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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
# Example for the use of the Python language (Burglar problem).
#
# Formulation of logical constraints.
#
# (C) 2018-2025 Fair Isaac Corporation
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.attributes.objval)
xsol = p.getSolution(x)
for i in Items:
print(" x(", i, "): ", xsol[i])
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| © Copyright 2025 Fair Isaac Corporation. |