Computer Network Questions and Answers

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.

Answer: B

Correct answer: the value of D just before the clock edge

A D flip-flop simply captures whatever is on its D input at the active clock edge and holds it until the next one. Its characteristic equation is \( Q_{next} = D \). The D stands for data or delay, and both readings are apt: the device stores one bit of data, and it delays a signal by exactly one clock period. Its virtue is that it has no forbidden or ambiguous input combination, unlike the SR flip-flop, and needs no input decoding, unlike the JK. That simplicity is why it is the workhorse of synchronous design and why registers, shift registers and pipeline stages are built from it. For the capture to be reliable the D input must be stable for the setup time before the edge and the hold time after it. Violating either can leave the output metastable - hovering between levels for an unpredictable period - which is why an asynchronous input is passed through two D flip-flops in series before being used.

Answer: B

Correct answer: RAM

RAM is volatile: it loses its contents the moment power is removed. It is used as working memory precisely because it is fast and freely writable, and its impermanence is the price of that. The other three are non-volatile and retain data without power. ROM is written once at manufacture; EPROM can be erased by ultraviolet light and reprogrammed; flash memory is electrically erasable and is what solid-state drives and memory cards are built from. RAM itself divides in two. SRAM stores each bit in a flip-flop, needs no refreshing, and is fast but expensive, so it is used for cache. DRAM stores each bit as a charge on a tiny capacitor, which leaks and must be refreshed thousands of times a second, but it is far denser and cheaper, so it is used for main memory. The name random access distinguishes both from sequential media such as tape, where access time depends on position.

Answer: A

Correct answer: 3

A multiplexer with \(2^n\) data inputs needs \(n\) select lines to address them. Since \(2^3 = 8\), three select lines are required to choose one of the eight inputs.

Answer: A

Correct answer: True

The statement is true. This is the complement law of Boolean algebra: a variable ORed with its own complement always gives 1. The reasoning is exhaustive. If \( A = 0 \) then \( \bar{A} = 1 \), and \( 0 + 1 = 1 \). If \( A = 1 \) then \( \bar{A} = 0 \), and \( 1 + 0 = 1 \). No third case exists in a two-valued algebra, so the result is 1 whatever A may be. Its dual is equally useful: \( A \cdot \bar{A} = 0 \), since a variable and its complement can never both be 1. The complement laws are among the basic identities used to simplify expressions, alongside the identity laws \( A + 0 = A \) and \( A \cdot 1 = A \), the null laws \( A + 1 = 1 \) and \( A \cdot 0 = 0 \), and the idempotent laws \( A + A = A \) and \( A \cdot A = A \). Note that in Boolean algebra + means OR, not arithmetic addition, which is why \( 1 + 1 = 1 \) rather than 2.