FCFS CPU Scheduling Algorithm

FCFS Scheduling Algorithm

First Come First Serve is a CPU scheduling algorithm where CPU execute each process according there appearance.

Code

/****Scheduling algorithm for FCFS****/
#include <stdio.h>
#include >malloc.h<

int main()
{
	int n,*b,*w,i,j,h;
	float avg =0;
	
	printf("\nEnter number of jobs:");
	scanf("%d",&n);
	
	//Create an dynamic array for specified job
	b=(int *)malloc(n*sizeof(int)); //Array for holding the burst time of the jobs
	w=(int *)malloc(n*sizeof(int)); //Array for holding the waiting time of the jobs
	
	//Taking the burst time from the user
	printf("\nEnter the burst time for corresponding jobs:");
	for(i=0;i<n;i++)
	{
		printf("\nProcess %d:",i+1);
		scanf("%d",&b[i]);
	}
	
	w[0]=0;
	printf("\nProcess 1 waiting time is 0"); //First process waiting time is always 0
	for(i=1;<i++)
	{
		w[i]=b[i-1]+w[i-1]; //Calculate the waiting time of the ith process
		printf("\nProcess %d waiting time is %d",i+1,w[i]); //Print the waiting time for the corresponding process
		avg+=w[i]; //Calculate the total waiting time
	}
	
	printf("\nTotal waiting time:%f",avg); //Print the total waiting time
	printf("\nThe average waiting time:%f",avg); //Print the average waiting time

	return 0;
}

Output

Enter number of jobs:5

Enter the burst time for corresponding jobs:
Process 1:3

Process 2:2

Process 3:5

Process 4:7

Process 5:2

Process 1 waiting time is 0
Process 2 waiting time is 3
Process 3 waiting time is 5
Process 4 waiting time is 10
Process 5 waiting time is 17
Total waiting time:35.000000
The average waiting time:7.000000

Shortest Job First CPU Scheduling | Operating System

SJF Scheduling Algorithm

Shortest Job First is a CPU scheduling algorithm where the CPU execute the process first which have the shorest burst time.

Code

/***Scheduling algorithm for SJF***/
#include <stdio.h>
#include <malloc.h>


int main()
{
 int n,*b,*w,*a,i,j,h,t,tt;
 float avg =0;
 
 printf("\nEnter the numbers of jobs:");
 scanf("%d",&n);
 
 b=(int *)malloc(n*sizeof(int)); //Array for holding the burst time of the jobs
 w=(int *)malloc(n*sizeof(int)); //Array for holding the waiting time of the jobs
 a=(int *)malloc(n*sizeof(int)); //Array for holding the jobs number according their appearance
 
 //Taking the burst time from the user
 printf("\nEnter the burst time for corresponding jobs:");
 for(i=1;i<=n;i++)
 {
  printf("\nProcess %d:",i);
  scanf("%d",&b[i]);
  a[i]=i;
 }
 
 for(i=1;i<=n;i++)
 {
  for(j=i;j<=n;j++)
  {
   //Shorting the process according to their burst time in ascending order
   if(b[i]>b[j]) 
   {
    t=b[i];tt=a[i];
    b[i]=b[j];a[i]=a[j];
    b[j]=t;a[j]=tt;
   }
   
   //If both process has same burst time then short them according their appearance
   if(b[i]==b[j]) 
   {
    if(a[i]>a[j])
    {
     t=a[i];
     a[i]=a[j];
     a[j]=t;
    }
   }
  }
 }
 
 w[1]=0;
 printf("\nProcess %d waiting time is 0",a[1]); //The first process waiting time is always 0
 for(i=2;i<=n;i++)
 {
  w[i]=b[i-1]+w[i-1]; //Calculate the waiting time of the i'th process
  printf("\nProcess %d waiting time is %d",a[i],w[i]); //Print the waiting time
  avg+=w[i]; //Calculate the total waiting time
 }
 
 printf("\nTotal waiting time :%f",avg); //Print the total waiting time
 printf("\nThe average waiting time :%f",avg/n);//Print the average waiting time

 return 0;
}

Output

Enter the numbers of jobs:5

Enter the burst time for corresponding jobs:
Process 1:4

Process 2:3

Process 3:6

Process 4:2

Process 5:1

Process 5 waiting time is 0
Process 4 waiting time is 1
Process 2 waiting time is 3
Process 1 waiting time is 6
Process 3 waiting time is 10
Total waiting time :20.000000
The average waiting time :4.000000