python - 在gekko中定义最大函数时出错

Question

我在 gekko-python 中定义 max 函数时遇到问题。

sum 函数工作正常但是当我创建另一个函数只是用 max 替换 sum 然后它抛出以下错误:

这是我使用的脚本的描述(包括模型、一些数据和结果)

#Model
import numpy as np
from gekko import GEKKO
import numpy as np
import pandas as pd
from pandas import ExcelWriter
from pandas import ExcelFile

m = GEKKO()  # Initialize gekko
m.options.SOLVER = 3  # IPOPT is an NLP solver
m.options.MAX_ITER = 10000 # maximum number of iterations
m.options.MAX_MEMORY = 6  # (2-10) memory allocation

R_sect_bin = {'W1': {'S1': 1}, 'W2': {'S1': 1, 'S2': 1, 'S4': 1}, 'W3': {'S1': 1, 'S2': 1, 'S3': 1, 'S4': 1, 'S5': 1, 'S6': 1}, 'W4': {'S4': 1}, 'W5': {'S4': 1, 'S5': 1, 'S6': 1}, 'W6': {'S6': 1}}
Input_Services_nonZero = {'S1': {'L1': 1, 'L3': 1}, 'S2': {'L2': 1}, 'S3': {'L4': 1}, 'S4': {'L1': 1}, 'S5': {'L3': 1}, 'S6': {'L1': 1, 'L2': 1}}

V = {}
for w in R_sect_bin:
    V[w] = {}
    for s in R_sect_bin[w]:
        V[w][s] = {}
        for l in Input_Services_nonZero[s]:
            V[w][s][l] = m.Var(value=10, lb=0, ub=100)

#functions:
sum_Vws1 = {}
def VWS1():
    global sum_Vws1
    for w in R_sect_bin:
        sum_Vws1[w] = {}
        for s in R_sect_bin[w]:
            sum_Vws1[w][s] = m.Intermediate(sum([V[w][s][l] for l in Input_Services_nonZero[s]]))
    return sum_Vws1
vws1 = VWS1()

sum_Vws2 = {}
def VWS2():
    global sum_Vws2
    for w in R_sect_bin:
        sum_Vws2[w] = {}
        for s in R_sect_bin[w]:
            sum_Vws2[w][s] = m.Intermediate(max([V[w][s][l] for l in Input_Services_nonZero[s]]))
    return sum_Vws2
vws2 = VWS2()

TypeError                                 Traceback (most recent call last)
<ipython-input-225-b48377242060> in <module>
     89             sum_Vws2[w][s] = m.Intermediate(max([V[w][s][l] for l in Input_Services_nonZero[s]]))
     90     return sum_Vws2
---> 91 vws2 = VWS2()
     92 
     93 #sum_Vws = {}

<ipython-input-225-b48377242060> in VWS2()
     87         sum_Vws2[w] = {}
     88         for s in R_sect_bin[w]:
---> 89             sum_Vws2[w][s] = m.Intermediate(max([V[w][s][l] for l in Input_Services_nonZero[s]]))
     90     return sum_Vws2
     91 vws2 = VWS2()

~\Anaconda3\lib\site-packages\gekko\gk_operators.py in __len__(self)
     23         return self.name
     24     def __len__(self):
---> 25         return len(self.value)
     26     def __getitem__(self,key):
     27         return self.value[key]

~\Anaconda3\lib\site-packages\gekko\gk_operators.py in __len__(self)
    132 
    133     def __len__(self):
--> 134         return len(self.value)
    135 
    136     def __getitem__(self,key):

TypeError: object of type 'int' has no len()

Answer 1

何登仁教授的回答:

您需要使用内置的 GEKKO max2 或 max3 函数。否则，Python 函数会创建一个不具有连续一阶或二阶导数的表达式，并且基于梯度的求解器可能无法找到解。

您需要使用 pip install gekko==0.2rc6 升级到最新的 gekko 版本 (>0.2rc5) 以使用 max2 或 max3 Gekko 函数。

下面是 max2 或 max3 的源代码，您也可以在 Gekko source 中找到它们.

def max2(self,x1,x2):
    """ Generates the maximum value with continuous first and
    second derivatives. The traditional method for max value (max) is not
    continuously differentiable and can cause a gradient-based optimizer
    to fail to converge.

    Usage: y = m.max2(x1,x2)
    Input: GEKKO variable, parameter, or expression
    Output: GEKKO variable
    """

    # verify that x1 and x2 are valid GEKKO variables or parameters
    if isinstance(x1,(GKVariable,GKParameter)):
        xin1 = x1

    else:
        # create input variable if it is an expression
        xin1 = self.Var()
        self.Equation(xin1==x1)

    if isinstance(x2,(GKVariable,GKParameter)):
        xin2 = x2

    else:
        # create input variable if it is an expression
        xin2 = self.Var()
        self.Equation(xin2==x2)

    # build max object with unique object name
    max_name = 'max2_' + str(len(self._objects) + 1)
    self._objects.append(max_name + ' = max')

    # add connections between x and max object attribute x
    self._connections.append(xin1.name + ' = ' + max_name+'.x[1]')
    self._connections.append(xin2.name + ' = ' + max_name+'.x[2]')

    # add connections between y and max object attribute y
    y = self.Var()
    self._connections.append(y.name + ' = ' + max_name+'.y')

    return y

def max3(self,x1,x2):
    """ Generates the maximum value with a binary switch variable.
    The traditional method for max value (max) is not continuously
    differentiable and can cause a gradient-based optimizer to fail
    to converge.

    Usage: y = m.max3(x1,x2)
    Input: GEKKO variable, parameter, or expression
    Output: GEKKO variable
    """

    # add binary (intb) and output (y) variable
    intb = self.Var(0,lb=0,ub=1,integer=True)
    y = self.Var()

    # add equations for switching conditions
    #  intb=0 when x1>x2 and y=x1
    #  intb=1 when x2>x1 and y=x2
    self.Equation((1-intb)*(x2-x1) <= 0)
    self.Equation(intb*(x1-x2) <= 0)
    self.Equation(y==(1-intb)*x1+intb*x2)

    # change default solver to APOPT (MINLP)
    self.options.SOLVER = 1

    return y 

max2 函数使用 MPCC，而 max3 使用二进制变量。这些函数仅通过使用函数调用就可以帮助使用这些更复杂的建模形式。您只能将两个值与 max2 或 max3 进行比较，因此如果您有一个列表，那么您需要执行以下操作:

y[0] = m.max3(x[0],x[1])
y[1] = m.max3(x[2],y[0])
y[2] = m.max3(x[3],y[1])

等等

这里是关于 why you need to use MPCCs or binary variables 的附加信息.还有一个 additional example here .