Computer Network Questions and Answers

Answer: C

Correct answer: \( A\bar{B} + \bar{A}B \)

The XOR function is 1 when the two inputs differ, which happens in exactly two cases: A is 1 and B is 0, or A is 0 and B is 1. Written as a sum of products that is \[ A \oplus B = A\bar{B} + \bar{A}B. \] The second option, \( AB + \bar{A}\bar{B} \), is 1 when the inputs are the same, and is therefore the complement - the XNOR function, also called the equivalence or coincidence gate. XOR has properties that make it unusually useful. \( A \oplus 0 = A \) and \( A \oplus 1 = \bar{A} \), so one input acts as a controlled inverter. \( A \oplus A = 0 \), which is why XOR finds the unpaired element in a list and why it underlies simple ciphers and the swap-without-a-temporary trick. In arithmetic it produces the SUM output of a half adder, and cascaded it generates and checks parity bits.

Answer: C

Correct answer: 3

A multiplexer with \( n \) select lines can choose among \( 2^{n} \) inputs, so for eight inputs \[ 2^{n} = 8 \quad\Rightarrow\quad n = 3. \] The three select lines form a binary address: 000 routes input 0 to the output, 001 routes input 1, and so on up to 111 for input 7. The same relation gives 2 select lines for a 4-to-1 multiplexer and 4 for a 16-to-1. It is the identical arithmetic that determines the number of flip-flops in a counter and the number of address lines for a memory, so it is worth being fluent in both directions. Multiplexers are used for far more than data routing. Because a \( 2^{n} \)-to-1 multiplexer can implement any Boolean function of \( n \) variables directly - by wiring the truth table values to its data inputs - it serves as a general-purpose logic element, which is why lookup tables in field-programmable gate arrays are built this way.

Answer: D

Correct answer: American Standard Code for Information Interchange

ASCII is the American Standard Code for Information Interchange, a character encoding published in 1963 that assigns a numeric value to each letter, digit and punctuation mark. It is a 7-bit code, giving 128 values: 0 to 31 are control characters such as carriage return and line feed, 32 is the space, 48 to 57 are the digits 0 to 9, 65 to 90 are the capital letters and 97 to 122 the small ones. Two consequences of that layout are worth knowing. The gap of exactly 32 between a capital and its small counterpart means the case of a letter is changed by flipping a single bit. And because the digits begin at 48, a numeric character is converted to its value by subtracting 48. Extended ASCII uses the eighth bit for a further 128 characters. Unicode superseded it for international text, but keeps ASCII as its first 128 code points, which is why the two remain compatible.

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.

Answer: C

Correct answer: 8

A decoder with \(n\) input lines activates one of \(2^n\) outputs, so a 3-to-8 decoder has eight outputs. Exactly one output is asserted at a time, corresponding to the binary value on the three inputs.

Answer: C

Correct answer: 100 Base-T4

100Base-T4 uses 8B/6T coding - eight bits mapped to six ternary symbols - which is what allows 100 Mbps over four pairs of ordinary Category 3 cable.

Answer: A

Correct answer: the D input just before the edge

The D flip-flop simply stores whatever is on its D input at the active clock edge, holding it until the next edge. This makes it the natural building block for registers and memory cells.

Answer: A

Correct answer: 30 kbps

This is Shannon's limit again, with a signal-to-noise ratio of 30 dB, that is 1000:

\[ C = 3000\log_{2}(1001) \approx 30{,}000\ \text{bps} \]

Note that the answer does not depend on how many levels the signalling uses; Shannon's bound holds whatever the modulation.

Answer: C

Correct answer: placing a telephone call prior to a conversion

A virtual circuit is set up before any data flows and torn down afterwards, so the route is fixed for the duration - exactly as dialling establishes a path before the conversation begins. Nothing is physically switched into place, which is what makes the circuit virtual.

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.