Unanswered Question
Nov 6th, 2009
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not done subnetting for a while, as a quick reminder, say I have

I know this means I have to networks, all the possible subnets are 4096, what would be thre start and end addresses of these subnets ? as I cannot find 4096, it it using all the combinations in the second octect, then spliting the third, ie, then, etc ?

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mike_guy29 Fri, 11/06/2009 - 03:54
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In the example you have posted - it would break down as follows - Network address - Broadcast address - First usable address - Last usable address

I think that was what you were asking. If I have misunderstood then please explain again and I shall try and help!



mike_guy29 Fri, 11/06/2009 - 05:11
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I'm not entirely sure what you mean but yes you would keep going up in subnets of 4096 addresses (4094 of which are usable). The last /20 subnet under the range would be... - Network address - Broadcast address - First usable address

10.255.2555.254 - Last usable address.

I think this answers your question?



viyuan700 Fri, 11/06/2009 - 07:22
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see if this helps you, your address is a Class A address so you have 12 subnet (20-8) bits and 12 (32-20) host bit. So you have 4096 subnet and 4094 addresses per subnet.

in binary


in first subnet i.e kept all my subnet bits 0) if i just keep increasing my 12 address bit by 1 my address range is

00001010.00000000.00000000.00000001 (



In next subnet i.e (i changed my first subnet bit)

00001010.00000000.00010000.00000000 ( network)

to get address range i kept increasing address bit by 1

00001010.00000000.00010000.00000001 (



Now i took my next subnet bit (


again keep increasing the address bit

you will get

Same way keep increasing the 12 subnet bit. Though i am changing the subnet from 0,1,2,3 but in octet its value is 0,16,32,48,64...


00000000.0001(second subnet but it value is 16 as last 4 bits in this octet are address )

00000000.0010 (third subnet but value is 32)

00000000.0011 (forth subnet but value is 48)

00000000.0100 (fifth subnet but value is 64 )

00000000.0101 (list goes on)

carl_townshend Fri, 11/06/2009 - 08:24
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hi there

can you tell me why you do 20-8 and 32-20 in the below ? would it always be something -8 and 32 minus the mask used for no matter what you use ?

" is a Class A address so you have 12 subnet (20-8) bits and 12 (32-20) host bit. So you have 4096 subnet and 4094 addresses per subnet"

viyuan700 Fri, 11/06/2009 - 08:57
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For Class A network you have 8 network bit & 24 host bit. Since it was /20 mask to get subnet bits you substract (20-8) and remaining (32-20) 12bits are for host. Total still is 24 bits. Only thing you divided a single network with 16777216 host into 4096 network and 4096 hots each network.

if i have network then 22-8=14 subnet bits and 32-22=10 host bit.

For class B example then i have 20-16=4 subnet bit and 32-20=12 host bit.

For class C, example i have 26-24=2 subnet bit and 32-26=6 host bit

You can take some example and apply the logic


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