transportation.py

__author__ = 'nick'

# Import PuLP modeler functions
from pulp import *

# Creates a list of all the supply nodes
warehouses = ['A', 'B']

# Creates a dictionary for the number of units of supply for each supply node
supply = {'A': 1000,
          'B': 4000}

# 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
        ]

# The cost data is made into a dictionary
costs = pulp.makeDict([warehouses, bars], costs,0)

# Creates the ’prob’ variable to contain the problem data
prob = pulp.LpProblem("Beer Distribution Problem", pulp.LpMinimize)

# Creates a list of tuples containing all the possible routes for transport
routes = [(w,b) for w in warehouses for b in bars]
print(routes)

# A dictionary called x is created to contain quantity shipped on the routes
x = pulp.LpVariable.dicts("route", (warehouses, bars), lowBound=0, cat=pulp.LpInteger)
print(x)

# The objective function is added to ’prob’ first
prob += sum([x[w][b]*costs[w][b] for (w, b) in routes]), "Sum_of_Transporting_Costs"

# Supply maximum constraints are added to prob for each supply node (warehouse)
for w in warehouses:
    prob += sum([x[w][b] for b in bars]) <= supply[w], "Sum_of_Products_out_of_Warehouse_%s"%w

# Demand minimum constraints are added to prob for each demand node (bar)
for b in bars:
    prob += sum([x[w][b] for w in warehouses]) >= demand[b], "Sum_of_Products_into_Bar%s"%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:", pulp.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 = ", prob.objective.value())