1

I will ask my question through an example.

I have a school with 2 campuses with the network address of the school to be 135.135.0.0/26 (host range: 135.135.0.1 - 135.135.0.62). Suppose I will subnet in order to give each campus its own hosts. So campus 1 will be for example 135.135.0.5 and campus 2 will have 135.135.50. Can I now repeat the subnet process for 135.135.0.5 and 135.135.0.50, separately, in order to get more hosts out of it in order for example to distribute them for computers in each campuses or printers or any other device?

2
  • 1
    Your network addresses will need to be a power of two, so 135.135.0.0/26 can be subnetted into 135.135.0.0/27 and 135.135.0.32/27, and one or both of those can be cut in half.
    – Ron Maupin
    Dec 5, 2016 at 21:54
  • Did any answer help you? If so, you should accept the answer so that the question doesn't keep popping up forever, looking for an answer. Alternatively, you could provide and accept your own answer.
    – Ron Maupin
    Aug 15, 2017 at 5:47

2 Answers 2

3

Your questions is a bit vague....

But if you have a /26 you indeed have 62 host IP addresses. If you want to split that subnet evenly between 2 sites you need to further subnet it to 2 x /27 subnets, in which each have 30 (IP) host addresses per Campus.

1

Implementing different subnets will separate your networks from each other thus the two campuses won't be able to communicate with each other unless there is a router between them. Sub-netting would not give you extra IP addresses if you still want them to communicate. But if this is not your concern yes you can divid your network using the 135.135.0.0/26 range ip addresses you have. You can use 135.135.0.0/27 which gives you 135.135.0.1-30 host addresses for network A and 135.135.0.32/27 which gives you 135.135.0.33-62 host addresses in network B. If you need 3 network you can continue the same pattern. For instance, 135.135.0.0/28 - 135.135.0.16/28 and leave the second part (network B) unchanged as 135.135.0.32/27.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.