בקטע הקודם מוסבר איך לפתור MIP באמצעות כמה משתנים ואילוצים, שמוגדרים בנפרד. בבעיות גדולות יותר, נוח יותר להגדיר את המשתנים והאילוצים על ידי חזרה על מערכים. הדוגמה הבאה ממחישה זאת.
דוגמה
בדוגמה הזאת נפתור את הבעיה הבאה.
מקסימום 7x1 + 8x2 + 2x3 + 9x4 + 6x5 בכפוף למגבלות הבאות:
- 5 x1 + 7 x2 + 9 x3 + 2 x4 + 1 x5 ≤ 250
- 18 x1 + 4 x2 - 9 x3 + 10 x4 + 12 x5 ≤ 285
- 4 x1 + 7 x2 + 3 x3 + 8 x4 + 5 x5 ≤ 211
- 5 x1 + 13 x2 + 16 x3 + 3 x4 - 7 x5 ≤ 315
כאשר x1, x2, ..., x5 הם מספרים שלמים לא שליליים.
בקטעים הבאים מוצגות תוכניות לפתרון הבעיה הזו. התוכנות משתמשות באותן שיטות כמו דוגמת ה-MIP הקודמת, אבל במקרה הזה הן מחילות אותן על ערכי מערך בלולאה.
מצהירים על הפותר
בכל תוכנית MIP, מתחילים בייבוא של ה-wrapper של הפותר הלינארי ובהצהרה על פותר MIP, כפי שמוצג בדוגמה הקודמת של ה-MIP.
יצירת הנתונים
הקוד הבא יוצר מערכים שמכילים את הנתונים לדוגמה: מקדמי המשתנים של האילוצים ושל פונקציית המטרה, והגבולות של האילוצים.
Python
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
C++
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;
};
Java
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;
}
C#
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;
}
יצירת מופע של הנתונים
הקוד הבא יוצר את מודל הנתונים.
Python
data = create_data_model()
C++
DataModel data;
Java
final DataModel data = new DataModel();
C#
DataModel data = new DataModel();
יצירת מיידי של הפותר
הקוד הבא יוצר את הפותר.
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;
}
הגדרת המשתנים
הקוד הבא מגדיר את המשתנים עבור הדוגמה בלולאה. בבעיות גדולות, קל יותר להגדיר את המשתנים בנפרד, כמו בדוגמה הקודמת.
Python
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())
C++
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();
Java
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());
C#
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());
מגדירים את האילוצים
הקוד הבא יוצר את האילוצים לדוגמה, באמצעות השיטה MakeRowConstraint (או וריאנט כלשהו, בהתאם לשפת התכנות). שני הארגומנטים הראשונים לשיטה הם הגבול התחתון והעליון של האילוץ. הארגומנט השלישי, שם האילוץ, הוא אופציונלי.
עבור כל אילוץ צריך להגדיר את המקדמים של המשתנים באמצעות השיטה SetCoefficient. השיטה מקצה את המקדם של המשתנה x[j] באילוץ i להיות הערך [i][j] של המערך constraint_coeffs.
Python
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])
C++
// 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();
Java
// 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());
C#
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());
מגדירים את היעד
הקוד הבא מגדיר את פונקציית היעד עבור הדוגמה. השיטה SetCoefficient מקצה את המקדמים הרצויים ליעד, ואילו השיטה SetMaximization מגדירה זאת כבעיה מקסימלית.
Python
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))
C++
// 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();
Java
MPObjective objective = solver.objective();
for (int j = 0; j < data.numVars; ++j) {
objective.setCoefficient(x[j], data.objCoeffs[j]);
}
objective.setMaximization();
C#
Objective objective = solver.Objective();
for (int j = 0; j < data.NumVars; ++j)
{
objective.SetCoefficient(x[j], data.ObjCoeffs[j]);
}
objective.SetMaximization();
קריאה לפותר
הקוד הבא מפעיל את הפותר.
Python
print(f"Solving with {solver.SolverVersion()}")
status = solver.Solve()
C++
const MPSolver::ResultStatus result_status = solver->Solve();
Java
final MPSolver.ResultStatus resultStatus = solver.solve();
C#
Solver.ResultStatus resultStatus = solver.Solve();
הצגת הפתרון
הקוד הבא מציג את הפתרון.
Python
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.")
C++
// 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();
}
Java
// 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.");
}
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("Optimal objective value = " + solver.Objective().Value());
for (int j = 0; j < data.NumVars; ++j)
{
Console.WriteLine("x[" + j + "] = " + x[j].SolutionValue());
}
הנה הפתרון לבעיה.
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
השלמה של תוכניות
אלה התוכניות המלאות.
Python
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()
C++
#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;
}
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;
/** 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() {}
}
C#
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");
}
}