Implementation of Shortest Job First Scheduling Algorithm SJF

Source code SJF Programming Algorithm
#include<stdio.h>
main()
{
int n,j,temp,temp1,temp2,
pr[10],b[10],t[10],w[10],p[10],i;
float att=0;awt=0;
for(i=0;i<10;i++)
{
b[i]=0;w[i]=0;
}
printf(“Enter the number of process:”);
scanf(“%d”,&n);
printf(“\n Enter the burst time”);
for(i=0;i<n;i++)
{
scanf(“%d”,&b[i]);
p[i]=i;
}
for(i=0;i<n;i++)
{
for(j=1;j<n;j++)
{
if(b[i]>b[j]
{
temp=b[i];
temp1=p[i];
b[i]=b[j];
p[i]=p[j];
b[j]=temp;
p[j]=temp1;
}
}
w[0]=0;
for(i=0;i<n;i++)
w[i+1]=w[i]+b[i];
for(i=0;i<n;i++)
{
t[i]=w[i]+b[i];
awt=aw+w[i]:
att=att+t[i];
}
awt=awt/n;
att=att/n;
printf(“\n\t Process \tWaitimgtime \t Turnaroundtime\n);
for(i=0;i<n;i++)
printf(“\t p[%d] \t %d\t\t %d\n”,p[i].w[i],t[i]);
printf(“\n The average waiting time is %t \n”,awt);
printf(“The average turnaround time is %f\n”,att);
return 1;
}

OUTPUT CS1252 Operating Systems lap
[redhat35@localhost Rhel5]$ cc sjf.c
[redhat35@localhost Rhel5]$ ./a.out
Enter the number of process:5
Enter the burst time:5
4
3
2
1

Process Waiting time Turnaround time
P[4] 0 1
P[0] 1 6
P[1] 6 10
P[2] 10 13
P[3] 13 15
The average waiting time is 6.000000
The average turnaround time is 9.000000