Help with Subnetting

Matt9300 · 29 Jun 2021 at 11:47

Given an available address space of 172.24.102.0/23, describe a subnetting scheme which would yield several smaller subnets, each of which must accommodate 6 hosts.

Is it possible to split the subnet down into further subnets? That's how I read this question.

So I have the following of

172.24.102.0/23 as the address space and subnet mask of 255.255.254.0 with 510 usable hosts with the following range of 172.24.102.1-103.254.

I can't figure out whether they mean use the next subnet of 172.24.104.1-172.25.105.254 or can it be split even further down?

ChrisD. · 29 Jun 2021 at 11:52

How many networks do you need in total? If you break it down into too many subnets you'll loose available host addresses as each subnet requires a network ID and broadcast address, plus a default gateway address.

I would just do two /24's to make it easy, or three /23's.
172.24.102.0/24
172.24.103.0/24

Or:
172.24.102.0/25
172.24.102.128/25
172.24.103.0/25
172.24.103.128/25

Or is this a CCNA type question? If so, go watch some videos and read some books.

Moley · 29 Jun 2021 at 11:58

A subnet mask of 255.255.255.248 will give you 8192 subnets of 6 hosts each.

VersionMonkey · 29 Jun 2021 at 14:27

Matt9300 said:
Given an available address space of 172.24.102.0/23, describe a subnetting scheme which would yield several smaller subnets, each of which must accommodate 6 hosts.

Is it possible to split the subnet down into further subnets? That's how I read this question.

So I have the following of

172.24.102.0/23 as the address space and subnet mask of 255.255.254.0 with 510 usable hosts with the following range of 172.24.102.1-103.254.

I can't figure out whether they mean use the next subnet of 172.24.104.1-172.25.105.254 or can it be split even further down?

They mean split that range down further - some authority has allocated you that address space, but you need to break it down into separate networks for whatever reason.

As per @Moley a mask of /29 will provide subnets of exactly 6 hosts and is probably the answer they are looking for (though as they don't say exactly 6 hosts, there is some ambiguity in the question).

Good luck with your exam

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