2

We're in the process of dividing a /24 network into a /25 network, a /26 network and a /27 network.

Our original plan was as follows:

10.97.21.0/24 (original network) to

10.97.21.0-31 /27

10.97.21.32-159 /25

10.97.21.160-223 /26

10.07.21.224-255 /27

However, I have been told that this won't work because it's imperative to always start with the largest network first. Why is that the case?

Fang
  • 264
  • 2
  • 9
  • 2
    You don't have to start with the largest, but it will help you to not make the mistake you made. Your /25 is not on a bit boundary, as @datagramnetwork explains in his answer. – Ron Trunk Feb 06 '17 at 12:52

2 Answers2

6

Is not gonna work as: 10.97.21.32/25 is part of 10.97.21.0/25 network which is overlapping with your 10.97.21.0/27

10.97.21.160/26 is part of 10.97.21.128/26 which is overlapping again with other of yours network.

That should work:

10.97.21.0/25

10.97.21.128/26

10.97.21.192/27

10.97.21.224/27

Datagram.Network
  • 1,587
  • 9
  • 17
3

It is not imperative to always start with the largest network but it is easier if you're not accustomed with sub-netting.

You can take it this way:

Divide your 10.97.21.0/24 network into 2 /25 networks:

10.97.21.0/25
10.97.21.128/25

Since you need a /25, keep one and further subnet the other one, once again in two networks:

10.97.21.0/25
10.97.21.128/26
10.97.21.192/26

You need a /26, so you keep the first one and divide the other one in two:

10.97.21.0/25
10.97.21.128/26
10.97.21.192/27
10.87.21.224/27

as you can see we get the solution given by @Datagram.Network, but if choose instead to divide the first subnet and keep the second we can also have

10.97.21.0/27
10.97.21.32/27
10.97.21.64/26
10.87.21.128/25

For more information you may find this answer to How do you calculate the prefix, network, subnet, and host numbers? interesting

JFL
  • 19,649
  • 1
  • 32
  • 64