我遇到了一些情况，Gekko 似乎陷入了局部最大值，并且想知道可以使用什么方法来解决这个问题或深入挖掘原因（包括下面的默认设置）。
例如，运行下面的场景会产生“-5127.34945104756”的目标
m = GEKKO(remote=False)
m.options.NODES = 3
m.options.IMODE = 3
m.options.MAX_ITER = 1000
m.options.SOLVER=1
#Limit max lnuc weeks
m.Equation(sum(x8)<=6)
m.Maximize(m.sum(simu_total_volume))
m.solve(disp = True)
#Objective : -5127.34945104756
现在如果我简单地改变“m.Equation(sum(x8))<=6)" to "m.Equation(sum(x8)==6)", it returns a better solution (-5638.55528892101):
m = GEKKO(remote=False)
m.options.NODES = 3
m.options.IMODE = 3
m.options.MAX_ITER = 1000
m.options.SOLVER=1
#Limit max lnuc weeks
m.Equation(sum(x8)==6)
m.Maximize(m.sum(simu_total_volume))
m.solve(disp = True)
# Objective : -5638.55528892101
鉴于“6”落在<=6, is there a reason why Gekko wouldn't try to go all the way up to 6 here? Posting the full code/values would also be difficult given size/scale of the problem, so appreciate any feedback based on this.
的范围内这是一个可以帮助解决局部与全局最小值这一重要主题的示例。以下脚本生成 (7,0,0) 的局部（非全局）解，目标为 951.0。
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}')
BARON、遗传算法、模拟退火等求解器中存在基于梯度的全局优化方法。一种简单的方法是通过网格搜索或智能地执行具有不同初始条件（猜测）的多启动方法如果初始猜测的数量很少，则可以使用贝叶斯方法进行更智能的搜索。
并行线程多启动
网格搜索很容易并行化，可以同时从多个位置开始。这是相同的优化问题，其中通过并行化 gekko 优化找到全局解决方案。
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}')
贝叶斯优化
另一种方法是通过将初始条件映射到优化解决方案的性能来进行智能搜索。它会搜索预期性能最佳的区域或尚未经过测试且不确定性较高的区域。
from gekko import GEKKO
from hyperopt import fmin, tpe, hp
from hyperopt import STATUS_OK, STATUS_FAIL
# Define the search space for the hyperparameters
space = {'x1': hp.quniform('x1', 0, 10, 3),
'x2': hp.quniform('x2', 0, 10, 3),
'x3': hp.quniform('x3', 0, 10, 3)}
def objective(params):
m = GEKKO(remote=False)
x = m.Array(m.Var,3,lb=0)
x1,x2,x3 = x
x1.value = params['x1']
x2.value = params['x2']
x3.value = params['x3']
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.options.SOLVER = 1
m.solve(disp=False,debug=False)
obj = m.options.objfcnval
if m.options.APPSTATUS==1:
s=STATUS_OK
else:
s=STATUS_FAIL
m.cleanup()
return {'loss':obj, 'status': s, 'x':x}
best = fmin(objective, space, algo=tpe.suggest, max_evals=50)
sol = objective(best)
print(f"Solution Status: {sol['status']}")
print(f"Objective: {sol['loss']:.2f}")
print(f"Solution: {sol['x']}")
两种多启动方法都找到了全局解：
Objective: 936.00
Solution: [[0.0] [0.0] [8.0]]
有关这些多启动方法的更多信息位于全局优化和求解器调整页面上的工程优化课程。