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Computer Science Engineering (CSE) Exam  >  Computer Science Engineering (CSE) Test  >  GATE Computer Science Engineering(CSE) 2027 Mock Test Series  >  Test: Data Link Layer & MAC-Sublayer- 2 - Computer Science Engineering (CSE) MCQ

GATE Computer Science Engineering(CSE) 2027 Test: Data Link Layer & MAC-Sublayer-


MCQ Practice Test & Solutions: Test: Data Link Layer & MAC-Sublayer- 2 (15 Questions)

You can prepare effectively for Computer Science Engineering (CSE) GATE Computer Science Engineering(CSE) 2027 Mock Test Series with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Data Link Layer & MAC-Sublayer- 2". These 15 questions have been designed by the experts with the latest curriculum of Computer Science Engineering (CSE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 45 minutes
  • - Number of Questions: 15

Sign up on EduRev for free to attempt this test and track your preparation progress.

Test: Data Link Layer & MAC-Sublayer- 2 - Question 1
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If an ethernet destination address is 07-01 -12 - 03 - 04 - 05, then that is a _____ address,

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

Detailed Solution: Question 2

Where n is maximum sequence number selective repeat

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Test: Data Link Layer & MAC-Sublayer- 2 - Question 3
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CRC can detect all bursts of upto m errors, if generator polynomial G(x) is of degree

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

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

Detailed Solution: Question 5

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

Test: Data Link Layer & MAC-Sublayer- 2 - Question 6
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In token ring, the tokens can be removed by

Detailed Solution: 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
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Who strips the data frame from the token ring network?

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

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

Detailed Solution: 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
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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: 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
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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: 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
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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: 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
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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: Question 13

Test: Data Link Layer & MAC-Sublayer- 2 - Question 14
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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: 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
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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: Question 15

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About this Computer Science Engineering (CSE) Practice Test

This test contains 15 MCQs on Data Link Layer & MAC-Sublayer- 2 with detailed solutions for each question. It is part of the GATE Computer Science Engineering(CSE) 2027 Mock Test Series course on EduRev, designed to align with the current Computer Science Engineering (CSE) syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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