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"""
The Beer Distribution Problem for the PuLP Modeller
Illustrates changing a constraint and solving
Authors: Antony Phillips, Dr Stuart Mitchell 2007
"""
# 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 = 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], "Sum_of_Products_out_of_Warehouse_%s"%w
# The demand minimum constraints are added to prob for each demand node (bar)
# These constraints are stored for resolve later
bar_demand_constraint = {}
for b in Bars:
constraint = lpSum([vars[w][b] for w in Warehouses])>=demand[b]
prob += constraint, "Sum_of_Products_into_Bar_%s"%b
bar_demand_constraint[b] = constraint
# The problem data is written to an .lp file
prob.writeLP("BeerDistributionProblem.lp")
for demand in range(500, 601, 10):
# reoptimise the problem by increasing demand at bar '1'
# note the constant is stored as the LHS constant not the RHS of the constraint
bar_demand_constraint['1'].constant = - demand
#or alternatively,
#prob.constraints["Sum_of_Products_into_Bar_1"].constant = - demand
# 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))
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