Computer Network Questions and Answers

Answer: C

Correct answer: Converts \(2^{n}\) input lines into an \(n\)-bit output code

An encoder takes \( 2^{n} \) input lines, of which one is active, and produces the \( n \)-bit binary code identifying which. A decimal-to-BCD encoder, for example, has ten inputs and four outputs. Its inverse is the decoder, described by the second option, which takes an \( n \)-bit code and activates one of \( 2^{n} \) outputs - used for address decoding and for driving displays. A plain encoder assumes only one input is active at a time and produces nonsense if two are. A priority encoder removes that limitation by responding to the highest-numbered active input and ignoring the rest, which is exactly what an interrupt controller needs when several devices request service at once. The third option describes a multiplexer and the fourth a flip-flop. Encoders and decoders together allow a keyboard, with one line per key, to send a compact code over few wires and have it expanded again at the other end.

Answer: A

Correct answer: CMOS

CMOS - complementary metal-oxide-semiconductor - consumes very little power, which is why it displaced every earlier family and is used in essentially all modern integrated circuits. The reason lies in its structure. Each gate pairs a p-channel and an n-channel transistor so that in either stable state one of the two is off, and almost no current flows from supply to ground. Power is drawn chiefly when the output changes, charging and discharging the capacitance of the next stage - which is why CMOS power consumption rises with clock frequency. TTL, transistor-transistor logic, is faster than early CMOS but draws current continuously. ECL, emitter-coupled logic, is the fastest of the three because its transistors never saturate, but it consumes the most power of all. RTL was among the earliest families and is obsolete. CMOS also tolerates a wide supply range and has high noise immunity, though it is sensitive to static discharge.

Answer: C

Correct answer: Carrier sense systems

Carrier sense schemes can be run either way: as plain contention, in which every station has equal standing, or with priority, in which higher-priority stations wait a shorter interval before transmitting and so win the medium. TDM and register insertion allocate access by position rather than by priority.

Answer: B

Correct answer: 20,000 bits

The signal takes

\[ \frac{200\times 10^{3}\ \text{m}}{200\ \text{m}/\mu\text{s}} = 1000\ \mu\text{s} \]

to go round, and at 20 Mbps that interval holds

\[ 20\times 10^{6}\times 10^{-3} = 20{,}000\ \text{bits} \]

Answer: C

Correct answer: 11010

Dividing repeatedly by two and reading the remainders upward gives the answer. 26 gives remainder 0, then 13 gives 1, then 6 gives 0, then 3 gives 1, then 1 gives 1 - reading upward, 11010. The check is quicker than the derivation: the place values from the left are 16, 8, 4, 2, 1, so 11010 is \[ 16 + 8 + 0 + 2 + 0 = 26. \] An alternative method subtracts the largest power of two that fits and repeats: 26 − 16 = 10, 10 − 8 = 2, 2 − 2 = 0, so the bits at 16, 8 and 2 are set. The distractors are all near misses of the kind produced by reversing the remainder order or dropping a bit, which is why reading the remainders in the right direction matters. In hexadecimal 26 is 1A, and grouping 11010 into fours from the right - 0001 1010 - gives 1 and A directly.

Answer: C

Correct answer: a small, very fast storage location inside the processor

A register is a small storage location built directly into the processor, typically holding one word - 32 or 64 bits - and accessible in a single cycle. Registers are the fastest storage in the machine and sit at the top of the memory hierarchy, above cache, main memory and disk. They are few in number precisely because they are expensive in silicon area, and a large part of what a compiler does is decide which values are worth keeping in them. Several have fixed roles. The program counter holds the address of the next instruction; the instruction register holds the instruction being decoded; the accumulator holds arithmetic results in simpler architectures; the memory address and memory buffer registers interface with memory; the stack pointer tracks the top of the stack; and the status or flag register holds the zero, carry, sign and overflow bits that conditional branches test. General-purpose registers hold whatever the program needs.

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.