lecture3logicallinklayer__2014_08_22_07_26_57.pptx

8/10/2019 Lecture3LogicalLinkLayer__2014_08_22_07_26_57.pptx

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Logical Link Layer

Prof. Hemang Kothari

Assistant Professor

Computer Engineering DepartmentMEFGI, Rajkot.

Email: [email protected]

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Function Provided By Logical Link La

•

Provide service interface to the network layer• Dealing with transmission errors

• Regulating data flow

– Slow receivers not swamped by fast senders



Content

•

Design Issue1. Framing.

2. Error control.

3. Flow control.

4. Error detection and correction.

• Elementary data link protocols: Simplex, Stop a

Sliding window protocol, HDLC.



Framing Methods

• Character Count.

• Starting & Ending Bytes (FLAG) With Byte Stuffi

• Starting & Ending Bit Pattern (Flag) With Bit Stu



Character Count - Framing

• This method specifies the number of characters

present in particular frame.

• This information is specified by using a special fi

header frame.



A character stream. (a) Without errors. (b) With on



Byte Stuffing – Framing

(a) A frame delimited by flag bytes.

(b) Four examples of byte sequences before and after stuffing.



Bit Stuffing

Bit stuffing

(a) The original data.

(b) The data as they appear on the line.

(c) The data as they are stored in receiver’s memory after d



Review

• In a data link protocol, the frame delimiter flag is give

0111. Assuming that bit stuffing is employed, the tran

sends the data sequence 01110110 as

(A) 01101011

(B) 011010110

(C) 011101100

(D) 0110101100



Review

• The following data fragment occurs in the middle of a

stream for which the bytestuffing algorithm describe

text is used: A B ESC C ESC FLAG FLAG D. What is the

after stuffing?

• Answer : A B ESC ESC C ESC ESC ESC FLAG ESC FLAG D



Review

• A bit string, 0111101111101111110, needs to be tran

at the data link layer. What is the string actually trans

after bit stuffing?

• Answer: The output is 011110111110011111010.



Flow Control & Error Control



Flow Control

• Flow control coordinates the amount of data that ca

before receiving acknowledgement

• Flow control is a set of procedures that tells the se

much data it can transmit before it must wa

acknowledgement from the receiver.

• Receiver has a limited speed at which it can process

data and a limited amount of memory in which

incoming data.



Flow Control

• Receiver must inform the sender before the limits ar

and request that the transmitter to send fewer fram

temporarily.

• Since the rate of processing is often slower than t

transmission, receiver has a block of memory (b

storing incoming data until they are processed.

Feedback is the breakfast of champions – Ken Blanch



Error Control

• Error control includes both error detection – corre

notification about error.

• It allows the receiver to inform the sender if a fram

damaged during transmission and coordin

retransmission of those frames by the sender.

• Error control in the data link layer is based on autom

request (ARQ). Whenever an error is detected,

frames are retransmitted.



Error and Flow Control Mechanism

• Stop-and-Wait

• Go-Back-N ARQ

• Selective-Repeat ARQ



Stop and Wait• Sender keeps a copy of the last f

receives an acknowledgement.

• For identification, both data acknowledgements (ACK) frames aalternatively 0 and 1.

• Sender has a control variable (S) tnumber of the recently sent frame.

• Receiver has a control variable (R) tnumber of the next frame expected

• Sender starts a timer when it sendan ACK is not received within a aperiod, the sender assumes that thlost or damaged and resends it

• Receiver send only positive ACK ifintact.

• ACK number always defines the nunext expected frame.



Stop-and-Wait ARQ, Corrupt fram

•

When a receivdamaged fram

it and keeps its

• After the tim

sender expir

copy of frame



Stop-and-Wait, lost ACK frame

• If the sender

damaged ACK, it

• When the tim

sender expires,

retransmits fram

• Receiver has

received framexpecting to rece

(R=0). Therefore

the second copy



Stop-and-Wait, delayed ACK frame

• The ACK can bethe receiver some problem

• It is receivedtimer for fraexpired.

• Sender retrancopy of frame 0

R =1 meanexpects to seeReceiver discduplicate frame

• Sender receivediscards the se



Disadvantage of Stop-and-Wait

• In stop-and-wait, at any point in time, there is only on

that is sent and waiting to be acknowledged.

• This is not a good use of transmission medium.

• To improve efficiency, multiple frames should be in tr

while waiting for ACK.

• Two protocol use the above concept,

– Go-Back-N ARQ

– Selective Repeat ARQ



Mathematical Review of Stop & Wa

• The Stop and Wait protocol is executed on a 10km wir

your modem and your ISPs edge router. The propagatioan electrical signal on the wire is 200,000 km/sec. The is 1000 bytes. The time to generate an acknowledgemreceiver is 4 microseconds. The channel capacity is (Mbps = megabit per second).

a) What is the propagation delay on the wire?

b) What is the packet transmission delay?

c) What is the maximum efficiency of the protocol maximum percentage of time is your modem actuadata)?




• What is the propagation delay on the wire?

• Answer: 50 microseconds, or 0.00005 seconds – Prop

delay is equal to the distance that the data must trav

by the speed that data travels on the wire, which in t

turns out to be 10 km / 200000 km/sec.




• What is the packet transmission delay?

• Answer: 80 microseconds, or 0.00008 seconds – Tran

delay is equal to the packet size divided by the chann

capacity, which in this case turns out to be (1000 * 8)

100,000,000. A common mistake was to forget to mu

packet size by 8 because it is in bytes not bits, and als

100 Mbps is equivalent to 100,000,000 bits per secon




• What is the maximum efficiency of the protocol (i.e., wh

maximum percentage of time is your modem actually sedata)?

• Answer: 43.48% - The time that the modem is actually seis equivalent to the packet transmission delay. The consists of the transmission delay at the modempropagation delay to get to the receiver plus the time create an acknowledgement plus the propagation delay to the sender.

• so we are left with the equation T / (T + 2P + Ack) = ((0.00008 + 2 * 0.00005 + 0.000004).



Sliding Window Protocols



Sliding Window

Sender

• Maintain sequence number of frames it is permitted to sSending window

Receiver

•

Maintain sequence number of frames it is expected to aReceiver window

Sliding window protocol allow us to send multiple packet



Go Back ARQ

• We can send up to W frames before worrying about ACK

• We keep a copy of these frames until the ACKs arrive.

• This procedure requires additional features to be added and-Wait ARQ.

– Frames from a sender are numbered sequentially.

– We need to set a limit since we need to include the sequence neach frame in the header.

– If the header of the frame allows m bits for sequence number,sequence numbers range from 0 to 2 m – 1. for m = 3, sequencare: 0,1, 2, 3, 4, 5, 6, 7.



Sender Sliding Window – Go Back A

• At the sending side, to hold the outstanding frames unt

acknowledged, we use the concept of a window.

• The size of the window is at most 2m -1 where m is the number o

sequence number.

• The window slides to include new unsent frames when the corr

received



Receiver Sliding Window – Go Back A

• Size of the window at the receiving site is always 1 in this pro

• Receiver is always looking for a specific frame to arrive in a sp

order.

• Any frame arriving out of order is discarded and needs to be

Receiver is waiting for frame 0 in part a.



Acknowledgement• Receiver sends positive ACK if a frame arrived safe and in order

• If the frames are damaged/out of order, receiver is silent and d

subsequent frames until it receives the one it is expecting.• The silence of the receiver causes the timer of the unacknowle

expire.

• Then the sender resends all frames, beginning with the one wittimer.

•

For example, suppose the sender has sent frame 6, but the timexpires (i.e. frame 3 has not been acknowledged), then the senand sends frames 3, 4, 5, 6 again. Thus it is called Go-Back-N-A

• The receiver does not have to acknowledge each frame receiveone cumulative ACK for several frames.



Go-Back-N ARQ, normal operation

• The sender keeps track of the outstanding frames and updat

variables and windows as the ACKs arrive.



Go-Back-N ARQ, lost frame• Frame 2

• When th

receives

discards

expectin

(accordi

• After the

frame 2

sender s

sends fr(go back



Go-Back-N ARQ, damaged/lost/delayed ACK

• If an ACK is damaged/lost, we can have two situations:

•

If the next ACK arrives before the expiration of any timer, therneed for retransmission of frames because ACKs are cumulativ

protocol.

• If ACK1, ACK2, and ACk3 are lost, ACK4 covers them if it arrive

the timer expires.

• If ACK4 arrives after time-out, the last frame and all the framethat are resent.

• Receiver never resends an ACK (i.e. no timer is maintain for AC

• A delayed ACK also triggers the resending of frames.

Go Back N ARQ sender window size



Go-Back-N ARQ, sender window size

• Size of the sender window must be less than 2 m. Size of the receiver is a

window size = 2 m – 1 = 3.

• Fig compares a window size of 3 and 4.



Selective Repeat ARQ

l l f



Selective Repeat ARQ, lost frame • Fra

acc

rec

are

spe

rec

for

• Re

to

ha

an

res

an

is iwin



I n Selective Repeat ARQ, the size of the send

receiver window must be at most one-half o

Selective Repeat ARQ sender window size



Selective Repeat ARQ, sender window size

• Size of the sender and receiver windows must be at most one-half of 2 m. If m = 2, windo

2 m /2 = 2. Fig compares a window size of 2 with a window size of 3. Window size is 3 and

sender sends duplicate of frame 0, window of the receiver expect to receive frame 0 (par

so accepts frame 0, as the 1st frame of the next cycle – an error.



Review

• Frames of 1000 bits are sent over a 1 Mbps satellite

propagation delay of 270 msec. Acknowledgements

piggybacked onto data frames.. Headers are very sh

bit sequence numbers are used. What is the

achievable channel utilization in frames per sec for

(a) stop-and-wait protocol

(b) protocol with go-back-n

(c) protocol with selective-repeat ?



Piggybacking



Solution

• Let t = 0 denote the start of transmission. At t = 1 mse

first frame has been fully transmitted. At t = 271 mseframe has fully arrived.

• At t = 272 msec, the frame acknowledging the first onbeen fully sent.

•

At t = 542 msec, the acknowledgement-bearing framarrived.

• Thus, the cycle is 542 msec. A total of k frames are semsec, for an efficiency of k/542.



Approx Solution

(a) k = 1, efficiency = 1/542 = 0.18%

(b) k = 7, efficiency = 7/542 = 1.29%

(c) k = 4, efficiency = 4/542 = 0.74%



Precise Solution

A. Channel utilization is 1 frame in 542 msec = 1

frames per sec

B. Pipeline 7 frames (3 bit seq number) with rou

548 msec giving channel utilization 7/548 = 1

frames per secC. pipeline 4 frames ( window size is sequence-s

in 545 msec giving channel utilization 4/545 =

frames per sec



Error Detection & Correction



Motivation

• Networks must be able to transfer data from one

another with complete accuracy.

• Data can be corrupted during transmission.

• For reliable communication, errors must be dete

corrected



Types of Error



Single Bit Error

Multiple Bits Error



Observation

• Single bit errors are the least likely type of errors in

transmission because the noise must have a vduration which is very rare. However this kind of happen in parallel transmission.

Example:

•

If data is sent at 1Mbps then each bit lasts only 1/1,0sec. or 1 μs.

• For a single-bit error to occur, the noise must have a dof only 1 μs, which is very rare.



Burst Error



Observation

• Burst error is most likely to happen in serial transmissio

duration of noise is normally longer than the duration of• The number of bits affected depends on the data rate an

of noise.

Example:

• If data is sent at rate = 1Kbps then a noise of 1/100 sec ca10 bits.(1/100*1000)

• If same data is sent at rate = 1Mbps then a noise of 1/10affect 10,000 bits.(1/100*106)



Error Detection

• Error detection means to decide whether the receive

correct or not without having a copy of the original m

• Error detection uses the concept of redundancy, whic

adding extra bits for detecting errors at the destinatio

To detect error you require some extra knowled



Redundancy



Error Detection Schemes

• Single Parity Check

• Two-Dimensional Parity Check

• Cyclic Redundancy Check

• Check Sum



Single Parity Check - VRC



Even Parity

• Add a parity bit to 7 bits of data to make an eve

of 1’s

• 0110100

• 1011010

• How many bits of error can be detected by a pa

• What’s the overhead?



Performance

• It can detect single bit error

• It can detect burst errors only if the total numb

errors is odd.



Two Dimension Parity Check

• Add one extra bit to a 7-bit code such

that the number of 1’s in the resulting 8bits is even (for even parity, and odd for

odd parity)

• Add a parity byte for the packet

• Example: five 7-bit character packet,even parity

0110100

1011010

0010110

1110101

1001011



Two Dimension Parity Check

0110100 1

1011010 0

0010110 1

1110101 1

1001011 0



Two Dimension Parity Check – Parity

0110100 1

1011010 0

0010110 1

1110101 1

1001011 0

___________________

1000110 1



All 1 Bit Error Will Be Detected



All 2 bit Error will be detected



Exceptional Case – Error Undetecte

0110100

1011010

0000111

1100100

1001011

_________________

1000110

0110100 1

1011010 0

0010110 1

1110101 1

1001011 0

___________________

1000110 1



Performance Summary

• If two bits in one data units are damaged and tw

exactly the same positions in another data unitdamaged, the 2-D checker will not detect an error.

• It can detect multiple bit and burst errors.



Review

• Assuming even parity, find the parity bit for ea

following data units.

a. 1001011

b. 0001100

c. 1000000d. 1110111



Review

• Suppose the information content of a packet is the b

1110 1011 1001 1101 and an even parity schemused. What would the value of the field containing

bits for the case of a two-dimensional parity sche

the input data should be represented by 4X4 2D array



Cyclic Redundancy Check



Cyclic Redundancy Check

• Given a k-bit frame or message, the transmitter gen

n-bit sequence, known as a frame check sequencethat the resulting frame, consisting of (k+n) bits,

divisible by some predetermined number.

• The receiver then divides the incoming frame by

number and, if there is no remainder, assumes that no error.



With & Without Error



Review - CRC

• Consider the following message M=1010001101. The

this message using the divisor polynomial x5 + x4 + x2

A. 01110

B. 01011

C. 10101

D. 10110



Example of Polynomial to Binary Fo



Review - CRC

GTU – 2012 / May

• Show the calculation polynomial code checksum for a

1101011011 using the generator x4 + x + 1

Check Sum



Check Sum



At Sender - Checksum

• The unit is divided into k sections, each of

• All sections are added together usin

complement to get the sum.

• The sum is complemented and becom

checksum.

• The checksum is sent with the data



At Receiver - Checksum

• The unit is divided into k sections, each of

• All sections are added together usin

complement to get the sum.

• The sum is complemented.

• If the result is zero, the data are a

otherwise, they are rejected.

Internet Checksum Example



Internet Checksum Example

• Note

–

When adding numbers, a carryout from the most significanneeds to be added to the result

• Example: add two 16-bit integers

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

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

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

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

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

Wraparound

Sum

Checksum



Review

• You are using the IP (Internet Protocol) Checksum for

Detection. Only, you do it based on 8 bit words, not 1in the real Internet. You must send the message 0011

1010 0101 1001.

• Show what sender does to create the checksum.

• Show what receiver does to verify no error occurs.

d k



Answer – Sender Work

1. Add zeros to get an integer number times 8 bits:

0011 1100

1010 0101

1001 0000

2. Add the three 8-bit ``words'' using one's complemen

0111 0010 .3. Take one's complement: 1000 1101 . That is the che

A i W k



Answer – Receiver Work

• The destination adds the four words

1000 1101

0011 1100

1010 0101

1001 0000

• Using (one's compl) and checks that the result is 1111and complement it gives 0000 0000. i.e. No Error



T f E C i



Types of Error Correction

• One is when an error is discovered; the receiver can

sender retransmit the entire data unit. This is backward error correction.

• In the other, receiver can use an error-correcting co

automatically corrects certain errors. This is known a

error correction.

E C i C



Error Correction Concept

• For correcting an error one has to know the exact p

error, i.e. exactly which bit is in error (to locate the in• For example, to correct a single-bit error in an ASCII

the error correction must determine which one of

bits is in error. To do this, we have to add some

redundant bits.• To calculate the numbers of redundant bits (r) re

correct d data bits. We use this relation 2r >= d + r +

H i C d B d E C t



Hamming Code Based - Error Correct

R d d t Bit C l l ti



Redundant Bit Calculation

R d d t Bit C l l ti



Redundant Bit Calculation

Example of Hamming Code Correcti



Example of Hamming Code Correcti

Si l Bit E O d



Single Bit Error Occurred





R i



Review

• Let us consider an example for 7-bit data. Assume t

transmission bit 5 has been changed from 1 to 0 asFigure. Find redundant bit and correct error.

Review



Review

• Sixteen-bit messages are transmitted using a Hammin

How many check bits are needed to ensure that the rcan detect and correct single-bit errors?

• Show the bit pattern transmitted for the message

• Assume that even parity is used in the Hamming code

Answer



Answer

• Parity bits are needed at positions 1, 2, 4, 8, and 16, s

messages that do not extend beyond bit 31 (includingparity bits) fit. Thus, 5 parity bits are sufficient.

• The bit pattern transmitted is 0110101100110011101



Extra Topics



High Level Data Link Control Protoc

HDLC Basic



HDLC Basic

• HDLC is a bit-oriented protocol.

•It specifies a packitization standard for serial links.

• It has been so widely implemented because it supportsduplex and full-duplex communication lines, point-to-to peer) and multi-point networks, and switched or nochannels.

• HDLC supports several modes of operation, includinsliding-window mode for reliable delivery. Since Internretransmission at higher levels (i.e., TCP), mosapplications use HDLC's unreliable delivery mode.

HDLC Stations and Configuration



HDLC Stations and Configuration

• Primary & Secondary Station: Primary has the respocontrolling all other stations on the link (usually stations).

• A primary issues commands and secondary issues Despite this important aspect of being on the link, tstation is also responsible for the organization of data flink.

• The secondary station is under the control of the primarhas no ability, or direct responsibility for controlling thonly activated when requested by the primary station.

• A combined station is a combination of a primary andstation.

HDLC Configuration



HDLC Configuration

• HDLC also defines three types of configurations for

types of stations. The word configuration referelationship between the hardware devices on a link.

• Following are the three configurations defined by HD

– Unbalanced Configuration

– Balanced Configuration – Symmetrical Configuration



HDLC Operation Modes



HDLC Operation Modes

• A mode in HDLC is the relationship between tw

involved in an exchange; the mode describes who colink. HDLC offers three different modes of operati

three modes of operations are:

– Normal Response Mode (NRM)

– Asynchronous Response Mode (ARM) – Asynchronous Balanced Mode (ABM)

Summary of Modes



Summary of Modes

HDLC Frames Types – I frame



HDLC Frames Types – I frame

• The function of the information command and

frame is to transfer sequentially numbered fram

containing an information field, across the data link.

HDLC Frames Types – S frame



HDLC Frames Types – S frame

• Supervisory (S) commands and responses frames are

perform numbered supervisory functions su

acknowledgment, polling, temporary suspension of inf

transfer, or error recovery. Frames with the S forma

field cannot contain an information field.

HDLC Frames Types – U frame



HDLC Frames Types – U frame

• The unnumbered format commands and responses fr

used to extend the number of data link control funct

unnumbered format frames have 5 modifier bits, wh

for up to 32 additional commands and 32 additional

functions.

Control Field of Frame



Control Field of Frame



Thank You

lecture3logicallinklayer__2014_08_22_07_26_57.pptx

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