Round robin process scheduling simulation
anomit | September 11, 2008No big deal. Everyone has to do it once for their UNIX concepts practical work I guess. What irked me was that the teacher and some of the ‘good students’ were hell bent on arguing that non-preemptive process scheduling simulations do not require any kind of data regarding the arrival time of processes in the ready queue. I still don’t understand how that simulates a real life scenario, that is, after the completion of a process, to which other process would the scheduler perform a context switch. Could someone point me out if I am missing something here?
Anyways, this is the code I wrote today during the practical class:
#include<stdio.h>
struct processinfo
{
int id;
int burst;
int arr;
};
typedef struct processinfo process;
/*checks if all the processes have finished execution*/
int check(process* p,int n)
{
int i=0,flag=0;
for(i=0;i<n;i++)
{
if(p[i].burst!=0)
{
flag=1;
break;
}
}
return flag;
}
int main()
{
int n;
printf("No. of processes:");
scanf("%d",&n);
process p[n];
int i=0,j=0;
int wt[n]/*the waiting time array*/,last[n]/*last[i] stores the last executed instant of the process i*/;
for(i=0;i<n;i++)
{
printf("Proccess id:");
scanf("%d",&p[i].id);
printf("Arrival time:");
scanf("%d",&p[i].arr);
printf("Burst time:");
scanf("%d",&p[i].burst);
wt[i]=0;
last[i]=0;
}
/*gah, bubble sort*/
/*anyways we won't be doing any sorting in a real scheduler, I guess*/
for(i=0;i<n-1;i++)
{
for(j=i+1;j<n;j++)
{
if(p[j-1].arr>p[j].arr)
{
process temp=p[j-1];
p[j-1]=p[j];
p[j]=temp;
}
}
}
int timequantum=2/*the time quantum*/,time=0/*running time*/;
while(1)
{
if(!check(p,n))
break;
int k=0;
for(k=0;k<n;k++)
{
if(p[k].burst!=0)
{
if(p[k].burst<timequantum)
{
wt[k]+=time-last[k];
time+=p[k].burst;
p[k].burst=0;
}
else
{
p[k].burst-=timequantum;
wt[k]+=time-last[k];
time+=timequantum;
}
last[k]=time;
}
}
}//end of while
for(i=0;i<n;i++)
printf("\nWaiting time for Process %d:%d",p[i].id,wt[i]);
return 0;
}
Well the arrival time of the process is of no
Ankit | October 8, 2008Well the arrival time of the process is of no use unless you are aiming for deadlines or preemption, or deferred scheduling. Linux kernel does not in fact use arrival time, except in RT scheduling.
Coming from you, I'd accept it for the time being.
anomit | October 8, 2008Coming from you, I’d accept it for the time being. I’d really like to find some time and learn more about this.
PS: Welcome to my blog!