Convert Quadratic Constraints to Second-Order Cone Constraints

This example shows how to convert a quadratic constraint to the second-order cone constraint form. A quadratic constraint has the form

$$x^{T}Qx+2q^{T}x+c\le 0.$$

Second-order cone programming has constraints of the form

$$\Vert A_{sc}(i)\cdot x-b_{sc}(i)\Vert \le d_{sc}(i)\cdot x-\gamma (i)$$.

The matrix $$Q$$ must be symmetric and positive semidefinite for you to convert quadratic constraints. Let $$S$$ be the square root of $$Q$$, meaning $$Q=S*S=S^T*S$$. You can compute $$S$$ using sqrtm. Suppose that there is a solution $$b$$ to the equation $$S^{T}b=–q$$, which is always true when $$Q$$ is positive definite. Compute $$b$$ using b = -S\q.

$$\begin{array}{ll}{x}^{T}Qx+2q^{T}x+c& =x^T{S}^{T}Sx-2{\left(S^{T}b\right)}^{T}x+c\\ & ={\left(Sx-b\right)}^T\left(Sx-b\right)-b^{T}b+c\\ & ={\Vert Sx-b\Vert }^2+c-b^{T}b.\end{array}$$

Therefore, if $$b^{T}b>c$$, then the quadratic constraint is equivalent to the second-order cone constraint with

• $$A_{sc}=S$$

• $$b_{sc}=b$$

• $$d_{sc}=0$$

• $$\gamma =-\sqrt{b^{T}b-c}$$

Numeric Example

Specify a five-element vector f representing the objective function $$f^{T}x$$.

f = [1;-2;3;-4;5];

Set the quadratic constraint matrix Q as a 5-by-5 random positive definite matrix. Set q as a random 5-element vector, and take the additive constant $$c=-1$$.

rng default % For reproducibility
Q = randn(5) + 3*eye(5);
Q = (Q + Q')/2; % Make Q symmetric
q = randn(5,1);
c = -1;

To create the inputs for coneprog, create the matrix S as the square root of Q.

S = sqrtm(Q);

Create the remaining inputs for the second-order cone constraint as specified in the first part of this example.

b = -S\q;
d = zeros(size(b));
gamma = -sqrt(b'*b-c);
sc = secondordercone(S,b,d,gamma);

Call coneprog to solve the problem.

[x,fval] = coneprog(f,sc)

Optimal solution found.

x = 5×1

   -0.7194
    0.2669
   -0.6309
    0.2543
   -0.0904

fval = 
-4.6148

Compare this result to the result returned by solving this same problem using fmincon. Write the quadratic constraint as described in Anonymous Nonlinear Constraint Functions.

x0 = randn(5,1); % Initial point for fmincon
nlc = @(x)x'*Q*x + 2*q'*x + c;
nlcon = @(x)deal(nlc(x),[]);
[xfmc,fvalfmc] = fmincon(@(x)f'*x,x0,[],[],[],[],[],[],nlcon)

Local 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.

xfmc = 5×1

   -0.7196
    0.2672
   -0.6312
    0.2541
   -0.0902

fvalfmc = 
-4.6148

The two solutions are nearly identical.