What is the C code for preemptive priority scheduling?

Answer 1

#include<stdio.h>

#include<conio.h>

int main()

{

char p[10][5],temp[5];

int i,j,pt[10],wt[10],totwt=0,pr[10],temp1,n;

float avgwt;

clrscr();

printf("enter no of processes:");

scanf("%d",&n);

for(i=0;i<n;i++)

{

printf("enter process%d name:",i+1);

scanf("%s",&p[i]);

printf("enter process time:");

scanf("%d",&pt[i]);

printf("enter priority:");

scanf("%d",&pr[i]);

}

for(i=0;i<n-1;i++)

{

for(j=i+1;j<n;j++)

{

if(pr[i]>pr[j])

{

temp1=pr[i];

pr[i]=pr[j];

pr[j]=temp1;

temp1=pt[i];

pt[i]=pt[j];

pt[j]=temp1;

strcpy(temp,p[i]);

strcpy(p[i],p[j]);

strcpy(p[j],temp);

}

}

}

wt[0]=0;

for(i=1;i<n;i++)

{

wt[i]=wt[i-1]+et[i-1];

totwt=totwt+wt[i];

}

avgwt=(float)totwt/n;

printf("p_name\t p_time\t priority\t w_time\n");

for(i=0;i<n;i++)

{

printf(" %s\t %d\t %d\t %d\n" ,p[i],pt[i],pr[i],wt[i]);

}

printf("total waiting time=%d\n avg waiting time=%f",tot,avg);

getch();

}

Answer 2

C program for Shortest Job First Preemptive algorithm.

#include

struct proc

{

int pid;

int at,bt,wt,tat,rbt;

int flag,flag1;

};

struct proc p1[10];

int i,j,k,n,no,m;

float atat=0.0,awt=0.0;

int tbt=0;

int minimum1();

int main()

{

int minv,locv,mins,locs;

printf("\nenter the number of processes:");

scanf("%d",&n);

printf("\nenter the proc information:");

printf("\npid at bt");

for(i=0;i

{

p1[i].wt=0;

p1[i].tat=0;

p1[i].flag=0;

p1[i].flag1=0;

scanf("%d%d%d",&p1[i].pid,&p1[i].at,&p1[i].bt);

tbt+=p1[i].bt;

p1[i].rbt=p1[i].bt;

}

printf("\nthe proc information:");

printf("\npid at bt");

for(i=0;i

{

printf("\n%d %d %d",p1[i].pid,p1[i].at,p1[i].bt);

}

minv=p1[0].at;

locv=0;

for(i=1;i

{

if(p1[i].at

{

locv=i; //tells min at process in locv

minv=p1[i].at;

}

}

for(i=0;i

{

if(p1[i].at==minv)

{

p1[i].flag1=1; //processes having same minimum at

}

}

mins=p1[0].bt;

locs=0;

for(i=0;i

{

if(p1[i].flag1==1&&p1[i].bt

{

mins=p1[i].bt; //gives process with minimum burst time

locs=i;

}

}

printf("\ngantt chart:");

for(i=minv;i

{

for(j=0;j

{

if(p1[j].rbt>0&&p1[j].at<=i)

{

p1[j].flag=1;

}

}

no=minimum1();

printf("%d p[%d]",i,p1[no].pid);

p1[no].rbt=p1[no].rbt-1;

for(k=0;k

{

if(p1[k].rbt>0&&p1[k].at<=i&&k!=no)

{

p1[k].wt++;

}

}

}

printf("%d",tbt+minv);

for(i=0;i

{

awt+=p1[i].wt;

}

awt=awt/n;

for(i=0;i

{

p1[i].tat=p1[i].wt+p1[i].bt;

atat+=p1[i].tat;

}

atat=atat/n;

printf("\n average wt=%f, average tat=%f",awt,atat);

printf("\nthe proc information:");

printf("\npid at bt wt tat");

for(i=0;i

{

printf("\n%d %d %d %d %d",p1[i].pid,p1[i].at,p1[i].bt,p1[i].wt,p1[i].tat);

}

}

int minimum1()

{

int loc,z;

int mini;

mini=99;

loc=-1;

for(z=0;z

{

if(p1[z].rbt>0&&p1[z].at<=i&&p1[z].rbt

{

mini=p1[z].rbt;

loc=z;

}

}

return loc;

}

Answer 3

APAT