"""
The Beer Distribution Problem with Extension for A Competitor Supply Node for the PuLP Modeller
"""
# Import PuLP modeler functions
from pulp import *
# Creates a list of all the supply nodes
Warehouses = ["A", "B", "C"]
# Creates a dictionary for the number of units of supply for each supply node
supply = {"A": 1000, "B": 4000, "C": 100}
# Creates a list of all demand nodes
Bars = ["1", "2", "3", "4", "5"]
# Creates a dictionary for the number of units of demand for each demand node
demand = {
"1": 500,
"2": 900,
"3": 1800,
"4": 200,
"5": 700,
}
# Creates a list of costs of each transportation path
costs = [ # Bars
# 1 2 3 4 5
[2, 4, 5, 2, 1], # A Warehouses
[3, 1, 3, 2, 3], # B
[0, 0, 0, 0, 0],
]
# The cost data is made into a dictionary
costs = makeDict([Warehouses, Bars], costs, 0)
# Creates the 'prob' variable to contain the problem data
prob = LpProblem("Beer Distribution Problem", LpMinimize)
# Creates a list of tuples containing all the possible routes for transport
Routes = [(w, b) for w in Warehouses for b in Bars]
# A dictionary called 'Vars' is created to contain the referenced variables(the routes)
vars = LpVariable.dicts("Route", (Warehouses, Bars), 0, None, LpInteger)
# The objective function is added to 'prob' first
prob += (
lpSum([vars[w][b] * costs[w][b] for (w, b) in Routes]),
"Sum_of_Transporting_Costs",
)
# The supply maximum constraints are added to prob for each supply node (warehouse)
for w in Warehouses:
prob += (
lpSum([vars[w][b] for b in Bars]) <= supply[w],
f"Sum_of_Products_out_of_Warehouse_{w}",
)
# The demand minimum constraints are added to prob for each demand node (bar)
for b in Bars:
prob += (
lpSum([vars[w][b] for w in Warehouses]) >= demand[b],
f"Sum_of_Products_into_Bar{b}",
)
# The problem data is written to an .lp file
prob.writeLP("BeerDistributionProblem.lp")
# The problem is solved using PuLP's choice of Solver
prob.solve()
# The status of the solution is printed to the screen
print("Status:", LpStatus[prob.status])
# Each of the variables is printed with it's resolved optimum value
for v in prob.variables():
print(v.name, "=", v.varValue)
# The optimised objective function value is printed to the screen
print("Total Cost of Transportation = ", value(prob.objective))