Banker’s Algorithm in Operating System

The banker’s algorithm is a resource allocation and deadlock avoidance algorithm that tests for safety by simulating the allocation for the 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.

Why Banker’s algorithm is named so?

Banker’s algorithm is named so because it is used in the banking system to check whether a loan can be sanctioned to a person or not. Suppose there are n number of account holders in a bank and the total sum of their money is S. If a person applies for a loan then the bank first subtracts the loan amount from the total money that the bank has and if the remaining amount is greater than S then only the loan is sanctioned. It is done because if all the account holders come to withdraw their money then the bank can easily do it.

In other words, the bank would never allocate its money in such a way that it can no longer satisfy the needs of all its customers. The bank would try to be in a safe state always.

The following Data structures are used to implement the Banker’s Algorithm:
Let ‘n’ be the number of processes in the system and ‘m’ be the number of resource types.
Available : 

• It is a 1-d array of size ‘m’ indicating the number of available resources of each type.
• Available[ j ] = k means there are ‘k’ instances of resource type R_j

Max :

• It is a 2-d array of size ‘n*m’ that defines the maximum demand of each process in a system.
• Max[ i, j ] = k means process P_i may request at most ‘k’ instances of resource type R_j.

Allocation :

• It is a 2-d array of size ‘n*m’ that defines the number of resources of each type currently allocated to each process.
• Allocation[ i, j ] = k means process P_i is currently allocated ‘k’ instances of resource type R_j

Need :

• It is a 2-d array of size ‘n*m’ that indicates the remaining resource need of each process.
• Need [ i,   j ] = k means process P_i currently needs ‘k’ instances of resource type R_j
• Need [ i,   j ] = Max [ i,   j ] – Allocation [ i,   j ]

Allocation specifies the resources currently allocated to process P_i and Need_i specifies the additional resources that process P_i may still request to complete its task.
Banker’s algorithm consists of a Safety algorithm and a Resource request algorithm.

Safety Algorithm

The algorithm for finding out whether or not a system is in a safe state can be described as follows: 

1) Let Work and Finish be vectors of length ‘m’ and ‘n’ respectively. 
Initialize: Work = Available 
Finish[i] = false; for i=1, 2, 3, 4….n
2) Find an i such that both 
a) Finish[i] = false 
b) Need_i <= Work 
if no such i exists goto step (4)
3) Work = Work + Allocation[i] 
Finish[i] = true 
goto step (2)
4) if Finish [i] = true for all i 
then the system is in a safe state 

Resource-Request Algorithm
Let Request_i be the request array for process P_i. Request_i [j] = k means process P_i wants k instances of resource type R_j. When a request for resources is made by process P_i, the following actions are taken:

1) If Request_i <= Need_i
Goto step (2) ; otherwise, raise an error condition, since the process has exceeded its maximum claim.
2) If Request_i <= Available 
Goto step (3); otherwise, P_i must wait, since the resources are not available.
3) Have the system pretend to have allocated the requested resources to process Pi by modifying the state as 
follows: 
Available = Available – Requesti 
Allocation_i = Allocation_i + Request_i
Need_i = Need_i– Request_i

Example:
Considering a system with five processes P₀ through P₄ and three resources of type A, B, C. Resource type A has 10 instances, B has 5 instances and type C has 7 instances. Suppose at time t₀ following snapshot of the system has been taken:

Question1. What will be the content of the Need matrix?
Need [i, j] = Max [i, j] – Allocation [i, j]
So, the content of Need Matrix is:

Question2.  Is the system in a safe state? If Yes, then what is the safe sequence?
Applying the Safety algorithm on the given system,

Question3. What will happen if process P₁ requests one additional instance of resource type A and two instances of resource type C?

We must determine whether this new system state is safe. To do so, we again execute Safety algorithm on the above data structures.

Hence the new system state is safe, so we can immediately grant the request for process  P_{1 .}
Code for Banker’s Algorithm

C

// Banker's Algorithm
#include <stdio.h>
int main()
{
    // P0, P1, P2, P3, P4 are the Process names here

    int n, m, i, j, k;
    n = 5; // Number of processes
    m = 3; // Number of resources
    int alloc[5][3] = { { 0, 1, 0 }, // P0    // Allocation Matrix
                        { 2, 0, 0 }, // P1
                        { 3, 0, 2 }, // P2
                        { 2, 1, 1 }, // P3
                        { 0, 0, 2 } }; // P4

    int max[5][3] = { { 7, 5, 3 }, // P0    // MAX Matrix
                      { 3, 2, 2 }, // P1
                      { 9, 0, 2 }, // P2
                      { 2, 2, 2 }, // P3
                      { 4, 3, 3 } }; // P4

    int avail[3] = { 3, 3, 2 }; // Available Resources

    int f[n], ans[n], ind = 0;
    for (k = 0; k < n; k++) {
        f[k] = 0;
    }
    int need[n][m];
    for (i = 0; i < n; i++) {
        for (j = 0; j < m; j++)
            need[i][j] = max[i][j] - alloc[i][j];
    }
    int y = 0;
    for (k = 0; k < 5; k++) {
        for (i = 0; i < n; i++) {
            if (f[i] == 0) {

                int flag = 0;
                for (j = 0; j < m; j++) {
                    if (need[i][j] > avail[j]){
                        flag = 1;
                         break;
                    }
                }

                if (flag == 0) {
                    ans[ind++] = i;
                    for (y = 0; y < m; y++)
                        avail[y] += alloc[i][y];
                    f[i] = 1;
                }
            }
        }
    }

      int flag = 1;

      for(int i=0;i<n;i++)
    {
      if(f[i]==0)
      {
        flag=0;
         printf("The following system is not safe");
        break;
      }
    }

      if(flag==1)
    {
      printf("Following is the SAFE Sequence\n");
      for (i = 0; i < n - 1; i++)
        printf(" P%d ->", ans[i]);
      printf(" P%d", ans[n - 1]);
    }


    return (0);

    // This code is contributed by Deep Baldha (CandyZack)
}

C++

// Banker's Algorithm
#include <iostream>
using namespace std;

int main()
{
    // P0, P1, P2, P3, P4 are the Process names here

  int n, m, i, j, k;
  n = 5; // Number of processes
  m = 3; // Number of resources
  int alloc[5][3] = { { 0, 1, 0 }, // P0 // Allocation Matrix
                     { 2, 0, 0 }, // P1
                     { 3, 0, 2 }, // P2
                     { 2, 1, 1 }, // P3
                     { 0, 0, 2 } }; // P4

  int max[5][3] = { { 7, 5, 3 }, // P0 // MAX Matrix
                   { 3, 2, 2 }, // P1
                   { 9, 0, 2 }, // P2
                   { 2, 2, 2 }, // P3
                   { 4, 3, 3 } }; // P4

  int avail[3] = { 3, 3, 2 }; // Available Resources

  int f[n], ans[n], ind = 0;
  for (k = 0; k < n; k++) {
    f[k] = 0;
  }
  int need[n][m];
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++)
      need[i][j] = max[i][j] - alloc[i][j];
  }
  int y = 0;
  for (k = 0; k < 5; k++) {
    for (i = 0; i < n; i++) {
      if (f[i] == 0) {

        int flag = 0;
        for (j = 0; j < m; j++) {
          if (need[i][j] > avail[j]){
            flag = 1;
            break;
          }
        }

        if (flag == 0) {
          ans[ind++] = i;
          for (y = 0; y < m; y++)
            avail[y] += alloc[i][y];
          f[i] = 1;
        }
      }
    }
  }

  int flag = 1;

  // To check if sequence is safe or not
  for(int i = 0;i<n;i++)
  {
        if(f[i]==0)
      {
        flag = 0;
        cout << "The given sequence is not safe";
        break;
      }
  }

  if(flag==1)
  {
    cout << "Following is the SAFE Sequence" << endl;
      for (i = 0; i < n - 1; i++)
        cout << " P" << ans[i] << " ->";
      cout << " P" << ans[n - 1] <<endl;
  }

    return (0);
}

Java

//Java Program for Bankers Algorithm
public class GfGBankers
{
    int n = 5; // Number of processes
    int m = 3; // Number of resources
    int need[][] = new int[n][m];
    int [][]max;
    int [][]alloc;
    int []avail;
    int safeSequence[] = new int[n];

    void initializeValues()
    {
    // P0, P1, P2, P3, P4 are the Process names here
    // Allocation Matrix
    alloc = new int[][] { { 0, 1, 0 }, //P0
                  { 2, 0, 0 }, //P1
                  { 3, 0, 2 }, //P2
                  { 2, 1, 1 }, //P3
                  { 0, 0, 2 } }; //P4

    // MAX Matrix
    max = new int[][] { { 7, 5, 3 }, //P0
             { 3, 2, 2 }, //P1
             { 9, 0, 2 }, //P2
             { 2, 2, 2 }, //P3
             { 4, 3, 3 } }; //P4

    // Available Resources
     avail = new int[] { 3, 3, 2 };
    }

    void isSafe()
    {
    int count=0;

    //visited array to find the already allocated process
    boolean visited[] = new boolean[n];
    for (int i = 0;i < n; i++)
    {
        visited[i] = false;
    }

    //work array to store the copy of available resources
    int work[] = new int[m];
    for (int i = 0;i < m; i++)
    {
        work[i] = avail[i];
    }

    while (count<n)
    {
        boolean flag = false;
        for (int i = 0;i < n; i++)
        {
            if (visited[i] == false)
             {
            int j;
            for (j = 0;j < m; j++)
            {
                if (need[i][j] > work[j])
                break;
            }
            if (j == m)
            {
               safeSequence[count++]=i;
               visited[i]=true;
               flag=true;

                for (j = 0;j < m; j++)
                {
                work[j] = work[j]+alloc[i][j];
                }
            }
             }
        }
        if (flag == false)
        {
            break;
        }
    }
    if (count < n)
    {
        System.out.println("The System is UnSafe!");
    }
    else
    {
        //System.out.println("The given System is Safe");
        System.out.println("Following is the SAFE Sequence");
                for (int i = 0;i < n; i++)
        {
            System.out.print("P" + safeSequence[i]);
            if (i != n-1)
            System.out.print(" -> ");
        }
    }
    }

    void calculateNeed()
    {
    for (int i = 0;i < n; i++)
    {
        for (int j = 0;j < m; j++)
         {
        need[i][j] = max[i][j]-alloc[i][j];
         }
    }
    }

    public static void main(String[] args)
    {
      int i, j, k;
      GfGBankers gfg = new GfGBankers();

      gfg.initializeValues();
      //Calculate the Need Matrix
      gfg.calculateNeed();

       // Check whether system is in safe state or not
      gfg.isSafe();
    }
}

Python3

# Banker's Algorithm

# Driver code:
if __name__=="__main__":

    # P0, P1, P2, P3, P4 are the Process names here
    n = 5 # Number of processes
    m = 3 # Number of resources

    # Allocation Matrix
    alloc = [[0, 1, 0 ],[ 2, 0, 0 ],
            [3, 0, 2 ],[2, 1, 1] ,[ 0, 0, 2]]

    # MAX Matrix
    max = [[7, 5, 3 ],[3, 2, 2 ],
            [ 9, 0, 2 ],[2, 2, 2],[4, 3, 3]]

    avail = [3, 3, 2] # Available Resources

    f = [0]*n
    ans = [0]*n
    ind = 0
    for k in range(n):
        f[k] = 0

    need = [[ 0 for i in range(m)]for i in range(n)]
    for i in range(n):
        for j in range(m):
            need[i][j] = max[i][j] - alloc[i][j]
    y = 0
    for k in range(5):
        for i in range(n):
            if (f[i] == 0):
                flag = 0
                for j in range(m):
                    if (need[i][j] > avail[j]):
                        flag = 1
                        break

                if (flag == 0):
                    ans[ind] = i
                    ind += 1
                    for y in range(m):
                        avail[y] += alloc[i][y]
                    f[i] = 1

    print("Following is the SAFE Sequence")

    for i in range(n - 1):
        print(" P", ans[i], " ->", sep="", end="")
    print(" P", ans[n - 1], sep="")

# This code is contributed by SHUBHAMSINGH10

C#

// C# Program for Bankers Algorithm
using System;
using System.Collections.Generic;

class GFG
{
static int n = 5; // Number of processes
static int m = 3; // Number of resources
int [,]need = new int[n, m];
int [,]max;
int [,]alloc;
int []avail;
int []safeSequence = new int[n];

void initializeValues()
{
    // P0, P1, P2, P3, P4 are the Process
    // names here Allocation Matrix
    alloc = new int[,] {{ 0, 1, 0 }, //P0
                        { 2, 0, 0 }, //P1
                        { 3, 0, 2 }, //P2
                        { 2, 1, 1 }, //P3
                        { 0, 0, 2 }};//P4

    // MAX Matrix
    max = new int[,] {{ 7, 5, 3 }, //P0
                        { 3, 2, 2 }, //P1
                      { 9, 0, 2 }, //P2
                      { 2, 2, 2 }, //P3
                      { 4, 3, 3 }};//P4

    // Available Resources
    avail = new int[] { 3, 3, 2 };
}

void isSafe()
{
    int count = 0;

    // visited array to find the
    // already allocated process
    Boolean []visited = new Boolean[n];
    for (int i = 0; i < n; i++)
    {
        visited[i] = false;
    }

    // work array to store the copy of
    // available resources
    int []work = new int[m];
    for (int i = 0; i < m; i++)
    {
        work[i] = avail[i];
    }

    while (count<n)
    {
        Boolean flag = false;
        for (int i = 0; i < n; i++)
        {
            if (visited[i] == false)
            {
                int j;
                for (j = 0; j < m; j++)
                {
                    if (need[i, j] > work[j])
                    break;
                }
                if (j == m)
                {
                    safeSequence[count++] = i;
                    visited[i] = true;
                    flag = true;
                    for (j = 0; j < m; j++)
                    {
                        work[j] = work[j] + alloc[i, j];
                    }
                }
            }
        }
        if (flag == false)
        {
            break;
        }
    }
    if (count < n)
    {
        Console.WriteLine("The System is UnSafe!");
    }
    else
    {
        //System.out.println("The given System is Safe");
        Console.WriteLine("Following is the SAFE Sequence");
        for (int i = 0; i < n; i++)
        {
            Console.Write("P" + safeSequence[i]);
            if (i != n - 1)
            Console.Write(" -> ");
        }
    }
}

void calculateNeed()
{
    for (int i = 0;i < n; i++)
    {
        for (int j = 0;j < m; j++)
        {
            need[i, j] = max[i, j] - alloc[i, j];
        }
    }
}

// Driver Code
public static void Main(String[] args)
{
    GFG gfg = new GFG();

    gfg.initializeValues();

    // Calculate the Need Matrix
    gfg.calculateNeed();

    // Check whether system is in
    // safe state or not
    gfg.isSafe();
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

  let n, m, i, j, k;
  n = 5; // Number of processes
  m = 3; // Number of resources
  let alloc = [ [ 0, 1, 0 ], // P0 // Allocation Matrix
                [ 2, 0, 0 ], // P1
                [ 3, 0, 2 ], // P2
                [ 2, 1, 1 ], // P3
                [ 0, 0, 2 ] ]; // P4

  let max = [ [ 7, 5, 3 ], // P0 // MAX Matrix
              [ 3, 2, 2 ], // P1
              [ 9, 0, 2 ], // P2
              [ 2, 2, 2 ], // P3
              [ 4, 3, 3 ] ]; // P4

  let avail = [ 3, 3, 2 ]; // Available Resources

  let f = [], ans = [], ind = 0;
  for (k = 0; k < n; k++) {
    f[k] = 0;
  }
  let need = [];
  for (i = 0; i < n; i++) {
      let need1 = [];
    for (j = 0; j < m; j++)
      need1.push(max[i][j] - alloc[i][j]);
    need.push(need1);
  }

  let y = 0;
  for (k = 0; k < 5; k++) {
    for (i = 0; i < n; i++) {
      if (f[i] == 0) {

        let flag = 0;
        for (j = 0; j < m; j++) {
          if (need[i][j] > avail[j]){
            flag = 1;
            break;
          }
        }

        if (flag == 0) {
          ans[ind++] = i;
          for (y = 0; y < m; y++)
            avail[y] += alloc[i][y];
          f[i] = 1;
        }
      }
    }
  }

  document.write("Following is the SAFE Sequence" + "<br>");
  for (i = 0; i < n - 1; i++)
    document.write(" P" + ans[i] + " ->");
  document.write( " P" + ans[n - 1] + "<br>");
</script>

Output

Following is the SAFE Sequence
 P1 -> P3 -> P4 -> P0 -> P2

• As the processes enter the system, they must predict the maximum number of resources needed which is not impractical to determine.
• In this algorithm, the number of processes remain fixed which is not possible in interactive systems.
• This algorithm requires that there should be a fixed number of resources to allocate. If a device breaks and becomes suddenly unavailable the algorithm would not work.
• Overhead cost incurred by the algorithm can be high when there are many processes and resources because it has to be invoked for every processes.