Bagian ini menampilkan contoh yang menunjukkan cara menyelesaikan soal tugas menggunakan pemecah masalah MIP dan pemecah masalah CP-SAT.
Contoh
Pada contoh, ada lima pekerja (bernomor 0-4) dan empat tugas (dinomori 0-3). Perhatikan bahwa ada satu pekerja lebih banyak daripada yang ada dalam contoh di Ringkasan.
Biaya menetapkan pekerja ke tugas ditampilkan dalam tabel berikut.
| Pekerja | Tugas 0 | Tugas 1 | Tugas 2 | Tugas 3 |
|---|---|---|---|---|
| 0 | 90 | 80 | 75 | 70 |
| 1 | 35 | 85 | 55 | 65 |
| 2 | 125 | 95 | 90 | 95 |
| 3 | 45 | 110 | 95 | 115 |
| 4 | 50 | 100 | 90 | 100 |
Masalahnya adalah menugaskan setiap pekerja ke maksimal satu tugas, tanpa dua pekerja yang melakukan tugas yang sama, sekaligus meminimalkan total biaya. Karena jumlah pekerja lebih banyak daripada tugas, satu pekerja tidak akan diberi tugas.
Solusi MIP
Bagian berikut menjelaskan cara menyelesaikan masalah menggunakan wrapper MPResolver.
Mengimpor library
Kode berikut mengimpor library yang diperlukan.
Python
from ortools.linear_solver import pywraplp
C++
#include <memory> #include <vector> #include "ortools/base/logging.h" #include "ortools/linear_solver/linear_solver.h"
Java
import com.google.ortools.Loader; import com.google.ortools.linearsolver.MPConstraint; import com.google.ortools.linearsolver.MPObjective; import com.google.ortools.linearsolver.MPSolver; import com.google.ortools.linearsolver.MPVariable;
C#
using System; using Google.OrTools.LinearSolver;
Membuat data
Kode berikut membuat data untuk masalah tersebut.
Python
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])
C++
const std::vector<std::vector<double>> costs{
{90, 80, 75, 70}, {35, 85, 55, 65}, {125, 95, 90, 95},
{45, 110, 95, 115}, {50, 100, 90, 100},
};
const int num_workers = costs.size();
const int num_tasks = costs[0].size();
Java
double[][] costs = {
{90, 80, 75, 70},
{35, 85, 55, 65},
{125, 95, 90, 95},
{45, 110, 95, 115},
{50, 100, 90, 100},
};
int numWorkers = costs.length;
int numTasks = costs[0].length;
C#
int[,] costs = {
{ 90, 80, 75, 70 }, { 35, 85, 55, 65 }, { 125, 95, 90, 95 }, { 45, 110, 95, 115 }, { 50, 100, 90, 100 },
};
int numWorkers = costs.GetLength(0);
int numTasks = costs.GetLength(1);
Array costs sesuai dengan tabel biaya
untuk menetapkan pekerja ke tugas, yang ditunjukkan di atas.
Mendeklarasikan pemecah MIP
Kode berikut mendeklarasikan pemecah MIP.
Python
# Create the mip solver with the SCIP backend.
solver = pywraplp.Solver.CreateSolver("SCIP")
if not solver:
return
C++
// Create the mip solver with the SCIP backend.
std::unique_ptr<MPSolver> solver(MPSolver::CreateSolver("SCIP"));
if (!solver) {
LOG(WARNING) << "SCIP solver unavailable.";
return;
}
Java
// Create the linear solver with the SCIP backend.
MPSolver solver = MPSolver.createSolver("SCIP");
if (solver == null) {
System.out.println("Could not create solver SCIP");
return;
}
C#
Solver solver = Solver.CreateSolver("SCIP");
if (solver is null)
{
return;
}
Membuat variabel
Kode berikut membuat variabel bilangan bulat biner untuk soal tersebut.
Python
# 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, "")
C++
// x[i][j] is an array of 0-1 variables, which will be 1
// if worker i is assigned to task j.
std::vector<std::vector<const MPVariable*>> x(
num_workers, std::vector<const MPVariable*>(num_tasks));
for (int i = 0; i < num_workers; ++i) {
for (int j = 0; j < num_tasks; ++j) {
x[i][j] = solver->MakeIntVar(0, 1, "");
}
}
Java
// x[i][j] is an array of 0-1 variables, which will be 1
// if worker i is assigned to task j.
MPVariable[][] x = new MPVariable[numWorkers][numTasks];
for (int i = 0; i < numWorkers; ++i) {
for (int j = 0; j < numTasks; ++j) {
x[i][j] = solver.makeIntVar(0, 1, "");
}
}
C#
// x[i, j] is an array of 0-1 variables, which will be 1
// if worker i is assigned to task j.
Variable[,] x = new Variable[numWorkers, numTasks];
for (int i = 0; i < numWorkers; ++i)
{
for (int j = 0; j < numTasks; ++j)
{
x[i, j] = solver.MakeIntVar(0, 1, $"worker_{i}_task_{j}");
}
}
Membuat batasan
Kode berikut membuat batasan untuk masalah.
Python
# 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)
C++
// Each worker is assigned to at most one task.
for (int i = 0; i < num_workers; ++i) {
LinearExpr worker_sum;
for (int j = 0; j < num_tasks; ++j) {
worker_sum += x[i][j];
}
solver->MakeRowConstraint(worker_sum <= 1.0);
}
// Each task is assigned to exactly one worker.
for (int j = 0; j < num_tasks; ++j) {
LinearExpr task_sum;
for (int i = 0; i < num_workers; ++i) {
task_sum += x[i][j];
}
solver->MakeRowConstraint(task_sum == 1.0);
}
Java
// Each worker is assigned to at most one task.
for (int i = 0; i < numWorkers; ++i) {
MPConstraint constraint = solver.makeConstraint(0, 1, "");
for (int j = 0; j < numTasks; ++j) {
constraint.setCoefficient(x[i][j], 1);
}
}
// Each task is assigned to exactly one worker.
for (int j = 0; j < numTasks; ++j) {
MPConstraint constraint = solver.makeConstraint(1, 1, "");
for (int i = 0; i < numWorkers; ++i) {
constraint.setCoefficient(x[i][j], 1);
}
}
C#
// Each worker is assigned to at most one task.
for (int i = 0; i < numWorkers; ++i)
{
Constraint constraint = solver.MakeConstraint(0, 1, "");
for (int j = 0; j < numTasks; ++j)
{
constraint.SetCoefficient(x[i, j], 1);
}
}
// Each task is assigned to exactly one worker.
for (int j = 0; j < numTasks; ++j)
{
Constraint constraint = solver.MakeConstraint(1, 1, "");
for (int i = 0; i < numWorkers; ++i)
{
constraint.SetCoefficient(x[i, j], 1);
}
}
Membuat fungsi tujuan
Kode berikut membuat fungsi objektif untuk soal ini.
Python
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))
C++
MPObjective* const objective = solver->MutableObjective();
for (int i = 0; i < num_workers; ++i) {
for (int j = 0; j < num_tasks; ++j) {
objective->SetCoefficient(x[i][j], costs[i][j]);
}
}
objective->SetMinimization();
Java
MPObjective objective = solver.objective();
for (int i = 0; i < numWorkers; ++i) {
for (int j = 0; j < numTasks; ++j) {
objective.setCoefficient(x[i][j], costs[i][j]);
}
}
objective.setMinimization();
C#
Objective objective = solver.Objective();
for (int i = 0; i < numWorkers; ++i)
{
for (int j = 0; j < numTasks; ++j)
{
objective.SetCoefficient(x[i, j], costs[i, j]);
}
}
objective.SetMinimization();
Nilai fungsi objektif adalah total biaya atas semua variabel yang diberi nilai 1 oleh pemecah masalah.
Memanggil pemecah masalah
Kode berikut memanggil pemecah.
Python
print(f"Solving with {solver.SolverVersion()}")
status = solver.Solve()
C++
const MPSolver::ResultStatus result_status = solver->Solve();
Java
MPSolver.ResultStatus resultStatus = solver.solve();
C#
Solver.ResultStatus resultStatus = solver.Solve();
Cetak solusi
Kode berikut mencetak solusi untuk masalah tersebut.
Python
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.")
C++
// Check that the problem has a feasible solution.
if (result_status != MPSolver::OPTIMAL &&
result_status != MPSolver::FEASIBLE) {
LOG(FATAL) << "No solution found.";
}
LOG(INFO) << "Total cost = " << objective->Value() << "\n\n";
for (int i = 0; i < num_workers; ++i) {
for (int j = 0; j < num_tasks; ++j) {
// Test if x[i][j] is 0 or 1 (with tolerance for floating point
// arithmetic).
if (x[i][j]->solution_value() > 0.5) {
LOG(INFO) << "Worker " << i << " assigned to task " << j
<< ". Cost = " << costs[i][j];
}
}
}
Java
// Check that the problem has a feasible solution.
if (resultStatus == MPSolver.ResultStatus.OPTIMAL
|| resultStatus == MPSolver.ResultStatus.FEASIBLE) {
System.out.println("Total cost: " + objective.value() + "\n");
for (int i = 0; i < numWorkers; ++i) {
for (int j = 0; j < numTasks; ++j) {
// Test if x[i][j] is 0 or 1 (with tolerance for floating point
// arithmetic).
if (x[i][j].solutionValue() > 0.5) {
System.out.println(
"Worker " + i + " assigned to task " + j + ". Cost = " + costs[i][j]);
}
}
}
} else {
System.err.println("No solution found.");
}
C#
// Check that the problem has a feasible solution.
if (resultStatus == Solver.ResultStatus.OPTIMAL || resultStatus == Solver.ResultStatus.FEASIBLE)
{
Console.WriteLine($"Total cost: {solver.Objective().Value()}\n");
for (int i = 0; i < numWorkers; ++i)
{
for (int j = 0; j < numTasks; ++j)
{
// Test if x[i, j] is 0 or 1 (with tolerance for floating point
// arithmetic).
if (x[i, j].SolutionValue() > 0.5)
{
Console.WriteLine($"Worker {i} assigned to task {j}. Cost: {costs[i, j]}");
}
}
}
}
else
{
Console.WriteLine("No solution found.");
}
Berikut adalah output program.
Total cost = 265.0 Worker 0 assigned to task 3. Cost = 70 Worker 1 assigned to task 2. Cost = 55 Worker 2 assigned to task 1. Cost = 95 Worker 3 assigned to task 0. Cost = 45
Selesaikan program
Berikut adalah program lengkap untuk solusi MIP.
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.")
if __name__ == "__main__":
main()
C++
#include <memory>
#include <vector>
#include "ortools/base/logging.h"
#include "ortools/linear_solver/linear_solver.h"
namespace operations_research {
void AssignmentMip() {
// Data
const std::vector<std::vector<double>> costs{
{90, 80, 75, 70}, {35, 85, 55, 65}, {125, 95, 90, 95},
{45, 110, 95, 115}, {50, 100, 90, 100},
};
const int num_workers = costs.size();
const int num_tasks = costs[0].size();
// Solver
// Create the mip solver with the SCIP backend.
std::unique_ptr<MPSolver> solver(MPSolver::CreateSolver("SCIP"));
if (!solver) {
LOG(WARNING) << "SCIP solver unavailable.";
return;
}
// Variables
// x[i][j] is an array of 0-1 variables, which will be 1
// if worker i is assigned to task j.
std::vector<std::vector<const MPVariable*>> x(
num_workers, std::vector<const MPVariable*>(num_tasks));
for (int i = 0; i < num_workers; ++i) {
for (int j = 0; j < num_tasks; ++j) {
x[i][j] = solver->MakeIntVar(0, 1, "");
}
}
// Constraints
// Each worker is assigned to at most one task.
for (int i = 0; i < num_workers; ++i) {
LinearExpr worker_sum;
for (int j = 0; j < num_tasks; ++j) {
worker_sum += x[i][j];
}
solver->MakeRowConstraint(worker_sum <= 1.0);
}
// Each task is assigned to exactly one worker.
for (int j = 0; j < num_tasks; ++j) {
LinearExpr task_sum;
for (int i = 0; i < num_workers; ++i) {
task_sum += x[i][j];
}
solver->MakeRowConstraint(task_sum == 1.0);
}
// Objective.
MPObjective* const objective = solver->MutableObjective();
for (int i = 0; i < num_workers; ++i) {
for (int j = 0; j < num_tasks; ++j) {
objective->SetCoefficient(x[i][j], costs[i][j]);
}
}
objective->SetMinimization();
// Solve
const MPSolver::ResultStatus result_status = solver->Solve();
// Print solution.
// Check that the problem has a feasible solution.
if (result_status != MPSolver::OPTIMAL &&
result_status != MPSolver::FEASIBLE) {
LOG(FATAL) << "No solution found.";
}
LOG(INFO) << "Total cost = " << objective->Value() << "\n\n";
for (int i = 0; i < num_workers; ++i) {
for (int j = 0; j < num_tasks; ++j) {
// Test if x[i][j] is 0 or 1 (with tolerance for floating point
// arithmetic).
if (x[i][j]->solution_value() > 0.5) {
LOG(INFO) << "Worker " << i << " assigned to task " << j
<< ". Cost = " << costs[i][j];
}
}
}
}
} // namespace operations_research
int main(int argc, char** argv) {
operations_research::AssignmentMip();
return EXIT_SUCCESS;
}
Java
package com.google.ortools.linearsolver.samples;
import com.google.ortools.Loader;
import com.google.ortools.linearsolver.MPConstraint;
import com.google.ortools.linearsolver.MPObjective;
import com.google.ortools.linearsolver.MPSolver;
import com.google.ortools.linearsolver.MPVariable;
/** MIP example that solves an assignment problem. */
public class AssignmentMip {
public static void main(String[] args) {
Loader.loadNativeLibraries();
// Data
double[][] costs = {
{90, 80, 75, 70},
{35, 85, 55, 65},
{125, 95, 90, 95},
{45, 110, 95, 115},
{50, 100, 90, 100},
};
int numWorkers = costs.length;
int numTasks = costs[0].length;
// Solver
// Create the linear solver with the SCIP backend.
MPSolver solver = MPSolver.createSolver("SCIP");
if (solver == null) {
System.out.println("Could not create solver SCIP");
return;
}
// Variables
// x[i][j] is an array of 0-1 variables, which will be 1
// if worker i is assigned to task j.
MPVariable[][] x = new MPVariable[numWorkers][numTasks];
for (int i = 0; i < numWorkers; ++i) {
for (int j = 0; j < numTasks; ++j) {
x[i][j] = solver.makeIntVar(0, 1, "");
}
}
// Constraints
// Each worker is assigned to at most one task.
for (int i = 0; i < numWorkers; ++i) {
MPConstraint constraint = solver.makeConstraint(0, 1, "");
for (int j = 0; j < numTasks; ++j) {
constraint.setCoefficient(x[i][j], 1);
}
}
// Each task is assigned to exactly one worker.
for (int j = 0; j < numTasks; ++j) {
MPConstraint constraint = solver.makeConstraint(1, 1, "");
for (int i = 0; i < numWorkers; ++i) {
constraint.setCoefficient(x[i][j], 1);
}
}
// Objective
MPObjective objective = solver.objective();
for (int i = 0; i < numWorkers; ++i) {
for (int j = 0; j < numTasks; ++j) {
objective.setCoefficient(x[i][j], costs[i][j]);
}
}
objective.setMinimization();
// Solve
MPSolver.ResultStatus resultStatus = solver.solve();
// Print solution.
// Check that the problem has a feasible solution.
if (resultStatus == MPSolver.ResultStatus.OPTIMAL
|| resultStatus == MPSolver.ResultStatus.FEASIBLE) {
System.out.println("Total cost: " + objective.value() + "\n");
for (int i = 0; i < numWorkers; ++i) {
for (int j = 0; j < numTasks; ++j) {
// Test if x[i][j] is 0 or 1 (with tolerance for floating point
// arithmetic).
if (x[i][j].solutionValue() > 0.5) {
System.out.println(
"Worker " + i + " assigned to task " + j + ". Cost = " + costs[i][j]);
}
}
}
} else {
System.err.println("No solution found.");
}
}
private AssignmentMip() {}
}
C#
using System;
using Google.OrTools.LinearSolver;
public class AssignmentMip
{
static void Main()
{
// Data.
int[,] costs = {
{ 90, 80, 75, 70 }, { 35, 85, 55, 65 }, { 125, 95, 90, 95 }, { 45, 110, 95, 115 }, { 50, 100, 90, 100 },
};
int numWorkers = costs.GetLength(0);
int numTasks = costs.GetLength(1);
// Solver.
Solver solver = Solver.CreateSolver("SCIP");
if (solver is null)
{
return;
}
// Variables.
// x[i, j] is an array of 0-1 variables, which will be 1
// if worker i is assigned to task j.
Variable[,] x = new Variable[numWorkers, numTasks];
for (int i = 0; i < numWorkers; ++i)
{
for (int j = 0; j < numTasks; ++j)
{
x[i, j] = solver.MakeIntVar(0, 1, $"worker_{i}_task_{j}");
}
}
// Constraints
// Each worker is assigned to at most one task.
for (int i = 0; i < numWorkers; ++i)
{
Constraint constraint = solver.MakeConstraint(0, 1, "");
for (int j = 0; j < numTasks; ++j)
{
constraint.SetCoefficient(x[i, j], 1);
}
}
// Each task is assigned to exactly one worker.
for (int j = 0; j < numTasks; ++j)
{
Constraint constraint = solver.MakeConstraint(1, 1, "");
for (int i = 0; i < numWorkers; ++i)
{
constraint.SetCoefficient(x[i, j], 1);
}
}
// Objective
Objective objective = solver.Objective();
for (int i = 0; i < numWorkers; ++i)
{
for (int j = 0; j < numTasks; ++j)
{
objective.SetCoefficient(x[i, j], costs[i, j]);
}
}
objective.SetMinimization();
// Solve
Solver.ResultStatus resultStatus = solver.Solve();
// Print solution.
// Check that the problem has a feasible solution.
if (resultStatus == Solver.ResultStatus.OPTIMAL || resultStatus == Solver.ResultStatus.FEASIBLE)
{
Console.WriteLine($"Total cost: {solver.Objective().Value()}\n");
for (int i = 0; i < numWorkers; ++i)
{
for (int j = 0; j < numTasks; ++j)
{
// Test if x[i, j] is 0 or 1 (with tolerance for floating point
// arithmetic).
if (x[i, j].SolutionValue() > 0.5)
{
Console.WriteLine($"Worker {i} assigned to task {j}. Cost: {costs[i, j]}");
}
}
}
}
else
{
Console.WriteLine("No solution found.");
}
}
}
Solusi CP SAT
Bagian berikut menjelaskan cara menyelesaikan soal menggunakan pemecah soal CP-SAT.
Mengimpor library
Kode berikut mengimpor library yang diperlukan.
Python
import io import pandas as pd from ortools.sat.python import cp_model
C++
#include <stdlib.h> #include <vector> #include "ortools/base/logging.h" #include "ortools/sat/cp_model.h" #include "ortools/sat/cp_model.pb.h" #include "ortools/sat/cp_model_solver.h"
Java
import com.google.ortools.Loader; import com.google.ortools.sat.CpModel; import com.google.ortools.sat.CpSolver; import com.google.ortools.sat.CpSolverStatus; import com.google.ortools.sat.LinearExpr; import com.google.ortools.sat.LinearExprBuilder; import com.google.ortools.sat.Literal; import java.util.ArrayList; import java.util.List; import java.util.stream.IntStream;
C#
using System; using System.Collections.Generic; using Google.OrTools.Sat;
Mendeklarasikan model
Kode berikut mendeklarasikan model CP-SAT.
Python
model = cp_model.CpModel()
C++
CpModelBuilder cp_model;
Java
CpModel model = new CpModel();
C#
CpModel model = new CpModel();
Membuat data
Kode berikut akan menyiapkan data untuk masalah tersebut.
Python
data_str = """
worker task cost
w1 t1 90
w1 t2 80
w1 t3 75
w1 t4 70
w2 t1 35
w2 t2 85
w2 t3 55
w2 t4 65
w3 t1 125
w3 t2 95
w3 t3 90
w3 t4 95
w4 t1 45
w4 t2 110
w4 t3 95
w4 t4 115
w5 t1 50
w5 t2 110
w5 t3 90
w5 t4 100
"""
data = pd.read_table(io.StringIO(data_str), sep=r"\s+")
C++
const std::vector<std::vector<int>> costs{
{90, 80, 75, 70}, {35, 85, 55, 65}, {125, 95, 90, 95},
{45, 110, 95, 115}, {50, 100, 90, 100},
};
const int num_workers = static_cast<int>(costs.size());
const int num_tasks = static_cast<int>(costs[0].size());
Java
int[][] costs = {
{90, 80, 75, 70},
{35, 85, 55, 65},
{125, 95, 90, 95},
{45, 110, 95, 115},
{50, 100, 90, 100},
};
final int numWorkers = costs.length;
final int numTasks = costs[0].length;
final int[] allWorkers = IntStream.range(0, numWorkers).toArray();
final int[] allTasks = IntStream.range(0, numTasks).toArray();
C#
int[,] costs = {
{ 90, 80, 75, 70 }, { 35, 85, 55, 65 }, { 125, 95, 90, 95 }, { 45, 110, 95, 115 }, { 50, 100, 90, 100 },
};
int numWorkers = costs.GetLength(0);
int numTasks = costs.GetLength(1);
Array costs sesuai dengan tabel biaya
untuk menetapkan pekerja ke tugas, yang ditunjukkan di atas.
Membuat variabel
Kode berikut membuat variabel bilangan bulat biner untuk soal tersebut.
Python
x = model.new_bool_var_series(name="x", index=data.index)
C++
// x[i][j] is an array of Boolean variables. x[i][j] is true
// if worker i is assigned to task j.
std::vector<std::vector<BoolVar>> x(num_workers,
std::vector<BoolVar>(num_tasks));
for (int i = 0; i < num_workers; ++i) {
for (int j = 0; j < num_tasks; ++j) {
x[i][j] = cp_model.NewBoolVar();
}
}
Java
Literal[][] x = new Literal[numWorkers][numTasks];
for (int worker : allWorkers) {
for (int task : allTasks) {
x[worker][task] = model.newBoolVar("x[" + worker + "," + task + "]");
}
}
C#
BoolVar[,] x = new BoolVar[numWorkers, numTasks];
// Variables in a 1-dim array.
for (int worker = 0; worker < numWorkers; ++worker)
{
for (int task = 0; task < numTasks; ++task)
{
x[worker, task] = model.NewBoolVar($"worker_{worker}_task_{task}");
}
}
Membuat batasan
Kode berikut membuat batasan untuk masalah.
Python
# Each worker is assigned to at most one task.
for unused_name, tasks in data.groupby("worker"):
model.add_at_most_one(x[tasks.index])
# Each task is assigned to exactly one worker.
for unused_name, workers in data.groupby("task"):
model.add_exactly_one(x[workers.index])
C++
// Each worker is assigned to at most one task.
for (int i = 0; i < num_workers; ++i) {
cp_model.AddAtMostOne(x[i]);
}
// Each task is assigned to exactly one worker.
for (int j = 0; j < num_tasks; ++j) {
std::vector<BoolVar> tasks;
for (int i = 0; i < num_workers; ++i) {
tasks.push_back(x[i][j]);
}
cp_model.AddExactlyOne(tasks);
}
Java
// Each worker is assigned to at most one task.
for (int worker : allWorkers) {
List<Literal> tasks = new ArrayList<>();
for (int task : allTasks) {
tasks.add(x[worker][task]);
}
model.addAtMostOne(tasks);
}
// Each task is assigned to exactly one worker.
for (int task : allTasks) {
List<Literal> workers = new ArrayList<>();
for (int worker : allWorkers) {
workers.add(x[worker][task]);
}
model.addExactlyOne(workers);
}
C#
// Each worker is assigned to at most one task.
for (int worker = 0; worker < numWorkers; ++worker)
{
List<ILiteral> tasks = new List<ILiteral>();
for (int task = 0; task < numTasks; ++task)
{
tasks.Add(x[worker, task]);
}
model.AddAtMostOne(tasks);
}
// Each task is assigned to exactly one worker.
for (int task = 0; task < numTasks; ++task)
{
List<ILiteral> workers = new List<ILiteral>();
for (int worker = 0; worker < numWorkers; ++worker)
{
workers.Add(x[worker, task]);
}
model.AddExactlyOne(workers);
}
Membuat fungsi tujuan
Kode berikut membuat fungsi objektif untuk soal ini.
Python
model.minimize(data.cost.dot(x))
C++
LinearExpr total_cost;
for (int i = 0; i < num_workers; ++i) {
for (int j = 0; j < num_tasks; ++j) {
total_cost += x[i][j] * costs[i][j];
}
}
cp_model.Minimize(total_cost);
Java
LinearExprBuilder obj = LinearExpr.newBuilder();
for (int worker : allWorkers) {
for (int task : allTasks) {
obj.addTerm(x[worker][task], costs[worker][task]);
}
}
model.minimize(obj);
C#
LinearExprBuilder obj = LinearExpr.NewBuilder();
for (int worker = 0; worker < numWorkers; ++worker)
{
for (int task = 0; task < numTasks; ++task)
{
obj.AddTerm((IntVar)x[worker, task], costs[worker, task]);
}
}
model.Minimize(obj);
Nilai fungsi objektif adalah total biaya atas semua variabel yang diberi nilai 1 oleh pemecah masalah.
Memanggil pemecah masalah
Kode berikut memanggil pemecah.
Python
solver = cp_model.CpSolver() status = solver.solve(model)
C++
const CpSolverResponse response = Solve(cp_model.Build());
Java
CpSolver solver = new CpSolver(); CpSolverStatus status = solver.solve(model);
C#
CpSolver solver = new CpSolver();
CpSolverStatus status = solver.Solve(model);
Console.WriteLine($"Solve status: {status}");
Cetak solusi
Kode berikut mencetak solusi untuk masalah tersebut.
Python
if status == cp_model.OPTIMAL or status == cp_model.FEASIBLE:
print(f"Total cost = {solver.objective_value}\n")
selected = data.loc[solver.boolean_values(x).loc[lambda x: x].index]
for unused_index, row in selected.iterrows():
print(f"{row.task} assigned to {row.worker} with a cost of {row.cost}")
elif status == cp_model.INFEASIBLE:
print("No solution found")
else:
print("Something is wrong, check the status and the log of the solve")
C++
if (response.status() == CpSolverStatus::INFEASIBLE) {
LOG(FATAL) << "No solution found.";
}
LOG(INFO) << "Total cost: " << response.objective_value();
LOG(INFO);
for (int i = 0; i < num_workers; ++i) {
for (int j = 0; j < num_tasks; ++j) {
if (SolutionBooleanValue(response, x[i][j])) {
LOG(INFO) << "Task " << i << " assigned to worker " << j
<< ". Cost: " << costs[i][j];
}
}
}
Java
// Check that the problem has a feasible solution.
if (status == CpSolverStatus.OPTIMAL || status == CpSolverStatus.FEASIBLE) {
System.out.println("Total cost: " + solver.objectiveValue() + "\n");
for (int i = 0; i < numWorkers; ++i) {
for (int j = 0; j < numTasks; ++j) {
if (solver.booleanValue(x[i][j])) {
System.out.println(
"Worker " + i + " assigned to task " + j + ". Cost: " + costs[i][j]);
}
}
}
} else {
System.err.println("No solution found.");
}
C#
// Check that the problem has a feasible solution.
if (status == CpSolverStatus.Optimal || status == CpSolverStatus.Feasible)
{
Console.WriteLine($"Total cost: {solver.ObjectiveValue}\n");
for (int i = 0; i < numWorkers; ++i)
{
for (int j = 0; j < numTasks; ++j)
{
if (solver.Value(x[i, j]) > 0.5)
{
Console.WriteLine($"Worker {i} assigned to task {j}. Cost: {costs[i, j]}");
}
}
}
}
else
{
Console.WriteLine("No solution found.");
}
Berikut adalah output program.
Total cost = 265 Worker 0 assigned to task 3 Cost = 70 Worker 1 assigned to task 2 Cost = 55 Worker 2 assigned to task 1 Cost = 95 Worker 3 assigned to task 0 Cost = 45
Selesaikan program
Berikut adalah program lengkap untuk solusi CP-SAT.
Python
import io
import pandas as pd
from ortools.sat.python import cp_model
def main() -> None:
# Data
data_str = """
worker task cost
w1 t1 90
w1 t2 80
w1 t3 75
w1 t4 70
w2 t1 35
w2 t2 85
w2 t3 55
w2 t4 65
w3 t1 125
w3 t2 95
w3 t3 90
w3 t4 95
w4 t1 45
w4 t2 110
w4 t3 95
w4 t4 115
w5 t1 50
w5 t2 110
w5 t3 90
w5 t4 100
"""
data = pd.read_table(io.StringIO(data_str), sep=r"\s+")
# Model
model = cp_model.CpModel()
# Variables
x = model.new_bool_var_series(name="x", index=data.index)
# Constraints
# Each worker is assigned to at most one task.
for unused_name, tasks in data.groupby("worker"):
model.add_at_most_one(x[tasks.index])
# Each task is assigned to exactly one worker.
for unused_name, workers in data.groupby("task"):
model.add_exactly_one(x[workers.index])
# Objective
model.minimize(data.cost.dot(x))
# Solve
solver = cp_model.CpSolver()
status = solver.solve(model)
# Print solution.
if status == cp_model.OPTIMAL or status == cp_model.FEASIBLE:
print(f"Total cost = {solver.objective_value}\n")
selected = data.loc[solver.boolean_values(x).loc[lambda x: x].index]
for unused_index, row in selected.iterrows():
print(f"{row.task} assigned to {row.worker} with a cost of {row.cost}")
elif status == cp_model.INFEASIBLE:
print("No solution found")
else:
print("Something is wrong, check the status and the log of the solve")
if __name__ == "__main__":
main()
C++
#include <stdlib.h>
#include <vector>
#include "ortools/base/logging.h"
#include "ortools/sat/cp_model.h"
#include "ortools/sat/cp_model.pb.h"
#include "ortools/sat/cp_model_solver.h"
namespace operations_research {
namespace sat {
void IntegerProgrammingExample() {
// Data
const std::vector<std::vector<int>> costs{
{90, 80, 75, 70}, {35, 85, 55, 65}, {125, 95, 90, 95},
{45, 110, 95, 115}, {50, 100, 90, 100},
};
const int num_workers = static_cast<int>(costs.size());
const int num_tasks = static_cast<int>(costs[0].size());
// Model
CpModelBuilder cp_model;
// Variables
// x[i][j] is an array of Boolean variables. x[i][j] is true
// if worker i is assigned to task j.
std::vector<std::vector<BoolVar>> x(num_workers,
std::vector<BoolVar>(num_tasks));
for (int i = 0; i < num_workers; ++i) {
for (int j = 0; j < num_tasks; ++j) {
x[i][j] = cp_model.NewBoolVar();
}
}
// Constraints
// Each worker is assigned to at most one task.
for (int i = 0; i < num_workers; ++i) {
cp_model.AddAtMostOne(x[i]);
}
// Each task is assigned to exactly one worker.
for (int j = 0; j < num_tasks; ++j) {
std::vector<BoolVar> tasks;
for (int i = 0; i < num_workers; ++i) {
tasks.push_back(x[i][j]);
}
cp_model.AddExactlyOne(tasks);
}
// Objective
LinearExpr total_cost;
for (int i = 0; i < num_workers; ++i) {
for (int j = 0; j < num_tasks; ++j) {
total_cost += x[i][j] * costs[i][j];
}
}
cp_model.Minimize(total_cost);
// Solve
const CpSolverResponse response = Solve(cp_model.Build());
// Print solution.
if (response.status() == CpSolverStatus::INFEASIBLE) {
LOG(FATAL) << "No solution found.";
}
LOG(INFO) << "Total cost: " << response.objective_value();
LOG(INFO);
for (int i = 0; i < num_workers; ++i) {
for (int j = 0; j < num_tasks; ++j) {
if (SolutionBooleanValue(response, x[i][j])) {
LOG(INFO) << "Task " << i << " assigned to worker " << j
<< ". Cost: " << costs[i][j];
}
}
}
}
} // namespace sat
} // namespace operations_research
int main(int argc, char** argv) {
operations_research::sat::IntegerProgrammingExample();
return EXIT_SUCCESS;
}
Java
package com.google.ortools.sat.samples;
import com.google.ortools.Loader;
import com.google.ortools.sat.CpModel;
import com.google.ortools.sat.CpSolver;
import com.google.ortools.sat.CpSolverStatus;
import com.google.ortools.sat.LinearExpr;
import com.google.ortools.sat.LinearExprBuilder;
import com.google.ortools.sat.Literal;
import java.util.ArrayList;
import java.util.List;
import java.util.stream.IntStream;
/** Assignment problem. */
public class AssignmentSat {
public static void main(String[] args) {
Loader.loadNativeLibraries();
// Data
int[][] costs = {
{90, 80, 75, 70},
{35, 85, 55, 65},
{125, 95, 90, 95},
{45, 110, 95, 115},
{50, 100, 90, 100},
};
final int numWorkers = costs.length;
final int numTasks = costs[0].length;
final int[] allWorkers = IntStream.range(0, numWorkers).toArray();
final int[] allTasks = IntStream.range(0, numTasks).toArray();
// Model
CpModel model = new CpModel();
// Variables
Literal[][] x = new Literal[numWorkers][numTasks];
for (int worker : allWorkers) {
for (int task : allTasks) {
x[worker][task] = model.newBoolVar("x[" + worker + "," + task + "]");
}
}
// Constraints
// Each worker is assigned to at most one task.
for (int worker : allWorkers) {
List<Literal> tasks = new ArrayList<>();
for (int task : allTasks) {
tasks.add(x[worker][task]);
}
model.addAtMostOne(tasks);
}
// Each task is assigned to exactly one worker.
for (int task : allTasks) {
List<Literal> workers = new ArrayList<>();
for (int worker : allWorkers) {
workers.add(x[worker][task]);
}
model.addExactlyOne(workers);
}
// Objective
LinearExprBuilder obj = LinearExpr.newBuilder();
for (int worker : allWorkers) {
for (int task : allTasks) {
obj.addTerm(x[worker][task], costs[worker][task]);
}
}
model.minimize(obj);
// Solve
CpSolver solver = new CpSolver();
CpSolverStatus status = solver.solve(model);
// Print solution.
// Check that the problem has a feasible solution.
if (status == CpSolverStatus.OPTIMAL || status == CpSolverStatus.FEASIBLE) {
System.out.println("Total cost: " + solver.objectiveValue() + "\n");
for (int i = 0; i < numWorkers; ++i) {
for (int j = 0; j < numTasks; ++j) {
if (solver.booleanValue(x[i][j])) {
System.out.println(
"Worker " + i + " assigned to task " + j + ". Cost: " + costs[i][j]);
}
}
}
} else {
System.err.println("No solution found.");
}
}
private AssignmentSat() {}
}
C#
using System;
using System.Collections.Generic;
using Google.OrTools.Sat;
public class AssignmentSat
{
public static void Main(String[] args)
{
// Data.
int[,] costs = {
{ 90, 80, 75, 70 }, { 35, 85, 55, 65 }, { 125, 95, 90, 95 }, { 45, 110, 95, 115 }, { 50, 100, 90, 100 },
};
int numWorkers = costs.GetLength(0);
int numTasks = costs.GetLength(1);
// Model.
CpModel model = new CpModel();
// Variables.
BoolVar[,] x = new BoolVar[numWorkers, numTasks];
// Variables in a 1-dim array.
for (int worker = 0; worker < numWorkers; ++worker)
{
for (int task = 0; task < numTasks; ++task)
{
x[worker, task] = model.NewBoolVar($"worker_{worker}_task_{task}");
}
}
// Constraints
// Each worker is assigned to at most one task.
for (int worker = 0; worker < numWorkers; ++worker)
{
List<ILiteral> tasks = new List<ILiteral>();
for (int task = 0; task < numTasks; ++task)
{
tasks.Add(x[worker, task]);
}
model.AddAtMostOne(tasks);
}
// Each task is assigned to exactly one worker.
for (int task = 0; task < numTasks; ++task)
{
List<ILiteral> workers = new List<ILiteral>();
for (int worker = 0; worker < numWorkers; ++worker)
{
workers.Add(x[worker, task]);
}
model.AddExactlyOne(workers);
}
// Objective
LinearExprBuilder obj = LinearExpr.NewBuilder();
for (int worker = 0; worker < numWorkers; ++worker)
{
for (int task = 0; task < numTasks; ++task)
{
obj.AddTerm((IntVar)x[worker, task], costs[worker, task]);
}
}
model.Minimize(obj);
// Solve
CpSolver solver = new CpSolver();
CpSolverStatus status = solver.Solve(model);
Console.WriteLine($"Solve status: {status}");
// Print solution.
// Check that the problem has a feasible solution.
if (status == CpSolverStatus.Optimal || status == CpSolverStatus.Feasible)
{
Console.WriteLine($"Total cost: {solver.ObjectiveValue}\n");
for (int i = 0; i < numWorkers; ++i)
{
for (int j = 0; j < numTasks; ++j)
{
if (solver.Value(x[i, j]) > 0.5)
{
Console.WriteLine($"Worker {i} assigned to task {j}. Cost: {costs[i, j]}");
}
}
}
}
else
{
Console.WriteLine("No solution found.");
}
Console.WriteLine("Statistics");
Console.WriteLine($" - conflicts : {solver.NumConflicts()}");
Console.WriteLine($" - branches : {solver.NumBranches()}");
Console.WriteLine($" - wall time : {solver.WallTime()}s");
}
}