data link layer - mississippi state universityweb.cse.msstate.edu/~ramkumar/dllayer.pdf · data...

Services Provides By DL LayerFraming

Error ControlData Link Protocols

Internet DL Protocols

Data Link Layer

Mahalingam RamkumarMississippi State University, MS

September 18, 2006

Ramkumar PL




Outline

1 Services Provides By DL LayerData Link Layer Issues

2 Framing

3 Error ControlError CorrectionError Detection

4 Data Link ProtocolsIssuesPrinciples of Reliable Data TransferEffective Data RateProtocols With Pipelining - Sliding Window Protocols

5 Internet DL ProtocolsPPP

Ramkumar PL




Data Link Layer Issues

Data Link Layer

Provide services to network layer

Types of services

Unacknowledged connectionless serviceAcknowledged connectionless serviceAcknowledged connection-oriented service

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Packet vs Frame

TrailerHeader Payload field

Frame

Sending machine

PacketPacket

Receiving machine

TrailerHeader Payload field

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Virtual / Actual Communication Path

4

3

2

1

4

3

2

1

4

3

2

1

4

3

2

1

Host 1 Host 2 Host 1 Host 2

Virtual data path

Actual data path

(a) (b)

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Purpose

Well defined service to network layer

Deal with transmission errors

Flow control

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Issues

All we need to do is send a sequence of bits from one end of awire to the other end.

How complicated can that be?

Synchronization

Logical organization of frames

Did the data actually reach the other end?

Integrity of the data received (remember noise?)

Type of service provided to higher layers.

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Framing

Organization of frames

Methods

Character countByte stuffingBit stuffingUsing physical layer “redundancy”

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Character Count

Errors may result in loss of synchronization

(b)

(a) 5 1 2 3 4 5 6 7 8 9 8 0 1 2 3 4 5 6 8 7 8 9 0 1 2 3

5 1 2 3 4 7 6 7 8 9 8 0 1 2 3 4 5 6 8 7 8 9 0 1 2 3

Character count One character

Error

Frame 1 5 characters

Frame 1


Frame 2 (Wrong)



Now a character count

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Byte Stuffing

Used in PPP. Only for byte oriented communications

A BESC FLAG

A BESC ESC

A ESC BESC ESC FLAG

A ESC BESC ESC ESC

BA FLAG

BA ESC

FLAGA ESC B

ESCA ESC B

FLAGTrailerFLAG Header Payload field

Original characters After stuffing

(a)

(b)

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Bit Stuffing

Simple algorithm - 01111110 used as flagAt transmitter - after 5 consecutive ones insert a zeroAt receiver - In a pattern of 111110 anywhere in the data, removethe trailing zero!

0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 0

0 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0 1 0 0 1 0

Stuffed bits

(a)

(b)

(c) 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 0

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Error CorrectionError Detection

Errors Happen!

Need to detect errors

If possible it will also be nice if we could correct errors

The key - redundancy

Some extra bits need to be transmitted for detecting /correcting errors

Example - parity bits

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Hamming Distance

A metric for measuring “distances” between two sequences ofbits

A = 1000110

B = 1010100

How many bits need to be flipped to get B or A (orvice-versa)?

Two bits - the Hamming distance between A and B is 2.

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Error Detection with Parity Bit

A = 1000110

With even parity A = 10001101

With odd parity A = 10001100

For all subsequent discussions we will only use even parity

If any of the eight (7+1) bits of A is flipped duringtransmission parity check will fail.

What happens if two bits get flipped?

Parity check fails to detect error...

With one redundant bit we can only detect one-bit errors.

Actually any odd number of bit-errors

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Redundancy

Need to transmit m bit symbols (2m possible symbols)

r redundant bits (for error detection / correction)

n = m + r bits actually transmitted.

Efficiency is mm+r = m

n

In general, more the redundancy, more the errors that can bedetected / corrected

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A Simple-Minded Approach

Just repeat every bit twice (efficiency 0.5) - any one bit errorcan be detected

Parity bit (just add one extra bit) - efficiency mm+1 - very high

for large m.

Why was the first method so inefficient?

Actually it is not - it can detect much more than “single biterrors”

How can we correct single bit errors automatically?

Repeat each bit three times (efficiency 0.33)

Are there better ways?

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Error Detection vs Correction

Obviously error correction needs more redundancy

Error detection is much more crucial than error correction -Why?

Typically error detection techniques address the problem ofdetecting multiple-bit errors.

Error correction is usually more often used only for correctionof single bit errors.

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Single Bit Errors are Statistically More Frequent!

Let us assume that the probability that any bit may beerroneously received is p = 10−9.

If N = 103 bits are transmitted, probabilty that there may bea bit error is roughly one in a million.

Probabilty that there may be two bit errors is roughly one in atrillion

We can just “ignore” two bit errors

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Our Focus

How do we correct one-bit errors?

If the receiver is able to correct errors, the sender does notneed to retransmit

How do we detect multiple-bit errors?

If error is detected, the receiver can request the sender toretransmit

Or simply not acknowledge reception

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One-bit Error Correction

m bit numbers need to be sent across a link

r redundant bits

We transmit n = m + r bits instead of m - efficiency mn

With n bits we can represent 2n unique sequences

We need to convey any of 2m unique sequences

Represent each of the 2m sequences with one n bit sequence

At the receiver there may be either one one-bit-error or noerrors

When a sender sends n bits, the receiver can receive upton + 1 different sequences

To detect error, the receiver should be able to get the correctn bit sequence from any of the n + 1 possible sequences.

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Partition!

Partition the 2n numbers into 2m non-overlapping partitionsEach partition should have n + 1 elementsEach partition has one sequence which is a representation ofone of the 2m symbols.Under no error or one bit error, the recieved sequence will stillbe in the same partitionHow much redundancy is needed?2n ≥ (n + 1)2m. Or 2m+r ≥ (m + r + 1)2m. Or2r ≥ m + r + 1.m ≤ 2r − r − 1.m = 1; r = 2; (1 = 4− 2− 1 = 1). This is a (1, 3) codem = 2; r = 3; (2 ≤ 8− 3− 1 = 4).m = 4; r = 3; (4 = 8− 3− 1 = 4). This is a (4, 7) code.r = 10, 2r − r − 1 = 1024− 10− 1 = 1013. (1013, 1024) code

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Hamming Codes

m = 2r − r − 1 - efficient codes. (4, 7), (11, 15)

Lets analyze the (4, 7) code: r = 3 parity bits, n = 7transmitted bits, m = 4 information bits.

Each bit position (1 to 7) can be represented in binary form

1 0 0 1 1 bits 1,2,4 are parity bits2 0 1 0 2 1 0 0 1 → 1 0 0 x 1 x x3 0 1 1 1 2 Even parity4 1 0 0 4 1 0 0 x 1 x x → 1 0 0 x 1 x 0

5 1 0 1 1 4 1 0 0 x 1 x 0 → 1 0 0 x 1 0 0

6 1 1 0 2 4 1 0 0 x 1 0 0 → 1 0 0 1 1 0 0

7 1 1 1 1 2 4 1001100 is the (4,7) code for 1001

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1001 → 1001100

Rx checks all parity bits and wites the result as xxx

If no error, all parities check out. So xxx = 000.

What happens if bit 6 is flipped? Rx gets 1101100

Parity bits 4 and 2 are wrong! So xxx = 110

If bit 5 is flipped parity bits 4 and 1 are wrong. xxx = 101

If bit 1 is flipped only parity bit 1 is affected. xxx = 001



xxx gives you the position of the error! Go ahead and correctit...

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Syndrome Decoding

A syndrome is a “collection of symptoms.”

Each result of parity check is a symptom.

The syndrome should unambiguously indicate the error.

With r redundant bits, we have 2r unique syndromes

We can detect 2r − 1 different “diseases” (one syndromecorresponds to “no illness”).

n ≤ 2r − 1

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Burst Errors

In practice, burst errors are more common (example uses (7,11)

Hamming code)

Char.

H a m m i n g c o d e

ASCII 1001000 1100001 1101101 1101101 1101001 1101110 1100111 0100000 1100011 1101111 1100100 1100101

Check bits 00110010000 10111001001 11101010101 11101010101 01101011001 01101010110 01111001111 10011000000 11111000011 10101011111 11111001100 00111000101 Order of bit transmissionRamkumar PL





Error Detection

Single error detection with parity bit

Burst error detection?

arrange M = w × h bits as a matrixParity bit for each row (h parity bits)Transmit column-wiseWe have w + 1 columns - total bits (w + 1)× h

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Multiple Error Detection

Double errors

arrange M = w × h bits as a matrix

Parity bit for each row (h parity bits) - we have (w + 1)× hmatrix

Parity bit for each columm - we end up with (w + 1)× (h + 1)matrix

Why does this work?

If 2 errors happen in the same row (column) they cannot be inthe same column (row)

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Cyclic Redundancy Checks

Need some background on polynomial arithmetic

Any bit string is considered as a polynomial in the field {0, 1}A polynomial - P(x) = x r + ar−1x

r−1 + · · ·+ a1x1 + a0

If P(x) is in the field of real numbers, x , a0 . . . ar−1 canassume any real value

P(x) in {0, 1} - x , a0 . . . ar−1 can only assume two values - 0or 1

Example - 110101 is represented as P(x) = x5 + x4 + x2 + 1.

Addition - 1 + 1 = 0 = 0 + 0, 1 + 0 = 0 + 1 = 1

Subtraction is the same as addition!

Multiplication - 1× 1 = 1, 0× x = 0.

P(0) = 1. P(1) = 1 + 1 + 1 + 1 = 0.

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Polynomial Arithmetic

P(x) = x5 + x4 + x2 + 1. Q(x) = x2 + x1

P(x) + Q(x) = x5 + x4 + x1 + 1

P(x)× Q(x) = (x5 + x4 + x2 + 1)(x2 + x1) =x7+x6+x4+x2+x6+x5+x3+x1 = x7+x5+x4+x3+x2+x1.

Polynomial division - what is P(x)/Q(x)

110 |110101 | 1001 (quotient)110---000101

110---011 (remainder)

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CRC

Choose a generator polynomial G (x) of degree r (r + 1 bitnumber)

Message to be sent M(x) (degree of M(x) not less than r)

Append r zeros to M(x). The result is x rM(x).

Evaluate the remainder of x rM(x)/G (x) (remainder will be apolynomial of degree less than r) - say R(x)

Transmit T (x) = x rM(x)− R(x) = x rM(x) + R(x)

Receiver receives T (x)

In case of error receiver gets T ′(x) = T (x) + E (x)

Receiver checks if remainder of T ′(x)/G (x) is zero

If remainder is not zero receiver decides that there is an error

As T (x)/G (x) = 0, T ′(x)/G (x) = E (x)/G (x)

Errors where E (x)/G (x) is zero goes undetected

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Good choice for G (x)

Both highest and lowest order bits should be 1 (1xx · · · xx1)

If G (x) has degree more than 1 all single bit errros will bedetected - E (x) = x i .

Two isolated single bit errors E (x) = x i + x j = x j(x i−j + 1)

Any G (x) that does not have xk + 1 as a factor (for allk ≤ m) is sufficient

If x + 1 is not a factor, then all odd nuber of errors can bedetected.

Any burst error of length ≤ r can be detected

32 bit polynomial specified in IEEE 802

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IssuesPrinciples of Reliable Data TransferEffective Data RateProtocols With Pipelining - Sliding Window Protocols

Basic Functions

Need to transmit sequence of packets from source todestination

Reliably (reception should be acknowledged)

Need to provide packets in the correct order to the higher layer

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Fundamental Limitations

Packet size (limited usually by channel error-rate)

Bits - number of bits - say P bitsDuration - depends on channel data rate. If rate is R bps thenpacket duration is τp = P

R seconds

Finite propagation time - depends on distance between sourceand destination, and velocity of propagation (electromagneticwaves) in the medium

If c is the velocity, and d is the distance, propagation time isτc = d

c .

Typically c ≈ 2.5× 108 m/s in copper wires.

Processing time τr - a finite amount of time needed forprocessing packets

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Example

100 Mbps channel, packet size 2000 bits

τp = 2000100000000 = 20 µs

d = 1000m. c = 2.5× 108 m/s. τc = 10002.5×108 = 4µs

Processing time - say 5µs

At t = 0 A starts transmission

At t = 4µs B senses the leading edge of the packet

At t = 24µs B senses trailing edge of the packet

At t = 29µs B starts sending the reply packet (say 2000 bits)

At t = 33µs A senses leading edge of reply

At t = 53µs A senses trailing edge of reply

Round trip time (RTT) - τp + τc + τr + τc + τp

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Reliable Data Transfer

Need to transmit a sequence of packets

Packets need to be acknowledged

Packets can get “lost”

Timeout for retransmission

But acknowledgements can get “lost” too!

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Lost Acknowledgements

Can both sender and receiver agree that P1′ is a retransmission ofP1 and not P2?

Timeout Interval

X(B)

ACK ACK

P1 P1’

(A)

Timeout Interval

ACK

P1 P2

ACK

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Packet Numbers!

Packets need to be numbered

Obviously numbers will be “recycled”

How many unique sequence numbers do we need?

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Ambiguous Acknowledgements

How does the sender know that the second acknowledgement is forP1 or P2?

Timeout Interval

X(C)

P1

ACK

P1’ P2

ACK

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Numbering

Acknowledgements need to be numbered too!

ABP protocol

One bit packets numbers - 0 or 1

One bit ACK numbers

Is this unambiguous?

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Alternating Bit Protocol (ABP)

Timeout Interval

0

1 2

1

1 2

Packet NoSeq No

Ack NoPacket No

0 1

(D)

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ABP is Unambiguous!

Timeout Interval

X(E)

1 1 2

0

0

0

Timeout Interval2

1 10

1 1 2

Timeout Interval

X(F)

10

10

0 0

2

2

1

Packet NoSeq No

Ack NoPacket No

Packet NoSeq No

Ack NoPacket No

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Effective Data Rate

100 Mbps link A ↔ B

d(A,B) = 2000m

packet size 100 bits

τp = 100100000000 = 1µs

τc = 20002.5×108 = 8µs

Assume τr = 1µs

What is the effective data rate between A and B when ABP isused?

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Sequence of Events

t = 0 - A starts transmission

t = τp + τc + τr = 10µ,

B starts acknowledgement packet to A

t = 10 + τp + τc + τr = 20µs

A starts sending packet 2 to B

t = 29µs - B completes reception of packet 2 from A

t = 30µs - B starts ACK for P2 to A.

t = 40µs - A is ready to send P3 to B.

Takes 40µs to send 2 packets from A to B

5 Mbps bit-rate achieved in a 100 Mbps line!

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Improving Throughput

What if A sends 8 packets without waiting foracknowledgement?

Window size 8.

Receiver could

Send ACK after 8 packetsSend ACK after every packet

Lets assume Rx sends ACK for every packet

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Improving Throughput

At t = 0, 1, 2, . . . , 8µs A starts sending packets 1 · · · 8 to B

At t = 10µs B starts ACK to A

At t = 11, 12, . . . 18µs B starts ACK to A for packets 2,3,...8

At t = 20µs - A starts sending packet 9

If everything goes well, at t = 40µs, A starts sending packet17.

Throughput - 20µs for 800 bits - or 40 Mbps

What if we used window size of 20?

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Pipelining

The secret to higher throughput

Assumes nodes can do “multiple” things at the same time.

ABP needs only half-duplex channels, for larger window sizeswe implicitly assumed full duplex channels

Nodes may need to buffer packets when things do not go well

What is the buffer size needed?

How do we number packets and acknowledgements

What if we do not acknowledge every packet (and sendcollective ACKs once in a while)?

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Selective Repeat Protocol

Similar to ABP, with window size W

Or ABP is SRP with window size 1

Pipelining

Buffer size of W packets

Numbering packets?

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SRP Rules

2W distinct numbers. If W = 4, numbers 0,1,...7.

P1 to P8 numbered 0 to 7 respectively

P9 / P17 are numbered 0. P10 / P18 are numbered 1...andso on.

If the lowest unacknowledged packet is i , sender is allowed tosend packets i through i +W −1 (sliding window of width W )

If sender has received ACKs for P1, P3 and P4 (or sender hasreceived ACKs with numbers 0, 2 and 3), sender can sendpackets P2 · · ·P5. Sender cannot send P6.

There can be no more than W outstanding unacknowledgedpackets.

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SRP: Receiver Logic

If the last acknowledged packet in an unbroken sequence is j ,the next received packet could be between j −W + 1 andj + W .

Receiver who has (received and) acknowledged packetsP1 · · ·P4 could receive (as the next packet) P1 · · ·P8 (or apacket numbered between 0 and 8). The next packet cannotbe P9.

If receiver has acknowledged P1 · · ·P5 and P7 (or packetswith numbers 0, 1, 2, 3, 4, and 6) the next packet could be inthe range P2 · · ·P9.

Cannot be P1. Cannot be P10.

If a packet with number 0 is received, is it P1 or P9?

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SRP, W = 4, Numbering 0,1,2,3

Timeout Interval

1 2 3 4 5 6 71 2

01 2

132

4 50

61

72

0 1 2 3 0

3

X

43

3

8 or 4 ?

Data Packet

Seq NoData Packet

Seq No

1)

2)

Data Packet Ack No

Data Packet Ack No

Timeout Interval

1 2 3 4 5 6 71 2

83

01 2

132

4 50

61

72

83

0 1 2 3 0

3

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SRP, W = 4, Numbering 0,1,2,3,4,5,6

Timeout Interval

1 2 3 4 5 6 75 6

80

01 2

132

4 54

65

76

80

0 1 2 3 4

3

Timeout Interval

1 2 3 4 5 6 75 6

80

01 2

132

40

0 1 2 3 4

3X X X

1 or 8?

Data Packet

Data Packet

Seq No

Seq No

3)

4)

Data Packet Ack No

Data Packet Ack No

Timeout Interval

1 2 3 4 10

01 2

132

4

0 1 2 3

3

X

1 or 8?

Data Packet Seq No

5)

Data Packet Ack No

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Negative Acknowledgements (NACK)

Say W = 100, and receiver did not receive packet 2 (butreceived packet 1 and 3).

Normally sender would wait for timeout for ACK for 2 beforehe knows that packet 2 was not received.

But an acknowledgement for 3 could indicate “negativeacknowledgement” for 2

In case sender has not sent all packets, he could stop sendingfurther packets and retransmit 2.

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Go Back N - GBN Protocol

Window size W

Receiver does not buffer packets

Peridically sends acknowledgement for the last recieved packetnumber in sequence

Out of sequence packets are discarded

Example, if Rx gets packets 1,2,3,4,5,7,8,9,10 the ACK is tillpacket 5. Sender gets no info about subsequent packets.

Sender retransmits all packets 6 and above.

If the last acknowledged packet is i sender can send packetsi + 1 to i + W .

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PPP

PPP in Dial-up Internet

PC

Routing process

Router

Modems

Internet provider's office

User's home

Client process using TCP/IP

Modem

Dial-up telephone line

TCP/IP connection using PPP

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PPP

Point-to-Point Protocol

RFC 1662, 1663

Features

FramingLink Control Protocol (LCP)Network Control Protocol (NCP)

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PPP

Overview of PPP

Byte oriented protocol

Byte stuffing for framing

Modem (client) makes a call to the modem of the serviceprovider

Connection established

LCP packets sent in PPP frames - negotiation for parameters(eg. data rate) to be used

Once a link is established, network layer parameters have tobe setup (NCP)

IP address - usually assigned by the service provider

Now the host has access to the Internet!

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PPP

PPP Frame Format

Flag byte 01111110

Address (recall that it is not needed for DL layer) always setto 11111111

Control field - default value indicates no ACKs.

During LCP parties can negotiate a decide to drop addressand control bytes

Protocol - specifies LCP, NCP and network protocols like IP,IPX, AppleTalk etc.

Payload

Checksum - 2 or 4 bytes (negotiated during LCP)

Flag 01111110

Flag 01111110

Address 11111111

ProtocolControl 00000011

Payload Checksum

Bytes 1 1 1 or 21 Variable 2 or 4 1

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PPP

PPP Phases

NetworkDead

Terminate Open

Establish Authenticate

Carrier detected

Both sides agree on options

Authentication successful

NCP configuration

Carrier dropped

Failed

Failed

Done

Ramkumar PL

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