1. Lecture on Data Link Control
Goal : Conversion of a virtual bit pipe into an error free link.
Functions
Error 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

2. 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.

3. 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)
data bits parity bit

4. Error Detection Method: Horizontal and Vertical Parity Check
Add one parity check for each row and one for each column. 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.
Order of transmission
Original
data

5. 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

6. 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 bits
S(0) S(1)…S(K-1) | X(0) X(1)…X(K+L-1) …
…………… …………….
………… …………….
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 patterns
of length K+L that can potentially be received.
2K information
patterns
2K code words

7. Automatic Repeat Request (ARQ)
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.

8. Time Space Three popular methods of ARQ: Stop and wait Go back n
Added bits
Packet
Three popular methods of ARQ:
Stop and wait
Go back n
Selective repeat
Each packet below has the form above. 1 2 3 4 5
Node A
Time
Space
Node B

9. 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
Sender A
Data+
Receiver B
Data+ is something that includes
{Ack or Nak} and parity check.

10. Examples of Trouble Scenario
timeout at A; repeat packet 0
Node A
Ack
Node B
Packet 0
Packet 0 or 1?
timeout at A;
repeat packet 0
ack received for 0;
But B thinks it’s for 1 1 2
Node A
for 0
for 0
Packet 1 is lost
Node B
Packet 0

11. - 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 ARQ
1. 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 working
For 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.

12. Go Back n ARQ
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

13. n = 4 Time out for packet 2 Window [0, 3] [2, 5] [4, 7] [5, 8] 1 2 3 4 5 2 4 5 SN
Node A
n = 4
Node B RN 0’ 1’ 2’ 3’ 4’ 5’ 6’
(piggy backed)

14. 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

15. Framing Decides where a frame begins and ends Three popular methods
Character-based
Bit-oriented
Length count

16. 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

17. 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)

18. Bit Oriented Framing
Eliminates the need of having integer multiple of characters.
Replace DLE ETX with a string called flag.
e.g., ( )
To prevent the accidental appearance of the flag, bit stuffing is needed.
→ Every becomes
→ 01j for j > 6 corresponds to an abnormal termination

19. Overhead in Bit Oriented Framing
Assumptions
Number 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
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Probability (stuff after bit i) =
Note: we ignore the probability of having 01n for

20. Overhead Calculation OV : Number of overhead bits
It is possible to find minimum Ek(OV) by varying j.
Optimal value of

21. 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.

22. 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

23. 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

24. Small kmax (assume j nodes j-1 links)
Source
Destination
Packet
transmission
time
Total packet
Delay over
both links
Time

25. - Reduces the transmission delay by pipelining T: total time
Half-packet
transmission
time
- Reduces the transmission delay by pipelining
T: total time
C: bits/second capacity on each link
Total delay for the
two half packets
over both links
Time

26. 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