""" 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))