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Test: Data Link Layer & MAC-Sublayer- 2 - Computer Science Engineering (CSE) MCQ


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15 Questions MCQ Test GATE Computer Science Engineering(CSE) 2025 Mock Test Series - Test: Data Link Layer & MAC-Sublayer- 2

Test: Data Link Layer & MAC-Sublayer- 2 for Computer Science Engineering (CSE) 2024 is part of GATE Computer Science Engineering(CSE) 2025 Mock Test Series preparation. The Test: Data Link Layer & MAC-Sublayer- 2 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Test: Data Link Layer & MAC-Sublayer- 2 MCQs are made for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Data Link Layer & MAC-Sublayer- 2 below.
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Test: Data Link Layer & MAC-Sublayer- 2 - Question 1

If an ethernet destination address is 07-01 -12 - 03 - 04 - 05, then that is a _____ address,

Detailed Solution for Test: Data Link Layer & MAC-Sublayer- 2 - Question 1

07-01 - 12 - 03 - 04 - 05. The ethernet has more than one destinations but not all. So it is a multicast address not a broadcast address.

Test: Data Link Layer & MAC-Sublayer- 2 - Question 2

If N is the maximum sequence number, then the window sizes in selective reject and Go-Back-N protocols are respectively:

Detailed Solution for Test: Data Link Layer & MAC-Sublayer- 2 - Question 2

Where n is maximum sequence number selective repeat

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Test: Data Link Layer & MAC-Sublayer- 2 - Question 3

CRC can detect all bursts of upto m errors, if generator polynomial G(x) is of degree

Detailed Solution for Test: Data Link Layer & MAC-Sublayer- 2 - Question 3

CRC guarantees that all burst error of length equal to the degree of the polynomials are detected and also burst errors affecting an odd number of bits are detected.

Test: Data Link Layer & MAC-Sublayer- 2 - Question 4

In a Go-back-N ARQ, if the window size is 63, what is the range of sequence numbers?

Detailed Solution for Test: Data Link Layer & MAC-Sublayer- 2 - Question 4

In Go back-N maximum window size 2N- 1
And range of sequence member 0 to 2N- 1
So, 0 to 63.

Test: Data Link Layer & MAC-Sublayer- 2 - Question 5

For stop and wait ARQ, for n data packets sent, ______________ acknowledgments are needed.

Detailed Solution for Test: Data Link Layer & MAC-Sublayer- 2 - Question 5

It n data packets sent we need less than n data packet acknowledgments.

Test: Data Link Layer & MAC-Sublayer- 2 - Question 6

In token ring, the tokens can be removed by

Detailed Solution for Test: Data Link Layer & MAC-Sublayer- 2 - Question 6

In token ring N number of system choose 1 monitor. The responsibility of this monitor is
(i) Lost of Token,
(ii) Garbage frame,
(iii) Orphan frame.
If a station is disabled the token generated by it is removed from the token rina,

Test: Data Link Layer & MAC-Sublayer- 2 - Question 7

Who strips the data frame from the token ring network?

Detailed Solution for Test: Data Link Layer & MAC-Sublayer- 2 - Question 7

The station who sending data strips the data frame from the token ring network.

Test: Data Link Layer & MAC-Sublayer- 2 - Question 8

Which of the following indicates the increasing order of accuracy in error detection?

Detailed Solution for Test: Data Link Layer & MAC-Sublayer- 2 - Question 8

Single parity can detect single bit error, while block sum check can more than one bit error and CRC can detect all errors present in data.
So. sinale Daritv < Block sum check < CRC.

Test: Data Link Layer & MAC-Sublayer- 2 - Question 9

A selective repeat ARQ is using 7 bits to represent the sequence numbers. What is the maximum size of the window?

Detailed Solution for Test: Data Link Layer & MAC-Sublayer- 2 - Question 9

With selective repeat ARQ of 7 bits. Maximum window size = 27-1 - 26 = 64 Maximum sequence number = (64 -1) = 63.

Test: Data Link Layer & MAC-Sublayer- 2 - Question 10

Which error detection method consists of a parity bit for each data unit as well as an entire data unit of parity bits?

Detailed Solution for Test: Data Link Layer & MAC-Sublayer- 2 - Question 10

Two dimensional parity check consist parity bits for each data unit as well as an entire data unit of parity bits.

Test: Data Link Layer & MAC-Sublayer- 2 - Question 11

The reference polynomial used in a CRC scheme  is x4 + x3 + 1. A data sequence 1010101010 is to be sent, Determine the actual bit string that is transmitted.

Detailed Solution for Test: Data Link Layer & MAC-Sublayer- 2 - Question 11

Reference polynomial x4 + r3 + 1 =11001
Datasequence = 1010101010
On dividing 1010101010 by 11001,
We get CRC (i.e. remainder) = 0010
On appending this CRC on data sequence, we get the actual message transmitted i.e. 10101010100010.

Test: Data Link Layer & MAC-Sublayer- 2 - Question 12

A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 μ sec/km.
The minimum number of bits required in the sequence number field of the packet is

Detailed Solution for Test: Data Link Layer & MAC-Sublayer- 2 - Question 12

First we find the Transmission Delay

= Packet size / Bandwidth

= 64 bytes / 1.536 Mbps

= (64 × 8 bits) / (1.536 × 106 bits per sec)

= 333.33 μsec

Then we find the Propagation Delay

For 1 km, propagation delay = 6 μsec

For 3000 km, propagation delay = 3000 x 6 μsec = 18000 μsec

Value of a is calculated by  :

= Tp / Tt

= 18000 μsec / 333.33 μsec

= 54

Bits required in sequence number field

= ⌈log2(1+2 x a)⌉

= ⌈log2(1 + 2 x 54)⌉

= ⌈log2(109)⌉

= ⌈6.76⌉

= 7 bits

Hence, option 2) is correct.

Test: Data Link Layer & MAC-Sublayer- 2 - Question 13

A 3000 km long trunk is used to transmit frames using a Go-Back-N protocol. The propagation speed is 6 jasec/ km and trunk data rate is 1.544 Mbps. We ignore the time taken to receive the bits in the acknowledgment. Frame size is 64 Bytes.

What is the maximum number of bits of the sequence number?

Detailed Solution for Test: Data Link Layer & MAC-Sublayer- 2 - Question 13

Bits for sequence no. = 2no of bits > Window size
=>    W = 110
27 = 128 > 110
Hence 7 bits used for sequence number.

Test: Data Link Layer & MAC-Sublayer- 2 - Question 14

A channel has bit rate of 1 Mbps and propagation delay of 270 msec. Frame size is of 125 bytes. Acknowledgment is always piggybacked onto data frames. Four bit sequence number is used, ignore header size.

What is the maximum achievable channel utilization for Go-Back-N?

Detailed Solution for Test: Data Link Layer & MAC-Sublayer- 2 - Question 14

Data rate = 1 Mbps
Tp = 270 msec
F = 125 bytes
= 125 x 8 bits
= 1000 bits

Four bit sequence number
For Go-back-N, W= 16-1 = 15

Test: Data Link Layer & MAC-Sublayer- 2 - Question 15

A channel has bit rate of 1 Mbps and propagation delay of 270 msec. Frame size is of 125 bytes. Acknowledgment is always piggybacked onto data frames. Four bit sequence number is used, ignore header size.

What is the maximum achievable channel utilization for Selective Repeat?

Detailed Solution for Test: Data Link Layer & MAC-Sublayer- 2 - Question 15

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