Maksimum akış sorunuyla yakından alakalı olan minimum maliyet (min. maliyet) grafikteki her yayın taşıma için birim maliyetine sahip olduğu akış problemi bir araya getirmektir. Sorun, en düşük toplam maliyete sahip bir akış bulmaktır.
Minimum maliyet akışı probleminde tedarik düğümleri veya talep adı verilen özel düğümler de vardır. ve havuzdaki kaynağa ve havuza benzer olan maks. akış problemi. Malzeme, tedarik düğümlerinden talep düğümlerine taşınır.
- Tedarik düğümünde, pozitif bir tutar (tedarik) akış. Örneğin, tedarik, o düğümdeki üretimi temsil edebilir.
- Talep düğümünde negatif bir tutar (talep) alınır biraz zorlanır. Talep, o düğümdeki tüketimi temsil edebilir. örneğine bakalım.
Kolaylık sağlaması açısından, tedarik veya talep düğümleri dışındaki tüm düğümlerin sıfır arz (ve talep) içeren olmaları.
Minimum maliyet akışı problemi için aşağıdaki akış koruma kuralımız vardır: proje yöneticisi de üretime gelecek
Aşağıdaki grafikte bir min. maliyet akışı sorunu gösterilmektedir. Yaylar çiftlerle etiketlenmiştir İlk sayı kapasite, ikinci sayı ise maliyeti belirtir. Düğümlerin yanındaki parantez içindeki sayılar, malzemeleri veya talepleri gösterir. Düğüm 0, 20 arz ile bir arz düğümüdür. Düğümler 3 ve 4 ise talep düğümleridir. için -5 ve -15'e karşılık gelmelidir.
Kitaplıkları içe aktarma
Aşağıdaki kod, gerekli kitaplığı içe aktarır.
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;
Çözücüyü açıklama
Sorunu çözmek için SimpleMinCostFlow çözücü olarak kullanılır.
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();
Verileri tanımlama
Aşağıdaki kod, soruna ilişkin verileri tanımlar. Bu durumda, başlangıç düğümleri, bitiş düğümleri, kapasiteler ve birim maliyetleri için dört dizi. Tekrar ediyorum, dizilerin uzunluğu, grafikteki yay sayısıdır.
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 };
Yayınları ekleme
Her başlangıç düğümü ve bitiş düğümü için başlangıç düğümünden bitiş düğümüne bir yay oluştururuz belirlenen kapasite ve birim maliyete göre, AddArcWithCapacityAndUnitCost.
Çözücünün SetNodeSupply yönteminde düğümler için bir tedarik vektörü oluşturulur.
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]); }
Çözücüyü çağır
Tüm yaylar tanımlandığına göre, geriye kalan tek şey
çözer ve sonuçları görüntüler. Solve() yöntemini çağırırız.
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();
Sonuçları görüntüle
Artık, akışı ve maliyeti her bir yayın boyunca görüntüleyebiliriz.
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); }
Python programının çıktısı şu şekildedir:
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
Programları tamamlama
Bir araya getirildiğinde programların tamamı burada verilmiştir.
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); } } }