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"""
The American Steel Problem for the PuLP Modeller
Authors: Antony Phillips, Dr Stuart Mitchell 2007
"""
# Import PuLP modeller functions
from pulp import *
# List of all the nodes
Nodes = ["Youngstown",
"Pittsburgh",
"Cincinatti",
"Kansas City",
"Chicago",
"Albany",
"Houston",
"Tempe",
"Gary"]
nodeData = {# NODE Supply Demand
"Youngstown": [10000,0],
"Pittsburgh": [15000,0],
"Cincinatti": [0,0],
"Kansas City": [0,0],
"Chicago": [0,0],
"Albany": [0,3000],
"Houston": [0,7000],
"Tempe": [0,4000],
"Gary": [0,6000]}
# List of all the arcs
Arcs = [("Youngstown","Albany"),
("Youngstown","Cincinatti"),
("Youngstown","Kansas City"),
("Youngstown","Chicago"),
("Pittsburgh","Cincinatti"),
("Pittsburgh","Kansas City"),
("Pittsburgh","Chicago"),
("Pittsburgh","Gary"),
("Cincinatti","Albany"),
("Cincinatti","Houston"),
("Kansas City","Houston"),
("Kansas City","Tempe"),
("Chicago","Tempe"),
("Chicago","Gary")]
arcData = { # ARC Cost Min Max
("Youngstown","Albany"): [0.5,0,1000],
("Youngstown","Cincinatti"): [0.35,0,3000],
("Youngstown","Kansas City"): [0.45,1000,5000],
("Youngstown","Chicago"): [0.375,0,5000],
("Pittsburgh","Cincinatti"): [0.35,0,2000],
("Pittsburgh","Kansas City"): [0.45,2000,3000],
("Pittsburgh","Chicago"): [0.4,0,4000],
("Pittsburgh","Gary"): [0.45,0,2000],
("Cincinatti","Albany"): [0.35,1000,5000],
("Cincinatti","Houston"): [0.55,0,6000],
("Kansas City","Houston"): [0.375,0,4000],
("Kansas City","Tempe"): [0.65,0,4000],
("Chicago","Tempe"): [0.6,0,2000],
("Chicago","Gary"): [0.12,0,4000]}
# Splits the dictionaries to be more understandable
(supply, demand) = splitDict(nodeData)
(costs, mins, maxs) = splitDict(arcData)
# Creates the boundless Variables as Integers
vars = LpVariable.dicts("Route",Arcs,None,None,LpInteger)
# Creates the upper and lower bounds on the variables
for a in Arcs:
vars[a].bounds(mins[a], maxs[a])
# Creates the 'prob' variable to contain the problem data
prob = LpProblem("American Steel Problem",LpMinimize)
# Creates the objective function
prob += lpSum([vars[a]* costs[a] for a in Arcs]), "Total Cost of Transport"
# Creates all problem constraints - this ensures the amount going into each node is at least equal to the amount leaving
for n in Nodes:
prob += (supply[n]+ lpSum([vars[(i,j)] for (i,j) in Arcs if j == n]) >=
demand[n]+ lpSum([vars[(i,j)] for (i,j) in Arcs if i == n])), "Steel Flow Conservation in Node %s"%n
# The problem data is written to an .lp file
prob.writeLP("AmericanSteelProblem.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))
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