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K.Balachandran

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To introduce the concepts, terminologies andtechnologies used in modern days data

communication and computer networking.

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To understand the concepts of datacommunications.

To study the functions of different layers. To introduce IEEE standards employed in

computer networking. To make the students to get familiarized with

different protocols and network components.

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Text book: Behrouz A. Forouzan, Data communication and

Networking, Tata McGraw-Hill, 2004. Reference books James F. Kurose and Keith W. Ross, Computer

Networking: A Top-Down Approach Featuring theInternet, Pearson Education, 2003.

Larry L.Peterson and Peter S. Davie, ComputerNetworks, Harcourt Asia Pvt. Ltd., Second Edition.

Andrew S. Tanenbaum, Computer Networks, PHI, FourthEdition, 2003.

William Stallings, Data and Computer Communication,Sixth Edition, Pearson Education, 2000.

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DATA COMMUNICATIONS DATA LINK LAYER

NETWORK LAYER TRANSPORT LAYER APPLICATION LAYER

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Components Direction of Data flow networks Components and Categories

types of Connections Topologies Protocolsand Standards ISO / OSI model Transmission Media Coaxial Cable FiberOptics Line Coding Modems RS232Interfacing sequences.

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Why study data communications? Data communication exchange of data

between two devices via a transmissionmedium

Effectiveness depends on:

Delivery, Accuracy, Timeliness, Jitter

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Sender Receiver

Message Medium Protocol

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Text represented as a bit pattern; codesoften used:

ASCII; Extended ASCII; Unicode; ISO Numbers represented by binary equivalent Images bit patterns representing pixels Audio Video

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Simplex unidirectional; one transmits, otherreceives

Half-duplex each can transmit/receive;communication must alternate

Full-duplex both can transmit/receivesimultaneously

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Set of devices (nodes) connected by media Distributed processing

Advantages

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Performance affected by # users, type ofmedium, HW/SW

Reliability measured by freq of failure,recovery time, catastrophe vulnerability Security protection from unauthorized

access, viruses/worms

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14

Point-to-point dedicated

Multipoint shared

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Physical or logical arrangement 4 basic types: mesh, star, bus, ring

May often see hybrid

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LAN MAN

WAN

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Protocols:

Syntax: structure and format of the data

Semantics: meaning of each section of bits Timing

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De facto: standards not approved by standardagencies

De jure: legislated and officially recognized

Standards Organizations: ISO

ITU- International Telecommunication Union

CCITT: Consultative committee for InternationalTelegraphy and Telephony

ANSI: American National Standards Institute

IEEE: institute of Electrical and Electronics Engineers

EIA: Electronics Industries Association

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OSI (Open Systems Interconnect) Network architecture based on a proposal developed by

ISO (International Standards Organization) to standardize the protocols used in various layers

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20

Dedicated point-to-pointlinks to every other device

Advantages Disadvantages

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21

Dedicated point-to-pointlinks to central controller(hub)

Controller acts as exchange Advantages Disadvantages

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22

Transmission Medium and Physical Layer

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24

Categories of unshielded twisted-pair cables

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25

(a)Category 3 UTP

(b)Category 5 UTP

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UTPconnector

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Physical description:

Each wire with copper conductor

Separately insulated wires

Twisted together to reduce cross talk Often bundled into cables of two or four twisted pairs

If enclosed in a sheath then is shielded twisted pair (STP)

otherwise often for home usage unshielded twisted pair (UTP).

Must be shield from voltage lines

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Application:

Common in building for digital signaling used at

speed of 10s Mb/s (CAT3) and 100Mb/s (CAT5)over 100s meters.

Common for telephone interconnection at home

and office buildings

Less expensive medium; limited in distance,bandwidth, and data rate.

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Category Maximum data

rate

Usual application

CAT 1 Less than 1 Mbps analog voice (plain old telephone

service) Integrated Services Digital

Network Basic Rate Interface in ISDN

Doorbell wiringCAT 2 4 Mbps Mainly used in the IBM Cabling System

for token ring networks

CAT 3 16 Mbps Voice and data on 10BASE-T Ethernet

(certify 16Mhz signal)

CAT 4 20 Mbps Used in 16Mbps Token Ring

Otherwise not used much

CAT 5 100 Mbps 100 Mbps TPDDI

155 Mbps asynchronous transfer

mode (certify 100 Mhz signal)

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Widely installed for use in business and corporationEthernet and other types of LANs.

Consists of inter copper insulator covered by claddingmaterial, and then covered by an outer jacket

Physical Descriptions

Covered by sheath material

Outer conductor is braided shielded (ground)

Separated by insulating material

Inner conductor is solid copper metal

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Applications:

TV distribution (cable tv); long distance telephone transmission;short run computer system links

Local area networks Transmission characteristics: Can transmit analog and digital signals

Usable spectrum for analog signaling is about 400 Mhz

Amplifier needed for analog signals for less than 1 Km and less

distance for higher frequency Repeater needed for digital signals every Km or less distance for

higher data rates

Operation of 100s Mb/s over 1 Km.

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Physical Description: Glass or plastic core of optical fiber = 2 to125 m

Cladding is an insulating material

Jacket is a protective cover

Laser or light emitting diode providestransmission light source

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Applications:

Long distance telecommunication

Greater capacity; 2 Gb/s over 10s of Km

Smaller size and lighter weight

Lower attenuation (reduction in strength of signal)

Electromagnetic isolation not effected by externalelectromagnetic environment. more privacy

Greater repeater spacing fewer repeaters, reduces lineregeneration cost

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multimode fiber is optical fiber that is designedto carry multiple light rays or modesconcurrently, each at a slightly different

reflection angle within the optical fiber core.used for relatively short distances because themodes tend to disperse over longer lengths (thisis called modal dispersion) .

For longer distances, single mode fiber(sometimes called monomode) fiber is used. Insingle mode fiber a single ray or mode of lightact as a carrier

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0.85, 1.30, 1.55 bands Bands are 25000 to 30000 GHz wide

Dispersion: Spreading out in length as theypropagate- Hyperbolic cosine Solitons: pulses which can travel 1000s of

KMs without any appreciable shape

dispersion

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Band Range Propagation Application

VLF 330 KHz Ground Long-range radio navigation

LF 30300 KHz GroundRadio beacons and

navigational locators

MF 300 KHz3 MHz Sky AM radio

HF 330 MHz SkyCitizens band (CB),

ship/aircraft communication

VHF 30300 MHzSky and

line-of-sight

VHF TV,

FM radio

UHF 300 MHz3 GHz Line-of-sightUHF TV, cellular phones,

paging, satellite

SHF 330 GHz Line-of-sight Satellite communication

EHF 30300 GHz Line-of-sight Long-range radio navigation

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43

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Process of converting binary data to a digital signal

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Residual direct-current(dc) components or zerofrequencies areundesirable

Some systems do not allow

passage of a dc component;

may distort the signal and

create output errors DC component is extra

energy and is useless

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46

Includes timinginformation in the databeing transmitted toprevent misinterpretation

Lack of synchronization

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Unipolar Polar

Bipolar

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Simplest method; inexpensive Uses only one voltage level Polarity is usually assigned to binary 1; a 0 is

represented by zero voltage

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Potential problems:

DC component

Lack of synchronization

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50

Uses two voltage levels, one positive and onenegative

Alleviates DC component Variations

Nonreturn to zero (NRZ)

Return to zero (RZ)

Manchester

Differential Manchester

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51

Value of signal is always positive or negative NRZ-L Signal level depends on bit represented; positive

usually means 0, negative usually means 1 Problem: synchronization of long streams of 0s or

1s NRZ-I (NRZ-Invert)

Inversion of voltage represents a 1 bit 0 bit represented by no change Allows for synchronization

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52

Uses an inversion at the middle of each bit interval forboth synchronization and bit representation

Negative-to-positive represents binary 1 Positive-to-negative represents binary 0

Achieves same level of synchronization with only 2levels of amplitude

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53

Inversion at middle of bit interval is used for synchronization Presence or absence of additional transition at beginning ofinterval identifies the bit

Transition means binary 0; no transition means 1 Requires two signal changes to represent binary 0; only one to

represent 1

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54

RS-232 is the Serial interface on the PC

Three major wires for the Serial interface:

Transmit -Pin 2

Receive -Pin 3

Ground -Pin 7 (25 pin connector)

- Pin 5 (9 pin connector)

Tx Tx

RxRx

GndGnd

Computer

Device

Transmit connects to Receive

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55

TD: transmitted data RD: received data DSR: data set ready

indicate whether DCE is powered on DTR: data terminal ready

indicate whether DTR is powered on

turning off DTR causes modem to hang up the line

RI: ring indicator ON when modem detects phone call

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DCD: data carrier detect ON when two modems have negotiated successfully and

the carrier signal is established on the phone line RTS: request to send

ON when DTE wants to send data Used to turn on and off modems carrier signal in multi-

point (i.e. multi-drop) lines

Normally constantly ON in point-to-point lines CTS: clear to send

ON when DCE is ready to receive data SG: signal ground

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Means to ask the transmitter to stop/resume sending indata

Required when: DTE to DCE speed > DCE to DCE speed

(e.g. terminal speed = 115.2kbps and line speed = 33.6kbps, inorder to benefit from modems data compression protocol)

without flow control, the buffer within modem will overflow sooner or later

the receiving end takes time to process the data and thus

cannot be always ready to receive

Position of the data-linklayer

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Data link layer duties

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LLC and MAC sublayers

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IEEE standards for LANs

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Error Detection

and

Correction

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Data can be corrupted during

transmission. For reliablecommunication, errors must be detected

and corrected.

Note:

T

f E

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Types of Error

Single-Bit Error

Burst Error

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In a single-bit error, only one bit in the

data unit has changed.

Note:

Single-bit error

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A burst error means that 2 or more bits

in the data unit have changed.

Note:

Burst error of length 5

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D t ti

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Detection

Redundancy

Parity Check

Cyclic Redundancy Check (CRC)

Checksum

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Error detection uses the concept of

redundancy, which means adding extrabits for detecting errors at the

destination.

Note:

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Detection methods

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Even-pari ty concept

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In parity check, a parity bit is added to

every data unit so that the total numberof 1s is even

(or odd for odd-parity).

Note:

Example 1

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Example 1

Suppose the sender wants to send the word world. In

ASCII the five characters are coded as

1110111 1101111 1110010 1101100 1100100

The following shows the actual bits sent

11101110 11011110 11100100 11011000 11001001

Example 2

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Example 2

Now suppose the word world in Example 1 is received by

the receiver without being corrupted in transmission.

11101110 11011110 11100100 11011000 11001001

The receiver counts the 1s in each character and comes up

with even numbers (6, 6, 4, 4, 4). The data are accepted.

Example 3

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Example 3

Now suppose the word world in Example 1 is corrupted

during transmission.

11111110 11011110 11101100 11011000 11001001

The receiver counts the 1s in each character and comes up

with even and odd numbers (7, 6, 5, 4, 4). The receiver

knows that the data are corrupted, discards them, and asks

for retransmission.

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Simple parity check can detect all

single-bit errors. It can detect bursterrors only if the total number of errors

in each data unit is odd.

Note:

Two-dimensional pari ty

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Example 4

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Example 4

Suppose the following block is sent:

10101001 00111001 11011101 11100111 10101010

However, it is hit by a burst noise of length 8, and some

bits are corrupted.

10100011 10001001 11011101 11100111 10101010

When the receiver checks the parity bits, some of the bits

do not follow the even-parity rule and the whole block is

discarded.

10100011 10001001 11011101 11100111 10101010

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In two-dimensional parity check, a block

of bits is divided into rows and aredundant row of bits is added to the

whole block.

Note:

CRC generator and checker

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Binary division in a CRC generator

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Binary division in CRC checker

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A polynomial representi ng a divisor

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Standardpolynomials

Name Polynomial Application

CRC-8 x8 +x2 +x+1 ATM header

CRC-10 x10+x9 +x5 +x4 +x2+1 ATM AAL

ITU-16 x16+x12+x5 + 1 HDLC

ITU-32x32+x26 +x23+x22+x16+x12+x11+x10

+x8+x7+x5+x4+x2+x +1LANs

Example 5

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p 5

It is obvious that we cannot choose x(binary 10) or x2+ x

(binary 110) as the polynomial because both are divisibleby x. However, we can choose x + 1(binary 11) because

it is not divisible by x, but is divisible by x + 1. We can

also choose x2+ 1(binary 101) because it is divisible by

x + 1 (binary division).

Example 6

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p

The CRC-12

x12+ x11+ x3+ x + 1

which has a degree of 12, will detect all burst errors

affecting an odd number of bits, will detect all bursterrors with a length less than or equal to 12, and will

detect, 99.97 percent of the time, burst errors with a

length of 12 or more.

Checksum

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Data uni t and checksum

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The sender follows these steps:

The unit is divided into k sections, each of n bits.

All sections are added using ones complement to get the sum.

The sum is complemented and becomes the checksum.

The checksum is sent with the data.

Note:

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The receiver follows these steps:

The unit is divided into k sections, each of n bits.

All sections are added using ones complement to get the sum.

The sum is complemented.

If the result is zero, the data are accepted: otherwise, rejected.

Note:

Example 7

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Suppose the following block of 16 bits is to be sent using a

checksum of 8 bits.10101001 00111001

The numbers are added using ones complement

10101001

00111001

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

Sum 11100010

Checksum 00011101

The pattern sent is 10101001 00111001 00011101

Example 8

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Now suppose the receiver receives the pattern sent in Example 7

and there is no error.10101001 00111001 00011101

When the receiver adds the three sections, it will get all 1s, which,

after complementing, is all 0s and shows that there is no error.

10101001

00111001

00011101

Sum 11111111

Complement 00000000 means that the pattern is OK.

Example 9

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Now suppose there is a burst error of length 5 that affects 4 bits.

10101111 11111001 00011101

When the receiver adds the three sections, it gets

10101111

11111001

00011101

Partial Sum 111000101

Carry 1

Sum 11000110

Complement 00111001 the pattern is corrupted.

Correction

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Correction

Retransmission

Forward Error Correction

Burst Error Correction

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Data and redundancy bits

Number of

data bits

m

Number of

redundancy bits

r

Total

bits

m + r

1 2 3

2 3 5

3 3 6

4 3 7

5 4 9

6 4 10

7 4 11

Positions of redundancy bits in Hamming code

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Redundancy bits calculation

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Example of redundancy bit calculation

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Er ror detection using Hamming code

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Burst error correction example

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How to find an error? Introducing redundancy -- using 2 bits message to send 1 bit

information in the previous example.

Message = information bits + redundant bits (checksum).

How to design codes that have errorcorrection/detection capability? Hamming distancebetween two code words: the

number of different bits between the two code words.

E.g 010101 and 111000? Hamming distance = ?

Hamming distanceof a complete code: the minimum

Hamming distance any of the two codewords in the code.

E.g 010101, 111000, 000111, 111111 Hamming_distance= ?

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Redundancy bits calculation

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10.16 Example of redundancy bit calculation

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Data transmission in one direction only Both Transmitting and receiving host are always

ready for operation Processing time can be ignored

Infinite buffer size available Communication channels between IMPs never

damages or loses frame No sequence Number, No acknowledgements Only event is FrameArrival arrival of undamaged

frame

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typedef enum {frme_arraival} event_type;#include protocol.hvoid sender1(void){ frame s; /*buffer for an outbound frame*/

packet buffer; /*buffer for an outbound packet*/while (true)

{from_network_layer(&buffer);s.info = buffer;

to_physical_layer(&s);}

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void receiver1(void){

frame r;/*buffers for the frames*/

event_type event;while (true){

wait_for_event(&event);from_physical_layer(&r);to_network_layer(&r.info);

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typedef enum {frme_arraival} event_type;#include protocol.hvoid sender2(void){ frame s; /*buffer for an outbound frame*/

packet buffer; /*buffer for an outbound packet*/

while (true){

from_network_layer(&buffer);s.info = buffer;

to_physical_layer(&s);wait_for_event(&event);}

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void receiver1(void){frame r,s;/*buffers for the frames*/event_type event;while (true)

{wait_for_event(&event);from_physical_layer(&r);to_network_layer(&r.info);

to_physical_layer(&s); /*sending dummy frame to awaken sender */}

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ACKs introduce a new issue how long doesreceiver wait before sending ONLY an ACKframe.

We need anACKTimer!!sender timeout periodneeds to set longer.

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Each outbound frame must contain a sequence number. With nbits for the sequence number field, the numbers range from 0to maxseq.

Sliding window :: sender has a windowof frames and maintainsa list of consecutive sequence numbers for frames that it ispermitted to send without waiting for ACKs.

receiver has a window that is a list of frame sequence numbersit is permitted to accept.

Notesending and receiving windows do NOT have to be the

same size.Windows can be fixed size or dynamically growing and

shrinking.

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Host is oblivious, message order at transportlevel is maintained.

senders window ::frames sent but not yet ACKed.

new packets from the Host cause the upper

edge inside sender window to be incremented.

ACKed frames from the receiver cause the

lower edge inside window to be incremented.

All frames in the senders window must be savedfor possible retransmission and we need onetimer per frame in the window.

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If the maximum sender window size is B,the sender needs B buffers.

If the sender windowgets full (i.e., reaches

its maximum window size, the protocolmust shut off the Host (the network layer)until buffers become available.

receiver window

Frames received with sequence numbers outsidethe receiver windoware not accepted.

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receiver window Frames received with sequence numbers

outside the receiver windoware not accepted.

The receiver window size is normally static.The set of acceptable sequence numbers isrotated as acceptable frames arrive.

a receiver window size = 1the protocol

only accepts frames in order.There is referred to as Go Back N.

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A sliding window of size 1, with a 3-bit sequence number.(a)Initially.(b)After the first frame has been sent.(c)After the first frame has been received.

(d)After the first acknowledgement hasbeen received.

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(a) Normal case. (b) Abnormal case.The notation is (seq, ack, packet number). Anasterisk indicates where a network layer accepts apacket.

Go Back N

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A

B

fr0

timefr1

fr2

fr3

fr4

fr5

fr6

fr3

ACK

1 error

Out-of-sequence frames

Go-Back-4: 4 frames are outstanding; so go back 4

fr5

fr6

fr4

fr7

fr8

fr9

ACK

2

ACK

3

ACK

4

ACK

5

ACK

6

ACK7

ACK

8

ACK

9

ACKing next frame expected

Go Back Nwith NAK error recovery

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A

B

fr0

timefr1

fr2

fr3

fr4

fr5

fr1

fr2

ACK

1

error

Out-of-sequenceframes

Go-Back-7:

fr4

fr5

fr3

fr6

fr7

fr0

NAK

1

ACK

3

ACK

4

ACK

5

ACK

6

ACK

7

ACK

2

Transmitter goes back to frame 1

Figure 5.17Leon-Garcia & Widjaja: Communication NetworksCopyright 2000 The McGraw Hill Companies

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A

B

fr0

timefr1

fr2

fr3

fr4

fr5

fr6

fr2

AC

K1 error

fr8

fr9

fr7

fr10

fr11

fr12

AC

K2

NA

K2

AC

K7

AC

K8

AC

K9

AC

K10

AC

K11

AC

K12

AC

K2

AC

K2

AC

K2

Figure 5.21

Selective Repeatwith NAK error recovery

Leon-Garcia & Widjaja: Communication NetworksCopyright 2000 The McGraw Hill Companies

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HDLC

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Con f igu rat ions and Trans fer Modes

Frames

Frame Format

Examples

Data Transparency

HDLC frame

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HDLC frame types

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I-frame

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S-frame control field in HDLC

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Table 11.1 U

-

frame control command and response

Command/response Meaning

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SNRM Set normal response mode

SNRME Set normal response mode (extended)

SABM Set asynchronous balanced mode

SABME Set asynchronous balanced mode (extended)

UP Unnumbered poll

UI Unnumbered information

UA Unnumbered acknowledgmentRD Request disconnect

DISC Disconnect

DM Disconnect mode

RIM Request information mode

SIM Set initialization mode

RSET Reset

XID Exchange ID

FRMR Frame reject

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Address Control Upper-layer data FCS

DSAP

SSAP

Control

Upperlayerdata

MACheader

MACpayload

FCS

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CS 640 137

Most popular packet-switched LAN technology Bandwidths: 10Mbps, 100Mbps, 1Gbps Max bus length: 2500m

500m segments with 4 repeaters

Bus and Star topologies are used to connect hosts

Hosts attach to network via Ethernet transceiver or hub or switch

Detects line state and sends/receives signals

Hubs are used to facilitate shared connections

All hosts on an Ethernet are competing for access to the medium

Switches break this model

Problem: Distributed algorithm that provides fair access

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CS 640 138

Ethernet by definition is a broadcastprotocol Any signal can be received by all hosts

Switching enables individual hosts tocommunicate Network layer packets are transmitted

over an Ethernet by encapsulating

Frame Format

Destaddr

64 48 32

CRCPreambleSrcaddr

Type Body

1648

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CS 640 139

Physical layer configurations are specified in three parts Data rate (10, 100, 1,000)

10, 100, 1,000Mbps Signaling method (base, broad)

Baseband Digital signaling

Broadband Analog signaling

Cabling (2, 5, T, F, S, L) 5 - Thick coax (original Ethernet cabling)

F Optical fiber

S Short wave laser over multimode fiber

L Long wave laser over single mode fiber

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CS 640 140

Developed in late 60s by Norm Abramson at Univ. of Hawaii(!!) for use with packet radio systems Any station can send data at any time

Receiver sends an ACK for data

Timeout for ACK signals that there was a collision What happens if timeout is poorly timed?

If there is a collision, sender will resend data after a random backoff

Utilization (fraction of transmitted frames avoiding collisionfor N nodes) was pretty bad

Max utilization = 18% Slotted Aloha (dividing transmit time into windows) helped Max utilization increased to 36%

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CS 640 141

In Aloha, decisions to transmit are made without paying

attention to what other nodes might be doing Ethernet uses CSMA/CD listens to line before/during sending If line is idle (no carrier sensed) send packet immediately upper bound message size of 1500 bytes

must wait 9.6us between back-to-back frames If line is busy (carrier sensed) wait until idle and transmit packet immediately

called 1-persistent sending If collision detected Stop sending and jam signal Try again later

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CS 640 142

Packet?

Sense

Carrier

Discard

Packet

Send Detect

Collision

Jam channel

b=CalcBackoff();wait(b);

attempts++;

No

Yes

attempts < 16

attempts == 16

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CS 640 143

A B

A B

Collisions are caused when two adaptors transmit at the samtime (adaptors sense collision based on voltage differences)

Both found line to be idleBoth had been waiting to for a busy line to become idle

A starts attime 0

Message almostthere at time T when

B starts collision!

How can we be sure A knows about the collision?

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CS 640 144

How can A know that a collision has taken place? There must be a mechanism to insure retransmission on collision As message reaches B at time T Bs message reaches A at time 2T So, A must still be transmitting at 2T

IEEE 802.3 specifies max value of 2T to be 51.2us This relates to maximum distance of 2500m between hosts At 10Mbps it takes 0.1us to transmit one bit so 512 bits (64B) take 51.2us to send So, Ethernet frames must be at least 64B long

14B header, 46B data, 4B CRC

Padding is used if data is less than 46B Send jamming signal after collision is detected to insure all hosts see

collision 48 bit signal

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CS 640 145

A B

A B

A B

time = 0

time = T

time = 2T

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CS 640 146

If a collision is detected, delay and try again Delay time is selected using binary exponential backoff 1st time: choose K from {0,1} then delay = K * 51.2us 2nd time: choose K from {0,1,2,3} then delay = K * 51.2us nthtime: delay = K x51.2us, for K=0..2n1

Note max value for k = 1023

give up after several tries (usually 16) Report transmit error to host

If delay were not random, then there is a chance that sourceswould retransmit in lock step

Why not just choose from small set for K

This works fine for a small number of hosts

Large number of nodes would result in more collisions

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CS 640 147

Senders handle all access control Receivers simply read frames with acceptable

address

Address to host

Address to broadcast

Address to multicast to which host belongs

All frames if host is in promiscuous mode

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CS 640 148

Fast Ethernet (100Mbps) has technology very similar to10Mbps Ethernet Uses different physical layer encoding (4B5B)

Many NICs are 10/100 capable Can be used at either speed

Gigabit Ethernet (1,000Mbps) Compatible with lower speeds

Uses standard framing and CSMA/CD algorithm

Distances are severely limited

Typically used for backbones and inter-router connectivity

Becoming cost competitive

How much of this bandwidth is realizable?

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a carrier-sense, multiple access with collisiondetection (CSMA/CD) protocol for a bustopology

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Token Bus protocol for a token-passingaccess method on a bus topology

The topology of the computer network can

include groups of workstations connectedby long trunk cables

token busis used in some manufacturing

environments, Ethernet and token ringstandards have become more prominent inthe office environment.

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Token Ring protocol which, like Token Bus, isanother token-passing access method, butfor a ring topology

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TOKEN RING NETWORK LIKE IEEE 802.5

TOKEN: A SPECIAL SEQUENCE OF BITS TOKEN CIRCULATES AROUND THE RING

A STATION REMOVES THE TOKEN FROM RINGBEFORE TRANSMISSION

AFTER TRANSMISSION, THE STATION

RETURNS THE TOKEN TO THE RING

COLLISIONS ARE PREVENTED AS THERE ISONLY ONE TOKEN IN THE RING

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DUAL-COUNTER-ROTATING TOKENRING ARCHITECTURE

ONE RING IS PRIMARY AND THEOTHER SECONDARY

UP TO 500 STATIONS WITH A

MAXIMUM DISTANCE OF 2 KMBETWEEN ANY PAIR OF STATIONS

FOR MULTIMODE FIBER

WITH SINGLE-MODE FIBER THE

DISTANCE CAN BE UP TO 40 KM

MAXIMUM RING LENGTH IS 100 KM(TOTAL FIBER LENGTH IS 200 KM

FOR TWO RINGS)

USES 4B/5B ENCODING

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154

FEATURES FDDI ETHERNET TOKEN RING

TRANSMISSION

RATE

125 MBAUD 20 MBAUD 8 & 32 MBAUD

DATA RATE 100 MBPS 10 MBPS 4 & 16 MBPS

SIGNALENCODING

4B/5B (80%EFFICIENT)

MANCHESTER(50%

EFFICIENT)

DIFFERENTIALMANCHESTER

(50% EFFICIENT)

MAXIMUM

COVERAGE

100 KM 2.5 KM CONFIGURATION

DEPENDENT

MAXIMUMNODES

500 1024 250

MAXIMUM

DISTANCE

BETWEEN

NODES

2 KM (MULTIMODE

FIBER)

40 KM (SINGLE-

MODE FIBER)

2.5 KM 300 M

(RECOMMENDED

100 M)

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802.11 is primarily concerned withthe lower layers of the OSI model.

Data Link Layer

Logical Link Control (LLC).

Medium Access Control (MAC).

Physical Layer

Physical Layer ConvergenceProcedure (PLCP).

Physical Medium Dependent (PMD).

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Well-supported, stable, and cost effective,but runs in the 2.4 GHz range that makes itprone to interference from other devices(microwave ovens, cordless phones, etc) andalso has security disadvantages.

Limits the number of access points in rangeof each other to three.

Has 11 channels, with 3 non-overlapping, and

supports rates from 1 to 11 Mbps, butrealistically about 4-5 Mbps max.

Uses direct-sequence spread-spectrumtechnology.

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Completely different from 11b and 11g. Flexible because multiple channels can be combined for

faster throughput and more access points can be co-located. Shorter range than 11b and 11g. Runs in the 5 GHz range, so less interference from other

devices. Has 12 channels, 8 non-overlapping, and supports rates from

6 to 54 Mbps, but realistically about 27 Mbps max Uses frequency division multiplexing

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Speed Slower than cable. Range Affected by various medium. Travels best through open space.

Reduced by walls, glass, water, etc Security Greater exposure to risks. Unauthorized access.

Compromising data.

Denial of service.

N t

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A bridge has a table used in filtering

decisions.

Note:

Bridge

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Networks: SONET 163

Digital carrier systems

The hierarchy of digital signals that the telephone networkuses.

Trunks and access links organized in DS (digital signal)hierarchy

Problem: rates are not multiples of each other.

In the 1980s Bellcore developed the Synchronous

Optical Network (SONET) standard. Previous efforts include: ISDNand BISDN.

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SONET:: encodes bit streams into optical signals propagatedover optical fiber. SONET defines a technology for carryingmany signals of different capacities through a synchronous,flexible, optical hierarchy.

A bit-way implementation providing end-to-end transportof bit streams.

All clocks in the network are locked to a common masterclock so that simple TDM can be used.

Multiplexing done by byte interleaving.

SONETis backward compatible to DS-1 and E-1 and forwardcompatible to ATM cells.

Demultiplexing is easy.

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Networks: SONET 165

SONET topology can be a mesh, but mostoften it is a dual ring.

Standard component of SONET ring is an

ADM (Add/Drop Multiplexer) Drop one incoming multiplexed stream and

replace it with another stream.

Used to make up bi-directional line switchingrings.

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Networks: SONET

a

b

c

d

a

b

c

d

(a) Dual ring (b) Loop-around in response to fault

Figure 4.12

ADM

ADM

ADM

ADM

Leon-Garcia & Widjaja: Communication NetworksCopyright 2000 The McGraw Hill Companies

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Networks: SONET

SONET Ring

(a)

STS

PTE

STS

PTE

SONET Architecture

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STE: Section Terminating Equipment, e.g. a repeaterLTE: Line Terminating Equipment, e.g. a STS-1 to STS-3 multiplexer

PTE: Path Terminating Equipment, e.g. an STS-1 multiplexer

optical

section

optical

section

optical

section

optical

section

line

optical

section

line

optical

section

line

path

optical

section

line

path

(b)

PTELTE

STE

STS-1 Path

STS Line

Section Section

Mux Muxreg reg reg

SONETTerminal

STE STELTE

SONETTerminal

Learning bridge

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Loop problem

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Prior to spanning tree application

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Position of network layer

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Network layer duties

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Internetwork

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Links in an internetwork

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Network layer in an internetwork

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Network layer at the source

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Note

:

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An IP address is a 32-bit address.

Note:

Note

:

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The IP addresses are unique

and universal.

Note:

Dotted-decimal notation

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Note

:

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The binary, decimal, and hexadecimal

number systems are reviewed in

Appendix B.

Note:

Example 1

Change the following IP addresses from binary notation to

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dotted-decimal notation.

a. 10000001 00001011 00001011 11101111

b. 11111001 10011011 11111011 00001111

Solution

We replace each group of 8 bits with its equivalent decimalnumber (see Appendix B) and add dots for separation:

a. 129.11.11.239b. 249.155.251.15

Example 2

Change the following IP addresses from dotted-decimal notation

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to binary notation.

a. 111.56.45.78

b. 75.45.34.78

SolutionWe replace each decimal number with its binary equivalent (seeAppendix B):

a. 01101111 00111000 00101101 01001110b. 01001011 00101101 00100010 01001110

Finding the address class

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Example 3

Find the class of each address:

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a. 00000001 00001011 00001011 11101111b. 11110011 10011011 11111011 00001111

Solution

See the procedure in Figure 19.11.

a. The first bit is 0; this is a class A address.b. The first 4 bits are 1s; this is a class E address.

Finding the class in decimal notation

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Example 4

Find the class of each address:

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a. 227.12.14.87b. 252.5.15.111

c. 134.11.78.56

Solutiona. The first byte is 227 (between 224 and 239); the class is D.

b. The first byte is 252 (between 240 and 255); the class is E.c. The first byte is 134 (between 128 and 191); the class is B.

Netid and hostid

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Blocks in class A

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Blocks in class B

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Blocks in class C

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Network address

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First Octet IP Address Characteristics

MostSignificant

BITS

ValueRanges

Addr.Class

Networkvs.

Host

#NETWORK

S# HOSTS

0000 0-126 A N.h.h.h 256 16,777,214

-- 127 - - Special - Local Loopback1000 128-191 B N.N.h.h 65,536 65,534

1100 192-223 C N.N.N.h 16,777,216 254

1110 224 - 239 D Special N/A N/A

1111 240 + E Special N/A N/A

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5-4

The Optimality Principle

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The Optimality Principle

Shortest Path Routing Flooding Distance Vector Routing

Link State Routing

Hierarchical Routing

Broadcast Routing

Multicast Routing

Routing for Mobile Hosts Routing in Ad Hoc Networks

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The first 5 steps used in computing the shortest path from A to D.The arrows indicate the working node.

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(a)A subnet. (b)Input from A, I, H, K, and the newrouting table for J.

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The count-to-infinity problem.

Each router must do the following:

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Each router must do the following:

1. Discover its neighbors, learn their network address.2. Measure the delay or cost to each of its neighbors.3. Construct a packet telling all it has just learned.

4. Send this packet to all other routers.

5. Compute the shortest path to every other router.

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(a)Nine routers and a LAN. (b)A graph modelof (a).

A subnet in which the East and West parts are

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A subnet in which the East and West parts areconnected by two lines.

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(a) A subnet. (b) The link state packets for thissubnet.

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The packet buffer for router B in the previousslide (Fig. 5-13).