Least-squares¶
Consider the following constrained least-squares problem
\[\begin{split}\begin{array}{ll}
\mbox{minimize} & \frac{1}{2} \|Ax - b\|_2^2 \\
\mbox{subject to} & 0 \leq x \leq 1
\end{array}\end{split}\]
The problem has the following equivalent form
\[\begin{split}\begin{array}{ll}
\mbox{minimize} & \frac{1}{2} y^T y \\
\mbox{subject to} & y = A x - b \\
& 0 \le x \le 1
\end{array}\end{split}\]
Python¶
import osqp
import numpy as np
import scipy as sp
from scipy import sparse
# Generate problem data
sp.random.seed(1)
m = 30
n = 20
Ad = sparse.random(m, n, density=0.7, format='csc')
b = np.random.randn(m)
# OSQP data
P = sparse.block_diag([sparse.csc_matrix((n, n)), sparse.eye(m)], format='csc')
q = np.zeros(n+m)
A = sparse.bmat([[Ad, -sparse.eye(m)],
[sparse.eye(n), None]], format='csc')
l = np.hstack([b, np.zeros(n)])
u = np.hstack([b, np.ones(n)])
# Create an OSQP object
prob = osqp.OSQP()
# Setup workspace
prob.setup(P, q, A, l, u)
# Solve problem
res = prob.solve()
Matlab¶
% Generate problem data
rng(1)
m = 30;
n = 20;
Ad = sprandn(m, n, 0.7);
b = randn(m, 1);
% OSQP data
P = blkdiag(sparse(n, n), speye(m));
q = zeros(n+m, 1);
A = [Ad, -speye(m);
speye(n), sparse(n, m)];
l = [b; zeros(n, 1)];
u = [b; ones(n, 1)];
% Create an OSQP object
prob = osqp;
% Setup workspace
prob.setup(P, q, A, l, u);
% Solve problem
res = prob.solve();
CVXPY¶
from cvxpy import *
import numpy as np
import scipy as sp
from scipy import sparse
# Generate problem data
sp.random.seed(1)
m = 30
n = 20
A = sparse.random(m, n, density=0.7, format='csc')
b = np.random.randn(m)
# Define problem
x = Variable(n)
objective = 0.5*sum_squares(A@x - b)
constraints = [x >= 0, x <= 1]
# Solve with OSQP
Problem(Minimize(objective), constraints).solve(solver=OSQP)
YALMIP¶
% Generate data
rng(1)
m = 30;
n = 20;
A = sprandn(m, n, 0.7);
b = randn(m, 1);
% Define problem
x = sdpvar(n, 1);
objective = 0.5*norm(A*x - b)^2;
constraints = [ 0 <= x <= 1];
% Solve with OSQP
options = sdpsettings('solver','osqp');
optimize(constraints, objective, options);