解決 MIP 相關問題

以下各節提供 MIP 問題的範例，並示範如何解決。問題如下：

盡可能提高 x + 10y 並遵循下列限制：

1. x + 7y ≤ 17.5
2. 0 ≤ x ≤ 3.5
3. 0 ≤ y
4. x，y 整數

由於限制為線性，因此這只是線性最佳化問題，解決方案必須為整數。下方圖表顯示問題可行地區的整數點。

請注意，這個問題與「解決 LP 問題」中所述的線性最佳化問題非常類似，但在本例中，我們要求解決方案必須為整數。

解決 MIP 問題的基本步驟

如要解決 MIP 問題，您的程式應包含下列步驟：

1. 匯入線性求解工具包裝函式。
2. 宣告 MIP 解析器
3. 您必須定義變數
4. 定義限制
5. 定義目標
6. 呼叫 MIP 解析器
7. 顯示解決方案

使用 MP 解題工具的解決方案

下一節展示了使用 MPSolver 包裝函式和 MIP 解析器解決問題的程式。

注意： 如要執行下列程式，您必須安裝 OR-Tools。

預設的 OR-Tools MIP 解析器是 SCIP。

匯入線性解題工具包裝函式

匯入 (或加入) OR-Tools 線性解析器包裝函式，這是 MIP 解答器和線性解題工具的介面，如下所示。

Python

from ortools.linear_solver import pywraplp

C++

#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;

宣告 MIP 解析器

下列程式碼會宣告問題的 MIP 解析器。這個範例使用第三方解析器 SCIP。

注意： 如果您從原始碼建構 OR-Tools，並想安裝並使用其他解決方案，請參閱所用平台的安裝指南。

Python

# Create the mip solver with the SCIP backend.
solver = pywraplp.Solver.CreateSolver("SAT")
if not solver:
    return

C++

// Create the mip solver with the SCIP backend.
std::unique_ptr<MPSolver> solver(MPSolver::CreateSolver("SCIP"));
if (!solver) {
  LOG(WARNING) << "SCIP solver unavailable.";
  return;
}

Java

// Create the linear solver with the SCIP backend.
MPSolver solver = MPSolver.createSolver("SCIP");
if (solver == null) {
  System.out.println("Could not create solver SCIP");
  return;
}

C#

// Create the linear solver with the SCIP backend.
Solver solver = Solver.CreateSolver("SCIP");
if (solver is null)
{
    return;
}

定義變數

以下程式碼會定義問題中的變數。

Python

infinity = solver.infinity()
# x and y are integer non-negative variables.
x = solver.IntVar(0.0, infinity, "x")
y = solver.IntVar(0.0, infinity, "y")

print("Number of variables =", solver.NumVariables())

C++

const double infinity = solver->infinity();
// x and y are integer non-negative variables.
MPVariable* const x = solver->MakeIntVar(0.0, infinity, "x");
MPVariable* const y = solver->MakeIntVar(0.0, infinity, "y");

LOG(INFO) << "Number of variables = " << solver->NumVariables();

Java

double infinity = java.lang.Double.POSITIVE_INFINITY;
// x and y are integer non-negative variables.
MPVariable x = solver.makeIntVar(0.0, infinity, "x");
MPVariable y = solver.makeIntVar(0.0, infinity, "y");

System.out.println("Number of variables = " + solver.numVariables());

C#

// x and y are integer non-negative variables.
Variable x = solver.MakeIntVar(0.0, double.PositiveInfinity, "x");
Variable y = solver.MakeIntVar(0.0, double.PositiveInfinity, "y");

Console.WriteLine("Number of variables = " + solver.NumVariables());

程式使用 MakeIntVar 方法 (或變化版本，視程式設計語言而定) 建立採用非負整數值的 x 和 y 變數。

定義限制條件

下列程式碼定義問題的限制條件。

Python

# x + 7 * y <= 17.5.
solver.Add(x + 7 * y <= 17.5)

# x <= 3.5.
solver.Add(x <= 3.5)

print("Number of constraints =", solver.NumConstraints())

C++

// x + 7 * y <= 17.5.
MPConstraint* const c0 = solver->MakeRowConstraint(-infinity, 17.5, "c0");
c0->SetCoefficient(x, 1);
c0->SetCoefficient(y, 7);

// x <= 3.5.
MPConstraint* const c1 = solver->MakeRowConstraint(-infinity, 3.5, "c1");
c1->SetCoefficient(x, 1);
c1->SetCoefficient(y, 0);

LOG(INFO) << "Number of constraints = " << solver->NumConstraints();

Java

// x + 7 * y <= 17.5.
MPConstraint c0 = solver.makeConstraint(-infinity, 17.5, "c0");
c0.setCoefficient(x, 1);
c0.setCoefficient(y, 7);

// x <= 3.5.
MPConstraint c1 = solver.makeConstraint(-infinity, 3.5, "c1");
c1.setCoefficient(x, 1);
c1.setCoefficient(y, 0);

System.out.println("Number of constraints = " + solver.numConstraints());

C#

// x + 7 * y <= 17.5.
solver.Add(x + 7 * y <= 17.5);

// x <= 3.5.
solver.Add(x <= 3.5);

Console.WriteLine("Number of constraints = " + solver.NumConstraints());

定義目標

下列程式碼定義了問題的 objective function。

Python

# Maximize x + 10 * y.
solver.Maximize(x + 10 * y)

C++

// Maximize x + 10 * y.
MPObjective* const objective = solver->MutableObjective();
objective->SetCoefficient(x, 1);
objective->SetCoefficient(y, 10);
objective->SetMaximization();

Java

// Maximize x + 10 * y.
MPObjective objective = solver.objective();
objective.setCoefficient(x, 1);
objective.setCoefficient(y, 10);
objective.setMaximization();

C#

// Maximize x + 10 * y.
solver.Maximize(x + 10 * y);

呼叫解題工具

下列程式碼會呼叫解題工具。

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();

顯示解決方案

下列程式碼顯示解決方案。

Python

if status == pywraplp.Solver.OPTIMAL:
    print("Solution:")
    print("Objective value =", solver.Objective().Value())
    print("x =", x.solution_value())
    print("y =", y.solution_value())
else:
    print("The problem does not have an optimal solution.")

C++

LOG(INFO) << "Solution:";
LOG(INFO) << "Objective value = " << objective->Value();
LOG(INFO) << "x = " << x->solution_value();
LOG(INFO) << "y = " << 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());

以下提供這項問題的解決方案。

Number of variables = 2
Number of constraints = 2
Solution:
Objective value = 23
x = 3
y = 2

目標函式的最佳值為 23，發生在 y = 2 x = 3 的時間點。

完成計畫

以下是完整的計畫。

Python

from ortools.linear_solver import pywraplp


def main():
    # Create the mip solver with the SCIP backend.
    solver = pywraplp.Solver.CreateSolver("SAT")
    if not solver:
        return

    infinity = solver.infinity()
    # x and y are integer non-negative variables.
    x = solver.IntVar(0.0, infinity, "x")
    y = solver.IntVar(0.0, infinity, "y")

    print("Number of variables =", solver.NumVariables())

    # x + 7 * y <= 17.5.
    solver.Add(x + 7 * y <= 17.5)

    # x <= 3.5.
    solver.Add(x <= 3.5)

    print("Number of constraints =", solver.NumConstraints())

    # Maximize x + 10 * y.
    solver.Maximize(x + 10 * y)

    print(f"Solving with {solver.SolverVersion()}")
    status = solver.Solve()

    if status == pywraplp.Solver.OPTIMAL:
        print("Solution:")
        print("Objective value =", solver.Objective().Value())
        print("x =", x.solution_value())
        print("y =", y.solution_value())
    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")
    print(f"Problem solved in {solver.nodes():d} branch-and-bound nodes")


if __name__ == "__main__":
    main()

C++

#include <memory>

#include "ortools/linear_solver/linear_solver.h"

namespace operations_research {
void SimpleMipProgram() {
  // Create the mip solver with the SCIP backend.
  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 integer non-negative variables.
  MPVariable* const x = solver->MakeIntVar(0.0, infinity, "x");
  MPVariable* const y = solver->MakeIntVar(0.0, infinity, "y");

  LOG(INFO) << "Number of variables = " << solver->NumVariables();

  // x + 7 * y <= 17.5.
  MPConstraint* const c0 = solver->MakeRowConstraint(-infinity, 17.5, "c0");
  c0->SetCoefficient(x, 1);
  c0->SetCoefficient(y, 7);

  // x <= 3.5.
  MPConstraint* const c1 = solver->MakeRowConstraint(-infinity, 3.5, "c1");
  c1->SetCoefficient(x, 1);
  c1->SetCoefficient(y, 0);

  LOG(INFO) << "Number of constraints = " << solver->NumConstraints();

  // Maximize x + 10 * y.
  MPObjective* const objective = solver->MutableObjective();
  objective->SetCoefficient(x, 1);
  objective->SetCoefficient(y, 10);
  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) << "Objective value = " << objective->Value();
  LOG(INFO) << "x = " << x->solution_value();
  LOG(INFO) << "y = " << y->solution_value();

  LOG(INFO) << "\nAdvanced usage:";
  LOG(INFO) << "Problem solved in " << solver->wall_time() << " milliseconds";
  LOG(INFO) << "Problem solved in " << solver->iterations() << " iterations";
  LOG(INFO) << "Problem solved in " << solver->nodes()
            << " branch-and-bound nodes";
}
}  // namespace operations_research

int main(int argc, char** argv) {
  operations_research::SimpleMipProgram();
  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;

/** Minimal Mixed Integer Programming example to showcase calling the solver. */
public final class SimpleMipProgram {
  public static void main(String[] args) {
    Loader.loadNativeLibraries();
    // Create the linear solver with the SCIP backend.
    MPSolver solver = MPSolver.createSolver("SCIP");
    if (solver == null) {
      System.out.println("Could not create solver SCIP");
      return;
    }

    double infinity = java.lang.Double.POSITIVE_INFINITY;
    // x and y are integer non-negative variables.
    MPVariable x = solver.makeIntVar(0.0, infinity, "x");
    MPVariable y = solver.makeIntVar(0.0, infinity, "y");

    System.out.println("Number of variables = " + solver.numVariables());

    // x + 7 * y <= 17.5.
    MPConstraint c0 = solver.makeConstraint(-infinity, 17.5, "c0");
    c0.setCoefficient(x, 1);
    c0.setCoefficient(y, 7);

    // x <= 3.5.
    MPConstraint c1 = solver.makeConstraint(-infinity, 3.5, "c1");
    c1.setCoefficient(x, 1);
    c1.setCoefficient(y, 0);

    System.out.println("Number of constraints = " + solver.numConstraints());

    // Maximize x + 10 * y.
    MPObjective objective = solver.objective();
    objective.setCoefficient(x, 1);
    objective.setCoefficient(y, 10);
    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");
    System.out.println("Problem solved in " + solver.nodes() + " branch-and-bound nodes");
  }

  private SimpleMipProgram() {}
}

C#

using System;
using Google.OrTools.LinearSolver;

public class SimpleMipProgram
{
    static void Main()
    {
        // Create the linear solver with the SCIP backend.
        Solver solver = Solver.CreateSolver("SCIP");
        if (solver is null)
        {
            return;
        }

        // x and y are integer non-negative variables.
        Variable x = solver.MakeIntVar(0.0, double.PositiveInfinity, "x");
        Variable y = solver.MakeIntVar(0.0, double.PositiveInfinity, "y");

        Console.WriteLine("Number of variables = " + solver.NumVariables());

        // x + 7 * y <= 17.5.
        solver.Add(x + 7 * y <= 17.5);

        // x <= 3.5.
        solver.Add(x <= 3.5);

        Console.WriteLine("Number of constraints = " + solver.NumConstraints());

        // Maximize x + 10 * y.
        solver.Maximize(x + 10 * 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");
        Console.WriteLine("Problem solved in " + solver.Nodes() + " branch-and-bound nodes");
    }
}

比較線性和整數最佳化

現在，讓我們比較解決方案和上圖的整數最佳化問題，以及解決方案與對應的線性最佳化問題，問題會移除整數限制。您可能猜測出整數問題的解決方式，會是最接近線性解決方案的可行區域中的整數點，也就是 x = 0、y = 2 點。在接下來的畫面裡 就並非如此

您只要進行以下變更，即可輕鬆修改上一節的程式來解決線性問題：

• 替換 MIP 解析器

  Python

  # Create the mip solver with the SCIP backend.
  solver = pywraplp.Solver.CreateSolver("SAT")
  if not solver:
      return

  C++

  // Create the mip solver with the SCIP backend.
  std::unique_ptr<MPSolver> solver(MPSolver::CreateSolver("SCIP"));
  if (!solver) {
    LOG(WARNING) << "SCIP solver unavailable.";
    return;
  }

  Java

  // Create the linear solver with the SCIP backend.
  MPSolver solver = MPSolver.createSolver("SCIP");
  if (solver == null) {
    System.out.println("Could not create solver SCIP");
    return;
  }

  C#

  // Create the linear solver with the SCIP backend.
  Solver solver = Solver.CreateSolver("SCIP");
  if (solver is null)
  {
      return;
  }

  註解解題工具

  Python

  # Create the linear solver with the GLOP backend.
  solver = pywraplp.Solver.CreateSolver("GLOP")
  if not solver:
      return

  C++

  // Create the linear solver with the GLOP backend.
  std::unique_ptr<MPSolver> solver(MPSolver::CreateSolver("GLOP"));

  Java

  // Create the linear solver with the GLOP backend.
  MPSolver solver = MPSolver.createSolver("GLOP");
  if (solver == null) {
    System.out.println("Could not create solver SCIP");
    return;
  }

  C#

  // Create the linear solver with the GLOP backend.
  Solver solver = Solver.CreateSolver("GLOP");
  if (solver is null)
  {
      return;
  }

• 取代整數變數

  Python

  infinity = solver.infinity()
  # x and y are integer non-negative variables.
  x = solver.IntVar(0.0, infinity, "x")
  y = solver.IntVar(0.0, infinity, "y")
  
  print("Number of variables =", solver.NumVariables())

  C++

  const double infinity = solver->infinity();
  // x and y are integer non-negative variables.
  MPVariable* const x = solver->MakeIntVar(0.0, infinity, "x");
  MPVariable* const y = solver->MakeIntVar(0.0, infinity, "y");
  
  LOG(INFO) << "Number of variables = " << solver->NumVariables();

  Java

  double infinity = java.lang.Double.POSITIVE_INFINITY;
  // x and y are integer non-negative variables.
  MPVariable x = solver.makeIntVar(0.0, infinity, "x");
  MPVariable y = solver.makeIntVar(0.0, infinity, "y");
  
  System.out.println("Number of variables = " + solver.numVariables());

  C#

  // x and y are integer non-negative variables.
  Variable x = solver.MakeIntVar(0.0, double.PositiveInfinity, "x");
  Variable y = solver.MakeIntVar(0.0, double.PositiveInfinity, "y");
  
  Console.WriteLine("Number of variables = " + solver.NumVariables());

  搭配連續變數

  Python

  infinity = solver.infinity()
  # Create the variables x and y.
  x = solver.NumVar(0.0, infinity, "x")
  y = solver.NumVar(0.0, infinity, "y")
  
  print("Number of variables =", solver.NumVariables())

  C++

  const double infinity = solver->infinity();
  // Create the variables x and y.
  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;
  // Create the variables x and y.
  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#

  // Create the variables x and y.
  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());

進行這些變更並再次執行程式後，您會得到下列輸出內容：

Number of variables = 2
Number of constraints = 2
Objective value = 25.000000
x = 0.000000
y = 2.500000

線性問題的解決方法發生在 x = 0 的 y = 2.5 點，其中目標函式等於 25。這張圖表顯示線性和整數問題的解決方法。

請注意，相較於可行區域內的大多數其他整數點，整數解決方案不接近線性解。一般而言，線性最佳化問題的解決方案，以及對應的整數最佳化問題可能很相似。因此，這兩種問題的解決方案都需要不同的方法。