Networking Basics Questions and Answers

Answer: D

The maximum bit rate is 12,000 bps.

For a noiseless channel, Nyquist's formula gives the maximum bit rate from the bandwidth B and the number of signal levels L:

\[ \begin{aligned} R &= 2B \log_2 L \\ &= 2 \times 3000 \times \log_2 4 \\ &= 6000 \times 2 = 12000\ \text{bps} \end{aligned} \]

6,000 bps is the signalling (baud) rate 2B, forgetting that each level carries 2 bits. 8,000 and 10,000 bps do not follow from the formula.

Answer: A

The minimum Hamming distance is 7.

To detect s errors a code needs \(d_{min} = s + 1\); to correct t errors it needs \(d_{min} = 2t + 1\).

\[ \begin{aligned} \text{detect 6:}\ d_{min} &= 6 + 1 = 7 \\ \text{correct 2:}\ d_{min} &= 2(2) + 1 = 5 \end{aligned} \]

A distance of 7 meets both requirements, while 5 allows correcting two errors but detects only up to four. 3 and 1 are too small for either.

Answer: D

A full-speed burst can last 25 msec.

During a burst at the maximum rate M, the bucket of capacity C empties while being refilled at the token rate \(\rho\), so the burst length S satisfies \(C + \rho S = M S\):

\[ \begin{aligned} S &= \frac{C}{M - \rho} = \frac{10^{6}}{(50 - 10) \times 10^{6}} \\ &= \frac{1}{40}\ \text{s} = 25\ \text{ms} \end{aligned} \]

Ignoring the refill and dividing 1 MB by 50 MB/s gives 20 ms, which is not offered; 25 sec and 50 sec are off by a factor of 1,000, and 50 msec does not follow from the formula.

Answer: B

The mechanism is Selective Repeat ARQ.

In Selective Repeat ARQ the receiver buffers correctly received frames that arrive out of order and the sender retransmits only the frames that were damaged or lost. On noisy channels, where errors are frequent, this avoids resending good frames and makes better use of the link.

Go-Back-N resends the damaged frame and every frame after it, and Stop-and-Wait sends one frame at a time and waits for each acknowledgement, so neither restricts retransmission to the damaged frames.

Answer: B

The propagation time is 50 ms.

Propagation time depends only on the distance and the signal speed, not on the size of the message or the bandwidth:

\[ \begin{aligned} t_p &= \frac{12000 \times 10^{3}\ \text{m}}{2.4 \times 10^{8}\ \text{m/s}} \\ &= 0.05\ \text{s} = 50\ \text{ms} \end{aligned} \]

The 2.5 kbyte attachment and the 1 Gbps bandwidth decide the transmission time instead: \(2.5 \times 8000 / 10^{9} = 20\ \mu s\), which is negligible here. None of 200, 208 or 600 ms follows from the given distance and speed.

Answer: B

The throughput is 2 Mbps.

Throughput is the rate at which data actually passes, whatever the bandwidth:

\[ \begin{aligned} T &= \frac{12000 \times 10000\ \text{bits}}{60\ \text{s}} \\ &= 2{,}000{,}000\ \text{bits/s} = 2\ \text{Mbps} \end{aligned} \]

So the link carries only a fifth of its 10 Mbps bandwidth. 20 Mbps would exceed the bandwidth itself, 0.2 Mbps is off by a factor of ten, and 0.83 Mbps comes from no consistent use of the figures.

Answer: C

The only layer that adds no header is the Physical layer.

As data passes down the OSI stack, each layer from application to data link adds its own control information: the transport layer adds a segment header with ports, the network layer a packet header with logical addresses, and the data link layer both a header and a trailer to form a frame.

The physical layer simply turns the bits of the frame into signals on the medium; it does not add a header of its own. So the application, network and transport layers are wrong.

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.

Answer: A

Correct answer: 1

A switch breaks up collision domains, not broadcast domains: a broadcast frame arriving on one port is flooded out of all the others, so all twelve ports remain in a single broadcast domain by default. Splitting that domain requires either a router or VLANs configured on the switch, each VLAN forming a broadcast domain of its own. The complementary question - how many collision domains - gives twelve.

Answer: A

Correct answer: 1

Spanning tree elects exactly one root bridge per broadcast domain, however many switches there are, and every other switch computes its shortest path to that root. The election is by lowest bridge ID - priority first, then MAC address as the tiebreaker - which is why the priority is normally set deliberately on the switch that should win, rather than left to the accident of which switch has the lowest MAC address.