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Basic MIP tasks: binary variables; logic constraints Description We wish to choose among items of different value and
weight those that result in the maximum total value for
a given weight limit. Further explanation of this example: 'Xpress Python Reference Manual'
Source Files By clicking on a file name, a preview is opened at the bottom of this page.
Data Files burglarl.py '''******************************************************* * Python Example Problems * * * * file burglarl.py * * Example for the use of the Python language * * (Burglar problem) * * -- Formulation of logical constraints -- * * * * (c) 2018-2024 Fair Isaac Corporation * *******************************************************''' from __future__ import print_function import xpress as xp Items = set(["camera", "necklace", "vase", "picture", "tv", "video", "chest", "brick"]) # Index set for items WTMAX = 102 # Max weight allowed for haul VALUE = {"camera": 15, "necklace": 100, "vase": 90, "picture": 60, "tv": 40, "video": 15, "chest": 10, "brick": 1} WEIGHT = {"camera": 2, "necklace": 20, "vase": 20, "picture": 30, "tv": 40, "video": 30, "chest": 60, "brick": 10} p = xp.problem() x = p.addVariables(Items, vartype=xp.binary) # 1 if we take item i; 0 otherwise # Objective: maximize total value p.setObjective(xp.Sum(VALUE[i]*x[i] for i in Items), sense=xp.maximize) # Weight restriction p.addConstraint(xp.Sum(WEIGHT[i]*x[i] for i in Items) <= WTMAX) # *** Logic constraint: # *** Either take "vase" and "picture" or "tv" and "video" # (but not both pairs). # * Values within each pair are the same p.addConstraint(x["vase"] == x["picture"]) p.addConstraint(x["tv"] == x["video"]) # * Choose exactly one pair (uncomment one of the 3 formulations A, B, or C) # (A) MIP formulation # p.addConstraint(x["tv"] == 1 - x["vase"]) # (B) Logic constraint # Note: Xpress Python interface doesn't use xor. # Need to introduce extra variable y = p.addVariable(vartype=xp.binary) # (C) Alternative logic formulation p.addIndicator(y == 1, x["tv"] + x["video"] >= 2) p.addIndicator(y == 0, x["vase"] + x["picture"] >= 2) p.addConstraint(x["tv"] + x["video"] + x["vase"] + x["picture"] <= 3) p.optimize() # Solve the MIP-problem # Print out the solution print("Solution:\n Objective: ", p.getObjVal()) for i in Items: print(" x(", i, "): ", p.getSolution(x[i])) | |||||||||||||||||
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