ज़्यादा से ज़्यादा फ़्लो की समस्या से काफ़ी हद तक कम से कम लागत (कम से कम लागत) फ़्लो की समस्या मिलती है. इसके तहत, ग्राफ़ में दिए गए हर चाप में, एक ही जगह पर सामग्री को एक जगह से दूसरी जगह ले जाने की एक इकाई होती है. समस्या सबसे कम कुल लागत वाला फ़्लो ढूंढना है.
कम से कम लागत के फ़्लो की समस्या में कुछ खास नोड भी होते हैं. इन्हें सप्लाई नोड या डिमांड नोड कहते हैं. ये नोड मैक्स फ़्लो समस्या में सोर्स और सिंक की तरह ही होते हैं. मटीरियल को सप्लाई नोड से डिमांड नोड में ले जाया जाता है.
- सप्लाई नोड पर फ़्लो में एक पॉज़िटिव रकम — सप्लाई — जोड़ दी जाती है. उदाहरण के लिए, सप्लाई उस नोड पर प्रोडक्शन को दिखा सकती है.
- मांग के नोड में, नेगेटिव रकम — मांग — को फ़्लो से हटा दिया जाता है. उदाहरण के लिए, मांग उस नोड में खपत को दिखा सकती है.
आपकी सुविधा के लिए, हम मान लेंगे कि सप्लाई या डिमांड नोड के अलावा, दूसरे सभी नोड की सप्लाई ज़ीरो (और डिमांड) होती है.
कम से कम लागत के फ़्लो की समस्या के लिए, हमने फ़्लो के संरक्षण का यह नियम तय किया है. इसमें, सप्लाई और मांग का ध्यान रखा जाता है:
नीचे दिया गया ग्राफ़ कम से कम लागत के फ़्लो से जुड़ी समस्या दिखाता है. चापों पर संख्याओं के जोड़े का लेबल होता है: पहली संख्या क्षमता और दूसरी संख्या लागत होती है. नोड के बगल में ब्रैकेट में दी गई संख्याएं सप्लाई या मांग को दिखाती हैं. नोड 0 एक सप्लाई नोड है, जिसकी सप्लाई 20 है, जबकि नोड 3 और 4 डिमांड नोड हैं, जहां डिमांड -5 और -15 हैं.
लाइब्रेरी इंपोर्ट करना
नीचे दिया गया कोड, ज़रूरी लाइब्रेरी को इंपोर्ट करता है.
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;
सॉल्वर का एलान करें
इस सवाल को हल करने के लिए, हम SimpleMinCostFlow सॉल्वर का इस्तेमाल करते हैं.
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();
डेटा तय करना
नीचे दिया गया कोड, समस्या से जुड़े डेटा के बारे में बताता है. इस मामले में, स्टार्ट नोड, एंड नोड, क्षमता, और इकाई की लागत के लिए चार कैटगरी होती हैं. फिर से, सरणियों की लंबाई, ग्राफ़ में मौजूद चापों की संख्या होती है.
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 };
चापों (आर्क) को जोड़ना
हर स्टार्ट नोड और एंड नोड के लिए, हम AddArcWithCapacityAndUnitCost के तरीके का इस्तेमाल करके, तय की गई क्षमता और यूनिट की लागत के साथ स्टार्ट नोड से आखिरी नोड तक एक आर्क बनाते हैं.
सॉल्वर का SetNodeSupply तरीका, नोड के लिए सप्लाई का एक वेक्टर बनाता है.
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]);
}
सॉल्वर को शुरू करें
अब सभी चापों को परिभाषित करने के बाद, बस सॉल्वर को शुरू करना और परिणाम दिखाना है. हम Solve() तरीके को शुरू करते हैं.
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();
नतीजे दिखाएं
अब, हम हर आर्क में फ़्लो और कीमत दिखा सकते हैं.
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);
}
यहां Python प्रोग्राम का आउटपुट दिया गया है:
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
प्रोग्राम पूरे करना
इन सबको एक साथ जोड़कर, यहां पूरे प्रोग्राम दिए गए हैं.
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);
}
}
}