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