1
Automatic Repeat Request 2 The Problem
- data in data out → → Tx - Rx -
Transmission of data requires series of PDUs to be sent PDU may be lost
PDU may be corrupted 3 Automatic Repeat Request
I - N ACK N data in data out → I - → N+1 Tx Rx NAK N+1
I - N+1 ACK N+1 automatic repeat request (ARQ) Data link layer (transport layer, application layer, . . . ) Number PDUs, and add error checking Data in I-frames, Acknowledgement in ACK-frames 4 Two Approaches to ARQ idle RQ Tx waits after each I(N) until it receives ACK(N) or NAK(N) or times out also called stop-and-wait or synchronous continuous RQ Tx sends continuous stream of I-frames can send I(N+1) before receiving ACK(N) also called asynchronous 5 Idle RQ: Tx and Rx Actions
T=N =N+1
I Tx N
⇒ time
Rx
R=N 6 Idle RQ: Tx and Rx Actions
T=N =N+1
I Tx N
⇒ time
N I Rx N
R=N =N+1 7 Idle RQ: Tx and Rx Actions
T=N =N+1
I Tx N
⇒ time
N I ACK Rx N N
R=N =N+1 8 Idle RQ: Tx and Rx Actions
T=N =N+1
I ACK Tx N N
⇒ time
N I ACK Rx N N
R=N =N+1 9 Idle RQ: Tx and Rx Actions
T=N =N+1 =N+2
I ACK I Tx N N N+1
⇒ time
N I ACK Rx N N
R=N =N+1 10 Idle RQ: Tx and Rx Actions
T=N =N+1 =N+2 =N+3
I ACK I ACK I Tx N N N+1 N+1 N+2
⇒ time
N N N I ACK I ACK I ACK Rx N N N+1 N+1 N+2 N+2
R=N =N+1 =N+2 =N+3 11 Worksheet: Automatic Repeat Request t ? Tx
Rx 12 Idle RQ: Stop-and-wait efficiency
T R -
I ACK I ACK I Tx N N N+1 N+1 N+2
⇒ time
N N N I ACK I ACK I ACK Rx N N N+1 N+1 N+2 N+2 - ---- TP TI TD TA TP
TR = TI + 2TP + TD + TA
TI ρs = TI + 2TP + TD + TA 13 Idle RQ: Improving performance
TI ρs = TI + 2TP + TD + TA
Increase ρs?
reduce TP by decreasing the distance between Tx and Rx usually not possible
increase TI by decreasing the bit rate would not want to do this tells us Idle RQ more efficient on slower channels
increase TI by increasing the number of bits per I-frame larger I-frames more likely to suffer errors 14 Idle RQ: Example Protocol Efficiency Calculation
Compute the maximum data throughput of Idle RQ 10Mbs−1communications system I-frame size is 92 bits and the ACK-frame size is 8 bits via a satellite in orbit at 37,500km above the Earth’s surface
propagation speed 300 × 106ms−1
92 −6 TI = 10×106 = 9.2 × 10 s 8 −6 TA = 10×106 = 0.8 × 10 s assume TD = 0 2×37.5×106 TP = 300×106 = 0.25s 9.2×10−6 −6 ρs = 9.2×10−5+2×0.25+0.8×10−6 = 18.4 × 10 −6 −1 −1 throughput = ρsB = 18.4 × 10 × 10Mbs = 184bs 15 Overall Efficiency of Idle RQ
Three factors cause the raw bit-rate of a channel to be wasted (i=no I-frame bits, c=control bits in I-frame, a=no ACK-frame bits)
Wastage due to stop-and-wait protocol
TI ρs = TI + 2TP + TA + TD Wastage due to errors forcing retransmission 1 ρ = (1 − BER)i+a ρ ≈ e e 1+ p
Wastage due to control information
i − c ρ = c i + a
Overall efficiency ρi = ρsρeρc 16 Worksheet: Overall Efficiency of Idle RQ
I-frames of 10000 bits; hosts separated by 500km over a 1Mbs−1 link. The I-frames consist of 1000 bits of header and 9000 bits of data. The signal propagation speed is 200 × 106ms−1, and the BER is 10−5 1. Calculate the probability that an I-frame is lost due to an error.
2. What is the error efficiency of the system ρe?
3. What is the stop-and-wait efficiency of the system ρs?
4. What is the control information efficiency of the system ρc? 5. What is the overall efficiency of the Idle RQ protocol? 17 Idle RQ Error Handling: Loss of I-frame
T=N =N+1 =N+2 T O -
I ACK I I Tx N N N+1 N+1
⇒ time
N N I ACK I ACK Rx N N N+1 N+1
R=N =N+1 =N+2
Timeout TO must be set at Tx
TO >TR 18 Idle RQ Error Handling: Corrupt I-frame
T=N =N+1 =N+2
I ACK I NAK I Tx N N N+1 N+1 N+1
⇒ time
N N N I ACK I NAK I ACK Rx N N N+1 (error) N+1 N+1 N+1
R=N =N+1 =N+2
NAK-frame speeds up retransmission of I-frame Do not have to use NAK-frames 19 Idle RQ Error Handling: Loss of ACK-frame
T=N =N+1 =N+2 T O -
I ACK I I Tx N N N+1 N+1
⇒ time
N N N I ACK I ACK I ACK Rx N N N+1 N+1 N+1 N+1
R=N =N+1 =N+2
Rx receives I(N+1) twice Really do have to number I-frames! 20 Continuous RQ
I I I I I Retransmission N+1 N+2 N+3 N+4 N+5 List I I I I I I N N N+1 N+2 N+3 N+4 T=N =N+1 =N+2 =N+3 =N+4 =N+5 =N+6
I I I I I I I Tx N N+1 N+2 N+3 N+4 N+5 N+6 ⇒ ACK ACK ACK ACK ACK time N N+1 N+2 N+3 N+4 ^ ^ ^ ^ ^ ^ I I I I I I Rx N N+1 N+2 N+3 N+4 N+5 R=N =N+1 =N+2 =N+3 =N+4 =N+5 =N+6 21 Go-Back-N: Loss of I-frame
I I N+4 N+4 I I I I I Retransmission N+1 N+2 N+3 N+3 N+3 List I I I I I I N N N+1 N+2 N+2 N+2 T=N =N+1 =N+2 =N+3 =N+4 =N+5 =N+5
I I I I I I I Tx N N+1 N+2 N+3 N+4 N+2 N+3 ⇒ ACK ACK NAK time N N+1 N+2 ^ ^ ^ ^ ^ I I I I I Rx N N+1 N+3 (ignore) N+4 (ignore) N+2 R=N =N+1 =N+2 =N+2 =N+2 =N+2 =N+3 22 Go-Back-N: Loss of NAK-frame
I N+5 I I N+4 N+4 I I I I I Retransmission N+1 N+2 N+3 N+3 N+3 List I I I I I I N N N+1 N+2 N+2 N+2 T=N =N+1 =N+2 =N+3 =N+4 =N+5 =N+5
I I I I I I I Tx N N+1 N+2 N+3 N+4 N+5 N+2 ⇒ ACK ACK NAK time N N+1 N+2 ^ ^ ^ ^ ^ I I I I I Rx N N+1 N+3 (ignore) N+4 (ignore) N+5 (ignore) R=N =N+1 =N+2 =N+2 =N+2 =N+2 =N+2 23 Idle RQ: Rough Formula for Efficiency r TR = number for frames sent in T = ⌈ TI ⌉ R p= probability frame suffers error error efficiency:
1 ρ ≈ e 1+ p protocol efficiency:
1 ρ = s r combined efficiency:
1 ρ = r + rp 24 Go-back-N: Efficiency
ρs = 1 send r TR I-frames for each erroneous I-frame = ⌈ TI ⌉ 1 ρ ≈ e 1+ pr
ρc same as idle RQ 25 Example: Continuous RQ Go-Back-N protocols
If Continuous RQ Go-Back-N is being used, calculate the efficiency and through- put of a 10Mbs−1 satellite communication system with BER of 10−6, a satellite height of 37,500km, propogation speed 300 × 106ms−1, I-frames 100 bits. p = 1 − 0.9999 = 10−4 100 −5 TI = 10×106 = 10 s 2×37.5×106 TP = 300×106 = 0.25s assume TD = 0,TA = 0 −5 TR = 10 + 2 × 0.25 = 0.5s 0.5 3 r = 10−5 = 50 × 10 1 1 ρg = 1+10−4×50×103 = 6 = 0.17 throughput = 1.7 Mbs−1 26 Worksheet: Efficiency of Continuous RQ Go-Back-N
Go-Back-N system uses I-frames of 10000 bits; hosts separated by 500km over a 10Mbs−1 link. Signal propagation speed is 200 × 106ms−1. 1. If no errors occur, what is the throughput we can expect on the link?
2. If the BER is 10−5, what is the probability of any one frame being lost? 3. What is the throughput with this BER?
4. What would have been the throughput if Idle RQ had been used? 27 Go-Back-N: Efficiency
1 r = 1 r = 10 0.8 r = 100 r = 1000
0.6 ρe 0.4
0.2
0 0 0.02 0.04 0.06 0.08 0.1 p 28 Selective Repeat: Explicit Request
I N+4 I I I I I Retransmission N+1 N+2 N+3 N+3 N+4 List I I I I I I N N N+1 N+2 N+2 N+2 T=N =N+1 =N+2 =N+3 =N+4 =N+5 =N+5
I I I I I I I Tx N N+1 N+2 N+3 N+4 N+2 N+5 ⇒ ACK ACK NAK ACK ACK time N N+1 N+2 N+3 N+4 ^ ^ ^ ^ ^ I I I I I Rx N N+1 N+3 N+4 N+2 R=N =N+1 =N+2 =N+2 =N+2 =N+2 =N+5
I I Receiver N+3 N+4 Buffer I N+3 29 Selective Repeat: Implicit Retransmission
I I N+4 N+5 I I I I I Retransmission N+1 N+2 N+3 N+3 N+4 List I I I I I I N N N+1 N+2 N+2 N+2 T=N =N+1 =N+2 =N+3 =N+4 =N+5 =N+6
I I I I I I I Tx N N+1 N+2 N+3 N+4 N+5 N+2 ⇒ ACK ACK ACK ACK time N N+1 N+3 N+4 ^ ^ ^ ^ ^ I I I I I Rx N N+1 N+3 N+4 N+5 R=N =N+1 =N+2 =N+2 =N+2 =N+2 =N+2
I I Receiver N+3 N+4 Buffer I N+3 30 Selective Repeat: efficiency protocol efficieny ρs = 1 error efficiency as idle request
1 ρ ≈ e 1+ p
ρc same as idle RQ 31 Example: Continuous RQ Selective-Repeat protocols
If Continuous RQ selective repeat is being used, calculate the efficiency and throughput of a 10Mbs−1 satellite communication system with BER of 10−6, a satellite height of 37,500km, propogation speed 300 × 106ms−1, I-frames 100 bits. p = 1 − (1 − 10−6)100 = 10−4 1 ρs = 1+10−4 = 0.9999 throughput = 10.0 Mbs−1 32 Comparison of ARQ Protocols
protocol error combined
ρs ρe ρsρe 1 1 1 Idle RQ r 1+p r+rp 1 1 Go-Back-N 1 1+rp 1+rp Selective 1 1 Repeat 1 1+p 1+p For all, control efficiency
i − c ρ = c i + a 33 Overall Efficiency for Go-Back-N
1 r = 1 r = 10 0.8
0.6 ρeρc 0.4
0.2
0 0 100 200 300 400 500 number of data bits per frame r = 1 graph corresponds to that for Selective Repeat 34 Duplex Operation
I I ACK I ACK Tx N N+1 M N+2 M+1 I I ACK I ACK Rx M M+1 N M+2 N+1 ⇒ time ^ ^ ^ I I ACK I ACK Tx M M+1 N M+2 N+1 I I ACK I ACK Rx N N+1 M N+2 M+1
Can send ACK-frames for I-frame by piggybacking it to I-frame in opposite direction 35 Idle RQ: Window Size
T=N =N+1 =N+2 =N+3
I ACK I ACK I Tx N N N+1 N+1 N+2
⇒ time
N N N I ACK I ACK I ACK Rx N N N+1 N+1 N+2 N+2
R=N =N+1 =N+2 =N+3 Tx and Rx only need buffer one I-frame 36 Continuous RQ Go-Back-N: Window Size
I I I I I Retransmission N+1 N+2 N+3 N+4 N+5 List I I I I I I N N N+1 N+2 N+3 N+4 T=N =N+1 =N+2 =N+3 =N+4 =N+5 =N+6
I I I I I I I Tx N N+1 N+2 N+3 N+4 N+5 N+6 ⇒ ACK ACK ACK ACK ACK time N N+1 N+2 N+3 N+4 ^ ^ ^ ^ ^ ^ I I I I I I Rx N N+1 N+2 N+3 N+4 N+5 R=N =N+1 =N+2 =N+3 =N+4 =N+5 =N+6
Tx must buffer r I-frames 37 Continuous RQ Selective Repeat: Window Size
I N+4 I I I I I Retransmission N+1 N+2 N+3 N+3 N+4 List I I I I I I N N N+1 N+2 N+2 N+2 T=N =N+1 =N+2 =N+3 =N+4 =N+5 =N+5
I I I I I I I Tx N N+1 N+2 N+3 N+4 N+2 N+5 ⇒ ACK ACK NAK ACK ACK time N N+1 N+2 N+3 N+4 ^ ^ ^ ^ ^ I I I I I Rx N N+1 N+3 N+4 N+2 R=N =N+1 =N+2 =N+2 =N+2 =N+2 =N+5
I I Receiver N+3 N+4 Buffer I N+3
Tx and Rx must buffer r I-frames 38 Sliding Windows
Error Send Receive S: Maximum sequence protocol window size window size number (counting from 1) Idle RQ 1 1 2 Go-back-N r 1 r + 1 Selective repeat r r 2r N does not need to be a infinite range Use modulo S