Computer netwrks

1
LECTURE NOTES
ON
COMPUTER NETWORKS
2018 – 2019
III B. Tech I Semester (JNTUA-R15)
Mr. K Munivara Prasad, Associate Professor
CHADALAWADA RAMANAMMA ENGINEERING COLLEGE
(AUTONOMOUS)
Chadalawada Nagar, Renigunta Road, Tirupati – 517 506
Department of Computer Science and Engineering

Computer Networks K Munivara Prasad
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY ANANTAPUR
B. Tech III - I sem (Common to CSE)
(15A05502) COMPUTER NETWORKS
III Year B.Tech. I Sem. Course SYLLABUS
Unit I
Introduction: Networks, Network Types, Internet History, Standards and Administration, Network
Models: Protocol Layering, TCP/IP Protocol Suite, The ISO Model. The Physical layer: Data and
Signals, Transmission impairment, Data rate limits, Performance, Transmission media:
Introduction, Guided Media, Unguided Media, Switching: Introduction, Circuit Switched
Networks, Packet switching.
Unit II
The Data Link Layer: Introduction, Link layer addressing, Error detection and Correction: Cyclic
codes, Checksum, Forward error correction, Data link control: DLC Services, Data link layer
protocols, HDLC, Point to Point Protocol, Media Access control: Random Access, Controlled
Access, Channelization, Connecting devices and virtual LANs: Connecting Devices.
Unit III
The Network Layer: Network layer design issues, Routing algorithms, Congestion control
algorithms, Quality of service, Internetworking, The network layer in the Internet: IPV4
Addresses, IPV6, Internet Control protocol, OSPF, BGP, IP, ICMPv4, IGMP.
Unit IV
The Transport Layer: The Transport Service, Elements of Transport Protocols, Congestion
Control, The internet transport protocols: UDP, TCP, Performance problems in computer
networks, Network performance measurement.
Unit V
The Application Layer: Introduction, Client Server Programming, WWW and HTTP, FTP, e-mail,
TELNET, Secure Shell, Domain Name System, SNMP.
Text Books:
1. ―Data communications and networkingǁ, Behrouz A. Forouzan, Mc Graw Hill Education, 5th
edition, 2012.
2. ―Computer Networksǁ, Andrew S. Tanenbaum, Wetherall, Pearson, 5th edition, 2010.
L T C
3 1 3

UNIT- I
Introduction
An interconnected collection of autonomous computers is called a computer network. Two
computers are said to be interconnected if they are able to exchange the information. If one
computer can forcibly start, stop and control another one, the computers are not autonomous. A
system with one control unit and many slaves is not a network, nor is a large computer with
remote printers and terminals.
In a Distributed system, the existence of multiple autonomous computers is transparent(i.e., not
visible) to the user. He can type a command to run a program and it runs. It is up to the operating
system to select the best processor, find and transport all the files to that processor, and put the
results in the appropriate place.
The user of a distributed system is not aware of that there are multiple processors; it looks like a
virtual uniprocessor. Allocation of jobs to processors and files to disks, movement of files between
where they are stored and where they are needed, and all system function are automatic.
With a network, users must explicitly log onto one machine, explicitly submit jobs remotely,
explicitly move files around and generally handle all the network management personally. The
distinction between Network and distributed system lies with software (OS) rather than software.
In network user invokes, in distributed system the system invokes.
A network is a set of devices connected by media links. Anode can be a computer, printer or any
other device capable of sending and receiving data generated by other nodes on the network.
The links connecting the devices are often called communication channels.
Networks use Distributed processing, in which a task is divided among multiple computers.
Advantages of Distributed processing are
 Security/ Encapsulation
 Distributed data bases
 Faster problem solving
 Security through Redundancy
 Collaborative processing
Network Criteria

Performance:
The performance can be measured in many ways and depends on number of factors.
 Number of users
 Type of transmission medium
 Hardware
 Software
Reliability
This is measured by the following factors
 Frequency of failure
 Recovery time of a network after a failure.
 Catastrophe.
Security
Network security issues include protecting data from the following
 Unauthorized access
 Viruses
Applications
 Accessing Remote databases
 Accessing Remote programs
 Value added communication facility
 Marketing and sales
 Financial services
 Manufacturing
 Electronic message
 Directory services
 Information services
 Teleconferencing
 Cellular telephone
 Cable television
Network Criteria
Performance Reliability Security

Network Structure
Hosts
IMP
Subnet
The end systems are called the HOSTS. The hosts are connected through a
communication subnet or simply Subnet as shown in fig.
The subnet consists of two parts: a) Transmission lines b) Switching elements.
The Transmission lines transmit the raw bits. The Switching elements are specialized
computers, which switches packets. This is called Interface Message Processor (IMP) or
Router or data switching exchanges or packet switching nodes.
The data can be transmitted through the subnet in two ways. They are
a) Point to point or store and forward
b) Broad casting
Network Architecture
To reduce the design complexity, most networks are organized as a series of layers or
levels, each built upon on the one below it. The number of layers, the name of each layer
,the contents of each layer ,and the function of each layer differ from network to network
However, in all networks the purpose of each layer is to offer certain services to the higher
layers ,shielding those layers from the details of how the offered services are actually
implemented.

Layer n on one machine carries on a conversation with layer n on another machine. The
rules and conventions used in this conversation are collectively known as the layer n
Protocol.
The entities comprising the corresponding layers on different machines are called Peers.
Layer 5 protocol
Layer 4/5 interface
Layer 4 protocol
Layer 3/4 interface
Layer 3 protocol
Layer 2/3 interface
Layer 2 protocol
Layer 1/2 interface
Layer 1 protocol
Layers, protocols and interfaces.
The interface defines which primitive operation and services the lower layer offers to the
upper one.
A set of layers and protocol is called network architecture.
Data transfer methods:
a. Simplex communication:
Data will be transferred in one direction only.
Simplex transmission
Ex: Keyboards, Monitors
b. Half -- duplex communication:
Data will be transferred in both the directions, but not simultaneously.
Layer 5
Layer 3
Layer 2
Layer 1
Physical Medium
Layer 5
Layer 4
Layer 3
Layer 2
Layer 1
Layer 4
A B

Half duplex transmission
Ex: One way bridge with two directional traffic.
c. Full – duplex communication:
Data will be transferred in both the directions simultaneously.
Ex: Two-way road, where traffic will be there in both the directions.
REFERENCE MODELS
The ISO OSI REFERENCE MODEL
In 1947, the International Standards Organization (ISO) proposed a network model that covers all
network communications .This model is called Open Systems Interconnection (OSI) model. An
open system is a model that allows any two different systems to communicate regardless of their
underlying architecture.
The OSI model is built of seven layers: Physical (layer 1), Data link (layer 2), Network
(layer 3), Transport (layer 4), Session (layer 5), Presentation (layer 6) and Application layers
(layer 7).
Within a single machine, each layer calls upon the services of the layer just below it.layer 3,for
example, uses the services provided by layer 2 and provides for layer 4.Between machines layer
on one machine communicates with layer x on another machine. This communication is governed
by protocols. The processes on each machine that communicate at a given layer are called peer
–to – peer processor.
At the physical layer, communicate is direct: Machine A sends a stream of bits to machine B. At
the higher layers, however, communication must move down through the layers on machine A,
over to machine B, and then back up through the layers. Each layer in the sending machine adds
its own information to the message it receives from the layer just above it and passes the whole
package to the layer just below it. This information is added in the form of headers or trailers.
Headers are added to the message at layers 6, 5, 4, 3, and 2.At layer 1 the entire message
converted to a form that can be transferred to the receiving machine. At the receiving machine,
A B
A B

the message is unwrapped layer by layer, with each process receiving and removing the data
meant for it.
Organization of the layers:
The seven layers can be thought of as belonging to three subgroups. Layers 1, 2, 3 –are the
network support layers; they deal with the physical aspects of moving data from onr machine to
another. Layers 5, 6, 7—can be thought of as user support layers: they allow interoperability
among unrelated software systems. Layer 4,the transport layer, ensures end to end reliable
transmission while layer 2 ensures reliable transmission on a single link. The upper layers are
implemented almost always in software; lower layers are a combination of hard ware and
software, where as physical layer is mostly hardware.
The OSI reference model
Application
Presentation
Session
Transport
Network
Data link
Physical
Application
Presentation
Session
Transport
Network
Data link
Physical
Network Network
Data link Data link
Physical Physical
Application protocol
Presentation protocol
Session protocol
Transport protocol
Layer
Name of unit
exchanged
Host A Host B
3
1
2
4
5
6
7
Interface

FUNCTIONS OF LAYERS:
Functions of the Layers
Physical Layer :
 Physical characteristics of interfaces and media
 Representation of bits.
 Data rate
 Synchronisation of bits
 Line configuration (point to point or multipoint)
 Transmission Mode
 Physical Topology
Data Link Layer :
 Framing
 Physical addressing
 Error control
 Flow control
 Access control
Network Layer :
- Routing
- Congestion control
- Billing
Transport Layer :
- Service – Point addressing
- Segmentation and reassembly
- Flow control
- Error control
Session Layer :
-Dialog control
-Synchronization
Presentation Layer :
- Data encoding
- Encryption
- Compression

Application Layer :
- File Transfer
- Mail services
- Directory services
TCP/IP reference model
Switching Methods
Two different types of switching methods are used: Circuit switching and Packet switching.
Circuit Switching
In this switching there are three phases
a. Circuit establishment b. Data transfer c. Circuit disconnection
Represents the node
The network station
A
1
2
4 6
3
5
7
B
C
D
Application
Session
Transport
Network
Data link
Physical
Application
Transport
Network
Data link
Physical
Not present in
the model
OSI TCP/IP
Presentation

Suppose if we want to send the data, say, from A to D. before sending the data a circuit will be
established between A to D as shown in fig with doted lines. All the data will follow the same path.
After data is transferred the circuit will be disconnected.
Circuit switching Virtual packet switching
Propagation delay
Propagation delay
call accept signal
DATA
Ack signal
Ack signal
1 2 3 4
1 2 3 4
1 2 3 4
Packet switching
call accept signal
P3
P2
P1
P3
P2
P1

Packet switching will be done in two ways.
1. Virtual Packet switching 2. Data gram Packet switching
Circuit switching Data gram packet Virtual packet
Dedicated transmission No dedicated path No dedicated path
Continuous transmission of
data Transmission of packets Transmission of packets
Messages are not stored Packets are stored
Packets are stored until
delivered
Path will be established for
entire conversation
Route will be established for
each packet.
Route will be established for
entire conversation
Fixed bandwidth transmission Dynamic use of bandwidth Dynamic use of bandwidth
X.25
X.25 is a packet switching wide area network.
It is an interface between DCE and DTE for terminal operation in the packet mode on public data
networks.
It defines how a packet- mode terminal connected to a packet network for the exchange of data.
It describes the procedures for establishing maintaining and terminating connections.
X.25 is known as a subscriber network interface (SNI).
It defines how the user’s DTE communicates with the network and how packets are sent over that
network using DCE’s.
X.25 has three layers:
 Physical layer
 Frame layer and
 Packet layer
Physical Layer:
At the physical layer, X.25 specifies a protocol called X.21.
This is similar to other physical layer protocols.

Frame Layer:
X.25 provides data link control using a bit oriented protocol called link access procedure balanced
(LAPB).
Packet Layer:
The Network layer in x.25 is called the Packet Layer Protocol (PLP).
 This layer is responsible for establishing the connection, transferring data and terminating
the connection.
 It is also responsible for creating the virtual circuits and negotiating network services
between two DTEs.
 The Frame layer is responsible for making a connection between a DTE and DCE, the
Packet layer is responsible for making a connection between two DTEs.
 End-to-End flow and error control between two DTEs are under the jurisdiction of the
Packet Layer.
Examples of Networks
NOVEL NETWARE
The most popular network in pc world system is novel netware.it was designed to be used by
companies from a mainframes to a network of PCs.
1. In this system, each user has a desk top PC functioning as a client.
2. Some number of power full PCs operate as servers providing file services ,data base
services and other services to a collection of clients it uses a proprietary protocol.
3. It is based an old Xerox network system, XNS with various modifications. Because of five-
layers, it looks much like TCP/IP than ISO OSI.
4. Physical and data link layer can choose an Ethernet, IBM token ring and ARC net protocols.
5. The network layer runs an unreliable connectionless Internet work protocol called ARC net
protocols.
6. It passes packets from source to destination transparently; even both are of different
networks.
7. Application layer uses SAP (Service Advertising protocol), to broadcast a packet and tell
what

service it offered. These packets are collected by special agents of a process running on the
router machine. With this information they construct databases of which server are running
where.
8. When client machine is booted, it broadcast s a request asking where the nearest server is.
The agent on the local router sees, looks into the database of servers and matches up the
request with the best server; with this the client can now establish a NCP connection and act
like client-server model in all aspects.
Integrated Services Digital Network (ISDN)
ISDN was developed by ITU- T in 1976.It is a set of protocols that combines digital telephony and
data transport services. The whole idea is to digitize the telephone network to permit the
transmission of audio, video, and text over existing telephone lines.
The goal of isdn is to form a wide network that provides universal end –to – end connectivity over
digital media. This can be done by integrating all of the separate transmission services into one
without adding links or subscriber lines.
HISTORY
Voice Communication over Analog Networks
Initially, telecommunications networks were entirely analog networks and were used for the
transmission of analog information in the form of voice.
Voice and Data Communication over Analog Networks
With the advent of digital processing, subscribers needed to exchange data as well as voice.
Modems were developed to allow digital exchange over analog lines.
Analog and Digital services to Subscribers
To reduce cost and improve performance, the telephone companies gradually began to add
digital technologies while continuing their analog services to their customers.

Integrated Digital Network (IDN)
Next, customers began to require access to a variety of networks, such as packet- switched
networks and circuit –switched networks. To meet these needs the telephone companies created
Integrated Digital Network (IDN). An IDN is a combination of networks available for different
purposes.
Integrated Services Digital Network (ISDN)
The ISDN integrates customer service with the IDN. With ISDN all customers’ services become
digital rather than analog and will allow the customers services to be made available on demand.
SERVICES
The purpose of the ISDN is to provide fully integrated digital services to users. These services fall
in to three categories: bearer services, teleservices, and supplementary services.
Bearer service
Bearer services provide the means to transfer information (voice, data, and voice) between users
without the network manipulating the content of information.
Tele Service
In teleservices the network may change or process the contents of the data. These services
correspond to layers 4 – 7 of the OSI ISO model. this service include telephony,telefax,videotex,
telex and teleconferencing.
Supplementary service
Supplementary services are those services that provide additional functionality to the bearer
service and teleservices. These services include call waiting, reverse charging, and message
handling.

Supplementary
services
Tele Services





- 
Bearer Services
SUBSCRIBER ACCESS TO THE ISDN
To allow flexibility, digital pipes between customers and the ISDN office are organized into
multiple channels of different sizes. The ISDN standard defines three channel types, each with a
different transmission rate: bearer channels, data channels, and hybrid channels
Channel Rates
Channel Data Rate(Kbps)
Bearer (B)
Data (D)
Hybrid (H)
64
16,64
384,1536,1920
B Channel
A B channel is defined at a rate of 64 Kbps .It is the basic user channel and can carry any type of
digital information in full duplex mode as long as the required transmission rate does not exceed
64 Kbps. A B channel can be used to carry digital data, digitized voice, or other low data – rate
information.
D Channel
A D channel can be either 16 or 64 Kbps, depending on the need of the user. The primary
function of a D channel is to carry control signaling for the B channels. A D channel carries the
    
  
Telephony Telefax Teletex
Telex
Teleconferencing
Call
waiting
Reverse
charging
Message
Handling
Circuit
switching
Packet
switching.
Cell
switching

control signaling for all the channels in a given path, using a method called common – channel
(Out – of – band) signaling.
Less common uses for the D channel include low- rate data transfer and applications such as
telemetry and alarm transmission.
H Channel
H Channels are available with data rates of 384 Kbps (HO), 1536 Kbps (H11), or 1920(H12).
These e rates suit for high data rate applications such as video, teleconferencing and so on.
Integrated Services Digital Network
I S D N
Integrated services digital network
I D N
Digital
Pipes
Office
User Interfaces
Digital subscriber loops are two types: basic rate interface (BRI ) and primary rate interface
(PRI ) .Each type is suited to a different level of customer needs .Both include one D channel and
some number of either B or H channels.
B R I
The basic rate interface specifies a digital pipe consisting of two B channels and one 16Kbps D
channel.
To ISDN office
64(B1) + 64 (B2) +16( D) +48(over head) = 192Kbps
Circuit
switching
………….
Packet
switched
Subscriber loops
I S D N

BRI requires a digital pipe of 192 Kbps as shown in the fig. Conceptually, the BRI service is like a
large pipe that contains three smaller pipes, two for the B channels and one for D channel. The
remainder of the space inside the large pipe carries overhead bits required for its operation.
The BRI is designed to meet the needs of residential and small – office customers.
P R I
The usual PRI specifies a digital pipe with 23 B channels and one 64 Kbps D channel.
To ISDN office
P R I 1.544 Mbps
64 x 23 (B1 – B23) + 64 (D) + 8 (overhead)
PRI requires a digital pipe of 1.544 Mbps. Conceptually, the PRI services is like a large pipe
containing 24 smaller pipes, 23 for the B channels and for the D channel. The rest of the pipe
carries the overhead bits.
One PRI can provide full – duplex transmission between as many as 23 sources and
receiving nodes. The individual transmission are collected from their source and multiplexed on to
a single path for sending to the ISDN office.
Functional Grouping
Functional Grouping used at the subscriber’s premises includes network terminations, terminal
equipment and terminal adapters, enables users to access the services of the BRI and PRI.
Network Teriminator 1 ( NT1)
An NT1 device controls the physical and electrical termination of the ISDN at user’s internal
system to the digital subscriber loop. These functions are comparable to those defined for the OSI
physical layer.
An NT1 organizes the date stream from connected subscribers into frames that can be sent
over the digital pipe, and translates the frame received from the network into a format usable by
the subscriber’s device.
Network Teriminator 2 ( NT2)
A NT1 device performs functions at the physical layer, data link, and net work layers of the OSI
model.NT2 provide multiplexing (layer 1),flow control (layer 2), and packetzing (layer 3).An NT2
provides intermediate signal processing between the Data – generating devices and an
NT1.There must be a point to point connection between an NT1 and NT1 ..NT2s are used
primarily to interface between a multi-user system and an NT1 in a PRI.
NT2s can be implemented by a variety of equipment types like a private branch exchange
(digital PBX), a LAN can function as an NT2.

Terminal Equipment 1 (TE1)
The TE is used by ISDN in the same manner as DTE in other protocol. Examples of TE1 are
digital telephones, integrated voice/data terminals, digital facsimiles.
Terminal Equipment 2 (TE2)
To provide backward compatibility with a customer’s existing equipment, the ISDN standard
defines a second level of terminal equipment called Terminal Equipment 1 ( TE1 ).This is a non
ISDN device, such as terminal, workstation or regular telephone. This can be used with the help
of another device called a terminal adapter (TA).
Reference Points
This refers to the label used to identify individual interface between two elements of an ISDN
installation. There are four reference points that defines the interface between a subscriber’s
equipment and the network. They are R, S, T and U.
Reference Point R defines the connection between a TE2 and a Ta. Reference Point S defines
the connection between a TE1 or TA and an NT1 or NT2. Reference Point T defines the interface
between an NT2 and NT1. Reference Point U defines the interface between an NT1 and the
ISDN office.
R S U
To ISDN
Office
S
S U
To ISDN
Office
S T U
To ISDN
Office
TE2
NT1
TA
TE1
NT1
TE1
NT1
NT
2

Reference Points
The ISDN Layers
Since the ISDN specifies two different channels ( B and D) with different functionalities ,unlike the
ISO OSI (which has seven layers) the ISDN is defined in three separate planes : the user plane,
the control panel ,and the management plane. All three planes are divided into seven layers
that correspond to the OSI model.
B channel D channel
Layers 4,5,6,7
Network
Data link
Physical
Simplified layers of ISDN
Physical Layer
The ISDN physical layer specifications are defined by two ITU-T standards: L430 for BRI access
and I.431 for PRI access. These standards define all aspects of the BRI and PRI. Of these
aspects, four are of primary importance:
 The mechanical and electrical specifications of interfaces R, S, T and U.
 Encoding
 Multiplexing channels to make them carriable by the BRI and PRI digital pipe.
 Power supply
Data Link Layer
B and D channels use different data link protocols. B channels use LAPB protocol. The D channel
uses link access procedure for D channel (LAPD).
Network Layer
Once a connection has been established by the D channel, the B channel sends data using circuit
switching, X.25,or other similar protocols. The network layer packet is called a message.
User’s choice
X.25 and others
LAPB and others
Call control
Q.931
LAPD
BRI (I.430) & PRI (I.431)

BROAD BAND ISDN
When the ISDN was originally designed data rates of 64 Kbps to 1.544 Mbps were sufficient to
handle all existing transmission needs. As applications using the telecommunications network
advanced, however, these rates proved inadequate to support many applications. In addition, the
original bandwidths provided too narrow to carry the large numbers of concurrent signals
produced by a growing industry of digital services providers.
To provide for the needs of the next generation of technology, an extension of ISDN,
called broadband ISDN is under study. B-ISDN provides subscribers to the network with data
rates in the range of 600 Mbps, almost 400 times faster than the PRI rate.
Services
B-ISDN provides two types of services: interactive and distributive.
Interactive services
Interactive services are those that require two-way exchanges between either two subscribers or
between a subscriber and a service provider. These services are of three types: conversational,
messaging and retrieval.
Conversational
These services are those, such as telephone calls, that support real time exchanges. These real
time services can be used for telephoning, video telephoning, and video conferencing, data
transfer.
Messaging
These services stored and forward exchanges. These are bi-directional, meaning that all parties
in an exchange can use them at same time. These services include voice mail, data mail, video
mail.
Retrieval
These services are those used to retrieve information from a central source called an information
center. These services are like libraries; they must allow public access to retrieve information on
demand. And information is not distributed unless asked for.
Distributive services
These services are unidirectional services sent from a provider to subscribers without the
subscriber having to transmit a request each time a service is desired. These services can be
without or with user control.
Access method
B-ISDN defines three access methods to provide for three levels of user needs. They are:
 155.520 Mbbs full duplex symmetrical
 155.520 Mbbs asymmetrical
 622.080 Mbbs full duplex
* * * * *

Expected Questions
1. Describe ISO OSI model with a neat sketch.
2. Compare and contrast the ISO OSI model with TCP/IP model.
3. Explain Novel NetWare and the ARPANET.
4. Explain the working of X.21
5. Explain this network structure
6. Explain different types of transmission media used in data communication
7. Give advantages and disadvantages of fiber optic cable over metallic cable
8. Draw the B-ISDN reference model and explain the functions of each layer.
9. Explain about ATM. What are the advantages of using fixed length cells.
10. Explain X.25 network.
11. What are the applications of networks?
12. Write the difference between B-ISDN and N-ISDN
13. Explain the following items
a) IMP
b) HOST
c) Subnet
14. What are the advantages of layered architecture
15. Explain the goals of a computer network.
***************************

Review Questions
1.Which OSI layers are the network support layers?
2. Which OSI layers are the user support layers?
3.What is the difference between network layer delivery and transport layer delivery?
4.List the layers of the OSI model.
5.What is a peer-to-peer process?
6.How does information get passed from one layer to the next?
7.What are the concerns of the Physical Layer?
8.What are the responsibilities of the data link layer?
9. What are the responsibilities of the network layer?
10. What are the responsibilities of the transport layer?
11.The transport layer creates a connection between the source and destination. What are the
three events involved in a connection?
12. What are the responsibilities of the session layer?
13. What are the responsibilities of the presentation layer?
14. What are the services provided by the application layer?
15.Name two categories of transmission media.
16.How do guided media differ from unguided?
17.What are the three major classes of guided media.
18.What is the major advantage of shielded twisted pair over unshielded twisted pair?
19.Why is coaxial cable superior to twisted-pair cable?
20.Name the advantages of optical over twisted pair and coaxial cable.
21. What are the disadvantages of optical fiber as a transmission medium?
22.How an IDN differs from ISDN?
23.What type of information can a B channel transmit?
24. What type of information can a D channel transmit?
25. What type of information can a H channel transmit?
26.What is the difference between in band signaling and out of band signaling?
27.What is NT1?
28. What is NT2 ?
29.What is TE1?
30. What is TE2?
31.What are X.25 layers?
32.How does the X.25 layers relate to the OSI model?
33.Name the X.25 frame types?

34.What are the functions of X.25 frame types?
35.What are the frame layer phases involved in the communication between a DTE and a DCE ?
36.How are packets associated with the virtual circuit on which they travel?
37.What is the purpose of an LCN?
38.What type of virtual circuits does X.25 use?
39.List of the fields of a PLP packet types?
Multiple Choice Questions
1.The OSI model consists of ----------- layers.
a. three b. five c. seven d. ten
2.The------layer decides the location of synchronization points.
a.transport b.session. c.physical d.application
3.the end-to-end delivery of the entire message is the responsibility of the------
a.network b. transport c.net work d.data link
4.The -------layer is the closest to the transmission medium.
a.physical b.datalink c.session d.application
5.In the ------layer the data unit is called a frame
a. data link b.network c. application.d. Network
6.Decryption and encryption of data are the responsibility of the ----- layer
a. physical b. data link c. session d. presentation
7.Dialog control is a function of the ----- layer
8.Mail services and directory services are available to network users through the ----layer.
a. transport b. session. c. physical d. application
9.Node –to-Node delivery of the data units is the responsibility of the ----- layer.
a. data link b. network c. application. d. Network
10.AS the data packets move from the lower to the upper layers, headers are-------
a. added b. subtracted c. rearranged d. modified
11. AS the data packets move from the upper to the lower layers, headers are-------
a. added b. subtracted c. rearranged d. modified
12.When data are transmitted from device A to device B, the header from A’s layer 5 is read by
B’s --------layer.
a. physical b. transport c. session d. presentation
13.In ----layer, translation from one character code to another occur.
14.The ----- layer changes bits into electromagnetic signals.

15.The ----layer can use the trailer of the frame for error detection.
16. The physical layer is concerned with the transmission of --------over the physical medium.
a. programs b. dialogs c. protocols d.bits.
17.Which of the following is an application layer service?
a. network virtual terminal b. file transfer c. mail service d. all of the above
18.Transmission media are usually categorized as------
a. fixed or unfixed b. guided media and unguided c. determinate or in determinate
d. metallic and nonmetallic
19.In fiber optics, the signal source is ------- waves.
a. light b. radio c. infrared d. very low frequency.
20.Which of the following is not a guided medium?
a. twisted pair b. coaxial cable c. fiber optic cable d. atmosphere
21.X.25 protocol uses ---- for end to end transmission.
a. message switching b. circuit switching. C. the datagram approach to packet switching
d. the virtual circuit approach.
22. The X.25 protocol operates in the -----of the OSI model.
a. physical layer b. data link layer c.net work layer d. all the above.
23.The physical layer protocol directly specified for the X.25 protocol is------
a. RS- 232 b. X.21 c. DB-15 d. DB- 37
24. The PLP packet is a product of the ------layer in the X.25 standard.
a. physical b. frame c. packet d. transport
25. The PLP ------1s used to transport data from upper layers in the X.25 standard
a. S-packet b. data packet c. C-packet d. P-packet
26. X.25 protocol requires error checking at the------layer.
a. Physical b. frame c. packet d. b and c
27. X.25 is ------- protocol.
a. a UNI b. an SNI c. AN NNI d. an SSN
28. ISDN is an acronym for ---------------------
a. Information services for digital network b. Internet work system for data networks
c. Integrated signals digital network d. Integrated services digital network
29.The ------channel is used for telemetry and alarms.
a. B b. C c. D d. H
30. ----------- is a group of non—ISDN equipment.

a.TE1 b. TE2 c. Tex d.T3
Expected Questions
1. Define Computer Network? Give the difference between a network and distributed system?
2. Discuss the applications and goals of the computer networks
3. Explain briefly the functions of different layers of the OSI reference model
4. Give the difference between ISO OSI and TCP /IP model.
5. Discuss the difference between connection –oriented and connections-less services.
6. Give the advantage and disadvantage of frame relay over a leased telephone line.
7. Why does ATM used small, fixed length cells? Explain ATM layers.
8. Explain ISDN design? What are the services that can be provided by the ISDN ?What are
the different ISDN phases?
9. What are the advantages of using layered architecture?
10. Briefly explain about the Novel NetWare and ARPANET
11. Explain X.21 digital interface?
12. Explain the following terms
a) HOST b) IMP c) Subnet d) Protocol e) Interface f) PEER Processor
13. Distinguish between guided and unguided transmission media.
14. Briefly explain the different types of transmission medias?
15. Give the advantages and disadvantages of using fiber optic cable over metallic cable.
* * * * * * * *

UNIT –II
DATALINK LAYER
Introduction
The Data Link Layer break the bit stream into discrete frames and compute the checksum
for each frame. When a Frame arrives at the destination, the checksum is recomputed. If
the newly computed checksum is different from one computed contained in the frame, the
data link layer knows that an error has occurred and takes steps to deal with it.
FRAMING METHODS
1. CHARATER COUNT METHOD
2. STARTING AND ENDING CHARACTERS, WITH CHARATER STUFFING
3. STARTING AND ENDING FLAGS, WITH BIT STUFFING
CHARATER COUNT METHOD:
In this method a field in the header will be used to specify the number of
CHARACTERS in the frame. When data link layer at the destination sees the character
count, it knows how many characters follow and hence where the end of the frame is.
The trouble with this algorithm is that the count can be garbed by a transmission error
resulting the destination will get out of synchronization and will be unable to locate the
start of the next frame. There is no way of telling where the next frame starts. For this
reason this method is rarely used.

(a) 5 1 2 3 4 5 6 7 8 9 8 0 1 2 3 4 5 6 8 7 8 9 0 1 2 3
Frame 1 Frame 2
5 characters 5 characters 8 characters 8 characters
5 1 2 3 4 7 6 7 8 9 8 0
1 2 3 4 5 6 8 7 8 9 0 1 2 3 5
Frame 1 Frame 2
(Wrong) Now a character count
A Character Stream (a) Without errors (b) With one error
CHARATER STUFFING METHOD:
In this method each frame will start with a FLAG and ends with a FLAG.
The starting flag is DLE STX ---- Data Link Escape Start of Text
The ending flag is DLE ETX ---- Data link Escape End of Text.
Character count
Error
Frame 3 Frame 4

Ex 1. The given Data ABRFCXDGJHKK12435ASBGXRR
The Data will be sent
DLE STX ABRFCXDGJHKK12435ASBGXRR DLE STX
Ex 2. The given Data ASHGTRDXZBNHG DLE STX %$#54378
The data will be sent as
DLE STX ASHGTRDXZBNHG DLE DLE STX %$#54378 DLE ETX
Dis Adv:
1.24 bits are unnecessarily stuffed.
2. Transmission delay.
BIT STUFFING METHOD
In this method every frame will start with a flag 01111110.
In the data if there are FIVE consecutive ONE ‘s are there then a ZERO will be
stuffed.
Ex. The given data is 01111000011111110101001111110 01111101100
The data will be sent as
01111110 0111100001111101101010011111010 011111001100
Advantages:
1. Only one bit is stuffed.
2. No transmission delay
Stuffed bits

ERROR – CORRECTING AND DETECTING CODES
Network designers have developed two basics strategies for dealing with errors. One way
is to include enough redundant information along with each block of data sent, to enable
the receiver to deduce what the transmitted data must have been .The other way is to
include only enough redundancy to allow the receiver to deduce that an error occurred,
but not which error, and have it request a retransmission. The former strategy uses Error
– correcting codes and the latter uses Error- detecting codes.
The Error – correcting and Error- detecting methods are
1. PARITY METHOD
2. LRC METHOD (Longitudinal redundancy check)
3. CRC METHOD (Cyclic redundancy check)
4. HAMMING CODE METHOD
PARITY METHOD
- appends a parity bit to the end of each word in the frame
- Even parity is used for asynchronous Transmission
- Odd parity is used for synchronous Transmission
Ex 1. Character code even parity odd parity
1100100 1100100 1 1100100 0
2. 0011000 0011000 0 0011000 1
If one bit or any odd no bits is erroneously inverted during Transmission, the Receiver
will detect an error. How ever if two or even no of bits are inverted an undetected error
occurs.

Ex 3. The Transmitted data is 10011010. The received data is 11011010.
Let both the transmitter and receiver are agreed on EVEN parity.
Now an error will be detected, since the no of ones received are ODD
4. The Transmitted data is 10011010. The received data is 01011010
The received data is wrong even though the no of ones are EVEN.
Science two bits are inverted error can’t be detected.
Longitudinal Redundancy Check(LRC)
The frame is viewed as a block of characters arranged in 2-dimensions. To each
character is appended a parity bit. In addition a parity bit is generated for each bit position
across all characters i.e., an additional character is generated in which the Ith
bit of the
character is parity bit for the Ith
bit of all other characters in the block. This can be
expressed mathematically using exclusive OR(+) operation. The parity bit at the end of
each character of row parity
Rj =b1j + b2j +------bnj
Where Rj=Parity bit of jth character
bij=ith bit in jth character
This equation generates even parity.
Ci=bi1+bi2+-----+bin
Where Ci=ith bit of parity check character
m=number of characters in a frame
In this format the parity bits at the end of each character are referred to as

The Vertical Redundancy Check (VRC) and the Parity check character is referred to as
the Longitudinal Redundancy Check (LRC).
1 0 1 1 0 1 1 1
1 1 0 1 0 1 1 1
0 0 1 1 1 0 1 0
1 1 1 1 0 0 0 0
1
0 0 0 1 0 1 1
0 1 0 1 1 1 1 1
CRC Method
1. The frame is expressed in the form of a Polynomial F(x).0 1 1 1 1 1 1 0
2. Both the sender and receiver will agree upon a generator polynomial G(x) in
advance.
3. Let ‘r’ be the degree of G(x).Append ‘r’ zero bits to the lower – order end of
frame now it contains m+r bits.
4. Divide the bit string by G(x) using Mod 2 operation.
5. Transmitted frame [T(x)] = frame + remainder
6. Divide T(x) by G(x) at the receiver end. If the result is a zero, then the frame is
transmitted correctly. Ex. Frame: 1101011011
Generator: 10011
Message after appending 4 zero bits: 11010110000
b21
bn1 R1
b22
bnm
c1
cn
bn2 R2
Rm
cn+1
bit
1
bit
2
bit
n
Parity
bit
Character 1
Character 2
Character m
Parity check
character
LRC
VRC
b11
b1m
b12
c2
b2m bnm

1100001010
10011 11010110 110 0 0 0
10011
10011
10011
00001
00000
00010
00000
00101
00000
01011
00000
10110
10011
01010
00000
10100
10011
01110 Remainder
00000
1110
Transmitted frame: 11010110111110

1100001010
10011 11010110 111 1 1 0
10011
10011
10011
00001
00000
00010
00000
00101
00000
01011
00000
10111
10011
01001
00000
10011
10011
00000 Remainder
00000
0000
Since the remainder is zero there is no error in the transmitted frame.

HAMMING CODES
Hamming codes provide another method for error correction. Error bits, called Hamming
bits, are inserted into message bits at random locations. It is believed that the
randomness of their locations reduces the odds that these Hamming bits themselves
would be in error. This is based on a mathematical assumption that because there are so
many more message bits compared with Hamming bits, there is a greater chance for a
message bit to be in error than for a Hamming bit to be wrong. Determining the
placement and binary value of the Hamming bits can be implemented using hardware,
but it is often more practical to implement them using software. The number of bits in a
message (M) are counted and used to solve the following equation to determine the
number of Hamming bits (H) to be used:
2H
≥ M + H + 1
Once the number of Hamming bits is determined, the actual placement of the bits into the
message is performed. It is important to note that despite the random nature of the
Hamming bit placements, the exact sample placements must be known and used by both
the transmitter and receiver. Once the Hamming bits are inserted into their positions, the
numerical values of the bit positions of the logic 1 bits in the original message are listed.
The equivalent binary numbers of these values are added in the same manner as used in
previous error methods by discarding all carry results. The sum produced is used as the
states of the Hamming bits in the message. The numerical difference between the
Hamming values transmitted and that produced at the receiver indicates the bit position
that contains a bad bit, which is then inverted to correct it.
Ex. The given data
10010001100101(14- bits)
The number of hamming codes
2H
≥ M + H + 1
H = ? M = 14 to satisfy this equation H should be 5 i.e. 5 hamming code
bits should be incorporated in the data bits.
1 0 0 1 0 0 0 1 1 0 H 0 H 1 H 0 H 1 H
Now count the positions where binary 1’s are present. Add using mod 2 operation (Ex-OR). The
result will give the Hamming code at the transmitter end.
1’s position Binary equivalent

2 - 0 0 0 1 0
6 - 0 0 1 1 0
11 - 0 1 0 1 1
12 - 0 1 1 0 0
16 - 1 0 0 0 0
19 - 1 0 0 1 1
Hamming code = 0 0 0 0 0
This Hamming code will be incorporated at the places of ‘H’ in the data bits and the data
will be transmitted.
How to find out there is an error in the data?
Let the receiver received the 12th
bit as zero. The receiver also finds out the Hamming
code in the same way as transmitter.
1’s position Binary equivalent
2 - 0 0 0 1 0
6 - 0 0 1 1 0
11 - 0 1 0 1 1
16 - 1 0 0 0 0
19 - 1 0 0 1 1
0 1 1 0 0
Hamming code at the Tx 0 0 0 0 0
Hamming code at the Rx 0 1 1 0 0
0 1 1 0 0
The decimal equivalent for the binary is 12 so error is occurred at 12th
place.
Hamming
code at the
receiver

Data Link Protocols
1. Unrestricted Simplex Protocol:
In this the following assumptions are made
a. Data transmission is simplex i.e. transmitted in one direction only.
b. Both transmitting and receiving network layers are ready.
c. Processing time is ignored.
d. Infinite buffer space is available.
e. An error free channel.
This is an unrealistic protocol, which has a nickname “Utopia”.
2. A simplex stop and wait protocol:
The following assumptions are made
a. Error free channel.
b. Data transmission simplex.
A B
1. Sends a frame 1. Receives the frame
Receives ack and and sends Ack
2. Sends next frame and 2. Receives frame sends
so on Ack
Since the transmitter waits for Δt time for an Ack this protocol is called stop and wait
protocol.

3. A simplex protocol for a noisy channel
A B
1. Sends a frame 1. Receives a frame
2. Receives Ack and sends Ack
and sends next frame damaged while no frame and no
Transmitting
no Ack waits Ack
Δt time sends duplicate Receives the next frame
When this protocol fails?
A B
1. Sends a frame 1. Received by B
2. Ack is not received Ack damaged while 2.And sends Ack
Transmitting
wait some time and sends Received duplicate duplicate
(same frame)
At this situation protocol fails because the receiver receives a duplicate frame and there is
no way to find out whether the receiver frame is original or duplicate. So the protocol fails
at this situation.
Now what is needed is some way for the Rx to distinguish a frame and a duplicate. To
achieve this, the sender has to put a sequence number in the header of each frame it
sends. The Rx can check the sequence number of each arriving frame to see if it is a new
frame or a duplicate.

Here a question arises: What is the minimum number of bits needed for the sequence
number? The ambiguity is between a frame and its successor. A
1-bit sequence number (0 or 1) is therefore sufficient. At each instant of time, the receiver
excepts a particular sequence number next. Any arriving frame containing wrong
sequence number is rejected as a duplicate. When a frame containing the correct
sequence number arrives, it is accepted, passed to the network layer and then expected
sequence number is incremented i.e. 0 becomes 1 and one becomes 0. Protocols in
which a sender waits for a positive ack before advancing to the next data item are often
called PAR (positive ack with retransmission) or ARQ (automatic repeat request).
When this protocol fails?
A B
1. A Sends a frame F0 1. B receives it and
sends Ack AF0
2. Receives AF0 send 2. B receives F1 and sends
next frame F1 Ack AF1 (while transmitting
3. A’s time out and sends Ack delayed and reached late).
duplicate frame F1
4. In between received the 3. Receives F1 and
delayed Ack AF1 and sends sends Ack AF1
frame F0 (while transmitting
F0 damaged.)
5. Now the duplicate frame 4.B does not get frame and no
Ack AF1 received Ack.
6. Now A thinks that the Ack received is the ack of new frame F0 and A sends next
frame F1. So a frame F0 is missed. At this situation this protocol fails.

PIGGY BACKING
In most practical situations there is a need of transmitting data in both directions. This can
be achieved by full duplex transmission. If this is done we have two separate physical
circuits each with a ‘forward ‘ and ‘reverse’ channel. In both cases, the reverse channel is
almost wasted. To overcome this problem a technique called piggy backing is used.
The technique of temporarily delaying outgoing acknowledgements so that they can be
hooked onto the next outgoing data frame is known as piggy backing.
However, piggybacking introduces a complication not present with separate
acknowledgements. How long should the data link layer wait longer than the sender’s
timeout period, the frame will be retransmitted, defeating the whole purpose of having
acknowledgements. Of course, the data link layer cannot foretell the future, so it must
resort to some ad hoc scheme, such as waiting a fixed number of milli seconds. If a new
packet arrives quickly, the acknowledgement is piggy backed onto it; otherwise, if no new
packet has arrived by the end of this time period, the data link layer just sends a separate
acknowledgement frame.
SLIDING WINDOW PROTOCOLS
In all sliding window protocols, each outbound frame contains a sequence number,
ranging from 0 up to some maximum. The maximum is usually 2n
–1 so the sequence
number fits nicely in an n-bit field. The stop-and-wait sliding window protocol uses n=1,
restricting the sequence numbers to 0 and 1, but more sophisticated versions can use
arbitrary n.
The essence of all sliding window protocols is that at any instant of time, the sender
maintains a set of sequence numbers corresponding to frames it is permitted to send.
These frames are said to fall with in the sending window. Similarly the receiver also
maintains a receiving window corresponding to the set of frames it is permitted to accept.
The sender’s window and the receiver’s window need not have the same lower and upper
limits, or even have the same size. In some protocols they are fixed in size, but in others
they can grow or shrink as frames are sent and received.
The sequence numbers with in the sender’s window represent frames sent but as yet not
acknowledged. Whenever a new packet arrives from the network layer, it is given the next

highest sequence number, and the upper edge of the window is advanced by one. When
an acknowledgement comes in, the lower edge is advanced by one. In this way the
continuously maintains a list of unacknowledged frames.
Sender
Receiver
(a) (b) (c) (d)
(a) Initially (b) After the first frame has been sent c) After the first frame has been
received. d) After the first acknowledgement has been received.
PIPELINING
1. Upto now we made the assumption that the transmission time required for a frame to
arrive at the receiver plus the transmission time for the ack to come back is negligible.
2. Sometimes this is not true, when there is a long round trip propagation time is there.
3. In these cases round trip propagation time can have important implications for the
efficiency of the bandwidth utilization.
0
7
6
5
4 3
2
1
0
7
6
5
4 3
2
1
0
7
6
5
4 3
2
1
0
7
6
5
4 3
2
1
0
7
6
5
4 3
2
1
0
7
6
5
4 3
2
1
0
7
6
5
4 3
2
1
0
7
6
5
4 3
2
1

Consider the below example.
Let the channel capacity b = 50Kbps.
round trip propagation delay = 500ms
Frame size = 1000bits
Without considering the round trip propagation delay
For one frame the time taken will be = 1000/500 ms
= 20 ms
Considering the round trip propagation delay
250 ms + 20 ms
250 ms
For one frame the time taken will be = 500 ms +20 ms
= 270 ms
The channel utilization = (20/520)*100 = 4%
i.e. We are wasting 96% of channel time. To overcome this problem we will go for a
technique called PIPELIING.
In this technique, the sender is allowed to transmit upto ‘w ‘ frames before blocking,
instead of just 1.With an appropriate choice of w the sender will be able to continuously
A B

transmit frames for a time equal to the round trip transmit time without filling up the
window.
In the above example w would be at least 26 frames. (520/20 = 26 frames)
By the time it has finished sending 26 frames, at t=520 ms, the ack for frame 0 will have
just arrived. Thereafter ack will arrive every 20 ms, so the sender always gets permission
to continue just when it needs it.
Hence, we can say the sender window size is 26.
Derivation:
Let the channel capacity = b Bps
Let the frame size = l bits
Let the round trip delay = R secs
To send one frame the time will be l/b secs
Due to round trip delay the time taken will be (l/b + R) Sec = l+Rb/b Sec
The channel utilization is l/b (l/b + R) Sec = (l / l + Rb) Sec
If l > bR the efficiency will be greater than 50%.
If l < bR the efficiency will be less than 50%.
If l = bR the efficiency will be 50%.
Ex 1. A channel has a bit rate of 4 kbps and a propagation delay of 20msec.For what
rage of frame sizes does stop and wait give an efficiency of at least 50 % ?

2.consider an error free 64 kbps channel used to send 512 –byte data frames in one
direction, with very short acknowledgements coming back the other way. What is the
maximum throughput for window sizes of 1,7,15,and 127?
Pipelining frames over an unreliable channel raises some serious issues.
First, what happens if a frame in the middle of a long stream is damaged or lost?
When a damaged frame arrives at the receiver, it obviously should discarded, but what
should the receiver do with all the correct frames following it?
There are two basic approaches to dealing with errors .
1. Go Back ‘n’
2. Selective repeat or Selective Reject
One way called in go back n, the receiver simply to discard all subsequent frames,
sending no acknowledgements for the discard frames. In the other words, the data link
layer refuses to accept any frame except the next one it must give to the network layer.
Selective Repeat:
The receiving data link layer store all the correct frames following the bad frame, not all its
successors. If the second try succeeds the receiving data link layer will now have many
correct frames in sequence, so they can all be handed off to the network layer quickly and
the highest number acknowledged. This strategy corresponds to a receiver window larger
than 1.
(a) Go-back-N
0 1 7
3 4 5 6
2 0
0 1 6
5
4
3
2
2 3 4 5
E D D D
Error Discarded frames

MEDIUM ACCESS CONTROL SUBLAYER (MAC)
Networks can be categories in to two ways
a) Point to point b) Broad cast channel
- In broadcast network, the key issue is how to share the channel among
several users.
- Ex a conference call with five people
-Broadcast channels are also called as multi-access channels or random access
channels.
-Multi-access channel belong to a sublayer at the DL layer called the MAC sublayer.
The Channel Allocation problem:
a) Static channel allocation in LANs & MANs
i) FDM ii) TDM
Drawbacks: -1) Channel is wasted if one or more stations do not send data.
2) If users increases this will not support.
b) Dynamic channel allocation
i) Pure ALOHA & Slotted ALOHA
CSMA/CD
ii) CSMA
CSMA/CA
0 1 3 4 5
2
0 1 6
5
4
3
2 6 7
E
Error Buffered by
receiver
2
Frames 2-5
released
b) Selective reject
0

Pure ALOHA
-1970’s Norman Abramson end his colleagues devised this method, used ground –based
radio broad costing. This is called the ALOHA system.
-The basic idea, many users are competing for the use of a single shared channel.
-There are two versions of ALOHA: Pure and Slotted.
-Pure ALOHA does not require global time synchronization, where as in slotted ALOHA
the time is divided into discrete slots into which all frames must fit.
-Let users transmit whenever they have data to be sent.
-There will be collisions and all collided frames will be damaged.
-Senders will know through feedback property whether the frame is destroyed or not by
listening channel.
[-With a LAN it is immediate, with a satellite, it will take 270m sec.]
-If the frame was destroyed, the sender waits random amount of time and again sends
the frame.
-The waiting time must be random otherwise the same frame will collide over and over.
USER
A
B
C
D
TIME

Frames are transmitted at completely arbitrary times
-Whenever two frames try to occupy the channel at the same time, there will be a collision
and both will be destroyed.
-We have to find out what is the efficiency of an ALOHA channel?
-Let us consider an infinite collection of interactive users sitting at their systems (stations).
-A user will always in two states typing or waiting.
-Let the ‘Frame time’ denotes the time required to transmit one fixed length frame.
-Assume that infinite populations of users are generating new frames according to
possion distribution with mean N frames per frame time.
-If N>1 users are generating frames at a higher rate than the channel can handle.
-For reasonable throughput 0<N<1.
-In addition to new frames, the station also generates retransmission of frames.
-Old and new frames are G per frame time.
-G> N
-At low load there will be few collisions, so G ~ N
-Under all loads, the throughput S = GPo, where Po is the probability that a frame does not
suffer a collision.
-A frame will not suffer a collision if no other frames are sent with one frame time of its
start.
-Let ‘t’ be the time required to send a frame.
-If any other user has generated a frame between time to and to+t, the end of that frame
will collide with the beginning of the shaded frame.
-Similarly, any other frame started b/w to+t and to+2t will bump into the end of the shaded
frame.
-The probability that ‘k’ frames are generated during a given frame time is given by the
possion distribution:
Pr[k] = Gk
e-G
k!
-The probability of zero frames is just e-G
-In an interval two frame times long, the mean number at frames generated is 2G.
-The probability at no other traffic being initiated during the entire vulnerable period is
given by
Po = e-2G
S= Ge-2G
[S=GPo]
The Maximum through put occurs at G=0.5 with S=1/2e = 0.184
The channel utilization at pure ALOHA =18%.

Vulnerable period for the shaded frame
Throughput versus offered traffic for ALOHA systems
Slotted ALOHA
-In 1972, Roberts’ devised a method for doubling the capacity of ALOHA system.
-In this system the time is divided into discrete intervals, each interval corresponding to
one frame.
Collides
with the
start of the
shaded
frame
Collides
with the
end of the
shaded
frame
t
to+t to+2t
to+3t Time
to
Vulnerable
S
(throughput
per
frame
time)
Slotted ALOHA : S = Ge-G
Pure ALOHA : S = Ge-G
0.5 1.0
0.184
0.368
G (attempts per packet time)

-One way to achieve synchronization would be to have one special station emit a pip at
the start of each interval, like a clock.
-In Roberts’ method, which has come to be known as slotted ALOHA, in contrast to
Abramson’s pure ALOHA; a computer is not permitted to send whenever a carriage return
is typed.
-Instead, it is required to wait for the beginning of the next slot.
-Thus the continuous pure ALOHA is turned into a discrete one.
-Since the vulnerable period is now halved, the of no other traffic during the same slot as
our test frame is e-G
which leads to
S = Ge –G
- At G=1, slotted ALOHA will have maximum throughput.
- So S=1/e or about 0.368, twice that of pure ALOHA.
- The channel utilization is 37% in slotted ALOHA.
Carrier Sense Multiple Access Protocols
Protocols in which stations listen for a carrier (transmission) and act accordingly are
called carries sense protocols.
Persistent CSMA
When a station has data to send, it first listens to the channel to see if any one else is
transmitting at that moment. If the channel is busy, the station waits until it become idle.
When the station detects an idle channel, it transmits a frame. If a collision occurs, the
station waits a random amount of time and starts all over again. The protocol is called 1-
persistent also because the station transmits with a probability of 1 when it finds the
channel idle.
The propagation delay has an important effect on the performance of the protocol. The
longer the propagation delay the worse the performance of the protocol.
Even if the propagation delay is zero, there will be collisions. If two stations listen the
channel, that is idle at the same, both will send frame and there will be collision.

Non persistent CSMA
In this, before sending, a station sense the channel. If no one else is sending, the station
begins doing so it self. However, if the channel is busy, the station does not continually
sense it but it waits a random amount of time and repeats the process.
This algorithms leads to better channel utilization but longer delays then
1-persistent CSMA.
With persistent CSMA, what happens if two stations become active when a third station is
busy? Both wait for the active station to finish, then simultaneously launch a packet,
resulting a collision. There are two ways to handle this problem.
a) P-persistent CSMA b) exponential backoff.
P-persistent CSMA
The first technique is for a waiting station not to launch a packet immediately when the
channel becomes idle, but first toss a coin, and send a packet only if the coin comes up
heads. If the coin comes up tails, the station waits for some time (one slot for slotted
CSMA), then repeats the process. The idea is that if two stations are both waiting for the
medium, this reduces the chance of a collision from 100% to 25%. A simple
generalization of the scheme is to use a biased coin, so that the probability of sending a
packet when the medium becomes idle is not 0.5, but p, where 0< p < 1. We call such a
scheme P-persistent CSMA. The original scheme, where p=1, is thus called 1-persitent
CSMA.
Exponential backoff
The key idea is that each station, after transmitting a packet, checks whether the packet
transmission was successful. Successful transmission is indicated either by an explicit
acknowledgement from the receiver or the absence of a signal from a collision detection
circuit. If the transmission is successful, the station is done. Otherwise, the station
retransmits the packet, simultaneously realizing that at least one other station is also
contending for the medium. To prevent its retransmission from colliding with the other
station’s retransmission, each station backs off (that is, idles) for a random time chosen
from the interval

[0,2*max-propagation_delay] before retransmitting its packet. If the retransmission also
fails, then the station backs off for a random time in the interval [0,4*
max_propagation_delay], and tries again. Each subsequent collision doubles the backoff
interval length, until the retransmission finally succeeds. On a successful transmission,
the backoff interval is reset to the initial value. We call this type of backoff exponential
backoff.
CSMA/CA
In many wireless LANS, unlike wired LANS, the station has no idea whether the packet
collided with another packet or not until it receives an acknowledgement from receiver. In
this situation, collisions have a greater effect on performance than with CSMA/CD, where
colliding packets can be quickly detected and aborted. Thus, it makes sense to try to
avoid collisions, if possible. CSMA/CA is basically p-persistence, with the twist that when
the medium becomes idle, a station must wait for a time called the interframe spacing or
IFS before contending for a slot. A station gets a higher priority if it is allocated smaller
inter frame spacing.
When a station wants to transmit data, it first checks if the medium is busy. If it is, it
continuously senses the medium, waiting for it to become idle. When the medium
becomes idle, the station first waits for an interframe spacing corresponding to its priority
level, then sets a contention timer to a time interval randomly selected in the range
[0,CW], where CW is a predefined contention window length. When this timer expires, it
transmits a packet and waits for the receiver to send an ack. If no ack is received, the
packet is assumed lost to collision, and the source tries again, choosing a contention
timer at random from an interval twice as long as the one before(binary exponential
backoff). If the station senses that another station has begun transmission while it was
waiting for the expiration of the contention timer, it does not reset its timer, but merely
freezer it, and restarts the countdown when the packet completes transmission. In this
way, stations that happen to choose a longer timer value get higher priority in the next
round of contention.
Collision-Free Protocols
A Bit-Map Protocol
In the basic bit-map method, each contention period consists of exactly N slots. If station
0 has a frame to send, it transmits a 1 bit during the zeroth slot. No other station is
allowed to transmit during this slot. Regardless of what station 0 does, station 1 gets the

opportunity to transmit a 1during slot 1, but only if it has a frame queued. In general,
station j may announce the fact that it has a frame to send by inserting a 1 bit into slot j.
after all N slots have passed by, each station has complete knowledge of which stations
with to transmit.
0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7
1 1 1 1 3 7 1 1 1 5
The basic bit-map protocol
Since everyone agrees on who goes next, there will never be any collisions. After the last
ready station has transmitted its frame, an event all stations can easily monitor, another N
bit contention period is begun. If a station becomes ready just after its bit slot has passed
by, it is out of luck and must remain silent until every station has had a chance and the bit
map has come around again. Protocols like this in which the desire to transmit is
broadcast before the actual transmission are called reservation protocols.
Binary Countdown
A problem with the basic bit-map protocol is that the overhead is 1 bit per station. A
station wanting to use the channel now broadcasts its address as a binary bit string,
starting with the high-order bit. All addresses are assumed to be the same length. The
bits in each address position from different stations are BOOLEAN ORed together. We
will call this protocol binary countdown. It is used in Datakit.
As soon as a station sees that a high-order bit position that is 0 in its address has been
overwritten with a 1, it gives up. For example, if station 0010,0100,1001, and 1010 are all
trying to get the channel, in the first bit time the stations transmit 0,0,1, and 1,
respectively. Stations 0010 and 0100 see the 1 and know that a higher-numbered station
is competing for the channel, so they give up for the current round. Stations 1001 and
1010 continue.
8 Contention slots
Frames
8 Contention slots

The next bit is 0, and both stations continue. The next bit is 1, so station 1001 gives up.
The winner is station 1010, because it has the highest address. After winning the bidding,
it may now transmit a frame, after which another bidding cycle starts.
The binary countdown protocol. A dash indicates silence
IEEE Standard 802 for LANS and MANS
The IEEE 802.3 is for a 1-persistent CSMA/CD LAN. Xerox built a 2.94 Mbps CSMA/CD
system to connect over 100 personal workstations on 1-Km cable. This system was
called Ethernet through which electromagnetic radiation was once thought to propagate.
Xerox DEC and Intel came with another standard for 100 Mbps Ethernet. This differs from
old one that it runs at speeds from 1 to 10 Mbps on various media. The second difference
between these two is in one header (802.3 length field is used for packet type in
Ethernet).
0 0 1 0
0 1 0 0
1 0 0 1
1 0 1 0
0 - - -
0 - - -
1 0 1 0
1 0 0 -
1 0 1 0
Result
Stations 0010
and 0100 see
this 1 and give
up
Station 1001
sees this 1
and gives up
0 1 2 3
Bit time

10Base5, 10Base2 10 Broad 36
10Base-T, 1Base5
100 Base-T
802.3 Cabling
Five types of cabling are commonly used, 10Base5 cabling called thick Ethernet, came
first. It resembles a yellow garden hose, with markings every 2.5 m to show where the
taps go. Connections to it are generally made using vampire taps, in which a pin is
carefully forced halfway into the coaxial cable’s core. The notation 10Base5 means that it
operates at 10 Mbps, uses baseband signaling, and can support segments of up to 500m.
Name Cable Max. segment Nodes/seg. Advantages
10Base5 Thick coax 500 m 100 Good for backbones
10Base2 Thin coax 200 m 30 Cheapest system
10Base-T Twisted pair 100 m 1024 Easy maintenance
10Base-F Fiber optics 2000 m 1024 Best between buildings
The second cable type was 10Base2 or thin Ethernet, which, in contrast to the garden-
hose-like thick Ethernet, bends easily. Connections to it are made using industry standard
BNC connectors to form T-junctions, rather than using vampire taps. These are easier to
use and more reliable. Thin Ethernet is much cheaper and easier to install, but it can run
for only 200m and can handle only 30 machines per cable segment.
Cable breaks, bad taps, or loose connectors can be detected by a devise called time
domain reflectometry.
For 10Base5, a transceiver is clamped securely around the cable so that its tap makes
contact with the inner core. The transceiver contains the electronics that handle carrier
detection and collision detection. When a collision is detected, the transceiver also puts a
802.3
Base band
Digital
(Manchester)
Broad band
Analog
( PSK)

special invalid signal on the cable to ensure that all other transceivers also realize that a
collision has occurred.
The transceiver cable terminates on an interface board inside the computer. The interface
board contains a controller chip that transmits frames to, and receives frames from, the
transceiver. The controller is responsible for assembling the data into the proper frame
format, as well as computing checksums on outgoing frames and verifying them on
incoming frames.
With 10Base2, the connection to the cable is just a passive BNC T-junction connector.
The transceiver electronics are on the controller board, and each station always has its
own transceiver.
With 10Base-T, there is no cable at all, just the hub (a box full of electronics). Adding or
removing a station is simple in this configuration, and cable breaks can be detected
easily. The disadvantage of 10Base-T is that the maximum cable run from the hub is only
100m, may be 150m if high-quality (category 5) twisted pairs are used. 10Base-Tis
becoming steadily more popular due to the ease of maintenance. 10Base-F, which uses
fiber optics. This alternative is expensive due to the cost of the connectors and
terminators, but it has excellent noise immunity and is the method of choice when running
between buildings or widely separated hubs.
Each version of 802.3 has a maximum cable length per segment. To allow larger
networks, multiple cables can be connected by repeaters. A repeater is a physical layer
device. It receives, amplifies, and retransmits signals in both directions. As far as the
software is concerned, a series of cable segments connected by repeaters is no different
than a single cable (except for some delay introduced by the repeater). A system may
contain multiple cable segments and multiple repeaters, but no two transceivers may be
more than 2.5km apart and no path between any two transceivers any traverse more than
four repeaters.

802.3 uses Manchester Encoding and differential Manchester Encoding
Bit stream 1 0 0 0 0 1 0 1 1 1 1
Binary encoding
Manchester encoding
Differential Manchester encoding
Transition here Lack of transition here
indicates a 0 indicates a 1
C
Trap
D
A B
B
Backbone
A B C D
Repeater

7 1 2 or 6 2 or 6 2 0-1500 0-46 4
Preamble
Destination
address
Source
address Data Pad Checksum
The 802.3 MAC sub layer protocol:
I) Preamble:
Each frame start with a preamble of 7 bytes each containing a bit pattern 10101010.
II) Start of frame byte:
It denotes the start of the frame itself. It contains 10101011.
III) Destination address:
This gives the destination address. The higher order bit is zero for ordinary address and
1for group address (Multi casting). All bits are 1s in the destination field frame will be
delivered to all stations (Broad casting).
The 46th
bit (adjacent to the high-order bit) is used to distinguish local from global
addresses.
IV) Length field:
This tells how many bytes are present in the data field from 0 to 1500.
V) Data field:
This contains the actual data that the frame contains.
VI) Pad:
Valid frame must have 64 bytes long from destination to checksum. If the frame size less
than 64 bytes pad field is used to fill out the frame to the minimum size.
VII) Checksum:
It is used to find out the receiver frame is correct or not. CRC will be used here.
Bytes
Start of frame
delimiter
Length of
data field

Other Ethernet Networks
Switched Fast Gigabit
Switched Ethernet:
- 10 Base-T Ethernet is a shared media network.
- The entire media is involved in each transmission.
- The HUB used in this network is a passive device. (not intelligent).
- In switched Ethernet the HUB is replaced with switch. Which is a active device
(intelligent )
Fast Ethernet
Gigabit Ethernet
100 Base_x
100 Base_Fx
100 Base_Tx 100 Base_T4
2 pairs of
UTP or STP
4 pairs of
UTP
2 optical
fibers
100 Base_Sx
(optical fiber
multi mode)
100 Base_Lx
(optical fiber
multi or single
mode)
1000 Base_Cx
STP
1000 Base_T
UTP

IEEE 802.4 (Token Bus)
802.3 frames do not have priorities, making them unsuited for real-time systems in which
important frames should not be held up waiting for unimportant frames. A simple system
with a known worst case is a ring in which the stations take turns sending frames. If there
are n stations and it takes T sec to send a frame, no frame will ever have to wait more
than nT sec to be sent.
This standard, 802.4, describes a LAN called a token bus. Physically, the token bus is a
linear or tree-shaped cable onto which the stations are attached. Logically, the stations
are organized into a ring, with each station knowing the address of the station to its “left”
and “right.” When the logical ring is initialized, the highest numbered station may send
the first frame. After it is done, it passes permission to its immediate neighbor by sending
the neighbor a special control frame called a token. The token propagates around the
logical ring, with only the token holder being permitted to transmit frames. Since only one
station at a time holds the token, collisions do not occur.
Since the cable is inherently a broadcast medium, each station receives each frame,
discarding those not addressed to it. When a station passes the token, it sends a token
frame specifically addressed to its logical neighbor in the ring, irrespective of where that
station is physically located on the cable. It is also worth noting that when stations are first
powered on, they will not be in the ring, so the MAC protocol has provisions for adding
stations to, and deleting stations from, the ring. For the physical layer, the token bus uses
the 75-ohm broadband coaxial cable used for cable television. Both single and dual-cable
systems are allowed, with or without head-ends.
1 1 2 or 6 2 or 6 0-8182 4 1
Bytes >
1
Logical ring
This station not
currently in the
logical ring
Direction of
token motion
Broadband
coaxial cable
17
14
20
13 11 7 19

Destination
address
Source
address Data Checksum
The frame control field is used to distinguish data frames from control frames. Fro data
frames, it carries the frame’s priority. It can also carry an indicator requiring the
destination station to acknowledge correct or incorrect receipt of the frame.
For control frames, the frame control field is used to specify the frame type.
The allowed types include token passing and various ring maintenance frames,
including the mechanism for letting new stations enter the ring, the mechanism for
allowing stations to leave the ring, and so on.
Connecting devices
Connecting devices and the OSI model
Start of delimiter End
delimiter
Preamble
Frame control
Connecting
devices
Internetworking
devices
Networking
devices
Repeaters Bridges Routers Gateways

Repeate
Bridge
Router
Gateway
Bridges
LANS can be connected by devices called bridges, which operate in the data link layer.
Bridges do not examine the network layer header and can thus copy IP, IPX, and OSI
packets equally well.
The various reasons why the bridges are used.
1) Many university and corporate departments have their own LANS, primarily to connect
their own personal computers, workstations, and servers. Since the goals of the various
departments differ, different departments choose different LANS, without regard to what
other departments are doing. Sooner or later, there is a need for interaction, so bridges
are needed.
2) The organization may be geographically spread over several buildings separated by
considerable distances. It may be cheaper to have separate LANS in each building and
connect them with bridges and infrared links than to run a single coaxial cable over the
entire site.
3) It may be necessary to split what is logically a single LAN into separate LANS to
accommodate the load. Putting all the workstations on a single LAN- the total bandwidth
needed is far too high. Instead multiple LANS connected by bridges are used.
4) In some situations, a single LAN would be adequate in terms of the load, but the
physical distance between the most distant machines is too great (e.g., more than 2.5km
for 802.3). Even if laying the cable is easy to do, the network would not work due to the
Application
Presentation
Session
Transport
Network
Data link
Physical
Application
Presentation
Session
Transport
Network
Data link
Physical

excessively long round-trip delay. Only solution is to partition the LAN and install bridges
between the segments.
5) There is the matter of reliability. On a single LAN, a defective node that keeps
outputting a continuous stream of garbage will cripple the LAN. Bridges can be inserted at
critical places, to prevent a single node which has gone berserk from bringing down the
entire system.
6) And last, bridges can contribute to the organization’s security. By inserting bridges at
various places and being careful not to forward sensitive traffic, it is possible to isolate
parts of the network so that its traffic cannot escape and fall into the wrong hands.
Types of Bridges
Simple Bridge
Simple bridges are the most primitive and least expensive type of bridge. A simple bridge
links two segments and contains a table that lists the addresses of all the stations
included in each of them. Before a simple bridge can be used, an operator must sit down
and enter the addresses of every station. Whenever a new station is added, the table
must be modified. If a station is removed, the newly invalid address must be deleted.
Installation and maintenance of simple bridges are time-consuming and potentially more
trouble than the cost savings are worth.
Transparent Bridge
A transparent, or learning, bridge builds its table of station addresses on its own as it
performs its bridge functions. When the transparent bridge is first installed, its table is
empty .As it encounters each packet, it looks at both the destination and the source
addresses. It checks the destination to decide where to send the packet. If it does not yet
recognize the destination address, it relays the packet to all of the stations on both
segments. It uses the source address to build its table. As it reads the source address, it
notes which side the packet came from and associates that address with the segment to
which it belongs. By continuing this process even after the table is complete, a
transparent bridge is also self-updating.
This bridge uses flooding and backward landing algorithms.
The routing procedure for an incoming frame depends on the LAN it arrives on (the
source LAN) and the LAN its destination is on (the destination LAN), as follows.
1) If destination and source LANS are the same, discard the frame.
2) If the destination and source LANS are different, forward the frame.

3) If the destination LAN is unknown, use flooding.
Two Parallel transparent bridges
Spanning Tree Algorithm
Bridges are normally installed redundantly, which means that two LANs may be
connected by more than one bridge. In this case, if the bridges are transparent bridges,
they may create a loop, which means a packet may be going round and round, from one
LAN to another and back again to the first LAN. To avoid this situation, bridges today use
what is called the spanning tree algorithm.
F1 F2
B1 B2
F
Bridge
Frame copied
by B2
Frame copied
by B1
Initial frame
LAN 2
LAN 1

Expected Questions
1 Briefly explain the functions of Data Link Layer.
2 Discuss different types of Framing Methods.
3 Discuss various Error detecting and correcting methods.
4 Explain CRC method with your own example.
5 Explain Hamming code method with your own example.
6 What is meant by Flow control? Discuss different flow control methods.
7 Discuss various stop and wait protocols.
8 Explain Piggy backing, pipelining techniques.
9 Explain sliding window protocol.
10 Explain Go back ‘n’ and selective repeat protocols.
11 Prove that the channel utilization is 18% in ALOHA and 37% slotted ALOHA.
12 Discuss the CSMA protocols.
13 Explain different STOP and WAIT protocols.
14 Explain persistence and non-persistence CSMA protocol.
15 Discuss HDLC in detail.
16 What is the difference information fields in an HDLC I – frame and an
U –frame.
17 Explain IEEE 802.3
20 Compare IEEE 802.3, IEEE802.4, IEEE802.5
21 Explain the operation of token bus
22 Explain the operation of token ring.
23 Prove that the channel utilization will be 37% in slotted ALOHA system.
24 Compare and Contrast pure ALOHA and Slotted ALOHA systems.
25 Write short note on CSMA/CD protocol.
26 Explain the collision free protocols.
27 Compare the ALOHA and CSMA protocols.
28 Explain Token bus and its frame format.
29 Explain token ring and its frame format.
30 Explain 802.3 frame format.

31 Explain different connecting devices.
32 Explain different types of Bridges.
33 Give different reasons why a bridge is required.
***********
Quiz Questions
1. What is the principle involved in the ALOHA?
2. What is the principle involved in Slotted ALOHA?
3. What is the principle involved in CSMA?
4. What is meant by Non Persistence CSMA?
5. What is meant by Persistence CSMA?
6. What is meant by P-Persistence CSMA?
7. The channel utilization in slotted ALOHA is ________________ and in pure
ALOHA _______________.
8. What is meant by exponential back-off algorithm?
9. What is meant by Vulnerable period w.r.t ALOHA?
10.What is meant by bit map protocol?
11.What is meant by binary count down protocol?
12.What is the function of Data Link Layer?
13.IEEE 802.3 is called ____________________.
14.IEEE 802.3 will use _________________ protocol.
15.10 Base 5 is called _________________.
16.What is meant by preamble in 802.3 frame format?
17.Why pad filed is used in 802.3 frame format?
18.Which purpose the time domain reflectometry is used?
19.In transparent bridge which algorithms are used to fill the routing table.
20.Why Spanning tree bridges are used?

21.Gateways are used in which layer?
22.IEEE 802.4 is called _______________.
23.IEEE 802.5 is called _______________.
24.How frames will be transmitted in IEEE 802.4?
25.Give 802.3 frame format?
26.Give 802.4 frame format?
27.What are different types of Ethernet available?
28.802.3 uses ____________ type of encoding.
29.To join two or more segments of Ethernet ______________will be used.
30.What is a token w.r.t 802.4.
31.How the cable breaks problem can be solved in 802.5?
32.Give the 802.5 Frame format ?
33.Give an example for character count framing method and explain the fields.
34.What is the draw back with the above method?
35.What are characters that are used in character stuffing method at the beginning
and ending of a frame?
36.Expand the above characters?
37.For the data given below : how it will be sent using character stuffing method?
abcdefghijka DLE 123456789RAMARAO DLE
38.What is the draw back with this method?
39.In bit stuffing method how a frame will start?
40.For the given data using bit stuffing method. How data will be sent?
011011111011111101111101111111011
41.Find out the LRC & VRC for the below characters?
0 0 1 1 0 1
1 1 0 1 0 0
0 1 1 0 1 0
0 0 1 0 1 1
42.What is the draw back with the LRC method?
43.How many zero bits will be append for the frame in CRC method?
44.What is meant by ‘piggy backing’ ?
45.What is meant by stop and wait protocol?
46.What is meant by pipelining?
47.What is the essence of sliding window protocol?
48.A channel has a propagation delay of 20m sec and has a frame size of 80bits. For
what range of band width does stop and wait give an efficiency of at least 50
percent?
49.How will you find out how many Hamming code bits should be incorporated?

UNIT – III
NETWORK LAYER
Functions of Net Work layer
1. Routing 2. Congestion Control
Routing algorithms
The main function of the network layer is routing packets from the source machine to the
destination machine. Routing algorithm can be grouped into two major classes. Nonadaptive and
Adaptive algorithms.
Non adaptive Adaptive
1) Routing decisions are not based on
measurements or estimates of the
current traffic and topology.
1) Routing decisions are based on
measurements of the current traffic and
topology.
2) The route is computed well in
advance.
2) The route is computed depends on
situation.
3) When the network is booted the
routers are downloaded.
3) The routers are not downloaded.
4) This is a static routing. 4) This is a dynamic routing.
Shortest Path Routing:
This is a static routing algorithm. The idea is to build a graph of the subnet, with each node of the
graph representing a router and each arc of the graph representing a communication line. To
choose a route between a given pair of routers, the algorithm just finds the shortest path between
them on the graph.

2
A
B
E
G H
D
C
F
2
7
2
6 4
2
2
3
1
B(2,A)
E(∞,-)
G(6,A) H(∞,-)
D(∞,-)
C(∞,-)
F(∞,-)
A
B(2,A)
E(4,B)
G(6,A) H(∞,-)
D(∞,-)
C(9,B)
F(∞,-)
A
B(2,A)
E(4,B)
G(5,E) H(∞,-)
D(∞,1)
C(9,B)
F(6,E)
A
B(2,A)
E(4,B)
G(5,E) H(9,G)
D(∞,-)
C(9,B)
F(6,E)
A
B(2,A)
E(4,B)
G(5,E) H(8,F)
D(∞,-)
C(9,B)
F(6,E)
A
(b)
(a)
(c)
(d)
(e) (f)
3

One-way of measuring path length is the number of hops. Using this metric, the paths
ABC and ABE are equally long. (Two hops).
Another metric is the Geographic distance in Kilometers. ABC is clearly longer than ABE.
Many other metrics are also possible besides hops and physical distance. Each are could
be labeled with the mean queuing and transmission delay for some standard test packets as
determined by hourly test runs. With this graph labeling, the shortest path is the fastest path,
rather than the path with the fewest arc or kilometers.
In most general case, the labels on the arcs could be computed as a function of the
distance, bandwidth, average traffic, communication cost, mean queue length, measured delay
and other factors.
The shortest path can be calculated using Dijkstra method. Each node is labeled with its
distance from the source along the best known path. Initially, no paths are known, so all nodes
are labeled with infinity. As the algorithm proceeds and paths are found, the labels may change,
reflecting better paths. Initially all labels are tentative. When it is discovered that a label
represents the shortest path from the source to that node, it is made permanent and never
changed thereafter.
In the above diagram, let the weights represents the distance. To find out the shortest path
from A to D. We start by marking A as permanent. The examine each one with the distance to A,
relabeling each one with the distance to A. Whenever a node is relabeling also label it with the
node from which the probe was made. After examing each of the nodes adjacent to A, examine
all the tentatively labeled nodes in the whole graph and make the one with the smallest label
permanent. This one becomes the new working node.
The same procedure is adopted to all the nodes and the shortest path is found.
Flooding:
This is a static algorithm. In this, every incoming packet is sent out on every outgoing line except
the one it arrived on. Flooding will generate vast numbers of duplicate packets, some measures
have to take to dump the duplicate packets. One such measure is to have a hop counter
contained in the header of each packet, which is decremented at each hop, with the packet being
discarded when the counter reaches zero. The hop counter should be initialized to the length of
the path from source to destination. If the sender does not know how long the path is it can
initialize the counter to full diameter of the subnet.
A variation of flooding is ‘Selective Flooding’. In this the routers do not send every
incoming packet on every line, instead only on those lines that are going approximately in the
right direction which leads to the destination.

Advantages
1) In military applications, where large numbers of routers are blown, flooding is desirable.
2) In Distributed database applications, it is some times necessary to update all the databases
concurrently, in which flooding is useful.
3) It is used as a metric against which other routing algorithms are compared.
4) Flooding chooses the shortest path, because it chooses all possible path in parallel.
Flow-based Routing:
The flooding and shortest path algorithm takes the topology in to account. Flow based
routing algorithm uses both topology and load for routing.
In some networks the mean data flow between each pair of nodes is relatively stable and
predictable. The average traffic is known in advance and to a reasonable approximation, constant
in time, it is possible to analyze the flows mathematically to optimize routing.
The basic idea behind this, for a given line, if the capacity and average flow are known, it
is possible to compute the mean packet delay on that line. From the mean delays on all the
lines, the mean packet delay for the whole subnet.
To use this technique, certain information must be known in advance. 1) About the subnet
topology 2) about the traffic 3) the line capacity 4) a routing algorithm
The fig.(b) gives the information packets /sec go from source i to destination j. Given this
information, it is easy to calculate the total in line for i i.e. i . Using the traffic from the source to
destination the mean number of packets/sec on each line. ci can be calculated, assuming a
mean packet size 1/ . The mean delay for each line can be derived where 1/ is the mean
packet size in bits,  is the mean flow in packets/sec.
With a capacity c = 25 packets/sec and an actual flow  = 14 packets/sec, the mean delay is 91
m sec. When  =0, the mean delay is 40m sec. With this example we can say the delay depends
on both queuing and service time.
The mean delay time for the entire subnet can be calculated as the sum of each of the eight lines,
with the weight being the fraction of the total traffic using that line.

Distance Vector Routing:
This is a dynamic routing algorithm. This algorithm operates by having each router
maintain a table (i.e. a vector) giving the best known distance to each destination and which line
to use. These tables are updated by exchanging information with the neighbors.
The routing table indexed by and containing one entry for each router in the subnet. This
entry contains two parts: The preferred outgoing line to use for the destination and an estimate of
time or distance to that destination. The metric used might be number of hops, time delay in
msec, total number of packets queued along the path or something similar.
The router is assumed to know the distance to each of its neighbors. If the metric is hops,
the distance is just one hop. If the metric is queue length, the router examines each queue. If the
metric is delay the router can measure it directly with a special ECHO packets.
Consider an example, in which the delay is used as metric and the router knows the delay
to each of its neighbors. Once every T msec each router send to each neighbor a list of its
estimated delays to each destination. It also receives a similar list from each neighbor. Let xi
being x’s estimate of how long it takes to get router ‘i’. If the router knows that the delay to x is ‘m’
m sec. To get router i via x is (xi +m) m sec. By performing this calculation for each neighbor, a
router can find out which estimate is the best and use that estimate and the corresponding line in
its new routing table.
A
B
C
D
E
F
A B C D E F
(a)
20
10
10
50
20
20
20
20
A
B C
D
E
F
3
BFD
8
BC
2
BFE
4
BF
9
AB
4
CBA
9
AB
1
ABFD
4
ABC
7
AE
4
AEF
1
DFBA
7
EA
4
FEA
4
FB
4
FEC
4
FD
4
FE
4
EF
4
DF
2
CEF
3
CE
3
CD
8
CB
3
DFB
2
EFB
3
DC
3
EC
3
ECD
3
DCE
Destination
Source
(b)

A I H K Line
C D
H
J K L
Subnet
JA JI JH JK
delay delay delay delay New
is is is is routing
8 10 12 6 table for J
Input from A,I,H, K and new routing table for J
Fig.(a) shows the subnet and fig.(b) shows the vectors of J for its neighbors. Fig.(c) shows the
new routing table for J. Let JA delay is 8, JI delay is 10, JH is 12, JK is 6.
The new route to G from J can be calculated as follows.
J can get A in 8 m sec.
A can get G in 18 m sec(from table)
 J can get G in (8+18) 26 m sec.
Similarly the delay to G via I,H and K is (31 +10) 41, (6+12)18, (31+6)37 m sec.
The best of these values is 18, so it makes an entry in its routing table that the delay to G is 18 m
sec and that route is via H.
0
12
25
40
14
23
18
17
21
9
24
29
24
36
18
27
7
20
31
20
0
11
22
33
20
31
19
8
30
19
6
0
14
7
22
9
21
28
36
24
22
40
31
19
22
10
0
9
8 A
20 A
28 I
20 H
17 I
30 I
18 H
12 H
10 I
0 -
6 K
15 K
A
B
C
D
E
F
G
H
I
J
K
L
E
G
F
A B
I
New estimated
delay from J
Vectors received from
J’s four neighbors
Router

2A 2B
2C 2D
3A
3B
4A
4B 4C
5B 5C
5E
1B
1A 1C
Hierarchical Routing:
As network grow in size, the router routing tables grow proportionally. Not only more
memory consumed by ever increasing tables, but more CPU time is needed to scan them more
bandwidth is needed to send status reports about them. At a certain point the network may grow
to the point where it is no longer feasible for every router to have an entry for every other router,
so the routing will have to be done hierarchically.
When hierarchical routing is used, the routers are divided into ‘Regions’. Each router
knows all the details about how to route packets to destinations within its own region, but doesn’t
know the internal structure of other regions. For huge networks, a two-level hierarchy may be
insufficient, it may be necessary to divide the regions into clusters, clusters into zones and zones
into groups and soon.
Consider a two level hierarchy with five regions as shown in fig. one router needs 17
entries for one table. The network contains 17 routers. So the total no. of entries will be 17 x 17.
This is for when we are not using hierarchy.
When routing is done hierarchically a router will consists of entries for all the local routers
and regions only.
For Ex. The router 1A consists of entries as shown in fig.(c). Hierarchical routing has reduced the
table from 17 to 7 entries.
Full table for 1A
Dest. Line Hops Dest. Line Hops
(a) (b) (c)
Disadvantages:
1A - -
1B 1B 1
1C 1C 1
2A 1B 2
2B 1B 3
2C 1B 3
3A 1C 4
3B 1C 3
3C 1C 2
4A 1C 3
4B 1C 4
4C 1C 4
5A 1C 4
5B 1C 5
5C 1B 5
5D 1C 6
5E 1C 5
1A - -
1B 1B 1
1C 1C 1
2 1B 2
3 1C 2
4 1C 3
5 1C 4
Region 1 Region 2
Region 3 Region 4 Region 5
Hierarchical table for 1A

Using hierarchy the path lengths will be increased. For Ex. The best path from 1A to 5C is via
region 2. But using hierarchy routing all traffic to region 5 goes via region 3, because it is the best
for most destination in region 5.
Ex. Consider a 720 routers subnet.
Without hierarchy each router required 720 entries. Total entries will be
720 x 720, with hierarchy, if the subnet is portioned into 24 regions and
30 routers/region, then each router needs 30+23 = 53 entries only.
If a 3-level hierarchy used, with 8 clusters, each contains 9 regions and
10 routers/region. Each router needs 10 + 8 + 7 = 25 entries
For a ‘N’ router subnet the optimal number of routers = lnN
The total no. of entries /router = elnN
Broadcast Routing:
In some applications, hosts need to send messages to many or all other hosts. For Ex.
Weather reports, stock market, updates etc. Sending a packet to all destinations simultaneously is
called ‘Broad Casting’.
1) One method is sending distinct packet to each destination by the source. This method wastes
the bandwidth and also requires the source to have a complete list of all destinations.
2) The second is using Flooding technique. This generates too many packets and consumes too
much bandwidth.
3) Another method is multi destination routing. In this each packet contains either a list of
destinations or a bit map indicating the desired destinations. When a packet arrives at a router,
the router checks all the destinations to determine the set of output lines that will be needed. The
router generates a new copy of the copy for each output line to be used and includes in each
packet only those destinations that are to use the line. This routing is like separately addressed
packets except that several packets must follow the same route.
4) Another algorithm which uses the spanning tree. A spanning tree is a subnet of the subnet that
includes all the routers but contains no loop. If each router knows which of its lines belong to the
spanning tree, it can copy an incoming broadcast packet onto all spanning tree lines except the
one it arrived on. This method makes use of bandwidth excellently and generates minimum no. of
packets necessary to do the job. The only disadvantage is that each router must have knowledge
of some spanning tree.
5) One more algorithm is an attempt to approximate the behavior of the previous one, even when
the routers do not know anything at all about spanning trees. The idea is remarkably simple once
it has been pointed out. When a broadcast packet arrives at a router, the router checks to see if
the packet arrived on that line that is normally used for sending packets to the source of the
broadcast. If so, there is an excellent chance that the broadcast packet itself followed the best

route from the router and is therefore the first copy to arrive at the router. This being the case, the
router forwards copies of it onto all lines except the one it arrived on. If, however, the broadcast
packet arrived on a line other than the preferred one for reaching the source the packet is
discarded as a likely duplicate.
B
A C
F D (a)
I
G
L N J
K
O
M
B
A C
E
F D
I
G
H L
J
K N
O
M
H
E

(b)
(c)
Multicast Routing :
For some applications, it is necessary for one process to send a message to all other members of
the group. If the group is small, it can just send each other member a point-to-point message. If
the group is large this strategy is expensive. Some times broad casting is used, but using broad
casting is used, but using broadcasting to inform 1000 machines on a million node network is
inefficient because most receivers are not interested in the message. Thus it is needed to send
message to well-defined groups. Sending message to such a group is called ‘multicasting’.
To do multicasting, group management is required. Some way is needed to create and
destroy groups and for processes to join and leave groups. When process joins a group, it informs
its host of this fact. It is important that routers know which of their hosts belong to which group.
Either hosts most inform their routers about change in group membership or routers must query
their hosts periodically. Routers tell their neighbors, so the information propagates through the
subnet.
To do multicast routing, each router computes a spanning tree covering all other routers in the
subnet. When a process sends a multicast packet, to a group, the first router examines its
spanning tree and prunes it, removing all lines that do not lead to hosts that are members in the
group. Multicast packets are forwarded. Only along the appropriate spanning tree.
A
E
I
F H J N
D
C
B
M
K
L
O
K
E O
G
B
L
H
G

Congestion Control Algorithms
What is Congestion?
When too many packets are present in the subnet performance degrades. This situation is called
Congestion.
The number of packets dumped into the subnet are within its carrying capacity, they are all
delivered.
However, if the traffic increases too far, the routers are unable to cope and begin losing packets.
At very high traffic, performance collapse completely and almost no packets are delivered.
What factors will lead to congestion?
1. Three or four input lines and only one output line queue will build up.
If there is insufficient memory to hold all of them, packets will lost.
Adding infinite memory congestion gets worse, because by the time packets get to the
front of the queue, the time out and duplicates have been sent.
2. Slow processors (routers) can cause congestion.
 A slow processor perform the book keeping tasks very slow, queues will build up.
3. Low band-width lines also cause congestion
 Upgrading lines but not changing the processor and vice-versa shifts the bottleneck.
This problem will persist until all components are in balance.
What is the difference between Congestion control and Flow control?
Congestion control is global issue and flow control is local issue.
Ex:
 Consider a network with a capacity of 1000Gbps on which a super computer is trying to
transfer a file to a personal computer at 1Gbps.Here a flow control is needed.
 Consider a network with 1Mbps lines and 1000 large computers, more than half are trying
to transfer files a 100kbps to the other half. The problem is here is the total offered traffic
exceeds than the network handle.

General Principles Of Congestion Control:
The congestion control can be done by two methods
Open loop
Closed loop
Open loop: These solutions attempt to solve the problem by good design, to make sure that it
does not occur in the first place. Once the system is up and running, midcourse corrections are
not made.
Tools for doing open-loop control include deciding when to accept new traffic, deciding
when to discard packets and which ones, and making scheduling decisions at various points in
the network.
In contrast, closed loop solutions are based on the concept of a feedback loop. This approach has
three parts when applied to congestion control:
1. Monitor the system to detect when and where congestion occurs.
2. Pass this information to places where action can be taken.
3. Adjust system operation to correct the problem.
Congestion Prevention Policy :
Traffic Shaping:
One of the main causes of congestion is that traffic is often bursty. If hosts could be made to
transmit at a uniform rate, congestion would be less common. Another open loop method to help
manage congestion is forcing the packets to be transmitted at a more predictable rate. This
approach to congestion management is widely used in ATM networks and is called traffic
shaping.
Monitoring a traffic flow is called traffic policing. Agreeing to a traffic shape and policing it
afterward are easier with virtual circuit subnet than with datagram subnets.
Leaky Bucket Algorithm
Imagine a bucket with a small hole in the bottom. No matter at what rate water enters the bucket,
the outflow is at a constant rate, , when there is any water in the bucket, and zero when the
bucket is empty. Also, once the bucket is full, any additional water entering it spills over the sides
and is lost.

Packet
Unregulated
Flow
Regulated
flow
Network
The same idea can be applied to packets, as shown in fig. Conceptually, each host is connected
to the network by an interface containing a leaky bucket, that is, a finite internal queue. If a packet
arrives at the queue when it is full, the packet is discarded. In other words, if one or more
processes within the host try to send a packet when the maximum numbers are already queued,
the new packet is unceremoniously discarded. This arrangement can be built into the hardware
interface or simulated by the host operating system.
The host is allowed to put one packet per clock tick onto the network. Again, this can be enforced
by the interface card or by the operating system. This mechanism turns an uneven flow of packets
from the user processes inside the host into an even flow of packets onto the network, smoothing
out bursts and greatly reducing the chances of congestion.
Implementing the original leaky bucket algorithm is easy. The leaky bucket consists of a finite
queue. When a packet arrives, if there is room on the queue it is appended to the queue;
otherwise, it is discarded. At every clock tick, one packet is transmitted (unless the queue is
empty).
Host
Computer
Interface
Containing
a leaky bucket
The bucket
holds
packets

The Token Bucket Algorithm:
The leaky bucket algorithm enforces a rigid output pattern at the average rate, no matter how
bursty the traffic is. For many applications, it is better to allow the output to speed up somewhat
when large bursts arrive, so a more flexible algorithm is needed, preferably one that never loses
data. One such algorithm is the token bucket algorithm.
Networks (a) Networks (b)
In this algorithm, the leaky bucket holds tokens, generated by a clock at the rate of one token ever
T sec. In figure (a), we see a bucket holding three tokens, with five packets waiting to be
transmitted. For a packet to be transmitted, it must capture and destroy one token. In figure(b), we
see that three of the five packets have gotten through, but the other two are stuck waiting for two
more tokens to be generated.
The leaky bucket algorithm does not allow idle hosts to save up permission to send large bursts
later. The token bucket algorithm does allow saving, up to the maximum size of the bucket, n.
Another difference between the two algorithms is that the token bucket algorithm throws away
tokens when the bucket fills up but never discards packets. In contrast, the leaky bucket algorithm
discards packets when the bucket fills up.
Host
Computer
One token
is added
to the bucket
every T
The bucket
holds
tokens
Host
Computer

Congestion Control in Datagram Subnets:
Each router can easily monitor the utilization of its output lines and other resources. It can
estimate each line about the recent utilization of that line (u). Periodically a sample at the
instantaneous line utilization (f) can be mad and u updated.
unew = a uold + (1-a)f
Where a is constant determines how fast the router forgets recent history.
Whenever u moves above the threshold, the output line enters a ‘warning’ state. Each new
arriving packet is checked if its output line is warning state. If it is some action is taken.
The Warning Bit:
When the output line reaches to warning state it is signaled by setting a special bit in the packet’s
header. When the packet arrived at its destination, the transport entity copied the bit into the next
acknowledgement sent back to source. The source then cot back on traffic. As long as the router
was in warning state, it continued to set warning bit. As long as the warning bits continued to flow
in, the source continued to decrease its transmission rate.
Choke packets:
In this algorithm, the router sends a choke packet back to the source host. The original packet is
tagged so that it will not generate any more choke packets farther along the path and is then
forwarded in the usual way.
When the source host gets the choke packet, it is required to reduce the traffic sent to the
specified destination by X percent. Since other packets aimed at the same destination are
probably already under way and will generate yet more choke packets, the host should ignore
choke packets referring to that destination for a fixed time interval. After that period has expired,
the host listens for more choke packets for another interval. If one arrives, the line is still
congested, so the host reduces the flow still more and begins ignoring choke packets again. If no
choke packets arrive during the listening period, the host may increase the flow again.
The first choke packet causes the data rate to be reduced to 0.50 of its previous rate, the next
one causes a reduction to 0.25, and so on. Increases are done in smaller increments to prevent
congestion from reoccurring quickly.
Hop by Hop choke packets:
For example, let the host A is sending packets to D. as shown in fig.(1). If D runs out of buffers, it
will take sometime for a choke packet to reach A to tell it to slow down. This is shown in fig
2,3,4.In this time another packets will be sent. Only after some more time the router D will be
noticing a slower flow (fig.7).
In other approach, as soon as choked packet reaches to F it cuts down the flow to D and D will
get immediate relief. (like a headache remedy in a TV). In the next set up, when choke reaches to
E it also cuts down the flow to F which in turn gives relief to F. Finally, when the choke packet
richer A and the flow genuinely slows down.

B C B C
A D A D
E F E F
(b)
Ch-choke
(a)
Flow is still at
maximum rate
Reduced
flow
Heavy
flow
Ch
Ch
Ch
Ch
Flow is reduced
(a) A choke packet that affects only the source (b) A choke packet that
affects each hop it passes through.

Load Shedding:
Load Shedding is a fancy way of saying that when routers are being inundated by packets that
they cannot handle, they just throw them away.
A router drowning in packets can just pick packets at random to drop, but usually it can do better
than that. Which packet to discard may depend on the applications running. For file transfer, an
old packet is worth more than a new one because dropping packet 6 and keeping packets 7
through 10 will cause a gap at the receiver that may force packets 6 through 10 to be
retransmitted (if the receiver routinely discards out-of –order packets). In a 12-packet file,
dropping 6 may require 7 through 12 to be retransmitted, whereas dropping 10 may require only
10 through 12 to be retransmitted. In contrast, for multimedia, a new packet is more important
than an old one. The former policy (old is better than new) is often called wine and the latter (new
is better than old) is often called milk.
A step above this in intelligence requires cooperation from the senders. For many applications,
some packets are more important than others. For example, certain algorithms for compressing
video periodically transmit an entire frame and then send subsequent frames as differences from
the last full frame. In this case, dropping a packet that is part of a difference is preferable to
dropping one that is part of a full frame. As another example, consider transmitting a document
containing ASCII text and pictures. Losing a line of pixels in some image is far less damaging
than losing a line of readable text.
Internetworking
When two or more networks are connected it is called Internet. There will be a variety of different
networks will always be around, for the following reasons.
1) Different networks will use different technologies like personal computers run TCP/IP,
mainframes run on IBM’s SNA.
2) As computers and networks get cheaper, the place where decisions get made moves
downwards in organizations.
3) As new hardware developments occur, new software will be created to fit the new hardware.
The purpose of interconnecting all these networks is to allow users on any of them to
communicate with users don all the other ones to allow users on any of them to access data on
any of them.
Networks differ in many ways. In the network layer the following differences can occur (fig.5.43).

How networks can be connected:
Networks can inter connected by different devices. In the physical layer networks are connected
by repeaters or hubs, which just move the bits from one network to an identical network.
At the Data Link layer bridges and switches are used. In network layer, routers are used to
connect two network layers, the router may be able to translate between the packet formats. A
router that can handle multiple protocols is called a ‘ multi protocol’ router.
In a switched network the entire frame is transported on the basis of its MAC address. With
router, the packet is extracted from the frame and the address in the packet is used for deciding
where to send it. Switches do not have to understand the network layer protocol used to switch
packets. Routers do.
Two ways of internetworking is possible.(i) a connection-oriented concatenated virtual subnets
and (ii) datagram internet. In the past most networks were connection oriented. Then with the
rapid acceptance of the Internet, datagrams became more popular. With growing importance of
multimedia networking, it is likely that connection-orientation is back in one form or another since
it is easier to guarantee quality of service with connections than without them.
In the concatenated virtual-circuit model a sequence of virtual circuit is set up from the source
through one or more gateways to the destination. Each gateway maintains tables telling which
virtual circuit pass through it, where they are to be routed, and what the new virtual –circuit
number.
In datagrams from one host to other host the packets will be routed in different routes through the
inter network. A routing decision is made separately for each packet, possibly depending on the
traffic at the moment the packet is sent. This strategy can use multiple routes and thus achieve a
higher bandwidth than the concatenated virtual circuit model.
Tunneling
The source and destination hosts are on the same type of network but there is a different network
in between ‘Tunneling’ will be used. An ex, think of an organization with TCP/IP based Ethernet at
one place and a TCP/IP base Ethernet in otherplace, and a PTT WAN in between as shown in fig.
Consider an example a person driving his car from one place to other under its own power. Let in
between he has to cross a river, which has no bridge. Hence his car has to be kept on a boat and
transported to other end. From there the car continues to move under its own power. Tunneling of
packets through a foreign network works the same way.
To send an IP packet to host 2, host 1, constructs the packet containing IP address of host 2,
inserts it into an Ethernet frame addressed to the multi protocol router, and puts it on the Ethernet.
When the multiprotocol router gets the frame, it removes the IP packet, inserts it in the payload
field of the WAN network layer packet, and addresses the latter to the WAN address of the other

multi protocol router to the other. Only the multiprotocol router has to understand IP and WAN
packets.
The IP Protocol
At the network layer, the Internet can be viewed as a collection of subnet-works or Autonomous
systems that are connected together. The network layer protocol that used for Internet is Internet
Protocol (IP). Its job is to provide a best-efforts way to transport datagrams from source to
destination, without regard to whether or not these machines are on the same network or not
these are other networks in between them.
Communication in the Internet works as follows. Each datagram is transmitted, after getting from
Transport layer, through the Internet, possibly being fragmented into smaller units as it goes.
When all pieces finally get to the destination machine, they are reassembled by the network layer
into the original datagram.
Internetwork protocol
At the network layer, TCP/IP supports the internetwork protocol .IP, in turn, contains four
supporting protocols:ARP ,RARP ,ICMP,and IGMP.
IP is the transmission mechanism used by the TCP/IP protocols. It is an un –reliable and
connectionless datagram protocol – a best effort delivery service. This is like a post office service.
The post office does its best to deliver the mail but does not always succeed. If an unregistered
letter is lost. it is up to the sender or would recipient to discover the loss and rectify the problem.
The post office itself does not keep track of every letter and cannot notify a sender of loss or
damage. An example of a situation similar to pairing IP with a protocol that contains reliability
functions is a self addressed ,stamped postcard included in a letter mailed through the post office.
when the letter is delivered , the receiver mails the postcard back to the sender to indicate
success. If the sender never receives the postcard, he or she assumes the letter was lost and
sends out another copy.
IP transports data in packets called Datagrams,each of which is transported separately.
Datagrams may travel along different routes and may arrive out of sequence or duplicated. IP
does not create virtual circuits for delivery.
Datagram
Packets in IP layer are called Datagrams. A Datagram is a variable length packet(upto 65,536
bytes) consisting of two parts : Header and Data. The header can be from 20 to 60 bytes and
contains information essential to routing and delivery.

Version The first field defines the version number of the IP. The current version is 4(IPv4),with
binary value 0100.
Header length (HLEN) The HLEN field defines the length of the header in multiples of four
bytes .The four bits can represent a number between 0 to 15,which,when multiplied by 4,gives a
maximum of 60 bytes.
Service Type. The service type field defines how datagram should be handled. It includes bits
that define the priority of the datagram. It also contains bits that specify the type of service the
sender desires such as the level of throughput, reliability, and delay.
Total Length The total length field defines the total length of the IP datagram. It is a two-byte
field (16 bits) and can define up to 65,535 bytes.
Identification The identification field is used in fragmentation. A datagram, when passing through
different networks, may be divided into fragments to match the network frame size. When this
happens, each fragment is identified with a sequence number in this field.
Flags The bits in the flags field deal with fragmentation (the datagram can or can not be
fragmented; can be first, middle, or last fragment; etc.).
Fragmentation offset The fragmentation offset is a pointer that shows the offset of the data in
the original datagram (if it is fragmented).
Time to live The time to live field defines the number of hops a datagram can travel before it is
discarded. The source host, when it creates the datagram, sets this field to an initial value. Then,
as the datagram travels through the Internet, router by router, each router decrements this value
by 1. If this value becomes 0 before the datagram reaches its final destination, the datagram is
discarded. This prevents a datagram from going back and forth forever between routers.
Protocol The protocol field defines which upper-layer protocol data are encapsulated in datagram
(TCP, UDP, ICMP etc.).
Header Checksum This is a 16-bit field used to check the integrity of the header, not the rest of
the packet.
Source address The source address field is a four-byte (32-bit) Internet address. It identifies the
original source of the datagram.
Destination address The destination address field is a four-byte (32-bit) Internet address. It
identifies the final destination of the datagram.
Options The options field gives more functionality to IP datagram. It can carry fields that control
routing, timing, management, and alignment.
ADDRESSING
In addition to the physical address the internet requires an additional addressing convention : an
address that identifies the connection of a host to its network.

Each Internet address consists of 4 bytes defining three fields : class type,netid,and hosted.
These parts are varying lengths depending on the class of the address.
CLASSES
There are currently five different classes:
They are Class A, Class B, Class C, Class D, Class E
Class A :
This can accommodate more hosts since 3 bytes are reserved for HOSTID. Class A will begin
with 0 .
Class B :
This will start with 10 and Host id will have 2 bytes length.
Class C :
This will start with 110 and Hostid will have 1 byte length.
Class D:
This will start with 1110 . This is reserved for Multicast addresses.
Class E :
This is reserved for feature use and will start with 1111 .
Ex. for classes
1. 01111011 10001111 11111100 11001111
The address is starting with 0 .Hence it is Class A.
2. 10011101 1000111111 11111100 11001111
The address is starting with 10 .Hence it is Class B.

3. 11011101 10001111 11111100 11001111
The address is starting with 110 .Hence it is Class C
4. 11101011 10001111 11111100 11001111
The address is starting with 1110 .Hence it is Class D.
5. 11110101 10001111 11111100 11001111
The address is starting with 1111. Hence it is Class E.
CLASS A :
Byte1 Byte 2 Byte3 Byte 4
000
CLASS B:
000
Class C :
000
Class D :
000
Class E :
000
0 Netid Hostid
10 Netid Hostid
110 Netid Hostid
11110 Multicast address
1111 Reserved for future use

Dotted-Decimal Notation:
To make 32 bit form shorter and easier to read, Internet addresses are usually written in decimal
form with decimal points separating the bytes – dotted – decimal notation.
IP addresses in decimal notation
Class ranges of Internet address
From To
Class A
From To
12 Class B
From To
Class C
From To
Class D
From To
Class E
10000000 00001011 00000011 00011111
128.11.3.31
0.0.0.0 127.255.255.2555
128.0.0.0 191.255.255.255
192.0.0.0 223.255.255.255
224.0.0.0 239.255.255.255
240.0.0.0 255.255.255.2555

Example:
Find the class of each address
a. 4.23.145.90
b. 227.34.78.7
c. 246.7.3.8
d. 129.6.8.4
e. 198.76.9.23
The first byte defines the class.
a. The binary equivalent for 4 is 0000 0100. Since the first bit is 0 ,it is CLASS A.
b. The binary equivalent for 227 is 11100011 . Since it is starting with 111 ,it is CLASS D.
c. The binary equivalent for 246 is 11110110 . Since it is starting with 111 ,it is CLASS E.
d. The binary equivalent for 129 is 10000001 . Since it is starting with 10,it is
CLASS B.
e. The binary equivalent for 198 is 11000110 . Since it is starting with 110 , it is CLASS C.
Example:
Find the netid and hosted for each address :
a. 4.23.145.90
b. 227.34.78.7
c. 246.7.3.8
d. 129.6.8.4
e. 198.76.9.23
First find the class and then netid and hosted.
a. Class A , netid :4 , hosted : 23.145.90
b. Class D , no netid or hosted;
c. Class E , no netid , or hosted;
d. Class B , netid :129.6 ,hosted :8.4 ;
e. Class C netid : 198.76.9 hostid 23 ;
TCP/IP supports four other protocols in the network layer :ARP,RARP,ICMP,and IGMP.

Address resolution protocol (ARP)
The address resolution Protocol associates an ip address with physical address. On a typical
physical network, such as a LAN, each device on a link is identified by a physical or station
address usually imprinted on the network interface card.(NIC)
Physical address have local jurisdiction and can be changed easily. For example, if the NIC on
a particular machine fails, the physical address changes. The IP address, on the other hand ,have
universal jurisdiction and cannot be changed. ARP is used to find the physical address of the
node when its Internet address is known.
Anytime a host or a router needs to find the physical address of another host on its
network, it formats an ARP query packet that includes the IP address and broadcast it over the
network. Every host on the network receives and processes the ARP packet, but only the
intended recipient recognizes its internet address and sends back its physical address. The host
both to its cache memory and to the datagram header, then sends the datagram on its way.
Reverse Address resolution protocol(RARP)
The RARP allows a host to discover its internet address when it knows only its physical
address. The question here is ,why do we need RARP? A host is supposed to have its internet
address stored on its hard disk !
RARP works much like ARP. The host wishing to retrieve its internet address broadcasts an
RARP query packet that contains its physical address to every host on its physical network. A
server on the network recognizes the RARP packet and returns the host’s internet address.
Internet Control Message Protocol (ICMP)
The Internet control message protocol is a mechanism used by hosts and routers to send
notification of datagram problems back to the sender.
IP is an unreliable and connectionless protocol. ICMP allows IP to inform a sender if a
datagram is undeliverable. A datagram travels from router to router until it reaches one that can
deliver it to its final destination. If a router is unable to route or deliver the datagram because of
unusual conditions or due to congestion, ICMP allows it to inform the original source.
ICMP uses echo test/reply to test whether a destination is reachable and responding. It
also handles both control and error message, but its sole function is tio report problems, not
correction them. A datagram carries only source and destination address. For this reason ICMP
can send message only to the source, not to an intermediate router.
Internet Group Message Protocol (IGMP)

Ip addressing supports multicasting. All 32-bit IP addresses that start with 1110(class D) are
multicast addresses. With 28 bits remaining for the group address, more than 250 million
addresses are available for assignment. Some of these addresses are permanently assigned.
The IGMP has been designed to help a multicast router identify the hosts in a lan that are
members of a multicast group. It is a companion to the IP protocol.
Expected questions
1.What is the difference between the adaptive and non-adaptive routing algorithms.
2.Explain the shortest path routing algorithm.
3.Explain the services that are provided by the network layer.
4.Explain Flooding routing algorithm.
5.Explain the Distance Vector Routing algorithm.
6.What is the count – to – infinity problem?
7.Explain link state routing algorithm.
8.Explain the Hierarchical Routing algorithm.
9.Explain Broadcast Routing and Multicast Routing.
10.What is congestion? Give the general principles of congestion control?
11.Explain Open loop and Close loop solutions for congestion.
12.How traffic shaping will be done to control congestion?
13.Explain The Leaky Bucket algorithm.
14. Explain the Token Bucket algorithm.
15.How the congestion can be controlled in Virtual Circuits?
16.What is a Choke packet? Explain when a choke packet is used.
17.Expalin the IP protocol.
18What is meant by Load shedding and Jitter control?
19. Explain the ICMP and ARP.
20.Explain the different IP address formats. For a hierarchical routing with 4800 routers, what
region and cluster sizes should be chosen to minimize the size of routing table for a three-layer
hierarchy?
* * * * *

Review Questions
2. How is a repeater different from an amplifier?
3. What is the difference between a simple bridge and transparent bridge?
4. What is the function of router?
5. How does a router differ from a bridge?
6. Why is adaptive routing is superior to non-adaptive routing?
7. What is the function of a gateway?
8. How does a multiprotocol router differ from a traditional single – protocol router?
1. HDLC is the acronym for--------
a. High-duplex line communication. b.high level data link control c.half-duplex
digital link combination d. none of the above
2.HDLC is a ------ protocol
a. Character-oriented b.bit-oriented c.byte – oriented d. count-oriented
3.The HDLC ----field defines the beginning and of a frame
a. Flag bladders c. control d. FCS
4.Polling and selecting are functions of the ----in HDLC protocol
a. P/F bit b.N( R ) c.N ( S ) d.code bits
5. Which of the following is not an internetworking device?
a. bridge b. gateway c. router d. all of them
6. Which of following uses the greatest number of layers in the OSI model?
a. bridge b. gateway c. router d .repeater
7. A simple bridge does the following
a. filters a data packet b. forwards a data packet c. extends Lans
d. all the above
8.The shortest path in routing can refer to --------
a. the least expensive path b the latest distant path
b. the path with the smallest number of hops d. any or combination of the above
9 In distance vector routing ,each router receives vectors from----
a. every router in the network b. every router less than two units away
b. a table stored by the software d. its neighbor only.
10.If there are five routers and six networks in an internet work using link
state routing, how many routing tables are there?
a. 1 b. 5 c. 6 d. 11
11. If there are five routers and six networks in an internet work using link state routing,
how many data bases are there?

a. 1 b. 5 c. 6 d. 1
12. Gateways function in which OSI layers?
a.the lower three b. the upper four c. all seven d. all but the physical layer
13. Repeaters function in the --- layer
a. physical b . datalink c. net work d. a and b
14.Bridges function in the -------- layer(s)
a. physical b . datalink c. net work d. a and b
Quiz Questions
1. What are the functions of Network Layer?
2. Give two differences between Adaptive and Nonadaptive routing algorithm.
3. In Shortest path routing algorithm what is the first step?
4. How the labels are measured in the above algorithm?
5. What is flooding algorithm?
6. What is meant by selective flooding?
7. What is the disadvantage with flooding algorithm?
8. Give one method to overcome the disadvantage with flooding?
9. When the flooding algorithm will be used?
10. When we will go for hierarchical routing?
11. How the hierarchy is divided?
12. In distance vector routing, what will be the contents of table?
13. In the hierarchical routing for a N router subnet the optimal no. of routers is
__________________________.
14. The total no. of entries / routers = _________________.
15. Give two examples of broadcast routing.
16.Give two methods of broadcast routing.
17. What is congestion?
18. Name the two congestion control algorithm.
19.What is the difference between a physical address and a logical address?
20.What are the advantages of using UDP over TCP?
21. What are the data packet at each TCP/IP protocol suite layer called?
22. Name the protocols at the network layer of the TCP/IP protocol suite.
23.What is the purpose of the time to live field in the datagram header?
24. Given an IP address in decimal – dotted notation, how can its class be determined?
25. How can a device have more than one IP address?

26. What is a Hostid ?
27. What is a Netid?
28. How does a netid differ from a network address?
29. What is the purpose of subnetting?
30. What is the purpose of ARP?
31. What is the purpose of RARP?
32. What is the purpose of ICMP?
33.What is the purpose of IGMP?
34.What is the diffrence between a logical address and port address?
35.Change the following IP address from dotted-decimal notation to binary notation
114.34.2.8
36. Change the following IP address from dotted-decimal notation to binary notation
208.34.54.12
37. Change the following binary notation Ip address from binary notation to dotted-decimal
notation .
01111111 11110000 01100111 01111101
Multiple choice Questions
1.Which OSI layer corresponds to the TCP-UDP layer?
a. physical b. data link c.network d. transport
2.Which OSI layer corresponds to the IP layer?
a. physical b. data link c.network d. transport
3. Which OSI layer(S) corresponds to the TCP/IP application layer?
a. application b. session c. presentation d.all of the above
4.Which IP address class has few hosts per network?
a. A b.B c.C d. D
5.For what does the data link layer look for as it sends a frame from one link to another?
a. hosted b.Ip address c. domain name d.station address
6.The purpose of ARP on a network is to find the ------- given the ---------
a. Intenet address, domain name
b. Intenet address , station address
c. Intenet address ,netid
d. station address ,Intenet address
7.Which of the following apply to UDP?
a. is unreliable and connectionless
b. contains destination and source port address
c. reports certain errors

d. all the above
8.Which of the following applies to both UDP and TCP?
a. transport layer protocols
b. port – to –port communication
c. services of IP layer used
e. all the above
9.Which of the following is a class A host address?
a. 128.4.5.6
b. 117.4.5.1
c. 117.0.0.0
d. 117.8.0.0
10.which of the following is a lass b host address?
a.233.0.0.0
b.130.4.5.6
c.230.0.0.0
d.30.4.5.6
11.Which of the following is a class C host address?
a.230.0.0.0
b. 130.4.5.6
c.230.0.0.0
d.30.4.5.6
12.The data unit in the TCP/IP data link layer is called a ------
a.message
b.segment
c.datagram
d.frame
13. The data unit in the TCP/IP layer is called a ------
a.message
b.segment
c.datagram
d.frame
14. The data unit from the transport layer that uses UDP is called a -----
a. user datagram
b. message
c. segment
d. frame

15.when a host knows its physical address but not its IP address,it can use-----
a.ICMP b.IGMP c. ARP d. RARP
16.This transport layer protocol is connectionless.
a.UDP b.TCP c. FTP d.NVT
17. This transport layer protocol requires acknowledgement.
a. UDP b.TCP c. FTP d.NVT
18.Which of the following is default mask for the address 98.0.46.201?
a.255.0.0.0
b.255.255.0.0
c.255.255.255.0
d.255.255.255.255
19. Which of the following is default mask for the address 98.0.46.201
a.255.0.0.0
b.255.255.0.0
c.255.255.255.0
d.255.255.255.255
20. Which of the following is default mask for the address 190.0.46.201?
a.255.0.0.0
b.255.255.0.0
c.255.255.255.0
d.255.255.255.255
21.Change the following IP address from dotted-decimal notation to binary notation
a. 114.34.2.8 b.129.14.6.8 c.208.34.54.12 d.238.34.2.1 e. 241.34.2.8
22.Find the class of the following IP address
a.208.34.54.12 b.238.34.2.1 c. 129.14.6.8 d.241.34.2.8
23.Find the netid and hosted for the following IP address.
a.114.34.2.8 b.19.34.21.5 c. 23.67.12.1 d. 126.23.4.0
24. Find the netid and hosted for the following IP address.
a.129.14.6.8 b.132.56.8.6 c.23.67.12.1 d.190.12.67.9

25. Find the netid and hosted for the following IP address.
a. 192.8.56.2 b. 220.34.8.9 c.208.34.54.12 d. 205.23.67.8
26.Find the network address of the following IP addresses.
a. 114.34.2.8 b. 171.34.14.8 c. 192.8.56.2 d.226.7.34.5
f.226.7.34.5 f. 225.23.6.7 g. 245.34.21.5
27.Write the following masks in binary notation:
a.255 .255.255.0 b.255.255.224.0 c.255.255.255.240
28. Find the subnet work address for the following
IP address : 125.34.12.56 Mask 255.255.0.0
29. Find the subnet work address for the following:
IP address : 120.14.22.16 Mask : 255.255.128.0
30. Find the subnet work address for the following:
Ip address 140.11.36.22 Mask : 255.255.255.0
* * * * *

UNIT – IV
TRANSPORT LAYER
Introduction
The transport layer is the core of the OSI model. Protocols at this layer oversee the delivery of
data from an application program on one device to an application program on another device.
They act as a liaison between the upper-layer protocols (session, presentation, and application)
and the services provided by the lower layers.
Duties of the transport layer:
The services provided are similar to those of the data link layer. The data link layer, however, is
designed to provide its services within a single network, while the transport layer provides these
services across an internetwork made of many networks. While the transport layer controls all
three of the lower layers.
The services provided by transport layer protocols can be divided into five broad categories: end-
to-end deliver, addressing, reliable delivery, flow control, and multiplexing.
Quality of Service
The transport protocol improves the QoS (Quality of Service) provided by the network layer.
Following are the QoS parameters:
Connection establishment delay:
The connection establishment delay is the amount of time elapsing between a transport
connection being requested and the confirmation being received by the user of the transport
service. It includes the processing delay in the remote transport entity. As with all parameters
measuring a delay, the shorter the delay, the better the service.
Connection establishment failure probability:
The connection establishment failure probability is the chance of a connection not being
established within the maximum establishment delay time, for example, due to network
congestion, lack of table space somewhere, or other internal problems.
Throughput:
The throughput parameter measures the number of bytes of user data transferred per second,
measured over some time interval. The throughput is measured separately for each direction.
Transit delay:
The transit delay measures the time between a message being sent by the transport user on the
source machine and its being received by the transport user on the destination machine. As with
throughput, each direction is handled separately.
The Residual error ratio :
Measures the number of lost or garbled messages as a fraction of the total sent. In theory, the
residual error rate should be zero, since it is the job of the transport layer to hide all network layer
errors. In practice it may have some (small) finite value.
The Protection parameter provides a way for the transport user to specify interest in having the
transport layer provide protection against unauthorized third parties (wiretappers) reading or
modifying the transmitted data.
The Priority parameter provides a way for a transport user to indicate that some of its
connections are more important than other ones, and in the event of congestion, to make sure
that the high-priority connections get serviced before the low-priority ones.
Finally, the Resilience parameter gives the probability of the transport layer itself spontaneously
terminating a connection due to internal problems or congestion.

The QoS parameters are specified by the transport user when a connection is requested. Both
the desired and minimum acceptable values can be given. In some cases, upon seeing the QoS
parameters, the transport layer may immediately realize that some of them are unachievable, in
which case it tells the caller that the connection attempt failed, without even bothering to contact
the destination. The failure report specifies the reason for the failure.
The transport layer knows it cannot achieve the desired goal (e.g.600 Mbps throughput), but it
can achieve a lower, but still acceptable rate (e.g.150 Mbps). It then sends the lower rate and the
minimum acceptable rate to the remote machine, asking to establish a connection. If the remote
machine cannot handle the proposed value, but it can handle a value above the minimum, it may
make a counteroffer. If it cannot handle any value above the minimum, it rejects the connection
attempt. Finally, the originating transport user is informed of whether the connection was
established or rejected, and if it was established, the values of the parameters agreed upon. This
process is called option negotiation.
End –to –end delivery
The network layer oversees the end-to-end delivery of individual packets but does not see any
relationship between those packets, even those belonging to a single message.
It treats each as an independent entity. The transport layer, on the other hand, makes sure that
the entire message (not just a single packet) arrives intact. Thus, it oversees the end-to-end
(source –to-destination) delivery of an entire message.
Addressing
The transport layer interacts with the functions of the session layer. However, many protocols (or
protocol stacks, meaning groups of protocols that interact at different levels) combine session,
presentation, and application level protocols into a single packages, called an application. In these
cases, delivery to the session layer functions is, in effect, delivery to the application. In these cases,
delivery to the session layer functions is, in effect, delivery to the application. So communication
occurs not just from end machine to end machine but from end application to end application. Data
generated by an application on one machine must be received not just by the other machine but by
the correct application on that other machine.
To ensure accurate delivery from service access point to service access point, we need another
level of addressing in addition to those at the data link and network levels. Data link level
protocols need to know which two computers within a network are communicating. Network level
protocols need to know which two computers within an internet are communicating. But at the
transport level, the protocol needs to know which upper-layer protocols are communicating.
Reliable Delivery
At the transport layer, reliable delivery has four aspects: error control, sequence control, loss
control, and duplication control.
Transport layer
duties
End –to-End
delivery Addressing
Flow control
Reliable
delivery
Multiplexing

Error Control
When transferring data, the primary goal of reliability is error control.
But if we already have error handling at the data link layer, why do we need it at the transport
layer? Data link layer functions guarantee error-free delivery node-to-node for each link. However,
node-to-node reliability does not ensure end-to-end reliability.
Sequence Control
The second aspect of reliability implemented at the transport layer is sequence control. On the
sending end, the transport layer is responsible for ensuring that data units received from the
upper layers are usable by the lower layers. On the receiving end, it is responsible for ensuring
that the various pieces of a transmission are correctly reassembled.
Segmentation and Concatenation
When the size of the data unit received from the upper layer is too long for the network layer
datagram or data link layer frame to handle, the transport protocol divides it into smaller, usable
blocks. The dividing process is called segmentation. When, on the other hand, the size of the data
units belonging to a single session are so small that several can fit together into a single
datagram or frame, the transport protocol combines them into a single data unit. The combing
process is called concatenation.
Sequence Numbers
Most transport layer services add a sequence number at the end of each segment. If a longer data
unit has been segmented, the numbers indicate the order for reassembly. If several shorter units
have been concatenated, the numbers indicate the end of each submit and allow them to be
separated accurately at the destination. In addition, each segment carries a field that indicates
whether it is the final segment of a transmission or a middle segment with more still to come.
Loss Control
The third aspect of reliability covered by the transport layer is loss control. The transport layer
ensures that all pieces of a transmission arrive at the destination, not just some of them. When data
have been segmented for delivery, some segments may be lost in transit. Sequence numbers allow
the receiver’s transport layer protocol to identify any missing segments and request redelivery.
Duplication Control
The fourth aspect of reliability covered by the transport layer is duplication control. Transport layer
functions must guarantee that no pieces of data arrive at the receiving system duplicated. Just as
they allow identification of lost packets, sequence numbers allow the receiver to identify and
discard duplicate segments.
Reliable
delivery
Error
control Sequence
control
Duplication
Loss
control

Flow Control
Like the data link layer, the transport layer is responsible for flow control. However, flow control at
this layer is performed end-to-end rather than across a single link. Transport layer flow control
also uses a sliding window protocol. However, the window at the transport layer can vary in size
to accommodate buffer occupancy.
Multiplexing
To improve transmission efficiency, the transport layer has the option of multiplexing. Multiplexing
at this layer occurs two ways: upward, meaning that multiple transport layer connections use the
same network connection, or downward, meaning that one transport-layer connection uses
multiple network connections.
Upward
The transport layer uses virtual circuits based on the services of the lower three layers.
Normally, the underlying networks charge for each virtual circuit connection. To make more cost-
effective use of an established circuit, the transport layer can send several transmissions bound
for the same destination along the same path by upward multiplexing. This means if the
underlying network protocol has a high throughput, for example in the range of 1 Gbps, and the
user can create data only in the range of Mbps, then several users can share one network
connection.
Downward
Downward multiplexing allows the transport layer to split a single connection among several
different paths to improve throughput (speed of delivery). This option is useful when the
underlying networks have low or slow capacity. For example, some network layer protocols have
restrictions on the sequence numbers that can be handled. X.25 uses a three-bit numbering code,
so sequence numbers are restricted to the range of 0 to 7 (only eight packets may be sent before
acknowledgment is required). In this case, throughput can be unacceptably low. To counteract
this problem, the transport layer can opt to use more than one virtual circuit at the network layer to
improve throughput. By sending several data segments at once, delivery is faster.
Transport Transport
layer layer
layer
Upward Downward
One virtual circuit
Upper
layers
stack
Upper
layers
stack
Upper
layers
stack
1 2 3
Three virtual circuits
Upper
layers
stack
2

TCP Protocol
ARPA established a packet-switching network of computers linked by point-to-point leased lines
called Advanced Research Project Agency Network(ARPANET) that provided a basics for early
research into networking. The conventions developed by ARPA to specify how individual
computers could communicate across that network became TCP/IP.
The transport layer is represented in TCP/IP by two protocols:TCP and UDP. Of these, UDP is
similar; it provides nonsequenced transport functionality when reliability and security are less
important than size and speed.
The transport protocols of the TCP/IP suite define a set of conceptual connections to individual
process called protocol ports or, more simply, ports. A protocol port is a destination point (usually
a buffer) for storing data for use by a particular process.
The IP is a host-to-host protocol, meaning that it can deliver a packet from one physical device to
another. TCP/IP’s transport level protocols are port-to-port protocols that work on top of the IP
protocols to deliver the packet from the originating port to the IP services at the start of a
transmission, and from the IP services to the destination port at the start end.
Figure Port addresses
Each port is defined by a positive integer address carried in the header of a transport layer
packet. An IP datagram uses the host’s 32-bit Internet address. A frame at the transport level
uses the process port address of 16 bits, enough to allow the support of up to 65,536(0 to 65535)
ports.
USER DATAGRM PROTOCOL(UDP)
TELNET
(server)
(23)
IP
Data Link
Physical
TELNET
(client)
(51001)
IP
Data Link
Physical
TCP OR UDP
TCP or UDP

The user datagram protocol (UDP) is the simpler of the two standard TCP/IP transport protocols.
It is an end-to-end transport level protocol that adds only port addresses, check sum error control,
and length information to the data from the upper layer. The packet produced by the UDP is
called a user datagram .
 Source port address. The source port address is the address of the application program
that has created the message.
 Destination port address. The destination port address is the address of the application
program that will receive the message.
 Total length. The total length field defines the total length of the user datagram in bytes.
 Check sum. The check sum is a 16-bit field used in error detection.
Figure UDP datagram format
UDP provides only the basic functions needed for end-to-end delivery of a transmission. It does
not provide any sequencing or recording functions and cannot specify the damaged packet when
reporting an error (for which it must be paired with ICMP). UDP can discover that an error has
occurred; ICMP can then inform the sender that a user datagram has been damaged and
discarded. Neither, however, has the ability to specify which packet has been lost. UDP contains
only a checksum; it does not contain an ID or sequencing number for a particular data segment.
Transmission Control Protocol(TCP)
Variable
8 bytes
Data
Source port address Destination port address
16 bits 16 bits
Total length Checksum
16 bits 16 bits
H
eader

The Transmission Control Protocol (TCP) provides full transport layer services to applications.
TCP is a reliable stream transport port-to-port protocol. The term stream, in this context, means
connection-oriented: a connection must be established between both ends of a transmission
before either may transmit data. By creating this connection, TCP generates a virtual circuit
between sender and receiver that is active for the duration of a transmission.(connections for the
duration of an entire exchange are different, and are handled by session functions in individual
applications.) TCP begins each transmission by altering the receiver that datagrams are on their
way (connection establishment) and ends each transmission with a connection termination. In this
way, the receiver knows to expect the entire transmission rather than a single packet.
IP and UDP treat multiple datagrams belonging to a single transmission as entirely separate units,
unrelated to each other. The arrival of each datagram at the destination is therefore a separate
event, unexpected by the receiver. TCP, on the other hand, as a connection-oriented service, is
responsible for the reliable delivery of the entire stream of bits contained in the message originally
generated by the sending application. Reliability is ensured by provision for error detection and
retransmission of damaged frames; all segments must be received and acknowledged before the
transmission is considered complete and the virtual circuit is discarded.
At the sending end of each transmission, TCP divides long transmissions into smaller data units
and packages each into a frame called a segment. Each segment includes a sequencing number
for reordering after receipt, together with an acknowledgement ID number and a window-size field
for sliding window ARQ. Segments are carried across network links inside of IP datagrams as it
comes in and reorders the transmission based on sequence numbers.
The TCP Segment
The scope of the services provided by TCP requires that the segment header be extensive. A
comparison of the TCP segment format with that of a UDP user datagram shows the differences
between the two protocols. TCP provides a comprehensive range of reliability functions but
sacrifices speed (connections must be established, acknowledgments waited for , etc.).Because
of its smaller frame size, UDP is much faster than TCP, but at the expense of reliability. A brief
description of each field is in order.

Figure TCP Segment format
 Source port address. The source port address defines the application program in the
source computer.
 Destination port address. The destination port address defines the application program
in the destination computer.
 Sequence number. A stream of data from the application program may be divided into
two or more TCP segments. The sequence number field shows the position of the data in
the original data stream.
 Acknowledgement number. The 32-bit acknowledgement number is used to
acknowledge the receipt of data from the other communicating device. This number is
valid only if the ACK bit in the control field(explained later) is set. In this case, it defines the
byte sequence number that is next expected.
 Header Length (HLEN). The four-bit HLEN field indicates the number of 32-bit (four-byte)
words in the TCP header. The four bits can define a number up to 15.This is multiplied by
4 to give the total number of bytes in the header. Therefore, the size of the header can be
a maximum of 60 bytes (4x15).Since the minimum required size of the header is 20 bytes,
40 bytes are thus available for the options section.
 Reserved. A six-bit field is reserved for future use.
 Control. Each bit of the six-bit control field functions individually and independently. A bit
can either define the use of a segment or serve as a validity check for other fields. The
Source port address 16 bits Destination port address 16 bits
Sequencing number 32 bits
Acknowledgement number 32 bits
HLEN Reserved U A P R S T Window size
4bits 6 bits R C S S Y I 16 bits
G K H T N N
Check sum 16 bits Urgent pointer 16 bits
Options & Padding
Data
Header

urgent bit, when set, validates the urgent pointer field. Both this bit and the pointer indicate
that the data in the segment are urgent. The ACK bit, when set, validates the
acknowledgement number field. Both are used together and have different functions,
depending on the segment type. The PSH bit is used to inform the sender that a higher
throughput is needed. If possible, data must be pushed through paths with higher
throughput. The reset bit is used to reset the connection when there is confusion in the
sequence numbers. The SYN bit is used for sequence number synchronization in three
types of segments: connection request, connection confirmation (with the ACK bit set),
and confirmation acknowledgement (with the ACK bit set). The FIN bit is used in
connection termination in three types of segments: termination request, termination
confirmation (with the ACK bit set), and acknowledgement of termination confirmation
(with the ACK bit set).
 Window size. The window is a 16-bit field that defines the size of the sliding window.
 Checksum. The checksum is a 16-bit field used in error detection.
 Urgent pointer. This is the last required field in the header. Its value is valid only if the
URG bit in the control field is set. In this case, the sender is informing the receiver that
there are urgent data in the data portion of the segment. This pointer defines the end of
urgent data and the start of normal data.
 Options and padding. The remainder of the TCP header defines the optional fields. They
are used to convey additional information to the receiver or for alignment purposes.

UNIT 5
Application Layer
Network Security
Security Attacks
Attacks on the security of a computer system or network are best characterized by viewing the
function of the computer system as providing information.
There are four general categories of attack:
 Interruption: An asset of the system is destroyed or becomes unavailable or unusable.
This is an attack on availability. Examples include destruction of a piece of hardware, such
as a hard disk, the cutting of a communication line, or the disabling of the file management
system.
 Interception: An unauthorized party gains access to an asset. This is an attack on
confidentiality. The unauthorized party could be a person, a program, or a computer.
Examples include wiretapping to capture data in a network, and the illicit copying of files or
programs.
 Modification: An unauthorized party not only gains access to but tampers with an asset.
This is an attack or integrity. Examples include changing values in a data file, altering a
program so that it performs differently, and modifying the content of messages being
transmitted in a network.
 Fabrication: An unauthorized party inserts counterfeit objects into the system. This is an
attack on authenticity. Examples include the insertion of spurious messages in a network
or the addition of records to a file.

Information Information
Source destination
(a) Normal flow
(b) Interruption (C) Interception
(d) Modification (e) Fabrication
CONVENTIONAL ENCRYPTION MODEL
The original intelligible message, referred to as plaintext, is converted into apparently
random nonsense, referred to as ciphertext. The encryption process consists of an algorithm and
a key. The key is a value independent of the plaintext. The algorithm will produce a different
output depending on the specific key being used at he time. Changing the key changes the output
of the algorithm.
Once the ciphertext is produced, it may be transmitted. Upon reception, the ciphertext can
be transformed back to the original plaintext by using a decryption algorithm and the same
key that was used for encryption.
The security of conventional encryption depends on the secrecy of the key, not the
secrecy of the algorithm. We do not need to keep the algorithm secret; we need to keep only the
key secret. A source produces a message in plaintext, X = [X1, X2, …….,XM ]. For encryption, a
key of the form K = [K1, K2,……..,KJ] is generated. If the key is generated at the message source,
then it must also be provided to the destination by means of some secure channel. Alternatively, a
third party could generate the key and securely deliver it to both source and destination.

With the message X and the encryption key K as input, the encryption algorithm forms the
ciphertext Y = [ Y1, Y2,……., YN]. We can write this as
Y = EK(X)
This notation indicates that Y is produced by using encryption algorithm E as a function of the
plaintext X, with the specific function determined by the value of the key K.
The intended receiver, in possession of the key, is able to invert the transformation:
X = DK(Y)
Substitution Techniques
A substitution technique is one in which the letters of plaintext are replaced by other letters or by
numbers or symbols. If the plaintext is viewed as a sequence of bits, then substitution involves
replacing plaintext bit patterns with ciphertext bit patterns.
Caesar Cipher
The earliest known use of a substitution cipher, and the simplest, was by Julius Caesar. The Caesar
cipher involves replacing each letter of the alphabet with the letter standing three places further
down the alphabet. For example,
plain : meet me after the toga party
cipher : PHHW PH DIWHU WKH WRJD SDUWB
Note that the alphabet is wrapped around, so that the letter following Z is A. We can define the
transformation by listing all possibilities, as follows:
plain: a b c d e f g I j k l m n o p q r s t u v w x y z
cipher: D E F G H I J K L M N O P Q R S T U V W X Y Z A B C
If we assign a numerical equivalent to each letter (a =1, b = 2, etc.), then the algorithm can be
expressed as follows. For each plaintext letter p, substitute the ciphertext letter C:
C = E(p) = (p + 3) mod (26)
A shift may be of any amount, so that the general Caesar algorithm is
C = E(p) = (p + k) mod (26)
Where k takes on a value in the range 1 to 25. The decryption algorithm is simply
P = D(c) = (C - k) mod (26)
Playfair Cipher
The bet-known multiple-letter encryption cipher is the Playfair, which treats digrams in the
plaintext as single units and translates these units into ciphertext digrams.
The Playfair algorithm is based on the use of a 5 X 5 matrix of letters constructed using a
keyword. Here is an example, solved by Lord Peter Wimsey in Dorothy Sayers’s Have His
carcase.
M O N A R
C H Y B D
E F G I/J K
L P Q S T
U V W X Z
In this case, the keyword is monarchy. The matrix is constructed by filling in the letters of the
keyword from left to right and from top to bottom, and then filling in the remainder of the matrix

with the remaining letters in alphabetic order. The letters I and J count as one letter. Plaintext is
encrypted two letters at a time, according to the following rules:
1. Repeating plaintext letters that would fall in the same pair are separated with a filler letter,
such as x, so that balloon would be enciphered as ba lx lo on.
2. Plaintext letters that fall in the same column are each replaced by the letter beneath, with
the top element of the row circularly following the last. For example, mu is encrypted as
CM.
3. Otherwise, each plaintext letter is replaced by the letter that lies in its own row and the
column occupied by the other plaintext letter. Thus, hs becomes BP and ea becomes IM
(or JM, as the encipherer wishes).
Simplified DES
The S-DES decryption algorithm takes an 8-bit block of plaintext (example: 10111101) and a 10-
bit key as input and produces an 8-bit block of ciphertext as output. The S-DES decryption
algorithm takes an 8-bit block of ciphertext and the same 10-bit key used to produce that
ciphertext as input and produces the original 8-bit block of plaintext.
The encryption algorithm involves five functions: an initial permutation (IP); a complex
function labeled fk, which involves both permutation substitution operations and depends on a key
input; a simple permutation function that switches (SW) the two halves of the data; the function fk
again, and finally a permutation function that is the inverse of the initial permutation (IP-1
).
The function fk takes as input not only the data passing through the encryption algorithm,
but also an 8-bit key. The algorithm could have been designed work with a 16-bit key, consisting
of two 8-bit subkeys, one used for each occurrence of fk. Alternatively, a single 8-bit key could
have been used, with the same key used twice in the algorithm. A compromise is to use a 10-bit
key from which two 8-bit subkeys are generated, as depicted in fig. In this case, the key is first
subjected to a permutation (P10). Then a shift operation is performed. The output of the shift
operation then passes through a permutation function that produces an 8-bit output (P8) for the
first subkey (K1). The output of the shift operation also feeds into another shift and another
instance of P8 to produce the second subkey (K2).

ENCRYPTION DECRYPTION
Simplified DES Scheme
P10
IP
Shift
IP-1
fk fk
SW SW
fk fk
IP-1
IP
Shift
P8
K1 K1
K2 K2
8-bit plaintext
8-bit plaintext
8-bit ciphertext 8-bit ciphertext
10 - bit key
P8

We can concisely express the encryption algorithm as a composition of functions:
IP-1 0
fk2
)
SW )
fk1
0
IP
Which can also be written as
ciphertext = IP-1
(fk ( SW ( fk ( IP ( plaintext ) ) ) ) )
Where
K1 = P8 (Shift (P10 (key) ) )
K2 = P8 (Shift (Shift (P10 (key) ) ) )
10 –bit key
10
5 5
5 5
8
K1
5 5
K2 8
Key Generation for Simplified DES
P8
P10
LS-1 LS-1
LS-1 LS-1
P8

Decryption is also shown in fig. and is essentially the reverse encryption:
plaintext = IP-1
( fk1 (SW (fk2 ( IP (ciphertext ) ) ) ) )
We now examine the elements of S-DES in more detail.
S-DES Key Generation
S-DES depends on the use of a 10-bit key shared between sender and receiver. From this
key, two 8-bit subkeys are produced for use in particular stages of the encryption and decryption
algorithm. Figure depicts the stages followed to produce the subkeys.
First, permute the key in the following fashion. Let the 10-bit key be designated as ( k1, k2,
k3, k4, k5, k6, k7, k8, k9, k10 ). Then the permutation P10 is defined as P10 (k1, k2, k3, k4, k5, k6, k7,
k8, k9, k10 ) = ( k3, k5, k2, k7, k4, k10, k1, k9, k8, k6 )
P10 can be concisely defined by the display:
P10
3 5 2 7 4 10 1 9 8 6
This table is read from left to right; each position in the table gives the identity of the input bit that
produces the output bit in that position. So the first output bit is bit 3 of the input; the second
output bit is bit 5 of the input, and so on. For example, the key (1010000010) is permuted to
(1000001100). Next, perform a circular left shift (LS-1), or rotation, separately on the first five bits
and the second five bits. In our example, the result is (00001 11000).
Next we apply P8, which picks out and permutes 8 of the 10 bits according to the following
rule:
P8
6 3 7 4 8 5 10 9
The result is subkey 1 (K1). In our example, this yields (10100100).
We then go back to the pair of 5-bit strings produced by the two LS-1 functions and
perform a circular left shift of 2 bit positions on each string. In our example, the value (00001
11000) becomes (00100 00011). Finally, P8 is applied again to produce K2. In our example, the
result is (01000011).
The RSA Algorithm
Description of the Algorithm
The scheme developed by Rivest, Shamir, and Adleman makes use of an expression with
exponentials. Plaintext is encrypted in blocks, with each block having a binary value less than
some number n. That is, the block size must be less than or equal to log2(n); in practice, the block
size is 2k
bits, where 2k
< n  2k+1
. Encryption and decryption are of the following form, for some
plaintext block M and ciphertext block C:
C = Me
mod n
M =Cd
mod n = (Me
)d
mod n = Med
mod n
Both sender and receiver must know the value of n. The sender knows the value of e, and only
the receiver knows the value of d. Thus, this is a public-key encryption algorithm with a public key
of KU = {e, n} and a private key of KR = {d, n}. For this algorithm to be satisfactory for public – key
encryption, the following requirements must be met:
1. It is possible to find values of e, d, n such that Med
= M mod n for all M < n.
2. It is relatively easy to calculate Me
and Cd
for all values of M< n.
3. It is infeasible to determine d given e and n.
4.

The RSA Algorithm (a)
Fig (a) summarizes the RSA algorithm.
Example 1:
Select two prime numbers, p=7 and q = 17.
1. Calculate n = pq = 7 X 17 = 119
2. Calculate (n) = (p-1) (q-1) = 96
3. Select e such that e is relatively prime to (n) = 96 and less than (n); in this case, e = 5.
4. Determine d such that de = 1 mod 96 and d < 96. The correct value is d = 77, because 77
X 5 = 385 = 4 X 96 +1.
The resulting keys are public key KU = {5,119} and private key KR = { 77,119}. The example
shows the use of these keys for a plaintext input of M = 19. For
ciphertext
66
KU = 5, 119 KR = 77,119
Example of RSA algorithm (b)
Key Generation
Select p,q p and q both prime
Calculate n = p x q
Calculate (n) = (p-1)(q-1)
Select integer e gcd((n), e) =1; 1< e < (n)
Calculate d d = e-1
mod (n)
Public key KU = {e,n}
Private key KR = {d,n}
Encryption
Plaintext: M < n
Ciphertext: C = Me
(mod n)
Decryption
Plaintext: C
Ciphertext: M = Cd
( mod n)
5
19 = 2476099 = 20807 with a
119 remainder of
66
Plaintext
77
66 = 1.27….x 10140
1.06 …..x10138
with
119 = a remainder of
19
Plaintext
19

Encryption, 19 is raised to the fifth power, yielding 2476099. Upon division by 119, the remainder
is determined to be 66. Hence 195
 66 mod 119, and the ciphertext is 66. For decryption, it is
determined that 6677
 19 mod 119.
Example 2 :
p = 3, q = 11, d = 17
assume plaintext symbol M = 5
n = p*q = 33, z = = (3-1) (11 – 1) = 20
Find e such that e * d = 1 mod z (z+1)
[ d = e-1
mod z ] k * z+1 (k =1 here)
e = 3 3 X 7 = 1 mod 20
public key = { e,n} = { 3, 33}
private key = { d, n} = { 7, 33}
Encryption M =5
C = Me
mod n
= 5e
mod 33 = 125 /33 = 3
with reminder 26
ciphertext = 26
decryption c = 26
p =M = Cd
mod n = 267
mod 33
= 8031810176/33 = 243388187
with reminder 5
plain text = 5
Example 3:
P = 17, q = 31, e = 7, m = 2
N = 17 X 31 = 527
z = (17-1) (31 – 1) = 16 x 30 = 480
e =7
Finding d such that e * d = 1 mod 480
and d < 480 = k * z + i
e = 7
the value obtained is 343 1/7 x (480 x k +1)
publickey = { 7, 527} private key = { 343, 527 }
ciphertext = 27
mod 527
= 128 mod 527 = 0

 with reminder = 128
ciphertext = 128
Decryption
128343
mod 527
2 is reminder
 plaintext =2
(a) Encryption
(b) Authentication

Public – Key Encryption
Conventional Encryption Public-Key Encryption
Needed to work: Needed to work:
1. The same algorithm with the same key
is used for encryption and decryption.
1. One algorithm is used for encryption
and decryption with a pair of keys, one
for encryption and one for decryption.
2. The sender and receiver must share
the algorithm and the key.
2. The sender and receiver must each
have one of the matched pair of keys
(not the same one).
Need for Security: Need for Security:
1.The key must be kept secret. 1. One of the two keys must be kept
secret.
2.It must be impossible or at least
impractical to decipher a message if no
other information is available.
2. It must be impossible or at least
impractical to decipher a message if no
other information is available.
3. Knowledge of the algorithm plus
samples of ciphertext must be insufficient
to determine the key.
3. Knowledge of the algorithm plus one
of the keys plus samples of ciphertext
must be insufficient to determine the
other key.
X
KRb
Source A Destination B
X Y X
KUb KRb
Cryptanalyst
Encryption
algorithm
Decryption
algorithm
Message
source Destination
Key pair
Source

E-mail :
E-mail system consists of two subsystems
- the user agent, and
- the message transfer agents
- User Agents :
They allow people to read and send e-mail they are local programs that provide a
command based, menu based, or graphical method for interacting with e-mail system.
- Message transfer agents :
They are responsible for moving the messages from the source to the destination. They
are typically system daemons that run in the background and move e-mail through the
system.
Typically, e-mail system support five basic functions given below.
(i) Composition :
It refers to the process of creating messages and answers.
(ii) Transfers :
it refers to moving messages from the originator to the recipient. This requires, establishing a
connection to the destination (or) some intermediate machine, outputting the message and
releasing the connection.
(iii) Reporting :
It informs the originator about the status of the message, whether it is delivered, rejected(or) lost.
(iv) Displaying :
These provides the incoming messages to be read by the people. Simple conversions and
formatting is performed.
(v) Disposition :
It is the final step and concerns what the recipient does with the message after receiving it.
Other Services of E-mail include:
Mailboxes :
Used for storing incoming E-mail.
Mailing List = List of e-mail addresses to whom, identical copies of messages need to be
sent.
Registered E-mail = It allows the originator to know that his mail has arrived.
High priority E-mail = Secret E-mail etc.

User Agent :
A user agent is normally a program that accepts a variety of commands for composing, receiving
and replying to messages as well as manipulating mail boxes.
Sending E-mail :
To send an e-mail a user must provide the message, the destination address and some other
parameters. The message can be produced in any text editor (or) the one built in user agent. The
destination address must be in the format that the user agent can deal with i.e., either DNS
address (or) X.400 address. Most e-mail systems support mailing list, so that a user can send the
same message to a list of people with a single command.
Reading E-mail :
When a user agent is started up, it will look at the user’s mailbox for incoming e-mail before
displaying anything on the screen. It then announces the number of messages in the mail box(or )
a one line summary of each one.
In a sophisticated system the user can specify the fields to be displayed by providing the display
format.
Eg:
1. Message numbers
2. Flag etc.
Message format:
Message consist of a primitive envelope, some number of header field, blank line followed by
message body. In normal usage, the user agent builds a message and passes it. To the message
transfer agent which then uses some of the header fields to construct the actual envelope.
Principal header include:
To :
DNS address of primary recipient.
CC :
DNS address of secondary recipient.
In terms of delivery there is no distinction between primary and secondary (carbon copies).
BCC :
Similar to CC, allows people to send copies to third parties without primary and secondary
knowing it.
From :
Who wrote the message.
Sender :
The one who sent the message.
Received :
Added by each message transfers agent along the way used for finding bugs in routing system.
Return path :
Added by final message transfer agent intended to tell how to get back to the sender etc.
Explain how e-mail works?
SMTP :

E-mail is delivered by having the source machine establish a TCP connection to destination.
Listening to this port is an E-mail daemon that speaks SMTP. This daemon accepts incoming
connections and copies messages from them to appropriate mail boxes. If the message cannot
be delivered, an error message is given.
After establishing a TCP connection, the sending machine operates as a client and waits
for receiving entity to talk first. The server starts by giving its identity and informing whether (or)
not it is prepared to receive mail. If it is not, the client releases the connection.
If the server is ready, the client announces whom the E-mail is coming from and whom it is
going to. If the recipient exists, the server gives a go-ahead to send the message. Then the client
sends the message and the server acknowledges it. When the E-mail has been exchanged then
the connection is released.
E-mail Gateways :
SMTP does not work, when both sender and receiver are not on internet. In order to overcome
this difficulty E-mail gateways are used.
Here the sender establishes a TCP connection to the gateway and then uses SMTP to transfer
the message. The daemon on the gateway then puts the message in a buffer of messages
destined for host2. Late TPU (similar to TCP) is established with host2 and the message is
transmitted.
Final Delivery :
Post Office Protocol (POP)
Used to fetch e-mail from a remote mail box, has commands for user to logon, logout, fetch and
delete messages. If fetches the mail and stores it in local system.
Interactive mail access protocol (IMAP)
This protocol is used by a person having multiple systems (office, residence, car, etc). Here the E-
mail server maintains a central repository that can be accessed from any machine. IMAP does not
copy E-mail as POP.
DOMAIN NAME SYSTEM
Generally host names, mailboxes and other resources are represented by using ASCII sting such
as rgm@vsnl.net.in.But the network itself only understands binary address i.e., the address
written in the binary form. So we need some mechanism to convert the ASCII strings to network
addresses in binary. It is easy to maintain the host names and their IP addresses in file for a
network of few hundred hosts. For a network of thousand hosts it is very difficult.
The Domain Name System, DNS is a distributes data that is used by TCP/IP application to
map between host names and IP addresses, and to provide electronic mail routing information.
We use the term distributed because no single site on the Internet knows all the information. Each
site maintains its own data base information and runs a server program that other systems
(clients) across the Internet can query. It is a good example of a TCP/IP client-server application.

The DNS provides the protocol that allows client and server to communicate with each other.
DNS is defined in RFC’s 1034 and 1035.
The DNS identifies each host on the internet with a unique name that identifies it as
unambiguously as its IP address as follows. To map a name onto an IP address, an application
program calls a library procedure called the resolver, passing it the name as a parameter. The
‘resolver’ sends a UDP packet to a local DNS server, which then looks up the name and returns
the IP address to the resolver, which then returns it to the caller. To create names that are unique
and at the same time decentralized and easy to change, the TCP/IP designers have chosen a
hierarchical system made up of a number of labels separated by dots.
THE DNS NAME SPACE
Internet is divided it several hundred top level domains, where each domain covers many hosts.
Each domain is partitioned into sub domains, these are further partitioned and so on. Thus DNS is
implemented using a tree in which each node represents one possible label of up to 63 characters.
The root of the tree is a special node with new label as shown in fig. Any comparison of label
considers uppercase and lower-case characters the same i.e., Domain names are case insensitive.
The leaves of the tree represent a company/organization and contain thousands of hosts.
Each domain is named by the path from it to the unnamed root. The components in the
name are separated by periods (dots), that is domain name of any node in the tree is the list of
labels starting at the node, working up to the root using the period (dot ) separate the labels.
The domain names that ends with a period is called an absolute domain name or fully
qualified domain name(FQDN).An example is vax.ugc,central.edu.
If domain does not end with a period, it is assumed that the name needs to be completed. How the
name is completed on the DNS software being used. If the incomplete names consist of two or
more labels, it might be considered to be complete. Otherwise, local addition might be added to the
right of the name. The name vax might be completed by adding the local suffix.ugc.central.edu.
The right most label in the name corresponds to the level of the tree closest to the root
(lowest), and left-most to the level farthest from the root(highest).The tree is divided into three
domains: generic, country and reverse as shown in fig.

Root-level
Un-named root
Second level
Domains united Arab
Emirates
Vax.ugc.control.edu ece.rmg.jntu.in
Generic domain Country
Domain
164.45.3 4.61.in-addr arpa.
Reverse
Domain
DOMAIN NAME SYSTEM
Generic Domain: The generic domain is also called the organization domain, divides registered
hosts according to their generic behaviour. Generic domain names, read left to the right , start
with the most specific information about the host(e.g. the name of the workstation) and become
more and more general with each label until they reach the rightmost label, which describes the
broadcast affiliation of the normal host i.e., the nature of the organization.
arpa com edu gov int mil net us
org ae in
In-addr
164
45
34
61
control
rgm
vax
ugc aict
e
JNTU
rgm
ece cse

The first level of the generic domain convention allows seven possible three character labels
describing organization type.
1. Com. commercial organization.
2. edu.: educational institution .
3. gov.: government institution.
4. int.: international organization.
5. mil.: military group.
6. net.: Network support center.
7. org. organizations other than listed above.
Each domain name corresponds to a particular IP address. To find the address, the resolution
application begins searching with the first level. As a much is found, a pointer leads to the next
level and finally to the associated IP address.
Country Domain: The country domain convention follows the same format as generic domain,
but uses two character country abbreviation in place of three character organizational
abbreviations at the first level shown in table. Second level labels can be organizational or they
can be more specific national designations.
Table: SOME DOMAIN NAME SYSTEM COUNTRY CODE
Country Code Country Name Country Code Country Name
AE
AU
BE
CA
CH
DE
DK
ES
FI
GR
Arubeme rates
Australia
Belgium
Canada
Switzerland
Germany
Denmark
Spain
Finland
Greece
IN
IT
JP
KW
NL
NO
NZ
SE
US
India
Italy
Japan
Kuwait
Netherlands
Norway
Newzeland
Sweden
United States of
America
Reverse Domain: If we have the IP address and need the domain name, you can reverse
domain the functions of DNS.
The domain can be inserted onto the tree in two ways. For example ugc.control.edu could equally
be listed under the country domain as cs.yale.ct.us.

To create a new domain, permission is required of the domain in which it will be included. For
example, rgm group was started under aicte and is known as rgm.aicte.control.edu. It needs
permission from which use manages aicte.control.edu. Naming follows organizational boundaries,
not physical networks.
RESOURCE RECORDS
Every domain in the DNS tree maintains a set of Resource Records, which are connected to it.
For a leaf node i.e., single host, the most common resource record is its IP address. When a
resolver gives a name to DNS, it gets back called as resource records associated with that name.
The original function of a DNS is to map domain names on to the resource records.
A resource record is a five tuple, in ASCII text they are represented as
Domain-name Time-to live type class value.
 The domain-name tells the domain to which this record belongs. This is the primary
search key used to satisfy queries.
 The time-to live field gives information regarding the stability of the record. A large value
such as 86-400(number of seconds in one day) indicates that the information is highly
stable. The small value such as 60(1 minute) indicates that the information is highly
volatile.
 The type of field tells what kind of record it is, some of the type records are listed in table
5.3.
S.No Type Meaning Value
1.
2.
3.
4.
5.
6.
7.
SoA
A
Mx
NS
CNAME
PTR
TXT
Start of Authority
IP address of a host
Mail Exchange
Name Server
Canonical name
Pointer
Text
Parameter for this zone
32 bit integer
Priority
Name of the server for this domain
Domain Name
Alias for an IP address
Uninterpreted ASCII text
1. The SOA record provides name of the primary source of information about (a) name
servers zone (b) e-mail address of its administration (c) various flags and (d) various time
outs.
2. The record A, holds a 32 bit IP address of the host. If a host connects two or more
networks, each case it has one type of a resource record per network connection.
3. The MX record specifies the name of domain prepared to accept e-mail for the specified
domain. It allows the host that is not on the internet to receive e-mail from internet sites.
4. NS record specifies Name server.

5. CNAME record specifies allows the aliases to be created.
6. PTR is a regular DNS data type whose interpretation depends on the context on which it is
found.
7. The TXT record allows domains to identify themselves in arbitrary way i.e., it is for user
convenience.
 The fourth field in the general structure of resource record is the class. It may be
Internet information, used IN and for non-internet information, other codes are
used.
 The value field can be number, domain name or an ASCII string.
NAME SERVERS
The Inter network Information center (Inter NIC) manages the top level domain names. The Inter
NIC delegates responsibility for assigning names to different organizations. Each organization is
responsible for a specific portion of the DNS tree structure. Internet professionals refer to these
areas of responsibilities as zones.
Alternatively, the Inter NIC delegates responsibility for assigning names with in a specific zone to
specific organizations. Each zone contains some part of the tree and also contains name servers
holding the authoritative information about the zone. Each zone contains one primary name
server and one or more secondary name servers. Primary name server and one or more
secondary name servers. Primary name server gets its information from a file on its disk, the
secondary name server and get their information from the primary name server. One or more
servers are located outside the zone, for each zone, for reliability. The number of name servers
needed in a zone depends on the zone boundaries.
Let us consider an example shown in fig connected with another domain. here a resolver on
“ece.rgm.jntu.in” wants to know the IP address of the host “rgm.aicte.control.edu” can be
explained in 8 steps.
Step 1: It sends a query to the local name server rgm.jntu.in.This query asks a record of type A
and the class IN.
Step 2: If the local name server had no such domain and knows nothing about it, it may ask a few
other near by name servers if none of them know, it sends a UDP packet to the server for “edu”
given in its database (see fig) edu.server.net.
Step 3: It forwards the request to the name server control.edu.

Step 4: And in turn this forwards the request aicte.control.edu, which has authoritative resource
records.
This is the request from client to a server, the resource record requested will work its way
back in step 5 to step 8.Once these records get back to rgm.jntu.in name server, they will be
entered into a cache/memory. However this information is not authoritative, since changes made
at aicte.control.edu will not be propagated to all the memories in the world. For this reason cache
should not live too long, so time-to-live field is used in each resource record. It tells the name
server how long to cache records.
WORKING OF A RESOLVER FOR A DOMAIN IN 8 STEPS
ELECTRONIC MAIL
Electronic mail or E-mail as it is popularly called, is a system that allows a person or a group to
electronically communicate with each other through a netork. Presently people can now receive
and send e-mail to:
 nearly any country in the world.
 one of millions of computer users.
 many users at once.
 computer programs.
The first e-map systems consisted of file transfer protocols, with the convention that the first line
of each message contained the recipient address. Some of the complaints at that time were
1. Sending a message to a group of people was inconvenient.
2. Messages had no internal structure, making computer processing difficult.
3. The sender never knew if a message arrived or not.
jntu.rgm edu name control name aicte name
Source domain name server server server server
Resource record
ece.rgm.
jntu.in
rgm.jntu. in
edu.server.
net
Control. edu aicte.control.
edu

4. It is difficult to forward the mails.
5. It is not possible to create and send messages containing a mixer of text, drawing
facsimile and voice.
After a decade of competition, email systems based on RFC822 are widely used, where all the
above problems are solved.
BASIC FUNCTIONS
Email systems support five basic functions, which are: Composition, Transfer, Reporting,
Displaying and Disposition.
1. Composition is a process for creating the messages and answers. This can be done by
text editor, outside the mailer, the system will provide assistance in addressing and
numerous header fields attached to each message. For eample:when answering a
message, the e mail system can extract the originator’s address from the incoming e-mail
and automatically insert it into the address space in reply.
2. Transfer refers to moving of messages from the source to the recipent. In some cases,
connection establishment is needed with the destination, outputting the message and
releasing the connection. The e-mail system should do automatically this.
3. Reporting is used to indicate the originator what happened to the message i.e.,
confirmation of the message delivery. Was it delivers successfully? Was it rejected? Was
it lost? Did errors occur?
4. Displaying It refers to read the incoming e-mail by the person. Sometimes conversion is
required or a special viewer must be invoked.
5. Disposition It concerns what the recipient does with the message after receiving it. The
possibilities are
(a) Throwing it away before reading
(b) Throwing it away after reading.
(c) Saving it and so on. It is also possible to forward them or process them in other
ways.
In addition to these basic services, most of e-mail systems provide a large variety of advanced
features such as
(a) It allows to create a mailbox to store incoming e-mail.
(b) It allows to have a mailing list, to which the e-mail messages have to send.
(c) Carbon copies, high priority email, secret email, registered email etc.

THE USER AGENT
The user agent is a program that allows users to read reply to, forward, save and compose
messages. User agents for electronic mail are sometimes called mail readers. Some user agents
have menu or icon driven interface that requires a mouse, some other requires only 1 character
command from keyboard.
Sending e-mail: To send an email message the user must provide
(a) message
(b) destination address and
(c) priority or security levels (options).
 Message can be produced with a free standing text editor, a word processing
program or by using a text editor built into the user agents. The format of an e-mail
message is similar to that of a conventional letter.
There are two main parts: Header and body.
The header contains out name and address, the name and address of the person it’s
being sent to, the name and address of the person who is being sent a copy, the date of the
message and the subject when we receive an e-mail from someone, the header tells us where it
came from, what it is about, how it was sent and when.
The body is the place where we write the contents of what we want to communicate. The
message sent should be simple and direct. Body is entirely for human recipient.
 The designation address must be in a format that the user agent can deal with. The basic
form of e-mail address is
User name @host name.subdomain.domain.
The text before the sign @(pronounced “at”) specifies the user name of the individual, the text
after the @ sign indicates how the computer system can locate that individual’s mailboxes.
For example
mvs@cs.colorado.edu
Here cs is a sub domain of Colorado is a sub domain of edu.the edu specifies the top-level
domain name.
The number of periods (pronounced as dots) varies from e-mail address.
Reading e-mail: On connecting to the net, the first thing a user usually does is check his mail, it’s
like checking the mailbox when we go home. The display like fig 5.28 appears on the screen.

Each line refers to one message. In the fig, the mailbox contains 4 (four) messages. The display
line contains several fields, which provides user profile.
AN EXAMPLE OF THE CONTENTS OF THE MAIL BOX
 The first field is the message number.
 The second field is flags, can contain,
K-means that, message was read previously and kept in mail box.
A-means the message has already answered and
F-message has been forwarded to someone else.
 The third field indicates the length of the message in bytes.
 Fourth field tells who sent the message, this field is simple extracted from the message, so
this field contains initials, log in name, first name etc.
 The last field is a ‘subject field’ gives brief summary of the message.
MESSAGE FORMATS
The e-mail message format was defined in RFC 822.There are two types: ASCII
e-mail and multimedia extensions.
ASCII e-mails using RFC 822: The e-mail message consists of a primitive envelope, some
number of header fields, a blank like and then message body.
Each header field consists of a single line of ASCII text containing the field name, a colon, and
a value of RFC.
The list of header fields related to message transport are
 A recipient’s address or “To”
 A sender’s address or “From”
 A subject.
The email header may additionally contain.
 A List of “Cd”: This is a list of e-mail or ‘carbon copies’ addresses to whom a copy of the
message is to be delivered. Multiple e-mail addresses in the “Cc” field are separated by a
comma.
S.No Flag Bytes Sender Subject
1.
2.
3.
4.
K
KA
KF
1000
2000
4000
1536
n / p
Smer
Vimicro
hiq
Got the job
Request for MP
Repair of controller
Enquiry of the book

 A List of “Bc”: This is same as “Cc” except that this is a carbon copy. The list of recipients
is not visible to the person who receives this message.
 Attached: This is a convenient method to share both data and programs. These files may
be attached or enclosed with an e-mail message.
 Signature: It contains sender’s full name and address or whatever information the sender
wishes to send.
Instead of creating a message from the scratch, we may choose to reply or forward the
messages.
 Replying: When we reply a message, the sender’s address is automatically put in the
“To” header and subject of the original message is reduced proceeded by Re, for the
reply.
 Forwarding: When we forward a message, the subject of the original message is reused,
with prefix “FW”.We must specify the e-mail address of the recipient of the forward
message.
 Redirecting: Some e-mail programs allow to redirect messages. It is similar to forwarding
a message, except that the message retains the original sender in the form header and
adds a notation that the message comes through you.
Multipurpose Internet Mail Extensions(MIME):
This is the solution defined in 1341 and updated in 1521 for the following problems.
1. Messages in languages with accents.
2. Messages in non Latin alphabets.
3. Messages in languages with out alphabets.
4. Messages not containing text at all.
The basic idea of MIME is to continue the use of RFC 822 format, but to add structure to the
message body defined encoding rules for non ASCII formats. The MIME messages can be sent
using the existing mail programs, and protocols.
The MIME defines five new message header
 MIME-Version: It tells the use agent receiving the message that it is dealing with a MIME
message, and which version of MIME it uses.
 Content-Description: It tells what is there in the message, this header helps the recipient
whether it is worth decoding and reading the message.

 Content-Transfer Encoding: It tells how the body is wrapped for transmission through a
network that may object to most characters other than letters, numbers and punctuation
marks.
 Content-Type: It specifies the nature of the message body. Seven types are defined in
RFC 1521, each of which has one or more sub types. The type and sub type are
separated by a slash. The sub type must be given explicitly in the header, no defaults are
provided. Table 5.4 shows the list of types and sub types.
TYPE AND SUB TYPE FIELDS DEFINED IN RFC 1521
S.No Type Sub Type Meaning
1. Text Plain
HTML
Rich text
Unformated text
Hyper text mark up language
Allows a simple mark up language to the
included in the text (standardized general mark
up language (SGML)
2. Image GIF
JPEG
PNG
To transmit still pictures in GIF format
To transmit still pictures in JPEG format
To transmit still pictures in portable network
graphics
3. Audio au
Basic
aiff
Sun micro systems sound
Audiable sound
Apple sound
4. Video sgi.movie
MPEG
avi
Silicon graphics movie
Visual information, the video format is
moving picture experts group MPEG
Microsoft audio video interleaved
5. Application Octet stream
Post Script
tex
It is a sequence of uninterrupted bytes
Which refers the postscript language
produced by Adobe systems and widely
used for describing printed pages.
TEX document.
6. Message RFC822
Partial
External
A MIME RFC-822 message (ASCII
characters message)
Break and encapsulated message up into
pieces and send them separately.
Used for very long message (i.e., video
films)
7. Multiport Mixed
Alternative
Parallel
Digest
Each part to be different with no additional
structure imposed
Each part must contain the same message
but expressed in a different medium or
encoding.
All parts must be viewed simultaneously
Many messages are packed together into
composite message.

MESSAGE TRANSFER
The message Transfer system, MTS is concerned with relaying messages from originator to the
recipent.The simplest way to do this is to establish a transport connection from source machine to
the destination machine and just transfer the message.
Mail servers are from the core of the e-mail infrastructure.Each recipient has a mail box, located
in one of the mail servers.A typical message starts its journey in the sender’s user agent, travels
to the sender’s main server, and then travels to the recipient mail server where it is deposited in
the recipient mail box.
A mail server needs to be running all the time, waiting for e-mail messages and routing them
approximately.If a mail server crashes or down for an extended period(3-4 days), e-mail can be
lost.There may be a limitation on the size of mail box.Generally once this limit is reached, new
incoming messages are refused until you free up space by deleting some messages.
SIMPLE MAIL TRANSFER PROTOCOL-SMTP
The simple mail transfer protocol (SMTP) is the principal application layer protocol for internet e-
mail. It is simple ASCII protocol. It uses the reliable data transfer service of TCP to transfer mail
from the sender’s mail server to the recipient’s mail server. In most application protocols SMTP
has two sides: a client side, which executes on the sender’s mail server and a server side-which
executes on the recipient mail server. When a mail server sends a mail (to other mail server), it
acts as a client SMTP.When a mail server receives a mail (from other mail server), it acts as an
SMTP server.
The SMTP defined in RF821, is at the heart of Internet e-mail.SMTP is much older than HTTP.To
illustrate the basic operation of SMTP, let’s walkthrough a common scenario. Suppose Ramu
wants to send Raju a simple ASCII message.
 Ramu invokes his user agent for e-mail, provides Raju’s e-mail address(example
Raju@some school.edu) composes a message, and instructs the user agent to send the
message.
 Ramu’s user agent sends the message to his mail server, where it is placed in a message
queue.
 The client side of SMTP, running on Ramu’s mail server, sees the message in the
message queue.It opens a TCP connection to a SMTP running Raju’s mail server.
 After some initial SMTP hand shaking, the SMTP client sends Ramu’s message into the
TCP connection.
 At Raju’s mail server host, the server side of SMTP recives the message.Raju’s mail
server then places the message in Raju’s mail box.
 Raju invokes his user agent to read the message at his convenience.
The scenario is summarized in fig.5.29

RAMU’S MAIL SERVER TRANSFERS RAMU’S MESSAGE TO RAJU’S MAIL SERVER
Let us now take closer look at how SMTP transfers a message from a sending mail server to a
receiving mail server.
We will see that the SMTP protocol has many similarities with protocols that are used for face-to-
face human interaction.
 The client SMTP has TCP to establish a connection on port 25 to server SMTP.If server is
down, the clients tries again later. Once the connection is established, the server and
client perform some application layer handshaking. During this SMTP handshaking phase,
the SMTP client indicates the e-mail address of the sender and the e-mail address of the
recipient. Once the SMTP client and server have introduced themselves to each other, the
client sends the message, SMTP can count on the reliable data transfer service of TCP to
get the message to the server without errors. The client then repeats this process over the
same TCP connection if it has other message to send to the server; otherwise it instructs
TCP to close the connection.
Even though the SMTP protocol is well defined, a few problems can still arise. These are.
1. Related to the Message Length : Some older implementations cannot handle messages
exceeding 64kB.
2. Related to Time Outs : If the client and server have different time-outs, one of them may
give up while the other is still busy, unexpectedly terminating the connection.
3. Infinite mail storms can be triggered .
To get around some of these problems, extended SMTP (ESMTP) has been defined in
RFC1425.
E-mail Gateways: E-mail using SMTP works best when both the sender and receiver on the
internet and can support TCP connections between sender and receiver.However many
SMTP
Internal
Ramu’s mail
server
Raju’s mail
server

machines that are not on the internet)because of security problem) still want to send and recive e-
mail from internet sites.
Another problem occurs when the sender speaks only RFC822 and the receiver speaks only
X.400 or some proprietary vendor specific mail protocol.
Both these problems can be solved using application layer e-mail gateways fig.5.30 shows the
gateway.
USE OF EMAIL GATEWAY
 Here Host1 speaks only TCP/IP and RFC822, where as host 2 speaks only OSITP4 ans
X.400.
They can exchange e-mail using an e-mail gateway.
Procedure:
1. Host 1 establishes a TCP connection to gateway and then use SMTP to transfer message
there.
2. The gateway then puts the message in a buffer of messages destined to host 2.
3. A TP4 connection is established between host 2 an the gateway.
4. The message is transferred using OSI equivalent of SMTA.
The problems here are
(a) The Internet address and X.400 address are totally different. Need of elaborating
mapping mechanism between them.
(b) Envelope and header fields are present in one system and are not present in the
other.
MAIL ACCESS PROTOCOL
Till now we have assumed that an users work on machines that are capable of sending and
receiving e-mail. Sometimes this situation is false. For example in an organization, users work on
desktop PCs that are no in the internet and are capable of sending and receiving e-mail from
outside. Instead the organization has one or more e-mail servers that can send and receive e-
Host 1 Gate way Message buffer Host 2
Mail Box Mail
Box
TCP connection TCP connection
Network
1 1 2 2

mail. To sned and receive e-mails, a PC must talk to an e-mail server using some kind of delivery
protocol.
There are currently two popular mail access protocols:POP3(Post office Protocol version3 ) and 1
MAP (internet mail access protocl)
POP3 : POP3 defined in RFC 1939, it is an extremely simple amil access protocol.POP3 begins
when the user agents (clients) opens a TCP connection to the mail server (the server) on port
100.With the TCP with TCP connection established, POP3 progress through three phases.
1. Authorization: The user agent sends a user name and a password to authenticate the
user downloading the mail.
2. Transaction: The user agent receives messages. In this phase the user agent can also
mark messages for deletion, remove deletion marks, and obtain mail statistics.
3. Update: During the third phase, update occurs after the client has issued the quit
command, ending the POP3 session. This time the mail server deletes the messages that
were marked for deletion.
IMAP: The Internet Mail Access Protocol (IMAP), is defined in RFC 2060.It has many features
than POP3 , but it is also significantly maore complex. It was designed to help the user whi uses
multiple computers, perhaps a workstation in the office, a PC at a home and laptop on the
road.The basic idea behind IMAP is for the e-mail server to maintain a central reposition that can
be accessed from any machine.Thus unlike POP3 , IMAP does not copy email to the user’s
personal machine because the user may have several.
The IMAP has many features.
a) It has commands that permit a user agents to obtain components of messages. This feature is
useful when there is a low bandwidth connection between the user agent and mail server.
b) An IMAP session consists of a client command, server data and a server completion result
response.
The IMAP server has four states.
1. Non Authenticated State: Initial state whenthe connection begins, the user must supply a
user name and password before most commands will be permitted.
2. Authenticated State: The user must select a folder before sending commands that affect
messages.
3. Selected State: The user can issue commands that affect messages.
4. Log Out State: Here the session is terminated.
* * * * *

REVIEW QUESTIONS
1.What is a secure communication and what are its features?
2.What is the need of cryptography?
3.Define the following terms
a. plain text b.cipher text c.encryption d.decrytion e.crypt analysis f.cryptology
4.With the help of suitable examples, explain transposition cipher and substitutional cipher.
5.Draw the block diagram of cryptography. Explain it.
6.Write about the services provided by the application layer.
6. Explain DES algorithm.
7.Explain the public key cryptography. Explain the MIT algorithm for public key encryption.
8.Distinguish between private and public key.
9. What is the diffrence between authentication protocol and digital signature?Give an example
for each.
10. What is the purpose of DNS, explain?
11. explain e-mail system
12.What are the different types of messages formats used in e-mail?
13.Write short notes on world wide web.
14.Write short notes on multimedia.
16.Write short notes on network security and privacy.
17.Using RSA algorithm with A = 1, B = 2 , ………,Z = 26
taking a.) p =5,q =11 and d =27,find ‘e’ and encrypt ‘ECE’

Quiz Questions
1.What are the properties of secure communication?
2.An intruder who can listen to and record the content and data messages on the channel is
called -----------
3.An intruder who can remove message from the channel and/or add message into the channel is
called -----------
4.The key which is available to every one is called --------------
5.The message to be encrypted is known as ----------------
6.The output of encryption process is known as --------------
7.In a------- each letter or group of letters is replaced by another letter or group of letters to
disguise it.
8.In ---------- type of cryptography, the messages do not disguise.
* * * * *

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