Python constrained non-linear optimization

Question

What's the recommended package for constrained non-linear optimization in python ?

The specific problem I'm trying to solve is this:

I have an unknown X (Nx1), I have M (Nx1) u vectors and M (NxN) s matrices.

max [5th percentile of (ui_T*X), i in 1 to M]
st 
0<=X<=1 and
[95th percentile of (X_T*si*X), i in 1 to M]<= constant

When I started out the problem I only had one point estimate for u and s and I was able to solve the problem above with cvxpy.

I realized that instead of one estimate for u and s, I had the entire distribution of values so I wanted to change my objective function so that I could use the entire distribution. The problem description above is my attempt to include that information in a meaningful way.

cvxpy cannot be used to solve this, I've tried scipy.optimize.anneal, but I can't seem to set bounds on the unknown values. I've looked at pulp too but it doesnt allow nonlinear constraints.

Answer 1

While the SLSQP algorithm in scipy.optimize.minimize is good, it has a bunch of limitations. The first of which is it's a QP solver, so it works will for equations that fit well into a quadratic programming paradigm. But what happens if you have functional constraints? Also, scipy.optimize.minimize is not a global optimizer, so you often need to start very close to the final results.

There is a constrained nonlinear optimization package (called mystic) that has been around for nearly as long as scipy.optimize itself -- I'd suggest it as the go-to for handling any general constrained nonlinear optimization.

For example, your problem, if I understand your pseudo-code, looks something like this:

import numpy as np

M = 10
N = 3
Q = 10
C = 10

# let's be lazy, and generate s and u randomly...
s = np.random.randint(-Q,Q, size=(M,N,N))
u = np.random.randint(-Q,Q, size=(M,N))

def percentile(p, x):
    x = np.sort(x)
    p = 0.01 * p * len(x)
    if int(p) != p:
        return x[int(np.floor(p))]
    p = int(p)
    return x[p:p+2].mean()

def objective(x, p=5): # inverted objective, to find the max
    return -1*percentile(p, [np.dot(np.atleast_2d(u[i]), x)[0] for i in range(0,M-1)])


def constraint(x, p=95, v=C): # 95%(xTsx) - v <= 0
    x = np.atleast_2d(x)
    return percentile(p, [np.dot(np.dot(x,s[i]),x.T)[0,0] for i in range(0,M-1)]) - v

bounds = [(0,1) for i in range(0,N)]

So, to handle your problem in mystic, you just need to specify the bounds and the constraints.

from mystic.penalty import quadratic_inequality
@quadratic_inequality(constraint, k=1e4)
def penalty(x):
  return 0.0

from mystic.solvers import diffev2
from mystic.monitors import VerboseMonitor
mon = VerboseMonitor(10)

result = diffev2(objective, x0=bounds, penalty=penalty, npop=10, gtol=200, \
                 disp=False, full_output=True, itermon=mon, maxiter=M*N*100)

print result[0]
print result[1]

The result looks something like this:

Generation 0 has Chi-Squared: -0.434718
Generation 10 has Chi-Squared: -1.733787
Generation 20 has Chi-Squared: -1.859787
Generation 30 has Chi-Squared: -1.860533
Generation 40 has Chi-Squared: -1.860533
Generation 50 has Chi-Squared: -1.860533
Generation 60 has Chi-Squared: -1.860533
Generation 70 has Chi-Squared: -1.860533
Generation 80 has Chi-Squared: -1.860533
Generation 90 has Chi-Squared: -1.860533
Generation 100 has Chi-Squared: -1.860533
Generation 110 has Chi-Squared: -1.860533
Generation 120 has Chi-Squared: -1.860533
Generation 130 has Chi-Squared: -1.860533
Generation 140 has Chi-Squared: -1.860533
Generation 150 has Chi-Squared: -1.860533
Generation 160 has Chi-Squared: -1.860533
Generation 170 has Chi-Squared: -1.860533
Generation 180 has Chi-Squared: -1.860533
Generation 190 has Chi-Squared: -1.860533
Generation 200 has Chi-Squared: -1.860533
Generation 210 has Chi-Squared: -1.860533
STOP("ChangeOverGeneration with {'tolerance': 0.005, 'generations': 200}")
[-0.17207128  0.73183465 -0.28218955]
-1.86053344078

mystic is very flexible, and can handle any type of constraints (e.g. equalities, inequalities) including symbolic and functional constraints. I specified the constraints as "penalties" above, which is the traditional way, in that they apply a penalty to the objective when the constraint is violated. mystic also provides nonlinear kernel transformations, which constrain solution space by reducing the space of valid solutions (i.e. by a spatial mapping or kernel transformation).

As an example, here's mystic solving a problem that breaks a lot of QP solvers, since the constraints are not in the form of a constraints matrix. It's optimizing the design of a pressure vessel.

"Pressure Vessel Design"

def objective(x):
    x0,x1,x2,x3 = x
    return 0.6224*x0*x2*x3 + 1.7781*x1*x2**2 + 3.1661*x0**2*x3 + 19.84*x0**2*x2

bounds = [(0,1e6)]*4
# with penalty='penalty' applied, solution is:
xs = [0.72759093, 0.35964857, 37.69901188, 240.0]
ys = 5804.3762083

from mystic.symbolic import generate_constraint, generate_solvers, simplify
from mystic.symbolic import generate_penalty, generate_conditions

equations = """
-x0 + 0.0193*x2 <= 0.0
-x1 + 0.00954*x2 <= 0.0
-pi*x2**2*x3 - (4/3.)*pi*x2**3 + 1296000.0 <= 0.0
x3 - 240.0 <= 0.0
"""
cf = generate_constraint(generate_solvers(simplify(equations)))
pf = generate_penalty(generate_conditions(equations), k=1e12)


if __name__ == '__main__':

    from mystic.solvers import diffev2
    from mystic.math import almostEqual
    from mystic.monitors import VerboseMonitor
    mon = VerboseMonitor(10)

    result = diffev2(objective, x0=bounds, bounds=bounds, constraints=cf, penalty=pf, \ 
                     npop=40, gtol=50, disp=False, full_output=True, itermon=mon)

    assert almostEqual(result[0], xs, rel=1e-2)
    assert almostEqual(result[1], ys, rel=1e-2)

Find this, and roughly 100 examples like it, here: https://github.com/uqfoundation/mystic.

I'm the author, so I am slightly biased. However, the bias is very slight. mystic is both mature and well-supported, and is unparalleled in capacity to solve hard constrained nonlinear optimization problems.

Answer 2

As others have commented as well, the SciPy minimize package is a good place to start. I've included an example below (Hock Schittkowski #71 benchmark) that includes an objective function, equality constraint, and inequality constraint.

import numpy as np
from scipy.optimize import minimize

def objective(x):
    return x[0]*x[3]*(x[0]+x[1]+x[2])+x[2]

def constraint1(x):
    return x[0]*x[1]*x[2]*x[3]-25.0

def constraint2(x):
    sum_eq = 40.0
    for i in range(4):
        sum_eq = sum_eq - x[i]**2
    return sum_eq

# initial guesses
n = 4
x0 = np.zeros(n)
x0[0] = 1.0
x0[1] = 5.0
x0[2] = 5.0
x0[3] = 1.0

# show initial objective
print('Initial SSE Objective: ' + str(objective(x0)))

# optimize
b = (1.0,5.0)
bnds = (b, b, b, b)
con1 = {'type': 'ineq', 'fun': constraint1} 
con2 = {'type': 'eq', 'fun': constraint2}
cons = ([con1,con2])
solution = minimize(objective,x0,method='SLSQP',\
                    bounds=bnds,constraints=cons)
x = solution.x

# show final objective
print('Final SSE Objective: ' + str(objective(x)))

# print solution
print('Solution')
print('x1 = ' + str(x[0]))
print('x2 = ' + str(x[1]))
print('x3 = ' + str(x[2]))
print('x4 = ' + str(x[3]))

There is also a more comprehensive discussion thread on nonlinear programming solvers for Python if SLSQP can't solve your problem. My course material on Engineering Design Optimization is available if you need additional information on the solver methods.

Answer 3

Typically for fitting you can use scipy.optimize functions, or lmfit which simply extends the scipy.optimize package to make it easier to pass things like bounds. Personally, I like using kmpfit, part of the kapteyn library and is based on the C implementation of MPFIT.

scipy.optimize.minimize() is probably the most easy to obtain and is commonly used.

Answer 4

scipy has a spectacular package for constrained non-linear optimization.

You can get started by reading the optimize doc, but here's an example with SLSQP:

minimize(func, [-1.0,1.0], args=(-1.0,), jac=func_deriv, constraints=cons, method='SLSQP', options={'disp': True})