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Gekko 约束中的最大方程长度限制

[英]Max equation length limit in Gekko constraint

 原文  2020-06-14 12:25:27  5   1    python/ numpy/ gekko

问题描述

I am working on optimizing a long list of equations under a constraint.我正在努力在约束下优化一长串方程。 The objective function can be split into several objective functions, but i am unable to split my constraint function in any way that i can find.目标 function 可以拆分为多个目标函数，但我无法以我能找到的任何方式拆分我的约束 function。 I have tried to solve this problem in several different ways over the last month, using both sympy and scipy.optimize, but both take ages to run.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，但两者都需要很长时间才能运行。 In this example i have limited the length to one of the shortest where i get the error, but i would like to be able to expand this as much as i want by increasing constraintmonth, nummonthstodo and adding to db1permonth.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。 How can i get around the length limit error: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...

In this code: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)

Thanks in advance for any help you can give, as a student and a beginner in data science this has really been giving me headaches for some time now.提前感谢您提供的任何帮助，作为一名学生和数据科学的初学者，这确实让我头疼了一段时间。

1 个解决方案

解决方案1
 1  2020-06-17 00:34:59

You can overcome the maximum equation length by declaring your products as a list and then use m.sum() .您可以通过将产品声明为列表来克服最大方程长度，然后使用m.sum() 。 Otherwise, it builds a very large expression that exceeds the limit of 15,000 characters per equation.否则，它会构建一个非常大的表达式，超过每个方程 15,000 个字符的限制。 You could also use m.Intermediate() variables to reduce the expression size.您还可以使用m.Intermediate()变量来减小表达式大小。 Here is a new version of your program that solves successfully.这是成功解决的程序的新版本。

import numpy as np
from gekko import GEKKO    
from math import e

m = GEKKO(remote=False)

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),lb=0,ub=1500)
for i,xi in enumerate(x):
    xi.value = ig[i]

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)]

prod1 = [db1s[j]*eo14s[i] for j in range(constraintmonth-i) \
                         for i in range(constraintmonth)]
db1month = m.sum(prod1)

prod2 = [cacs[i]*eo14s[i] for i in range(constraintmonth)]
spend = m.sum(prod2)

alleo14 = eo14(0,len(ig))

m.Equation(db1month - spend >= constraint)

for i in range(len(ig)):
   m.Maximize((np.multiply(db1life-x,alleo14))[i])

m.solve()

print(x)

This gives the solver output:这给出了求解器 output：

Number of Iterations....: 9

                                   (scaled)                 (unscaled)
Objective...............: -1.7974777841412724e+003  -2.8845129361943263e+006
Dual infeasibility......:  1.3421365394256974e-009   2.1538014234551661e-006
Constraint violation....:  0.0000000000000000e+000   0.0000000000000000e+000
Complementarity.........:  2.1861799117816599e-010   3.5082849379387629e-007
Overall NLP error.......:  1.3421365394256974e-009   2.1538014234551661e-006


Number of objective function evaluations             = 10
Number of objective gradient evaluations             = 10
Number of equality constraint evaluations            = 10
Number of inequality constraint evaluations          = 0
Number of equality constraint Jacobian evaluations   = 10
Number of inequality constraint Jacobian evaluations = 0
Number of Lagrangian Hessian evaluations             = 9
Total CPU secs in IPOPT (w/o function evaluations)   =      0.066
Total CPU secs in NLP function evaluations           =      0.001

EXIT: Optimal Solution Found.

 The solution was found.

 The final value of the objective function is  -2884512.9361943263

 ---------------------------------------------------
 Solver         :  IPOPT (v3.12)
 Solution time  :  0.127 sec
 Objective      :  -2884512.9361943263
 Successful solution
 ---------------------------------------------------

And the solution:和解决方案：

[[1359.5263445] [561.97176237] [203.39330742] [1359.5263445]
 [561.97176237] [203.39330742] [1359.5263445] [561.97176237]
 [203.39330742] [1359.5263445] [561.97176237] [203.39330742]
 [1359.5263445] [561.97176237] [203.39330742] [1359.5263445]
 [561.97176237] [203.39330742] [1359.5263445] [561.97176237]
 [203.39330742] [1359.5263445] [561.97176237] [203.39330742]
 [1359.5263445] [561.97176237] [203.39330742] [1359.5263445]
 [561.97176237] [203.39330742] [1359.5263445] [561.97176237]
 [203.39330742] [1359.5263445] [561.97176237] [203.39330742]
 [1359.5263445] [561.97176237] [203.39330742] [1359.5263445]
 [561.97176237] [203.39330742] [1359.5263445] [561.97176237]
 [203.39330742] [1359.5263445] [561.97176237] [203.39330742]
 [1359.5263445] [561.97176237] [203.39330742] [1359.5263445]
 [561.97176237] [203.39330742] [1359.5263445] [561.97176237]
 [203.39330742] [1359.5263445] [561.97176237] [203.39330742]]

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