Is there any quadratic programming function that can have both lower and upper bounds - Python

Question

Normally I have been using GNU Octave to solve quadratic programming problems.

I solve problems like

x = 1/2x'Qx + c'x

With subject to

A*x <= b
lb <= x <= ub

Where lb and ub are lower bounds and upper bounds, e.g limits for x

My Octave code looks like this when I solve. Just one simple line

U = quadprog(Q, c, A, b, [], [], lb, ub);

The square brackets [] are empty because I don't need the equality constraints

Aeq*x = beq,

So my question is: Is there a easy to use quadratic solver in Python for solving problems

x = 1/2x'Qx + c'x

With subject to

A*x <= b
lb <= x <= ub

Or subject to

b_lb <= A*x <= b_ub
lb <= x <= ub

Answer 1

You can write your own solver based scipy.optimize, here is a small example on how to code your custom python quadprog():

# python3
import numpy as np
from scipy import optimize

class quadprog(object):

    def __init__(self, H, f, A, b, x0, lb, ub):
        self.H    = H
        self.f    = f
        self.A    = A
        self.b    = b
        self.x0   = x0
        self.bnds = tuple([(lb, ub) for x in x0])
        # call solver
        self.result = self.solver()

    def objective_function(self, x):
        return 0.5*np.dot(np.dot(x.T, self.H), x) + np.dot(self.f.T, x)

    def solver(self):
        cons = ({'type': 'ineq', 'fun': lambda x: self.b - np.dot(self.A, x)})
        optimum = optimize.minimize(self.objective_function, 
                                    x0          = self.x0.T,
                                    bounds      = self.bnds,
                                    constraints = cons, 
                                    tol         = 10**-3)
        return optimum

Here is how to use this, using the same variables from the first example provided in matlab-quadprog:

# init vars
H  = np.array([[ 1, -1],
               [-1,  2]])

f  = np.array([-2, -6]).T

A  = np.array([[ 1, 1],
               [-1, 2],
               [ 2, 1]])

b  = np.array([2, 2, 3]).T
x0 = np.array([1, 2])
lb = 0
ub = 2

# call custom quadprog
quadprog  = quadprog(H, f, A, b, x0, lb, ub)
print(quadprog.result)

The output of this short snippet is:

     fun: -8.222222222222083
     jac: array([-2.66666675, -4.        ])
 message: 'Optimization terminated successfully.'
    nfev: 8
     nit: 2
    njev: 2
  status: 0
 success: True
       x: array([0.66666667, 1.33333333])

For more information on how to use scipy.optimize.minimize please refer to the docs.