IMPLEMENTATION OF SJF SCHEDULING

                  

AIM:

To write a C program to perform SJF scheduling in Linux.

ALGORITHM:

1.      Stop the program.

2.      Read the number of request

3.      Read the servicing time for each process

4.      Sort the process by servicing time(i.e. the job having least burst time is executed first)

5.      Calculate the waiting time for all the process.

6.      Calculate the average waiting time.

7.      Draw the gantt chart and print the waiting time, average waiting time, turnaround time and average turnaround time

8.      Stop the process

PROGRAM:

#include<stdio.h>

int  main()

{

int p[5],b[5],w[5],f[5],n,tb,t[5],i,j,tw,tt,temp,wt,c[10];

float aw,at;

printf("\n Enter the no. of process:");

scanf("%d",&n);

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

{

printf("Enter the burst time:");

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

p[i]=i+1;

}

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

for(j=i+1;j<n;j++)

if(b[i]>b[j])

{

temp=b[i];

b[i]=b[j];

b[j]=temp;

temp=p[i];

p[i]=p[j];

p[j]=temp;

}

tw=w[0]=tt=t[0]=tb=0;

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

{

w[i]=tb;

tb=tb+b[i];

t[i]=b[i]+w[i];

tw=tw+w[i];

tt=tt+t[i];

}

aw=(float)tw/n;

at=(float)tt/n;

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

c[i]=w[i];

c[n]=t[n-1];

printf("\n Process       Burst time       Waiting time Turnaround time\n");

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

printf("\np%d\t\t%d\t\t%d\t\t%d\n",p[i],b[i],w[i],t[i]);

printf("\nGantt chart\n");

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

for(j=i+1;j<=n;j++)

if(c[i]>c[j])

{

temp=c[i];

c[i]=c[j];

c[j]=temp;

}

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

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

printf("\n");

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

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

printf("\n The total waiting time=%d",tw);

printf("\n Total turnaround time is=%d",tt);

printf("\n Average waiting time=%f",aw);

printf("\n Average turnaround time=%f",at);

printf("\n");

}

}

OUTPUT:

 [root@localhost ~]# ./p5

Enter the no. of process:4

Enter the burst time:3

Enter the burst time:5

Enter the burst time:6

Enter the burst time:1

 Process       Burst time       Waiting time      Turnaround time

P4                     1                        0                               1

P1                     3                        1                               4

P2                    5                         4                               9

P3                    6                         9                              15

Gantt chart

                                        P4      P1        P2      P3

                                        0        1        4       9       15

 The total waiting time=14

 Total turnaround time is=29

 Average waiting time=3.500000

 Average turnaround time=7.250000