File: SpongeRollProblem1.py

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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 = [[0, 2, 2], [1, 1, 0], [1, 0, 1]]  # A B C  # 5  # 7  # 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))