Warm Start quadprog
This example shows how a warm start object increases the speed of the solution in a large, dense quadratic problem. Create a scaled problem with N
variables and 10N
linear inequality constraints. Set N
to 1000.
rng default % For reproducibility N = 1000; rng default A = randn([10*N,N]); b = 5*ones(size(A,1),1); f = sqrt(N)*rand(N,1); H = (4+N/10)*eye(N) + randn(N); H = H + H'; Aeq = []; beq = []; lb = -ones(N,1); ub = -lb;
Create a warm start object for quadprog
, starting from zero.
opts = optimoptions('quadprog','Algorithm','active-set'); x0 = zeros(N,1); ws = optimwarmstart(x0,opts);
Solve the problem, and time the result.
tic [ws1,fval1,eflag1,output1,lambda1] = quadprog(H,f,A,b,Aeq,beq,lb,ub,ws);
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. <stopping criteria details>
toc
Elapsed time is 9.221035 seconds.
The solution has several active linear inequality constraints, and no active bounds.
nnz(lambda1.ineqlin)
ans = 211
nnz(lambda1.lower)
ans = 0
nnz(lambda1.upper)
ans = 0
The solver takes a few hundred iterations to converge.
output1.iterations
ans = 216
Change one random objective to twice its original value.
idx = randi(N); f(idx) = 2*f(idx);
Solve the problem with the new objective, starting from the previous warm start solution.
tic [ws2,fval2,eflag2,output2,lambda2] = quadprog(H,f,A,b,Aeq,beq,lb,ub,ws1);
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. <stopping criteria details>
toc
Elapsed time is 1.490214 seconds.
The solver takes much less time to solve the new problem.
The new solution has about the same number of active constraints.
nnz(lambda2.ineqlin)
ans = 214
nnz(lambda2.lower)
ans = 0
nnz(lambda2.upper)
ans = 0
The new solution is near the previous solution.
norm(ws2.X - ws1.X)
ans = 0.0987
norm(ws2.X)
ans = 2.4229
The difference in speed is largely due to the solver taking many fewer iterations.
output2.iterations
ans = 29