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python - 如何解决@error：GEKKO 中的方程定义

问题描述

我想解决python中的优化问题。
这是代码。认为 ：

W = [0.010858969983152403,0.15750120163876366,0.14594534721059332,0.08588827233293823,0.14391967610026943,0.17068447485608854,0.17510026127394213,0.11010179660425223]

C = np.array([[0.99365367, 0.97892888, 1.01870907, 1.00434405, 0.99742434,
        0.98994678, 1.00610998, 1.0014477 ],
       [0.99065144, 1.00254236, 0.97842508, 0.93742212, 0.99908661,
        0.99232329, 0.99406251, 0.99902616],
       [0.99355243, 0.9896095 , 1.00603939, 1.01114646, 1.00859356,
        1.00421901, 0.9994433 , 0.96580307],
       [0.99310202, 1.00188421, 1.01455517, 0.99027971, 0.99445973,
        0.99638549, 0.98567891, 1.00278336],
       [0.98696926, 0.99425696, 1.01039431, 1.0066784 , 0.99775556,
        0.99873331, 0.99854812, 1.00948166]])

现在由 GEKKO 优化：

import numpy as np
from gekko import GEKKO

nd = 5
qw = GEKKO()
x = qw.Array(qw.Var,nd,value=1/nd,lb=0,ub=1) # x.shape --> (5, )

qw.Equation(sum(x) == 1)

ww = np.array(W) # ww.shape --> (8, )

def Log_Caculator(Array):
    '''
    final goal of This function is to Calculate Logarithm of every element of the 'Array'
    and return the new Array
    '''
    for j in range(len(Array)):
        Array[j] = qw.log10(Array[j])
    
    return Array

qw.Maximize(ww * Log_Caculator(np.dot(x.T ,  C)))
qw.solve(disp=True)
for i,xi in enumerate(x):
    print(i+1,xi.value)

输出：

Exception:  @error: Equation Definition
 Equation without an equality (=) or inequality (>,<)
 ((0.15750120163876366)*(log10(((((((v1)*(0.97892888))+((v2)*(1.00254236)))+((v3
 )*(0.9896095)))+((v4)*(1.00188421)))+((v5)*(0.99425696))))))
 STOPPING...

通过 Visual Studio Code 的调试功能，我得到了这些：
在执行之前qw.solve(disp=True)：

在此处输入图像描述

执行后qw.solve(disp=True)：

在此处输入图像描述

如果您尝试比较它们，您会发现x它发生了变化！这意味着已经找到了最优解！我认为优化是由算法完成的。
但它仍然向我展示了我在输出部分提到的错误。
我应该如何解决这个问题？

标签： pythongekko

解决方案

这个问题可以通过改变目标函数来解决：

qw.Maximize(np.dot(ww,Log_Caculator(np.dot(x.T, C))))

m.Maximize()使用或m.Minimize()必须是标量（单个）值定义的目标函数。附加的 np.dot() 函数是制作ww * Log_Caculator(np.dot(x.T , C))标量的一种方法。

这是完整的脚本：

import numpy as np
from gekko import GEKKO

W = [0.010858969983152403,0.15750120163876366,0.14594534721059332,\
     0.08588827233293823,0.14391967610026943,0.17068447485608854,\
     0.17510026127394213,0.11010179660425223]
ww = np.array(W)

C = np.array([[0.99365367, 0.97892888, 1.01870907, 1.00434405, \
               0.99742434, 0.98994678, 1.00610998, 1.0014477 ],
              [0.99065144, 1.00254236, 0.97842508, 0.93742212, \
               0.99908661, 0.99232329, 0.99406251, 0.99902616],
              [0.99355243, 0.9896095 , 1.00603939, 1.01114646, \
               1.00859356, 1.00421901, 0.9994433 , 0.96580307],
              [0.99310202, 1.00188421, 1.01455517, 0.99027971, \
               0.99445973, 0.99638549, 0.98567891, 1.00278336],
              [0.98696926, 0.99425696, 1.01039431, 1.0066784 , \
               0.99775556, 0.99873331, 0.99854812, 1.00948166]])

nd = 5
qw = GEKKO()
x = qw.Array(qw.Var,nd,value=1/nd,lb=0,ub=1)
qw.Equation(sum(x) == 1)

def Log_Caculator(Array):
    for j in range(len(Array)):
        Array[j] = qw.log10(Array[j])    
    return Array

qw.Maximize(np.dot(ww,Log_Caculator(np.dot(x.T, C))))
qw.solve(disp=True)
for i,xi in enumerate(x):
    print(i+1,xi.value)

这是解决方案：

EXIT: Optimal Solution Found.
 
 The solution was found.
 
 The final value of the objective function is  -5.540853005129416E-004
 
 ---------------------------------------------------
 Solver         :  IPOPT (v3.12)
 Solution time  :   9.600000004866160E-003 sec
 Objective      :  -5.540853005129416E-004
 Successful solution
 ---------------------------------------------------
 
1 [3.7304341888e-05]
2 [4.8543858174e-06]
3 [3.9756289011e-05]
4 [2.5331754827e-05]
5 [0.99989275323]

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