Subnetting question

Sorry if this sounds really dumb or anything but I found that subnetting made easy guide and so far so good I'm really catching things but there is this question that messes with my brain
What is the valid host range of the 1st subnet of 172.16.0.0/17?
So the answer is 172.16.0.1-172.16.127.254?
I know it may sound dumb but since /17 means it is on the 3rd octet shouldn't it be 172.16.1.0 to 172.16.126.0?

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esr0159

172.16.0 000 0000.0000 0000 will be your host portion *binary notation*
So the first usable will be 172.16.0 000 0000.0000 0001 = 172.16.0.1
the last usable address will be 172.16.0 111 1111.1111 1110= 172.16.127.254

_Gonzalo_

egibrajshori wrote: »

I know it may sound dumb but since /17 means it is on the 3rd octet shouldn't it be 172.16.1.0 to 172.16.126.0?

There are no dumb questions.

First of all, when you are doing subnetting, you are using a given network and making equal size pieces out of it. Let´s say that, for your example, you started with 172.16.0.0/16

You want to get a /17 out of that, so you need to keep in mind that you will get two equal sized subnets, as you will be using one more bit for your network part. Hence, your original network will be split in two.

Starting with:
172.16.0.0/17, which is the first /17
Now, as we are in the third octet and splitting it in two, next/other one will be:
172.16.128.0/17

Now, about usable addresses. The first is the network address and the last is the broadcast address, as you already know. Usable addresses are the ones in between. Sticking to the first example:

172.16.0.0/17
First address: 172.16.0.0
Last address: 172.16.127.255 (because the next one is already the network address of the next /17 portion)

I hope I´m not confusing you further! I tried to keep summarization out of this while hinting at it.

Switch1

egibrajshori wrote: »

Sorry if this sounds really dumb or anything but I found that subnetting made easy guide and so far so good I'm really catching things but there is this question that messes with my brain
What is the valid host range of the 1st subnet of 172.16.0.0/17?
So the answer is 172.16.0.1-172.16.127.254?
I know it may sound dumb but since /17 means it is on the 3rd octet shouldn't it be 172.16.1.0 to 172.16.126.0?

Remember, just because /17 falls into the 3rd octet, that doesn't mean the 4th octet remains untouched!

In fact,
everything after /17 is used to determine the valid host ranges, NNNNNNNN.NNNNNNNN.NHHHHHHH.HHHHHHHH.

N: Network Bits
H: Host Bits

Just keep spending time with subnetting everyday and you'll start to see how predictable it really is