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"""
The Simplified Sponge Roll Problem for the PuLP Modeller
Authors: Antony Phillips, Dr Stuart Mitchell 2007
"""
# Import PuLP modeler functions
from pulp import *
# A list of all the roll lengths is created
LenOpts = ["5","7","9"]
# A dictionary of the demand for each roll length is created
rollDemand = {"5":150,
"7":200,
"9":300}
# A list of all the patterns is created
PatternNames = ["A","B","C"]
# Creates a list of the number of rolls in each pattern for each different roll length
patterns = [#A B C
[0,2,2],# 5
[1,1,0],# 7
[1,0,1] # 9
]
# The cost of each 20cm long sponge roll used
cost = 1
# The pattern data is made into a dictionary
patterns = makeDict([LenOpts,PatternNames],patterns,0)
# The problem variables of the number of each pattern to make are created
vars = LpVariable.dicts("Patt",PatternNames,0,None,LpInteger)
# The variable 'prob' is created
prob = LpProblem("Cutting Stock Problem",LpMinimize)
# The objective function is entered: the total number of large rolls used * the fixed cost of each
prob += lpSum([vars[i]*cost for i in PatternNames]),"Production Cost"
# The demand minimum constraint is entered
for i in LenOpts:
prob += lpSum([vars[j]*patterns[i][j] for j in PatternNames])>=rollDemand[i],"Ensuring enough %s cm rolls"%i
# The problem data is written to an .lp file
prob.writeLP("SpongeRollProblem.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("Production Costs = ", value(prob.objective))
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