असाइनमेंट से जुड़ी समस्या हल करना

इस सेक्शन में एक उदाहरण दिया गया है, जिसमें असाइनमेंट के सवाल को कैसे हल करें इसके लिए, हम MIP सॉल्वर और CP-SAT सॉल्वर, दोनों का इस्तेमाल करते हैं.

उदाहरण

इस उदाहरण में, पांच वर्कर (संख्या 0-4) और चार टास्क (नंबरों के साथ) हैं 0 से 3 तक). ध्यान दें कि उदाहरण में दिए गए कर्मचारी से एक और कर्मचारी है खास जानकारी.

टास्क के लिए कर्मचारियों को असाइन करने की कीमत नीचे दी गई टेबल में शामिल करें.

वर्कर	टास्क 0	कार्य 1	कार्य 2	कार्य 3
0	90	80	75	70
1	35	85	55	65
2	125	95	90	95
3	45	110	95	115
4	50	100	90	100

समस्या यह है कि हर कर्मचारी को ज़्यादा से ज़्यादा एक ही काम दिया जाए, कोई भी दो कर्मचारी काम नहीं कर रहे हैं और कुल लागत को कम से कम करते समय, एक ही टास्क करने में मदद मिलेगी. क्योंकि वहां से ज़्यादा कर्मचारी हैं काम है, तो एक कर्मचारी को काम असाइन नहीं किया जाएगा.

एमआईपी सलूशन

नीचे दिए सेक्शन में, MPSolver रैपर.

लाइब्रेरी इंपोर्ट करें

यहां दिया गया कोड ज़रूरी लाइब्रेरी इंपोर्ट करता है.

Python

from ortools.linear_solver import pywraplp

C++

#include <memory>
#include <vector>

#include "ortools/base/logging.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;

डेटा बनाना

यह कोड समस्या के लिए डेटा बनाता है.

Python

costs = [
    [90, 80, 75, 70],
    [35, 85, 55, 65],
    [125, 95, 90, 95],
    [45, 110, 95, 115],
    [50, 100, 90, 100],
]
num_workers = len(costs)
num_tasks = len(costs[0])

C++

const std::vector<std::vector<double>> costs{
    {90, 80, 75, 70},   {35, 85, 55, 65},   {125, 95, 90, 95},
    {45, 110, 95, 115}, {50, 100, 90, 100},
};
const int num_workers = costs.size();
const int num_tasks = costs[0].size();

Java

double[][] costs = {
    {90, 80, 75, 70},
    {35, 85, 55, 65},
    {125, 95, 90, 95},
    {45, 110, 95, 115},
    {50, 100, 90, 100},
};
int numWorkers = costs.length;
int numTasks = costs[0].length;

C#

int[,] costs = {
    { 90, 80, 75, 70 }, { 35, 85, 55, 65 }, { 125, 95, 90, 95 }, { 45, 110, 95, 115 }, { 50, 100, 90, 100 },
};
int numWorkers = costs.GetLength(0);
int numTasks = costs.GetLength(1);

costs कलेक्शन, लागत की टेबल से मेल खाता है कर्मचारियों को टास्क में असाइन करने के लिए किया जा सकता है.

MIP सॉल्वर के बारे में बताएं

यहां दिया गया कोड, MIP सॉल्वर के बारे में बताता है.

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#

Solver solver = Solver.CreateSolver("SCIP");
if (solver is null)
{
    return;
}

वैरिएबल बनाना

नीचे दिया गया कोड, इस सवाल के लिए बाइनरी इंटीजर वैरिएबल बनाता है.

Python

# x[i, j] is an array of 0-1 variables, which will be 1
# if worker i is assigned to task j.
x = {}
for i in range(num_workers):
    for j in range(num_tasks):
        x[i, j] = solver.IntVar(0, 1, "")

C++

// x[i][j] is an array of 0-1 variables, which will be 1
// if worker i is assigned to task j.
std::vector<std::vector<const MPVariable*>> x(
    num_workers, std::vector<const MPVariable*>(num_tasks));
for (int i = 0; i < num_workers; ++i) {
  for (int j = 0; j < num_tasks; ++j) {
    x[i][j] = solver->MakeIntVar(0, 1, "");
  }
}

Java

// x[i][j] is an array of 0-1 variables, which will be 1
// if worker i is assigned to task j.
MPVariable[][] x = new MPVariable[numWorkers][numTasks];
for (int i = 0; i < numWorkers; ++i) {
  for (int j = 0; j < numTasks; ++j) {
    x[i][j] = solver.makeIntVar(0, 1, "");
  }
}

C#

// x[i, j] is an array of 0-1 variables, which will be 1
// if worker i is assigned to task j.
Variable[,] x = new Variable[numWorkers, numTasks];
for (int i = 0; i < numWorkers; ++i)
{
    for (int j = 0; j < numTasks; ++j)
    {
        x[i, j] = solver.MakeIntVar(0, 1, $"worker_{i}_task_{j}");
    }
}

कंस्ट्रेंट बनाएं

यह कोड समस्या के लिए पाबंदियां बनाता है.

Python

# Each worker is assigned to at most 1 task.
for i in range(num_workers):
    solver.Add(solver.Sum([x[i, j] for j in range(num_tasks)]) <= 1)

# Each task is assigned to exactly one worker.
for j in range(num_tasks):
    solver.Add(solver.Sum([x[i, j] for i in range(num_workers)]) == 1)

C++

// Each worker is assigned to at most one task.
for (int i = 0; i < num_workers; ++i) {
  LinearExpr worker_sum;
  for (int j = 0; j < num_tasks; ++j) {
    worker_sum += x[i][j];
  }
  solver->MakeRowConstraint(worker_sum <= 1.0);
}
// Each task is assigned to exactly one worker.
for (int j = 0; j < num_tasks; ++j) {
  LinearExpr task_sum;
  for (int i = 0; i < num_workers; ++i) {
    task_sum += x[i][j];
  }
  solver->MakeRowConstraint(task_sum == 1.0);
}

Java

// Each worker is assigned to at most one task.
for (int i = 0; i < numWorkers; ++i) {
  MPConstraint constraint = solver.makeConstraint(0, 1, "");
  for (int j = 0; j < numTasks; ++j) {
    constraint.setCoefficient(x[i][j], 1);
  }
}
// Each task is assigned to exactly one worker.
for (int j = 0; j < numTasks; ++j) {
  MPConstraint constraint = solver.makeConstraint(1, 1, "");
  for (int i = 0; i < numWorkers; ++i) {
    constraint.setCoefficient(x[i][j], 1);
  }
}

C#

// Each worker is assigned to at most one task.
for (int i = 0; i < numWorkers; ++i)
{
    Constraint constraint = solver.MakeConstraint(0, 1, "");
    for (int j = 0; j < numTasks; ++j)
    {
        constraint.SetCoefficient(x[i, j], 1);
    }
}
// Each task is assigned to exactly one worker.
for (int j = 0; j < numTasks; ++j)
{
    Constraint constraint = solver.MakeConstraint(1, 1, "");
    for (int i = 0; i < numWorkers; ++i)
    {
        constraint.SetCoefficient(x[i, j], 1);
    }
}

मकसद फ़ंक्शन बनाना

यह कोड समस्या के लिए मकसद फ़ंक्शन बनाता है.

Python

objective_terms = []
for i in range(num_workers):
    for j in range(num_tasks):
        objective_terms.append(costs[i][j] * x[i, j])
solver.Minimize(solver.Sum(objective_terms))

C++

MPObjective* const objective = solver->MutableObjective();
for (int i = 0; i < num_workers; ++i) {
  for (int j = 0; j < num_tasks; ++j) {
    objective->SetCoefficient(x[i][j], costs[i][j]);
  }
}
objective->SetMinimization();

Java

MPObjective objective = solver.objective();
for (int i = 0; i < numWorkers; ++i) {
  for (int j = 0; j < numTasks; ++j) {
    objective.setCoefficient(x[i][j], costs[i][j]);
  }
}
objective.setMinimization();

C#

Objective objective = solver.Objective();
for (int i = 0; i < numWorkers; ++i)
{
    for (int j = 0; j < numTasks; ++j)
    {
        objective.SetCoefficient(x[i, j], costs[i, j]);
    }
}
objective.SetMinimization();

मकसद फ़ंक्शन की वैल्यू, असाइन किए गए सभी वैरिएबल की कुल लागत होती है वैल्यू 1 को सॉल्वर से हल करें.

सॉल्वर आज़माएं

यह कोड, सॉल्वर को शुरू करता है.

Python

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

C++

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

Java

MPSolver.ResultStatus resultStatus = solver.solve();

C#

Solver.ResultStatus resultStatus = solver.Solve();

समाधान को प्रिंट करें

निम्न कोड समस्या का समाधान प्रिंट करता है.

Python

if status == pywraplp.Solver.OPTIMAL or status == pywraplp.Solver.FEASIBLE:
    print(f"Total cost = {solver.Objective().Value()}\n")
    for i in range(num_workers):
        for j in range(num_tasks):
            # Test if x[i,j] is 1 (with tolerance for floating point arithmetic).
            if x[i, j].solution_value() > 0.5:
                print(f"Worker {i} assigned to task {j}." + f" Cost: {costs[i][j]}")
else:
    print("No solution found.")

C++

// Check that the problem has a feasible solution.
if (result_status != MPSolver::OPTIMAL &&
    result_status != MPSolver::FEASIBLE) {
  LOG(FATAL) << "No solution found.";
}

LOG(INFO) << "Total cost = " << objective->Value() << "\n\n";

for (int i = 0; i < num_workers; ++i) {
  for (int j = 0; j < num_tasks; ++j) {
    // Test if x[i][j] is 0 or 1 (with tolerance for floating point
    // arithmetic).
    if (x[i][j]->solution_value() > 0.5) {
      LOG(INFO) << "Worker " << i << " assigned to task " << j
                << ".  Cost = " << costs[i][j];
    }
  }
}

Java

// Check that the problem has a feasible solution.
if (resultStatus == MPSolver.ResultStatus.OPTIMAL
    || resultStatus == MPSolver.ResultStatus.FEASIBLE) {
  System.out.println("Total cost: " + objective.value() + "\n");
  for (int i = 0; i < numWorkers; ++i) {
    for (int j = 0; j < numTasks; ++j) {
      // Test if x[i][j] is 0 or 1 (with tolerance for floating point
      // arithmetic).
      if (x[i][j].solutionValue() > 0.5) {
        System.out.println(
            "Worker " + i + " assigned to task " + j + ".  Cost = " + costs[i][j]);
      }
    }
  }
} else {
  System.err.println("No solution found.");
}

C#

// Check that the problem has a feasible solution.
if (resultStatus == Solver.ResultStatus.OPTIMAL || resultStatus == Solver.ResultStatus.FEASIBLE)
{
    Console.WriteLine($"Total cost: {solver.Objective().Value()}\n");
    for (int i = 0; i < numWorkers; ++i)
    {
        for (int j = 0; j < numTasks; ++j)
        {
            // Test if x[i, j] is 0 or 1 (with tolerance for floating point
            // arithmetic).
            if (x[i, j].SolutionValue() > 0.5)
            {
                Console.WriteLine($"Worker {i} assigned to task {j}. Cost: {costs[i, j]}");
            }
        }
    }
}
else
{
    Console.WriteLine("No solution found.");
}

ये रहे प्रोग्राम का आउटपुट.

Total cost =  265.0

Worker 0 assigned to task 3.  Cost = 70
Worker 1 assigned to task 2.  Cost = 55
Worker 2 assigned to task 1.  Cost = 95
Worker 3 assigned to task 0.  Cost = 45

प्रोग्राम पूरे करें

MIP समाधान के लिए संपूर्ण प्रोग्राम यहां दिए गए हैं.

Python

from ortools.linear_solver import pywraplp


def main():
    # Data
    costs = [
        [90, 80, 75, 70],
        [35, 85, 55, 65],
        [125, 95, 90, 95],
        [45, 110, 95, 115],
        [50, 100, 90, 100],
    ]
    num_workers = len(costs)
    num_tasks = len(costs[0])

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

    if not solver:
        return

    # Variables
    # x[i, j] is an array of 0-1 variables, which will be 1
    # if worker i is assigned to task j.
    x = {}
    for i in range(num_workers):
        for j in range(num_tasks):
            x[i, j] = solver.IntVar(0, 1, "")

    # Constraints
    # Each worker is assigned to at most 1 task.
    for i in range(num_workers):
        solver.Add(solver.Sum([x[i, j] for j in range(num_tasks)]) <= 1)

    # Each task is assigned to exactly one worker.
    for j in range(num_tasks):
        solver.Add(solver.Sum([x[i, j] for i in range(num_workers)]) == 1)

    # Objective
    objective_terms = []
    for i in range(num_workers):
        for j in range(num_tasks):
            objective_terms.append(costs[i][j] * x[i, j])
    solver.Minimize(solver.Sum(objective_terms))

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

    # Print solution.
    if status == pywraplp.Solver.OPTIMAL or status == pywraplp.Solver.FEASIBLE:
        print(f"Total cost = {solver.Objective().Value()}\n")
        for i in range(num_workers):
            for j in range(num_tasks):
                # Test if x[i,j] is 1 (with tolerance for floating point arithmetic).
                if x[i, j].solution_value() > 0.5:
                    print(f"Worker {i} assigned to task {j}." + f" Cost: {costs[i][j]}")
    else:
        print("No solution found.")


if __name__ == "__main__":
    main()

C++

#include <memory>
#include <vector>

#include "ortools/base/logging.h"
#include "ortools/linear_solver/linear_solver.h"

namespace operations_research {
void AssignmentMip() {
  // Data
  const std::vector<std::vector<double>> costs{
      {90, 80, 75, 70},   {35, 85, 55, 65},   {125, 95, 90, 95},
      {45, 110, 95, 115}, {50, 100, 90, 100},
  };
  const int num_workers = costs.size();
  const int num_tasks = costs[0].size();

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

  // Variables
  // x[i][j] is an array of 0-1 variables, which will be 1
  // if worker i is assigned to task j.
  std::vector<std::vector<const MPVariable*>> x(
      num_workers, std::vector<const MPVariable*>(num_tasks));
  for (int i = 0; i < num_workers; ++i) {
    for (int j = 0; j < num_tasks; ++j) {
      x[i][j] = solver->MakeIntVar(0, 1, "");
    }
  }

  // Constraints
  // Each worker is assigned to at most one task.
  for (int i = 0; i < num_workers; ++i) {
    LinearExpr worker_sum;
    for (int j = 0; j < num_tasks; ++j) {
      worker_sum += x[i][j];
    }
    solver->MakeRowConstraint(worker_sum <= 1.0);
  }
  // Each task is assigned to exactly one worker.
  for (int j = 0; j < num_tasks; ++j) {
    LinearExpr task_sum;
    for (int i = 0; i < num_workers; ++i) {
      task_sum += x[i][j];
    }
    solver->MakeRowConstraint(task_sum == 1.0);
  }

  // Objective.
  MPObjective* const objective = solver->MutableObjective();
  for (int i = 0; i < num_workers; ++i) {
    for (int j = 0; j < num_tasks; ++j) {
      objective->SetCoefficient(x[i][j], costs[i][j]);
    }
  }
  objective->SetMinimization();

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

  // Print solution.
  // Check that the problem has a feasible solution.
  if (result_status != MPSolver::OPTIMAL &&
      result_status != MPSolver::FEASIBLE) {
    LOG(FATAL) << "No solution found.";
  }

  LOG(INFO) << "Total cost = " << objective->Value() << "\n\n";

  for (int i = 0; i < num_workers; ++i) {
    for (int j = 0; j < num_tasks; ++j) {
      // Test if x[i][j] is 0 or 1 (with tolerance for floating point
      // arithmetic).
      if (x[i][j]->solution_value() > 0.5) {
        LOG(INFO) << "Worker " << i << " assigned to task " << j
                  << ".  Cost = " << costs[i][j];
      }
    }
  }
}
}  // namespace operations_research

int main(int argc, char** argv) {
  operations_research::AssignmentMip();
  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 that solves an assignment problem. */
public class AssignmentMip {
  public static void main(String[] args) {
    Loader.loadNativeLibraries();
    // Data
    double[][] costs = {
        {90, 80, 75, 70},
        {35, 85, 55, 65},
        {125, 95, 90, 95},
        {45, 110, 95, 115},
        {50, 100, 90, 100},
    };
    int numWorkers = costs.length;
    int numTasks = costs[0].length;

    // Solver
    // 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;
    }

    // Variables
    // x[i][j] is an array of 0-1 variables, which will be 1
    // if worker i is assigned to task j.
    MPVariable[][] x = new MPVariable[numWorkers][numTasks];
    for (int i = 0; i < numWorkers; ++i) {
      for (int j = 0; j < numTasks; ++j) {
        x[i][j] = solver.makeIntVar(0, 1, "");
      }
    }

    // Constraints
    // Each worker is assigned to at most one task.
    for (int i = 0; i < numWorkers; ++i) {
      MPConstraint constraint = solver.makeConstraint(0, 1, "");
      for (int j = 0; j < numTasks; ++j) {
        constraint.setCoefficient(x[i][j], 1);
      }
    }
    // Each task is assigned to exactly one worker.
    for (int j = 0; j < numTasks; ++j) {
      MPConstraint constraint = solver.makeConstraint(1, 1, "");
      for (int i = 0; i < numWorkers; ++i) {
        constraint.setCoefficient(x[i][j], 1);
      }
    }

    // Objective
    MPObjective objective = solver.objective();
    for (int i = 0; i < numWorkers; ++i) {
      for (int j = 0; j < numTasks; ++j) {
        objective.setCoefficient(x[i][j], costs[i][j]);
      }
    }
    objective.setMinimization();

    // Solve
    MPSolver.ResultStatus resultStatus = solver.solve();

    // Print solution.
    // Check that the problem has a feasible solution.
    if (resultStatus == MPSolver.ResultStatus.OPTIMAL
        || resultStatus == MPSolver.ResultStatus.FEASIBLE) {
      System.out.println("Total cost: " + objective.value() + "\n");
      for (int i = 0; i < numWorkers; ++i) {
        for (int j = 0; j < numTasks; ++j) {
          // Test if x[i][j] is 0 or 1 (with tolerance for floating point
          // arithmetic).
          if (x[i][j].solutionValue() > 0.5) {
            System.out.println(
                "Worker " + i + " assigned to task " + j + ".  Cost = " + costs[i][j]);
          }
        }
      }
    } else {
      System.err.println("No solution found.");
    }
  }

  private AssignmentMip() {}
}

C#

using System;
using Google.OrTools.LinearSolver;

public class AssignmentMip
{
    static void Main()
    {
        // Data.
        int[,] costs = {
            { 90, 80, 75, 70 }, { 35, 85, 55, 65 }, { 125, 95, 90, 95 }, { 45, 110, 95, 115 }, { 50, 100, 90, 100 },
        };
        int numWorkers = costs.GetLength(0);
        int numTasks = costs.GetLength(1);

        // Solver.
        Solver solver = Solver.CreateSolver("SCIP");
        if (solver is null)
        {
            return;
        }

        // Variables.
        // x[i, j] is an array of 0-1 variables, which will be 1
        // if worker i is assigned to task j.
        Variable[,] x = new Variable[numWorkers, numTasks];
        for (int i = 0; i < numWorkers; ++i)
        {
            for (int j = 0; j < numTasks; ++j)
            {
                x[i, j] = solver.MakeIntVar(0, 1, $"worker_{i}_task_{j}");
            }
        }

        // Constraints
        // Each worker is assigned to at most one task.
        for (int i = 0; i < numWorkers; ++i)
        {
            Constraint constraint = solver.MakeConstraint(0, 1, "");
            for (int j = 0; j < numTasks; ++j)
            {
                constraint.SetCoefficient(x[i, j], 1);
            }
        }
        // Each task is assigned to exactly one worker.
        for (int j = 0; j < numTasks; ++j)
        {
            Constraint constraint = solver.MakeConstraint(1, 1, "");
            for (int i = 0; i < numWorkers; ++i)
            {
                constraint.SetCoefficient(x[i, j], 1);
            }
        }

        // Objective
        Objective objective = solver.Objective();
        for (int i = 0; i < numWorkers; ++i)
        {
            for (int j = 0; j < numTasks; ++j)
            {
                objective.SetCoefficient(x[i, j], costs[i, j]);
            }
        }
        objective.SetMinimization();

        // Solve
        Solver.ResultStatus resultStatus = solver.Solve();

        // Print solution.
        // Check that the problem has a feasible solution.
        if (resultStatus == Solver.ResultStatus.OPTIMAL || resultStatus == Solver.ResultStatus.FEASIBLE)
        {
            Console.WriteLine($"Total cost: {solver.Objective().Value()}\n");
            for (int i = 0; i < numWorkers; ++i)
            {
                for (int j = 0; j < numTasks; ++j)
                {
                    // Test if x[i, j] is 0 or 1 (with tolerance for floating point
                    // arithmetic).
                    if (x[i, j].SolutionValue() > 0.5)
                    {
                        Console.WriteLine($"Worker {i} assigned to task {j}. Cost: {costs[i, j]}");
                    }
                }
            }
        }
        else
        {
            Console.WriteLine("No solution found.");
        }
    }
}

सीपी एसएटी सलूशन

इन सेक्शन में CP-SAT सॉल्वर का इस्तेमाल करके, समस्या को हल करने का तरीका बताया गया है.

लाइब्रेरी इंपोर्ट करें

यहां दिया गया कोड ज़रूरी लाइब्रेरी इंपोर्ट करता है.

Python

import io

import pandas as pd

from ortools.sat.python import cp_model

C++

#include <stdlib.h>

#include <vector>

#include "ortools/base/logging.h"
#include "ortools/sat/cp_model.h"
#include "ortools/sat/cp_model.pb.h"
#include "ortools/sat/cp_model_solver.h"

Java

import com.google.ortools.Loader;
import com.google.ortools.sat.CpModel;
import com.google.ortools.sat.CpSolver;
import com.google.ortools.sat.CpSolverStatus;
import com.google.ortools.sat.LinearExpr;
import com.google.ortools.sat.LinearExprBuilder;
import com.google.ortools.sat.Literal;
import java.util.ArrayList;
import java.util.List;
import java.util.stream.IntStream;

C#

using System;
using System.Collections.Generic;
using Google.OrTools.Sat;

मॉडल का एलान करना

यहां दिया गया कोड, CP-SAT मॉडल के बारे में बताता है.

Python

model = cp_model.CpModel()

C++

CpModelBuilder cp_model;

Java

CpModel model = new CpModel();

C#

CpModel model = new CpModel();

डेटा बनाना

यह कोड समस्या के लिए डेटा सेट अप करता है.

Python

  data_str = """
worker  task  cost
    w1    t1    90
    w1    t2    80
    w1    t3    75
    w1    t4    70
    w2    t1    35
    w2    t2    85
    w2    t3    55
    w2    t4    65
    w3    t1   125
    w3    t2    95
    w3    t3    90
    w3    t4    95
    w4    t1    45
    w4    t2   110
    w4    t3    95
    w4    t4   115
    w5    t1    50
    w5    t2   110
    w5    t3    90
    w5    t4   100
"""

  data = pd.read_table(io.StringIO(data_str), sep=r"\s+")

C++

const std::vector<std::vector<int>> costs{
    {90, 80, 75, 70},   {35, 85, 55, 65},   {125, 95, 90, 95},
    {45, 110, 95, 115}, {50, 100, 90, 100},
};
const int num_workers = static_cast<int>(costs.size());
const int num_tasks = static_cast<int>(costs[0].size());

Java

int[][] costs = {
    {90, 80, 75, 70},
    {35, 85, 55, 65},
    {125, 95, 90, 95},
    {45, 110, 95, 115},
    {50, 100, 90, 100},
};
final int numWorkers = costs.length;
final int numTasks = costs[0].length;

final int[] allWorkers = IntStream.range(0, numWorkers).toArray();
final int[] allTasks = IntStream.range(0, numTasks).toArray();

C#

int[,] costs = {
    { 90, 80, 75, 70 }, { 35, 85, 55, 65 }, { 125, 95, 90, 95 }, { 45, 110, 95, 115 }, { 50, 100, 90, 100 },
};
int numWorkers = costs.GetLength(0);
int numTasks = costs.GetLength(1);

costs कलेक्शन, लागत की टेबल से मेल खाता है कर्मचारियों को टास्क में असाइन करने के लिए किया जा सकता है.

वैरिएबल बनाना

नीचे दिया गया कोड, इस सवाल के लिए बाइनरी इंटीजर वैरिएबल बनाता है.

Python

x = model.new_bool_var_series(name="x", index=data.index)

C++

// x[i][j] is an array of Boolean variables. x[i][j] is true
// if worker i is assigned to task j.
std::vector<std::vector<BoolVar>> x(num_workers,
                                    std::vector<BoolVar>(num_tasks));
for (int i = 0; i < num_workers; ++i) {
  for (int j = 0; j < num_tasks; ++j) {
    x[i][j] = cp_model.NewBoolVar();
  }
}

Java

Literal[][] x = new Literal[numWorkers][numTasks];
for (int worker : allWorkers) {
  for (int task : allTasks) {
    x[worker][task] = model.newBoolVar("x[" + worker + "," + task + "]");
  }
}

C#

BoolVar[,] x = new BoolVar[numWorkers, numTasks];
// Variables in a 1-dim array.
for (int worker = 0; worker < numWorkers; ++worker)
{
    for (int task = 0; task < numTasks; ++task)
    {
        x[worker, task] = model.NewBoolVar($"worker_{worker}_task_{task}");
    }
}

कंस्ट्रेंट बनाएं

यह कोड समस्या के लिए पाबंदियां बनाता है.

Python

# Each worker is assigned to at most one task.
for unused_name, tasks in data.groupby("worker"):
    model.add_at_most_one(x[tasks.index])

# Each task is assigned to exactly one worker.
for unused_name, workers in data.groupby("task"):
    model.add_exactly_one(x[workers.index])

C++

// Each worker is assigned to at most one task.
for (int i = 0; i < num_workers; ++i) {
  cp_model.AddAtMostOne(x[i]);
}
// Each task is assigned to exactly one worker.
for (int j = 0; j < num_tasks; ++j) {
  std::vector<BoolVar> tasks;
  for (int i = 0; i < num_workers; ++i) {
    tasks.push_back(x[i][j]);
  }
  cp_model.AddExactlyOne(tasks);
}

Java

// Each worker is assigned to at most one task.
for (int worker : allWorkers) {
  List<Literal> tasks = new ArrayList<>();
  for (int task : allTasks) {
    tasks.add(x[worker][task]);
  }
  model.addAtMostOne(tasks);
}

// Each task is assigned to exactly one worker.
for (int task : allTasks) {
  List<Literal> workers = new ArrayList<>();
  for (int worker : allWorkers) {
    workers.add(x[worker][task]);
  }
  model.addExactlyOne(workers);
}

C#

// Each worker is assigned to at most one task.
for (int worker = 0; worker < numWorkers; ++worker)
{
    List<ILiteral> tasks = new List<ILiteral>();
    for (int task = 0; task < numTasks; ++task)
    {
        tasks.Add(x[worker, task]);
    }
    model.AddAtMostOne(tasks);
}

// Each task is assigned to exactly one worker.
for (int task = 0; task < numTasks; ++task)
{
    List<ILiteral> workers = new List<ILiteral>();
    for (int worker = 0; worker < numWorkers; ++worker)
    {
        workers.Add(x[worker, task]);
    }
    model.AddExactlyOne(workers);
}

मकसद फ़ंक्शन बनाना

यह कोड समस्या के लिए मकसद फ़ंक्शन बनाता है.

Python

model.minimize(data.cost.dot(x))

C++

LinearExpr total_cost;
for (int i = 0; i < num_workers; ++i) {
  for (int j = 0; j < num_tasks; ++j) {
    total_cost += x[i][j] * costs[i][j];
  }
}
cp_model.Minimize(total_cost);

Java

LinearExprBuilder obj = LinearExpr.newBuilder();
for (int worker : allWorkers) {
  for (int task : allTasks) {
    obj.addTerm(x[worker][task], costs[worker][task]);
  }
}
model.minimize(obj);

C#

LinearExprBuilder obj = LinearExpr.NewBuilder();
for (int worker = 0; worker < numWorkers; ++worker)
{
    for (int task = 0; task < numTasks; ++task)
    {
        obj.AddTerm((IntVar)x[worker, task], costs[worker, task]);
    }
}
model.Minimize(obj);

मकसद फ़ंक्शन की वैल्यू, असाइन किए गए सभी वैरिएबल की कुल लागत होती है वैल्यू 1 को सॉल्वर से हल करें.

सॉल्वर आज़माएं

यह कोड, सॉल्वर को शुरू करता है.

Python

solver = cp_model.CpSolver()
status = solver.solve(model)

C++

const CpSolverResponse response = Solve(cp_model.Build());

Java

CpSolver solver = new CpSolver();
CpSolverStatus status = solver.solve(model);

C#

CpSolver solver = new CpSolver();
CpSolverStatus status = solver.Solve(model);
Console.WriteLine($"Solve status: {status}");

समाधान को प्रिंट करें

निम्न कोड समस्या का समाधान प्रिंट करता है.

Python

if status == cp_model.OPTIMAL or status == cp_model.FEASIBLE:
    print(f"Total cost = {solver.objective_value}\n")
    selected = data.loc[solver.boolean_values(x).loc[lambda x: x].index]
    for unused_index, row in selected.iterrows():
        print(f"{row.task} assigned to {row.worker} with a cost of {row.cost}")
elif status == cp_model.INFEASIBLE:
    print("No solution found")
else:
    print("Something is wrong, check the status and the log of the solve")

C++

if (response.status() == CpSolverStatus::INFEASIBLE) {
  LOG(FATAL) << "No solution found.";
}

LOG(INFO) << "Total cost: " << response.objective_value();
LOG(INFO);
for (int i = 0; i < num_workers; ++i) {
  for (int j = 0; j < num_tasks; ++j) {
    if (SolutionBooleanValue(response, x[i][j])) {
      LOG(INFO) << "Task " << i << " assigned to worker " << j
                << ".  Cost: " << costs[i][j];
    }
  }
}

Java

// Check that the problem has a feasible solution.
if (status == CpSolverStatus.OPTIMAL || status == CpSolverStatus.FEASIBLE) {
  System.out.println("Total cost: " + solver.objectiveValue() + "\n");
  for (int i = 0; i < numWorkers; ++i) {
    for (int j = 0; j < numTasks; ++j) {
      if (solver.booleanValue(x[i][j])) {
        System.out.println(
            "Worker " + i + " assigned to task " + j + ".  Cost: " + costs[i][j]);
      }
    }
  }
} else {
  System.err.println("No solution found.");
}

C#

// Check that the problem has a feasible solution.
if (status == CpSolverStatus.Optimal || status == CpSolverStatus.Feasible)
{
    Console.WriteLine($"Total cost: {solver.ObjectiveValue}\n");
    for (int i = 0; i < numWorkers; ++i)
    {
        for (int j = 0; j < numTasks; ++j)
        {
            if (solver.Value(x[i, j]) > 0.5)
            {
                Console.WriteLine($"Worker {i} assigned to task {j}. Cost: {costs[i, j]}");
            }
        }
    }
}
else
{
    Console.WriteLine("No solution found.");
}

ये रहे प्रोग्राम का आउटपुट.

Total cost = 265

Worker  0  assigned to task  3   Cost =  70
Worker  1  assigned to task  2   Cost =  55
Worker  2  assigned to task  1   Cost =  95
Worker  3  assigned to task  0   Cost =  45

प्रोग्राम पूरे करें

यहां CP-SAT के लिए सभी प्रोग्राम दिए गए हैं.

Python

import io

import pandas as pd

from ortools.sat.python import cp_model


def main() -> None:
    # Data
    data_str = """
  worker  task  cost
      w1    t1    90
      w1    t2    80
      w1    t3    75
      w1    t4    70
      w2    t1    35
      w2    t2    85
      w2    t3    55
      w2    t4    65
      w3    t1   125
      w3    t2    95
      w3    t3    90
      w3    t4    95
      w4    t1    45
      w4    t2   110
      w4    t3    95
      w4    t4   115
      w5    t1    50
      w5    t2   110
      w5    t3    90
      w5    t4   100
  """

    data = pd.read_table(io.StringIO(data_str), sep=r"\s+")

    # Model
    model = cp_model.CpModel()

    # Variables
    x = model.new_bool_var_series(name="x", index=data.index)

    # Constraints
    # Each worker is assigned to at most one task.
    for unused_name, tasks in data.groupby("worker"):
        model.add_at_most_one(x[tasks.index])

    # Each task is assigned to exactly one worker.
    for unused_name, workers in data.groupby("task"):
        model.add_exactly_one(x[workers.index])

    # Objective
    model.minimize(data.cost.dot(x))

    # Solve
    solver = cp_model.CpSolver()
    status = solver.solve(model)

    # Print solution.
    if status == cp_model.OPTIMAL or status == cp_model.FEASIBLE:
        print(f"Total cost = {solver.objective_value}\n")
        selected = data.loc[solver.boolean_values(x).loc[lambda x: x].index]
        for unused_index, row in selected.iterrows():
            print(f"{row.task} assigned to {row.worker} with a cost of {row.cost}")
    elif status == cp_model.INFEASIBLE:
        print("No solution found")
    else:
        print("Something is wrong, check the status and the log of the solve")


if __name__ == "__main__":
    main()

C++

#include <stdlib.h>

#include <vector>

#include "ortools/base/logging.h"
#include "ortools/sat/cp_model.h"
#include "ortools/sat/cp_model.pb.h"
#include "ortools/sat/cp_model_solver.h"

namespace operations_research {
namespace sat {

void IntegerProgrammingExample() {
  // Data
  const std::vector<std::vector<int>> costs{
      {90, 80, 75, 70},   {35, 85, 55, 65},   {125, 95, 90, 95},
      {45, 110, 95, 115}, {50, 100, 90, 100},
  };
  const int num_workers = static_cast<int>(costs.size());
  const int num_tasks = static_cast<int>(costs[0].size());

  // Model
  CpModelBuilder cp_model;

  // Variables
  // x[i][j] is an array of Boolean variables. x[i][j] is true
  // if worker i is assigned to task j.
  std::vector<std::vector<BoolVar>> x(num_workers,
                                      std::vector<BoolVar>(num_tasks));
  for (int i = 0; i < num_workers; ++i) {
    for (int j = 0; j < num_tasks; ++j) {
      x[i][j] = cp_model.NewBoolVar();
    }
  }

  // Constraints
  // Each worker is assigned to at most one task.
  for (int i = 0; i < num_workers; ++i) {
    cp_model.AddAtMostOne(x[i]);
  }
  // Each task is assigned to exactly one worker.
  for (int j = 0; j < num_tasks; ++j) {
    std::vector<BoolVar> tasks;
    for (int i = 0; i < num_workers; ++i) {
      tasks.push_back(x[i][j]);
    }
    cp_model.AddExactlyOne(tasks);
  }

  // Objective
  LinearExpr total_cost;
  for (int i = 0; i < num_workers; ++i) {
    for (int j = 0; j < num_tasks; ++j) {
      total_cost += x[i][j] * costs[i][j];
    }
  }
  cp_model.Minimize(total_cost);

  // Solve
  const CpSolverResponse response = Solve(cp_model.Build());

  // Print solution.
  if (response.status() == CpSolverStatus::INFEASIBLE) {
    LOG(FATAL) << "No solution found.";
  }

  LOG(INFO) << "Total cost: " << response.objective_value();
  LOG(INFO);
  for (int i = 0; i < num_workers; ++i) {
    for (int j = 0; j < num_tasks; ++j) {
      if (SolutionBooleanValue(response, x[i][j])) {
        LOG(INFO) << "Task " << i << " assigned to worker " << j
                  << ".  Cost: " << costs[i][j];
      }
    }
  }
}
}  // namespace sat
}  // namespace operations_research

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

Java

package com.google.ortools.sat.samples;
import com.google.ortools.Loader;
import com.google.ortools.sat.CpModel;
import com.google.ortools.sat.CpSolver;
import com.google.ortools.sat.CpSolverStatus;
import com.google.ortools.sat.LinearExpr;
import com.google.ortools.sat.LinearExprBuilder;
import com.google.ortools.sat.Literal;
import java.util.ArrayList;
import java.util.List;
import java.util.stream.IntStream;

/** Assignment problem. */
public class AssignmentSat {
  public static void main(String[] args) {
    Loader.loadNativeLibraries();
    // Data
    int[][] costs = {
        {90, 80, 75, 70},
        {35, 85, 55, 65},
        {125, 95, 90, 95},
        {45, 110, 95, 115},
        {50, 100, 90, 100},
    };
    final int numWorkers = costs.length;
    final int numTasks = costs[0].length;

    final int[] allWorkers = IntStream.range(0, numWorkers).toArray();
    final int[] allTasks = IntStream.range(0, numTasks).toArray();

    // Model
    CpModel model = new CpModel();

    // Variables
    Literal[][] x = new Literal[numWorkers][numTasks];
    for (int worker : allWorkers) {
      for (int task : allTasks) {
        x[worker][task] = model.newBoolVar("x[" + worker + "," + task + "]");
      }
    }

    // Constraints
    // Each worker is assigned to at most one task.
    for (int worker : allWorkers) {
      List<Literal> tasks = new ArrayList<>();
      for (int task : allTasks) {
        tasks.add(x[worker][task]);
      }
      model.addAtMostOne(tasks);
    }

    // Each task is assigned to exactly one worker.
    for (int task : allTasks) {
      List<Literal> workers = new ArrayList<>();
      for (int worker : allWorkers) {
        workers.add(x[worker][task]);
      }
      model.addExactlyOne(workers);
    }

    // Objective
    LinearExprBuilder obj = LinearExpr.newBuilder();
    for (int worker : allWorkers) {
      for (int task : allTasks) {
        obj.addTerm(x[worker][task], costs[worker][task]);
      }
    }
    model.minimize(obj);

    // Solve
    CpSolver solver = new CpSolver();
    CpSolverStatus status = solver.solve(model);

    // Print solution.
    // Check that the problem has a feasible solution.
    if (status == CpSolverStatus.OPTIMAL || status == CpSolverStatus.FEASIBLE) {
      System.out.println("Total cost: " + solver.objectiveValue() + "\n");
      for (int i = 0; i < numWorkers; ++i) {
        for (int j = 0; j < numTasks; ++j) {
          if (solver.booleanValue(x[i][j])) {
            System.out.println(
                "Worker " + i + " assigned to task " + j + ".  Cost: " + costs[i][j]);
          }
        }
      }
    } else {
      System.err.println("No solution found.");
    }
  }

  private AssignmentSat() {}
}

C#

using System;
using System.Collections.Generic;
using Google.OrTools.Sat;

public class AssignmentSat
{
    public static void Main(String[] args)
    {
        // Data.
        int[,] costs = {
            { 90, 80, 75, 70 }, { 35, 85, 55, 65 }, { 125, 95, 90, 95 }, { 45, 110, 95, 115 }, { 50, 100, 90, 100 },
        };
        int numWorkers = costs.GetLength(0);
        int numTasks = costs.GetLength(1);

        // Model.
        CpModel model = new CpModel();

        // Variables.
        BoolVar[,] x = new BoolVar[numWorkers, numTasks];
        // Variables in a 1-dim array.
        for (int worker = 0; worker < numWorkers; ++worker)
        {
            for (int task = 0; task < numTasks; ++task)
            {
                x[worker, task] = model.NewBoolVar($"worker_{worker}_task_{task}");
            }
        }

        // Constraints
        // Each worker is assigned to at most one task.
        for (int worker = 0; worker < numWorkers; ++worker)
        {
            List<ILiteral> tasks = new List<ILiteral>();
            for (int task = 0; task < numTasks; ++task)
            {
                tasks.Add(x[worker, task]);
            }
            model.AddAtMostOne(tasks);
        }

        // Each task is assigned to exactly one worker.
        for (int task = 0; task < numTasks; ++task)
        {
            List<ILiteral> workers = new List<ILiteral>();
            for (int worker = 0; worker < numWorkers; ++worker)
            {
                workers.Add(x[worker, task]);
            }
            model.AddExactlyOne(workers);
        }

        // Objective
        LinearExprBuilder obj = LinearExpr.NewBuilder();
        for (int worker = 0; worker < numWorkers; ++worker)
        {
            for (int task = 0; task < numTasks; ++task)
            {
                obj.AddTerm((IntVar)x[worker, task], costs[worker, task]);
            }
        }
        model.Minimize(obj);

        // Solve
        CpSolver solver = new CpSolver();
        CpSolverStatus status = solver.Solve(model);
        Console.WriteLine($"Solve status: {status}");

        // Print solution.
        // Check that the problem has a feasible solution.
        if (status == CpSolverStatus.Optimal || status == CpSolverStatus.Feasible)
        {
            Console.WriteLine($"Total cost: {solver.ObjectiveValue}\n");
            for (int i = 0; i < numWorkers; ++i)
            {
                for (int j = 0; j < numTasks; ++j)
                {
                    if (solver.Value(x[i, j]) > 0.5)
                    {
                        Console.WriteLine($"Worker {i} assigned to task {j}. Cost: {costs[i, j]}");
                    }
                }
            }
        }
        else
        {
            Console.WriteLine("No solution found.");
        }

        Console.WriteLine("Statistics");
        Console.WriteLine($"  - conflicts : {solver.NumConflicts()}");
        Console.WriteLine($"  - branches  : {solver.NumBranches()}");
        Console.WriteLine($"  - wall time : {solver.WallTime()}s");
    }
}