local area networks by r.s. chang, dept. csie, ndhu 1 chapter 3 the data link layer user a user b...

Local Area Networks by R.S. Chang, Dept. CSIE, NDHU 1

Chapter 3 The Data Link Layer

user A

user B

reliable transmission over unreliable physical link



3.1 Data Link Layer Design Issues

Providing a well-defined service interface to the network layer

Determining how the bits of the physical layer are grouped into frames

Dealing with transmission errors Regulating the flow of frames so

that slow receivers are not swamped by fast senders

Functions of Data Link Layer




Relationship between packets and frames




3.1.1 Services Provided to the Network layer

Model used





Three reasonable possibilities for data link services:1. Unacknowledged connectionless service2. Acknowledged connectionless service3. Acknowledged connection-oriented service

Why not unacknowledged connection-oriented service?





Unacknowledged connectionless service

1. The source sends independent frames to the destination without the destination acknowledging them.

2. No connection is established beforehand or released afterward

3. If a frame is lost due to noise on the line, no attempt is made to recover it in the data link layer

4. Appropriate when the error rate is very low5. Appropriate for real-time traffic, such as speech, in which

late data are worse than bad data6. Most LANs use unacknowledged connectionless service





Acknowledged connectionless service

Each frame sent is individually acknowledged. In this way, the sender knows whether or not a frame has arrived safely. If it has not arrived within a specified time interval, it can be sent again.

It is perhaps worth emphasizing that providing acknowledgements in the data link layer is just an optimization, never a requirement. The network layer can always send a packet and wait for it to be acknowledged.





Acknowledged connectionless service

The trouble with this strategy is that if the average packet is broken up into, say, 10 frames, and 20 percent of all frames are lost, it may take a very long time for the packet to get through. If individual frames are acknowledged and retransmitted, entire packet get through much faster.

On reliable channels, such as fiber, the overhead of a heavyweight data link protocol may be unnecessary, but on wireless channels it is well worth the cost due to their inherent unreliability.





Acknowledged connection-oriented service

1. The source and the destination establish a connection before and data are transferred.

2. Each frame sent over the connection is numbered3. The data link layer guarantees that each frame sent is

indeed received.4. Furthermore, it guarantees that each frame is received

exactly once and that all frames are received in the right order.

5. Connection-oriented service provides the network layer with the equivalent of a reliable bit stream.





Role of Data Link Protocol

The routing code frequently wants the job done right by data link protocol.




3.1.2 Framing

The data link uses the services provided by the physical layer. The physical layer will not be perfect. It is up to the data link layer to detect, and if necessary, correct errors.

The usual approach is for the data link layer to break the bit stream up into discrete frames and compute the checksum for each frame. When a frame arrives at the destination, the checksum is recomputed. If the newly computed checksum is different from the one contained in the frame, the data link layer knows that an error has occurred and takes steps to deal with it.




3.1.2 Framing

Breaking the bit stream up into frames is more difficult than it first appears.

Four methods:1. Character count.2. Flag bytes with byte stuffing.3. Flag bits with bit stuffing.4. Physical layer coding violations.




3.1.2 FramingCharacter count

Without error

With one error




3.1.2 Framing

Character count

Even if the checksum is incorrect so the destination knows that the frame is bad, it still has no way of telling where the next frame starts.

Sending a frame back to the source asking for a retransmission does not help either, since the destination does not know how many characters to skip over to get to the start of the retransmission.




3.1.2 FramingCharacter (or) byte-oriented




3.1.2 Framing

Starting and ending characters

A major disadvantage of using this framing method is that it is closely tied to 8-bit characters in general and the ASCII character code in particular.

We need a technique to allow arbitrary sized characters.




3.1.2 FramingFlags (01111110) at the beginning and end

Bit-oriented framing with bit stuffing




3.1.2 Framing

Physical layer coding violations

This method of framing is only applicable to networks in which the encoding on the physical medium contains some redundancy.

For example, some LANs encode 1 bit of data by using 2 physical bits. Normally, a 1 bit is a high-low pair and a 0 bit is a low-high pair. The combinations high-high and low-low are not used for data and they can be used as frame boundaries.




3.1.3 Error Control

The usual way to ensure reliable delivery is to provide the sender with some feedback about what is happening at the other end of the line.

Sender

Data Frame

Receiver Acknowledgement (ACK)

Or negativeAcknowledgement (NAK)




3.1.3 Error Control

An additional complication comes from the possibility that hardware troubles may cause a frame to vanish completely. In this case, the receiver will not react at all.

This possibility is dealt with by introducing timers into the data link layer.


retransmitif time-out

ACK if correct

correct and ordered

packeterror code

ack

error code

networklayer

networklayer

physicallayer

physicallayer

Sender Receiver



3.1.3 Error Control




3.1.3 Error Control

However, when frames may be transmitted multiple times there is a danger that the receiver will accept the same frame two or more times, and pass it to the network layer more than once. (When will this happen?)

To prevent this from happening, it is generally necessary to assign sequence numbers to outgoing frames, so that the receiver can distinguish retransmissions from originals.




3.1.4 Flow Control

What to do with a sender that systematically wants to transmit frames faster than the receiver can accept them?

The usual solution is to introduce flow control to throttle the sender into sending no faster than the receiver can handle the traffic. This throttling generally requires some kind of a feedback mechanism, so the sender can be made aware of whether or not the receiver is able to keep up.




3.1.4 Flow Control

Various flow control schemes are known, but most of them use the same basic principle. The protocol contains well-defined rules about when a sender may transmit the next frame.

These rules often prohibit frames from being sent until the receiver has granted permission, either implicitly or explicitly.

Feedback-based flow controlRate-based flow control



3.2 Error Detection and Correction

Transmission errors are going to be a fact of life for many years to come. (in local loop, in wireless communications)

As a result of the physical processes that generate them, errors on some media (e.g., radio) tend to come in bursts rather than singly.


error model: bursty errors

every packet would be in error

bursty error, only one packet is incorrect



Advantage:

Disadvantage: They are much harder to detect and correct than isolated errors.




3.2.1 Error-Correcting Codes

m data bits r check bits

codeword

There are 2n possible codewords and 2m possible data messages.Hamming distance between codewords: min d(C1,C2)=number of (same bit position) bits which differ

d(10010010,00010001)=3

n=m+r, code rate=m/n

If two codewords are a hamming distance d apart, it will require d single-bit errors to convert one into the other.





In most data transmission applications, all 2m possible data messages are legal, but due to the way the check bits are computed, not all 2n possible codewords are used.

Given the algorithm for computing the check bits, it is possible to construct a complete list of codewords, and from this list find the two codewords whose Hamming distance is minimum. This distance is the Hamming distance of the complete code.





To detect d errors, you need a distance d+1 code.

d+1

radius=d bits

distance<d+1 distance





<2d+1 2d+1radius=d bits

To correct d errors, you need a distance 2d+1 code.





Parity check is a code with distance 2. Therefore, it can detect single-bit error.

Consider a code with only 4 valid codewords:0000000000, 0000011111, 1111100000, 1111111111This code has a distance of 5, which means that it can detect double errors.

If the codeword 0000000111 arrives and we know at most two bits are in error, the receiver knows that the original must have been 0000011111. If, however, a triple error changes 0000000000 into 0000000111, the error will not be corrected properly.






In a block code, the r check bits are computed solely as a function of the m data bits with which they are associated.

In a systematic code, the m data bits are sent directly, along with the check bits, rather than being encoded themselves before they are sent.

In a linear code, the r check bits are computed as a linear function of the m data bits.




3.2.1 Error-Correcting CodesA 1-bit error correcting code (Hamming code)


n=m+rEach of the 2m legal messages has n illegal codewords at a distance 1 from it.Thus each of the 2m legal messages requires n+1 bit patterns dedicated to it.

r

rmm

nm

rm

rm

n

2)1(

22)1(

22)1(

Therefore,





Given m, this puts a lower limit on the number of check bits needed to correct single errors.This theoretical lower limit can, in fact, be achieved using a method due to Hamming.

The bits of codewords are numbered consecutively, starting with bit 1 at the left end. The bits that are power of 2 (1, 2, 4, 8, …) are check bits. The rest are filled up with the data bits. Each check bit forces the parity of some collection of bits, including itself, to be even (or odd).





Correct 12-bit burst errors by transmitting vertically.


0

0

0

0

111098

7654

1110732

1197531

bbbb

bbbb

bbbbb

bbbbbb




A 1-bit error correcting code (Hamming code)

When a codeword arrives, the receiver initializes a counter to zero. It then examines each check bit, k, to see if it has the correct parity. If not, it adds k to the counter. If the counter is zero afterwards, the codeword is accepted as valid. If the counter is nozero, it contains the number of the incorrect bit.

(Using even parity)





A 1-bit error correcting code (Hamming code)





Convolutional Code

Input=111, output=11 10 01 (assuming all internal states are 0)

Code rate=1/2, constraint length=7

Decoding using Viterbi algorithm





Reed-Solomon Code

RS codes are based on the fact that every n degree polynomial is uniquely determined by n+1 points.

For m bit symbols, the codewords are 2m-1 symbols long.m=8 for bytes, a codeword is then 255 bytes long.

(255,223) code: add 32 redundant symbols to 223 data symbols





Low-Density Parity Check (LDPC) Codecodeword

Parity checks

8 possible 6-bit strings corresponding to valid codewords: (000000, 011001, 110010, 101011, 111100, 100101, 001110, 010111)




3.2.2 Error-detecting Codes

error control: error detection or error correction?

assume error rate=1/106 data=1000 bitserror detecting check bits: 1 bit (e.g. parity check)error correcting check bits: 10 bits (e.g. Hamming code)

In 103 packets:error detecting overhead=103+1001=2001 bitserror correcting overhead=10*103 bits=10000 bits

retransmission

Therefore, error correction codes is used in some critical situation.For example, real-time memory system (in space shuttle).





Parity





Checksum





Checksum

479c+2=479e --->1’s complement --->b861

Computing checksum

Checking checksum

fffd+2=ffff --->1’s complement --->0000

An IP packet





CRC (Cyclic Redundancy Code)

Treat bit strings as polynomials with coefficient 0 and 1.(E.g. 10001001:x7+x3+1, 01010111=x6+x4+x2+x+1)

1. Sender and receiver agree upon a polynomial of degree r. (the generating polynomial, G(x))

).()()(mit Then trans

).(remainder with )()(

)( Assume 2.

xRxxMxT

xRxAxG

xxM

r

r

3. When receiver receives T(x), it divides it by G(x). If there is a remainder, then an error has occurred.(All the above operations are done in modulo 2.)





error detecting capabilities of CRC

E(x): error (E.g., E(x)=x5+x+1 means bits 0,1,6 are in error)

Errors can go undetected if T(x)+E(x) can be divided by G(x)with no remainder.

If there are k 1 bits in E(x), k single-bit errors have occurred. A single burst error is characterized by an initial 1, a mixture of 0s and 1s, and a final 1, with all other bits being 0.






1. single bit error E(x)=xi

if G(x) contains a constant term, then E(x)/G(x) will have remainder.

2. odd number of bits in error

Assume E(x)=G(x)Q(x). If we let G(x) has even number ofterms, then when x=1, E(1)=1 but G(1)=0, a contradiction.Or G(x) containing a factor of (x+1) will also do.

CRC-CCITT: x16+x12+x5+1






3. If there have been two isolated single-bit errors, E(x)=xi+xj, where i>j.

E(x)=xi+xj=xj(xi-j+1)

If we assume that G(x) is not divisible by x, a sufficient condition for all double errors to be detected is that G(x) does not divide xk+1 for any k up to the maximum value of i-j (i.e., up to the maximum frame length). Simple, low-degree polynomials that give protection to long frames are known. For example, x15+x14+1 will not divide xk+1 for any value of k below 32768.






Finally, and most important, a polynomial code with r check bits will detect all burst errors of length less than r.

If the burst length is r+1, the remainder of the division by G(x) will be zero if and only if the burst is identical to G(x), which has a probability of 1/2r-1 (excluding the first and the last bits).



Problems: 1, 2, 3, 4, 7, 9, 15, 16

local area networks by r.s. chang, dept. csie, ndhu 1 chapter 3 the data link layer user a user b...

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