【发布时间】：2020-10-03 22:27:05
【问题描述】：
我正在努力在约束条件下优化一长串方程。目标函数可以拆分为多个目标函数，但我无法以任何我能找到的方式拆分我的约束函数。在上个月，我尝试以几种不同的方式解决这个问题，同时使用 sympy 和 scipy.optimize，但都需要很长时间才能运行。在此示例中，我将长度限制为出现错误的最短长度之一，但我希望能够通过增加constraintmonth、nummonthstodo 并添加到db1permonth 来尽可能多地扩展它。我怎样才能绕过长度限制错误：
Exception: @error: Max Equation Length
Error with line number: 67
(((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((
[...lots of numbers here..]
0/(1.0+(2.718281828459045^(-(((v14-476.0))/(120)))))))))+((-2.02635827)*((200.0
/(1.0+(2.718281
APM model error: string > 15000 characters
Consider breaking up the line into multiple equations
The may also be due to only using newline character CR
instead of CR LF (for Windows) or LF (for MacOS/Linux)
To fix this problem, save APM file with appropriate newline characters
STOPPING...
在这段代码中：
import numpy as np
from gekko import GEKKO
from math import e
m = GEKKO()
nummonthstodo = 20
constraintmonth = 15
markets=3
fudgefactor = 1.18
constraint = -100000
ig = np.tile([1169.00,476.00,191.00],(nummonthstodo))
dynamic_scale = np.tile(np.array([100,120,80]),nummonthstodo)
turning_point =np.tile( np.array([1169.00,476.00,191.00]),nummonthstodo)
eo14_scale = np.tile(np.array([72.00,200.00,500.00]),nummonthstodo)
db1life = np.tile(np.array([2131.64415, 927.62553, 376.79813]),nummonthstodo)
db1permonth = np.array([-7.69469125e+01, -8.27515903e+01, -8.23699131e+01, 4.45215469e+02,
3.69705278e+02, 1.58977477e+02, -3.88706011e+00, -3.02636074e+00,
-2.79420475e+00, -7.67567831e-01, -3.03178864e+00, -2.79921627e+00,
6.82114433e+01, 5.11866268e+01, 7.41661986e+01, -3.22847133e+00,
-2.28929507e+00, -1.28846317e+00, -7.10922109e-01, -2.29483815e+00,
-1.29158293e+00, 5.33895986e+01, 3.53279803e+01, 4.70884129e+01,
-2.98307479e+00, -2.02448766e+00, -8.09906719e-01, -8.69398271e-01,
-2.02635827e+00, -8.10655068e-01, 4.40599327e+01, 2.74298296e+01,
3.40114832e+01, -2.83729295e+00, -1.90180770e+00, -5.82931770e-01,
-4.06643364e+01, -2.85105835e+01, -8.73890925e+00, 2.05687539e+02,
1.15368035e+02, 2.54844135e+01, -1.93419651e+00, -1.17779492e+00,
-4.36383453e-01, -3.54686842e-01, -1.17891974e+00, -4.36800210e-01,
3.28820412e+01, 1.73557733e+01, 2.03414999e+01, -1.85294103e+00,
-1.09751072e+00, -3.47833721e-01, -4.44622167e-01, -1.09856091e+00,
-3.48166558e-01, 2.90051018e+01, 1.46659871e+01, 1.69382174e+01])
x = m.Array(m.Var,(len(ig)))
# lower bounds
for i,xi in enumerate(x):
xi.value = ig[i]
xi.lower = 0
xi.upper = 1500
def eo14(a,b):
return np.divide(np.tile(eo14_scale[0:3],int((b-a)/3)),(np.ones(b-a)+e**(-np.divide((x[a:b]-np.tile(turning_point[0:3],int((b-a)/3))),np.tile(dynamic_scale[0:3],int((b-a)/3))))))
def db1(a,b):
return db1permonth[a:b]
def cac(a,b):
return x[a:b]
eo14s = [eo14(i*markets,(i+1)*markets) for i in range(nummonthstodo)]
db1s = [db1(i*markets,(i+1)*markets) for i in range(nummonthstodo)]
cacs = [cac(i*markets,(i+1)*markets) for i in range(nummonthstodo)]
db1month = 0
for i in range(constraintmonth):
for j in range(constraintmonth-i):
db1month = db1month + np.multiply(db1s[j],eo14s[i])
spend = 0
for i in range(constraintmonth):
spend = spend + np.multiply(cacs[i],eo14s[i])
alleo14 = eo14(0,len(ig))
m.Equation(m.sum(db1month - spend) >= constraint)
for i in range(len(ig)):
m.Obj(-(np.multiply(db1life-x,alleo14))[i])
m.solve()
print(x)
提前感谢您提供的任何帮助，作为数据科学的学生和初学者，这确实让我头疼了一段时间。
【问题讨论】：