Using NumPy and Xpress

#
# Example: use numpy to print the product of a matrix by a random vector.
#
# Uses xpress.Dot to on a matrix and a vector. Note that the NumPy dot operator
# works perfectly fine here.
#

from __future__ import print_function

import numpy as np
import xpress as xp

x = [xp.var() for i in range(5)]

p = xp.problem()
p.addVariable(x)
p.addConstraint(xp.Sum(x) >= 2)

p.setObjective(xp.Sum(x[i]**2 for i in range(5)))
p.solve()

A = np.array(range(30)).reshape(6, 5)  # A is a 6x5 matrix
sol = np.array(p.getSolution())  # suppose it's a vector of size 5
columns = A*sol         # not a matrix-vector product!
v = xp.Dot(A, sol)      # this is a matrix-vector product A*sol

print(v)

#
# An example of a problem formulation that uses the xpress.Dot() operator
# to formulate constraints simply. Note that the NumPy dot operator is not
# suitable here as the result is an expression in the Xpress variables.
#

import xpress as xp
import numpy as np

A = np.random.random(30).reshape(6, 5)  # A is a 6x5 matrix
Q = np.random.random(25).reshape(5, 5)  # Q is a 5x5 matrix
x = np.array([xp.var() for i in range(5)])  # vector of variables
x0 = np.random.random(5)  # random vector

Q += 4 * np.eye(5)  # add 5 * the identity matrix

# 6 constraints (rows of A)
Lin_sys = xp.Dot(A, x) <= np.array([3, 4, 1, 4, 8, 7])

# One quadratic constraint
Conv_c = xp.Dot(x, Q, x) <= 1

p = xp.problem()

p.addVariable(x)
p.addConstraint(Lin_sys, Conv_c)
p.setObjective(xp.Dot(x-x0, x-x0))

p.solve()