Lecture on Data Link Control
-
Upload
denise-carson -
Category
Documents
-
view
33 -
download
2
description
Transcript of 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