Nonlinear Optimization Using Python

Question

What is the recommended package for limited nonlinear optimization in python?

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

I have an unknown X(Nx1), I have vectors M(Nx1) uand M(NxN) s.

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 the problem, I had only one point estimate for uand s, and I was able to solve the problem above with cvxpy.

I realized that instead of a single evaluation for uand sI had the entire distribution of values, so I wanted to change my objective function so that I could use the entire distribution. The description of the problem above is my attempt to include this information in a meaningful way.

cvxpycan not be used to solve this problem, I tried scipy.optimize.anneal, but I can not set boundaries for unknown values. I also looked at pulp, but does not allow non-linear constraints.

Answer 1

scipy has an impressive package for limited nonlinear optimization.

You can start by reading the optimize doc , but here is an example with SLSQP:

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

Answer 2

SLSQP scipy.optimize.minimize , . QP, , . , ? , scipy.optimize.minimize , .

( mystic), , scipy.optimize - .

, , , :

import numpy as np

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

# let 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)]

, mystic, .

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]

:

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 (, , ), . "" , , , . mystic , (.. ).

, mystic , QP, . .

"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)

100 : https://github.com/uqfoundation/mystic.

, . . mystic , .

Answer 3

, SciPy - . (Hock Schittkowski # 71 benchmark), , .

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]))

Python, SLSQP . , .

Answer 4

scipy.optimize, lmfit, scipy.optimize, , . kmpfit, kapteyn C MPFIT.

scipy.optimize.minimize(), , .