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Subnetting - Next Usable Address Space

Posted on 2014-09-17
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Last Modified: 2014-09-18
I have a subnet of 172.16.63.0/20.  This would equate to a subnet mask of 255.255.240.0.  I am needing to create another subnet in the address space immediately after this subnet.  What would that subnet's address space look like?  At what address would I be able to create the next subnet at?  The subnet I would create would be a /25.
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Question by:marrj
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woolmilkporc earned 500 total points
ID: 40328147
Your subnet starts at 172.16.48.0 (network address) or 172.16.48.1 (first host address) and ends at 172.16.63.254 (last host address) or 172.16.63.255 (broadcast address).

So the next subnet starts at 172.16.64.0 (network address) or 172.16.64.1 (first host address).
If it's a /20 subnet it ends at 172.16.79.254 (last host address) or 172.16.79.255 (broadcast address), if it's a /25 subnet it ends at 172.16.64.126 (last host address) or 172.16.64.127 (broadcast address).

Each /20 subnet contains 4096 addresses, where 4094 can be used for hosts, and each /25 subnet contains 128 addresses, where 126 can be used for hosts.
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by:Spartan_1337
ID: 40328207
A /25 will only give you 126 hosts. So you have use 172.16.30.64/25 and your range is 172.16.64.1 - 126
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by:mikebernhardt
ID: 40328441
Here's a little trick for you: Subtract the partial mask (in this case 240) from 256 and you'll have the number of subnets in that octet. In this case, it's 16. So the starting range for 172.16.63.0/20 must be divisable by 16. That puts you 172.16.48.0/20 with 172.16.63.255 at the end.

Next available is 172.16.64.0/25 and the mask is 255.255.255.128. Again, subtract 128 from 256 and you get 128, so the end of the range will be 172.16.64.128.

That helps me figure it out quickly if my brain is running slow.
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by:John Braun
ID: 40331198
You can always use a subnetting calculator for these problems. Try http://subnettingpractice.com/vlsm.html for example.
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