1. Mixed Integer Programming

Easy

Description

Formulate and solve the following simple MIP model:

Maximize
    x + y + 2z

subject to constraints
    x + 2y + 3z <= 4
    x +  y      >= 1
    x, y, z binary

🚀 Test your code

Try to solve the problem below in the code editor before reading the solution.

View Answer

Objective value: 3.0
x: 1.0
y: 0.0
z: 1.0

Solutions

GurobiPulp

import gurobipy as gp
from gurobipy import GRB

# Create a new model
m = gp.Model("mip1")

# Create variables
x = m.addVar(vtype=GRB.BINARY, name="x")
y = m.addVar(vtype=GRB.BINARY, name="y")
z = m.addVar(vtype=GRB.BINARY, name="z")

# Set objective
m.setObjective(x + y + 2 * z, GRB.MAXIMIZE)

# Add constraint: x + 2 y + 3 z <= 4
m.addConstr(x + 2 * y + 3 * z <= 4, "c0")

# Add constraint: x + y >= 1
m.addConstr(x + y >= 1, "c1")

# Optimize model
m.optimize()

for v in m.getVars():
    print(f"{v.VarName} {v.X:g}")

print(f"Obj: {m.ObjVal:g}")

from pulp import *

def solve_mip_model():

    # Create the model
    model = LpProblem(name="maximize_xyz", sense=LpMaximize)

    # Define the decision variables
    x = LpVariable(name="x", cat=LpBinary)
    y = LpVariable(name="y", cat=LpBinary)
    z = LpVariable(name="z", cat=LpBinary)

    # Add the objective function
    model += lpSum([x, y, 2*z]), "objective_function"

    # Add the constraints
    model += (x + 2*y + 3*z <= 4), "constraint_1"
    model += (x + y >= 1), "constraint_2"

    # Solve the model
    model.solve()

    # Print the results
    print(f"Status: {model.status}")
    print(f"Objective value: {model.objective.value()}")
    for var in model.variables():
        print(f"{var.name}: {var.value()}")

if __name__ == "__main__":
    solve_mip_model()