Computer Network Questions and Answers

Answer: C

Correct answer: 255

With \( n \) bits there are \( 2^{n} \) distinct patterns, so eight bits give 256 of them - but they represent the values 0 to 255, since one pattern is used for zero. Hence the maximum is \[ 2^{8} - 1 = 255. \] The off-by-one is the whole point of the question, and 256 is the classic wrong answer: it is the count of values, not the largest of them. That is unsigned. If the eight bits hold a signed value in 2’s complement, the range becomes −128 to +127 - which is where the other two distractors come from. The asymmetry arises because 2’s complement has a single representation of zero, leaving one more pattern for the negatives than for the positives. The pattern 11111111 is therefore 255 when read as unsigned and −1 when read as signed. Nothing in the bits themselves says which is meant; it is the instruction operating on them that decides.

Answer: C

Correct answer: 3

An encoder performs the reverse of a decoder, turning one active input among \(2^n\) into an \(n\)-bit code. With 8 inputs, \(2^3 = 8\) gives three output lines.

Answer: C

Correct answer: stores one bit of information

A flip-flop is a bistable circuit: it has two stable states and remains in whichever it was last put into, so it stores exactly one bit. It is the basic element of all sequential logic and of static memory. The distinction from combinational logic is the essential one. A combinational circuit’s output depends only on its present inputs; a sequential circuit’s output depends on its inputs and on its stored state, which is what gives it memory. The common types are the SR flip-flop, which sets and resets but has a forbidden input combination; the D flip-flop, which simply captures the value on its data input at the clock edge; the JK, which resolves the SR ambiguity by toggling; and the T, which toggles on each clock pulse. Chained together, flip-flops make registers, shift registers and counters. A latch is the level-triggered relative of the edge-triggered flip-flop.

Answer: B

Correct answer: interface card

Every machine on a LAN needs a network interface card to put frames on the cable and take them off. A modem is for telephone lines, which is a wide-area connection.

Answer: C

Correct answer: both (a) and (b)

Slotted ALOHA divides time into slots equal to one frame time and allows transmission only at a slot boundary, which halves the vulnerable period and doubles the throughput. Both properties follow from the same requirement: every station must agree on where the slots begin, which needs global synchronisation.

Answer: D

Correct answer: All of these

A wide area network spans distances no single medium covers economically, so it uses whatever is available: leased telephone lines, terrestrial microwave links and satellite hops.

Answer: A

Correct answer: False

The statement is false. With both S and R at 1 the SR flip-flop enters its forbidden or invalid condition. The problem is twofold. While both inputs are held high, Q and its complement are driven to the same level, so the outputs are no longer complementary and the device is not storing a meaningful bit. Worse, when the inputs are released together the circuit settles into whichever state wins a race between its two gates, which depends on minute differences in propagation delay - so the resulting state is genuinely unpredictable. The JK flip-flop was designed to remove this defect. By feeding the outputs back into the input gating, it converts the J = K = 1 case into a useful toggle, and that toggle is what makes binary counters possible. The D flip-flop avoids the problem differently, by deriving both internal inputs from a single data line so the forbidden combination can never arise.

Answer: A

Correct answer: 15

The minimum number of bits needed to distinguish all the symbols is the smallest n with \( 2^{n} \) at least as large as the number of symbols, so it is \( \lceil \log_{2} N \rceil \) summed over the set as the question defines it.

Answer: D

Correct answer: Contention

In a contention scheme such as CSMA/CD a station transmits when it judges the medium free, and whether it succeeds depends on what other stations happen to do - so access is probabilistic. Polling and token passing grant access deterministically, by turn.

Answer: D

Correct answer: both (a) and (b)

The destination does not remove the frame; it copies the data and sets the address-recognised and frame-copied bits in the frame status field - four bits in all. The frame continues round to the sender, which removes it, and that is how the sender learns the frame was received.