Convert Quadratic Programming Problem to Second-Order Cone Program

This example show how to convert a positive semidefinite quadratic programming problem to the second-order cone form used by the coneprog solver. A quadratic programming problem has the form

$$\underset{x}{\min }\frac{1}{2}{x}^{T}Hx+f^{T}x$$,

possibly subject to bounds and linear constraints. coneprog solves problems in the form

$$\underset{x}{\min }{f}_{sc}^{T}x$$

such that

$$\Vert A_{sc}x-b_{sc}\Vert \le d_{sc}^{T}x-\gamma$$,

possibly subject to bounds and linear constraints.

To convert a quadratic program to coneprog form, first calculate the matrix square root of the matrix $$H$$. Assuming that $$H$$ is a symmetric positive semidefinite matrix, the command

A = sqrtm(H);

returns a positive semidefinite matrix A such that A'*A = A*A = H. Therefore,

$$x^{T}Hx=x^T{A}^{T}Ax=(Ax{)}^{T}Ax=\Vert Ax{\Vert }^2$$.

Modify the form of the quadratic program as follows:

$$\underset{x}{\min }\frac{1}{2}{x}^{T}Hx+f^{T}x=-\frac{1}{2}+\underset{x,t}{\min }[(t+1/2)+f^{T}x]$$

where $$t$$ satisfies the constraint

$$t+1/2\ge \frac{1}{2}{x}^{T}Hx$$.

Extend the control variable $$x$$ to $$u$$, which includes $$t$$ as its last element:

$\mathit{u}=\left[\begin{array}{c}\mathit{x}\\ \mathit{t}\end{array}\right]$.

Extend the second-order cone constraint matrices and vectors as follows:

$${\mathit{A}}_{\mathrm{sc}}=\left[\begin{array}{cc}\mathit{A}& 0\\ 0& 1\end{array}\right]$$

$${\mathit{b}}_{\mathrm{sc}}=\left[\begin{array}{c}0\\ ⋮\\ 0\end{array}\right]$$

$${\mathit{d}}_{\mathrm{sc}}=\left[\begin{array}{c}0\\ ⋮\\ 0\\ 1\end{array}\right]$$

$\gamma =-1$.

Extend the coefficient vector $$f$$ as well:

${\mathit{f}}_{\mathrm{sc}}=\left[\begin{array}{c}\mathit{f}\\ 1\end{array}\right]$.

In terms of the new variables, the quadratic programming problem becomes

$$\underset{u}{\min }\frac{1}{2}{u}^T{A}_{sc}^{2}u+f_{sc}^{T}u=-1/2+\underset{u}{\min }[1/2+f_{sc}^{T}u]$$

where

$$u(end)+1/2\ge \frac{1}{2}{u}^T{A}_{sc}u$$.

This quadratic constraint becomes a cone constraint through the following calculation, which uses the earlier definitions of $$A_{sc}$$, $$d_{sc}$$, and $$\gamma$$:

$$\frac{1}{2}\Vert Hx{\Vert }^2=\frac{1}{2}{u}^T{A}_{sc}^{2}u\le t+\frac{1}{2}$$

$$\Vert Hx{\Vert }^2\le 2t+1$$

$$\Vert Hx{\Vert }^2+t^2\le t^2+2t+1=(t+1{)}^2$$

$$\sqrt{\Vert Hx{\Vert }^2+t^2}\le |t+1|=\pm (t+1)$$

But $$\Vert Hx{\Vert }^2+t^2=\Vert A_{sc}u{\Vert }^2$$. So, recalling that $$\gamma =-1$$ and the definition of $$d_{sc}$$, the inequality becomes

$$\Vert A_{sc}u\Vert \le \pm (t-\gamma )$$

$$\Vert A_{sc}u\Vert \le \pm (d_{sc}^{T}u-\gamma )$$.

The quadratic program has the same solution as the corresponding cone program. The only difference is the added term $$-1/2$$ in the cone program.

Numerical Example

The quadprog documentation gives this example.

H = [1,-1,1
    -1,2,-2
    1,-2,4];
f = [-7;-12;-15];
lb = zeros(3,1);
ub = ones(size(lb));
Aineq = [1,1,1];
bineq = 3;
[xqp fqp] = quadprog(H,f,Aineq,bineq,[],[],lb,ub)

Minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.

xqp = 3×1

    1.0000
    1.0000
    1.0000

fqp = 
-32.5000

Referring to the description at the beginning of this example, specify the second-order cone constraint variables, and then call the coneprog function.

Asc = sqrtm(H);
Asc((end+1),(end+1)) = 1;
d = [zeros(size(f(:)));1];
gamma = -1;
b = zeros(size(d));
qp = secondordercone(Asc,b,d,gamma);
Aq = Aineq;
Aq(:,(end+1)) = 0;
lb(end+1) = -Inf;
ub(end+1) = Inf;
[u,fval,eflag] = coneprog([f(:);1],qp,Aq,bineq,[],[],lb,ub)

Optimal solution found.

u = 4×1

    1.0000
    1.0000
    1.0000
    1.0000

fval = 
-33.0000

eflag = 
1

The first three elements of the cone solution u are equal to the elements of the quadratic programming solution xqp, to display precision:

disp([xqp,u(1:(end-1))])

    1.0000    1.0000
    1.0000    1.0000
    1.0000    1.0000

The returned quadratic function value fqp is the returned cone value minus 1/2 when $$2t+1$$ is positive, or plus 1/2 when $$2t+1$$ is negative.

disp([fqp-sign(2*u(end)+1)*1/2 fval])

  -33.0000  -33.0000