Python 基于纸浆的仓库调度优化

我正在尝试使用纸浆模块优化商店时间表，但我面临第四个约束的问题 下文将详细阐述这些制约因素

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商店总需求量不应超过一天的容量（如果您希望两天之间的间隔为2，请将限制更改为

prob+=storeVars[d][s]+storeVars[d+1][s]+storeVars[d+2][s]==1

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