Lecture on Data Link Control

Lecture on Data Link ControlGoal : Conversion of a virtual bit pipe into an error free link.

FunctionsError Detection

Automatic Repeat Request (ARQ)

Framing

- Sending DLC receives packets from Network layer.- Receiving DLC delivers packets to Network Layer.- Packet ordering is checked at the receiving DLC.- Sending DLC must add additional bits to the beginning and end of each packet.

DLC Frame

Added bits Packet Added bits

Error Detection- Virtual bit pipe is unreliable due to transmission

errors.- Receiving DLC must detect errors.- If errors are detected, arrived packets are

discarded and retransmission is requested.

Note: Error correction is not usually done due to lack of knowledge on Physical Layer characteristics. Without specific knowledge on Physical Layer, error correction is not reliable.

Error Detection Method: Single Parity Check

- Add one parity check bit to the packet to make the number of ones even (or odd).

- Able to detect odd numbers of error.- Even numbered errors are not detectable.- Example : ASCII Code (7 bit data + 1 bit parity)

0 1 1 0 1 1 1 1

data bits parity bit

Add one parity check for each row and one for each column.

1 0 1 1 1

1 1 0 1 1

0 0 0 1 1

1 0 1 0 0

1 1 0 1 x

-x can be the parity check for the bottom row, the right most column, or the whole array of data including all previous parity bits.

Error Detection Method: Horizontal and Vertical Parity Check

Originaldata

Order of transmission

- An odd number of errors in any row or column can be detected.- Four errors confined in two rows and two columns in the following

manner cannot be detected

1 0 1 1 1

1 1* 0 1* 1

0 0 0 1 1

1 0* 1 0* 0

1 1 0 1 1

Error detection method : Coding- Given the information bits S(0) S(1)…S(K-1), generate parity bits C(0) C(1)…C(L-1)

with each C(i) depending on some information bits.

- Transmit X(0) X(1)…X(K+L-1) where X(0)=C(0) X(1)=C(1) ………… X(L-1)=C(L-1) X(L)=S(0) X(L+1)=S(1) ……….. X(K+L-1)=S(K-1)

Information bits Transmitted bitsS(0) S(1)…S(K-1) | X(0) X(1)…X(K+L-1) 0 1 … 1 0 0 ... 1 …………… ……………. ………….. …………….

2K code words2K information

patterns

Each correctly generated X(0) X(1)…X(K+L-1) is called a codeword. There are 2K code words in the code book. However, there are 2K+L random bit patternsof length K+L that can potentially be received.

- After an error is detected, the receiving DLC requests the sending DLC to retransmit.

- Assumptions Framing is done properly There exist a non zero probability that transmission is a

success over the link. FIFO on the communication link All errors are properly detected.

Automatic Repeat Request (ARQ)

Three popular methods of ARQ: Stop and wait Go back n Selective repeat

Added bits Packet Added bits

1 2 3 4 5

Node B

Node A

Each packet below has the form above.

TimeSpace

Stop and Wait ARQ

Ensures correct transmission before next transmission.

Possible scenarios- Data may get lost : nothing happens at B- Data arrive at B with error : B sends Nak in Data+- Data arrive at B without error : B sends Ack in Data+- Ack received at A : A sends the next data- Nak received at A: A repeats the current data- A times out : A repeats the current data

Data

Data+

Sender A

Receiver B

Data+ is something that includes {Ack or Nak} and parity check.

0 0 1 2Node A

Packet 0

for 0 for 0

0 0

timeout at A; repeat packet 0

Node A

Packet 0 Packet 0 or 1?

Node B

Ack

Examples of Trouble Scenario

timeout at A; repeat packet 0

ack received for 0;But B thinks it’s for 1

Packet 1 is lost Node B

- Stop and Wait ARQ needs a certain ID system for data, ack, and nak.- Add SN and RN to the header of packets

SN : sequence number for the packet being transmitted at the sender RN : sequence number of the next packet expected at the receiver

Correctness of stop and wait ARQ1. Safety : Does not do anything incorrect Packets are accepted only if they are error free Packets are accepted only if they are the next one expected Then, Stop and wait ARQ is safe

2. Liveliness : Never stops workingFor stop and wait ARQ, it is known that for the probability of transmission being a success greater than zero, we can show that

t(1) < t(2) < t(3) < t(1) : time when a packet is sent for the first time t(2) : time when RN is incremented t(3) : time when SN is updated Then, Stop and Wait ARQ is live.

- Sender does not wait. It can send n packets before acknowledgement is received.

- Receiver acknowledges the receipt of the next packet expected.

- Sender

SNmin : smallest index not acknowledged.

SNmax : smallest index not sent.- Receiver

RN : next packet expected

Go Back n ARQ

0 1 2 3 4 5

0 2 4 2SNNode A

n = 4

1 3 5

0’Node B

RN(piggy backed)

1’ 2’ 3’ 4’ 5’ 6’

Window [0, 3] [2, 5] [4, 7] [5, 8]

4 5

Time out for packet 2

Selective Repeat ARQ

In go back n, a single error causes round trip delay worth of retransmitted packets. Need to store round-trip worth of packets Probability of error in a packet is small.

Go back n ARQ is inefficient.

Selective repeat ARQ can achieve Sender : Sends SNmin to SNmin + n –1 Receiver : Accepts any RN to RN + n –1

success) of prob :p n,utilizatiolink :( 1 p

.1 achieveCan p

Framing

Decides where a frame begins and ends

Three popular methods- Character-based- Bit-oriented- Length count

Character Based Framing

- DLE is needed for differentiating accidental appearances of STX, ETX, and DLE itself DLE STX : Start * STX: binary string that happens to be the same as

STX DLE DLE STX : binary string that happens to be DLE

STX DLE DLE * : binary string that happens to be DLE *

SYN SYN DLE STX HEADER Packet DLE ETX CRC SYN SYN SYN

filler start Pay load EDD check filler

Disadvantages of Character-based Framing

The length of a packet must be integer multiple of character

High overhead Accidental appearance of DLE ETX causes a

packet to be terminated prematurely CRC does not check Loss of packet Probability of a random CRC being accepted

correctly = 2-L (L:length of CRC)

Bit Oriented Framing

Eliminates the need of having integer multiple of characters.

Replace DLE ETX with a string called flag. e.g., 0160 (01111110)

To prevent the accidental appearance of the flag, bit stuffing is needed. → Every 11111 becomes 111110.→ 01j for j > 6 corresponds to an abnormal termination

Overhead in Bit Oriented Framing

AssumptionsNumber of bits to be stuffed is small

0 or 1 happens with the equal probability

01j0 is the flag

bit1 bit j-1 bit k

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxProbability (stuff after bit i) =

Note: we ignore the probability of having 01n for

kjji

jifor

jiP

P

j

j

i

i

,...,1 , ,2

1 ,2

1 ,0)1(

12 jn

Overhead Calculation

OV : Number of overhead bits

It is possible to find minimum Ek(OV) by varying j.

Optimal value of

1 ,1

1 ),1(2)()(

)1(2)3(

)1(2121 )1(

jkj

jkjkEOVE

jjk

jOV

jk

j

flag

k

ji

jj

2

2

2

)(log)(,

)(log

kEOVEjoptimalFor

kEj

maxlog k2

Length Field Framing

Sends k (the number of bits in the frame) as part of the header

For a given kmax, needs bits for transmitting the length field.

)(log 2 kEj

.2)(log)( , 2 kEOVEjoptimalFor

Minimum Bits for Length Field

Let pk be the probability of a frame being of length k.

If pk is uniform, H=log2 kmax

If pk is geometric, H=log2 E(k)+log2 e

Example) Geometrically distributed k

Represent

Encode the length as 01i r where r is represented in a binary string of j bits.

If k=7 and j=2, i=1 and r=3, 0111

To implement this, you need bits for each k

kkk ppH

12log

jKEOVE j 12)()(

)/( jk 2

jj rrik 20 ,2

Effect of Maximum Frame Size

- M:The number of total bits in a message

- V:The number of overhead bits per frame

- :The number of bits in a full frame without overhead

- Large

Total bits

Number of frames

Processing overhead

maxk

VkM

framekM

maxM/kMsent bits Total

full) benot may onelast The full. are frames 1

max

max

/(

./

maxk

- Small kmax (assume j nodes j-1 links)

Source Destination

Packet transmission

time

Total packet Delay overboth links

Time

- Reduces the transmission delay by pipeliningT: total timeC: bits/second capacity on each link

Time

Half-packet transmission

time

Total delay for thetwo half packetsover both links

1

T, minimize then to0, toequalset and Take

21

1

j

VMEk

dkd

VkVMEMEjVkCTE

VkMMjVkCT

)(

/

//)()())(()(

/))((

max

max

maxmax

maxmax

Stream Type Traffic: Voice

- Delay calculation between the arrival of a bit to the delivery of this bit.- Assume the bits arrive as a packet of length k at bit rate of R/second

- This packet needs to traverse a series of links at Ci bits/second.

waiting time

before transmission

- Small k reduces T for finite queue

i iC

vk

R

kT

i

C

vk

R

k

i

,max