This set of Network Security Multiple Choice Questions & Answers (MCQs) focuses on “Designing Subnets – 1”.
1. N_sub = n + log_2(N/N_sub) is used to find the suffix length.
a) True
b) False
View Answer
Explanation: N_sub = n + log_2(N/N_sub) is used to find the prefix length.
2. An organization is granted the block 130.34.12.64/26. Find the number of addresses for the whole network.
a) 128
b) 32
c) 256
d) None of the mentioned
View Answer
Explanation: N = 2(32-36) = 64.
3. An organization is granted the block 130.34.12.64/26. What is the last address of the network?
a) 130.34.12.63/26
b) 130.34.12.64/26
c) 130.34.12.127/26
d) 130.34.12.128/28
View Answer
Explanation: Last address = 130.34.12.127/26.
This can be found via: Last address = (any address) OR [NOT (network mask)].
4. An organization is granted the block 130.34.12.64/26. The organization needs four subnetworks, each with an equal number of hosts. What is the number of addresses to each subnetwork?
a) 4
b) 16
c) 8
d) 32
View Answer
Explanation: N = 2^(32-36) = 64.
Nsub= N/No. of subnetworks = 64/4 = 16.
5. In the above question, what is the subnetwork mask for each network?
a) 27
b) 29
c) 28
d) 26
View Answer
Explanation: nsub1 = nsub2 = nsub3 = nsub4 = n + log2(N/Nsub) =26+ log2(64/16)= 28.
6. An organization is granted the block 130.34.12.64/26. The organization needs four subnetworks, each with an equal number of hosts. What is the first address for the 3rd subnetwork?
a) 130.34.12.64/28
b) 130.34.12.96/28
c) 130.34.12.96/26
d) 130.34.12.80/27
View Answer
Explanation: Each subnetwork has 16 addresses. 64+32 = 96. This is the starting address of the 3rd block.
An organization is granted a block of addresses with the beginning address 14.24.74.0/24. The organization needs to have 3 subblocks of addresses to use in its three subnets as shown below:
One subblock of 120 addresses
One subblock of 60 addresses
One subblock of 10 addresses
Answer the following questions based on the above information –
7. Number of addresses :
a) 64
b) 128
c) 256
d) 512
View Answer
Explanation: N = 2^(32-24) = 256.
8. Last Address –
a) 14.24.74.64/24
b) 14.24.74.127/24
c) 14.24.74.255/24
d) 14.24.74.256/24
View Answer
Explanation: Last Address = 14.24.74.255/24
Last address = (any address) OR [NOT (network mask)].
9. The subnet mask nsub1=
a) 23
b) 25
c) 26
d) 27
View Answer
Explanation: nsub1= n+log2(N/Nsub1) = 25.
10. First address of the second subblock –
a) 14.24.74.192/26
b) 14.24.74.128/26
c) 14.24.74.127/28
d) 14.24.74.67/27
View Answer
Explanation: First address = 14.24.74.128/26
First address = (any address) AND (network mask).
11. Nsub3 =
a) 16
b) 14
c) 12
d) 10
View Answer
Explanation: The number of addresses in the third subblock (10) is not a power of 2.
Thus, we allocate Nsub3 = 16 addresses.
12. First Address of the third subblock –
a) 14.24.74.128/28
b) 14.24.74.127 /28
c) 14.24.74.192/27
d) 14.24.74.192/28
View Answer
Explanation: First address = 14.24.74.192/28
First address = (any address) AND (network mask).
13. The subnet mask nsub3=
a) 28
b) 27
c) 26
d) 24
View Answer
Explanation: The subnet mask nsub3= n+log2(N/Nsub3) = 28.
Sanfoundry Global Education & Learning Series – Cryptography and Network Security.
To practice all areas of Cryptography and Network Security, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
- Apply for Computer Science Internship
- Practice Computer Science MCQs
- Practice Cyber Security MCQ
- Check Computer Science Books
- Check Cryptography and Network Security Books
