Global minimum versus local minima solution with Python Gekko

Question

A simple optimization example has 2 local minima at (0,0,8) with objective 936.0 and (7,0,0) with objective 951.0. What are techniques to use local optimizers in Python Gekko (APOPT,BPOPT,IPOPT) to find a global solution?

from gekko import GEKKO
m = GEKKO(remote=False)
x = m.Array(m.Var,3,lb=0)
x1,x2,x3 = x
m.Minimize(1000-x1**2-2*x2**2-x3**2-x1*x2-x1*x3)
m.Equations([8*x1+14*x2+7*x3==56,
             x1**2+x2**2+x3**2>=25])
m.solve(disp=False)
res=[print(f'x{i+1}: {xi.value[0]}') for i,xi in enumerate(x)]
print(f'Objective: {m.options.objfcnval:.2f}')

This produces a local minimum:

x1: 7.0
x2: 0.0
x3: 0.0
Objective: 951.00

There are solvers for a global optimum such as BARON, COCOS, GlobSol, ICOS, LGO, LINGO, and OQNLP, but what are some quick strategies that can be used with a local optimizer to search for a global solution? Some industrial applications have highly nonlinear models that haven't been fully tested for global solutions in control and design. Can the strategy be parallelized in Python?

Answer 1

Multi-start approaches (using random starting points) can be useful. No guarantee about global optimality, but at least you are protected a bit against some embarrassingly bad local solutions. Some local NLP solvers have this built-in (e.g. Knitro).

Here is Python code for the example using a multi-start method to get the global solution. It uses multi-threading to parallelize the search.

import numpy as np
import threading
import time, random
from gekko import GEKKO

class ThreadClass(threading.Thread):
    def __init__(self, id, xg):
        s = self
        s.id = id
        s.m = GEKKO(remote=False)
        s.xg = xg
        s.objective = float('NaN')

        # initialize variables
        s.m.x = s.m.Array(s.m.Var,3,lb=0)
        for i in range(3):
            s.m.x[i].value = xg[i]
        s.m.x1,s.m.x2,s.m.x3 = s.m.x

        # Equations
        s.m.Equation(8*s.m.x1+14*s.m.x2+7*s.m.x3==56)
        s.m.Equation(s.m.x1**2+s.m.x2**2+s.m.x3**2>=25)

        # Objective
        s.m.Minimize(1000-s.m.x1**2-2*s.m.x2**2-s.m.x3**2
                     -s.m.x1*s.m.x2-s.m.x1*s.m.x3)

        # Set solver option
        s.m.options.SOLVER = 1

        threading.Thread.__init__(s)

    def run(self):
        print('Running application ' + str(self.id) + '\n')
        self.m.solve(disp=False,debug=0) # solve
        # Retrieve objective if successful
        if (self.m.options.APPSTATUS==1):
            self.objective = self.m.options.objfcnval
        else:
            self.objective = float('NaN')
        self.m.cleanup()

# Optimize at mesh points
x1_ = np.arange(0.0, 10.0, 3.0)
x2_ = np.arange(0.0, 10.0, 3.0)
x3_ = np.arange(0.0, 10.0, 3.0)
x1,x2,x3 = np.meshgrid(x1_,x2_,x3_)

threads = [] # Array of threads

# Load applications
id = 0
for i in range(x1.shape[0]):
    for j in range(x1.shape[1]):
        for k in range(x1.shape[2]):
            xg = (x1[i,j,k],x2[i,j,k],x3[i,j,k])
            # Create new thread
            threads.append(ThreadClass(id, xg))
            # Increment ID
            id += 1

# Run applications simultaneously as multiple threads
# Max number of threads to run at once
max_threads = 8
for t in threads:
    while (threading.activeCount()>max_threads):
        # check for additional threads every 0.01 sec
        time.sleep(0.01)
    # start the thread
    t.start()

# Check for completion
mt = 10.0 # max time (sec)
it = 0.0  # time counter
st = 1.0  # sleep time (sec)
while (threading.active_count()>=3):
    time.sleep(st)
    it = it + st
    print('Active Threads: ' + str(threading.active_count()))
    # Terminate after max time
    if (it>=mt):
        break

# Initialize array for objective
obj = np.empty_like(x1)

# Retrieve objective results
id = 0
id_best = 0; obj_best = 1e10
for i in range(x1.shape[0]):
    for j in range(x1.shape[1]):
        for k in range(x1.shape[2]):
            obj[i,j,k] = threads[id].objective
            if obj[i,j,k]<obj_best:
                id_best = id
                obj_best = obj[i,j,k]
            id += 1

print(obj)
print(f'Best objective {obj_best}')
print(f'Solution {threads[id_best].m.x}')

It produces the global solution:

Best objective 936.0
Solution [[0.0] [0.0] [8.0]]