Esta seção apresenta um exemplo que mostra como resolver um problema de atribuição usando os solucionadores MIP e CP-SAT.
Exemplo
No exemplo, há cinco workers (numerados de 0 a 4) e quatro tarefas (numeradas 0 a 3). Observe que há um worker a mais do que no exemplo da Visão geral.
Os custos de atribuição de workers a tarefas são mostrados no tabela a seguir.
| Worker | Tarefa 0 | Tarefa 1 | Tarefa 2 | Tarefa 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 |
O problema é atribuir cada worker a no máximo uma tarefa, sem dois workers realizando a mesma tarefa, minimizando o custo total. Como há mais workers do que tarefas, um worker não receberá uma tarefa.
Solução MIP
As seções a seguir descrevem como resolver o problema usando a Wrapper MPSolver.
Importar as bibliotecas
O código a seguir importa as bibliotecas necessárias.
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;
Criar os dados
O código a seguir cria os dados para o 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);A matriz costs corresponde à tabela de custos.
para atribuir workers a tarefas, conforme mostrado acima.
Declarar o solucionador MIP
O código a seguir declara o solucionador 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; }
Criar as variáveis
O código a seguir cria variáveis inteiras binárias para o 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}"); } }
Criar as restrições
O código a seguir cria as restrições para o 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); } }
Criar a função objetiva
O código a seguir cria a função de objetivo para o 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();
O valor da função de objetivo é o custo total de todas as variáveis com o valor 1 pelo solucionador.
Invocar o solucionador
O código a seguir invoca o solucionador.
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();
Imprimir a solução
O código a seguir imprime a solução para o 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."); }
Esta é a saída do programa.
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
Programas completos
Estes são os programas completos para a solução 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."); } } }
Solução CP SAT
As seções a seguir descrevem como resolver o problema usando o solucionador CP-SAT.
Importar as bibliotecas
O código a seguir importa as bibliotecas necessárias.
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;
Declarar o modelo
O código a seguir declara o modelo CP-SAT.
Python
model = cp_model.CpModel()
C++
CpModelBuilder cp_model;
Java
CpModel model = new CpModel();
C#
CpModel model = new CpModel();
Criar os dados
O código a seguir define os dados para o 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);A matriz costs corresponde à tabela de custos.
para atribuir workers a tarefas, conforme mostrado acima.
Criar as variáveis
O código a seguir cria variáveis inteiras binárias para o 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}");
}
}Criar as restrições
O código a seguir cria as restrições para o 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); }
Criar a função objetiva
O código a seguir cria a função de objetivo para o 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);
O valor da função de objetivo é o custo total de todas as variáveis com o valor 1 pelo solucionador.
Invocar o solucionador
O código a seguir invoca o solucionador.
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($"Solve status: {status}");
Imprimir a solução
O código a seguir imprime a solução para o problema.
Python
if status == cp_model.OPTIMAL or status == cp_model.FEASIBLE: print(f"Total cost = {solver.objective_value}\n") selected = data.loc[solver.boolean_values(x).loc[lambda x: x].index] for unused_index, row in selected.iterrows(): print(f"{row.task} assigned to {row.worker} with a cost of {row.cost}") elif status == cp_model.INFEASIBLE: print("No solution found") else: print("Something is wrong, check the status and the log of the solve")
C++
if (response.status() == CpSolverStatus::INFEASIBLE) { LOG(FATAL) << "No solution found."; } LOG(INFO) << "Total cost: " << response.objective_value(); LOG(INFO); for (int i = 0; i < num_workers; ++i) { for (int j = 0; j < num_tasks; ++j) { if (SolutionBooleanValue(response, x[i][j])) { LOG(INFO) << "Task " << i << " assigned to worker " << j << ". Cost: " << costs[i][j]; } } }
Java
// Check that the problem has a feasible solution. if (status == CpSolverStatus.OPTIMAL || status == CpSolverStatus.FEASIBLE) { System.out.println("Total cost: " + solver.objectiveValue() + "\n"); for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { if (solver.booleanValue(x[i][j])) { System.out.println( "Worker " + i + " assigned to task " + j + ". Cost: " + costs[i][j]); } } } } else { System.err.println("No solution found."); }
C#
// Check that the problem has a feasible solution. if (status == CpSolverStatus.Optimal || status == CpSolverStatus.Feasible) { Console.WriteLine($"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."); }
Esta é a saída do programa.
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
Programas completos
Estes são os programas completos da solução 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"); } }