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Banker’s Algorithm in Operating System

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Prerequisite – Resource Allocation Graph (RAG), Banker’s Algorithm, Program for Banker’s Algorithm 
Banker’s Algorithm is a resource allocation and deadlock avoidance algorithm. This algorithm test for safety simulating the allocation for predetermined maximum possible amounts of all resources, then makes an “s-state” check to test for possible activities, before deciding whether allocation should be allowed to continue.

In simple terms, it checks if allocation of any resource will lead to deadlock or not, OR if is it safe to allocate a resource to a process and if not then resource is not allocated to that process. Determining a safe sequence(even if there is only 1) will assure that system will not go into deadlock.

Banker’s algorithm is generally used to find if a safe sequence exist or not. But here we will determine the total number of safe sequences and print all safe sequences.

How does the Banker’s Algorithm work?
The Banker’s Algorithm works by maintaining a matrix of resources and processes. Each row represents a process, and each column represents a resource. The matrix contains information about the current state of the system, including the maximum number of resources each process needs, the number of resources currently allocated to each process, and the number of resources available in the system.

: What is the purpose of the Banker’s Algorithm?
A: The purpose of the Banker’s Algorithm is to prevent deadlock, a situation where two or more processes are blocked, waiting for each other to release resources they need in order to proceed. Deadlock can cause a system to become unresponsive and may require a reboot to recover.

The data structure used are: 

  • Available vector
  • Max Matrix
  • Allocation Matrix
  • Need Matrix

Example:
Input: 


 

Output: Safe sequences are:
P2--> P4--> P1--> P3
P2--> P4--> P3--> P1
P4--> P2--> P1--> P3
P4--> P2--> P3--> P1

There are total 4 safe-sequences 


Explanation: 

Total resources are R1 = 10, R2 = 5, R3 = 7 and allocated resources are R1 = (0+2+3+2 =) 7, R2 = (1+0+0+1 =) 2, R3 = (0+0+2+1 =) 3. Therefore, remaining resources are R1 = (10 – 7 =) 3, R2 = (5 – 2 =) 3, R3 = (7 – 3 =) 4. 
Remaining available = Total resources – allocated resources 
and 
Remaining need = max – allocated
 

So, we can start from either P2 or P4. We can not satisfy remaining need from available resources of either P1 or P3 in first or second attempt step of Banker’s algorithm. There are only four possible safe sequences. 

These are : 
            P2--> P4--> P1--> P3 
            P2--> P4--> P3--> P1 
            P4--> P2--> P1--> P3 
            P4--> P2--> P3--> P1

Implementation:

C++

// C++ Program to Print all possible safe sequences using banker's algorithm
#include <iostream>
#include <string.h>
#include <vector>
// total number of process
#define P 4
// total number of resources
#define R 3
 
// total safe-sequences
int total = 0;
 
using namespace std;
 
// function to check if process
// can be allocated or not
bool is_available(int process_id, int allocated[][R],
                  int max[][R], int need[][R], int available[])
{
 
    bool flag = true;
 
    // check if all the available resources
    // are less greater than need of process
    for (int i = 0; i < R; i++) {
 
        if (need[process_id][i] > available[i])
            flag = false;
    }
 
    return flag;
}
 
// Print all the safe-sequences
void safe_sequence(bool marked[], int allocated[][R], int max[][R],
                   int need[][R], int available[], vector<int> safe)
{
 
    for (int i = 0; i < P; i++) {
 
        // check if it is not marked
        // already and can be allocated
        if (!marked[i] && is_available(i, allocated, max, need, available)) {
 
            // mark the process
            marked[i] = true;
 
            // increase the available
            // by deallocating from process i
            for (int j = 0; j < R; j++)
                available[j] += allocated[i][j];
 
            safe.push_back(i);
            // find safe sequence by taking process i
            safe_sequence(marked, allocated, max, need, available, safe);
            safe.pop_back();
 
            // unmark the process
            marked[i] = false;
 
            // decrease the available
            for (int j = 0; j < R; j++)
                available[j] -= allocated[i][j];
        }
    }
 
    // if a safe-sequence is found, display it
    if (safe.size() == P) {
 
        total++;
        for (int i = 0; i < P; i++) {
 
            cout << "P" << safe[i] + 1;
            if (i != (P - 1))
                cout << "--> ";
        }
 
        cout << endl;
    }
}
 
// Driver Code
int main()
{
 
    // allocated matrix of size P*R
    int allocated[P][R] = { { 0, 1, 0 },
                            { 2, 0, 0 },
                            { 3, 0, 2 },
                            { 2, 1, 1 } };
 
    // max matrix of size P*R
    int max[P][R] = { { 7, 5, 3 },
                      { 3, 2, 2 },
                      { 9, 0, 2 },
                      { 2, 2, 2 } };
 
    // Initial total resources
    int resources[R] = { 10, 5, 7 };
 
    // available vector of size R
    int available[R];
 
    for (int i = 0; i < R; i++) {
 
        int sum = 0;
        for (int j = 0; j < P; j++)
            sum += allocated[j][i];
 
        available[i] = resources[i] - sum;
    }
 
    // safe vector for displaying a safe-sequence
    vector<int> safe;
 
    // marked of size P for marking allocated process
    bool marked[P];
    memset(marked, false, sizeof(marked));
 
    // need matrix of size P*R
    int need[P][R];
    for (int i = 0; i < P; i++)
        for (int j = 0; j < R; j++)
            need[i][j] = max[i][j] - allocated[i][j];
 
    cout << "Safe sequences are:" << endl;
    safe_sequence(marked, allocated, max, need, available, safe);
 
    cout << "\nThere are total " << total << " safe-sequences" << endl;
    return 0;
}

                    

Java

// Java Program to Print all possible safe
// sequences using banker's algorithm
import java.util.*;
 
public class GFG
{
 
    // total number of process
    static int P = 4;
     
    // total number of resources
    static int R = 3;
     
    // total safe-sequences
    static int total = 0;
 
    // function to check if process
    // can be allocated or not
    static boolean is_available(int process_id, int allocated[][],
                    int max[][], int need[][], int available[])
    {
 
        boolean flag = true;
 
        // check if all the available resources
        // are less greater than need of process
        for (int i = 0; i < R; i++)
        {
 
            if (need[process_id][i] > available[i])
            {
                flag = false;
            }
        }
 
        return flag;
    }
 
    // Print all the safe-sequences
    static void safe_sequence(boolean marked[], int allocated[][], int max[][],
                        int need[][], int available[], Vector<Integer> safe)
    {
 
        for (int i = 0; i < P; i++)
        {
 
            // check if it is not marked
            // already and can be allocated
            if (!marked[i] && is_available(i, allocated, max, need, available))
            {
 
                // mark the process
                marked[i] = true;
 
                // increase the available
                // by deallocating from process i
                for (int j = 0; j < R; j++)
                {
                    available[j] += allocated[i][j];
                }
 
                safe.add(i);
                 
                // find safe sequence by taking process i
                safe_sequence(marked, allocated, max, need, available, safe);
                safe.removeElementAt(safe.size() - 1);
 
                // unmark the process
                marked[i] = false;
 
                // decrease the available
                for (int j = 0; j < R; j++)
                {
                    available[j] -= allocated[i][j];
                }
            }
        }
 
        // if a safe-sequence is found, display it
        if (safe.size() == P)
        {
 
            total++;
            for (int i = 0; i < P; i++)
            {
 
                System.out.print("P" + (safe.get(i) + 1));
                if (i != (P - 1))
                {
                    System.out.print("--> ");
                }
            }
 
            System.out.println("");;
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
         
        // allocated matrix of size P*R
        int allocated[][] = {{0, 1, 0},
        {2, 0, 0},
        {3, 0, 2},
        {2, 1, 1}};
 
        // max matrix of size P*R
        int max[][] = {{7, 5, 3},
        {3, 2, 2},
        {9, 0, 2},
        {2, 2, 2}};
 
        // Initial total resources
        int resources[] = {10, 5, 7};
 
        // available vector of size R
        int[] available = new int[R];
 
        for (int i = 0; i < R; i++)
        {
 
            int sum = 0;
            for (int j = 0; j < P; j++)
            {
                sum += allocated[j][i];
            }
 
            available[i] = resources[i] - sum;
        }
 
        // safe vector for displaying a safe-sequence
        Vector<Integer> safe = new Vector<Integer>();
 
        // marked of size P for marking allocated process
        boolean[] marked = new boolean[P];
 
        // need matrix of size P*R
        int[][] need = new int[P][R];
        for (int i = 0; i < P; i++)
        {
            for (int j = 0; j < R; j++)
            {
                need[i][j] = max[i][j] - allocated[i][j];
            }
        }
 
        System.out.println("Safe sequences are:");
        safe_sequence(marked, allocated, max, need, available, safe);
 
        System.out.println("\nThere are total " + total + " safe-sequences");
    }
}
 
/* This code contributed by PrinciRaj1992 */

                    

Python3

# Python3 program to print all
# possible safe sequences
# using banker's algorithm
 
# Total number of process
P = 4
 
# Total number of resources
R = 3
  
# Total safe-sequences
total = 0
  
# Function to check if process
# can be allocated or not
def is_available(process_id, allocated,
                 max, need, available):
                      
    flag = True
  
    # Check if all the available resources
    # are less greater than need of process
    for i in range(R):
        if (need[process_id][i] > available[i]):
            flag = False
  
    return flag
 
# Print all the safe-sequences
def safe_sequence(marked, allocated,
                  max, need, available, safe):
     
    global total, P, R
     
    for i in range(P):
         
        # Check if it is not marked
        # already and can be allocated
        if (not marked[i] and
            is_available(i, allocated, max,
                         need, available)):
                              
            # mark the process
            marked[i] = True
  
            # Increase the available
            # by deallocating from process i
            for j in range(R):
                available[j] += allocated[i][j]
  
            safe.append(i)
             
            # Find safe sequence by taking process i
            safe_sequence(marked, allocated, max,
                          need, available, safe)
            safe.pop()
  
            # unmark the process
            marked[i] = False
  
            # Decrease the available
            for j in range(R):
                available[j] -= allocated[i][j]
         
    # If a safe-sequence is found, display it
    if (len(safe) == P):
        total += 1
         
        for i in range(P):
            print("P" + str(safe[i] + 1), end = '')
            
            if (i != (P - 1)):
                print("--> ", end = '')
             
        print()
  
# Driver code       
if __name__=="__main__":
     
    # Allocated matrix of size P*R
    allocated = [ [ 0, 1, 0 ],
                  [ 2, 0, 0 ],
                  [ 3, 0, 2 ],
                  [ 2, 1, 1 ]]
  
    # max matrix of size P*R
    max = [ [ 7, 5, 3 ],
            [ 3, 2, 2 ],
            [ 9, 0, 2 ],
            [ 2, 2, 2 ] ]
  
    # Initial total resources
    resources = [ 10, 5, 7 ]
  
    # Available vector of size R
    available = [0 for i in range(R)]
     
    for i in range(R):
        sum = 0
         
        for j in range(P):
            sum += allocated[j][i]
  
        available[i] = resources[i] - sum
     
  
    # Safe vector for displaying a
    # safe-sequence
    safe = []
  
    # Marked of size P for marking
    # allocated process
    marked = [False for i in range(P)]
  
    # Need matrix of size P*R
    need = [[0 for j in range(R)]
               for i in range(P)]
     
    for i in range(P):
        for j in range(R):
            need[i][j] = (max[i][j] -
                    allocated[i][j])
     
    print("Safe sequences are:")
     
    safe_sequence(marked, allocated, max,
                  need, available, safe)
     
    print("\nThere are total " + str(total) +
          " safe-sequences")
 
# This code is contributed by rutvik_56

                    

C#

// C# Program to Print all possible safe
// sequences using banker's algorithm
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // total number of process
    static int P = 4;
     
    // total number of resources
    static int R = 3;
     
    // total safe-sequences
    static int total = 0;
 
    // function to check if process
    // can be allocated or not
    static Boolean is_available(int process_id,
                                int [,]allocated,
                                int [,]max, int [,]need,
                                int []available)
    {
        Boolean flag = true;
 
        // check if all the available resources
        // are less greater than need of process
        for (int i = 0; i < R; i++)
        {
 
            if (need[process_id, i] > available[i])
            {
                flag = false;
            }
        }
 
        return flag;
    }
 
    // Print all the safe-sequences
    static void safe_sequence(Boolean []marked,
                              int [,]allocated,
                              int [,]max, int [,]need,
                              int []available,
                              List<int> safe)
    {
        for (int i = 0; i < P; i++)
        {
 
            // check if it is not marked
            // already and can be allocated
            if (!marked[i] &&
                is_available(i, allocated, max,
                               need, available))
            {
 
                // mark the process
                marked[i] = true;
 
                // increase the available
                // by deallocating from process i
                for (int j = 0; j < R; j++)
                {
                    available[j] += allocated[i, j];
                }
 
                safe.Add(i);
                 
                // find safe sequence by taking process i
                safe_sequence(marked, allocated, max,
                               need, available, safe);
                safe.RemoveAt(safe.Count - 1);
 
                // unmark the process
                marked[i] = false;
 
                // decrease the available
                for (int j = 0; j < R; j++)
                {
                    available[j] -= allocated[i, j];
                }
            }
        }
 
        // if a safe-sequence is found,
        // display it
        if (safe.Count == P)
        {
            total++;
            for (int i = 0; i < P; i++)
            {
                Console.Write("P" + (safe[i] + 1));
                if (i != (P - 1))
                {
                    Console.Write("--> ");
                }
            }
            Console.WriteLine("");;
        }
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
         
        // allocated matrix of size P*R
        int [,]allocated = {{0, 1, 0},
                            {2, 0, 0},
                            {3, 0, 2},
                            {2, 1, 1}};
 
        // max matrix of size P*R
        int [,]max = {{7, 5, 3},
                      {3, 2, 2},
                      {9, 0, 2},
                      {2, 2, 2}};
 
        // Initial total resources
        int []resources = {10, 5, 7};
 
        // available vector of size R
        int[] available = new int[R];
 
        for (int i = 0; i < R; i++)
        {
            int sum = 0;
            for (int j = 0; j < P; j++)
            {
                sum += allocated[j,i];
            }
 
            available[i] = resources[i] - sum;
        }
 
        // safe vector for displaying a safe-sequence
        List<int> safe = new List<int>();
 
        // marked of size P for marking
        // allocated process
        Boolean[] marked = new Boolean[P];
 
        // need matrix of size P*R
        int[,] need = new int[P,  R];
        for (int i = 0; i < P; i++)
        {
            for (int j = 0; j < R; j++)
            {
                need[i, j] = max[i, j] - allocated[i, j];
            }
        }
 
        Console.WriteLine("Safe sequences are:");
        safe_sequence(marked, allocated, max,
                       need, available, safe);
 
        Console.WriteLine("\nThere are total " +
                            total + " safe-sequences");
    }
}
 
// This code is contributed by Rajput-Ji

                    

Javascript

// total number of process
const P = 4;
// total number of resources
const R = 3;
 
// total safe-sequences
let total = 0;
 
// function to check if process
// can be allocated or not
function is_available(process_id, allocated, max, need, available) {
  let flag = true;
 
  // check if all the available resources
  // are less greater than need of process
  for (let i = 0; i < R; i++) {
    if (need[process_id][i] > available[i]) {
      flag = false;
      break;
    }
  }
 
  return flag;
}
 
// Print all the safe-sequences
function safe_sequence(marked, allocated, max, need, available, safe) {
  for (let i = 0; i < P; i++) {
 
    // check if it is not marked
    // already and can be allocated
    if (!marked[i] && is_available(i, allocated, max, need, available)) {
 
      // mark the process
      marked[i] = true;
 
      // increase the available
      // by deallocating from process i
      for (let j = 0; j < R; j++)
        available[j] += allocated[i][j];
 
      safe.push(i);
      // find safe sequence by taking process i
      safe_sequence(marked, allocated, max, need, available, safe);
      safe.pop();
 
      // unmark the process
      marked[i] = false;
 
      // decrease the available
      for (let j = 0; j < R; j++)
        available[j] -= allocated[i][j];
    }
  }
 
  // if a safe-sequence is found, display it
  if (safe.length === P) {
    total++;
    let result = "";
    for (let i = 0; i < P; i++) {
      result += "P" + (safe[i] + 1);
      if (i !== P - 1) {
        result += "--> ";
      }
    }
    console.log(result);
  }
}
 
// allocated matrix of size P*R
const allocated = [
  [0, 1, 0],
  [2, 0, 0],
  [3, 0, 2],
  [2, 1, 1]
];
 
// max matrix of size P*R
const max = [
  [7, 5, 3],
  [3, 2, 2],
  [9, 0, 2],
  [2, 2, 2]
];
 
// Initial total resources
const resources = [10, 5, 7];
 
// available vector of size R
const available = [];
 
for (let i = 0; i < R; i++) {
  let sum = 0;
  for (let j = 0; j < P; j++) {
    sum += allocated[j][i];
  }
  available.push(resources[i] - sum);
}
 
// safe vector for displaying a safe-sequence
const safe = [];
 
// marked of size P for marking allocated process
let marked = new Array(P).fill(false);
 
// need matrix of size P*R
let need = new Array(P);
for (let i = 0; i < P; i++) {
need[i] = new Array(R);
for (let j = 0; j < R; j++) {
need[i][j] = max[i][j] - allocated[i][j];
}
}
 
console.log("Safe sequences are:");
 
safe_sequence(marked, allocated, max, need, available, []);
 
console.log(`\nThere are total ${total} safe-sequences`);
 
// This code is contributed by Prince Kumar

                    

Output
Safe sequences are:
P2--> P4--> P1--> P3
P2--> P4--> P3--> P1
P4--> P2--> P1--> P3
P4--> P2--> P3--> P1

There are total 4 safe-sequences

Time complexity: O(P*R)

Auxiliary Space: O(P*R)

FAQ:

Q1: What are the advantages of using the Banker’s Algorithm?
Employing the Banker’s Algorithm has various benefits such as averting a state of deadlock and enabling every process to finish its execution seamlessly. This could aid in boosting the overall stability and dependability of the system.

Q2: What are the limitations of the Banker’s Algorithm?
The Banker’s Algorithm has certain limitations as it assumes resources to be fixed in number that may not be feasible in several circumstances. Additionally, the algorithm supposes that a process’s maximum resource requirements are known beforehand, which may not hold true every time.

Q3: What are some real-world examples of where the Banker’s Algorithm is used?
The Banker’s Algorithm finds widespread usage across operating systems, resource allocation systems, manufacturing control systems, and airline reservation systems.



Last Updated : 05 May, 2023
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