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
Second-order cone programming has constraints of the form
The matrix must be symmetric and positive semidefinite for you to convert quadratic constraints. Let be the square root of , meaning . You can compute using
sqrtm. Suppose that there is a solution to the equation , which is always true when is positive definite. Compute using
b = -S\q.
Therefore, if , then the quadratic constraint is equivalent to the second-order cone constraint with
Specify a five-element vector
f representing the objective function .
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 .
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
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);
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.