For FCFS scheduling algorithm, read the number of processes/jobs in the system, their CPU burst times. The scheduling is performed on the basis of arrival time of the processes irrespective of their other parameters. Each process will be executed according to its arrival time. Calculate the waiting time and turnaround time of each of the processes accordingly.
Algorithm:
#include<stdio.h>
// #include<conio.h>
int main()
{
int bt[20], wt[20], tat[20], i, n;
float wtavg, tatavg;
// clrscr();
printf("\nEnter the number of processes -- ");
scanf("%d", &n);
for(i=0;i<n;i++)
{
printf("\nEnter Burst Time for Process %d -- ", i);
scanf("%d", &bt[i]);
}
wt[0] = wtavg = 0;
tat[0] = tatavg = bt[0];
for(i=1;i<n;i++)
{
wt[i] = wt[i-1] +bt[i-1];
tat[i] = tat[i-1] +bt[i];
wtavg = wtavg + wt[i];
tatavg = tatavg + tat[i];
}
printf("\t PROCESS \tBURST TIME \t WAITING TIME\t TURNAROUND TIME\n");
for(i=0;i<n;i++)
printf("\n\t P%d \t\t %d \t\t %d \t\t %d", i, bt[i], wt[i], tat[i]); printf("\nAverage Waiting Time -- %f", wtavg/n); printf("\nAverage Turnaround Time -- %f", tatavg/n);
// getch();
}Enter the number of processes -- 5
Enter Burst Time for Process 0 -- 13
Enter Burst Time for Process 1 -- 41
Enter Burst Time for Process 2 -- 99
Enter Burst Time for Process 3 -- 41
Enter Burst Time for Process 4 -- 34
PROCESS BURST TIME WAITING TIME TURNAROUND TIME
P0 13 0 13
P1 41 13 54
P2 99 54 153
P3 41 153 194
P4 34 194 228
Average Waiting Time -- 82.800003
Average Turnaround Time -- 128.39999
For SJF scheduling algorithm, read the number of processes/jobs in the system, their CPU burst times. Arrange all the jobs in order with respect to their burst times. There may be two jobs in queue with the same execution time, and then FCFS approach is to be performed. Each process will be executed according to the length of its burst time. Then calculate the waiting time and turnaround time of each of the processes accordingly.
Algorithm:
#include<stdio.h>
int main()
{
int p[20], bt[20], wt[20], tat[20], i, k, n, temp;
float wtavg, tatavg;
printf("\nEnter the number of processes -- ");
scanf("%d", &n);
for(i=0;i<n;i++)
{
p[i]=i;
printf("Enter Burst Time for Process %d -- ", i);
scanf("%d", &bt[i]);
}
for(i=0;i<n;i++)
for(k=i+1;k<n;k++)
if(bt[i]>bt[k])
{
temp=bt[i];
bt[i]=bt[k];
bt[k]=temp;
temp=p[i];
p[i]=p[k];
p[k]=temp;
}
wt[0] = wtavg = 0;
tat[0] = tatavg = bt[0];
for(i=1;i<n;i++)
{
wt[i] = wt[i-1] +bt[i-1];
tat[i] = tat[i-1] +bt[i];
wtavg = wtavg + wt[i];
tatavg = tatavg + tat[i];
}
printf("\n\t PROCESS \tBURST TIME \t WAITING TIME\t TURNAROUND TIME\n"); for(i=0;i<n;i++)
printf("\n\t P%d \t\t %d \t\t %d \t\t %d", p[i], bt[i], wt[i], tat[i]); printf("\nAverage Waiting Time -- %f", wtavg/n); printf("\nAverage Turnaround Time -- %f", tatavg/n);
}
Enter the number of processes -- 4
Enter Burst Time for Process 0 -- 12
Enter Burst Time for Process 1 -- 13
Enter Burst Time for Process 2 -- 15
Enter Burst Time for Process 3 -- 60
PROCESS BURST TIME WAITING TIME TURNAROUND TIME
P0 12 0 12
P1 13 12 25
P2 15 25 40
P3 60 40 100
Average Waiting Time -- 19.250000
Average Turnaround Time -- 44.250000
For round robin scheduling algorithm, read the number of processes/jobs in the system, their CPU burst times, and the size of the time slice. Time slices are assigned to each process in equal portions and in circular order, handling all processes execution. This allows every process to get an equal chance. Calculate the waiting time and turnaround time of each of the processes accordingly.
Algorithm:
#include<stdio.h>
int main(){
int i,j,n,bu[10],wa[10],tat[10],t,ct[10],max;
float awt=0,att=0,temp=0;
// clrscr();
printf("Enter the no of processes -- ");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("\nEnter Burst Time for process %d -- ", i+1);
scanf("%d",&bu[i]);
ct[i]=bu[i];
}
printf("\nEnter the size of time slice -- ");
scanf("%d",&t);
max=bu[0];
for(i=1;i<n;i++)
if(max<bu[i])
max=bu[i];
for(j=0;j<(max/t)+1;j++)
for(i=0;i<n;i++)
if(bu[i]!=0)
if(bu[i]<=t)
{
tat[i]=temp+bu[i];
temp=temp+bu[i];
bu[i]=0;
}
else
{
bu[i]=bu[i]-t;
temp=temp+t;
}
for(i=0;i<n;i++)
{
wa[i]=tat[i]-ct[i];
att+=tat[i];
awt+=wa[i];
}
printf("\nThe Average Turnaround time is -- %f",att/n);
printf("\nThe Average Waiting time is -- %f ",awt/n);
printf("\n\tPROCESS\t BURST TIME \t WAITING TIME\tTURNAROUND TIME\n");
for(i=0;i<n;i++)
printf("\t%d \t %d \t\t %d \t\t %d \n",i+1,ct[i],wa[i],tat[i]);
}Solution:
Enter the no of processes -- 4
Enter Burst Time for process 1 -- 12
Enter Burst Time for process 2 -- 14
Enter Burst Time for process 3 -- 59
Enter Burst Time for process 4 -- 31
Enter the size of time slice -- 23
The Average Turnaround time is -- 64.250000
The Average Waiting time is -- 35.250000
PROCESS BURST TIME WAITING TIME TURNAROUND TIME
1 12 0 12
2 14 12 26
3 59 57 116
4 31 72 103
For priority scheduling algorithm, read the number of processes/jobs in the system, their CPU burst times, and the priorities. Arrange all the jobs in order with respect to their priorities. There may be two jobs in queue with the same priority, and then FCFS approach is to be performed. Each process will be executed according to its priority. Calculate the waiting time and turnaround time of each of the processes accordingly.
Algorithm:
#include<stdio.h>
int main()
{
int p[20],bt[20],pri[20], wt[20],tat[20],i, k, n, temp;
float wtavg, tatavg;
printf("Enter the number of processes --- ");
scanf("%d",&n);
for(i=0;i<n;i++)
{
p[i] = i;
printf("Enter the Burst Time & Priority of Process %d --- ",i); scanf("%d %d",&bt[i], &pri[i]);
}
for(i=0;i<n;i++)
for(k=i+1;k<n;k++)
if(pri[i] > pri[k])
{
temp=p[i];
p[i]=p[k];
p[k]=temp;
temp=bt[i];
bt[i]=bt[k];
bt[k]=temp;
temp=pri[i];
pri[i]=pri[k];
pri[k]=temp;
}
wtavg = wt[0] = 0;
tatavg = tat[0] = bt[0];
{
wt[i] = wt[i-1] + bt[i-1];
tat[i] = tat[i-1] + bt[i];
wtavg = wtavg + wt[i];
tatavg = tatavg + tat[i];
}
printf("\nPROCESS\t\tPRIORITY\tBURST TIME\tWAITING TIME\tTURNAROUND TIME"); for(i=0;i<n;i++)
printf("\n%d \t\t %d \t\t %d \t\t %d \t\t %d ",p[i],pri[i],bt[i],wt[i],tat[i]);
printf("\nAverage Waiting Time is --- %f",wtavg/n);
printf("\nAverage Turnaround Time is --- %f",tatavg/n);
}solution:
Enter the number of processes --- 5
Enter the Burst Time & Priority of Process 0 --- 5
3
Enter the Burst Time & Priority of Process 1 --- 13 20
Enter the Burst Time & Priority of Process 2 --- 40 19
Enter the Burst Time & Priority of Process 3 --- 13 20
Enter the Burst Time & Priority of Process 4 --- 11 22
PROCESS PRIORITY BURST TIME WAITING TIME TURNAROUND TIME
0 3 5 0 5
2 19 40 0 32701
1 20 13 1700966438 0
3 20 13 0 0
4 22 11 1983920792 -1903357744
Average Waiting Time is --- 396784160.000000
Average Turnaround Time is --- -380671552.000000