Consider the Figure Below. a Sender Begins Sending Packetized Audio Periodically at T=1
17 April 2009
EE6412 Test
Total Marks = 100
Question 1 (25 marks)
Consider the figure below. A sender begins sending packetized audio periodically at t=1. The first packet arrives at the receiver at t= 6.
a)What are the delays (from sender to receiver, ignoring any playout delays) of packets 1 through 8? Note that each vertical and horizontal line segment in the figure has a length of 1, 2, or 3 time units. [5 marks]
b)If audio playout begins at t = 5, which of the first eight packet sent will not arrive in time for playout? [5 marks]
c)If audio playout begins at t= 6, which of the first eight packets sent will not arrive in time for playout? [5 marks]
d)If audio playout begins at t= 7, which of the first eight packets sent will not arrive in time for playout? [5 marks]
e)What is the minimum playout delay at the receiver that results in all of the first eight packets arriving in time for their playout? [5 marks]
a) The delay of packet 1 is 5 slots. The delay of packet 2 is 6 slots. The delay of packet 3 is 5 slots. The delay of packet 4 is 7 slots. The delay of packet 5 is 6 slots. The delay of packet 6 is 5 slots. The delay of packet 7 is 5 slots. The delay of packet 8 is = 4 slots.
b) Packets 1, 2, 3, 4, 5, 6, and 7 will not be received in time for their playout if playout begins at t= 5.
c) Packets 2, 4 and 5 will not be received in time for their playout if playout begins at t=6.
d) Packet 4 will not be received in time for their playout if playout time begins at t= 7.
e) The minimum playout delay at the receiver that results in all of the first eight packets arriving in time for their playout is 2 and therefore playout time begins at t= 8.
Question 2 (15 marks)
Suppose you want to implement an IP telephony systemusingvoice codec G.711. Assuming 10 msec sampling duration and 64 Kbps voice signal rate, discuss the transmission efficiency of the resulting system for the following cases: (i) one sample per packet (ii) twosamples per packet and (iii) three samples per packet. Please comment on the results.
(i) G.711 is used for encoding the 64 Kbps voice. Assuming 10 msec sampling frame size, the payload is 80 bytes,IP/UDP/RTP header is 40 bytes and the efficiency is 66.67%. Note that Transmission Efficiency = (Payload Size)/(Payload Size+IP/UDP/RTP Header Size).
(ii) For 20 msec frame size, payload is 160 bytes and efficiency is 80%.
(iii) For 30 msec frame size, payload is 240 bytes and efficiency is 85.71%.
Efficiency will increase if number of samples/ pkt increases(i.e. larger frame size is used). However, when number of samples/ pkt increases, delay increases.
Question 3. (30marks)
- Given that n = (N/2)1/2 and that the number of links between stages is equal to the number of the incoming trunks, design a three-stage network for connecting 120 incoming trunks to 120 outgoing trunks. If there are more than one solution, determine which one is better. Draw the end result. [10 marks]
- Assuming that 30 erlangs of traffic is offered to the above network, calculate the blocking probability if the above switching network is used as a route switch in a telephone network . [10 marks]
- For the offered 30 erlangs of traffic, what is the blocking probability if the numbers of links between stages and the number of secondary switches are increased by 25%. [10 marks]
1a.120/2 = 7.75. Therefore, we use n=6 and n=8.
i)If n=6, there are:
20 primary switches of size 66
6 secondary switches of size 2020
20 tertiary switches of size 66
The number of crosspoints is = 22066+62020 =3840
ii)If n=8, there are:
15 primary switches of size 88
8 secondary switches of size 1515
15 tertiary switches of size 88
The number of crosspoints is = 21588+81515 = 3720
Since network ii) has less crosspoints than network i), it is better. Moreover, the second network has more secondary switches. It therefore provides a greater number of paths between an incoming and an outgoing trunk and will exhibit less blocking.
1. b. a=b=c=30/120=0.25
For network ii):
B1 = [1 - (1 - a)(1-b)]g2
= [1- (1-0.25)(1-0.25)]8
= 1.34 x 10-3
B2 = [B1 + c(1-B1)]g3
= [0.000257 + 0.25(1-0.000257)]15
= 9.98 x 10-10
1. c. a=b = 30/150 = 0.2, c=30/120=0.25
B1 = [1 - (1 - a)(1-b)]g2
= [1 - (1 - 0.20)(1-0.20)]10
= 3.656 x 10-5
B2 = [B1 + c(1-B1)]g3
= [3.656 x 10-5+ 0.25(1-3.656 x 10-5)]15
= 9.335 x 10-10
Question 4. (30 marks)
(a) An S-T-S network has 16 incoming and 16 outgoing highways, each of which conveys 24 PCM channels. Between the incoming and outgoing space switches there are 20 links containing time switches. During the busy hour, the network is offered 200 E of traffic and it can be assumed that this is evenly distributed over the outgoing channels.
(i)Derive an equivalent space-division network.
(ii)Estimate the blocking probability as an expander.
(iii)Estimate the grade of service when an incoming call must be connected to a selected outgoing highway but may use any free channel on it.
[20 marks]
(a) (i) m = 16, n = 24, k = 20
(ii) For the equivalent space-division network shown in Figure 6.9: m = 16, n = 24, k = 20. The occupancy of a link is
b = 200/(24 20) = 0.417 E.
B1 = [1 (1 b )2]k = [1 (1 0.417 )2]20
= 0.66020 = 2.4710 -4
(iii) The occupancy of a highway is
c = 200/(24 16) = 0.521 E
B2 = [B1 + c (1 B1)]n
= [2.4710 -4 + 0.521(1 2.4710 -4 )] 24
= 0.52124 =1.6110 -7
(b) An STS switch has 32 incoming and outgoing highways, having 24 PCM channels and 32 time-switch links. Calculate the numbers of crosspoints and bytes of storage required for this network with the assumption that each memory location in the connection/speech store requires one byte memory. [10 marks]
For a time switch:
No. of crosspoints = 0
Storage required = 2 no. ofPCM channels (for both speech store and connection store)
For a space switch:
No. of crosspoints = no. of incoming highways no. of outgoing highways
Storage required = no. ofPCM channels (only for connection store since there is no speech store in a space switch)
For S–T–S switch,
Crosspoints required = 23232 = 2,048
Storage required = 2432 + 22432 + 2432 = 3,072 bytes
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