Ivan is planning his company network. He is going to use at least 400 subnets, with up to 1000
hosts on each subnet. Which class of addresses should he use?

A.
Class A.

B.
Class B.

C.
Class C.

D.
Class D.

Explanation:
One must be careful, the wrong way to do this is to take 400 times 1000 (which gives 400,000)
and see how many bits 400,000 can fit in – even if it leads to the correct answer, it won’t always.
The first step is to add 2 to each number, because there are two reserved host and network
addresses. Remember, the formula is: 2 n -2.
The next step is to round each number up to the next straight power of 2. For the networks we get
512 (400+2 raised up) and 1024 (1002 raised up).
Next, find the number of subnet mask bits required to represent each: 512 will require 9 network
bits, and 1024 will require 10 bits, for a total of 19 bits (10 + 9).
A Class A address gives us 8 bits of Networking, and 24 bits in Host addressing. We can carve out
our new network by using 19 of the 24 bits for the host.
Incorrect Answers:
B: A Class B address gives 16 bits of networking and 16 bits of host. We can’t fit 19 bits into a 16
bit number, and we can’t touch any of the first 16 bits, which is the assigned network address.
C: A Class C address gives 24-bits of networking and 8-bits of host. We can’t fit 19 bits into a 8-bit
number, and we can’t touch any of the first 24-bits, which is the assigned network address.
D: A Class D address is a Multicast address range, and can’t be used for network/host addressing.