Network Lab Manual(a & B)

8/2/2019 Network Lab Manual(a & B)

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Computer Networks Lab Manual

COMPUTER NETWORKS

LAB MANUAL

For

VII SEMESTER(ISE)

by:

Sreenath M,

Sr. Lecturer, Dept. of ISE,

Note: Student is required to solve one problem from PART-A and one problem from

PART-B. The questions are allotted based on lots. Both questions carry equal marks.

PART - A

SIMULATION EXERCISESThe following experiments shall be conducted using either NS / NCTUns or any other

suitable simulator.

Department Of Information Science and Engineering Page 1


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1. Simulate a three nodes point to point network with duplex links

between them. Set the queue size and vary the bandwidth and find the

number of packets dropped.

2. Simulate a four node point-to-point network with the links connected asfollows:n0 n2, n1 n2 and n2 n3. Apply TCP agent between n0-n3

and UDP between n1-n3. Apply relevant applications over TCP and UDP

agents changing the parameter and determine the number of packets

sent by TCP / UDP.

3. Simulate the different types of Internet traffic such as FTP and TELNET

over a network and analyze the throughput.

4. Simulate the transmission of ping messages over a network topologyconsisting of 6 nodes and find the number of packets dropped due to

congestion.

5. Simulate an Ethernet LAN using n nodes (6-10), change error rate and

data rate and compare throughput.

6. Simulate an Ethernet LAN using n nodes and set multiple traffic nodes

and determine collision across different nodes.

7. Simulate an Ethernet LAN using n nodes and set multiple traffic nodes

and plot congestion window for different source / destination.

8. Simulate simple ESS and with transmitting nodes in wire-less LAN by

simulation and determine the performance with respect to transmission

of packets.

PART - B

IMPLEMENT THE FOLLOWING BY C/C++

1. Write a program for error detecting code using CRC-CCITT (16- bits).

2. Write a program for frame sorting technique used in buffers.



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3. Write a program for distance vector algorithm to find suitable path for

transmission.

4. Using TCP/IP sockets, write a client server program to make the client send

the file name and to make the server send back the contents of the requestedfile if present.

5. Implement the above program using as message queues or FIFOs as IPC

channels.

6. Write a program for simple RSA algorithm to encrypt and decrypt the data.

7. Write a program for Hamming code generation for error detection and

correction.

8. Write a program for congestion control using leaky bucket algorithm.



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PART A Programs

The following experiments shall be conducted using NS2 or NCTUns-6.0 or other suitablesimulator

Steps to execute Part-A:

1.After grub loading select NCTUns kernel

2.others3.username:root

Password:root123

STEP1-open terminal1

[root@lab3~]#cd /usr/local/nctuns/bin

[root@lab3~bin]# ./dispatcher

STEP2-open terminal2

[root@lab3~]#cd /usr/local/nctuns/bin[root@lab3~bin]# ./coordinator

STEP3-open terminal3[root@lab3~]#cd /usr/local/nctuns/bin

[root@lab3~bin]# ./nctunsclient



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1. Simulate a three-node point-to-point network with a duplex link between them.

Set the queue size and vary the bandwidth and find the number of packets

dropped.

Procedure:

1. Draw the n/w topology as specified in the draw topology mode of the simulatorDraw toplogy mode-

Select the hub icon and drag it on the working window

Select the host icon on the toolbar and drag it on the working window

Repeat for another host

Select the link icon from the toolbar & make a connection b/w host1 and hub, host2 and

hub. Here the hub acts as node 3 in the point to point n/w topology

2. Go to edit mode and save the topology as a .tpl file

3. Double click on host1,click on the node editor & select MAC layer and then select half

duplex for hubs & in log statistics , select no of drop packets, no of collisions, &

throuput of outgoing packets & press ok. Select FIFO , set the queue size to 50 & pressok . click on command box & type the following command Shown after diagram

4. Double click on the link b/w host1 and hub to set the b/w to some initial value ,say10mbps

5. Save the changes made to the topology .now click on the R button to run the simulation

6. Now go to menu -> simulation -> run . execution

7. Play8. Go to g-settings , plot the graph -> fileopen ->.results

Topology:-

Sender:-

stg u 1024 100 1.0.1.2

Receiver:-

rtg u w log1

Parameters:-

Drop Packets and Collision Packets

2. Simulate a four-node point-to-point network and connect the link as follows: Apply

a TCP agent between n0 to n3 and apply a UDP agent between n1 and n3. Apply



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relevant applications over TCP and UDP agents changing the parameters and

determine the number of packets sent by two agents.

Procedure:

1. Create the toplogy of three nodes connect these nodes to a hub

2. Goto edit mode & save the topology3. Setup the tcp connection b/w n0-n3 and udp connection b/w n1-n3 as fallows using

the commands

TCP & UDP

Sender-Stcp p 7000 l 1024 1.0.1.3

Stg u 7000 l 100 1.0.1.3

Receiver-rtcp p 7000 l 1024

rtg u w log1

4. Set the i/p throuput for the receiver & the o/p throuput for the sender in the log file

statistics5. Back to edit mode

6. Run the simulation7. Play

8. For results of graph, -> g-settings

->.plot graph

->select parameters9. make a comparisons

Topology:-

Sender:-

Stcp p 7000 l 1024 1.0.1.3



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Stg u 7000 l 100 1.0.1.3

Receiver:-

rtcp p 7000 l 1024

rtg u w log1

Parameters:-

Throughput of incoming and outgoing Packets



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3. Simulate the different types of Internet traffic such as FTP, TELNET over a

network and analyze the throughput.

Procedure:

Steps to FTP connection

1. Create the topology of three nodesConnect these nodes to a hub

2. Go to edit mode and save the toplogy

3. Setup FTP connection b/w 1 and 3 using the following commands

Stcp p 21 l 1024 1.0.1.34. Set the i/p and o/p throughput logfile as described in the previous experiments

5. To view results, go to filename results folder

To setup TELNET connection

1. Create the topology of three nodes; connect these nodes to the hub

2. Go to edit mode and save the topology

3. Setup TELNET connection b/w node2 and node3Using the following command

Stcp p 23 l 1024 1.0.1.3Rtcp p 23 l 1024

4. Set the input and output throughput log file as described in the previous

experiments

5. To view the results, go filename results folder

Topology:-

Sender:-For FTP

stcp p 21 l 1024 1.0.1.3

For Telnetstcp p 23 l 1024 1.0.1.3



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Receiver:-

For FTP

rtcp p 21 l 1024For Telnet

rtcp p 23 l 1024

Parameters:-

Throughput of incoming and outgoing Packets



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4. Simulate the transmission of ping messages over a network topology consisting

of 6 nodes and find the number of packets dropped due to congestion.

Procedure:

1.click on the hub icon on the toolbar and drag it on the working window

2. select the 6 nodes and a suitable radius b/w the host and the hub3. a single host is made to act as a receiver and other as sender

4.run the simulation

5.in the run mode from sending node the receiving node is pinged using ping

1.0.1.6Note:the no of dropped packets are obtained only when the traffic is more is the

n/w

Command console:go to toolsSimulation time and change

Simulation time to 100

During run mode, doulble click host2 and then click command console and execute

Ping 1.0.1.6

Topology:-



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Sender:-

stcp p 2000 l 1024 1.0.1.4

Receiver:-rtcp p 2000 l 1024

Command Console:-

Goto tools-> simulation time and change Simulation time to 100. During run mode,

double click host 2 and then click command console. And execute the following

command.

ping 1.0.1.4

Parameters:-

Drop Packets and Collision Packets.



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5. Simulate an Ethernet LAN using N nodes (6-10), change error rate and data

rate and compare throughput.

Procedure:

1. Connect one set of host with a hub and another set of host also through a huband connect these two hubs with a switch , this forms Ethernet LAN

2. To setup a tcp connection b/w a host or one hub & host or another hub using

following commands

Stcp p portno l buffer size IP addressStcp p 2000 l 1024 1.0.1.4

rtcp p 2000 l 1024

3. Setup the error route , data route in the physical layer, i/p and o/p throughput inthe MAC layer

4. Change error route and data route and compare throughput

5. View the results in file name results

Note:double click on reciver link and change BER to 0.000001, run again

Topology:-

Sender:-

stcp p 2000 l 1024 1.0.1.4

Receiver:-

rtcp p 2000 l 1024

Double click on receiver link and change BER to 0.000001, Run Again.

Parameters:-Throughput of outgoing Packets



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6. Simulate an Ethernet LAN using N nodes and set multiple traffic nodes and

determine collisions across different nodes.

Procedure:

1. Connect one set of host with a hub and another set of host also through a hub and

connect these two hubs through a switch, these forms Ethernet LAN2. Setup multiple traffic connection b/w the hosts on one hub & hosts on another

hub using the following command

Stcp p 2000 l 1024 1.0.1.4

Rtcp p 2000 l 10243. Setup the collision log at the destination host in the MAC layer

4. Give the results in filename .results

Topology:-

Sender:-stcp p 2000 l 1024 1.0.1.4

Receiver:-

rtcp p 2000 l 1024

Parameters:-

Collision Packets and Drop Packets (Optional)



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7. Simulate an Ethernet LAN using N nodes and set multiple traffic nodes and

plot congestion window for different source/destination.

Procedure:1. Connect one set of host with a hub and another set of hosts also through a hub and

connect these two hubs through a switch this forms an Ethernet LAN

2. Setup multiple traffic connection b/w the hosts on one hub and host on another hubusing the following

Stcp p 2000 l 1024 1.0.1.4

Rtcp p 2000 l 1024

3. Setup the collision log at the destination host in the MAC layer4. Plot the congestion window , go to menu->tools->plotgraph->file->open filename

results-> filename .coll.log

5. Give the results in filename .results

Topology:-

Sender:-

stcp p 2000 l 1024 1.0.1.4

Receiver:-

rtcp p 2000 l 1024

Parameters:-

Receiver side Collision Packets and Drop Packets



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8. Simulate simple BSS and with transmitting nodes in wireless LAN by

simulation and determine the performance with respect to transmission of

packets.

Procedure:

1. Draw the topology consisting of a : route b:host c:WLAN d:WLAN accesspoint (two)

2. Go to tools menu, select the option wireless LAN -> mobile nodes

3. Generate infrastructure mobile nodes IP

4. A dialog box appear, entries to be done areSubnet ID 8 gateway IP

Subnetid=2 & gateway Ip= 1.0.1.0

5. Here mobile nodes are the senders & the host is the receiver6. For each mobile node, add a command

ttcp t -u s p 7000

ttcp t u s p 7001 for another node

7. For the receiver, add the following commandTtcp r u s p 7000

8. Run the simulation9. To see the results of the graph , go to g-settings ->plot graph ->file open-

>results, select the parameters required

Topology:-

Click on access point. Goto wireless interface and tick on show transmission range

and then click OK.

Double click on Router -> Node Editor and then

Left stack -> throughput of Incoming packets

Right stack -> throughput of Outgoing packets

Select mobile hosts and access points then click on.

Tools -> WLAN mobile nodes-> WLAN Generate infrastructure.



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Subnet ID: Port number of router (2)

Gateway ID: IP address of router

Mobile Host 1

ttcp t u s p 3000 1.0.1.1

Mobile Host 1

ttcp t u s p 3001 1.0.1.1

Host(Receiver)ttcp r u s p 3000

ttcp r u s p 3001

Run and then play to plot the graph.



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PART-BError detecting code using CRC-CCITT

Overview:The accurate implementations (long-hand and programmed) of the 16-bit

CRC-CCITT specification, is as follows:

Width = 16 bits

Truncated polynomial = 0x1021

Initial value = 0xFFFF

Input data is NOT reflected

Output CRC is NOT reflected

No XOR is performed on the output CRC

Theoretical Concepts:Important features of a standard CRC are that it:

Can be used to validate data

Is reproducible by others

The first feature above is easy to realize in a closed system if corruption

of data is infrequent (but substantial when it occurs). The term "closed

system" refers to a situation where the CRC need not be communicated

to others. A correct implementation of a 16-bit CRC will detect a

change in a single bit in a message of over 8000 bytes. An erroneous

CRC implementation may not be able to detect such subtle errors. Iferrors are usually both rare and large (affecting several bits), then a

faulty 16-bit CRC implementation may still be adequate in a closed

system. The second feature above -- that the CRC is reproducible by

others -- is crucial in an open system; that is, when the CRC must be

communicated to others. If the integrity of data passed between two

applications is to be verified using a CRC defined by a particular

standard, then the implementation of that standard must produce the

same result in both applications -- otherwise, valid data will be reported

as corrupt.

Reproducibility may be satisfied by even a botched implementation of a

standard CRC in most cases -- if everyone uses the same erroneous

implementation of the standard. But this approach:

Modifies the standard in ways that are both unofficial and

undocumented.



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Creates confusion when communicating with others who have not

adopted the botched implementation as the implied standard.

The CRC value for the 9-byte reference string, "123456789" is 0xE5CC.

The need to focus on the 16-bit CRC-CCITT (polynomial 0x1021) and notCRC16 (polynomial 0x8005),are as follows :

Is a straightforward 16-bit CRC implementation in that it doesn't

involve:

reflection of data

reflection of the final CRC value

Starts with a non-zero initial value -- leading zero bits can't affect the

CRC16 used by LHA, ARC, etc., because its initial value is zero.

It requires no additional XOR operation after everything else is done.

The CRC32 implementation used by Ethernet, Pkzip, etc., requires

this operation; less common 16-bit CRCs may require it as well.

The need to use a 16-bit CRC instead of a 32-bit CRC is as follows :

Can be calculated faster than a 32-bit CRC.

Requires less space than a 32-bit CRC for storage, display or printing.

Is usually long enough if the data being safeguarded is fewer than

several thousand bytes in length, e.g., individual records in a database.

Example:-

Calculation of the 16-bit CRC-CCITT for a one-byte

message consisting of the letter "A":

Quotient= 111100001110111101011001

poly=

------------------------------------------

10001000000100001)

1111111111111111010000010000000000000000

10001000000100001

-------------------------------------------------------

---

red bits are initial value

11101111110111111

bold bits are message

10001000000100001

blue bits are augmentation



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-------------------------------------------------------

----

11001111100111100

10001000000100001

-------------------------

-----

10001111000111010

10001000000100001

---------------------------------------

----

00001110000110110

00000000000000000

---------------------------

-----

00011100001101100

00000000000000000

---------------------------

-----

00111000011011000

00000000000000000

----------------------------

-----

01110000110110001 00000000000000000

----------------------------

-----

11100001101100010

10001000000100001

----------------------------

-----

11010011010000110

10001000000100001----------------------------

----

10110110101001110



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Program 1:

Write a program for error detecting code using CRC-CCITT (16-bits).

#include

#include

char t[128],cs[128],g[ ]="10001000000100001";

#define N strlen(g)

int a,e,c;

void xor()

{

for(c=1;c


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for(e=a;e


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FRAME SORTING TECHNIQUE USED IN BUFFERSOverview & Theoretical Concepts:

We need the frame sorter when the frame source is using TCP/IP so we can

run packet re-assemblers on other computers in an effort to balance the

workload.

If the frames are coming in over ISIS IP multicasts, we don't need a sorter:

Each re-assembler can receive all frames and discard those they don't need.

This way we get each frame on the network one time only, and the work in

each re-assembler to input and reject frames on the wrong virtual channel is

very little. If ISIS is not using true IP multicasting, then, once again, the

sorter is needed to reduce the network traffic.

There are side benefits to such a design. In cases like I&T where the frame

source is using TCP/IP, we need to sort the transfer frames by virtual

channel so we can run some packet re-assemblers on other workstations to

balance the compute load. But even so, we may re-assemble housekeeping

packets on the workstation doing the frame sorting. Why do it in a separate

process? This is especially true on Solaris where we have control over thread

binding, and therefore, thread concurrency.

In a multi-threaded environment, the frame sorter and packet re-assembler

both can be constructed from a collection of objects: input tasks, outputtasks, packet assembly tasks, pool objects, sorter objects, and queue objects.

The packet re-assembler might looks like this:

input --> queue --> pkt assy --> sorter --> queue --> output

Transfer frames enter an input. Input enqueues and possibly archives them.

The packet assembly task reads frames out of the queue and constructs

packets. It hands completed packets to a sorter which enqueues them for

output tasks and possibly archives them. Each output task reads packets

fromits queue and sends packets out.

Program 2:

Write a program for frame sorting technique used in the buffers.

#include

struct frame{

int fslno;



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char finfo[20];

};

struct frame arr[10];

int n;

void sort()

{

int i,j,ex;

struct frame temp;

for(i=0;i


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printf("-----------------------------------------------------------\n");

}

return 0;

}

Output :

Enter the number of frames: 3

Enter the frame sequence and frame content:

5 ink

3 pot

8 pen

The frame in sequence

3 pot

---------------------------------5 ink

---------------------------------

8 pen

---------------------------------

DISTANCE VECTOR ALGORITHM

Background: One of the most popular & dynamic routing algorithms, the

distance vector routing algorithms operate by using the concept of each

router having to maintain a table( ie., a vector ), that lets the router know the

best or shortest distance to each destination and the next hop to get to there.

Also know as Bellman Ford(1957) and Ford Fulkerson (1962). It was the

original ARPANET algorithm.

Algorithm Overview:Each router maintains a table containing entries for

and indexed by each other router in the subnet. Table contains two parts :

1.A preferred outing line.

2.Estimate of time or distance to that destination.

The metric used here is the transmission delay to each destination. Thismetric maybe number of hops, queue length, etc.

The router is assumed to know the distance metric to each of its

neighbours.Across the network the delay is calculated by sending echo

packets to each of neighbours.

Example:


x y z

x

y

z

0 2 7

from

cost to

from

from

x y z

x

y

z

0 2 3

from

cost tox y z

x

y

z

0 2 3

from

cost to

x y zx

y

z

cost to

x y zx

y

z

0 2 7

from

cost to

x y z

x

y

z

0 2 3

from

cost to

x y z

x

y

z

0 2 3

from

cost to

x y z

x

y

z

0 2 7

from

cost to

x y z

x

y

z 7 1 0

cost to

2 0 1

2 0 1

7 1 0

2 0 1

7 1 0

2 0 1

3 1 0

2 0 1

3 1 0

2 0 1

3 1 0

2 0 1

3 1 0

x z

12

7

y

Dx(y) = min{c(x,y) + D

y(y), c(x,z) + D

z(y)}

= min{2+0 , 7+1} = 2


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Program 3:

Write a program for distance vector algorithm to find suitable path for

transmission.

#include

#include

int p,a[10][10];

int i,j;

void path(int n)

{

printf("If no path exist,enter a value > 999 \n");

printf("enter the source node:");

scanf("%d",&p);

for(i=1;i


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{

int sma=9999,pos;

for(i=1;ia[1][i])

{sma==a[1][i];

pos=i;

}

for(j=i;j1)

{

int sma=minv(a,s);

s=s-1;

for(i=p;ia[p][sma]+a[sma][i])a[p][i]=a[p][sma]+a[sma][i];

}

}

}

void disp(int n)

{

for(i=1;i


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scanf("%d",&n);

path(n);

fsp(n);

disp(n);

}

Output:

Enter the number of node:3

If no path exists,enter a value >999

Enter the source node: 1

Enter the value of 1:2:::path

200

Enter the value of 1:3:::path255


256


643


235


2345The path from 1 to 1:0

The path from 1 to 2:200

The path from 1 to 3:255

CLIENT SERVER PROGRAM USING SOCKET PROGRAMMING

OVERVIEW:Unix sockets is just like two way FIFO's. All data

communication will take place through the socket's interface, instead ofthrough the file interface. Although unix socket's are a special file in the file

system(just like FIFO's), there's usage of socket(), bind(), recv(),etc and not

open(), read().

When programming with socket's, usually there's creation of server and

client programs. The server will sit listening for incoming connections from



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clients and handling. This is similar to the situation that exists with internet

sockets but with fine differences.

For instance, when describing which unix socket that has to be used(i.e the

path to the special file that is the socket). The structure struct sockaddr_un

has the following fields:struct sockaddr_un{

unsigned short sa_family; // Address family,AF_XXXX

char sa_data; // 14 bytes of protocol address

};

This is the structure you will be passing to the bind() function, which

associates a socket descriptor(a file descriptor) with a certain file(the name

for which is in the sun_path field).

The structure struct sockaddr_in is used when we need IP address and Port

number to be binded to the Sockets. It has following fields:

struct sockaddr_in {

short int sin_family; // Address family

unsigned short int sin_port; // Port number

struct in_addr sin_addr; // Internet address

unsigned char sin_zero[8] // Same size as struct

sockaddr

};

// Internet adress

struct in_addr {

unsigned long s_addr; // 32 bits or 4 bytes long};

BACKGROUND REQUIRED:

1. UNIX File I/O system calls

2. UNIX IPC system calls

3. UNIX socket programming

THEORETICAL CONCEPTS:

Most interprocess communication uses the client server model. These terms

refer to the two processes which will be communication with each other.

One of the two processes , the client , connects to the other process, the

server, typiceally to make a request for information. A good analogy is a

person who makes a phone call to another person.

Notice that the client needs to know of the existence of and the address of

the server, but the server does not need to know the adresss of(or even the

existence of) the client prior to the connection being established. Notice also



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that once a connection is established, both sides can send and receive

information.

The system calls for establishing a connection are somewhat different for the

client and the server, but both involve the basic construct of a socket. A

socket is one end of an interprocess communication channel. The twoprocesses each establish their own socket.

The steps involved in establishing a socket on the client side are as follows-

1. Create a socket with the socket() system call .

2. Connect the socket to the address of the server using the connect()

system call.

3. Send and receive data.There are a number of ways to do this, but the

simplest is to use the read() and write() systen calls.

The stepd involved in establishing a socket on the server side are as

follows-

1. Create a socket with the socket() system call.

2. Bind the socket to an address using the bind() system call. For a

server socket on the internet,an address consists of a port number on

the host machine.

3. Listen for connections with the listen() system call.

4. Acept a connection with the accept() system call. This call typically

blocks until a client connects with the server.

5. Send and receive the data.

Socket Types:When a socket is created, the program has to specify the address domain and

the socket type. Two processes can communicate with each other only if

their sockets are of the same type and in the same domain, in which two

processes running on any two hosts on the Internet communicate. Each of

these has it's own adress format.

The address of a socket in the Unix domain is a character string which is

basically an entry in the file system.

The address of a socket in the Internet domain consists of the Internet

address of the host machine (every computer on the Internet has a unique 32

bit address, often reffered to as it's IP address). In addition , each socket

needs a port number on that host. Port numbers are 16 bit unsigned integers.

The lower numbers are reserved in Unix for standard services. For eg, the

port number for the FTP server is 21.There are two widely used socket

types, stream sockets , and datgram sockets. Stream sockets treat

communications as a continuous stream of characters, while datagram



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sockets have to read entire messages at once. Each uses it's own

communications protocol. Stream sockets use TCP , which is a reliable ,

stream oriented protocol, and datagram sockets use UDP, which is unreliable

and message oriented.

The primary socket calls are as follows:-

1. socket() - Create a new socket and return it's descriptor.

2. bind() - Associate a socket with a port and address .

3. Listen() -Establish a queue for connection requests.

4. Accept()- Accepts a connection request.

5. Connect()- Initiate a connection to a remote host.

6. Recv() - Receive data from a socket descriptor.

7. Send() - Send data to a socket descriptor.

8. Close() - one-way close of a socket descriptor,

The other system calls used are as follows:-

1. gethostbyname- given a hostname , returns a structure which specifies

it's DNS name(s) and IP address(es).

2. getservbyname given service name and protocol , returns a structure

which specifies its name(s) and its port address.

The socket utility functions are as follows:-

1. htons/ntohl- convert short/long from host byte order to network byte

order.

2. inet_ntoa/inet_addr- converts 32 bit IP address (network byte orderto/from a dotted decimal string)

The header files used in order are:-

1. -prerequisite typedefs.

2. -names for erno values (error numbers)

3. - struct sock addr ;system prototypes and constants .

4. - network info lookup prototypes and structures.

5. - struct sockaddr_in; byte ordering macros.

6. - utility function prototypes.

Program 4

Using TCP/IP sockets, write a client-server program to make client

sending the file name and the server to send back the contents of the

requested file if present.



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Client program:

#include

#include

#include

#include#include

#include

#include

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

{

int sockfd,portno,n;

char buffer[256],c[20000];

struct sockaddr_in serv_addr;

if(argc


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}

Server program:

#include#include

#include

#include

#include

#include

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

{

int sockfd,newsockfd,portno,clilen,n;

char buffer[256],c[2000],cc[20000];struct sockaddr_in serv_addr,cli_addr;

FILE *fd;

if(argc


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strcat(cc,c);

write(newsockfd,cc,strlen(cc));

}else{

printf("server:file not found\n");

n=write(newsockfd,"file not found\n",15);}

return 0;

}

Output:

cc server.c

./a.out 8888

Server waiting for client:

(in another terminal)

cc client.c

./a.out 8888

Enter the path:

(in that folder jst create a new document and type some text inside the

document)

Take the path of that folder and type in the client terminal.

Enter the path : /nwlab/ns/ns 2.33/a

Client : displaying contents of /nwlab/ns/ns 2.33/a

EXECUTION INSTRUCTIONS:

There are two programs(Server-side & Client-side) to be typed on an editor

(eg vi or kdevelop), exit the editor after saving the progarm, complile the

programs on different terminals using the command cc .c

-o and then the command ./ is



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entered if no errors are encountered(else correct the errors and then perform

the above).

1. Compile and execute the Server side program specifying an unsed port

no. after the ouput filename, then in the other terminal compile and

execute the Client sideprogram with IP address of Server system andcommon port no.

2. Enter the pathname of any file on the Server system as a request to the

Server and wait for reply.

3. Now switch to Server terminal and check if the file requested is present or

not and the status of Server. If the file is found then switch to the Client

terminal and watch the contents of file displayed on the terminal. Else it

displays that file is not found with pathname.

MESSAGE QUEUES / FIFOs AS IPC CHANNELS

Client Server communication using FIFO as IPC

OVERVIEW:

FIFO's is used to send data between a Clientand Server. If we have a server

that is contacted by numerous clients, each client can write its request to a

well-known(i.e, the pathname is known by all clients) FIFO that the server

creates. The figure below shows this arrangement. Since there are multiple

writers for the FIFO, the requests sent by the clients to the server need to be



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less than PIPE_BUF bytes in size. This prevents any interleaving of the

client writes.

The problem in using FIFO's for this type of client-server communication is

how to send replies back from the server to each client. A single FIFO can't

be used, as the clients would never know when to read their response, versusresponses for other clients. One solution is for each client to send its process

ID with the request. The server then creates a unique FIFO for each client,

using a pathname based on the client's process ID.

BACKGROUND REQUIRED:

1. UNIX File I/O system calls

2. UNIX IPC system calls

ALGORITHM: There are two programs- Server side & Client side.Server side:

1. Create a well-known FIFO using mknod command.

Ex: mknod(, S_IFIFO | 0666 , 0)

2. Open the above FIFO in readonly mode to accept requests from

the clients.

3. When the client opens the other end of FIFO in writeonly mode

then read the contents and store the request(pathname of file) in

a buffer.

4. Now create another FIFO(client-specific) in writeonly mode to

send the reply(contents of the file requested)

5. Open the file requested by client and write the contents into the

client-speicfic FIFO and terminate the connection.

Client side:

1. Open the well-known Server FIFO in write mode.


Well-known

FIFO

Client Client

Server

Client-specific

FIFOClient-specific

FIFO


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2. Write the pathname of file into this FIFO and send as a request.

3. Now open the Client-specific FIFO in read mode and wait for

reply from server.

4. When the contents of file are available on this FIFO display it

on the terminal.

Program 5

Implement the above program using as message queues of FIFOs as

IPC channel.

server program:

#include

#include

#include

#include#include

#include

#include

#include

#define FIFO1_NAME "fifo1"


int main()

{

char p[100],f[100],c[3000];

int num,num2,f1,fd,fd2,in;

mknod(FIFO1_NAME,S_IFIFO | 0666,0);


printf("SERVER ONLINE\n");



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fd = open(FIFO1_NAME,O_RDONLY);

printf("Client online \n waiting for request \n");

while(1){

if((num = read(fd, p, 100))==-1)

perror("Read Error\n");else{

p[num] ='\0';

if((f1=open(p,O_RDONLY))


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int main()

{char p[1000],f[1000],c[3000];

int num,num2,f1,fd,fd2;



printf("Waiting for server\n");

fd=open(FIFO1_NAME,O_WRONLY);

printf("server online\n Client:Enter the Path\n");

while(gets(p)!=feof(stdin)){

if((num= write(fd,p,strlen(p)))==-1)

perror("write Error\n");

else

{

printf("waiting for reply\n");

fd2=open(FIFO2_NAME,O_RDONLY);

if((num2=read(fd2,c,300))==-1)

perror("Transfer error\n");

else{

printf("File Recevied \n displaying the contents \n");

if(fputs(c,stdout)==EOF)perror("Print error\n");

exit(1);

}

}

}

}

Output:

clientterminal Server terminal



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cc client.c

./a.out

Waiting for server

Server online

Client:enter the path:/ns/nwlab/ns2.33/file

Waiting for reply

File received

Displaying the contents:

Bangalore college of engineering

cc server.c

./a.out

Server online

Client online, waiting for the request

Waiting for requestServer /ns/nwlab/ns2.33/file found

Transferring the contents

Server:transfer completed size=33

EXECUTION INSTRUCTIONS:

There are two programs(Server-side & Client-side) to be typed on an editor,

exit the editor after saving the program, compile the programs on different

terminals using the command cc .c and then the

command ./a.out is entered if no errors are encountered(else correct theerrors and then perform the above).

1. Compile and execute the Server side program first then in the other

terminal compile and execute the client side program.

2. Enter the pathname of any file on the system as a request to the Server

and wait for reply.

3. Now switch to Server terminal and check if the file requested is present or

not and the status of Server. If the file is found then switch to the Client

terminal and watch the contents of file displayed on the terminal. Else it

displays that file is not found with pathname.



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RSA ALGORITHM

Overview:

The RSA Public key algorithm was invented in 1977 by Ron Rivest, Adi

Shamir and Leonard Adelman (RSA). The algorithm supports encryption

and digital signatures. It is the most widely used public key algorithm. RSA

gets its security from the integer factorization problem. The algorithm is

relatively easy to understand and implement. It has been patent free since

2000.RSA is used in security protocols such as IPSEC/IKE -IP data security,

TLS/SSL -transport data security (web), PGP -email security, SSH -terminal

connection security, SILC -conferencing service security.

Theoretical Concepts:

RSA gets its security from the factorization problem. The difficulty of

factoring large numbers is the basis of security of RSA.

The Integer factorization problem (finding a number's prime factors):

For a positive integer n, find its prime factors:

n = p1 p2 ... pi where pi is positive distinct prime number

Example: 257603 = 41 * 61 * 103

Factorization algorithms can be used to factor faster than brute forcing.

Some of them are Trial division, Pollard's rho, Pollard's p-1, Quadratic sieve,

elliptic curve factorization, Random square factoring, Number field sieve,

etc. A Prime number is a positive integer and is divisible only by itself and

1. Prime numbers are found

With primality testing; an algorithm which tests a probable prime for

primality. If primality testing returns false prime numbers the cryptographic

algorithm may be insecure (or will not function correctly). RSA depends on

prime numbers in key generation. It also uses strong primes, numbers



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whose factors of the prime are also primes.

The RSA algorithm:

Key generation:

1)Select random prime numbers p and q, and check that p != q2)Compute modulus n = pq

3)Compute phi = (p - 1)(q - 1)

4)Select public exponent e, 1 < e < phi such that gcd(e,phi) = 1

5)Compute private exponent d = e^-1 mod phi

6)Public key is {n, e}, private key is d

Encryption: c = m*e mod n, decryption: m = c*d mod n

The selected public exponent e, which is used as public key with n. It is used

to encrypt messages and to verify digital signatures. The e is stored for later

with n. The e is usually small number but it can be 1 < e < phi . The e must

be relatively prime to phi , hence gcd(e, phi ) = 1.

(gcd = greatest common divisor, using the Euclidean algorithm)

The private exponent d, is the actual RSA private key. The d must not be

disclosed at any time or the security of the RSA is compromised. The d is

found by computing the multiplicative inverse d = e^-1 mod phi . The

extended Euclidean algorithm is commonly used to compute inverses. The d

exponent is used to decrypt messages and to compute digital signatures.Implementations try to

find as small d as possible to make decryption faster. This is fine as long as

it is assured that d is about the same size as n. If it is only one quarter of size

it is not considered safe to be used. It is possible to find a smaller d by using

lcm(p-1,q-1) instead of phi (lcm = least common multiple, lcm(p-1,q-1) =

phi /gcd(p-1,q-1) ).

Example of RSA with small numbers:

p = 47, q = 71, compute n = p*q = 3337

Compute phi = 46 * 70 = 3220

Let e be 79, compute d = 79-1 mod 3220 = 1019



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Public key is n and e, private key d, discard p and q.

Encrypt message m = 688, 68879 mod 3337 = 1570 = c.

Decrypt message c = 1570, 15701019 mod 3337 = 688 = m.

Program 6:

Write a program for simple RSA algorithm to encrypt and decrypt the

data.

#include

#include

int gcd(int a,int b)

{

int c;

while(a!=b)

{

if(a


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}

int main()

{

int p,q,n,e,m,c,d,x,z;int en[100],de[100],j=0;

printf("\nenter the value of p & q\n");

scanf("%d%d",&p,&q);

n=p*q;

z=(p-1)*(q-1);

for(e=1;e


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printf("\nchoose D\n");

scanf("%d",&d);

x=mod(c,d,n);

printf("after decryption=%d\n",x);

return 0;}

Output:

Enter the value of p& q: 3 5

1 3 5 7

Choose e: 7Enter the character to be encrypted: 3

Before encryption : 3

After encryption: 12

7

Choose d: 7

After decryption : 3

HAMMING CODES

Overview:

Developed by 1947 Richard. W. Hamming for detecting and correcting

single bit errors in transmitted data. This technique requires that three parity

bits (or check bits) be transmitted with every four data bits. The algorithm is



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called a (7, 4) code, because it requires seven bits to encoded four bits of

data.

Theoretical Concepts :

Background Concepts Required:1. matrix multiplication

2. modulo 2 arithmetic

3. parity.

Parity:A parity bit is an extra bit that forces a binary string to have a specific

parity.

Two types:

1.Even(Even number of 1's i.e., the modulo 2 sum of the bits is 0)

2.Odd (Odd number of 1's i.e., the modulo 2 sum of the bits is 1)

The table below lists all possible three bit values and value of a parity bit

required to create a four bit sequence with even parity.

3 Bit String Parity Bit Verification

000 0 0 + 0 + 0 + 0 = 0

001 1 0 + 0 + 1 + 1 = 0

010 1 0 + 1 + 0 + 1 = 0

011 0 0 + 1 + 1 + 0 = 0100 1 1 + 0 + 0 + 1 = 0

101 0 1 + 0 + 1 + 0 = 0

110 0 1 + 0 + 1 + 0 = 0

111 1 1 + 1 + 1 + 1 = 0

Table 1. Even Parity

It is very common for communication protocols to specify that a block of

bits will be transmitted with a specific parity. If a block of data arrives at its

intended destination with a parity other than the specified parity, it must be

the case that at least one of the bits has been corrupted. A single parity bit is

not sufficient to identify an even number of bits with errors, nor is it

sufficient to allow the receiver to correct an error. Hamming codes use

multiple parity bits to allow for error correction.



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Encoding

Traditional Hamming codes are (7, 4) codes, encoding four bits of data into

seven bit blocks (a Hamming code word). The extra three bits are parity bits.

Each of the three parity bits are parity for three of the four data bits, and no

two parity bits are for the same three data bits. All of the parity bits are evenparity.

Example:

Given: data bits d1, d2, d3, and d4

A (7, 4) Hamming code may define parity bits p1, p2, and p3 as

p1 = d2 + d3 + d4

p2 = d1 + d3 + d4

p3 = d1 + d2 + d4

There's a fourth equation for a parity bit that may be used in Hamming

codes:

p4 = d1 + d2 + d3

Valid Hamming codes may use any three of the above four parity bit

definitions.

Valid Hamming codes may also place the parity bits in any location within

the block of 7 data and parity bits. Two Hamming codes with different parity

bits or parity bits in a different bit position are considered equivalent. Theywill produce different results, but they are still Hamming codes.

One method for transforming four bits of data into a seven bit Hamming

code word is to use a 47 generator matrix [G].

Define d to be the 14 vector [d1 d2 d3 d4]

It's possible to create a 47 generator matrix [G] such that the product

modulo 2 of d and [G] (d[G]) is the desired 17 Hamming code word. Here's

how it's done:

Step 1: Represent each data bit with a column vector as follows:

| 1 |

d1 = | 0 |

| 0 |

| 0 |



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| 0 |

d2 = | 1 || 0 |

| 0 |

| 0 |

d3 = | 0 |

| 1 |

| 0 |

| 0 |

d4 = | 0 |

| 0 |

| 1 |

Step 2:Represent each parity bit with a column vector containing a 1 in the

row corresponding to each data bit included in the computation and a zero in

all other rows. Using the parity bit definitions from the example above:

| 0 |p1 = | 1 |

| 1 |

| 1 |

| 1 |

p2 = | 0 |

| 1 |

| 1 |

| 1 |

p3 = | 1 |

| 0 |

| 1 |



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Step 3:Arrange the column vectors from the previous steps into a 47 matrix

such that the columns are ordered to match their corresponding bits in a code

word.

Using the vectors from the previous steps, the following will produce code

words of the form [p1 p2 p3 d1 d2 d3 d4]

| 0 1 1 1 0 0 0 |

G = | 1 0 1 0 1 0 0 |

| 1 1 0 0 0 1 0 |

| 1 1 1 0 0 0 1 |

Arranging the columns in any other order will just change the positions of

bits in the code word.

Example:

Encode the data value 1010 using the Hamming code defined by the matrix

G (above).

| (1 0) + (0 1) + (1 1) + (0 1) |

| 1 |

| (1 1) + (0 0) + (1 1) + (0 1) || 0 |

| 0 1 1 1 0 0 0 | | (1 1) + (0 1) + (1 0) + (0 1) |

| 1 |

|1010 | | 1 0 1 0 1 0 0 | = | (1 1) + (0 0) + (1 0) + (0 0) |

= | 1 |

| 1 1 0 0 0 1 0 | | (1 0) + (0 1) + (1 0) + (0 0) |

| 0 |

| 1 1 1 0 0 0 1 | | (1 0) + (0 0) + (1 1) + (0 0) |

| 1 |

| (1 0) + (0 0) + (1 0) + (0 1) |

| 0 |

So 1010 encodes to 1011010. Equivalent Hamming codes represented by

different generator matrices will produce different results.



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Decoding

In a world without errors decoding a Hamming code word would be very

easy. Just throw out the parity bits. The encoding example produced a 7 bit

code word. Its parity bits are 101 and its data bits are 1010. If you receive a

1011010, just decode it as 1010. But what happens if you receive a codeword with an error and one or more of the parity bits are wrong.

Suppose the Hamming code defined by the matrix G in the example above is

being used and the code word 1011011 is received. How is that word

decoded? The first step is to check the parity bits to determine if there is an

error.

Arithmetically, parity may be checked as follows:

p1 = d2 + d3 + d4 = 0 + 1 + 1 = 0

p2 = d1 + d3 + d4 = 1 + 1 + 1 = 1

p3 = d1 + d2 + d4 = 1 + 0 + 1 = 0

In this case every parity bit is wrong. p1, p2, and p3 should have been 010,

but we received 101.

Parity may also be validated using matrix operations. A 37 parity check

matrix [H] may be constructed such that row 1 contains 1s in the position of

the first parity bit and all of the data bits that are included in its parity

calculation. Row 2 contains 1s in the position of the second parity bit and all

of the data bits that are included in its parity calculation. Row 3 contains 1sin the position of the third parity bit and all of the data bits that are included

in its parity calculation.

Example:

Using the code from example above, the matrix H may be defined as

follows:

| 1 0 0 0 1 1 1 |

H = | 0 1 0 1 0 1 1 |

| 0 0 1 1 1 0 1 |

Multiplying the 37 matrix [H] by a 71 matrix representing the encoded

data produces a 31 matrix called the "syndrome". There are two useful

proprieties of the syndrome. If the syndrome is all zeros, the encoded data is



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error free. If the syndrome has a non-zero value, flipping the encoded bit that

is in the position of the column matching the syndrome will result in a valid

code word.

Example:Using the parity check matrix from the example above we can correct and

verify the code word 1011011.

| 1 |

| 0 |

| 1000111 | | 1 | |(11)+(00)+(01)+(01)+(10)+(11) +(11) | |

1 |

| 0101011 | | 1 | = |(00)+(10)+(01)+(11)+(00)+(11)+(11) | = |

1 |

| 0011101 | | 0 | |(01)+(00)+(11)+(11)+(10)+(01)+(11) | |

1 |

| 1 |

| 1 |

A column of all 1s is not the column of all 0s, so there is a parity error.

Looking back at the matrix [H], you will see that the seventh column is all

1s, so the seventh bit is the bit that has an error. Changing the seventh bit

produces the code word 1011010.

| 1 |

| 0 |

|1000111| | 1 | |(11)+(00)+(01)+(01)+(10)+(11)+(10) |

| 0 |

|0101011| | 1 | = |(00)+(10)+(01)+(11)+(00)+(11)+(10) |

= | 0 |

|0011101| | 0 | |(01)+(00)+(11)+(11)+(10)+(01)+(10) |

| 0 |

| 1 |

| 0 |

Sure enough 1011010 is a valid code word. As I stated at the top of this

section remove the parity bits to get the encoded value. In this case 1011011



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was likely transmitted as 1011010, which encodes 1010.

Program 7:

Write a program for Hamming Code generation for error detection

and correction.

#include

#include

#include

int main()

{

int i,j;

char data[4];

int d1[5],d2[5],d3[5],d4[5];int p1[5],p2[5],p3[5];

int gmatrix[4][8];

int encode[7]={0,0,0,0,0,0,0};

int edata[7]={0,0,0,0,0,0,0};

int errdig;

int syn[3]= {0,0,0};

int hmatrix[3][7] = { 1,0,0,0,1,1,1,

0,1,0,1,0,1,1,

0,0,1,1,1,0,1

};

int choice;

system("clear");

for(;;)

{

printf("\n 1.ENCODE ");



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printf("\n 2.DECODE ");

printf("\n 3.EXIT ");

printf("\n Enter your choice : ");

scanf("%d",&choice);

switch(choice){

case 1:

printf("\n Enter 4 data bits\n");

scanf("%s",data);

for(i=0;i


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for(i=0;i


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}

Output:

cc prgm8.c

./a.out1.Encode

2.Decode

3.Exit

Enter your choice : 1

Enter the 4 bits : 1101

The generator matrix is

0 1 1 1 0 0 0

1 0 1 0 1 0 0

1 1 0 0 0 1 0

1 1 1 0 0 0 1

The encoded data is:

0011101

1.Encode

2.Decode

3.Exit


Enter the encoded data :

0 0 1 1 1 0 1Error free data

The actual data error free data: 1101

1.Encode

2.Decode

3.Exit


LEAKY BUCKET ALGORITHM

The leaky-bucket implementation is used to control the rate at which trafficis sent to the network. A leaky bucket provides a mechanism by which

bursty traffic can be shaped to present a steady stream of traffic to the

network, as opposed to traffic with erratic bursts of low-volume and high-

volume flows.



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Traffic analogy An appropriate analogy for the leaky bucket is a scenario in

which four lanes of automobile traffic converge into a single lane. A

regulated admission interval into the single lane of traffic flow helps the

traffic move. The benefit of this approach is that traffic flow into the major

arteries (the network) is predictable and controlled. The major liability is

that when the volume of traffic is vastly greater than the bucket size, in

conjunction with the drainage-time interval, traffic backs up in the bucket

beyond bucket capacity and is discarded.

The Leaky-bucket algorithm

The algorithm can be conceptually understood as follows:

Arriving packets (network layer PDUs) are placed in a bucket with a

hole in the bottom.

The bucket can queue at most b bytes. If a packet arrives when the



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bucket is full, the packet is discarded.

Packets drain through the hole in the bucket, into the network, at a

constant rate ofrbytes per second, thus smoothing traffic bursts.

The size b of the bucket is limited by the available memory of the system.

Sometimes the leaky bucket and token algorithms are lumped together under

the same name.

Program 8:

Write a program for Congestion control using the leaky bucket

algorithm.

#include

#define min(x,y) ((x) < (y) ? (x) : (y))



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#define MAX 25

int main()

{

int cap, oprt, nsec, cont, i=0,dr=0, inp[MAX]={0}, ch, x;printf("\n\nLEAKY BUCKET ALGORITHM\n");

printf("\nEnter bucket size : ");

scanf("%d",&cap);

printf("\nEnter output rate (no..of pkts/sec) : ");

scanf("%d",&oprt);

do{ printf("\nEnter the no..of packets entering at second %d :

",i+1);

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

printf("Enter 1 to insert packets or 0 to quit : ");

scanf("%d",&ch);

} while(ch);

nsec=i;

printf("\nSecond : Packets sent : Packets recvd : In bucket:

Dropped\n");

for(cont=i=0; cont || (i0)

if(x>cap)

{

cont=cap;

dr=x-cap;

}

else

{

cont=x;



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dr=0;

}

else

cont=0;

printf(" %d\t:",cont); printf(" %d\n",dr);

}

return 0;

}

Output :

Leaky bucket algorithm:

Enter bucket size: 3

Enter the output rate(No. of packets/sec): 2

Enter the no. of packets sent at second 1: 6

Enter 1 to insert packets or 0 to quit:0

Second : packet sent : packet recvd : In bucket : Dropped

(1) (6) (2) (3) (1)

(2) (0) (2) (1) (0)

(3) (0) (1) (0) (0)

Viva Questions



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1. What are functions of different layers in OSI And TCP/IP model?2. Differentiate between TCP/IP Layers and OSI Layers

3. Why header is required?

4. What is the use of adding header and trailer to frames?

5. What is encapsulation?

6. Why fragmentation requires?7. What is MTU?

8. Which layer imposes MTU?

9. Differentiate between flow control and congestion control.

10. Differentiate between Point-to-Point Connection and End-to-End connections.

11. Which are protocols running in different layers?12. What is Protocol Stack?

13. Differentiate between TCP and UDP.

14. Differentiate between Connectionless and connection oriented connection.

15. Why frame sorting is required?

16. What is meant by subnet?

17. What is meant by Gateway?

18. What is an IP address?19. What is MAC address?

20. Why IP address is required when we have MAC address?

21. What is meant by port?

22. What are ephemerical port number and well known port numbers?

23. What is a socket?24. What are the parameters of socket()?

25. Describe bind(), listen(), accept(),connect(), send() and recv().

26. What are system calls? Mention few of them.

27. What is IPC? Name three techniques.

28. Explain mkfifo(), open(), close() with parameters.

29. What is meant by file descriptor?

30. What is meant by traffic shaping?31. How do you classify congestion control algorithms?

32. Differentiate between Leaky bucket and Token bucket.

33. How do you implement Leaky bucket?

34. How do you generate busty traffic?

35. What is the polynomial used in CRC-CCITT?

36. What are the other error detection algorithms?

37. What is difference between CRC and Hamming code?38. Why Hamming code is called 7,4 code?

39. What is odd parity and even parity?

40. What is meant by syndrome?

41. What is generator matrix?42. What is spanning tree?

43. Where Pirms algorithm does finds its use in Networks?

44. Differentiate between Prims and Kruskals algorithm.

45. What are Routing algorithms?46. How do you classify routing algorithms? Give examples for each.

47. What are drawbacks in distance vector algorithm?

48. How routers update distances to each of its neighbor?

49. How do you overcome count to infinity problem?50. What is cryptography?



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51. How do you classify cryptographic algorithms?

52. What is public key?

53. What is private key?

54. What are key, ciphertext and plaintext?

55. What is simulation?

56. What are advantages of simulation?57. Differentiate between Simulation and Emulation.

58. What is meant by router?

59. What is meant by bridge?

60. What is meant by switch?

61. What is meant by hub?

62. Differentiate between route, bridge, switch and hub.

63. What is ping and telnet?

64. What is FTP?

65. What is BER?

66. What is meant by congestion window?

67. What is BSS?

68. What is incoming throughput and outgoing throughput?

69. What is collision?

70. How do you generate multiple traffics across different sender-receiver pairs?

71. How do you setup Ethernet LAN?

72. What is meant by mobile host?

73. What is meant by NCTUns?74. What are dispatcher, coordinator and nctunsclient?

75. Name few other Network simulators

76. Differentiate between logical and physical address.

77. Which address gets affected if a system moves from one place to another place?

78. What is ICMP? What are uses of ICMP? Name few.

79. Which layer implements security for data?



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Guidelines For installation of NCTUns Network Simulator

Follow the following steps carefully,

Please don't skip any steps that have been mentioned below

1. Install LinuxRecommended Fedora 12 with dvd(gcc compiler should be installed if it not installed

after fedora installation)

2. After installation Boot into Linux as root.

3. Copy the .tgz installation file of NCTUns that you got from college to the folder /bin/localPlease don't change any folder name in this folder that is created after unzipping the above

file. Dont even change the Case of the folder that is created

4. Now unzip the .tgz file by opening the terminal and changing the directory to /bin/localbythe command :-

[root@localhost ~] cd /bin/local

To unzip use the following command:-

[root@localhost local] tar xvzf [the file name].tar

(Note there is no '-' before xvzf)

This will create a folder called NCTUns in the directory /bin/local...

5. Now disable the Secure Linux option by running the following command :-

[root@localhost local]vi /etc/selinux/config

When the file opens, there is a line similar to -- SELINUX=enforcing

Change the "enforcing" to "disabled" (Note without quotes)

6. From the directory/bin/localchange the current working directory to NCTUns by following

command :-[root@localhost local]cd NCTUns

7. Now from here execute the installation shell script that will do the required compilations

and settings for you:-

[root@localhost local]./install.sh

During this part it will ask for installation of tunnel files. Please type yes and Enter to

continue

8. If the installation is successful, it will display the success message at the end. Now restartyour computer. You will find a new entry in GRUB Menu "NCTUns kernel login". Boot

into Linux using this entry.

9. before using simulator, please execute the following command

[root@localhost ~]iptables -F

10. Run the simulator using three commands where each command should be executed in

different window.

[root@localhost ~]dispatcher

[root@localhost ~]coordinator



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[root@localhost ~]nctunsclient

11. In the NCTUns window Settings Dispatcher. Provide the username and password of the

user account u created in step 11. Then Click OK.

Documents

Network Lab Manual(a & B)