# 5 1 Victor S. Frost Dan F. Servey Distinguished Professor Electrical Engineering and Computer...

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: frost@eecs.ku.edu

http://www.ittc.ku.edu/

How to cope with last hop impairments?Part 2ARQ #5

How to cope with last hop impairments?

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

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

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

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

• For access networks the main issue is error control

Error and Flow Control Protocol 1: The Unrestricted Simplex Protocol

• Assumptions– One directional information flow– Infinite buffers– No errors– Network Layer always has a packet to

NetworkNetworkLayerLayer

Data LinkData LinkLayerLayer

Error and Flow Control Protocol 2:

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

Error and Flow Control Protocol 2: The Simplex Stop & Wait Protocol

ACKACKChannelChannel

Assume Network Layer always has data to send

Error and Flow Control Protocol 3: 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

– When to retransmit– What to retransmit

# 5 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

# 5 11

Error and Flow Control Protocol 3: The Simplex Protocol for a Noisy

Channel

–Timeout interval too short•Duplicate packets

–Timeout interval too long •Reduced throughput

# 5 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

# 5 13

Error and Flow Control Protocol 3: One Simplex Protocol for a Noisy Channel

Event Transmitter Action Receiver Action

Packet arrived to DLL from NL

Store in Buffer.Insert s.seq.Transmit Frame & starttimer.

Frame arrive at receiver Get frame from physical layer.If in error then dump it.If out of sequence then dump it.If in-sequence and error free then, send to NL, increment Frame Expected,Send Acknowledgement

ReceiveAcknolwedgement

Remove from bufferGet next packet from NLIncrement s.seqStore packet in Buffer.Insert s.seq.Transmit Frame & starttimer.

# 5 14

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

# 5 15

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, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 32, 3, 0, 1, 2, 3

Receive ack and advance window

# 5 16

Error and Flow Control Sliding Window Protocols:

A to B Data TrafficA to B Data TrafficB to A Ack TrafficB to A Ack Traffic

B to A Data TrafficB to A Data TrafficA to B Ack TrafficA to B Ack Traffic

# 5 17

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)

# 5 18

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

# 5 19

Error and Flow Control

Sliding Window Protocols: How many frames can be pipelined: Problem if max # frames in

pipeline = 2n

• Assume that # frames in pipeline 2n

• Assume n = 3, Node A sends 0...7 (8 frames)• Node B receives 0...7 ok and sends Ack• First Ack gets lost• Packet 0 of Node A times out• Node B receives another packet 0, expects a

packet 0, but this is a duplicate

• Thus: # frames in pipline <= 2n-1

# 5 20

Error and Flow Control Sliding Window Protocols: How many frames can be

pipelined (1)

• Now with # frames in pipline = N= 2n-1

– 0....6 (7 frames)• Node A sends 0...6• Node B receives 0...6 ok• Node B sends Ack• Ack gets lost

# 5 21

Error and Flow Control Sliding Window Protocols: How many frames can be pipelined

• Node A times out• Node A retransmitts 0...6•But Node B is expecting

frame #7• Node ignores 0...6 (often will

send a RR frame explicitly telling Node A it is expecting Frame #7)

# 5 22

Error and Flow Control

• Focus on which frames to retransmit

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

# 5 23

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– 1000m/c = 3.33 us– 2 R= 1000 bits

# 5 24

• 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

# 5 25

• Case 2: Stop and Wait (N=1)– Reduce capacity to 1.5 Mb/s– 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

# 5 26

• 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

# 5 27

• 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 R = 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%

# 5 28

6.66us 3333us

Frame # 1,000

Ack for Frame 1

# 5 29

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

# 5 30

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

Example: Transmit 1,2,3,4,5 and frame 2 is in error then 3, 4, and 5 are received out of sequence and

retransmit 2,3,4,5

# 5 31

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

# 5 32

Error and Flow Control Other Enhancements

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

received the receiver sends a NAK frame to the transmitter, the NAK frame 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.

# 5 33

Error and Flow Control Sliding Window Protocols: Piggyback ACKS

A to B Data TrafficA to B Data TrafficB to A Ack TrafficB to A Ack Traffic

B to A Data TrafficB to A Data TrafficA to B Ack TrafficA to B Ack Traffic

Reverse traffic is used to Piggyback Reverse traffic is used to Piggyback ACKSACKS

# 5 34

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.

# 5 35

Error and Flow Control Performance

• Definition for effective rate

protocol thegiven Bits transfer toTime

Delivered Bits#effR

# 5 36

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 = • Processing time (in receiver and

transmitter) = tproc

# 5 37

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

# 5 38

• Reff = (nf - no)/ to

• Efficiency = Reff/R =

# 5 39

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

procproc

0 so )3

0 so 1)

Assuming

Define 2tR =Define 2tR =Delay-Bandwidth ProductDelay-Bandwidth Product

For fixed DLL parametersFor fixed DLL parametersAs Delay-Bandwidth Product As Delay-Bandwidth Product Efficiency Efficiency

# 5 40

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

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

# 5 41

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:

# 5 42

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

get first ack• Or Nnf / R > 2 + nf / R

• Then channel is always busy sending packets

• Efficiency = (nf - no)/ nf

# 5 43

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

time to get first ack•Or Nnf / R < 2 + nf / R

•Then channel is Not always busy

sending packets: Time is wasted waiting for an Ack

# 5 44

• Time to send one window = Nnf / R

• Number of bits sent = Nnf

• Time to send Nnf bits = 2 + nf / R

• Effective rate = Nnf /(2 + nf / R )

• Efficiency = Nnf /(2R + nf )= N /(2R/ nf + 1)

# 5 45

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

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

Note you can control the rate by changing NNote you can control the rate by changing N

# 5 46

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) n

• If p << 1 then Pf = pnf

# 5 47

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 2 transmission to send a frame = Pf Pf(1- Pf)

1)P1(kP ions transmissofnumber Average

# 5 48

• Without errors to = 2 + tf + tack + 2 tproc = 2( + tproc ) + (na + nf)/R ~ 2 + nf/R

• With Errors to = (2 + nf/R)/ (1- Pf)

• with errors(1- Pf)without errors

# 5 49

• Stop and wait• Capacity = 2 Mb/s • Frame size= 512

bits• Propagation time =

10us• Efficiency without

errors = 0.92

Performance of Stop and Wait with Errors

0.000001 0.00001 0.0001 0.001 0.01

Bit error rate

# 5 50

Performance of Sliding Window

• Assume Go-Back-N• Assume large window case

• 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

fffffGBN

NttPPtt

Modified from: Leon-Garcia & Widjaja: Communication Networks

# 5 51

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/(

# 5 52

Error and Flow Control With Errors

ARQ Efficiency Comparison

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

- LOG(p)

SelectiveRepeat

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

Delay-Bandwidth product = 10, 100

# 5 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

# 5 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 decrease BER, e.g., adapt the modulation scheme

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

# 5 55

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.com Sep 24, 2002

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

Mode IEEE802.11bMb/s

IEEE802.11aMb/s

3 5.5 12

4 11 18

# 5 56

Hybrid ARQ: Incremental Redundancy

• Goal: transmit a packet of J information bits

• Add CRC to the J information bits to create a J + c packet

• Use a 1/n convolution code get n(J+c) bits

• Puncture the 1/n convolutional code* to get a rate 1 code (save all n bits)

• Send the J+c bits*Need to use special Convolutional codes: Rate Compatible Codes (RCPC)

# 5 57

• At receiver pad with random bits to get the n(J+c) bits

• Use Viterbi decoding • Check CRC

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

• The sender now sends transmitters next J+c bits

• At receiver now have rate 2/n code• Use Viterbi decoding • Check CRC

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

• Continue until sent all n(J+c) bits

# 5 58

• Hybrid ARQ or Incremental Redundancy

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

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

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

# 5 59

• Advantage: If mobile is close to BS then lost of power and 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.

# 5 60

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

• More….

# 5 1 Victor S. Frost Dan F. Servey Distinguished Professor Electrical Engineering and Computer...

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