6.6 Geometric Programming

Geometric programs (GP) are a particular class of optimization problems which can be expressed in special polynomial form as positive sums of generalized monomials. More precisely, a geometric problem in canonical form is

(6.17)$$\begin{array}{r}\begin{array}{lll}\text{minimize}& f_0(x)& \\ \text{subject to}& f_i(x)\le 1,& i=1,\dots ,m,\\ & x_j>0,& j=1,\dots ,n,\end{array}\end{array}$$

where each $f_0,\dots ,f_m$ is a posynomial, that is a function of the form

$$f(x)=\sum _k{c}_k{x}_1^{{\alpha }_{k1}}{x}_2^{{\alpha }_{k2}}\cdots x_n^{{\alpha }_{kn}}$$

with arbitrary real ${\alpha }_{ki}$ and $c_k>0$. The standard way to formulate GPs in convex form is to introduce a variable substitution

$$x_i=\exp (y_i).$$

Under this substitution all constraints in a GP can be reduced to the form

(6.18)$$\log (\sum _k\exp (a_k^{T}y+b_k))\le 0$$

involving a log-sum-exp bound. Moreover, constraints involving only a single monomial in $x$ can be even more simply written as a linear inequality:

$$a_k^{T}y+b_k\le 0$$

We refer to the MOSEK Modeling Cookbook and to [BKVH07] for more details on this reformulation. A geometric problem formulated in convex form can be entered into MOSEK with the help of exponential cones.

6.6.1 Example GP1

The following problem comes from [BKVH07]. Consider maximizing the volume of a $h\times w\times d$ box subject to upper bounds on the area of the floor and of the walls and bounds on the ratios $h/w$ and $d/w$:

(6.19)$$\begin{array}{r}\begin{array}{rrl}\text{maximize}& hwd& \\ \text{subject to}& 2(hw+hd)& \le A_{\mathrm{wall}},\\ & wd& \le A_{\mathrm{floor}},\\ & \alpha & \le h/w\le \beta ,\\ & \gamma & \le d/w\le \delta .\end{array}\end{array}$$

The decision variables in the problem are $h,w,d$. We make a substitution

$$h=\exp (x),w=\exp (y),d=\exp (z)$$

after which (6.19) becomes

(6.20)$$\begin{array}{r}\begin{array}{rl}\text{maximize}& x+y+z\\ \text{subject to}& \log (\exp (x+y+\log (2/A_{\mathrm{wall}}))+\exp (x+z+\log (2/A_{\mathrm{wall}})))\le 0,\\ & y+z\le \log (A_{\mathrm{floor}}),\\ & \log (\alpha )\le x-y\le \log (\beta ),\\ & \log (\gamma )\le z-y\le \log (\delta ).\end{array}\end{array}$$

Next, we demonstrate how to implement a log-sum-exp constraint (6.18). It can be written as:

(6.21)$$\begin{array}{r}\begin{array}{l}{u}_k\ge \exp (a_k^{T}y+b_k),(\mathrm{equiv}.(u_k,1,a_k^{T}y+b_k)\in K_{\exp }),\\ \sum _k{u}_k=1.\end{array}\end{array}$$

This presentation requires one extra variable $u_k$ for each monomial appearing in the original posynomial constraint. In this case the affine conic constraints (ACC, see Sec. 6.2 (From Linear to Conic Optimization)) take the form:

$$\begin{array}{r}\left[\begin{array}{ccccc}0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0\\ 1& 1& 0& 0& 0\\ 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0\\ 1& 0& 1& 0& 0\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\\ u_1\\ u_2\end{array}\right]+\left[\begin{array}{c}0\\ 1\\ \log (2/A_{\mathrm{wall}})\\ 0\\ 1\\ \log (2/A_{\mathrm{wall}})\end{array}\right]\in K_{\exp }\times K_{\exp }.\end{array}$$

As a matter of demonstration we will also add the constraint

$$u_1+u_2-1=0$$

as an affine conic constraint. It means that to define the all the ACCs we need to produce the following affine expressions (AFE) and store them:

$$u_1,u_2,x+y+\log (2/A_{\mathrm{wall}}),x+z+\log (2/A_{\mathrm{wall}}),1.0,u_1+u_2-1.0.$$

We implement it by adding all the affine expressions (AFE) and then picking the ones required for each ACC:

Listing 6.7 Implementation of log-sum-exp as in (6.21). Click here to download.

      // Affine expressions appearing in affine conic constraints
      // in this order:
      // u1, u2, x+y+log(2/Awall), x+z+log(2/Awall), 1.0, u1+u2-1.0
      long numafe    = 6;
      int u1 = 3, u2 = 4;     // Indices of slack variables
      long[]   afeidx = {0, 1, 2, 2, 3, 3, 5, 5};
      int[]    varidx = {u1, u2, x, y, x, z, u1, u2};
      double[] fval   = {1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0};
      double[] gfull  = {0, 0, Math.log(2/Aw), Math.log(2/Aw), 1.0, -1.0};

      // New variables u1, u2
      task.appendvars(2);
      task.putvarboundsliceconst(u1, u2+1, boundkey.fr, -inf, inf);

      // Append affine expressions
      task.appendafes(numafe);
      task.putafefentrylist(afeidx, varidx, fval);
      task.putafegslice(0, numafe, gfull);

      // Two affine conic constraints
      long expdom = task.appendprimalexpconedomain();

      // (u1, 1, x+y+log(2/Awall)) \in EXP
      task.appendacc(expdom, new long[]{0, 4, 2}, null);

      // (u2, 1, x+z+log(2/Awall)) \in EXP
      task.appendacc(expdom, new long[]{1, 4, 3}, null);     

      // The constraint u1+u2-1 \in \ZERO is added also as an ACC
      task.appendacc(task.appendrzerodomain(1), new long[]{5}, null);

We can now use this function to assemble all constraints in the model. The linear part of the problem is entered as in Sec. 6.1 (Linear Optimization).

Listing 6.8 Source code solving problem (6.20). Click here to download.

  public static double[] max_volume_box(double Aw, double Af, 
                                        double alpha, double beta, double gamma, double delta)
  {
    // Basic dimensions of our problem
    int numvar    = 3;  // Variables in original problem
    int x=0, y=1, z=2;  // Indices of variables
    int numcon    = 3;  // Linear constraints in original problem

    // Linear part of the problem involving x, y, z
    double[] cval  = {1, 1, 1};
    int[]    asubi = {0, 0, 1, 1, 2, 2};
    int[]    asubj = {y, z, x, y, z, y};
    double[] aval  = {1.0, 1.0, 1.0, -1.0, 1.0, -1.0};
    boundkey[] bkc = {boundkey.up, boundkey.ra, boundkey.ra};
    double[] blc   = {-inf, Math.log(alpha), Math.log(gamma)};
    double[] buc   = {Math.log(Af), Math.log(beta), Math.log(delta)};

    try (Task task = new Task()) 
    {
      // Directs the log task stream to the user specified
      // method task_msg_obj.stream
      task.set_Stream(
        streamtype.log,
        new Stream()
      { public void stream(String msg) { System.out.print(msg); }});

      // Add variables and constraints
      task.appendvars(numvar);
      task.appendcons(numcon);

      // Objective is the sum of three first variables
      task.putobjsense(objsense.maximize);
      task.putcslice(0, numvar, cval);
      task.putvarboundsliceconst(0, numvar, boundkey.fr, -inf, inf);

      // Add the linear constraints
      task.putaijlist(asubi, asubj, aval);
      task.putconboundslice(0, numcon, bkc, blc, buc);

      // Affine expressions appearing in affine conic constraints
      // in this order:
      // u1, u2, x+y+log(2/Awall), x+z+log(2/Awall), 1.0, u1+u2-1.0
      long numafe    = 6;
      int u1 = 3, u2 = 4;     // Indices of slack variables
      long[]   afeidx = {0, 1, 2, 2, 3, 3, 5, 5};
      int[]    varidx = {u1, u2, x, y, x, z, u1, u2};
      double[] fval   = {1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0};
      double[] gfull  = {0, 0, Math.log(2/Aw), Math.log(2/Aw), 1.0, -1.0};

      // New variables u1, u2
      task.appendvars(2);
      task.putvarboundsliceconst(u1, u2+1, boundkey.fr, -inf, inf);

      // Append affine expressions
      task.appendafes(numafe);
      task.putafefentrylist(afeidx, varidx, fval);
      task.putafegslice(0, numafe, gfull);

      // Two affine conic constraints
      long expdom = task.appendprimalexpconedomain();

      // (u1, 1, x+y+log(2/Awall)) \in EXP
      task.appendacc(expdom, new long[]{0, 4, 2}, null);

      // (u2, 1, x+z+log(2/Awall)) \in EXP
      task.appendacc(expdom, new long[]{1, 4, 3}, null);     

      // The constraint u1+u2-1 \in \ZERO is added also as an ACC
      task.appendacc(task.appendrzerodomain(1), new long[]{5}, null);

      // Solve and map to original h, w, d
      task.optimize();
      double[] xyz = task.getxxslice(soltype.itr, 0, numvar);
      double[] hwd = new double[numvar];     
      for(int i = 0; i < numvar; i++) hwd[i] = Math.exp(xyz[i]);
      return hwd;
    }
  }

Given sample data we obtain the solution $h,w,d$ as follows:

Listing 6.9 Sample data for problem (6.19). Click here to download.

  public static void main(String[] args)
  {
    double Aw    = 200.0;
    double Af    = 50.0;
    double alpha = 2.0;
    double beta  = 10.0;
    double gamma = 2.0;
    double delta = 10.0;
    
    double[] hwd = max_volume_box(Aw, Af, alpha, beta, gamma, delta);

    System.out.format("h=%.4f w=%.4f d=%.4f\n", hwd[0], hwd[1], hwd[2]);
  }