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Data Communication ChannelsRate and Delay

SwitchingReliable Data Transfer

Reliable and Efficient Data Transfer

Data Communication Principles

Mahalingam RamkumarMississippi State University, MS

August 17, 2015

Ramkumar CSE 4153 / 6153




1 Data Communication ChannelsData Rate of a Communication ChannelTransmission MediaModulationMultiplexing

2 Rate and DelayRepeatersRound Trip Time

3 SwitchingPacket Switching

4 Reliable Data TransferError ControlError CorrectionError DetectionARQ ProtocolsAlternating Bit Protocol

5 Reliable and Efficient Data TransferSliding Window Protocols

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Data Rate of a Communication ChannelTransmission MediaModulationMultiplexing

Fundamental Factors

Channel bandwidth H

Signal power S

Noise power N

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Channel Capacity R

Achievable bit-rate R

Shannon’s Capacity Limit

R = H log2(1 + S/N) bps. (1)

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S/N in Decibels (dB)

Signal to Noise Ratio (SNR)

Usually expressed in dB

(S/N)dB = 10log10(S/N)

20 dB → S/N = 100;

30 dB → S/N = 1000;

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Example

What is the limit on achievable rate R in a channel withbandwidth 1 MHz and SNR 30 dB?

30dB → S/N = 1000

R = 1× 106 × log2(1 + 1000) ≈ 1× 106 × 10 = 10Mbps (2)

To double the rate to 20 Mbps, we can

Double the bandwidth, orincrease SNR to 60 dB (or increase S/N ≈ 1 million)log2(1, 000, 000) ≈ 20

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Nyquist Sampling Theorem

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 uniquelyreconstructed from discrete samples taken at a rate 2H persecond

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Nyquist Capacity

Assume each sample can take V possible values

log2(V ) bits per sample

Nyquist capacity C = 2H log2(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).

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Transmission Media

Two basic types

Guided

Twisted pairCoaxial cablesFiber opticsMagnetic media

Unguided

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Twisted Pair

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)

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Coaxial Cables

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

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Fiber Optics

Extremely high bandwidth

Single mode and multimode

Single mode fibers can handle upto 50GHz over 100km!

Total internal reflection.

Air/silica boundary

Light sourceSilica

Air

(a) (b)

β1 β2 β3

α1 α2 α3

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Fiber vs Copper Wire

Weight vs bandwidth

Cost

Security

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Electromagnetic Spectrum

100 102 104 106 108 1010 1012 1014 1016 1018 1020 1022 1024

Radio Microwave Infrared UV X-ray Gamma ray

f (Hz)

Visible light

104 105 106 107 108 109 1010 1011 1012 1013 1014 1015 1016

f (Hz)

Twisted pairCoax

Satellite

TV

Terrestrial microwave

Fiber

optics

MaritimeAM

radioFM

radio

Band LF MF HF VHF UHF SHF EHF THF

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Allocation of Spectrum

ITU-R, FCCISM (Industrial, Scientific, Medical) Bands

Freq.

Bandwidth

902 MHz

928 MHz

26 MHz

2.4 GHz

2.4835 GHz

5.735 GHz

5.860 GHz

. . . . . .

. . . . . .83.5 MHz

125 MHz

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Satellites

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 areasPoint-to-point vs BroadcastMobile communications“Right of way” for laying cables

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Modulation

Modulation: information signal + carrier signal → modulatedsignal

Speech signals have frequencies between 300 and 3000 HzSignals cannot propagate directly over all mediumsSpeech 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 themodulated signal

Types of modulation: Amplitude (AM), Phase (PM),Frequency (FM)

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0 1 0 1 1 0 0 1 0 0 1 0 0

(a)

(b)

(c)

(d)

Phase changes

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Bandwidth of Modulated Signals

The same as the bandwidth of the “information” signal

The bandwidth is now around a “center” frequency — thefrequency of the “carrier”

Many signals can coexist simultaneously in the medium.

By using different carrier frequencies

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Modulation for FAX/Dial-up Internet

Send bits over telephone cable using Modem (Modulator -Demodulator)

Cable meant for carrying voice signals between 300-3000 Hz

Amplitude + phase modulation - series of sinusoidal pulses

Modulator: Bits are chunked, represented using sinusoidalpulses

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

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QPSK, QAM-16, QAM-64

Quadrature Phase Shift keying (2 bits per sample), QuadratureAmplitude Modulation

270

(a)

90

0 180

270

(b)

90

0

270

(c)

90

0 180

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Telephone/FAX Modem

Each pulse represents multiple bits

If we use 16 different types of pulses (eg., QAM 16) we cansend 4 bits per pulse (16 = 24)

Bit rate and Baud rate

Baud rate - number of sinusoidal pulses per secondFixed at 2400 for telephone linesBit rate = Baud rate x number of bits per pulseHow do we get 56 K with baud rate of 2400?

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Trellis Coded Modulation

Add more bits per sampleAdd extra bits to each sample for error correctionV.32 (4+1, 32) - 9600 bps, V.32 bis (6 + 1, 128) 14,400 bpsV.34 - 28,800 bps, V.34 bis - 36,600 bps

270

(b)

90

270

(c)

90

0 180 0 180

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Modern Modems

Full duplex - transmissions possible in both directions at thesame 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 whileonline

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Actual Capacity of a Telephone Line

50

40

20

30

10

00 1000 2000 3000 4000

Meters5000 6000

Mpb

s

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ADSL

We have been sitting on a gold-mine!

Telephones filter out higher frequencies to suppress noise

1.1 MhZ spectrum divided into 256 channels - 4312 Hz each(DMT - Discrete Multitone)

First channel used for POTS (plain old telephone service)

Pow

er

Voice Upstream Downstream

256 4-kHz Channels

0 25 1100 kHz

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ADSL Equipment

DSLAM

Splitter

Codec

Splitter

Telephone

To ISP

ADSL modem

Ethernet

Computer

Telephone line

Telephone company end office Customer premises

Voice switch

NID

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Multiplexing and Demultiplexing

Multiplexing - many to one

Many signal mixed together to produce one signal. Themultiplexed signal is transmitted over a channel

Demultiplexing - one to many

Multiplexed signal is split into individual components

Types: FDM, TDM, CDM

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Frequency division multiplexing (FDM)

Telephone - 4000Hz per channel (450 + 3100 + 450)

300 3100

Channel 3

Channel 2

Channel 1

1

1

1

Atte

nuat

ion

fact

or

64

Frequency (kHz)

(c)

Channel 1 Channel 3Channel 2

68 72

60 64

Frequency (kHz)

(b)

Frequency (Hz)

(a)

68 72

60

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Wavelength Division Multiplexing (WDM)

Spectrum on the shared fiber

Pow

er

λ

Fiber 4 spectrum

Pow

er

λ

Fiber 3 spectrum

Pow

er

λ

Fiber 2 spectrum

Pow

er

λ

λ1

λ1+λ2+λ3+λ4

Fiber 1 spectrum

Pow

er

λ

Fiber 1λ2

Fiber 2λ3

Fiber 3Combiner Splitter

Long-haul shared fiber

λ4

λ2

λ4

λ1

λ3Fiber 4

Filter

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Time Division Multiplexing (TDM)

N signals, each having M samples per secInterleaved together in a channel which can carry NM samplesper secPOTS signal (speech or clipped sinusoids) sampled at 8000samples per sec (sent as 64000 bps - each sample isrepresented using 8 bits)T1 lines - 1.544 Mbps, 24 channels

Channel 1

Channel 2

Channel 3

Channel 4

Channel 24

193-bit frame (125 µsec)

7 Data bits per channel

per sample

Bit 1 is a framing code

Bit 8 is for signaling

0

1

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TDM in Telephone Network

Telephone

End

office

Toll

office

Intermediate switching office(s)

Telephone

End

office

Toll

office

Local loop

Toll connecting

trunk

Very high bandwidth

intertoll trunks

Toll connecting

trunk

Local loop

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T1, T2, T3, T4

6 5 4 3 2 1 05 1

4 0

6 2

7 3

6:17:14:1

4 T1 streams in

1 T2 stream out

6.312 Mbps

T2

1.544 Mbps

T1

44.736 Mbps

T3

274.176 Mbps

T4

7 T2 streams in 6 T3 streams in

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Code Division Multiple Access (CDMA)

TDM - different channels can overlap in frequencies, but notin 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.

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CDMA

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 tosend a zero

x .x = 1, x .(−x) = −1

x .y = 0 if y is orthogonal to x .

In practice difficult to obtain strict orthogonality

x .y can be made very small compared to 1

“Tolerable” interference with quasi-orthogonal sequences

CDMA is also more efficient (closer to Shannon’s limit)

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Why CDMA?

If n chips per bit, then chip frequency n times higher thanbit-frequency

Energy to sent a bit spread over a larger frequency

Useful during war-time as transmission virtuallyindistinguishable for noise in any narrow band.

CDMA is also more efficient

Higher frequency and lower S/N takes us closer to Shannon’slimit

Capacity increases linearly with bandwidth (onlylogarithmically with S/N)

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RepeatersRound Trip Time

Characterizing Data Transmission

Bit-rate R and bit duration,

Packet duration τp

Propagation speed c , propagation delay τc

One-way delay

Round trip time (RTT)

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Packet Duration

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.

Packet duration is bit-duration times packet size (number ofbits in the packet )

10 Mbps bit-rate: duration of a 100 byte packet isτp = 100× 8× 0.1 = 80µs.

1 Gbps bit-rate: duration of a 100 byte packet isτp = 100× 8 = 800ns = 0.8µs.

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One-way Delay

propagation at the speed c (light speed in the medium)

c = 3× 108 m/s in free-space and air, c = 2.5× 108 m/s incopper.

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 taketo receive a 100 byte packet?

propagation delay is τc = 10µsassume transmission started at time t = 0leading edge of the packet sent at t = 0; trailing edge sent attime τp = 80µsleading edge arrives at t = 10µs; trailing edge arrives att = 90µs

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One-way Delay

Pictorial Representation

t

τp

τc

distance

One-way Delay

One-way delay = Propagationdelay τc + packet duration τp

y axis is time

x axis represents distance

the slope represents speed ofpropagation

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Repeaters

Used in long transmission lines

Converts electrical signals to bits, and back to electricalsignals, for retransmission

Repeaters placed strategically to ensure that signal strengthreceived at repeaters is strong enough to eliminate errors

The conversion process eliminates the effects of noise betweenrepeaters or between repeaters and end-points.

Repeaters introduce additional propagation delay

For a line of length 2500 m, with four repeaters (assume eachrepeater introduces a delay of 4µs) the propagation delay is10 + 4× 4 = 26µs.

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Repeater Delay

Repeater Delay

Repeater Delay

Repeaters

packet duration does not affectrepeater delay

repeaters do not have to receivethe entire packet before startingthe conversion

Why not repeaters for analogcommunications?

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Processing Delay

It takes a finite amount of time for the receiver to “process”the received packet (say, τr )

Data-link frames typically have a cyclic redundancy check(CRC) that will be verified (if inconsistent drop the corruptedpacket)

Routers will need time to look-up routing tables to determinethe 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 outgoinginterface for the packet at the head of the queue is busy.

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Round Trip Time

Typically, every packet is acknowledged.

Round trip time (RTT) = Packet duration τp + propagationdelay τc + processing delay τr + duration of acknowledgementτ ′p + ack propagation delay τc + ack processing delay τ ′r .

A decent approximation is RTT = 2× (τp + τc + τr )

What is the RTT over a 10 Mbps line of length 2500 m, if thepacket size (both sent and ACK) is 100 bytes, and theprocessing delay is 5 µs?

For the same length, packet size, and processing delay, what isthe RTT if the bit-rate is 1 Gbps?

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Multi-hop Propagation

One way propagation delay for a channel with n hops isn(τc + τp + τr )

τp

τr

τc

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A Simpler Representation of Delays

Representation

t

τc

τp

τr

τ

τ ′c

τ ′p

τ ′r

A

B

C

Simpler Representation

Forward path AB, reverse pathBC

τ = τc + τp + τr total forwardpath delay (τ ′ = τ ′c + τ ′p + τ ′rtotal reverse path delay)

In the simpler representationonly the lines AB and BC willbe depicted

Also often rotated by 90 degrees(time in X -axis)

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Packet Switching

Switching

Circuit Switching

Cut-through Switching

Store and Forward Switching

Circuit switching was used in Telephone networks (not anylonger)

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Packet Switching

Circuit Switching

Physical path established over multiple links

Circuit established at the physical layer level

Delay for multi-hop propagation is the sum of propagationdelay over each link

Was used in Telephone networks (not any longer)

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Packet Switching

Cut-through Switching

Begin forwarding a packet to next hop even before the entirepacket has been received

Leading bytes in the packet will indicate destination

Useful when look-up for destination can be accomplished veryfast

Delay for multi-hop propagation similar to the situation whenrepeaters are used.

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Packet Switching

Store and Forward or Packet Switching

Entire packet received before the packet can be sent to nexthop

Packet duration influences the overall delay betweenend-points.

Two types: Virtual Circuits and Datagram Switching

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Packet Switching

Datagram Switching

Datagram Subnet

A E F Process P2

LAN

Router

1

Carrier's equipment

Process P1

B

H1 H2

D

C

Packet

3

4

2

A –

B B

initially

C C

D B

E C

F C

Dest.

A –

B B

laterA's table

C C

D B

E B

F B

A A

B A

C's table

C –

D D

E E

F E

A C

B D

E's table

C C

D D

E –

F F

Line

Datagram Switching

Routers maintain a routingtable indicating best next hopfor each destination

Routing tables are dynamic

Every packet routedindividually

Paths may change dynamically

Packets not guaranteed to bereceived in the same order.

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Packet Switching

Virtual Circuit Switching

VC Subnet

E FProcess P2

LAN

Router

1

Carrier's equipment

Process P1

Process P3

B

H1

H3

H2

D

C

3

42

H1 1

H3 1

C 1

C 2

A's table

A 1

A 2

E 1

E 2

C's table

C 1

C 2

F 1

F 2

E's table

A

In Out

Virtual Circuit

A path is established beforeeven the first packet can besent

Path accepted by all routers inthe path

Packets marked with a pathidentifier

Path identifier helps the routerdetermine how to forward thenext packet

Size of routing tables in arouter depends on number ofactive paths through therouter.

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Packet Switching

Datagram vs VC

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Packet Switching

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 higherlevel (transport layer).

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Packet Switching

Latency Example

Audio channel, 100 kbps (bit duration is 10µs)

10 hops, each 500 Km. τc = 20 ms (total for all 10 hops)

Assume low processing delay of 1µs at each hop (totalτr = 10µ s)

If packet size is 1000 bytes, packet duration τp = 80ms

Total delay 20 + 0.01 + 10 ∗ 80 ≈ 820ms

What if packet size is 100 bytes?

Total delay 20 + 0.01 + 10 ∗ 8 ≈ 100ms

What if channel rate is 100 Mbps? delay due to packetduration becomes negligible.

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Packet Switching

To realize low latency channels

Circuit switching — only propagation delay.

Cut-through switching

Store and forward (packet switching) vs cut-throughswitching — error checking at every hop

For circuit and cut-through switching packet duration isirrelevant.

For packet switching use smaller packet sizes in low-bit ratechannels

In high bit-rate links packet duration can be negligible(compared to propagation delay).

With ever increasing bit-rates the advantages of CircuitSwitching and Cut-through Switching become less relevant.

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Error ControlError CorrectionError DetectionARQ ProtocolsAlternating Bit Protocol

Reliable Data Transfer

Need to 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, beduplicated, dropped, · · · .How does the sender know if the receiver got a specificpacket?

How does the receiver know that the sender knows? (ARQProtocols)

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Errors Are Unavoidable

Need to detect errors

correct errors / request retransmission

The key: redundancy — extra bits need to be transmitted fordetecting / correcting errors

Example: parity bits for error detection

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Hamming Distance

A metric for measuring “distances” between two sequences ofbits

A = 1000110

B = 1010100

How many bits need to be flipped to get B or A (orvice-versa)?

Two bits — the Hamming distance between A and B is 2.

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Error Detection with Parity Bit

A = 1000110

With even parity A = 10001101

With odd parity A = 10001100

For all subsequent discussions we will only use even parity

If any of the eight (7+1) bits of A is flipped duringtransmission 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

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Redundancy

Need to transmit m bit symbols (2m possible symbols)

r redundant bits (for error detection / correction)

n = m + r bits actually transmitted.

Efficiency is mm+r = m

n

In general, more the redundancy, more the errors that can bedetected / corrected

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A Simple-Minded Approach

Just repeat every bit twice (efficiency 0.5) - any one bit errorcan be detected

Parity bit (just add one extra bit) - efficiency mm+1 is high for

large m.

How can we correct single bit errors automatically?

Repeat each bit two more times (efficiency 0.33)

Are there better ways?

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Error Detection vs Correction

Error correction needs more redundancy

Error detection is much more crucial than error correction —why?

Practical error detection techniques address the problem ofdetecting multiple-bit errors.

Error correction is usually used only for correction of single biterrors.

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Single Bit Errors are Statistically More Frequent!

Let us assume that the probability that any bit may beerroneously received is p = 10−9.

If N = 103 bits are transmitted, probability that there may bea bit error is roughly one in a million.

Probability that there may be two errors is roughly one in atrillion (million x million)

We need to detect even rare errors — but not a big deal if wecan not correct it (request re-Tx).

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Our Focus

How do we correct one-bit errors?

If the receiver is able to correct errors, the sender does notneed to retransmit

How do we detect multiple-bit errors?

If error is detected, the receiver can request the sender toretransmit

Or simply not acknowledge reception

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One-bit Error Correction

Problem statement:

X is transmitted n-bit messageY is received n-bit messageH(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 , orother n cases where only one bit is flipped (first or second or· · · or nth)

Example, X = 1011101 (n = 7), Y has eight possible values

Y ∈[

1011101 1011100 1011111 10110011010101 1001101 1111101 0011101

]

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For each X there is a Y -set (n + 1 possible Y s)

Each set includes X , and n values one Hamming distanceaway away from X

What is the maximum number of non-overlapping sets?2n/(n + 1).

If X is a 7 bit number, then 2n/(n + 1) = 128/8 = 16

If X is a 10 bit numbers then 210/11 = 1024/11 = 93.09 (or93 non-overlapping sets)

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If 2m ≤ 2n/(n + 1) we have a non overlapping set of Y s forevery m-bit number

For every m-bit number there is a set with i) an n-bit X andii) n numbers one Hamming distance away from X

The X s are 2m unique codes represented using n = m + r bits.

If any of the 2m X s are sent, we can readily identify which onewas sent even if a bit was flipped in the channel.

To send an m-bit number over a noisy channel (in which notmore than one bit can get flipped) we send a n = m + r -bitcode

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Efficient Choice of n and m

If 2n/(n + 1) is a power of 2 then we have 2m = 2n/(n + 1).

Else we have to choose 2m < 2n/(n + 1) (which is notefficient 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 21 = 23/4.

m = 2; n = 5 r = 3; — not efficient as 25/6 = 32/6 > 22 = 4

(m = 4, n = 7) (r = 3) is an efficient code(27/(7 + 1) = 128/8 = 16)

For efficient codes n = 2r − 1, m = n − r

r = 8, n = 256− 1 = 255, m = 255− 8 = 247: (247, 255)code

r = 10, n = 1024− 1 = 1023. m = 1023− 10 = 1013:(1013, 1023) code

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Hamming Codes

n = 2r − 1: efficient codes like(4, 7), (11, 15), . . . (1013, 1023), . . .

(4, 7) code: r = 3 parity bits, n = 7 transmitted bits, m = 4information bits.

Each code of the form b7 b6 b5 b4 b3 b2 b1

The bold bits are r = 3 parity bits. Other m = 4 bits are theactual information bits.

What is special about positions 1,2, and 4?

Powers of 2. (20, 21, 22).

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Hamming Code (4, 7)

b7 b6 b5 b4 b3 b2 b1

b1 is the parity bits for bits b7, b5, b3 (all odd positions7,5,3,1)

b2 is the parity for bits b3, b6, b7 (positions 2,3,6,7)

b4 is the parity for bits b5, b6, b7 (positions 4,5,6,7)

Assume we want to send m = 4 bits 1011

b7 = 1 b6 = 0 b5 = 1 b4 =? b3 = 1b2 =? b1 =?

b1 = 1, b2 = 0, b4 = 0 are the correct parities

The code for 1011 is 1010101

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Hamming Code Example

The receiver checks parity bits and writes down the results ofthe check as x4x2x1

x4 = 0 if parity b4; x4 = 1 if parity b4 is wrong.

If no bit is flipped by the channel then all parity bits will becorrect (result 000)

If b7 is flipped all three parities will be wrong (result 111)

If b3 is flipped result is 011

If b5 is flipped result is 101

See a pattern? The result gives the position of the erroneousbit.

Go ahead and flip it back.

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Hamming Code Example

Code X = 1010101

If Y = 1010101 (no error) result 000

If Y = 1010100 (pos 1 error) result 001







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(11, 15) Hamming Code

Code b15 · · · b1b1, b2, b4, b8 are parity bitsExample 0 1 0 1 0 1 0 ? 0 0 1 ? 0 ? ? (11-bit information01010100010)b15 b14 b13 b12 b11 b10 b9 b8 b7 b6 b5 b4 b3 b2 b10 1 0 1 0 1 0 ? 0 0 1 ? 0 ? ?

0 1 0 1 0 1 0 ? 0 0 1 ? 0 ? 10 1 0 1 0 1 0 ? 0 0 1 ? 0 0 10 1 0 1 0 1 0 ? 0 0 1 1 0 0 10 1 0 1 0 1 0 1 0 0 1 1 0 0 1

Result written as x8x4x2x1.If bit 13 is flipped result is 1101

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Syndrome Coding

A syndrome is a “collection of symptoms.”

Each result of parity check is a symptom: each bit of theresult (for example 1011 if r = 4 is a symptom)

The syndrome should unambiguously indicate the error.

With r redundant bits, we have 2r unique syndromes

We can detect 2r − 1 different “diseases” (one syndromecorresponds to “no illness”).

n ≤ 2r − 1

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Burst Errors

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

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Interleaving

40 bits need to be sent

Use (7,4) Hamming code to encode 4 bits at a time: 40 bitsbecomes 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 mosterror in each of the ten codes

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Error Detection

Single error detection with parity bit

Burst error detection: Interleaving

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Multi-bit Error Detection

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 inthe same column (row)

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Cyclic Redundancy Checks (CRC)

A polynomial - y = P(x) = x r + ar−1xr−1 + · · ·+ a1x

1 + a0

If P(x) is in the field of real numbers, x , a0 . . . ar−1, y can beany real number

Any bit string can be seen coefficients of a polynomial in thefinite field {0, 1}P(x), x , a0 . . . ar−1 can only be 0 or 1

Example: 110101 is P(x) = x5 + x4 + x2 + x0.

Addition: 1 + 1 = 0 = 0 + 0, 1 + 0 = 0 + 1 = 1

Subtraction is the same as addition!

Multiplication: 1× 1 = 1, 0× x = 0.

P(0) = 1. P(1) = 1 + 1 + 1 + 1 = 0.

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Polynomial Arithmetic

P(x) = x5 + x4 + x2 + 1. Q(x) = x2 + x1

P(x) + Q(x) = x5 + x4 + x1 + 1

P(x)× Q(x) = (x5 + x4 + x2 + 1)(x2 + x1) =x7+x6+x4+x2+x6+x5+x3+x1 = x7+x5+x4+x3+x2+x1.

Polynomial division - what is P(x)/Q(x)

110 |110101 | 1001 (quotient)

110

---

000101

110

---

011 (remainder)

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CRC

Choose a generator polynomial G (x) of degree r (r + 1 bitnumber)

Message to be sent M(x)

Append r zeros to M(x). The result is x rM(x).

Evaluate the remainder of x rM(x)/G (x) (remainder will be apolynomial of degree less than r) - say R(x)

Transmit T (x) = x rM(x)− R(x) = x rM(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

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Good choice for G (x)

Both highest and lowest order bits should be 1 (1xx · · · xx1)

If G (x) has degree more than 1 all single bit errors will bedetected - 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 xk + 1 as a factor (for allk ≤ m) is sufficient

If x + 1 is not a factor, then all odd number of errors can bedetected.

Any burst error of length ≤ r can be detected

32 bit polynomial specified in IEEE 802x

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ARQ (Automatic Repeat Request) Protocols

Packets need to be acknowledged

Time-out for retransmission

But acknowledgements can get lost too!

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Lost Acknowledgements

In (B) ACK was lost. So sender retransmits packet P1. Can bothsender and receiver agree that P1′ is a retransmission of P1 andnot P2?

Timeout Interval

X(B)

ACK ACK

P1 P1’

(A)

Timeout Interval

ACK

P1 P2

ACK

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Packet Numbers!

Packets need to be numbered

Obviously numbers will be “recycled”

How many unique sequence numbers do we need?

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Ambiguous Acknowledgements

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 forP1 or P2?

Timeout Interval

X(C)

P1

ACK

P1’ P2

ACK

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Numbering

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?

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Alternating Bit Protocol (ABP)

Timeout Interval

0

1 2

1

1 2

Packet NoSeq No

Ack NoPacket No

0 1

(D)

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ABP is Unambiguous!

Timeout Interval

X(E)

1 1 2

0

0

0

Timeout Interval2

1 10

1 1 2

Timeout Interval

X(F)

10

10

0 0

2

2

1

Packet NoSeq No

Ack NoPacket No

Packet NoSeq No

Ack NoPacket No

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ABP Sender Rules

If the last packet sent was numbered 0,

allowed to send the next packet with number 1 only after ACKfor 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 ACKfor 1 is received(else retransmit 1)

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ABP Receiver Rules

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)

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Sliding Window Protocols

Effective Data Rate

100 Mbps link A↔ B (R = 100x106), L = 2000m, packetsize F = 100 bits, τr = 1µs

P = τp = 100100000000 = 1µs, τc = 2000

2.5×108 = 8µs

What is the effective data rate between A and B when ABP isused?

RTT = 2τp + 2τc + 2τr = 2 + 16 + 2 = 20.

In each 20 µs interval transmission occurs only for 1 µs

Effective data rate is 5 Mbps (P/RTT x R)

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Improving Throughput

During the first micro second sender sends the first packet

Sender does nothing for 19 µs. ACK for first packet receivedafter 20µs

What if sender is allowed to send 20 packets before receivingthe ACK for the first packet?

Only after the ACK for the first packet is received the sendercan send packet 21

Only after the ACK for the second packet is received can thesender send packet 22, and so on

Window size of 20 packets

More generally, Window size ≥ RTT/packet duration

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Pipelining Example, W ≥ RTT/τp = 13

0

0

RTT

1 2 3 4 5 6 7 8 9 10 11 12

13 packets (0 to 12) sent in one RTT

ACK for 0 processed; ready to send 13

13

1

ACK for 1 rcd; ready to send 14

14

2

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0

0

RTT

1

2 3 4 5 6 7 8 9 10 11 12



13

1


14

2

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0

0

RTT

1 2

3 4 5 6 7 8 9 10 11 12



13

1


14

2

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0

0

RTT

1 2 3 4 5 6 7 8 9 10 11

12



13

1


14

2

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0

0

RTT

1 2 3 4 5 6 7 8 9 10 11 12



13

1


14

2

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Pipelining

The secret to higher throughput

ABP needs only half-duplex channels, for pipelining weimplicitly 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 ?

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Selective Repeat Protocol

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?)

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SRP Protocol

Sender Window (size 5)

Sent and ACKed

ACK pending

Not yet sent

3 4 5 6 7 8 9 0 1 2 3 4

SRP Sender Rules

Legal to send any packet within the window

Head of the queue blocked by an outstandingACK.

Sender window can not advance until ACK isreceived for the blocking packet.

Normally next unsent packet (blue) is sent.

On time-out for the blocking red packetresend the packet.

Sender may have to buffer up to W packets.

2W unique packet numbers — 0 to 2W − 1.

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SRP Window

Rx. Window

received

pending

p.w f.w

3 4 5 6 7 8 9 0 1 2 3 4

Receiver Rules

Past Window and Future Window; each of size W .

Packets left of the Past Window: Receipt of ACKconfirmed by sender

Packets in the Past Window: ACKs not confirmed bysender

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 windowposition. Reception of 8 → ACK for 3 confirmed.

Future window is the upper bound on sender windowposition

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SRP Window

Rx. Window

received

pending

3 4 5 6 7 8 9 0 1 2 3 4

Receiver Rules

Up to W packets may need to be buffered

2W possibilities for next packet. Using 2W distinctnumbers eliminates ambiguities.

If next received packet is numbered 4, 5, 6, 7 or 8 ACKand discard.

If next received packet is numbered 9, 0, 1, 2 or 3 ACKand store.

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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 aNACK for 2?

ACK for 3 is generally received well before time-out for 2.

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Go Back N - GBN Protocol

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 receivedpacket number

Example, if Rx gets packets 1,2,3,4,5,7,8,9 the ACK for 5 issent. 7,8,9 not ACKed.

Sender retransmits 6,7,8,9...

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Windowing Protocols in DL vs TL

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 oneRTT”)

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 τp, τc , τr , RTT etc.,are known at design time. (Easy to compute W ≥ RTT/τp).

In TL how well do we know the “virtual link” parameters?How do we choose window size?

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