Lecturer: Tae-Hyong Kim (D132) [email protected] Ch.2 Direct Link Networks Computer Networks.

Lecturer: Tae-Hyong Kim (D132)[email protected]

Ch.2 Direct Link Networks

Computer Networks

2

How to connect hosts physically? Suitable medium Framing Encoding Media access control Error Detection Reliable Delivery

Problem: Physically Problem: Physically Connecting HostsConnecting Hosts

3

Signals Sine Wave Ti me & Frequency Domain Composite Signals Digital Signals Shannon Capacity

Hardware Building Blocks Encoding Framing Error Detection Reliable Transmission Project #1 Ethernet Wireless Assignment #2

ContentsContents

4

Analog and digital signals Analog signal: infinitely many levels of intensity

continuous Digital signal: only a limited number of defined values

discrete

Periodic and Nonperiodic Signals Period? What signals can be used for data transmission?

SignalsSignals

5

Sine WaveSine Wave

Sine wave s(t) = Asin(2ft +

) s : instantaneous

amplitude A : peak amplitude f : frequency : phase

Period = ?

6

Period and Frequency Period: the amount of time a signal needs to complete

one cycle

Frequency: the number of periods in one second

they are inverses of each other T = 1/f

Ex. 100 ms = 10-1 s 1/(10-1) Hz = 10Hz = 10-2KHz

Sine WaveSine Wave

7

More about Frequency Frequency: the rate of change wrt. time.

Change in a short span of time high frequency Change over a long span of time low frequency

Two Extremes A signal does not change at all frequency = 0 (DC) A signal changes instantaneously frequency = infinite

Phase Position of the waveform relative to time zero

Sine WaveSine Wave

8

Distance a simple signal can travel in one period Usually used for the transmission of light in an optical fiber

Wavelength() = propagation speed(c) period(T) = propagation speed(c) / frequency(f)

Depend on both the frequency of a signal and the medium = cT = c/f

Ex.1 the wavelength of red light (f = 41014 Hz) in air: = c/f = (3108)/(41014) = 0.75m

WavelengthWavelength

9

Time-domain plot instantaneous amplitude with respect to time.

Frequency-domain plot maximum amplitude with respect to frequency Analog signals are best represented in the freq. domain

Relation

Time and Frequency Time and Frequency DomainDomain

10

Relation The frequency domain is more compact and useful

when we are dealing with more than one sine wave

Time and Frequency Time and Frequency DomainDomain

11

Usage of a single sine wave Carry of electric energy (power) single tone not useful in data communications to make signals that can carry information, we

have to add several different sine waves (composite signals)

Composite Signals A periodic signal decomposed into two or more

sine waves. Fourier Analysis (Transform) is used to decompose

a composite signal into its components

Composite SignalsComposite Signals

12

Fourier Analysis Any composite signal can be represented as a

combination of simple sine waves with different frequencies, phases, and amplitudes

An example: a square wave


13

Fourier Series Periodic time domain signals discrete

frequency domain signals


14

Fourier Transform Nonperiodic time domain signals

continuous frequency domain signals


15

Fourier Analysis An example: a square wave

First three harmonics : f, 3f, 5f

Adding first three harmonics


16

Fourier Analysis An example: a square wave

Frequency spectrum comparison


17

Fourier Analysis An example: nonperiodic composite signal

e.g., voice level (microphone)


18

Definition: The range of frequencies contained in a composite

signal the difference between the highest and the lowest

frequencies

BandwidthBandwidth

19

Example: Which signal has a wider bandwidth, a

sine wave with a frequency of 100Hz or a sine wave with a frequency of 200Hz?

BandwidthBandwidth

20

Digital signals with different signal levels if a signal has L levels each level needs log2L bits

e.g., 8 levels no. of bits per level = log28 = 3

Digital SignalsDigital Signals

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the number of bit sent in 1 second (bps) Ex. 100 page (24line*80col) text per minute

100*24*80*8 = 1,636,000 bps = 1.636 Mbps Ex. HDTV : 1920*1080, refresh rate : 30/s,

24 bit color depth 1920*1080*30*24 = 1,492,992,000 ≈ 1.5Gbps

Bit RateBit Rate

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A digital signal is a composite analog signal

Bandwidth = ?

Bandwidth of Digital Bandwidth of Digital SignalsSignals

23

Two Approaches Baseband transmission: digital ( digital) Broadband transmission: digital analog

BW of Physical Medium The frequency BW that medium can pass

Digital Signal TransmissionDigital Signal Transmission

24

The required BW for the given bit rate Nyquist theorem (Noiseless assumption)

Bit Rate (n) 2BW log2L (L = # of signal levels) Why?

Ex.1 Consider the same noiseless channel, transmitting a signal with four signal levels. The maximum bit rate is:

BitRate = 2 3000 log24 = 12,000bps

Ex.2 We need to send 256 kbps over a noiseless channel with a BW of 20kHz. how many signal levels do we need?

256,000 = 2 20,000 log2L

log2L = 6.625, L = 26.625 = 98.7 levels 128 levels

Bit Rate and BW Bit Rate and BW

25

Theoretical highest data rate for a noisy channel: Capacity = BW log2(1+SNR) (bps)

SNR (Signal-to-Noise Ratio) = Signal Power/Noise Power

Ex.2 What is the theoretical highest bit rate of a regular telephone line? (BW:3000 hz, SNR : 3162 (35 dB)) C=3000log2(1+3162)=3000log23163=34,680 bps

How are 56kbps modems possible?

Shannon CapacityShannon Capacity

26

Using Both LimitsUsing Both Limits

Shannon capacity the upper limit Nyquist formula no. of signal levels Ex.1 We have a channel with a 1MHz BW.

The SNR for this channel is 63; what is the appropriate bit rate and signal level? Upper limit: by Shannon formula

Capacity = B log2(1+SNR) = 106 log2 (1+63) = 106 log2 64

= 6Mbps

Let’s choose 4Mbps for better performance Then, the number of signal levels: by Nyquist formula

4M bps = 2 1 MHz log2L

L = 4

27

Modulation : digital signal analog signal

Why? Examples?

Broadband TransmissionBroadband Transmission

28

Speed of Electromagnetic waves = Speed of light 3108m/s

Electromagnetic SpectrumElectromagnetic Spectrum

29

Signals Hardware Building Blocks

Nodes Links

Encoding Framing Error Detection Reliable Transmission Project #1 Ethernet Wireless Assignment #2

ContentsContents

30

Workstations Hosts, switches, routers

NodesNodes

31

Network adaptor

NodesNodes

32

Why throughput < BW? Nodes

Sender Network Receiver

Node design Memory, bus, …

Congestion Where?

NodesNodes

33

Common types of cables

CablesCables

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Common BW (service) available from the carriers

Leased LinesLeased Lines

Line

T1

T3

OC-1

OC-3

OC-12

OC-24

OC-48

35

Common services available to connect your home

VDSL

Last-Mile LinksLast-Mile Links

Service BW

ADSL ~12Mbps / ~1.3Mbps

VDSL ~55Mbps / ~2.3Mbps

Cable Modem 10Mbps

Metro Optical Ethernet 100Mbps, 1Gbps

36

Radio band should be licensed ISM(Industrial, Scientific, Medical)

Bands Ex. 900MHs, 2.4GHz, 5.8GHz

Wireless LinksWireless Links

Service BW

Wi-Fi (802.11G/A) 54Mbps

Wi-Fi(802.11N) 150, 300, 600Mbps

Bluetooth (2.0EDR) 2.1Mbps

HSDPA (3.5G) 14Mbps

WiMax(802.16) 75Mbps

LTE(3.9G) 100Mbps

LTE-Advanced(4G) 1Gbps

37

Signals Hardware Building Blocks Encoding

NRZI, RZ, Manchester, Block coding Decoding Problems

Framing Error Detection Reliable Transmission Project #1 Ethernet Wireless Assignment #2

ContentsContents

Encoding Data Code (Bits? Signals?)

Signal encoding is required for physical transmission cf. Modulation

Digital (data) to digital (signal) encoding Line coding

Bit-by-bit encoding NRZ, NRZI, Manchester, …

block coding (Prior to line coding) Block-by-block encoding 4B/5B, 8B/10B, …

What is Encoding?What is Encoding?

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NRZ(NonReturn to Zero)

Decoding problems? (consecutive 1’s or 0’s)

Baseline wander Average level of the encoded signal?

Clock recoveryHow to decode correctly when clock difference or

signal delay change?

Basic Encoding: NRZBasic Encoding: NRZ

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Baseline Wander How to make average = 0?

Clock Recovery Separate clock transmission?

Encoding Design Problem No Baseline Wander Self Clock Recovery Signal BW , bit rate Reliability (special features)

How to Solve Decoding How to Solve Decoding Problems?Problems?

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41

Bit Rates and Baud RatesBit Rates and Baud Rates

Signal element vs. Data Element r = no. of data elements carried by each signal element

42

Bit Rates and Baud RatesBit Rates and Baud Rates

Data rate vs. Signal rate data rate (N): no. of data elements (bits) sent in 1s (bit rate) signal rate (S): no. of signal elements sent in 1s (baud rate,

symbol rate) BW is related to Bit rate? or Baud Rate?

How to increase bit rate while decreasing baud rate? S = c ×N × 1/r (baud)

c: case factor (0≤c≤1), depends on no. of 0's and 1's usually BW S

Ex.1 A signal is carrying data in which one data element is encoded as one signal element (r=1). If the bit rate is 100 kbps, what is the average value of the baud rate if c is between 0 and 1?

caverage = 1/2, S = c×N×1/r = 1/2×100,000×1=50,000=50k baud

NRZI(NRZ Inverted)

r =? Consecutive 1’s? Consecutive 0’s? Signal BW (baud rate)?

NRZINRZI

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44

RZRZ

RZ (Return-to-Zero) uses three values (positive, negative, zero)

1 : positive-to-zero 0 : negative-to-zero

Two signal changes for a bit BW? Baseline Wander? Clock Recovery?

r = 1/2, BW(RZ) = 2 BW(NRZ)

Manchester & Differential Manchester

Baseline wander? Clock Recovery? Signal BW (baud rate)? (r=?)

ManchesterManchester

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4B/5B 1 Redundant bit (cost), why? No more than 3 consecutive 0’s are

encountered How to solve consecutive 1’s? NRZI

Unused code(25-24)? cf. 8B/10B

Block CodingBlock Coding

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Data Code Data Code0000 11110 1000 100100001 01001 1001 100110010 10100 1010 101100011 10101 1011 101110100 01010 1100 110100101 01011 1101 110110110 01110 1110 111000111 01111 1111 11101

47

Signals Hardware Building Blocks Encoding Framing

Byte-Oriented vs. Bit-Oriented Framing Approaches

Error Detection Reliable Transmission Project #1 Ethernet Wireless Assignment #2

ContentsContents

Break bitstream into a frame Typically implemented by network

adaptor

Design problem in framing Point-to-point links Multipoint links

What is Framing?What is Framing?

48

Frame a collection of bytes (characters)

Examples BISYNC(Binary Synchronous Comm.) (IBM,

1967)

DDCMP(Digital Data Comm. Message Prot.)(DECNET)

PPP(Point-to-Point Protocol) (RFC1661)

Byte-Oriented ProtocolsByte-Oriented Protocols

49

Frame a collection of bits Examples

HDLC(High-level Data Link Control)(ISO)

Ethernet, LLC, … How to construct a frame?

Frame starting? Header? Trailer?

Bit-Oriented ProtocolsBit-Oriented Protocols

50

Sentinel-based Delineate frame with special character or

pattern BISYNC, DDCMP: SYN character PPP, HDLC: Flag(01111110)

Problem: special char. or pattern appears in the payload

Charater stuffing (Byte-Oriented Protocols) Extra characters are inserted (ESC character)

Bit stuffing (Bit-Oriented Protocols) Sender: insert 0 after five consecutive 1’s (why?) Receiver: delete 0 that follows five consecutive 1’s

Framing ApproachesFraming Approaches

51

Counter-based (Byte-Counting) When payload length is variable Include payload length in header E.g., DDCMP

Problem: count field corrupted Solution: catch when CRC fails

Cf. CRC(Cyclic Redundancy Check)


52

Clock-based Each frame is a specific time long E.g., SONET(Synchronous Optical Network)

(125s) STS-n multiplexing (STS-1 = 51.84Mbps)


53

concatenated

54

Signals Hardware Building Blocks Encoding Framing Error Detection

Error Detection and Error Correction 2D parity, Checksum, CRC

Reliable Transmission Project #1 Ethernet Wireless Assignment #2

ContentsContents

Single-bit error The least likely type of error in serial data transmission

Burst error two or more bits in the data unit have changed Does not necessary mean that the errors occur in

consecutive bits

Type of ErrorsType of Errors

55

56

Detection Vs. CorrectionDetection Vs. Correction

Redundancy Adding extra bits and Sending

Error Detection Has any error occurred?

Yes, or No

Error Correction the number of bits corrupted the location of corrupted bits ex. the number of bits: 10 bits, 1000 bit message

# of possibility: 1000C10 = 1000×999×…×991/(10×9×…×1)=?

Error correcting code (FEC) vs. Retransmission

57

Error Detecting CodeError Detecting Code

How do we create redundant bits? Cost? Performance?

Parity Check One bit is added (even-parity bit, odd-parity

bit) Performance?

Two-Dimensional Parity CRC (Cyclic Redundancy Check) Checksum

58

Modular ArithmeticModular Arithmetic

Features a limited range of integers are used : 0 ~ N-1 an upper limit: N Modulus Modulo-N arithmetic

No carry when you add two digits in a column No carry when you subtract one digit from another in a column

Modulo-2 arithmetic: N = 2 (0~1) Adding: 0+0 = 0, 0+1=1, 1+0=1, 1+1=0 Subtracting: 0−0 = 0, 0−1=1, 1−0=1, 1−1=0 same result same as XOR operation

Cost? Added bits?

Performance? One-bit error? 2-bit, 3-bit error? 4-bit error?

Two-Dimensional ParityTwo-Dimensional Parity

59

Sum (1's complement arithmetic wrapped sum) If the number has more than n bits, the extra leftmost

bits is added to the n rightmost bits (wrapping)

Checksum Inverting all bits of sum (negative value)

ChecksumChecksum

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Routine in C (16-bit checksum)

Cost & Performance?

ChecksumChecksum

61

u_short cksum(u_short *buf, int count){ register u_long sum = 0; while (count--) { sum += *buf++; if (sum & 0xFFFF0000) { /* carry occurred, so wrap around */ sum &= 0xFFFF; sum++; } } return ~(sum & 0xFFFF);}

Cyclic code If a codeword is cyclically shifted (rotated), the

result is another codeword E.g., CRC code with C(7,4)

CRC(Cyclic Redundancy CRC(Cyclic Redundancy Check)Check)

62

Encoding concept


63

Decoding concept


64

Polynomial representation

CRC divisor using polynomial

CRC PolynomialCRC Polynomial

65

66

CRC PerformanceCRC Performance

Some definitions Message: M(x) Transmitted Message: P(x) Divisor: C(x) Error: E(x)

In a cyclic code: If P(x)/C(x) ≠ 0, one or more bits is corrupted If P(x)/C(x) = 0, either

No bit is corrupted, or Some bits are corrupted, but the decoder failed to detect them

Received codeword = P(x) + E(x) Received codeword/g(x) = P(x)/C(x) + E(x)/C(x) E(x) errors that are divisible by C(x) are not caught

67


Single-bit error E(x) = xi (i: the position of the bit) If a single-bit error is caught, xi is not divisible by C(x) C(x) has more than one term and the coefficient of x0 is 1 all single-bit errors can be caught

e.g., C(x) = x3+1

Two isolated single-bit error E(x) = xj + xi (j – i = the distance between the two errors)

E(x) = xi (xj-i + 1), j-i>1, i≥0 If C(x) has more than one term and one term is x0 it cannot divide

xj. And if C(x) cannot divide xt+1 (2≤t≤n-1) (at least 3 terms) all

isolated double errors can be detected (n=degree of P(x)+1)

68


Odd number of errors C(x) containing (x+1) can detect all odd numbers of errors

Proof? e.g., x4+x2+x+1 = (x+1)(x3+x2+1)

Burst errors E(x) = xj+…+xi (two terms or more) E(x) = xi(xj-i+…+1) If C(x) can detect a single error, it cannot divide xi

the remainder of (xj-i+…+1)/(xr+…+1) must not be zero (C(x)=xr+…+1) If (j – i < r), the remainder can never be zero j – i = L – 1 (L=error length) (L – 1 < r) (L < r+1)

(L ≤ r)

All burst errors with L≤r will be detected

69

CRC Summary (1)CRC Summary (1)

Characteristics of Good Polynomial Divisor1. It should have at least three terms2. The coefficient of the term x0 should be 13. It should have the factor x+1

Standard Polynomials

70

CRC Summary (2)CRC Summary (2)

Advantages of CRC Very good performance in detecting single-bit errors,

double errors, an odd number of errors, and burst errors

can be easily implemented in hardware

71

Signals Hardware Building Blocks Encoding Framing Error Detection Reliable Transmission

Stop-and Wait Sliding Windows Go-Back-N, Selective Repeat

Project #1 Ethernet Wireless Assignment #2

ContentsContents

Flow control Restrict the amount of data that the sender can send

before waiting for acknowledgment

Error Control Based on Automatic Repeat reQuest (ARQ): the

retransmission of data

Protocols Stop-and-Wait ARQ Sliding Window

Go-Back-N ARQ Selective Repeat ARQ

Flow and Error ControlFlow and Error Control

72

Basic Stop-and-Wait Scenarios (1) No error and frame lost

Stop-and-WaitStop-and-Wait

73

Basic Stop-and-Wait Scenarios (2) How to solve the scenario (d)?


74

SAW with sequence number (1-bit) Can this solve scenario (d)? Cost? Ack number rule?

Link Utilization (U) ttrans = x, tprop = y, a=y/x Perceived latency 2y + x U = x/(2y+x)=1/(2a+1) U L/(RTTBW)=x/2y=1/(2a)

E.g. 1.5Mbps link, RTT=45ms , frame size = 1KB, U=?

How to increase U?


75

76

frame frame prop

The time for transmission of the frame in the case that a frame lost or that its ACK

is lost. Two transmission attempts are required for succesful transmission

Timeout 2

.

assume that

T T T T

frame prop

frame prop

f

prop Timeout value is equal to twice (In fact, slightly longer)

: the average of

2( 2 )

( 2

times each frame must be transmitted

normalized throughput (link utilizat

)

ion)

r

r

T

N

T T T

T N T T

TU

rame

frame prop

1

( 2 ) (1 2 )r rN T T N a


Link Utilization (at error condition)

77

1

Pr[exactly attempts] Pr[( 1) unsuccessful attempts] Pr[successful atte

the probability : probability

mpts]

that a single frame is

in error

(1

E

)k

r

k

N

p

k -

p p

1

1

1

[transmissions] ( Pr[ transmissions])

1 ( (1

normalized throughput (link utilizat

))1

1 1

(

ion)

1 2

) 2

1

i

i

i

r

i i

ip pp

pU

N a a


Link Utilization (at error condition)

78


Link Utilization (at error condition) Performance of stop-and-wait protocol (p=10-3)

1 1

(1 2 ) 1 2r

pU

N a a

How to improve the efficiency of SAW? Concept in timeline

Sequence number modulo-2m (m: the size of the sequence number field in

bits) e.g., m=3 0,1,2,3,4,5,6,7,0,1,2, ...

How to manage frames to be sent?

Sliding WindowSliding Window

79

Sliding window on sender

LFS – LAR SWS

Sliding window on receiver

LAF-LFR RWS

Notations SWS: Send Window Size, RWS: Receive Window Size LAR: Last Ack Received, LFS: Last Frame Sent LFR: Last Frame Received, LAF: Last Acceptable Frame


80

The receiver, when receiving a frame with

SeqNum if (SeqNum ≤ LFR) or (SeqNum > LAF) (out of valid range)

Discard the frame?

Let’s consider in more detail later

If (LFR < SeqNum ≤ LAF) (within range) Accept the frame!

How to send an ACK? Ack Number? SeqNumToAck

all frames upto SeqNumToAck have been received

ACK may be sent for each frame or cumulative frames

Set LFR SeqNumToAck and LAF LFR+RWS


81


82


83

Design Issues When the receiver receives the frame with

(LFR<SeqNum≤LAF) but (SeqNum ≠ LFR+1) ACK? AckNum? Negative ACK(NAK)? Selective ACK(SACK)? Pros and Cons?

When the timer for the lost frame expires, Send the frame only? Send the frame and the subsequent frames that

have been sent together? Pros and Cons?


84

Well-known sliding window ARQ protocols Go-Back-N

Retransmit all subsequent frames sent if frame error RWS = 1 Motivation? Effect?

Selective Repeat (Selective Reject) Retransmit the error frame only RWS = SWS Motivation? Effect?

What if RWS > SWS?


85

Sequence numbers m bits in header 0 … 2m-1

Sequence numbers and sliding window size If sequence numbers are 0 … MaxSeqNum-1,

what is the maximum size of SWS (to increase channel utilization)?

SWS > MaxSeqNum, possible? SWS = MaxSeqNum, possible? SWS ≤ MaxSeqNum – 1, sufficient?

When Go-Back-N ARQ (RWS=1)? When Selective Repeat ARQ (RWS=SWS)?

Sequence Numbers and Sequence Numbers and Sliding WindowSliding Window

86

Go-Back-N ARQ (MaxSeqNum = 4)


87

Selective Repeat ARQ (MaxSeqNum = 4) SWS < (MaxSeqNum+1)/2


88

89

Go-Back-N ARQGo-Back-N ARQ

Performance Timing of sliding window mechanism (error-free)

≥

90


Performance Timing of sliding window mechanism (error-free)

91


Performance The utilization of error-free sliding window mechanism

1 2 1

2 12 1

W aU W

W aa

W=2n-1 W=1: stop and wait W=7: many case W=127: high speed

WANs

92

1

1

E[number of transmitted frames to succesfully transmit one frame]

( ) (1 ) where ( ) is the number of frames transmitted

if the

r

i

i

N

f i p p f i

1 1

1 1

original frame must be transmitted times.

( ) 1 ( 1) (1 ) ( : number of frames to be retransmitted)

1(1 ) (1 ) (1 ) 1

1 1

is approximately equal to

i ir

i i

i

f i i K K Ki K

K p KpN K p p K ip p K

p p

K

No error

No error

(2 1) for (2 1), and for (2 1)

1/ 2 1

1 2

(1 )/ 2 1

(2 1)(1 )

r

r

a w a K W W a

pU U N W a

ap

W pU U N W a

a p Wp

Performance The utilization of Go-back-N ARQ


93

average number of transmitted frames to succesfully transmit one frame

1 ( : probability that a single frame is in error)

1

' ( ' error-free utilization of sliding window mechanism)

1 2'

r

r

r

N

N pp

UU U

N

WU

1

2 12 1

1 2 1

(1 )2 1

2 1

a

WW a

a

p W aU W p

W aa

Performance The utilization of selective repeat ARQ

Selective Repeat ARQSelective Repeat ARQ

94



95



Protocol stack (assumption)

Frame header

Implementation of Sliding Implementation of Sliding WindowWindow

96

HLP

LINK

HLP

LINK

SWP SWP

State of SW protocol


97

sendSWP() send(SWP, packet)


98

deliverSWP()


99


100

swpInWindow()


101

Piggybacking In full-duplex transmission Frame can carry both user data with SeqNum and AckNum Example: HDLC


102

To reliably deliver frames across an unreliable link

To preserve the order in which frames are transmitted

To support flow control A feedback mechanism by which the

receiver is able to throttle the sender

The Role of Sliding Window The Role of Sliding Window AlgorithmAlgorithm

103

104

Signals Hardware Building Blocks Encoding Framing Error Detection Reliable Transmission Project #1 Ethernet Wireless Assignment #2

ContentsContents

Mandatory Implement Selective Repeat ARQ (SWS=RWS)

algorithm in C Refer to Code in Textbook

Selective (1) Implement real-time Simulator of Selective

Repeat ARQ (2) Implement visual trace-based Simulator of

Selective Repeat ARQ (3) Implement performance-based Simulator

of Selective Repeat ARQ

Project #1Project #1

105

Requirements and Materials Discrete Event-based Simulator Engine will be

provided Study Discrete Event-Driven Simulation With

Slides Lecture movie

Configure Simulation Environment Design Simulator UI Show simulation information and/or

performance result Detailed guideline will be provided

Project #1Project #1

106

107

Signals Hardware Building Blocks Encoding Framing Error Detection Reliable Transmission Project #1 Ethernet

IEEE Standards Physical Properties Frame Format Transmitter Algorithm Evolution of Ethernet

Wireless Assignment #2

ContentsContents

108

IEEE StandardsIEEE Standards

IEEE Standard for LAN: Project 802

109


Data Link Layer Logical Link Control (LLC)

in charge of flow control, error control, framing (partly)

110


Data Link Layer Media Access Control (MAC)

Physical Layer dependent on the implementation and type of physical

media used Ethernet evolution through four

generations

111

Physical PropertiesPhysical Properties

Categories of standard Ethernet

10Base5 (Thick Ethernet)

10Base2 (Thin Ethernet)


112

10Base-T: Twisted Pair Ethernet


113

Maximum link length 2500m with repeaters

Collision domain Hosts are competing for access


114

802.3 MAC frame

Ethernet V2 frame

115

Frame FormatFrame Format

~1500B

PDUType ~1500B

116


Preamble (7B) for synchronization of receiver’s H/W with the incoming signal bit pattern : 10101010….. Added at the physical layer (not formal part of the frame)

SFD (1B: 10101011) indicates a last chance for synchronization

Destination/Source address (6B) serial number on the NIC (unique) Broadcast address (only for DA) : all 6 bytes set to 1

Length/Type (2B) < 1518 (802.3): length field length of data field > 1536 (V2): the type of the PDU packet encapsulated in the

frame Data (46-1500B) CRC (4B): CRC-32

117


Frame length: minimum and maximum

The maximum length (1518B): historical reasons by reducing size of buffer, for preventing monopolizing the

medium The minimum length (64B): for CSMA/CD

If there is a collision before the physical layer sends a whole frame, it must be heard by all stations

64 bytes for 10Mbps Ethernet (collision domain = 2500m) If the upper-layer packet is less than 46 bytes, padding is

added to make up the difference

118


Addressing 6-byte physical address on network interface card

(NIC)

Source address unicast (only one station)

Destination address Unicast: 1-to-1 Multicast: 1-to-many Broadcast: 1-to-all (FFFFFFFFFFFF)

Who, When can send data? Random Access

CSMA/CD, CSMA/CA Controlled Access

Token Bus, Token Ring Reservation-based Polling-based

Channelization (Data Link Layer Techniques) FDMA, TDMA, CDMA

Access Control MethodsAccess Control Methods

119

120

CSMA(Carrier Sense CSMA(Carrier Sense Multiple Access)Multiple Access)

Approach Minimize the chance of collision increase the

performance Each station first listen to the medium before sending

(CS) Reduces the possibility of collision but not eliminate

due to propagation delay Collision in CSMA

121

CSMACSMA

Persistence Method The procedure for a station that senses a busy

medium 1-persistent, nonpersistent, p-persistent

122

CSMACSMA

Persistence Method 1-persistent, nonpersistent, p-persistent (cont.)

if p=1

123

CSMA/CD (Collision CSMA/CD (Collision Detection)Detection)

Approach A station monitors the medium after it sends

a frame to see if the transmission was successful, If so, the station is finished, otherwise, the frame is sent again

124

CSMA/CDCSMA/CD

Procedure

125

CSMA/CDCSMA/CD

Minimum frame size Before sending the last bit of the frame, the sending

station must detect a collision frame transmission time ≥ 2 × max. propagation time L/R ≥ 2dmax/V L ≥ 2Rdmax/V

126

Evolution of EthernetEvolution of Ethernet

Four Generation Ethernet

127

Fast EthernetFast Ethernet

Goals Upgrade the data rate to 100 Mbps Make it compatible with Standard Ethernet Keep the same 48-bit address Keep the same frame format Keep the same minimum and maximum frame length

Access Method Half duplex: CSMA/CD (collision domain = 250m WHY?) Full duplex: no CSMA/CD The implementations keep CSMA/CD for backward compatibility

Minimum and maximum frame size: same as those of Ethernet

128

Fast EthernetFast Ethernet

Autonegotiation Allows two devices to negotiate the mode or data rate

of operation In order to allow incompatible devices to connect one

another In order to allow one device to have multiple capabilities In order to allow a station to check a hub’s capabilities

Implementations

129

Gigabit EthernetGigabit Ethernet

Goals Upgrade the data rate to 1Gbps Make it compatible with Standard or Fast Ethernet Keep the same 48-bit address Keep the same frame format Keep the same minimum and maximum frame length To support autonegotiation as defined in Fast Ethernet

Usage Backbone, high-speed links

Implementation

130

Gigabit EthernetGigabit Ethernet

Access method Full-duplex Mode

No collision CSMA/CD is not used the maximum length of the cable is determined by

the signal attenuation in the cable Half-duplex Mode (1000BaseT)

Traditional Slot time for Gigabit Ethernet: 5.12 bit = 0.512μsCollision domain = 25m too short

Frame Bursting not efficient Jumbo frame: up to 9Kbytes

131

10-Gigabit Ethernet10-Gigabit Ethernet

Goals Upgrade the data rate to 10Gbps Make it compatible with Standard, Fast, and Gigabit Ethernet Keep the same 48-bit address Keep the same frame format Keep the same minimum and maximum frame length Allow the interconnection of existing LANs into a metropolitan

are network (MAN) or a wide are network (WAN) Make Ethernet compatible with technologies such as Frame

Relay and ATM

Usage Backbone, high-speed links

Only Full-duplex mode with optical fiber

Capture packets in Ethernet LAN Check the version of Ethernet Check the fields of Ethernet

frame Framing sequences Addresses Size/Type CRC

Packet Capture with Packet Capture with WiresharkWireshark

132

133

Signals Hardware Building Blocks Encoding Framing Error Detection Reliable Transmission Project #1 Ethernet Wireless

Overview Bluetooth (802.15.1) Wi-Fi (802.11): Physical, CSMA/CA, Architecture, Frame Format WiMax Cell Phone Technologies

Assignment #2

ContentsContents

Leading wireless technologies

OverviewOverview

134

Bluetooth802.15.1

Wi-Fi802.11

WiMax802.16

3G Cellular

Typical Link

Length10m 100m 10km Tens of km

Typical BW

21.Mbps(shared)

54Mbps(shared)

70Mbps(shared)

384+Kbps(per conn)

Typical Use

Link a peripheral

to a notebook computer

Link a notebook computer to a wired

base

Link a building to

a wired tower

Link a cell phone to a

wired tower

Wired Analogy

USB EthernetCoaxial cable

DSL

Wireless network using a base station

OverviewOverview

135

Wireless ad hoc or mesh network

OverviewOverview

136

137

BluetoothBluetooth

Features WLAN technology designed to connect devices of different

functions such as telephones, notebooks, computers, cameras, printers, coffee makers, and so on.

A Bluetooth LAN is an ad hoc network formed spontaneously Applications

Peripheral devices: wireless mouse or keyboard Monitoring devices: sensor devices, home security devices

Origin of name Harald Blaatand, king of Denmark (a project by the Ericsson Co.)

Standard IEEE 802.15.1

138

BluetoothBluetooth

Two types of networks piconet (small net)

can have up to eight active stations (1 primary, the rest secondaries)

the communication between the primary and the secondary can be one-to-one or one-to-many

communication is only between the primary and a secondary/secondaries

an additional 8 secondaries can be in parked state cannot take part in communication

139

ArchitectureArchitecture

Two types of networks scatternet

Piconets can be combined to form a scatternet A secondary station in one piconet can be the primary in

another piconet This station can receive messages from the primary in the

first piconet and deliver to the secondaries in the second piconet

A station can be a member of two piconets

140

BluetoothBluetooth

Bluetooth devices has built-in short range radio transmitter data rate: 1Mbps with a 2.4-GHz band

possible interference between IEEE 802.11b WLAN and Bluetooth LANs

Two sublayers

Wi-FiWi-Fi

141

142

Wi-FiWi-Fi

Physical Properties

ISM (Industrial, Scientific, and Medical) band

143

Wi-FiWi-Fi

Wireless Problem In wireless applications, stations must be able to share

air medium without interception by an eavesdropper and without being subject to jamming from a malicious intruder

Solution approach : Spread spectrum spread the original spectrum needed for each station

BSS >> B (the required BW)

Frequency hopping spread spectrum (FHSS) Direct Sequence Spread Spectrum (DSSS)

FHSS concept

Wi-FiWi-Fi

144

DSSS Concept

Wi-FiWi-Fi

145

OFDM? Orthogonal Frequency Division

Multiplexing FDM and OFDM

Wi-FiWi-Fi

146

147

Wi-Fi MACWi-Fi MAC

DCF (Distributed Coordination Function) Access method: CSMA/CA The reasons WLAN cannot implement CSMA/CD

For collision detection, a station must be able to send data and receive collision signals at the same time costly station and increased BW requirements

The distance between stations can be great. Signal fading could prevent a station at one end from hearing a collision at the other end

Even Carrier Sense may not be possible because of the hidden station problem

148

Wi-Fi MACWi-Fi MAC

CSMA/CA

149

Wi-Fi MACWi-Fi MAC

CSMA/CA and NAV

150

Wi-Fi MACWi-Fi MAC

Point Coordination Function (PCF) Optional access method for an infrastructure network implemented on top of the DCF used mostly for time-sensitive transmission centralized, contention-free polling access method

AP performs polling for stations that are capable of being polled The stations are polled one after another, sending any data they

have to the AP To give priority to PCF over DCF, another set of interframe

spaces has neen defined: PIFS (PCF IFS) < DIFS

151

Wi-Fi MACWi-Fi MAC

Hidden Node Problem

Solution: RTS/CTS handshaking

152

Wi-Fi MACWi-Fi MAC

Exposed Node Problem a station refrains from using a channel when it is available(BA (no

i/f) CD) waste the capacity of the channel

CTS/RTS cannot help in this case: half-duplex cannot hear when sending

153

Wi-Fi MACWi-Fi MAC

CTS/RTS shortcoming Case 1) (AB (no interference) CD)

A B C DRTS

CTS CTS RTS

Collisionpacket

transmission CTS

RTS

Collision

CTS

time

154

Wi-Fi MACWi-Fi MAC

CTS/RTS shortcoming Case 2)

A B C DRTS

CTSCTS RTS

packettransmission

A B

CTS

time

packettransmission

C D

packettransmission

C DCollision

C did not hear B’s CTS since it was

transmitting its own RTS to D

155

Wi-Fi ArchitectureWi-Fi Architecture

Basic Service Set (BSS) Made of stationary or mobile wireless stations and an

optional central base station (AP) Adhoc network: a BSS without an AP Intrastructure network: a BSS with an AP

156


Extended Service Set (ESS) Made of two or more BSSs with Aps BSSs are connected through a distribution system

Distribution system connects the APs in the BSSs A mobile station can belong to more than one BSS at

the same time

157


Station Types Based on mobility

No-transition mobility Stationary : not moving Moving only inside a BSS

BSS-transition mobility Moving from one BSS to another inside one BSS Inter-BSS Handover

ESS-transition mobility Moving from one ESS to another Inter-ESS Handover

cf. Handover (Handoff) Issues Seamless HO, Smooth HO, Fast HO Soft HO, Hard HO Vertical HO

158

Wi-Fi Frame FormatWi-Fi Frame Format

Frame format

FC: Frame Control type of frame and some control information

D: NAV, or ID of the frame SC: Sequence Control Seq. no for flow control

159


Frame format Subfields in FC field

160


Frame Types Management frames

for the initial communication between stations and APs Control frames

for accessing the channel and acknowledging frames

values of subfields in control frames

Data frames for carrying data and control information

161


Four cases in addresses use 'To DS' and 'From DS' flags in the FC field

Address 1: the address of the next device Address 2: the address of the previous device Address 3: the address of the final station if it is not

defined by address 1 Address 4: the address of the original source if it is not

the same as address 2

162


Four cases in addresses

Capture packets in Wireless LAN Check Wireless LAN technology Check the fields of WLAN frame

Non-security frame Security frame

Consider WLAN Security Issues Eavesdropping Hacking

Packet Capture with Packet Capture with WiresharkWireshark

163

What is WiMax(Worldwide Interoperability for Microwave Access)? For the delivery of last mile wireless broadband access as

an alternative to cable and DSL. Provides fixed, nomadic, portable and, eventually, mobile

wireless broadband connectivity without the need for direct line-of-sight (LOS) with a base station.

In a typical cell radius deployment of three to ten kilometers, WiMAX Forum Certified™ systems can be expected to deliver capacity of up to 40 Mbps per channel, for fixed and portable access applications.

Mobile network deployments are expected to provide up to 15 Mbps of capacity within a typical cell radius deployment of up to three kilometers.

WiMaxWiMax

164

WiMax Standard

WiMaxWiMax

165

802.16 802.16a 802.16-2004

802.16e-2005

Date Completed

December 2001

January 2003

June 2004

December 2005

Spectrum 10-66 GHz < 11 GHz < 11 GHz

< 6 GHz

Operation

LOS Non-LOS Non-LOS Non-LOS and Mobile

Bit Rate 32-134 Mbps

Up to 75 Mbps

Up to 75 Mbps

Up to 15 Mbps

Cell Radius

1-3 miles 3-5 miles 3-5 miles 1-3 miles

Usage

WiMaxWiMax

166

Overview

Cell Phone TechnologiesCell Phone Technologies

167

168


Provides communications between two moving units (Mobile Stations) or between one mobile unit and one stationary unit

A service provider must be able to: Locate and track a caller Assign a channel to the call Transfer the channel from BS (Base Station) to BS as caller moves out

of range Cells: small regions each cellular service area is divided

into Contains an antenna (uses own range of frequency) Controlled by a small office (BS) controlled by a switching office (Mobile Switching Center)

169


Message Switching Center (MSC) Coordinates communication between all BSs and the telephone

central office Telephone central office

Connects calls, records call information, and bills Shape of cells

Square Hexagon: for equidistance antennas Cell size

Depends on the population of the area Typically 1~12 miles in radius The transmission power of each cell is kept low to prevent its

signal from interfering with those of other signals

170


Handoff(Handover) Hard handoff (using one BS) : early model

First communication must be broken with the previous BS and then communication can be established with the new one

Soft handoff (using two BSs; Seamless handoff) During handoff MS may continue with the new BS before

breaking from the old one

BSi

Signal strength due

to BSj

E

X

1

Signal strength due

to BSi

BSjX

3

X

4

X

2

X

5

Xth

MSPmin

Pi(x)

Pj(x)

171


Roaming Problem A user can have access to

communication or can be reached where there is coverage

But a service provider usually has limited coverage

Solution approach: roaming Neighboring service providers can

provide extended coverage through a roaming contract

172


Satellite Network telephony

Exercises Calculate: 1, 3, 7, 19, 31, 32, 33, 34, 41, 43, 44, 47, 48 Analyze: 13, 24, 27, 28, 35, 52, 66

Experiments Wireshark experiments

Capture HTTP traffic from your PC with capture filter Screenshots: capture filter, captured traffic and information

5-Slide Survey Wireless mesh networks Motivation (why?), Problem (what?), Technique (how?)

Use PPT slides and upload at the Report board

Assignment #2Assignment #2

173

Lecturer: Tae-Hyong Kim (D132) [email protected] Ch.2 Direct Link Networks Computer Networks.

Documents

frequency analog signals

relationthe frequency

componentscomposite

harmonicscomposite signals

simple signal

square wavecomposite

signal needs

fourier analysisan example