> I guess you'll need at least two statements:
>
> 192.168.1.0/24 is the complete C-subnet
> 192.168.1.0/25 is the first half of 255.255.255.128
> 192.168.1.0/26 " " " quarter of 255.255.255.192
>
> so to get from 64 to the last you'd add the second half:
>
> 192.168.1.128/25 (from 129-254 usable)
Would the above allow 192.168.1.191 and 192.168.1.192 to be valid host IPs
?. Then would require the following two routes.
route add -net 192.168.1.0 netmask 255.255.255.192
route add -net 192.168.1.64 netmask 255.255.255.192
route add -net 192.168.1.128 netmask 255.255.255.128
correct me if I am wrong.
Irfan Akber
> plus the second quarter of the /26:
> 192.168.1.64/26 (65-127)
>
> kr=
>
> \\\___///
> \\ - - //
> ( @ @ )
> +---------------oOOo-(_)-oOOo-------------+
> | kris carlier - kris@iguana.be |
> | Hiroshima 45, Tsjernobyl 86, Windows 95 |
> | Linux, the choice of a GNU gener8ion |
> | SMS: +32-75-61.43.05 |
> +------------------------Oooo-------------+
> oooO ( )
> ( ) ) /
> \ ( (_/
> \_)
>
>
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