Data Communication Principles

Transcription

1 Data Communication Principles Mahalingam Ramkumar Mississippi State University, MS August 15, 2016 Ramkumar CSE 4153 / 6153

2 1 Data Communication Channels Data Rate of a Communication Channel Transmission Media Modulation Multiplexing 2 Repeaters Round Trip Time 3 Packet 4 Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol 5 2 / 103

3 Fundamental Factors Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Channel bandwidth H Hz Signal power S Watts Noise power N Watts 3 / 103

4 Channel Capacity R Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Achievable bit-rate R Shannon s Capacity Limit R = H log 2 (1 + S/N) bps. (1) 4 / 103

5 S/N in Decibels (db) Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Signal to Noise Ratio (SNR) Usually expressed in db (S/N) db = 10log 10 (S/N) 20 db S/N = 100; 30 db S/N = 1000; 5 / 103

6 Example Data Communication Channels Data Rate of a Communication Channel Transmission Media Modulation Multiplexing What is the limit on achievable rate R in a channel with bandwidth 1 MHz and SNR 30 db? 30dB S/N = 1000 R = log 2 ( ) = 10Mbps (2) To double the rate to 20 Mbps, we can Double the bandwidth, or increase SNR to 60 db (or increase S/N 1 million) log 2 (1, 000, 000) 20 6 / 103

7 Nyquist Sampling Theorem Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Any signal emerging from a medium is band-width limited Bandwidth - H hertz (cycles/sec) A band-limited signal (limited to H Hz) can be uniquely reconstructed from discrete samples taken at a rate 2H per second 7 / 103

8 Nyquist Capacity Data Communication Channels Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Assume each sample can take V possible values log 2 (V ) bits per sample Nyquist capacity C = 2H log 2 (V ) No of distinct recognizable values V is determined by noise. If noise is zero, V, so C Every medium has inherent noise Even at absolute zero (0 Kelvin). 8 / 103

9 Transmission Media Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Two basic types Guided Twisted pair Coaxial cables Fiber optics Magnetic media Unguided 9 / 103

10 Twisted Pair Data Communication Channels Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Telephone cables Any wire is an antenna... Twisting substantially reduces interference Waves from different twists cancel out Number of twists per cm Categories 3 (16MHz), 5 (100 MHz), 6 (250), and 7(600) Ethernet cables (a) (b) 10 / 103

11 Coaxial Cables Data Communication Channels Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Up to 1 GHz Twisted pair cables are catching up! Better noise immunity than twisted pairs Used for TV signals, cable TV... Copper core Insulating material Braided outer conductor Protective plastic covering 11 / 103

12 Fiber Optics Data Communication Channels Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Extremely high bandwidth Single mode and multimode Single mode fibers can handle upto 50GHz over 100km! Air/silica boundary Air β 1 β 2 β 3 Total internal reflection. α 1 α 2 α 3 Silica Light source (a) (b) 12 / 103

13 Fiber vs Copper Wire Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Weight vs bandwidth Cost Security 13 / 103

14 Electromagnetic Spectrum Data Rate of a Communication Channel Transmission Media Modulation Multiplexing f (Hz) Radio Microwave Infrared UV X-ray Gamma ray Visible light f (Hz) Twisted pair Satellite Fiber Coax Terrestrial optics microwave Maritime AM radio FM radio TV Band LF MF HF VHF UHF SHF EHF THF 14 / 103

15 Allocation of Spectrum Data Rate of a Communication Channel Transmission Media Modulation Multiplexing ITU-R, FCC ISM (Industrial, Scientific, Medical) Bands Bandwidth 26 MHz 83.5 MHz MHz Freq MHz MHz GHz GHz GHz GHz 15 / 103

16 Satellites Data Communication Channels Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Geo-stationary (about 36,000 km) Medium Earth Orbit (18,000 km, 6 hr orbits) Low Earth Orbit (less than 1000 km) Satellite vs Fiber Remote / hostile areas Point-to-point vs Broadcast Mobile communications Right of way for laying cables 16 / 103

17 Modulation Data Communication Channels Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Modulation: information signal + carrier signal modulated signal Speech signals have frequencies between 300 and 3000 Hz Signals cannot propagate directly over all mediums Speech signals s(t) can be sent directly over a copper cable: but not over air or free-space Carrier signals are used to carry the information signals Modulated signal transmitted over a medium Demodulation: extracting information signal from the modulated signal Types of modulation: Amplitude (AM), Phase (PM), Frequency (FM) 17 / 103

18 Data Rate of a Communication Channel Transmission Media Modulation Multiplexing (a) (b) (c) (d) Phase changes 18 / 103

19 Bandwidth of Modulated Signals Data Rate of a Communication Channel Transmission Media Modulation Multiplexing The same as the bandwidth of the information signal The bandwidth is now around a center frequency the frequency of the carrier Many signals can coexist simultaneously in the medium. By using different carrier frequencies 19 / 103

20 Modulation for FAX/Dial-up Internet Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Send bits over telephone cable using Modem (Modulator - Demodulator) Cable meant for carrying voice signals between Hz Amplitude + phase modulation - series of sinusoidal pulses Modulator: Bits are chunked, represented using sinusoidal pulses Demodulator: Convert received sinusoidal pulse to bits Clipped sinusoid is the basic signal Variations achieved by modifying amplitude and phase Represented on a constellation diagram 20 / 103

21 QPSK, QAM-16, QAM-64 Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Quadrature Phase Shift keying (2 bits per sample), Quadrature Amplitude Modulation (a) 270 (b) 270 (c) 21 / 103

22 Telephone/FAX Modem Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Each pulse represents multiple bits If we use 16 different types of pulses (eg., QAM 16) we can send 4 bits per pulse (16 = 2 4 ) Bit rate and Baud rate Baud rate - number of sinusoidal pulses per second Fixed at 2400 for telephone lines Bit rate = Baud rate x number of bits per pulse How do we get 56 K with baud rate of 2400? 22 / 103

23 Trellis Coded Modulation Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Add more bits per sample Add extra bits to each sample for error correction V.32 (4+1, 32) bps, V.32 bis (6 + 1, 128) 14,400 bps V.34-28,800 bps, V.34 bis - 36,600 bps (b) 270 (c) 23 / 103

24 Modern Modems Data Communication Channels Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Full duplex - transmissions possible in both directions at the same time Half-duplex - Both directions, but not at the same time Simplex - Only one direction V.90, V.92 - full duplex Dedicated uplink and downlink channel 56 kbps downlink, 33.6 kbps uplink (V.90) 48kbps uplink (V.92) + facility to detect incoming calls while online 24 / 103

25 Actual Capacity of a Telephone Line Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Mpbs Meters / 103

26 ADSL Data Communication Channels Data Rate of a Communication Channel Transmission Media Modulation Multiplexing We have been sitting on a gold-mine! Telephones filter out higher frequencies to suppress noise 1.1 MhZ spectrum divided into 256 channels Hz each (DMT - Discrete Multitone) First channel used for POTS (plain old telephone service) kHz Channels Power khz Voice Upstream Downstream 26 / 103

27 ADSL Equipment Data Communication Channels Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Voice switch Telephone Codec Splitter Telephone line NID Splitter Computer DSLAM To ISP Telephone company end office ADSL modem Ethernet Customer premises 27 / 103

28 Multiplexing and Demultiplexing Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Multiplexing - many to one Many signal mixed together to produce one signal. The multiplexed signal is transmitted over a channel Demultiplexing - one to many Multiplexed signal is split into individual components Types: FDM, TDM, CDM 28 / 103

29 Frequency division multiplexing (FDM) Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Telephone Hz per channel ( ) 1 Channel 1 Attenuation factor Channel 2 1 Channel 3 1 Channel 2 Channel 1 Channel Frequency (khz) (c) Frequency (Hz) (a) Frequency (khz) (b) 29 / 103

30 Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Wavelength Division Multiplexing (WDM) Fiber 1 spectrum Fiber 2 spectrum Fiber 3 spectrum Fiber 4 spectrum Spectrum on the shared fiber Power Power Power Power Power λ λ λ λ λ Filter Fiber 1 λ 1 λ 2 Fiber 2 Fiber 3 Fiber 4 λ 2 λ 3 λ 4 Combiner λ 1 +λ 2 +λ 3 +λ 4 Long-haul shared fiber Splitter λ 4 λ 1 λ 3 30 / 103

31 Time Division Multiplexing (TDM) Data Rate of a Communication Channel Transmission Media Modulation Multiplexing N signals, each having M samples per sec Interleaved together in a channel which can carry NM samples per sec POTS signal (speech or clipped sinusoids) sampled at 8000 samples per sec (sent as bps - each sample is represented using 8 bits) T1 lines Mbps, 24 channels 193-bit frame (125 µsec) Channel 1 Channel 2 Channel 3 Channel 4 Channel Bit 1 is 7 Data Bit 8 is for 31 / 103

32 TDM in Telephone Network Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Telephone End office Toll office Intermediate switching office(s) Toll office End office Telephone Local loop Toll connecting trunk Very high bandwidth intertoll trunks Toll connecting trunk Local loop 32 / 103

33 T1, T2, T3, T4 Data Communication Channels Data Rate of a Communication Channel Transmission Media Modulation Multiplexing 4 T1 streams in 7 T2 streams in 6 T3 streams in T2 stream out : :1 6: Mbps Mbps Mbps Mbps T1 T2 T3 T4 33 / 103

34 Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Code Division Multiple Access (CDMA) TDM - different channels can overlap in frequencies, but not in time FDM - no overlap in frequencies, overlap in time CDMA - overlap in both time and frequency Separable by orthogonality of codes Two vectors are orthogonal if their inner-product is zero. 34 / 103

35 CDMA Data Communication Channels Data Rate of a Communication Channel Transmission Media Modulation Multiplexing Each bit is converted to a vector of chips Different users assigned different orthogonal chip vectors A user assigned vector x transmits x to send a one or x to send a zero x. x = 1, x.( x) = 1 x.ȳ = 0 if ȳ is orthogonal to x. In practice difficult to obtain strict orthogonality x.ȳ can be made very small compared to 1 Tolerable interference with quasi-orthogonal sequences CDMA is also more efficient (closer to Shannon s limit) 35 / 103

36 Why CDMA? Data Communication Channels Data Rate of a Communication Channel Transmission Media Modulation Multiplexing If n chips per bit, then chip frequency n times higher than bit-frequency Energy to sent a bit spread over a larger frequency Useful during war-time as transmission virtually indistinguishable for noise in any narrow band. CDMA is also more efficient Higher frequency and lower S/N takes us closer to Shannon s limit Capacity increases linearly with bandwidth (only logarithmically with S/N) 36 / 103

37 Characterizing Data Transmission Repeaters Round Trip Time Bit-rate R, transmission delay τ p Propagation speed c, propagation delay τ c Round trip time (RTT) 37 / 103

38 Packet Duration Data Communication Channels Repeaters Round Trip Time Bit duration is the inverse of the bit-rate R If bit rate is 1 Mbps, the bit duration is 1 µs 1 Kbps 1 ms, 1 Gbps 1 ns. Transmission delay is bit-duration times packet size (number of bits in the packet ) 10 Mbps bit-rate, 100 byte packets: τ p = = 80µs. 1 Gbps bit-rate, 100 byte packet: τ p = = 800ns = 0.8µs. 38 / 103

39 Total Delay Data Communication Channels Repeaters Round Trip Time propagation at the speed c (light speed in the medium) c = m/s in free-space and air, c = m/s in copper. Distance traveled in 1 µs is 250 m (25cm in 1 ns) Over a 10 Mbps line of length 2500 m, how long does it take to receive a 100 byte packet? propagation delay is τ c = 10µs bit duration is 0.1µs; transmission delay τ p = = 80µs at time t = 0 leading edge of the packet sent at t = 0; trailing edge sent at time 80µs leading edge arrives at t = 10µs; trailing edge arrives at t = 90µs 39 / 103

40 Total Delay Data Communication Channels Repeaters Round Trip Time Pictorial Representation t τ c τ p Total Delay Total delay = Propagation delay τ c + transmission delay τ p y axis is time x axis represents distance the slope represents speed of propagation distance 40 / 103

41 Repeaters Data Communication Channels Repeaters Round Trip Time Used in long transmission lines Converts electrical signals to bits, and back to electrical signals, for retransmission Repeaters placed strategically to ensure that signal strength received at repeaters is strong enough to eliminate errors The conversion process eliminates the effects of noise between repeaters or between repeaters and end-points. Repeaters introduce additional propagation delay For a line of length 2500 m, with four repeaters (assume each repeater introduces a delay of 4µs) the propagation delay is = 26µs. 41 / 103

42 Repeater Delay Data Communication Channels Repeaters Round Trip Time Repeater Delay Repeater Delay Repeaters repeaters do not affect transmission delay repeaters do not have to receive the entire packet before starting the conversion Why not repeaters for analog communications? 42 / 103

43 Processing Delay Data Communication Channels Repeaters Round Trip Time It takes a finite amount of time for the receiver to process the received packet (say, τ r ) Data-link frames typically have a cyclic redundancy check (CRC) that will be verified (if inconsistent drop the corrupted packet) Routers will need time to look-up routing tables to determine the best next hop. There may also be queuing delay at routers Routers may have several incoming and outgoing interfaces A queue in an incoming interface may be stuck if the outgoing interface for the packet at the head of the queue is busy. 43 / 103

44 Round Trip Time Data Communication Channels Repeaters Round Trip Time Typically, every packet is acknowledged. Round trip time (RTT) = trans. delay τ p + propagation delay τ c + processing delay τ r + trans. delay for ack τ p + ack propagation delay τ c + ack processing delay τ r. A decent approximation is RTT = 2 (τ p + τ c + τ r ) What is the RTT over a 10 Mbps line of length 2500 m, if the packet size (both sent and ACK) is 100 bytes, and the processing delay is 5 µs? For the same length, packet size, and processing delay, what is the RTT if the bit-rate is 1 Gbps? 44 / 103

45 Multi-hop Propagation Repeaters Round Trip Time One way propagation delay for a channel with n hops is n(τ c + τ p + τ r ) τ c τ p τ r 45 / 103

46 Repeaters Round Trip Time A Simpler Representation of Delays Representation A t τ c τ p τ r C τ c τ p τ r B τ Simpler Representation Forward path AB, reverse path BC τ = τ c + τ p + τ r total forward path delay (τ = τ c + τ p + τ r total reverse path delay) In the simpler representation only the lines AB and BC will be depicted Also often rotated by 90 degrees (time in X -axis) 46 / 103

47 Data Communication Channels Packet Circuit Cut-through Store and Forward Circuit switching was used in Telephone networks (not any longer) 47 / 103

48 Circuit Data Communication Channels Packet Physical path established over multiple links Circuit established at the physical layer level Propagation delay is the sum of propagation delays over each link Transmission delay is not affected Was used in Telephone networks (not any longer) 48 / 103

49 Cut-through Packet Begin forwarding a packet to next hop even before the entire packet has been received Leading bytes in the packet will indicate destination Useful when look-up for destination can be accomplished very fast Delay for multi-hop propagation similar to the situation when repeaters are used. 49 / 103

50 Packet Store and Forward or Packet Entire packet received before the packet can be sent to next hop delay at each hop is transmission delay+propagation delay+ processing delay total delay is the delays at each hop packet duration has a significant impact on total delay. Two types: Virtual Circuits and Datagram 50 / 103

51 Datagram Packet Datagram Subnet Datagram H1 Packet Router Carrier's equipment B D 4 1 A E F H2 Process P2 Routers maintain a routing table indicating best next hop for each destination Process P1 3 A's table initially later A A B B B B C C C C 2 C C's table A A B A C E's table A C B D C C LAN Routing tables are dynamic Every packet routed individually Paths may change dynamically D B E C F C Dest. Line D E F B B B D E F D E E D D E F F Packets not guaranteed to be received in the same order. 51 / 103

52 Virtual Circuit Packet Virtual Circuit VC Subnet A path is established before even the first packet can be sent H3 Process P3 B Router D Carrier's equipment Path accepted by all routers in the path H1 Process P1 H1 1 H3 1 In A A's table C 1 C 2 Out 4 3 A 1 A 2 C C's table 2 E 1 E 2 C 1 C 2 E E's table 1 F 1 F 2 F LAN H2 Process P2 Packets marked with a path identifier Path identifier helps the router determine how to forward the next packet Size of routing tables in a router depends on number of active paths through the router. 52 / 103

53 Datagram vs VC Data Communication Channels Packet 53 / 103

54 Packet Connectionless vs Connection-Oriented Connectionless: Snail mail, telegram, etc. Establishing connections makes the job of end points easier. Circuit switching established a physical layer connection Virtual circuits are established at the network layer level In datagram-routing connection can be established at a higher level (transport layer). 54 / 103

55 Latency Example Data Communication Channels Packet Audio channel, 100 kbps (bit duration is 10µs) 10 hops, each 500 Km. τ c = 20 ms (total for all 10 hops) Assume low processing delay of 1µs at each hop (total τ r = 10µ s) If packet size is 1000 bytes, packet trans delay τ p = 80ms Total delay ms What if packet size is 100 bytes? Total delay ms What if channel rate is 100 Mbps? delay due to packet duration becomes negligible. 55 / 103

56 To realize low latency channels Packet For circuit switching and cut-through switching packet duration is irrelevant. For packet switching use smaller packet sizes - especially in low-bit rate channels In high bit-rate links packet duration (transmission delay) can be negligible (compared to propagation delay). With ever increasing bit-rates the advantages of Circuit and Cut-through become less relevant. 56 / 103

57 Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Reliably transmit a sequence of packets over a link. This is a requirement for data-link and transport layers Link can be real (data link layer) or virtual (transport layer) Packets can be corrupted (errors) Over a virtual link packets can also arrive out of order, be duplicated, dropped,. How does the sender know if the receiver got a specific packet? How does the receiver know that the sender knows that the receiver knows? (ARQ Protocols) 57 / 103

58 Errors Are Unavoidable Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Need to detect errors correct errors / request retransmission The key: redundancy extra bits need to be transmitted for detecting / correcting errors Example: parity bits for error detection 58 / 103

59 Hamming Distance Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol A metric for measuring distances between two sequences of bits A = B = How many bits need to be flipped to get B or A (or vice-versa)? Two bits the Hamming distance between A and B is / 103

60 Error Detection with Parity Bit Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol A = With even parity A = With odd parity A = For all subsequent discussions we will only use even parity If any of the eight (7+1) bits of A is flipped during transmission parity check will fail. What happens if two bits get flipped? Parity check fails to detect error... With one redundant bit we can only detect one-bit errors. Actually any odd number of bit-errors 60 / 103

61 Redundancy Data Communication Channels Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Need to transmit m bit symbols (2 m possible symbols) r redundant bits (for error detection / correction) n = m + r bits actually transmitted. Efficiency is m m+r = m n In general, more the redundancy, more the errors that can be detected / corrected 61 / 103

62 A Simple-Minded Approach Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Just repeat every bit twice (efficiency 0.5) - any one bit error can be detected Parity bit (just add one extra bit) - efficiency large m. m m+1 How can we correct single bit errors automatically? Repeat each bit two more times (efficiency 0.33) Are there better ways? is high for 62 / 103

63 Error Detection vs Correction Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Error correction needs more redundancy Error detection is much more crucial than error correction why? Practical error detection techniques address the problem of detecting multiple-bit errors. Error correction is usually used only for correction of single bit errors. 63 / 103

64 Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Single Bit Errors are Statistically More Frequent! Let us assume that the probability that any bit may be erroneously received is p = If N = 10 3 bits are transmitted, probability that there may be a bit error is roughly one in a million. Probability that there may be two errors is roughly one in a trillion (million x million) We need to detect even rare errors but not a big deal if we can not correct it (request re-tx). 64 / 103

65 Our Focus Data Communication Channels Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol How do we correct one-bit errors? If the receiver is able to correct errors, the sender does not need to retransmit How do we detect multiple-bit errors? If error is detected, the receiver can request the sender to retransmit Or simply not acknowledge reception 65 / 103

66 One-bit Error Correction Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Problem statement: X is transmitted n-bit message Y is received n-bit message H(X, Y ) 1 (Hamming distance is at most 1) For any X there is a set consisting of n + 1 possible Y s Y = X, or other n cases where only one bit is flipped (first or second or or n th ) Example, X = (n = 7), Y has eight possible values [ ] Y / 103

67 One-bit Error Correction Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol For each X there is a Y -set (n + 1 possible Y s) Each set includes X, and n values one Hamming distance away away from X What is the maximum number of non-overlapping sets? 2 n /(n + 1). If X is a 7 bit number, then 2 n /(n + 1) = 128/8 = 16 If X is a 10 bit numbers then 2 1 0/11 = 1024/11 = (or 93 non-overlapping sets) 67 / 103

68 One-bit Error Correction Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol If 2 m 2 n /(n + 1) we have a non overlapping set of Y s for every m-bit number For every m-bit number there is a set with i) an n-bit X and ii) n numbers one Hamming distance away from X The X s are 2 m unique codes represented using n = m + r bits. If any of the 2 m X s are sent, we can readily identify which one was sent even if a bit was flipped in the channel. To send an m-bit number over a noisy channel (in which not more than one bit can get flipped) we send a n = m + r-bit code 68 / 103

69 Efficient Choice of n and m Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol If 2 n /(n + 1) is a power of 2 then we have 2 m = 2 n /(n + 1). Else we have to choose 2 m < 2 n /(n + 1) (which is not efficient as we waste some usable codes) Efficient codes are represented as (m, n) (Hamming codes) m = 1; n = 3; r = 2; an efficient code as 2 1 = 2 3 /4. m = 2; n = 5 r = 3; not efficient as 2 5 /6 = 32/6 > 2 2 = 4 (m = 4, n = 7) (r = 3) is an efficient code (2 7 /(7 + 1) = 128/8 = 16) For efficient codes n = 2 r 1, m = n r r = 8, n = = 255, m = = 247: (247, 255) code r = 10, n = = m = = 1013: (1013, 1023) code 69 / 103

70 Hamming Codes Data Communication Channels Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol n = 2 r 1: efficient codes like (4, 7), (11, 15),... (1013, 1023),... (4, 7) code: r = 3 parity bits, n = 7 transmitted bits, m = 4 information bits. Each code of the form b 7 b 6 b 5 b 4 b 3 b 2 b 1 The bold bits are r = 3 parity bits. Other m = 4 bits are the actual information bits. What is special about positions 1,2, and 4? Powers of 2. (2 0, 2 1, 2 2 ). 70 / 103

71 Hamming Code (4, 7) Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol b 7 b 6 b 5 b 4 b 3 b 2 b 1 b 1 is the parity bits for bits b 7, b 5, b 3 (all odd positions 7,5,3,1) b 2 is the parity for bits b 3, b 6, b 7 (positions 2,3,6,7) b 4 is the parity for bits b 5, b 6, b 7 (positions 4,5,6,7) Assume we want to send m = 4 bits 1011 b 7 = 1 b 6 = 0 b 5 = 1 b 4 =? b 3 = 1b 2 =? b 1 =? b 1 = 1, b 2 = 0, b 4 = 0 are the correct parities The code for 1011 is / 103

72 Hamming Code Example Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol The receiver checks parity bits and writes down the results of the check as x 4 x 2 x 1 x 4 = 0 if parity b 4 ; x 4 = 1 if parity b 4 is wrong. If no bit is flipped by the channel then all parity bits will be correct (result 000) If b 7 is flipped all three parities will be wrong (result 111) If b 3 is flipped result is 011 If b 5 is flipped result is 101 See a pattern? The result gives the position of the erroneous bit. Go ahead and flip it back. 72 / 103

73 Hamming Code Example Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Code X = If Y = (no error) result 000 If Y = (pos 1 error) result 001 If Y = (pos 2 error) result 010 If Y = (pos 3 error) result 011 If Y = (pos 4 error) result 100 If Y = (pos 5 error) result 101 If Y = (pos 6 error) result 110 If Y = (pos 7 error) result / 103

74 (11, 15) Hamming Code Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Code b 15 b 1 b 1, b 2, b 4, b 8 are parity bits Example ? 0 0 1? 0?? (11-bit information ) b 15 b 14 b 13 b 12 b 11 b 10 b 9 b 8 b 7 b 6 b 5 b 4 b 3 b 2 b ? 0 0 1? 0?? ? 0 0 1? 0? ? 0 0 1? ? Result written as x 8 x 4 x 2 x 1. If bit 13 is flipped result is / 103

75 Syndrome Coding Data Communication Channels Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol A syndrome is a collection of symptoms. Each result of parity check is a symptom: each bit of the result (for example 1011 if r = 4 is a symptom) The syndrome should unambiguously indicate the error. With r redundant bits, we have 2 r unique syndromes We can detect 2 r 1 different diseases (one syndrome corresponds to no illness ). n 2 r 1 75 / 103

76 Burst Errors Data Communication Channels Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Burst errors affect a consecutive sequence of bits Power surges, explosions Hamming codes can only detect one-bit errors Can we handle burst errors? Yes: by interleaving 76 / 103

77 Interleaving Data Communication Channels Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol 40 bits need to be sent Use (7,4) Hamming code to encode 4 bits at a time: 40 bits becomes 70 bits at positions (1, 2,..., 70) 10 independent codes at positions (1, 2, 3, 4, 5, 6, 7), (8, 9, 10, 11, 12, 13, 14) (64, 65, 66, 67, 68, 69, 70) Reorder the sequence of 70 bits by interleaving (1,8,15,... 64), (2,9,...,65), (3,10,...,66) (7,14,... 70) At the receiver: de-interleave and then decode Any burst sequence up to 10 bits long will produce at most error in each of the ten codes 77 / 103

78 Error Detection Data Communication Channels Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Single error detection with parity bit Burst error detection: Interleaving 78 / 103

79 Multi-bit Error Detection Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Double errors arrange n = w h bits as a matrix Add parity bit for each row (we now have (w + 1) h matrix) Parity bit for each column we end up with a (w + 1) (h + 1) matrix If 2 errors happen in the same row (column) they cannot be in the same column (row) 79 / 103

80 Cyclic Redundancy Checks (CRC) Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol A polynomial - y = P(x) = x r + a r 1 x r a 1 x 1 + a 0 If P(x) is in the field of real numbers, x, a 0... a r 1, y can be any real number Any bit string can be seen coefficients of a polynomial in the finite field {0, 1} P(x), x, a 0... a r 1 can only be 0 or 1 Example: is P(x) = x 5 + x 4 + x 2 + x 0. Addition: = 0 = 0 + 0, = = 1 Subtraction is the same as addition! Multiplication: 1 1 = 1, 0 x = 0. P(0) = 1. P(1) = = / 103

81 Polynomial Arithmetic Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol P(x) = x 5 + x 4 + x Q(x) = x 2 + x 1 P(x) + Q(x) = x 5 + x 4 + x P(x) Q(x) = (x 5 + x 4 + x 2 + 1)(x 2 + x 1 ) = x 7 +x 6 +x 4 +x 2 +x 6 +x 5 +x 3 +x 1 = x 7 +x 5 +x 4 +x 3 +x 2 +x 1. Polynomial division - what is P(x)/Q(x) (quotient) (remainder) 81 / 103

82 CRC Data Communication Channels Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Choose a generator polynomial G(x) of degree r (r + 1 bit number) Message to be sent M(x) Append r zeros to M(x). The result is x r M(x). Evaluate the remainder of x r M(x)/G(x) (remainder will be a polynomial of degree less than r) - say R(x) Transmit T (x) = x r M(x) R(x) = x r M(x) + R(x) Receiver receives T (x) In case of error receiver gets T (x) = T (x) + E(x) Receiver checks if remainder of T (x)/g(x) is zero If remainder is not zero receiver decides that there is an error As T (x)/g(x) = 0, T (x)/g(x) = E(x)/G(x) Errors where E(x)/G(x) is zero goes undetected 82 / 103

83 Good choice for G(x) Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Both highest and lowest order bits should be 1 (1xx xx1) If G(x) has degree more than 1 all single bit errors will be detected - E(x) = x i. Two isolated single bit errors E(x) = x i + x j = x j (x i j + 1): any G(x) that does not have x k + 1 as a factor (for all k m) is sufficient If x + 1 is not a factor, then all odd number of errors can be detected. Any burst error of length r can be detected 32 bit polynomial specified in IEEE 802x 83 / 103

84 Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol ARQ (Automatic Repeat Request) Protocols Packets need to be acknowledged Time-out for retransmission But acknowledgements can get lost too! 84 / 103

85 Lost Acknowledgements Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol In (B) ACK was lost. So sender retransmits packet P1. Can both sender and receiver agree that P1 is a retransmission of P1 and not P2? Timeout Interval P1 P2 (A) ACK ACK Timeout Interval (B) P1 P1 X 85 / 103

86 Packet Numbers! Data Communication Channels Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Packets need to be numbered Obviously numbers will be recycled How many unique sequence numbers do we need? 86 / 103

87 Ambiguous Acknowledgements Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol The ACK was P1 was delayed. Sender retransmits P1. Soon after, ACK for P1 is received and sender sends P2 which gets lost. How does the sender know that the second acknowledgement is for P1 or P2? Timeout Interval P1 P1 P2 (C) X ACK ACK 87 / 103

88 Numbering Data Communication Channels Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Acknowledgements need to be numbered too! ABP protocol (alternating bit protocol) One bit packets numbers - 0 or 1 One bit ACK numbers Also called Stop and Wait Protocol (SWP) Is ABP/SWP unambiguous? 88 / 103

89 Alternating Bit Protocol (ABP) Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol (D) Timeout Interval Packet No Seq No Ack No Packet No 89 / 103

90 ABP is Unambiguous! Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol Timeout Interval Timeout Interval Packet No Seq No 1 (E) X 0 0 Ack No 1 1 Packet No 2 Timeout Interval Packet No Seq No (F) X 0 0 Ack No 2 Packet No 90 / 103

91 ABP Sender Rules Data Communication Channels Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol If the last packet sent was numbered 0, allowed to send the next packet with number 1 only after ACK for 0 is received (else retransmit 0) If the last packet sent was numbered 1, allowed to send the next packet with number 0 only after ACK for 1 is received (else retransmit 1) 91 / 103

92 ABP Receiver Rules Error Control Error Correction Error Detection ARQ Protocols Alternating Bit Protocol If last packet received was numbered 0 If 1 received ACK and keep (new packet) If 0 received ACK and drop (retransmission of previous packet) If last packet received was numbered 1 If 0 received ACK and keep (new packet) If 1 received ACK and drop (retransmission of previous packet) 92 / 103

93 Effective Data Rate Sliding Window Protocols 100 Mbps link A B (R = 100x10 6 ), L = 2000m, packet size F = 100 bits, τ r = 1µs P = τ p = = 1µs, τ c = = 8µs What is the effective data rate between A and B when ABP is used? RTT = 2τ p + 2τ c + 2τ r = = 20. In each 20 µs interval transmission occurs only for 1 µs Effective data rate is 5 Mbps (P/RTT x R) 93 / 103

94 Improving Throughput Sliding Window Protocols During the first micro second sender sends the first packet Sender does nothing for 19 µs. ACK for first packet received after 20µs What if sender is allowed to send 20 packets before receiving the ACK for the first packet? Only after the ACK for the first packet is received the sender can send packet 21 Only after the ACK for the second packet is received can the sender send packet 22, and so on Window size of 20 packets More generally, Window size RTT/packet duration 94 / 103

95 Sliding Window Protocols Pipelining Example, W RTT /τ p = RTT

96 Sliding Window Protocols Pipelining Example, W RTT /τ p = RTT

97 Sliding Window Protocols Pipelining Example, W RTT /τ p = RTT

98 Sliding Window Protocols Pipelining Example, W RTT /τ p = RTT

99 Sliding Window Protocols Pipelining Example, W RTT /τ p = RTT 13 packets (0 to 12) sent in one RTT ACK for 0 processed; ready to send 13

100 Sliding Window Protocols Pipelining Example, W RTT /τ p = RTT 13 packets (0 to 12) sent in one RTT ACK for 0 processed; ready to send 13 ACK for 1 rcd; ready to send 14

101 Sliding Window Protocols Pipelining Example, W RTT /τ p = RTT 13 packets (0 to 12) sent in one RTT ACK for 0 processed; ready to send 13 ACK for 1 rcd; ready to send / 103

102 Pipelining Data Communication Channels Sliding Window Protocols The secret to higher throughput ABP needs only half-duplex channels, for pipelining we implicitly assume full duplex channels ACKs and packets cross each other Nodes may need to buffer packets when things do not go well What is the buffer size needed? How do we number packets and acknowledgements? 96 / 103

103 Selective Repeat Protocol Sliding Window Protocols Similar to ABP, with window size W packets Or ABP is SRP with window size 1 packet Buffer size (W. Why?) Numbering packets? (2W unique numbers. Why?) 97 / 103

104 SRP Protocol Data Communication Channels Sliding Window Protocols Sender Window (size 5) Not yet sent ACK pending Sent and ACKed SRP Sender Rules Legal to send any packet within the window Head of the queue blocked by an outstanding ACK. Sender window can not advance until ACK is received for the blocking packet. Normally next unsent packet (blue) is sent. On time-out for the blocking red packet resend the packet. Sender may have to buffer up to W packets W unique packet numbers 0 to 2W / 103

105 SRP Window Data Communication Channels Sliding Window Protocols Receiver Rules Past Window and Future Window; each of size W. Rx. Window received pending p.w f.w Packets left of the Past Window: Receipt of ACK confirmed by sender Packets in the Past Window: ACKs not confirmed by sender Future Window blocked by earliest pending packet. Next packet received can be from either window If packet from past window ACK and drop packet If packet from future window ACK and store packet Past Window is the lower bound on sender window position. Reception of 8 ACK for 3 confirmed. 99 / 103

106 SRP Window Data Communication Channels Sliding Window Protocols Rx. Window received pending Receiver Rules Up to W packets may need to be buffered 2W possibilities for next packet. Using 2W distinct numbers eliminates ambiguities. If next received packet is numbered 4, 5, 6, 7 or 8 ACK and discard. If next received packet is numbered 9, 0, 1, 2 or 3 ACK and store. 100 / 103

107 Sliding Window Protocols Negative Acknowledgments (NACK) Assume ACK for packet number 2 not received Normally sender would wait for a timeout period What if an ACK for 3 (while 2 is missing) is interpreted as a NACK for 2? ACK for 3 is generally received well before time-out for / 103

108 Go Back N - GBN Protocol Sliding Window Protocols Window size W Sender logic is the same Receiver does not buffer packets (no window) Out of sequence packets are simply discarded Periodically sends acknowledgement for the last received packet number Example, if Rx gets packets 1,2,3,4,5,7,8,9 the ACK for 5 is sent. 7,8,9 not ACKed. Sender retransmits 6,7,8, / 103

109 Windowing Protocols in DL vs TL Sliding Window Protocols Window size need not always be expressed in packets If expressed in packets we assume equal duration packets (window size is number of packet durations in one RTT ) It can be in bits too ( number of bits that can be sent in one RTT ) Or bytes ( number of bytes that can be sent in one RTT ) In transport layer window size is expressed in bytes In data link layer, the link parameters like τ p, τ c, τ r, RTT etc., are known at design time. (Easy to compute W RTT /τ p ). In TL how well do we know the virtual link parameters? How do we choose window size? 103 / 103