Error Detection and Correction MCQ in Data Link Layer

1. Which one of the following is not error detection technique…

  • A. Hamming code
  • B. Cyclic Redundancy Check
  • C. Checksum
  • D. Stop and Wait ARQ
  • E. None of the above
Error can be detect by using Hamming code, Cyclic Redundancy Check, Checksum but Stop And Wait technique is used in flow control.

2. Which one of the following is error correction technique…

  • A. Hamming code
  • B. Cyclic Redundancy Check
  • C. Checksum
  • D. Stop and Wait ARQ
  • E. None of the above

3. Hamming code technique is used to correct how many bit error…

  • A. One bit
  • B. Two bit
  • C. Three bit
  • D. Four bit
  • E. None of the above

4. What is the minimum hamming distance of the string 110110,001001,011101 and 110101…

  • A. One
  • B. Two
  • C. Three
  • D. Four
  • E. None of the above
We have to given four binary string
110110
001001
011101
110101
After performing XOR between all string we find minimum hamming distance two.

5. Minimum Hamming distance (Data link layer) required to correct ‘m’ bit error in given string…

  • A. 2m+1
  • B. m+1
  • C. 2m+3
  • D. m+3
  • E. None of the above

6. Minimum Hamming distance (Data link layer) required to detect ‘m’ bit error in given string…

  • A. 2m+1
  • B. m+1
  • C. 2m+3
  • D. m+3
  • E. None of the above

7. How many bit error will be correct if we have minimum hamming distance is 5.

  • A. 2
  • B. 4
  • C. 3
  • D. 5
  • E. None of the above

Solution:- Minimum Hamming distance (Data link layer) required to correct ‘m’ bit error is 2m+1

in above question we have given minimum hamming distance is 5 put this value in the formula

5=2m+1 => m=2

8. How many bit error will be detect if we have minimum hamming distance is 5.

  • A. 2
  • B. 4
  • C. 3
  • D. 5
  • E. None of the above

Solution:- Minimum Hamming distance (Data link layer) required to Detect ‘m’ bit error is m+1

in above question we have given minimum hamming distance is 5 put this value in the formula

5=m+1 => m=4

9. Each original block string of k bit in error detection technique is called…

  • A. Codeword
  • B. Syndrom
  • C. Dataword
  • D. Generator
  • E. None of the above

10. In a error detection technique we add r redundant bit to data block now the sending string is called

  • A. Codeword
  • B. Syndrom
  • C. Dataword
  • D. Generator
  • E. None of the above

11. Difference of bits between sending and receiving bits in strings is called…

  • A. Hamming code
  • B. Cyclic Redundancy Check
  • C. Checksum
  • D. Hamming distance
  • E. None of the above

12. What is the hamming distance of detection 15 bit error.

  • A. 16
  • B. 17
  • C. 31
  • D. 15
  • E. None of the above

Solution:- Minimum Hamming distance (Data link layer) required to Detect ‘m’ bit error is m+1

in above question we have given minimum hamming distance is 5 put this value in the formula

Minimum Hamming distance=m+1 => 15+1 =>m = 16

13 . What is the minimum hamming distance of the string 110110,001001,011101 and 110101 and how many bit error will be detect by this technique..

  • A. One, One
  • B. Two , One
  • C. Three, two
  • D. Four, One
  • E. None of the above

We have to given four binary string
110110
001001
011101
110101
After performing XOR between all string we find minimum hamming distance two.

Minimum Hamming distance (Data link layer) required to Detect ‘m’ bit error is m+1

in above question we have given minimum hamming distance is 2 put this value in the formula

Minimum Hamming distance=m+1 => 2 = m+1 =>m=1

14. Simple parity check technique is used to detect how many bit error.

  • A. Odd bit error
  • B. Even bit error
  • C. One bit error
  • D. Two bit error
  • E. None of the above

15. In which technique if single bit is rotated than a new codeword is found…

  • A. Hamming code
  • B. Cyclic
  • C. Checksum
  • D. Stop and Wait ARQ
  • E. None of the above

16. In error detection technique we are using block coding technique. if n=4 and k=3 than invalid codewords are…

  • A. 6
  • B. 7
  • C. 8
  • D. 16
  • E. None of the above
In a block coding technique we add r redundant bit in original datagram and send this datagram as codeword.
here n is total code words which is 2n and and divide each string in k bit block datagram
Total codeword is 2n and total blocks are 2k
Total invalid codewords are = 2n-2k
Now put the value we found 24-23=8

17. Three linear codeword are given as follows…

  1. 0011
  2. 0101
  3. 1010

which one of the following is not a linear code.

  • A. 1111
  • B. 0001
  • C. 1001
  • D. 0110
  • E. None of the above

Solution:- In error detection technique the sum of two codeword is a valid code word.

Here we have given three codeword is are

code 1 is 0011

code 2 is 0101

code 3 is 1010

when XOR any of these two codewords we found a valid code word.

code 1 XOR code 2=> 0011 XOR 0101=0110

code 1 XOR code 3=> 0011 XOR 1010=1001

code 2 XOR code 3=> 0101 XOR 1010=1111

so 0001 is not valid linear code word

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