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1.Shell programming

#!/bin/bash

echo "Enter First Number:"read x

echo "Enter Second Number:"

read y

(( sum=x+y ))

echo "Result of addition:$sum"

Output

chezhian@chezhian-desktop:~$ bash ex2.sh

Enter First Number : 25

Enter Second Number : 25

Result of addition : 50

2.(A)CPU Scheduling: Round Robin

#include

main()

{

int i,j,n,bu[10],wa[10],tat[10],t,ct[10],max;

float awt=0,att=0,temp=0;

printf("Enter the no of processes:"); scanf("%d",&n);

for(i=0;i

{

printf("\n Enter Burst Time for process %d:", i+1); scanf("%d",&bu[i]);

ct[i]=bu[i];

}

printf("\n Enter the size of time slice:"); scanf("%d",&t);

max=bu[0];

for(i=1;i

if(max

max=bu[i];

for(j=0;j<(max/t)+1;j++)

for(i=0;i

if(bu[i]!=0)

if(bu[i]<=t)

{

tat[i]=temp+bu[i];

temp=temp+bu[i];

bu[i]=0;

}

else

{

bu[i]=bu[i]-t;

temp=temp+t;

}

for(i=0;i

{

wa[i]=tat[i]-ct[i];

att+=tat[i];

awt+=wa[i];

}

printf("\n Average Turnaround time is:%f",att/n); printf("\n Average Waiting time is:%f ",awt/n);

printf("\n\tPROCESS\t BURST TIME \t WAITING TIME\tTURNAROUND TIME\n");

for(i=0;i

printf("\t%d \t %d \t\t %d \t\t %d \n",i+1,ct[i],wa[i],tat[i]); getch();

}

Output

Enter the no of processes:3

Enter Burst Time for process 0: 24

Enter Burst Time for process 1: 3

Enter Burst Time for process 2: 3

Enter the size of time slice:3

PROCESS BURST TIME WAITING TIME TURNAROUND

P0 24 6 30

P1 3 4 7

P2 3 7 10

Average Turnaround time is : 15.666667

Average Waiting time is : 5.666667

2.(B)CPU Scheduling:Sortest job first

#include

int main()

{

int

n,j,temp,temp1,temp2,pr[10],b[10],t[10],w[10],p[10],i; float att=0,awt=0; for(i=0;i<10;i++)

{

b[i]=0;w[i]=0;

}

printf("enter the number of process"); scanf("%d",&n);

printf("enter the burst times");

for(i=0;i

{

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

p[i]=i;

}

for(i=0;i

{

for(j=i;j

{

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

{

temp=b[i];

temp1=p[i];

b[i]=b[j];

p[i]=p[j];

b[j]=temp;

p[j]=temp1;

}

w[0]=0;

for(i=0;i

w[i+1]=w[i]+b[i];

for(i=0;i

{

t[i]=w[i]+b[i];

awt=awt+w[i];

att=att+t[i];

}

awt=awt/n;

att=att/n;

printf("\n\t process \t waiting time \t turn around time \n");

for(i=0;i

printf("\t p[%d] \t %d \t\t %d \n",p[i],w[i],t[i]);

printf("the average waitingtimeis %f\n",awt);

printf("the average turn around time is %f\n",att); return 1;

}

Output

Enter the number of process:5

Enter the burst times:

2 4 5 6 8

PROCESS WAITING TIME TURNAROUND

P[0] 0 2

P[1] 2 6

P[2] 6 11

P[3] 11 17

P[4] 17 25

Average Turnaround time is : 7.200000

Average Waiting time is : 12.200000

2.(c)CPU scheduling: First Come First Serve

#include

int main()

{

int n,a[10],b[10],t[10],w[10],g[10],i,m;

float att=0,awt=0;

for(i=0;i<10;i++)

{

a[i]=0; b[i]=0; w[i]=0; g[i]=0;

}

printf("Enter number of process:")

scanf("%d",&n);

printf("Enter burst times:");

for(i=0;i

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

printf("\n Enter arrival times:");

for(i=0;i

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

g[0]=0;

for(i=0;i<10;i++)

g[i+1]=g[i]+b[i];

for(i=0;i

{

w[i]=g[i]-a[i];

t[i]=g[i+1]-a[i];

awt=awt+w[i];

att=att+t[i];

}

awt =awt/n;

att=att/n;

printf("\n\tProcess\tWaiting time\t Turn arround time\n");

for(i=0;i

{

printf("\tp%d\t\t%d\t\t%d\n",i,w[i],t[i]);

}

printf("Average waiting time: %f\n",awt); printf("Average turn around time:%f\n",att);

}

Output

Enter number of process: 4

Enter burst times:

4 9 8 3

Enter arrival times:

0 2 4 3

Process Waiting time Turn arround time

p0 0

p1 2 11

p2 9 17

p3 18 21

Average waiting time : 7.250000

Average turn around time : 13.250000

2.(D)Priority Scheduling

#include

int main()

{

int

bt[20],p[20],wt[20],tat[20],pr[20],i,j,n,total=0,pos,temp,avg_wt,av g_tat;

printf("Enter Total Number of Process:");

scanf("%d",&n);

printf("\nEnter Burst Time and Priority\n"); for(i=0;i

{

printf("\nP[%d]\n",i+1);

printf("Burst Time:");

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

printf("Priority:");

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

p[i]=i+1;

}

for(i=0;i

{

pos=i;

for(j=i+1;j

{

if(pr[j]

pos=j;

}

temp=pr[i];

pr[i]=pr[pos];

pr[pos]=temp;

temp=bt[i];

bt[i]=bt[pos];

bt[pos]=temp;

temp=p[i];

p[i]=p[pos];

p[pos]=temp;

}

wt[0]=0;

for(i=1;i

{

wt[i]=0;

for(j=0;j

wt[i]+=bt[j];

total+=wt[i];

}

avg_wt=total/n;

total=0;

printf("\nProcess\t Burst Time \tWaiting Time\tTurnaround Time");

for(i=0;i

{

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

total+=tat[i];

printf("\nP[%d]\t\t%d\t\t%d\t\t\t%d",p[i],bt[i],wt[i],tat[i]);

}

avg_tat=total/n; //average turnaround time

printf("\n\nAverage Waiting Time=%d",avg_wt);

printf("\nAverage Turnaround Time=%d\n",avg_tat);

return 0;

}

Output

Enter number of process: 4

Enter burst times:

4 9 8 3

Enter arrival times:

0 2 4 3

Process Waiting time Turn arround time

p0 0 4

p1 2 11

p2 9 17

p3 18 21

Average waiting time : 7.250000

Average turn around time : 13.250000

3.(A)Sequential Method

#include

int main()

{

int n,i,j,b[20],sb[20],t[20],x,c[20][20];

printf("Enter no.of files:");

scanf("%d",&n);

for(i=0;i

{

printf("Enter no. of blocks occupied by file%d",i+1); scanf("%d",&b[i]);

printf("Enter the starting block of file%d",i+1);

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

t[i]=sb[i];

for(j=0;j

c[i][j]=sb[i]++;

}

printf("Filename\tStart lock\tlength\n");

for(i=0;i

printf("%d\t %d \t%d\n",i+1,t[i],b[i]);

printf("Enter file name:");

scanf("%d",&x);

printf("File name is:%d",x);

printf("length is:%d",b[x-1]);

printf("blocks occupied:");

for(i=0;i

printf("%4d",c[x-1][i]);

return 0;

}

Output

Enter no.of files: 2

Enter no. of blocks occupied by file1 : 4

Enter the starting block of file1 : 2

Enter no. of blocks occupied by file2 : 10

Enter the starting block of file2 : 5

Filename Start block length

1 2 4

2 5 10

Enter file name: Sai

File name is:12803 length is:0 blocks occupied

3.(B)Indexed Method

#include

int main()

{

int n,m[20],i,j,ib[20],b[20][20];

printf("Enter no. of files:");

scanf("%d",&n);

for(i=0;i

{

printf("Enter index block :",i+1);

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

printf("Enter blocks occupied by file%d:",i+1); scanf("%d",&m[i]);

printf("enter blocks of file%d:",i+1);

for(j=0;j

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

}

printf("\nFile\t index\tlength\n");

for(i=0;i

printf("%d\t%d\t%d\n",i+1,ib[i],m[i]);

printf("blocks occupiedare:");

for(i=0;i

{

printf("fileno%d",i+1);

for(j=0;j

printf("\t%d--->%d\n",ib[i],b[i][j]);

printf("\n");

}

return 0;

}

Output

Enter no. of files :2

Enter index block :3

Enter blocks of file1:

Enter index block :5

File index length

1 3 4

2 5 2

blocks occupied are:fileno1

3--->9

3--->4

3--->6

3--->7

Fileno2

5--->10

5--->8

3.(C)Indexed Method

#include

struct file

{

char fname[10];

int start,size,block[10];

}f[10];

int main()

{

int i,j,n;

printf("Enter no. of files:");

scanf("%d",&n);

for(i=0;i

{

printf("Enter file name:");

scanf("%s",&f[i].fname);

printf("Enter starting block:");

scanf("%d",&f[i].start);

f[i].block[0]=f[i].start;

printf("Enter no.of blocks:");

scanf("%d",&f[i].size);

printf("Enter block numbers:");

for(j=1;j<=f[i].size;j++)

{

scanf("%d",&f[i].block[j]);

}

printf("File\tstart\tsize\tblock\n");

for(i=0;i

{

printf("%s\t%d\t%d\t",f[i].fname,f[i].start,f[i].size);

for(j=1;j<=f[i].size-1;j++)

printf("%d--->",f[i].block[j]);

printf("%d",f[i].block[j]);

printf("\n");

}

return 0;

}

Output

Enter no. of files:2

Enter file name:venkat

Enter starting block:20

Enter no.of blocks:6

Enter block numbers: 4 12 15 45 32 25

Enter file name:rajesh

Enter starting block:12

Enter no.of blocks:5

Enter block numbers:6 5 4 3 2

File start size block

venkat 20 6 4--- >12--- >15 --- >45--- >32--- >25

rajesh 12 5 6--- >5--- >4--- >3 --->2

4.Semaphores

#include

sem_t mutex;

void* thread(void* arg)

{

//wait

sem_wait(&mutex);

printf("\nEntered thread\n");

//critical section

sleep(4);

//signal

printf("\n Exit thread\n");

sem_post(&mutex);

}

int main()

{

sem_init(&mutex, 0, 1);

pthread_t t1,t2;

pthread_create(&t1,NULL,thread,NULL);

sleep(2);

pthread_create(&t2,NULL,thread,NULL);

pthread_join(t1,NULL);

pthread_join(t2,NULL);

sem_destroy(&mutex);

return 0;

}

Output

Output 1: (Have semaphore mean)

Entered Thread

Exit thread

Entered Thread

Exit thread

Output 2: (Don’t Have semaphore mean)

Entered Thread

Exit thread

5.(A)Single level directory

#include

main()

{

intmaster,s[20];

char f[20][20][20];

char d[20][20];

inti,j;

printf("\n Enter number of directories:");

scanf("%d",&master);

printf("\n Enter name of directories:");

for(i=0;i

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

printf("\n Enter the size of directories:");

for(i=0;i

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

printf("\n Enter the file names:");

for(i=0;i

for(j=0;j

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

printf("\n");

printf("\n Directory\tsize\tfilenames");

printf("\n ********************************");

for(i=0;i

{

printf("\n %s\t\t%d\t",d[i],s[i]);

for(j=0;j

printf("%s\n\t\t\t",f[i][j]);

printf("\n");

}

printf("\t\n");

}

Output

Enter number of directories

: 3

Enter name of directories

Enter the size of directories :

Enter the file names :

Directory size filenames

************************************

1 3 9

2 2 6

3 1 4

5.(B)Two level directory

#include

struct st

{

char dname[10];

char sdname[10][10];

char fname[10][10][10];

int ds,sds[10];

}

dir[10];

int main()

{

int i,j,k,n;

printf("enter number of directories:");

scanf("%d",&n);

for(i=0;i

{

printf("enter directory %d names:",i+1);

scanf("%s",&dir[i].dname);

printf("enter size of directories:");

scanf("%d",&dir[i].ds);

for(j=0;j

{

printf("enter subdirectory name and size:");

scanf("%s",&dir[i].sdname[j]);

scanf("%d",&dir[i].sds[j]);

for(k=0;k

{

printf("enter file name:");

scanf("%s",&dir[i].fname[j][k]);

}

printf("\nDirname\t\tSize\tSubdirname\tSize\tFiles");

printf("\n*\n");

for(i=0;i

{

printf("%s\t\t%d",dir[i].dname,dir[i].ds);

for(j=0;j

{

printf("\t%s\t\t%d\t",dir[i].sdname[j],dir[i].sds[j]);

for(k=0;k

printf("%s\t",dir[i].fname[j][k]);

printf("\n\t\t");

}

printf("\n");

}

return 0;

}

Output

Enter number of directories : 1

Enter directories 1 names : Kavin

Enter size of directories : 1

Enter subdirectory name and size : Sai 1

Enter file name : Tamil

Dirname Size Subdirname Size Files ******************************************************

Kavin 1 Sai 1 Tamil

5.(C)Hierarchical directory`

#include

struct tree_element

{

char name[20];

int x, y, ftype, lx, rx, nc, level;

struct tree_element *link[5];

};typedef struct tree_element node;

int main()

{

int gd=DETECT,gm;

node *root;

root=NULL;

create(&root,0,"root",0,639,320);

initgraph(&gd,&gm,"c:\tc\BGI");

display(root);

closegraph();

}

create(node **root,int lev,char *dname,int lx,int rx,int x)

{

int i, gap; if(*root==NULL)

{

(*root)=(node *)malloc(sizeof(node));

printf("Enter name of dir/file(under %s) : ",dname);

fflush(stdin); gets((*root)->name);

printf("Enter 1 for Dir/2 for file :");

scanf("%d",&(*root)->ftype);

(*root)->level=lev;

(*root)->y=50+lev*50;

(*root)->x=x;

(*root)->lx=lx;

(*root)->rx=rx;

for(i=0;i<5;i++)

(*root)->link[i]=NULL;

if((*root)->ftype==1)

{

printf("No of sub directories/files(for %s):",(*root)->name);

scanf("%d",&(*root)>nc);

if((*root)->nc==0)

gap=rx-lx;

else

gap=(rx-lx)/(*root)->nc;

for(i=0;i<(*root)->nc;i++)

create(&((*root)>link[i]),lev+1,(*root)>name,lx+gap*i,lx+gap*i+gap,lx+gap*i+gap/2);

}

else

(*root)->nc=0;

}

display(node *root)

{

int i; settextstyle(2,0,4);

settextjustify(1,1);

setfillstyle(1,BLUE);

setcolor(14);

if(root !=NULL)

{

for(i=0;inc;i++)

line(root->x,root->y,root->link[i]->x,root->link[i]->y);

if(root->ftype==1)

bar3d(root->x-20,root->y-10,root->x+20,root>y+10,0,0);

else

fillellipse(root->x,root->y,20,20);

outtextxy(root->x,root->y,root->name);

for(i=0;inc;i++)

display(root->link[i]);

}

Output

Enter Name of dir/file(under root) : ROOT

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for ROOT) : 2

Enter Name of dir/file(under ROOT) : USER1

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for USER1) : 1

Enter Name of dir/file(under USER1) : SUBDIR1

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for SUBDIR1): 2

Enter Name of dir/file(under USER1) : JAVA

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for JAVA) : 0

Enter Name of dir/file(under SUBDIR1) : VB

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for VB) : 0

Enter Name of dir/file(under ROOT) : USER2

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for USER2) : 2

Enter Name of dir/file(under ROOT) : A

Enter 1 for Dir/2 for File : 2

Enter Name of dir/file(under USER2) : SUBDIR2

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for SUBDIR2): 2

Enter Name of dir/file(under SUBDIR2) : PPL

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for PPL) : 2

Enter Name of dir/file(under PPL) : B

Enter 1 for Dir/2 for File : 2

Enter Name of dir/file(under PPL) : C

Enter 1 for Dir/2 for File : 2

Enter Name of dir/file(under SUBDIR) : AI

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for AI) : 2

Enter Name of dir/file(under AI) : D

Enter 1 for Dir/2 for File : 2

Enter Name of dir/file(under AI) : E

Enter 1 for Dir/2 for File : 2

5.(D)DAG

#include

struct tree_element

{

char name[20];

int x,y,ftype,lx,rx,nc,level;

struct tree_element *link[5];

};

typedef struct tree_element node;

typedef struct

{

char from[20];

char to[20];

}

link;

Link L[10]; Int nofl;

node * root;

void main()

{

int gd=DETECT,gm; root=NULL;

create(&root,0,"root",0,639,320);

read_links();

clrscr();

initgraph(&gd,&gm,"c:\\tc\\BGI");

draw_link_lines();

display(root);

getch();

closegraph();

}

read_links()

{

int i;

printf("how many links");

scanf("%d",&nofl);

for(i=0;i

{

printf("File/dir:");

fflush(stdin);

gets(L[i].from);

printf("user name:");

fflush(stdin);

gets(L[i].to);

}

draw_link_lines()

{

int i,x1,y1,x2,y2;

for(i=0;i

{

search(root,L[i].from,&x1,&y1);

search(root,L[i].to,&x2,&y2);

setcolor(LIGHTGREEN);

setlinestyle(3,0,1);

line(x1,y1,x2,y2);

setcolor(YELLOW);

setlinestyle(0,0,1);

}

search(node *root,char *s,int *x,int *y)

{

int i;

if(root!=NULL)

{

if(strcmpi(root->name,s)==0)

{

*x=root->x;

*y=root->y;

return;

}

else

{

for(i=0;inc;i++) search(root->link[i],s,x,y);

}

create(node **root,int lev,char * dname,int lx,int rx,int x)

{

int i,gap; if(*root==NULL)

{

(*root)=(node *)malloc(sizeof(node));

printf("enter name of dir/file(under %s):",dname);

fflush(stdin);

gets((*root)->name);

printf("enter 1 for dir/ 2 for file:");

scanf("%d",&(*root)- >ftype);

(*root)->level=lev;

(*root)->y=50+lev*50;

(*root)->x=x;

(*root)->lx=lx;

(*root)->rx=rx;

for(i=0;i<5;i++) (*root)->link[i]=NULL;

if((*root)->ftype==1)

{

printf("no of sub directories /files (for %s):",(*root)->name);

scanf("%d",&(*root)->nc);

if((*root)->nc==0) gap=rx-lx;

else

gap=(rx-lx)/(*root)->nc;

for(i=0;i<(*root)->nc;i++)

create( & ( (*root)->link[i] ) , lev+1 , (*root)->name,lx+gap*i,lx+gap*i+gap,lx+gap*i+gap/2);

}

else

(*root)->nc=0;

}

display(node *root)

{

int i;

settextstyle(2,0,4);

settextjustify(1,1);

setfillstyle(1,BLUE);

setcolor(14);

ot!=NULL)

{

for(i=0;inc;i++)

{

line(root->x,root->y,root->link[i]->x,root->link[i]->y);

}

if(root->ftype==1) bar3d(root->x-20,root->y-10,root->x+20,root->y+10,0,0);

else

fillellipse(root->x,root->y,20,20);

outtextxy(root->x,root->y,root->name);

for(i=0;inc;i++)

{

display(root->link[i]);

}

Output

Enter Name of dir/file(under root) : ROOT

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for ROOT) : 2

Enter Name of dir/file(under ROOT) : USER1

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for USER1) : 1

Enter Name of dir/file(under USER1) : SUBDIR1

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for SUBDIR1): 2

Enter Name of dir/file(under USER1) : JAVA

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for JAVA) : 0

Enter Name of dir/file(under SUBDIR1) : VB

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for VB) : 0

Enter Name of dir/file(under ROOT) : USER2

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for USER2) : 2

Enter Name of dir/file(under ROOT) : A

Enter 1 for Dir/2 for File : 2

Enter Name of dir/file(under USER2) : SUBDIR2

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for SUBDIR2): 2

Enter Name of dir/file(under SUBDIR2) : PPL

Enter 1 for Dir/2 for File : 1

No of subdirectories/files(for PPL) : 2

Enter Name of dir/file(under PPL) : B

6.Deadlock Avoidance

#include

int main()

{

int n, m, i, j, k;

n = 5;

m = 3;

int alloc[5][3] = { { 0, 1, 0 },

{ 2, 0, 0 }, { 3, 0, 2 }, { 2, 1, 1 }, { 0, 0, 2 }

};

int max[5][3] = { { 7, 5, 3 },

{ 3, 2, 2 }, { 9, 0, 2 }, { 2, 2, 2 }, { 4, 3, 3 } };

int avail[3] = { 3, 3, 2 };

int f[n], ans[n], ind = 0;

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

{

f[k] = 0;

}

int need[n][m];

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

{

for (j = 0; j < m; j++)

need[i][j] = max[i][j] - alloc[i][j];

}

int y = 0;

for (k = 0; k < 5; k++) { for (i = 0; i < n; i++)

{

if (f[i] == 0)

{

int flag = 0;

for (j = 0; j < m; j++)

{

if (need[i][j] > avail[j])

{

flag = 1;

break;

}

if (flag == 0)

{

ans[ind++] = i;

for (y = 0; y < m; y++)

avail[y] += alloc[i][y];

f[i] = 1;

}

int flag = 1;

for(int i=0;i

{

if(f[i]==0)

{

flag=0;

printf("The following system is not safe");

break;

}

if(flag==1)

{

printf("Following is the SAFE Sequence\n");

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

printf(" P%d ->", ans[i]);

printf(" P%d", ans[n - 1]);

}

return (0);

}

Output

Following is the SAFE Sequence

P1 -> P3 -> P4 -> P0 -> P2

7.Deadlock detection

#include

int main()

{

int

found,flag,l,p[4][5],tp,tr,c[4][5],i,j,k=1,m[5],r[5],a[5],temp[5],sum=0;

printf("Enter total no of processe

scanf("%d",&tp);

printf("Enter total no of resources");

scanf("%d",&tr);

printf("Enter claim (Max. Need) matrix\n");

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

{

printf("process %d:\n",i);

for(j=1;j<=tr;j++)

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

}

printf("Enter allocation matrix\n");

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

{

printf("process %d:\n",i);

for(j=1;j<=tr;j++)scanf("%d",&p[i][j]);

}

printf("Enter resource vector (Total resources):\n");

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

{

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

}

printf("Enter availability vector (available resources):\n");

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

{

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

temp[i]=a[i];

}

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

{

sum=0;

for(j=1;j<=tr;j++)

{

sum+=p[i][j];

}

if(sum==0)

{

m[k]=i; k++;

}

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

{

for(l=1;l

if(i!=m[l])

{

flag=1; for(j=1;j<=tr;j++)

if(c[i][j]

{

flag=0; break;

}

if(flag==1)

{

m[k]=i; k++;

for(j=1;j<=tr;j++)

temp[j]+=p[i][j];

}

printf("Deadlock causing processes are:");

for(j=1;j<=tp;j++)

{

found=0;

for(i=1;i

{

if(j==m[i])

found=1;

}

if(found==0)

printf("%d\t",j);

}

Output

Enter total no. of processes : 4

Enter total no. of resources : 5

Enter claim (Max. Need) matrix :

0 0 0 0 1

1 0 1 0 1

Enter allocation matrix :

1 0 1 1 0

1 1 0 0 0

0 0 0 1 0

0 00 0 0

Enter resource vector (Total resources) :

2 1 1 2 1

Enter availability vector (available resources) :

0 00 0 1

deadlock causing processes are : 2 3

8.(A)FIFO

#include

int fr[3];

void main()

{

void display();

int i,j,page[12]=

{2,3,2,1,5,2,4,5,3,2,5,2};

intflag1=0,flag2=0,pf=0,frsize=3,top=0;

for(i=0;i<3;i++)

{

fr[i]=-1;

}

for(j=0;j<12;j++)

{

flag1=0;

flag2=0;

for(i=0;i<12;i++)<\p>

{

if(fr[i]==page[j])

{

flag1=1;

flag2=1;

break;

}

if(flag1==0)

{

for(i=0;i

{

if(fr[i]==-1)

{

fr[i]=page[j];

flag2=1;

break;

}

if(flag2==0)

{

fr[top]=page[j];

top++;

pf++;

if(top>=frsize)

top=0;

}

display();

}

printf("Number of page faults : %d ",pf+frsize); }

void display()

{

int i;

printf("\n"); for(i=0;i<3;i++);

printf("%d\t",fr[i]);

}

Output

2 -1 -1

2 3 -1

2 3 1

5 3 1

5 2 1

5 2 4

3 2 4

3 5 4

3 5 2

8.(B)LRU

#include

int fr[3];

void main()

{

void display();

int p[12]={2,3,2,1,5,2,4,5,3,2,5,2},i,j,fs[3];

int

index,k,l,flag1=0,flag2=0,pf=0,frsize=3;

for(i=0;i<3;i++)

{

fr[i]=-1;

}

for(j=0;j<12;j++)

{

flag1=0; flag2=0;

for(i=0;i<3;i++)

{

if(fr[i]==p[j])

{

flag1=1; flag2=1;

break;

}

if(flag1==0)

{

for(i=0;i<3;i++)

{

if(fr[i]==-1)

{

fr[i]=p[j];

flag2=1;

break;

}

if(flag2==0)

{

for(i=0;i<3;i++) fs[i]=0;

for(k=j-1,l=1;l<=frsize-1;l++,k--)

{

for(i=0;i<3;i++)

{

if(fr[i]==p[k])

fs[i]=1;

}

for(i=0;i<3;i++)

{

if(fs[i]==0)

index=i;

}

fr[index]=p[j];

pf++;

}

display();

}

printf("\n no of page faults :%d",pf+frsize);

}

void display()

{

int i;

for(i=0;i<3;i++)

printf("\t%d",fr[i]);

}

Output

2 -1 -1

2 3 -1

2 3 1

2 5 1

2 5 4

3 5 4

3 5 2

No of page faults: 7

8.(C)LFU

#include

int fr[3], n, m;

void display();

void main()

{

int i,j,p ge[20],fs[10];

int max,found=0,lg[3],index,k,l,flag1=0,flag2=0,pf=0; float pr;

clrscr();

printf("Enter length of the reference string: ");

scanf("%d",&n);

printf("Enter the reference string: ");

for(i=0;i

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

printf("Enter no of frames: ");

scanf("%d",&m);

for(i=0;i

pf=m;

for(j=0;j

{

flag1=0;

flag2=0;

for(i=0;i

{

if(fr[i]==page[j])

{

flag1=1;

flag2=1;

break;

}

if(flag1==0)

{

for(i=0;i

{

if(fr[i]==-1)

{

fr[i]=page[j];

flag2=1;

break;

}

if(flag2==0)

{

for(i=0;ilg[i]=0;

for(i=0;i{

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

{

if(fr[i]==page[k])

{

lg[i]=k-j;

}

found=0;

for(i=0;i

{

if(lg[i]==0)

{

}

if(found==0)

{

max=lg[0];

index=0;

for(i=0;i

{

if(max

{

max=lg[i];

index=i;

}

fr[index]=page[j];

pf++;

}

display();

}

printf("Number of page faults : %d\n",); pr=(float)pf/n*100;

rintf("Page fault rate = %f \n", pr);

}

void display()

int i;

for(i=0;i

printf("%d\t",fr[i]);

}

output

Enter length of the reference string: 12

Enter the reference string: 1 2 3 4 1 2 5 1 2 3 4 5

Enter no of frames: 3

1 -1 -1

1 2 -1

1 2 3

1 2 4

1 2 5

3 2 5

4 2 5

Number of page faults : 7

Page fault rate = 58.333332

9.Shared Memory and IPC

#include

main()

{

int pid,pfd[2],n,a,b,c;

if(pipe(pfd)==-1)

{

printf("\n Error in pipe onnection\n");

exit(1);

}

pid=fork();

if(pid>0)

{

printf("\n Parent Process");

printf("\n\n\t Fibonacci Series");

printf("\n Enter the limit for the

series:");

scanf("%d",&n);

close(pfd[0]);

write(pfd[1],&n,sizeof(n));

close(pfd[1]);

exit(0);

}

else

{

close(pfd[1]);

read(pfd[0],&n,sizeof(n));

printf("\n Child process");

a=0;b=1;

close(pfd[0]);

printf("\n Fibonacci series is:");

printf("\n\n%d\n%d",a,b);

while(n>2)

{

c=a+b;

printf("\n%d",c);

a=b;b=c;

n--;

}

Output

Parent Process

Fibonacci Series

Enter the limit for the series : 5

Child Process

Fibonacci Series is:

10.Paging Technique

#include

int main()

{

int np,ps,i;

int *sa;

printf("Enter how many pages\n");

scanf("%d",&np);

printf("Enter the page size \n");

scanf("%d",&ps);

sa=(int*)malloc(2*np);

for(i=0;i

{

sa[i]=(int)malloc(ps);

printf("Page%d\t Address %u\n",i+1,sa[i]);

return 0;

}

Output

Enter how many pages: 5

Enter the page size: 4

Page1 Address: 1894

Page2 Address: 1902

Page3 Address: 1910

Page4 Address: 1918

Page5 Address: 1926

11.Thread

#include

pthread_t tid[2];

int counter;

void* doSomeThing(void *arg)

{

unsigned long i = 0;

counter += 1;

printf("\n Job %d started\n",

counter);

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

printf("\n Job %d finished\n",

counter);

return NULL;

}

int main(void)

{

int i = 0;

int err;

while(i < 2)

{

int counter;

void* &doSomeThing, NULL);

if (err != 0)

printf ("\ncan't create thread :[%s]", strerror(err));

i++;

}

pthread_join(tid[0], NULL);

pthread_join(tid[1], NULL);

urn 0;

}

Output

Job 1 started

Job 2 started

Job 1 finished

Job 2 finished