최소 비용 흐름 솔버를 사용하여 할당 문제의 특수한 사례를 해결할 수 있습니다.
실제로 최소 비용 흐름은 MIP 또는 CP-SAT 솔버보다 더 빠르게 솔루션을 반환할 수 있습니다. 그러나 MIP 및 CP-SAT는 최소 비용 흐름보다 더 큰 종류의 문제를 해결할 수 있으므로 대부분의 경우 MIP 또는 CP-SAT가 최선의 선택입니다.
다음 섹션에서는 최소 비용 흐름 솔버를 사용하여 다음 할당 문제를 해결하는 Python 프로그램을 보여줍니다.
선형 할당 예
이 섹션에서는 선형 대입 문제 해결사 섹션에 설명된 이 예를 최소 비용 흐름 문제로 해결하는 방법을 보여줍니다.
라이브러리 가져오기
다음 코드는 필요한 라이브러리를 가져옵니다.
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;
문제 해결사 선언
다음 코드는 최소 비용 흐름 솔버를 만듭니다.
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();
데이터 만들기
이 문제의 흐름 다이어그램은 소스와 싱크가 추가된 비용 매트릭스의 이분자 그래프 (약간 다른 예는 할당 개요 참고)로 구성됩니다.
데이터에는 시작 노드, 종료 노드, 용량 및 문제의 비용에 해당하는 다음 4개의 배열이 포함됩니다. 각 배열의 길이는 그래프의 호의 수입니다.
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 };
데이터 설정 방법을 명확히 하기 위해 각 배열은 3개의 하위 배열로 나뉩니다.
- 첫 번째 배열은 소스 밖으로 이어지는 호에 해당합니다.
- 두 번째 배열은 작업자와 태스크 간의 호에 해당합니다.
costs의 경우 벡터로 평면화된 비용 행렬(선형 할당 솔버에서 사용됨)입니다. - 세 번째 배열은 싱크로 이어지는 호에 해당합니다.
데이터에는 각 노드에서 공급을 제공하는 벡터 supplies도 포함됩니다.
최소 비용 흐름 문제가 할당 문제를 나타내는 방식
위의 최소 비용 흐름 문제는 할당 문제를 어떻게 나타내나요? 첫째, 모든 원호의 용량은 1이므로 소스에서 4개의 호가 공급되면 작업자로 이어지는 네 개의 호가 각각 흐름이 1이 됩니다.
다음으로 Flow-in-equals-flow-out 조건은 각 작업자의 흐름이 1이 되도록 강제합니다. 가능한 경우 솔버는 각 작업자로부터 이어지는 최소 비용 원호를 가로질러 해당 흐름을 전달할 것입니다. 하지만 솔버는 두 작업자의 흐름을 단일 태스크로 보낼 수 없습니다. 이렇게 하면 해당 작업에 2의 결합된 흐름이 생성되어 용량 1의 단일 호를 통해 작업에서 싱크로 전송될 수 없습니다. 즉, 문제 해결사는 할당 문제의 필요에 따라 단일 작업자에게만 태스크를 할당할 수 있습니다.
마지막으로 Flow-in-equals-flow-out 조건은 각 태스크의 유출이 1이 되도록 강제하므로 일부 작업자가 각 태스크를 수행합니다.
그래프 및 제약조건 만들기
다음 코드는 그래프와 제약 조건을 만듭니다.
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]);
}
솔버 호출
다음 코드는 솔버를 호출하고 솔루션을 표시합니다.
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();
이 솔루션은 작업자와 작업 사이의 호로 구성되며, 이 호는 솔버에 의해 1 흐름이 할당됩니다. 소스나 싱크에 연결된 아크는 솔루션의 일부가 아닙니다.
프로그램은 각 원호를 확인하여 흐름 1이 있는지 확인하고, 흐름 1이 있으면 할당의 작업자 및 작업에 해당하는 원호의 Tail (시작 노드)과 Head (종료 노드)를 출력합니다.
프로그램 출력
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);
}
프로그램의 출력은 다음과 같습니다.
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
결과는 선형 할당 솔버와 동일합니다 (작업자 수와 비용이 다른 경우 제외). 선형 할당 솔버는 최소 비용 흐름보다 약간 빠릅니다(0.000147초이지만 0.000458초).
전체 프로그램
전체 프로그램은 아래에 표시되어 있습니다.
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);
}
}
}
작업자 팀의 배정
이 섹션에서는 보다 일반적인 할당 문제를 다룹니다. 이 문제에서 6명의 작업자는 두 팀으로 나뉩니다. 문제는 작업자에게 4개의 태스크를 할당하여 팀 간에 워크로드가 균등하게 균형을 이루도록 하여 각 팀이 두 가지 태스크를 수행하도록 하는 것입니다.
이 문제에 대한 MIP 솔버 솔루션은 작업자팀으로 할당을 참조하세요.
다음 섹션에서는 최소 비용 흐름 솔버를 사용하여 문제를 해결하는 프로그램을 설명합니다.
라이브러리 가져오기
다음 코드는 필요한 라이브러리를 가져옵니다.
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;
문제 해결사 선언
다음 코드는 최소 비용 흐름 솔버를 만듭니다.
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();
데이터 만들기
다음 코드는 프로그램의 데이터를 만듭니다.
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 };
작업자는 노드 1 - 6에 해당합니다. 팀 A는 작업자 1, 3, 5로 구성되고 팀 B는 작업자 2, 4, 6으로 구성됩니다. 작업 번호는 7~10입니다.
소스와 작업자 사이에는 11과 12, 두 개의 새 노드가 있습니다. 노드 11은 팀 A의 노드에 연결되고 노드 12는 용량 1의 호가 있는 팀 B의 노드에 연결됩니다. 아래 그래프는 소스에서 작업자로의 노드와 호만 보여줍니다.
워크로드 균형의 핵심은 소스 0이 용량 2의 원호로 노드 11과 12에 연결된다는 것입니다. 즉, 노드 11과 12 (따라서 팀 A와 B팀)의 최대 흐름은 2입니다. 따라서 각 팀은 최대 두 가지 작업을 수행할 수 있습니다.
제약조건 만들기
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]);
}
솔버 호출
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();
프로그램 출력
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);
}
다음은 프로그램의 출력을 보여줍니다.
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
A팀에는 작업 9와 10이 할당되고 B팀에는 작업 7과 8이 할당됩니다.
이 문제의 경우 최소 비용 흐름 솔버는 약 0.006초가 걸리는 MIP 솔버보다 더 빠릅니다.
전체 프로그램
전체 프로그램은 아래에 표시되어 있습니다.
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);
}
}
}