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Round Robin Scheduling Method C Program

#include<stdio.h>
#include<conio.h>

int z[10],b[10],n,m[50],r,q,e=0,avg=0,i,j;
float f;

main()
{
 clrscr();
 printf("\n\tJOB SCHEDULING ALGORITHM[RR]");
 printf("\n\t*******************************************************\n");
 printf("\nEnter how many jobs:");
 scanf("%d",&n);
 printf("\nEnter burst time for corresponding job...\n");
 for(i=1;i<=n;i++)
 {
  printf("\nProcess %d: ",i);
  scanf("%d",&b[i]); z[i]=b[i];
 }

 printf("\nENTER THE TIME SLICE VALUE:");
 scanf("%d",&q);

 rr();

 average();

 getch();
 return 0;
}

rr()
{
 int max=0;
 max=b[1];
    for(j=1;j<=n;j++)
       if(max<=b[j])
    max=b[j];

 if((max%q)==0)
   r=(max/q);
 else
  r=(max/q)+1;
   for(i=1;i<=r;i++)
   {
      printf("\nround %d",i);
      for(j=1;j<=n;j++)
    {
      if(b[j]>0)
        {
         b[j]=b[j]-q;

         if(b[j]<=0)
        {
        b[j]=0;
         printf("\nprocess %d is completed",j);
        }
         else
         printf("\nprocess %d remaining time is %d",j,b[j]);
        }
     }
     delay(1000);
   }
   return 0;
}

average()
{
 for(i=1;i<=n;i++)
 {
    e=0;
    for(j=1;j<=r;j++)
     {
    if(z[i]!=0)
    {
      if(z[i]>=q)
      {
      m[i+e]=q;  z[i]-=q;
      }
      else
      {
      m[i+e]=z[i];  z[i]=0;
      }
    }
    else
      m[i+e]=0;
       e=e+n;
     }
   }
  for(i=2;i<=n;i++)
     for(j=1;j<=i-1;j++)
      avg=avg+m[j];
  for(i=n+1;i<=r*n;i++)
  {
    if(m[i]!=0)
    {
      for(j=i-(n-1);j<=i-1;j++)
      avg=m[j]+avg;
    }
  }
 f=avg/n;
 printf("\nTOTAL WATING:%d",avg);
 printf("\n\nAVERAGE WAITING TIME:%f\n",f);
    for(i=1;i<=r*n;i++)
     { if(m[i]!=0)
       if(i%n==0){
       printf("P%d",(i%n)+(n));      }


       else
       printf("P%d",(i%n));
       for(j=1;j<=m[i];j++)
        printf("%c",22);
     }
printf("\n");
 getch();
 return 0;
}
Output:
Enter how many jobs:4
Enter burst time for corresponding job...
Process 1: 7
Process 2: 5
Process 3: 4
Process 4: 2
ENTER THE TIME SLICE VALUE:2
round 1
process 1 remaining time is 5
process 2 remaining time is 3
process 3 remaining time is 2
process 4 is completed
round 2
process 1 remaining time is 3
process 2 remaining time is 1
process 3 is completed
round 3
process 1 remaining time is 1
process 2 is completed
round 4
process 1 is completed
TOTAL WAITING : 39
AVERAGE WAITING TIME:9.000000

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