7.3 Power Cone Optimization

The structure of a typical conic optimization problem is

\[\begin{split}\begin{array}{lccccl} \mbox{minimize} & & & c^T x+c^f & & \\ \mbox{subject to} & l^c & \leq & A x & \leq & u^c, \\ & l^x & \leq & x & \leq & u^x, \\ & & & Fx+g & \in & \D, \end{array}\end{split}\]

(see Sec. 12 (Problem Formulation and Solutions) for detailed formulations). Here we discuss how to set-up problems with the primal/dual power cones.

MOSEK supports the primal and dual power cones, defined as below:

• Primal power cone:

  \[\POW_n^{\alpha_k} = \left\{ x\in \real^n ~:~ \prod_{i=0}^{n_\ell-1} x_i^{\beta_i} \geq \sqrt{\sum_{j=n_\ell}^{n-1}x_j^2},\ x_0\ldots,x_{n_\ell-1}\geq 0 \right\}\]

  where \(s = \sum_i \alpha_i\) and \(\beta_i = \alpha_i / s\), so that \(\sum_i \beta_i=1\).

• Dual power cone:

  \[(\POW_n^{\alpha_k}) = \left\{ x\in \real^n ~:~ \prod_{i=0}^{n_\ell-1} \left(\frac{x_i}{\beta_i}\right)^{\beta_i} \geq \sqrt{\sum_{j=n_\ell}^{n-1}x_j^2},\ x_0\ldots,x_{n_\ell-1}\geq 0 \right\}\]

  where \(s = \sum_i \alpha_i\) and \(\beta_i = \alpha_i / s\), so that \(\sum_i \beta_i=1\).

Perhaps the most important special case is the three-dimensional power cone family:

\[\POW_3^{\alpha,1-\alpha} = \left\lbrace x \in \real^3: x_0^\alpha x_1^{1-\alpha}\geq |x_2|,\ x_0,x_1\geq 0 \right\rbrace.\]

which has the corresponding dual cone:

For example, the conic constraint \((x,y,z)\in\POW_3^{0.25,0.75}\) is equivalent to \(x^{0.25}y^{0.75}\geq |z|\), or simply \(xy^3\geq z^4\) with \(x,y\geq 0\).

For other types of cones supported by MOSEK, see Sec. 14.8 (Supported domains) and the other tutorials in this chapter. Different cone types can appear together in one optimization problem.

7.3.1 Example POW1

Consider the following optimization problem which involves powers of variables:

(7.3)\[\begin{split}\begin{array} {lrcl} \mbox{maximize} & x_0^{0.2}x_1^{0.8} + x_2^{0.4} - x_0 & & \\ \mbox{subject to} & x_0+x_1+\frac12 x_2 & = & 2, \\ & x_0,x_1,x_2 & \geq & 0. \end{array}\end{split}\]

We convert (7.3) into affine conic form using auxiliary variables as bounds for the power expressions:

(7.4)\[\begin{split}\begin{array} {lrcl} \mbox{maximize} & x_3 + x_4 - x_0 & & \\ \mbox{subject to} & x_0+x_1+\frac12 x_2 & = & 2, \\ & (x_0,x_1,x_3) & \in & \POW_3^{0.2,0.8}, \\ & (x_2,1.0,x_4) & \in & \POW_3^{0.4,0.6}. \end{array}\end{split}\]

The two conic constraints shown in (7.4) can be expressed in the ACC form as shown in (7.5):

(7.5)\[\begin{split}\left[\begin{array}{ccccc}1&0&0&0&0\\0&1&0&0&0\\0&0&0&1&0\\0&0&1&0&0\\0&0&0&0&0\\0&0&0&0&1\end{array}\right] \left[\begin{array}{c}x_0\\x_1\\x_2\\x_3\\x_4\end{array}\right] + \left[\begin{array}{c}0\\0\\0\\0\\1\\0\end{array}\right] \in \POW^{0.2,0.8}_3 \times \POW^{0.4,0.6}_3.\end{split}\]

We start by creating the optimization model:

      using (Model M = new Model("pow1"))
      {

We then define the variable x corresponding to the original problem (7.3), and auxiliary variables appearing in the conic reformulation (7.4).

        Variable x  = M.Variable("x", 3, Domain.Unbounded());
        Variable x3 = M.Variable();
        Variable x4 = M.Variable();

The linear constraint is defined using the dot product operator Expr.Dot:

        // Create the linear constraint
        double[] aval = new double[] {1.0, 1.0, 0.5};
        M.Constraint(Expr.Dot(x, aval), Domain.EqualsTo(2.0));

The primal power cone is referred to via Domain.InPPowerCone with an appropriate list of variables or expressions in each case.

        // Create the exponential conic constraint
        // Create the conic constraints
        M.Constraint(Var.Vstack(x.Slice(0,2), x3), Domain.InPPowerCone(0.2));
        M.Constraint(Expr.Vstack(x.Index(2), 1.0, x4), Domain.InPPowerCone(0.4));      

We only need the objective function:

        // Set the objective function
        double[] cval = new double[] {1.0, 1.0, -1.0};
        M.Objective(ObjectiveSense.Maximize, Expr.Dot(cval, Var.Vstack(x3, x4, x.Index(0))));

Calling the Model.Solve method invokes the solver:

        M.Solve();

The primal and dual solution values can be retrieved using Variable.Level, Constraint.Level and Variable.Dual, Constraint.Dual. Here we just display the primal solution

        double[] solx = x.Level();
        Console.WriteLine("x,y,z = {0}, {1}, {2}", solx[0], solx[1], solx[2]);

which is

[ 0.06389298  0.78308564  2.30604283 ]

Listing 7.3 Fusion implementation of model (7.3). Click here to download.

using System;
using mosek.fusion;

namespace mosek.fusion.example
{
  public class pow1
  {
    public static void Main(string[] args)
    {
      using (Model M = new Model("pow1"))
      {

        Variable x  = M.Variable("x", 3, Domain.Unbounded());
        Variable x3 = M.Variable();
        Variable x4 = M.Variable();

        // Create the linear constraint
        double[] aval = new double[] {1.0, 1.0, 0.5};
        M.Constraint(Expr.Dot(x, aval), Domain.EqualsTo(2.0));

        // Create the exponential conic constraint
        // Create the conic constraints
        M.Constraint(Var.Vstack(x.Slice(0,2), x3), Domain.InPPowerCone(0.2));
        M.Constraint(Expr.Vstack(x.Index(2), 1.0, x4), Domain.InPPowerCone(0.4));      

        // Set the objective function
        double[] cval = new double[] {1.0, 1.0, -1.0};
        M.Objective(ObjectiveSense.Maximize, Expr.Dot(cval, Var.Vstack(x3, x4, x.Index(0))));

        // Solve the problem
        M.Solve();

        // Get the linear solution values
        double[] solx = x.Level();
        Console.WriteLine("x,y,z = {0}, {1}, {2}", solx[0], solx[1], solx[2]);
      }
    }
  }
}