Match the following OSI layers with protocols :(a) Layer 7 (Application)(i) IPv4(b) Layer 4 (Transport)(ii) UPP(c) Layer 3 (Network)(iii) HTTP(d) Layer 2 (Data Link)(iv) Ethernet, WiFi (802.11)

The correct matching is a-(iii), b-(ii), c-(i), d-(iv).Layer 7, application: HTTP.Layer 4, transport: UDP (printed 'UPP').Layer 3, network: IPv4.Layer 2, data link: Ethernet and Wi-Fi (802.11).Options A, B and C all put Ethernet and Wi-Fi at the application layer and HTTP or IPv4 at the wrong layers, so only D is right.

Networking Basics Questions and Answers

Q: A noiseless channel with bandwidth of 3000 Hz transmitting a signal with four signal levels (for each level we send 2 bits). Then the maximum bit rate is :

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.

Q: What is the minimum Hamming distance for detection of six errors or correction of two errors ?

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.

Q: Imagine a flow specification that has a maximum packet size of 1000 bytes, a token bucket rate of 10 million bytes/sec, a token bucket size of 1 million bytes, and a maximum transmission rate of 50 million bytes/sec. How long can a burst at maximum speed last ?

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.

Q: Which ARQ mechanism deals with the transmission of only damaged or lost frames despite the other multiple frames by increasing the efficiency and its utility in noisy channels ?

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.

Q: Suppose we have to send an E-mail with an attachment of 2.5 kbyte with 1 Gbps bandwidth. Assume that the distance between the sender and the receiver is 12000 km and the light travels at 2.4 × 108 m/s. What will be the propagation time ?

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.

Q: We have a LAN with 10 Mbps bandwidth and average 12000 frames per minutes and size of each frame is 10000 bits. Then its throughput will be ................... .

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.

Q: Which is the only layer of OSI model that prevents itself from adding its own header to the data during the data transmission process ?

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.

Q: Suppose we have to send an E-mail with an attachment of 2.5 kbyte with 1 Gbps bandwidth. Assume that the distance between the sender and the receiver is 12000 km and that light travels at \(2.4 \times 10^{8}\) m/s. What will be the transmission time ?

The transmission time is 0.020 ms.Transmission time is the time to put all the bits on the link, message size divided by bandwidth:\[ \begin{aligned} T_t &= \frac{2500 \times 8}{10^{9}} = \frac{20000}{10^{9}} \text{ s} \\ &= 20\ \mu\text{s} = 0.020 \text{ ms} \end{aligned} \]The distance and the speed of light give the propagation time instead, which is the trap:\[ \begin{aligned} T_p &= \frac{12 \times 10^{6}}{2.4 \times 10^{8}} \\ &= 0.05 \text{ s} = 50 \text{ ms} \end{aligned} \]So 50 ms answers a different question, and 2 ms and 0.2 ms are off by factors of 100 and 10.

Q: If the frame to be transmitted is 1101011011 and the CRC polynomial to be used for generating checksum is \(x^{4} + x + 1\), then what is the transmitted frame ?

The transmitted frame is 11010110111110.The generator \(x^{4} + x + 1\) is the bit pattern 10011, of degree 4. Append four 0s to the frame and divide by 10011 using modulo-2 (XOR) division:\[ \begin{aligned} &11010110110000 \div 10011 \\ &\text{remainder} = 1110 \end{aligned} \]Replacing the four appended 0s with the remainder gives the transmitted frame 1101011011 1110. The receiver divides it by 10011 and gets remainder 0 if no error occurred.The other options keep the original frame but attach the wrong check bits (1011, 1101 or 1001), so their division by 10011 would not leave remainder 0.

21.

A noiseless channel with bandwidth of 3000 Hz transmitting a signal with four signal levels (for each level we send 2 bits). Then the maximum bit rate is :

A.6,000 bps
B.8,000 bps
C.10,000 bps
D.12,000 bps

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

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22.

What is the minimum Hamming distance for detection of six errors or correction of two errors ?

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

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23.

Imagine a flow specification that has a maximum packet size of 1000 bytes, a token bucket rate of 10 million bytes/sec, a token bucket size of 1 million bytes, and a maximum transmission rate of 50 million bytes/sec. How long can a burst at maximum speed last ?

A.50 sec
B.50 msec
C.25 sec
D.25 msec

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

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24.

Which ARQ mechanism deals with the transmission of only damaged or lost frames despite the other multiple frames by increasing the efficiency and its utility in noisy channels ?

A.Go-Back-N ARQ
B.Selective Repeat ARQ
C.Stop-and-Wait ARQ
D.All of the above

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

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25.

Suppose we have to send an E-mail with an attachment of 2.5 kbyte with 1 Gbps bandwidth. Assume that the distance between the sender and the receiver is 12000 km and the light travels at 2.4 × 10⁸ m/s. What will be the propagation time ?

A.208 ms
B.50 ms
C.200 ms
D.600 ms

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

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.

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26.

We have a LAN with 10 Mbps bandwidth and average 12000 frames per minutes and size of each frame is 10000 bits. Then its throughput will be ................... .

A.0.83 Mbps
B.2 Mbps
C.0.2 Mbps
D.20 Mbps

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

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.

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27.

Which is the only layer of OSI model that prevents itself from adding its own header to the data during the data transmission process ?

A.Application layer
B.Network layer
C.Physical layer
D.Transport layer

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

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.

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28.

Suppose we have to send an E-mail with an attachment of 2.5 kbyte with 1 Gbps bandwidth. Assume that the distance between the sender and the receiver is 12000 km and that light travels at \(2.4 \times 10^{8}\) m/s. What will be the transmission time ?

A.50 ms
B.2 ms
C.0.020 ms
D.0.2 ms

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The transmission time is 0.020 ms.

Transmission time is the time to put all the bits on the link, message size divided by bandwidth:

\[ \begin{aligned} T_t &= \frac{2500 \times 8}{10^{9}} = \frac{20000}{10^{9}} \text{ s} \\ &= 20\ \mu\text{s} = 0.020 \text{ ms} \end{aligned} \]

The distance and the speed of light give the propagation time instead, which is the trap:

\[ \begin{aligned} T_p &= \frac{12 \times 10^{6}}{2.4 \times 10^{8}} \\ &= 0.05 \text{ s} = 50 \text{ ms} \end{aligned} \]

So 50 ms answers a different question, and 2 ms and 0.2 ms are off by factors of 100 and 10.

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29.

Match the following OSI layers with protocols :

(a) Layer 7 (Application)	(i) IPv4
(b) Layer 4 (Transport)	(ii) UPP
(c) Layer 3 (Network)	(iii) HTTP
(d) Layer 2 (Data Link)	(iv) Ethernet, WiFi (802.11)

A.a-(iv), b-(iii), c-(i), d-(ii)
B.a-(iv), b-(iii), c-(ii), d-(i)
C.a-(iv), b-(ii), c-(iii), d-(i)
D.a-(iii), b-(ii), c-(i), d-(iv)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The correct matching is a-(iii), b-(ii), c-(i), d-(iv).

Layer 7, application: HTTP.
Layer 4, transport: UDP (printed 'UPP').
Layer 3, network: IPv4.
Layer 2, data link: Ethernet and Wi-Fi (802.11).

Options A, B and C all put Ethernet and Wi-Fi at the application layer and HTTP or IPv4 at the wrong layers, so only D is right.

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30.

If the frame to be transmitted is 1101011011 and the CRC polynomial to be used for generating checksum is \(x^{4} + x + 1\), then what is the transmitted frame ?

A.11010110111110
B.11010110111011
C.11010110111101
D.11010110111001

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The transmitted frame is 11010110111110.

The generator \(x^{4} + x + 1\) is the bit pattern 10011, of degree 4. Append four 0s to the frame and divide by 10011 using modulo-2 (XOR) division:

\[ \begin{aligned} &11010110110000 \div 10011 \\ &\text{remainder} = 1110 \end{aligned} \]

Replacing the four appended 0s with the remainder gives the transmitted frame 1101011011 1110. The receiver divides it by 10011 and gets remainder 0 if no error occurred.

The other options keep the original frame but attach the wrong check bits (1011, 1101 or 1001), so their division by 10011 would not leave remainder 0.

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