使用 Pulp 优化商店计划

Question

我正在尝试使用 PuLP 模块优化商店计划，但我遇到了第 4 个约束的问题下面将详细说明约束条件

Store_demand一天的总和不能超过容量（<=容量）
每个商店将根据他们的天数分配给工作日（Store_Days）

例如：“S4”应该只安排在三天后

应该在 3 天内删除的商店有一个单独的约束“每隔一天”条件来获得一天的差距

EX：“S4”商店

如果第一天安排在 SAT，其他日子将是 MON 和周三
如果第一天安排在星期日，其他日子将是星期二和星期四

应该在2天内掉落的商店在下一次掉落之前应该有两天的差距

EX：“S8”商店

如果第一天安排在 SAT，那么另一天将是 TUE
如果它的第一天安排在星期日，那么另一天将是星期三
如果第一天安排在星期一，那么另一天将是星期四

我得到了一个最优解，虽然这不是我需要的结果，因为输出显示连续两天，所以我想我有一个定位问题

例如： enter image description here

1表示当天掉线
0表示不会掉线

我想要显示的结果如下table 商店 ROUTE Carton SAT SUN MON TUE 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

如果您希望天数间隔为 2，请将限制更改为 prob += storeVars[d][s] + storeVars[d+1][s] + storeVars[d+2][s] == 1。

使用 Pulp 优化商店计划

store schedule optimization with Pulp

python

pandas

pulp