Questa sezione presenta un esempio che mostra come risolvere un problema di assegnazione utilizzando sia il risolutore MIP sia quello CP-SAT.
Esempio
Nell'esempio ci sono cinque worker (numerati da 0 a 4) e quattro attività (numerati da 0 a 3). Tieni presente che è presente un solo worker in più rispetto all'esempio nella Panoramica.
I costi per l'assegnazione dei worker alle attività sono mostrati nella tabella seguente.
| Worker | Attività 0 | Compito 1 | Compito 2 | Compito 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 |
Il problema è assegnare ogni worker al massimo a un'attività, senza che due worker eseguano la stessa attività, riducendo al minimo il costo totale. Poiché esistono più worker che attività, a un worker non verrà assegnata un'attività.
Soluzione MIP
Le seguenti sezioni descrivono come risolvere il problema utilizzando il wrapper MPSolver.
Importa le librerie
Il codice seguente importa le librerie richieste.
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;
Crea i dati
Il codice seguente crea i dati relativi al problema.
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);
L'array costs corrisponde alla tabella dei costi per l'assegnazione di worker alle attività, mostrata sopra.
Dichiara il risolutore MIP
Il codice seguente dichiara il risolutore 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;
}
Crea le variabili
Il codice seguente crea variabili con numeri interi binari per il problema.
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}");
}
}
Crea i vincoli
Il codice seguente crea i vincoli per il problema.
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);
}
}
Creare la funzione obiettivo
Il codice seguente crea la funzione obiettivo per il problema.
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();
Il valore della funzione obiettivo corrisponde al costo totale di tutte le variabili a cui il risolutore assegna il valore 1.
Richiama il risolutore
Il codice seguente richiama il risolutore.
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();
Stampa la soluzione
Il codice seguente stampa la soluzione al problema.
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.");
}
Ecco l'output del programma.
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
Completa i programmi
Di seguito sono riportati i programmi completi per la soluzione 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.");
}
}
}
Soluzione CP SAT
Le seguenti sezioni descrivono come risolvere il problema utilizzando il risolutore CP-SAT.
Importa le librerie
Il codice seguente importa le librerie richieste.
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;
Dichiara il modello
Il codice seguente dichiara il modello CP-SAT.
Python
model = cp_model.CpModel()
C++
CpModelBuilder cp_model;
Java
CpModel model = new CpModel();
C#
CpModel model = new CpModel();
Crea i dati
Il codice seguente configura i dati per il problema.
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);
L'array costs corrisponde alla tabella dei costi per l'assegnazione di worker alle attività, mostrata sopra.
Crea le variabili
Il codice seguente crea variabili con numeri interi binari per il problema.
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}");
}
}
Crea i vincoli
Il codice seguente crea i vincoli per il problema.
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);
}
Creare la funzione obiettivo
Il codice seguente crea la funzione obiettivo per il problema.
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);
Il valore della funzione obiettivo corrisponde al costo totale di tutte le variabili a cui il risolutore assegna il valore 1.
Richiama il risolutore
Il codice seguente richiama il risolutore.
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($"Solve status: {status}");
Stampa la soluzione
Il codice seguente stampa la soluzione al problema.
Python
if status == cp_model.OPTIMAL or status == cp_model.FEASIBLE:
print(f"Total cost = {solver.objective_value}\n")
selected = data.loc[solver.boolean_values(x).loc[lambda x: x].index]
for unused_index, row in selected.iterrows():
print(f"{row.task} assigned to {row.worker} with a cost of {row.cost}")
elif status == cp_model.INFEASIBLE:
print("No solution found")
else:
print("Something is wrong, check the status and the log of the solve")
C++
if (response.status() == CpSolverStatus::INFEASIBLE) {
LOG(FATAL) << "No solution found.";
}
LOG(INFO) << "Total cost: " << response.objective_value();
LOG(INFO);
for (int i = 0; i < num_workers; ++i) {
for (int j = 0; j < num_tasks; ++j) {
if (SolutionBooleanValue(response, x[i][j])) {
LOG(INFO) << "Task " << i << " assigned to worker " << j
<< ". Cost: " << costs[i][j];
}
}
}
Java
// Check that the problem has a feasible solution.
if (status == CpSolverStatus.OPTIMAL || status == CpSolverStatus.FEASIBLE) {
System.out.println("Total cost: " + solver.objectiveValue() + "\n");
for (int i = 0; i < numWorkers; ++i) {
for (int j = 0; j < numTasks; ++j) {
if (solver.booleanValue(x[i][j])) {
System.out.println(
"Worker " + i + " assigned to task " + j + ". Cost: " + costs[i][j]);
}
}
}
} else {
System.err.println("No solution found.");
}
C#
// Check that the problem has a feasible solution.
if (status == CpSolverStatus.Optimal || status == CpSolverStatus.Feasible)
{
Console.WriteLine($"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.");
}
Ecco l'output del programma.
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
Completa i programmi
Di seguito sono riportati i programmi completi per la soluzione 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");
}
}