Download the hong kong polytechnic university hong kong community

THE HONG KONG POLYTECHNIC UNIVERSITY
HONG KONG COMMUNITY COLLEGE
Subject Title
: Data Communications and
Networking
Session
: Semester 1, 2011/12
Subject Code
:
CC3204
Numerical Answers
Section B
Question B1
a)
No. of upstream channels: (41.5- 5.5)/3 = 12
c)
Bit rate = 2 x 3 x log2 32 = 30 Mbps
Alternate Answer
Assuming the baud rate = signal frequency band, Bit rate = 3 x log2 32 = 15 Mbps (Mark is
given only when this assumption is explicitly stated.)
d)
SNR = 5000/10 = 500
Maximum bit rate = (750-550) log2( 1+ 500) = 1793.7 Mbps
Page 1 of 6
Question B2
a)
go in binary 110 0111110 1111
Divisor 10011
append 110 0111110 1111 with 0000
1 1 0 1 1 0 0 1 0 0 0 0 1 1
1 0 0 1 1 1 1 0 0 1 1 1 1 1 0 1 1 1 1 0 0 0 0
1 0 0 1 1
1 0 1 0 1
1 0 0 1 1
1 1 0 1 1
1 0 0 1 1
1 0 0 0 1
1 0 0 1 1
1 0 0 1 1
1 0 0 1 1
1 1 0 0 0
1 0 0 1 1
1 0 1 1 0
1 0 0 1 1
0 1 0 1
CRC 0101
Resultant Data : 1 1 0 0 1 1 1 1 1 0 1 1 1 1 0 1 0 1
b)
0 1 1 1 1 1 1 0 1 1 0 0 1 1 1 1 1 0 0 1 1 1 1 0 1 0 1 0 1 1 1 1 1 1 0
Page 2 of 6
Question B3
a)
(i) Attenuation is 10 dB/Km
Total Attenuation : 3 x 10 = 30dB
10 log10 (PR / PT) = -30
log10 ( PR / 0.2) = -3
PR = 2 x 10-4 W or 0.2 mW
(ii)
Acceptable attenuation is 10 log10(8 x10-3/ 0.2 ) = -13.98 dB
Without amplifiers, the attenuation = 30 dB
If rounded, attenuation is = -14.0 dB
Gain in dB from amplifiers needed = 30 -14.0 = 16.0 dB
Minimum number of amplifiers needed = 16.0 / 4 = 4
Therefore 4 amplifiers are needed.
Alternate Answer
Gain in dB from amplifiers needed = 30 -13.98 = 16.02 dB
Minimum number of amplifiers needed = 16.02 / 4 = 4.005
Therefore ⎡ 4.005⎤ = 5 amplifiers are needed
Question B6
b)
Classful IP address 215.70.120.35
(a) the first three bits of 215 after converting to binary are 110
therefore the host belongs to class C
(b) Class C uses the first three bytes as net-id
therefore the network mask is 255.255.255.0
(c) netid is the first three bytes, i.e. 215.70.120
(d) hostid is the last byte, i.e. 35
(e) all bits of the host-id are “0” in the network address
therefore the network address is 215.70.120.0
(f) all bits of the host-id are “1” in the broadcast address
therefore the broadcast address is 215.70.120.255
(g) - For the same network mask, 24 bits are used for the netid,
- the classless IP address is 215.70.120.35/24
Page 3 of 6
Section C
Question C1
a)
For a round trip,
Transmission Time (Tx) = 1000 bit / (200x103) bps = 0.005s
Propagation Time (Tp) = 5000 x 103/ 2x 108 = 0.025s
Processing Time(Tc) = 0.002s
Transmission Time (Tx) for ACK= 100 bit / (200x103) bps = 0.0005s
Propagation Time (Tp) for ACK= 0.025s
For a round trip, delay = 0.005 + 0.025 + 0.0005 + 0.0025 + 0.002 x2= 0.0595s
No. of Round Trips = ⎡2 Mbit /(1000-1) bits ⎤ =2003
Total Delay = 2003 x 0.0595 = 119.1785s
Page 4 of 6
b)
Sender
Receiver (Time)
0
1
0.025
2
3
4
X
5
0.005
6
7
0.025
ACK 0
8
9
ACK 2
NAK 3
10
0.025
3
0.005
0.005
0.005
0.005
0.005
0.025
ACK 11
11
12
13
0.025
0.005
0.005
0.005
Size of the sliding window in the sender = 24/2 = 8
Correct identification of main ACK
Correct identification of NAK
Correct sequence of frames
Timing of delay components
Total delay : 0.17 sec.
(Alternate answer: last acknowledgement (Ack (14)) : Total delay = 0.17 + 0.025 = 0.195 sec )
Page 5 of 6
Question C2
b)
Datagram packet switching
i)
no. of data bits per packet = 1024 – 32 = 992
no. of packet required = 9000/992 = 9.073
therefore 10 packets are required
ii)
transmission time for one packet = 1024/5120
= 0.2 second
time for transmitting the 1st packet to the destination (via 5 hops and 4 nodes)
= 5 x (propagation delay + packet time) + 4 x nodal processing time
= 5 x (0.5 + 0.2) + 4 x 2
= 11.5 seconds
Therefore B will receive the first packet at 2:00:11.5 pm
iii)
time for the remaining 9 packets reaching the destination via the last hop by pipelining
= 9 x (nodal processing time + propagation delay + packet time)
= 9 x (2 + 0.5 + 0.2) = 24.3 s
end-to-end delay = 11.5 + 24.3
= 35.8 seconds
Page 6 of 6