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simple subnetting question

In the following site http://www.learntosubnet.com/Subnetting_Step2.htm

# Of Hosts per SN-ID

128-2 = 126 (Invalid)
64–2 = 62
32–2 = 30
16-2 = 14
8–2 = 6
4-2 = 2
2–2 = 0 (Invalid)
0 (Invalid)

I have no idea why 126, 0 are invalide number of hosts. I appreciate any help.

Thanks
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aja101498
Asked:
aja101498
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1 Solution
 
cf_rootCommented:
I am not sure why they are listing 126 as invalid. Because technically with that subnet mask you have 126 available address. The 2 unusable addresses are for your broadcast and network. That is why the "2-2" is not valid. One is the network the other is the broadcast address therefore no hosts can reside on this mask. Please let me know if you have questions.
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chicagoanCommented:
/25 is a valid mask
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MaBCommented:
They are probably listing 126 as invalid because with a class C net subnetted into two 128 subnets with 126 possible hosts in each subnet that would mean that the first subnets network adress would be X.X.X.0 which is the same as the class C nets network address which is "no good".
Similar reason with the second subnet which will have X.X.X.128 as network adress which is "legal" but X.X.X.255 as broadcast adress which is the class C nets broadcast address thus also "no good".

You will always lose the first and last subnet as "no good" when subnetting.

See more extensive description in http://www.experts-exchange.com/Networking/Q_20816228.html#9874506
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MaBCommented:
0 is invalid because you wont have need for a subnet that can't have ane hostes attached.
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chicagoanCommented:
This is a limitation of the windoze ip implementation, not because some other subnet class uses them as a network address or broadcast address. Using that logic a .252 mask would invalidate 124 addresses in a .255 network.
Supernetting always yields 0 and 255 host addresses, and these should be reserved in your allocation table for that reason.
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