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

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

Second-order cone programming has constraints of the form

$‖{A}_{sc}\left(i\right)\cdot x-{b}_{sc}\left(i\right)‖\le {d}_{sc}\left(i\right)\cdot x-\gamma \left(i\right)$.

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.

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

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.