7.2 Conic Quadratic Optimization

Conic optimization is a generalization of linear optimization, allowing constraints of the type

\[x^t \in \K_t,\]

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 quadratic optimization problems of the form

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

where the domain restriction, \(x \in \K\), implies that all variables are partitioned into convex cones

\[x = (x^0, x^1, \ldots , x^{p-1}),\quad \mbox{with } x^t \in \K_t \subseteq \real^{n_t}.\]

In this tutorial we describe how to use the two types of quadratic cones defined as:

• Quadratic cone:

  \[\Q^n = \left\lbrace x \in \real^n: x_0 \geq \sqrt{\sum_{j=1}^{n-1} x_j^2} \right\rbrace.\]

• Rotated quadratic cone:

  \[\Qr^n = \left\lbrace x \in \real^n: 2 x_0 x_1 \geq \sum_{j=2}^{n-1} x_j^2,\quad x_0\geq 0,\quad x_1 \geq 0 \right\rbrace.\]

For other types of cones supported by MOSEK see Sec. 7.3 (Power Cone Optimization), Sec. 7.4 (Conic Exponential Optimization), Sec. 7.5 (Semidefinite Optimization). See Domain for a list and definitions of available cone types. Different cone types can appear together in one optimization problem.

For example, the following constraint:

\[(x_4, x_0, x_2) \in \Q^3\]

describes a convex cone in \(\real^3\) given by the inequality:

\[x_4 \geq \sqrt{x_0^2 + x_2^2}.\]

In Fusion the coordinates of a cone are not restricted to single variables. They can be arbitrary linear expressions, and an auxiliary variable will be substituted by Fusion in a way transparent to the user.

7.2.1 Example CQO1

Consider the following conic quadratic problem which involves some linear constraints, a quadratic cone and a rotated quadratic cone.

(7.2)\[\begin{split} \begin{array}{ll} \minimize & y_1 + y_2 + y_3 \\ \st & x_1 + x_2 + 2.0 x_3 = 1.0,\\ & x_1,x_2,x_3 \geq 0.0,\\ & (y_1,x_1,x_2) \in \Q^3,\\ & (y_2,y_3,x_3) \in \Qr^3. \end{array}\end{split}\]

We start by creating the optimization model:

with Model('cqo1') as M:

We then define variables x and y. Two logical variables (aliases) z1 and z2 are introduced to model the quadratic cones. These are not new variables, but map onto parts of x and y for the sake of convenience.

    x = M.variable('x', 3, Domain.greaterThan(0.0))
    y = M.variable('y', 3, Domain.unbounded())

    # Create the aliases
    #      z1 = [ y[0],x[0],x[1] ]
    #  and z2 = [ y[1],y[2],x[2] ]
    z1 = Var.vstack(y.index(0), x.slice(0, 2))
    z2 = Var.vstack(y.slice(1, 3), x.index(2))

The linear constraint is defined using the dot product:

    # Create the constraint
    #      x[0] + x[1] + 2.0 x[2] = 1.0
    M.constraint("lc", Expr.dot([1.0, 1.0, 2.0], x), Domain.equalsTo(1.0))

The conic constraints are defined using the logical views z1 and z2 created previously. Note that this is a basic way of defining conic constraints, and that in practice they would have more complicated structure.

    # Create the constraints
    #      z1 belongs to C_3
    #      z2 belongs to K_3
    # where C_3 and K_3 are respectively the quadratic and
    # rotated quadratic cone of size 3, i.e.
    #                 z1[0] >= sqrt(z1[1]^2 + z1[2]^2)
    #  and  2.0 z2[0] z2[1] >= z2[2]^2
    qc1 = M.constraint("qc1", z1, Domain.inQCone())
    qc2 = M.constraint("qc2", z2, Domain.inRotatedQCone())

We only need the objective function:

    # Set the objective function to (y[0] + y[1] + y[2])
    M.objective("obj", ObjectiveSense.Minimize, Expr.sum(y))

Calling the Model.solve method invokes the solver:

    M.solve()
    M.writeTask('cqo1.opf')

The primal and dual solution values can be retrieved using Variable.level, Constraint.level and Variable.dual, Constraint.dual, respectively:

    # Get the linear solution values
    solx = x.level()
    soly = y.level()

    # Get conic solution of qc1
    qc1lvl = qc1.level()
    qc1sn = qc1.dual()

Listing 7.2 Fusion implementation of model (7.2). Click here to download.

from mosek.fusion import *

with Model('cqo1') as M:

    x = M.variable('x', 3, Domain.greaterThan(0.0))
    y = M.variable('y', 3, Domain.unbounded())

    # Create the aliases
    #      z1 = [ y[0],x[0],x[1] ]
    #  and z2 = [ y[1],y[2],x[2] ]
    z1 = Var.vstack(y.index(0), x.slice(0, 2))
    z2 = Var.vstack(y.slice(1, 3), x.index(2))

    # Create the constraint
    #      x[0] + x[1] + 2.0 x[2] = 1.0
    M.constraint("lc", Expr.dot([1.0, 1.0, 2.0], x), Domain.equalsTo(1.0))

    # Create the constraints
    #      z1 belongs to C_3
    #      z2 belongs to K_3
    # where C_3 and K_3 are respectively the quadratic and
    # rotated quadratic cone of size 3, i.e.
    #                 z1[0] >= sqrt(z1[1]^2 + z1[2]^2)
    #  and  2.0 z2[0] z2[1] >= z2[2]^2
    qc1 = M.constraint("qc1", z1, Domain.inQCone())
    qc2 = M.constraint("qc2", z2, Domain.inRotatedQCone())

    # Set the objective function to (y[0] + y[1] + y[2])
    M.objective("obj", ObjectiveSense.Minimize, Expr.sum(y))

    # Solve the problem
    M.solve()
    M.writeTask('cqo1.opf')

    # Get the linear solution values
    solx = x.level()
    soly = y.level()
    print('x1,x2,x3 = %s' % str(solx))
    print('y1,y2,y3 = %s' % str(soly))

    # Get conic solution of qc1
    qc1lvl = qc1.level()
    qc1sn = qc1.dual()
    print('qc1 levels                = %s' % str(qc1lvl))
    print('qc1 dual conic var levels = %s' % str(qc1sn))