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:
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;
}
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- Categories
- Coding, GNU/Linux
- Tags
- scheduler, unix
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I think doing a md5 hash of the resulting ciphertext would add an extra layer of security. What do the others think of this idea? Of course, the safe storage and transmission of the XOR key becomes an issue. Check the code and see if you could come up with suggestions to optimize it. I particularly don't like the exponential order for loop at the very end. The XOR key is stored in the xor-key file in the same directory where the code is run.
