# 7.5 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

(7.8)$\begin{split}\begin{array}{lll} \minimize & f_0(x) & \\ \st & f_i(x) \leq 1, &i=1,\ldots,m, \\ & x_j>0, &j=1,\ldots,n, \end{array}\end{split}$

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

$f(x) = \sum_k c_kx_1^{\alpha_{k1}}x_2^{\alpha_{k2}}\cdots x_n^{\alpha_{kn}}$

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

(7.9)$\log(\sum_k\exp(a_k^Ty+b_k)) \leq 0$

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:

$a_k^Ty+b_k\leq 0$

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.

## 7.5.1 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 $$d/w$$:

(7.10)$\begin{split}\begin{array}{rrl} \maximize & hwd & \\ \st & 2(hw + hd) & \leq A_{\mathrm{wall}}, \\ & wd & \leq A_{\mathrm{floor}}, \\ & \alpha & \leq h/w \leq \beta, \\ & \gamma & \leq d/w \leq \delta. \end{array}\end{split}$

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 (7.10) becomes

(7.11)$\begin{split}\begin{array}{rll} \maximize & x+y+z \\ \st & \log(\exp(x+y+\log(2/A_{\mathrm{wall}}))+\exp(x+z+\log(2/A_{\mathrm{wall}}))) \leq 0, \\ & y+z \leq \log(A_{\mathrm{floor}}), \\ & \log(\alpha) \leq x-y \leq \log(\beta), \\ & \log(\gamma) \leq z-y \leq \log(\delta). \end{array}\end{split}$

Next, we demonstrate how to implement a log-sum-exp constraint (7.9). It can be written as:

(7.12)$\begin{split}\begin{array}{l} u_k\geq \exp(a_k^Ty+b_k),\quad (\mathrm{equiv.}\ (u_k,1,a_k^Ty+b_k)\in\EXP),\\ \sum_k u_k = 1. \end{array}\end{split}$

This presentation requires one extra variable $$u_k$$ for each monomial appearing in the original posynomial constraint.

Listing 7.5 Implementation of log-sum-exp as in (7.12). Click here to download.
  // Models log(sum(exp(Ax+b))) <= 0.
// Each row of [A b] describes one of the exp-terms
public static void logsumexp(Model      M,
double[][] A,
Variable   x,
double[]   b)
{
int k = A.length;
Variable u = M.variable(k);
M.constraint(Expr.sum(u), Domain.equalsTo(1.0));
M.constraint(Expr.hstack(u,
Expr.constTerm(k, 1.0),
}


We can now use this function to assemble all constraints in the model. The linear part of the problem is entered as in Sec. 7.1 (Linear Optimization).

Listing 7.6 Source code solving problem (7.11). Click here to download.
  public static double[] max_volume_box(double Aw, double Af,
double alpha, double beta, double gamma, double delta)
throws SolutionError
{
Model M = new Model("max_vol_box");
try {
Variable xyz = M.variable(3);
M.objective("Objective", ObjectiveSense.Maximize, Expr.sum(xyz));

logsumexp(M,
new double[][] {{1,1,0}, {1,0,1}},
xyz,
new double[] {Math.log(2.0/Aw), Math.log(2.0/Aw)});

M.constraint(Expr.dot(new double[] {0,1,1}, xyz), Domain.lessThan(Math.log(Af)));
M.constraint(Expr.dot(new double[] {1,-1,0}, xyz), Domain.inRange(Math.log(alpha),Math.log(beta)));
M.constraint(Expr.dot(new double[] {0,-1,1}, xyz), Domain.inRange(Math.log(gamma),Math.log(delta)));

M.setLogHandler(new java.io.PrintWriter(System.out));
M.solve();

double[] xyzVal = xyz.level();
double[] hwdVal = new double[3];
for(int i=0; i<3; i++) hwdVal[i] = Math.exp(xyzVal[i]);

return hwdVal;
} finally {
M.dispose();
}
}


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

Listing 7.7 Sample data for problem (7.10). Click here to download.
  public static void main(String[] args)
throws SolutionError
{
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]);
}