Как написать итерационные переменные LP, функцию obj и ограничения, используя PuLP? - PullRequest
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Как написать итерационные переменные LP, функцию obj и ограничения, используя PuLP?

0 голосов
/ 26 апреля 2020

Переменные решения (x1, x2, ..., x8) в приведенном ниже коде являются повторяющимися, которые я хочу повторить, и поэтому я хочу знать, как будут изменены уравнения ограничения 1, 2, 3 и целевая функция. Я попытался перебрать параметр EH, и с этим код в настоящее время работает нормально, но я хочу убедиться, что это правильно, или мне нужно что-то там изменить. Спасибо 

import time
start_time = time.time()
from pulp import *
from openpyxl import load_workbook

N=8
Time=16
Days=7

I=list(range(Days))
J=list(range(Time))

wb1=load_workbook(filename= 'Employer Interface.xlsx')
wse=wb1['Sheet1']

MH=[[0 for j in range(Time)] for i in range(Days)]

for i in range(Days):
    for j in range(Time):
        MH[i][j]=wse.cell(row=i+3, column=j+3).value

wb2=load_workbook(filename= 'Employee Interface.xlsx')

def employeeIF(sheet):
    return wb2.worksheets[sheet]

EH=[[0 for j in range(Time)] for i in range(Days)]

def EH(eh):
    for eh in list(range(1,9,1)):
        for i in range(Days):
            for j in range(Time):
                return employeeIF(eh-1).cell(row=i+3, column=j+3).value

prob= LpProblem("EL-NR", LpMinimize)
x1= LpVariable.matrix("x1", (I,J), 0, 1, LpBinary)
x2= LpVariable.matrix("x2", (I,J), 0, 1, LpBinary)
x3= LpVariable.matrix("x3", (I,J), 0, 1, LpBinary)
x4= LpVariable.matrix("x4", (I,J), 0, 1, LpBinary)
x5= LpVariable.matrix("x5", (I,J), 0, 1, LpBinary)
x6= LpVariable.matrix("x6", (I,J), 0, 1, LpBinary)
x7= LpVariable.matrix("x7", (I,J), 0, 1, LpBinary)
x8= LpVariable.matrix("x8", (I,J), 0, 1, LpBinary)

prob+= lpSum([lpSum([x1[i][j]*EH(1) for j in J]) for i in I])+\
    lpSum([lpSum([x2[i][j]*EH(2) for j in J]) for i in I])+\
    lpSum([lpSum([x3[i][j]*EH(3) for j in J]) for i in I])+\
    lpSum([lpSum([x4[i][j]*EH(4) for j in J]) for i in I])+\
    lpSum([lpSum([x5[i][j]*EH(5) for j in J]) for i in I])+\
    lpSum([lpSum([x6[i][j]*EH(6) for j in J]) for i in I])+\
    lpSum([lpSum([x7[i][j]*EH(7) for j in J]) for i in I])+\
    lpSum([lpSum([x8[i][j]*EH(8) for j in J]) for i in I])    #Objective Function

for i in I:
    for j in J:
        prob+=MH[i][j]==x1[i][j]*EH(1)+\
        x2[i][j]*EH(2)+x3[i][j]*EH(3)+\
        x4[i][j]*EH(4)+x5[i][j]*EH(5)+\
        x6[i][j]*EH(6)+x7[i][j]*EH(7)+x8[i][j]*EH(8) #First constraint

prob+=lpSum([lpSum([x1[i][j] for j in J]) for i in I])<=37.5 #Second constraint
prob+=lpSum([lpSum([x2[i][j] for j in J]) for i in I])<=37.5 #Second constraint
prob+=lpSum([lpSum([x3[i][j] for j in J]) for i in I])<=37.5 #Second constraint
prob+=lpSum([lpSum([x4[i][j] for j in J]) for i in I])<=37.5 #Second constraint
prob+=lpSum([lpSum([x5[i][j] for j in J]) for i in I])<=37.5 #Second constraint
prob+=lpSum([lpSum([x6[i][j] for j in J]) for i in I])<=37.5 #Second constraint
prob+=lpSum([lpSum([x7[i][j] for j in J]) for i in I])<=37.5 #Second constraint
prob+=lpSum([lpSum([x8[i][j] for j in J]) for i in I])<=37.5 #Second constraint

for i in I:
    for j in J:
        prob+=EH(1)>=x1[i][j] #Third constraint
        prob+=EH(2)>=x2[i][j] #Third constraint
        prob+=EH(3)>=x3[i][j] #Third constraint
        prob+=EH(4)>=x4[i][j] #Third constraint
        prob+=EH(5)>=x5[i][j] #Third constraint
        prob+=EH(6)>=x6[i][j] #Third constraint
        prob+=EH(7)>=x7[i][j] #Third constraint
        prob+=EH(8)>=x8[i][j] #Third constraint

prob.solve()
print(prob.objective.value())
...