recitation 5

Recitation 5: ARQ Schemes

Hung-Bin (Bing) Chang and Yu-Yu Lin

Electrical Engineering Department University of California (UCLA), USA,[email protected] and [email protected]

Prof. Izhak Rubin (UCLA) EE 132B 2014 Fall 1 / 15

Outline

1 Stop and WaitExample 1: Stop and Wait

2 Go back NExample 2: Go back N

3 Selective RepeatExample 3: Selective Repeat


Stop and Wait

Stop and Wait

Half duplexIt is important to number the packet and ACK

PKT 1

ACK 1

PKT 3

ACK 3

transmitter receiver

Send PKT 1

Time expires

PKT 2

ACK 2

PKT 4 (error)

PKT 4

ACK 4

tto

Figure : ARQ scheme: Stop and waitProf. Izhak Rubin (UCLA) EE 132B 2014 Fall 3 / 15

Stop and Wait Example 1: Stop and Wait

Parameter setup

Transmission rate R = 32 KbpsEach packet contains np characters.

150 data characters (i.e., nd = 150).10 header characters (i.e., nh = 10).160 characters in a packet (i.e., np = nd + nh = 160).

Acknowledgements consist of 10 characters (i.e., ns = 10).There are 8 bits per character (i.e., nb = 8).tp = tta = 0.02 sec (half duplex, line-turnaround time).trr = ttr = 0.02 sec (full duplex, receiver reaction time, transmitterreaction time).Bit error rate (BER) = p = 104.



Example 1 - Contd

Question 1-1What is the packet error rate?

PE = 1 (1 p)Nbp = 1 (1 104)(150+10)8 = 0.1202,

where Npb = np nb is the number of bits in a packet.Question 1-2

If packets are sent without error, what is the time betweensuccessive packets by using stop and wait?

T =(150+ 10) 8

32K+

10 832K

+ 2 (0.02+ 0.02) = 0.1225.



Example 1 - Contd

Question 1-3Now, with errors occurring, what is the average number oftransmissions required for successful reception?

P(NT = n) = Pn1E (1 PE),n = 1,2, . . . ,E [NT ] =

11 PE =

11 0.1202 = 1.1366.

Question 1-4What is the effective throughput rate () for stop and wait?

=NdbTp

=150 8

1.1366 0.1225 = 8618 bps,

where Ndb is the number of data bits per packet and Tp is the timeto successfully transmit one packet.



Example 1 - Contd

Question 1-5What is the normalized throughput for stop and wait scheme?

R=

861832K

= 26.7%,

where R is transmission rate and can be also regarded as maxthroughout rate.The general formula for stop and wait scheme throughout is givenby

=Ndb

11 PE =E [NT ]

(NpbR

+NabR

+ 2(tp + tta))

=T

=Ndb(1 PE)R

Npb + Nab + 2R(tp + tta),

where Npb = np nb, Ndb = nd nb and Nab = na nb.


Go back N

Go back N

Full duplex, pipelining and ensure the oldest frame transmittedsuccessfully

ACK 4

1

ACK 1


Time expires

23

ACK 2

ACK 3

456

Ignored by receiver

(out of sequence

packets)45 6

5 6

ACK 5

ACK 6

Figure : ARQ scheme: Go back N


Go back N Example 2: Go back N

Example 2 - Contd

Question 2-1Now consider Go Back N. If there are no errors, what is the time totransmit one packet?

NpbR

=(150+ 10) 8

32Ksec.

Question 2-2If errors occur, what is the average number of retransmissions?

P(NR = n) = PnE(1 PE),n = 0,1,2, . . . ,E [NR] =

PE1 PE .



Example 2 - Contd

Question 2-3What is the optimal value of N for Go Back N?

Select N such that time-out duration (tto) is given by

tto = (N 2)NpbR ,

where

N =

2tp + trr + ttr + NabR

NpbR

+ 2

=

2 0.02+ 0.02+ 0.02+ 10832K

160832K

+ 2 = 5,

where tto = 2tp + trr + ttr + NabR .



Example 2 - Contd

Question 2-4How many packets must be transmitted, on average, in order tosuccessfully receive one packet?

For each retransmission happening, we need to send N packets.

N E [NR] + 1 = N(

PE1 PE

)+ 1

=NPE + 1 PE

1 PE=

PE(N 1)1 PE

=0.1202 4+ 11 0.1202 = 1.6831.



Example 2 - Contd

Question 2-5What is the throughput for go back N scheme?

=NdbTp

=Ndb

E [NT ] TR=

150 81.6831 160832K

= 17824 bps (55.7%),

where Ndb = nd nb is the number of data bits in a packet, E [NT ] isthe average number of packet transmissions for one successfulreception and Tp is the time to transmit one packet.The general formula for Go back N is given by

=Ndb

NpbRTp

((N 1)PE + 1

1 PE

)

E [NT ]

.


Selective Repeat

Selective Repeat

Full duplex,Receiver has the ability to receive out of sequence packetWe could have window.

1

ACK 1


23

ACK 2

ACK 3

456

ACK 5

ACK 6

Time expires

ACK 4

4

Figure : ARQ scheme: Selective repeat


Selective Repeat Example 3: Selective Repeat

Example 3

Question 3What is the throughput for selective repeat?

=Ndb

NpbR E [NT ]

=Ndb

NpbR

(1

1PE

)=

150 8160832K

(1

10.1202) = 26394 bps (82.5%),

Note: Under a special case of no error (i.e., p = 0), Go back Nscheme is same with Selective repeat scheme.


Selective Repeat Example 3: Selective Repeat

Q&A


Stop and WaitExample 1: Stop and Wait

Go back NExample 2: Go back N

Selective RepeatExample 3: Selective Repeat

recitation 5

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