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

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1 How to cope with last hop impairments? Part 2 ARQ #5 Victor S. Frost Dan F. Servey Distinguished Proessor Electrical Engineering and Computer Science University o Kansas 2335 Irving Hill Dr. Lawrence, Kansas Phone: (785) FAX:(785) rost@eecs.ku.edu All material copyright 2006 Victor S. Frost, All Rights Reserved # 5 1 How to cope with last hop impairments? Techniques or coping with noise Forward error detection/correction coding Automatic Repeat request (ARQ) Co-existance or modiications to end-to-end protocols Techniques or coping with multipath ading ading mitigation techniques, e.g., Equalizers Diversity RAKE receivers OFDM # 5 2

2 ARQ ARQ is used or both Flow control Error control For access networks the main issue is error control # 5 3 Protocol 1: The Unrestricted Simplex Protocol Assumptions One directional inormation low Ininite buers No errors Network Layer always has a packet to send Network Layer Data Link Layer Data Link Layer Network Layer # 5 4

3 Protocol 2: The Simplex Stop & Wait Protocol: Assumptions One directional inormation low No errors Network Layer always has a packet to send Finite receive buers Finite buer means that there must be some way to stop the transmitter rom sending when the buer is ull # 5 5 Protocol 2: The Simplex Stop & Wait Protocol Assume Network Layer always has data to send Network Layer Data Link Layer ACK Channel Data Link Layer Network Layer # 5 6

4 Protocol 2: The Simplex Stop & Wait Protocol Assume Network Layer always has data to send Network Layer Data Link Layer ACK Channel Data Link Layer Network Layer # 5 7 Protocol 2: The Simplex Stop & Wait Protocol Assume Network Layer always has data to send Network Layer Data Link Layer ACK Channel Data Link Layer Network Layer # 5 8

5 Protocol 3: The Simplex Protocol or a Noisy Channel Assumptions One directional inormation low Network Layer always has a packet to send Finite receive buers Allow errors Data link protocols must address What to transmit (in access networks) When to retransmit What to retransmit # 5 9 Access Networks-What to Transmit Scheduling based on Requests or upstream transmission opportunities Channel conditions CoS QoS # 5 10

6 Protocol 3: The Simplex Protocol or a Noisy Channel 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 I no acknowledgment received in 2.1 ms then, Packet in error Acknowledgment lost # 5 11 Protocol 3: The Simplex Protocol or a Noisy Channel Timeout interval too short Duplicate packets Timeout interval too long Reduced throughput # 5 12

7 Protocol 3: The Simplex Protocol or a Noisy Channel Sequence numbers are used to determine what to retransmitt Transmitter assigns a number to each rame Receiver keeps track o the expected rame number How to deal with out o sequence rames, i.e., i the received sequence number does not match what is expected, The rame is dumped (or some DLC protocols) Frame stored Again aced with the issue o how to schedule the retransmission wrt current state o buers and channels. # 5 13 Protocol 3: One Simplex Protocol or a Noisy Channel Event Transmitter Action Receiver Action Packet arrived to DLL rom NL Frame arrive at receiver Receive Acknolwedgement Store in Buer. Insert s.seq. Transmit Frame & start timer. Remove rom buer Get next packet rom NL Increment s.seq Store packet in Buer. Insert s.seq. Transmit Frame & start timer. Get rame rom physical layer. I in error then dump it. I out o sequence then dump it. I in-sequence and error ree then, send to NL, increment Frame Expected, Send Acknowledgement # 5 14

8 Sliding Window Protocols: Assumptions Two directional inormation low Network Layer always has a packet to send Finite receive buers Finite number o bits/sequence number Piggybacking Put Acknowledgments in reverse traic low Increases protocol eiciency Reduces interrupts # 5 15 Sliding Window Protocols: Advantage pipeline Why called sliding window Assume 2 bits/sequence number Possible rame 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 # 5 16

9 Sliding Window Protocols: A A to B Data Traic B to A Ack Traic B to A Data Traic A to B Ack Traic B # 5 17 Sliding Window Protocols Transmitter keeps a list o sequence # s it can use Sending window Receiver keeps a list o sequence # s it will accept Receiving window n = # bits/(sequence number) # 5 18

10 Sliding Window Protocols Sequence numbers in range n -1 This allows N=2 n -1 packets to be sent beore getting and acknowledgment Requires N=2 n -1 packets buers # 5 19 Sliding Window Protocols: How many rames can be pipelined: Problem i max # rames in pipeline = 2 n Assume that # rames in pipeline 2 n Assume n = 3, Node A sends (8 rames) Node B receives ok and sends Ack First Ack gets lost Packet 0 o Node A times out Node B receives another packet 0, expects a packet 0, but this is a duplicate Thus: # rames in pipline <= 2 n -1 # 5 20

11 Sliding Window Protocols: How many rames can be pipelined (1) Now with # rames in pipline = N= 2 n (7 rames) Node A sends Node B receives ok Node B sends Ack Ackgets lost # 5 21 Sliding Window Protocols: How many rames can be pipelined (2) Node A times out Node A retransmitts But Node B is expecting rame #7 Node ignores (oten will send a RR rame explicitly telling Node A it is expecting Frame #7) # 5 22

12 Focus on which rames to retransmit Pipeline: send up to N rames beore receiving an acknowledgment # 5 23 Example Distance between nodes = 1 km Frame length = 1000 bits Capacity = 150 Mb/s No errors Delay-bandwidth product Assume ree space τ =1000m/c = 3.33 us 2 τr= 1000 bits # 5 24

13 Example Case 1: Stop and Wait (N=1) Frame transmission time = 6.66us Propagation time = 3.33us Transmit rame at t=0, At 6.66 us us rame received At 6.66us us the acknowledgment is received, thereore transmitted 1000 bits in 6.66us us Eective transmission rate is 1000/13.3us ~ 75Mb/s Eiciency: (75Mb/s)/(150Mb/s) ~ 50.0% eicient # 5 25 Example 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 rame at t=0, At 666 us us rame received At 666us us the acknowledgment is received, thereore transmitted 1000 bits in 666us us Eective transmission rate is 1000/672us ~ Mb/s Eiciency: (1.488Mb/s)/(1.50Mb/s) ~ 99.2% eicient # 5 26

14 Example Case 3: Stop and Wait (N=1) Capacity to 150 Mb/s Frame transmission time = 6.66us WAN: D=1000km Propagation time = 3333us Transmit rame at t=0, At 6.66 us us rame received At 6.66us us the acknowledgment is received, thereore transmitted 1000 bits in 6.66us us Eective transmission rate is 1000/6672us ~.149Mb/s Eiciency: (.149Mb/s)/(150Mb/s) ~ 0.1% eicient # 5 27 Example Case 4: Sliding window (N=1023) Capacity to 150 Mb/s Frame transmission time = 6.66us WAN: D=1000km Propagation time = 3333us Transmit rame at t=0, Note 2 τr = 999,900 bits or in rames rames Since time to transmit 1023 rames > Always have a sequence number to use Never have to wait or ACK Eiciency 100% # 5 28

15 Example 6.66us 3333us Frame # 1,000 Ack or Frame 1 # 5 29 Go-Back-N Protocol (1) Problem: I there is an error or lost rame then what rules are used to determine the rames to retransmit. Go-back-N Retransmit all rames transmitted ater the erred rame The receiver ignores all out-o sequence rames # 5 30

16 Go-Back-N Protocol (2) Example: Transmit 1,2,3,4,5 and rame 2 is in error then 3, 4, and 5 are received out o sequence and retransmit 2,3,4,5 # 5 31 Selective Repeat Receiver accepts out o sequence rames Requires buers in receiver and transmitter Requires extra processing to deliver packets in order to the Network Layer # 5 32

17 Other Enhancements Negative Acknowledgment When an out-o-sequence rame is received the receiver sends a NAK rame to the transmitter, the NAK rame contains the sequence number o the expected data rame. NAK enables aster error recovery, without a NAK time-out must be used to learn about errors. # 5 33 Sliding Window Protocols: Piggyback ACKS Reverse traic is used to Piggyback ACKS A A to B Data Traic B to A Ack Traic B to A Data Traic A to B Ack Traic B # 5 34

18 Other Enhancements: Acknowledgment timer I there is light (or no) reverse traic then ACKS may not be sent. An acknowledgment timer is used to insure ACKS are sent. Upon receipt o a rame an AckTimer is started. I reverse traic arrives beore the AckTimer ires then piggyback the ACK. I the AckTimer ires then send a supervisory ACK rame. # 5 35 Perormance Deinition or eective rate R e = Time to # Bits Delivered transer Bits given the protocol # 5 36

19 Perormance Length o data packet (bits) = D Number o overhead bits/packet = n o Link Rate(b/s) = R Length o Ack Packet (bits)= n a Frame size = n = D+ n o One-way propagation delay = τ Processing time (in receiver and transmitter) = t proc # 5 37 Perormance-Stop & Wait Eective rate and eiciency or simplex stop-and-wait protocol t ack = n a /R t = n /R Time to transmit one rame = t o t o = 2 τ + t + t ack + 2 t proc = 2(τ + t proc ) + (n a + n )/R # 5 38

20 Perormance-Stop & Wait R e = (n - n o )/ t o Eiciency = R e /R = ηo = 1+ n n a no 1 n 2R( τ + t + n proc ) # 5 39 Perormance-Stop & Wait: Limiting Case Assuming 1) n 2) t a proc 3) n n o then n 1 ηo = 2τ R 1+ n τ so t n so n a proc n so n o 0 + τ τ 0 Deine 2tR = Delay-Bandwidth Product For ixed DLL parameters As Delay-Bandwidth Product Eiciency # 5 40

21 Perormance-Stop & Wait Example Frame size = 1024 bytes Overhead = Ack= 8 bytes τ = 50 ms Case 1: R=30 Kb/s Eiciency = 73% Case 2: R=1.5 Mb/s Eiciency = 5% # 5 41 Perormance-Sliding Window Protocol Case 1: Large window Window Size = N Transmit N packet and wait or Ack Making the same assumption as beore First Ack arrives at sender at: 2τ + n R # 5 42

22 Perormance-Sliding Window Protocol Case 1: Large window I time to transmit N packets > time to get irst ack Or Nn / R > 2τ + n / R Then channel is always busy sending packets Eiciency = (n -n o )/ n # 5 43 Perormance-Sliding Window Protocol Case 2: Small Window I time to transmit N packets < time to get irst ack Or Nn / R < 2τ + n / R Then channel is Not always busy sending packets: Time is wasted waiting or an Ack # 5 44

23 Perormance-Sliding Window Protocol Time to send one window = Nn / R Number o bits sent = Nn Time to send Nn bits = 2τ + n / R Eective rate = Nn /(2τ + n / R ) Eiciency = Nn /(2τR + n ) = N /(2τR/ n + 1) # 5 45 Perormance-Sliding Window Protocol Example: Frame size = 1024 bytes Overhead = Ack= 0 bytes τ = 1 ms Rate = 40 Mb/s Case 1: N = 12 Eiciency = 100% 40 Mb/s Case 2: N = 8 Eiciency = ~75% 30 Mb/s Case 3: N = 4 Eiciency = ~ 37% 15 Mb/s Note you can control the rate by changing N # 5 46

24 With bit errors Let p = Probability o a bit error Assume bits errors are random Let P = Probability o a rame error P = 1 - (1-p) n I p << 1 then P = pn # 5 47 Stop and Wait Perormance The Probability that it takes 1 transmission to send a rame = 1- P The Probability that it takes 2 transmission to send a rame = P (1- P ) The Probability that it takes 3 transmission to send a rame = P P (1- P ) Average number o transmissions = kp k= 1 k-1 1 (1 P ) = (1 P ) # 5 48

25 Stop and Wait Perormance Without errors t o = 2 τ + t + t ack + 2 t proc = 2(τ + t proc ) + (n a + n )/R ~ 2 τ + n /R With Errors t o = (2 τ + n /R)/ (1- P ) η with errors = (1- P )η without errors # 5 49 Stop and Wait Perormance Stop and wait Capacity = 2 Mb/s Frame size= 512 bits Propagation time = 10us Eiciency without errors = 0.92 Eiciency Perormance o Stop and Wait with Errors Bit error rate # 5 50

26 Perormance o Sliding Window Assume Go-Back-N (GBN) Assume large window case with W s = window size Assume P rame loss probability, then time to deliver a rame is: t i irst rame transmission succeeds (1 P ) t + W s t /(1-P ) i the irst transmission does not succeed P t η GBN GBN = t (1 P ) + P { t = n n t R GBN o no 1 n = 1+ ( N 1) P Nt + } = t 1 P (1 P ) Nt + P 1 P and Modiied rom: Leon-Garcia & Widjaja: Communication Networks # 5 51 Perormance o Sliding Window Assume Selective repeat Assume P rame loss probability, then number o transmissions required to deliver a rame is: t / (1-P ) n n η SR = t o /(1 P R ) no = (1 n )(1 P ) Modiied rom: Leon-Garcia & Widjaja: Communication Networks # 5 52

27 With Errors Eiciency ARQ Eiciency Com parison Selective Repeat LOG(p) p Delay-Bandwidth product = 10, 100 Go Back N 10 Stop and Wait 100 Go Back N 100 Stop and Wait 10 Modiied rom: Leon-Garcia & Widjaja: Communication Networks # 5 53 Impact o ARQ Increase delay wait or successul retransmission Increase the delay variation Reduce the eective throughput Signiicantly reduces the probability o an errored packet reaching upper layers especially when combined with FEC. Practical issue: The link layer will only try to send a packet or so long, eventually it will time out and let high layers deal with the problem; thereore higher layers can still see losses # 5 54

28 Other actions to cope with errors Adaptive power control In CDMA system this increases the intererence to other users Used in ixed 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 or retransmission # 5 55 Adaptive Bit Rate Rate vs S/N 800 Date Rate (kb S/N (db) + S/N obtained via measurements + Measurement called channel state inormation + Rate change by: - Modulation - Number codes assigned - Number o time slots assigned # 5 56

29 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 * GPRS=General Packet Radio Service GSM=Global System or Mobile Communications EDGE= Enhanced date rates or GSM Evolution CS=Coding Scheme MCS=Modulation and Coding Scheme Mode IEEE b Mb/s IEEE a Mb/s *From: Dealing with the EDGE Evolution, Terry Locke, Si Nguyen, and Dominique Moreuil, Nortel Networks CommsDesign.com Sep 24, 2002 # 5 57 Hybrid ARQ: Incremental Redundancy Goal: transmit a packet o J inormation bits Add c error detection bits to the J inormation 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) # 5 58

30 Hybrid ARQ: Incremental Redundancy 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 I 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 I checks then accept packet and send ACK Else save received bits and send NACK Continue until sent all K(J+c) bits # 5 59 Hybrid ARQ: Incremental Redundancy Each retransmission adds redundancy that will improve the probability o correct reception Modiied From: Incremental Redundancy in EGPRS, Feb, # 5 60

31 Hybrid ARQ: Incremental Redundancy Rate 1/3 means 3 coded bits or each inormation bit Modiied From: Incremental Redundancy in EGPRS, Feb, # 5 61 Hybrid ARQ: Incremental Redundancy Advantage: I mobile is close to BS then lots o power and only irst packet (rate =1) will be needed As the mobile moves away rom 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 o CSI # 5 62

32 Reerences 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, (8): p 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 Mercy, P., Incremental Redundancy in EGPRS. Wireless Design & Development, 2005 p Mukhtar, R., et al., A model or the perormance evaluation o packet transmissions using type-ii hybrid ARQ over a correlated error channel Wirel. Netw., (1): p Nanda, S., K. Balachandran, and S. Kumar, Adaptation techniques in wireless packet data services. Communications Magazine, IEEE, (1): p Porat, D., Incremental Redundancy. 2002, Communication & Signal Processing Ltd. p # 5 63