Computer Network Questions and Answers

Answer: C

Correct answer: one variable

Adjacent cells in a Karnaugh map differ in exactly one variable, and that single property is what makes the map work. When two adjacent cells both contain 1, the variable that differs between them can be eliminated. Algebraically this is the identity \[ A B + A \bar{B} = A(B + \bar{B}) = A. \] Combining a pair removes one variable, a group of four removes two, a group of eight removes three. The arrangement is achieved by labelling the axes in Gray code, in which successive values change one bit at a time: 00, 01, 11, 10. Plain binary order would place 01 next to 10, which differ in two bits, and the grouping would be invalid. The map also wraps round, so the first and last columns are adjacent, and in a four-variable map the four corner cells form a valid group of four. Missing that wraparound is the commonest error in using the method.

Answer: B

Correct answer: 8

A byte is 8 bits. The convention was settled by the IBM System/360 in the 1960s and has been universal since; a byte of eight bits can represent 256 distinct values, which was enough for a character set with room to spare. A group of four bits is a nibble, and it corresponds exactly to one hexadecimal digit - which is why a byte is written as two hexadecimal characters, from 00 to FF. Above the byte the units run kilobyte, megabyte, gigabyte and terabyte. There is a real ambiguity here worth knowing: in the decimal convention a kilobyte is 1,000 bytes, while in the binary convention it is \( 2^{10} = 1{,}024 \). The standard prefixes for the binary values are kibibyte, mebibyte and gibibyte, though they are rarely used outside standards documents. Storage manufacturers quote decimal values and operating systems often report binary ones, which is why a drive sold as 1 TB shows as roughly 931 GB.

Answer: C

Correct answer: 4

A counter built from \(n\) flip-flops passes through \(2^n\) states, so a MOD-16 counter needs \(n = 4\) since \(2^4 = 16\). It counts 0 through 15 and then returns to 0.

Answer: B

Correct answer: 802.5

IEEE 802.5 is Token Ring, and one station on the ring is designated the active monitor, whose duties include removing orphaned frames and ensuring exactly one token circulates. 802.3 is Ethernet, which has no token.

Answer: D

Correct answer: all of above

Broadband means a channel of wide bandwidth, and coaxial cable, optical fibre and microwave links all qualify - each carries far more than a single voice channel.

Answer: B

Correct answer: 6 kbps

Nyquist again, with two levels:

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

Answer: D

Correct answer: 13

Each position carries a power of two: \(1101_2 = 8 + 4 + 0 + 1 = 13\). The zero in the twos place is what separates it from 1111, which would be 15.

Answer: C

Correct answer: NOT

The NOT gate is the inverter. It has a single input and a single output, and the output is always the complement of the input: 0 becomes 1 and 1 becomes 0. Its Boolean expression is \( Y = \bar{A} \). On a circuit diagram it is drawn as a triangle with a small circle, or bubble, at its output. That bubble is the general symbol for inversion, which is why a NAND gate is drawn as an AND gate with a bubble and a NOR gate as an OR gate with one. The NOT gate is the only basic gate with one input; AND, OR, NAND, NOR and XOR all take two or more. It is easily built from a universal gate: tying both inputs of a NAND or a NOR together produces an inverter, which is part of what makes those gates universal. Two inverters in series cancel, giving \( \bar{\bar{A}} = A \) - the involution law of Boolean algebra.

Answer: B

Correct answer: 13%

Of every 8-bit sample T1 devotes 1 bit to signalling, and each 193-bit frame adds a framing bit:

\[ \frac{1}{8} = 12.5\% \quad\text{plus}\quad \frac{1}{193} \approx 0.5\% \]

which together come to about 13 per cent.

Answer: B

Correct answer: (A · B)’ = A’ + B’

De Morgan’s second theorem states that the complement of a product equals the sum of the complements: (A · B)’ = A’ + B’. In gate terms, a NAND is the same as an OR gate fed by inverted inputs.