For stop-and-wait flow control, for n data packets sent, how many acknowledgments are needed ?
Answer: A
Correct answer: n
In stop-and-wait the sender transmits one frame and waits for its acknowledgement before sending the next, so n frames require n acknowledgements. Grouping them is what the sliding-window protocols allow.
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The collision domain of tranditional Ethernet and Fast Ethernet respectively are
Answer: C
Correct answer: 2500m, 250m
Traditional 10 Mbps Ethernet allows a collision domain of about 2500 m. Fast Ethernet runs ten times faster, so a frame occupies the wire for a tenth of the time and the domain must shrink by the same factor to 250 m if collisions are still to be detected before the frame ends.
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In a crossbar with 1000 crosspoints, how many statistically are in use at any time ?
Answer: B
Correct answer: 250
Crossbars are dimensioned on the observation that only about a quarter of the crosspoints are needed at once, since not every input is active and each active one uses a single crosspoint. For 1000 crosspoints that is 250.
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Which of the following is a time - division switch?
Answer: D
Correct answer: both (a) and (b)
A time-slot interchanger and a TDM bus both switch by moving samples between time slots rather than by making a physical path, so both are time-division switches. A crosspoint makes a path and is space-division.
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In which routing method do all the routers have a common database ?
Answer: B
Correct answer: Link state
In a link state protocol every router floods a description of its own links to all the others, so each ends with the same map of the network and computes its own shortest paths from it. Distance vector routers exchange only their own distance tables and never share a common database.
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A terminal multiplexer has six 1200 bps terminals and n 300 bps terminals connected to it. The outgoing line is 9600 bps. What is the maximum value of n?
Answer: C
Correct answer: 8
The 300 bps terminals may use whatever the 1200 bps terminals leave:
\[ 9600 - 6\times 1200 = 2400\ \text{bps} \qquad n = \frac{2400}{300} = 8 \]
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Data link layer retransmits the damaged frames in most networks. If probability of a frame's being damaged is p, then what is the mean number of transmissions required to send a frame if acknowledgements are never lost ?
Answer: B
Correct answer: 1 / K - P
If each transmission fails with probability p independently, the number of attempts is geometric and its mean is \( 1/(1-p) \). The expression given is that quantity written with the question's own symbols.
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If a designer wants to design a point-to-point subnetwork with 10 routers of full duplex line, then total number of lines among them would be
Answer: D
Correct answer: 45
A full-duplex line carries both directions, so one line serves each pair of routers:
\[ \binom{10}{2} = \frac{10\times 9}{2} = 45 \]
It would take 90 lines only if each were simplex and a separate one were needed for each direction.
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Maximum data rate of a noiseless 4 kHz channel using T1 PCM system is
Answer: B
Correct answer: 56 kbps
T1 PCM samples at 8000 samples per second and uses seven of the eight bits of each sample for data, the eighth being taken for signalling:
\[ 8000\times 7 = 56{,}000\ \text{bps} \]
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Phase shift keying (PSK.) method is used to modulate digital signals at 9600 bps using 16 levels. The line signals speed (i.e. modulation rate) will be
Answer: A
Correct answer: 2400 bands
Sixteen levels carry \( \log_{2}16 = 4 \) bits per symbol, so
\[ \text{baud} = \frac{9600}{4} = 2400 \]
The baud rate is the symbol rate; it equals the bit rate only when each symbol carries one bit.
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