EC8551 COMMUNICATION NETWORKS

KARPAGAM INSTITUTE OF TECHNOLOGY,
COIMBATORE - 105
Course Code with Name : EC8551 COMMUNICATION NETWORKS
Staff Name / Designation : Dr. M.S.Gowtham /AP
Department :ECE
Year / Semester :III/05
Department of ECE
Karpagam Institute of Technology 1

COURSE SYLLABUS
EC8551 COMMUNICATION NETWORKS L T P C 3 0 0 3
UNIT I FUNDAMENTALS & LINK LAYER 9
Overview of Data Communications- Networks – Building Network and its types– Overview of Internet - Protocol Layering -
OSI Mode – Physical Layer – Overview of Data and Signals - introduction to Data Link Layer - Link layer Addressing- Error
Detection and Correction
UNIT II MEDIA ACCESS & INTERNETWORKING 9
Overview of Data link Control and Media access control - Ethernet (802.3) - Wireless LANs – Available Protocols – Bluetooth
– Bluetooth Low Energy – WiFi – 6LowPAN–Zigbee - Network layer services – Packet Switching – IPV4 Address – Network
layer protocols ( IP, ICMP, Mobile IP)
UNIT III ROUTING
Routing - Unicast Routing – Algorithms – Protocols – Multicast Routing and its basics – Overview of Intradomain and
interdomain protocols – Overview of IPv6 Addressing – Transition from IPv4 to IPv6
UNIT IV TRANSPORT LAYER 9
Introduction to Transport layer –Protocols- User Datagram Protocols (UDP) and Transmission Control Protocols (TCP) –
Services – Features – TCP Connection – State Transition Diagram – Flow, Error and Congestion Control - Congestion
avoidance (DECbit, RED) – QoS – Application requirements
UNIT V APPLICATION LAYER 9
Application Layer Paradigms – Client Server Programming – World Wide Web and HTTP - DNS- Electronic Mail (SMTP,
POP3, IMAP, MIME) – Introduction to Peer to Peer Networks – Need for Cryptography and Network Security – Firewalls.
TOTAL: 45 PERIODS

COURSE OBJECTIVES
• The student should be made:
• Understand the division of network functionalities into layers
• Be familiar with the components required to build different types of
networks
• Be exposed to the required functionality at each layer
• Learn the flow control and congestion control algorithms
• COURSE OUTCOMES (COs)
• Upon completion of the course, the students should be able to:
• Identify the components required to build different types of
networks
• Choose the required functionality at each layer for given application
• Identify solution for each functionality at each layer
• Trace the flow of information from one node to another node in the
network
3Karpagam Institute of Technology

PROGRAM SPECIFIC OUTCOMES
1. To analyze, design and develop solutions by applying
foundational concepts of electronics and
communication engineering.
2. To apply design principles and best practices for
developing quality products for scientific and
business applications.
3. To adapt to emerging information and
communication technologies (ICT) to innovate ideas
and solutions to existing/novel problems

COURSE OUTCOME Vs. PROGRAM OUTCOMES AND
PROGRAM SPECIFIC OUTCOMES MAPPING
Course Outcomes PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12 PSO2 PSO2 PSO3
C504.1 3 3 3 3 3 3 - - - - 2 2 3 3 3
C504.2 3 3 3 3 3 3 - - - - 2 1 3 3 3
C504.3 3 3 3 3 3 2 - - - - 1 1 3 3 3
C504.4 3 3 3 3 3 2 - - - - 2 2 3 3 3
C504.5 3 3 3 3 3 3 - - - - 2 2 3 3 3

UNIT 1
FUNDAMENTALS AND LINK LAYER

Definition
A communication Network is a collection of methods
that users employ to pass on valuable information. The
communication network is the sum of all the means and
methods that an organization employs to communicate.
Applications
• Most people know about the Internet (a computer
network) through applications
• World Wide Web,Email,Online Social Network,Streaming
Audio Video,File Sharing,Instant Messaging
7
Karpagam Institute of Technology

Application Protocol
• URL: Uniform resource locater
• HTTP: Hyper Text Transfer Protocol
• TCP: Transmission Control Protocol
• 17 messages for one URLrequest
• 6 to find the IP (Internet Protocol) address
• 3 for connection establishment of TCP
• 4 for HTTP request and acknowledgement
• Request: I got your request and I will send the data
• Reply: Here is the data you requested; I got the data
• 4 messages for tearing down TCP connection
8

Requirements
• Application Programmer
• List the services that his application needs: delay
bounded delivery of data
• Network Designer
• Design a cost-effective network with sharable
resources
• Network Provider
• List the characteristics of a system that is easy to
manage
9

Connectivity
Need to understand
the following
terminologies
Scale
Link
Nodes
Point-to-point
Multiple access
Switched Network
Circuit Switched
Packet Switched
Packet, message
Store-and-forward
(a) Point-to-point
(b) Multiple access
10

Connectivity
Terminologies (contd.)
Cloud
Hosts
Switches
internetwork
Router/gateway
Host-to-host connectivity
Address
Routing
Unicast/broadcast/multicast(a) A switchednetwork
(b) Interconnection of networks
(a)
(b)
11

Cost-Effective Resource Sharing
Resource: links and
nodes
How to share a link?
Multiplexing
De-multiplexing
Synchronous Time-
division Multiplexing
Time
transmitted
slots/data
in
predetermined slots
Multiplexing multiple logicalflows
over a single physical link
12

Cost-Effective Resource Sharing
FDM: Frequency
Division Multiplexing
Statistical Multiplexing
Data is transmitted based
on demand of each flow.
What is a flow?
Packets vs. Messages
FIFO, Round-Robin,
Priorities (Quality-of-
Service (QoS))
Congested?
LAN, MAN, WAN
SAN (System Area
Networks
A switch multiplexing packetsfrom
multiple sources onto one
shared link
13

Support for Common Services
Logical Channels
Application-to-Application communication path or a pipe
Process communicating over an abstractchannel
14

Common Communication Patterns
• Client/Server
• Two types of communication
channel
1. Request/Reply Channels
2. Message Stream Channels
15

Reliability
• Network should hide the errors
• Bits are lost
• Bit errors (1 to a 0, and vice versa)
• Burst errors – several consecutive errors
• Packets are lost (Congestion)
• Links and Node failures
• Messages are delayed
• Messages are delivered out-of-order
• Third parties eavesdrop
16

Network Architecture
Example of a layered network
system
Layered system with alternative
abstractions available at a given
layer
17

Protocols
Protocol defines the interfaces between the layers
in the same system and with the layers of peer
system
Building blocks of a network architecture
Each protocol object has two different interfaces
• service interface: operations on this protocol
• peer-to-peer interface: messages exchanged with peer
Term ―protocol‖ is overloaded
• specification of peer-to-peer interface
• module that implements this interface
18

Interfaces
Service and Peer Interfaces
Protocol Specification: prose, pseudo-code, state
transition diagram
19
 Interoperable: when two or more protocols
implement the specification accurately
 IETF: Internet Engineering Task Force
that

Protocol Graph
Example of a
protocol graph
nodes are the
protocols and links
the ―depends-on‖
relation
20

Encapsulation
High-level messages are encapsulated
inside of low-level messages
21

OSIArchitecture
The OSI 7-layer Model
OSI – Open Systems Interconnection
22

Description of Layers
Physical Layer
Handles the transmission of raw bits over a communication link
Data Link Layer
Collects a stream of bits into a larger aggregate called a frame
Network adaptor along with device driver in OS implement the
protocol in this layer
Frames are actually delivered to hosts
Network Layer
Handles routing among nodes within a packet-switched network
Unit of data exchanged between nodes in this layer is calleda
packet
The lower three layers are implemented on all network nodes
23

Description of Layers
Transport Layer
Implements a process-to-process channel
Unit of data exchanges in this layer is called a message
Session Layer
Provides a name space that is used to tie together the potentially
different transport streams that are part of a single application
Presentation Layer
Concerned about the format of data exchanged between peers
Application Layer
Standardize common type of exchanges
The transport layer and the higher layers typically run onlyon
end-hosts and not on the intermediate switches and routers
24

InternetArchitecture
Internet Protocol Graph
Alternative view of the
Internet architecture.The
―Network‖ layer shown
here is sometimes referred
to as the ―sub-network‖ or
―link‖layer.
25

InternetArchitecture
Defined by IETF
Three main features
1. Does not imply strict layering. The application is free to
bypass the defined transport layers and to directly use IP
or other underlying networks
2. An hour-glass shape – wide at the top, narrow in the
middle and wide at the bottom. IP serves as the focal
point for the architecture
3. In order for a new protocol to be officially included in the
architecture, there needs to be both a protocol
specification and at least one (and preferably two)
representative implementations of the specification
26

Application Programming Interface
Interface exported by the network
Since most network protocols are implemented
(those in the high protocol stack) in software and
nearly all computer systems implement their
network protocols as part of the operating
system, when we refer to the interface ―exported
by the network‖, we are generally referring to the
interface that the OS provides to its networking
subsystem
The interface is called the network Application
Programming Interface (API)
27

Application Programming Interface
(Sockets)
Socket Interface was originally provided
by the Berkeley distribution of Unix
Now supported in virtually all operating
systems
Each protocol provides a certain set of
services, and the API provides a syntax
by which those services can be invoked
in this particular OS
28

Socket
What is a socket?
The point where a local application process attaches
to the network
An interface between an application and the network
An application creates the socket
The interface defines operations for
Creating a socket
Attaching a socket to the network
Sending and receiving messages through the socket
Closing the socket
29

Socket
Socket Family
PF_INET denotes the Internet family
PF_UNIX denotes the Unix pipe facility
PF_PACKET denotes direct access to the network
interface (i.e., it bypasses the TCP/IP protocol stack)
Socket Type
SOCK_STREAM is used to denote a byte stream
SOCK_DGRAM is an alternative that denotes a
message oriented service, such as that provided by
UDP
30

Creating a Socket
int sockfd = socket(address_family, type, protocol);
The socket number returned is the socket
descriptor for the newly created socket
int sockfd = socket (PF_INET,SOCK_STREAM, 0);
int sockfd = socket (PF_INET, SOCK_DGRAM, 0);
The combination of PF_INET and
SOCK_STREAM implies TCP
31

Client Server Model with TCP
Server
Passive open
Prepares to accept connection, does not actually establish
a connection
Server invokes
int bind(int socket,struct sockaddr *address, int addr_len)
int listen (int socket, int backlog)
int accept(int socket,struct sockaddr *address,int *addr_len)
32

 Bind
Binds the newly created socket to the specified address i.e. the network
address of the local participant (the server)
Address is a data structure which combines IP and port
 Listen
Defines how many connections can be pending on the specified socket
 Accept
Carries out the passive open
Blocking operation
Does not return until a remote participant has established a
connection
When it does, it returns a new socket that corresponds to the new
established connection and the address argument contains the
remote participant’s address
33

Client
Application performs active open
It says who it wants to communicate with
Client invokes
int connect(int socket,struct sockaddr *address,int addr_len)
Connect
Does not return until TCP has successfully established a
connection at which application is free to begin sending
data
Address contains remote machine’s address
34

 In practice
The client usually specifies only remote participant’s address
and let’s the system fill in the local information
Whereas a server usually listens for messages on a well-known
port
A client does not care which port it uses for itself, the OSsimply
selects an unused one
 Once a connection is established, the application process invokes
two operation
 int send (int socket, char *msg, int msg_len, int flags);
 int recv (int socket, char *buff, int buff_len, int flags);
35

Performanc
e
Bandwidth
Width of the frequency band
Number of bits per second that can be transmitted
over a communication link
1 Mbps: 1 x 106 bits/second = 1x220 bits/sec
1 x 10-6 seconds to transmit each bit or imagine that a
timeline, now each bit occupies 1 micro second space.
On a 2 Mbps link the width is 0.5 micro second.
Smaller the width more will be transmission per unit
time.
36

Bandwidth
Bits transmitted at a particular bandwidth can be regarded
as having some width:
(a) bits transmitted at 1Mbps (each bit 1 μs wide);
(b) bits transmitted at 2Mbps (each bit 0.5 μs wide).
37

Performance
• Latency = Propagation + transmit + queue
• Propagation = distance/speed of light
• Transmit = size/bandwidth
• One bit transmission => propagation is important
• Large bytes transmission => bandwidth is important
38

Delay X Bandwidth
 We think the channel between a pair of processes as a hollow pipe
 Latency (delay) length of the pipe and bandwidth the width of the pipe
 Delay of 50 ms and bandwidth of 45 Mbps
 50 x 10-3 seconds x 45 x 106 bits/second
 2.25 x 106 bits = 280 KB data.
Network as a pipe
 Relative importance of bandwidth and latency depends on application
For large file transfer, bandwidth is critical
For small messages (HTTP, NFS, etc.), latency is critical
Variance in latency (jitter) can also affect some applications (e.g.,
audio/video conferencing)
39

Delay X Bandwidth
How many bits the sender must transmit before the first bit arrives
at the receiver if the sender keeps the pipe full
Takes another one-way latency to receive a response from the
receiver
If the sender does not fill the pipe—send a whole delay
bandwidth product’s worth of data before it stops to wait for a
signal—the sender will not fully utilize the network
 Infinite bandwidth
RTT dominates
Throughput = TransferSize / TransferTime
TransferTime = RTT + 1/Bandwidth x TransferSize
 Its all relative
1-MB file to 1-Gbps link looks like a 1-KB packet to 1-Mbps link
40

Relationship bet. bandwidth & latency
A 1-MB file would fill the 1-Mbps link 80 times, but only fill the 1-Gbpslink
1/12 of one time
41

Link layer Services
42

Perspectives on Connecting
An end-user’s view of the Internet
43

Link Capacity & Shannon-Hartley
Theorem
Gives the upper bound to the capacity of a link in
terms of bits per second (bps) as a function of
signal-to-noise ratio of the link measured in
decibels (dB).
C = Blog2(1+S/N)
Where B = 3300 – 300 = 3000Hz, S is the signal power,
N the average noise.
The signal to noise ratio (S/N) is measured in decibels is
related to dB = 10 x log10(S/N). If there is 30dB of
noise then S/N = 1000.
Now C = 3000 x log2(1001) = 30kbps.
How can we get 56kbps?
44

Links
 All practical links rely on some sort of electromagnetic
radiation propagating through a medium or, in some cases,
through free space
 One way to characterize links, then, is by the medium they use
Typically copper wire in some form (as in Digital Subscriber
Line (DSL) and coaxial cable),
Optical fiber (as in both commercial fiber-to-the home
services and many long-distance links in the Internet’s
backbone), or
Air/free space (for wireless links)
 Another important link characteristic is the frequency
Measured in hertz, with which the electromagnetic waves
oscillate
45

Links
 Distance between the adjacent pair of maxima or minima of a wave
measured in meters is called wavelength
 Speed of light divided by frequency gives the wavelength.
 Frequency on a copper cable range from 300Hz to 3300Hz; Wavelength for
300Hz wave through copper is speed of light on a copper /frequency
 2/3 x 3 x 108 /300 = 667 x 103meters.
Electromagnetic
spectrum
46

Links
Common services available to connect yourhome
Placing binary data on a signal is called encoding.
Modulation involves modifying the signals in terms of
their frequency, amplitude, and phase.
47

Encoding
Signals travel between signaling components; bits flow between adaptors
NRZ encoding of a bit stream
48

Encoding
Problem with NRZ
Baseline wander
The receiver keeps an average of the signals it has seen so far
Uses the average to distinguish between low and high signal
When a signal is significantly low than the average, it is 0, else it is1
Too many consecutive 0’s and 1’s cause this average to change, making it
difficult to detect
Clock recovery
Frequent transition from high to low or vice versa are necessary toenable
clock recovery
Both the sending and decoding process is driven by a clock
Every clock cycle, the sender transmits a bit and the receiver recovers a bit
The sender and receiver have to be precisely synchronized
49

Encoding
NRZI
Non Return to Zero Inverted
Sender makes a transition from the current signal to encode 1
and stay at the current signal to encode 0
Solves for consecutive 1’s
Manchester encoding
Merging the clock with signal by transmitting Ex-OR of the
NRZ encoded data and the clock
Clock is an internal signal that alternates from low to high, a
low/high pair is considered as one clock cycle
In Manchester encoding
0: low high transition
1: high low transition
5
0

Encoding
Different encoding strategies
51

Encoding
 Problem with Manchester encoding
 Doubles the rate at which the signal transitions are made on thelink
Which means the receiver has half of the time to detect each pulse of the
signal
 The rate at which the signal changes is called the link’s baud rate
 In Manchester the bit rate is half the baud rate
 4B/5B encoding
 Insert extra bits into bit stream so as to break up the long sequence of 0’s and
1’s
 Every 4-bits of actual data are encoded in a 5- bit code that is transmitted to
the receiver
 5-bit codes are selected in such a way that each one has no more than one
leading 0(zero) and no more than two trailing0’s.
 No pair of 5-bit codes results in more than three consecutive0’s
52

Encoding
 4B/5B encoding
• 0000  11110
• 0001  01001
• 0010  10100
• ..
• ..
• 1111  11101
16 left
11111 – when the line is idle
00000 – when the line is dead
00100 – to mean halt
13 left : 7 invalid, 6 for various
control signals
53

Framing
 We are focusing on packet-switched networks, which means that
blocks of data (called frames at this level), not bit streams, are
exchanged between nodes.
 It is the network adaptor that enables the nodes to exchange
frames.
 When node A wishes to transmit a frame to node B, it tells its
adaptor to transmit a frame from the node’s memory. This results in
a sequence of bits being sent over the link.
Bits flow between adaptors,
frames between hosts
54

Framing
The adaptor on node B then collects together the
sequence of bits arriving on the link and deposits the
corresponding frame in B’s memory.
Recognizing exactly what set of bits constitute a frame—
that is, determining where the frame begins and ends—is
the central challenge faced by the adaptor
Byte-oriented Protocols
To view each frame as a collection of bytes (characters) rather
than bits
BISYNC (Binary Synchronous Communication) Protocol
Developed by IBM (late 1960)
DDCMP (Digital Data Communication Protocol)
Used in DECNet
55

Framing
 BISYNC – sentinel approach
 Frames transmitted beginning with leftmost field
 Beginning of a frame is denoted by sending a special SYN (synchronize)
character
 Data portion of the frame is contained between special sentinel characterSTX
(start of text) and ETX (end of text)
 SOH : Start of Header DLE : Data Link Escape CRC: Cyclic Redund.Check
 Recent PPP which is commonly run over Internet links uses sentinelapproach
 Special start of text character denoted as Flag
0 1 1 1 1 1 1 0
 Address, control : default numbers
 Protocol for demux : IP / IPX
 Payload : negotiated (1500 bytes)
 Checksum : for error detection
55
5
6

Framing
 Byte-counting approach
DDCMP
count : how many bytes are contained in the frame body
If count is corrupted
Framing error
 Bit-oriented Protocol
HDLC : High Level Data Link Control
Beginning and Ending Sequences
0 1 1 1 1 1 1 0
On the sending side, any time five consecutive 1’s have been
transmitted from the body of the message (i.e. excluding when the
sender is trying to send the distinguished 01111110sequence)
The sender inserts 0 before transmitting the nextbit
57

Framing
HDLC Protocol
On the receiving side
5 consecutive 1’s
Next bit 0 : Stuffed, so discard it
1 : Either End of the frame marker Or Error has been
introduced in the bitstream
 Look at the next bit
 If 0 ( 01111110 )  End of the frame marker
 If 1 ( 01111111 )  Error, discard the whole frame
The receiver needs to wait for next 01111110 before it can start
receiving again
58

Error Detection
 Bit errors are introduced into frames
Because of electrical interference and thermal noises
 Detecting Error and Correction Error
 Two approaches when the recipient detects an error
Notify the sender that the message was corrupted, so the sender can send
again.
If the error is rare, then the retransmitted message will be error-free
Using some error correct detection and correction algorithm, the receiver
reconstructs the message
 Common technique for detecting transmission error
CRC (Cyclic Redundancy Check)
Used in HDLC, DDCMP, CSMA/CD, Token Ring
Other approaches
Two Dimensional Parity (BISYNC)
Checksum (IP) 55
5
9

Error Detection
 Basic Idea of Error Detection
To add redundant information to a frame that can be used to
determine if errors have been introduced
Imagine (Extreme Case)
Transmitting two complete copies of data
Identical  No error
Differ  Error
Poor Scheme ???
n bit message, n bit redundant information
Error can go undetected
In general, we can provide strong error detection technique
k redundant bits, n bits message, k << n
In Ethernet, a frame carrying up to 12,000 bits of data
requires only 32-bit CRC 59
5
6

Error Detection
 Extra bits are redundant
 They add no new information to the message
 Derived from the original message using some algorithm
 Both the sender and receiver know the algorithm
Sender Receiver
 Receiver computes r using m
 If they match, no error
m rm r
61

Two-dimensional parity
Two-dimensional parity is exactly what the name suggests
It is based on ―simple‖ (one-dimensional) parity, which
usually involves adding one extra bit to a 7-bit code to
balance the number of 1s in the byte. For example,
Odd parity sets the eighth bit to 1 if needed to give an odd
number of 1s in the byte, and
Even parity sets the eighth bit to 1 if needed to give an even
number of 1s in the byte
Two-dimensional parity does a similar calculation
for each bit position across each of the bytes
contained in the frame
62

Two-dimensional parity
This results in an extra parity
byte for the entire frame, in
addition to a parity bit for each
byte
Two-dimensional parity catches
all 1-, 2-, and 3-bit errors and
most 4-bit errors
63

Internet ChecksumAlgorithm
Not used at the link level
Add up all the words that are transmitted and then transmit
the result of that sum
The result is called the checksum
The receiver performs the same calculation on the received
data and compares the result with the received checksum
If any transmitted data, including the checksum itself, is
corrupted, then the results will not match, so the receiver
knows that an error occurred
Consider the data being checksummed as a sequence of 16-
bit integers.
64

Add them together using 16-bit ones complement
arithmetic (explained next slide) and then take the ones
complement of the result.
That 16-bit number is the checksum
In ones complement arithmetic, a negative integer −x is
represented as the complement of x;
Each bit of x is inverted.
When adding numbers in ones complement arithmetic, a
carryout from the most significant bit needs to be added
to the result.
65

Consider, for example, the addition of −5 and −3 in
ones complement arithmetic on 4-bit integers
+5 is 0101, so −5 is 1010; +3 is 0011, so −3 is 1100
If we add 1010 and 1100 ignoring the carry, we get
0110
In ones complement arithmetic, the fact that this
operation caused a carry from the most significant bit
causes us to increment the result, giving 0111, which is
the ones complement representation of −8 (obtained by
inverting the bits in 1000), as we would expect
66

Cyclic Redundancy Check (CRC) Reduce the number of extra bits and maximize protection
 Given a bit string 110001 we can associate a polynomial on a single
variable x for it.
1.x5+1.x4+0.x3+0.x2+0.x1+1.x0 = x5+x4+1 and the degree is5.
A k-bit frame has a maximum degree of k-1
 Let M(x) be a message polynomial and C(x) be a generator
polynomial.
 Let M(x)/C(x) leave a remainder of 0.
 When M(x) is sent and M’(x) is received we have M’(x) =
M(x)+E(x)
 The receiver computes M’(x)/C(x) and if the remainder is nonzero,
then an error has occurred.
 The only thing the sender and the receiver should know is C(x).
67
6
7

Cyclic Redundancy Check (CRC)
 Polynomial Arithmetic Modulo2
 Any polynomial B(x) can be divided by a divisor polynomial C(x) if B(x) is
of higher degree than C(x).
 Any polynomial B(x) can be divided once by a divisor
polynomial C(x) if B(x) is of the same degree as C(x).
 The remainder obtained when B(x) is divided by C(x) is obtained by
subtracting C(x) from B(x).
 To subtract C(x) from B(x), we simply perform the exclusive- OR (XOR)
operation on each pair of matching coefficients.
 Let M(x) be a frame with m bits and let the generator polynomial have less than
m bits say equal to r.
 Let r be the degree of C(x). Append r zero bits to the low-order end of the
frame, so it now contains m+r bits and corresponds to the polynomial xrM(x).
68
6
8

Divide the bit string corresponding to xrM(x) by the bit s
corresponding to C(x) using modulo 2 division.
Subtract the remainder (which is always r or fewer bits) from
xrM(x) using modulo
subtraction (addition
the string corresponding to
2
and
subtraction are the same in
modulo 2).
 The result
checksummed
transmitted. Call
is the
frame to be
it
polynomial M’(x).
CRC Calculation using
Polynomial Long Division
656
9

Cyclic Redundancy Check (CRC) Properties of Generator Polynomial
 Let P(x) represent what the sender sent and P(x) + E(x) is the received string. A 1 in
E(x) represents that in the corresponding position in P(x) the message the bit is flipped.
 We know that P(x)/C(x) leaves a remainder of 0, but if E(x)/C(x) leaves a
remainder of 0, then either E(x) = 0 or C(x) is factor ofE(x).
 When C(x) is a factor of E(x) we have problem; errors gounnoticed.
 If there is a single bit error then E(x) = xi, where i determines the bit in error. If C(x)
contains two or more terms it will never divide E(x), so all single bit errors will be
detected.
 In general, it is possible to prove that the following types of errors can be detected by a C(x)
with the stated properties
 All single-bit errors, as long as the xk and x0 terms have nonzerocoefficients.
 All double-bit errors, as long as C(x) has a factor with at least threeterms.
 Any odd number of errors, as long as C(x) contains the factor(x+1).
 Any ―burst‖error (i.e., sequence of consecutive error bits) for which the length of the
burst is less than k bits. (Most burst errors of larger than k bits can also
69
7
0

Six generator polynomials that have become
international standards are:
CRC-8 = x8+x2+x+1
CRC-10 = x10+x9+x5+x4+x+1
CRC-12 = x12+x11+x3+x2+x+1
CRC-16 = x16+x15+x2+1
CRC-CCITT = x16+x12+x5+1
CRC-32 =
x32+x26+x23+x22+x16+x12+x11+x10+x8+x7+x5+x4+x2+x+1
71

Reliable Transmission
 CRC is used to detect errors.
 Some error codes are strong enough to correct errors.
 The overhead is typically too high.
 Corrupt frames must be discarded.
 A link-level protocol that wants to deliver frames reliably must
recover from these discarded frames.
 This is accomplished using a combination of two fundamental
mechanisms
Acknowledgements and Timeouts
 An acknowledgement (ACK for short) is a small control frame that
a protocol sends back to its peer saying that it has received the
earlier frame.
A control frame is a frame with header only (no data).
72

Reliable Transmission
The receipt of an acknowledgement indicates to the
sender of the original frame that its frame was
successfully delivered.
If the sender does not receive an acknowledgment after a
reasonable amount of time, then it retransmits the original
frame.
The action of waiting a reasonable amount of time is
called a timeout.
The general strategy of using acknowledgements and
timeouts to implement reliable delivery is sometimes
called Automatic Repeat reQuest (ARQ).
73

Stop and Wait Protocol
Idea of stop-and-wait
protocol is
straightforward
After transmitting one
frame, the sender waits
for an acknowledgement
before transmitting the
next frame.
If the acknowledgement
does not arrive after a
certain period of time,
the sender times out and
retransmits the original
frame
Timeline showing four different scenarios for the stop-
and-wait algorithm. (a) The ACK is received before the
timer expires; (b) the original frame is lost; (c) the
ACK is lost; (d) the timeout fires too soon
7
4

 If the acknowledgment is lost or delayed
in arriving
 The sender times out and retransmits the
original frame, but the receiver will think
that it is the next frame since it has correctly
received and acknowledged the first frame
 As a result, duplicate copies of frames will
be delivered
 How to solve this
 Use 1 bit sequence number (0 or 1)
 When the sender retransmits frame 0, the
receiver can determine that it is seeing a
second copy of frame 0 rather than the first
copy of frame 1 and therefore can ignore it
(the receiver still acknowledges it, in case
the first acknowledgement was lost)
Timeline for stop-and-wait
with 1-bit sequence
number
75

 The sender has only one outstanding frame on the link at a
time
 This may be far below the link’s capacity
 Consider a 1.5 Mbps link with a 45 ms RTT
 The link has a delay bandwidth product of 67.5 Kb or approximately
8 KB
 Since the sender can send only one frame per RTT and assuming a
frame size of 1 KB
 Maximum Sending rate
 Bits per frame Time per frame = 1024 8 0.045 = 182 Kbps
 Or about one-eighth of the link’s capacity
 To use the link fully, then sender should transmit up to eight frames
before having to wait for an acknowledgement
76

Sliding Window Protocol
Timeline for Sliding Window Protocol
 Sender assigns a sequence
number denoted as SeqNum to
each frame.
 Assume it can grow infinitely large
 Sender maintains three variables
 Sending Window Size (SWS)
Upper bound on the number of
outstanding (unacknowledged)
frames that the sender can transmit
 Last AcknowledgementReceived
(LAR)
Sequence number of the last
77
acknowledgement received
 Last Frame Sent (LFS)
Sequence number of the last frame
sent

 Sender also maintains the following invariant
 LFS – LAR ≤ SWS
Sliding Window
on Sender
 When an acknowledgement arrives
 the sender moves LAR to right, thereby allowing the sender to transmit
another frame
 Also the sender associates a timer with each frame it transmits
 It retransmits the frame if the timer expires before the ACK isreceived
 Note that the sender has to be willing to buffer up to SWS
frames
 WHY?
78

 Receiver maintains three variables
 Receiving Window Size (RWS)
Upper bound on the number of out-of-order frames that the receiver is willing to
accept
 Largest Acceptable Frame (LAF)
Sequence number of the largest acceptable frame
 Last Frame Received (LFR)
Sequence number of the last frame received
 Receiver also maintains the following invariant
LAF – LFR ≤ RWS
Sliding Window
on Receiver
79

 When a frame with sequence number SeqNum arrives, what does the
receiver do?
 If SeqNum ≤ LFR or SeqNum > LAF
Discard it (the frame is outside the receiver window)
 If LFR < SeqNum ≤ LAF
Accept it
Now the receiver needs to decide whether or not to send anACK
 Let SeqNumToAck
Denote the largest sequence number not yet acknowledged, such that all frames
with sequence number less than or equal to SeqNumToAck have been received
 The receiver acknowledges the receipt of SeqNumToAck even if high-
numbered packets have been received
This acknowledgement is said to be cumulative.
 The receiver then sets
LFR = SeqNumToAck and adjusts
LAF = LFR + RWS
8
0

Issues with Sliding Window Protocol
 When timeout occurs, the amount of data in transit decreases
 Since the sender is unable to advance its window
 When the packet loss occurs, this scheme is no longer keeping the
pipe full
 The longer it takes to notice that a packet loss has occurred, the more severe
the problem becomes
 How to improve this
 Negative Acknowledgement(NAK)
 AdditionalAcknowledgement
 SelectiveAcknowledgement
 Negative Acknowledgement (NAK)
 Receiver sends NAK for frame 6 when frame 7 arrive (in the previous example)
 However this is unnecessary since sender’s timeout mechanism will be sufficient to catch
the situation
 AdditionalAcknowledgement
 Receiver sends additional ACK for frame 5 when frame 7 arrives
 Sender uses duplicate ACK as a clue for frame loss
8
1

References
• TEXT BOOK:
• T1: Behrouz A. Forouzan, ―“Data communication and
Networking”, Fifth Edition, Tata McGraw – Hill, 2013 (UNIT I –V)
• REFERENCES:
• R1 James F. Kurose, Keith W. Ross, “Computer Networking - A
Top-Down Approach Featuring the Internet”,
• seventh Edition, Pearson Education, 2016.
• R2 Nader. F. Mir, “Computer and Communication Networks”,
Pearson Prentice Hall Publishers, 2nd Edition, 2014.
• R3 Ying-Dar Lin, Ren-Hung Hwang, Fred Baker, “Computer
Networks: An Open Source Approach”, Mc Graw Hill
• Publisher, 2011.
• R4 Larry L. Peterson, Bruce S. Davie, “Computer Networks: A
Systems Approach”, Fifth Edition, Morgan Kaufmann
• Publishers, 2011.
8
2

SUMMARY
• The data link layer is the protocol layer in a program that handles the
moving of data into and out of a physical link in a network.
• Data bits are encoded, decoded and organized in the data link layer,
before they are transported as frames between two adjacent nodes
on the same LAN or WAN

ASSIGNMENT QUESTIONS
S.No. Topic K Level
1. Analyze the cyclic codes to find its capabilities by using
polynomials.
K3
2. Evaluate the following:
(i) Hardware Implementation of Cyclic codes
(ii) Checksum
K2
8
4

Video Tutorials
S.No. Topic LINK
1. Network Protocol Layers https://www.youtube.com/
watch?v=_9QayISruzo
2 Error Detection and Correction https://www.youtube.com/
watch?v=aNqiTCZ-nko
8
5

QUIZ QUESTIONS1.Several computers linked to a server to share programs and storage space.
A. Library
B. Network
C. Grouping
D. Integrated system
2.Which of the following device is used to connect two systems, especially if the
systems use different protocols?
A. Repeater
B. Gateway
C. Bridge
D. Hub
3.The collection of links throughout the Internet creates an interconnected
network called the
A. WWW
B. Web
8
6

4.Network architecture has a stack of layers. Which of the following is not true for
this architecture?
A. Different layers can be developed separately
B. Layers internals are independent
C. Network protocols cannot work with multiple layers
D. A device can handle one or more layers as per requirement
5.Who defines the Internet architecture?
A. IETF
B. IEEE
C. ACM
D. None
6.What are the end network devices?
A. Servers, smartphones and computers
B. Transceivers, NICs
C. Routers and servers
D. Smartphones, cell towers
8
7

7.P2P is a _____ application architecture.
A. Client/server
B. Distributed
C. Centralized
D. 1-tier
8._______ is the most important/powerful computer in a typical network.
A. Desktop
B. Network server
C. Network client
D. Network switch
9. IP is responsible for packet switching
True B. False
10. OSI Model consists of _______layers
A.5 B.7 C.9 D.6 8
8

11. Which network topology requires a central controller or hub?
a) Star
b) Mesh
c) Ring
d) Bus
Answer: a
12. Data communication system spanning states, countries, or the whole world is ________
a) LAN
b) WAN
c) MAN
d) PAN
Answer: b
13. _____ is the multiplexing technique that shifts each signal to a different carrier frequency.
a) FDM
b) TDM
c) Both FDM & TDM
d) PDM
Answer: a
14. Which multiplexing technique used to transmit digital signals?
a) FDM
b) TDM
c) WDM
d) FDM & WDM
Answer: b
15. Which of the following delay is faced by the packet in travelling from one end system to another?
a) Propagation delay
b) Queuing delay
c) Transmission delay
d) All of the mentioned
Answer: d

17. Which of the following is an example of a bounded medium?
(A) coaxial cable
(B) wave guide
(C) fiber optic cable
(D) all of these
Answer . D
18. LAN can use ________ architecture.
A. Client and server
B. Peer–to–peer
C. Both a and b
D. Neither A or B
ANS: C
19. Which sublayer of the data link layer performs data link functions that depend upon the type of medium?
a) logical link control sublayer
b) media access control sublayer
c) network interface control sublayer
d) error control sublayer
Answer: b
20. Transport layer aggregates data from different applications into a single stream before passing it to ____________
a) network layer
b) data link layer
c) application layer
d) physical layer
Answer: a

22.Which of the following transport layer protocolss is used to support electronic mail?
(A) SMTP
(B) IP
(C) TCP
(D) UDP
Answer :(C)
23. Which address is used to identify a process on a host by the transport layer?
a) physical address
b) logical address
c) port address
d) specific address
Answer: c
24.Bits can be sent over guided and unguided media as analog signal by ___________
a) digital modulation
b) amplitude modulation
c) frequency modulation
d) phase modulation
Answer: a
25.The technique of temporarily delaying outgoing acknowledgements so that they can be hooked onto the next outgoing data frame is called ____________
a) piggybacking
b) cyclic redundancy check
c) fletcher’s checksum
d) parity check
Answer: a
26.Multiplexing is used in _______
a) Packet switching
b) Circuit switching
c) Data switching
d) Packet & Circuit switching
Answer: b
27.Fiber optics posses following properties __________
a) Immune electromagnetic interference
b) Very less signal attenuation
c) Very hard to tap
d) All of the mentioned
Answer: d

46. What is the benefit of the Networking?
A. File Sharing
B. Easier access to Resources
C. Easier Backups
D. All of the Above
Answer: D
47. Which of the following is the fastest communication channel?
A. Micro wave
B. Optical fiber
C. Radio wave
D. None of the above
Answer: A
48. Encryption and decryption are the function of
A. Session layer
B. Presentation layer
C. Transport layer
D. None of the above
Answer: B
49. An inter–company network which used to distribute information documents files and database is called as
A. MAN
B. WAN
C. LAN
D. Switch
Answer: C Karpagam Institute of Technology 92

Unit - II
MEDIAACCESS &INTERNETWORKING

Ethernet
 Most successful local area networking technology of last 20 years.
 Developed in the mid-1970s by researchers at the Xerox Palo Alto
Research Centers (PARC).
 Uses CSMA/CD technology
 Carrier Sense Multiple Access with CollisionDetection.
 A set of nodes send and receive frames over a sharedlink.
 Carrier sense means that all nodes can distinguish between an idle and a busy
link.
 Collision detection means that a node listens as it transmits and can therefore
detect when a frame it is transmitting has collided with a frame transmitted by
another node.
 Uses ALOHA (packet radio network) as the rootprotocol
 Developed at the University of Hawaii to support communication across the
Hawaiian Islands.
 For ALOHA the medium was atmosphere, for Ethernet the medium is a coax
cable.
94

Ethernet
 DEC and Intel joined Xerox to define a 10-Mbps Ethernet
standard in 1978.
 This standard formed the basis for IEEE standard 802.3
 More recently 802.3 has been extended to include a 100-Mbps
version called Fast Ethernet and a 1000-Mbps version called
Gigabit Ethernet.
 An Ethernet segment is implemented on a coaxial cable of up to 500 m.
 This cable is similar to the type used for cable TV except that it typically
has an impedance of 50 ohms instead of cable TV’s 75 ohms.
 Hosts connect to an Ethernet segment by tapping into it.
 A transceiver (a small device directly attached to the tap) detects when
the line is idle and drives signal when the host is transmitting.
 The transceiver also receives incoming signal.
95

Ethernet
 The transceiver is connected to an
Ethernet adaptor which is plugged
into the host.
 The protocol is implemented on the
adaptor.
 Multiple Ethernet segments can
be joined together by repeaters.
 A repeater is a device that
forwards digital signals.
 No more than four repeaters may
be positioned between any pair of
hosts.
 An Ethernet has a total reach of only
2500 m.
Ethernet transceiver and adaptor
96

Ethernet
Ethernet repeater
Any signal
97
placed on the
host isby a
over the entire
Ethernet
broadcast
network
Signal is propagated in both
directions.
Repeaters forward the signal
on all outgoing segments.
Terminators attached to the
end of each segment absorb
the signal.
Ethernet uses Manchester
encoding scheme.

Ethernet
New Technologies in Ethernet
 Instead of using coax cable, an Ethernet can be constructed from athinner
cable known as 10Base2 (the original was 10Base5)
10 means the network operates at 10 Mbps
Base means the cable is used in a baseband system
2 means that a given segment can be no longer than 200 m
 Another cable technology is 10BaseT
T stands for twisted pair
Limited to 100 m in length
 With 10BaseT, the common configuration is to have several point to point
segments coming out of a multiway repeater, called Hub
Ethernet Hub
98

Access Protocol for Ethernet
 The algorithm is commonly called Ethernet’s Media Access Control
(MAC).
 It is implemented in Hardware on the network adaptor.
 Frame format
 Preamble (64bit): allows the receiver to synchronize with the signal
(sequence of alternating 0s and 1s).
 Host and Destination Address (48biteach).
 Packet type (16bit): acts as demux key to identify the higher level protocol.
 Data (up to 1500 bytes)
Minimally a frame must contain at least 46 bytes of data.
Frame must be long enough to detect collision.
 CRC (32bit)
99

Ethernet Addresses
 Each host on an Ethernet (in fact, every Ethernet host in the world)
has a unique EthernetAddress.
 The address belongs to the adaptor, not the host.
 It is usually burnt into ROM.
 Ethernet addresses are typically printed in a human readable format
 As a sequence of six numbers separated by colons.
 Each number corresponds to 1 byte of the 6 byte address and is given by a pair
of hexadecimal digits, one for each of the 4-bit nibbles in thebyte
 Leading 0s are dropped.
 For example, 8:0:2b:e4:b1:2 is
00001000 00000000 00101011 11100100 10110001 00000010
 To ensure that every adaptor gets a unique address, each
manufacturer of Ethernet devices is allocated a different prefix that
must be prepended to the address on every adaptor they build
AMD has been assigned the 24bit prefix 8:0:20
100

Ethernet Addresses
 Each frame transmitted on an Ethernet is received by every adaptor
connected to that Ethernet.
 Each adaptor recognizes those frames addressed to its address and
passes only those frames on to the host.
 In addition, to unicast address, an Ethernet address consisting of all
1s is treated as a broadcast address.
 All adaptors pass frames addressed to the broadcast address up to the host.
 Similarly, an address that has the first bit set to 1 but is not the
broadcast address is called a multicast address.
 A given host can program its adaptor to accept some set of
multicast addresses. To summarize, an Ethernet adaptor receives all
frames and accepts
 Frames addressed to its own address
 Frames addressed to the broadcast address
 Frames addressed to a multicast addressed if it has been instructed
101

Ethernet TransmitterAlgorithm
 When the adaptor has a frame to send and the line is idle, it transmits
the frame immediately.
 The upper bound of 1500 bytes in the message means that the adaptor can
occupy the line for a fixed length of time.
 When the adaptor has a frame to send and the line is busy, it waits
for the line to go idle and then transmits immediately.
 The Ethernet is said to be 1-persistent protocol because an adaptor
with a frame to send transmits with probability 1 whenever a busy
line goes idle.
 Since there is no centralized control it is possible for two (or more)
adaptors to begin transmitting at the same time,
 Either because both found the line to be idle,
 Or, both had been waiting for a busy line to become idle.
 When this happens, the two (or more) frames are said to be collide
on the network.
102

 Since Ethernet supports collision detection, each sender is able
to determine that a collision is in progress.
 At the moment an adaptor detects that its frame is colliding
with another, it first makes sure to transmit a 32-bit jamming
sequence and then stops transmission.
 Thus, a transmitter will minimally send 96 bits in the case of collision
64-bit preamble + 32-bit jamming sequence
 One way that an adaptor will send only 96 bit (called a runt
frame) is if the two hosts are close to each other.
 Had they been farther apart,
 They would have had to transmit longer, and thus send more bits,
before detecting the collision.
103

The worst case scenario happens when the two hosts are at
opposite ends of the Ethernet.
To know for sure that the frame its just sent did not collide
with another frame, the transmitter may need to send as
many as 512 bits.
Every Ethernet frame must be at least 512 bits (64 bytes) long.
14 bytes of header + 46 bytes of data + 4 bytes of CRC
Why 512 bits?
Why is its length limited to 2500 m?
The farther apart two nodes are, the longer it takes for a
frame sent by one to reach the other, and the network is
vulnerable to collision during this time
104

 A begins transmitting a frame at time t
 d denotes the one link latency
 The first bit of A’s frame arrives at B at
time t + d
 Suppose an instant before host A’s frame
arrives, host B begins to transmit its own
frame
 B’s frame will immediately collide with A’s
frame and this collision will be detected by
host B
 Host B will send the 32-bit jamming
sequence
 Host A will not know that the collision
occurred until B’s frame reaches it, which
will happen at t + 2 * d
 Host A must continue to transmit until this
time in order to detect the collision
Host A must transmit for 2 * d to be
sure that it detects all possible
collisions
Worst-case scenario: (a) A sends a frame
at time t; (b) A’sframe arrives at B attime
t + d; (c) B begins transmitting at time t +
d and collides with A’sframe; (d) B’s runt
(32-bit) frame arrives at Aat time t + 2d.
14Karpagam Institute of Technology 105

Consider that a maximally configured Ethernet is 2500 m
long, and there may be up to four repeaters between any two
hosts, the round trip delay has been determined to be 51.2
s
Which on 10 Mbps Ethernet corresponds to 512 bits
The other way to look at this situation,
We need to limit the Ethernet’s maximum latency to a fairly small
value (51.2 s) for the access algorithm to work
Hence the maximum length for the Ethernet is on the order of 2500m.
 Once an adaptor has detected a collision, and stopped its transmission,
it waits a certain amount of time and tries again.
 Each time the adaptor tries to transmit but fails, it doubles the amount
of time it waits before trying again.

 This strategy of doubling the delay interval between each
retransmission attempt is known as Exponential Backoff.
 The adaptor first delays either 0 or 51.2 s, selected at random.
s (selected If this effort fails, it then waits 0, 51.2, 102.4, 153.6
randomly) before trying again;
 This is k * 51.2 for k = 0, 1, 2, 3
 After the third collision, it waits k * 51.2 for k = 0…23 – 1 (again
selected at random).
 In general, the algorithm randomly selects a k between 0 and 2n –
1 and waits for k * 51.2 s, where n is the number of collisions
experienced so far.

Experience with Ethernet
 Ethernets work best under lightly loaded conditions.
 Under heavy loads, too much of the network’s capacity is wasted by
collisions.
 Most Ethernets are used in a conservative way.
 Have fewer than 200 hosts connected to them which is far fewer than the
maximum of 1024.
 Most Ethernets are far shorter than 2500m with a round-trip
delay of closer to 5 s than 51.2 s.
 Ethernets are easy to administer and maintain.
 There are no switches that can fail and no routing and configuration tables
that have to be kept up-to-date.
 It is easy to add a new host to the network.
 It is inexpensive.
 Cable is cheap, and only other cost is the network adaptor on each host.

Wireless Links
 Wireless links transmit electromagnetic signals
 Radio, microwave, infrared
 Wireless links all share the same “wire” (so to speak)
 The challenge is to share it efficiently without unduly interfering with eachother
 Most of this sharing is accomplished by dividing the “wire” alongthe
dimensions of frequency and space
 Exclusive use of a particular frequency in a particular geographicarea
may be allocated to an individual entity such as a corporation
 Devices that use license-exempt frequencies are still subject to certain
restrictions
 The first is a limit on transmission power
 This limits the range of signal, making it less likely to interfere with another
signal
 For example, a cordless phone might have a range of about 100 feet.

Wireless Links
These allocations are determined by government agencies
such as FCC (Federal Communications Commission) in
USA
Specific bands (frequency) ranges are allocated to certain
uses.
Some bands are reserved for government use
Other bands are reserved for uses such as AM radio, FM radio,
televisions, satellite communications, and cell phones
Specific frequencies within these bands are then allocated to
individual organizations for use within certain geographical areas.
Finally, there are several frequency bands set aside for “license
exempt” usage
Bands in which a license is not needed

Wireless Links
The second restriction requires the use of Spread
Spectrum technique
Idea is to spread the signal over a wider frequency band
So as to minimize the impact of interference from other devices
Originally designed for military use
Frequency hopping
Transmitting signal over a random sequence of frequencies
First transmitting at one frequency, then a second, then a third…
The sequence of frequencies is not truly random, instead computed
algorithmically by a pseudorandom number generator
The receiver uses the same algorithm as the sender, initializes it with
the same seed, and is
Able to hop frequencies in sync with the transmitter to correctly receive
the frame

Wireless Links
 A second spread spectrum technique called Direct sequence
 Represents each bit in the frame by multiple bits in the transmitted signal.
 For each bit the sender wants to transmit
It actually sends the exclusive OR of that bit and n random bits
 The sequence of random bits is generated by a pseudorandom number
generator known to both the sender and the receiver.
 The transmitted values, known as an n-bit chipping code, spread the
signal across a frequency band that is n times wider
Example 4-bit chipping sequence

Wireless Links
 Wireless technologies differ in a variety of dimensions
 How much bandwidth they provide
 How far apart the communication nodes can be
 Four prominent wireless technologies: Bluetooth, Wi-Fi (more formally
known as 802.11), WiMAX (802.16), 3G cellularwireless
Overview of leading wireless technologies

Wireless Links
widely used
links
 Mostly
wireless
are
today
usually
asymmetric
 Two end-points are
usually different kinds
of nodes
 One end-point usually
has no mobility, but
has wired connection
to the Internet
base(known as
station)
 The node at the other
end of the link is often
mobile
A wireless network using a basestation

Wireless Links
 Wireless communication supports point-to-multipoint communication
 Communication between non-base (client) nodes is routed via the base
station
 Three levels of mobility for clients
 No mobility: the receiver must be in a fix location to receive a directional
transmission from the base station (initial version of WiMAX)
 Mobility is within the range of a base (Bluetooth)
 Mobility between bases (Cell phones and Wi-Fi)
 Mesh or Ad-hoc network
 Messages may be forwarded via a chain of peer nodes
 Nodes are peers

IEEE 802.11
 Also known as Wi-Fi
 Like its Ethernet and token ring siblings, 802.11 is designed for use
in a limited geographical area (homes, office buildings, campuses)
 Primary challenge is to mediate access to a shared communication medium –
in this case, signals propagating throughspace
 802.11 supports additional features
 power management and
 security mechanisms
 Original 802.11 standard defined two radio-based physical layerstandard
 One using the frequency hopping
 Over 79 1-MHz-wide frequency bandwidths
 Second using direct sequence
 Using 11-bit chipping sequence
 Both standards run in the 2.4-GHz and provide up to 2 Mbps

IEEE 802.11
Then physical layer standard 802.11b was added
Using a variant of direct sequence 802.11b provides up to
11 Mbps
Uses license-exempt 2.4-GHz band
Then came 802.11a which delivers up to 54 Mbps
using OFDM
802.11a runs on license-exempt 5-GHz band
Most recent standard is 802.11g which is backward
compatible with 802.11b
Uses 2.4 GHz band, OFDM and delivers up to 54 Mbps

IEEE 802.11 – Collision Avoidance
 Consider the situation in the following figure where each of four
nodes is able to send and receive signals that reach just the nodes to
its immediate left and right
 For example, B can exchange frames with A and C, but it cannot reachD
 C can reach B and D but notA
Example of a wireless network

 Suppose both A and C want
to communicate with B and
so they each send it a frame.
 A and C are unaware of each
other since their signals do not
carry that far
 These two frames collide with
each other at B
But unlike an Ethernet,
neither A nor C is aware of
this collision
 A and C are said to hidden
nodes with respect to each
other
The “Hidden Node” Problem.Although
A and C are hidden from each
other, their signals can collide at B. (B’s
reach is not shown.)

 Another problem called exposed
node problem occurs
 Suppose B is sending to A. Node
C is aware of this communication
because it hears B’s transmission.
 It would be a mistake for C to
conclude that it cannot transmit to
anyone just because it can hear B’s
transmission.
 Suppose C wants to transmit to
node D.
 This is not a problem since C’s
transmission to D will not interfere
with A’sability to receive from B.
Exposed Node Problem. Although B and
C are exposed to each other’s signals,
there is no interference if B transmits to A
while C transmits to D. (A and D’s reaches
are not shown.)

802.11 addresses these two problems with an algorithm
called Multiple Access with Collision Avoidance(MACA).
Key Idea
Sender and receiver exchange control frames with each other
before the sender actually transmits any data.
This exchange informs all nearby nodes that a transmission is
about to begin
Sender transmits a Request to Send (RTS) frame to the receiver.
The RTS frame includes a field that indicates how long the sender wants to
hold the medium
- Length of the data frame to be transmitted
Receiver replies with a Clear to Send (CTS) frame
This frame echoes this length field back to the senderKarpagam Institute of Technology

 Any node that sees the CTS frame knows that
it is close to the receiver, therefore
cannot transmit for the period of time it takes to send a fram
the specified length
Any node that sees the RTS frame but not the CTS frame
is not close enough to the receiver to interfere with it, and
so is free to transmit

 Using ACK in MACA
 Proposed in MACAW: MACA for WirelessLANs
 Receiver sends an ACK to the sender after successfully receiving a
frame
 All nodes must wait for this ACK before trying to transmit
 If two or more nodes detect an idle link and try to transmit an RTS
frame at the same time
 Their RTS frame will collide with each other
 802.11 does not support collision detection
 So the senders realize the collision has happened when they do not receive the
CTS frame after a period of time
 In this case, they each wait a random amount of time before trying again.
 The amount of time a given node delays is defined by the same exponential
backoff algorithm used on the Ethernet.

IEEE 802.11 – Distribution System
802.11 is suitable for an ad-hoc configuration of nodes that
may or may not be able to communicate with all other
nodes.
Nodes are free to move around
The set of directly reachable nodes may change over time
To deal with this mobility and partial connectivity,
802.11 defines additional structures on a set of nodes
Instead of all nodes being created equal,
some nodes are allowed to roam
 some are connected to a wired network infrastructure
they are called Access Points (AP) and they are connected to each other by a
so-called distribution system

 Following figure illustrates a distribution system that connects three accesspoints,
each of which services the nodes in the same region
 Each of these regions is analogous to a cell in a cellular phone system withthe
APIs playing the same role as a base station
 The distribution network runs at layer 2 of the ISO architecture
 Although two nodes can communicate directly with each other if they are within
reach of each other, the idea behind this configuration is
 Each nodes associates itself with one access point
 For node A to communicate with node E, A first sends a frame to its AP-1 which
forwards the frame across the distribution system to AP-3, which finally transmits the
frame to E
Access points connected
to a distribution network

 How do the nodes select their access points
 How does it work when nodes move from one cell to another
 The technique for selecting an AP is calledscanning
 The node sends a Probe frame
 All APs within reach reply with a Probe Responseframe
 The node selects one of the access points and sends that AP an Association
Request frame
 The AP replies with an Association Responseframe
 A node engages this protocol whenever
 it joins the network, as well as
 when it becomes unhappy with its currentAP
This might happen, for example, because the signal from its current AP has
weakened due to the node moving away from it
Whenever a node acquires a new AP, the new AP notifies the old AP of the
change via the distribution system

 Consider the situation shown in the following figure when node C moves fromthe
cell serviced by AP-1 to the cell serviced byAP-2.
 As it moves, it sends Probe frames, which eventually result in ProbeResponses
fromAP-2.
 At some point, C prefersAP-2 overAP-1 , and so it associates itself with that
access point.
 This is called active scanning since the node is actively searching for an access point
 APs also periodically send a Beacon frame that advertises the capabilities of the
access point; these include the transmission rate supported by the AP
 This is called passive scanning
 A node can change to this AP based on the Beacon frame simply by sending it an
Association Request frame back to the access point.
Node Mobility

IEEE 802.11 – Frame Format
Source and Destinations addresses: each 48 bits
Data: up to 2312 bytes
CRC: 32 bit
Control field: 16 bits
Contains three subfields (of interest)
6 bit Type field: indicates whether the frame is an RTS or CTS frame or
being used by the scanning algorithm
A pair of 1 bit fields : called ToDS and FromDS
Frame Format

 Frame contains four addresses
 How these addresses are interpreted depends on the settings of
the ToDS and FromDS bits in the frame’s Control field
 This is to account for the possibility that the frame had to be
forwarded across the distribution system which would mean
that,
 the original sender is not necessarily the same as the most recent
transmitting node
 Same is true for the destination address
 Simplest case
 When one node is sending directly to another, both the DS bits are 0,
Addr1 identifies the target node, andAddr2 identifies the source node

Most complex case
Both DS bits are set to 1
Indicates that the message went from a wireless node onto the
distribution system, and then from the distribution system to
another wireless node
With both bits set,
Addr1 identifies the ultimate destination,
Addr2 identifies the immediate sender (the one that forwarded the
frame from the distribution system to the ultimatedestination)
Addr3 identifies the intermediate destination (the one that accepted
the frame from a wireless node and forwarded across the
distribution system)
Addr4 identifies the original source
Addr1: E,Addr2:AP-3,Addr3:AP-1,Addr4:A

Bluetooth
 Used for very short range
communication between mobile
phones, PDAs, notebook
computers and other personal or
peripheral devices
 Operates in the license-exempt
band at 2.45 GHz
 Has a range of only 10 m
 Communication
typically belong to
devices
one
individual or group
 Sometimes categorized as Personal
Area Network (PAN) A BluetoothPiconet

Bluetooth
 Version 2.0 provides speeds up to 2.1 Mbps
 Power consumption is low
 Bluetooth is specified by an industry consortium called the
Bluetooth Special Interest Group
 It specifies an entire suite of protocols, going beyond the link layer
to define application protocols, which it calls profiles, for a range of
applications
 There is a profile for synchronizing a PDA with personalcomputer
 Another profile gives a mobile computer access to a wired LAN
 The basic Bluetooth network configuration is called a piconet
 Consists of a master device and up to seven slave devices
 Any communication is between the master and a slave
 The slaves do not communicate directly with each other
 A slave can be parked: set to an inactive, low-powerstate

ZigBee
ZigBee is a new technology that competes with
Bluetooth
Devised by the ZigBee alliance and standardized as IEEE
802.15.4
It is designed for situations where the bandwidth
requirements are low and power consumption must be
very low to give very long battery life
It is also intended to be simpler and cheaper than
Bluetooth, making it financially feasible to incorporate in
cheaper devices such as a wall switch that wirelessly
communicates with a ceiling-mounted fan

Switching and Forwarding
Switch
A mechanism that allows us to
interconnect links to form a
large network
A multi-input, multi-output
device which transfers packets
from an input to one or more
outputs
A switch is connected to a set of
links and for each of these links,
runs the appropriate data link
protocol to communicate with
that node
Adds the star topology
to the links

 A switch’s primary job is to receive incoming packets on one of its
links and to transmit them on some other link
This function is referred as switching or forwarding
According to OSI architecture this is the main function of the
network layer
 How does the switch decide which output port to place each packet
on?
It looks at the header of the packet for an identifier that it uses to
make the decision
Two common approaches
Datagram or Connectionless approach
Virtual circuit or Connection-oriented approach
A third approach source routing is less common

packet contains
 Datagrams
Key Idea
Every
enough information to
enable any switch to decide
how to get it to destination
Every packet contains
the complete
destination address
 To decide how to forward a
consults a
forwarding
packet, a switch
table (sometimes
called a routing table)An example network
Dest Port
-------------------
A 3
B 0
C 3
D 3
E 2
F 1
G 0
H 0
ForwardingTable
for Switch2

 Characteristics of Connectionless (Datagram) Network
 A host can send a packet anywhere at any time, since any packet that turns
up at the switch can be immediately forwarded using the forwarding table
 When a host sends a packet, it does NOT know if the network is capable
of delivering it or if the destination host is even up and running
 Each packet is forwarded independently of previous packets that might
have been sent to the same destination.
Thus two successive packets from host A to host B may follow
completely different paths
 A switch or link failure might not have any serious effect on
communication if it is possible to find an alternate route around the failure
and update the forwarding table accordingly
 Virtual Circuit Switching (connection-oriented)
 Uses the concept of virtual circuit (VC)
 First set up a virtual connection from the source host to the destination host
and then send the data

Two-stage process
Connection setup
Data Transfer
Host A wants tosend
packets to host B
Connection setup
Establish “connection state” in each of the switches
between the source and destination hosts
The connection state for a single connection consists of an
entry in the “VC table” in each switch through which the
connection passes

 Characteristics of VC
 Since host A has to wait for the connection request to reach the far side of the network
and return before it can send its first data packet, there is at least one RTT of delay
before data is sent
 While the connection request contains the full address for host B (which might be quite
large, being a global identifier on the network), each data packet contains only a small
identifier, which is only unique on one link.
 Thus the per-packet overhead caused by the header is reduced relative to the datagram model
 If a switch or a link in a connection fails, the connection is broken and a new one will
need to be established.
 Also the old one needs to be torn down to free up table storage space in the switches
 The issue of how a switch decides which link to forward the connection request on has
similarities with the function of a routing algorithm
 Comparison with the Datagram Model
 Datagram network has no connection establishment phase and each switch processes
each packet independently
 Each arriving packet competes with all other packets for buffer space
 If there are no buffers, the incoming packet must be dropped

 Good Properties of VC
 By the time the host gets the go-ahead to send data, it knows quite a lot about the
network-
For example, that there is really a route to the receiver and that the receiver is
willing to receive data
 It is also possible to allocate resources to the virtual circuit at the time it is established
X.25 network ( an early virtual-circuit-based networking technology but now
largely obsolete) allocates buffers per VC
 In VC, we could imagine providing each circuit with a different quality of service (QoS)
 The network gives the user some kind of performance related guarantee
Switches set aside the resources they need to meet this guarantee
For example, a percentage of each outgoing link’s bandwidth
Delay tolerance on each switch
 Most popular examples of VC technologies are X.25, Frame Relay andATM
 However, with the success of the Internet’s connection-less model, none of them enjoys
great popularity today

 ATM (Asynchronous Transfer Mode)
 Most well-known VC-based networkingtechnology
 Somewhat pasts its peak in terms of deployment
 Was important in the 1980s and early 1990s
 High-speed switching technology
 Was thought of to take over the world
 Connection-oriented packet-switched network
 Packets are called cells
 5 byte header + 48 byte payload
 Fixed length packets are easier to switch in hardware
 Simpler to design
 ATM
VPI: Virtual Path Identifier
CLP: Cell Loss Priority
 GFC: Generic Flow Control (not used)
 VCI: Virtual Circuit Identifier
 (VPI + VCI together makes the VC number we talked about)
 Type: management, congestion control HEC: Header Error Check (CRC-8)

 Source Routing
 All the information about network topology that is
required to switch a packet across the network is
provided by the source host
 Notes on Source Routing
Assumes that the source host knows
enough about the topology of the network
Analogous the problem of building the
forwarding tables in datagram networks or
figuring out where to send a setup packet in a virtual
circuit network
 We can not predict how the header needs to be (# of
switches in the path)
 Can be used in both datagram and virtual circuit
networks
 For example, IP, which is a datagram protocol includes a
source route option that allows selected packets to be
source routed.

Bridges and LAN Switches
 Bridges and LAN Switches
 Class of switches that is used to forward packets between shared-media LANs such as
Ethernets
 Known as LAN switches
 Referred to as Bridges
 Suppose you have a pair of Ethernets that you want to interconnect
 One approach is put a repeater in between them
 It might exceed the physical limitation of theEthernet
 No more than four repeaters between any pair of hosts
 No more than a total of 2500 m in length isallowed
 An alternative would be to put a node between the two Ethernets and have the node forward
frames from one Ethernet to the other
 This node is called a Bridge
 A collectionof LANs connected by one or more bridges is usually said to form an Extended LAN
 Simplest Strategy for Bridges
 Accept LAN frames on their inputs and forward them out to all other outputs
 Used by early bridges
 Learning Bridges
 Observe that there is no need to forward all the frames that a bridge receives

 Consider the following figure
 When a frame from host A that is addressed to host B arrives on port 1, there
is no need for the bridge to forward the frame out over port 2.
 How does a bridge come to learn on which port the various hosts reside?
 Solution
 Download a table into the bridge
 Who does the download?
 Human
 Too much work for maintenance
Host Port
--------------------
A 1
B 1
C 1
X 2
Y 2
Z 2

 Can the bridge learn this information by itself?
 Yes
 How
 Each bridge inspects the source address in all the frames it receives
 Record the information at the bridge and build the table
 When a bridge first boots, this table is empty
 Entries are added over time
 A timeout is associated with eachentry
 The bridge discards the entry after a specified period oftime
To protect against the situation in which a host is moved from one network to
another
 If the bridge receives a frame that is addressed to host not
currently in the table
 Forward the frame out on all other ports

 Strategy works fine if the extended LAN does not have a loop
in it
 Why?
Frames potentially loop through the extended LAN forever
Bridges B1, B4, and B6 form a loop

How does an extended LAN come to have a loop in it?
Network is managed by more than one administrator
For example, it spans multiple departments in an
organization
It is possible that no single person knows the entire
configuration of the network
A bridge that closes a loop might be added without anyoneknowing
to provideLoops are built into the network
redundancy in case of failures
Solution
Distributed Spanning TreeAlgorithm
14

Spanning TreeAlgorithm
 Think of the extended LAN as being represented by a graph that
possibly has loops (cycles)
 A spanning tree is a sub-graph of this graph that covers all the
vertices but contains no cycles
Spanning tree keeps all the vertices of the original graph but
throws out some of the edges
Example of (a) a cyclic graph; (b) a corresponding spanning tree.
14

Spanning TreeAlgorithm
 Developed by Radia Perlman at Digital
A protocol used by a set of bridges to agree upon a spanning
tree for a particular extended LAN
IEEE 802.1 specification for LAN bridges is based on this
algorithm
Each bridge decides the ports over which it is and is not willing
to forward frames
In a sense the extended LAN is reduced to an acyclic tree
Details are NOT required for the exam purposes
Take point: Spanning Tree Algorithm removes the cycles/loops
from the extended (bridged) LANs
14

Limitation of Bridges
 Do not scale
 Spanning tree algorithm does not scale
 Broadcast does not scale
Nodes get bothered with too many broadcasts that the bridges forward
ALLnodes
 Do not accommodate heterogeneity
Ethernet with Ethernet, Wi-Fi with Wi-Fi, etc.
 A solution
Virtual LAN (VLAN)
15

Virtual LANs (VLANs)
Allow a single extended LAN to be partitioned into
several logical LANs
Each VLAN is assigned an ID (or color)
Frames can only be travel between LANs segments
within the same VLAN
Partially solves the broadcast problem in the extended
LAN
One Attractive feature of VLANs is
We can change the logical topology of the extended
LAN without moving/changing any wire or addresses
Just change the Bridge configuration
15

Virtual LANs (VLANs)
 When a frame from X arrives at bridge B2, the bridge observes that
it came in a port that was configured as being in VLAN 100, so it
inserts a VLAN header (has the VLAN ID) between the Ethernet
header and its payload
 When the frame arrives at B1, it will only forward it to the port of
VLAN 100 and not to VLAN 200
 The link between B1 and B2 is considered to be in both VLANs
15

Internetworking
 What is internetwork
 An arbitrary collection of networks interconnected to provide some sortof
host-to-host packet delivery service
 A simple internetwork where H represents hosts and R represents routers
15

Internetworking
 What is IP
 IP stands for InternetProtocol
 Key tool used today to build scalable, heterogeneousinternetworks
 It runs on all the nodes in a collection of networks and defines the
infrastructure that allows these nodes and networks to function as a single
logical internetwork
 A simple internetwork showing the protocollayers
15

IP Service Model
Two parts
Global Addressing Scheme
Provides a way to identify all hosts in the network
Datagram (Connectionless) model for data delivery
Best-effort delivery (unreliable service)
packets are lost
packets are delivered out of order
duplicate copies of a packet are delivered
packets can be delayed for a long time
15

Packet Format
 Version (4 bits):
 currently 4 or 6.
 Also called IPv4 and IPv6
 Hlen (4 bits):
 number of 32-bit words in header
 usually 5 32-bit words with no options
 TOS (8 bits):
 type of service (not widely used)
 Length (16 bits):
 number of bytes in this datagram including the
header
 Ident (16 bits) and Flags/Offset (16 bits):
 used by fragmentation
 Protocol (8 bits):
 demux key (TCP=6, UDP=17)
 Checksum (16 bits):
 of the header only
15

Packet Format
 TTL (8 bits):
 number of hops/routers this packet can travel
 discard the looping packets
 Originally based on time, but changed to a hop-countbased
 Each router decrements it by 1
 Discard the packet when it becomes 0
 Default is 64
 Problems
 Setting it too high the packet will loop a lot
 Setting it too low the packet will not reach the destination
 DestAddr & SrcAddr (32 bits)
 The key for datagram delivery
 Every packet contains a full destination address
 Forwarding/routing decisions are made at each router
 The source address is for the destination to know the sender and if it wants
to reply to it
15

IP Fragmentation and Reassembly
Each network has some MTU (Maximum Transmission
Unit)
Ethernet (1500 bytes), FDDI (4500 bytes)
IP packets need to fit in the payload of link-layerframe
Solutions
Make all packet size small enough to fit all
Or fragment the large packets into smaller ones and reassembles them
later
Strategy
Fragmentation occurs in a router when it receives a datagram that
it wants to forward over a network which has (MTU < datagram)
Reassembly is done at the receiving host
All the fragments carry the same identifier in the Ident field
Fragments are self-contained datagrams
15

IP Fragmentation and Reassembly
+Suppose PPP has MTU of 532-byte packet (20 header
512 payload)
IP datagrams traversing the sequence of physicalnetworks
Header fields used in IP fragmentation.
(a) Unfragmented packet; (b) fragmented packets.
 MTU path discovery is a good strategy to avoid fragmentation
Send some packets first just discover the MTUs on the path to D
15

Global Addresses
 IP addresses Properties
 globally unique
 hierarchical: network + host
 Network part: identifies the network the host is attached to
 Host: identifies a unique host on that network
 Ethernet addresses, even globally unique, are flat (no structure and thus no meaning) and can
not be use for routing
 Note that a router is attached to at least two networks, so it must have anIP
address on each port/interface
 Thus it is more precise to think of IP addresses as belonging to interfaces rather than to hosts
 Approximately, 4 Billion IPaddress,
half are A type, ¼ is B type, and 1/8 is Ctype
(a) ClassA (b) Class B (c) Class C
16

GlobalAddresses
 Class A was intended for Wide AreaNetworks
 Thus there should a very few of them
 Class B was intended for a modest size networks (like a campus)
 Class C is for the large number of LANs
and today’s IP However, these classifications are not flexible
addresses are normally “classless” as we will see
 Format
 4 bytes, each byte is represented by a decimal number
 Dot notation
 10.3.2.4
 128.96.33.81
 192.12.69.77
16

IP Datagram Forwarding
 Strategy
 every datagram contains destination's address
 if directly connected to destination network, then forward tohost
 if not directly connected to destination network, then forward to somerouter
 forwarding table maps network number into next hop
 each host has a default router
 each router maintains a forwarding table
 Example (router R2)
Algorithm
if (NetworkNum of destination = NetworkNum of one of my interfaces)then
deliver packet to destination over thatinterface
else
if (NetworkNum of destination is in my forwarding table)then
deliver packet to NextHop router
else
deliver packet to default router
For a host with only one interface and only a default router in its forwarding table, this simplifies to
if (NetworkNum of destination = my NetworkNum)then
deliver packet to destinationdirectly
else
deliver packet to default router
16

Subnetting
 The network number part was designed to uniquely identify exactly one physical network
 However, this approach has some problems
 A network with only 2 hosts has to have at least a class C network!!
 A network with only 256 hosts has to have at least a class B network!!
 Thus, we will waste our valuable IP address space
 Solution: Subnetting
 Key Idea
 Allocate a single network number and use it for several physical networks - subnets
 Several things need to be done
Subnets need to be physically close to each other
From the Internet point of view, they all look ONE network
A perfect situation to use subnetting is for large campus or corporation
Configure all nodes on each subnet with a subnet mask
It masks the network part
Introduces the subnet number
All nodes on the same subnet have the same subnet number and the same mask
 The IP address of a nodes ANDed with the subnet mask give the subnet number
 IP AND subnet mask  subnet number
16

Subnetting
Increases the number of
networks and reducesthe
number of hosts
 When a host wants to send a packet to a certain IPaddress
 First, it does the bitwise AND between its own subnet mast and destination IP address
 If the result equals the subnet number of the sender, then the destination host is on the same
subnet so the packet can be delivered directly (without a router)
 Else, the packet will be forwarded to another subnet (through a router)
16

Subnetting
Forwarding Table at Router R1
ForwardingAlgorithm
D = destination IPaddress
for each entry < SubnetNum, SubnetMask, NextHop>
D1 = SubnetMask & D
if D1 = SubnetNum
if NextHop is an interface
deliver datagram directly to destination
else
deliver datagram to NextHop (a router)

ClasslessAddressing
 Subnetting has a counterpart, sometimes called supernetting, but
often called Classless Interdomain Routing, CIDR (pronounced
cider)
 Address assignment efficiency
 A network with 256 hosts needs a class Baddress
 Address assignment efficiency = 256/65535 = 0.39
 Solutions
 Use subnetting
 Only give class C networks
 Give class B only with a proof of that the network has more than 64K hosts
Problem with this solution
Excessive storage requirement at the routers.

ClasslessAddressing
 If a single site has, say 16 class C network numbers assigned to it,
Every Internet backbone router needs 16 entries in its routing tables
for that site (too much for one site)
This is true, even if the path to every one of these networks is the same
 If we had assigned a class B address to theAS
The same routing information can be stored in one entry
 Efficiency = number of hosts / number of available addresses =16
255 / 65, 536 = 6.2% (not efficient)
 CIDR tries to balance the desire to minimize the number of routes that a
router needs to know against the need to hand out addresses efficiently.
 CIDR uses aggregate routes
Uses a single entry in the forwarding table to tell the router how to
reach a lot of different networks
Breaks the rigid boundaries between address classes
16

ClasslessAddressing
 Consider a site with 16 class C network numbers.
 Instead of handing out 16 addresses at random, hand out a block of
contiguous class C addresses
 Suppose we assign the class C network numbers from 192.4.16
through 192.4.31
 Observe that top 20 bits of all the addresses in this range are the
same (11000000 00000100 0001)
 We have created a 20-bit network number (which is in between class B
network number and class C number)
 Requires to hand out blocks of class C addresses that share a
common prefix
 The convention is to place a /X after the prefix where X is the prefix
length in bits
16

ClasslessAddressing
 For example, the 20-bit prefix for all the networks 192.4.16
through 192.4.31 is represented as 192.4.16/20
 By contrast, if we wanted to represent a single class C network
number, which is 24 bits long, we would write it 192.4.16/24
Route aggregation with CIDR
16

Different Protocols
ARP (Address Resolution Protocol)
DHCP (Dynamic Host Configuration
Protocol)
ICMP (Internet Control Message
Protocol)
17

Address Translation Protocol (ARP)
Map IP addresses into physical addresses
ARP (Address Resolution Protocol)
table of IP to physical address bindings
The router broadcasts a request (who-has / tell)
if the required IP address not in the ARPtable
Ex., who-has 192.168.0.29 tell 192.168.0.1
target machine (with IP 192.168.0.29 in the
example) responds with its physical address (its
MAC)
17

Host IP Configurations
Most host Operating Systems provide a way to
manually configure the IP information for the
host
Drawbacks of manual configuration
A lot of work to configure all the hosts in a large
network
Configuration process is error-prune
Automated Configuration Process is required
Using the DHCP protocol
17

Dynamic Host Configuration Protocol
 DHCP server is responsible for providing configuration
information to hosts
 There is at least one DHCP server for an administrative domain
 DHCP server maintains a pool/set of available addresses
 Newly booted or attached host sends DHCPDISCOVER message
to a special IP address (255.255.255.255)
 DHCP relay agent unicasts the message to DHCP server and waits
for the response
17

Internet Control Message Protocol
Defines a collection of error messages that are
sent back to the source host whenever a router or
host is unable to process an IP datagram
successfully
Destination host unreachable due to link /node failure
Reassembly process failed
TTL had reached 0 (so datagrams don't cycle forever)
IP header checksum failed
ICMP-Redirect
From router to a source host
With a better route information
17

Routing
Forwarding versus Routing
 Forwarding:
 to select an output port based on destination address and routing
table
 Routing:
 routing table is built
 Network as a Graph
 The basic problem of routing is to find the lowest-cost path between anytwo
nodes
Where the cost of a path equals the sum of the costs of all the edgesthat
make up the path
17

Routing
• For a simple network, we can calculate all shortest paths and
load them into some nonvolatile storage on each node.
• Such a static approach has several shortcomings
• It does not deal with node or link failures
• It does not consider the addition of new nodes or links
• It implies that edge costs cannot change
• What is the solution?
• Need a distributed and dynamic protocol
• Two main classes of protocols
• Distance Vector
• Link State
Details are not required for
the exam purposes
17

SUMMARY
• A media access control is a network data transfer policy that
determines how data is transmitted between two computer terminals
through a network cable. The media access control policy involves
sub-layers of the data link layer 2 in the OSI reference model
• An internetwork is a collection of individual networks, connected by
intermediate networking devices, that functions as a single large
network. Internetworking refers to the industry, products, and
procedures that meet the challenge of creating and administering
internetworks.

TOPIC LEVEL
Compose controlled access methods with packet formats k2
Validate the network layer protocols and its types. k3
Compile IPv4 addresses with classful and classless addressing formats. k3

VIDEO TUTORIAL
Media Access Control https://www.coursera.org/lecture/peer-to-peer-
protocols-local-area-networks/medium-access-
control-nWWWd

QUIZ QUESTIONS
1. The Media access control layer frame has -------------------------------
a) 5 Fields
b) 7 Fields
c) 9 Fields
d) 11 Fields
1. In Third generation of cellular telephony , the data rate for access in a moving vehicle is ------------------
a) 244 kbps
b) 238 kbps
c) 150 kbps
d) 144 kbps
1. The field of the MAC frame that alter the receive and enables it to synchronize is known as -------------------------
a) SFD
b) Preamble
c) Source Address
d) Designation Address
1. The value of the frame body field of Media Access Control frame is in between----------------------
a) 0 and 512 bytes
b) 0 and 1214 bytes
c) 0 to 2312 bytes
d) 0 to 3450 bytes
•
1. The duration field D of Media Access Control layer frame contains -----------------
a) 2 bytes
b) 4 bytes
c) 6 bytes
d) 8 bytes.

1. In ---------------------- methods, no station is superior to another station and none is assigned the control over another.
a) Random access
b) Controlled access
c) Channelization
d) None of the above
1. In----------------------- the chance, of collision can be reduced if a station sense the medium before trying to use it.
a) MA
b) CSMA
c) FDMA
d) CDMA.
1. ------------------------- requires that each station first listen to the medium before sending.
a) MA
b) CSMA
c) CDMA
d) FDMA
1. Media Access Control is the sub layer of --------------
a) LLC
b) IEEE
c) ANSI
d) Both a and c.
1. -------------------------- augments the CSMA algorithm to detect collision
a) CSMA/CD
b) CSMA/CA
c) Either (a) or(b)
d) Both (a) and (b)

1. Hardware address is known as-----------------
a) MAC address
b) IP Address
c) Network Interface Card
d) Address Resolution Protocol.
1. What translates IP address into MAC address? ------------------------
a) Organizationally Unique Identifier
b) Address Resolution Protocol
c) Network Interface Card
d) Burned In Address
1. Networking Hardware Address is referred with----------------------
a) IP address
b) MAC address
c) NIC
d) Organizationally Unique Identifier
1. Does MAC address contain characters.--------------------
a) True
b) False.
1. On wireless networks--------------------- filtering is the security measure.
a) OUI
b) IP
c) NIC
d) MAC.

1. MAC addresses are used as----------------------------
a) Network addresses
b) IP address
c) Hardware address
d) Burned in address.
1. IEEE standards for Institute of Electrical and Electronics Engineers----------------
a) False
b) True
1. The original IEEE 802 MAC address comes from--------------------
a) MAC address
b) IP address
c) Ethernet address
d) Http.
1. MAC addresses are very useful in diagnosing network issues--------------
a) True
b) False.
1. What is the size of MAC address? -------------
a) 16-bits
b) 32-bits
c) 48-bits
d) 64-bits.

1. 802.11 wireless networking uses what method as the media access method? ---------------
a) CSMA/CD
b) CTS/RTS
c) CSMA/CA
d) CSCD/CA
1. In ---------------------- each station sends a frame whenever it has a frame to send.
a) Pure ALOHA
b) Slotted ALOHA
c) Both (a) and (b)
d) Neither (a) nor (b).
1. In pure ALOHA, the vulnerable time is ----------------------- the frame transmission time.
a) The same as
b) Two times
c) Three times.
d) None of the above.
1. The maximum throughput for pure ALOHA is ------------------ percent
a) 12.2
b) 18.4
c) 36.8
1. In -----------------------method a station cannot send unless it has been authorized by other stations.
a) Random access
b) Controlled access
c) Channelization

1. The vulnerable time for CSMA is the --------------------------- propagation time.
a) The same as
b) Two times
c) Three times
1. In the ------------------------ method a station needs to make a reservation before sending data. Time is divided into intervals.
a) Reservation
b) Polling
c) Token passing
1. The channelization protocols---------------------- method.
a) Two
b) Three
c) Four
1. The popular controlled access -------------------- methods
a) Two
b) Three
c) Four
1. In--------------------- the available bandwidth is divided into frequency bands
a) FDMA
b) CDMA
c) TDMA

Unit - III
ROUTING

Intra-AS Routing
also known as interior gateway protocols
(IGP)
most common intra-AS routing protocols:
RIP: Routing Information Protocol
OSPF: Open Shortest Path First
IGRP: Interior Gateway Routing Protocol
(Cisco proprietary)
1

RIP ( Routing Information Protocol)
 included in BSD-UNIX distribution in 1982
 distance vector algorithm
 distance metric: # hops (max = 15 hops), each link has cost1
 DVs exchanged with neighbors every 30 sec in response message(aka
advertisement)
 each advertisement: list of up to 25 destination subnets (in IP addressing
sense)
DC
u v
A B w
x
y
z
subnet hops
u 1
v 2
w 2
x 3
y 3
z 2
from router A to destinationsubnets:
1

RIP: example
w x
z
y
A
C
routing table in router D
D B
destination subnet
w
y
next router
A
B
# hops to dest
2
2
z B 7
x -- 1
…. …. ....
1

w x
z
y
A D B
C
routing table in router D
destination subnet
w
next router #
A
hops to dest
2
y B
A
2 5
z B 7
x -- 1
…. …. ....
dest next hops
w
x
z
….
- 1
- 1
C 4
… ...
A-to-D advertisement
RIP: example
1

RIP: link failure, recovery
if no advertisement heard after 180 sec -->
neighbor/link declared dead
routes via neighbor invalidated
new advertisements sent to neighbors
neighbors in turn send out new advertisements (if
tables changed)
link failure info quickly (?) propagates to entire net
poison reverse used to prevent ping-pong loops
(infinite distance = 16 hops)
1

RIP table processing
RIP routing tables managed by application-
level process called route-d (daemon)
physical
link
network
(IP)
forwarding
table
transport
(UDP)
advertisements sent
periodically repeated
routed
physical
link
network
(IP)
transprt
(UDP)
in UDP packets,
routed
forwarding
table
192

OSPF (Open Shortest Path First)
“open”: publicly available
uses link state algorithm
LS packet dissemination
topology map at each node
route computation using Dijkstra’s algorithm
OSPF advertisement carries one entry per neighbor
advertisements flooded to entire AS
carried in OSPF messages directly over IP (rather than
TCP or UDP
IS-IS routing protocol: nearly identical to OSPF
193

OSPF “advanced” features (not in RIP)
security: all OSPF messages authenticated (to prevent
malicious intrusion)
multiple same-cost paths allowed (only one path in RIP)
for each link, multiple cost metrics for different TOS (e.g.,
satellite link cost set “low” for best effort ToS; high for real
time ToS)
integrated uni- and multicast support:
Multicast OSPF (MOSPF) uses same topology data base as OSPF
hierarchical OSPF in large domains.
194

Hierarchical OSPF
boundary router
backbone router
area 1
area 2
area 3
backbone
area
border
routers
internal
routers
195

two-level hierarchy: local area, backbone.
link-state advertisements only in area
each nodes has detailed area topology; only know
direction (shortest path) to nets in other areas.
area border routers: “summarize” distances to nets
in own area, advertise to other Area Border
routers.
backbone routers: run OSPF routing limited to
backbone.
boundary routers: connect to otherAS’s.
Hierarchical OSPF
196

Internet inter-AS routing: BGP
 BGP (Border Gateway Protocol): the de facto inter-domain routing
protocol
 “glue that holds the Internet together”
 BGP provides each AS a means to:
 eBGP: obtain subnet reachability information from neighboring
ASs.
 iBGP: propagate reachability information to all AS-internal
routers.
 determine “good” routes to other networks based on
reachability information and policy.
 allows subnet to advertise its existence to rest of Internet: “I am
here”
197

BGP basics
AS2
3b
3a
AS3 1c
1a
AS1 1d
1b
2a
2c
2b
other
networks
other
networks
 BGP session: two BGP routers (“peers”) exchange BGP
messages:
 advertising paths to different destination network prefixes (“pathvector”
protocol)
 exchanged over semi-permanent TCPconnections
 when AS3 advertises a prefix toAS1:
 AS3 promises it will forward datagrams towards that prefix
 AS3 can aggregate prefixes in its advertisement
3c
BGP
message
198

BGP: distributing path information
AS2
3b
3a
AS3
1c
1a
AS1 1d
1b
2a
2c
2b
other
networks
other
networks
using eBGP session between 3a and 1c, AS3 sendsprefix
reachability info toAS1.
1c can then use iBGP do distribute new prefix info to all
routers inAS1
1b can then re-advertise new reachability info to AS2 over1b-
to-2a eBGP session
when router learns of new prefix, it creates entry for
prefix in its forwarding table.
eBGP session
iBGP session
199

Path attributes and BGP routes
 advertised prefix includes BGP attributes
prefix + attributes = “route”
 two important attributes:
AS-PATH: contains ASs through which prefix advertisement
has passed: e.g., AS 67, AS 17
NEXT-HOP: indicates specific internal-AS router to next-hop
AS. (may be multiple links from current AS to next-hop-AS)
 gateway router receiving route advertisement uses import policy
to accept/decline
e.g., never route through AS x
policy-based routing
200

BGP route selection
router may learn about more than 1 route to
destination AS, selects route based on:
 local preference value attribute: policy
decision
 shortestAS-PATH
 closest NEXT-HOP router: hot potato routing
 additional criteria
201

BGP messages
BGP messages exchanged between peers over TCP
connection
BGP messages:
OPEN: opens TCP connection to peer and
authenticates sender
UPDATE: advertises new path (or withdraws old)
KEEPALIVE: keeps connection alive in absence of
UPDATES; also ACKs OPEN request
NOTIFICATION: reports errors in previous msg; also
used to close connection
202

BGP routing policy
 A,B,C are provider networks
 X,W,Y are customer (of provider networks)
 X is dual-homed: attached to two networks
X does not want to route from B via X to C
.. so X will not advertise to B a route to C
A
B
C
W
X
Y
legend:
customer
network:
provider
network
203

BGP routing policy (2)
A
B
C
W
X
Y
 A advertises pathAW to B
 B advertises path BAW to X
 Should B advertise path BAW to C?
 No way! B gets no “revenue” for routing CBAW since neither W norC
are B’s customers
 B wants to force C to route to w viaA
 B wants to route only to/from its customers!
legend:
customer
network:
provider
network
204

Why different Intra-, Inter-AS routing ?
 policy:
 inter-AS: admin wants control over how its traffic
routed, who routes through its net.
 intra-AS: single admin, so no policy decisions needed
 scale:
 hierarchical routing saves table size, reduced update
traffic
 performance:
 intra-AS: can focus on performance
 inter-AS: policy may dominate over performance
205

The Global Internet
The tree structure of the Internet in 1990
206

The Global Internet
A simple multi-providerInternet
207

Interdomain Routing (BGP)
Internet is organized as autonomous systems (AS) each of
which is under the control of a single administrative
entity
Autonomous System (AS)
corresponds to an administrative domain
examples: University, company, backbone network
A corporation’s internal network might be a single AS, as
may the network of a single Internet service provider
208

Interdomain Routing
A network withtwo
autonomous system
209

Route Propagation
Idea: Provide an additional way to hierarchically
aggregate routing information is a large internet.
Improves scalability
Divide the routing problem in two parts:
Routing within a single autonomous system
Routing between autonomous systems
in the Another name for autonomous systems
Internet is routing domains
Two-level route propagation hierarchy
Inter-domain routing protocol (Internet-widestandard)
Intra-domain routing protocol (each AS selects itsown)
210

EGP and BGP
Inter-domain Routing Protocols
Exterior Gateway Protocol (EGP)
Forced a tree-like topology onto the Internet
Did not allow for the topology to become general
Tree like structure: there is a single backbone and autonomous systems
are connected only as parents and children and not aspeers
Border Gateway Protocol (BGP)
Assumes that the Internet is an arbitrarily interconnected set of
ASs.
Today’s Internet consists of an interconnection of multiple
backbone networks (they are usually called service provider
networks, and they are operated by private companies rather
than the government)
Sites are connected to each other in arbitrary ways
211

BGP
Some large corporations connect directly to one
or more of the backbone, while others connect to
smaller, non-backbone service providers.
Many service providers exist mainly to provide
service to “consumers” (individuals with PCs in
their homes), and these providers must connect to
the backbone providers
Often many providers arrange to interconnect
with each other at a single “peering point”
212

BGP-4: Border Gateway Protocol
Assumes the Internet is an arbitrarily interconnected set
ofAS's.
Define local traffic as traffic that originates at or
terminates on nodes within an AS, and transit traffic as
traffic that passes through anAS.
We can classify AS's into three types:
 Stub AS: an AS that has only a single connection to one other AS;
such an AS will only carry local traffic (small corporation in the figure of
the previous page).
 Multihomed AS: an AS that has connections to more than one other AS, but
refuses to carry transit traffic (large corporation at the top in the figure of
the previous page).
 Transit AS: an AS that has connections to more than one other AS, and is
designed to carry both transit and local traffic (backbone providers in the
figure of the previous page).
213

The goal of Inter-domain routing is to
find any path to the intended destination
that is loop free
We are concerned with reachability than
optimality
Finding path anywhere close to optimal is
considered to be a great achievement
Why?
BGP
214

Scalability: An Internet backbone router must be able to
forward any packet destined anywhere in the Internet
Having a routing table that will provide a match for any valid IP
address
Autonomous nature of the domains
It is impossible to calculate meaningful path costs for a path that
crosses multipleASs
A cost of 1000 across one provider might imply a great path but it
might mean an unacceptable bad one from another provid
Issues of trust
Provider A might be unwilling to believe certain advertisements
from provider B
BGP
215

Each AS has:
One BGP speaker that advertises:
local networks
other reachable networks (transit AS only)
gives path information
In addition to the BGP speakers, the AS has one or
more border “gateways” which need not be the same
as the speakers
The border gateways are the routers through which
packets enter and leave theAS
BGP
216

BGP does not belong to either of the two main classes of
routing protocols (distance vectors and link-state
protocols)
BGP advertises complete paths as an enumerated lists of
ASs to reach a particular network
BGP
Example
217

BGP Example
Speaker for AS 2 advertises reachability to Pand
Q
Network 128.96, 192.4.153, 192.4.32, and 192.4.3,
can be reached directly from AS 2.
Speaker for backbone network then advertises
Networks 128.96, 192.4.153, 192.4.32, and 192.4.3
can be reached along the path <AS 1, AS 2>.
Speaker can also cancel previously advertised
paths
218

BGP Issues
It should be apparent that the AS
benumbers carried in BGP need to
unique
For example, AS 2 can only recognize
itself in the AS path in the example if no
other AS identifies itself in the same way
AS numbers are 16-bit numbers
assigned by a central authority
219

Integrating Interdomain & Intradomain Routing
All routers run iBGP and an intradomain routing protocol.
Border routers (A, D, E) also run eBGP to otherASs
220

Integrating Interdomain & Intradomain Routing
BGP routing table, IGP routing table, and combined table at routerB
221

RoutingAreas
A domain divided intoarea
router
Backbone area
Area border
(ABR)
222

Next Generation IP
(IPv6)
223

Major Features
 128-bit addresses
 Multicast
 Real-time service
 Authentication and security
 Auto-configuration
 End-to-end fragmentation
Enhanced routing functionality, including support for
mobile hosts
224

IPv6 Addresses
Classless addressing/routing (similar to CIDR)
Notation: x:x:x:x:x:x:x:x(x= 16-bit hex number)
contiguous 0s are compressed: 47CD::A456:0124
IPv6 compatible IPv4 address: ::128.42.1.87
Address assignment
provider-based
geographic
225

IPv6 Header
40-byte “base” header
Extension headers (fixed order, mostly fixed
length)
fragmentation
source routing
authentication and
 security
other options
226

Internet
Multicast
227

Overview
IPv4
class D addresses
demonstrated with MBone
uses tunneling
Integral part of IPv6
problem is making it scale
One-to-many
Radio station broadcast
Transmitting news, stock-price
Software updates to multiple hosts 228

Overview
 Many-to-many
Multimedia teleconferencing
Online multi-player games
Distributed simulations
 Without support for multicast
A source needs to send a separate packet with the identical data
to each member of the group
This redundancy consumes more bandwidth
Redundant traffic is not evenly distributed, concentrated near
the sending host
Source needs to keep track of the IP address of each member in
the group
Group may be dynamic

Overview
To support many-to-many and one-to-many IP provides
an IP-level multicast
Basic IP multicast model is many-to-many based
on multicast groups
Each group has its own IP multicast address
Hosts that are members of a group receive copies of
any packets sent to that group’s multicast address
A host can be in multiple groups
A host can join and leave groups

Overview
Using IP multicast to send the identical packet to
each member of the group
A host sends a single copy of the packet addressed to
the group’s multicast address
The sending host does not need to know the individual
unicast IP address of each member
Sending host does not send multiple copies of the
packet
IP’s original many-to-many multicast has been
supplemented with support for a form of one-to-
many multicast 23Karpagam Institute of Technology 231

Overview
 One-to-many multicast
Source specific multicast (SSM)
A receiving host specifies both a multicast group and aspecific
sending host
 Many-to-many model
Any source multicast (ASM)
 A host signals its desire to join or leave a multicast groupby
communicating with its local router using a special protocol
In IPv4, the protocol is Internet Group Management Protocol
(IGMP)
In IPv6, the protocol is Multicast Listener Discovery (MLD)
 The router has the responsibility for making multicast behave
correctly with regard to the host
232

Multicast Routing
A router’s unicast forwarding tables indicate for any IP
address, which link to use to forward the unicast packet
To support multicast, a router must additionally have
multicast forwarding tables that indicate, based on
multicast address, which links to use to forward the
multicast packet
Unicast forwarding tables collectively specify a set of
paths
Multicast forwarding tables collectively specify a set of
trees
Multicast distribution trees
233

Multicast Routing
To support source
multicast forwarding
specific multicast, the
tables must indicate
• which links to use based on the combination of multicast address
the unicast IP address of the source
Multicast routing is the process by which multicast distribution t
are determined
234

Distance Vector Multicast
Each router already knows that shortest path to
source S goes through router N.
When receive multicast packet from S, forward on
all outgoing links (except the one on which the
packet arrived), iff packet arrived from N.
onlyEliminate duplicate broadcast packets by
letting
“parent” for LAN (relative to S) forward
shortest path to S (learn via distance vector)
smallest address to break ties
235

Reverse Path Broadcast (RPB)
Goal: Prune networks that have no hosts in group G
Step 1: Determine of LAN is a leaf with no members in G
leaf if parent is only router on the LAN
determine if any hosts are members of G using IGMP
Step 2: Propagate “no members of G here” information
augment <Destination, Cost> update sent to neighbors with set
of groups for which this network is interested in receiving
multicast packets.
only happens when multicast address becomes active.
Distance Vector Multicast
236

Protocol Independent Multicast (PIM)
Shared Tree
Source specific
tree
237

Protocol Independent Multicast (PIM)
Delivery of a packet along a shared tree. R1 tunnels the
packet to the RP, which forwards it along the shared tree to
R4 and R5.
238

Routing for Mobile Hosts
 Mobile IP
 home agent
Router located on the home network of the mobile hosts
 home address
The permanent IP address of the mobile host.
Has a network number equal to that of the home network and thus of the home
agent
 foreign agent
Router located on a network to which the mobile node attaches itself when it is
away from its home network
239

Problem of delivering a packet to the mobile
node
How does the home agent intercept a packet that is
destined for the mobile node?
ProxyARP
How does the home agent then deliver the packet to
the foreign agent?
IP tunnel
Care-of-address
How does the foreign agent deliver the packet to the
mobile node?
240

Route optimization in Mobile IP
The route from the sending node to mobile node can be
significantly sub-optimal
One extreme example
The mobile node and the sending node are on the same network, but the
home network for the mobile node is on the far side of the Internet
Triangle Routing Problem
Solution
Let the sending node know the care-of-address of the mobile node. The
sending node can create its own tunnel to the foreignagent
Home agent sends binding update message
The sending node creates an entry in the bindingcache
The binding cache may become out-of-date
The mobile node moved to a different network
Foreign agent sends a binding warning message
241

SUMMARY
• Routers are physical devices that join multiple wired or wireless
networks together. Technically, a wired or wireless router is a Layer 3
gateway, meaning that the wired/wireless router connects networks
(as gateways do), and that the router operates at the network layer of
the OSI model.

TOPIC LEVEL
Elaborate in detail about the Address Space and Address Space Allocation
and IPv6 addressing.
k3
Develop the transition from IPv4 to IPv6 with its strategies and uses. k2
(i) Evaluate the working of Protocol Independent Multicast (PIM) in detail.
(ii) Outline the need of DVMRP
k3

VIDEO TUTORIAL
Introduction to Routing https://www.youtube.com/watch?v=Am
lOSGYkKXc
Routing 1 https://www.youtube.com/watch?v=u0B
TdiqhVRI

• QUIZ QUESTIONS:
1. When a router cannot route a datagram, the datagram is discarded and send a message to source-----------------------
a) Destination Unreachable
b) Destination Unverified
c) Destination Unavailable
d) Destination no-entry.
1. When a router needs to send a packet destined for another network ,it must know the -----------------
a) Datagram
b) Medium
c) Path flow
d) IP address
1. An option which is used to record the time of datagram processing by a router is called----------------
a) Timestamp
b) Time frame
c) Time delay
d) Time wrap
1. A host or router can join a group by sending a --------------------
a) Initiative report
b) Joining report
c) Membership report message
d) Membership Leave report
1. In network addresses, each router has two---------------
a) Addresses
b) Masks
c) Blocks
d) Segments.

1. Which of the following is not the requirement of routing function?-----------------------
a) Correctness
b) Robustness
c) Delaytime
d) Stability
1. The-----------------protocol allows the administrator to assign a cost,called the metric, to eachroute.
a) OSPF
b) RIP
c) BGP
d) BBGP
1. If there is only one routing sequence for each source destination pair, the scheme is known as -------------------------
a) static routing
b) fixed alternative routing
c) standard routing
d) dynamic routing
1. The OpenShortest Path First(OSPF) protocol is an intra domain routing protocol basedon ------------------routing.
a) distance vector
b) link state
c) path vector
d) non distance vector
1. A ---------------------routing scheme is designed to enable switches to react to changing traffic patterns on the network.
a) static routing
b) fixed alternative routing
c) standard routing
d) dynamic routing.

1. The Routing Information Protocol(RIP) is an intra domain routing basedon ---------------routing.
a) distance vector
b) link state
c) path vector
d) distance code
1. The term ------------- refers to which node or nodes in the network are responsible for the routing decision.
a) decision place
b) routing place
c) node place
d) switching place
1. In ------------------ routing the least cost route betweenany two nodes is the minimum distance.
a) path vector
b) distance vector
c) link state
d) switching
1. For centralized routing the decision is made by some designated node called-----------------
a) designated center
b) control center
c) network center
d) network control center
1. For purposesof routing, the Internet is divided into ----------------
a) wide area networks
b) autonomous networks
c) local area networks
d) autonomous system

1. In -------------------- a route is selectedfor eachdestination pair of nodes in the network
a) Flooding
b) variable routing
c) fixed routing
d) random routing.
1. To createa neighborhood relationship, a router running BGP sendsan -----------------message.
a) Open
b) Update
c) keep alive
d) Close
1. The technique which requires no network information required is---------------------
a) Flooding
b) variable routing
c) fixed routing
d) random routing
1. An area is-------------------
a) partof an AS
b) composed of at least two AS
c) another term for an AS
d) composed more than two AS
1. Which of the following produces high traffic network? -----------------
a) Variable routing
b) Flooding
c) Fixed routing
d) Random routing

1. In ------------------ routing, we assume that there is one node (or more) in each autonomous system that acts on behave of the entire autonomous system.
a) distant vector
b) path vector
c) link state
d) multipoint
1. When a direct delivery is made, both the deliverer and receiver have the same---------------
a) routing table
b) host id
c) IP address
d) Netid
1. In OSPF, a ------------------ link is a network with several routers attached to it.
a) point-to-point
b) transient
c) Stub
d) Multipoint
1. In ---------------------- routing, the mask and the destination addressare both 0.0.0.0in routing table.
a) next-hop
b) host-specific
c) network-specific
d) default
1. In--------------------------- the router forwards the receive packet through only one of its interfaces.
a) unicasting
b) Multicasting
c) Broadcasting
d) point to point

1. An example of the routing algorithm is--------------------
a) TELNET
b) TNET
c) ARPANET
d) ARNET
1. ----------------------- protocol is a popular example of a link-state routing protocol.
a) SPF
b) BGP
c) RIP
d) OSPF
1. Alternate and adaptive routing algorithm belongs to ---------------------
a) static routing
b) permanent routing
c) standard routing
d) dynamic routing
1. The Enhanced Interior Gateway Routing Protocol(EIGRP) is categorized as a------------------
a) Distance vector routing protocols
b) Link state routing protocols
c) Hybrid routing protocols
d) Automatic state routing protocols
1. In ----------------routing, the routing table hold the address of just the next hop instead of complete route information.
a) next-hop
b) host-specific
c) network-specific
default.

• TEXT BOOK:
• T1: Behrouz A. Forouzan, ―“Data communication and Networking”, Fifth Edition, Tata
McGraw – Hill, 2013 (UNIT I –V)
• REFERENCES:
• R1 James F. Kurose, Keith W. Ross, “Computer Networking - A Top-Down Approach
Featuring the Internet”,
• seventh Edition, Pearson Education, 2016.
• R2 Nader. F. Mir, “Computer and Communication Networks”, Pearson Prentice Hall
Publishers, 2nd Edition, 2014.
• R3 Ying-Dar Lin, Ren-Hung Hwang, Fred Baker, “Computer Networks: An Open Source
Approach”, Mc Graw Hill
• Publisher, 2011.
• R4 Larry L. Peterson, Bruce S. Davie, “Computer Networks: A Systems Approach”, Fifth
Edition, Morgan Kaufmann
• Publishers, 2011.

THANK YOU

EC8551 COMMUNICATION NETWORKS

More Related Content

What's hot (20)

Similar to EC8551 COMMUNICATION NETWORKS (20)

Recently uploaded (20)

EC8551 COMMUNICATION NETWORKS