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Chapter 5

Chapter 5. Reliable Data Transfer and ARQ Protocols. Reliable Data Transfer and ARQ Protocols. Introduction. Reliability & Sequencing. Reliability : Are messages or information stream delivered error-free and without loss or duplication?

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Chapter 5

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  1. Chapter 5 Reliable Data Transfer and ARQ Protocols

  2. Reliable Data Transfer and ARQ Protocols Introduction

  3. Reliability & Sequencing • Reliability: Are messages or information stream delivered error-free and without loss or duplication? • Sequencing: Are messages or information stream delivered in order? • ARQ protocols combine error detection, retransmission, and sequence numbering to provide reliability & sequencing • Examples: TCP and High Level Data Link Control (HDLC) Protocol

  4. End-to-End vs. Hop-by-Hop • A service feature can be provided by implementing a protocol • End-to-end across the network • Across every hop in the network • Example: • Perform error control at every hop in the network or only between the source and destination? • The same question for flow control and congestion control. • We next consider the tradeoffs between the two approaches

  5. Data Link operates over wire-like, directly-connected systems Frames can be corrupted, but arrive in order Data link performs error-checking & retransmission Ensures error-free packet transfer between two systems Packets Packets Data link layer Data link layer Frames A B Physical layer Physical layer 1 1 2 2 3 3 1 2 1 2 2 1 1 2 Medium 2 1 B A 1 Physical layer entity 2 Data link layer entity 3 Network layer entity Error Control in Data Link Layer (a) (b)

  6. Messages Messages Segments Transport layer Transport layer Network layer Network layer Network layer Network layer Data link layer Data link layer Data link layer Data link layer End system A End system B Physical layer Physical layer Physical layer Physical layer Network Error Control in Transport Layer • Transport layer protocol (e.g. TCP) sends segments across network and performs end-to-end error checking & retransmission • Underlying network is assumed to be unreliable

  7. 3 2 1 2 1 C 2 2 1 1 End System α End System β 1 1 1 2 2 2 3 3 3 2 4 1 4 2 2 2 1 2 1 2 1 1 1 3 Medium 2 1 B A Network 3 Network layer entity 4 Transport layer entity • Segments can experience long delays, can be lost, or arrive out-of-order because packets can follow different paths across network • End-to-end error control protocol more difficult

  8. Hop-by-Hop Data Data Data Data 1 2 3 4 5 ACK/NAK ACK/NAK ACK/NAK ACK/NAK Hop-by-Hop Vs. End-to-End Hop-by-hop cannot ensure E2E correctness Initiate error recovery more quickly End-to-End Simple inside the network ACK/NAK More scalable if complexity at the edge 1 5 2 3 4 Data Data Data Data

  9. ARQ Protocols and Reliable Data Transfer Automatic Repeat Request (ARQ)

  10. Automatic Repeat Request (ARQ) • Purpose: to ensure that the sequence of information packets/ frames is delivered in order and without errors or duplications (despite transmission errors & losses) • We will look at: • Stop-and-Wait ARQ • Go-Back N ARQ • Selective Repeat ARQ • Basic elements of ARQ: • Error-detecting code with high error coverage • ACKs (positive acknowledgments) • NAKs (negative acknowledgments) • Time-out mechanism

  11. Packet Information frame Control frame Header CRC Information packet Header CRC Control frame: ACKs Information frame Stop-and-Wait (S&W) ARQ Transmit a frame, wait for ACK Error-free packet Receiver (Process B) Transmitter (Process A) Timer set after each frame transmission

  12. (a)Frame 1 Lost Time-out Time A Frame 0 Frame 1 Frame 1 Frame 2 ACK ACK B (b) ACK Lost Time-out Time A Frame 0 Frame 1 Frame 1 Frame 2 ACK ACK ACK B Need for Sequence Numbers • In cases (a) & (b) the transmitting station A acts the same way • But in case (b) the receiving station B accepts Frame 1 twice Question: How is the receiver to know the second Frame is also Frame 1? • Answer: Add frame sequence number in header • Slast is sequence number of most recent transmitted frame

  13. The transmitting station A misinterprets duplicate ACKs Incorrectly assumes second ACK acknowledges Frame 1 Question: How A knows that the second ACK is for Frame 0? Answer: Add frame sequence number in ACK header Rnext is sequence number of next frame expected by the receiver Implicitly acknowledges receipt of all prior frames Time-out Time A Frame 0 Frame 0 Frame 2 Frame 1 ACK ACK B Sequence Numbers (c) Premature Time-out

  14. 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 Rnext Slast Timer Slast Receiver B Transmitter A Rnext 1-Bit Sequence Numbering Suffices Global State: (Slast, Rnext) Error-free Frame 0 arrives at receiver (0,0) (0,1) ACK for Frame 0 arrives at transmitter ACK for Frame 1 arrives at transmitter Error-free Frame 1 arrives at receiver (1,0) (1,1)

  15. Transmitter Ready state Await request from higher layer for packet transfer When request arrives, transmit frame with updated Slast and CRC Go to Wait State Wait state Wait for ACK or timer to expire If timeout expires: retransmit frame and reset timer If ACK received: If sequence number is incorrect or if errors detected: ignore ACK If sequence number is correct (Rnext = Slast +1) and no errors: accept frame, go to Ready state Receiver Always in Ready State Wait for arrival of new frame When frame arrives, check for errors If no errors detected and sequence number is correct (Slast=Rnext), then accept frame, update Rnext, send ACK frame with Rnext, deliver packet to higher layer. If no errors detected and wrong sequence number discard frame send ACK frame with Rnext.. If errors detected discard frame (send ACK frame with Rnext) or (let timer to expire). Stop-and-Wait ARQ

  16. Last frame bit enters channel ACK arrives First frame bit enters channel Channel idle while transmitter waits for ACK t A B t Receiver processes frame and prepares ACK First frame bit arrives at receiver Last frame bit arrives at receiver Stop-and-Wait Efficiency • 10,000 bit frame @ 1 Mbps takes 10 ms to transmit • If wait for ACK = 1 ms, then efficiency = 10/11= 91% • If wait for ACK = 20 ms, then efficiency =10/30 = 33%

  17. t0 = total time to transmit 1 frame A tproc B frame tftime tprop tprop tproc tack Stop-and-Wait Model bits/info frame bits/ACK frame channel transmission rate

  18. S&W Efficiency in Error-Free Channel Bits for header & CRC Effective transmission rate: Transmission efficiency: Effect of frame overhead& CRC Effect of ACK frame Effect of Delay-Bandwidth Product or Effect of Reaction Time

  19. Example: Impact of Delay-Bandwidth Product nf=1250 bytes = 10000 bits, na=no=25 bytes = 200 bits Stop-and-Wait does not work well for very high speeds or long propagation delays.

  20. Effect of frame loss S&W Efficiency in Channel with Errors • Let 1 – Pf = probability frame arrives without errors • Avg. # of transmissions to first correct arrival is then 1/ (1–Pf ), if frame transmission errors are independent. • “If 1-in-10 gets through without error, then avg. 10 tries to success.” • S&W on average requires t0 / (1 – Pf) seconds to get a frame through.

  21. Example: Impact Bit Error Rate nf=1250 bytes = 10000 bits, na=no=25 bytes = 200 bits Find efficiency for random bit errors with p=0, 10-6, 10-5, 10-4 Bit errors impact performance as nfp approaches 1

  22. Applications of S&W ARQ • IBM Binary Synchronous Communications protocol (Bisync): character-oriented data link control. It has been replaced by DLC based on HDLC. • Xmodem: modem file transfer protocol. • Trivial File Transfer Protocol (RFC 1350): simple protocol for file transfer over UDP.

  23. Go-Back-N ARQ • Improve Stop-and-Wait by not waiting! • Keep channel busy by continuing to send frames • Allow a window of up to Ws (=N) outstanding frames • Use m-bit sequence numbering • If ACK for oldest frame arrives before window is exhausted, we can continue transmitting • If window is exhausted and ACK not received, pull back and retransmit all outstanding frames • Alternative: Use time-out

  24. 4 frames are outstanding (Ws=4); so go back 4 Go-Back-4: Time fr 0 fr 1 fr 2 fr 3 fr 4 fr 5 fr 3 fr 4 fr 5 fr 6 fr 6 fr 7 fr 8 fr 9 A B out of sequence frames ACK1 ACK2 ACK4 ACK5 ACK3 ACK7 ACK6 ACK9 ACK8 Rnext 0 1 2 3 3 4 5 6 7 8 9 Go-Back-N ARQ • Frame transmission are pipelined to keep the channel busy • Frame with errors and subsequent out-of-sequence frames are ignored • Transmitter is forced to go back when window of 4 is exhausted

  25. Window size long enough to cover round trip time Time-out expires Stop-and-Wait ARQ Time fr 1 fr 0 fr 0 A B ACK1 Receiver is looking for Rnext=0 Four frames are outstanding; so go back 4 Go-Back-N ARQ fr 0 fr 1 fr 2 fr 3 fr 0 fr 1 fr 2 fr 3 fr 4 fr 5 fr 6 Time A B ACK1 ACK5 ACK2 ACK6 ACK4 ACK3 Receiver is looking for Rnext=0 Out-of-sequence frames

  26. Go-Back-N with Timeout • Problem with Go-Back-N as presented: • If frame is lost and source does not have frame to send, then window will not be exhausted and recovery will not commence • Use a timeout with each frame • When timeout expires, resend all outstanding frames

  27. Receiver Transmitter Send Window Receive Window ... Frames transmitted and ACKed Slast Srecent Slast+Ws-1 Frames received Buffers Rnext oldest un-ACKed frame Slast Timer Slast+1 Timer ... most recent transmission Srecent Timer ... max Seq # allowed Slast+Ws-1 Go-Back-N Transmitter & Receiver Receiver will only accept a frame that is error-free and that has sequence number Rnext When such frame arrives Rnext is incremented by one, so the receive window slides forward by one

  28. Transmitter Send Window m-bit Sequence Numbering ... Frames transmitted and ACKed 0 Slast 2m – 1 1 Srecent Slast+Ws-1 2 Slast send window i i + 1 i + Ws – 1 Sliding Window Operation Transmitter waits for error-free ACK frame with sequence number Slast When such ACK frame arrives, Slast is incremented by one, and the send window slides forward by one

  29. M = 22 = 4, Go-Back-3: Transmitter goes back 3 fr 0 fr 1 fr 0 fr 2 fr 2 fr 1 Time A ACK2 ACK3 ACK1 B Receiver has Rnext= 3 , so it rejects the old frame 0 Rnext 0 1 2 3 Maximum Allowable Window Size is Ws = 2m-1 Transmitter goes back 4 M=22 = 4, Go-Back - 4: fr 0 fr 2 fr 3 fr 1 fr 1 fr 2 fr 3 Time fr 0 A B ACK1 ACK 0 ACK2 ACK3 Receiver has Rnext= 0, but it does not know whether its ACK for frame 0 was received, so it does not know whether this is the old frame 0 or a new frame 0 Rnext 0 1 2 30

  30. SArecent RAnext Receiver Transmitter Transmitter Receiver SBrecent RBnext “A” Receive Window “B” Receive Window RAnext RBnext “A” Send Window “B” Send Window ... ... SAlast SBlast SAlast+WAs-1 SBlast+WBs-1 Buffers Buffers SAlast SBlast Timer Timer Note: Out-of-sequence error-free frames discarded after Rnext examined SAlast+1 SBlast+1 Timer Timer ... ... SArecent Timer Timer SBrecent ... ... SAlast+WAs-1 SBlast+WBs-1 Timer Timer ACK Piggybacking in Bidirectional GBN

  31. tout tproc tprop tf tprop tf Required Timeout & Window Size(for ACK piggybacking) • The timeout value should allow for two propagation times +1 processing time (2 tprop + tproc), the frame that begins transmission right before our frame arrives tf, and the next frame which carries the ACK, tf. tout= 2 tprop + tproc+2tf • The window size Ws should be large enough to keep the channel busy for tout. Assumption: a frame begins transmission right before our frame arrives, and the next frame carries the ACK.

  32. Required Window Size (Ws) forDelay-Bandwidth Product (we use control frames )

  33. Efficiency of Go-Back-N • GBN is completely efficient, if Ws large enough to keep channel busy, and if channel is error-free • Assume Pfframe loss probability. With probability Pf the first frame is in error, and so, additional retransmissions are required. Each time a retransmission is required, Go-Back N transmits Ws frames, each of duration tf = nf /R, and the average number of retransmissions is 1/(1- Pf ). The total average time required to transmit a frame in GBN, and the Efficiency are respectively given as: Delay-bandwidth product determines Ws

  34. Example: Impact of the Bit Error Rate on GBN nf =1250 bytes = 10,000 bits, na=no=25 bytes = 200 bits Compare S&W with GBN efficiency for random bit errors with p = 0, 10-6, 10-5, 10-4 and R = 1 Mbps & 100 ms 1 Mbps x 100 ms = 100,000 bits = 10 frames→ Use Ws= 11 • GBN significant improvement over S&W for large delay-bandwidth product. • GBN becomes inefficient as error rate increases.

  35. Applications of Go-Back-N ARQ • HDLC (High-Level Data Link Control): • Variants: • Link Access Procedure-Data (LAPD) , in Integrated Services Digital Network (ISDN). • Mobile Data Link Protocol (MDLP) in cellular digital packet data systems. • V.42 Modem: error control over telephone modem links

  36. Selective Repeat (SR) ARQ • In channels with high error rate, GBN protocol is inefficient because of the need to retransmit the frame in error and all subsequent frames. • Selective Repeat (SR) retransmits only an individual frame • Timeout causes individual corresponding frame to be resent. • NAK causes retransmission of oldest un-acked frame. • Receiver maintains a receive window (=WR) of sequence numbers that can be accepted • Error-free, but out-of-sequence frames with sequence numbers within the receive window are buffered. • Arrival of frame with Rnext causes window to slide forward by 1 or more.

  37. fr 1 fr 2 fr 0 fr 3 fr 4 fr 5 fr 2 fr 7 fr 8 fr 11 fr 6 fr 9 fr 10 fr 12 Time A B ACK2 NAK2 ACK2 ACK2 ACK7 ACK8 ACK9 ACK10 ACK11 ACK12 ACK1 ACK2 Selective Repeat ARQ

  38. Receiver Transmitter Receive Window Send Window ... Frames transmitted and ACKed Frames received Rnext Rnext + Wr-1 Slast Srecent Slast+ Ws-1 Buffers Buffers Slast Rnext+ 1 Timer Slast+ 1 Rnext+ 2 Timer ... ... Srecent Timer max seq # accepted Rnext+ Wr- 1 ... Slast+ Ws - 1 Selective Repeat ARQ

  39. 0 0 2m-1 1 2m-1 1 2 2 Rnext Slast j receive window send window i j i + 1 i + Ws– 1 j + Wr– 1 Moves k forward when ACK arrives withRnext = Slast + k k = 1, …, Ws-1 Moves forward by 1 or more when frame arrives with Seq. # = Rnext Send & Receive Windows Transmitter Receiver

  40. Frame 0 resent fr2 fr0 fr0 fr1 A Time B ACK2 ACK1 ACK3 Receive Window {1,2,3} {0,1,2} {2,3,0} {3,0,1} Old frame 0 accepted as a new frame because it falls in the receive window What size Ws and Wr allowed? • Example: M=22=4, Ws=3, Wr=3

  41. Frame 0 resent fr0 fr0 fr1 A Time B ACK2 ACK1 Receive Window {0,1} {1,2} {2,3} Old frame 0 rejected because it falls outside the receive window Ws + Wr = 2mis maximum allowed • Example: M=22=4, Ws=2, Wr=2

  42. Transmitter sends frames 0 to Ws-1; send window empty All arrive at receiver All ACKs lost Window slides forward to {Ws,…,Ws+Wr-1} Why Ws + Wr = 2m works ? • Receiver window : {0, …, Wr} • Receiver rejects frame 0 because it is outside receive window • Transmitter resends frame 0 0 0 2m-1 1 2m-1 1 Ws +Wr-1 Slast 2 2 Rnext receive window Ws send window Ws-1

  43. Applications of SR ARQ • TCP (Transmission Control Protocol) : transport layer protocol uses variation of selective repeat to provide reliable stream service. • SSCOP (Service Specific Connection Oriented Protocol) : error control for signaling messages in ATM networks. This was originally designed for high-speed satellite links (large delay-bandwidth product). It is suitable to ATM networks, which also exhibit large delay-bandwidth product.

  44. Decreases with the frame loss probability Efficiency of Selective Repeat • Assume Pf theframe loss probability. Then, the number of transmissions required to deliver a frame is 1/ (1-Pf) and the average transmission time is tf / (1-Pf). Note that tf =nf /R. • The efficiency of the SR ARQ is:

  45. Comparison of ARQ Efficiencies Assume na and noare negligible relative to nf, and L = 2(tprop+tproc)R/nf=(Ws-1), then Selective-Repeat: For Pf≈0, SR & GBN same Go-Back-N: For Pf→1, GBN & SW same Stop-and-Wait:

  46. Example: Impact of the Bit Error Rate on SR nf=1250 bytes = 10,000 bits, na=no=25 bytes = 200 bits Compare S&W, GBN & SR efficiency for random bit errors with p=0, 10-6, 10-5, 10-4 and R= 1 Mbps & 100 ms, Ws=11 • Selective Repeat outperforms GBN and S&W, but efficiency drops as error rate increases

  47. ARQ Efficiency Comparison Selective Repeat 1.5 Go Back N L=10 1 Efficiency Stop and Wait 0.5 L=10 0 Go Back N L=100 -9 -8 -7 -6 -5 -4 -3 -2 -1 10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 Stop and Wait - log(p) p L=100 Delay-Bandwidth Product = 10, 100 times, and nf=10,000 bits, i.e., L=10,100 ARQ Efficiencies

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