This example shows the value of using sparse arithmetic when you have a sparse problem. The matrix has n
rows, where you choose n
to be a large value, and a few nonzero diagonal bands. A full matrix of size n
-by-n
can use up all available memory, but a sparse matrix presents no problem.
The problem is to minimize x'*H*x/2 + f'*x
subject to
x(1) + x(2) + ... + x(n) <= 0
,
where f = [-1;-2;-3;...;-n]
. H
is a sparse symmetric banded matrix.
Create a symmetric circulant matrix based on shifts of the vector [3,6,2,14,2,6,3]
, with 14 being on the main diagonal. Have the matrix be n
-by-n
, where n = 30,000
.
n = 3e4;
H2 = speye(n);
H = 3*circshift(H2,-3,2) + 6*circshift(H2,-2,2) + 2*circshift(H2,-1,2)...
+ 14*H2 + 2*circshift(H2,1,2) + 6*circshift(H2,2,2) + 3*circshift(H2,3,2);
View the matrix structure.
spy(H)
The linear constraint is that the sum of the solution elements is nonpositive. The objective function contains a linear term expressed in the vector f
.
A = ones(1,n); b = 0; f = 1:n; f = -f;
Solve the quadratic programming problem using the 'interior-point-convex'
algorithm. To keep the solver from stopping prematurely, set the StepTolerance
option to 0
.
options = optimoptions(@quadprog,'Algorithm','interior-point-convex','StepTolerance',0); [x,fval,exitflag,output,lambda] = ... quadprog(H,f,A,b,[],[],[],[],[],options);
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>
On many computers you cannot create a full n
-by-n
matrix when n
= 30,000. So you can run this problem only by using sparse matrices.
View the objective function value, number of iterations, and Lagrange multiplier associated with linear inequality.
fprintf('The objective function value is %d.\nThe number of iterations is %d.\nThe Lagrange multiplier is %d.\n',... fval,output.iterations,lambda.ineqlin)
The objective function value is -3.133073e+10. The number of iterations is 7. The Lagrange multiplier is 1.500050e+04.
Because there are no lower bounds, upper bounds, or linear equality constraints, the only meaningful Lagrange multiplier is lambda.ineqlin
. Because lambda.ineqlin
is nonzero, you can tell that the inequality constraint is active. Evaluate the constraint to see that the solution is on the boundary.
fprintf('The linear inequality constraint A*x has value %d.\n',A*x)
The linear inequality constraint A*x has value 9.150244e-08.
The sum of the solution components is zero to within tolerances.
The solution x
has three regions: an initial portion, a final portion, and an approximately linear portion over most of the solution. Plot the three regions.
subplot(3,1,1) plot(x(1:60)) title('x(1) Through x(60)') subplot(3,1,2) plot(x(61:n-60)) title('x(61) Through x(n-60)') subplot(3,1,3) plot(x(n-59:n)) title('x(n-59) Through x(n)')