Showing posts with label OPERATING SYSTEM LAB PROGRAMS. Show all posts
Showing posts with label OPERATING SYSTEM LAB PROGRAMS. Show all posts

Monday, 2 September 2013

PRIORITY BASED CPU SCHEDULING C PROGRAM



/*CSEMATTERBLOGSPOT.IN

Write a program in C to implement priority based CPU scheduling algo
calculate avg. waiting time & average turn around time*/

#include<stdio.h>
#include<conio.h>
#include<graphics.h>
int main()
{
    int gd=DETECT,gm,i,j,k,n,b,count=0,arr[20],prio[20],bt[20],wt[20],tat[20];
    float avg_wt,avg_tat,sum_wt=0.0,sum_tat=0.0;
    initgraph(&gd,&gm,"C:\TC\BIN");
    printf("Enter the no. of processes  ");
    scanf("%d",&n);
    printf("Enter the burst time of processes \n");
    for(i=0;i<n;i++)
        scanf("%d",&bt[i]);
    printf("Enter priority of processes \n");
    for(i=0;i<n;i++)
    {    scanf("%d",&prio[i]);
        arr[i]=prio[i];}
    for(i=0;i<=(n-2);i++)
    {for(j=0;j<=((n-2)-i);j++)
      {   if(arr[j]>arr[j+1])
          {
              b=arr[j];
              arr[j]=arr[j+1];
              arr[j+1]=b;
          }} }
    wt[0]=0;
    tat[0]=0;
    for(i=0;i<n;i++)
    {   for(j=0;j<n;j++)
        {      if(arr[i]!=prio[j])
               count=count+1;
           else
               break;}
        wt[i+1]=bt[count]+wt[i];
        tat[i+1]=bt[count]+tat[i];
        count=0; }
    for(i=0;i<n;i++)
    sum_wt=sum_wt+wt[i];
    for(i=0;i<(n+1);i++)
    sum_tat=sum_tat+tat[i];
    avg_wt=(sum_wt/(float)n);
    avg_tat=(sum_tat/(float)n);
    printf("Process_no. Burst_Time Priority Waiting_Time Turn_Around_Time \n");
    for(i=0;i<n;i++)
    {
    printf("P%d\t\t%d\t%d\t\t%d\t\t%d\n",i,bt[i],prio[i],wt[i],tat[i+1]);}
    printf("Average waiting time is: %f \n",avg_wt);
    printf("Average turn around time is: %f",avg_tat);
    printf("\n\n \t GANTT CHART \n");
    for(i=0;i<n;i++)
        printf("P%d \t",i);
    printf("\n");
    for(i=0;i<n;i++)
    {
        printf("%d \t",wt[i]);
    }
    rectangle(2,300,270,320);
    getch();
    closegraph();
}
/* OUTPUT
Enter the no. of processes 5
Enter the burst time of processes
12 8 4 6 10
Enter priority of processes
5 1 2 4 3

Process_no.  Burst_Time  Priority  Waiting_Time  Turn_Around_Time
P1 8    1 0 8
P2 4    2 8 12
P4 10    3 12 22
P3 6    4 22 28
P0 12    5 28 40

Average waiting time is 14.00000
Average turn around time is 22.00000

GANTT CHART
 ||P1|| ||P2||  ||P4||   ||P3||   ||P0||
0      8       12 22 28  40*/

PREEMPTIVE SCHEDULING C PROGRAM


/* CSEMATTER.BLOGSPOT.IN
C Program for preemptive cpu scheduling*/
#include<stdio.h>
#include<conio.h>
int main()
{
    int min(int, int []);
    int temp,time,temp4,c,n,i,j,a1[20],wt[20],pr[20],bt[20],gc[20],arr_ti[20],tat[20],sum_bt=0,a=0,p=0,temp1,count;
    printf("Enter the no. of proceses");
    scanf("%d",&n);
    printf("Enter the burst time of processes");
    for(i=0; i<n; i++)
    {scanf("%d",&bt[i]);
      sum_bt=sum_bt+bt[i];
       a1[i]=bt[i];}
    for(i=0; i<n; i++)
    pr[i]=i+1;
    printf("Enter the arrival time");
    for(i=0; i<n; i++)
        scanf("%d",&arr_ti[i]);
    for(i=0; i<(n-2); i++)
    {  for(j=i; j<((n-2)-i); i++)
        {
            if(arr_ti[j]>arr_ti[j+1])
            {   temp=arr_ti[j];
                arr_ti[j]=arr_ti[j+1];
                arr_ti[j+1]=temp;
                temp=a1[j];
                a1[j]=a1[j+1];
                a1[j+1]=temp;
                temp=pr[j];
                pr[j]=pr[j+1];
                pr[j+1]=temp;
            }}}
    for(i=0; i<n; i++)
    {
        wt[i]=0;
        tat[i]=0;
    }
    for(i=0; i<=sum_bt; i++)
    {
        a++;
        if(a<5)
        {
            c=min(a,a1)+1;
            gc[i]=c;
        }
        else
        {
            c=min(5,a1)+1;
            gc[i]=c;
        }  }
    for(i=0; i<n; i++)
    {
        temp=i+1;
        time=0;
        temp1=0;
        for(j=0; j<sum_bt; j++)
        {time++;
            if(temp==gc[j])
            {
             if(temp1==0)
            {
            wt[i]=wt[i]+time-(i+1);
            temp1=j+2;
            }
            else
            {
            wt[i]=wt[i]+(j+1)-temp1;
            temp1=j+2;
            }}}}

    for(i=0; i<n; i++)
    {
        temp=0;
        for(j=0; j<sum_bt; j++)
        {
            if(i+1==gc[j])
            {
                if(temp==0)
                {
                    tat[i]=tat[i]+bt[i]+j-arr_ti[i];
                    temp=j+1;
                }
                else
                {
                    tat[i]=tat[i]+j-temp;
                    temp=j+1;
                }  }}}
     //printing the output

    printf("DATA IS AS FOLLOWS: \n");
    printf("Process\t Burst Time\t Arrival Time\t Waiting Time\t Turn Around Time\n");
    for(i=0;i<n;i++)
    {
        printf("%d\t %d\t\t\t %d\t\t %d\t\t %d\n",pr[i],bt[i],arr_ti[i],wt[i],tat[i]);
    }
    printf("\n \n");
    printf("GANTT CHART: \n");
    c=gc[0];
    printf("  ||P%d\t",gc[0]);
    for(i=1;i<sum_bt;i++)
    {
        if(gc[i]!=c)
        {printf(" ||P%d|| ",gc[i]);
        c=gc[i];}
    }
    printf("\n0\t");
    c=gc[0];
    for(i=1;i<=sum_bt;i++)
    {    p=p+1;
         if(gc[i]!=c)
        {printf("%d \t",p);
        c=gc[i];
        }
        }
    getch();
}


int min(int a,int a1[20])
{
    int i,count=0;
    int temp=a1[0];
    if(a==1)
    { temp--;}
    else
    {
        for(i=0; i<a; i++)
        {
            if(temp>a1[i])
            {
                temp=a1[i];
                count=i;}}
        temp--;}
    if(temp==0)
    {temp=500;}
    a1[count]=temp;
    return(count);  /* returning the index of array whose value is minimum*/
}

/*OUTPUT:
Enter the no. of proceses 5
Enter the burst time of processes
5   2   2   3   6
Enter the arrival time
0   1   2   3   4
DATA IS AS FOLLOWS:
Process  Burst Time  ArrivalTime  Waiting Time Turn AroundTime
1           5           0           7               12
2           2           1           0               2
3           2           2           1               3
4           3           3           2               5
5           6           4           8               14
GANTT CHART:
  ||P1||    ||P2||  ||P3||  ||P4||  ||P1||  ||P5||
0        1        3        5       8      12      18*/

SJF CPU SCHEDULING C PROGRAM


/*CSEMATTER.BLOGSPOT.IN
program in C to implement shortest job first CPU scheduling algo
calculate avg. waiting time & average turn around time*/

#include<stdio.h>
#include<conio.h>
int main()
{
int p[10],temp;
int tot=0,wt[10],pt[10],tat[10],i,j,n,temp1,tot1=0;
float avg=0,avg1=0;
printf("enter no of processes:");
scanf("%d",&n);
printf("enter process time");
for(i=0;i<n;i++)
{
scanf("%d",&pt[i]);
p[i]=i;
}
for(i=0;i<n-1;i++)
{
for(j=i+1;j<n;j++)
{
if(pt[i]>pt[j])
{
temp1=pt[i];
pt[i]=pt[j];
pt[j]=temp1;
temp=p[i];
p[i]=p[j];
p[j]=temp;

}
}
}
wt[0]=0;
for(i=1;i<=n;i++)
{
wt[i]=wt[i-1]+pt[i-1];

tot=tot+wt[i];
}
for(i=0;i<n;i++)
   {tat[i]=wt[i]+pt[i];
    tot1=tot1+tat[i];}
avg=(float)tot/n;
avg1=(float)tot1/n;
printf("p_no.\t P_time\t Waiting_Time\t  Turn_Around_Time\n");
for(i=0;i<n;i++)
printf("%d\t%d\t%d\t %d\n",p[i],pt[i],wt[i],tat[i]);
printf("avg waiting time=%f\n",avg);
printf("avg turn around time=%f",avg1);
printf("\n\n \t GANTT CHART \n");
    for(i=0;i<n;i++)
        printf("P%d \t",i);
    printf("\n");
    for(i=0;i<n;i++)
    {
        printf("%d \t",wt[i]);
    }
        getch();
}
/* OUTPUT
Enter the no. of processes 5
Enter the burst time of processes
12 8 4 6 10
P_no.  P_Time  Waiting_Time  Turn_Around_Time
2 4     0 4
3 6     4 10
1 8     10 18
4       10        18 28
0 12     28 40

Average waiting time is 20.000000
Average turn around time is 20.000000

GANTT CHART
P0 P1 P2 P3 P4
0 4 10 18 28*/

ROUND ROBIN OS C PROGRAM

/*CSEMATTERBLOG
PROGARM TO IMPLEMENT ROUND ROBIN ALGO*/
#include<stdio.h>
#include<conio.h>
int main()
{
    int bt[20],gc[20],wt[20],tat[20],bt1[20],st[20],ts,n,i,j,k,count=0,count1,sum_bt=0,tq;
    int swt=0,stat=0,temp,sq=0,c,p=0;
float awt=0.0,atat=0.0;
    printf("Enter the nuo. of processs");
    scanf("%d",&n);
    printf("Enter the burst time");
    for(i=0; i<n; i++)
        {scanf("%d",&bt[i]);
        bt1[i]=bt[i];
        st[i]=bt[i];}
    printf("Enter the time slice");
    scanf("%d",&ts);
    tq=ts;
    for(i=0; i<n; i++)
    sum_bt=sum_bt+bt[i];
        for(k=0; k<n; k++)
    {
       do
        {
            for(i=0; i<n; i++)
            {
                if(bt[i]>=ts)
                {
                    for(j=count; j<(count+ts); j++)
                        gc[j]=i+1;
                    count+=ts;
                    bt[i]=bt[i]-ts;
                }
                else
                {
                    for(j=count; j<=(count+bt[i]); j++)
                        gc[j]=i+1;
                    count+=bt[i];
                    bt[i]=0;
                }
            }
        }while(bt[k]!=0);
    }

 while(1)
{
       for(i=0,count=0;i<n;i++)
       {
       temp=tq;
       if(st[i]==0)
      {
        count++;
        continue;
       }
if(st[i]>tq)
st[i]=st[i]-tq;
else
if(st[i]>=0)
{
temp=st[i];
st[i]=0;
}
sq=sq+temp;
tat[i]=sq;
}
if(n==count)
break;
}
for(i=0;i<n;i++)
{
wt[i]=tat[i]-bt1[i];
swt=swt+wt[i];
stat=stat+tat[i];
}
awt=(float)swt/n;
atat=(float)stat/n;
printf("Process_no Burst time Wait time Turn around time\n");
for(i=0;i<n;i++)
printf("%d %d %d %d\n",i+1,bt1[i],wt[i],tat[i]);
printf("Avg wait time is %f Avg turn around time is %f",awt,atat);
 printf("\n \n");
    printf("GANTT CHART: \n");
    c=gc[0];
    printf("  ||P%d\t",gc[0]);
    for(i=1;i<sum_bt;i++)
    {
        if(gc[i]!=c)
        {printf(" ||P%d|| ",gc[i]);
        c=gc[i];}
    }
    printf("\n0\t");
    c=gc[0];
    for(i=1;i<=sum_bt;i++)
    {    p=p+1;
         if(gc[i]!=c)
        {printf("%d \t",p);
        c=gc[i];
        }}
   getch();
}
/*OUTPUT:
Enter the no. of proceses 5
Enter the burst time of processes
5   1   2   2   3
Enter the time slice 2
Process  Burst Time   Waiting Time Turn AroundTime
1           5           8               13
2           1           2               3
3           2           3               5
4           2           5               7
5           3           9               12
GANTT CHART:
  ||P1||    ||P2||  ||P3||  ||P4||  ||P5||  ||P1||  ||P5||  ||P1||
0        2        3        5      7       9       11      12      13*/

PRODUCER CONSUMER C PROGRAM

/*CSEMATTER.BLOGSPOT.IN
C Program to implement producer consumer problem*/
#include<stdio.h>
#include<conio.h>
int main()
{
int producer(int [],int,int);
int consumer(int [],int,int);
int buffer[20],max,n,a=0,a1=0,ch,cons=0;
 printf("\n\n\n\tEnter Stack Size :");
 scanf("%d",&max);
do
 {
   printf("\n\t\tCHOICES\n\t\t\n\t1.Producer\n\t2.Consumer\n\t3.Exit\nEnter your choice :\n ");
   scanf("%d",&ch);
   switch(ch)
   {
    case 1:
  {
   if(a==max & cons==0)
   {printf("STACK FULL...\n");
            break;}
     else if(cons==2)
     {
         printf("Consumer has not consumed all item yet");
         break;
    }
    else
    {a=producer(buffer,max,a);
     break;}
    }
    case 2:
  {if(a!=max)
  {printf("STACK NOT FULL YET...PRODUCER TURNS TO PRODUCE\n");
        break;}
      if(a==max)
     {a1=consumer(buffer,max,a1);
      if(a1==max)
      {a=0;
      cons=0;}
      else
      cons=2;}
     break;}
     case 3:
     break;
    }
 }
while(ch!=3);
getch();
}
 int producer(int buffer[20],int max,int a)
{
int i,n;
int counter=a;
printf("Enter the no. of items to be produced");
scanf("%d",&n);
for(i=a;i<a+n;i++)
{
printf("Enter the item to be produced");
scanf("%d",&buffer[i]);
counter=counter+1;
}
return(counter);
}
int consumer(int buffer[20],int max,int a1)
{
int i,n;
int counter1=a1;
printf("Enter the no. of items to be consumed");
scanf("%d",&n);
for(i=a1;i<a1+n;i++)
{
printf("Consumed item is: %d \n",buffer[i]);
buffer[i]=0;
counter1=counter1+1;
}
return(counter1);}
/*OUTPUT
     Enter Stack Size :5
                CHOICES
        1.Producer
        2.Consumer
        3.Exit
Enter your choice :
 2
STACK NOT FULL YET...PRODUCER TURNS TO PRODUCE
                CHOICES
        1.Producer
        2.Consumer
        3.Exit
Enter your choice :
 1
Enter the no. of items to be produced3
Enter the item to be produced1
Enter the item to be produced2
Enter the item to be produced3
                CHOICES
        1.Producer
        2.Consumer
        3.Exit
Enter your choice :
 2
STACK NOT FULL YET...PRODUCER TURNS TO PRODUCE
                CHOICES
        1.Producer
        2.Consumer
        3.Exit
Enter your choice :
 1
Enter the no. of items to be produced2
Enter the item to be produced4
Enter the item to be produced5
                CHOICES
        1.Producer
        2.Consumer
        3.Exit
Enter your choice :
 1
STACK FULL...
                CHOICES
        1.Producer
        2.Consumer
        3.Exit
Enter your choice :
 2
Enter the no. of items to be consumed3
Consumed item is: 1
Consumed item is: 2
Consumed item is: 3
                CHOICES
        1.Producer
        2.Consumer
        3.Exit
Enter your choice :
 1
Consumer has not consumed all item yet
                CHOICES
        1.Producer
        2.Consumer
        3.Exit
Enter your choice :
 2
Enter the no. of items to be consumed2
Consumed item is: 4
Consumed item is: 5
                CHOICES
        1.Producer
        2.Consumer
        3.Exit
Enter your choice :
3 */

READER WRITER C PROGRAM

/*CSEMATTER.BLOGSPOT.IN
C Program to implement reader writer problem*/
#include<stdio.h>
#include<conio.h>
int ch,write=0,readcount=0;
int main()
{
void writerin();
void writerout();
void readerin();
void readerout();
 do
 {
   printf("\n\t\tCHOICES\n\t\t\n\t1 Writer in\n\t2 Writer out\n\t3.Reader in");
   printf("\n\t3..Reader out\n\t5.exit\nEnter your choice :\n ");
   scanf("%d",&ch);
   switch(ch)
   {
    case 1:
    {
     writerin();
    }
    case 2: 
    {
       writerout();   
     }
     case3:
 {
         readerin();
 }
     case4:
       {readerout();
       }
     case 5:
     break;
    }
 }
while(ch!=5);
getch();
}
void readerout()
{
if(write==1 && readcount>0)
      {
        readcount--;
 }
 else if(write==1 && readcount==0)
        {
      write=0;
         }
  else
 printf("no reader is reading");
}
void readerin()
{
if(write==0 && readcount==0)
      {
      readcount++;
      write=1;
      }
else if(write==1 && readcount>0)
{
readcount++;
}
else
{
printf("writer is writing");
}
}

void writerin()
{
if(readcount==0)
{
write=1;
printf("writer is writing\n");
}
else
{
printf("%d reader are reading..wait for finish\n",readcount);
}
}
void writerout()
{
if(write==1)
{
write=0;
printf("writer is exiting\n");
}
else
{
printf("no writer is writing\n");}
}
}

FIFO PAGE REPLACEMENT C PROGRAM


/* CSEMATTERBLOG

C PROGRAM TO IMPLEMENT FIFO PAGE REPLACEMENT ALGO*/
#include<stdio.h>
#include<conio.h>
void main()
{
int frame[20],pages[20],n,i,j,k,fr,count=0,avail;
clrscr();
printf("enter the no. of pages");
scanf("%d",&n);
printf("enter the page sequence");
for(i=0;i<n;i++)
scanf("%d",&pages[i]);
printf("enter the no. of frames");
scanf("%d",&fr);
for(i=0;i<fr;i++)
frame[i]=-1;
j=0;
printf("ref string \t page frame\n");
for(i=0;i<n;i++)
{
printf("%d\t\t",pages[i]);
avail=0;
for(k=0;k<fr;k++)
if(frame[k]==pages[i])
avail=1;
if(avail==0)
{
frame[j]=pages[i];
j=(j+1)%fr;
count++;
for(k=0;k<fr;k++)
printf("%d\t",frame[k]);
}
printf("\n");
}
printf("page fault is :%d",count);
getch();
}
/*OUTPUT
enter the no. of pages12
enter the page sequence2
3
4
5
2
3
6
2
3
4
5
6
enter the no. of frames3
ref string       page frame
2               2       -1      -1
3               2       3       -1
4               2       3       4
5               5       3       4
2               5       2       4
3               5       2       3
6               6       2       3
2
3
4               6       4       3
5               6       4       5
6
page fault is :9
*/

FCFS DISK SCHEDULING C PROGRAM

/*CSEMATTER.BLOGSPOT.IN

C PROGRAM TO IMPLEMENT FCFS DISK SCHEDULING  ALGO*/
#include<stdio.h>
#include<conio.h>
void main()
{
int queue[20],n,head,i,j,k,seek=0,max,diff;
float aver;
clrscr();
printf("enter the max range of disk");
scanf("%d",&max);
printf("enter the size of queue request");
scanf("%d",&n);
printf("enter the queue");
for(i=1;i<=n;i++)
{scanf("%d",&queue[i]);}
printf("enter the initial head position");
scanf("%d",&head);
queue[0]=head;
for(j=0;j<=n-1;j++)
{
diff=abs(queue[j+1]-queue[j]);
seek+=diff;
printf("move is from %d to %d with seek %d\n",queue[j],queue[j+1],diff);
}
printf("total seek time is%d\n",seek);
aver=seek/(float)n;
printf("avrage seek time is %f\n",aver);
getch();
}
/*OUTPUT:
enter the max range of disk180
enter the size of queue request8
enter the queue87
170
40
150
36
72
66
15
enter the initial head position60
move is from 60 to 87 with seek 27
move is from 87 to 170 with seek 83
move is from 170 to 40 with seek 130
move is from 40 to 150 with seek 110
move is from 150 to 36 with seek 114
move is from 36 to 72 with seek 36
move is from 72 to 66 with seek 6
move is from 66 to 15 with seek 51
total seek time is557
avrage seek time is 69.625000
*/