Python PuLP“if”只符合第一个条件。

Question

我试图用纸浆来优化一个系统，使它的成本最小化。我使用多个If，问题是它总是满足第一个条件。这是我的密码。我只是希望有人能帮我学语言。在

import numpy as np
import pandas as pd
from pulp import *

idx = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23]
d = {
'day': pd.Series(['01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14'], index=idx),
'hour':pd.Series(['00:00:00', '01:00:00', '02:00:00', '03:00:00', '04:00:00', '05:00:00', '06:00:00', '07:00:00', '08:00:00', '09:00:00', '10:00:00', '11:00:00', '12:00:00', '13:00:00', '14:00:00', '15:00:00', '16:00:00', '17:00:00', '18:00:00', '19:00:00', '20:00:00', '21:00:00', '22:00:00', '23:00:00'], index=idx),
'output':pd.Series([0,0,0,0.087,0.309,0.552,0.682,0.757,0.783,0.771,0.715,0.616,0.466,0.255,0.022,0,0,0,0,0,0,0,0,0], index=idx)}
cfPV = pd.DataFrame(d)


idx = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23]
d1 = {
'day': pd.Series(['01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14'], index=idx),
'hour':pd.Series(['00:00:00', '01:00:00', '02:00:00', '03:00:00', '04:00:00', '05:00:00', '06:00:00', '07:00:00', '08:00:00', '09:00:00', '10:00:00', '11:00:00', '12:00:00', '13:00:00', '14:00:00', '15:00:00', '16:00:00', '17:00:00', '18:00:00', '19:00:00', '20:00:00', '21:00:00', '22:00:00', '23:00:00'], index=idx),
'output':pd.Series([0.528,0.512,0.51,0.448,0.62,0.649,0.601,0.564,0.541,0.515,0.502,0.522,0.57,0.638,0.66,0.629,0.589,0.544,0.506,0.471,0.448,0.438,0.443,0.451], index=idx)}
cfWT = pd.DataFrame(d1)


prob = LpProblem ("System", LpMinimize)

CPV = LpVariable ("PVCapacity",0) #PV Capacity in kW
CWT = LpVariable ("WTurCapacity",0) #WT Capacity in kW
CBA = LpVariable ("BatteryCapacity",0) #Battery Capacity kW

prob+= 63.128*CPV + 88.167*CWT + 200*CBA, "TotalCostSystem"

xEne = 0
xREin = 0
xBin = 0
xBout = 0
SOCB = 0
xPEMin = 0
xOvEn = 0
xSum = 0

CPEM = 230

for i in idx:

    xEne = (CPV*cfPV['output'][i]+CWT*cfWT['output'][i])

    #Low limit for Variables
    prob += (CPV*cfPV['output'][i]+CWT*cfWT['output'][i]) >= 0
    prob += xREin >= 0
    prob += xBin >= 0
    prob += xBout >= 0
    prob += SOCB >= 0
    prob += xPEMin >= 0
    prob += xOvEn >= 0
    prob += xSum >= 0
    prob += CBA >= SOCB
    prob += xBin <= (CBA - SOCB)
    prob += xBout <= SOCB

    #Cases

    #Case 1 xEne > CPEM
    if xEne >= CPEM:

        xREin = CPEM
        xBout = 0
        xOvEn = xEne - CPEM 

        #Case 1.1 xOvEn < CBA - SOCB
        if (value(xOvEn) <= (CBA - value(SOCB))): 
            xBin = xOvEn

        #Case 1.2 xOvEn > CBA -SOCB
        else: 
            xBin = CBA - SOCB 

    #Case 2 xEne < CPEM
    else:
        xREin = xEne
        xBin = 0 
        xOvEn = 0

        #Case 2.1 SOCB > CPEM - xREin
        if (value(SOCB) >= (CPEM - value(xREin))):
            xBout = (CPEM - xREin)

        #Case 2.2 SOCB < CPEM - xREin 
        else:

            xBout = SOCB 

    SOCB = SOCB + xBin - xBout
    xPEMin = xREin + xBout 

    xSum += xPEMin

prob += xSum >= 5000


prob.writeLP("PVWTBattSyste.lp")

prob.solve()

给出的解总是满足第一个条件。此外，当条件不满足时（例如，将CPEM更改为50000000000000），if将按真的那样工作。在

我用“elif”得到了同样的结果。如果我改变了条件的顺序（也就是说如果第一个条件是xEne<；=CPEM，那么结果将只符合第一个新条件。在

提前谢谢你！在

卡斯蒂略普里西拉

Answer 1

if语句不起作用，因为变量在优化之后才有定义的值。在

在lp问题中，你建立了lp变量之间的关系，优化者找到了与之匹配的解决方案。在

在您的例子中，混合LpVariables和python整数的方式是不正确的。在

也许看看 http://www.yzuda.org/Useful_Links/optimization/if-then-else-02.html

或者

https://ocw.mit.edu/courses/sloan-school-of-management/15-053-optimization-methods-in-management-science-spring-2013/lecture-notes/MIT15_053S13_lec11.pdf

或者

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.464.6182&rep=rep1&type=pdf

寻求帮助

斯图