Networking Basics Questions and Answers

91.

The message 11001001 is to be transmitted using the CRC polynomial \(x^{3} + 1\) to protect it from errors. The message that should be transmitted is

A.110010011001
B.11001001
C.110010011001001
D.11001001011

Exam Year(s)

UGC NET - 2013,

Answer: D

The transmitted message is 11001001011. The generator \(x^{3} + 1\) is 1001, of degree 3, so three zeros are appended to the message, giving 11001001000. Dividing by 1001 using modulo-2 arithmetic leaves the remainder 011.

\[ 11001001000 \div 1001 \Rightarrow \text{remainder } 011 \]

The remainder is placed in the place of the appended zeros, so the frame is 11001001011. Option B is the unprotected message and options A and C do not have 3 check bits appended.

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92.

What is the bit rate for transmitting uncompressed \(800 \times 600\) pixel colour frames with 8 bits/pixel at 40 frames/second ?

A.2.4 Mbps
B.15.36 Mbps
C.153.6 Mbps
D.1536 Mbps

Exam Year(s)

UGC NET - 2013,

Answer: C

The bit rate is 153.6 Mbps. Each frame has \(800 \times 600 = 480000\) pixels at 8 bits per pixel, giving 3,840,000 bits per frame. At 40 frames per second:

\[ 3\,840\,000 \times 40 \]

\[ = 153\,600\,000 \text{ bits/s} \]

That is 153.6 Mbps. The figure 15.36 Mbps is ten times too small, 1536 Mbps is ten times too large, and 2.4 Mbps does not follow from the data.

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93.

Which layer of OSI reference model is responsible for decomposition of messages and generation of sequence numbers to ensure correct re-composition from end to end of the network ?

A.Physical
B.Data-link
C.Transport
D.Application

Exam Year(s)

UGC NET - 2013,

Answer: C

The transport layer is responsible for breaking messages into segments, generating sequence numbers and reassembling them correctly from end to end. The physical layer carries raw bits, the data-link layer handles frames between adjacent nodes, and the application layer provides services to user programs.

Only the transport layer provides this end-to-end segmentation and sequencing between the two hosts.

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94.

A client-server system uses a satellite network, with the satellite at a height of 40,000 kms. What is the best-case delay in response to a request ? (Note that the speed of light in air is 3,00,000 km/second).

A.133.33 m sec
B.266.67 m sec
C.400.00 m sec
D.533.33 m sec

Exam Year(s)

UGC NET - 2013,

Answer: D

The best-case delay is 533.33 msec. A request goes from the client up to the satellite (40,000 km) and down to the server (40,000 km), and the response makes the same trip, so the signal travels \(4 \times 40\,000 = 160\,000\) km.

\[ t = \frac{160\,000}{300\,000} = 0.5333 \text{ s} \]

\[ t = 533.33 \text{ msec} \]

133.33 msec is a single 40,000 km hop, 266.67 msec is a one-way trip of 80,000 km, and 400 msec corresponds to 120,000 km, so none of these covers the full round trip.

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95.

The start and stop bits are used in serial communication for

A.error detection
B.error correction
C.synchronization
D.slowing down the communication

Exam Year(s)

UGC NET - 2013,

Answer: C

Start and stop bits are used in serial (asynchronous) communication for synchronization. The start bit tells the receiver that a character is beginning, so it can start sampling at the right moment, and the stop bit marks the end of the character and returns the line to idle.

They are not used for error detection or correction (parity bits do detection), and they do not slow the communication down; in fact they add a little overhead.

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96.

_________ is a type of transmission impairment in which the signal looses strength due to the resistance of the transmission medium.

A.Attenuation
B.Distortion
C.Noise
D.Decibel

Exam Year(s)

UGC NET - 2013,

Answer: A

The impairment in which the signal loses strength because of the resistance of the transmission medium is attenuation. Energy is converted to heat as the signal travels, so its amplitude falls with distance, and amplifiers or repeaters are used to restore it.

Distortion is a change in the shape of the signal, noise is unwanted energy added to the signal, and the decibel is only a unit used to measure gain or loss.

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97.

In a fully connected mesh network with n devices, there are ________ physical channels to link all devices.

A.n(n-1)/2
B.n(n+1)/2
C.2n
D.2n+1

Exam Year(s)

UGC NET - 2013,

Answer: A

In a fully connected mesh network with n devices, every device is linked to each of the other n - 1, and each link is shared by two devices, so the number of physical channels is \(\frac{n(n - 1)}{2}\). The answer is (A).

The count \(\frac{n(n + 1)}{2}\) in (B) would include a link from each device to itself, and 2n and 2n + 1 in (C) and (D) grow only linearly, which is the number for a ring or a star with a spare link, not for a full mesh.

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98.

The baud rate of a signal is 600 baud/second. If each signal unit carries 6 bits, then the bit rate of a signal is ________.

A.3600
B.100
C.6/600
D.None of the above

Exam Year(s)

UGC NET - 2013,

Answer: A

The bit rate is the baud rate multiplied by the number of bits carried by each signal unit:

\[ 600 \times 6 = 3600 \text{ bps} \]

so option (A) is the answer. The value 100 in (B) is 600 divided by 6, which is what the baud rate would be if the bit rate were 600, and 6/600 in (C) is the inverse of the baud rate per bit and is not a rate in bits per second. So (D) 'none of the above' is also wrong.

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99.

Match the following :

a. Data link layer	i. Flow control
b. Network layer	ii. Node to node delivery
c. Transport layer	iii. Mail services
d. Application layer	iv. Routing

Codes : a b c d

A.ii i iv iii
B.ii iv i iii
C.ii i iii iv
D.ii iv iii i

Exam Year(s)

UGC NET - 2013,

Answer: B

The correct matching is the data link layer with node to node delivery (ii), the network layer with routing (iv), the transport layer with flow control (i) and the application layer with mail services (iii). This is code ii iv i iii, option (B).

Options (A) and (C) give the network layer flow control or mail services, and option (D) gives the transport layer mail services. Routing is the work of the network layer, so only (B) is right on the second pair.

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100.

An image is 1024∗800 pixels with 3 bytes/pixel. Assume the image is uncompressed. How long does it take to transmit it over a 10-Mbps Ethernet ?

A.196.6 seconds
B.19.66 seconds
C.1.966 seconds
D.0.1966 seconds

Exam Year(s)

UGC NET - 2013,

Answer: C

The image has \(1024 \times 800\) pixels of 3 bytes each, so its size is

\[ 1024 \times 800 \times 3 = 24{,}57{,}600 \text{ bytes} \]

that is 196,60,800 bits. At 10 Mbps the time taken is

\[ \frac{19{,}660{,}800}{10 \times 10^{6}} \]

\[ = 1.966 \text{ seconds} \]

This is option (C). The other values are out by a factor of 10 or 100, which is what comes from taking the wrong number of bits for a byte or the wrong power of ten for the rate.

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