> Well, it is Monday morning and what the heck. I'm not sure how it
> goes, but this is how I think it goes, which someone can correct if
> wrong.
> 1 block
> 192.168.1.0-255 netmask 255.255.255.0
correct
> 2 blocks
> 192.168.1.0-127 netmask 255.255.255.128
> 192.168.1.128-255 netmask 255.255.255.128
correct, but we're not supposed to use the top and bottom nets, are we? :)
> 4 blocks
> 192.168.1.0-63 netmask 255.255.255.64
> 192.168.1.64-127 netmask 255.255.255.64
> 192.168.1.128-191 netmask 255.255.255.64
> 192.168.1.192-255 netmask 255.255.255.64
no, the netmask for this subnetting arrangement is 255.255.255.192 (/26).
its simple to work out the netmask... remember all we are doing is borrowing
bits from the host addresses to give to the networks.
e.g.
Netmask 255.255.255.192 /26 (11111111.11111111.11111111.11000000)
4 subnets
x.x.x.0 x.x.x.63
x.x.x.64 x.x.x.127
x.x.x.128 x.x.x.191
x.x.x.192 x.x.x.255
taking the last byte, 11000000 = 192.
and for the next network, 8 subnets... we're using a last byte in the netmask
of 11100000 right? :)
so you get a netmask of 255.255.255.224
this gives us 8 networks ( - 2 :)... /27
hope this helps.
Cheers,
Nigel
--
Nigel Maddock | I'm out of my mind,
[EMAIL PROTECTED] | but feel free to leave a message...
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