একটি মডেল সংজ্ঞায়িত করতে অ্যারে ব্যবহার করা

পূর্ববর্তী বিভাগটি দেখিয়েছে কিভাবে একটি MIP সমাধান করা যায় মাত্র কয়েকটি ভেরিয়েবল এবং সীমাবদ্ধতার সাথে, যা পৃথকভাবে সংজ্ঞায়িত করা হয়েছে। বড় সমস্যাগুলির জন্য, অ্যারেগুলির উপর লুপ করে ভেরিয়েবল এবং সীমাবদ্ধতাগুলি সংজ্ঞায়িত করা আরও সুবিধাজনক। পরবর্তী উদাহরণ এটি ব্যাখ্যা করে।

উদাহরণ

এই উদাহরণে আমরা নিম্নলিখিত সমস্যার সমাধান করব।

সর্বাধিক 7 x 1 + 8 x 2 + 2 x 3 + 9 x 4 + 6 x 5 নিম্নলিখিত সীমাবদ্ধতা সাপেক্ষে:

  1. 5 x 1 + 7 x 2 + 9 x 3 + 2 x 4 + 1 x 5 ≤ 250
  2. 18 x 1 + 4 x 2 - 9 x 3 + 10 x 4 + 12 x 5 ≤ 285
  3. 4 x 1 + 7 x 2 + 3 x 3 + 8 x 4 + 5 x 5 ≤ 211
  4. 5 x 1 + 13 x 2 + 16 x 3 + 3 x 4 - 7 x 5 ≤ 315

যেখানে x 1 , x 2 , ..., x 5 অ-ঋণাত্মক পূর্ণসংখ্যা।

নিম্নলিখিত বিভাগগুলি এই সমস্যার সমাধান করে এমন প্রোগ্রামগুলি উপস্থাপন করে। প্রোগ্রামগুলি পূর্ববর্তী MIP উদাহরণের মতো একই পদ্ধতি ব্যবহার করে, কিন্তু এই ক্ষেত্রে সেগুলিকে একটি লুপে অ্যারের মানগুলিতে প্রয়োগ করে৷

সমাধানকারী ঘোষণা করুন

যেকোন MIP প্রোগ্রামে, আপনি লিনিয়ার সলভার র্যাপার আমদানি করে এবং MIP সল্ভার ঘোষণা করে শুরু করেন, যেমনটি আগের MIP উদাহরণে দেখানো হয়েছে।

ডেটা তৈরি করুন

নিম্নলিখিত কোডটি উদাহরণের জন্য ডেটা ধারণকারী অ্যারে তৈরি করে: সীমাবদ্ধতা এবং উদ্দেশ্য ফাংশনের জন্য পরিবর্তনশীল সহগ এবং সীমাবদ্ধতার জন্য সীমা।

পাইথন

def create_data_model():
    """Stores the data for the problem."""
    data = {}
    data["constraint_coeffs"] = [
        [5, 7, 9, 2, 1],
        [18, 4, -9, 10, 12],
        [4, 7, 3, 8, 5],
        [5, 13, 16, 3, -7],
    ]
    data["bounds"] = [250, 285, 211, 315]
    data["obj_coeffs"] = [7, 8, 2, 9, 6]
    data["num_vars"] = 5
    data["num_constraints"] = 4
    return data

সি++

struct DataModel {
  const std::vector<std::vector<double>> constraint_coeffs{
      {5, 7, 9, 2, 1},
      {18, 4, -9, 10, 12},
      {4, 7, 3, 8, 5},
      {5, 13, 16, 3, -7},
  };
  const std::vector<double> bounds{250, 285, 211, 315};
  const std::vector<double> obj_coeffs{7, 8, 2, 9, 6};
  const int num_vars = 5;
  const int num_constraints = 4;
};

জাভা

static class DataModel {
  public final double[][] constraintCoeffs = {
      {5, 7, 9, 2, 1},
      {18, 4, -9, 10, 12},
      {4, 7, 3, 8, 5},
      {5, 13, 16, 3, -7},
  };
  public final double[] bounds = {250, 285, 211, 315};
  public final double[] objCoeffs = {7, 8, 2, 9, 6};
  public final int numVars = 5;
  public final int numConstraints = 4;
}

সি#

class DataModel
{
    public double[,] ConstraintCoeffs = {
        { 5, 7, 9, 2, 1 },
        { 18, 4, -9, 10, 12 },
        { 4, 7, 3, 8, 5 },
        { 5, 13, 16, 3, -7 },
    };
    public double[] Bounds = { 250, 285, 211, 315 };
    public double[] ObjCoeffs = { 7, 8, 2, 9, 6 };
    public int NumVars = 5;
    public int NumConstraints = 4;
}

তথ্য ইনস্ট্যান্ট

নিম্নলিখিত কোড তথ্য মডেল instantiates.

পাইথন

data = create_data_model()

সি++

DataModel data;

জাভা

final DataModel data = new DataModel();

সি#

DataModel data = new DataModel();

সমাধানকারীকে ইনস্ট্যান্টিয়েট করুন

নিম্নলিখিত কোড সমাধানকারীকে তাৎক্ষণিকভাবে নির্দেশ করে।

পাইথন

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

সি++

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

জাভা

// 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;
}

সি#

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

ভেরিয়েবলের সংজ্ঞা দাও

নিম্নলিখিত কোডটি একটি লুপে উদাহরণের জন্য ভেরিয়েবলকে সংজ্ঞায়িত করে। বড় সমস্যাগুলির জন্য, পূর্ববর্তী উদাহরণের মতো, পৃথকভাবে ভেরিয়েবলগুলিকে সংজ্ঞায়িত করার চেয়ে এটি সহজ।

পাইথন

infinity = solver.infinity()
x = {}
for j in range(data["num_vars"]):
    x[j] = solver.IntVar(0, infinity, "x[%i]" % j)
print("Number of variables =", solver.NumVariables())

সি++

const double infinity = solver->infinity();
// x[j] is an array of non-negative, integer variables.
std::vector<const MPVariable*> x(data.num_vars);
for (int j = 0; j < data.num_vars; ++j) {
  x[j] = solver->MakeIntVar(0.0, infinity, "");
}
LOG(INFO) << "Number of variables = " << solver->NumVariables();

জাভা

double infinity = java.lang.Double.POSITIVE_INFINITY;
MPVariable[] x = new MPVariable[data.numVars];
for (int j = 0; j < data.numVars; ++j) {
  x[j] = solver.makeIntVar(0.0, infinity, "");
}
System.out.println("Number of variables = " + solver.numVariables());

সি#

Variable[] x = new Variable[data.NumVars];
for (int j = 0; j < data.NumVars; j++)
{
    x[j] = solver.MakeIntVar(0.0, double.PositiveInfinity, $"x_{j}");
}
Console.WriteLine("Number of variables = " + solver.NumVariables());

সীমাবদ্ধতা সংজ্ঞায়িত করুন

নিম্নলিখিত কোডটি MakeRowConstraint পদ্ধতি ব্যবহার করে উদাহরণের জন্য সীমাবদ্ধতা তৈরি করে (অথবা কিছু বৈকল্পিক, কোডিং ভাষার উপর নির্ভর করে)। পদ্ধতির প্রথম দুটি আর্গুমেন্ট হল সীমাবদ্ধতার জন্য নিম্ন এবং উপরের সীমানা। তৃতীয় যুক্তি, সীমাবদ্ধতার একটি নাম, ঐচ্ছিক।

প্রতিটি সীমাবদ্ধতার জন্য, আপনি SetCoefficient পদ্ধতি ব্যবহার করে ভেরিয়েবলের সহগ নির্ধারণ করেন। পদ্ধতিটি সীমাবদ্ধতার মধ্যে x[j] ভেরিয়েবলের সহগ নির্ধারণ i অ্যারের constraint_coeffs এর [i][j] এন্ট্রি হতে।

পাইথন

for i in range(data["num_constraints"]):
    constraint = solver.RowConstraint(0, data["bounds"][i], "")
    for j in range(data["num_vars"]):
        constraint.SetCoefficient(x[j], data["constraint_coeffs"][i][j])
print("Number of constraints =", solver.NumConstraints())
# In Python, you can also set the constraints as follows.
# for i in range(data['num_constraints']):
#  constraint_expr = \
# [data['constraint_coeffs'][i][j] * x[j] for j in range(data['num_vars'])]
#  solver.Add(sum(constraint_expr) <= data['bounds'][i])

সি++

// Create the constraints.
for (int i = 0; i < data.num_constraints; ++i) {
  MPConstraint* constraint = solver->MakeRowConstraint(0, data.bounds[i], "");
  for (int j = 0; j < data.num_vars; ++j) {
    constraint->SetCoefficient(x[j], data.constraint_coeffs[i][j]);
  }
}
LOG(INFO) << "Number of constraints = " << solver->NumConstraints();

জাভা

// Create the constraints.
for (int i = 0; i < data.numConstraints; ++i) {
  MPConstraint constraint = solver.makeConstraint(0, data.bounds[i], "");
  for (int j = 0; j < data.numVars; ++j) {
    constraint.setCoefficient(x[j], data.constraintCoeffs[i][j]);
  }
}
System.out.println("Number of constraints = " + solver.numConstraints());

সি#

for (int i = 0; i < data.NumConstraints; ++i)
{
    Constraint constraint = solver.MakeConstraint(0, data.Bounds[i], "");
    for (int j = 0; j < data.NumVars; ++j)
    {
        constraint.SetCoefficient(x[j], data.ConstraintCoeffs[i, j]);
    }
}
Console.WriteLine("Number of constraints = " + solver.NumConstraints());

উদ্দেশ্য সংজ্ঞায়িত করুন

নিম্নলিখিত কোড উদাহরণের জন্য উদ্দেশ্য ফাংশন সংজ্ঞায়িত করে। SetCoefficient পদ্ধতিটি উদ্দেশ্যের জন্য সহগ নির্ধারণ করে, যখন SetMaximization এটিকে সর্বাধিকীকরণ সমস্যা হিসাবে সংজ্ঞায়িত করে।

পাইথন

objective = solver.Objective()
for j in range(data["num_vars"]):
    objective.SetCoefficient(x[j], data["obj_coeffs"][j])
objective.SetMaximization()
# In Python, you can also set the objective as follows.
# obj_expr = [data['obj_coeffs'][j] * x[j] for j in range(data['num_vars'])]
# solver.Maximize(solver.Sum(obj_expr))

সি++

// Create the objective function.
MPObjective* const objective = solver->MutableObjective();
for (int j = 0; j < data.num_vars; ++j) {
  objective->SetCoefficient(x[j], data.obj_coeffs[j]);
}
objective->SetMaximization();

জাভা

MPObjective objective = solver.objective();
for (int j = 0; j < data.numVars; ++j) {
  objective.setCoefficient(x[j], data.objCoeffs[j]);
}
objective.setMaximization();

সি#

Objective objective = solver.Objective();
for (int j = 0; j < data.NumVars; ++j)
{
    objective.SetCoefficient(x[j], data.ObjCoeffs[j]);
}
objective.SetMaximization();

সমাধানকারীকে কল করুন

নিম্নলিখিত কোড সমাধানকারী কল.

পাইথন

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

সি++

const MPSolver::ResultStatus result_status = solver->Solve();

জাভা

final MPSolver.ResultStatus resultStatus = solver.solve();

সি#

Solver.ResultStatus resultStatus = solver.Solve();

সমাধান প্রদর্শন করুন

নিম্নলিখিত কোড সমাধান প্রদর্শন করে.

পাইথন

if status == pywraplp.Solver.OPTIMAL:
    print("Objective value =", solver.Objective().Value())
    for j in range(data["num_vars"]):
        print(x[j].name(), " = ", x[j].solution_value())
    print()
    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")
else:
    print("The problem does not have an optimal solution.")

সি++

// 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();

for (int j = 0; j < data.num_vars; ++j) {
  LOG(INFO) << "x[" << j << "] = " << x[j]->solution_value();
}

জাভা

// Check that the problem has an optimal solution.
if (resultStatus == MPSolver.ResultStatus.OPTIMAL) {
  System.out.println("Objective value = " + objective.value());
  for (int j = 0; j < data.numVars; ++j) {
    System.out.println("x[" + j + "] = " + x[j].solutionValue());
  }
  System.out.println();
  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");
} else {
  System.err.println("The problem does not have an optimal solution.");
}

সি#

// 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("Optimal objective value = " + solver.Objective().Value());

for (int j = 0; j < data.NumVars; ++j)
{
    Console.WriteLine("x[" + j + "] = " + x[j].SolutionValue());
}

এখানে সমস্যার সমাধান।

Number of variables = 5
Number of constraints = 4
Objective value = 260.0
x[0]  =  10.0
x[1]  =  16.0
x[2]  =  4.0
x[3]  =  4.0
x[4]  =  3.0

Problem solved in 29.000000 milliseconds
Problem solved in 315 iterations
Problem solved in 13 branch-and-bound nodes

সম্পূর্ণ প্রোগ্রাম

এখানে সম্পূর্ণ প্রোগ্রাম আছে.

পাইথন

from ortools.linear_solver import pywraplp


def create_data_model():
    """Stores the data for the problem."""
    data = {}
    data["constraint_coeffs"] = [
        [5, 7, 9, 2, 1],
        [18, 4, -9, 10, 12],
        [4, 7, 3, 8, 5],
        [5, 13, 16, 3, -7],
    ]
    data["bounds"] = [250, 285, 211, 315]
    data["obj_coeffs"] = [7, 8, 2, 9, 6]
    data["num_vars"] = 5
    data["num_constraints"] = 4
    return data



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

    infinity = solver.infinity()
    x = {}
    for j in range(data["num_vars"]):
        x[j] = solver.IntVar(0, infinity, "x[%i]" % j)
    print("Number of variables =", solver.NumVariables())

    for i in range(data["num_constraints"]):
        constraint = solver.RowConstraint(0, data["bounds"][i], "")
        for j in range(data["num_vars"]):
            constraint.SetCoefficient(x[j], data["constraint_coeffs"][i][j])
    print("Number of constraints =", solver.NumConstraints())
    # In Python, you can also set the constraints as follows.
    # for i in range(data['num_constraints']):
    #  constraint_expr = \
    # [data['constraint_coeffs'][i][j] * x[j] for j in range(data['num_vars'])]
    #  solver.Add(sum(constraint_expr) <= data['bounds'][i])

    objective = solver.Objective()
    for j in range(data["num_vars"]):
        objective.SetCoefficient(x[j], data["obj_coeffs"][j])
    objective.SetMaximization()
    # In Python, you can also set the objective as follows.
    # obj_expr = [data['obj_coeffs'][j] * x[j] for j in range(data['num_vars'])]
    # solver.Maximize(solver.Sum(obj_expr))

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

    if status == pywraplp.Solver.OPTIMAL:
        print("Objective value =", solver.Objective().Value())
        for j in range(data["num_vars"]):
            print(x[j].name(), " = ", x[j].solution_value())
        print()
        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")
    else:
        print("The problem does not have an optimal solution.")


if __name__ == "__main__":
    main()

সি++

#include <memory>
#include <vector>

#include "ortools/linear_solver/linear_solver.h"

namespace operations_research {
struct DataModel {
  const std::vector<std::vector<double>> constraint_coeffs{
      {5, 7, 9, 2, 1},
      {18, 4, -9, 10, 12},
      {4, 7, 3, 8, 5},
      {5, 13, 16, 3, -7},
  };
  const std::vector<double> bounds{250, 285, 211, 315};
  const std::vector<double> obj_coeffs{7, 8, 2, 9, 6};
  const int num_vars = 5;
  const int num_constraints = 4;
};

void MipVarArray() {
  DataModel data;

  // 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[j] is an array of non-negative, integer variables.
  std::vector<const MPVariable*> x(data.num_vars);
  for (int j = 0; j < data.num_vars; ++j) {
    x[j] = solver->MakeIntVar(0.0, infinity, "");
  }
  LOG(INFO) << "Number of variables = " << solver->NumVariables();

  // Create the constraints.
  for (int i = 0; i < data.num_constraints; ++i) {
    MPConstraint* constraint = solver->MakeRowConstraint(0, data.bounds[i], "");
    for (int j = 0; j < data.num_vars; ++j) {
      constraint->SetCoefficient(x[j], data.constraint_coeffs[i][j]);
    }
  }
  LOG(INFO) << "Number of constraints = " << solver->NumConstraints();

  // Create the objective function.
  MPObjective* const objective = solver->MutableObjective();
  for (int j = 0; j < data.num_vars; ++j) {
    objective->SetCoefficient(x[j], data.obj_coeffs[j]);
  }
  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();

  for (int j = 0; j < data.num_vars; ++j) {
    LOG(INFO) << "x[" << j << "] = " << x[j]->solution_value();
  }
}
}  // namespace operations_research

int main(int argc, char** argv) {
  operations_research::MipVarArray();
  return EXIT_SUCCESS;
}

জাভা

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;

/** MIP example with a variable array. */
public class MipVarArray {
  static class DataModel {
    public final double[][] constraintCoeffs = {
        {5, 7, 9, 2, 1},
        {18, 4, -9, 10, 12},
        {4, 7, 3, 8, 5},
        {5, 13, 16, 3, -7},
    };
    public final double[] bounds = {250, 285, 211, 315};
    public final double[] objCoeffs = {7, 8, 2, 9, 6};
    public final int numVars = 5;
    public final int numConstraints = 4;
  }

  public static void main(String[] args) throws Exception {
    Loader.loadNativeLibraries();
    final DataModel data = new DataModel();

    // 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;
    MPVariable[] x = new MPVariable[data.numVars];
    for (int j = 0; j < data.numVars; ++j) {
      x[j] = solver.makeIntVar(0.0, infinity, "");
    }
    System.out.println("Number of variables = " + solver.numVariables());

    // Create the constraints.
    for (int i = 0; i < data.numConstraints; ++i) {
      MPConstraint constraint = solver.makeConstraint(0, data.bounds[i], "");
      for (int j = 0; j < data.numVars; ++j) {
        constraint.setCoefficient(x[j], data.constraintCoeffs[i][j]);
      }
    }
    System.out.println("Number of constraints = " + solver.numConstraints());

    MPObjective objective = solver.objective();
    for (int j = 0; j < data.numVars; ++j) {
      objective.setCoefficient(x[j], data.objCoeffs[j]);
    }
    objective.setMaximization();

    final MPSolver.ResultStatus resultStatus = solver.solve();

    // Check that the problem has an optimal solution.
    if (resultStatus == MPSolver.ResultStatus.OPTIMAL) {
      System.out.println("Objective value = " + objective.value());
      for (int j = 0; j < data.numVars; ++j) {
        System.out.println("x[" + j + "] = " + x[j].solutionValue());
      }
      System.out.println();
      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");
    } else {
      System.err.println("The problem does not have an optimal solution.");
    }
  }

  private MipVarArray() {}
}

সি#

using System;
using Google.OrTools.LinearSolver;

public class MipVarArray
{
    class DataModel
    {
        public double[,] ConstraintCoeffs = {
            { 5, 7, 9, 2, 1 },
            { 18, 4, -9, 10, 12 },
            { 4, 7, 3, 8, 5 },
            { 5, 13, 16, 3, -7 },
        };
        public double[] Bounds = { 250, 285, 211, 315 };
        public double[] ObjCoeffs = { 7, 8, 2, 9, 6 };
        public int NumVars = 5;
        public int NumConstraints = 4;
    }
    public static void Main()
    {
        DataModel data = new DataModel();

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

        Variable[] x = new Variable[data.NumVars];
        for (int j = 0; j < data.NumVars; j++)
        {
            x[j] = solver.MakeIntVar(0.0, double.PositiveInfinity, $"x_{j}");
        }
        Console.WriteLine("Number of variables = " + solver.NumVariables());

        for (int i = 0; i < data.NumConstraints; ++i)
        {
            Constraint constraint = solver.MakeConstraint(0, data.Bounds[i], "");
            for (int j = 0; j < data.NumVars; ++j)
            {
                constraint.SetCoefficient(x[j], data.ConstraintCoeffs[i, j]);
            }
        }
        Console.WriteLine("Number of constraints = " + solver.NumConstraints());

        Objective objective = solver.Objective();
        for (int j = 0; j < data.NumVars; ++j)
        {
            objective.SetCoefficient(x[j], data.ObjCoeffs[j]);
        }
        objective.SetMaximization();

        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("Optimal objective value = " + solver.Objective().Value());

        for (int j = 0; j < data.NumVars; ++j)
        {
            Console.WriteLine("x[" + j + "] = " + x[j].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");
    }
}