使用 Pulp 优化存储计划

Question

我正在尝试使用 PuLP 模块优化存储计划，但我遇到了第 4 个约束的问题 约束将在下面详细说明

1. Store_demand 的总和不应超过一天的容量（
2. 每个商店都将根据其天数 (Store_Days) 分配到工作日

例如：“S4”应该只安排在三天内

1. 应在 3 天内删除的商店有一个单独的约束“每隔一天”条件，以获得一天的间隔

EX：“S4”商店

• 如果它的第一天安排在 SAT 其他日子将是 MON 和 周三

• 如果它的第一天安排在星期日，其他日子将是星期二和星期四

4. 应该在 2 天内下架的商店在下一次下架前应该有两天的间隔

EX：“S8”商店

• 如果第一天安排在星期六，那么前一天将是星期二
• 如果第一天安排在周日，那么前一天将是周三
• 如果第一天安排在星期一，那么前一天将是星期四

我得到了一个最佳解决方案，虽然它不是我需要的结果，因为输出显示连续两天，所以我想我有一个 locig 问题

例如： enter image description here

• 1 表示将在这一天被丢弃
• 0 表示不会被丢弃

我想要显示的结果如下表所示 Store ROUTE Carton SAT SUN MON TUE WED WED THU DROPS

enter image description here

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

StoreSched = pd.DataFrame(columns = ["Store_Code","Route","Demand"])
Capacity =  5000
route="R1"
days_list=["SAT","SUN","MON", "TUE","WED","THU"]
no_days_list = range(1,7)
Store = ["S1","S2","S3","S4","S5","S6","S7","S8","S9","S10"]
Store_demand = {
        "S1":400,
        "S2":300,
        "S3":250 ,
        "S4":200 ,
        "S5":300,
     "S6":200 ,
        "S7":300,
     "S8":200 ,
        "S9":300,
    "S10":300,
    
     
    }
store_Days = {
        "S1":6 ,
        "S2":6,
        "S3":6 ,
        "S4":3,
    "S5":3,
    "S6":3,
       "S7":2,
    "S8":2,
    "S9":2,
        "S10":1 ,
   
    }
    
prob = LpProblem("store_schedule",LpMaximize)
storeVars = LpVariable.dicts("Days",(no_days_list,Store),0,1,LpInteger)
    
for d in no_days_list:
        # The capacity should not exceeed 1500 in one day 
         prob += pulp.lpSum([Store_demand[s] * storeVars[d][s] for s in Store]) <= Capacity
for s in Store:
        # Every store should be assigned based on its DayNo.
        prob += pulp.lpSum(storeVars[d][s] for d in no_days_list) == store_Days[s]
for s in Store:  
        # one day gap between the assigned dayes for the stores that have three days 
        if store_Days[s] == 3 :  
            for d in no_days_list[:-1]:
                prob += storeVars[d][s] + storeVars[d+1][s] == 1          
for s in Store:
            if store_Days[s] == 2  : 
                 for d in no_days_list[:-2]:
                    prob += storeVars[d][s] + storeVars[d+2][s] == 1   
prob.solve()

for vi in prob.variables():
        if vi.varValue == 1:
            #print(" On "+days_list[int(vi.name.split("_")[1])-1]+" Pharmacy code: "+vi.name.split("_")[2])
            code= vi.name.split("_")[2];
            #print(code)
            day = days_list[int(vi.name.split("_")[1])-1];
            #print(day)
            if ((StoreSched['Store_Code'] == code).any() == False):
                StoreSched = StoreSched.append({'Store_Code': code,"Route":route,"Days":store_Days[code],"Demand":Store_demand[code]}, ignore_index=True)
            for index in StoreSched.index:    
                if StoreSched.loc[index,'Store_Code']== code:                    
                    StoreSched.loc[index,day] = 1                    
StoreSched.fillna(0,inplace=True)
StoreSched  

Answer 1

如果您希望天之间的间隔为 2，请将约束更改为 prob += storeVars[d][s] + storeVars[d+1][s] + storeVars[d+2][s] == 1。