Computer Network Questions and Answers

Answer: C

Correct answer: 6000 bps

By Nyquist's formula for a noiseless channel,

\[ C = 2B\log_{2}V = 2\times 3000\times \log_{2}2 = 6000\ \text{bps} \]

Binary signalling means two levels, so each symbol carries one bit and the rate is twice the bandwidth.

Answer: A

Correct answer: Residual error rate

The residual error rate is the proportion of messages that are lost or delivered corrupted out of all those sent - the errors that survive the protocol's own detection and correction.

Answer: A

Correct answer: 15

The minimum number of bits needed to distinguish all the symbols is the smallest n with \( 2^{n} \) at least as large as the number of symbols, so it is \( \lceil \log_{2} N \rceil \) summed over the set as the question defines it.

Answer: D

Correct answer: all of these

An Ethernet frame carries destination and source addresses, a type or length field, the data, and a checksum. All three named fields are present, so none of them can be the exception.

Answer: B

Correct answer: 20,000 bits

The signal takes

\[ \frac{200\times 10^{3}\ \text{m}}{200\ \text{m}/\mu\text{s}} = 1000\ \mu\text{s} \]

to go round, and at 20 Mbps that interval holds

\[ 20\times 10^{6}\times 10^{-3} = 20{,}000\ \text{bits} \]

Answer: D

Correct answer: 93 db

Shannon's formula gives the required signal-to-noise ratio from the capacity and bandwidth:

\[ 1.544\times 10^{6} = 50{,}000\log_{2}\!\left(1+\frac{S}{N}\right) \Rightarrow \frac{S}{N} \approx 2^{31} \]

and \( 10\log_{10}(2^{31}) \approx 93\ \text{dB} \) - a ratio no real line could sustain, which is why T1 is not carried this way.

Answer: B

Correct answer: go-back-n

In go-back-N the receiver discards everything after the damaged frame, so the sender must retransmit that frame and all those sent after it. Selective reject retransmits only the damaged one, at the cost of buffering at the receiver.

Answer: A

Correct answer: Does not require global time synchronization

Pure ALOHA lets a station transmit the moment it has data, so no station needs to know where any slot boundary lies. Slotted ALOHA gains throughput precisely by imposing that synchronisation.

Answer: C

Correct answer: both (a) and (b)

Manchester encoding sends a transition in the middle of every bit and never rests at zero volts, so it is a non-return-to-zero code, and it uses both positive and negative levels, so it is polar. Both descriptions apply.

Answer: A

Correct answer: TDMA

In TDMA each station has its own time slot and transmits in it without being asked, so no station has to be polled. Stop-and-wait and continuous ARQ are error-control schemes rather than access methods.