Network Security Subcategories

Networking Overview I	Network Hardware
Freshers	Socket Secure Layer I
Network Security	Classful Addressing
Experienced	Secure Shell I
Designing Subnets I

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

Explanation: N = 2(32-36) = 64.

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

Explanation: Last address = 130.34.12.127/26. This can be found via: Last address = (any address) OR [NOT (network mask)].

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

Explanation: N = 2^(32-36) = 64. Nsub= N/No. of subnetworks = 64/4 = 16.

In the above question, what is the subnetwork mask for each network?

A. 27
B. 29
C. 28
D. 26

Explanation: nsub1 = nsub2 = nsub3 = nsub4 = n + log2(N/Nsub) =26+ log2(64/16)= 28.

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

Explanation: Each subnetwork has 16 addresses. 64+32 = 96. This is the starting address of the 3rd block.

Number of addresses :

A. 64
B. 128
C. 256
D. 512

Explanation: N = 2^(32-24) = 256.

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

Explanation: Last Address = 14.24.74.255/24 Last address = (any address) OR [NOT (network mask)].

The subnet mask nsub1=

A. 23
B. 25
C. 26
D. 27

Explanation: nsub1= n+log2(N/Nsub1) = 25.

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

Explanation: First address = 14.24.74.128/26 First address = (any address) AND (network mask).

Nsub3 =

A. 16
B. 14
C. 12
D. 10

Explanation: The number of addresses in the third subblock (10) is not a power of 2. Thus, we allocate Nsub3 = 16 addresses.