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[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
[13] Nup 07 5
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[13] Nup 07 5

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  • 1. Protokolle der OSI-Schicht 2 Error Control, Link Layer Control Kapitel 7.5 Netze und Protokolle Dr.-Ing. J. Steuer Institut für Kommunikationstechnik www.ikt.uni-hannover.de
  • 2. FEC vs ARQ FEC, Forward Error Correction, see Musmann, Kaiser errors must be detected and corrected high amount of redundant information applied if application is time critical and/or error rate is high ARQ, Automatic Repeat Request errors need to detected only less amount of redundant information applied with low error rate and sufficient time to apply for data repetition repetition is initiated by a negative acknowledgement or a missing acknowledgement and a time out (2)
  • 3. Methods for error detection & error correction one dimensional parity bit two dimensional parity bit codes convolutional codes BCH codes Reed Solomon codes For details refer to lectures of Prof. Musmann and Prof. Kaiser (3)
  • 4. Verletzung des OSI-Modells Hybride ARQ Typ2 Verfahren (Kontrolle Schicht 2, Daten der Schicht 1) 2 2 1 1 1.) Übertragung von gering geschützten Daten 2.) Nach Fehler Sendung weiterer Redundanz auf Schicht 1 (4)
  • 5. Process of error detection in L2 network layer network layer (3) (3) link layer (2): link layer (2): 1.take info from layer 3 1.in case of positive ack hand 2.calculate redundant over L3 to L7 and user info information, using an to layer 3, otherwise wait for appropriate coding repetition scheme 4. performe FEC or give positive 3.add redundant ar negative acknowledgment information to the 3.compare calculated redundance layer 3-info with received redundance 4. continue with other 2. calculate redundant info from tasks of the link layer L3 to L7 and user info 5. hand over enhanced 1.seperate L3 to L7 anduser info information to layer 1 network physical layer physical layer (1) (1) L1 L3 L2:Redundance, others L4 L5 L6 L7 user info (5)
  • 6. ARQ-Protocols differentiation: positive/negative acknowledgment (ack/nack) positive acknowledgment only, (be careful:timing) stop & wait ARQ continous ARQ pigy-packed ARQ question: would a pure negative ack be a sensible solution? (6)
  • 7. ARQ-Protocols, Stop and Wait using positive and negative acknowledgement or positive acknowledgment and timeout: frame n-1 frame n-1 positive positive acknowledg- acknowledgement ment frame n frame n corruption timeout corruption negative acknowledg- repetition of ment frame n repetiton of frame n time time remark: the ACK could be lost as well (7)
  • 8. ARQ-Protocols, Stop and Wait with lost frames (packets) frame n-1 frame n-1 positive positive acknowledg- acknowledgement ment frame n frame n timeout lost timeout lost repetiton of repetition of frame n frame n time time Question: Is it necessary to distinguish frame n from the repeated frame n at the receiver? (8)
  • 9. ARQ-Protocols, Stop and Wait with lost frames (packets) and Enquiry frame n-1 frame n-1 positive positive acknowledg- acknowledgement ment frame n frame n positive timeout lost timeout acknowledg- ment lost ENQ ENQ repetition of positive repetition acknowledgment of positive acknowledg- time ment time • Why not just repeating the frame which is not acknowledged? • how to differentiate which ACK is received? The transmitter does not know which frame is lost, its own or the ACK! The way out is the odd and even numbering of the ACK´s! (9)
  • 10. ARQ-Protocols, Continuous stop and wait is time consuming, especially on links with high propagation delay time saving: continuous transmission without waiting for ACK problem: propagation delay might result in late information from the transmitter on transmission errors, consequently a number of packets need to be retransmitted solution: sliding window mechanism (10)
  • 11. Continuous ARQ-Protocols for full duplex channels frame 1 frame 1 frame 2 ACK 1 frame 2 ACK 1 frame 3 frame 3 ACK 2 ACK 2 frame 4 frame 4 NACK 3 NACK 3 frame 5 frame 5 ACK 4 frame 3 frame 3 ACK 5 frame 4 ACK 3 ACK 3 frame 5 ACK 4 ACK 5 Go back n selective time time repeat Go back n is easier to implement, selective repeat is more efficient! (11)
  • 12. piggy packing of ACK frame 1 frame 2 frame 1,ACK 1 frame 3 frame2,ACK 2 frame 4, ACK1 frame3, NACK 3 frame 5,ACK2 frame 3,ACK3 frame 4 frame4, ACK 3 frame 5 frame5, ACK 4 ACK 4 frame6, ACK 5 ACK 5 ACK 6 time Go back n •The acknoledgement is partly transported with communication frames •Sequence count required! (12)
  • 13. Communications Efficiency of ARQ- stop&wait (1) Performance of DLC − Pr otocols requires consideration of retransmissions : def . : Re := effective data transmission rate number of user data bits accepted by the destination Re = Total time for accep tan ce K − nh Re = (6.5.1) K + na + 2(t p + tta )) N( R def . : K := total bits per frame nh := total bits per header in a frame N := average number of attempts made to get accep tan ce na := total bits per acknowledgment R := transmission bit rate t p := propagation time on the link tta := turn around time to switch the direction of communication (13)
  • 14. Communications Efficiency of ARQ- stop&wait (2) Next we have to solve the average number of transmissions N: • consider P to be the probability of an error on any frame transmission • errors on different frames are independent • j transmissions will be required to be successful, this means we need j-1 erroneous transmissions followed by one successful: Pj = P j −1 (1 − P ) ( 6 .5 .2 ) • the average of j is the average number of transmissions N: ∞ N = ∑ jP j −1 (1 − P ) = (1 − P )[1 + 2 P + 3P 2 + ... + nP ∞ −1 ] (6.5.3) j =1 • with Bronstein 1.61: 1 − (∞ + 1) P ∞ + ∞P ∞ +1 1 ∞ −1 [1 + 2 P + 3P + ... + nP ] = = 2 (6.5.4) (1 − P ) (1 − P ) 2 2 (1 − P ) 1 No, we still miss the (6.5.5) N= = (1 − P ) 1 − P 2 Ready? frame error rate P (14)
  • 15. Communications Efficiency of ARQ- stop&wait (3) To get the average number of transmissions we have to find the probability for disturbed frames P. The most common approach is to use the bit error rate p and the number of bits per frame K (bit errors shall be independent of each other): P = 1 − (1 − p ) K ( 6 .5 .6 ) Which effect do we neglect with the given approach above? considering p as burst error rate and combining 6.5.6 with 6.5.1 we get: K − nh K − nh Re = = K + na 1 K + na + 2(t p + tta )) + 2(t p + tta )) N( ( 1− P R R (1 − p ) K R ( K − nh ) Re = ( 6 .5 .7 ) K + na + 2 R (t p + t a ) (15)
  • 16. R:=4800bit/s; nh:=48 bit; (1 − p ) R ( K − nh ) K ta:=0.1s; tp:=0.01s; Re = note: K + na + 2 R (t p + t a ) na:=48bit; • optimal frame length • effective frame rate drops, if too many repetitions are p1:=10^(-5); required Re:=effektive transmission rate in bit/s; K := frame length in Byte p2:=10^(-4); Which error rate do you need, in order to run IEEE802.3 efficient? (16)
  • 17. Communications Efficiency of error free continuous ARQ in an error free environment the time for acceptance is just the serialization time: K/R def . : Re := effective data transmission rate number of user data bits accepted by the destination Re = Total time for accep tan ce K − nh Re = (6.5.8) K R def . : K := total bits per frame nh := total bits per header in a frame time R := transmission bit rate (17)
  • 18. Communications Efficiency for continuous ARQ and Go Back n (1) each corrupted frame will cause a step of n frames backwards. The first frame can be correct. This means we transmit one frame to be successful. This causes an average of one frame to be transmitted. The first frame can be corrupted and request a repetition of n frames. The probability for that is: P (1 − P ) with the average transmitted frames : n * P (1 − P ) • The second frame can be corrupted as well and request a repetition of 2n frames. The probability for that is: P 2 (1 − P ) with the average transmitted frames : 2n * P 2 (1 − P ) • The third frame can be corrupted as well and request a repetition of 3n frames. The probability for that is: P 3 (1 − P ) with the average transmitted frames : 3n * P 3 (1 − P ) (18)
  • 19. Communications Efficiency for continuous ARQ and Go Back n (2) • The average number N of frames to be transmitted is the sum of the average frames of the individual cases from the previous slide: N = 1 + n * P (1 − P ) + 2n * P 2 (1 − P ) + 3n * P 3 (1 − P ) + ... 6.5.9 • again using Bronstein 1.61: 1 + (n − 1) P P N = 1+ n * = 6.5.10 (1 − P ) 1− P • if we multiply now the average transmitted frames with the time to serialize one frame, we get the „total time to get the bits accepted“ and thus the effective transmission rate: ( K − nh ) (1 − P)( K − nh ) R (1 − [1 − (1 − p ) K ])( K − nh ) R Re = = = 6.5.11 1 + (n − 1) P K [1 + (n − 1) P ]K [1 + (n − 1)[1 − (1 − p ) K ]]K 1− P R (19)
  • 20. R:=4800bit/s; (1 − [1 − (1 − p ) ])( K − nh ) R K nh:=48 bit; Re = [1 + (n − 1)[1 − (1 − p ) K ]]K n:=5 p1:=10^(-5); Go Back n p1:=10^(-5); stop and wait p2:=10^(-4); Go Back n p2:=10^(-4); stop and wait (20)
  • 21. Fenstermechanismus Sliding Window unterer Fensterrand (hier:0) unterer Fensterrand: Sendelaufnummer des ältesten noch nicht bestätigten Datenpaketes 0 7 w=3 1 oberer Fensterrand: 6 Sendelaufnummer des ersten noch nicht zur 2 5 Sendung freigegebenen 4 3 Datenpaketes oberer Fensterrand (hier:3) (21)
  • 22. Sliding Window direction A direction B lower window edge P0 (here:0) P1 P0 P2 P1 P0 P2 P1 Q0 0 7 w=3 1 P2 Q0 Q1 6 Q0 Q1 Q2 2 5 P3 Q1 Q2 4 3 P4 P3 Q2 upper window edge P5 P4 P3 (here:3) P5 P4 Q3 1 P5 Q3 Q4 time time Q3 Q4 Q5 Q4 Q5 (22)
  • 23. demonstration of the sliding window one packet two packets 7 packets under under under window window window control control control (23)
  • 24. sliding window mechanismn (see notes) (24)
  • 25. performance of the sliding window -Acknowledgment at end of window- (1) Tt : time to serialize τ : propagation time Tt τ Tp : time to serialize the Acknowledgment WTt fram e1 fram e2 fram The delay d from the start of transmission d e3 until the complete receipt of the acknowledgment is Tp Ack1 d = W * Tt + Tp + 2τ 6.5.12 fram e1 fram e2 with the data rate R we get the ratio: time fram (efficient data rate Re): e3 Re W * Tt = window size: W=3 6.5.13 R d (25)
  • 26. performance of the sliding window -Acknowledgment at end of window- (2) combining equation 6.5.12 with 6.5.13: W * Tt Re = R 6.5.14 W * Tt + Tp + 2τ the packet length is: K= R *Tt 6.5.15 6.5.15 in 6.5.14: K W* R Re = R 6.5.16 K W * + Tp + 2τ R W *K Re = 6.5.15 W *K + Tp + 2τ R (26)
  • 27. performance of the sliding window -Acknowledgment at end of window- (3) W *K Re = 6.5.15 W *K + Tp + 2τ R W=5 parameters: W=3 W=1 R=4800bit/sek Tp=20ms τ =20ms (27)
  • 28. performance of the sliding window -Acknowledgment after receipt of frame-(1) Tt : time to serialize τ : propagation time Tt τ Tp : time to serialize the Acknowledgment d WTt fram e1 The delay d from the start of transmission fram e2 until the complete receipt of the acknowledgment is: fram Ack1 Tp e3 d = Tt + Tp + 2τ 6.5.16 fram e4 with the data rate R we get the ratio: fram e5 (efficient data rate Re): fram Ack1 e6 ⎧W * Tt ⎫ Re = min ⎨ ,1⎬ 6.5.17 time ⎩d ⎭ fram R e7 Equ. 6.5.17 this expresses the continous data flow window size: W=3 for: W * Tt ≥ d 6.5.18 (28)
  • 29. performance of the sliding window -Acknowledgment after receipt of frame- vs. - Acknowledgment at end of window- W=5 W=3 W=5 parameters: W=3 R=4800bit/sek Tp=20ms W=1 τ =20ms ⎧W * Tt ⎫ Re = min ⎨ ,1⎬ 6.5.17 ⎩d ⎭ R (29)
  • 30. Fragen (2) Wieviele Pakete können maximal ins Netz geschickt werden, bevor ein acknowledge (Bestätigung) für den Empfang kommen muß? Wie können N(R) und N(S) benutzt werden, um eine Fehlerfall-Steuerung zu realisieren? Auf welchen Leitungen werden Sie die Fenstergröße groß und auf welchen klein machen? (30)
  • 31. Ab hier nicht mehr drucken (31)
  • 32. ARQ-Protocols differentiation: positive/negative acknowledgment positive acknowledgment only, (be careful:timing) question: would a pure negative ack be a sensible solution? Could be done, but it takes time to wait for the timer! (32)
  • 33. ARQ-Protocols, Stop and Wait with lost frames (packets) frame n-1 frame n-1 positive positive acknowledg- acknowledgement timeout timeout ment frame n frame n lost lost repetiton of repetition of time time frame n frame n Question: Is it necessary to distinguish frame n from the repeated frame n at the receiver? answer: Of course it is. The receiver does not know the content of the frame. The content should not be delivered twice at layer 3. next Question: Propose the most simple method to differentiate both. answer: add simply one bit to the header, which is set to one for the original message and zero for the repetition. When does this method fail? (33)
  • 34. Communications Efficiency of ARQ-Protocols (3) To get the average number of transmissions we have to find the probability for disturbed frames P. The most common approach is to use the bit error rate p and the number of bits per frame K (bit errors shall be independent of each other): P = 1 − (1 − p ) K ( 6 .5 .6 ) Which effect do we neglect with the given approach above? The error probability for the bits is (depending on the transmission media) is often not independent of each other. Very often, especially in radio systems we have burst errors. In such cases it is more realistic to consider burst errors as independent rather than the bit errors! An approximation from experience is that the burst error rate is 1/3 to 1/10 of the bit error rate. (34)
  • 35. R:=4800bit/s; nh:=48 bit; (1 − p ) R ( K − nh ) K ta:=0.1s; tp:=0.01s; Re = note: K + na + 2 R (t p + t a ) na:=48bit; • optimal frame length • effective frame rate p1:=10^(-5); drops, if too many repetitions are required Re:=effektive transmission rate in bit/s; p2:=10^(-4); K := frame length in Byte Which error rate do you need, in order to run IEEE802.3 efficient? Better Than p1, because the max. Frame length is about 1500 octetts; for that using p2 Re is already down. (35)

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