Shortest job first scheduling algorithm : Implementation in C

Shortest remaining time First Algorithm

In this scheduling algorithm, the process with the smallest amount of time remaining until completion is selected to execute. Since the currently executing process is the one with the shortest amount of time remaining by definition, and since that time should only reduce as execution progresses, processes will always run until they complete or a new process is added that requires a smaller amount of time.

Shortest remaining time is advantageous because short processes are handled very quickly. The system also requires very little overhead since it only makes a decision when a process completes or a new process is added, and when a new process is added the algorithm only needs to compare the currently executing process with the new process, ignoring all other processes currently waiting to execute.

Implementation  Of SRTF in C

# include <stdio.h> 

# include <conio.h> 

void main() 

{ 

 int i,p[20],a[20],b[20],br[20],tat[20],gantt[100][2],n1,n2,n3,c,d,e=0,f=0,g,time=0,t1,t2,t3,t4,t5,t6; 

 float v1=0.0,v2=0.0; 

 FILE *fp; 

 clrscr(); 

 fp=fopen("sjf.txt","r"); 

 while (fscanf(fp,"%d%d%d",&n1,&n2,&n3)!=EOF) 

 { 

   p[e]=n1; 

   a[e]=n2; 

   b[e]=n3; 

   br[e++]=n3; 

 } 

 while (1) 

 { 

  for (i=1;i<e;i++) 

   if ((br[0]==0 || time<a[0]?1:br[i]<br[0] || (br[i]==br[0] && a[i]<a[0])) && time>=a[i] && br[i]!=0) 

   { 

    t1=p[0]; 

    t2=a[0]; 

    t3=b[0]; 

    t4=br[0]; 

    t5=tat[0]; 

    p[0]=p[i]; 

    a[0]=a[i]; 

    b[0]=b[i]; 

    br[0]=br[i]; 

    tat[0]=tat[i]; 

    p[i]=t1; 

    a[i]=t2; 

    b[i]=t3; 

    br[i]=t4; 

    tat[i]=t5; 

   } 

  if (p[0]>time || br[0]==0) 

  { 

   if (f==0 || gantt[f-1][0]!=-1) 

   { 

    gantt[f][0]=-1; 

    gantt[f++][1]=time; 

   } 

  } 

  else 

  if (br[0]!=0) 

  { 

   if (f==0 || gantt[f-1][0]!=p[0]) 

   { 

    gantt[f][0]=p[0]; 

    gantt[f++][1]=time; 

   } 

   if (--br[0]==0) 

     tat[0]=time+1; 

  } 

  time++; 

  c=0; 

  for (i=0;i<e;i++) 

   if (br[i]!=0) 

    c=1; 

  if (c==0) 

   break; 

 } 

 gantt[f][1]=time; 

 printf ("%8s%8s%8s%8s%8s\n\n","P","A","B","W","TAT"); 

 for (i=0;i<e;i++) 

  printf ("%8d%8d%8d%8d%8d\n",p[i],a[i],b[i],tat[i]-b[i],tat[i]); 

 printf("\n\nAverage waiting time is:"); 

  for(i=0;i<e;i++) 

    v1+=tat[i]-b[i]; 

printf("%f",v1/e); 

printf("\n\nAverage turnaround time is:"); 

  for(i=0;i<e;i++) 

    v2+=tat[i]; 

printf("%f",v2/e); 

printf("\n\n\n"); 

 printf ("\n\n "); 

 for (i=0;i<f;i++) 

  printf ("--- "); 

 printf ("\n|"); 

 for (i=0;i<f;i++) 

  printf (gantt[i][0]==-1?"   |":" %d |",gantt[i][0]); 

 printf ("\n "); 

 for (i=0;i<f;i++) 

  printf ("--- "); 

 printf ("\n0"); 

 for (i=1;i<=f;i++) 

  printf ("  %2d",gantt[i][1]); 

 getch(); 

}