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Il problema del bin packing

Come nel caso degli zaini multipli, anche il problema dello imballaggio degli zaini comporta imballare gli articoli in cestini. Tuttavia, il problema del bin packing presenta obiettivo: trovare il minor numero di bin che conterrà tutti gli elementi.

Di seguito vengono riepilogate le differenze tra i due problemi:

Problema con più zaini: imballa un sottoinsieme di articoli in un numero fisso di bin, con capacità diverse, in modo che il valore totale degli articoli imballati è il massimo.
Problema di bin packing: dato il numero necessario di bin con una capacità comune, trovare il minor numero di elementi che conterrà tutti gli elementi. In questo problema, gli elementi non sono assegnati valori, perché l'obiettivo non implica valore.

Il prossimo esempio mostra come risolvere un problema di bin packing.

Esempio

In questo esempio, articoli di vari pesi devono essere imballati in un set di cestini con una capacità comune. Supponendo che ci siano abbastanza bin per contenere tutti i elementi, il problema è trovare il minor numero di elementi sufficiente.

Le sezioni seguenti presentano programmi per risolvere questo problema. Per l'intera vedi Programmi completati.

In questo esempio viene utilizzato il wrapper MPSolver.

Importa le librerie

Il codice riportato di seguito importa le librerie richieste.

Python

from ortools.linear_solver import pywraplp

C++

#include <iostream>
#include <memory>
#include <numeric>
#include <ostream>
#include <vector>

#include "ortools/linear_solver/linear_expr.h"
#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;

crea i dati

Il codice seguente crea i dati per l'esempio.

Python

def create_data_model():
    """Create the data for the example."""
    data = {}
    weights = [48, 30, 19, 36, 36, 27, 42, 42, 36, 24, 30]
    data["weights"] = weights
    data["items"] = list(range(len(weights)))
    data["bins"] = data["items"]
    data["bin_capacity"] = 100
    return data

C++

struct DataModel {
  const std::vector<double> weights = {48, 30, 19, 36, 36, 27,
                                       42, 42, 36, 24, 30};
  const int num_items = weights.size();
  const int num_bins = weights.size();
  const int bin_capacity = 100;
};

Java

static class DataModel {
  public final double[] weights = {48, 30, 19, 36, 36, 27, 42, 42, 36, 24, 30};
  public final int numItems = weights.length;
  public final int numBins = weights.length;
  public final int binCapacity = 100;
}

C#

class DataModel
{
    public static double[] Weights = { 48, 30, 19, 36, 36, 27, 42, 42, 36, 24, 30 };
    public int NumItems = Weights.Length;
    public int NumBins = Weights.Length;
    public double BinCapacity = 100.0;
}

I dati includono:

weights: un vettore contenente i pesi degli elementi.
bin_capacity: un singolo numero che indica la capacità dei bin.

Non ci sono valori assegnati agli elementi perché l'obiettivo di ridurre al minimo di bin non comporta alcun valore.

Tieni presente che num_bins è impostato sul numero di elementi. Questo perché se se esiste una soluzione, il peso di ogni elemento deve essere minore o uguale a fino alla capacità del bin. In questo caso, il numero massimo di bin che potresti aver bisogno è il numero di articoli, perché potevi sempre mettere ogni articolo in una fascia separata.

Dichiara il risolutore

Il seguente codice dichiara il risolutore.

Python

  # Create the mip solver with the SCIP backend.
  solver = pywraplp.Solver.CreateSolver("SCIP")

  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;
      }

Crea le variabili

Il seguente codice crea le variabili per il programma.

Python

# Variables
# x[i, j] = 1 if item i is packed in bin j.
x = {}
for i in data["items"]:
    for j in data["bins"]:
        x[(i, j)] = solver.IntVar(0, 1, "x_%i_%i" % (i, j))

# y[j] = 1 if bin j is used.
y = {}
for j in data["bins"]:
    y[j] = solver.IntVar(0, 1, "y[%i]" % j)

C++

std::vector<std::vector<const MPVariable*>> x(
    data.num_items, std::vector<const MPVariable*>(data.num_bins));
for (int i = 0; i < data.num_items; ++i) {
  for (int j = 0; j < data.num_bins; ++j) {
    x[i][j] = solver->MakeIntVar(0.0, 1.0, "");
  }
}
// y[j] = 1 if bin j is used.
std::vector<const MPVariable*> y(data.num_bins);
for (int j = 0; j < data.num_bins; ++j) {
  y[j] = solver->MakeIntVar(0.0, 1.0, "");
}

Java

MPVariable[][] x = new MPVariable[data.numItems][data.numBins];
for (int i = 0; i < data.numItems; ++i) {
  for (int j = 0; j < data.numBins; ++j) {
    x[i][j] = solver.makeIntVar(0, 1, "");
  }
}
MPVariable[] y = new MPVariable[data.numBins];
for (int j = 0; j < data.numBins; ++j) {
  y[j] = solver.makeIntVar(0, 1, "");
}

C#

Variable[,] x = new Variable[data.NumItems, data.NumBins];
for (int i = 0; i < data.NumItems; i++)
{
    for (int j = 0; j < data.NumBins; j++)
    {
        x[i, j] = solver.MakeIntVar(0, 1, $"x_{i}_{j}");
    }
}
Variable[] y = new Variable[data.NumBins];
for (int j = 0; j < data.NumBins; j++)
{
    y[j] = solver.MakeIntVar(0, 1, $"y_{j}");
}

Come nell'esempio con più zaini, si definisce un array di variabili x[(i, j)], il cui valore è 1 se l'elemento i viene inserito nella fascia j e 0 negli altri casi.

Per il bin packing, devi definire anche un array di variabili, y[j], il cui valore è 1 se viene utilizzata la fascia j (ossia, se al suo interno sono imballati degli articoli) e 0 negli altri casi. La somma di y[j] sarà il numero di bin utilizzati.

Definisci i vincoli

Il seguente codice definisce i vincoli per il problema:

Python

# Constraints
# Each item must be in exactly one bin.
for i in data["items"]:
    solver.Add(sum(x[i, j] for j in data["bins"]) == 1)

# The amount packed in each bin cannot exceed its capacity.
for j in data["bins"]:
    solver.Add(
        sum(x[(i, j)] * data["weights"][i] for i in data["items"])
        <= y[j] * data["bin_capacity"]
    )

C++

// Create the constraints.
// Each item is in exactly one bin.
for (int i = 0; i < data.num_items; ++i) {
  LinearExpr sum;
  for (int j = 0; j < data.num_bins; ++j) {
    sum += x[i][j];
  }
  solver->MakeRowConstraint(sum == 1.0);
}
// For each bin that is used, the total packed weight can be at most
// the bin capacity.
for (int j = 0; j < data.num_bins; ++j) {
  LinearExpr weight;
  for (int i = 0; i < data.num_items; ++i) {
    weight += data.weights[i] * LinearExpr(x[i][j]);
  }
  solver->MakeRowConstraint(weight <= LinearExpr(y[j]) * data.bin_capacity);
}

Java

double infinity = java.lang.Double.POSITIVE_INFINITY;
for (int i = 0; i < data.numItems; ++i) {
  MPConstraint constraint = solver.makeConstraint(1, 1, "");
  for (int j = 0; j < data.numBins; ++j) {
    constraint.setCoefficient(x[i][j], 1);
  }
}
// The bin capacity contraint for bin j is
//   sum_i w_i x_ij <= C*y_j
// To define this constraint, first subtract the left side from the right to get
//   0 <= C*y_j - sum_i w_i x_ij
//
// Note: Since sum_i w_i x_ij is positive (and y_j is 0 or 1), the right side must
// be less than or equal to C. But it's not necessary to add this constraint
// because it is forced by the other constraints.

for (int j = 0; j < data.numBins; ++j) {
  MPConstraint constraint = solver.makeConstraint(0, infinity, "");
  constraint.setCoefficient(y[j], data.binCapacity);
  for (int i = 0; i < data.numItems; ++i) {
    constraint.setCoefficient(x[i][j], -data.weights[i]);
  }
}

C#

for (int i = 0; i < data.NumItems; ++i)
{
    Constraint constraint = solver.MakeConstraint(1, 1, "");
    for (int j = 0; j < data.NumBins; ++j)
    {
        constraint.SetCoefficient(x[i, j], 1);
    }
}

for (int j = 0; j < data.NumBins; ++j)
{
    Constraint constraint = solver.MakeConstraint(0, Double.PositiveInfinity, "");
    constraint.SetCoefficient(y[j], data.BinCapacity);
    for (int i = 0; i < data.NumItems; ++i)
    {
        constraint.SetCoefficient(x[i, j], -DataModel.Weights[i]);
    }
}

I vincoli sono i seguenti:

Ogni articolo deve essere posizionato esattamente in un'unica cassa. Questo vincolo è impostato da richiedendo che la somma di x[i][j] in tutti i bin j sia uguale a 1. Nota che questo è diverso dal problema degli zaini multipli, in cui la somma è deve essere minore o uguale a 1, perché non tutti gli elementi devono essere pacchettizzati.
Il peso totale imballato in ogni bin non può superare la sua capacità. Questo è il dello stesso vincolo del problema con più zaini, ma in questo caso moltiplica la capacità del bin sul lato destro delle disuguaglianze per y[j].

Perché moltiplicare per y[j]? Perché obbliga y[j] a essere uguale a 1 se un elemento è confezionato in contenitore j. Questo è dovuto al fatto che se y[j] fossero 0, il lato destro di la disuguaglianza è pari a 0, mentre il peso bin a sinistra sarebbe maggiore di 0, violando il vincolo. In questo modo collegherai le variabili y[j] all'obiettivo del problema, per ora il risolutore cercherà di ridurre al minimo numero di bin per i quali y[j] è 1.

Definisci l'obiettivo

Il seguente codice definisce la funzione obiettivo per il problema.

Python

# Objective: minimize the number of bins used.
solver.Minimize(solver.Sum([y[j] for j in data["bins"]]))

C++

// Create the objective function.
MPObjective* const objective = solver->MutableObjective();
LinearExpr num_bins_used;
for (int j = 0; j < data.num_bins; ++j) {
  num_bins_used += y[j];
}
objective->MinimizeLinearExpr(num_bins_used);

Java

MPObjective objective = solver.objective();
for (int j = 0; j < data.numBins; ++j) {
  objective.setCoefficient(y[j], 1);
}
objective.setMinimization();

C#

Objective objective = solver.Objective();
for (int j = 0; j < data.NumBins; ++j)
{
    objective.SetCoefficient(y[j], 1);
}
objective.SetMinimization();

Poiché y[j] è 1 se si utilizza bin j e 0 altrimenti, la somma di y[j] è il numero di bin utilizzati. L'obiettivo è minimizzare la somma.

Chiama il risolutore e stampa la soluzione

Il codice seguente chiama il risolutore e stampa la soluzione.

Python

print(f"Solving with {solver.SolverVersion()}")
status = solver.Solve()
if status == pywraplp.Solver.OPTIMAL:
    num_bins = 0
    for j in data["bins"]:
        if y[j].solution_value() == 1:
            bin_items = []
            bin_weight = 0
            for i in data["items"]:
                if x[i, j].solution_value() > 0:
                    bin_items.append(i)
                    bin_weight += data["weights"][i]
            if bin_items:
                num_bins += 1
                print("Bin number", j)
                print("  Items packed:", bin_items)
                print("  Total weight:", bin_weight)
                print()
    print()
    print("Number of bins used:", num_bins)
    print("Time = ", solver.WallTime(), " milliseconds")
else:
    print("The problem does not have an optimal solution.")

C++

const MPSolver::ResultStatus result_status = solver->Solve();
// Check that the problem has an optimal solution.
if (result_status != MPSolver::OPTIMAL) {
  std::cerr << "The problem does not have an optimal solution!";
  return;
}
std::cout << "Number of bins used: " << objective->Value() << std::endl
          << std::endl;
double total_weight = 0;
for (int j = 0; j < data.num_bins; ++j) {
  if (y[j]->solution_value() == 1) {
    std::cout << "Bin " << j << std::endl << std::endl;
    double bin_weight = 0;
    for (int i = 0; i < data.num_items; ++i) {
      if (x[i][j]->solution_value() == 1) {
        std::cout << "Item " << i << " - Weight: " << data.weights[i]
                  << std::endl;
        bin_weight += data.weights[i];
      }
    }
    std::cout << "Packed bin weight: " << bin_weight << std::endl
              << std::endl;
    total_weight += bin_weight;
  }
}
std::cout << "Total packed weight: " << total_weight << std::endl;

Java

final MPSolver.ResultStatus resultStatus = solver.solve();
// Check that the problem has an optimal solution.
if (resultStatus == MPSolver.ResultStatus.OPTIMAL) {
  System.out.println("Number of bins used: " + objective.value());
  double totalWeight = 0;
  for (int j = 0; j < data.numBins; ++j) {
    if (y[j].solutionValue() == 1) {
      System.out.println("\nBin " + j + "\n");
      double binWeight = 0;
      for (int i = 0; i < data.numItems; ++i) {
        if (x[i][j].solutionValue() == 1) {
          System.out.println("Item " + i + " - weight: " + data.weights[i]);
          binWeight += data.weights[i];
        }
      }
      System.out.println("Packed bin weight: " + binWeight);
      totalWeight += binWeight;
    }
  }
  System.out.println("\nTotal packed weight: " + totalWeight);
} else {
  System.err.println("The problem does not have an optimal solution.");
}

C#

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($"Number of bins used: {solver.Objective().Value()}");
double TotalWeight = 0.0;
for (int j = 0; j < data.NumBins; ++j)
{
    double BinWeight = 0.0;
    if (y[j].SolutionValue() == 1)
    {
        Console.WriteLine($"Bin {j}");
        for (int i = 0; i < data.NumItems; ++i)
        {
            if (x[i, j].SolutionValue() == 1)
            {
                Console.WriteLine($"Item {i} weight: {DataModel.Weights[i]}");
                BinWeight += DataModel.Weights[i];
            }
        }
        Console.WriteLine($"Packed bin weight: {BinWeight}");
        TotalWeight += BinWeight;
    }
}
Console.WriteLine($"Total packed weight: {TotalWeight}");

La soluzione mostra il numero minimo di bin necessari per imballare tutti gli articoli. Per ogni bin utilizzato, la soluzione mostra gli articoli imballati al suo interno e peso totale bin.

Output del programma

Quando esegui il programma, viene visualizzato l'output seguente.

Bin number 0
  Items packed: [1, 5, 10]
  Total weight: 87

Bin number 1
  Items packed: [0, 6]
  Total weight: 90

Bin number 2
  Items packed: [2, 4, 7]
  Total weight: 97

Bin number 3
  Items packed: [3, 8, 9]
  Total weight: 96


Number of bins used: 4.0

Completa i programmi

Di seguito sono riportati i programmi completi per il problema del bin packing.

Python

from ortools.linear_solver import pywraplp


def create_data_model():
    """Create the data for the example."""
    data = {}
    weights = [48, 30, 19, 36, 36, 27, 42, 42, 36, 24, 30]
    data["weights"] = weights
    data["items"] = list(range(len(weights)))
    data["bins"] = data["items"]
    data["bin_capacity"] = 100
    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

    # Variables
    # x[i, j] = 1 if item i is packed in bin j.
    x = {}
    for i in data["items"]:
        for j in data["bins"]:
            x[(i, j)] = solver.IntVar(0, 1, "x_%i_%i" % (i, j))

    # y[j] = 1 if bin j is used.
    y = {}
    for j in data["bins"]:
        y[j] = solver.IntVar(0, 1, "y[%i]" % j)

    # Constraints
    # Each item must be in exactly one bin.
    for i in data["items"]:
        solver.Add(sum(x[i, j] for j in data["bins"]) == 1)

    # The amount packed in each bin cannot exceed its capacity.
    for j in data["bins"]:
        solver.Add(
            sum(x[(i, j)] * data["weights"][i] for i in data["items"])
            <= y[j] * data["bin_capacity"]
        )

    # Objective: minimize the number of bins used.
    solver.Minimize(solver.Sum([y[j] for j in data["bins"]]))

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

    if status == pywraplp.Solver.OPTIMAL:
        num_bins = 0
        for j in data["bins"]:
            if y[j].solution_value() == 1:
                bin_items = []
                bin_weight = 0
                for i in data["items"]:
                    if x[i, j].solution_value() > 0:
                        bin_items.append(i)
                        bin_weight += data["weights"][i]
                if bin_items:
                    num_bins += 1
                    print("Bin number", j)
                    print("  Items packed:", bin_items)
                    print("  Total weight:", bin_weight)
                    print()
        print()
        print("Number of bins used:", num_bins)
        print("Time = ", solver.WallTime(), " milliseconds")
    else:
        print("The problem does not have an optimal solution.")


if __name__ == "__main__":
    main()

C++

#include <iostream>
#include <memory>
#include <numeric>
#include <ostream>
#include <vector>

#include "ortools/linear_solver/linear_expr.h"
#include "ortools/linear_solver/linear_solver.h"

namespace operations_research {
struct DataModel {
  const std::vector<double> weights = {48, 30, 19, 36, 36, 27,
                                       42, 42, 36, 24, 30};
  const int num_items = weights.size();
  const int num_bins = weights.size();
  const int bin_capacity = 100;
};

void BinPackingMip() {
  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;
  }

  std::vector<std::vector<const MPVariable*>> x(
      data.num_items, std::vector<const MPVariable*>(data.num_bins));
  for (int i = 0; i < data.num_items; ++i) {
    for (int j = 0; j < data.num_bins; ++j) {
      x[i][j] = solver->MakeIntVar(0.0, 1.0, "");
    }
  }
  // y[j] = 1 if bin j is used.
  std::vector<const MPVariable*> y(data.num_bins);
  for (int j = 0; j < data.num_bins; ++j) {
    y[j] = solver->MakeIntVar(0.0, 1.0, "");
  }

  // Create the constraints.
  // Each item is in exactly one bin.
  for (int i = 0; i < data.num_items; ++i) {
    LinearExpr sum;
    for (int j = 0; j < data.num_bins; ++j) {
      sum += x[i][j];
    }
    solver->MakeRowConstraint(sum == 1.0);
  }
  // For each bin that is used, the total packed weight can be at most
  // the bin capacity.
  for (int j = 0; j < data.num_bins; ++j) {
    LinearExpr weight;
    for (int i = 0; i < data.num_items; ++i) {
      weight += data.weights[i] * LinearExpr(x[i][j]);
    }
    solver->MakeRowConstraint(weight <= LinearExpr(y[j]) * data.bin_capacity);
  }

  // Create the objective function.
  MPObjective* const objective = solver->MutableObjective();
  LinearExpr num_bins_used;
  for (int j = 0; j < data.num_bins; ++j) {
    num_bins_used += y[j];
  }
  objective->MinimizeLinearExpr(num_bins_used);

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

  // Check that the problem has an optimal solution.
  if (result_status != MPSolver::OPTIMAL) {
    std::cerr << "The problem does not have an optimal solution!";
    return;
  }
  std::cout << "Number of bins used: " << objective->Value() << std::endl
            << std::endl;
  double total_weight = 0;
  for (int j = 0; j < data.num_bins; ++j) {
    if (y[j]->solution_value() == 1) {
      std::cout << "Bin " << j << std::endl << std::endl;
      double bin_weight = 0;
      for (int i = 0; i < data.num_items; ++i) {
        if (x[i][j]->solution_value() == 1) {
          std::cout << "Item " << i << " - Weight: " << data.weights[i]
                    << std::endl;
          bin_weight += data.weights[i];
        }
      }
      std::cout << "Packed bin weight: " << bin_weight << std::endl
                << std::endl;
      total_weight += bin_weight;
    }
  }
  std::cout << "Total packed weight: " << total_weight << std::endl;
}
}  // namespace operations_research

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

/** Bin packing problem. */
public class BinPackingMip {
  static class DataModel {
    public final double[] weights = {48, 30, 19, 36, 36, 27, 42, 42, 36, 24, 30};
    public final int numItems = weights.length;
    public final int numBins = weights.length;
    public final int binCapacity = 100;
  }

  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;
    }

    MPVariable[][] x = new MPVariable[data.numItems][data.numBins];
    for (int i = 0; i < data.numItems; ++i) {
      for (int j = 0; j < data.numBins; ++j) {
        x[i][j] = solver.makeIntVar(0, 1, "");
      }
    }
    MPVariable[] y = new MPVariable[data.numBins];
    for (int j = 0; j < data.numBins; ++j) {
      y[j] = solver.makeIntVar(0, 1, "");
    }

    double infinity = java.lang.Double.POSITIVE_INFINITY;
    for (int i = 0; i < data.numItems; ++i) {
      MPConstraint constraint = solver.makeConstraint(1, 1, "");
      for (int j = 0; j < data.numBins; ++j) {
        constraint.setCoefficient(x[i][j], 1);
      }
    }
    // The bin capacity contraint for bin j is
    //   sum_i w_i x_ij <= C*y_j
    // To define this constraint, first subtract the left side from the right to get
    //   0 <= C*y_j - sum_i w_i x_ij
    //
    // Note: Since sum_i w_i x_ij is positive (and y_j is 0 or 1), the right side must
    // be less than or equal to C. But it's not necessary to add this constraint
    // because it is forced by the other constraints.

    for (int j = 0; j < data.numBins; ++j) {
      MPConstraint constraint = solver.makeConstraint(0, infinity, "");
      constraint.setCoefficient(y[j], data.binCapacity);
      for (int i = 0; i < data.numItems; ++i) {
        constraint.setCoefficient(x[i][j], -data.weights[i]);
      }
    }

    MPObjective objective = solver.objective();
    for (int j = 0; j < data.numBins; ++j) {
      objective.setCoefficient(y[j], 1);
    }
    objective.setMinimization();

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

    // Check that the problem has an optimal solution.
    if (resultStatus == MPSolver.ResultStatus.OPTIMAL) {
      System.out.println("Number of bins used: " + objective.value());
      double totalWeight = 0;
      for (int j = 0; j < data.numBins; ++j) {
        if (y[j].solutionValue() == 1) {
          System.out.println("\nBin " + j + "\n");
          double binWeight = 0;
          for (int i = 0; i < data.numItems; ++i) {
            if (x[i][j].solutionValue() == 1) {
              System.out.println("Item " + i + " - weight: " + data.weights[i]);
              binWeight += data.weights[i];
            }
          }
          System.out.println("Packed bin weight: " + binWeight);
          totalWeight += binWeight;
        }
      }
      System.out.println("\nTotal packed weight: " + totalWeight);
    } else {
      System.err.println("The problem does not have an optimal solution.");
    }
  }
  private BinPackingMip() {}
}

C#

using System;
using Google.OrTools.LinearSolver;

public class BinPackingMip
{
    class DataModel
    {
        public static double[] Weights = { 48, 30, 19, 36, 36, 27, 42, 42, 36, 24, 30 };
        public int NumItems = Weights.Length;
        public int NumBins = Weights.Length;
        public double BinCapacity = 100.0;
    }
    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.NumItems, data.NumBins];
        for (int i = 0; i < data.NumItems; i++)
        {
            for (int j = 0; j < data.NumBins; j++)
            {
                x[i, j] = solver.MakeIntVar(0, 1, $"x_{i}_{j}");
            }
        }
        Variable[] y = new Variable[data.NumBins];
        for (int j = 0; j < data.NumBins; j++)
        {
            y[j] = solver.MakeIntVar(0, 1, $"y_{j}");
        }

        for (int i = 0; i < data.NumItems; ++i)
        {
            Constraint constraint = solver.MakeConstraint(1, 1, "");
            for (int j = 0; j < data.NumBins; ++j)
            {
                constraint.SetCoefficient(x[i, j], 1);
            }
        }

        for (int j = 0; j < data.NumBins; ++j)
        {
            Constraint constraint = solver.MakeConstraint(0, Double.PositiveInfinity, "");
            constraint.SetCoefficient(y[j], data.BinCapacity);
            for (int i = 0; i < data.NumItems; ++i)
            {
                constraint.SetCoefficient(x[i, j], -DataModel.Weights[i]);
            }
        }

        Objective objective = solver.Objective();
        for (int j = 0; j < data.NumBins; ++j)
        {
            objective.SetCoefficient(y[j], 1);
        }
        objective.SetMinimization();

        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($"Number of bins used: {solver.Objective().Value()}");
        double TotalWeight = 0.0;
        for (int j = 0; j < data.NumBins; ++j)
        {
            double BinWeight = 0.0;
            if (y[j].SolutionValue() == 1)
            {
                Console.WriteLine($"Bin {j}");
                for (int i = 0; i < data.NumItems; ++i)
                {
                    if (x[i, j].SolutionValue() == 1)
                    {
                        Console.WriteLine($"Item {i} weight: {DataModel.Weights[i]}");
                        BinWeight += DataModel.Weights[i];
                    }
                }
                Console.WriteLine($"Packed bin weight: {BinWeight}");
                TotalWeight += BinWeight;
            }
        }
        Console.WriteLine($"Total packed weight: {TotalWeight}");
    }
}