使用 Python Gekko 的全局最小值与局部最小值解决方案

一个简单的优化示例有 2 个局部最小值(0,0,8)有目标的936.0 and (7,0,0)有目标的951.0。在 Python Gekko 中使用本地优化器的技术有哪些（APOPT,BPOPT,IPOPT）寻找全局解决方案？

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}')

这会产生局部最小值：

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

有一些用于全局最优解的求解器，例如BARON, COCOS, GlobSol, ICOS, LGO, LINGO, and OQNLP，但是有哪些快速策略可以与局部优化器一起使用来搜索全局解决方案？一些工业应用具有高度非线性模型，这些模型尚未针对控制和设计方面的全局解决方案进行全面测试。该策略可以在 Python 中并行化吗？

多启动方法（使用随机起点）可能很有用。不能保证全局最优，但至少你可以免受一些令人尴尬的糟糕的本地解决方案的影响。一些本地 NLP 求解器具有此内置功能（例如 Knitro）。

以下是使用多启动方法获取全局解的示例的 Python 代码。它使用多线程来并行化搜索。

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}')

它产生了全局解决方案：

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