Puedes usar el solucionador de flujo de costos mínimo para resolver casos especiales del problema de asignación.
De hecho, el flujo de costo mínimo a menudo puede mostrar una solución más rápido que el solucionador de MIP o CP-SAT. Sin embargo, MIP y CP-SAT pueden resolver una clase más grande de problemas que el flujo de costo mínimo, por lo que en la mayoría de los casos MIP o CP-SAT son las mejores opciones.
En las siguientes secciones, se presentan programas de Python que resuelven los siguientes problemas de asignación con el solucionador de flujo de costo mínimo:
- Un ejemplo de asignación lineal mínimo.
- Un problema de asignación con los equipos de trabajadores.
Ejemplo de asignación lineal
En esta sección, se muestra cómo resolver el ejemplo descrito en la sección Resolutor de asignaciones lineales como un problema de flujo de costos mínimo.
Importa las bibliotecas
Con el siguiente código, se importa la biblioteca requerida.
Python
from ortools.graph.python import min_cost_flow
C++
#include <cstdint> #include <vector> #include "ortools/graph/min_cost_flow.h"
Java
import com.google.ortools.Loader; import com.google.ortools.graph.MinCostFlow; import com.google.ortools.graph.MinCostFlowBase;
C#
using System; using Google.OrTools.Graph;
Cómo declarar el solucionador
Con el siguiente código, se crea el solucionador de flujo de costo mínimo.
Python
# Instantiate a SimpleMinCostFlow solver. smcf = min_cost_flow.SimpleMinCostFlow()
C++
// Instantiate a SimpleMinCostFlow solver. SimpleMinCostFlow min_cost_flow;
Java
// Instantiate a SimpleMinCostFlow solver. MinCostFlow minCostFlow = new MinCostFlow();
C#
// Instantiate a SimpleMinCostFlow solver. MinCostFlow minCostFlow = new MinCostFlow();
Crea los datos
El diagrama de flujo del problema consiste en el grafo bipartito para la matriz de costos (consulta la descripción general de la asignación para obtener un ejemplo ligeramente diferente), con una fuente y un receptor agregados.
Los datos contienen los siguientes cuatro arreglos que corresponden a los nodos de inicio, los nodos finales, las capacidades y los costos del problema. La longitud de cada arreglo es la cantidad de arcos en el gráfico.
Python
# Define the directed graph for the flow.
start_nodes = (
[0, 0, 0, 0] + [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4] + [5, 6, 7, 8]
)
end_nodes = (
[1, 2, 3, 4] + [5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8] + [9, 9, 9, 9]
)
capacities = (
[1, 1, 1, 1] + [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] + [1, 1, 1, 1]
)
costs = (
[0, 0, 0, 0]
+ [90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115]
+ [0, 0, 0, 0]
)
source = 0
sink = 9
tasks = 4
supplies = [tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks]
C++
// Define four parallel arrays: sources, destinations, capacities,
// and unit costs between each pair. For instance, the arc from node 0
// to node 1 has a capacity of 15.
const std::vector<int64_t> start_nodes = {0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2,
3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8};
const std::vector<int64_t> end_nodes = {1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8,
5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9};
const std::vector<int64_t> capacities = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
const std::vector<int64_t> unit_costs = {0, 0, 0, 0, 90, 76, 75, 70,
35, 85, 55, 65, 125, 95, 90, 105,
45, 110, 95, 115, 0, 0, 0, 0};
const int64_t source = 0;
const int64_t sink = 9;
const int64_t tasks = 4;
// Define an array of supplies at each node.
const std::vector<int64_t> supplies = {tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks};
Java
// Define four parallel arrays: sources, destinations, capacities, and unit costs
// between each pair.
int[] startNodes =
new int[] {0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8};
int[] endNodes =
new int[] {1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9};
int[] capacities =
new int[] {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
int[] unitCosts = new int[] {
0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 0, 0, 0, 0};
int source = 0;
int sink = 9;
int tasks = 4;
// Define an array of supplies at each node.
int[] supplies = new int[] {tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks};
C#
// Define four parallel arrays: sources, destinations, capacities, and unit costs
// between each pair.
int[] startNodes = { 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8 };
int[] endNodes = { 1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9 };
int[] capacities = { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 };
int[] unitCosts = { 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65,
125, 95, 90, 105, 45, 110, 95, 115, 0, 0, 0, 0 };
int source = 0;
int sink = 9;
int tasks = 4;
// Define an array of supplies at each node.
int[] supplies = { tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks };
Para dejar en claro cómo se configuran los datos, cada array se divide en tres subarreglos:
- El primer array corresponde a arcos que salen del origen.
- El segundo array corresponde a los arcos entre los trabajadores y las tareas.
Para
costs, esto es solo la matriz de costos (que usa el solucionador de asignaciones lineales), acoplada en un vector. - El tercer array corresponde a los arcos que conducen al receptor.
Los datos también incluyen el vector supplies, que proporciona el suministro en cada nodo.
Cómo representa un problema de asignación un problema de flujo de costos mínimo
¿De qué manera el problema de flujo de costo mínimo anterior representa un problema de asignación? En primer lugar, como la capacidad de cada arco es de 1, la oferta de 4 en la fuente obliga a cada uno de los cuatro arcos que conducen a los trabajadores a tener un flujo de 1.
A continuación, la condición flujo-en-igual-flujo-salida obliga al flujo de cada trabajador a ser 1. Si es posible, el solucionador dirigirá ese flujo a través del arco de costo mínimo que sale de cada trabajador. Sin embargo, el solucionador no puede dirigir los flujos de dos trabajadores diferentes a una sola tarea. Si así fuera, habría un flujo combinado de 2 en esa tarea, que no se podría enviar a través del arco único con capacidad de 1 desde la tarea hasta el receptor. Esto significa que el solucionador solo puede asignar una tarea a un solo trabajador, según lo requiera el problema de asignación.
Por último, la condición flujo-igual-flujo-salida obliga a cada tarea a tener una salida de 1, por lo que algún trabajador realiza cada tarea.
Crea el gráfico y las restricciones
El siguiente código crea el gráfico y las restricciones.
Python
# Add each arc.
for i in range(len(start_nodes)):
smcf.add_arc_with_capacity_and_unit_cost(
start_nodes[i], end_nodes[i], capacities[i], costs[i]
)
# Add node supplies.
for i in range(len(supplies)):
smcf.set_node_supply(i, supplies[i])
C++
// Add each arc.
for (int i = 0; i < start_nodes.size(); ++i) {
int arc = min_cost_flow.AddArcWithCapacityAndUnitCost(
start_nodes[i], end_nodes[i], capacities[i], unit_costs[i]);
if (arc != i) LOG(FATAL) << "Internal error";
}
// Add node supplies.
for (int i = 0; i < supplies.size(); ++i) {
min_cost_flow.SetNodeSupply(i, supplies[i]);
}
Java
// Add each arc.
for (int i = 0; i < startNodes.length; ++i) {
int arc = minCostFlow.addArcWithCapacityAndUnitCost(
startNodes[i], endNodes[i], capacities[i], unitCosts[i]);
if (arc != i) {
throw new Exception("Internal error");
}
}
// Add node supplies.
for (int i = 0; i < supplies.length; ++i) {
minCostFlow.setNodeSupply(i, supplies[i]);
}
C#
// Add each arc.
for (int i = 0; i < startNodes.Length; ++i)
{
int arc =
minCostFlow.AddArcWithCapacityAndUnitCost(startNodes[i], endNodes[i], capacities[i], unitCosts[i]);
if (arc != i)
throw new Exception("Internal error");
}
// Add node supplies.
for (int i = 0; i < supplies.Length; ++i)
{
minCostFlow.SetNodeSupply(i, supplies[i]);
}
Cómo invocar el solucionador
El siguiente código invoca el solucionador y muestra la solución.
Python
# Find the minimum cost flow between node 0 and node 10. status = smcf.solve()
C++
// Find the min cost flow. int status = min_cost_flow.Solve();
Java
// Find the min cost flow. MinCostFlowBase.Status status = minCostFlow.solve();
C#
// Find the min cost flow. MinCostFlow.Status status = minCostFlow.Solve();
La solución consiste en los arcos entre los trabajadores y las tareas a las que el solucionador asigna un flujo de 1. (Los arcos conectados a la fuente o el receptor no forman parte de la solución).
El programa verifica cada arco para ver si tiene el flujo 1 y, de ser así, imprime el Tail (nodo de inicio) y el Head (nodo final) del arco, que corresponden a un trabajador y una tarea en la asignación.
Resultado del programa
Python
if status == smcf.OPTIMAL:
print("Total cost = ", smcf.optimal_cost())
print()
for arc in range(smcf.num_arcs()):
# Can ignore arcs leading out of source or into sink.
if smcf.tail(arc) != source and smcf.head(arc) != sink:
# Arcs in the solution have a flow value of 1. Their start and end nodes
# give an assignment of worker to task.
if smcf.flow(arc) > 0:
print(
"Worker %d assigned to task %d. Cost = %d"
% (smcf.tail(arc), smcf.head(arc), smcf.unit_cost(arc))
)
else:
print("There was an issue with the min cost flow input.")
print(f"Status: {status}")
C++
if (status == MinCostFlow::OPTIMAL) {
LOG(INFO) << "Total cost: " << min_cost_flow.OptimalCost();
LOG(INFO) << "";
for (std::size_t i = 0; i < min_cost_flow.NumArcs(); ++i) {
// Can ignore arcs leading out of source or into sink.
if (min_cost_flow.Tail(i) != source && min_cost_flow.Head(i) != sink) {
// Arcs in the solution have a flow value of 1. Their start and end
// nodes give an assignment of worker to task.
if (min_cost_flow.Flow(i) > 0) {
LOG(INFO) << "Worker " << min_cost_flow.Tail(i)
<< " assigned to task " << min_cost_flow.Head(i)
<< " Cost: " << min_cost_flow.UnitCost(i);
}
}
}
} else {
LOG(INFO) << "Solving the min cost flow problem failed.";
LOG(INFO) << "Solver status: " << status;
}
Java
if (status == MinCostFlow.Status.OPTIMAL) {
System.out.println("Total cost: " + minCostFlow.getOptimalCost());
System.out.println();
for (int i = 0; i < minCostFlow.getNumArcs(); ++i) {
// Can ignore arcs leading out of source or into sink.
if (minCostFlow.getTail(i) != source && minCostFlow.getHead(i) != sink) {
// Arcs in the solution have a flow value of 1. Their start and end nodes
// give an assignment of worker to task.
if (minCostFlow.getFlow(i) > 0) {
System.out.println("Worker " + minCostFlow.getTail(i) + " assigned to task "
+ minCostFlow.getHead(i) + " Cost: " + minCostFlow.getUnitCost(i));
}
}
}
} else {
System.out.println("Solving the min cost flow problem failed.");
System.out.println("Solver status: " + status);
}
C#
if (status == MinCostFlow.Status.OPTIMAL)
{
Console.WriteLine("Total cost: " + minCostFlow.OptimalCost());
Console.WriteLine("");
for (int i = 0; i < minCostFlow.NumArcs(); ++i)
{
// Can ignore arcs leading out of source or into sink.
if (minCostFlow.Tail(i) != source && minCostFlow.Head(i) != sink)
{
// Arcs in the solution have a flow value of 1. Their start and end nodes
// give an assignment of worker to task.
if (minCostFlow.Flow(i) > 0)
{
Console.WriteLine("Worker " + minCostFlow.Tail(i) + " assigned to task " + minCostFlow.Head(i) +
" Cost: " + minCostFlow.UnitCost(i));
}
}
}
}
else
{
Console.WriteLine("Solving the min cost flow problem failed.");
Console.WriteLine("Solver status: " + status);
}
Este es el resultado del programa.
Total cost = 265 Worker 1 assigned to task 8. Cost = 70 Worker 2 assigned to task 7. Cost = 55 Worker 3 assigned to task 6. Cost = 95 Worker 4 assigned to task 5. Cost = 45 Time = 0.000245 seconds
El resultado es el mismo que el de la resolución de asignaciones lineales (excepto por la diferente cantidad de trabajadores y los costos). La resolución de asignaciones lineales es un poco más rápida que el flujo de costos mínimo: 0.000147 segundos frente a 0.000458 segundos.
Todo el programa
A continuación, se muestra todo el programa.
Python
"""Linear assignment example."""
from ortools.graph.python import min_cost_flow
def main():
"""Solving an Assignment Problem with MinCostFlow."""
# Instantiate a SimpleMinCostFlow solver.
smcf = min_cost_flow.SimpleMinCostFlow()
# Define the directed graph for the flow.
start_nodes = (
[0, 0, 0, 0] + [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4] + [5, 6, 7, 8]
)
end_nodes = (
[1, 2, 3, 4] + [5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8] + [9, 9, 9, 9]
)
capacities = (
[1, 1, 1, 1] + [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] + [1, 1, 1, 1]
)
costs = (
[0, 0, 0, 0]
+ [90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115]
+ [0, 0, 0, 0]
)
source = 0
sink = 9
tasks = 4
supplies = [tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks]
# Add each arc.
for i in range(len(start_nodes)):
smcf.add_arc_with_capacity_and_unit_cost(
start_nodes[i], end_nodes[i], capacities[i], costs[i]
)
# Add node supplies.
for i in range(len(supplies)):
smcf.set_node_supply(i, supplies[i])
# Find the minimum cost flow between node 0 and node 10.
status = smcf.solve()
if status == smcf.OPTIMAL:
print("Total cost = ", smcf.optimal_cost())
print()
for arc in range(smcf.num_arcs()):
# Can ignore arcs leading out of source or into sink.
if smcf.tail(arc) != source and smcf.head(arc) != sink:
# Arcs in the solution have a flow value of 1. Their start and end nodes
# give an assignment of worker to task.
if smcf.flow(arc) > 0:
print(
"Worker %d assigned to task %d. Cost = %d"
% (smcf.tail(arc), smcf.head(arc), smcf.unit_cost(arc))
)
else:
print("There was an issue with the min cost flow input.")
print(f"Status: {status}")
if __name__ == "__main__":
main()
C++
#include <cstdint>
#include <vector>
#include "ortools/graph/min_cost_flow.h"
namespace operations_research {
// MinCostFlow simple interface example.
void AssignmentMinFlow() {
// Instantiate a SimpleMinCostFlow solver.
SimpleMinCostFlow min_cost_flow;
// Define four parallel arrays: sources, destinations, capacities,
// and unit costs between each pair. For instance, the arc from node 0
// to node 1 has a capacity of 15.
const std::vector<int64_t> start_nodes = {0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2,
3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8};
const std::vector<int64_t> end_nodes = {1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8,
5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9};
const std::vector<int64_t> capacities = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
const std::vector<int64_t> unit_costs = {0, 0, 0, 0, 90, 76, 75, 70,
35, 85, 55, 65, 125, 95, 90, 105,
45, 110, 95, 115, 0, 0, 0, 0};
const int64_t source = 0;
const int64_t sink = 9;
const int64_t tasks = 4;
// Define an array of supplies at each node.
const std::vector<int64_t> supplies = {tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks};
// Add each arc.
for (int i = 0; i < start_nodes.size(); ++i) {
int arc = min_cost_flow.AddArcWithCapacityAndUnitCost(
start_nodes[i], end_nodes[i], capacities[i], unit_costs[i]);
if (arc != i) LOG(FATAL) << "Internal error";
}
// Add node supplies.
for (int i = 0; i < supplies.size(); ++i) {
min_cost_flow.SetNodeSupply(i, supplies[i]);
}
// Find the min cost flow.
int status = min_cost_flow.Solve();
if (status == MinCostFlow::OPTIMAL) {
LOG(INFO) << "Total cost: " << min_cost_flow.OptimalCost();
LOG(INFO) << "";
for (std::size_t i = 0; i < min_cost_flow.NumArcs(); ++i) {
// Can ignore arcs leading out of source or into sink.
if (min_cost_flow.Tail(i) != source && min_cost_flow.Head(i) != sink) {
// Arcs in the solution have a flow value of 1. Their start and end
// nodes give an assignment of worker to task.
if (min_cost_flow.Flow(i) > 0) {
LOG(INFO) << "Worker " << min_cost_flow.Tail(i)
<< " assigned to task " << min_cost_flow.Head(i)
<< " Cost: " << min_cost_flow.UnitCost(i);
}
}
}
} else {
LOG(INFO) << "Solving the min cost flow problem failed.";
LOG(INFO) << "Solver status: " << status;
}
}
} // namespace operations_research
int main() {
operations_research::AssignmentMinFlow();
return EXIT_SUCCESS;
}
Java
package com.google.ortools.graph.samples;
import com.google.ortools.Loader;
import com.google.ortools.graph.MinCostFlow;
import com.google.ortools.graph.MinCostFlowBase;
/** Minimal Assignment Min Flow. */
public class AssignmentMinFlow {
public static void main(String[] args) throws Exception {
Loader.loadNativeLibraries();
// Instantiate a SimpleMinCostFlow solver.
MinCostFlow minCostFlow = new MinCostFlow();
// Define four parallel arrays: sources, destinations, capacities, and unit costs
// between each pair.
int[] startNodes =
new int[] {0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8};
int[] endNodes =
new int[] {1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9};
int[] capacities =
new int[] {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
int[] unitCosts = new int[] {
0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 0, 0, 0, 0};
int source = 0;
int sink = 9;
int tasks = 4;
// Define an array of supplies at each node.
int[] supplies = new int[] {tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks};
// Add each arc.
for (int i = 0; i < startNodes.length; ++i) {
int arc = minCostFlow.addArcWithCapacityAndUnitCost(
startNodes[i], endNodes[i], capacities[i], unitCosts[i]);
if (arc != i) {
throw new Exception("Internal error");
}
}
// Add node supplies.
for (int i = 0; i < supplies.length; ++i) {
minCostFlow.setNodeSupply(i, supplies[i]);
}
// Find the min cost flow.
MinCostFlowBase.Status status = minCostFlow.solve();
if (status == MinCostFlow.Status.OPTIMAL) {
System.out.println("Total cost: " + minCostFlow.getOptimalCost());
System.out.println();
for (int i = 0; i < minCostFlow.getNumArcs(); ++i) {
// Can ignore arcs leading out of source or into sink.
if (minCostFlow.getTail(i) != source && minCostFlow.getHead(i) != sink) {
// Arcs in the solution have a flow value of 1. Their start and end nodes
// give an assignment of worker to task.
if (minCostFlow.getFlow(i) > 0) {
System.out.println("Worker " + minCostFlow.getTail(i) + " assigned to task "
+ minCostFlow.getHead(i) + " Cost: " + minCostFlow.getUnitCost(i));
}
}
}
} else {
System.out.println("Solving the min cost flow problem failed.");
System.out.println("Solver status: " + status);
}
}
private AssignmentMinFlow() {}
}
C#
using System;
using Google.OrTools.Graph;
public class AssignmentMinFlow
{
static void Main()
{
// Instantiate a SimpleMinCostFlow solver.
MinCostFlow minCostFlow = new MinCostFlow();
// Define four parallel arrays: sources, destinations, capacities, and unit costs
// between each pair.
int[] startNodes = { 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8 };
int[] endNodes = { 1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9 };
int[] capacities = { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 };
int[] unitCosts = { 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65,
125, 95, 90, 105, 45, 110, 95, 115, 0, 0, 0, 0 };
int source = 0;
int sink = 9;
int tasks = 4;
// Define an array of supplies at each node.
int[] supplies = { tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks };
// Add each arc.
for (int i = 0; i < startNodes.Length; ++i)
{
int arc =
minCostFlow.AddArcWithCapacityAndUnitCost(startNodes[i], endNodes[i], capacities[i], unitCosts[i]);
if (arc != i)
throw new Exception("Internal error");
}
// Add node supplies.
for (int i = 0; i < supplies.Length; ++i)
{
minCostFlow.SetNodeSupply(i, supplies[i]);
}
// Find the min cost flow.
MinCostFlow.Status status = minCostFlow.Solve();
if (status == MinCostFlow.Status.OPTIMAL)
{
Console.WriteLine("Total cost: " + minCostFlow.OptimalCost());
Console.WriteLine("");
for (int i = 0; i < minCostFlow.NumArcs(); ++i)
{
// Can ignore arcs leading out of source or into sink.
if (minCostFlow.Tail(i) != source && minCostFlow.Head(i) != sink)
{
// Arcs in the solution have a flow value of 1. Their start and end nodes
// give an assignment of worker to task.
if (minCostFlow.Flow(i) > 0)
{
Console.WriteLine("Worker " + minCostFlow.Tail(i) + " assigned to task " + minCostFlow.Head(i) +
" Cost: " + minCostFlow.UnitCost(i));
}
}
}
}
else
{
Console.WriteLine("Solving the min cost flow problem failed.");
Console.WriteLine("Solver status: " + status);
}
}
}
Asignación con equipos de trabajadores
En esta sección, se presenta un problema de asignación más general. En este problema, seis trabajadores se dividen en dos equipos. El problema es asignar cuatro tareas a los trabajadores para que la carga de trabajo esté equitativamente equilibrada entre los equipos, es decir, que cada equipo realice dos de las tareas.
Si deseas obtener una solución de resolución de MIP para este problema, consulta Asignación con equipos de trabajadores.
En las siguientes secciones, se describe un programa que resuelve el problema con el solucionador de flujo de costos mínimos.
Importa las bibliotecas
Con el siguiente código, se importa la biblioteca requerida.
Python
from ortools.graph.python import min_cost_flow
C++
#include <cstdint> #include <vector> #include "ortools/graph/min_cost_flow.h"
Java
import com.google.ortools.Loader; import com.google.ortools.graph.MinCostFlow; import com.google.ortools.graph.MinCostFlowBase;
C#
using System; using Google.OrTools.Graph;
Cómo declarar el solucionador
Con el siguiente código, se crea el solucionador de flujo de costo mínimo.
Python
smcf = min_cost_flow.SimpleMinCostFlow()
C++
// Instantiate a SimpleMinCostFlow solver. SimpleMinCostFlow min_cost_flow;
Java
// Instantiate a SimpleMinCostFlow solver. MinCostFlow minCostFlow = new MinCostFlow();
C#
// Instantiate a SimpleMinCostFlow solver. MinCostFlow minCostFlow = new MinCostFlow();
Crea los datos
El siguiente código crea los datos para el programa.
Python
# Define the directed graph for the flow.
team_a = [1, 3, 5]
team_b = [2, 4, 6]
start_nodes = (
# fmt: off
[0, 0]
+ [11, 11, 11]
+ [12, 12, 12]
+ [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6]
+ [7, 8, 9, 10]
# fmt: on
)
end_nodes = (
# fmt: off
[11, 12]
+ team_a
+ team_b
+ [7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10]
+ [13, 13, 13, 13]
# fmt: on
)
capacities = (
# fmt: off
[2, 2]
+ [1, 1, 1]
+ [1, 1, 1]
+ [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
+ [1, 1, 1, 1]
# fmt: on
)
costs = (
# fmt: off
[0, 0]
+ [0, 0, 0]
+ [0, 0, 0]
+ [90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95]
+ [0, 0, 0, 0]
# fmt: on
)
source = 0
sink = 13
tasks = 4
# Define an array of supplies at each node.
supplies = [tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks]
C++
// Define the directed graph for the flow.
const std::vector<int64_t> team_A = {1, 3, 5};
const std::vector<int64_t> team_B = {2, 4, 6};
const std::vector<int64_t> start_nodes = {
0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3,
3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 8, 9, 10};
const std::vector<int64_t> end_nodes = {
11, 12, 1, 3, 5, 2, 4, 6, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8,
9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 13, 13, 13, 13};
const std::vector<int64_t> capacities = {2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
const std::vector<int64_t> unit_costs = {
0, 0, 0, 0, 0, 0, 0, 0, 90, 76, 75, 70,
35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115,
60, 105, 80, 75, 45, 65, 110, 95, 0, 0, 0, 0};
const int64_t source = 0;
const int64_t sink = 13;
const int64_t tasks = 4;
// Define an array of supplies at each node.
const std::vector<int64_t> supplies = {tasks, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, -tasks};
Java
// Define the directed graph for the flow.
// int[] teamA = new int[] {1, 3, 5};
// int[] teamB = new int[] {2, 4, 6};
int[] startNodes = new int[] {0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3,
4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 8, 9, 10};
int[] endNodes = new int[] {11, 12, 1, 3, 5, 2, 4, 6, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7,
8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 13, 13, 13, 13};
int[] capacities = new int[] {2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
int[] unitCosts = new int[] {0, 0, 0, 0, 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95,
90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95, 0, 0, 0, 0};
int source = 0;
int sink = 13;
int tasks = 4;
// Define an array of supplies at each node.
int[] supplies = new int[] {tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks};
C#
// Define the directed graph for the flow.
int[] teamA = { 1, 3, 5 };
int[] teamB = { 2, 4, 6 };
// Define four parallel arrays: sources, destinations, capacities, and unit costs
// between each pair.
int[] startNodes = { 0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3,
3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 8, 9, 10 };
int[] endNodes = { 11, 12, 1, 3, 5, 2, 4, 6, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8,
9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 13, 13, 13, 13 };
int[] capacities = { 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 };
int[] unitCosts = { 0, 0, 0, 0, 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95,
90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95, 0, 0, 0, 0 };
int source = 0;
int sink = 13;
int tasks = 4;
// Define an array of supplies at each node.
int[] supplies = { tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks };
Los trabajadores corresponden al nodo del 1 al 6. El equipo A está compuesto por los trabajadores 1, 3 y 5, y el equipo B está compuesto por los trabajadores 2, 4 y 6. Las tareas están numeradas del 7 al 10.
Hay dos nodos nuevos, 11 y 12, entre la fuente y los trabajadores. El nodo 11 está conectado a los nodos del equipo A y el nodo 12 está conectado a los nodos del equipo B, con arcos de capacidad 1. En el siguiente gráfico, solo se muestran los nodos y arcos de la fuente a los trabajadores.
La clave para balancear la carga de trabajo es que la fuente 0 está conectada a los nodos 11 y 12 mediante arcos de capacidad 2. Esto significa que los nodos 11 y 12 (y, por lo tanto, los equipos A y B) pueden tener un flujo máximo de 2. Como resultado, cada equipo puede realizar como máximo dos de las tareas.
Crea las restricciones
Python
# Add each arc.
for i in range(0, len(start_nodes)):
smcf.add_arc_with_capacity_and_unit_cost(
start_nodes[i], end_nodes[i], capacities[i], costs[i]
)
# Add node supplies.
for i in range(0, len(supplies)):
smcf.set_node_supply(i, supplies[i])
C++
// Add each arc.
for (int i = 0; i < start_nodes.size(); ++i) {
int arc = min_cost_flow.AddArcWithCapacityAndUnitCost(
start_nodes[i], end_nodes[i], capacities[i], unit_costs[i]);
if (arc != i) LOG(FATAL) << "Internal error";
}
// Add node supplies.
for (int i = 0; i < supplies.size(); ++i) {
min_cost_flow.SetNodeSupply(i, supplies[i]);
}
Java
// Add each arc.
for (int i = 0; i < startNodes.length; ++i) {
int arc = minCostFlow.addArcWithCapacityAndUnitCost(
startNodes[i], endNodes[i], capacities[i], unitCosts[i]);
if (arc != i) {
throw new Exception("Internal error");
}
}
// Add node supplies.
for (int i = 0; i < supplies.length; ++i) {
minCostFlow.setNodeSupply(i, supplies[i]);
}
C#
// Add each arc.
for (int i = 0; i < startNodes.Length; ++i)
{
int arc =
minCostFlow.AddArcWithCapacityAndUnitCost(startNodes[i], endNodes[i], capacities[i], unitCosts[i]);
if (arc != i)
throw new Exception("Internal error");
}
// Add node supplies.
for (int i = 0; i < supplies.Length; ++i)
{
minCostFlow.SetNodeSupply(i, supplies[i]);
}
Cómo invocar el solucionador
Python
# Find the minimum cost flow between node 0 and node 10. status = smcf.solve()
C++
// Find the min cost flow. int status = min_cost_flow.Solve();
Java
// Find the min cost flow. MinCostFlowBase.Status status = minCostFlow.solve();
C#
// Find the min cost flow. MinCostFlow.Status status = minCostFlow.Solve();
Resultado del programa
Python
if status == smcf.OPTIMAL:
print("Total cost = ", smcf.optimal_cost())
print()
for arc in range(smcf.num_arcs()):
# Can ignore arcs leading out of source or intermediate, or into sink.
if (
smcf.tail(arc) != source
and smcf.tail(arc) != 11
and smcf.tail(arc) != 12
and smcf.head(arc) != sink
):
# Arcs in the solution will have a flow value of 1.
# There start and end nodes give an assignment of worker to task.
if smcf.flow(arc) > 0:
print(
"Worker %d assigned to task %d. Cost = %d"
% (smcf.tail(arc), smcf.head(arc), smcf.unit_cost(arc))
)
else:
print("There was an issue with the min cost flow input.")
print(f"Status: {status}")
C++
if (status == MinCostFlow::OPTIMAL) {
LOG(INFO) << "Total cost: " << min_cost_flow.OptimalCost();
LOG(INFO) << "";
for (std::size_t i = 0; i < min_cost_flow.NumArcs(); ++i) {
// Can ignore arcs leading out of source or intermediate nodes, or into
// sink.
if (min_cost_flow.Tail(i) != source && min_cost_flow.Tail(i) != 11 &&
min_cost_flow.Tail(i) != 12 && min_cost_flow.Head(i) != sink) {
// Arcs in the solution have a flow value of 1. Their start and end
// nodes give an assignment of worker to task.
if (min_cost_flow.Flow(i) > 0) {
LOG(INFO) << "Worker " << min_cost_flow.Tail(i)
<< " assigned to task " << min_cost_flow.Head(i)
<< " Cost: " << min_cost_flow.UnitCost(i);
}
}
}
} else {
LOG(INFO) << "Solving the min cost flow problem failed.";
LOG(INFO) << "Solver status: " << status;
}
Java
if (status == MinCostFlow.Status.OPTIMAL) {
System.out.println("Total cost: " + minCostFlow.getOptimalCost());
System.out.println();
for (int i = 0; i < minCostFlow.getNumArcs(); ++i) {
// Can ignore arcs leading out of source or intermediate nodes, or into sink.
if (minCostFlow.getTail(i) != source && minCostFlow.getTail(i) != 11
&& minCostFlow.getTail(i) != 12 && minCostFlow.getHead(i) != sink) {
// Arcs in the solution have a flow value of 1. Their start and end nodes
// give an assignment of worker to task.
if (minCostFlow.getFlow(i) > 0) {
System.out.println("Worker " + minCostFlow.getTail(i) + " assigned to task "
+ minCostFlow.getHead(i) + " Cost: " + minCostFlow.getUnitCost(i));
}
}
}
} else {
System.out.println("Solving the min cost flow problem failed.");
System.out.println("Solver status: " + status);
}
C#
if (status == MinCostFlow.Status.OPTIMAL)
{
Console.WriteLine("Total cost: " + minCostFlow.OptimalCost());
Console.WriteLine("");
for (int i = 0; i < minCostFlow.NumArcs(); ++i)
{
// Can ignore arcs leading out of source or into sink.
if (minCostFlow.Tail(i) != source && minCostFlow.Tail(i) != 11 && minCostFlow.Tail(i) != 12 &&
minCostFlow.Head(i) != sink)
{
// Arcs in the solution have a flow value of 1. Their start and end nodes
// give an assignment of worker to task.
if (minCostFlow.Flow(i) > 0)
{
Console.WriteLine("Worker " + minCostFlow.Tail(i) + " assigned to task " + minCostFlow.Head(i) +
" Cost: " + minCostFlow.UnitCost(i));
}
}
}
}
else
{
Console.WriteLine("Solving the min cost flow problem failed.");
Console.WriteLine("Solver status: " + status);
}
A continuación, se muestra el resultado del programa.
Total cost = 250 Worker 1 assigned to task 9. Cost = 75 Worker 2 assigned to task 7. Cost = 35 Worker 5 assigned to task 10. Cost = 75 Worker 6 assigned to task 8. Cost = 65 Time = 0.00031 seconds
Al equipo A se le asignan las tareas 9 y 10, mientras que al equipo B se le asignan las tareas 7 y 8.
Ten en cuenta que el solucionador de flujo de costo mínimo es más rápido para este problema que el solucionador de MIP, que tarda alrededor de 0.006 segundos.
Todo el programa
A continuación, se muestra todo el programa.
Python
"""Assignment with teams of workers."""
from ortools.graph.python import min_cost_flow
def main():
"""Solving an Assignment with teams of worker."""
smcf = min_cost_flow.SimpleMinCostFlow()
# Define the directed graph for the flow.
team_a = [1, 3, 5]
team_b = [2, 4, 6]
start_nodes = (
# fmt: off
[0, 0]
+ [11, 11, 11]
+ [12, 12, 12]
+ [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6]
+ [7, 8, 9, 10]
# fmt: on
)
end_nodes = (
# fmt: off
[11, 12]
+ team_a
+ team_b
+ [7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10]
+ [13, 13, 13, 13]
# fmt: on
)
capacities = (
# fmt: off
[2, 2]
+ [1, 1, 1]
+ [1, 1, 1]
+ [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
+ [1, 1, 1, 1]
# fmt: on
)
costs = (
# fmt: off
[0, 0]
+ [0, 0, 0]
+ [0, 0, 0]
+ [90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95]
+ [0, 0, 0, 0]
# fmt: on
)
source = 0
sink = 13
tasks = 4
# Define an array of supplies at each node.
supplies = [tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks]
# Add each arc.
for i in range(0, len(start_nodes)):
smcf.add_arc_with_capacity_and_unit_cost(
start_nodes[i], end_nodes[i], capacities[i], costs[i]
)
# Add node supplies.
for i in range(0, len(supplies)):
smcf.set_node_supply(i, supplies[i])
# Find the minimum cost flow between node 0 and node 10.
status = smcf.solve()
if status == smcf.OPTIMAL:
print("Total cost = ", smcf.optimal_cost())
print()
for arc in range(smcf.num_arcs()):
# Can ignore arcs leading out of source or intermediate, or into sink.
if (
smcf.tail(arc) != source
and smcf.tail(arc) != 11
and smcf.tail(arc) != 12
and smcf.head(arc) != sink
):
# Arcs in the solution will have a flow value of 1.
# There start and end nodes give an assignment of worker to task.
if smcf.flow(arc) > 0:
print(
"Worker %d assigned to task %d. Cost = %d"
% (smcf.tail(arc), smcf.head(arc), smcf.unit_cost(arc))
)
else:
print("There was an issue with the min cost flow input.")
print(f"Status: {status}")
if __name__ == "__main__":
main()
C++
#include <cstdint>
#include <vector>
#include "ortools/graph/min_cost_flow.h"
namespace operations_research {
// MinCostFlow simple interface example.
void BalanceMinFlow() {
// Instantiate a SimpleMinCostFlow solver.
SimpleMinCostFlow min_cost_flow;
// Define the directed graph for the flow.
const std::vector<int64_t> team_A = {1, 3, 5};
const std::vector<int64_t> team_B = {2, 4, 6};
const std::vector<int64_t> start_nodes = {
0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3,
3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 8, 9, 10};
const std::vector<int64_t> end_nodes = {
11, 12, 1, 3, 5, 2, 4, 6, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8,
9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 13, 13, 13, 13};
const std::vector<int64_t> capacities = {2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
const std::vector<int64_t> unit_costs = {
0, 0, 0, 0, 0, 0, 0, 0, 90, 76, 75, 70,
35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115,
60, 105, 80, 75, 45, 65, 110, 95, 0, 0, 0, 0};
const int64_t source = 0;
const int64_t sink = 13;
const int64_t tasks = 4;
// Define an array of supplies at each node.
const std::vector<int64_t> supplies = {tasks, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, -tasks};
// Add each arc.
for (int i = 0; i < start_nodes.size(); ++i) {
int arc = min_cost_flow.AddArcWithCapacityAndUnitCost(
start_nodes[i], end_nodes[i], capacities[i], unit_costs[i]);
if (arc != i) LOG(FATAL) << "Internal error";
}
// Add node supplies.
for (int i = 0; i < supplies.size(); ++i) {
min_cost_flow.SetNodeSupply(i, supplies[i]);
}
// Find the min cost flow.
int status = min_cost_flow.Solve();
if (status == MinCostFlow::OPTIMAL) {
LOG(INFO) << "Total cost: " << min_cost_flow.OptimalCost();
LOG(INFO) << "";
for (std::size_t i = 0; i < min_cost_flow.NumArcs(); ++i) {
// Can ignore arcs leading out of source or intermediate nodes, or into
// sink.
if (min_cost_flow.Tail(i) != source && min_cost_flow.Tail(i) != 11 &&
min_cost_flow.Tail(i) != 12 && min_cost_flow.Head(i) != sink) {
// Arcs in the solution have a flow value of 1. Their start and end
// nodes give an assignment of worker to task.
if (min_cost_flow.Flow(i) > 0) {
LOG(INFO) << "Worker " << min_cost_flow.Tail(i)
<< " assigned to task " << min_cost_flow.Head(i)
<< " Cost: " << min_cost_flow.UnitCost(i);
}
}
}
} else {
LOG(INFO) << "Solving the min cost flow problem failed.";
LOG(INFO) << "Solver status: " << status;
}
}
} // namespace operations_research
int main() {
operations_research::BalanceMinFlow();
return EXIT_SUCCESS;
}
Java
package com.google.ortools.graph.samples;
import com.google.ortools.Loader;
import com.google.ortools.graph.MinCostFlow;
import com.google.ortools.graph.MinCostFlowBase;
/** Minimal Assignment Min Flow. */
public class BalanceMinFlow {
public static void main(String[] args) throws Exception {
Loader.loadNativeLibraries();
// Instantiate a SimpleMinCostFlow solver.
MinCostFlow minCostFlow = new MinCostFlow();
// Define the directed graph for the flow.
// int[] teamA = new int[] {1, 3, 5};
// int[] teamB = new int[] {2, 4, 6};
int[] startNodes = new int[] {0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3,
4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 8, 9, 10};
int[] endNodes = new int[] {11, 12, 1, 3, 5, 2, 4, 6, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7,
8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 13, 13, 13, 13};
int[] capacities = new int[] {2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
int[] unitCosts = new int[] {0, 0, 0, 0, 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95,
90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95, 0, 0, 0, 0};
int source = 0;
int sink = 13;
int tasks = 4;
// Define an array of supplies at each node.
int[] supplies = new int[] {tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks};
// Add each arc.
for (int i = 0; i < startNodes.length; ++i) {
int arc = minCostFlow.addArcWithCapacityAndUnitCost(
startNodes[i], endNodes[i], capacities[i], unitCosts[i]);
if (arc != i) {
throw new Exception("Internal error");
}
}
// Add node supplies.
for (int i = 0; i < supplies.length; ++i) {
minCostFlow.setNodeSupply(i, supplies[i]);
}
// Find the min cost flow.
MinCostFlowBase.Status status = minCostFlow.solve();
if (status == MinCostFlow.Status.OPTIMAL) {
System.out.println("Total cost: " + minCostFlow.getOptimalCost());
System.out.println();
for (int i = 0; i < minCostFlow.getNumArcs(); ++i) {
// Can ignore arcs leading out of source or intermediate nodes, or into sink.
if (minCostFlow.getTail(i) != source && minCostFlow.getTail(i) != 11
&& minCostFlow.getTail(i) != 12 && minCostFlow.getHead(i) != sink) {
// Arcs in the solution have a flow value of 1. Their start and end nodes
// give an assignment of worker to task.
if (minCostFlow.getFlow(i) > 0) {
System.out.println("Worker " + minCostFlow.getTail(i) + " assigned to task "
+ minCostFlow.getHead(i) + " Cost: " + minCostFlow.getUnitCost(i));
}
}
}
} else {
System.out.println("Solving the min cost flow problem failed.");
System.out.println("Solver status: " + status);
}
}
private BalanceMinFlow() {}
}
C#
using System;
using Google.OrTools.Graph;
public class BalanceMinFlow
{
static void Main()
{
// Instantiate a SimpleMinCostFlow solver.
MinCostFlow minCostFlow = new MinCostFlow();
// Define the directed graph for the flow.
int[] teamA = { 1, 3, 5 };
int[] teamB = { 2, 4, 6 };
// Define four parallel arrays: sources, destinations, capacities, and unit costs
// between each pair.
int[] startNodes = { 0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3,
3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 8, 9, 10 };
int[] endNodes = { 11, 12, 1, 3, 5, 2, 4, 6, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8,
9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 13, 13, 13, 13 };
int[] capacities = { 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 };
int[] unitCosts = { 0, 0, 0, 0, 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95,
90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95, 0, 0, 0, 0 };
int source = 0;
int sink = 13;
int tasks = 4;
// Define an array of supplies at each node.
int[] supplies = { tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks };
// Add each arc.
for (int i = 0; i < startNodes.Length; ++i)
{
int arc =
minCostFlow.AddArcWithCapacityAndUnitCost(startNodes[i], endNodes[i], capacities[i], unitCosts[i]);
if (arc != i)
throw new Exception("Internal error");
}
// Add node supplies.
for (int i = 0; i < supplies.Length; ++i)
{
minCostFlow.SetNodeSupply(i, supplies[i]);
}
// Find the min cost flow.
MinCostFlow.Status status = minCostFlow.Solve();
if (status == MinCostFlow.Status.OPTIMAL)
{
Console.WriteLine("Total cost: " + minCostFlow.OptimalCost());
Console.WriteLine("");
for (int i = 0; i < minCostFlow.NumArcs(); ++i)
{
// Can ignore arcs leading out of source or into sink.
if (minCostFlow.Tail(i) != source && minCostFlow.Tail(i) != 11 && minCostFlow.Tail(i) != 12 &&
minCostFlow.Head(i) != sink)
{
// Arcs in the solution have a flow value of 1. Their start and end nodes
// give an assignment of worker to task.
if (minCostFlow.Flow(i) > 0)
{
Console.WriteLine("Worker " + minCostFlow.Tail(i) + " assigned to task " + minCostFlow.Head(i) +
" Cost: " + minCostFlow.UnitCost(i));
}
}
}
}
else
{
Console.WriteLine("Solving the min cost flow problem failed.");
Console.WriteLine("Solver status: " + status);
}
}
}