Network Security Interview Subcategories

Freshers	Network Security Interview
Network Hardware	Experienced
Classful Addressing

What are the bits transmitted for the return to zero system?

A. 1001
B. 10110
C. 10010
D. 1101

Explanation: First half low – 0, First Half High – 1.

What are the bits transmitted in the differential Manchester coding?

A. 1101
B. 101100
C. 110010
D. 10011

Explanation: -ve to +ve transition – 1 bit, +ve to –ve – 0 bit.

What are the bits transmitted for the AMI and Pseudoternary System?

A. 101101
B. 101100
C. 10010
D. None of the mentioned

Explanation: Pseudoternary Has alternating high and low levels as 0 and neutral level as logic 1. AMI has Alternating high levels as logic 1 and neutral level as logic 0.

Five channels, each with a 100-kHz bandwidth, are to be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10 kHz between the channels to prevent interference?

A. 4600 khz
B. 500 khz
C. 540 khz
D. 580 khz

Explanation: For five channels, we need at least four guard bands. This means that the required bandwidth is at least 5 × 100 + 4 × 10 = 540 kHz.

Find the checksum byte for the fallowing data words: 10110011, 10101011, 01011010, 11010101

A. 10001101
B. 1110010
C. 10110101
D. 1110010

Explanation: Do binary addition to compute the result.

Generate the CRC codeword for the message x3+1 using the generator polynomial x3+x+1.

A. 1001101
B. 1001001
C. 1001110
D. 1001111

Explanation: M(x) = 1001. G(x) = 1011. x3*M(x) = 1001000. On Dividing we get remainder as 110. Therefore, Codeword is 1001110.

What is the hamming distance between these 2 codes: 10010010 and 11011001?

A. 3
B. 4
C. 6
D. 2

Explanation: Hamming distance is number of dissimilar bits between 2 streams.

What is the hamming code for the data: 1001101?

A. 10011100101
B. 11010000101
C. 10001100101
D. 11111001011

Explanation: Find the 1st 2nd 4th and 8th bits using the hamming algorithm and thus proceed to get the hamming code.

10010100101 is the code received. Find the error bit.

A. 7
B. 5
C. 2
D. 3

Explanation: Error occurs in bit 7 and can be found via hamming code.

The channel capacity is 100Mbps, the frame length is 10000 bits and the arrival rate is 5000 frames/sec. Calculate the mean time delay.

A. 400 μsec
B. 20 msec
C. 2000 μsec
D. 200 μsec

Explanation: T = 200 µsec from T= 1/(μC-λ).