Category: Cryptography & Network Security Objective Questions

Network Security Inteview Questions freshers on “Layers”.

1. What is the hamming distance between these 2 codes: 10010010 and 11011001?
a) 3
b) 4
c) 6
d) 2

Answer: b
Clarification: Hamming distance is number of dissimilar bits between 2 streams.

2. What is the hamming code for the data: 1001101?
a) 10011100101
b) 11010000101
c) 10001100101
d) 11111001011

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

3. 10010100101 is the code received. Find the error bit.
a) 7
b) 5
c) 2
d) 3

Answer: a
Clarification: Error occurs in bit 7 and can be found via hamming code.

4. 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

Answer: d
Clarification: T = 200 µsec from T= 1/(μC-λ).

5. 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

Answer: c
Clarification: 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.

6. Find the checksum byte for the fallowing data words: 10110011, 10101011, 01011010, 11010101
a) 10001101
b) 01110010
c) 10110101
d) 01110010

Answer: a
Clarification: Do binary addition to compute the result.

7. Generate the CRC codeword for the message x³+1 using the generator polynomial x³+x+1.
a) 1001101
b) 1001001
c) 1001110
d) 1001111

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

8. Generate the CRC codeword for the data word 1101011011 using generator 10011. Also write both in the polynomial form.
a) 11010110110011
b) 11010110110110
c) 11010110111100
d) 11010110111110

Answer: d
Clarification: Solve using polynomial division and then appending the remainder to the divisor.

Network Security online quiz on “IEEE 802.11 WLAN”.

1. Which layer in the IEEE 802.11 protocol stack has the function of flow control and error control?
a) Physical Layer
b) Logic Link Control Layer
c) Medium Access Layer
d) None of the mentioned

Answer: b
Clarification: Logic Link Layer has the function of flow control and error control.

2. With respect to IEEE 802.11 Wireless LAN, MSDU stands for-
a) MAC service data unit.
b) Main server data user
c) Multiframe service datagram usage
d) MAC server device usage

Answer: a
Clarification: MSDU stands for MAC service data unit.

3. Frequency band definition and Wireless signal encoding are functions of which layer?
a) Physical Layer
b) Logic Link Control Layer
c) Medium Access Layer
d) None of the mentioned

Answer: a
Clarification: Frequency band definition and Wireless signal encoding are functions of the Physical Layer.

4. The correct order of the of the MAC header is-
a) MAC Control, Destination MAC Address, Source MAC Address
b) Destination MAC Address, Source MAC Address, MAC Control
c) Source MAC Address, Destination MAC Address, MAC Control
d) none of the mentioned

Answer: a
Clarification: The correct order of arrangement is MAC Control, Destination MAC Address, Source MAC Address.

5. CRC is a component of the MAC trailer.
a) True
b) False

Answer: a
Clarification: CRC is a component of the MAC trailer. The statement is true.

6. Reliable data delivery and Wireless access control protocols are functions of which layer?
a) Physical Layer
b) Logic Link Control Layer
c) Medium Access Layer
d) None of the mentioned

Answer: c
Clarification: Reliable data delivery and Wireless access control protocols are functions of the Medium Access Layer.

7. Which layer keeps track of the frames that have been transmitted and received?
a) Physical Layer
b) Logic Link Control Layer
c) Medium Access Layer
d) None of the mentioned

Answer: b
Clarification: The LLC keeps track of the frames that have been transmitted and received.

8. The smallest building block of a wireless LAN is –
a) Unit server set
b) Unit service set
c) Basic server set
d) Basic service set

Answer: d
Clarification: The smallest building block of a wireless LAN is Basic service set (BSS).

9. In an IBSS system all communications are done via access points (APs).
a) True
b) False

Answer: b
Clarification: IBSS stands for Independent Basic Service Set.

Cryptography Inteview Questions freshers on “The AES Algorithm”.

1. How many step function do Round 1 and 2 each have in S-AES?
a) 4 and 3
b) Both 4
c) 1 and 4
d) 3 and 4

Answer: a
Clarification: Round 1 has four step functions whereas Round 2 has three step functions.

2. For a key 25D5 and PT input A479 what is the output we obtain after the “add round key” function?
a) F34D
b) 81AC
c) 79DF
d) 327D

Answer: b
Clarification: Simply apply XOR to the state matrix PT with the key matrix to obatain the output which in this case is 81AC.

3. The output of the previous question, on passing through “nibble substitution” gets us the output
a) 3267
b) 1344
c) 64C0
d) CA37

Answer: c
Clarification: 81AC after passing through the “nibble substitution” round produces an output 64C0. A corresponding substitution is referred to in this step.

4. The output of the previous question on passing through the “shift row” step function gives us the output
a) C046
b) 0C64
c) 64C0
d) 640C

Answer: b
Clarification: 64C0 after passing through the “shift row” round produces an output 640C. A one nibble shift is performed in the second row in this step.

5. The output of the previous question on passing through the “mix columns” step function gives us the output
a) 3252
b) 3743
c) 3425
d) 3473

Answer: d
Clarification: 64C0 after passing through the “mix columns” round produces an output 3473. This is obtained by multiplying this matrix with the matrix 1441.

6. Which function can be used in AES multiplication
a) m(x)=x⁷+x⁴+x³
b) m(x)=x⁸+x⁴+x³+x+1
c) m(x)=x⁸+x³+x²+x+1
d) m(x)=x⁸+x⁵+x³+x

Answer: b
Clarification: m(x)=x⁸+x⁴+x³+x+1 stands for 100011011 which is an irreducible polynomial. Others are not irreducible polynomials.

7. The inverse transformation matrix times the forward transformation matrix equals the identity matrix.
a) True
b) False

Answer: c
Clarification: The statement is true. The inverse transformation matrix times the forward transformation matrix does equal the identity matrix.

8. Which of the following are true
i) In the AES key expansion algorithm, the function g() operates on w i+3
ii) Perform a one-byte right circular rotation on the argument 4-byte word
iii) The round constant follows the formula RC[j] = RC[j − 1]

a) i) ii)
b) ii) only
c) i) only
d) All

Answer: c
Clarification: Only the first option is true. There is a left circular rotation in AES key expansion. RC[j] = 2 × RC [j − 1] is the formula for the round constant.

9. How many round keys are generated in the AES algorithm?
a) 11
b) 10
c) 8
d) 12

Answer: a
Clarification: 11 round keys are generated. One for each of the 10 rounds and one of the initial permutations (Round 0).

Cryptography Multiple Choice Questions on “Number Theory”.

1. ᶲ(231)=
a) 230
b) 60
c) 80
d) 120

Answer: d
Clarification: ᶲ(231) = ᶲ(3) x ᶲ(7) x ᶲ(11) = 2 x 6 x 10 = 120.

2. n is prime if and only if n divides (2ⁿ – 2).
a) True
b) False

Answer: b
Clarification: This isn’t true for all cases. Take for example 341 which is non prime.

3. Find x for the CRT when x= 2 mod 3; x= 3 mod 5; x = 2 mod 7
a) 33
b) 22
c) 23
d) 31

Answer: c
Clarification: We have M = 3 x 5 x 7 = 105; M/3 = 35; M/5 = 21; M/7 = 15.
The set of linear congruences
35 x b1 = 1 (mod 3); 21 x b2 = 1 (mod 5); 15 x b3 = 1 (mod 7)
has the solutions b1 = 2; b2 = 1; b3 = 1. Then,
x = 2 x 2 c 35 + 3 x 1 x 21 + 2 x 1 x 15 = 233 (mod 105) = 23.

4. Consider a function: f(n) = number of elements in the set {a: 0 <= a < n and gcd(a,n) = 1}. What is this function?
a) Primitive
b) Totient
c) Primality
d) All of the mentioned

Answer: b
Clarification: Such a set is known as Totient.

5. The inverse of 49 mod 37 is –
a) 31
b) 23
c) 22
d) 34

Answer: d
Clarification: 49^-1 mod 37 = 34.

6. Six teachers begin courses on Monday Tuesday Wednesday Thursday Friday and Saturday, respectively, and announce their intentions of lecturing at intervals of 2,3,4,1,6 and 5 days respectively. Sunday lectures are forbidden. When first will all the teachers feel compelled to omit a lecture? Use CRT.
a) 354
b) 371
c) 432
d) 213

Answer: b
Clarification: Use CRT to get the answer as 371.

7. How many primitive roots are there for 25?
a) 4
b) 5
c) 7
d) 8

Answer: d
Clarification: 2, 3, 8, 12, 13, 17, 22, 23 are the primitive roots of 25.

Given 2 as a primitive root of 29, construct a table of discrete algorithms and solve for x in the following –

8. 17 x² = 10 ( mod 29 )
a) x = 3, 22 (mod 29)
b) x = 7, 28 (mod 29)
c) x = 2, 27 (mod 29)
d) x = 4, 28 (mod 29)

Answer: c
Clarification: On solving we get x = 2, 27 (mod 29).

9. x – 4x – 16 = 0 (mod 29)
a) x = 6, 24 (mod 29)
b) x = 9, 24 (mod 29)
c) x = 9, 22 (mod 29)
d) x = 6, 22 (mod 29)
View Answer

Answer: b
Clarification: On solving we get x = 9, 24 (mod 29).

10. x⁷ = 17 (mod 29)
a) x = 8, 9, 12, 13, 15, 24, 28 (mod 29)
b) x = 8, 10, 12, 15, 18, 26, 27 (mod 29)
c) x = 8, 10, 12, 15, 17, 24, 27 (mod 29)
d) x = 8, 9, 13, 15, 17, 24, 28 (mod 29)

Answer: b
Clarification: On solving we get x = 8, 10, 12, 15, 18, 26, 27 (mod 29).

11. The inverse of 37 mod 49 is –
a) 23
b) 12
c) 4
d) 6

Answer: c
Clarification: 37^-1 mod 49 = 4.

12. How many primitive roots are there for 19?
a) 4
b) 5
c) 3
d) 6

Answer: d
Clarification: 2, 3, 10, 13, 14, 15 are the primitive roots of 19.

Cryptography Multiple Choice Questions on “Whirlpool Algorithm”.

1. Which of the following is not true for whirlpool hashing function?
a) It is a two way hashing function
b) It is based on a 512 bit block cipher
c) The block cipher is similar to AES structure
d) It is resistant to differential attacks

Answer: a
Clarification: The whirlpool algorithm is a one way hashing algorithm.

2. The expected workload of generating a collision is of the order of ____________ executions of Whirlpool.
a) 2^(n/2)
b) 2^(n-1)
c) 2⁽ⁿ⁾
d) 2⁽²ⁿ⁾

Answer: a
Clarification: The expected workload of generating a collision is 2^(n/2).

3. Given a message, the expected workload of finding a message that hashes to that value is of the order ____________
a) 2^(n/2)
b) 2^(n-1)
c) 2⁽ⁿ⁾
d) 2⁽²ⁿ⁾

Answer: c
Clarification: The expected workload would be 2ⁿ in such a case.

4. Given a message and its n-bit hash result, the expected workload of finding a message that hashes to the same value is of the order _______________
a) 2^(n/2)
b) 2^(n-1)
c) 2⁽ⁿ⁾
d) 2⁽²ⁿ⁾

Answer: c
Clarification: When the message and hash value are provided, the workload turns out to be 2ⁿ.

5. The message is padded to
a) 1024 bits
b) 512 bits
c) Odd multiple of 256 bits
d) Odd multiple of 512 bits

Answer: c
Clarification: The message in Whirlpool is padded to an odd multiple of 512 bits.

6. The length of the message is appended to the message. What is the size of this length?
a) 256 bits
b) 512 bits
c) 128 bits
d) 1024 bits

Answer: a
Clarification: The length is expressed as a 256 bit unsigned integer, with most significant byte being the leftmost.

7. The hash matrix is of the size
a) 16 × 8
b) 16 × 4
c) 8 × 8
d) 4 × 16

Answer: c
Clarification: The hash matrix has a 8 × 8 matrix.

8. The 8 × 8 hash matrix holds values of size
a) 16 bits
b) 32 bits
c) 8 bits
d) 4 bits

Answer: c
Clarification: 8 × 8 × 8 = 512 bits. 512 bits is the size of the message block. Thus we require 8 bits only.

9. The hash matrix of 8 × 8 is preferred over a 16 × 4 matrix because
a) it is symmetric
b) it is smaller in size
c) it is faster in computation
d) none of the mentioned

Answer: c
Clarification: 8 × 8 is preferred because it is faster.

10. How many rounds are there in one Whirlpool Iteration?
a) 3
b) 4
c) 6
d) 8

Answer: b
Clarification: There are 4 rounds in each Whirlpool step.

11. Which one of these is not a Whirlpool function?
a) Add Key
b) Substitute Bytes
c) Mix Rows
d) Shift Rows

Answer: d
Clarification: Shift columns is a valid operation. Rows are not shifted.

tricky Network Security questions and answers on “Layers”.

1. A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the requirement to make this frame collision-free?
a) 2msec
b) 4msec
c) 2sec
d) 4sec

Answer: a
Clarification: Average frame transmission time
Tfr = 200 bits/200 kbps or 1 ms
Vulnerable time = 2x Tfr = 2 × 1 ms = 2 ms.

2. A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the throughput if the system (all stations together) produces 1000 frames per second?
a) 150 frames
b) 80 frames
c) 135 frames
d) 96 frames

Answer: c
Clarification: Frame transmission time
Tfr= 200/200 kbps or 1 ms.
If the system creates 1000 frames per second, or 1 frame
per millisecond, then G = 1
S = G × e−2G = 0.135 (13.5 percent)
This means that,
Throughput =1000 × 0.135 = 135 frames.

3. A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the throughput if the system (all stations together) produces 500 frames per second?
a) 146 frames
b) 92 frames
c) 38 frames
d) 156 frames

Answer: b
Clarification: If the system creates 500 frames per second, or 1/2 frames per millisecond.
Then G = 1/2.
S = G × e−2G = 0.184 (18.4 percent)
This means that
Throughput = 500 × 0.184 = 92
Only 92 frames out of 500 will probably survive
This is the maximum throughput case, percentage-wise.

4. A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the throughput if the system (all stations together) produces 250 frames per second?
a) 38 frames
b) 48 frames
c) 96 frames
d) 126 frames

Answer: a
Clarification: If the system creates 250 frames per second, or 1/4 frames per millisecond, then G = ¼.
S = G × e−2G = 0.152 (15.2 percent).
This means that
Throughput = 250 × 0.152 = 38.
Only 38 frames out of 250 will probably survive.

5. A slotted ALOHA network transmits 200-bit frames using a shared channel with a 200-kbps bandwidth. Find the throughput if the system (all stations together) produces 1000 frames per second.
a) 92 frames
b) 368 frames
c) 276 frames
d) 151 frames

Answer: b
Clarification: G =1 S=G×e−G=0.368 (36.8%)
Throughput = 1000 × 0.0368 = 368 frames.

6. A slotted ALOHA network transmits 200-bit frames using a shared channel with a 200-kbps bandwidth. Find the throughput if the system (all stations together) produces 500 frames per second.
a) 92 frames
b) 368 frames
c) 276 frames
d) 151 frames

Answer: d
Clarification: G=1/2 S = G × e−G = 0.303 (30.3 percent)
Throughput is 500 × 0.0303 = 151.

7. A network using CSMA/CD has a bandwidth of 10 Mbps. If the maximum propagation time (including the delays in the devices and ignoring the time needed to send a jamming signal, as we see later) is 25.6 μs, what is the minimum size of the frame?
a) 128 bytes
b) 32 bytes
c) 16 bytes
d) 64 bytes

Answer: d
Clarification: The minimum frame transmission time is Tfr = 2 × Tp = 51.2 μs. This means, in the worst case, a station needs to transmit for a period of 51.2 μs to detect the collision.
The minimum size of the frame is 10 Mbps × 51.2 μs = 512 bits or 64 bytes. This is actually the minimum size of the frame for Standard Ethernet.

8. In collision free protocol channel efficiency is given by-
a) d/(d + log2(N))
b) d*(d + log2(N))
c) log2(N)
d) (d + log2(N))

Answer: a
Clarification: In collision free protocol channel efficiency is given by d/(d + log2(N)).

9. After performing bit stuffing on the following stream : 01101111111111111110010, the output is-
a) 01101111101111101111100010
b) 01101111111111111110010111
c) 10010000000000000000001101
d) 01101111111111111111110010

Answer: a
Clarification: Bit stuffing involves adding a 0 after every five 1s during transmission.