
            it is down to 19 minutes. Do you remember last time it took 20 minutes for me to write up my answer... anyway it is small but PROGRESS...
            $endgroup$
            – Oguz Toragay
            9 hours ago




            2




            2




            $begingroup$
            @Oguz Toragay Sometimes someone else posts basically the same answer as mine while I am composing. When that happens, I just delete my post.
            $endgroup$
            – Mark L. Stone
            9 hours ago




            $begingroup$
            @Oguz Toragay Sometimes someone else posts basically the same answer as mine while I am composing. When that happens, I just delete my post.
            $endgroup$
            – Mark L. Stone
            9 hours ago












            $begingroup$
            I will do the same then. Thank you for letting me know what happens when two people post exactly the same answer.
            $endgroup$
            – Oguz Toragay
            9 hours ago




            $begingroup$
            I will do the same then. Thank you for letting me know what happens when two people post exactly the same answer.
            $endgroup$
            – Oguz Toragay
            9 hours ago











            2












            $begingroup$

            Some of the reasons are convention (and the reason $1$ is the convention is that after substituting $x_i = e^y_i$ and taking logs of both sides, we get a convex program where all right-hand sides are $0$). But the primary practical reason to put the constraints in a standard form - that is, $g_i(mathbf x) le 1$ - is that then the dual of a geometric program is slightly more convenient to state.



            I'm going to be lazy and link to my own notes on the geometric programming dual. The dual geometric program is described on the first page. The description is, uh, kind of long. The example at the bottom of page $3$ might be more helpful.



            There is a dual variable $delta_i$ for each term $C_i x_1^alpha_i1 x_2^alpha_i2 dotsm x_m^alpha_im$ that appears in the geometric program, wherever it appears: in the objective function, or in one of the constraints. The key point to highlight for your question is that, provided your geometric program is in standard form, the dual objective function has a $left(fracC_idelta_iright)^delta_i$ factor.



            If we're working with a geometric program that's not in standard form, and the $i^textth$ term comes from a constraint with right-hand side $B$, then this factor should look like $left(fracC_i/Bdelta_iright)^delta_i$ instead. Effectively, we are forced to divide through by $B$ before we can take the dual.




            I would still consider programs that have the more general constraints of the form $textposynomial le B$ (for $B>0$) to be geometric programs. They're not in standard form, but depending on what you're about to do with them, you may not care.



            In fact, we could even allow the right-hand side to be any monomial with a positive coefficient. For example, the constraint $$x_1^2 + x_2^2 le x_3^2$$ is not any harder to deal with. The reason, of course, is that we can divide through by such a monomial to get a posynomial constraint $$x_1^2 x_3^-2 + x_2^2 x_3^-2 le 1$$ in standard form. I would still call programs with such constraints geometric programs, though somewhat less neatly stated ones.






            share|improve this answer








            New contributor



            Misha Lavrov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.





            $endgroup$



















              2












              $begingroup$

              Some of the reasons are convention (and the reason $1$ is the convention is that after substituting $x_i = e^y_i$ and taking logs of both sides, we get a convex program where all right-hand sides are $0$). But the primary practical reason to put the constraints in a standard form - that is, $g_i(mathbf x) le 1$ - is that then the dual of a geometric program is slightly more convenient to state.



              I'm going to be lazy and link to my own notes on the geometric programming dual. The dual geometric program is described on the first page. The description is, uh, kind of long. The example at the bottom of page $3$ might be more helpful.



              There is a dual variable $delta_i$ for each term $C_i x_1^alpha_i1 x_2^alpha_i2 dotsm x_m^alpha_im$ that appears in the geometric program, wherever it appears: in the objective function, or in one of the constraints. The key point to highlight for your question is that, provided your geometric program is in standard form, the dual objective function has a $left(fracC_idelta_iright)^delta_i$ factor.



              If we're working with a geometric program that's not in standard form, and the $i^textth$ term comes from a constraint with right-hand side $B$, then this factor should look like $left(fracC_i/Bdelta_iright)^delta_i$ instead. Effectively, we are forced to divide through by $B$ before we can take the dual.




              I would still consider programs that have the more general constraints of the form $textposynomial le B$ (for $B>0$) to be geometric programs. They're not in standard form, but depending on what you're about to do with them, you may not care.



              In fact, we could even allow the right-hand side to be any monomial with a positive coefficient. For example, the constraint $$x_1^2 + x_2^2 le x_3^2$$ is not any harder to deal with. The reason, of course, is that we can divide through by such a monomial to get a posynomial constraint $$x_1^2 x_3^-2 + x_2^2 x_3^-2 le 1$$ in standard form. I would still call programs with such constraints geometric programs, though somewhat less neatly stated ones.






              share|improve this answer








              New contributor



              Misha Lavrov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.





              $endgroup$

















                2












                2








                2





                $begingroup$

                Some of the reasons are convention (and the reason $1$ is the convention is that after substituting $x_i = e^y_i$ and taking logs of both sides, we get a convex program where all right-hand sides are $0$). But the primary practical reason to put the constraints in a standard form - that is, $g_i(mathbf x) le 1$ - is that then the dual of a geometric program is slightly more convenient to state.



                I'm going to be lazy and link to my own notes on the geometric programming dual. The dual geometric program is described on the first page. The description is, uh, kind of long. The example at the bottom of page $3$ might be more helpful.



                There is a dual variable $delta_i$ for each term $C_i x_1^alpha_i1 x_2^alpha_i2 dotsm x_m^alpha_im$ that appears in the geometric program, wherever it appears: in the objective function, or in one of the constraints. The key point to highlight for your question is that, provided your geometric program is in standard form, the dual objective function has a $left(fracC_idelta_iright)^delta_i$ factor.



                If we're working with a geometric program that's not in standard form, and the $i^textth$ term comes from a constraint with right-hand side $B$, then this factor should look like $left(fracC_i/Bdelta_iright)^delta_i$ instead. Effectively, we are forced to divide through by $B$ before we can take the dual.




                I would still consider programs that have the more general constraints of the form $textposynomial le B$ (for $B>0$) to be geometric programs. They're not in standard form, but depending on what you're about to do with them, you may not care.



                In fact, we could even allow the right-hand side to be any monomial with a positive coefficient. For example, the constraint $$x_1^2 + x_2^2 le x_3^2$$ is not any harder to deal with. The reason, of course, is that we can divide through by such a monomial to get a posynomial constraint $$x_1^2 x_3^-2 + x_2^2 x_3^-2 le 1$$ in standard form. I would still call programs with such constraints geometric programs, though somewhat less neatly stated ones.






                share|improve this answer








                New contributor



                Misha Lavrov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.





                $endgroup$



                Some of the reasons are convention (and the reason $1$ is the convention is that after substituting $x_i = e^y_i$ and taking logs of both sides, we get a convex program where all right-hand sides are $0$). But the primary practical reason to put the constraints in a standard form - that is, $g_i(mathbf x) le 1$ - is that then the dual of a geometric program is slightly more convenient to state.



                I'm going to be lazy and link to my own notes on the geometric programming dual. The dual geometric program is described on the first page. The description is, uh, kind of long. The example at the bottom of page $3$ might be more helpful.



                There is a dual variable $delta_i$ for each term $C_i x_1^alpha_i1 x_2^alpha_i2 dotsm x_m^alpha_im$ that appears in the geometric program, wherever it appears: in the objective function, or in one of the constraints. The key point to highlight for your question is that, provided your geometric program is in standard form, the dual objective function has a $left(fracC_idelta_iright)^delta_i$ factor.



                If we're working with a geometric program that's not in standard form, and the $i^textth$ term comes from a constraint with right-hand side $B$, then this factor should look like $left(fracC_i/Bdelta_iright)^delta_i$ instead. Effectively, we are forced to divide through by $B$ before we can take the dual.




                I would still consider programs that have the more general constraints of the form $textposynomial le B$ (for $B>0$) to be geometric programs. They're not in standard form, but depending on what you're about to do with them, you may not care.



                In fact, we could even allow the right-hand side to be any monomial with a positive coefficient. For example, the constraint $$x_1^2 + x_2^2 le x_3^2$$ is not any harder to deal with. The reason, of course, is that we can divide through by such a monomial to get a posynomial constraint $$x_1^2 x_3^-2 + x_2^2 x_3^-2 le 1$$ in standard form. I would still call programs with such constraints geometric programs, though somewhat less neatly stated ones.







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