Banker’s algorithm in operating system (OS) -Advantages – Disadvantages

What is Banker’s algorithm

Banker’s algorithm is an algorithm to avoid deadlock and to allocate resources to the processes safely. Banker’s algorithm helps the operating system to successfully share the resources among all the processes.

Let’s discuss with an example;

Examples of bankers algorithm

Example 1

bankers algorithm Operating systems Step by step — **Figure:** bankers algorithm Operating systems – Step by step

Example 2

Process	Total instances of Resource A	Total instances of Resource B	Total instances of Resource C
P0	0	1	0
P1	2	0	0
P2	3	0	2
P3	2	1	1
P4	0	0	2

Table: Showing resources already processes occupies

Note: PCB manage the memory by keeping a record of the allocated resources and free resources.

Explanation of the table

Process P0 already has 0 resource instances of A, 1 resource instance of B, and similarly 0 resource instance of C.

Process P1 already has 2 resource instances of A, 0 resource instance of B, and similarly 0 resource instance of C.

Process P2 already have 3 resource instances of A, 0 resource instance of B, and similarly 2 resource instance of C.

Process P3 already has 2 resource instances of A, 1 resource instance of B, and similarly 1 resource instance of C.

Process P4 already have 0 resource instances of A, 0 resource instance of B, and similarly 2 resource instance of C.

Process	Total instances of Resource A	Total instances of Resource B	Total instances of Resource C
P0	7	5	3
P1	3	2	2
P2	9	0	2
P3	2	2	2
P4	4	3	3

Table: Showing maximum resources required for complete execution.

Explanation of the table

Process P0 requires a maximum of 7 resource instances of A, 51 resource instances of B, and similarly 3 resource instances of C for the successful execution.
Process P1 requires a maximum of 3 resource instances of A, 2 resource instance of B, and similarly 2 resource instances of C for the successful execution.
Process P2 requires a maximum of 9 resource instances of A, 0 resource instance of B, and similarly 2 resource instances of C for the successful execution.
Process P3 requires a maximum of 2 resource instances of A, 2 resource instance of B, and similarly 2 resource instances of C for the successful execution.
Process P4 requires a maximum of 4 resource instances of A, 3 resource instance of B, and similarly 3 resource instances of C for the successful execution.

Process	Total instances of Resource A	Total instances of Resource B	Total instances of Resource C
P0	7	4	3
P1	1	2	2
P2	6	0	0
P3	0	1	1
P4	4	3	1

Table: Showing needed resources to each process for execution.

We can get the needed resources with the help of the following formula;

Needed resources = maximum resources – allocated resources

Process	Total instances of Resource A	Total instances of Resource B	Total instances of Resource C
P0	3	3	2
P1	5	3	2
P2	5	3	2
P3	7	4	3
P4	7	4	5
P0	7	5	5
P2	10	5	7

Table: Showing available free resources for each process.

Now compare available free resources of each process with its needed resources. The needed matrix should less than/equal as compared to the available matrix for process execution.

First iteration:

P0=finish[p0]=False can’t execute because if we assign free resources

to P0, and we add the free resources of P0 with

needed resources of P0, it still does not fulfill the

the requirement of maximum resources needed to P0 for

it’s execution.

P1=FINISH[P1]=TRUE can execute

P2=FINISH[P2]=False can’t execute

P3=FINSH[P3]=TRUE can execute

P4=FINISH[P4]=TRUE can execute

2nd iteration:

P0=FINISH[P0]=TRUE can execute

P2=FINISH[P2]=TRUE can execute

Example 3

Allocated Resources

	A	B	C
P1	2	0	1
P2	5	2	6
P3	3	1	5
P4	4	0	9

Maximum Resources

	A	B	C
P1	3	5	4
P2	9	2	8
P3	5	7	5
P4	8	6	10

Needed Resources

	A	B	C
P1	1	5	3
P2	4	0	2
P3	2	6	0
P4	4	6	1

Free Resources

A	B	C
3	7	2

P1= fails ( as the number of needed resources are greater than the free resources)

P2=fails ( as the number of needed resources are greater than the free resources)

P3= works and executes

After the execution of P3, it releases its allocated resources as well so that the number of free resources increased.

Free Resources

A	B	C
6	8	7

P4= works and executes

After the execution of P4, it releases its allocated resources to increased the number of free resources.

Free ResourceP1

A	B	C
10	8	16

P1= works and executes

After the execution of P1, it releases its allocated resources to increased the number of free resources.

Free Resources

A	B	C
12	8	17

P2= works and executes

After the execution of P2, it releases its allocated resources to increased the number of free resources.

Free Resources

A	B	C
15	10	23

These are the free resources when all the processes execute successfully.

Click Below to Download Implementation file

Bankers algorithm implementation Operating Systems (OS)

Advantages of Banker’s algorithm

Banker’s algorithm Avoids deadlock and it is less restrictive than deadlock prevention.

Disadvantages of Banker’s algorithm

It only works with a fixed number of resources and processes.

Banker Algorithm code in C

Here, i am sharing with you the banker algorithm code in C.

#include <iostream>
#include <stdio.h>
int  Resources_Available[5];
int Resources_Required[5][5];
int Resources_Already_Having[5][5];
int allocation[5] = {0, 0, 0, 0, 0};
int maxres[5], running[5], safe = 0;
int counter = 0, i, j, Exection, resources, processes, k = 1;
 
int main()
{
	printf("\nEnter number of processes: ");
    	scanf("%d", &processes);
 
    	for (i = 0; i < processes; i++) 
	{
        	running[i] = 1;
        	counter++;
    	}
 
    	printf("\nEnter number of resources: ");
    	scanf("%d", &resources);
 
    	printf("\nEnter Claim Resources:");
    	for (i = 0; i < resources; i++) 
	{ 
	        scanf("%d", &maxres[i]);
    	}
 
   	printf("\nEnter Allocated Resource Table:\n");
    	for (i = 0; i < processes; i++) 
	{
	        for(j = 0; j < resources; j++) 
		{
  			scanf("%d", &Resources_Already_Having[i][j]);
        	}
    	}
 
    	printf("\nEnter Maximum Claim Table:\n");
    	for (i = 0; i < processes; i++) 
	{
        	for(j = 0; j < resources; j++) 
		{
            		scanf("%d", &Resources_Required[i][j]);
        	}
    	}
 
	printf("\nThe Claim Resources is: ");
    	for (i = 0; i < resources; i++) 
	{
	        printf("\t%d", maxres[i]);
	}
 
    	printf("\nThe Allocated Resource Table:\n");
    	for (i = 0; i < processes; i++) 
	{
	        for (j = 0; j < resources; j++) 
		{
            		printf("\t%d", Resources_Already_Having[i][j]);
        	}
		printf("\n");
    	}
 
    	printf("\nThe Maximum Claim Table:\n");
    	for (i = 0; i < processes; i++) 
	{
        	for (j = 0; j < resources; j++) 
		{
		        printf("\t%d", Resources_Required[i][j]);
        	}
        	printf("\n");
    	}
 
    	for (i = 0; i < processes; i++) 
	{
        	for (j = 0; j < resources; j++) 
		{
            		allocation[j] += Resources_Already_Having[i][j];
        	}
    	}
 
    	printf("\nAllocated resources:");
    	for (i = 0; i < resources; i++) 
	{
        	printf("\t%d", allocation[i]);
    	}
 
    	for (i = 0; i < resources; i++) 
	{
	        Resources_Available[i] = maxres[i] - allocation[i];
	}
 
    	printf("\nAvailable resources:");
    	for (i = 0; i < resources; i++) 
	{
        	printf("\t%d", Resources_Available[i]);
    	}
    	printf("\n");
 
    	while (counter != 0) 
	{
        	safe = 0;
        	for (i = 0; i < processes; i++) 
		{
            		if (running[i]) 
			{
                		Exection = 1;
                		for (j = 0; j < resources; j++) 
				{
                    			if (Resources_Required[i][j] - Resources_Already_Having[i][j] > Resources_Available[j]) 
					{
                        			Exection = 0;
                        			break;
                    			}
                		}
                		if (Exection) 
				{
                    			printf("\nProcess%d is Execting\n", i + 1);
                    			running[i] = 0;
                    			counter--;
                    			safe = 1;
 
                    			for (j = 0; j < resources; j++) 
					{
                        			Resources_Available[j] += Resources_Already_Having[i][j];
                    			}
			                break;
                		}
            		}
        	}
        	if (!safe) 
		{
            		printf("\nThe processes are in unsafe state.\n");
            		break;
        	} 
		else 
		{
            		printf("\nThe process is in safe state");
            		printf("\nAvailable Resources:");
 
            		for (i = 0; i < resources; i++) 
			{
                		printf("\t%d", Resources_Available[i]);
            		}
 
		        printf("\n");
        	}
    	}
    	return 0;
}

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#include <iostream>

#include <stdio.h>

int Resources_Available[5];

int Resources_Required[5][5];

int Resources_Already_Having[5][5];

int allocation[5] = {0, 0, 0, 0, 0};

int maxres[5], running[5], safe = 0;

int counter = 0, i, j, Exection, resources, processes, k = 1;

int main()

{

printf("\nEnter number of processes: ");

scanf("%d", &processes);

for (i = 0; i < processes; i++)

{

running[i] = 1;

counter++;

}

printf("\nEnter number of resources: ");

scanf("%d", &resources);

printf("\nEnter Claim Resources:");

for (i = 0; i

< resources; i++)

{

scanf("%d", &maxres[i]);

}

printf("\nEnter Allocated Resource Table:\n");

for (i = 0; i < processes; i++)

{

for(j = 0; j < resources; j++)

{

scanf("%d", &Resources_Already_Having[i][j]);

}

printf("\nEnter Maximum Claim Table:\n");

for (i = 0; i < processes; i++)

{

for(j = 0; j < resources; j++)

{

scanf("%d", &Resources_Required[i][j]);

}

printf("\nThe Claim Resources is: ");

for (i = 0; i < resources; i++)

{

printf("\t%d", maxres[i]);

}

printf("\nThe Allocated Resource Table:\n");

for (i = 0; i < processes; i++)

{

for (j = 0; j <

resources; j++)

{

printf("\t%d", Resources_Already_Having[i][j]);

}

printf("\n");

}

printf("\nThe Maximum Claim Table:\n");

for (i = 0; i < processes; i++)

{

for (j = 0; j < resources; j++)

{

printf("\t%d", Resources_Required[i][j]);

}

printf("\n");

}

for (i = 0; i < processes; i++)

{

for (j = 0; j < resources; j++)

{

allocation[j] += Resources_Already_Having[i][j];

}

printf("\nAllocated resources:");

for (i = 0; i < resources; i++)

{

printf("\t%d", allocation[i]);

}

for (i = 0; i < resources; i++)

{

Resources_Available[i] = maxres[i] - allocation[i];

}

printf("\nAvailable resources:");

for (i = 0; i < resources; i++)

{

printf("\t%d", Resources_Available[i]);

}

printf("\n");

while (counter != 0)

{

safe = 0;

for (i = 0; i < processes; i++)

{

if (running[i])

{

Exection = 1;

for (j = 0; j < resources; j++)

{

if (Resources_Required[i][j] - Resources_Already_Having[i][j] > Resources_Available[j])

{

Exection = 0;

break;

}

if (Exection)

{

printf("\nProcess%d is Execting\n", i + 1);

running[i] = 0;

counter--;

safe = 1;

for (j = 0; j < resources; j++)

{

Resources_Available[j] += Resources_Already_Having[i][j];

}

break;

}

if (!safe)

{

printf("\nThe processes are in unsafe state.\n");

break;

}

else

{

printf("\nThe process is in safe state");

printf("\nAvailable Resources:");

for (i = 0; i < resources; i++)

{

printf("\t%d", Resources_Available[i]);

}

printf("\n");

}

return 0;

}

Output

Frequently Asked Questions (FAQ)

Who developed the banker’s algorithm?
Edsger Dijkstra developed the banker’s algorithm.

Excercise

Answer the following questions using the bankers algorithm.

System is in safe state or not?
How you will apply bankers algorithm on the given data.

	Allocation	Max	Available
	A B C D	A B C D	A B C D
P1	4111	4414	7741
P1	7141	5454
P4	4117	4716
P7	1714	1444
P4	1474	7665

Banker’s algorithm in operating system (OS) -Advantages – Disadvantages
By: Prof. Dr. Fazal Rehman | Last updated: March 3, 2022

What is Banker’s algorithm

Examples of bankers algorithm

Advantages of Banker’s algorithm

Disadvantages of Banker’s algorithm

Banker Algorithm code in C

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