Back to Or Tools

assignment_mip

examples/notebook/linear_solver/assignment_mip.ipynb

2016-063.1 KB
Original Source
Copyright 2025 Google LLC.

Licensed under the Apache License, Version 2.0 (the "License"); you may not use this file except in compliance with the License. You may obtain a copy of the License at

http://www.apache.org/licenses/LICENSE-2.0

Unless required by applicable law or agreed to in writing, software distributed under the License is distributed on an "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the License for the specific language governing permissions and limitations under the License.

assignment_mip

<table align="left"> <td> <a href="https://colab.research.google.com/github/google/or-tools/blob/main/examples/notebook/linear_solver/assignment_mip.ipynb">Run in Google Colab</a> </td> <td> <a href="https://github.com/google/or-tools/blob/main/ortools/linear_solver/samples/assignment_mip.py">View source on GitHub</a> </td> </table>

First, you must install ortools package in this colab.

python
%pip install ortools

MIP example that solves an assignment problem.

python
from ortools.linear_solver import pywraplp



def main():
    # Data
    costs = [
        [90, 80, 75, 70],
        [35, 85, 55, 65],
        [125, 95, 90, 95],
        [45, 110, 95, 115],
        [50, 100, 90, 100],
    ]
    num_workers = len(costs)
    num_tasks = len(costs[0])

    # Solver
    # Create the mip solver with the SCIP backend.
    solver = pywraplp.Solver.CreateSolver("SCIP")

    if not solver:
        return

    # Variables
    # x[i, j] is an array of 0-1 variables, which will be 1
    # if worker i is assigned to task j.
    x = {}
    for i in range(num_workers):
        for j in range(num_tasks):
            x[i, j] = solver.IntVar(0, 1, "")

    # Constraints
    # Each worker is assigned to at most 1 task.
    for i in range(num_workers):
        solver.Add(solver.Sum([x[i, j] for j in range(num_tasks)]) <= 1)

    # Each task is assigned to exactly one worker.
    for j in range(num_tasks):
        solver.Add(solver.Sum([x[i, j] for i in range(num_workers)]) == 1)

    # Objective
    objective_terms = []
    for i in range(num_workers):
        for j in range(num_tasks):
            objective_terms.append(costs[i][j] * x[i, j])
    solver.Minimize(solver.Sum(objective_terms))

    # Solve
    print(f"Solving with {solver.SolverVersion()}")
    status = solver.Solve()

    # Print solution.
    if status == pywraplp.Solver.OPTIMAL or status == pywraplp.Solver.FEASIBLE:
        print(f"Total cost = {solver.Objective().Value()}\n")
        for i in range(num_workers):
            for j in range(num_tasks):
                # Test if x[i,j] is 1 (with tolerance for floating point arithmetic).
                if x[i, j].solution_value() > 0.5:
                    print(f"Worker {i} assigned to task {j}." + f" Cost: {costs[i][j]}")
    else:
        print("No solution found.")


main()