Shortest Job First (SJF)

  • Best approach to minimize waiting time.

  • Impossible to implement

  • Processer should know in advance how much time process will take.



Example:

Process Arrival TimeExecute TimeService Time
P0 0 5 0
P1 1 3 3
P2 2 8 8
P3 3 6 16


P1

P0

P3

P2

0 3 8 16 22


Wait time of each process is following

ProcessWait Time : Service Time - Arrival Time
P03 - 0 = 3
P10 - 0 = 0
P216 - 2 = 14
P38 - 3 = 5

Average Wait Time: (3+0+14+5) / 4 = 5.50





C code for SJF


#include<stdio.h>
#include<conio.h>

struct process
{
 int arr_time;
 int burst_time;
 int no;
 int rem_time;
 int priority;
};


struct process read(int i)
{
 struct process p;
 printf("\n\n The process no.:%d.\n",i);
 p.no=i;
 printf("Enter the arrival time:");
 scanf("%d",&p.arr_time);
 printf("Enter the burst time:");
 scanf("%d",&p.burst_time);
 p.rem_time=p.burst_time;
 return p;
}



struct process readp(int i)
{
 struct process p;
 printf("\n\n The process no.:%d.\n",i);
 p.no=i;
 printf("Enter the arrival time:");
 scanf("%d",&p.arr_time);
 printf("Enter the burst time:");
 scanf("%d",&p.burst_time);
 p.rem_time=p.burst_time;
 printf("Enter the priority:");
 scanf("%d",&p.priority);
 return p;
}



void swap(struct process *i, struct process *j)
{
 struct process *t;
 i=t;
 i=j;
 j=t;
}



//SHORTEST JOB FIRST SCHEDULING ALGO.

int main()
{
 int  n;        //To hold the no. of processes.
 struct process p[10],tmp;     //To hold the details of the processes.
 int i,j;
 int ready[10];       //List of ready processes  index
 int running;       //Running process index
 int t;        //Time variable
 int last,min;
 int time;
 printf("Enter the number if processes you want to enter:");
 scanf("%d",&n);


 for(i=0;i<n;i++)
  p[i]=read(i);      //Read the details of the processes

 t=0;
 last=-1;
 min=0;
 time=0;

 do
 {
  last=-1;
  for(i=0;i<n;i++)
  {
   if(p[i].arr_time<=time && p[i].rem_time>0)
   {
    ready[++last]=i;   //update the ready queue
    
   }

  }

  min = ready[0];
  if(last<0)
   continue;
  for(i=0;i<=last;i++)
  {
   if(p[ready[i]].rem_time<p[min].rem_time)
    min=ready[i];
  }
  running = min;      //Schedule a process and print the process that was scheduled
  printf("\n\nTime:%d to time: %d Running Process: %d",time,time+p[running].rem_time,running);
  
  time = time + p[running].rem_time;
  p[running].rem_time=0;     //Remove the process from the ready queue
  

   
 }while(last>=0);
 printf("\n"); 
 getch();
 return 0;
}