C program for shortest job first scheduling algorithm


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IMPLEMENTATION OF SHORTEST JOB FIST SCHEDULING ALGORITHM

AIM

To implement the c program for shortest job first scheduling algorithm



ALGORITHM


1. Start the process

2. Declare the array size

3. Get the number of elements to be inserted

4. Select the process which have shortest burst will execute first

5. If two process have same burst length then FCFS scheduling algorithm used

6. Make the average waiting the length of next process

7.  Start with the first process from it’s selection as above and  let other process to be in
     queue

6. Calculate the total number of burst time

7.  Display the values

8. Stop the process


PROGRAM:

#include<stdio.h>
int 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("enter the burst times");

for(i=0;i<n;i++)
            {
                        scanf("%d",&b[i]);
                        p[i]=i;
            }
for(i=0;i<n;i++)
            {
                        for(j=i;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=awt+w[i];
            att=att+t[i];
        }
awt=awt/n;
att=att/n;
printf("\n\t process \t waiting time \t turn around time \n");
for(i=0;i<n;i++)
printf("\t p[%d] \t %d \t\t %d \n",p[i],w[i],t[i]);
printf("the average waitingtimeis %f\n",awt);
printf("the average turn around time is %f\n",att);
return 1;
}


OUTPUT:
enter the number of process 5
enter the burst times
2  4  5  6  8

         process         waiting time    turn around time
         p[0]                     0                                   2
         p[1]                     2                                   6
         p[2]                     6                                   11
         p[3]                     11                                  17
         p[4]                     17                                  25
the average waitingtime is 7.200000
the average turn around time is 12.200000

7 comments:

raje said...

can yu pls post the program for the processes which has same burst time...

Anonymous said...

can you pls post the prog for the sjf in which the processes have the same burst time...

sugan said...

do you plz give separate program for pre emptive and non pre emptive sjf.....

Anonymous said...

Earlist deadline first



#include
#include
#include
void main()
{
char p[10][5],temp[5];
int tot=0,wt[10],pt[10],i,j,n,temp1;
float avg=0;
clrscr();
printf("enter no of processes:");
scanf("%d",&n);
for(i=0;ipt[j])
{
temp1=pt[i];
pt[i]=pt[j];
pt[j]=temp1;
strcpy(temp,p[i]);
strcpy(p[i],p[j]);
strcpy(p[j],temp);
}
}
}
wt[0]=0;
for(i=1;i<n;i++)
{
wt[i]=wt[i-1]+et[i-1];
tot=tot+wt[i];
}
avg=(float)tot/n;
printf("p_name\t P_time\t w_time\n");
for(i=0;i<n;i++)
printf("%s\t%d\t%d\n",p[i],et[i],wt[i]);
printf("total waiting time=%d\n avg waiting time=%f",tot,avg);
getch();

}


OUTPUT:

enter no of processes: 5
Enter process1 name: aaa
enter process time: 4
enter process2 name: bbb
enter process time: 3
enter process3 name: ccc
enter process time: 2
enter process4 name: ddd
enter process time: 5
enter process5 name: eee
enter process time: 1

p_name P_time w_time
eee 1 0
ccc 2 1
bbb 3 3
aaa 4 6
ddd 5 10
total waiting time=20
avg waiting time=4.00

Anonymous said...

please send program for shortest job first preemptive with arrival time

Anonymous said...

here is your solution



#include

struct proc
{
int pid;
int at,bt,wt,tat,rbt;
int flag,flag1;
};
struct proc p1[10];
int i,j,k,n,no,m;
float atat=0.0,awt=0.0;
int tbt=0;
int minimum1();

int main()
{
int minv,locv,mins,locs;
printf("\nenter the number of processes:");
scanf("%d",&n);
printf("\nenter the proc information:");
printf("\nprocess id arrival time burst time");
for(i=0;i0&&p1[j].at<=i)
{
p1[j].flag=1;
}
}
no=minimum1();


printf("%d p[%d]",i,p1[no].pid);
p1[no].rbt=p1[no].rbt-1;

for(k=0;k0&&p1[k].at<=i&&k!=no)
{
p1[k].wt++;
}
}
}

printf("%d",tbt+minv);

for(i=0;i0&&p1[z].at<=i&&p1[z].rbt<mini)
{
mini=p1[z].rbt;
loc=z;
}
}
return loc;
}

praveen said...

in the sorting of shortest job,,,can u try this ex program and practical
the burst time for 8 process were 20,4,3,3,2,4,1,2 respectively

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