問題不可行時的紙漿約束

Question

我正在嘗試使用 Python 中的 Pulp 解決線性優化問題。

這是代碼：

import pandas as pd
import pulp

D_XB = 20
D_XP = 0
D_XC = 0

Available_Time = 1440 #in minutes

test = [['A1', 'A2', 'A3', 'A4', 'A5'], [1,2,1,0,3], [16,32,0,16,32], [10,10,10,10,10], [120,210,180,180,350]]

Cycles = pd.DataFrame(test, index=['Cycles', 'QTA1', 'QTA2', 'QTA3', 'T_TOT']).T

A1 = pulp.LpVariable("Cycle_A1", lowBound=0, cat='Integer')
A2 = pulp.LpVariable("Cycle_A2", lowBound=0, cat='Integer')
A3 = pulp.LpVariable("Cycle_A3", lowBound=0, cat='Integer')
A4 = pulp.LpVariable("Cycle_A4", lowBound=0, cat='Integer')
A5 = pulp.LpVariable("Cycle_A5", lowBound=0, cat='Integer')
    
# Defining the problem as a minimization problem (Minimize Storage)
problem_5 = pulp.LpProblem("Storage_Minimization_3", pulp.LpMinimize)

S_XB = pulp.lpSum((Cycles.iloc[0]["QTA1"])*A1 + (Cycles.iloc[1]["QTA1"])*A2 + (Cycles.iloc[2]["QTA1"])*A3 + (Cycles.iloc[3]["QTA1"])*A4 + (Cycles.iloc[4]["QTA1"])*A5)
S_XP = pulp.lpSum((Cycles.iloc[0]["QTA2"])*A1 + (Cycles.iloc[1]["QTA2"])*A2 + (Cycles.iloc[2]["QTA2"])*A3 + (Cycles.iloc[3]["QTA2"])*A4 + (Cycles.iloc[4]["QTA2"])*A5)
S_XC = pulp.lpSum((Cycles.iloc[0]["QTA3"])*A1 + (Cycles.iloc[1]["QTA3"])*A2 + (Cycles.iloc[2]["QTA3"])*A3 + (Cycles.iloc[3]["QTA3"])*A4 + (Cycles.iloc[4]["QTA3"])*A5)

Tot_Time = pulp.lpSum((Cycles.iloc[0]["T_TOT"])*A1 + (Cycles.iloc[1]["T_TOT"])*A2 + (Cycles.iloc[2]["T_TOT"])*A3 + (Cycles.iloc[3]["T_TOT"])*A4 + (Cycles.iloc[4]["T_TOT"])*A5)

Stock_XB = (S_XB - D_XB)
Stock_XP = (S_XP - D_XP)
Stock_XC = (S_XC - D_XC)

problem_5 += Stock_XB + Stock_XP + Stock_XC
    
# Constraints: Time constraint present
problem_5 += S_XB >= D_XB
problem_5 += S_XP >= D_XP
problem_5 += S_XC >= D_XC
problem_5 += A1 >= 0
problem_5 += A2 >= 0
problem_5 += A3 >= 0
problem_5 += A4 >= 0
problem_5 += A5 >= 0
problem_5 += Tot_Time <= Available_Time
    
# Solving the probelm
status = problem_5.solve()

result = pd.DataFrame({'A1':[pulp.value(A1)], 'A2':[pulp.value(A2)], 'A3':[pulp.value(A3)], 'A4':[pulp.value(A4)], 'A5':[pulp.value(A5)], 
                       'Demand XB':[D_XB], 'Demand XP':[D_XP], 'Demand XC':[D_XC],  'Minimum storage':[pulp.value(problem_5.objective)], 
                       'Stock_XB':[pulp.value(Stock_XB)], 'Stock_XP':[pulp.value(Stock_XP)], 'Stock_XC':[pulp.value(Stock_XC)], 
                       'Total Time needed':[pulp.value(Tot_Time)]})

術語：S_* 是某個項目的生產，D_* 是該項目的需求。 這些是在問題定義之前定義的。

這個問題並不總是可行的，因為有時生產會超過可用時間。 在這種情況下，我希望遵守對周期的限制，並且應該打破按時的限制以解決問題。

我怎樣才能做到這一點？

謝謝，卡洛塔。

Answer 1

他們這樣做的方法是將松弛變量添加到時間限制約束中，如下所示：

首先添加一個新變量：

#(...)
slack_time = pulp.LpVariable("slack_time", lowBound=0, cat=pulp.LpContinuous)
#(...)

然后你在目標 function 中懲罰它：

#(...)
big_enough_number = 10000
problem_5 += Stock_XB + Stock_XP + Stock_XC + slack_time*big_enough_number
#(...)

最后，您將其添加到您的時間限制約束中：

#(...)
problem_5 += Tot_Time - slack_time <= Available_Time
#(...)

然后，您將獲得一個盡可能不違反時間限制的解決方案。 如果您選擇了足夠好的big_enough_number ，則 model 只會在沒有其他選項的情況下違反時間限制。