The only way to understand it quite well is doing lots of exercises. But to summarize, there are a few concepts:
- In practice, there are 3 IP address types (A, B and C). Each of them have their base netmask, which you can subnet afterwards. For IP addresses of type A, netmask is
255.0.0.0, and the private addresses are in the format 10.0.0.0 to 10.255.255.255. For B class, the netmask is 255.255.0.0 and the private addresses go from 172.16.0.0 to 172.31.0.0. And for class C, the base netmask is 255.255.255.0 and the base IP address goes from 192.168.0.0 to192.168.255.0`. You have more info here.
- On the netmask, you have to be aware how much zeros you have and how many ones. The zeros represent number of nets that you have, while the zeros represent the number of hosts that can have each of the nets.
A very common question in this kind of exams are in the form:
What is the minimal netmask that should be used for a subnet to have 10 hosts on it, using class C private adresses?
It's important that to know that, you need to apply the 2^n - 2 formula, where n are the bits of the host, to know how much hosts you can have within a subnet. The - 2 part is because there are two IP addresses that are not usable: the net address (the first one) and the broadcast address (the last one).
So as said, class C addresses have their netmask in this form: 255.255.255.0. In binary, that is written that way:
11111111.11111111.11111111.00000000
If you don't know how to pass a decimal number to binary, have a look here.
On that binary subnet mask, let's center on the zeros as we said we're seeking the number of **hosts*.
- What would happen if we take 1 zero bit? We would have
2^1 - 2 = 0, which is less than the 10 requested hosts. It's not enough.
- What would happen if we take 2 zero bits? We would have
2^2 - 2 = 2, which is less than the 10 requested hosts. It's not enough.
- What would happen if we take 3 zero bits? We would have
2^3 - 2 = 6, which is less than the 10 requested hosts. It's not enough.
- What would happen if we take 4 zero bits? We would have
2^4 - 2 = 14, which is more than 10. It's enough!
The question asks for the minimal subnet mask, so from the last operation we now know that we need only 4 zero bytes (remember, zeros represent the number of hosts). In practice that means that the netmask would be:
11111111.11111111.11111111.11110000
Synonyms for this netmask are:
Look that you have four ones as well. That means that you can have: 2^4 = 16 nets. So starting with 192.168.1.0 (I'm taking .1. as the third octet, but you could take any other), the 16 nets would be (divide 256 between the number of nets):
192.168.1.0 - 15192.168.1.16 - 31192.168.1.32 - 45
...
What are the network and broadcast addresses of each of these subnets? The first and last, respectively:
- Network address for the first subnet:
192.168.1.0
- Network address for the second subnet:
192.168.1.16
- ...
- Broadcast address for the first subnet:
192.168.1.15
- Broadcast address for the second subnet:
192.168.1.31
- ...
This is just one exercise but in essence it's the base of any other. There are a few more useful links which explain this subject in deep, but remember, the secret is doing lots and lots of exercises and this way you'll acquire the needed knowledge to do anything.
