store schedule optimization with Pulp

Question

I am trying to optimize the store schedule using the PuLP module, but I am facing a problem with the 4th constraint The constraints will be elaborated on below

1. The Total of Store_demand should not exceed the capacity in one day (<= capacity)
2. Every store will be assigned to weekdays based on their Days No. (Store_Days)

Ex: "S4" should be scheduled in three days only

1. The store that should be dropped in 3 days has a separate constraint "Every other day" condition to get one day gap

EX:" S4" store

• If its first day was scheduled on SAT the other days will be MON and WED

• If its first day was scheduled on SUN the other days will be TUE and THU

4. The store that should be dropped in 2 days should have two days gaps before the next drop

EX:" S8" store

• If its first day was scheduled on SAT the other day will be TUE
• If its first day was scheduled on SUN the other day will be WED
• If its first day was scheduled on MON the other day will be THU

I got an optimal solution although it is not the result that I need as the output shows two consecutive days, so I guess I have a locig problem

Ex: enter image description here

• 1 means it will be dropped on this day
• 0 means it will not be dropped

The Result that I want to be shown as the below table Store 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

Change the constraint to prob += storeVars[d][s] + storeVars[d+1][s] + storeVars[d+2][s] == 1 if you want a gap of 2 between the days.