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Lets say I have to create 256 subnets (it could be any number), how would I find the first and last (network/broadcast addresses) for a randomly chosen subnet? For instance, subnet 139.

I'll give an example. Network IP: 145.0.0.0/8 This IP needs to be subsetted into 300 subnets. Network mask: 255.0.0.0 Network address (first address):145.0.0.0/8 Broadcast address: 145.255.255.255/8 Number of hosts: 2^(32-8) = 16777216 Addresses/subnet: 65536 (16777216/256) Subnet mask: /16

I'm not really concerned about usable addresses, just trying to understand a concept.

The first subnet would be: Network address: 145.0.0.0/16 Broadcast Address: 145.0.255.255

So, is there a formula or method to finding a randomly picked subnet's network/broadcast address?

Jeremy
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1 Answers1

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Take the subnet number, subtract one (the first one is 0), and put that in the subnet part of the address. In your example, 145.0.0.0/8 will have 256 /16 subnets. The 139th subnet will be 145.138.0.0/16. You only need the subnet and the mask to figure out anything else.

To do this correctly (otherwise you can make some serious mistakes with non-octet bounded networks), you need to do it in binary. See this answer for how to do all the calculations.

Community
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Ron Maupin
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  • Ok thanks! That makes sense. So what if the number of subnets is more than 256? Lets say I need 512 subnets. I'm assuming I would borrow from the right. So subnet 405, would be 145.255.149.0? (255 for the second octet, then the remaining [405-256] for the third octet? – Jeremy Feb 05 '17 at 19:35
  • You need to calculate the subnet mask. To get 512 subnets from a `/8` network the subnets need to be `/17`. That gets you off the octet boundaries, and you need to do this in binary. Get `404` in binary (`110010100`) and place it into the subnet part of the address. You get `14.202.0.0/17`. Forget octets, they are only to make an address easier for humans to read, and they really play no part in addressing. IPv4 addresses are simply 32-bit binary numbers. – Ron Maupin Feb 05 '17 at 19:39