以下のセクションでは、LP 問題の例と解決方法について説明します。問題は次のとおりです。
次の制約に従って 3x + 4y
を最大化します。
x + 2y
: 14 以下3x - y
: 0 以上x - y
~ 2
目的関数 3x + 4y
と制約の両方が一次式によって与えられるため、これは線形問題になります。
制約によって実現可能領域(内部を含む以下の三角形)が定義されます。
LP に関する問題を解決するための基本的な手順
LP の問題を解決するため、プログラムに次のステップを含める必要があります。
- 線形ソルバーのラッパーをインポートします。
- LP ソルバーを宣言して、
- 変数を定義し、
- 制約を定義し、
- 目標を定義し
- LP ソルバーを呼び出す
- 解答を表示する
MPSolver を使用した解答
次のセクションでは、MPSolver ラッパーと LP ソルバーを使用して問題を解決するプログラムを示します。
注:以下のプログラムを実行するには、OR-Tools をインストールする必要があります。
OR-Tools の主要な線形最適化ソルバーは、Google の社内線形プログラミング ソルバーである Glop です。高速、メモリ効率、数値安定です。
線形ソルバーのラッパーをインポートする
以下に示すように、OR-Tools の線形ソルバー(MIP ソルバーと線形ソルバーのインターフェース)をインポートまたは追加します。
Python
from ortools.linear_solver import pywraplp
C++
#include <iostream> #include <memory> #include "ortools/linear_solver/linear_solver.h"
Java
import com.google.ortools.Loader; import com.google.ortools.linearsolver.MPConstraint; import com.google.ortools.linearsolver.MPObjective; import com.google.ortools.linearsolver.MPSolver; import com.google.ortools.linearsolver.MPVariable;
C#
using System; using Google.OrTools.LinearSolver;
LP ソルバーを宣言する
MPsolver
は、Glop を含むさまざまなソルバーのラッパーです。以下のコードは GLOP ソルバーを宣言しています。
Python
solver = pywraplp.Solver.CreateSolver("GLOP") if not solver: return
C++
std::unique_ptr<MPSolver> solver(MPSolver::CreateSolver("SCIP")); if (!solver) { LOG(WARNING) << "SCIP solver unavailable."; return; }
Java
MPSolver solver = MPSolver.createSolver("GLOP");
C#
Solver solver = Solver.CreateSolver("GLOP"); if (solver is null) { return; }
注: GLOP
を PDLP
に置き換えて、別の LP ソルバーを使用します。ソルバーの選択の詳細については、高度な LP ソルバーをご覧ください。サードパーティのソルバーのインストールについては、インストール ガイドをご覧ください。
変数を作成する
まず、値が 0 ~無限大の変数 x と y を作成します。xx
Python
x = solver.NumVar(0, solver.infinity(), "x") y = solver.NumVar(0, solver.infinity(), "y") print("Number of variables =", solver.NumVariables())
C++
const double infinity = solver->infinity(); // x and y are non-negative variables. MPVariable* const x = solver->MakeNumVar(0.0, infinity, "x"); MPVariable* const y = solver->MakeNumVar(0.0, infinity, "y"); LOG(INFO) << "Number of variables = " << solver->NumVariables();
Java
double infinity = java.lang.Double.POSITIVE_INFINITY; // x and y are continuous non-negative variables. MPVariable x = solver.makeNumVar(0.0, infinity, "x"); MPVariable y = solver.makeNumVar(0.0, infinity, "y"); System.out.println("Number of variables = " + solver.numVariables());
C#
Variable x = solver.MakeNumVar(0.0, double.PositiveInfinity, "x"); Variable y = solver.MakeNumVar(0.0, double.PositiveInfinity, "y"); Console.WriteLine("Number of variables = " + solver.NumVariables());
制約を定義する
次に、変数に制約を定義します。各制約に一意の名前(constraint0
など)を付けて、制約の係数を定義します。
Python
# Constraint 0: x + 2y <= 14. solver.Add(x + 2 * y <= 14.0) # Constraint 1: 3x - y >= 0. solver.Add(3 * x - y >= 0.0) # Constraint 2: x - y <= 2. solver.Add(x - y <= 2.0) print("Number of constraints =", solver.NumConstraints())
C++
// x + 2*y <= 14. MPConstraint* const c0 = solver->MakeRowConstraint(-infinity, 14.0); c0->SetCoefficient(x, 1); c0->SetCoefficient(y, 2); // 3*x - y >= 0. MPConstraint* const c1 = solver->MakeRowConstraint(0.0, infinity); c1->SetCoefficient(x, 3); c1->SetCoefficient(y, -1); // x - y <= 2. MPConstraint* const c2 = solver->MakeRowConstraint(-infinity, 2.0); c2->SetCoefficient(x, 1); c2->SetCoefficient(y, -1); LOG(INFO) << "Number of constraints = " << solver->NumConstraints();
Java
// x + 2*y <= 14. MPConstraint c0 = solver.makeConstraint(-infinity, 14.0, "c0"); c0.setCoefficient(x, 1); c0.setCoefficient(y, 2); // 3*x - y >= 0. MPConstraint c1 = solver.makeConstraint(0.0, infinity, "c1"); c1.setCoefficient(x, 3); c1.setCoefficient(y, -1); // x - y <= 2. MPConstraint c2 = solver.makeConstraint(-infinity, 2.0, "c2"); c2.setCoefficient(x, 1); c2.setCoefficient(y, -1); System.out.println("Number of constraints = " + solver.numConstraints());
C#
// x + 2y <= 14. solver.Add(x + 2 * y <= 14.0); // 3x - y >= 0. solver.Add(3 * x - y >= 0.0); // x - y <= 2. solver.Add(x - y <= 2.0); Console.WriteLine("Number of constraints = " + solver.NumConstraints());
目的関数を定義する
次のコードでは、目的関数 3x + 4y
を定義し、これが最大化問題であることを指定しています。
Python
# Objective function: 3x + 4y. solver.Maximize(3 * x + 4 * y)
C++
// Objective function: 3x + 4y. MPObjective* const objective = solver->MutableObjective(); objective->SetCoefficient(x, 3); objective->SetCoefficient(y, 4); objective->SetMaximization();
Java
// Maximize 3 * x + 4 * y. MPObjective objective = solver.objective(); objective.setCoefficient(x, 3); objective.setCoefficient(y, 4); objective.setMaximization();
C#
// Objective function: 3x + 4y. solver.Maximize(3 * x + 4 * y);
ソルバーを呼び出す
次のコードはソルバーを呼び出します。
Python
print(f"Solving with {solver.SolverVersion()}") status = solver.Solve()
C++
const MPSolver::ResultStatus result_status = solver->Solve(); // Check that the problem has an optimal solution. if (result_status != MPSolver::OPTIMAL) { LOG(FATAL) << "The problem does not have an optimal solution!"; }
Java
final MPSolver.ResultStatus resultStatus = solver.solve();
C#
Solver.ResultStatus resultStatus = solver.Solve();
解答を表示する
次のコードは解答を示しています。
Python
if status == pywraplp.Solver.OPTIMAL: print("Solution:") print(f"Objective value = {solver.Objective().Value():0.1f}") print(f"x = {x.solution_value():0.1f}") print(f"y = {y.solution_value():0.1f}") else: print("The problem does not have an optimal solution.")
C++
LOG(INFO) << "Solution:"; LOG(INFO) << "Optimal objective value = " << objective->Value(); LOG(INFO) << x->name() << " = " << x->solution_value(); LOG(INFO) << y->name() << " = " << y->solution_value();
Java
if (resultStatus == MPSolver.ResultStatus.OPTIMAL) { System.out.println("Solution:"); System.out.println("Objective value = " + objective.value()); System.out.println("x = " + x.solutionValue()); System.out.println("y = " + y.solutionValue()); } else { System.err.println("The problem does not have an optimal solution!"); }
C#
// Check that the problem has an optimal solution. if (resultStatus != Solver.ResultStatus.OPTIMAL) { Console.WriteLine("The problem does not have an optimal solution!"); return; } Console.WriteLine("Solution:"); Console.WriteLine("Objective value = " + solver.Objective().Value()); Console.WriteLine("x = " + x.SolutionValue()); Console.WriteLine("y = " + y.SolutionValue());
すべてのプログラム
すべてのプログラムを以下に示します。
Python
from ortools.linear_solver import pywraplp def LinearProgrammingExample(): """Linear programming sample.""" # Instantiate a Glop solver, naming it LinearExample. solver = pywraplp.Solver.CreateSolver("GLOP") if not solver: return # Create the two variables and let them take on any non-negative value. x = solver.NumVar(0, solver.infinity(), "x") y = solver.NumVar(0, solver.infinity(), "y") print("Number of variables =", solver.NumVariables()) # Constraint 0: x + 2y <= 14. solver.Add(x + 2 * y <= 14.0) # Constraint 1: 3x - y >= 0. solver.Add(3 * x - y >= 0.0) # Constraint 2: x - y <= 2. solver.Add(x - y <= 2.0) print("Number of constraints =", solver.NumConstraints()) # Objective function: 3x + 4y. solver.Maximize(3 * x + 4 * y) # Solve the system. print(f"Solving with {solver.SolverVersion()}") status = solver.Solve() if status == pywraplp.Solver.OPTIMAL: print("Solution:") print(f"Objective value = {solver.Objective().Value():0.1f}") print(f"x = {x.solution_value():0.1f}") print(f"y = {y.solution_value():0.1f}") else: print("The problem does not have an optimal solution.") print("\nAdvanced usage:") print(f"Problem solved in {solver.wall_time():d} milliseconds") print(f"Problem solved in {solver.iterations():d} iterations") LinearProgrammingExample()
C++
#include <iostream> #include <memory> #include "ortools/linear_solver/linear_solver.h" namespace operations_research { void LinearProgrammingExample() { std::unique_ptr<MPSolver> solver(MPSolver::CreateSolver("SCIP")); if (!solver) { LOG(WARNING) << "SCIP solver unavailable."; return; } const double infinity = solver->infinity(); // x and y are non-negative variables. MPVariable* const x = solver->MakeNumVar(0.0, infinity, "x"); MPVariable* const y = solver->MakeNumVar(0.0, infinity, "y"); LOG(INFO) << "Number of variables = " << solver->NumVariables(); // x + 2*y <= 14. MPConstraint* const c0 = solver->MakeRowConstraint(-infinity, 14.0); c0->SetCoefficient(x, 1); c0->SetCoefficient(y, 2); // 3*x - y >= 0. MPConstraint* const c1 = solver->MakeRowConstraint(0.0, infinity); c1->SetCoefficient(x, 3); c1->SetCoefficient(y, -1); // x - y <= 2. MPConstraint* const c2 = solver->MakeRowConstraint(-infinity, 2.0); c2->SetCoefficient(x, 1); c2->SetCoefficient(y, -1); LOG(INFO) << "Number of constraints = " << solver->NumConstraints(); // Objective function: 3x + 4y. MPObjective* const objective = solver->MutableObjective(); objective->SetCoefficient(x, 3); objective->SetCoefficient(y, 4); objective->SetMaximization(); const MPSolver::ResultStatus result_status = solver->Solve(); // Check that the problem has an optimal solution. if (result_status != MPSolver::OPTIMAL) { LOG(FATAL) << "The problem does not have an optimal solution!"; } LOG(INFO) << "Solution:"; LOG(INFO) << "Optimal objective value = " << objective->Value(); LOG(INFO) << x->name() << " = " << x->solution_value(); LOG(INFO) << y->name() << " = " << y->solution_value(); } } // namespace operations_research int main(int argc, char** argv) { operations_research::LinearProgrammingExample(); return EXIT_SUCCESS; }
Java
package com.google.ortools.linearsolver.samples; import com.google.ortools.Loader; import com.google.ortools.linearsolver.MPConstraint; import com.google.ortools.linearsolver.MPObjective; import com.google.ortools.linearsolver.MPSolver; import com.google.ortools.linearsolver.MPVariable; /** Simple linear programming example. */ public final class LinearProgrammingExample { public static void main(String[] args) { Loader.loadNativeLibraries(); MPSolver solver = MPSolver.createSolver("GLOP"); double infinity = java.lang.Double.POSITIVE_INFINITY; // x and y are continuous non-negative variables. MPVariable x = solver.makeNumVar(0.0, infinity, "x"); MPVariable y = solver.makeNumVar(0.0, infinity, "y"); System.out.println("Number of variables = " + solver.numVariables()); // x + 2*y <= 14. MPConstraint c0 = solver.makeConstraint(-infinity, 14.0, "c0"); c0.setCoefficient(x, 1); c0.setCoefficient(y, 2); // 3*x - y >= 0. MPConstraint c1 = solver.makeConstraint(0.0, infinity, "c1"); c1.setCoefficient(x, 3); c1.setCoefficient(y, -1); // x - y <= 2. MPConstraint c2 = solver.makeConstraint(-infinity, 2.0, "c2"); c2.setCoefficient(x, 1); c2.setCoefficient(y, -1); System.out.println("Number of constraints = " + solver.numConstraints()); // Maximize 3 * x + 4 * y. MPObjective objective = solver.objective(); objective.setCoefficient(x, 3); objective.setCoefficient(y, 4); objective.setMaximization(); final MPSolver.ResultStatus resultStatus = solver.solve(); if (resultStatus == MPSolver.ResultStatus.OPTIMAL) { System.out.println("Solution:"); System.out.println("Objective value = " + objective.value()); System.out.println("x = " + x.solutionValue()); System.out.println("y = " + y.solutionValue()); } else { System.err.println("The problem does not have an optimal solution!"); } System.out.println("\nAdvanced usage:"); System.out.println("Problem solved in " + solver.wallTime() + " milliseconds"); System.out.println("Problem solved in " + solver.iterations() + " iterations"); } private LinearProgrammingExample() {} }
C#
using System; using Google.OrTools.LinearSolver; public class LinearProgrammingExample { static void Main() { Solver solver = Solver.CreateSolver("GLOP"); if (solver is null) { return; } // x and y are continuous non-negative variables. Variable x = solver.MakeNumVar(0.0, double.PositiveInfinity, "x"); Variable y = solver.MakeNumVar(0.0, double.PositiveInfinity, "y"); Console.WriteLine("Number of variables = " + solver.NumVariables()); // x + 2y <= 14. solver.Add(x + 2 * y <= 14.0); // 3x - y >= 0. solver.Add(3 * x - y >= 0.0); // x - y <= 2. solver.Add(x - y <= 2.0); Console.WriteLine("Number of constraints = " + solver.NumConstraints()); // Objective function: 3x + 4y. solver.Maximize(3 * x + 4 * y); Solver.ResultStatus resultStatus = solver.Solve(); // Check that the problem has an optimal solution. if (resultStatus != Solver.ResultStatus.OPTIMAL) { Console.WriteLine("The problem does not have an optimal solution!"); return; } Console.WriteLine("Solution:"); Console.WriteLine("Objective value = " + solver.Objective().Value()); Console.WriteLine("x = " + x.SolutionValue()); Console.WriteLine("y = " + y.SolutionValue()); Console.WriteLine("\nAdvanced usage:"); Console.WriteLine("Problem solved in " + solver.WallTime() + " milliseconds"); Console.WriteLine("Problem solved in " + solver.Iterations() + " iterations"); } }
最適なソリューション
以下に示すように、プログラムは問題の最適な解を返します。
Number of variables = 2
Number of constraints = 3
Solution:
x = 6.0
y = 4.0
Optimal objective value = 34.0
このグラフは、この解答を示しています。
緑色の破線は目的関数を最適値の 34 に設定することで定義されます。方程式が 3x + 4y = c
という形式の線は破線と平行です。34 は、線が実行可能領域を交差する c の最大値です。
線形最適化の問題を解決する。高度な LP の解法をご覧ください。