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deb-python-pulp/examples/SpongeRollProblem4.py
2014-01-23 18:05:08 +08:00

138 lines
4.6 KiB
Python

"""
The Full Sponge Roll Problem using Classes for the PuLP Modeller
Authors: Antony Phillips, Dr Stuart Mitchell 2007
"""
def calculatePatterns(totalRollLength,lenOpts,head):
"""
Recursively calculates the list of options lists for a cutting stock problem. The input
'tlist' is a pointer, and will be the output of the function call.
The inputs are:
totalRollLength - the length of the roll
lenOpts - a list of the sizes of remaining cutting options
head - the current list that has been passed down though the recusion
Returns the list of patterns
Authors: Bojan Blazevic, Dr Stuart Mitchell 2007
"""
if lenOpts:
patterns =[]
#take the first option off lenOpts
opt = lenOpts[0]
for rep in range(int(totalRollLength/opt)+1):
#reduce the length
l = totalRollLength - rep*opt
h = head[:]
h.append(rep)
patterns.extend(calculatePatterns(l, lenOpts[1:], h))
else:
#end of the recursion
patterns = [head]
return patterns
def makePatterns(totalRollLength,lenOpts):
"""
Makes the different cutting patterns for a cutting stock problem.
The inputs are:
totalRollLength : the length of the roll
lenOpts: a list of the sizes of cutting options as strings
Authors: Antony Phillips, Dr Stuart Mitchell 2007
"""
# calculatePatterns is called to create a list of the feasible cutting options in 'tlist'
patternslist = calculatePatterns(totalRollLength,lenOpts,[])
# The list 'PatternNames' is created
PatternNames = []
for i in range(len(patternslist)):
PatternNames += ["P"+str(i)]
#Patterns = [0 for i in range(len(PatternNames))]
Patterns = []
for i,j in enumerate(PatternNames):
Patterns += [Pattern(j, patternslist[i])]
# The different cutting lengths are printed, and the number of each roll of that length in each
# pattern is printed below. This is so the user can see what each pattern contains.
print("Lens: %s" %lenOpts)
for i in Patterns:
print(i, " = %s"%[i.lengthsdict[j] for j in lenOpts])
return Patterns
class Pattern:
"""
Information on a specific pattern in the SpongeRoll Problem
"""
cost = 1
trimValue = 0.04
totalRollLength = 20
lenOpts = [5, 7, 9]
def __init__(self,name,lengths = None):
self.name = name
self.lengthsdict = dict(zip(self.lenOpts,lengths))
def __str__(self):
return self.name
def trim(self):
return Pattern.totalRollLength - sum([int(i)*self.lengthsdict[i] for i in self.lengthsdict])
# Import PuLP modeler functions
from pulp import *
rollData = {#Length Demand SalePrice
5: [150, 0.25],
7: [200, 0.33],
9: [300, 0.40]}
# The pattern names and the patterns are created as lists, and the associated trim with each pattern
# is created as a dictionary. The inputs are the total roll length and the list (as integers) of
# cutting options.
Patterns = makePatterns(Pattern.totalRollLength,Pattern.lenOpts)
# The rollData is made into separate dictionaries
(rollDemand,surplusPrice) = splitDict(rollData)
# The variable 'prob' is created
prob = LpProblem("Cutting Stock Problem",LpMinimize)
# The problem variables of the number of each pattern to make are created
pattVars = LpVariable.dicts("Patt",Patterns,0,None,LpInteger)
# The problem variables of the number of surplus rolls for each length are created
surplusVars = LpVariable.dicts("Surp",Pattern.lenOpts,0,None,LpInteger)
# The objective function is entered: (the total number of large rolls used * the cost of each) - (the value of the surplus stock) - (the value of the trim)
prob += lpSum([pattVars[i]*Pattern.cost for i in Patterns]) - lpSum([surplusVars[i]*surplusPrice[i] for i in Pattern.lenOpts]) - lpSum([pattVars[i]*i.trim()*Pattern.trimValue for i in Patterns]),"Net Production Cost"
# The demand minimum constraint is entered
for j in Pattern.lenOpts:
prob += lpSum([pattVars[i]*i.lengthsdict[j] for i in Patterns]) - surplusVars[j] >= rollDemand[j],"Ensuring enough %s cm rolls"%j
# The problem data is written to an .lp file
prob.writeLP("SpongeRollProblem.lp")
# The problem is solved
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))