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. 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].
Create the parameter n and the utility
matrix T. The matrix T is a
sparse circulant matrix that is simply a helper for creating the sparse
positive-definite quadratic matrix H.
n = 30000; % Adjust n to a large value
T = spalloc(n,n,n); % make a sparse circulant matrix
r = 1:n-1;
for m = r
T(m,m+1)=1;
end
T(n,1) = 1;Create a sparse vector v. Then create
the matrix H by shifted versions of v*v'.
The matrix T creates shifts of v.
v(n) = 0; v(1) = 1; v(2) = 2; v(4) = 3;
v = (sparse(v))';
% Make a banded type of matrix
H = spalloc(n,n,7*n);
r = 1:n;
for m = r
H = H + v*v';
v = T*v;
endTake a look at the structure of H:
spy(H)

Create the problem vector f and linear
constraint.
f = -r; % linear term A = ones(1,n); b = 0;
Solve the quadratic programming problem with the interior-point-convex algorithm.
Set the StepTolerance option to a very low value
so that the algorithm does not stop too early.
options = optimoptions(@quadprog,'Algorithm','interior-point-convex','StepTolerance',1e-15);
[x,fval,exitflag,output,lambda] = ...
quadprog(H,f,A,b,[],[],[],[],[],options);View the solution value, number of iterations, and Lagrange multiplier associated with linear inequalities:
fval,output.iterations,lambda.ineqlin
fval =
-3.1331e+10
ans =
6
ans =
1.5000e+04Since there are no lower bounds or upper bounds, all the values
in lambda.lower and lambda.upper are 0.
The inequality constraint is active, since lambda.ineqlin is
nonzero.
On many computers you cannot create a full n-by-n matrix
when n = 30000. So you can run this problem only
using sparse matrices.