CSIS0234A Computer and Communication Networks

Direct Link Network Point-to-point

Point-to-point

1

Readings  Peterson

et. al, Computer Networks – A System Approach, 3rd edition, Morgan Kaufmann. 

Chapter 2

Stallings, Data and Computer Communications, 7th Edition, Prentice Hall

 W.



Sections 3.2 and 3.4

 B.A.

Forouzan, Data Communications and Networking, 3rd Edition, McGraw-Hill 

Point-to-point

Sections 10.3, 11.3 and 11.4

2

Terminology  Direct 

link

No intermediate devices

 Simplex 

 Point-to-point 

Direct link



Only 2 devices share link



 Half 

 Signals 

Electric or electromagnetic representations of data

 Transmission 

Point-to-point

Communication of data by propagation and processing of signals

One direction

duplex

Either direction, but only one way at a time 

 Full 

e.g. Television

e.g. police radio

duplex

Both directions at the same time 

e.g. telephone 3

Signals  Physical

medium carries signals by which data are propagated 

electromagnetic waves

 Signals 

Analog signal – Continuously variable 

Signal intensity varies in a “smooth and continuous” fashion over time 



Point-to-point

can be analog or digital

e.g. Sine Wave

Digital signal – Discrete values 

Maintains a constant level then changes to another constant level



Use two DC components 4

Analog & Digital Transmission  Analog

signal transmitted without regard to content

 Maybe

analog/digital data

 May

become weaker (attenuated) over distance 

Use amplifiers to boost signal

 Also 



Point-to-point

amplifies noise

audio or video data – quality deteriorate but still “usable”

 Concerned

with content

 Integrity

endangered by noise, attenuation etc.

 Repeaters

used



Repeater receives signal



Recovers bit pattern



Retransmits the digital signal

 Attenuation  Noise

is overcome

is not amplified

digital data – introduce errors

5

Advantages of Digital Transmission  Data

integrity



Using repeaters remove the effects of noise



Longer distances over lower quality lines

 Capacity

utilization



High degree of multiplexing easier with digital techniques



Using statistical multiplexing rather that frequency division

 Security 

& Privacy

Encryption

 Integration 

Can treat analog and digital data digitally



Economics of scale as integration of voice, video, and digital data

Point-to-point

6

Channel Capacity  Data 

rate - In bits per second

Rate at which data can be communicated

 Bandwidth 

Any transmission system has a limited band of frequencies 



Constrained by transmitter and medium

Bandwidth Wc is range of frequencies passed by the communication channel

 Noise 

Average level of noise over the communication channel

 Error 

Point-to-point

rate

The rate at which errors occur 7

Spectrum

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8

Nyquist Bandwidth A 

noise free channel



If rate of signal transmission is 2Wc then signal with frequencies no greater than Wc is sufficient to carry signal rate



Bit Rate = 2 x 3000 = 6000 bps

Given binary signal 

Two signal levels



Data rate supported by Wc Hz is 2Wc bps

Point-to-point

Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. The maximum bit rate can be calculated as

 For

a given bandwidth, can data rate be increased? 

How?

9

Multilevel Signaling  With

two signal levels, each pulse carries one bit of information



Bit rate = 2Wc bits/second



 With

M = 2m signal levels, each pulse carries m bits Bit rate =



2Wc pulses/s * m bits/pulse = 2Wc m = 2 Wc log2 M bps



Example of multilevel signals 

Four levels {-1, -1/3, 1/3, +1} for {00, 01,10,11}



Waveform for 11,10, 01 sends +1, +1/3, -1/3

Consider the same noiseless channel, transmitting a signal with four signal levels (for each level, we send two bits). The maximum bit rate can be calculated as:

Bit Rate = 2 x 3000 x log2 4 = 12,000 bps Point-to-point

10

Practical limitation 

Receiver needs to distinguish M (2m) possible signal elements during each signal time



Error rate depends on relative value of noise amplitude and spacing between signal levels 

If transmitted power is limited, then as M increases, spacing between levels decreases



Presence of noise at receiver causes more frequent errors to occur as M is increased

Point-to-point

11

Signal-to-Noise Ratio Presence of noise limits accuracy of measurement of received signal amplitude signal

signal + noise

noise

High SNR

noise

signal

virtually error-free signal + noise

Low SNR error-prone Bit Error Rate (BER) increases with decreasing signal-to-noise ratio Point-to-point

SNR =

Average Signal Power Average Noise Power SNR (dB) = 10 log10 SNR

12

Shannon Channel Capacity 



Consider the relationship among data rate, noise and error rate Faster data rate “shortens” each bit so burst of noise affects more bits 



Signal to noise ratio (in decibels) 



At given noise level, high data rate means higher error rate



Arbitrarily reliable communications is possible if the transmission rate R < C.



If R > C, then arbitrarily reliable communications is not possible.



“Arbitrarily reliable” means the BER can be made arbitrarily small through sufficiently complex coding.



C can be used as a measure of how close a system design is to the best achievable performance.

SNRdb = 10 log10 (signal/noise)

Capacity C = Wc log2(1+SNR)

Point-to-point

13

Example We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is usually 3162. For this channel the capacity is calculated as

C = Wc log2 (1 + SNR) = 3000 log2 (1 + 3162) = 3000 log2 (3163) = 34,860 bps

Point-to-point

14

Example We have a channel with a 1 MHz bandwidth. The SNR for this channel is 24dB; what is the appropriate bit rate and signal level? First, we use the Shannon formula to find our upper limit. SNRdB = 24 dB = 10 log10(SNR) SNR = 251 C = 1 MHz log2 (1 + 251) = 106 log2 252 ≈ 8 Mbps

This is a theoretical limit, is unlikely to be reached. But assume we can achieve the limit. Then we use the Nyquist formula to find the number of signal levels. 8 Mbps = 2 × 1 MHz × log2 M  M = 16 Point-to-point

15

Line Coding  Signals 

How to represent the 1s and 0s by the electrical or light pulses

 Digital 

propagate over a physical medium signal

Discrete, discontinuous voltage pulses 

sent as a series of "square waves" of either positive or negative voltage. 

e.g. Voltages vary between +3/-3 to +24/-24, depending on the circuit.



Each pulse is a signal element



We are encoding the binary data by a signal element or a group of signal elements

 Each

digital transmission standard defines



What voltage levels correspond to a bit value of 0 or 1.



Which encoding scheme it is using

Point-to-point

16

Non-Return-to-Zero (NRZ)  Two

different voltages for 0 and 1 bits

 Voltage 

remain constant during “bit interval” – no transition

i.e. no return to zero voltage

 Examples: 

Unipolar NRZ – Absence of voltage for zero, constant positive voltage for one



Polar NRZ – Negative voltage for one value and positive for the other

 Long

strings of 1 or 0



Poor timing – unable to recover clock



Low-frequency content B its

Point-to-point

NR Z

0 0 1

0 1

1

1

1

0 1

0 0 0 0 1

0

17

Differential Coding  The

information is encoded as a change of signals between successive data symbols rather than the signal levels

 Non-return 

Makes a transition from the current signal to encode a one, and stay at the current signal to encode a zero 



Point-to-point

to Zero Inverted (NRZI)

Data encoded as presence or absence of signal transition at beginning of bit time or at the middle of bit time

Solves the problem of consecutive ones

18

Manchester code  Every

interval has transition in middle - mid-bit transition. 

Transition serves as clock and data representation One is Zero is

Pros

Cons Self clocking Error detection Absence of expected transition

Double the bandwidth

 Used

in 10-Mbps Ethernet & other LAN standards

Point-to-point

19

4B/5B coding 

Every 4 bits of data encoded in a 5-bit code



5-bit codes selected to have no more than one leading 0 and no more than two trailing 0s 



resulting 5-bit codes are transmitted using NRZI 



No problem with consecutive 1s

achieves 80% efficiency 



Never get more than 3 consecutive 0s

Using 5 signal elements to encode 4 bits

Used by 100Base-TX and FDDI

Point-to-point

20

Framing  The

physical layer is moving data as a sequence of bits

 Upper

layers work with a block of data, not bit streams

 The

data link layer needs to find some way to break the sequence of bits into a frame

 Typically

implemented by network adaptor B its

Node A

A daptor

A daptor

Node B

Frame s

Point-to-point

21

Approaches (1)  Sentinel-based 

Delineate frame with special pattern 

Byte-Oriented: e.g. PPP 



Frames consist of integer number of bytes

Bit-Oriented: e.g. HDLC 

a collection of bits, is not concerned with byte boundaries



Problem: special pattern appears in the payload



Solution 

Byte-Oriented: byte stuffing

Point-to-point

22

Approaches (2) 

Problems with byte stuffing ✖ ✖



Size of frame varies unpredictably due to byte insertion Malicious users can inflate bandwidth by inserting those control/escape characters

Bit-Oriented: bit stuffing, e.g. 01111110 

sender: insert 0 after five consecutive 1s



receiver: when it detects a 0 that follows five consecutive 1s, delete it ✖

if after five 1s, next bits are 10: frame marker ☛

If next bits are 11: error

Data received

01111110000111011111011111011001111110 Data to be sent

0110111111111100

After destuffing and deframing

*000111011111-11111-110*

After stuffing and framing

0111111001101111101111100001111110 Point-to-point

23

Approaches (3)  Counter-based 

include payload length in header 

e.g. DECNET's DDCMP 8

8

8

14

42

Count

H e ade r

16 B ody

 problem:

count field corrupted

 solution:

catch when CRC fails

Point-to-point

CR C

24

Question  Suppose

the following sequence of bits arrives over a

link: 1101011111010111110010111110110 Show the resulting frame after any stuffed bits have been removed.

Point-to-point

25

Error Control  How 

to make sure the integrity of all received frames ?

Error control 

For reliable communication, errors must be detected and corrected



Error types include corrupted data and missing data



Error control is concerned with: 1) detecting AND 2) correcting errors.



Point-to-point



Tell the sender if error occur -- feedback



Sender cautiously waits for feedback -- timing



Which frames we are currently handling -- labeling/sequencing

Two basic approaches: 

Error detection & retransmission (ARQ)



Forward error correction (FEC) 26

Errors  Single

bit errors



One bit altered



Adjacent bits not affected

 Burst

errors



Length B



Contiguous sequence of B bits in which first last and any number of intermediate bits in error



Impulse noise



Fading in wireless



Effect is greater at higher data rates

Point-to-point

27

Key Idea  All

transmitted data blocks (“codewords”) satisfy a pattern

 If

received block doesn’t satisfy pattern, it is in error

 Redundancy:

the amount of information that is transmitted is over and above the required minimum

 Blindspot:

when channel transforms a codeword into another codeword All inputs to channel satisfy pattern or condition

User Encoder information Point-to-point

Channel

Channel output

Pattern checking

Deliver user information or set error alarm

28

Error Detection Process At the receiving end, the error detection value is recalculated and checked against the received value If they differ, an error has occurred

Error detection uses the concept of redundancy, which means adding extra bits for detecting errors at the destination

An error detection value is first calculated by the sender and transmitted along with the data

Point-to-point

29

Single Parity Check  Parity

checking is one of the oldest and simplest error detection techniques

 Append

an overall parity check to k information bits



For even parity, the sum of the bits (including the parity bit) remains even



For odd parity, the sum remains odd

 Receiver 

All error patterns that change an odd # of bits are detectable

 Parity 

Point-to-point

checks to see if # of 1s is even or odd

checking is simple, but doesn't catch all errors.

If two (or an even number of) bits have been changed at the same time, the parity check appears to be correct even though an error has occurred.

30

Two-Dimensional Parity Check  More

parity bits to improve coverage

 Adds

an additional character of parity checks, called a block control character (BCC) to each block of data.

 The

BCC parity calculations are based on the 1st bit, 2nd bit, etc.

 Catching

over 98% of all errors, but still misses a 1 0 0 1 0 0 significant number of errors. 0 1 0 0 0 1 1 0 0 1 0 0 1 1 0 1 1 0

Last column consists of check bits for each row

1 0 0 1 1 1 Bottom row consists of

Point-to-point

check bit for each column

31

Error-detecting capability 1 0 0 1 0 0

1 0 0 1 0 0

0 0 0 0 0 1

0 0 0 0 0 1

1 0 0 1 0 0

One error

1 0 0 1 0 0

1 1 0 1 1 0

1 0 0 1 1 0

1 0 0 1 1 1

1 0 0 1 1 1

1 0 0 1 0 0

1 0 0 1 0 0

0 0 0 1 0 1

0 0 0 1 0 1

1 0 0 1 0 0 Three errors 1 0 0 1 1 0

1 0 0 1 0 0

1 0 0 1 1 1

1 0 0 1 1 1

1 0 0 0 1 0

Arrows indicate failed check bits

Point-to-point

Two errors

1, 2, or 3 errors can always be detected; Not all patterns >4 errors can be detected Four errors (undetectable)

32

Internet Checksum 



Several Internet protocols (e.g. IP, TCP, UDP) use check bits to detect errors in the header 

A checksum is calculated and included in a special field



Algorithm selected for ease of implementation in software

View message M as a sequence of 16-bit integers 

Add all these integers using 16-bit ones complement arithmetic X = ( b0 + b1 + b2+ ...+ bL-1 ) 

Ones complement addition, carryout from most significant bit is added to result 1010 1100 10110 0111

Point-to-point

33

Verifying Internet Checksum 

Then take the ones complement of the result bL = - X 

In ones complement, negative number is represented each bit inverted 0101 → 1010



Resulting 16-bit number is the checksum

 On 

the receiver side

Do the same checksum operation on the received message M' 0 = - ( b0 + b1 + b2+ ...+ bL-1 + bL )

Point-to-point

34

Example Suppose the following block of 16 bits is to be sent using a checksum of 8 bits. 10101001 00111001 The numbers are added using one’s complement 10101001 169 00111001 57 -----------Sum 11100010 226 Checksum 00011101 29 The pattern sent is 10101001 00111001 00011101 Now suppose the receiver receives the pattern and there is no error. 10101001 00111001 00011101 When the receiver adds the three sections, it will get all 1s, which, after complementing, is all 0s and shows that there is no error. 10101001 169 00111001 57 00011101 29 Sum 11111111 255 Complement 00000000 means that the pattern is OK.

Point-to-point

Now suppose there is a burst error of length 5 that affects 4 bits. 10101111 11111001 00011101 When the receiver adds the three sections, it gets 10101111 175 11111001 249 00011101 29 Partial Sum 111000101 453 Carry 1 Sum 11000110 198 Complement 00111001 57 the pattern is corrupted.

Usually, Internet checksum uses a 16-bit word, not 8-bit

35

Cyclic Redundancy Check (CRC) 

Add k bits of redundant data to an n-bit message 

We want k n+1 > m+r+1;

e.g., for m=7, r >= 4



Useful for one way transmissions or when transmission times are very long (e.g., communications to spacecraft)



Hamming Code - one example of FEC techniques

Point-to-point

39

Hamming Code  Extra

parity values are calculated so that each data bit figures into at least two parity bit calculations 

This means that if any one bit, either parity or data, gets changed in transmission, the change in the received data can be detected and corrected



The secret of error correction is to locate the invalid bit

 This

technique, however, only works for one bit errors



Point-to-point

The bits that are powers of 2 are check bits 40

Hamming Code (2)

Point-to-point

A bit (at pos k) is checked by those check bits which sum equal to k. e.g., 11 = 1 + 2 + 8

41

Example Data - 1001101

Code - 10011100101 Point-to-point

42

Error Correction via Retransmission  Detect

error, correct by retransmission

 The

process of requesting that a data transmission be resent is called an Automatic Repeat Request or ARQ 

Basic elements of ARQ: 

Error-detecting code with high error coverage



Acknowledgment



 Main



ACKs (positive acknowledgments)



NAKs (negative acknowledgments)**

Timeout mechanism

ARQ protocols are:



Stop-and-Wait



Sliding Window - Go-Back N & Selective Repeat

Point-to-point

43

Stop-and-Wait  Source

transmits single

frame 

Wait for ACK

 Destination

Replies with ACK

ACK damaged/lost,



Transmitter will retransmit



Problem: Receiver gets two copies of frame



Question: How to differentiate?

receives

frame 

 If

 If

received frame damaged, discard it 

Transmitter has timeout, retransmit



Question: How long would the sender wait?

Point-to-point

44

Need for Sequence Numbers (a) Frame 1 lost

A B (b) ACK lost

A B     

Time-out

Frame Frame 0 1 ACK

Frame Frame 1 2 ACK

Time-out Frame Frame Frame Frame 0 1 1 ACK 2 ACK ACK

Time

Time

In cases (a) & (b) the transmitting station A acts the same way But in case (b) the receiving station B accepts frame 1 twice Question: How is the receiver to know the second frame is also frame 1? Answer: Add frame sequence number (1-bit) in header Put the current sequence number in the header

Point-to-point

45

Sequence Numbers (c) Premature Time-out

Is this acknowledgment for Frame 1 ??

Time-out A

Frame 0 ACK

B

Time Frame 0

ACK

Frame 1

Frame 2



The transmitting station A misinterprets duplicate ACKs



Incorrectly assumes second ACK acknowledges Frame 1



Question: How is the receiver to know second ACK is for frame 0?



Answer: Add frame sequence number in ACK header



the sequence number of next frame expected by the receiver

Point-to-point

46

Stop-and-Wait Model

A

t0 = total time to transmit 1 frame  



B

 tprop

tproc

 frame time tf

tproc

tack

tprop bits/info frame

    

t 0 = 2t prop + 2t proc + t f + t ack bits/ACK frame nf na = 2t prop + 2t proc + + R R

First bit enters channel Last bit enters channel First bit arrives at receiver Last bit arrives at receiver Channel idle while transmitter waits for ACK

Point-to-point

channel transmission rate

47

Transmission Efficiency bits for header & CRC

Effective transmission rate: R eff =

n f −n0 Number of information bits delivered to destination = Total time required to deliver the information bits t0

Transmission efficiency: n f − no

η0=

Point-to-point

Reff R

=

t0 R

= 1+

na + nf

Extra overhead in carrying ACK frame

no 1− nf . 2(t prop + t proc ) R

Effect of frame overhead – loss in transmission efficiency

nf Effect of Delay-Bandwidth Product

48

Example nf=1250 bytes = 10000 bits, na=no=25 bytes = 200 bits 2xDelayxBW Efficiency

1 ms

10 ms

100 ms

1 sec

1 Mbps

200 km 103

2000 km 104

20000 km 105

200000 km 106

1 Gbps

88% 106

49% 107

9% 108

1% 109

1%

0.1%

0.01%

0.001%

Stop-and-Wait does not work well for very high speeds or long propagation delays Point-to-point

49

Sliding Window  Improve  Keep

channel busy by continuing to send frames

 Allow 

Stop-and-Wait by not waiting!

a window of up to window un-ACKed frames

Frame transmission are pipelined to keep the channel busy

 Assign

sequence number to each frame

 If

receive ACKs before window is exhausted, we can Se nde r R e ce ive r continue transmitting

Point-to-point

50

Sender's Sliding Window Frames transmitted and ACKed

LAR 

...

LFS

LAR + SWS

LFS - LAR