Lecture Datalink

8/6/2019 Lecture Datalink

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1

Automatic Repeat Request


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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


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3

Automatic Repeat Request

data in

data out

Tx

-

-

-

I

N

ACK

N

I

N+1

NAK

N+1

I

N+1

ACK

N+1

Rx

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


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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


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5

Idle RQ: Tx and Rx Actions

Rx

TxI

N

time

R=N

T=N =N+1


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6


Rx

TxI

N

I

NN

time

R=N =N+1

T=N =N+1


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7


Rx

TxI

N

I

NACK

NN

time

R=N =N+1

T=N =N+1


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Rx

TxI

NACK

N

I

NACK

NN

time

R=N =N+1

T=N =N+1


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Rx

TxI

NACK

NI

N+1

I

NACK

NN

time

R=N =N+1

T=N =N+1 =N+2


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10


Rx

TxI

NACK

NI

N+1ACK

N+1I

N+2

I

NACK

NI

N+1ACK

N+1I

N+2ACK

N+2N

N

N

time

R=N =N+1 =N+2 =N+3

T=N =N+1 =N+2 =N+3


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11

Worksheet: Automatic Repeat Request

Rx

Tx?

t


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12

Idle RQ: Stop-and-wait efficiency

Rx

TxI

NACK

NI

N+1ACK

N+1I

N+2

I

NACK

NI

N+1ACK

N+1I

N+2ACK

N+2

N

N

N

time

TR -

TP

-TI

-TD

-TA

-TP

-

TR = TI + 2TP + TD + TA

s =TI

TI + 2TP + TD + TA


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13

Idle RQ: Improving performance

s =TI

TI + 2TP + TD + TA

Increase s?

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


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14

Idle RQ: Example Protocol Efficiency Calculation

Compute the maximum data throughput of Idle RQ

10Mbs1communications 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 Earths surface

propagation speed300 106

ms1

TI =92

10106= 9.2 106s

TA =8

10106= 0.8 106s

assume TD = 0

TP = 237.5106300106

= 0.25s

s =9.2106

9.2105+20.25+0.8106= 18.4 106

throughput = sB = 18.4 106 10M bs1 = 184bs1

5


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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

s =TI

TI + 2TP + TA + TD

Wastage due to errors forcing retransmission

e = (1 BER)i+a e

1

1 + p

Wastage due to control information

c =i c

i + a

Overall efficiency i = sec

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16

Worksheet: Overall Efficiency of Idle RQ

I-frames of 10000 bits; hosts separated by 500km over a 1Mbs1 link. The

I-frames consist of 1000 bits of header and 9000 bits of data. The signalpropagation speed is 200 106ms1, and the BER is 105

1. Calculate the probability that an I-frame is lost due to an error.

2. What is the error efficiency of the system e?

3. What is the stop-and-wait efficiency of the system s?

4. What is the control information efficiency of the system c?

5. What is the overall efficiency of the Idle RQ protocol?

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17

Idle RQ Error Handling: Loss of I-frame

Rx

TxI

NACK

NI

N+1I

N+1

I

NACK

NI

N+1ACK

N+1N

N

time

TO -

R=N =N+1 =N+2

T=N =N+1 =N+2

Timeout TO must be set at Tx

TO > TR

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Idle RQ Error Handling: Corrupt I-frame

Rx

TxI

NACK

NI

N+1I

N+1

I

NACK

NI

N+1 (error)NAK

N+1

NAK

N+1

I

N+1ACK

N+1N

N N

time

R=N =N+1 =N+2

T=N =N+1 =N+2

NAK-frame speeds up retransmission of I-frame

Do not have to use NAK-frames

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Idle RQ Error Handling: Loss of ACK-frame

Rx

TxI

NACK

NI

N+1I

N+1

I

NACK

NI

N+1ACK

N+1I

N+1ACK

N+1N

N N

time

TO -

R=N =N+1 =N+2

T=N =N+1 =N+2

Rx receives I(N+1) twice

Really do have to number I-frames!

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Continuous RQ

Rx

TxI

NI

N+1I

N+2I

N+3I

N+4I

N+5I

N+6

I

NI

N+1I

N+2I

N+3I

N+4I

N+5

^ ^ ^ ^ ^ ^

ACK

N

ACK

N+1

ACK

N+2

ACK

N+3

ACK

N+4 time

R=N

T=N =N+1 =N+2 =N+3 =N+4 =N+5 =N+6

=N+1 =N+2 =N+3 =N+4 =N+5 =N+6

RetransmissionList

I

N

I

N+1I

N

I

N+2I

N+1

I

N+3I

N+2

I

N+4I

N+3

I

N+5I

N+4

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21

Go-Back-N: Loss of I-frame

Rx

TxI

NI

N+1I

N+2I

N+3I

N+4I

N+2I

N+3

I

NI

N+1

I

N+3 (ignore)I

N+4 (ignore)I

N+2

^ ^ ^ ^ ^

ACK

N

ACK

N+1

NAK

N+2 time

R=N

T=N =N+1 =N+2 =N+3 =N+4 =N+5 =N+5

=N+1 =N+2 =N+2 =N+2 =N+2 =N+3

RetransmissionList

I

N

I

N+1I

N

I

N+2I

N+1

I

N+3I

N+2

I

N+4IN+3

I

N+2

I

N+4IN+3

I

N+2

22


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22

Go-Back-N: Loss of NAK-frame

Rx

TxI

NI

N+1I

N+2I

N+3I

N+4I

N+5I

N+2

I

NI

N+1

I

N+3 (ignore)I

N+4 (ignore)I

N+5 (ignore)

^ ^ ^ ^ ^

ACK

N

ACK

N+1

NAK

N+2 time

R=N

T=N =N+1 =N+2 =N+3 =N+4 =N+5 =N+5

=N+1 =N+2 =N+2 =N+2 =N+2 =N+2

RetransmissionList

I

N

I

N+1I

N

I

N+2I

N+1

I

N+3I

N+2

I

N+4IN+3

I

N+2

I

N+5I

N+4IN+3

I

N+2

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23

Idle RQ: Rough Formula for Efficiency

r = TRTI= number for frames sent in TR

p= probability frame suffers error

error efficiency:

e 1

1 + p

protocol efficiency:

s =1

r

combined efficiency:

=1

r + rp

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24

Go-back-N: Efficiency

s = 1

send r = TRTI I-frames for each erroneous I-frame

e 1

1 + pr

c same as idle RQ

25


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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 10Mbs1 satellite communication system with BER of 106 , a satelliteheight of 37,500km, propogation speed 300 106ms1, I-frames 100 bits.

p = 1 0.9999 = 104

TI =100

10106= 105s

TP =237.5106

300106= 0.25s

assume TD = 0, TA = 0

TR = 105 + 2 0.25 = 0.5s

r = 0.5105 = 50 10

3

g =1

1+10450103= 1

6= 0.17

throughput = 1.7 Mbs1

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Worksheet: Efficiency of Continuous RQ Go-Back-N

Go-Back-N system uses I-frames of 10000 bits; hosts separated by 500km over a

10Mbs1

link. Signal propagation speed is 200 106

ms1

.

1. If no errors occur, what is the throughput we can expect on the link?

2. If the BER is 105, 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?

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Go-Back-N: Efficiency

0

0.2

0.4

0.6

0.8

1

0 0.02 0.04 0.06 0.08 0.1

e

p

r = 1r = 10r = 100

r = 1000

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8

Selective Repeat: Explicit Request

Rx

TxI

NI

N+1I

N+2I

N+3I

N+4I

N+2I

N+5

I

NI

N+1I

N+3I

N+4I

N+2

^ ^ ^ ^ ^

ACKN

ACKN+1

NAKN+2

ACKN+3

ACKN+4 time

R=N

T=N =N+1 =N+2 =N+3 =N+4 =N+5 =N+5

=N+1 =N+2 =N+2 =N+2 =N+2 =N+5

RetransmissionList

I

N

I

N+1I

N

I

N+2I

N+1

I

N+3I

N+2

I

N+4I

N+3I

N+2

I

N+4I

N+2

ReceiverBuffer

I

N+3I

N+4I

N+3

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Selective Repeat: Implicit Retransmission

Rx

TxI

NI

N+1I

N+2I

N+3I

N+4I

N+5I

N+2

I

NI

N+1I

N+3I

N+4I

N+5

^ ^ ^ ^ ^

ACKN

ACKN+1

ACKN+3

ACKN+4 time

R=N

T=N =N+1 =N+2 =N+3 =N+4 =N+5 =N+6

=N+1 =N+2 =N+2 =N+2 =N+2 =N+2

RetransmissionList

I

N

I

N+1I

N

I

N+2I

N+1

I

N+3I

N+2

I

N+4I

N+3I

N+2

I

N+5I

N+4I

N+2

ReceiverBuffer

I

N+3I

N+4I

N+3

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Selective Repeat: efficiency

protocol efficieny s = 1

error efficiency as idle request

e 1

1 + p

c same as idle RQ

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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, asatellite height of 37,500km, propogation speed 300 106ms1 , I-frames 100

bits.

p = 1 (1 106)100 = 104

s =1

1+104 = 0.9999throughput = 10.0 Mbs1

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Comparison of ARQ Protocols

protocol error combineds e se

Idle RQ 1r

1

1+p1

r+rp

Go-Back-N 1 11+rp

1

1+rp

Selective

Repeat 1 11+p 11+p

For all, control efficiency

c =i c

i + a

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Overall Efficiency for Go-Back-N

0

0.2

0.4

0.6

0.8

1

0 100 200 300 400 500

ec

number of data bits per frame

r = 1r = 10

r = 1 graph corresponds to that for Selective Repeat

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Duplex Operation

Rx

Tx

Rx

TxI

NI

N+1ACK

MI

N+2ACK

M+1

IM

IM+1

ACKN

IM+2

ACKN+1

I

NI

N+1ACK

MI

N+2ACK

M+1

I

M

I

M+1

ACK

N

I

M+2

ACK

N+1

^ ^ ^

time

Can send ACK-frames for I-frame by piggybacking it to I-frame in opposite

direction

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Idle RQ: Window Size

Rx

TxI

NACK

NI

N+1ACK

N+1I

N+2

I

NACK

NI

N+1ACK

N+1I

N+2ACK

N+2N

N

N

time

R=N =N+1 =N+2 =N+3

T=N =N+1 =N+2 =N+3

Tx and Rx only need buffer one I-frame

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Continuous RQ Go-Back-N: Window Size

Rx

TxI

NI

N+1I

N+2I

N+3I

N+4I

N+5I

N+6

I

NI

N+1I

N+2I

N+3I

N+4I

N+5

^ ^ ^ ^ ^ ^

ACK

N

ACK

N+1

ACK

N+2

ACK

N+3

ACK

N+4 time

R=N

T=N =N+1 =N+2 =N+3 =N+4 =N+5 =N+6

=N+1 =N+2 =N+3 =N+4 =N+5 =N+6

RetransmissionList

I

N

I

N+1I

N

I

N+2I

N+1

I

N+3I

N+2

I

N+4I

N+3

I

N+5I

N+4

Tx must buffer r I-frames

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Continuous RQ Selective Repeat: Window Size

Rx

TxI

NI

N+1I

N+2I

N+3I

N+4I

N+2I

N+5

I

NI

N+1I

N+3I

N+4I

N+2

^ ^ ^ ^ ^

ACK

N

ACK

N+1

NAK

N+2

ACK

N+3

ACK

N+4 time

R=N

T=N =N+1 =N+2 =N+3 =N+4 =N+5 =N+5

=N+1 =N+2 =N+2 =N+2 =N+2 =N+5

RetransmissionList

I

N

I

N+1I

N

I

N+2I

N+1

I

N+3I

N+2

I

N+4I

N+3I

N+2

I

N+4I

N+2

ReceiverBuffer

I

N+3I

N+4I

N+3

Tx and Rx must buffer r I-frames

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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 an infinite range

Use modulo S

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