Operating system Lab

Ex. No:1

UNIX BASIC COMMANDS

S.NO	COMMAND	SYNTAX	DESCRIPTION
		General Purpose Utilities
		$ date	Displays current month, date and
			time
		$ date +%D	Displays date in the format
			mm/dd/yy
		$date +%H	Display hour
1	DATE	$date +%M	Display minutes
		$date +%$	Display seconds
		$date +%h	Month name
		$date +%m	Month no
		$date +%T	Time in am /pm
		$date +%y	Last two digits of the year
		$cal	Displays calendar for the current
			month
2	cal : calendar	$cal monthno year	Displays calendar for the
			corresponding month in the year
		$cal year	Displays the calendar for the
			whole year
3	who : Login details	$ who	Gives the details of who all have
			logged into the system currently.
		$who -Hu	Displays the details with header
4	who am I : who is on the		Tells who are you with the
	system		terminal name, date & time
5	type : to know the	$type command	To know the location of the
	execution of the command		program the system will execute
6	echo : to display messaae	$echo message
		$echo 'message'
7	man : online help	$man command	Online help about a particular
			command
	uname : know your	$uname -r	Display the version number
8		$uname -n	Display the machine name
	machine details
		$tty	Tells you the terminal number
9	clear & tput clear	$tput clear	Clears the screen
		$clear
		Usage of Directory Commands
10	pwd: present working	$pwd
	Directory
		$ls	Lists all the files
			Lists files with their mode, no of
	ls: displays the list of files	$ls -l	links, owner, file size,modification
			date and time and file name.
11	in the current working
			Lists in order of last modification
	directory.	$ls -t
			time.
		Sis -a	Lists all entries including the
			hidden files.

S.NO	COMMAND	SYNTAX	DESCRIPTION
		$ls -d	Lists directory files instead of its
			contents.
		$ls -p	Puts a slash after each directory
		$is -u	Lists in order of last access time
		$ls -x	To produce a multi-column output
		$ls -F	To identify directories & files . / -
			dir, * - exe files
		$ls -r	The files will be displayed in the
			reverse order
		$ls -R	Recursive listing of all files in sub
			directories
		$ls -I	Shows inode no. of a file
		$ls -id dirname	Listing the attributes of a particular
			directory
12	mkdir: making directory	$mkdir dirname	Creates home directory
13	cd : change directory	$cd dirname	Change directory
14	rmdir: removing	$rmdir dirname
	directories
15	HOME: to know the user	$echo SHOME
	directory
		File Commands
		$cat > filename	Type the contents & press ctrl-d to
			end
16	cat: displaying & creating	$cat filename or $cat <	Displays the contents of the file.
	files	filename
		$cat >> filename	To append the file content.Type
			the contents & press ctrl-d to end
		$cp fl f2	Fl is copied into f2.
			Interactive copy (warns the user
		$cp -I f1 f2	before overwriting the destination
17	cp : copying a file		file)
			Copying directory structures.(
		S cp - r dirl dir2	copies all files & subdirectories in
			dirl to dir2)
		$rm filename	To delete a file
		$rm *	To delete all files
18	rm : deleting files	$rm -I f1	Interactive deletion
		$rm -f f1	Forcibly remove write-protected
			file also
19	mv : renaming files	$mv f1 f2	Rename the file f1 as f2. (f1 no
			longer exist)
		$more f1	Allows the user to view the file f1,
			one screen at a time.(q- exit)
20	more : paging output
		f	To forward one screen
		b	To scroll back
21	file : know the file types	$file f1	tell whether f1 is a
			ordinary,directory or device file.
22	wc : line,word and	$wc f1	Displays no.of lines,words and
	character counting		characters

S.NO	COMMAND	SYNTAX	DESCRIPTION
		$wc -l f1	tells no.of lines only
		$wc -w f1	No of words only
		$wc -c f1	No of characters only
		$od f1	display the file contents in octal
			form
23	od : displaying the data in	$od -b f1	Displays the octal value of each
	octal		character in the file
		$od -bc f1	Displays each & every character
			along with their octal equivalent.
			Splits into no. of files containing
	split : spliting a file into	$split f1	'n' number of lines in each file.
24			(filenames xaa,xab,..xyz)
	multiple files
		$split f1 f2	creates resultant files in the names
			of f2aa, f2ab,..
			Files are compared byte by byte I
			and the location of the 1-st mistake
		$cmp fl f2	is echoed. (if 2 files are same, cmp
			displays no message, simply
25	cmp : comparing two files		returns S prompt)
			Gives a detailed list of the byte no.
		$cmp -l fl f2	& the directory bytes in octal for
			each character that differ in both
			files.
	comm : finding what is		Compares each line of fl with its
26	common (compares two	$comm f1 f2
			corresponding line in f2
	sorted files)
27	diff: to display the file	diff fl £2	display line-by-line differences
	differences		between pairs of text files
	chmod : changing file	$chmod category operation
	permission (u - user, g -	permission filename
	group o - others r- read, w-		Assigns execute permission to user
28	write & x- execute +	$chmod u+x,go+r fl	and read permission to group and
	addition of rights -		others.
	removal of rights = assign		Assigns read & write permissions
	permission in octal r-4 w	$chmod 644 fl	to user, read permission to group
	—2 x-1)		& others
		Environment Variables
		$set	Displays a complete set of all the
			environment variables
		$echo $PATH	It tells the route to locate any
			executable command
29
		$echo $HOME	Displays the user directory
		$echo $LOGNAME	Tells the login user
		$echo $TERM	Tells the terminal type
		$echo $Ps1	Tells about the shell
		Simple Filters
30	head : displaying the	$head filename	Displays the first few lines of the
	beginning of a file		file'
		$head -n filename	Displays the first 'n' no. of lines

S.NO	COMMAND	SYNTAX	DESCRIPTION
31	tail: displaying the end of a	$tail -n filename	Displays 'n' no. of lines from the
	file		end of a file.
		$tail +n filename	n- th line onwards, upto the end
32	cut: slitting a file vertically	$cut -c range,range filename	To extract specific columns from
			the file.
		$cut -c 1-5,10-15 fl
33	paste: pasting files	$paste fl f2	To merge two files
		$sort filename	Sorts the contents based on ASCII.
	sort : ordering a file (		Alphabetical order
34	sorting is in the order -	$sort -r filename	Sorts in the reverse order
	numerals, uppercase letters	$sort -n filename	n- numerical sort
	& lowercase letters)	$sort -c filename	checks if file is sorted
		$sort -u filename	removes duplicate records
	uniq : locating repeated		It fetches one copy of each record
35	lines(it requires sorted	$uniq filename
			& writes it to the standard output.
	files as input)
36	nl: line numbering	$nl filename	It numbers only logical lines.
	tr: translating characters (tr
	doesn't accept a filename		Lower case to uppercase
37	as an argument, it takes its	$cat filename | tr '[a-z]' '[A-Z]'
			conversion
	input through redirection
	or a pipe)
		Pattern Searching
		$grep pattern filename	Displays the lines that containing
			the pattern
		$grep-v pattern filename	Displays the lines which are not
			having that pattern
	grep: searching for a
38			Displays the lines that containing
	pattern	$grep -n pattern filename
			the pattern along with the line no.
		$grep -i pattern filename	Ignoring the case
		$grep -c pattern filename	Displays count of number of
			occurrences
	eg rep : extending grep		Searches the two patterns in the
39	(searches more than one	$egrep 'pattern1|pattern2' f1
			filefl
	pattern in a file)
40	fgrep : multiple string	$fgrep -f filel file2	Searches pattern from more than
	searching		one file
		Process
41	$$ : process no. of the sh	$echo $$	Displays the process number of the
	shell		current shell
		$ps	Displays the attributes of a process
			such as PID TTY TTME CMD
42	ps : process status		Displaying process ancestry such
		$ps -f	as UID PID PPID C STTME TTY
			TIME CMD
		$ps -a	Lists the processes of all users
		$ps -e	Displays all system processes

S.NO	COMMAND	SYNTAX	DESCRIPTION
			Nohup permits execution of the
43	nohup : log outsafely	$nohup command filename &	process even after the user has
			logged out. It sends the default
			output to a file nohup.out.
44	kill: premature termination	$kill processidno	Terminates the job with the PID
	of a process
	nice : job execution with		To run a job with a low priority
45		$nice command filename	(nice reduces the priority of any
	low priority
			process by 10 units)
46	at & batch: execute later	$at time command filename	at the particular time, that
		$batch < filename
		Interprocess Communication
			Any user can read any message
47	news : the bulletin board	$news	that is sent by the system
			administrator
	write : two way		The message appears on user's
			terminal provided he is logged
48	communication (o - over	$write username message 0
			in.If he is not, the system responds
	oo - over & out)
			with an error message.
49	mesg : your willingness to	$mesg	User can receive messages (
	talk		default option will be displayed)
		$mesg n	to prevent from others intrusions.
	talk : an alternative to	$talk username message ctrl-c	It splits the screen horizontally into
50			2 windows; one is for receiving
	write	(or) press delete key to end)
			and the other is for transmission
			It is used for later viewing, and the
51	mail : the universal mailer	$mail username message ctrl-d	user doesn't need to be logged in
			for the mail to reach her.
		Miscellaneous Command
		$finger	It produces a list of all logged
	finger or W: details of		users.
52
	users	$finger username	Provide details of a particular user
		$ x=vl y=v2 $expr $x op $y	evaluate arguments as an
53	expr : computation	(or) $z=`expr $x op $y` $echo
			expression
		$z (op - any operator)
			Displays the result of the
54	be : the calculator	$bc any calculation ctrl-d	calculation.(it is also used for base
			conversion)
55	spell: check your spellings	$spell filename	Lists all misspelled words

S.NO	COMMAND	SYNTAX	DESCRIPTION
		$touch mm/dd/hh mm file	Touch changes both modification
		name	time and access time
56	touch : changing the time	$touch - m mm dd hh min file	Only changes modification time
	stamps	name
		$touch - a mm dd hh min file	Only access time
		name
57	ln : linking of files	$ln f1 f2	The changes in fl also reflected in
			f2
		$chown owner name filename	Gives away the ownership of a file
	chown & chgrp:changing		to any user
58
	file ownership	$chgrp  group name filename	Changes the group ownership of
			the file
		$cc filename	To compile a C programs
59	cc : compiling C programs	$cc file1.c - o file2 or $cc -o	Compiles file1.c and places the
			resulting executable program in
		file2 file1.c
			the file2 rather than a. out

SHELL PROGRAMMING

EX NO: 2(a)                 TO FIND GREATEST AMONG THREE NUMBERS

Date :

PRE LAB:

AIM:

            To find the greatest of three numbers using shell programming.

IN LAB:

ALGORITHM:

 Step 1: Start the program.

 Step 2: Read the three numbers a,b,c.

 Step 3: If a is greater than b and c.

 Step 4: Print “A is greatest”.

 Step 5: If b is greater than a and c.

 Step 6: Print “B is greater”.

 Step 7: Else print “C is greater”.

 Step 8: Stop the program.

PROGRAM:

echo "Enter three no"

read a

read b

read c

if test $a -gt $b

then

if test $a -gt $c

then

echo "a is greatest"

fi

fi

if test $b -gt $c

then

echo "b is greatest"

else

echo "c is greatest"

fi

OUTPUT :

[cse@unixSERVER ~]$ sh great.sh

 Enter three no

 10

 12

 13

 C is greatest

RESULT:

       Thus the program to find greatest of three numbers using shell programming was         executed and verified

EX NO: 2(b)                     TO PERFORM ARITHMETIC OPERATION                  

Date :

           

PRE LAB:

AIM:

      To perform arithmetic operations using shell programming.

IN LAB:

ALGORITHM:

 Step 1: Start the program.

 Step 2: Read the two numbers a and b.

 Step 3: Select the arithmetic operation to be performed.

 Step 4: Perform “Addition” if case 1 is selected.

 Step 5: Perform “Subtraction” if case 2 is selected.

 Step 6: Perform “Multiplication” if case 3 is selected.

 Step 7: Perform “Division” if case 4 is selected.

 Step 8: Print the required result.

 Step 9: Stop the program.

PROGRAM:

echo "1.add"

echo "2.subtract"

echo "3.multiply"

echo "4.divide"

echo "Enter your choice"

read i

echo "Enter two numbers"

read a

read b

case $i in

1) x=`expr $a + $b`

 echo "The sum is $x";;

2) x=`expr $a - $b`

 echo "The result is $x";;

3) x=`expr $a \* $b`

 echo "The product is $x";;

4) x=`expr $a / $b`

 echo "The result is $x";;

esac

OUTPUT:

 [cse@unixSERVER ~]$ sh arit.sh

 1.add

 2.subtract

 3.multiply

 4.divide

 Enter your choice

 1

 Enter two numbers

 2

 3

 The sum is 5

 [cse@unixSERVER ~]$ sh arit.sh

 1.add

 2.subtract

 3.multiply

 4.divide

 Enter your choice

 2

 Enter two numbers

 5

 1

 The result is 4

 [cse@unixSERVER ~]$ sh arit.sh

 1.add

 2.subtract

 3.multiply

 4.divide

 Enter your choice

 3

 Enter two numbers

 6

 5

 The product is 30

 [cse@unixSERVER ~]$ sh arit.sh

 1.add

 2.subtract

 3.multiply

 4.divide

 Enter your choice

 4 

Enter two numbers

 20

 5

 The result is 4

RESULT:

       Thus the program to perform arithmetic operations using shell programming was      executed and verified.

Ex.no:2(c)                    FACTORIAL OF A GIVEN NUMBER

Date  :

PRE LAB:

AIM:

      To find the factorial of a given number using shell programming.

IN LAB:

ALGORITHM:

  Step 1: Start the program.

  Step 2: Declare a variable a.

  Step 3: Assume n is equal to a.

  Step 4: If a is o, then print the factorial value as 1.

  Step 5: Else, subtract the value of a by 1.

  Step 6: Multiply with the previous value of a.

  Step 7: Print the required result.

  Step 8: Stop the program.

PROGRAM:

echo "Factorial of the given number"

echo "Enter the number"

read a

fact=1

n=$a

if [ $a -eq 0 ]

then

echo "Factorial of 0 is 1"

else

while [ $a -gt 0 ]

do

fact=`expr $fact \* $a`

a=`expr $a - 1`

done

echo "Factorial of $n is $fact"

fi

OUTPUT:

 [cse@unixSERVER ~]$ sh fact.sh

 Factorial of the given number

 Enter the number

 3

 Factorial of 3 is 6

RESULT:

      Thus the program to find the factorial of a given number using shell programming was executed and verified.

CPU SCHEDULING ALGORITHMS

EX.NO:3 (a)                   ROUND ROBIN CPU SCHEDULING ALGORITHM

DATE:

PRE LAB:

AIM:

          To perform the round robin scheduling using a c program.

ALGORITHM:

 Step1: Start the program.

 Step2: First make a 2-dimensional arry.

 Step3: Enter the no of process and quantum.

 Step4: Enter the processno,servicetime and arrivaltime.

 Step5: In round robin scheduling we execute the process for the given time quantum.

 Step6: Calculate the waiting time and turn around time using the giventime quantum.

 Step7: Now calculate the average waiting time and average turn around time.

 Step8: Display the process and respective time.

 Step9: Display the average waiting time and turn around time.

 Step10: Stop the program.

IN LAB:

PROGRAM:

#include<stdio.h>

main()

{

struct process

{

float at,tt,st,ft,bt,wt;

int pno;

}p[20],t[20],temp;

float stat=0,time=0,q=1,avgwt=0,avgtt=0;

int i,n;

int arr[50],arr1[50],k=1;

printf("ROUND ROBIN SCHEDULING\n");

printf("Enter the no of process:");

scanf("%d",&n);

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

{

printf("Enter the burst time of %d process:",i);

scanf("%f",&p[i].bt);

t[i].bt=p[i].bt;

p[i].pno=i+1;

p[i].at=0;

}

while(stat<n)

{

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

{

if(t[i].bt>q)

{

p[i].ft=time+q;

t[i].bt=t[i].bt-q;

time=time+q;

arr[k]=p[i].pno;

k++;

}

else if(t[i].bt>0&&t[i].bt<=q)

{

p[i].ft=t[i].bt+time;

arr1[k]=t[i].bt;

time=time+t[i].bt;

arr[k]=p[i].pno;

 k++;

t[i].bt=0;

stat++;

}

}

}

printf("\n\npno\tat\tst\tbt\tft\twt\ttt\n");

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

{

p[i].wt=p[i].ft-p[i].bt;

p[i].tt=p[i].wt+p[i].bt;

avgwt=avgwt+p[i].wt;

avgtt=avgtt+p[i].tt;

printf("%d\t2.2f\t%2.2f\t%2.2f\t%2.2f\t%2.2f\t%2.2f\n",p[i].pno,p[i].at,p[i].st,p[i].bt,p[i].ft,p[i].wt,p[i].tt);

}

avgwt=(avgwt/n);

avgtt=(avgtt/n);

printf("\naverage waiting time:%2.2f",avgwt);

printf("\naverage turn around time:%2.2f",avgtt);

}

      

OUTPUT:

[cse@unixSERVER ~]$ vi roro.c

[cse@unixSERVER ~]$ cc roro.c

[cse@unixSERVER ~]$ ./a.out

ROUND ROBIN SCHEDULING

Enter the no of process: 3

Enter the burst time of 0 process: 5

Enter the burst time of 1 process: 7

Enter the burst time of 2 process: 3

pno     at      st      bt      ft      wt      tt

1       2.2f    0.00    0.00    5.00    12.00   7.00

2       2.2f    0.00    0.00    7.00    15.00   8.00

3       2.2f    0.00    0.00    3.00    9.00    6.00

average waiting time: 7.00

average turn around time: 12.00

RESULT:

          Thus the program for round robin scheduling has been executed and output is

verified successfully.

POST LAB:

VIVA VOCE:

1. What is time slice?

2. How turnaround time is calculated?

3. Define arrival time.

4. What is the unit of average waiting time?

5. Define entry section and exit section

                                                                                                      

EX.NO:3 (b)                  SJF CPU SCHEDULING ALGORITHM

DATE:

PRE LAB:

AIM:

        To write a   program for implementing shortest job first using c programming.

IN LAB:

ALGORITHM:

 Step 1 :Start the program.

 Step 2:Declare the variables.

 Step 3:Initialize total waiting time and total as 0.

 Step 4:Pint the no of process.

 Step 5:Use two dimensional array to print burst time, arrival time in a matrix form.

 Step 6:Calculate average waiting time and turn around time.

 Step 7:Print the result.

 Step 8:Stop.

PROGRAM:

#include<stdio.h>

int main()

{

int m,a[10][6],i,j,k,te,p,q;

int twt=0,tot=0;

float awt,tat;

printf("SJF\n");

printf("Enter the no.of process:");

scanf("%d",&m);

printf("Enter %d *3 rowwise  \n process no \nburst time \narrival time\n",m);

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

{

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

{

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

}

}

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

{

for(j=i+1;j<3;j++)

{

if(a[i+1][1]<a[i][1])

{

te=a[i][0];a[i][0]=a[i+1][0];a[i+1][0]=te;

q=a[i][1];a[i][1]=a[i+1][1];a[i+1][1]=q;

p=a[i][2];a[i][2]=a[i+1][2];a[i+1][2]=p;

}

}

}

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

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

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

printf("\nP.NO\tB.T\tA.T\tW.T\tR.T\tTAT\n");

a[0][3]=0;

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

{

a[i][3]=a[i-1][3]+a[i-1][1];

twt+=a[i][3];

}

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

{

a[i][4]=a[i][3]-a[i][2];

a[i][5]=a[i][1]+a[i][4];

tot+=a[i][5];

}

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

{

for(j=0;j<6;j++)

{

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

}

awt=twt/m;

tat=tot/m;

printf("\n");

}

printf("avg waiting time is %f\n",awt);

printf("avg turn around time is %f\n",tat);

}

OUTPUT:

[cse@unixSERVER ~]$ cc sjf.c

[cse@unixSERVER ~]$ ./a.out

 SJF

Enter the no of process 3

Enter 3*3 rowwise

process no      burst time      arrival time

1 3  5

2 4  6

3 5  7

1       3       5

2       4       6

3       5       7

PNO     BT      AT      WT      RT      TAT

1       3       5       0       5       8

2       4       6       3       9       13

3       5       7       7       14      19

average waiting time is 3.000000

average turnaround time is 13.000000[7133@unixSERVER ~]$SJF

RESULT:

          Thus the program for SJFscheduling has been executed and output is verified successfully.

POST LAB:

VIVA VOCE:

1. Define CPU scheduling

2. What is a Dispatcher?

3. Classify two types of SJF.

4. What is response time?

5. Delineate Gantt chart.

EX.NO:3 (c)                   FCFS CPU SCHEDULING ALGORITHM

DATE:

PRE LAB:

AIM:

       To write a program for implementing first come first serve using  C programming.

IN LAB:

ALGORITHM:

Step1: Start the program.

Step2: Declare the variables.

Step3: Initialize   average waiting time and average turn around time as 0.

Step4: Print the no of process.

Step5: Based on the process print the burst time and arrival time.

Step6: Check the arrival time of process I and j.

Step7: If process I is greater than j, then swap.

Step8: Calculate average waiting time and average turn around time.

Step9: Display the result.

Step10: Stop

PROGRAM:

#include<stdio.h>

main()

{

struct process

{

float at,tt,st,ft,bt,wt;

}p[20],temp;

float avgwt=0.0,avgtt=0.0;

int n,i,j;

printf("FCFS\n");

printf("\nEnter the no of process:");

scanf("%d",&n);

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

{

printf("\nEnter the arrival time and burst time of %d proce

scanf("%f%f",&p[i].at,&p[i].bt);

}

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

{

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

{

temp=p[i];

p[i]=p[j];

p[j]=temp;

}

}

}

p[0].st=p[0].at;

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

{

p[i].ft=p[i].st+p[i].bt;

p[i+1].st=p[i].ft;

if(p[i].st>p[i].at)

p[i].wt=p[i].st-p[i].at;

else

p[i].wt=0;

p[i].tt=p[i].wt+p[i].bt;

avgwt=avgwt+p[i].wt;

avgtt=avgtt+p[i].tt;

}

printf("pno \t at\t  bt\t  wt\t tt\n");

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

{

printf("%d\t%2.2f\t%2.2f\t%2.2f\t%2.2f\t%2.2f\t%2.2f\n",i

t,p[i].tt);

}

avgwt=(avgwt/n);

avgtt=(avgtt/n);

printf("avgwt is %2.2f\t%2.2f",avgwt);

printf("avgtt is %2.2f\t%2.2f",avgtt);

}

OUTPUT:

[cse@unixSERVER ~]$ cc fcfs.c

[cse@unixSERVER ~]$ ./a.out

FCFS

Enter the no of process: 3

Enter the arrival time and burst time of 0 process 1 4

Enter the arrival time and burst time of 1 process 2 5

Enter the arrival time and burst time of 2 process 4 7

Pno      at       bt      wt     tt

0       1.00    4.00    0.00    4.00    0.00    0.00

1       2.00    5.00    3.00    8.00    0.00    0.00

2       4.00    7.00    6.00    13.00   0.00    0.00

avgwt is 3.00   0.00 avgtt is 8.33       0.00

RESULT:

       Thus the program for implementing first come first serve has been executed and verified.           

POST LAB:

VIVA VOCE:

1. Whether FCFS is preemptive or non preemptive?

2. Define dispatcher and its functions.

3. Delineate context switch.

4. What is the default arrival time?

5. Define dispatch latency?

EX.NO:3 (d)              PRIORITY CPU SCHEDULING ALGORITHM

DATE:

PRE LAB:

AIM:
        To write a c program to implement the Priority Scheduling (PR).

IN LAB:

ALGORITHM:
Step 1:  Get the number of process from the user.
Step 2:  Get also CPU service time for each process from the user.
Step 3:  Sort the CPU time of process according to the process priority in ascending order
Step 4:  Now for each process the waiting time is equivalent to CPU time of previous process.
Step 5:  The ratio of waiting time of all the process to the number of process will give the average waiting time.
Step 6:  Display the result.

PROGRAM:
# include<stdio.h>
main()
{
int j,i,k=0,ptime[25],n,s=0,I,sum=0;
char name[25][25];
int t,p,time[10],pos[10],priority[25];
float avg;
printf ("enter the no. of process: \t");
scanf ("%d",&n);
for(i=0;i<n;i++)
{
printf("enter the name for processes: \t");
printf("%d \t",i+1);
scanf("%s",name[i]);
}
printf("\n \n");

for(i=0;i<n;i++)
{
printf("enter the process time: \t");
printf("%s \t",name[i]);
scanf("%d",&ptime[i]);
}
printf("\n \n");
for(i=0;i<n;i++)
{
printf("enter the priority for process: \t");
printf("%d \t",i+1);
scanf("%d",&priority[i]);
}
printf("\n process – name \t process – time \n");
for(i=0;i<n;i++)
{
printf("\t %s \t \t %d \t\t %d\n",name[i],ptime[i],priority[i]);
}
printf("\n \n PRIORITY SCHEDULING \n \n");
for(i=0;i<n;i++)
{
pos[i]=i;
time[i]=ptime[i];
}
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
p=time[i];
time[i]=time[j];
time[j]=p;
t=pos[i];
pos[i]=pos[j];
pos[j]=t;
p=priority[i];
priority[i]=priority[j];
priority[j]=p;
}
}

for(i=0;i<n;i++)
{
printf("process %s from %d to %d \n",name,pos[i],k,(k+time[i]));
k+=time[i];
}
for(i=0;i<(n-1);i++)
{
s+=time[i];
sum+=s;
}
avg=(float)sum/n;
printf("\n\n average waiting time: \t");
printf("%2fmsec",avg);
sum=avg=s=0;
for(i=0;i<n;i++)
{
s+=ptime[i];
sum+=s;
}
avg=(float)sum/n;
printf("\n turn around time is \t");
printf("%2fmsec",avg);
return 0;
}

RESULT :

Thus the program to implement the Priority Scheduling  was executed successfully and output was verified.

     

POST LAB:

VIVA VOCE:

1. What are privileged instructions?

2. How is the protection for memory provided?

3. How can a user program disrupt the normal operations of a system?

4. Justify, how priority is considered for job scheduling?

5. Write the formula to calculate waiting time.

IMPLEMENT ALL FILE ALLOCATION STRATEGIES

EX.NO:4(A)                           SEQUENTIAL FILE ALLOCATION

DATE:

PRE LAB:

AIM:

       To implement the sequential file allocation using C programming.

IN LAB:

ALGORITHM:

Step 1: Start the program.

Step 2: Get the number of memory partition and their sizes.

Step 3: Get the number of processes and values of block size for each process.

Step 4: First fit algorithm searches all the entire memory block until a hole which is    

              big enough is encountered. It allocates that memory block for the requesting

              process.

Step 5: Best-fit algorithm searches the memory blocks for the smallest hole which can

              be allocated to requesting process and allocates if.

Step 6: Worst fit algorithm searches the memory blocks for the largest hole and

               allocates it to the process.

Step 7: Analyses all the three memory management techniques and display the best

            algorithm which utilizes the memory resources effectively and efficiently.

Step 8: Stop the program.

 PROGRAM:

#include<stdio.h>

#include<conio.h>

main()

{

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();

}

OUTPUT:

enter starting block & length of files

4

5

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

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

POST LAB:

RESULT :

Thus the program to implement the sequential file allocation  was executed successfully and output was verified.

     

EX.NO:4(B)                                       INDEXED FILE ALLOCATION

DATE:

PRE LAB:

AIM:

       To implement the indexed file allocation using C programming.

IN LAB:

ALGORITHM:

Step 1: Start the program.

Step 2: Get the number of memory partition and their sizes.

Step 3: Get the number of processes and values of block size for each process.

Step 4: First fit algorithm searches all the entire memory block until a hole which is    

              big enough is encountered. It allocates that memory block for the requesting

              process.

Step 5: Best-fit algorithm searches the memory blocks for the smallest hole which can

              be allocated to requesting process and allocates if.

Step 6: Worst fit algorithm searches the memory blocks for the largest hole and

               allocates it to the process.

Step 7: Analyses all the three memory management techniques and display the best

            algorithm which utilizes the memory resources effectively and efficiently.

Step 8: Stop the program.

 PROGRAM:

#include<stdio.h>

#include<conio.h>

main()

{

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

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( );

}

OUTPUT:

enter index block 3

enter no of files on index

4 5 6 7 8

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

POST LAB:

RESULT :

Thus the program to implement the indexed file allocation was executed successfully and output was verified.

EX.NO:4(C)                                       LINKED FILE ALLOCATION

DATE:

PRE LAB:

AIM:

       To implement the linked file allocation using C programming.

IN LAB:

ALGORITHM:

Step 1: Start the program.

Step 2: Get the number of memory partition and their sizes.

Step 3: Get the number of processes and values of block size for each process.

Step 4: First fit algorithm searches all the entire memory block until a hole which is    

              big enough is encountered. It allocates that memory block for the requesting

              process.

Step 5: Best-fit algorithm searches the memory blocks for the smallest hole which can

              be allocated to requesting process and allocates if.

Step 6: Worst fit algorithm searches the memory blocks for the largest hole and

               allocates it to the process.

Step 7: Analyses all the three memory management techniques and display the best

            algorithm which utilizes the memory resources effectively and efficiently.

Step 8: Stop the program.

PROGRAM:

#include<stdio.h>

#include<conio.h>

main()

{

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( );

}

OUTPUT:

enter how many blocks already allocated 5

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

POST LAB:

RESULT :

Thus the program to implement the linked file allocation was executed successfully and output was verified.

     

EX.NO:5            PRODUCER – CONSUMER PROBLEM USING SEMAPHORES

DATE:

PRE LAB:

AIM:

       To implement the producer- consumer problem using semaphores in C programming.

IN LAB:

ALGORITHM:

1. Start the program

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.

PROGRAM :

#include<stdio.h>

#include<stdlib.h>

#include<sys/types.h>

#include<sys/ipc.h>

#include<sys/sem.h>

#include<unistd.h>

#define num_loops 2

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

{

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);

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:

semaphore set created

semaphore set id ‘327690’

producer: ‘0’

consumer:’0’

producer:’1’

consumer:’1’

POST LAB:

RESULT :

Thus the program to implement the producer-consumer problem was executed successfully and output was verified.

     Ex.No:7                     BANKERS ALGORITHM FOR DEAD LOCK AVOIDANCE

Date:

PRE LAB:

AIM:

To implement the bankers Algorithm for Dead Lock Avoidance

IN LAB:

PROGRAM:

#include<stdio.h>
int alloc[10][10];
int max[10][10];
int cneed[10][10];
int a[10];
int total[10];
void main()
{
int r,p,i,j,k,flag,f,count,x,flag1;
count=0;
flag1=0;
printf(“\nEnter number of resources:”);
scanf(“%d”,&r);
f=r;
printf(“\nEnter total memory of given resources sequentially:”);
for(i=0;i<r;i++)
scanf(“%d”,&total[i]);
printf(“\nEnter number of processes:”);
scanf(“%d”,&p);
printf(“\nEnter Allocated and Maxneed memory…”);
getch();
for(i=0;i<p;i++)
{
for(j=0;j<r;j++)
{
printf(“\nFor Process %d=”,i);
scanf(“%d%d”,&alloc[i][j],&max[i][j]);
//calculating current need
cneed[i][j]=max[i][j]-alloc[i][j];
}
}
printf(“\nSequence of execution is…\n”);
k=0;
for(j=0;j<r;j++)
{
for(i=0;i<p;i++)
{
a[k]+=alloc[i][j];
}
a[k]=total[k]-a[k];
k++;
}
while(count!=p)
{
for(i=0;i<p;i++)
{
flag=0;
if(cneed[i][f]!=1)
{
for(j=0;j<r;j++)
{
total[j]=a[j]-cneed[i][j];
}
for(k=0;k<r;k++)
{
if(total[k]<0)
{
flag=1;
}
}
}
if((flag==0)&&(cneed[i][f]!=1))
break;
}
x=i;
cneed[x][f]=1;
printf(“P%d->”,x);
count++;
for(i=0;i<r;i++)
{
total[i]+=max[x][i];
a[i]=total[i];
}
for(i=0;i<p;i++)
{
if((cneed[i][0]<a[0])&&(cneed[i][f]==0))
flag1=1;
}
if(flag1==0)
break;
}
if(flag1==0)
printf(“\nUnsafe…”);
else
printf(“\nSafe…”);
getchar();
}

 OUTPUT:

//For Safe condition

Enter number of resources:3

Enter total memory of given resources sequentially:10 5 7

Enter number of processes:5

Enter Allocated and Maxneed memory...

For Process 0=0 7

For Process 0=1 5

For Process 0=0 3

For Process 1=2 3

For Process 1=0 2

For Process 1=0 2

For Process 2=3 9

For Process 2=0 0

For Process 2=2 2

For Process 3=2 2

For Process 3=1 2

For Process 3=1 2

For Process 4=0 4

For Process 4=0 3

For Process 4=2 3

Sequence of execution is...

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

Safe...

//For Unsafe condition

Enter number of resources:3

Enter total memory of given resources sequentially:6 5 4

Enter number of processes:3

Enter Allocated and Maxneed memory...

For Process 0=0 7

For Process 0=1 5

For Process 0=0 3

For Process 1=2 3

For Process 1=0 2

For Process 1=0 2

For Process 2=3 9

For Process 2=0 0

For Process 2=2 2

Sequence of execution is...

P1->Unsafe...

Ex.No:8                     BANKERS ALGORITHM FOR DEAD LOCK DETECTION

Date:

PRE LAB:

AIM:

To write a C program to implement the bankers Algorithm for Dead Lock Detection

IN LAB:

PROGRAM:

#include <stdio.h>;
#include <conio.h>;
void 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;
 clrscr();
 printf("Enter total no of processes");
 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<k;l++)
  if(i!=m[l])
  {
   flag=1;
   for(j=1;j<=tr;j++)
   if(c[i][j]<temp[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<k;i++)
  {
   if(j==m[i])
   found=1;
  }
  if(found==0)
  printf("%d\t",j);
 }
 getch();
}

OUTPUT:
Enter total no. of processes : 4
Enter total no. of resources : 5
Enter claim (Max. Need) matrix :
0 1 0 0 1
0 0 1 0 1
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 0 0 0 0
Enter resource vector (Total resources) :
2 1 1 2 1
Enter availability vector (available resources) :
0 0 0 0 1
deadlock causing processes are : 2 3

POSTLAB:

RESULT:

            Thus the program to implement the bankers Algorithm for Dead Lock Detection has been executed successfully.

Ex.No:9(a)      IMPLEMENTATION OF FIFO PAGE REPLACEMENT ALGORITHM

Date:

PRE LAB:

Aim

To write a program to implement FIFO page replacement algorithm

 INLAB:

.Algorithm

1. Start the process

2. Declare the size with respect to page length

3. Check the need of replacement from the page to memory

4. Check the need of replacement from old page to new page in memory

5. Forma queue to hold all pages

6. Insert the page require memory into the queue

7. Check for bad replacement and page fault

8. Get the number of processes to be inserted

9. Display the values

10. Stop the process

Program:

#include<stdio.h>

int main()

{

int i,j,n,a[50],frame[10],no,k,avail,count=0;

            printf("\n ENTER THE NUMBER OF PAGES:\n");

scanf("%d",&n);

            printf("\n ENTER THE PAGE NUMBER :\n");

            for(i=1;i<=n;i++)

            scanf("%d",&a[i]);

            printf("\n ENTER THE NUMBER OF FRAMES :");

            scanf("%d",&no);

for(i=0;i<no;i++)

            frame[i]= -1;

                        j=0;

                        printf("\tref string\t page frames\n");

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

                        {

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

                                    avail=0;

                                    for(k=0;k<no;k++)

if(frame[k]==a[i])

                                                avail=1;

                                    if (avail==0)

                                    {

                                                frame[j]=a[i];

                                                j=(j+1)%no;

                                                count++;

                                                for(k=0;k<no;k++)

                                                printf("%d\t",frame[k]);

}

                                    printf("\n");

}

                        printf("Page Fault Is %d",count);

                        return 0;

}

Output

ENTER THE NUMBER OF PAGES:  20

ENTER THE PAGE NUMBER :       7 0 1 2 0 3 0 4 2 3 0 3 2 1 2 0 1 7 0 1

ENTER THE NUMBER OF FRAMES :3

      ref string       page frames

7               7       -1      -1

0               7       0       -1

1               7       0       1

2               2       0       1

0

3               2       3       1

0               2       3       0

4               4       3       0

2               4       2       0

3               4       2       3

0               0       2       3

3

2

1               0       1       3

2               0       1       2

0

1

7               7       1       2

0               7       0       2

1               7       0       1

Page Fault Is 15

POSTLAB:

Result

The program for FIFO page replacement was implemented and hence verified.

Ex.No:9(b)     IMPLEMENTATION OF LRU PAGE REPLACEMENT ALGORITHM

Date:

PRELAB:

Aim

To write a program a program to implement LRU page replacement algorithm

INLAB:

Algorithm

1. Start the process

2. Declare the size

3. Get the number of pages to be inserted

4. Get the value

5. Declare counter and stack

6. Select the least recently used page by counter value

7. Stack them according the selection.

8.  Display the values

9. Stop the process

Program

#include<stdio.h>

main()

{

int q[20],p[50],c=0,c1,d,f,i,j,k=0,n,r,t,b[20],c2[20];

printf("Enter no of pages:");

scanf("%d",&n);

printf("Enter the reference string:");

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

            scanf("%d",&p[i]);

printf("Enter no of frames:");

scanf("%d",&f);

q[k]=p[k];

printf("\n\t%d\n",q[k]);

c++;

k++;

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

            {

                        c1=0;

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

                        {

                                    if(p[i]!=q[j])

                                    c1++;

                        }

                        if(c1==f)

                        {

                                    c++;

                                    if(k<f)

                                    {

                                                q[k]=p[i];

                                                k++;

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

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

                                                printf("\n");

                                    }

                                    else

                                    {

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

                                                {

                                                            c2[r]=0;

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

                                                            {

                                                            if(q[r]!=p[j])

                                                            c2[r]++;

                                                            else

                                                            break;

                                                }

                                    }

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

                                     b[r]=c2[r];

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

                                    {

                                                for(j=r;j<f;j++)

                                                {

                                                            if(b[r]<b[j])

                                                            {

                                                                        t=b[r];

                                                                        b[r]=b[j];

                                                                        b[j]=t;

                                                            }

                                                }

                                    }

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

                                    {

                                                if(c2[r]==b[0])

                                                q[r]=p[i];

                                                printf("\t%d",q[r]);

                                    }

                                    printf("\n");

                        }

            }

}

printf("\nThe no of page faults is %d",c);

}

POSTLAB:

Output:

Enter no of pages:10

Enter the reference string:7 5 9 4 3 7 9 6 2 1

Enter no of frames:3

        7

        7       5

        7       5       9

        4       5       9

        4       3       9

        4       3       7

        9       3       7

        9       6       7

        9       6       2

        1       6       2

The no of page faults is 10

POSTLAB:

Result:

The program for LRU page replacement was implanted and hence verified.

Ex.No:9(c)                OPTIMAL(LFU) PAGE REPLACEMENT ALGORITHM

Date:

PRELAB:

AIM:  

To implement page replacement algorithms Optimal (The page which is not used for longest time) Optimal algorithm. Here we select the page that will not be used for the longest period of time.

INLAB:

ALGORITHM:

OPTIMAL:

Step 1:  Create a array

Step 2:  When the page fault occurs replace page that will not be used for the longest

                period of time

PROGRAM:

#include<stdio.h>

#include<conio.h>

int i,j,nof,nor,flag=0,ref[50],frm[50],pf=0,victim=-1;

int recent[10],optcal[50],count=0;

int optvictim();

void main()

{

   clrscr();

   printf("\n OPTIMAL PAGE REPLACEMENT ALGORITHN");

   printf("\n.................................");

   printf("\nEnter the no.of frames");

   scanf("%d",&nof);

   printf("Enter the no.of reference string");

   scanf("%d",&nor);

   printf("Enter the reference string");

   for(i=0;i<nor;i++)

       scanf("%d",&ref[i]);

   clrscr();

   printf("\n OPTIMAL PAGE REPLACEMENT ALGORITHM");

   printf("\n................................");

   printf("\nThe given string");

   printf("\n....................\n");

   for(i=0;i<nor;i++)

       printf("%4d",ref[i]);

   for(i=0;i<nof;i++)

   {

       frm[i]=-1;

       optcal[i]=0;

   }

   for(i=0;i<10;i++)

       recent[i]=0;

   printf("\n");

   for(i=0;i<nor;i++)

   {

      flag=0;

      printf("\n\tref no %d ->\t",ref[i]);

      for(j=0;j<nof;j++)

      {

            if(frm[j]==ref[i])

            {

               flag=1;

               break;

            }

      }

      if(flag==0)

      {

            count++;

            if(count<=nof)

                victim++;

            else

                victim=optvictim(i);

            pf++;

            frm[victim]=ref[i];

            for(j=0;j<nof;j++)

               printf("%4d",frm[j]);

      }

   }

   printf("\n Number of page faults: %d",pf);

   getch();

}

int optvictim(int index)

{

   int i,j,temp,notfound;

   for(i=0;i<nof;i++)

   {

      notfound=1;

      for(j=index;j<nor;j++)

            if(frm[i]==ref[j])

            {

                notfound=0;

                optcal[i]=j;

                break;

            }

      if(notfound==1)

               return i;

   }

   temp=optcal[0];

   for(i=1;i<nof;i++)

       if(temp<optcal[i])

               temp=optcal[i];

   for(i=0;i<nof;i++)

       if(frm[temp]==frm[i])

               return i;

 return 0;

}

OUTPUT:    

OPTIMAL PAGE REPLACEMENT ALGORITHM

Enter no.of Frames....3

Enter no.of reference string..6

Enter reference string..6 5 4 2 3 1

OPTIMAL PAGE REPLACEMENT ALGORITHM

 The given reference string:

          ………………….  6   5   4   2   3   1

         Reference NO 6->          6  -1  -1

         Reference NO 5->          6   5  -1

         Reference NO 4->          6   5   4

         Reference NO 2->          2   5   4

         Reference NO 3->          2   3   4

         Reference NO 1->          2   3   1

         No.of page faults...6

POST LAB:

RESULT :

Thus the program to implement the paging technique of memory management was executed successfully and output was verified.

     

Ex.No:10                    Implement Shared memory and IPC

Date:

PRELAB:

AIM:

To write a c program to develop an application using Inter process Communication (IPC)

using pipes.

INLAB:

ALGORITHM:

1. Create the child process using Fork()

2. Create the pipe structure using pipe()

3. Now close the read end of the parent process using close()

4. Write the data in the pipe using write()

5. Now close the write end of child process using close()

6. Read y the data in the pipe using read()

7. Display the string.

PROGRAM:

#include<stdio.h>

int main()

{

int fd[2],child;

char a[10],b[10];

printf("\nEnter the string to enter into the pipe");

scanf("%s",a);

pipe(fd);

child=fork();

if(!child)

{

close(fd[0]);

write(fd[1],a,5);

wait(0);

}

else

{

close(fd[1]);

read(fd[0],b,5);

printf("\nThe string retrived from the pipe is %s\n",b);

}

SENDER:

#include<stdio.h>

#include<sys/shm.h>

#include<sys/ipc.h>

#define size 32

int main()

{

int shmid;

char *s[100],*str;

printf("\nipc message passing using shared memory sender");

shmid=shmget(60,size,IPC_CREAT|0666);

str=shmat(shmid,0,0);

printf("\neneter the message to be sent");

gets(s);

strcpy(str,s);

printf("\nyour mesage has been sent");

return 0;

}

RECEIVER:

#include<stdio.h>

#include<sys/shm.h>

#include<sys/ipc.h>

#define size 32

int main()

{

printf("\nipc message passing using shared memory-receiver");

int shmid;

char *str;

shmid=shmget(60,size,IPC_CREAT|0666);

str=shmat(shmid,0,0);

printf("\nreceived message is....");

puts(str);

return 0;

}

OUTPUT:

Sender

ipc message passing using shared memory sender

enter the message to be sentsuccess

your mesage has been sent

Receiver

ipc message passing using shared memory-receiver

received message is....success

POSTLAB:

RESULT:

The C program to develop an application using Inter process Communication (IPC)

using pipes is implemented.

Ex.No:11         Implement Paging Technique of memory management

Date:        

PRELAB:

AIM:  

To implement the Memory management policy- Paging.

INLAB:

ALGORITHM:

Step 1: Read all the necessary input from the keyboard.

Step 2: Pages - Logical memory is broken into fixed - sized blocks.

Step 3: Frames – Physical memory is broken into fixed – sized blocks.

Step 4: Calculate the physical address using the following

             Physical address = ( Frame number * Frame size ) + offset

Step 5: Display the physical address.

Step 6: Stop the process.

PROGRAM :

#include <stdio.h>

struct pstruct

{

            int fno;

            int pbit;

}ptable[10];

int pmsize,lmsize,psize,frame,page,ftable[20],frameno;

void info()

{

            printf("\n\nMEMORY MANAGEMENT USING PAGING\n\n");

            printf("\n\nEnter the Size of Physical memory: ");

            scanf("%d",&pmsize);

            printf("\n\nEnter the size of Logical memory: ");

            scanf("%d",&lmsize);

            printf("\n\nEnter the partition size: ");

            scanf("%d",&psize);

            frame = (int) pmsize/psize;

            page = (int) lmsize/psize;

            printf("\nThe physical memory is divided into %d no.of frames\n",frame);

            printf("\nThe Logical memory is divided into %d no.of pages",page);

}

void assign()

{

            int i;

            for (i=0;i<page;i++)

            {

            ptable[i].fno = -1;

            ptable[i].pbit= -1;

            }

            for(i=0; i<frame;i++)

                        ftable[i] = 32555;

            for (i=0;i<page;i++)

            {

            printf("\n\nEnter the Frame number where page %d must be placed: ",i);

                        scanf("%d",&frameno);

                        ftable[frameno] = i;

                        if(ptable[i].pbit == -1)

                        {

                                    ptable[i].fno = frameno;

                                    ptable[i].pbit = 1;

                        }

            }

                        printf("\n\nPAGE TABLE\n\n");

            printf("PageAddress  FrameNo. PresenceBit\n\n");

            for (i=0;i<page;i++)

                        printf("%d\t\t%d\t\t%d\n",i,ptable[i].fno,ptable[i].pbit);

            printf("\n\n\n\tFRAME TABLE\n\n");

            printf("FrameAddress   PageNo\n\n");

            for(i=0;i<frame;i++)

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

}

void cphyaddr()

{

            int laddr,paddr,disp,phyaddr,baddr;

            printf("\n\n\n\tProcess to create the Physical Address\n\n");

            printf("\nEnter the Base Address: ");

            scanf("%d",&baddr);

            printf("\nEnter theLogical Address: ");

            scanf("%d",&laddr);

            paddr = laddr / psize;

            disp = laddr % psize;

             if(ptable[paddr].pbit == 1 )

                        phyaddr = baddr + (ptable[paddr].fno*psize) + disp;

            printf("\nThe Physical Address where the instruction present: %d",phyaddr);

}

void main()

{

            info();

            assign();

            cphyaddr();

            }

OUTPUT:

MEMORY MANAGEMENT USING PAGING

 Enter the Size of Physical memory: 16

Enter the size of Logical memory: 8

Enter the partition size: 2

The physical memory is divided into 8 no.of frames

The Logical memory is divided into 4 no.of pages

Enter the Frame number where page 0 must be placed: 5

Enter the Frame number where page 1 must be placed: 6

Enter the Frame number where page 2 must be placed: 7

Enter the Frame number where page 3 must be placed: 2

PAGE TABLE

PageAddress            FrameNo.                  PresenceBit

         0                           5                                   1

         1                           6                                   1

         2                           7                                   1

         3                           2                                   1

FRAME TABLE

  FrameAddress               PageNo

        0                           32555

        1                           32555

        2                              3

        3                           32555

        4                           32555

        5                              0

        6                              1

        7                              2

Process to create the Physical Address

   Enter the Base Address: 1000

  Enter theLogical Address: 3

The Physical Address where the instruction present: 1013

POSTLAB:

RESULT:

The Memory management policy using Paging is implemented.

Ex.No:12                    Implement Threading & Synchronization Applications

Date:        

PRELAB:

AIM:

               To write a program to implement threading and synchronization applications.

INLAB:

ALGORITHM:        

Step 1: Start the program.

Step 2: Enter the logical memory address.

Step 3: Enter the page table which has offset and page frame.

Step 4: The corresponding physical address can be calculate by,

PA = [ pageframe* No. of page size ] + Page offset.

Step 5: Print the physical address for the corresponding logical address.

Step 6: Terminate the program.

PROGRAM:

$gcc thread.c -lpthread

#include<stdio.h>

#include<string.h>

#include<pthread.h>

#include<stdlib.h>

#include<unistd.h>

pthread_t tid[2];

void* doSomeThing(void *arg)

{

    unsigned long i = 0;

    pthread_t id = pthread_self();

    if(pthread_equal(id,tid[0]))

    {

        printf("\n First thread processing\n");

    }

    else

    {

        printf("\n Second thread processing\n");

    }

    for(i=0; i<(0xFFFFFFFF);i++);

    return NULL;

}

int main(void)

{

    int i = 0;

    int err;

    while(i < 2)

    {

        err = pthread_create(&(tid[i]), NULL, &doSomeThing, NULL);

        if (err != 0)

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

        else

            printf("\n Thread created successfully\n");

        i++;

    }

    sleep(5);

    return 0;

}

OUTPUT:

First thread processing 1

Thread created successfully

Second thread processing 2

can't create thread

POSTLAB:

RESULT:

            Thus the program has been executed and verified successfully.