Computer Network Questions and Answers

Answer: D

Correct answer: 10

With \( n \) address lines a processor can select \( 2^{n} \) distinct locations. Since 1 KB in the binary sense is 1,024 bytes, \[ 2^{n} = 1024 \quad\Rightarrow\quad n = 10. \] The powers of two are worth having by heart for this family of questions: \( 2^{10} = 1\text{K} \), \( 2^{20} = 1\text{M} \), \( 2^{30} = 1\text{G} \). So 1 MB needs 20 address lines and 4 GB needs 32 - which is precisely why a 32-bit processor can address no more than 4 GB without extensions. The distractor 1024 confuses the number of locations with the number of lines needed to select among them. The width of the data bus is a separate matter: it determines how many bits are transferred at once, not how many locations exist. A memory of 1K × 8 has ten address lines and eight data lines.

Answer: B

Correct answer: 20

A simplex line carries traffic in one direction only, so each pair of routers needs two of them - one each way. With 5 routers there are \( 5\times 4 = 20 \) ordered pairs, and so 20 lines. Full-duplex lines would need only \( \binom{5}{2}=10 \).

Answer: C

Correct answer: 100 m

A twisted pair segment is limited to 100 m, which is why structured cabling is planned around that figure: 90 m of fixed cable and 10 m of patch leads.

Answer: D

Correct answer: 2000 m

Fibre carries a signal furthest of the media in the list - 2000 m per segment for the 100Base-FX and 10Base-FL standards - because glass attenuates light far less than copper attenuates a current.

Answer: A

Correct answer: equilizer

In a synchronous modem the digital-to-analog converter's output goes to the equalizer, which compensates for the distortion the telephone line will impose before the signal is put on the line.

Answer: C

Correct answer: simplify Boolean expressions

A Karnaugh map is a graphical method of minimising a Boolean expression. It arranges the cells of a truth table so that any two adjacent cells differ in exactly one variable, which lets a redundant variable be spotted and cancelled by grouping. That adjacency is achieved by labelling the axes in Gray code - 00, 01, 11, 10 rather than 00, 01, 10, 11 - and the map wraps at its edges, so the leftmost and rightmost columns are also adjacent. Groups must contain a power of two cells: 1, 2, 4, 8. Each doubling of a group removes one variable from the resulting term, so the aim is to cover every 1 with as few and as large groups as possible. Overlapping is allowed. Cells marked as don’t care may be treated as 1 or 0, whichever makes the groups larger. The map is practical up to four variables; beyond that the Quine–McCluskey algorithm is used instead.

Answer: B

Correct answer: selects one of several inputs and routes it to a single output

A multiplexer, or data selector, has several data inputs, one output, and a set of select lines that decide which input reaches the output. A multiplexer with \( n \) select lines can choose among \( 2^{n} \) inputs, so a 4-to-1 multiplexer needs two select lines and an 8-to-1 needs three. Its inverse is the demultiplexer, described by the second option, which takes one input and routes it to one of several outputs. The two are used together to send many signals over a single shared line and separate them again at the far end - the principle behind time-division multiplexing in communications. A decoder, the third option, converts an \( n \)-bit code into one of \( 2^{n} \) active outputs and is used for address decoding in memory; an encoder does the reverse. Multiplexers can also implement arbitrary logic functions directly, which makes them useful beyond data routing.

Answer: C

Correct answer: Go-back-N has better line utilization

Selective repeat retransmits only the frame that failed, whereas go-back-N retransmits it and everything sent after it, so selective repeat makes the better use of the line. It pays for that with more buffering at the receiver and more complex logic, which is what the other statements record.