6.4 Power Cone Optimization¶
Conic optimization is a generalization of linear optimization, allowing constraints of the type
where \(x^t\) is a subset of the problem variables and \(\K_t\) is a convex cone. Since the set \(\real^n\) of real numbers is also a convex cone, we can simply write a compound conic constraint \(x\in \K\) where \(\K=\K_1\times\cdots\times\K_l\) is a product of smaller cones and \(x\) is the full problem variable.
MOSEK can solve conic optimization problems of the form
where the domain restriction, \(x \in \K\), implies that all variables are partitioned into convex cones
In this tutorial we describe how to use the power cone. The primal power cone of dimension \(n\) with parameter \(0<\alpha<1\) is defined as:
In particular, the most important special case is the three-dimensional power cone family:
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 \(x,y\geq 0\).
MOSEK also supports the dual power cone:
For other types of cones supported by MOSEK see Sec. 6.3 (Conic Quadratic Optimization), Sec. 6.5 (Conic Exponential Optimization), Sec. 6.6 (Semidefinite Optimization). Different cone types can appear together in one optimization problem.
Furthermore, each variable may belong to one cone at most. The constraint \(x_i - x_j = 0\) would however allow \(x_i\) and \(x_j\) to belong to different cones with same effect.
6.4.1 Example POW1¶
Consider the following optimization problem which involves powers of variables:
With \((x,y,z)=(x_0,x_1,x_2)\) we convert it into conic form using auxiliary variables as bounds for the power expressions:
The linear constraints are specified as if the problem was a linear problem. The cone elements are specified using two index lists cones.subptr
and cones.sub
. Cone-type identifiers appear in cones.type
, and the cone parameters \(\alpha\) are in cones.conepar
. The elements of all the cones are listed in cones.sub
, and cones.subptr
specifies the index of the first element in cones.sub
for each cone.
Listing 6.8 demonstrates how to solve the example (6.7) using MOSEK. The solution is
[ 0.06389298 0.78308564 2.30604283 ]
function pow1()
clear prob;
[r, res] = mosekopt('symbcon');
% Specify the non-conic part of the problem.
prob.c = [-1 0 0 1 1 0];
prob.a = [1 1 0.5 0 0 0];
prob.blc = [2.0];
prob.buc = [2.0];
prob.blx = [-inf -inf -inf -inf -inf 1.0];
prob.bux = [ inf inf inf inf inf 1.0];
% Specify the cones.
prob.cones.type = [res.symbcon.MSK_CT_PPOW res.symbcon.MSK_CT_PPOW];
prob.cones.conepar= [0.2 0.4];
prob.cones.sub = [1 2 4 3 6 5];
prob.cones.subptr = [1 4];
% The field 'type' specifies the cone types, in this case power cones.
%
% The fields 'sub' and 'subptr' specify the members of the cones,
% i.e., the above definitions imply that
% (x(1), x(2), x(4)) and (x(3), x(6), x(5))
% are cones.
%
% The field 'conepar' specifies the alpha cone parameters (exponents)
% Optimize the problem.
[r,res]=mosekopt('maximize',prob);
% Display the primal solution.
res.sol.itr.xx'