Datornätverk A – lektion 8

1. Datornätverk A – lektion 8 Kapitel 11: Flow control and Error control. (Kapitel 12: Point-to-point access PPP. Översiktligt.)

2. 11.1 Flow and Error Control Flow Control (Flödesstyrning) Error Control (Felhantering) • Båda dessa funktioner hanteras av vissa datalänkprotokoll (lager 2), i LLC-sublagret, t.ex. vid trådlös kommunikation eller vid modem. • End-to-end flödesstyrning och felkontroll hanteras av transportprotokollet TCP (lager 4).

3. Flow control Necessary when data is being sent faster than it can be processed by receiver to avoid that the receiver’s buffer is overwhelmed.

4. Felhantering med hjälp av felrättande koder FEC = Forward Error Correction. Baseras på felrättande istället för felupptäckande koder. Kräver ingen backkanal. Två typer: • Faltningskoder (convolutional codes). Ex:Vid Faltningskod med kodtakt (code rate) 1/3 infogas två redundanta bitar mellan varje bit i nyttomeddelandet. Dessa felrättande bitar beräknas kontinuerligt för varje inkommande bit i nyttomeddelandet. • Blockkoder (block codes) Ex: I digital-TV-systemet används en s.k. Read Salomon-kod med beteckningen RS(204, 188, 8). Det innebär att nyttoinformationen delas upp i 188 byte stora block. För varje block beräknas en felrättande kod, som läggs till blocket så att blocket blir 204 byte. Redundanden är alltså 204 – 188 = 16 byte. Koden klarar 8 felaktiga byte.

5. Felhantering med hjälp av felupptäckande koder Alternativ 1: Bortkastning av felaktiga paket. Alternativ 2: ARQ = Automatic Repeat reQuest = automatisk omsändning av paket vid bitfel, eller om paketet inte når fram. I fortsättning kommer vi med begreppet ”error control” eller ”felkontroll” att avse ARQ.

6. Protocols to be presented • Stop-and-wait ARQ • Sliding Window Flow Control • Go-back-N ARQ • Selective Repeat ARQ Sliding WindowProtocols

7. The Stop-and-Wait Protocol The simplest protocol for error and flow control How the protocol operates: • Source may not send a new frame until the receiver acknowledges previous one. • The receiver sends only positive acknowledgements (ACK) to notify the sender that the frame was received. • If the frame 0 was received, the ACK 1 is sent. In that way the sender is notified that the receiver is expecting frame 1. • The ID of the frame is called a sequence number. • 1 bit sequence numbers is sufficient. Sequence: 0 1 0 1 0 ... .

8. 11.1Normal operation ACK n = Acknowledgement. Expecting frame number n

9. 11.2Stop-and-Wait ARQ, lost frame

10. Lost or Damaged Frame • The sender starts a timer when it sends each frame • If the ACK is not received before the timer expires, the sender resends the same frame again

11. 11.3Stop-and-Wait ARQ, lost ACK frame

12. Lost or damaged ACK • Lost ACK causes duplicate frames • A duplicate frame is recognized by the sequence number and is discarded • The receiver sends the same ACK again

13. 11.4Stop-and-Wait ARQ, delayed ACK

14. Note: Numbered acknowledgments are needed if an acknowledgment is delayed and the next frame is lost.

15. Piggybacking • Usually the communication is in both ways – this means that the sender is a receiver and the receiver is the sender, too. (both send and receive data) • To save on the processing and bandwidth the short ACKs messages are not sent as separate frames. Instead, they may be included in the frames with data. • This technique is called piggybacking

16. 11.5Piggybacking

17. Efficiency of Stop-and-Wait • Very inefficient, having in mind that most of the time the sender is idle • Example: 40 km copper cable, 10 Mbps rate, 1000 bit frame, • Signal in copper propagates at 2 x 108 m/sec • Transmission time is 1000/10000000 (Takes 0.1 msec to transmit frame) • Propagation time is 40000/ 2 x 108 (0.2 msec delay to begin arriving at the receiver) • Total time is 0.3 msec. to get to the receiver • ACK transmission time is approximately 0 (assuming the ACK is very short (length  0) • 0.2 msec is the time for the ACK to arrive at the sender • Total time is 0.5 msec before the sender can transmit again • 0.5 ms for 0.1 msec frame or efficiency is 20%

18. Sliding-Window flow control • Several frames can be sent without acknowledgement being received • N is the window size – the maximum number of frames that can be sent and not being acknowledged. • The receiver must be able to buffer N frames. • Sequence numbers are used to identify each frame. They are carried in the header. • The number of different sequence numbers must be at least N+1. • If the field for sequence numbers allows m bits, the number of different sequence number is 2m and the sequence numbers range from 0 to 2m-1. In that case the maximum window size is N = 2m-1.

19. 11.6Sender sliding window The sender window is the the set of frames that may be transmitted before an ACK. It slides when the senderhas received an ACK and sent next frame.

20. 11.7Receiver sliding window The receiver window is the the set of frames that may be accepted before the buffer is full. While the buffer is full, the receiver sends no ACK. The window of a stuffed receiver slides when the receiver has ”consumed” a frame and thus sent an ACK.

21. Stop-and-Wait vs. Sliding Window Window size N=3. Sender Sender Receiver Receiver Transmission + propagation time for the packet Frame 0 Frame 0 propa-gation time Transmission time for the packet Frame 1 ACK 0 Frame 2 . . . ACK 1 Frame consump-tiondelay Time Frame 1 ACK 2 Transmission + propagation time for the ACK ACK 1 Time ACK 3 Frame 0 Frame consumptiondelay ACK 0 . . . . . . Sequence numbers are 1 bit long (0 or 1) Sequence numbers from 0 to 2m-1. m-bit field for the seq. num.

22. Sender and Receiver Prospective The window size is 7

23. 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 F0 F1 ACK1 F2 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 ACK3 ACK2 F3 F4 ACK5 ACK4 F5 ACK6 Sliding Window Flow Control N=6

24. ARQ with Sliding Window Problems arise when some of the frames are discarded (errors or lost frames). Two strategies are developed to deal with this problem: • Go-back-N strategy • The reciever simply discards all frames after the damaged frame without sending acknowledgement. • Selective repeat strategy • The receiver keeps all the frames after the damaged frame. It sends negative acknowledgement (NACK) for the damaged frame. When the sender finaly notice that something is wrong it retransmits the bad frame. The two strategies are trade-offs between bandwidth and data-link buffer space.

25. Go-Back-N Strategy • If a frame is lost, the lost frame and all the frames sent after it are sent again. • Sending window of size N, receiving window of size 1. • The sender has to buffer N frames • Bandwidth is wasted.

26. 11.9Go-Back-N ARQ, normal operation

27. 11.10Go-Back-N ARQ, lost frame

28. 11.11Go-Back-N ARQ: sender window size

29. Selective Repeat Strategy • Only retransmit the frames that are in error • Both sending and receiving window are of size N

30. 11.13Selective Repeat ARQ, lost frame

31. 11.14Selective Repeat ARQ, sender window size

32. Bandwidth – Delay Product • The product of the bit rate (bandwidth expressed as bits per seconds) and the propagation time gives the number of bits that can be on the channel and thus can give orientation about the window size • When propagation time is high (for example in satellite channels), the window size need to be larger

33. Example 1 In a Stop-and-Wait ARQ system, the bandwidth of the line is 1 Mbps, and 1 bit takes 20 ms to make a round trip. What is the bandwidth-delay product? If the system data frames are 1000 bits in length, what is the utilization percentage of the link? Solution The bandwidth-delay product is 1  106 20  10-3 = 20,000 bits The system can send 20,000 bits during the time it takes for the data to go from the sender to the receiver and then back again. However, the system sends only 1000 bits. We can say that the link utilization is only 1000/20,000, or 5%. For this reason, for a link with high bandwidth or long delay, use of Stop-and-Wait ARQ wastes the capacity of the link.

34. Example 2 What is the utilization percentage of the link in Example 1 if the link uses Go-Back-N ARQ with a 15-frame sequence? Solution The bandwidth-delay product is still 20,000. The system can send up to 15 frames or 15,000 bits during a round trip. This means the utilization is 15,000/20,000, or 75 percent. Of course, if there are damaged frames, the utilization percentage is much less because frames have to be resent.

35. High-level Data Link Control Protocol • HDLC is one of the first protocols that implements mechanisms of ARQ • Supports half-duplex and full-duplex mode on point-to-point links • Uses three types of frames: information (I-frames), supervisory (S-frames) and unnumbered (U-frames) • Only I frames carry information, S frames carry transport control information and U frames are used for managing the link

36. Flag Address Control Information FCS Flag HDLC Frame Structure • Flag: 01111110, at start and end • Physical Address: secondary station (for multidrop configurations) • Information: the data to be transmitted • Frame check sequence (FCS): 16- or 32-bit CRC • Control: purpose or function of frame • Information frames: contain user data • Supervisory frames: flow/error control (ACK/ARQ) • Unnumbered frames: variety of control functions (see p.220)

37. 11.18HDLC frame types

38. The Need for Bit Stuffing • The flags show the receiver the start and the end of frame • There is a problem if the flag appears in the middle of the frame as a part of data • The receiver will ”think” it is the end of frame • A technique called “bit stuffing” is used to resolve this problem

39. Bit Stuffing • The sender stuffs redundant 0s • Every time it encounters five 1s in a row, it inserts a redundant 0 • The redundant 0 tells the receiver that the sequence is not a flag • The receiver removes all redundant 0s to restore the original frame • Example: Bit stuff the following data: 0001111111110111100011111011 000111110111101111000111110011 Redundant 0s

40. 11.24Bit stuffing and removal

41. 11.25Bit stuffing in HDLC

42. PPP (Point-to-Point Protocol) • Based upon HDLC • Used for point-to-point access • Common protocol used for connecting home users to the Internet (via dial-up, DSL or cable modem or leased line) • Defines the negotiation for establishment of the link • Defines the protocol carried on the network layer • Includes authentication and a field about the type of network protocol carried within the frame

43. PPP Frame Format Number of bytes in a field variable 1 1 1 1 1 or 2 2 or 4 Flag 01111110 Flag 01111110 Address 11111111 Control 00000011 Payload Protocol CRC • Physical Address field with all 1s indicate broadcasting, i.e. that all stations accept the frame • Since the Address and Control fields are constant, the two parties can negotiate to omit them, thus saving 2 bytes • Protocol field defines what is carried in the payload field (user data or other information) • CRC bits are error control bits