subnetting

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Nov 6th, 2009
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Hi

not done subnetting for a while, as a quick reminder, say I have 10.1.0.0/20

I know this means I have 10.1.0.0 to 10.1.15.0 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

10.0.0.0, then 10.0.16.0, 10.0.32.0 etc ?

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mike_guy29 Fri, 11/06/2009 - 03:54
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Hi,


In the example you have posted - 10.1.0.0/20 it would break down as follows


10.1.0.0 - Network address

10.1.15.255 - Broadcast address

10.1.0.1 - First usable address

10.1.15.254 - 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!


Regards

Mike

mike_guy29 Fri, 11/06/2009 - 05:11
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Hi,


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 10.0.0.0 range would be...


10.255.240.0 - Network address

10.255.255.255 - Broadcast address

10.255.240.1 - First usable address

10.255.2555.254 - Last usable address.


I think this answers your question?


Regards

Mike

viyuan700 Fri, 11/06/2009 - 07:22
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see if this helps you, your address

10.0.0.0/20 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

00001010.00000000.00000000.00000000

in first subnet i.e 10.0.0.0(i 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 (10.0.0.1

00001010.00000000.00001111.11111110

(10.0.15.254)


In next subnet i.e 10.0.16.0 (i changed my first subnet bit)

00001010.00000000.00010000.00000000 (10.0.16.0 network)


to get address range i kept increasing address bit by 1

00001010.00000000.00010000.00000001 (10.0.16.1)

00001010.00000000.00011111.11111110

(10.0.16.254)


Now i took my next subnet bit (10.0.32.0)

00001010.00000000.00100000.00000000


again keep increasing the address bit

you will get 10.0.32.1-10.0.32.254)


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...


00000000.0000

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 ?


"10.0.0.0/20 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 10.0.0.0/22 network then 22-8=14 subnet bits and 32-22=10 host bit.


For class B 172.16.0.0/20 example then i have 20-16=4 subnet bit and 32-20=12 host bit.


For class C, 192.168.1.0/26 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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