Computer Network Questions and Answers

Answer: C

Correct answer: 15

With n bits for sequence numbers a go-back-N window may be at most \( 2^{n}-1 \), one less than the number of distinct sequence numbers, so that a full window can never be mistaken for an empty one:

\[ 2^{4}-1 = 15 \]

Answer: C

Correct answer: Four

Binary Coded Decimal stores each decimal digit in its own group of four bits, so 27 becomes 0010 0111. The six combinations from 1010 to 1111 are unused, which makes BCD less compact than pure binary.

Answer: D

Correct answer: 3

A full adder has three inputs - the two bits to be added and a carry-in from the previous stage - and two outputs, the sum and the carry-out. The carry-in is what distinguishes it from a half adder, which has only two inputs and cannot be used beyond the least significant bit of a multi-bit addition. Its Boolean expressions are \[ S = A \oplus B \oplus C_{in}, \qquad C_{out} = AB + C_{in}(A \oplus B), \] and it can be built from two half adders and an OR gate. The sum output is 1 when an odd number of the three inputs is 1, which is why it is a cascade of XOR gates; the carry is 1 when at least two of the three are. Chaining \( n \) full adders gives a ripple-carry adder for \( n \)-bit numbers, simple but slow because each stage waits for the carry below it. Carry-lookahead adders compute the carries in parallel to remove that delay.

Answer: A

Correct answer: within each user session

An adaptive routing table is rebuilt as conditions change - as links fail or queues lengthen - so it may well change several times during one session. A table fixed at system generation is static routing.

Answer: B

Correct answer: 15

Octal is base 8, so each digit carries a place value that is a power of eight. For 17 that gives \[ 1 \times 8 + 7 = 15. \] The trap is reading the numeral as though it were decimal, which is why 17 is offered as a distractor. Octal uses only the digits 0 to 7; the digits 8 and 9 do not exist in it, so a numeral containing either is not a valid octal number at all - a quick way to spot a bad option. Octal is used as shorthand for binary because each octal digit corresponds to exactly three bits: 1 is 001 and 7 is 111, so 17 in octal is 001111, which is 15. Conversion is done by grouping bits in threes from the right, with no arithmetic. It was common on machines with word lengths divisible by three, and survives in Unix file permissions, where 755 and 644 are octal.

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.