computer networks by r.s. chang, dept. csie, ndhu1 chapter 3 the data link layer user a user b...

Computer 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 transport layer can always send a message and wait for it to be acknowledged.





Acknowledged connectionless service

The trouble with this strategy is that if the average message is broken up into, say, 10 frames, and 20 percent of all frames are lost, it may take a very long time for the message to get through. If individual frames are acknowledged and retransmitted, entire messages 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. Starting and ending characters, with character stuffing.3. Starting and ending flags, 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.



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.

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


error model: bursty errors every packet would be in error

bursty error, only one packet is incorrect



Advantage:




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

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.




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

m data bits r check bits

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.


1

1

1

1

11108

7654

1110732

1197531

bbb

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.




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).





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))2.

Assume with remainder

Then transmit

M x x

G xA x R x

T x M x x R x

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).



3.3 Elementary Data Link Protocols




3.3.1 An Unrestricted Simplex Protocol




3.3.2 A Simplex Stop-and-Wait Protocol




3.3.3 A Simplex Protocol for Noisy Channel

sequence numbers (to number packets and ACKs)

sender

receiver

P(1) P(2)

ACK

sender

receiver

P(1) P(1)

ACK

error

time-out

If there is no sequence number for packets, the receivercan not distinguish between these two cases.

new old





ACKs must also be numbered.

sender

receiver

P(1) P(1) P(2)

error

ACK ACK

time-out

The sender cannot tell whether this ACK is for P(1) or P(2) if not numbered.

A potential failure





Protocols in which the sender waits for a positive acknowledgement before advancing to the next data item are often called PAR (Positive Acknowledgement with Retransmission) or ARQ (Automatic Repeat reQuest).

Although it can handle lost frames by timing out, it requires the timeout interval to be long enough to prevent premature timeouts. If the sender times out early, while the acknowledgement is still on the way, it will send a duplicate.



3.4 Sliding Window Protocols

Piggybacking: The technique of temporarily delaying outgoing acknowledgements so that they can be hooked onto the next outgoing data frame.

Full-duplex operation: data and acknowledgements are bi-directional and intermixed

The principal advantage of using piggybacking over having distinct acknowledgement frames is a better use of the available channel bandwidth (because it saves the overhead of a distinct frame).




However, piggybacking introduces a complication not present with separate acknowledgements. How long should the data link layer wait for a packet onto which to piggyback the acknowledgements?

If the data link layer waits longer than the sender’s timeout period, the frame will be retransmitted, defeating the whole purpose of having acknowledgements.

Resort to some ad hoc schemes, such as waiting for a fixed number of milliseconds.




The ultimate purpose: To design a protocol that remained synchronized in the face of any combination of garbled frames, lost frames, and premature timeouts.

Sliding window protocols: The essence of all sliding window protocols is that at any instant of time, the sender maintains a set of sequence numbers corresponding to frames it is permitted to send. These frames are said to fall within the sending window.

Similarly, the receiver also maintains a receiving window corresponding to the set of frames it is permitted to accept.




The sender’s window and the receiver’s window need not have the same lower and upper limits, or even have the same size. In some protocols they are fixed in size, but in others they can grow or shrink as frames are sent and received.

The sequence numbers within the sender’s window represent frames sent but as yet not acknowledged. Whenever a new packet arrives from the network layer, it is given the next higher sequence number, and the upper edge of the window is advanced by one. When an acknowledgement comes in, the lower edge is advanced by one. In this way the window continuously maintains a list of unacknowledged frames.




A sliding window of size 1 with 3-bit sequence number.(a) Initially(b) After the first frame has been sent.




A sliding window of size 1 with 3-bit sequence number.(c) After the first frame has been received.(d) After the first acknowledgement has been received.




Since frames currently within the sender’s window may ultimately be lost or damaged in transmit, the sender must keep all these frames in its memory for possible retransmission.

Thus if the maximum window size is n, the senders need n buffers to hold the unacknowledged frames.

If the window ever grows to its maximum size, the sending data link layer must forcibly shut off the network layer until another buffer becomes free.




The receiving data link layer’s window corresponds to the frames it may accept. Any frame falling outside the window is discarded without comment. When a frame whose sequence number is equal to the lower edge of the window is received, it is passed to the network layer, an acknowledgement is generated, and the window is rotated by one.

Unlike the sender’s window, the receiver’s window always remains at its initial size. Note that a window size of 1 means that the data link layer only accepts frames in order, but for larger windows this is not so.

The network layer, in contrast, is always fed data in the proper order, regardless of the data link layer’s window size.




3.4.1 A One Bit Sliding Window Protocol

normal scenario

sender

receiver

0 1 0 1

ACK0 ACK1 ACK0 ACK1

sender time-out

sender

receiver

0 1

ACK0 ACK1 ACK1 ACK0

1time-out 0 ignored

Stop-and-Waitprotocol




3.4.1 A One Bit Sliding Window ProtocolACK lost or corrupted

sender

receiver

0 1

ACK0 ACK1 ACK1 ACK0

1time-out 0

Packet lost or corrupted

sender

receiver

0 1

ACK0 ACK1 ACK0

1time-out 0

NAK can be used here to speed up processing.

NAK





A normal scenario

(seq, ack, packet number)An asterisk indicates where a network layer accepts a packet





A peculiar scenario(half of the frames are duplicates)

(seq, ack, packet number)




3.4.2 A Protocol Using Go Back n

sender

receiver

TRANSP

PROP PROC TRANSA

PROP PROC

S=minimum time between successive packet transmissions

processing time

S=TRANSP+TRANSA+2(PROC+PROP)efficiency(without error)=TRANSP/S

Efficiency of sliding window of size 1 (stop and wait) protocol

propagation delay





Example 1: a transmission line of 500km at 56kbps, packet and ack

are 700 bits: TRANSA = TRANSP =700bits

56000bpss

PROP =500000m

m / ss. Consequently,

S 0.0125s + 2 (0.00217s) + 2 PROC + 0.0125s= 0.02934s + 2 PROC

efficiency0.0125s

0.02934s + 2 PROC

0.0125s

0.02934s

0 0125

2 3 100 00217

43%

8

.

..


Example 2: 500km optical fiber at 100Mbps, packet and ack

are 10000 bits: TRANSA = TRANSP =10000bits

10 bpss

PROP =500000m

m / ss. Assume PROC = 1ms.

Consequently, S 10 s + 2 (0.0025s) + 2 0.001s +10 s= 7.2 10 s

efficiency10 s

7.2 10 s

8

-4 -4

-3

-4

-3

10

2 100 0025

14%

4

8.

.








Example 3: Using satellite link

50-kbps satellite channel with a 500 msec round-trip delay

1000-bit frame, TRANSP=20 msec,Assume all other times are negligible.

Efficiency=20/520=4%

Clearly, the combination of a long transmit time, high bandwidth, and short frame length is disastrous in terms of efficiency.





The technique of pipelining: Filling the pipe with frames

For example, in the last satellite example, if the sender’s window size is 26, after sending the 26th frame (at time 520 msec), the acknowledgement of the first frame will be back. The 27th frame can be sent now without any idle time in the sender.





An important design parameter: How many unacknowledged outstanding packets are allowed?

It affects:•sequence number•buffer sizes in sender and receiver

sender has to buffer those sent but unacknowledged (for possible retransmission) (IF) receiver has to buffer those received but are out-of-

sequence (receiver has to deliver packets to network layer in order)

SN





Pipelining frames over an unreliable channel raises some serious issues. First, what happens if a frame in the middle of a long stream is damaged or lost. What should the receiver do with all the correct frames following it?

Two approaches:go back n: the receiver discards all frames (the receiver has a window of size 1)selective repeat: buffer all the correct frames





Go Back n





Selective Repeat





k-bit sequence number

packet numbering: 0,1,2,3,4, ..., 2k-1

maximum window size: 2k-1

(3 bits)

(0,1,2,3,4,5,6,7)

(7 buffers at sender)

packets

packets 0~7 received correctlyACK of 7 received correctlynew batch of 0~7 (correct)ACK of 7 received correctly

packets 0~7 received correctlyACK of 7 received correctlynew batch of 0~7 (lost)ACK of 7 received correctly

(can not be 2k, otherwise consider the following)





Because this protocol has multiple outstanding frames. It logically needs multiple timers, one per outstanding frame. All of these timers can easily be simulated in software using a single hardware clock that causes interrupts periodically.




3.4.3 A Protocol Using Selective Repeat

How many outstanding and unacknowledged frames are allowed?

sender’s window

receiver’swindow

Consider when the cases when acks are ok or when acks are lost.

All seven frames received correctly





How many outstanding and unacknowledged frames are allowed?

sender’s window

receiver’swindow

Make sure there is no overlap when receiver advances the window.





k-bit sequence number

packet numbering: 0,1,2,3,4, ..., 2k-1

maximum window size: 2k-1

(3 bits)

(0,1,2,3,4,5,6,7)

(4 buffers)

packets



3.5 Protocol Verification

Protocol correctness: How do you know that the protocol design is correct? For example, no deadlock, no loop, etc.

Protocol (implementation) conformance: How do you know that your implementation is correct?




3.5.1 Finite State Machine Models

A finite state machine model of a protocol can be regarded as a quadruple (S, M, I, T), where:

S is the set of states the processes and channel can be inM is the set of frames that can be exchanged over the channelI is the set of initial states of the processesT is the set of transitions between states




3.5.1 Finite State Machine Models

(a) State diagram for protocol 3. (b) Transmissions

SRC

Normal operation




3.5.2 Petri Net Models

A Petri net has four basic elements: places, transitions, arcs, and tokens.

1. A place represents a state which (part of) the system may be in.

2. A transition is indicated by a horizontal or vertical bar.3. A transition is enabled if there is at least one input token in

each of its places.4. Any enabled transition may fire at will, removing one

token from each input place and depositing a token in each output place.

Enabledtransition





Petri net model for protocol 3





Petri nets can be represented in a convenient algebraic form resembling a grammar.

1. BD→AC2. A→AC3. AD→BE4. B→BE5. C→6. D→7. E→8. CF→DF9. EG→DG10. CG→DF11. EF→DG

A, C, G have tokens. So rules 2, 5, 10 can fire.



3.6 Example data Link Protocols

3.6.1 HDLC—High-level Data Link Control

SDLC (Synchronous Data Link Control) ofIBM’s SNA (System Network Architecture)

ADCCP (Advanced Data Communication Control Procedure)

HDLC (High-level Data LinkControl)

ANSI ISO

LAP (Link Access Procedure)

LAPB (Link Access ProcedureBalanced)





Frame format (bit-oriented)





The P/F bit stands for Poll/Final. It is used when a computer is polling a group of terminals. When used as P, the computer is inviting the terminal to send data. All the frames sent by the terminal, except the final one, have the P/F bit set to P. The final one is set to F.





A supervisory frame

type=0 receiver ready (an acknowledgement frame)type=1 reject (negative acknowledgement frame)type=2 receiver not ready (stop sending data)type=3 selective reject (retransmission of only the frame specified)




3.6.2 The Data Link Layer in the Internet





PPP (Point-to-Point Protocol) (RFCs 1661, 1662, 1663)

PPP handles error detection, supports multiple protocols (not just IP), allows IP addresses to be negotiated at connection time, permits authentication, and has many other features.






PPP provides three things:

1. A framing method that unambiguously delineates the end of one frame and the start of the next one. The frame format also handles error detection.







2. A link control protocol for bringing lines up, testing them, negotiating options, and bring them down again gracefully when they are no longer needed. The protocol is called LCP (Link Control Protocol).







3. A way to negotiate network-layer options in a way that is independent of the network layer protocol to be used. The method chosen is to have a different NCP (Network Control Protocol) for each network layer supported.






Making a communication:1. PC calls the provider’s router by a modem2. Use LCP packets to select PPP parameters3. Use NCP to receive an IP address4. Send and receive IP packets5. Use NCP to tear down the network layer connection and free up

the IP address6. Use LCP to shutdown the data link connection7. The computer tells the modem to hang up






The PPP full frame format for unnumbered mode operation



1. Implement selective repeat protocol in Figure 3-19. (Simulate the send and receive using interprocess

communication mechanisms)

2. Page 243, Problem 2 Page 244, Problems 12, 15, 18 Page 245, Problem 29

computer networks by r.s. chang, dept. csie, ndhu1 chapter 3 the data link layer user a user b...

Documents