Cpu Scheduling Algorithms I

FIRST COME FIRST SERVED (FCFS)

Program:

#include<stdio.h>
struct process
{
int burst,wait;
}
p[20]={0,0};
int main()
{
int n,i,totalwait=0,totalturn=0;
printf("\nEnter The No Of Process :");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("Enter The Burst Time (in ms) For Process #%2d :",i+1);
scanf("%d",&p[i].burst);
}
printf("\nProcess \t Waiting Time TurnAround Time ");
printf("\n \t (in ms) (in ms)");
for(i=0;i<n;i++)
{
printf("\nProcess # %-12d%-15d%-15d",i+1,p[i].wait,p[i].wait+p[i].burst);
p[i+1].wait=p[i].wait+p[i].burst;
totalwait=totalwait+p[i].wait;
totalturn=totalturn+p[i].wait+p[i].burst;
}
printf("\n\nAVERAGE\n--------- ");
printf("\nWaiting Time : %f ms",totalwait/(float)n);
printf("\nTurnAround Time : %f ms\n\n",totalturn/(float)n);
return 0;
}

Output:

Enter The No Of Process :3
Enter The Burst Time (in ms) For Process # 1 :10
Enter The Burst Time (in ms) For Process # 2 :30
Enter The Burst Time (in ms) For Process # 3 :20
Process Waiting Time TurnAround Time
(in ms) (in ms)
Process # 1 0 10
Process # 2 10 40
Process # 3 40 60

AVERAGE
---------
Waiting Time : 16.666667 ms
TurnAround Time : 36.666667 ms

SHORTEST JOB FIRST(SJF)

Program:

#include<stdio.h>
struct process{
int burst,wait,no;
}p[20]={0,0};
int main()
{
int n,i,j,totalwait=0,totalturn=0;
printf("\nEnter The No Of Process :");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("Enter The Burst Time (in ms) For Process #%2d :",i+1);
scanf("%d",&p[i].burst);
p[i].no=i+1;
}
for(i=0;i<n;i++)
for(j=0;j<n-i-1;j++)
if(p[j].burst>p[j+1].burst){
p[j].burst^=p[j+1].burst^=p[j].burst^=p[j+1].burst;
p[j].no^=p[j+1].no^=p[j].no^=p[j+1].no;
}
printf("\nProcess \t Waiting Time TurnAround Time ");
for(i=0;i<n;i++){
printf("\nProcess # %-12d%-15d%-15d",p[i].no,p[i].wait,p[i].wait+p[i].burst);
p[i+1].wait=p[i].wait+p[i].burst;
totalwait=totalwait+p[i].wait;
totalturn=totalturn+p[i].wait+p[i].burst;
}
printf("\n\nAverage\n---------");
printf("\nWaiting Time : %f ms",totalwait/(float)n);
printf("\nTurnAround Time : %f ms\n\n",totalturn/(float)n);
return 0;
}

Output:

Enter The No Of Process :3
Enter The Burst Time (in ms) For Process # 1 :20
Enter The Burst Time (in ms) For Process # 2 :30
Enter The Burst Time (in ms) For Process # 3 :10
Process Waiting Time TurnAround Time
Process # 3 0 10
Process # 1 10 30
Process # 2 30 60
Average
---------
Waiting Time : 13.333333 ms
TurnAround Time : 33.333333 ms