Yang terkait erat dengan masalah alur maksimum adalah biaya minimum (biaya min) masalah alur, di mana setiap busur dalam grafik memiliki biaya satuan untuk pengangkutan di seluruh materi. Masalahnya adalah menemukan alur dengan total biaya paling rendah.
Masalah aliran biaya min juga memiliki node khusus, yang disebut node pasokan atau permintaan node, yang mirip dengan resource sumber dan sink masalah alur maks. Material diangkut dari node pasokan ke node permintaan.
- Pada node suplai, jumlah positif — pasokan — ditambahkan ke dengan aliran data. Sebagai contoh, pasokan dapat mewakili produksi pada node tersebut.
- Pada node permintaan, jumlah negatif — permintaan — diambil menjauh dari aliran. Permintaan dapat merepresentasikan konsumsi pada node tersebut, untuk contoh.
Untuk memudahkan, kita asumsikan bahwa semua {i>node<i}, selain node penawaran atau permintaan, memiliki penawaran (dan permintaan) nol.
Untuk masalah aliran biaya minimum, kita memiliki aturan konservasi aliran berikut, yang mempertimbangkan persediaan dan permintaan:
Grafik di bawah menunjukkan masalah alur biaya min. Busur tersebut diberi label pasangan angka: angka pertama adalah kapasitas dan angka kedua adalah biaya. Angka dalam tanda kurung di samping node mewakili persediaan atau permintaan. {i>Node<i} 0 adalah simpul pasokan dengan suplai 20, sementara simpul 3 dan 4 adalah simpul permintaan, dengan menuntut masing-masing -5 dan -15.
Mengimpor library
Kode berikut akan mengimpor library yang diperlukan.
Python
import numpy as np 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;
Mendeklarasikan pemecah
Untuk mengatasi masalah ini, kita menggunakan SimpleMinCostFlow pemecah masalah.
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();
Menentukan data
Kode berikut menentukan data untuk masalah tersebut. Dalam hal ini, ada empat array untuk node awal, node akhir, kapasitas, dan biaya satuan. Sekali lagi, panjang array adalah jumlah busur dalam grafik.
Python
# 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. start_nodes = np.array([0, 0, 1, 1, 1, 2, 2, 3, 4]) end_nodes = np.array([1, 2, 2, 3, 4, 3, 4, 4, 2]) capacities = np.array([15, 8, 20, 4, 10, 15, 4, 20, 5]) unit_costs = np.array([4, 4, 2, 2, 6, 1, 3, 2, 3]) # Define an array of supplies at each node. supplies = [20, 0, 0, -5, -15]
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. std::vector<int64_t> start_nodes = {0, 0, 1, 1, 1, 2, 2, 3, 4}; std::vector<int64_t> end_nodes = {1, 2, 2, 3, 4, 3, 4, 4, 2}; std::vector<int64_t> capacities = {15, 8, 20, 4, 10, 15, 4, 20, 5}; std::vector<int64_t> unit_costs = {4, 4, 2, 2, 6, 1, 3, 2, 3}; // Define an array of supplies at each node. std::vector<int64_t> supplies = {20, 0, 0, -5, -15};
Java
// 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. // Problem taken From Taha's 'Introduction to Operations Research', // example 6.4-2. int[] startNodes = new int[] {0, 0, 1, 1, 1, 2, 2, 3, 4}; int[] endNodes = new int[] {1, 2, 2, 3, 4, 3, 4, 4, 2}; int[] capacities = new int[] {15, 8, 20, 4, 10, 15, 4, 20, 5}; int[] unitCosts = new int[] {4, 4, 2, 2, 6, 1, 3, 2, 3}; // Define an array of supplies at each node. int[] supplies = new int[] {20, 0, 0, -5, -15};
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. // Problem taken From Taha's 'Introduction to Operations Research', // example 6.4-2. int[] startNodes = { 0, 0, 1, 1, 1, 2, 2, 3, 4 }; int[] endNodes = { 1, 2, 2, 3, 4, 3, 4, 4, 2 }; int[] capacities = { 15, 8, 20, 4, 10, 15, 4, 20, 5 }; int[] unitCosts = { 4, 4, 2, 2, 6, 1, 3, 2, 3 }; // Define an array of supplies at each node. int[] supplies = { 20, 0, 0, -5, -15 };
Menambahkan busur
Untuk setiap node awal dan node akhir, kita buat busur dari node awal ke node akhir dengan kapasitas dan biaya satuan yang ditentukan, menggunakan metode AddArcWithCapacityAndUnitCost.
Solusi SetNodeSupply membuat vektor suplai untuk node tersebut.
Python
# Add arcs, capacities and costs in bulk using numpy. all_arcs = smcf.add_arcs_with_capacity_and_unit_cost( start_nodes, end_nodes, capacities, unit_costs ) # Add supply for each nodes. smcf.set_nodes_supplies(np.arange(0, len(supplies)), supplies)
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]); }
Memanggil pemecah masalah
Setelah semua busur didefinisikan, yang tersisa hanyalah memanggil metode
pemecah masalah dan
menampilkan hasilnya. Kita akan memanggil metode Solve().
Python
# Find the min cost flow. 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();
Menampilkan hasil
Sekarang, kita dapat menampilkan alur dan biaya di setiap busur.
Python
if status != smcf.OPTIMAL: print("There was an issue with the min cost flow input.") print(f"Status: {status}") exit(1) print(f"Minimum cost: {smcf.optimal_cost()}") print("") print(" Arc Flow / Capacity Cost") solution_flows = smcf.flows(all_arcs) costs = solution_flows * unit_costs for arc, flow, cost in zip(all_arcs, solution_flows, costs): print( f"{smcf.tail(arc):1} -> {smcf.head(arc)} {flow:3} / {smcf.capacity(arc):3} {cost}" )
C++
if (status == MinCostFlow::OPTIMAL) { LOG(INFO) << "Minimum cost flow: " << min_cost_flow.OptimalCost(); LOG(INFO) << ""; LOG(INFO) << " Arc Flow / Capacity Cost"; for (std::size_t i = 0; i < min_cost_flow.NumArcs(); ++i) { int64_t cost = min_cost_flow.Flow(i) * min_cost_flow.UnitCost(i); LOG(INFO) << min_cost_flow.Tail(i) << " -> " << min_cost_flow.Head(i) << " " << min_cost_flow.Flow(i) << " / " << min_cost_flow.Capacity(i) << " " << cost; } } else { LOG(INFO) << "Solving the min cost flow problem failed. Solver status: " << status; }
Java
if (status == MinCostFlow.Status.OPTIMAL) { System.out.println("Minimum cost: " + minCostFlow.getOptimalCost()); System.out.println(); System.out.println(" Edge Flow / Capacity Cost"); for (int i = 0; i < minCostFlow.getNumArcs(); ++i) { long cost = minCostFlow.getFlow(i) * minCostFlow.getUnitCost(i); System.out.println(minCostFlow.getTail(i) + " -> " + minCostFlow.getHead(i) + " " + minCostFlow.getFlow(i) + " / " + minCostFlow.getCapacity(i) + " " + cost); } } 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("Minimum cost: " + minCostFlow.OptimalCost()); Console.WriteLine(""); Console.WriteLine(" Edge Flow / Capacity Cost"); for (int i = 0; i < minCostFlow.NumArcs(); ++i) { long cost = minCostFlow.Flow(i) * minCostFlow.UnitCost(i); Console.WriteLine(minCostFlow.Tail(i) + " -> " + minCostFlow.Head(i) + " " + string.Format("{0,3}", minCostFlow.Flow(i)) + " / " + string.Format("{0,3}", minCostFlow.Capacity(i)) + " " + string.Format("{0,3}", cost)); } } else { Console.WriteLine("Solving the min cost flow problem failed. Solver status: " + status); }
Berikut adalah output dari program Python:
Minimum cost: 150 Arc Flow / Capacity Cost 0 -> 1 12 / 15 48 0 -> 2 8 / 8 32 1 -> 2 8 / 20 16 1 -> 3 4 / 4 8 1 -> 4 0 / 10 0 2 -> 3 12 / 15 12 2 -> 4 4 / 4 12 3 -> 4 11 / 20 22 4 -> 2 0 / 5 0
Selesaikan program
Menyatukan semuanya, berikut ini program lengkapnya.
Python
"""From Bradley, Hax and Maganti, 'Applied Mathematical Programming', figure 8.1.""" import numpy as np from ortools.graph.python import min_cost_flow def main(): """MinCostFlow simple interface example.""" # Instantiate a SimpleMinCostFlow solver. smcf = min_cost_flow.SimpleMinCostFlow() # 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. start_nodes = np.array([0, 0, 1, 1, 1, 2, 2, 3, 4]) end_nodes = np.array([1, 2, 2, 3, 4, 3, 4, 4, 2]) capacities = np.array([15, 8, 20, 4, 10, 15, 4, 20, 5]) unit_costs = np.array([4, 4, 2, 2, 6, 1, 3, 2, 3]) # Define an array of supplies at each node. supplies = [20, 0, 0, -5, -15] # Add arcs, capacities and costs in bulk using numpy. all_arcs = smcf.add_arcs_with_capacity_and_unit_cost( start_nodes, end_nodes, capacities, unit_costs ) # Add supply for each nodes. smcf.set_nodes_supplies(np.arange(0, len(supplies)), supplies) # Find the min cost flow. status = smcf.solve() if status != smcf.OPTIMAL: print("There was an issue with the min cost flow input.") print(f"Status: {status}") exit(1) print(f"Minimum cost: {smcf.optimal_cost()}") print("") print(" Arc Flow / Capacity Cost") solution_flows = smcf.flows(all_arcs) costs = solution_flows * unit_costs for arc, flow, cost in zip(all_arcs, solution_flows, costs): print( f"{smcf.tail(arc):1} -> {smcf.head(arc)} {flow:3} / {smcf.capacity(arc):3} {cost}" ) if __name__ == "__main__": main()
C++
// From Bradley, Hax and Maganti, 'Applied Mathematical Programming', figure 8.1 #include <cstdint> #include <vector> #include "ortools/graph/min_cost_flow.h" namespace operations_research { // MinCostFlow simple interface example. void SimpleMinCostFlowProgram() { // 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. std::vector<int64_t> start_nodes = {0, 0, 1, 1, 1, 2, 2, 3, 4}; std::vector<int64_t> end_nodes = {1, 2, 2, 3, 4, 3, 4, 4, 2}; std::vector<int64_t> capacities = {15, 8, 20, 4, 10, 15, 4, 20, 5}; std::vector<int64_t> unit_costs = {4, 4, 2, 2, 6, 1, 3, 2, 3}; // Define an array of supplies at each node. std::vector<int64_t> supplies = {20, 0, 0, -5, -15}; // 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) << "Minimum cost flow: " << min_cost_flow.OptimalCost(); LOG(INFO) << ""; LOG(INFO) << " Arc Flow / Capacity Cost"; for (std::size_t i = 0; i < min_cost_flow.NumArcs(); ++i) { int64_t cost = min_cost_flow.Flow(i) * min_cost_flow.UnitCost(i); LOG(INFO) << min_cost_flow.Tail(i) << " -> " << min_cost_flow.Head(i) << " " << min_cost_flow.Flow(i) << " / " << min_cost_flow.Capacity(i) << " " << cost; } } else { LOG(INFO) << "Solving the min cost flow problem failed. Solver status: " << status; } } } // namespace operations_research int main() { operations_research::SimpleMinCostFlowProgram(); return EXIT_SUCCESS; }
Java
// From Bradley, Hax, and Maganti, 'Applied Mathematical Programming', figure 8.1. package com.google.ortools.graph.samples; import com.google.ortools.Loader; import com.google.ortools.graph.MinCostFlow; import com.google.ortools.graph.MinCostFlowBase; /** Minimal MinCostFlow program. */ public class SimpleMinCostFlowProgram { 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. For instance, the arc from node 0 to node 1 has a // capacity of 15. // Problem taken From Taha's 'Introduction to Operations Research', // example 6.4-2. int[] startNodes = new int[] {0, 0, 1, 1, 1, 2, 2, 3, 4}; int[] endNodes = new int[] {1, 2, 2, 3, 4, 3, 4, 4, 2}; int[] capacities = new int[] {15, 8, 20, 4, 10, 15, 4, 20, 5}; int[] unitCosts = new int[] {4, 4, 2, 2, 6, 1, 3, 2, 3}; // Define an array of supplies at each node. int[] supplies = new int[] {20, 0, 0, -5, -15}; // 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("Minimum cost: " + minCostFlow.getOptimalCost()); System.out.println(); System.out.println(" Edge Flow / Capacity Cost"); for (int i = 0; i < minCostFlow.getNumArcs(); ++i) { long cost = minCostFlow.getFlow(i) * minCostFlow.getUnitCost(i); System.out.println(minCostFlow.getTail(i) + " -> " + minCostFlow.getHead(i) + " " + minCostFlow.getFlow(i) + " / " + minCostFlow.getCapacity(i) + " " + cost); } } else { System.out.println("Solving the min cost flow problem failed."); System.out.println("Solver status: " + status); } } private SimpleMinCostFlowProgram() {} }
C#
// From Bradley, Hax, and Magnanti, 'Applied Mathematical Programming', figure 8.1. using System; using Google.OrTools.Graph; public class SimpleMinCostFlowProgram { static void Main() { // Instantiate a SimpleMinCostFlow solver. MinCostFlow minCostFlow = new MinCostFlow(); // 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. // Problem taken From Taha's 'Introduction to Operations Research', // example 6.4-2. int[] startNodes = { 0, 0, 1, 1, 1, 2, 2, 3, 4 }; int[] endNodes = { 1, 2, 2, 3, 4, 3, 4, 4, 2 }; int[] capacities = { 15, 8, 20, 4, 10, 15, 4, 20, 5 }; int[] unitCosts = { 4, 4, 2, 2, 6, 1, 3, 2, 3 }; // Define an array of supplies at each node. int[] supplies = { 20, 0, 0, -5, -15 }; // 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("Minimum cost: " + minCostFlow.OptimalCost()); Console.WriteLine(""); Console.WriteLine(" Edge Flow / Capacity Cost"); for (int i = 0; i < minCostFlow.NumArcs(); ++i) { long cost = minCostFlow.Flow(i) * minCostFlow.UnitCost(i); Console.WriteLine(minCostFlow.Tail(i) + " -> " + minCostFlow.Head(i) + " " + string.Format("{0,3}", minCostFlow.Flow(i)) + " / " + string.Format("{0,3}", minCostFlow.Capacity(i)) + " " + string.Format("{0,3}", cost)); } } else { Console.WriteLine("Solving the min cost flow problem failed. Solver status: " + status); } } }