72 lines
2.3 KiB
Python
72 lines
2.3 KiB
Python
"""
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The Beer Distribution Problem for the PuLP Modeller
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Authors: Antony Phillips, Dr Stuart Mitchell 2007
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"""
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# Import PuLP modeler functions
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from pulp import *
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# Creates a list of all the supply nodes
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Warehouses = ["A", "B"]
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# Creates a dictionary for the number of units of supply for each supply node
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supply = {"A": 1000,
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"B": 4000}
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# Creates a list of all demand nodes
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Bars = ["1", "2", "3", "4", "5"]
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# Creates a dictionary for the number of units of demand for each demand node
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demand = {"1":500,
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"2":900,
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"3":1800,
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"4":200,
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"5":700,}
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# Creates a list of costs of each transportation path
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costs = [ #Bars
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#1 2 3 4 5
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[2,4,5,2,1],#A Warehouses
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[3,1,3,2,3] #B
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]
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# The cost data is made into a dictionary
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costs = makeDict([Warehouses,Bars],costs,0)
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# Creates the 'prob' variable to contain the problem data
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prob = LpProblem("Beer Distribution Problem",LpMinimize)
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# Creates a list of tuples containing all the possible routes for transport
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Routes = [(w,b) for w in Warehouses for b in Bars]
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# A dictionary called 'Vars' is created to contain the referenced variables(the routes)
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vars = LpVariable.dicts("Route",(Warehouses,Bars),0,None,LpInteger)
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# The objective function is added to 'prob' first
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prob += lpSum([vars[w][b]*costs[w][b] for (w,b) in Routes]), "Sum_of_Transporting_Costs"
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# The supply maximum constraints are added to prob for each supply node (warehouse)
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for w in Warehouses:
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prob += lpSum([vars[w][b] for b in Bars])<=supply[w], "Sum_of_Products_out_of_Warehouse_%s"%w
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# The demand minimum constraints are added to prob for each demand node (bar)
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for b in Bars:
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prob += lpSum([vars[w][b] for w in Warehouses])>=demand[b], "Sum_of_Products_into_Bar%s"%b
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# The problem data is written to an .lp file
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prob.writeLP("BeerDistributionProblem.lp")
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# The problem is solved using PuLP's choice of Solver
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prob.solve()
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# The status of the solution is printed to the screen
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print("Status:", LpStatus[prob.status])
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# Each of the variables is printed with it's resolved optimum value
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for v in prob.variables():
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print(v.name, "=", v.varValue)
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# The optimised objective function value is printed to the screen
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print("Total Cost of Transportation = ", value(prob.objective))
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