#
# 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

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

p = xp.problem ()

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

p.solve ()