Data Link Layer

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)

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

Services to Network Layer Transferring data between network layers of machines

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

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. 802.11 (WiFi)

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

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

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

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

Example The following character encoding is used in a data link protocol: A: 01000111; B: 11100011; FLAG: 01111110; ESC: 11100000 Show the bit sequence transmitted (in binary) for the four-character frame: A B ESC FLAG When Character count framing method is used: Answer 00000101 01000111 11100011 11100000 01111110

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

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.

Example The following character encoding is used in a data link protocol: A: 01000111; B: 11100011; FLAG: 01111110; ESC: 11100000 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 01111110 01000111 11100011 11100000 11100000 11100000 01111110 01111110

Flag Bits with Bit Stuffing Each frame begins and ends with special bit pattern (flag byte): 01111110 Problem: 6 consecutive 1s in data Solution: Bit Stuffing: inserting a 0 after 5 consecutive 1s Original Data After Stuffing After received and destuffed

Example The following character encoding is used in a data link protocol: A: 01000111; B: 11100011; FLAG: 01111110; ESC: 11100000 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 01111110 01000111 11010001 11110000 0 011111010 0111110

Exercises The following character encoding is used in a data link protocol: A: 11010101; B: 10101001; FLAG: 01111110; ESC: 10100011 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) 01111110 11010101 10100011 10100011 10101001 10100011 10100011 10100011 01111110 01111110 b) 01111110 11010101 10100011 10101001 10100011 011111010 0111110

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, 0111101111101111110, needs to be transmitted at the data link layer. What is the string actually transmitted after bit stuffing? The output is 011110111110011111010

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

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

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

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

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

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

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

Types of Errors

Single-bit error

Burst error

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

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.

Error-Correcting Codes Hamming codes (our focus) Binary convolution codes GSM mobile phone system, satellite communication, 802.11 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, 802.11

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

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

Hamming Code Example 7-bit data word "0110101“(d -data bits, p -parity bits)

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

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

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 P1+1+0+0+0+1  P1 = 0 P2+1+0+0+0+1  P2 = 0 P4+0+0+0  P4 = 0 P8+0+0+1  P8 = 1

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

Use of a Hamming code to correct burst errors.

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

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 1011010, its codeword is ________ if even parity is used ________ if odd parity is used 10110100 10110101

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

Example Interleaving of parity bits to detect a burst error

Error-Detecting Code - CRC Bit stream is treated as polynomial w/ coefficients 0 and 1 Example: data: 10100111 polynomial: degree = 7 Modulo 2 arithmetic is used Example: 10011011 11110000 +11001010 -10100110 01010001 01010110 XOR

CRC generator and checker

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)!!

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

CRC - Example Generator: 0 0

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

k + 1 bit check sequence c, equivalent to a degree-k polynomial CRC – Example 1 G(x) = x3  x2  1 = 1101 Generator P(x) = x7  x4  x3  x = 10011010 Message 1101 k + 1 bit check sequence c, equivalent to a degree-k polynomial 10011010000 Message plus k zeros 1101 1001 1101 1000 Result: Transmit message followed by remainder: 10011010101 1101 1011 1101 1100 1101 1000 Remainder m mod c 1101 101

Example 1 CRC Checking (No Errors) G(x) = x3  x2  1 = 1101 Generator P(x) = x10  x7  x6  x4  x2  1 = 10011010101 Received Message 1101 k + 1 bit check sequence c, equivalent to a degree-k polynomial 10011010101 Received message, no errors 1101 1001 1101 1000 Result: CRC test is passed 1101 1011 1101 1100 1101 1101 Remainder m mod c 1101

Example 1:CRC Checking (With Errors) G(x) = x3  x2  1 = 1101 Generator P(x) = x10  x7  x5  x4  x2  1 = 10010110101 Received Message 1101 k + 1 bit check sequence c, equivalent to a degree-k polynomial 10010110101 Received message 1101 Two bit errors 1000 1101 1011 Result: CRC test failed 1101 1101 1101 0101 Remainder m mod c

Example 2 Calculating CRC

Example 2 :Checking CRC

Exercise For G=110011 and P=11100011, find the CRC or For P(x)=x7+x6+x5+x+1, G(x)=x5+x4+x+1 calculate CRC.

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

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

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

Implementation of Physical, Data Link, and Network Layers

Data Structures and Primitives

Data Structures and Primitives

Data Structures and Primitives

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

Unrestricted Simplex Protocol - Sender

Unrestricted Simplex Protocol - Receiver

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

Simplex Stop-and-Wait Protocol - Sender

Simplex Stop-and-Wait Protocol - Receiver

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)

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?

PAR Protocol – Sender

PAR Protocol – Sender (Cont’d)

PAR Protocol – Receiver

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

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

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

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

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

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

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

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

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

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

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

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.

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

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.

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: 500+20=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 !

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)

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

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.

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

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

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

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

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

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?

Selective Repeat with NAK Nak 2, lost

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 ?

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

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.