ارتباط نزدیک با مسئله حداکثر جریان، مسئله جریان حداقل هزینه ( حداقل هزینه ) است که در آن هر قوس در نمودار دارای هزینه واحد برای حمل مواد در سراسر آن است. مشکل پیدا کردن جریانی با کمترین هزینه کل است.
مسئله جریان هزینه حداقل نیز دارای گره های خاصی است که به آنها گره های عرضه یا گره های تقاضا گفته می شود که در مسئله حداکثر جریان مشابه منبع و فرورفته هستند. مواد از گره های عرضه به گره های تقاضا منتقل می شوند.
- در یک گره عرضه ، مقدار مثبت - عرضه - به جریان اضافه می شود. برای مثال یک عرضه می تواند تولید را در آن گره نشان دهد.
- در یک گره تقاضا ، مقدار منفی - تقاضا - از جریان برداشته می شود. برای مثال یک تقاضا می تواند نشان دهنده مصرف در آن گره باشد.
برای راحتی، فرض می کنیم که همه گره ها، به غیر از گره های عرضه یا تقاضا، عرضه (و تقاضا) صفر دارند.
برای مسئله جریان هزینه حداقل، قانون حفظ جریان زیر را داریم که عرضه ها و تقاضاها را در نظر می گیرد:
نمودار زیر یک مشکل جریان حداقل هزینه را نشان می دهد. قوس ها با جفت اعداد برچسب گذاری شده اند: عدد اول ظرفیت و عدد دوم هزینه است. اعداد داخل پرانتز در کنار گره ها نشان دهنده عرضه یا تقاضا هستند. گره 0 یک گره عرضه با عرضه 20 است، در حالی که گره های 3 و 4 گره های تقاضا هستند که به ترتیب تقاضاهای 5- و 15- دارند.
کتابخانه ها را وارد کنید
کد زیر کتابخانه مورد نیاز را وارد می کند.
پایتون
import numpy as np from ortools.graph.python import min_cost_flow
C++
#include <cstdint> #include <vector> #include "ortools/graph/min_cost_flow.h"
جاوا
import com.google.ortools.Loader; import com.google.ortools.graph.MinCostFlow; import com.google.ortools.graph.MinCostFlowBase;
سی شارپ
using System; using Google.OrTools.Graph;
حل کننده را اعلام کنید
برای حل مشکل از حل کننده SimpleMinCostFlow استفاده می کنیم.
پایتون
# Instantiate a SimpleMinCostFlow solver. smcf = min_cost_flow.SimpleMinCostFlow()
C++
// Instantiate a SimpleMinCostFlow solver. SimpleMinCostFlow min_cost_flow;
جاوا
// Instantiate a SimpleMinCostFlow solver. MinCostFlow minCostFlow = new MinCostFlow();
سی شارپ
// Instantiate a SimpleMinCostFlow solver. MinCostFlow minCostFlow = new MinCostFlow();
داده ها را تعریف کنید
کد زیر داده های مشکل را تعریف می کند. در این مورد، چهار آرایه برای گره های شروع، گره های پایانی، ظرفیت ها و هزینه های واحد وجود دارد. باز هم طول آرایه ها تعداد کمان های نمودار است.
پایتون
# 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};جاوا
// 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};سی شارپ
// 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 };کمان ها را اضافه کنید
برای هر گره شروع و گره پایانی، با استفاده از روش AddArcWithCapacityAndUnitCost ، یک قوس از گره شروع به گره انتهایی با ظرفیت و هزینه واحد داده شده ایجاد می کنیم.
متد SetNodeSupply حلگر یک بردار از منابع برای گره ها ایجاد می کند.
پایتون
# 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]);
}جاوا
// 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]);
}سی شارپ
// 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]);
}حل کننده را فراخوانی کنید
اکنون که تمام کمان ها تعریف شده اند، تنها چیزی که باقی می ماند فراخوانی حل کننده و نمایش نتایج است. متد Solve() فراخوانی می کنیم.
پایتون
# Find the min cost flow. status = smcf.solve()
C++
// Find the min cost flow. int status = min_cost_flow.Solve();
جاوا
// Find the min cost flow. MinCostFlowBase.Status status = minCostFlow.solve();
سی شارپ
// Find the min cost flow. MinCostFlow.Status status = minCostFlow.Solve();
نمایش نتایج
اکنون، می‌توانیم جریان و هزینه را در هر قوس نمایش دهیم.
پایتون
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;
}جاوا
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);
}سی شارپ
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);
}این هم خروجی برنامه پایتون:
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
برنامه های کامل
با کنار هم قرار دادن همه اینها، در اینجا برنامه های کامل وجود دارد.
پایتون
"""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;
}جاوا
// 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() {}
}سی شارپ
// 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);
}
}
}