マルチナップサック問題と同様に、ビンパッキング問題にも 商品をビンに梱包しますただし、ビンパッキングの問題には、 すべてのアイテムを保持する最小のビンを見つけることです。
2 つの問題の違いは次のとおりです。
複数のナップサック問題: アイテムのサブセットを固定数の さまざまな容量のビンに保管されるため、梱包されたアイテムの 上限です。
ビンパッキングの問題: 共通の容量を持つビンを必要な数だけ与えると、 最小のアイテムを見つけることですこの問題では、 目標には値が関係しないため、値が割り当てられません。
次の例は、ビンパッキングの問題を解く方法を示しています。
例
この例では、さまざまな重量の商品を一連のビンに梱包する必要があります 共通の容量を持つ ネットワークインターフェースですすべてのテーブルを保持するのに十分なビンがあることを 問題は、十分な数のアイテムを見つけることです。
以降のセクションでは、この問題を解決するプログラムについて説明します。完全な 完全なプログラムをご覧ください。
この例では、MPSolver ラッパーを使用しています。
ライブラリのインポート
以下のコードでは、必要なライブラリをインポートします。
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;
データを作成する
以下のコードは、この例のデータを作成します。
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; }
データには次のものが含まれます。
weights
: アイテムの重みを含むベクトル。bin_capacity
: ビンの容量を示す単一の数値。
商品数の上限を最小化することを目標に、商品に値は割り当てられていません。 値が含まれないことを意味します。
num_bins
はアイテム数に設定されています。なぜなら、
問題に解答がある場合、すべての項目の重みが
ビンの容量になります。その場合、必要となるビンの最大数は
アイテムの数を減らすことができます。
解法を宣言する
次のコードはソルバーを宣言しています。
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; }
変数を作成する
次のコードは、プログラムの変数を作成します。
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}"); }
複数のナップサックの例と同様に、変数 x[(i,
j)]
の配列を定義します。変数 i
がビン j
に配置されている場合は値が 1、それ以外の場合は 0 です。
ビンパッキングのために、値が 1 の変数の配列 y[j]
も定義します。
ビン j
が使用されている場合(ビンに梱包されているアイテムがある場合)、0
できません。y[j]
の合計が、使用されるビンの数になります。
制約を定義する
次のコードは、問題の制約を定義しています。
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]); } }
制約は次のとおりです。
- 各アイテムは 1 つのビンに入れる必要があります。この制約は、
すべてのビン
j
のx[i][j]
の合計が 1 に等しい必要があります。備考 マルチ ナップサック問題とは異なります。 1 以下であることのみが必要です。これは、すべてのアイテムですべてのアイテムが です。 各ビンに梱包する合計重量が容量を超えないようにする必要があります。これが 複数のナップサック問題と同じ制約が適用されますが、 不等式の右側のビンの容量に
y[j]
を掛けます。y[j]
を掛ける理由いずれかの項目が一致すると、y[j]
が強制的に 1 になるためです。 ビンj
に詰められています。これは、y[j]
が 0 の場合、 不等式は 0 になり、左側のビンの重みは 制約に違反しています。これにより、変数y[j]
が接続されます。 と言えます。今のところ、解法は単語の確率分布をy[j]
が 1 のビンの数。
目標を定義する
次のコードは、問題の目的関数を定義しています。
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();
ビン j を使用する場合は y[j]
が 1、それ以外の場合は 0 になるため、y[j]
の合計は
使用されるビンの数です。目標は合計を最小化することです。
解法を呼び出し、解答を出力する
次のコードはソルバーを呼び出し、解答を出力します。
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}");
この解答は、すべての商品を梱包するために必要な最小ビンの数を示しています。 ソリューションには、使用されるビンごとに、その中に梱包されたアイテムが表示され、 ビンの重みです。
プログラムの出力
プログラムを実行すると、次の出力が表示されます。
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
プログラムを完了する
ビンパッキング問題の完全なプログラムを以下に示します。
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}"); } }