Computer Network Questions and Answers

Answer: C

Correct answer: narrow band channel

A narrowband channel carries the least data per second, being defined by its small range of frequencies; voice-band channels are the next step up and wideband channels the fastest.

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.

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: B

Correct answer: Successive values differ in exactly one bit

In Gray code only one bit changes between consecutive values. That property matters in position encoders, where several bits changing at once could momentarily be read as a wildly wrong value.

Answer: C

Correct answer: Distributive law

This is the distributive law, which lets AND be distributed over OR just as multiplication distributes over addition in ordinary algebra. Boolean algebra also allows the reverse form, A + BC = (A + B)(A + C), which has no arithmetic counterpart.

Answer: C

Correct answer: 0.736

Two uplink slotted ALOHA channels each deliver at most \( 1/e \approx 0.368 \) of their capacity, and the downlink carries the sum of the two:

\[ 2\times 0.368 = 0.736 \]

Answer: B

Correct answer: True

The statement is true. This is the absorption law, and it is one of the most useful identities for simplifying expressions. The proof is short: \[ A + A\cdot B = A(1 + B) = A \cdot 1 = A, \] using the distributive law and then the null law \( 1 + B = 1 \), which holds because ORing anything with 1 gives 1. It can also be seen by inspection. If \( A = 1 \) the whole expression is 1 whatever B may be; if \( A = 0 \) then both terms are 0. So the value depends only on A, and B is absorbed. Its dual is \( A \cdot (A + B) = A \), and a related form is \( A + \bar{A}B = A + B \), which is not absorption and is worth keeping distinct. These identities matter because every gate removed from a design saves cost, power and propagation delay - the same purpose served by Karnaugh maps and the Quine–McCluskey method for larger expressions.

Answer: D

Correct answer: primary-to-secondary

Poll and select belong to a configuration with one primary station controlling several secondaries: the primary polls a secondary to invite it to send, and selects it to warn it that data is coming. Peers by definition need no such controller.

Answer: C

Correct answer: Address

In a poll/select discipline the primary addresses each transmission to a particular secondary, so every frame carries an address and each station acts only on the frames bearing its own.

Answer: A

Correct answer: in-service expansion

A token bus passes the token logically, by address, rather than physically round a cable, so a station can be added to or removed from the logical ring while the network keeps running. The number of stations and the span are both limited by the medium.