C Program to Simulate Round Robin CPU Sceduling Algorithm

The round robin CPU scheduling algorithm is simulated using C program. The time slice or time quantum is read from the user.

Program:

#include<stdio.h>
struct process{



int burst,wait,comp,f;
}p[20]={0,0};
int main(){
int
n,i,j,totalwait=0,totalturn=0,quantum,flag=1,
time=0;
printf("\nEnter The No Of Process
:");
scanf("%d",&n);
printf("\nEnter The Quantum time (in ms)
:");
scanf("%d",&quantum);
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].f=1;
}
printf("\nOrder Of Execution \n");
printf("\nProcess \tStarting\tEnding\tRemaining");
printf("\n\tTime \tTime\tTime");
while(flag==1){
flag=0;
for(i=0;i<n;i++){
if(p[i].f==1){
flag=1;
j=quantum;
if((p[i].burst-p[i].comp)>quantum){
p[i].comp+=quantum;
}
else{
p[i].wait=time-p[i].comp;
j=p[i].burst-p[i].comp;
p[i].comp=p[i].burst;
p[i].f=0;
}
printf("\nprocess # %-3d %-10d %-10d %-10d",i+1,time,time+j,p[i].burst-
p[i].comp);
time+=j;
}}}
printf("\n\n------------------");
printf("\nProcess \t Waiting Time  TurnAround Time ");
for(i=0;i<n;i++){
printf("\nProcess # %-12d%-15d%-15d",i+1,p[i].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;
}

C Program to Simulate Priority Scheduling CPU scheduling Algorithm

The following C program implements Simulation of  priority scheduling (CPU Scheduling algorithm). Each process or job is given a priority. The priority is represented using a positive integer. The priority increases as the value of the integer decreases.

Program:

#include<stdio.h>
struct process{



int burst,wait,no,priority;
}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);
printf("Enter The Priority For Process #%2d :",i+1);
scanf("%d",&p[i].priority);
p[i].no=i+1;
}
for(i=0;i<n;i++)
for(j=0;j<n-i-1;j++)
if(p[j].priority>p[j+1].priority)
{
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;
//Simple way to swap 2 var’s
p[j].priority^=p[j+1].priority^=p[j].priority^=p[j+1].priority;
//printf("j");
}
printf("\nProcess \t Starting\tEnding\tWaiting\tTurnAround ");
printf("\n\t Time\tTime\tTime\tTime ");
for(i=0;i<n;i++)
{
printf("\nProcess # %-11d%-10d%-10d%-10d%10d",p[i].no,p[i].wait,p[i].wait+p[i].burst,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 :30
Enter The Priority For Process # 1 :2
Enter The Burst Time (in ms) For Process #2 :20
Enter The Priority For Process # 2 :1
Enter The Burst Time (in ms) For Process #3 :40
Enter The Priority For Process # 3 :3

Process          Starting  Ending  Waiting  TurnAround
                          Time        Time      Time          Time
Process # 2        0             20           0                20
Process # 1       20            50          20              50
Process # 3       50            90          50              90

Average
---------
Waiting Time : 23.333333 ms
TurnAround Time : 53.333333 ms

Simulation of Shortest Job First (SJF) CPU scheduling algorithm in C Language

This is a C program to simulate Shortest Job First (SJF) CPU scheduling algorithm. In Shortest Job first CPU scheduling algorithm, the process with  shortest CPU burst time is executed first. The following program is to simulate non preemptive SJF scheduling. Preemptive SJF CPU scheduling algorithm will be discussed in another post.

#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

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