1. Data Link Layer

2. Data link Layer Design Issues
Services interface to the network layer.
Framing
Dealing with transmission errors.(Error Control)
Regulating the flow of data so that slow receivers are not swamped by fast senders.(Flow Control)

3. Functions of the Data Link Layer
Relationship between packets and frames.

4. Services to Network Layer
Transferring data between network layers of machines

5. Types of Services
1. Unacknowledged Connectionless Service. 2. Acknowledged connectionless Service. 3. Acknowledged connection-oriented Service.

6. Services Unacknowledged connectionless service
Real-time traffic, e.g., speech, video
Most LANs, such as Ethernet
Acknowledged connectionless service
Useful over unreliable channels
Each frame sent individually acknowledged
e.g., wireless systems, e.g (WiFi)

7. Services Acknowledged connection-oriented service Guarantees
Each frame sent is received without error
All frames sent are received in right order
Three phases:
Connection establishment
Variables and counters initialization
Frame transmission
Connection release
Variables, buffers, resources freed up

8. Framing Fact Data link layer detects/corrects errors
Raw bit stream delivered by physical layer is not error free
Data link layer detects/corrects errors
Framing
Computing checksum
Handling error if any

9. Framing Approaches Character Count. Flag bytes with byte stuffing.
Starting and ending flags, with bit stuffing.

10. Character Count
A field in header specifies number of characters in a frame.
Problem?

11. Example
The following character encoding is used in a data link protocol:
A: ;
B: ;
FLAG: ;
ESC:
Show the bit sequence transmitted (in binary) for the four-character frame: A B ESC FLAG
When Character count framing method is used:
Answer

12. Flag Bytes with Byte Stuffing
A frame delimited by flag bytes
Four examples of byte sequences before and after stuffing

13. Byte Stuffing / Character Stuffing
A serious problem may easily happen that the flag byte's bit pattern occurs in the data.
One way to solve this problem is to have the sender's data link layer insert a special escape byte (ESC) just before each ''accidental'' flag byte in the data.
The data link layer on the receiving end removes the escape byte before the data are given to the network layer.
This technique is called byte stuffing or character stuffing.
Thus, a framing flag byte can be distinguished from one in the data by the absence or presence of an escape byte before it.

14. Example
The following character encoding is used in a data link protocol:
A: ;
B: ;
FLAG: ;
ESC:
Show the bit sequence transmitted (in binary) for the four-character frame: A B ESC FLAG
When Flag bytes with byte stuffing framing method is used:
Answer

15. Flag Bits with Bit Stuffing
Each frame begins and ends with special bit pattern (flag byte):
Problem: 6 consecutive 1s in data
Solution: Bit Stuffing: inserting a 0 after 5 consecutive 1s
Original Data
After
Stuffing
After received
and destuffed

16. Example
The following character encoding is used in a data link protocol:
A: ;
B: ;
FLAG: ;
ESC:
Show the bit sequence transmitted (in binary) for the four-character frame: A B ESC FLAG
When Starting and ending flag bytes, with bit stuffing framing method is used:
Answer

17. Exercises
The following character encoding is used in a data link protocol:
A: ; B: ; FLAG: ; ESC:
Show the bit sequence transmitted (in binary) for the five-character frame: A ESC B ESC FLAG
when each of the following framing methods are used:
(a) Flag bytes with byte stuffing.
(b) Starting and ending flag bytes, with bit stuffing.
ANSWER: a) b)

18. Exercises
1.Given the output after byte-stuffing:
FLAG A B ESC ESC C ESC ESC ESC FLAG ESC FLAG D F FLAG.
What is the original data?
ANSWER:
A B ESC C ESC FLAG FLAG D F
2. Given the output after byte-stuffing:
FLAG A B ESC ESC C ESC ESC ESC FLAG ESC FLAG D FLAG.
A B ESC C ESC FLAG FLAG D
3. A bit string, , needs to be transmitted at the data link layer. What is the string actually transmitted after bit stuffing?
The output is

19. Error Control Using acknowledgement
Positive- If the sender receives a positive acknowledgement about a frame, it knows the frame has arrived safely.
Negative-On the other hand, a negative acknowledgement means that something has gone wrong, and the frame must be transmitted again.
Problem: In some cases, sender waits for acknowledgement forever if a frame is lost for ever.
Solution: Timer
Problem: Duplicate transmission (receiver will accept the same frame two or more times)
Solution: Sequence number

20. Positive Acknowledgement
Sender sends a message, waits for acknowledgement from receiver, and then sends next message

21. Reliability and Acknowledgement
Case 1: no error
Sender Receiver
Case 2: data lost
Sender Receiver
Time
Data
Time
Data X
Timeout
Ack.
Data
Ack.
Timeout and retransmission

22. Reliability and Acknowledgement
Case 3: data error
Sender Receiver
Case 4: ack. lost
Sender Receiver
Time
Data
Time
Data
Error
Timeout
Timeout X
Data
Data
Ack.
Ack.
Timeout and retransmission
New problem? Duplicate
Solution: Sequence number

23. Flow Control Needed Problem Solution
When frames are transmitted faster than receiver can accept, frames will be lost
Solution
Flow control by feedback mechanism
The protocol contains well-defined rules about when a sender may transmit the next frame

24. Introduction Transmission impairments (errors) Attenuation
Loss of energy as signal propagates
Delay Distortion
Components travel at different speeds
Noise
Unwanted energy from other sources

25. Attenuation, distortion, and noise
Delay
Distortion
Noise
The McGraw-Hill Companies, Inc., 2004

26. Types of Errors

27. Single-bit error

29. Burst error

30. Error Correcting/Detecting Codes
Error correction
Referred to as forward error correction
Detect and correct error
Error detection
Detect error and request retransmission
Redundancy added to message
Codeword (n bits)
message + redundancy
(m bits) (r bits)
n = m + r
Code rate = m / n

31. Error Correcting/Detecting Codes
Where are error correction/detection used?
Correcting: physical layer and higher layers
Detecting: data link, network, and transport
Error correcting or error detecting?
Circumstances, application, etc.

32. Error-Correcting Codes
Hamming codes (our focus)
Binary convolution codes
GSM mobile phone system, satellite communication,
Reed-Solomon codes
DSL, data over cable, satellite communications
CDs, DVDs, Blue-ray disks
Low-density parity check codes
Digital video broadcasting, 10 Gbps Ethernet, power-line networks,

33. Hamming Distance
The number of bit positions in which two codewords differ is called the Hamming distance .
Its significance is that if two codewords are a Hamming distance ‘d’ apart, it will require ‘d’ single-bit errors to convert one into the other

34. Error Correction Hamming Code Example: If m = 7, then r = 4
Linear systematic block code
For m-bit message we need r-bit redundancy, where (m +r + 1)  2r  Why?
Redundancy bits are placed in position of 2’s power
Example: If m = 7, then r = 4

35. Hamming Code Example
7-bit data word " “(d -data bits, p -parity bits)

36. Combination p1 : bits 1, 3, 5, 7, 9, 11.. p2 : bits 2,3,6,7,10,11..

37. Hamming Code Example (Cont.)
Assume the final bit gets corrupted and turned from 1 to 0
Flag each parity bit as 1 when the even parity check fails

38. Hamming Code Example How to construct codeword message: m = 7 bits
1. find r using the formula  r = 4
2. place data bits in codeword
3. calculate P bits: even parity
Sum should be even
P  P1 = 0
P  P2 = 0
P  P4 = 0
P  P8 = 1

39. Hamming Code Example (Cont’d)
How to correct error and retrieve message
1. compute syndrome similar to codeword construction  error position
2. flip the bit in the error position
3. remove P bits from codeword
Example of an (11, 7) Hamming code correcting a single-bit error

40. Use of a Hamming code to correct burst errors.

41. Error-Detecting Codes
Parity
Checksums
Internet
Cyclic Redundancy Checks (CRCs)
They are all linear systematic block code

42. Error-Detecting Codes
Parity bit
Added to data so that number of 1 bits in codeword is
Even (even parity)
Odd (odd parity)
E.g., ASCII of ‘H’ is , its codeword is
________ if even parity is used
________ if odd parity is used

43. Use Parity to Detect Burst Errors
Organize a block into (k x n) matrix
One parity for each column
 one row of parities at the bottom
Transmit one row at a time
Can detect burst errors of length  n

44. Example
Interleaving of parity bits to detect a burst error

45. Error-Detecting Code - CRC
Bit stream is treated as polynomial w/ coefficients 0 and 1
Example:
data:
polynomial:
degree = 7
Modulo 2 arithmetic is used
Example:
XOR

46. CRC generator and checker

48. Error-Detecting Code - CRC
Use generator polynomial G(x) to calculate checksum R(x)
Frame: P(x) generator: G(x)
degree of G(x) = d
Transmitted: checksummed frame P(x)·xd + R(x)
It’s guaranteed that P(x)·xd + R(x) is divisible by G(x)!!

49. Error-Detecting Code - CRC
Receiver
Divides checksummed frame by G(x)
If remainder is zero
No error, CRC is removed
Otherwise
Error, the frame is discarded

50. CRC - Example
Generator:
0 0

51. CRC - Example Generator: Error during transmission ≠ 0 
Received
Error during transmission 1
CRC - Example
Generator:
0 0
1 0
≠ 0 
Frame discarded
0 0
1 0

52. k + 1 bit check sequence c, equivalent to a degree-k polynomial
CRC – Example 1
G(x) = x3  x2  = Generator
P(x) = x7  x4  x3  x = Message
1101
k + 1 bit check sequence c, equivalent to a degree-k polynomial
Message plus k zeros
1101
1001
1101
1000
Result:
Transmit message followed by remainder:
1101
1011
1101
1100
1101
1000
Remainder
m mod c
1101
101

53. Example 1 CRC Checking (No Errors)
G(x) = x3  x2  1 = Generator
P(x) = x10  x7  x6  x4  x2  1 = Received Message
1101
k + 1 bit check sequence c, equivalent to a degree-k polynomial
Received message, no errors
1101
1001
1101
1000
Result:
CRC test is passed
1101
1011
1101
1100
1101
1101
Remainder
m mod c
1101

54. Example 1:CRC Checking (With Errors)
G(x) = x3  x2  1 = 1101 Generator
P(x) = x10  x7  x5  x4  x2  1 = Received Message
1101
k + 1 bit check sequence c, equivalent to a degree-k polynomial
Received message
1101
Two bit errors
1000
1101
1011
Result:
CRC test failed
1101
1101
1101
0101
Remainder
m mod c

55. Example 2 Calculating CRC

56. Example 2 :Checking CRC

57. Exercise
For G= and P= , find the CRC or
For P(x)=x7+x6+x5+x+1, G(x)=x5+x4+x+1 calculate CRC.

58. CRC Properties Single error detection
Double error detection w/ carefully chosen G(x)
Odd number error detection if (x + 1) is a factor of G(x)
Detect burst error length  r for r check bits
Can be implemented in hardware using simple shift register circuit

59. Standard CRC Examples CRC-8 (for ATM header): x8 + x2 + x + 1
CRC-32 (for LANs):
x32 + x26 + x23 + x22 + x16 + x12 + x11 +
x10 + x8 + x7 + x5 + x4 + x2 + x + 1
CRC-16-(for Bluetooth, etc.):
x16 + x12 + x5 + 1

60. Elementary Data Link Protocols
Key Assumptions
Network, data link, and physical layers are independent processes communicating by sending messages
Machine A wants to send a long stream of data to machine B over a reliable, connection-oriented service

61. Implementation of Physical, Data Link, and Network Layers

62. Data Structures and Primitives

63. Data Structures and Primitives

64. Data Structures and Primitives

65. Unrestricted Simplex Protocol
Utopia protocol
Assumptions
Unidirectional data transmission
Transmitting/receiving network layers are always ready
Processing time is ignored
Infinite buffer space
No errors

66. Unrestricted Simplex Protocol - Sender

67. Unrestricted Simplex Protocol - Receiver

68. Simplex Stop-and-Wait Protocol
Assumptions
Unidirectional data transmission
Transmitting/receiving network layers are always ready
Finite processing speed
Finite buffer capacity
No errors
Problem: Sender sends too fast
Stop-and-wait
Senders sends one frame and then waits for an acknowledgement before processing

69. Simplex Stop-and-Wait Protocol - Sender

70. Simplex Stop-and-Wait Protocol - Receiver

71. Protocol for noisy channel Simplex PAR Protocol
Positive acknowledgement with retransmission (PAR)
Sender waits for a positive acknowledgement before advancing to the next data item
Also Known as ARQ (Automatic Repeat reQuest)

72. PAR Protocol Assumptions Timer + sequence number
Unidirectional data transmission
Transmitting/receiving network layers are always ready
Finite processing speed
Finite buffer capacity
Errors, can be detected
Timer + sequence number
Size (i.e., # bits) of sequence number?

73. PAR Protocol – Sender

74. PAR Protocol – Sender (Cont’d)

75. PAR Protocol – Receiver

76. Sliding window protocols
Data Frame transmission
Unidirectional assumption in previous elementary protocols
 Not general
Full-duplex - approach 1
Two separate communication channels
Forward channel for data
Reverse channel for acknowledgement
 Problems: 1. reverse channel bandwidth wasted
2. cost

77. Sliding Window Protocols
Full-duplex - approach 2
Same circuit for both directions
Data and acknowledgement are intermixed
How do we tell acknowledgement from data?
"kind" field telling data or acknowledgement
Approach 3
Attaching acknowledgement to outgoing data frames
 Piggybacking

78. Piggybacking
Temporarily delaying transmission of outgoing acknowledgement so that they can be hooked onto the next outgoing data frame
Advantage: higher channel bandwidth utilization
Complication:
How long to wait for a packet to piggyback?
If longer than sender timeout period then
sender retransmit
 Purpose of acknowledgement is lost

79. Piggybacking Solution for timing complexion
If a new packet arrives quickly
 Piggybacking
If no new packet arrives after a receiver ack timeout
 Sending a separate acknowledgement frame

80. Sliding Window Protocol
We are going to study three bidirectional sliding window protocols (max sending window size, receiving window size)
One-bit sliding window protocol (1, 1)
Go back N (>1, 1)
Selective repeat (>1, >1)
Differ in efficiency, complexity, and buffer requirements

81. Sliding Window Protocol
Each outbound frame contains an n-bit sequence number
Range: 0 - MAX_SEQ (MAX_SEQ = 2n - 1)
At any instance of time
Sender maintains a set of sequence numbers of frames permitted to send
These frames fall within sending window
Receiver maintains a set of sequence numbers of frames permitted to accept
These frames fall within receiving window

82. Sliding Window Protocol
Lower limit, upper limit, and size of two windows need not be the same
Fixed or variable size
Requirements
Packets delivered to the receiver's network layer must be in the same order that they were passed to the data link layer on the sending machine
Frames must be delivered by the physical communication channel in the order in which they were sent

83. Sending Window
Contains frames can be sent or have been sent but not yet acknowledged – outstanding frames
When a packet arrives from network layer
Next highest sequence number assigned
Upper edge of window advanced by 1
When an acknowledgement arrives
Lower edge of window advanced by 1

84. Sending Window
If the maximum window size is n, n buffers is needed to hold unacknowledged frames
Window full (maximum window size reached)
 shut off network layer

85. Receiving Window Contains frames may be accepted
Frame outside the window  discarded
When a frame's sequence number equals to lower edge
Passed to the network layer
Acknowledgement generated
Window rotated by 1

86. Receiving Window Contains frames may be accepted
Always remains at initial size (different from sending window)
Size
=1 means frames only accepted in order
>1 not so
Again, the order of packets fed to the receiver’s network layer must be the same as the order packets sent by the sender’s network layer

87. A sliding window of size 1, with a 3-bit sequence number.
Actually, 1-bit sequence number is enough for this example. The purpose of using 3-bit is to demonstrate the idea of sliding window.
A sliding window of size 1, with a 3-bit sequence number.
(a) Initially.
(b) After the first frame has been sent.
(c) After the first frame has been received.
(d) After the first acknowledgement has been received.
In many textbooks, an array of boxes are used to represent the window.

89. One Bit Sliding Window Protocol
Sending window size = receiving window size = 1
Stop-and-wait
Acknowledgement =
Sequence number of last frame received w/o error*
Problem of sender and receiver send simultaneously
*: some protocols define the acknowledgement to be the sequence number expected to receive

90. Case 2: simultaneous start
Case 1: normal case
Case 2: simultaneous start
(a) Case 1: Normal case. (b) Case 2: Abnormal case The notation is (seq, ack, packet number). An asterisk indicates where a network layer accepts a packet.

91. Problem with stop and wait protocols
Example: 
50 Kbps bandwidth (satellite channel), 1000 bit frames, 500 msec round-trip propagation delay
Sender
Receiver
first packet bit transmitted, t = 0
500
first packet bit arrives: 500/2=250
ACK arrives: =520
last packet bit transmitted, t = 20
500/2=250
1000 bits/50,000 bps=20 msec
Line utilization = 20/520=3.85% !!  sender is blocked 96.15% of time !

92. Pipelining Solution for poor channel utilization:
Use larger frames, but the maximum size is limited by the bit error rate of the channel. The larger the frame, the higher the probability that it will become damaged during transmission.
 Allowing the sender to send more than 1 frame, w frames send before blocking (Pipelining)
In Pipelining:
Sender does not wait for each frame to be ACK'ed. Rather it sends many frames with the assumption that they will arrive. Must still get back ACKs for each frame.
 Provides more efficient use of transmit bandwidth, but error handling is more complex.
Value of w? times equal to round-trip time,
For previous example, W=26 (Because 520/20=26)

93. Pipelining In Pipelining:
Sender after sending the wth frame, will get the ACK of first frame.
If error occurred:
What if 26 frames transmitted, and the second has an error. Frames 3-26 will be ignored at receiver side? Sender will have to retransmit. What are the possibilities?
 Two strategies as solutions (depends on receive Window size)
Go-Back-N
Selective-Repeat

94. Go-Back-N Go-Back-N: Equivalent to receiver's window size = 1.
If receiver sees bad frames or missing sequence numbers, subsequent frames are discarded. No ACKs for discarded frames.

95. Go Back n Protocol
Receiver discards all subsequent frames following an error one, and send no acknowledgement for those discarded
Receiving window size = 1 (i.e., frames must be accepted in the order they were sent)
Sending window might get full
If so, re-transmitting unacknowledged frames
Wasting a lot of bandwidth if error rate is high

96. Go Back n Protocol Implementation
Sender has to buffer unacknowledged frames
Acknowledge n means frames n,n-1,n-2, ... are acknowledged (i.e., received correctly) and those buffers can be released
One timer for each outstanding frame in sending window

97. Pipelining and error recovery
Pipelining and error recovery. Effect of an error when (a) receiver's window size is 1 and (b) receiver's window size is large

98. Selective-Repeat Selective-Repeat:
Receiver's window size larger than one.
Store all received frames after the bad one. ACK only last one received in sequence.
Use NAK (Negative ACK) when an error detected: checksum error or out of sequence

99. Select Repeat Protocol
Receiver stores correct frames following the bad one
Sender retransmits the bad one after noticing
Receiver passes data to network layer and acknowledge with the highest number
Receiving window > 1 i.e., any frame within the window may be accepted and buffered until all the preceding one passed to the network layer
Might need large memory

100. Negative Acknowledgement (NAK)
SRP is often combined with NAK
When error is suspected by receiver, receiver request retransmission of a frame
Arrival of a damaged frame
Arrival of a frame other than the expected
Does receiver keep track of NAK?
What if NAK gets lost?

101. Selective Repeat with NAK
Nak 2, lost

102. Select Repeat Protocol Implementation
Receiver has a buffer for each sequence number within receiving window
Each buffer is associated with an "arrived" bit
Check whether sequence number of an arriving frame within window or not
If so, accept and store
Maximum window size = ? Can it be MAX_SEQ ?

103. Select Repeat Protocol - Window Size
Problem is caused by new and old windows overlapped
Solution
Window size=(MAX_SEQ+1)/2
E.g., if 4-bit window is used, MAX_SEQ = 15
 window size = (15+1)/2 = 8
Number of buffers needed
= window size

104. Select Repeat Protocol
(a) Initial situation with a window size seven.
(b) After seven frames sent and received, but not acknowledged.
(c) Initial situation with a window size of four.
(d) After four frames sent and received, but not acknowledged.