The Data Link Layer Data Link Layer Design Issues

The Data Link Layer Chapter p 3

Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Data Link Layer Design Issues • • • •

Network layer services Framing E Error control t l Flow control


Packets and Frames

Relationship between packets and frames frames. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Network Layer Services

(a) Virtual communication. communication (b) Actual communication. communication Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Possible Services Offered 1 Unacknowledged 1. U k l d d connectionless ti l service. i  Ex: Ethernet  Appropriate for low error rate and real-time traffic

2. Acknowledged connectionless service.  Ack/Timer/Resend  Useful in unreliable channels, WiFi

3. Acknowledged connection-oriented service.  Guarantee frames are received exactly once and in the right order pp p over long, g unreliable links such as a satellite  Appropriate channel or a long-distance telephone circuit Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Framing Methods 1 Byte count: specify the number of bytes in the 1. frame 2 Flag bytes with byte stuffing: Flag byte is 2. used as both the starting and ending delimiter. delimiter 3. Flag bits with bit stuffing: Each frame begins and d ends d with ith a special i l bit pattern tt (ex:01111110) 4. Physical layer coding violations: use some reserved signals to indicate the start and end of frames. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Framing (1)

Problem: the count can be garbled!! No way of telling where the next frame starts. Retransmission also does not help.

A byte stream stream. (a) Without errors errors. (b) With one error error. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Framing (2)

a))

Problem: flag f bytes may happens in the data, especially when binary data such as photographs or songs are transmitted. – Solve by inserting a special escape byte before the flag byte in the data

b)

Four examples of byte sequences before and after byte stuffing. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Framing (3)

Stuff a 0 bit in the outgoing bit stream t whenever h it encounters t five 1s in the data.

Bit stuffing. (a) The original data. (b) The data as they appear on the line. line (c) The data as they are stored in the receiver receiver’s s memory after destuffing. If the user data contain the flag pattern, 01111110, this flag is transmitted as 011111010. The side effect: the length g of a frame depends p on the contents of the data it carries.


Error Detection Codes (1)

1. 2. 3. 4.

Hamming codes. Binary convolutional codes. Reed-Solomon codes. Low-Density Parity Check codes.


Codes  All these codes add redundancy to the information that is sent.  A fframe consists i t off m data d t (message) ( ) bits bit andd r redundant d d t (i.e. check) bits  Block code (區塊碼): r check bits are computed solely as a function of m data bits, as though the m bits were looked up in a large table to find r check bits.  Linear code (線性碼): the r check bits are computed as a linear function (ex: XOR) of the m data bits.  Let L t the th total t t l length l th off a block bl k be b n (i.e., (i n=m+r). + ) We W describe d ib this thi as (n,m) code.  codeword: An n-bit n bit unit containing data and check bits is referred to as an n-bit codeword.  code rate: the fraction of the codeword that carries information that is not redundant, or m/n. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Codes  Hamming distance: 兩個 codewords 之間相異的位元數.  For example, the Hamming distance between v = (1 0 0 1 0 1 1) and w = (0 1 0 0 0 1 1) is 3

codeword

 Two words in this figure are connected by an edge if their Hamming distance is 1. 用另一個角度看，Hamming 個角度看 Hamming distance 代表是某代表是某一個codeword 個codeword  用另需要轉換幾個 bit，才能轉換到另一個 codeword. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011



Codes 一個碼 (code) 的最小Hamming distance (minimum hamming distance) 就是所有的碼字組(codewords) 之間的最小 Hamming distance d =min dist(x, y), where the minimization is over all pairs of distinct codewords (x, y). 001

010

100

111

001

X

2

2

2

010

2

X

2

2

100

2

2

X

2

111

2

2

2

X

Distance is 2. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Codes   





Can detect one bit error, but cannot correction even for one bit! 要偵測 d errors, 一個碼它的距離必須至少為 d+1. 即正確的 codewords 間距離必須大於 d. 要糾正 d errrors,一個碼它的距離必須至少為 2d+1. 這樣錯誤的 codeword 才能清楚判定它離哪一個正確的 codeword 最接近。 Ex: An error-correction code, consider a code with four valid codewords:0000000000, d d 0000000000 0000011111 0000011111, 1111100000 1111100000, 1111111111。The code has a distance 5, which means it can correct double errors or detect quadruple errors errors. If the codeword 0000000111 arrives and we expect only single- or double single double-bit bit errors, the receiver will know that the original data must have been 0000011111. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Codes   

If a triple error changes 0000000000 into 0000000111, the error will not be corrected properly. Finding the legal codeword that is closest to the received codeword. Design a code with m message bits and r check bits that will allow all single error to be corrected. corrected Each of the 2m legal messages has n illegal codewords at a distance of 1 from it. There are n bits and the total number of bit pattern is 2n.

 n  1 2m  2n ,  n  m  r ,   m  r  1  2r 

假設我們的資料位元 m = 4. 那  4  r  1  2r 表示 r 要大於等於 3才行. 才行這表示我們要對一個4個bit的資料字組構建一個可更正單一位元錯誤的碼字組時就要增加三個檢查位元. 元錯誤的碼字組時, 就要增加三個檢查位元 Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Hamming Code  



The bits that are power of 2 (1,2,4,8,16,….) are check bits. The rest are filled up with the m data bits. 假設我們的資料長度 m = 8,  8  r  1  2r 表示 r 必需大於或等於 4. 這表示我們要對一個8個bit的資料字組構建一個可更正單一位元錯誤的碼字組時, 就要增加4個檢查位元, 所以整體碼字組長度為12. 那如何決定這些檢查碼呢?


Hamming Code 



當碼字組的長度為12時, 我們可看到每個數字, 從1 到12, 都可以表示成2的次方相加. 所以: 1 = 20 5 = 22 + 20 9 = 23 + 20 2 = 21 6 = 22 + 21 10 = 23 + 21 3 = 21 + 20 7 = 22 + 21 + 20 11 = 23 + 21 + 20 4 = 22 8 = 23 12 = 23 + 22 1 (= 20) 出現在所有的奇數數字. 2 ((= 21) 出現在, 2,, 3,, 6,, 7,, 10,, and 11. 4 (= 22) 出現在, 4, 5, 6, 7, 12. 8 ((= 23) 出現在, 8,, 9,, 10,, 11,, 12. . . . 以此類推 . . . 我們利用這個概念來產生檢查碼. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Hamming Code  

 

以我們用長度為12的碼字組為例, 從1開始對最低位元順序標上號碼. 這些檢查位元就會包含每個位元數字和中的同位位元.

因為2 (= 21) 會出現在數字, 2, 3, 6, 7, 10, and 11. 所以位置 2 會包含 3, 6, 7, 10, 和11的同位位元. 當我們使用偶同位時, 就是這些位元的模同餘2之和.如下所示, 第二個位元的位置會產生0的同位元值.

?+0+1+0+0+1=0 (mod 2) ∴?=0 Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Hamming Code 

那麼其他同位元呢?

?+1+1+0+1=0 (mod 2) ∴?=1



When a codeword arrives arrives, the receiver redoes the check bit computation including the value of the received check bits. If the check bits are correct,, for even parity p y sums,, each check result should be zero.


Error Detection Codes (2) 0 1 0 0 0 1 0( d2) 0+1+0+0+0+1=0(mod2)

0+1+1+0+0+1=1(mod2) Means 5th bit is error

Example of an (11, 7) Hamming code correcting a single single-bit bit error error. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Convolutional Code    



In a convolutional code, an encoder processes a sequence of input bits and generates a sequence of output bits. The output depends on the current and previous input bits. That is, the encoder has memory. The number of previous bits on which the output depends is called the constraint length of the code. A binary convolutional code is denoted by a three-tuple (n, k, m). n output bits are generated whenever k input bits are received m designates the number of previous kk-bit received. bit input blocks that must be memorized in the encoder. m is called the memoryy order of the convolutional code. Widely used in deployed networks (ex: GSM, satellite communications, 802.11)) Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Error Detection Codes (3)

The NASA binary convolutional code used in 802.11.


Error Detecting Codes (1) Error-Detecting Linear, systematic block codes 1. Parity bit: A single parity bit is appended to the data. Ex: 1011010->10110100 with even parity. Problem: only odd number of errors can be detected. 2. Checksums: a group of check bits associated with a message, regardless dl off how h they h are calculated. l l d Usually U ll placed l d at the h endd of the message. Errors can be detected by summing the entire received word, word both data bits and checksum. checksum If the result is zero, zero no errors has been detected. 3. Cyclic Redundancy Checks (CRCs): treating bit strings as representations of polynomials with coefficients of 0s and 1s. Ex: 110001 represents 1x5+1x4+0x3+0x2+0x1+1x0.


Error-Detecting Codes (2)

Parity bit is chosen so that the number of 1 bits in the codeword is even (or odd)

Interleaving of parity bits to detect a burst error error. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Error-Detecting Codes (2) 1. 2. 3.

Let r be the degree of G(x). Append r zero bits to the loworder end of the frame,, so it now contains m+r bits. Divide the bit string corresponding to G(x) using modulo 2 division. Subtract the remainder from the bit string using modulo 2 subtraction.


Error-Detecting Codes (3)

Example calculation of the CRC Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Elementary Data Link Protocols (1) •

Utopian Simplex Protocol • • • • •

•

Data are transmitted in one direction only Both the transmitting and receiving network layers are ready Processing time ignored I fi i bbuffer Infinite ff space Communication channel between the data link layer never damages or loses frames

Simplex Stop-and-Wait Protocol •

The receiver has finite buffer length and needs processing time •Stop and wait protocol

• Error-Free Channel • Simplex Stop-and-Wait Protocol • Noisy Channel – Frame Lost and Damaged •ARQ (Automatic Repeat Request) / PAR ( Positive Acknowledgement with Retransmission) Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Elementary Data Link Protocols (2)

Implementation of the physical physical, data link link, and network layers layers. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011


... Some definitions needed in the protocols to follow follow. These definitions are located in the file protocol.h. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011


... Some definitions needed in the protocols to follow follow. These definitions are located in the file protocol.h. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011


Some definitions needed in the protocols to follow follow. These definitions are located in the file protocol.h. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Utopian Simplex Protocol (1)

... A utopian simplex protocol. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Utopian Simplex Protocol (2)

A utopian simplex protocol protocol. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Simplex Stop-and-Wait Protocol for a Noisy Channel (1)

... A simplex stop-and-wait protocol. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Simplex Stop-and-Wait Protocol for a Noisy Channel (2)

A simplex stop stop-and-wait and wait protocol protocol. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Sliding Window Protocols (1)

... A positive acknowledgement with retransmission protocol protocol. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011


... A positive acknowledgement with retransmission protocol protocol. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011


A positive acknowledgement with retransmission protocol protocol. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

PiggyBacking & Sliding Window    

The Ack costs only a few bits and every frame needs a header. Better use of channel bandwidth Delay the ACK. The acknowledgement is attached to the next outgoing frame to send back. Problem: wait for a packet to piggyback the acknowledgement, but how long should the receiver wait? 



Too long: Retransmission

F ll D Full Duplex l and d Pi Pipelined li d T Transmission i i


Sliding Window  

Go-Back-N and Selective Repeat Sliding Window



A sliding window of size 1 1, with a 3-bit 3 bit sequence number number. (a) Initially. (b) After the first frame has been sent. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011


A sliding window of size 1, with a 3-bit sequence number (c) After the first frame has been received received. (d) After the first acknowledgement has been received. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

One-Bit Sliding Window Protocol (1)

... A 1-bit sliding window protocol. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011


... A 1-bit sliding window protocol. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011


A 1-bit sliding window protocol. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011


Two scenarios for protocol 4. (a) Normal case. (b) Abnormal case. The notation is (seq, ack, packet number). An asterisk indicates where a network layer accepts a packet Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Go-Back-N 

Sending window: The sender maintains a set of sequence permitted to send. numbers p 

  

Receiving window Sliding window Receiver only sends cumulative acks: ACKs all frames up to including seq # n - “cumulative to, cumulative ACK ACK”    



Sender can have up to N unacked frames in pipeline

always send ACK for correctly-received frames with highest in-order sequence number may generate duplicate ACKs Doesn’t ack frame if there’s a gap out-of-order frame: discard (don’t (don t buffer) -> no receiver buffering!

Sender has timer for oldest unacked frame:   

In theory, timer for each in-flight frame If timer expires, retransmit all unacked frames may receive duplicate ACKs (see receiver)


Protocol Using Go-Back-N (1)

Pipelining and error recovery recovery. Effect of an error when (a) receiver’s window size is 1 Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011


Pipelining and error recovery recovery. Effect of an error when (b) receiver’s window size is large. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011


... A sliding window protocol using go-back-n go back n. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011












A sliding window protocol using go-back-n go back n. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011


Simulation of multiple timers in software. (a) The queued timeouts (b) The situation after the first timeout has expired. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Selective Repeat • •

Sender can have up to N unacked frames in pipeline Receiver acks individual frames • Buffers frames, as needed, for eventual in-order delivery to upper layer • S d maintains Sender i t i titimer ffor each h unacked k d fframe –

When timer expires, retransmit only unack frame


Selective Repeat


Selective repeat: dilemma Example: • seq #’s: 0, 1, 2, 3 • window size=3 • receiver sees no difference in two scenarios! i ! • incorrectly passes duplicate data as new in (a) Q: what relationship between seq # size and window size? Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Protocol Using Selective Repeat (1)

... A sliding window protocol using selective repeat repeat. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011






...

A sliding window protocol using selective repeat repeat. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011


...



...









a) b) c) d)

Initial situation with a window of size7 After 7 frames sent and received but not acknowledged. Initial situation with a window size of 4. After 4 frames sent and received but not acknowledged.


Example Data Link Protocols

1. Packet over SONET 1 2. ADSL (Asymmetric Digital Subscriber Loop)


Packet over SONET (1)

Packet over SONET. SONET (a) A protocol stack. stack (b) Frame relationships Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Packet over SONET (2) PPP Features 1. Separate packets, error detection ) bringing g g lines up, p testing g 2. Link Control Protocol ((LCP): them, bringing them down when no longer needed. 3. Negotiate network layer: Network Control Protocol (NCP)


Packet over SONET (3)

IP or other network layer protocols

Broadcast frame

Unnumbered frame

Negotiated using LCP during line setup

The PPP full frame format for unnumbered mode operation Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

Packet over SONET (4) Select PPP options for the link

Establish a physical later connection

Start here

Start data transport Configure the network layer

State diagram for bringing a PPP link up and down Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

ADSL (Asymmetric Digital Subscriber Loop) (1)

ADSL protocol stacks stacks. Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

ADSL (Asymmetric Digital Subscriber Loop) (1)

AAL5 frame carrying PPP data Computer Networks, Fifth Edition by Andrew Tanenbaum and David Wetherall, © Pearson Education-Prentice Hall, 2011

End Chapter p 3
