Adapt the code with the new information provided We have feedback from our drivers that we need to allow an extra $15$ minutes between pairs of farms on a milk run. Additionally, we need to allow $60$ minutes cleaning time between milk runs made by the same tanker$.$

Taking these times into account, which of the milk runs should we now use? Please provide us with the minimum total cost of collections.

The previuos code is:

from gurobipy import $*$

from Milkruns import Milkruns

# Farms

farms $=[\text{'}$Cowbell$\text{'},$ 'Creamy Acres', 'Milky Way', 'Happy Cows', 'Udder Delight', 'Fresh Pail', 'Cowabunga', 'Utopia', 'Moo Meadows', 'Bluebell', 'Harmony', 'Velvet Valley', 'Moonybrook', 'Cloven Hills', 'Midnight Moo', 'Willows Bend', 'Moosa Heads', 'Dreamy Dairies', 'Happy Hooves', 'Highlands'$]$

F$=$ range$($len$($farms$))$

# Supply from each farm

Supply $=[9100,6100,6500,9400,6800,6900,4300,3000,5000,4700,3600,3000,3700,3300,3300,3200,4700,3400,4000,3400]$

# Processing facilities

facilities $=[\text{'}$P$1\text{'},\text{'}$P$2\text{'},\text{'}$P$3\text{'}]$

P$=$range$($len$($facilities$))$

Tankers $=$ range$(5)$

#Data Milkruns

#R$[0]-$ facility tanker must leave from

#R$[1]-$ list of farms visited

#R$[2]-$ total time taken$($mins$)$

#R$[3]-$ total cost of milk run$($$$)$

R $=$ range$($len$($Milkruns$))$

# Processing facility capacities

PMin $=[19000,25000,19000]$

PMax $=[35000,43000,43000]$

# Create a new model

m $=$ Model$("$Teal Cow Dairy"$)$

#VARIABLES

#binary variable$-1$ if milkrun r is assigned to tanker k at facility j

X $=\{($j$,$k$,$r$)$: m$.$addVar$($vtype$=$ GRB$.$BINARY$)$ for r in R for j in P for k in Tankers$\}$

#binary variable $-1$ if tanker k is part of the fleet at processing facility j

Z $=\{($j$,$k$)$: m$.$addVar$($vtype$=$GRB$.$BINARY$)$ for k in Tankers for j in P$\}$

#OBJECTIVE: Minimise total cost of collections

m$.$setObjective$($quicksum$($Milkruns$[$r$][3]*$ X$[$j$,$ k$,$ r$]$ for r in R for j in P for k in Tankers$)+$

quicksum$((500-30*$ k$)*$ Z$[$j$,$ k$]$ for k in Tankers for j in P$),$ GRB$.$MINIMIZE$)$

#Constraints

#each milkrun is assigned to one tanker at a processing facility

for r in R:

m$.$addConstr$($quicksum$($X$[$j$,$k$,$r$]$ for j in P for k in Tankers if j $!=$ Milkruns$[$r$][0])==0)$

#each farm i is srvicied by exactly one milkrun

for i in F:

m$.$addConstr$($quicksum$($X$[$j$,$k$,$r$]$ for j in P for k in Tankers for r in R if i in Milkruns$[$r$][1])==1)$

for j in P:

#Processing capacity constraint

m$.$addConstr$($quicksum$($Supply$[$i$]*$X$[$j$,$k$,$r$]$ for r in R for k in Tankers for i in Milkruns$[$r$][1])>=$ PMin$[$j$])$

m$.$addConstr$($quicksum$($Supply$[$i$]*$X$[$j$,$k$,$r$]$ for r in R for k in Tankers for i in Milkruns$[$r$][1])<=$ PMax$[$j$])$

for k in Tankers:

#each tanker can only be used for $10$ hours

m$.$addConstr$($quicksum$($Milkruns$[$r$][2]*$X$[$j$,$k$,$r$]$ for r in R$)<=600*$Z$[$j$,$k$])$

#unlock tankers in order and check if total timeis within the limits

if k$>0$:

m$.$addConstr$($Z$[$j$,$k$]<=$ Z$[$j$,$k$-1])$

m$.$addConstr$($quicksum$($Milkruns$[$r$][2]*$X$[$j$,$k$-1,$r$]$ for r in R$)>=600*$Z$[$j$,$k$])$

#each milkrun r is assigned to at most one processing facility and a tanker

for r in R:

m$.$addConstr$($quicksum$($X$[$j$,$k$,$r$]$ for j in P for k in Tankers$)<=1)$

m$.$optimize$()$

print$("$Min total cost of collection: $",$ m$.$ObjVal$)$

Consider this info :

When the questions states 'extra $15$ minutes between pairs of farms on a milk run', does this mean if there are $2$ farms, it will take $15$ minutes and if there are $3$ farms, it will take $30$ minutes to unload the milk

So a tanker doing three milk runs would require $120$ minutes of cleaning, while a tanker doing a single run would require no cleaning during the day $($it can be done at night, before the next day$).$