Subnetting question problem

Hi Friends,
can some please explain to me in detail how to subnet the following two examples:

1) 172.16.0.0/16
2) 172.16.0.0/24

Thanks in advance

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101010

http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html <- Have you checked out this post? It contains a very thorough and, in my opinion, easy subnetting process.

If you have, or are using some other method, what are you having issues with? If you can show us where you are hitting the wall, we may be able to point you in the right direction, as opposed to reinventing the wheel.

urvi

Pls don't take it otherwise,but your answer was not comprehensive.
Instead of giving links ,had you posted at least the answer to the question ,that wud have made me happy. It is not possible to browse through all those webpages of Subnetting Made Easy to find the answer.
I already know the answer to the above problems. But I just want to confirm the methods.
The Odom book show how to subnet with exactly 8 subnet bits in the 3rd octet for a class B address but doesn't show how to subnet a class B address with a /16 mask.
My prob is mainly with the first one-172.16.0.0/16.
I know the ans will be 1 subnet (2^0=1 subnet) and 65,534 hosts(2^16-2 host) but by the Wendell Odom mentioned method,which will be the interesting octet?
2)In the 2nd case (/24) the 172.16.0.0 the 3rd octet will be interesting octet as it contains the subnet bits.
So answer will be :
172.16.0.0
172.16.1.0
172.16.2.0
and so on...
Total 2^8=256 subnets
and 2^8-2=254 hosts.

I answered my own question.
Now will anyone tell me -
how to find the interesting octet in the first case i.e. -172.16.0.0/16 ?

bbarrick

The first one is what it is. Technically from what I understand there is no subnetting there. A class B address has a default of 16 bits for the network which is the 172.16 and the rest are host bits. So a 172.16.0.0 /16 is the easiest of the two. Now to actually subnet it you'd turn your prefix into something other than /16. Such as /18 or /24 so that it would be a subnetted address and not technically a Class B anymore.

Basically there is no interesting octet.

This is just my rudimentary understanding so far in my studies.

101010

urvi wrote: »

how to find the interesting octet in the first case i.e. -172.16.0.0/16 ?

I believe you have already answered your own question:

urvi wrote: »

In the 2nd case (/24) the 172.16.0.0 the 3rd octet will be interesting octet as it contains the subnet bits.

The "Interesting Octet" is simply the last octet that contains network bits.

So in your second example: 172.16.0.0/24

The /24 indicates that there are 24 network bits, and the "Interesting Octet" will be the last one containing network bits as shown below:

/24 = 24 Network Bits

NNNNNNNN.NNNNNNNN.NNNNNNNN.HHHHHHHH [3rd octet will be incremented]

So you then know which octet to begin your incrementation. I in this case it would be an increment of 1, as you had calculated:

urvi wrote: »

I know the ans will be 1 subnet (2^0=1 subnet)

You will simply follow the same logic in your first example: 172.16.0.0 /16

/16 = 16 Network Bits

NNNNNNNN.NNNNNNNN.HHHHHHHH.HHHHHHHH [2nd Octet will be incremented]

As in your second example the increment will be 1 [2^0 = 1].

So your networks will be:

172.16.0.0
172.17.0.0
172.18.0.0
.......
172.255.0.0

Here is another example so you can see the method: 172.16.0.0 /22

/22 = 22 Network Bits

NNNNNNNN.NNNNNNNN.NNNNNNHH.HHHHHHHH [3rd Octet will be incremented]

Incrementation will be 4 [2^2 = 4]

Networks will be:

172.16.0.0
172.16.4.0
172.16.8.0
........
172.16.252.0

Also, as bbarrick had noted, your first example is just a pure Class B address. You should make note of the /8, /16 and /24 masks to save yourself some trouble in the future.

I hope this helps.