Computer Networking Questions and Answers

Answer: B

R receives the whole file at 8.008 s.

The file is \(10^{6}\) bytes = \(8 \times 10^{6}\) bits, in 1000 chunks of 8000 bits. Each chunk takes \(8000/10^{6} = 0.008\) s to transmit on either link. P sends the chunks back to back, so the last chunk leaves P at

\[ \begin{aligned} t_{P} &= \frac{8 \times 10^{6}}{10^{6}} = 8.000\ \text{s} \\ t_{R} &= 8.000 + 0.008 = 8.008\ \text{s} \end{aligned} \]

Q forwards each chunk as soon as it has received it, while receiving the next, so the two links work in a pipeline and only the last chunk's extra hop adds time. 8.000 ignores that hop, and 16.000 treats the transfer as store-and-forward of the whole file, sending it twice in sequence; 15.992 has no basis in the pipeline.

Answer: B

The correct matching is (a)-(I), (b)-(III), (c)-(II).

• The network layer routes packets from source to destination across networks, using logical (IP) addresses: (a)-(I).
• The transport layer provides end-to-end, host-to-host (process-to-process) communication, with segmentation, flow control and reliability as in TCP: (b)-(III).
• The data link layer groups bits into frames and detects errors on a single link, for example with a CRC: (c)-(II).

Option A gives framing to the transport layer, and C and D give routing to the transport or data link layer, so each misplaces at least two layers.

Answer: B

The minimum window is 21 frames.

The transmission time of one frame and the round-trip time are:

\[ \begin{aligned} T_{t} &= \frac{1000}{100\times10^{3}} = 10\ \text{ms} \\ 2T_{p} &= 200\ \text{ms} \end{aligned} \]

For 100% utilisation the sender must keep transmitting until the first acknowledgement returns, which takes one frame time plus the round trip:

\[ W \ge \frac{T_{t}+2T_{p}}{T_{t}} = \frac{10+200}{10} = 21 \]

20 counts only the round-trip time and leaves the link idle for one frame time per cycle. 10 and 11 use the one-way delay instead of the round trip.

Answer: D

The true statement is the exponential backoff mechanism reduces the probability of collision on retransmissions.

After a collision, each station waits a random number of slot times chosen from a range that doubles after each successive collision (binary exponential backoff). Spreading the retries over a growing range makes it less likely that the same stations choose the same slot again.

The others describe CSMA/CD wrongly. A station keeps sensing the channel while transmitting; that is how it detects a collision. The jamming signal tells all stations a collision occurred; padding short frames is a separate rule for the minimum frame size. And a station stops transmitting the frame once it detects a collision, after sending the jam signal.

Answer: B

The true statement is The destination IP address is the IP address of Y.

IP addresses are end to end: the packet carries Y's IP address as its destination all the way, and router R forwards it without changing that field. MAC addresses are hop by hop: X sends the frame to the next IP hop on its own network, which is router R, so the destination MAC address is R's.

R's IP address would be the destination only if the packet were addressed to the router itself. An Ethernet switch is transparent at layer 2 and does not appear as a frame's destination, so S's MAC address is wrong, and Y's MAC address is unknown to X because Y is on another network.

Answer: A

The channel utilization is about 88.23%.

In stop-and-wait the sender transmits one frame and then waits a full round trip for the acknowledgement. With negligible ACK size and processing times:

\[ \begin{aligned} T_{t} &= \frac{3000}{2000} = 1.5\ \text{s} \\ T_{p} &= 100\ \text{ms} = 0.1\ \text{s} \\ U &= \frac{T_{t}}{T_{t} + 2T_{p}} = \frac{1.5}{1.7} = 0.8824 \end{aligned} \]

That is 88.23% to two decimal places (88.235 truncated). 93.75% uses only one propagation delay (1.5/1.6), forgetting that the acknowledgement must travel back. 66.67% and 85.44% match no consistent use of these figures.

Answer: B

The transfer takes 64 seconds.

With no per-packet overhead and negligible delays, the data streams through the path, and the throughput is limited by the slowest link, the bottleneck of 500 kbps:

\[ \begin{aligned} \text{Size} &= 4 \times 10^{6} \times 8 = 32 \times 10^{6}\ \text{bits} \\ t &= \frac{32 \times 10^{6}}{500 \times 10^{3}} = 64\ \text{s} \end{aligned} \]

8 s divides the size in bytes, not bits, by the bottleneck rate, and 16 s is the time over the 2 Mbps link alone. 731 matches no consistent calculation. The faster links do not add to the time, because the stream flows through them while the bottleneck is still sending.

Answer: D

The remainder appended is 100.

The generator 1001 has degree 3, so append three zeros to the data, 110001011 000, and divide by 1001 using modulo-2 (XOR) arithmetic. Each step XORs 1001 under the leading 1:

110001011000
1001
010101011000
 1001
000111011000
   1001
000011111000
    1001
000001101000
     1001
000000100000
      1001
000000000100

The last three bits, 100, are the remainder, and the transmitted frame is 110001011100. A check: that frame divided by 1001 leaves remainder 000.

000 would mean the data were already a multiple of the generator, and 011 and 101 come from slips in the XOR steps.

Answer: B

Utilisation is lowest with a longer link length and higher transmission rate.

In Stop-and-Wait the sender transmits one frame and then idles for a round trip until the acknowledgement returns. With transmission time \(T_{t} = L/R\) and propagation delay \(T_{p} = d/v\):

\[ \begin{aligned} U &= \frac{T_{t}}{T_{t} + 2T_{p}} = \frac{1}{1 + 2a} \\ a &= \frac{T_{p}}{T_{t}} \end{aligned} \]

Utilisation falls as a grows. A longer link raises \(T_{p}\), and a higher transmission rate shrinks \(T_{t}\); both increase a, so together they give the lowest utilisation. A shorter link with a lower rate is the best case, and the two mixed options lie in between.

Answer: B

Correct answer: Every 2 seconds

Bridge protocol data units are sent every two seconds by default, which is the spanning tree hello time. They carry the bridge ID and path cost information from which every switch determines the root bridge, its own root port and any blocking ports. The interval matters because convergence times are derived from it: max age is twenty seconds, ten missed hellos, and forward delay fifteen.