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# 7 1

Victor S. FrostDan F. Servey Distinguished Professor

Electrical Engineering and Computer ScienceUniversity of Kansas2335 Irving Hill Dr.

Lawrence, Kansas 66045Phone: (785) 864-4833 FAX:(785) 864-7789

e-mail: [email protected]://www.ittc.ku.edu/

How to cope with last hop impairments?Part 2ARQ #7

All material copyright 2006Victor S. Frost, All Rights Reserved

# 7 2

How to cope with last hop impairments?

• Techniques for coping with multipath fading fading mitigation techniques, e.g., – Equalizers– Diversity– RAKE receivers– OFDM

• Techniques for coping with noise– Forward error detection/correction coding

– Automatic Repeat reQuest (ARQ)– Co-existence or modifications to end-to-end protocols

# 7 3

ARQ

• ARQ is used for both– Flow control– Error control

• For access networks the main issue is error control

# 7 4

Error and Flow Control The Simplex Stop & Wait Protocol: Assumptions

• One directional information flow• No errors• Network Layer always has a packet to

send• Finite receive buffers

– Finite buffer means that there must be some way to stop the transmitter from sending when the buffer is full

# 7 5

Error and Flow ControlThe Simplex Stop & Wait Protocol

NetworkLayer

NetworkLayer

Data LinkLayer

Data LinkLayer

ACKChannel

Assume Network Layer always has data to send

# 7 6


NetworkLayer

NetworkLayer

Data LinkLayer

Data LinkLayer

ACKChannel


# 7 7


NetworkLayer

NetworkLayer

Data LinkLayer

Data LinkLayer

ACKChannel


# 7 8

Error and Flow Control The Simplex Protocol for a Noisy Channel

• Assumptions– One directional information flow– Network Layer always has a packet to send– Finite receive buffers– Allow errors

• Data link protocols must address – What to transmit (in access networks)– When to retransmit– What to retransmit

# 7 9

Error and Flow Control Access Networks-What to Transmit

• Scheduling based on – Requests for upstream transmission

opportunities– Channel conditions– CoS– QoS

# 7 10


• Timeout to determine when to retransmitt• Example:

– Assume a 1 ms propagation time– Assume a .1 ms receiver packet processing time – Timeout interval >2.1 ms

• If no acknowledgment received in 2.1 ms then,

– Packet in error– Acknowledgment lost

# 7 11


–Timeout interval too short•Duplicate packets

–Timeout interval too long •Reduced throughput

# 7 12


• Sequence numbers are used to determine what to retransmitt– Transmitter assigns a number to each frame– Receiver keeps track of the expected frame number– How to deal with out of sequence frames, i.e., if the

received sequence number does not match what is expected,

• The frame is dumped (for some DLC protocols)• Frame stored

• Again faced with the issue of how to schedule the retransmission wrt current state of buffers and channels.

# 7 13

Error and Flow Control Sliding Window Protocols: Assumptions

• Two directional information flow• Network Layer always has a packet to send• Finite receive buffers• Finite number of bits/sequence number• Piggybacking

– Put Acknowledgments in reverse traffic flow– Increases protocol efficiency– Reduces interrupts

# 7 14

Error and Flow Control Sliding Window Protocols:

• Advantage pipeline• Why called sliding window

– Assume 2 bits/Sequence number– Possible frame numbers 0, 1, 2, 3

0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3

Receive ack and advance window

# 7 15

Error and Flow Control Sliding Window Protocols:

A B

A to B Data TrafficB to A Ack Traffic

B to A Data TrafficA to B Ack Traffic

# 7 16

Error and Flow Control Sliding Window Protocols

• Transmitter keeps a list of sequence #’s it can use

– Sending window• Receiver keeps a list of

sequence #’s it will accept– Receiving window

• n = # bits/(sequence number)

# 7 17

Error and Flow Control Sliding Window Protocols

• Sequence numbers in range 0...2n-1

• This allows N=2n-1 packets to be sent before getting and acknowledgment

• Requires N=2n-1 packets buffers

# 7 18

Error and Flow Control

• Focus on which frames to retransmit

• Pipeline: send up to N frames before receiving an acknowledgment

# 7 19

Error and Flow Control Example

• Distance between nodes = 1 km

• Frame length = 1000 bits• Capacity = 150 Mb/s• No errors• Delay-bandwidth product

– Assume free space– τ =1000m/c = 3.33 us– 2 τR= 1000 bits

# 7 20


• Case 1: Stop and Wait (N=1)– Frame transmission time = 6.66us – Propagation time = 3.33us– Transmit frame at t=0, – At 6.66 us + 3.33us frame received – At 6.66us + 6.66us the acknowledgment is

received, therefore transmitted 1000 bits in 6.66us + 6.66us

– Effective transmission rate is 1000/13.3us ~ 75Mb/s

– Efficiency: (75Mb/s)/(150Mb/s) ~ 50.0% efficient

# 7 21


• Case 2: Stop and Wait (N=1)– Reduce capacity to 1.5 Mb/s (like a 3G rate)– Frame transmission time = 666us – Propagation time = 3.33us – Transmit frame at t=0, – At 666 us + 3.33us frame received – At 666us + 6.66us the acknowledgment is

received, therefore transmitted 1000 bits in 666us + 6.66us

– Effective transmission rate is 1000/672us ~ 1.488 Mb/s

– Efficiency: (1.488Mb/s)/(1.50Mb/s) ~ 99.2% efficient

# 7 22


• Case 3: Stop and Wait (N=1)– Capacity to 150 Mb/s– Frame transmission time = 6.66us – WAN: D=1000km Propagation time = 3333us– Transmit frame at t=0, – At 6.66 us + 3333us frame received – At 6.66us + 6666us the acknowledgment is

received, therefore transmitted 1000 bits in 6.66us + 6666us

– Effective transmission rate is 1000/6672us ~ .149Mb/s

– Efficiency: (.149Mb/s)/(150Mb/s) ~ 0.1% efficient

# 7 23


• Case 4: Sliding window (N=1023)– Capacity to 150 Mb/s– Frame transmission time = 6.66us – WAN: D=1000km Propagation time = 3333us– Transmit frame at t=0, – Note 2 τR = 999,900 bits or in frames 999.9 frames– Since time to transmit 1023 frames > 999.9

• Always have a sequence number to use• Never have to wait for ACK

– Efficiency 100%

# 7 24

Error and Flow Control Go-Back-N Protocol (1)

• Problem:If there is an error or lost frame then what rules are used to determine the frames to retransmit.

• Go-back-N– Retransmit all frames transmitted after

the erred frame– The receiver ignores all out-of sequence

frames

# 7 25

Error and Flow Control Go-Back-N Protocol (2)

Example:Transmit 1,2,3,4,5 and

frame 2 is in error then3, 4, and 5 are received out

of sequence andretransmit 2,3,4,5

# 7 26

Error and Flow Control Selective Repeat

• Receiver accepts out of sequence frames

• Requires buffers in receiver and transmitter

• Requires extra processing to deliver packets in order to the Network Layer

# 7 27

Error and Flow Control Other Enhancements

• Negative Acknowledgment– When an out-of-sequence frame is

received the receiver sends a NAKframe to the transmitter, the NAKframe contains the sequence number of the expected data frame.

– NAK enables faster error recovery, without a NAK time-out must be used to learn about errors.

# 7 28

Error and Flow ControlSliding Window Protocols: Piggyback ACKS

A B

A to B Data TrafficB to A Ack Traffic

B to A Data TrafficA to B Ack Traffic

Reverse traffic is used to Piggyback ACKS

# 7 29

Error and Flow Control Other Enhancements: Acknowledgment timer

• If there is light (or no) reverse traffic then ACKS may not be sent.

• An acknowledgment timer is used to insure ACKS are sent.

• Upon receipt of a frame an AckTimer is started. If reverse traffic arrives before the AckTimer fires then piggyback the ACK. If the AckTimer fires then send a supervisory ACK frame.

# 7 30

Error and Flow Control Performance

• Definition for effective rate

protocol thegiven Bits transfer toTime Delivered Bits#

=effR

# 7 31

Error and Flow Control Performance

• Length of data packet (bits) = D• Number of overhead bits/packet = no

• Link Rate (b/s) = R• Length of Ack Packet (bits)= na

• Frame size = nf = D+ no

• One-way propagation delay = τ• Processing time (in receiver and

transmitter) = tproc

# 7 32

Error and Flow Control Performance-Stop & Wait

• Effective rate and efficiency for simplex stop-and-wait protocol– tack = na/R– tf = nf/R– Time to transmit one frame = to

to = 2 τ + tf + tack + 2 tproc= 2(τ + tproc ) + (na + nf)/R

# 7 33


• Reff = (nf - no)/ to• Efficiency = Reff/R =

f

proc

f

a

f

o

o

ntR

nn

nn

)(21

1

+++

−= τη

# 7 34

Error and Flow Control Performance-Stop & Wait: Limiting Case

f

f

ofo

procproc

f

afa

nR

nnnn

tt

nnnn

τη

τττ

21

1then

0 so )3

so )2

0 so 1)

Assuming

o

+=

⎯→⎯⟨⟨

≈+⟨⟨

⎯→⎯⟨⟨ Define 2tR =Delay-Bandwidth Product

For fixed DLL parametersAs Delay-Bandwidth Product ↑Efficiency ↓

# 7 35


• Example– Frame size = 1024 bytes– Overhead = Ack = 8 bytes– τ = 50 ms

• Case 1: R=30 Kb/s →Efficiency = 73%• Case 2: R=1.5 Mb/s →Efficiency = 5%

# 7 36

Error and Flow Control Performance-Sliding Window Protocol

• Case 1: Large window– Window Size = N– Transmit N packet and wait for Ack– Making the same assumption as before– First Ack arrives at sender at:

Rn f+τ2

# 7 37


• Case 1: Large window– If time to transmit N packets > time to get

first ack• Or Nnf / R > 2τ + nf / R • Then channel is always busy sending packets• Efficiency = (nf - no)/ nf

# 7 38


• Case 2: Small Window– If time to transmit N packets <

time to get first ack• Or Nnf / R < 2τ + nf / R • Then channel is Not always busy sending

packets: Time is wasted waiting for an Ack

# 7 39


• Time to send one window = Nnf / R • Number of bits sent = Nnf• Time to send Nnf bits = 2τ + nf / R • Effective rate = Nnf /(2τ + nf / R )• Efficiency = Nnf /(2τR + nf )

= N /(2τR/ nf + 1)

# 7 40


• Example:– Frame size = 1024 bytes– Overhead = Ack = 0 bytes– τ = 1 ms– Rate = 40 Mb/s

• Case 1: N = 12 → Efficiency = 100% → 40 Mb/s• Case 2: N = 8 → Efficiency = ~75% → 30 Mb/s• Case 3: N = 4 → Efficiency = ~ 37% →15 Mb/s

Note you can control the rate by changing N

# 7 41

With bit errors

• Let p = Probability of a bit error• Assume bits errors are random• Let Pf = Probability of a frame error• Pf = 1 - (1-p) nf• If p << 1 then Pf = pnf

# 7 42

Stop and Wait Performance

• The Probability that it takes 1 transmission to send a frame = 1- Pf

• The Probability that it takes 2 transmission to send a frame = Pf(1- Pf)

• The Probability that it takes 3 transmission to send a frame = Pf Pf(1- Pf)

)P1(1)P1(kP ions transmissofnumber Average

ff

1k

1-kf −

=−= ∑∞

=

# 7 43


• Without errorsto = 2 τ + tf + tack + 2 tproc

= 2(τ + tproc ) + (na + nf)/R~ 2 τ + nf/R

• With Errorsto = (2 τ + nf/R)/ (1- Pf)

• ηwith errors= (1- Pf)ηwithout errors

# 7 44


• Stop and wait• Capacity = 2 Mb/s • Frame size= 512 bits• Propagation time = 10us• Efficiency without

errors = 0.92• Note to keep high

normalized throughput need to keep BER<0.0001

Performance of Stop and Wait with Errors

0.5

0.55

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

0.000001 0.00001 0.0001 0.001 0.01

Bit error rate

Effi

cien

cy

# 7 45

Stop and Wait Performance with Adaptive Modulation

• Suppose Rb=2Mb/s with QPSK is used • Assume Eb /No= 8.75dB• Then BER = 0.0001 and efficiency with

errors ~ 0.90 ~1.8Mb/s• Then assume the Eb /No= 5.75dB (e.g.,

move away from BS)– BER= 0.01– Pf=0.99– Efficiency with errors ~ 0.01– Broken link……

# 7 46


• Now switch to BPSK @ 1Mb/s • Gain 3 dB• Eb /No= 8.75dB• BER = 0.0001 and efficiency with

errors ~ 0.90 0.9Mb/s

# 7 47


From: Channel-Adaptive Technologies and Cross-Layer Designs for Wireless Systems with Multiple Antennas: Theory and Applications, Vincent K.N. Lau and Yu-Kwong Ricky Kwok

# 7 48


BER constant over a 15dB range

Modified from: Channel-Adaptive Technologies and Cross-Layer Designs for Wireless Systems with Multiple Antennas: Theory and Applications, Vincent K.N. Lau and Yu-Kwong Ricky Kwok

# 7 49

Performance of Sliding Window• Assume Go-Back-N (GBN)• Assume large window case with Ws = window size • Assume Pf frame loss probability, then time to deliver

a frame is:– tf if first frame transmission succeeds (1 – Pf )– tf + Wstf /(1-Pf) if the first transmission does not succeed Pf

)1()1(1

1

and 1

}1

{)1(

ff

f

o

GBN

of

GBN

f

fff

f

fffffGBN

PPN

nn

Rt

nn

PNt

PtP

NttPPtt

−−+

−=

−

=

−+=

−++−=

η

Modified from: Leon-Garcia & Widjaja: Communication Networks

# 7 50

Performance of Sliding Window

• Assume Selective repeat• Assume Pf frame loss probability, then number

of transmissions required to deliver a frame is:– tf / (1-Pf)

)1)(1()1/(

ff

off

of

SR Pnn

RPtnn

−−=−−

=η


# 7 51

Error and Flow Control With Errors

ARQ Efficiency Comparison

0

0.5

1

1.5

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

- LOG(p)

Effic

ienc

ySelectiveRepeat

Go Back N 10

Stop and Wait100

Go Back N 100

Stop and Wait10

10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1

BER

Delay-Bandwidth product = 10, 100


# 7 52

BER for QAM

Eb/No

From: Modern Digital and Analog Communications Systems, B. P. Lathi, Bryden Press 1989

20 dB32-PSK

14.8 dB16-PSK

4.5 dB8-PSK

3 dBQPSK

RefBPSK

Relative Eb/No forBER 10-4

Modulation

# 7 53

Impact of ARQ

• Increase delay wait for successful retransmission

• Increase the delay variation• Reduce the effective throughput• Significantly reduces the probability of an errored

packet reaching upper layers especially when combined with FEC.

• Practical issue: The link layer will only try to send a packet for so long, eventually it will time out and let high layers deal with the problem; therefore higher layers can still see losses

# 7 54

Other actions to cope with errors

• Adaptive power control– In CDMA system this increases the interference to

other users– Used in fixed wireless deployments

• Adaptive bit rate: via measurements the system knows to reduce its transmission rate to keep BER (PER) constant, e.g., adapt the modulation scheme

• With constant PER the time in ARQ retransmissions is ~constant

• Combine FEC with ARQ: Save the errored packet and ask for retransmission

# 7 55

Adaptive Bit RateRate vs S/N

0

200

400

600

800

-10 -5 0 5 10 15 20

S/N (dB)

Date

Rat

e (k

b

+ S/N obtained via measurements+ Measurement called “channel state

information”+ Rate change by:

- Modulation- Number codes assigned- Number of time slots assigned

# 7 56

Adaptive Bit Rate: IEEE 802.16

From: IEEE 802 Wireless Systems: Protocols, Multi-Hop Mesh/Relaying: Performance and Spectrum Coexistence, Bernhard H. Walke, Stefan Mangold, Lars Berlemann, Wiley, 2006

# 7 57

Rate adaptation techniques• Basic idea: change the

modulation and coding scheme during the communication to adapt to the changing link conditions, e.g., radio environment

*From: Dealing with the EDGE Evolution, Terry Locke, Si Nguyen, and Dominique Moreuil, Nortel Networks CommsDesign.comSep 24, 2002

* GPRS=General Packet Radio ServiceGSM=Global System for Mobile CommunicationsEDGE= Enhanced date rates for GSM EvolutionCS=Coding SchemeMCS=Modulation and Coding Scheme

568

487

366

245

18114

125.53

922

611

IEEE802.1aMb/s

IEEE802.1bMb/s

Mode

# 7 58

Hybrid ARQ: Incremental Redundancy

• Goal: transmit a packet of J information bits• Add c error detection bits to the J

information bits to create a J + c packet• Use a convolutional code with 1/K redundancy

ratio• Use a 1/K convolution code get K(J+c) bits• Puncture the 1/K convolutional code* to get a

rate 1 code and save all K(J+c) bits• Send the J+c bits *Need to use special Convolutional codes:

Rate Compatible Codes (RCPC)

# 7 59


• At receiver recive only J+C bits • Pad J+C bits with random bits to get the K(J+c)

bits• Use Viterbi decoding • Check CRC

– If checks then accept packet and send ACK done– Else save received bits and send NACK

• The sender now sends transmitters next J+c bits• At receiver now have higher rate code 1/2• Use Viterbi decoding • Check CRC

– If checks then accept packet and send ACK– Else save received bits and send NACK

• Continue until sent all K(J+c) bits

# 7 60


Modified From: IncrementalRedundancy in EGPRS, Feb, 2005www.wirelessdesignmag.com

Each retransmission adds redundancy that will improve the probability of correct reception

# 7 61


Modified From: IncrementalRedundancy in EGPRS, Feb, 2005www.wirelessdesignmag.com

Rate 1/3 means 3 coded bits for each information bit

# 7 62


• Advantage: If mobile is close to BS then lots of power and only first packet (rate =1) will be needed

• As the mobile moves away from the BS there is less power and this scheme automatically compensates by sending additional redundancy bits.

• Process automatically matches coding rate to the channel S/N without explicit measurement of CSI

# 7 63

References

• Leon-Garcia & Widjaja: Communication Networks, McGraw Hill, 2004• Dealing with the EDGE Evolution, Terry Locke, Si Nguyen, and Dominique

Moreuil, Nortel Networks CommsDesign.com Sep 24, 2002• Cianca, E., et al., Channel-adaptive techniques in wireless communications: an

overview. Wireless Communications and Mobile Computing, 2002. 2(8): p. 799-813.

• Falahati, S., et al., Convolutional Coding in Hybrid Type-II ARQ Schemes on Wireless Channels, in Proceedings Radiovetenskap och Kommunikations. 1999: Karlskrona, Sweden.

• Langton, C., Coding adn decoding with convolutional codes. p. 1-29.• Mercy, P., Incremental Redundancy in EGPRS. Wireless Design &

Development, 2005 p. 30-35.• Mukhtar, R., et al., A model for the performance evaluation of packet

transmissions using type-II hybrid ARQ over a correlated error channel Wirel. Netw. , 2004. 10 (1): p. 7-16.

• Nanda, S., K. Balachandran, and S. Kumar, Adaptation techniques in wireless packet data services. Communications Magazine, IEEE, 2000. 38(1): p. 54-64.

• Porat, D., Incremental Redundancy. 2002, Communication & Signal Processing Ltd. p. 1-7.

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