JYOTHISHMATHI INSTITUTE OF TECHNOLOGY & SCIENCE NUSTULAPUR, KARIMNAGAR

JYOTHISHMATHI INSTITUTE OF TECHNOLOGY & SCIENCE

1

JYOTHISHMATHI INSTITUTE OF TECHNOLOGY & SCIENCE NUSTULAPUR, KARIMNAGAR


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LAB MANUAL OF UNIX & OS LAB (IT)

(III B.Tech. II SEM)


3

INDEX

SNO PROGRAM NAME PAGE

NO

1 Write a shell script to print the multiplication

table for the given number. 1

2 Write a shell script that copies multiple files

into a directory. 2

3 Write a shell script to find no. of words and

characters in a given file. 3

4 Write a shell script to display all files in a

given directory. 4

5 Write a shell script to perform simple calculator. 5

6 Write a C program to count no.of blanks,characters

a) Using standard i/o function. b) Using system

calls.

6-7

7 Write a C program to illustrate the following

Command using system Calls a) cat b) ls c)

mv

8-10

8 Write a C program to illustrate the stat system

call. 11

9 Write a C program that illustrates the creation of

child process using fork() system call. 12

10 Write a C program that illustrates file locking. 13-14

11 Write a C program that implements producer

consumer problem using semaphore system calls. 15-16

12 Write a C program to illustrate inter process

communication using shared memory system calls. 17-18

13 Write a C program to (a) create message queue

(b) write to message queue (c) read from

message queue

19-20


4

O.S. PROGRAMMES

S.NO PROGRAM NAME PAGE

NO

1 Write a C program for ROUND ROBIN CPU scheduling

algorithm 20-21

2 Write a C program for SJF CPU scheduling algorithm 22-23

3 Write a C program for FCFS CPU scheduling

algorithm 24

4 Write a C Program for priority CPU scheduling

algorithm.

25-26

5 Write a C Program for sequential file allocation. 27-28

6 Program for indexed file allocation strategy. 29-30

7 Program for linked file allocation strategy. 31-32

8 Write a C Program for MVT first fit. 33

9 Write a C Program for MFT . 34

10 Write a C Program for MVT best fit. 35-36

11 Program for bankers algorithm 37-39

12 Write a C program for FIFO page allocation

algorithm.

40-41

13 Write a C program for LRU page replacement

algorithm.

42-44


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Program 1:Program 1:Program 1:Program 1: AIM: AIM: AIM: AIM: Write a shell script to print the multiplication table for the Write a shell script to print the multiplication table for the Write a shell script to print the multiplication table for the Write a shell script to print the multiplication table for the given number.given number.given number.given number.

Algorithm:Algorithm:Algorithm:Algorithm:

1.Start.

2.Read n.

3.i=1.

4.f = n * i.

5.i = i + 1.

6.Display f.

7.Repeat 4,5,6 steps until i = 10.

8.End.

Source Code:Source Code:Source Code:Source Code: #! /bin/sh

echo “enter the number”

read n

for i in 1 2 3 4 5 6 7 8 9 10

do

x= èxpr $n \* $i`

done

Input: Input: Input: Input:

enter the number

6

Output:Output:Output:Output:

6 * 1 = 6

6 * 2 = 12


6

6 * 3 = 18

6 * 4 = 24

6 * 5 = 30

6 * 6 = 36

6 * 7 = 42

6 * 8 = 48

6 * 9 = 54

6 * 10 = 60

Program 2:Program 2:Program 2:Program 2:

AIM: AIM: AIM: AIM: Write a shell script that copies multiple files into a directory.Write a shell script that copies multiple files into a directory.Write a shell script that copies multiple files into a directory.Write a shell script that copies multiple files into a directory.


1.Start.

2.Read *.c files into i.

3.Copy $i to the root directory.

4.Repeat 2,3 steps until all values of $i..

5.End.


for i in `ls *.c`

do

cp $i /root

done

echo “file are copied into root”

Output:Output:Output:Output: file are copied into root”

Program 3 :Program 3 :Program 3 :Program 3 :

AIM: AIM: AIM: AIM: Write a shell script to find no. of words and characters in a Write a shell script to find no. of words and characters in a Write a shell script to find no. of words and characters in a Write a shell script to find no. of words and characters in a given file.given file.given file.given file. Algorithm:Algorithm:Algorithm:Algorithm:

1. Start.


7

2. cc = 0,w=0. 3. give file name as commandline argument (i.e $1). 4. for i in `cat $1`. 5. w = w + 1. 6. c = expr “$i” : “*”. 7. cc = cc + c. 8. Repeat 4,5,6,7 steps until eof. 9. End.

Source Code:Source Code:Source Code:Source Code:

# /bin/sh

w=0

cc=0

for i in ‘cat $1’

do

j= $i

echo $j

w=èxpr $w + 1`

c=èxpr “$j”:”.*”`

cc = èxpr $cc + $c`

done

echo “no.of characters “ $cc

echo “no.of words” $w

Input:Input:Input:Input:. ./a.out sun


no. of characters 56

no. of words 11


8

Program 4:Program 4:Program 4:Program 4: AIM: AIM: AIM: AIM: Write a shell script to display all files in a given directory.Write a shell script to display all files in a given directory.Write a shell script to display all files in a given directory.Write a shell script to display all files in a given directory. Algorithm:Algorithm:Algorithm:Algorithm: 1.Start.

2.Read i value from ls sss.

3.Display i value .

4.Repeat above 2 steps until end of directory.

5.End.


for i in `ls sss`

do

echo $i

done

Output:Output:Output:Output: abc.txt

ss.c


9

Program 5:Program 5:Program 5:Program 5: AIM:AIM:AIM:AIM: Write a shell script to perform simple calculator.Write a shell script to perform simple calculator.Write a shell script to perform simple calculator.Write a shell script to perform simple calculator. Algorithm:Algorithm:Algorithm:Algorithm:

1.Start

2.Read a and b

3.Read op

4.Depending on value of operator perform case

Operation.

5.End


#! /bin/sh

echo ‘enter the value for a’

read a

echo ‘enter the value for b’

read b


10

echo ‘enter operator’

read op

case $op in

+) c=èxpr $a + $b`;;

-) c = èxpr $a -$b`;;

\*) c = èxpr $a \* $b`;;

/) c = èxpr $a / $b ;;

esac

echo $c

Input:Input:Input:Input:

enter the value for a

3

enter the value for b

6

enter the operator

*


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Program 6a:Program 6a:Program 6a:Program 6a: AIM: AIM: AIM: AIM: Write a C program to count no. of blanks, characters, lines Write a C program to count no. of blanks, characters, lines Write a C program to count no. of blanks, characters, lines Write a C program to count no. of blanks, characters, lines using standard i/o function.using standard i/o function.using standard i/o function.using standard i/o function. Algorithm:Algorithm:Algorithm:Algorithm:

1. Start. 2. open the file using fopen( ) function in “r” mode 3. ch=fgetc(fp) (to read character by character) 4. if ch = ‘ ‘ or ‘\n’ b=b+1.

5. if ch = ‘\n’ l=l+1.

6. c=c+1. 7. Repeat 3,4,5&6 steps until ch = eof 8. End

Source Code:Source Code:Source Code:Source Code: #include<stdio.h>

int main( )

{


11

file *fp:

int b=0,nl=0,c=0;

char ch;

fp=fopen(“text.txt”,”r”);

while(ch!=eof)

{

ch=fgetc(fp);

if(ch==’ ‘)

b++;

if(ch==’\n’)

nl++;

c++;

}

printf(“no.of blanks %d”,b);

printf(“no.of lines %d”,nl);

printf(“no.of characters %d”,c);

}

Input:Input:Input:Input: ./a.out sss.txt


no.of blanks 5

n.of lines 2

no.of characters 36

Program 6b:Program 6b:Program 6b:Program 6b: AIM:AIM:AIM:AIM: Write a C program to count no. of blanks, characters, lines Write a C program to count no. of blanks, characters, lines Write a C program to count no. of blanks, characters, lines Write a C program to count no. of blanks, characters, lines using system calls.using system calls.using system calls.using system calls. Algorithm:Algorithm:Algorithm:Algorithm:

1. Start. 2. open the file using open( ) system call in O_RDONLY 3. rc=read(fd,buf,5) (to read characters ) 4. if ch = ‘ ‘ or ‘\n’ w=w+1.

5. if ch = ‘\n’ l=l+1.

6. c=c+1. 7. Repeat 3,4,5&6 steps until ch = eof 8. End


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int main( )

{

int fd:

int b=0,nl=0,c=0;

char ch[1];

fd=open(“text.txt”,O_RDONLY);

while((rc=read(fd,ch,1))>0)

{

if(ch==’ ‘)

b++;

if(ch==’\n’)

nl++;

c++;

}

printf(“no.of blanks %d”,b);

printf(“no.of lines %d”,nl);

printf(“no.of characters %d”,c);

}

Input:Input:Input:Input: ./a.out sss.txt

Output:Output:Output:Output: no. of blanks 5

no. of lines 2

no. of characters 36

Program 7a:Program 7a:Program 7a:Program 7a: AIM:AIM:AIM:AIM: Write a C program to illustrate the mv command using Write a C program to illustrate the mv command using Write a C program to illustrate the mv command using Write a C program to illustrate the mv command using system Callssystem Callssystem Callssystem Calls Algorithm:Algorithm:Algorithm:Algorithm:

1. open one existed file and one new open file using open( ) system call

2. read the contents from keyboard using read( ) 3. write these contents into file using write() 4. repeat 2,3 steps until eof 5. close 2 file using fclose( ) system call

6. delete existed file using using unlink( ) system


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#include<stdio.h>

#include<unistd.h>

#include<sys/types.h>

int main()

{

int fd,fd1,rc;

char ch[5];

fd1=open(“sss”,O_WRONLY | O_CREATE);

while((rc=read(stdin,ch,5))>0)

write(fd1,ch,rc);

close fd1;

fd1=open(“sss”,O_RDONLY);

while((rc=read(fd1,ch,5))>0)

write(stdout,ch,rc);

close fd1;

}

Input:Input:Input:Input: hai hello A friend in need is a friend in deed

^z

Output:Output:Output:Output: hai hello A friend in need is a friend in dee

Program 7b :Program 7b :Program 7b :Program 7b : AIM: AIM: AIM: AIM: Write a program to illustrate “ls” command using system callsWrite a program to illustrate “ls” command using system callsWrite a program to illustrate “ls” command using system callsWrite a program to illustrate “ls” command using system calls Algorithm:Algorithm:Algorithm:Algorithm:

1. Start. 2. open directory using opendir( ) system call. 3. read the directory using readdir( ) system call. 4. print dp.name and dp.inode . 5. repeat above step until end of directory. 6. End


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#include<stdio.h>

#include<dirent.h>

int main(int argc,char *argv[])

{

dir *dp;

struct dirent *d;

if(((dp=opendir(“sss”)))==null)

printf(“cannodt open\n”);

while((d=readdir(dp)!=null))

{

printf(“%s\t”,d->d_name);

printf(“%d\n”,d->d_ino);

}

closedir(dirp);

exit(0);

}

output:output:output:output:

mul.sh 12344

division.c 12345

process.c 12346

Program 7c:Program 7c:Program 7c:Program 7c: AIM:AIM:AIM:AIM: Write a C program to illustrate the mv command using Write a C program to illustrate the mv command using Write a C program to illustrate the mv command using Write a C program to illustrate the mv command using system calls.system calls.system calls.system calls. Algorithm:Algorithm:Algorithm:Algorithm:

1. Start 2. open an existed file and one new open file using open()


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system call

3. read the contents from existed file using read( ) system call

4. write these contents into new file using write system call using write( ) system call

5. repeat above 2 steps until eof 6. close 2 file using fclose( ) system call

7. delete existed file using using unlink( ) system 8. End.


#include<stdio.h>

#include<unistd.h>


int main()

{

int fd,fd1,rc;

char ch[5];

fd=open(“sss”,O_RDONLY );

fd1= open(“rrr”,O_WRONLY | O_CREATE);

while((rc=read(fd,ch,5))>0)

write(fd1,ch,rc);

close fd1;

close fd;

unlink(fd);

printf(“ file is copied “);

}

Output:Output:Output:Output: file is copied

Program 8:Program 8:Program 8:Program 8: AIM:AIM:AIM:AIM: Write a C program to illustrate the stat system callsWrite a C program to illustrate the stat system callsWrite a C program to illustrate the stat system callsWrite a C program to illustrate the stat system calls Algorithm:Algorithm:Algorithm:Algorithm:

1. Declare s and st of type struct stat. 2. Use stat system call by specifying sss.c file


16

3. stat returns s (s may be 0 or -1) 4. if s<0 then

display there is no file with the name sss.c.

5. Display information about sss.c file 6. Display inode number of sss.c. 7. Display last access time of file sss.c. 8. Display size of the file sss.c.


#include<stdio.h>

#include<unistd.h>


#include<sys/stat.h>

int main()

{

int s;

struct stat st;

s=stat(“sss.c”,&st);

if(s<0)

{

printf(“use of stat() is unsuccessful “);

exit(0);

}

printf(“inode number of sss.c file is %d“,st.st_ino);

printf(“no.of links to sss.c is %d“, st.st_nlink );

printf(“file permissions sss.c is %d“,st.st_mode);

printf(“size of sss.c file is “,st.st_size);

printf(“last access time of sss.c %d”,st.st_atime );

}

OutputOutputOutputOutput:::: inode number of sss.c file is 12345 no.of links to sss.c file is 2

file permissions of sss.c file is 666

size of sss.c file is 8

last access time of sss.c file is 7:40

Program 9:Program 9:Program 9:Program 9: AIM: AIM: AIM: AIM: Write a C program thWrite a C program thWrite a C program thWrite a C program that illustrates the creation of child at illustrates the creation of child at illustrates the creation of child at illustrates the creation of child process using fork( ) system callprocess using fork( ) system callprocess using fork( ) system callprocess using fork( ) system call


17


1. Start 2. Declare pid 3. create new process using fork( ) system call 4. If pid!=0 then 5. Display parent process getpid(),getppid(). 6. Else 7. Display child process getpid().getppid(). 8. End

Source code:Source code:Source code:Source code:

#include<stdio.h>

int main( )

{

printf(“original process with pid %d ppid %d\n”,

getpid() ,getppid());

pid=fork();

if(pid!=0)

printf(“parent process with pid %d ppid %d \n”,

getpid(),getppid());

else

{

sleep(5);

printf(“child process with pid %d ppid %d\n”,

getpid(),getppid());

}

printf(“ pid %d terminates “,getpid());

}


original process with pid 3456 and ppid 3525

child process with pid 3457 and ppid 3456

pid 3457 terminates

parent process with pid 3456 and ppid 3525

pid 3456 terminates


AIM:AIM:AIM:AIM: Write a C program that illustrates file locking.Write a C program that illustrates file locking.Write a C program that illustrates file locking.Write a C program that illustrates file locking.


18

AlAlAlAlgorithm:gorithm:gorithm:gorithm:

1. Start 2. Flock is predefined data structure for locking techniques.

3. Opening rr.txt in read and write mode. 4. l_type is read lock 5. By pressing return it reads that character through getchar() function.

6. It gets the lock. 7. Next we are sending unlock type into l_type. 8. By pressing return it gets unlocked.

9. End


#include<stdio.h>

#include<stdlib.h>

#include<fcntl.h>

#include<unistd.h>

int main(int argc,char* argv[])

{

struct flock f1={f_wrlck,seek_set,0,0,0);

int fd;

f1.l_type=f_rdlck;

if((fd=open(‘rr.txt”,o_rdwr))==-1)

{

perror(“open”);

exit(1);

}

printf(“press<return> to try to get lock:”);

getchar();

printf(“trying to get lock”):

if(fnctl(fd,f_setlkw,&f1)==-1)

{

perror(“fcntl”);

exit(1);

}

printf(“got lock \n”);

printf(“press <return> to release lock:”);

getchar( );

f1.l_type=f_unlck;

if(fcntl(fd,f_setlk,&f1)==-1)

{

perror(“fcntl”);

exit(1);

}


19

printf(“unlocked\n”);

close(fd);

}


press<return> to try to get lock

trying to get lock

press <return> to release lock

unlocked


20

Program 11:Program 11:Program 11:Program 11: AIM:AIM:AIM:AIM:

Write a C program that implements producer consumerWrite a C program that implements producer consumerWrite a C program that implements producer consumerWrite a C program that implements producer consumer problem using semaphore system callsproblem using semaphore system callsproblem using semaphore system callsproblem using semaphore system calls


1. Start 2. create semaphore using semget( ) system call 3. if successful it returns positive value 4. create two new processes 5. first process will produce 6. until first process produces second process cannot

consume

7. End.

Source code:Source code:Source code:Source code:

#include<stdio.h>

#include<stdlib.h>


#include<sys/ipc.h>

#include<sys/sem.h>

#include<unistd.h>

#define num_loops 2


{

int sem_set_id;

int child_pid,i,sem_val;

struct sembuf sem_op;

int rc;

struct timespec delay;

clrscr();

sem_set_id=semget(ipc_private,2,0600);

if(sem_set_id==-1)

{

perror(“main:semget”);

exit(1);

}

printf(“semaphore set created,semaphore setid‘%d’\n ”,

sem_set_id);


21

child_pid=fork();

switch(child_pid)

{

case -1:

perror(“fork”);

exit(1);

case 0:

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

{

sem_op.sem_num=0;

sem_op.sem_op=-1;

sem_op.sem_flg=0;

semop(sem_set_id,&sem_op,1);

printf(“producer:’%d’\n”,i);

fflush(stdout);

}

break;

default:

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

{

printf(“consumer:’%d’\n”,i);

fflush(stdout);

sem_op.sem_num=0;

sem_op.sem_op=1;

sem_op.sem_flg=0;

semop(sem_set_id,&sem_op,1);

if(rand()>3*(rano_max14));

{

delay.tv_sec=0;

delay.tv_nsec=10;

nanosleep(&delay,null);

}

}

break;

}

return 0;

}

Output:Output:Output:Output: semaphore set created

semaphore set id ‘327690’

producer: ‘0’

consumer:’0’

producer:’1’


22

consumer:’1’

Program 12:Program 12:Program 12:Program 12: AIM: AIM: AIM: AIM: Write a C program to illustrate inter process communication Write a C program to illustrate inter process communication Write a C program to illustrate inter process communication Write a C program to illustrate inter process communication using shared memory system calls.using shared memory system calls.using shared memory system calls.using shared memory system calls. Algorithm:Algorithm:Algorithm:Algorithm:

1. Start 2. create shared memory using shmget( ) system call 3. if success full it returns positive value 4. attach the created shared memory using shmat( ) system

call

5. write to shared memory using shmsnd( ) system call 6. read the contents from shared memory using shmrcv( )

system call

7. End .


#include<stdio.h>

#include<stdlib.h>

#include<sys/ipc.h>


#include<string.h>

#include<sys/shm.h>

#define shm_size 1024

int main(int argc,char * argv[])

{

key_t key;

int shmid;

char *data;

int mode;

if(argc>2)

{

fprintf(stderr,”usage:stdemo[data_to_writte]\n”);

exit(1);

}

if((shmid=shmget(key,shm_size,0644/ipc_creat))==-1)

{

perror(“shmget”);

exit(1);


23

}

data=shmat(shmid,(void *)0,0);

if(data==(char *)(-1))

{

perror(“shmat”);

exit(1);

}

if(argc==2)

printf(writing to segment:\”%s”\”\n”,data);

if(shmdt(data)==-1)

{

perror(“shmdt”);

exit(1);

}

return 0;

}


#./a.out swarupa


writing to segment swarupa


24

Program 13:Program 13:Program 13:Program 13: AIM: AIM: AIM: AIM:

Write a C program to Write a C program to Write a C program to Write a C program to ( a ) create message queue( a ) create message queue( a ) create message queue( a ) create message queue ( b ) write to message queue( b ) write to message queue( b ) write to message queue( b ) write to message queue ( c ) read from message queue( c ) read from message queue( c ) read from message queue( c ) read from message queue Algorithm:Algorithm:Algorithm:Algorithm:

1. Start 2. create message queue using msgget( ) system call 3. while creating message queue it returns id into file descriptor

4. print the message queue id 5. if successful rite into message queue using msgsnd()

system call

6. read the contents from message queue using msgrcv( ) system call

7. End.


#include<stdio.h>

#include<sys/ipc.h>

#include<sys/msg.h>



{

int q_id;

if(q_id=msgget(ipc_private,0600)==-1)

perror(msgget”);

struct msgbuf

{

long mtype;

char mtext[20];

}

struct msgbuf* msg;


25

struct msgbuf* recv_msg;

int rc;

printf(“message queue created %d\n”,q_id);

msg=(struct msgbuf*)malloc(sizeof(struct msgbuf) +

strlen(“hello world”));

msg->mtype=1;

strcpy(msg->mtext,helloworld”);

free(msg);

printf(“message placed on queue successfully\n”);

recv_msg=(struct msgbuf*)malloc(sizeof(struct msgbuf)+

strlen(“hello world”));

if(rc==-1)

{

perror(“main:msgrcv”);

exit(1);

}

printf(“msgrcv:received message:\nmtype’%d’\n; mtext

’%s’\n”,recv_msg->mtype,recv_msg->mtext);

return 0;

}


message queue created 0

message placed on the queue successfully

msgrcv:received message:

mtype: 0

mtext: hello world


26

LAB MANUAL OFLAB MANUAL OFLAB MANUAL OFLAB MANUAL OF

OPERATING SYSTEM LABOPERATING SYSTEM LABOPERATING SYSTEM LABOPERATING SYSTEM LAB


27

Program 1:Program 1:Program 1:Program 1: AIM: AIM: AIM: AIM: Write a C program for ROUND ROBIN CPU scheduling algorithm.Write a C program for ROUND ROBIN CPU scheduling algorithm.Write a C program for ROUND ROBIN CPU scheduling algorithm.Write a C program for ROUND ROBIN CPU scheduling algorithm. Source Code:Source Code:Source Code:Source Code: #include<stdio.h>

main()

{

int s[10],p[10],n,i,j,wi=0,w[10],t[10], st[10],tq,tst=0;

int tt=0,tw=0;

float aw at;

printf("enter no.of process");

scanf("%d",&n);

printf("\n enter time quanum");

scanf("%d",&tq);

printf("\n enter process&service time");

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

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


28

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

st[i]=s[i];

tst=tst+s[i];

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

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

{

if(s[i]>tq)

{

s[i]=s[i]-tq;

w1=w1+tq;

t[i]=w1;

w[i]=t[i]-st[i];

}

else if(s[i]!=0)

{

w1=w1+tq;

t[i]=w1;

w[i]=t[i]-st[i];

s[i]=s[i]-tq;

}

}

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

{

tw=tw+w[i];

tt=tt+t[i];

}

aw=tw/n;

at=tt/n;

printf("process\tst\twt\ttt");

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

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

printf("awt=%d",aw);

printf("att=%d",at);

}


enter no of process 3

enter time quantum 2

enter process&service time

1

4

2

6

3


29

2

Output:Output:Output:Output: process st wt tt

1 4 4 8

2 6 6 12

3 2 4 6

Awt = 4.000000

att = 8.000000

Program 2:Program 2:Program 2:Program 2: AIM:AIM:AIM:AIM: Write a C program for SJF CPU scheduling algorithmWrite a C program for SJF CPU scheduling algorithmWrite a C program for SJF CPU scheduling algorithmWrite a C program for SJF CPU scheduling algorithm Source Code:Source Code:Source Code:Source Code:

#include<stdio.h>

main()

{

int i,j,bt[10],n,pt[10],wt[10],tt[10],t,k,l,w1=0,t1=0;

float at,aw;

printf(“enter no of jobs”);

scanf(“%d”,&n);

printf(“enter burst time”);

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

scanf”(%d”,&bt[i]);

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

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


30

if(bt[i]<bt[j])

{

t=bt[i];

bt[i]=bt[j];

bt[j]=t;

}

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

{

wt[i+1]=bt[i]+wt[i];

tt[i+1]=tt[i]+bt[i];

w1=w1+wt[i];

t1=t1+tt[i];

}

aw=w1/n;

at=t1/n;

printf(“\nbt\twt\ttt”);

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

printf(“%d\t%d\t%d\n”,bt[i],wt[i],tt[i]);

printf(“aw=%d\nat=%d”,aw,at);

getch();

}

Input:Input:Input:Input: Enter no of jobs

4

Enter burst time

5

12

8

20

Output:Output:Output:Output: Bt wt tt

5 0 5

12 5 13

8 13 25

20 25 45


31

aw=10.75000

at=22.000000


AIM:AIM:AIM:AIM: Write a C program for FCFS CPU scheduling algorithmWrite a C program for FCFS CPU scheduling algorithmWrite a C program for FCFS CPU scheduling algorithmWrite a C program for FCFS CPU scheduling algorithm

Source Code:Source Code:Source Code:Source Code: include<stdio.h>

main()

{


float at,aw;


scanf(“%d”,&n);


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



32

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

{



w1=w1+wt[i];

t1=t1+tt[i];

}

aw=w1/n;

at=t1/n;

printf(“\nbt\twt\ttt”);

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

printf(“%d\t%d\t%d\n”,bt[i],wt[i],tt[i]);


getch();

}

Input:Input:Input:Input: enter no of jobs

3

enter bursttime

12

8

20

output:output:output:output:

bt wt tt

12 0 12

8 12 20

20 20 40

aw=10.666670

at=24.00000


AIM:AIM:AIM:AIM: Write a C PrograWrite a C PrograWrite a C PrograWrite a C Program for priority CPU scheduling algorithm.m for priority CPU scheduling algorithm.m for priority CPU scheduling algorithm.m for priority CPU scheduling algorithm. Source Code:Source Code:Source Code:Source Code:

#include<stdio.h>

main()

{


float at,aw;



33

scanf(“%d”,&n);


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


printf(“enter priority values”);

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

scanf(“%d”,&pt[i]);

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

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

if(pt[i]<pt[j])

{

t=pt[i];

pt[i]=pt[j];

pt[j]=t;

k=bt[i];

bt[i]=bt[j];

bt[j]=k;

}

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

{



w1=w1+wt[i];

t1=t1+tt[i];

}

aw=w1/n;

at=t1/n;

printf(“\nbt\tprority\twt\ttt”);

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

printf(“%d\t%d\t%d\t%d\n”,bt[i],pt[i],wt[i],tt[i]);


getch();

}

Input:Input:Input:Input: Enter no of jobs

4

Enter bursttime

10

2

4

7

Enter priority values

4


34

2

1

3

Output:Output:Output:Output: Bt priority wt tt

4 1 0 4

2 2 4 6

7 3 6 13

10 4 13 23

aw=5.750000

at=12.500000


AIM:AIM:AIM:AIM: Write a C Program for sequential file allocation.Write a C Program for sequential file allocation.Write a C Program for sequential file allocation.Write a C Program for sequential file allocation. Source Code:Source Code:Source Code:Source Code:

#include<stdio.h>

#include<conio.h>

main()

{


35

int f[50],i,st,j,len,c,k,count=0;

for(i=0;i<50;i++)

f[i]=0;

X:

printf("\n enter starting block & length of files");

scanf("%d%d",&st,&len);

printf("\n file not allocated(yes-1/no-0)");

for(k=st;k<(st+len);k++)

if(f[k]==0)

count++;

if(len==count)

{

for(j=st;j<(st+len);j++)

if(f[i]==0)

{

f[j]=1;

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

if(j==(st+len-1))

printf("\n the file is allocated to disk");

}

}

else

printf("file is not allocated");

count=0;

printf("\n if u want to enter more files(y-1/n-0)");

scanf("%d",&c);

if(c==1)

goto X;

else

exit();

getch();

}


enter starting block & length of files

4

5

OutputOutputOutputOutput:

file not allocated (y-1/n-0)


36

4 1

5 1

6 1

7 1

8 1

file is allocated to disk

if u want to enter more files(y-1/n-0)

0


AIM:AIM:AIM:AIM: Write a C Program for indexed file allocation strategy.Write a C Program for indexed file allocation strategy.Write a C Program for indexed file allocation strategy.Write a C Program for indexed file allocation strategy.


#include<stdio.h>

#include<conio.h>

main()

{

int f[50],i,k,j,index[50],n,c,count=0;


37

for(i=0;i<50;i++)

f[i]=0;

X:

printf("enter index block");

scanf("%d",&i);

if(f[i]!=1)

{

f[i]=1;

printf("enter no of files on index");

scanf("%d",&n);

}

y:

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

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

if(f[index[i]==0)

count++;

if(count==n)

{

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

f[index[j]=1;

printf("\nallocated");

printf("\n file indexed");

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

printf("\n%d->%d:%d",i,index[k],f[index[k]);

}

else

{

printf("\n file in the index already allocation");

printf("\nenter another file indexed");

goto y;

}

printf("\n index is already allocated");

count=0;

printf("\n if u enter one more block(1/0)");

scanf("%d",&c);

if(c==1)

goto x;

getch( );

}

Input:Input:Input:Input: enter index block

3

enter no of files on index


38

4

5

6

7

8

Output:Output:Output:Output: Allocated

file indexed

4->5:1

4->6:1

4->7:1

4->8:1

index is already allocated

if u enter one more block(1/0)

0

Program 7:Program 7:Program 7:Program 7: AIM:AIM:AIM:AIM: Write a C Program for linked file allocation strategy.Write a C Program for linked file allocation strategy.Write a C Program for linked file allocation strategy.Write a C Program for linked file allocation strategy.


#include<conio.h>

main()


39

{

int f[50],p,i,j,k,a,st,len,n;

for(i=0;i<50;i++)

f[i]=0;

printf("enter how many blocks already allocated");

scanf("%d",&p);

printf("\nenter the blocks nos");

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

{

scanf("%d",&a);

f[a]=1;

}

X:

printf("enter index sarting block & length");

scanf("%d%d",&st,&len);

k=len;

if(f[st]==0)

{

for(j=st;j<(k+st);j++)

{

if(f[j]==0)

{

f[j]=1;

printf("\n%d->%d",j,f[j]);

}

else

{

printf("\n %d->file is already allocated",j);

k++;

}

}

}

else

printf("\nif u enter one more (yes-1/no-0)");

scanf("%d",&c);

if(c==1)

goto X;

else

exit();

getch( );

}

Input:Input:Input:Input: enter how many blocks already allocated

5


40

enter block nos

3

7

9

10

14

enter index starting block & length

4 10


4->1 5->1 6->1 7->file is already allocated 8->1 9->file is already allocated 10->file is already allocated 11->1 12->1 13->1 14->file is already allocated 15->1 16->1 17->1

Program 8:Program 8:Program 8:Program 8: AIM:AIM:AIM:AIM: Write a C Program for MVT first fit.Write a C Program for MVT first fit.Write a C Program for MVT first fit.Write a C Program for MVT first fit.


#include<conio.h>


41

main()

{

int i,j,t1=0,k,n;

int p[10],h[10];

static int t[10];

printf(“enter no of blocks”);

scanf(“%d”,&n);

printf(“enter block sizes”);

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

scanf(“%d”,&h[j]);

printf(“enter no of process”);

scanf(“%d”,&k);

printf(“enter each process size”);

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

scanf(“%d”,&p[i]);

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

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

if(p[i]<=h[j])

{

t[i]=h[j]=h[j]-p[i];

break;

}

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

t1+=t[i];

printf(“total fragmentation=%d”,t1);

getch();

}

InputInputInputInput:

Enter no of blocks3

Enter block size100

200

300

Enter no of process2

Enter process123

243

OutputOutputOutputOutput:

Total fragmentation=134


AIM:AIM:AIM:AIM: Write a C Program for MFT .Write a C Program for MFT .Write a C Program for MFT .Write a C Program for MFT .


42


#include<stdio.h>

#include<conio.h>

main( )

{

int a[10],b[10],c[10],i,j,p,s,n,t=0;

printf(“enter no of process”);

scanf(“%d”,&n);

printf(“enter size of each process”);

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

scanf(“%d”,&a[i]);

printf(“enter size of memory”);

scanf(“%d”,&s);

printf(“enter no of partition”);

scanf(“%d”,&p);

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

b[i]=s/n;

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

c[i]=b[i]-a[i];

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

t=t+c[i];

printf(“\nprocess\tmemorysize\tprocessizefragementation”);

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

printf(“%d\t%d\t%d\t%d\n”,t,b[i],a[i],c[i]);

printf(“total fragementation=%d”,t);

getch();

}

InputInputInputInput: Enter no of process3

Enter size of process100

200

300

Enter size of memory900

Enter no of partitions3

outputoutputoutputoutput:

Process memorysize processize fragmentation

1 300 100 200

2 300 200 100

3 300 300 0

Total fragmentation=300


AIM:AIM:AIM:AIM: Write a C Program for MVT best fit.Write a C Program for MVT best fit.Write a C Program for MVT best fit.Write a C Program for MVT best fit.


43


#include<conio.h>

main()

{

int n,d,p[10],b[10],i,j,ts;

printf(“enter no of blocks”);

scanf(“%d”,&d);

printf(“enter size of block”);

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

scanf(“%d”,&b[i]);

printf(“\n enter no of process”);

scanf(“%d”,&n);

printf(“enter size of each process”);

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

scanf(“%d”,&p[i]);

sort(&b,d);

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

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

if(p[i]<=b[j])

{

b[j]=b[j]-p[i];

break;

}

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

ts=ts+b[i];

printf(“total fragmentation=%d”,ts);

getch();

}

sort(int b[10],int n)

{

int i,j,t;

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

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

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

{

t=b[i];

b[i]=b[j];

b[j]=t;

}

}

Input:Input:Input:Input: enter no of blocks

3


44

enter size of block

43

43

21

enter no of process

3

enter size of each process

32

40

20


total fragmentation 15


45


AIM:AIM:AIM:AIM: Program for bankers algorithmProgram for bankers algorithmProgram for bankers algorithmProgram for bankers algorithm


#include<stdio.h>

#include<conio.h>

main()

{

int

i,j,a=0,b=0,c=0,f[10],t[10][10],al[10][10],ta[10][10];

int a1[10][10], max[10][10], n[10][10], n1,p,k=0;

printf(“\n enter no.of resources”);

scanf(“%d”,n1);

printf(“\nenter the max no .of resources for each type”);

for(i=0;i<n1;i++)

scanf(“%d”,&t[b][i]);

printf(“\nenter no .of process”);

scanf(“%d”,&p);

printf(“\nenter allocation resources”);

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

{

f[i]=0;

for(j=0;j<n1;j++)

scanf(“%d”,&a1[i][j]);

}

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

for(j=0;j<n1;j++)

{

if(a1[i][j]<=t[b][j])

{

t[b][j]+=a1[i][j];

continue;

}

else

printf(“\n wrong resourcesallocation”);

printf(“\n chance of deadlock occurrence after

allocation”);

for(j=0;j<n1;j++)

printf(“%d”,a1[b][j]);

printf(“\n enter the max resources for every process”);

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


46

for(j=0;j<n1;j++);

{

scanf(“%d”,&max[i][j]);

n[i][j]=max[i][j]-a1[i][j];

}

j=0;

printf(“\n needed resources for every process to start

execution”);

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

printf(“\n%d %d%d”,n[i][j],n[i][j+1],n[i][j+2]);

printf(“\n safe sequence the sequence of process to

compute

their execution”);

for(a=0;a<(p-c);)

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

{

j=0;

b=0;

if(f[i]==0)

{

if(n[i][j]<=a1[b][j]&&n[i][j+1]<=a1[b][j+1]&&

n[i][j+2]<=a1[b][j+2])

{

printf(“\n process %d execution started and

completed”,i+1);

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

a1[b][k]+=a1[i][k];

f[i]=1;

c++;

}

else

f[i]=0;

}

}

getch();

}


47


enter no.of resources

3

enter the max no .of resources for each type

10 5 7

enter the no .of process

5

enter allocation of resources

0 1 0 2 0 0 3 0 2 2 1 1 0 0 2


total available resources after allocation

3 3 2

enter the max resources for every process

7 5 3 3 2 2 9 0 2 2 2 2 4 3 3

needed resources for every process to start execution

7 4 3

1 2 2

6 0 0

0 1 1

4 3 1

Safe sequence ,the sequence of process to complete their

execution

Procee 2 execution started & completed






48

Program 12:Program 12:Program 12:Program 12: AIM:AIM:AIM:AIM:

Write a C program for FIFO page allocation algorithm.Write a C program for FIFO page allocation algorithm.Write a C program for FIFO page allocation algorithm.Write a C program for FIFO page allocation algorithm.


#include<stdio.h>

#include<conio.h>

main()

{

int a[10],b[5],n,t,i,c=0,k=0,j=0,m,x;

printf("\nenter no of requests");

scanf("%d",&n);

printf("\nenter no of frames");

scanf("%d",&t);

printf("enter requests one by one");

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

{

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

if(k<t)

{

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

if(b[x]==a[i])

break;

if(x==k)

{

b[k]=a[i];

k++;

m=i;

}

}

}

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

{

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

if(b[x]==a[i])

break;

if(x==3)

{


49

c++;

b[j]=a[i];

j++;

if(j==t)

j=0;

}

}

printf("no of pagefaults=%d",c);

getch( );

}

Input:Input:Input:Input: enter no of requests

13

enter no of frames

3

enter requests one by one

2

5

3

2

0

1

3

5

4

2

2

3

4

Output:Output:Output:Output: no of pagefaults

6


50

Program 13:Program 13:Program 13:Program 13: AIM:AIM:AIM:AIM: Write a C program for LRU page replacement algorithm.Write a C program for LRU page replacement algorithm.Write a C program for LRU page replacement algorithm.Write a C program for LRU page replacement algorithm. Source Code:Source Code:Source Code:Source Code:

#include<stdio.h>

#include<conio.h>

main()

{

int a[50],b[40],c[15],n,f,i,y,s,p=0,k=0,l=0,m,x,g;

printf("\nenter no of requests");

scanf("%d",&n);

printf("\nenter no of frames");

scanf("%d",&f);

printf("enter requests one by one");

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

{

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

if(k<f)

{

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

if(b[x]==a[i])

break;

if(x==k)

{

b[k]=a[i];

c[k]=a[i];

k++;

m=i;

}

}

}

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

{

printf("%d",a[i]);

for(x=0;x<f;x++)


51

printf("%4d",b[x]);

printf("---------");

for(x=0;x<f;x++)

if(b[x]==a[i])

break;

if(x==f)

{

p++;

l=0;

for(g=i-1;g>0;g--)

for(x=0;x<f&&l<=f;x++)

if(a[g]==b[x]&&c[x]==b[x])

{

y=c[x];

c[x]=-1;

l++;

}

if(l==f)

{

for(s=0;s<f;s++)

if(y==b[s])

b[s]=a[i];

}

if(l<f)

for(s=0;s<f;s++)

if(c[s]!=-1)

{

b[s]=a[i];

break;

}

for(s=0;s<f;s++)

c[s]=b[s];

}

for(s=0;s<f;s++)

printf("%4d",b[s]);

printf("\n");

}

printf("\n no of pagefaults=%d",p);

getch();

}


52

Input:Input:Input:Input: enter no of requests

13

enter no of frames

3

enter requests one by one

2 5 3 2 0 1 3 5 4 2 2 3 4

Output:Output:Output:Output: 3 2 5 3 ------------ 2 5 3

2 2 5 3 ------------ 2 5 3

0 2 5 3 ------------ 2 0 3

1 2 0 3 ------------ 2 0 1

3 2 0 1 ------------ 2 0 1

5 2 0 1 ------------ 2 5 1

4 2 5 1 ------------ 2 5 4

2 2 5 4 ------------ 2 5 4

2 2 5 4 ------------ 2 5 4

3 2 5 4 ------------ 2 3 4

4 2 3 4 ------------ 2 3 4

No of pagefaults=7

JYOTHISHMATHI INSTITUTE OF TECHNOLOGY & SCIENCE NUSTULAPUR, KARIMNAGAR

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