>From Subnetworking mini Howto:
******--------------------------------
Network Broadcast Netmask Hosts
192.168.1.0 192.168.1.63 255.255.255.192 62
192.168.1.64 192.168.1.127 255.255.255.192 62
182.168.1.128 192.168.1.255 255.255.255.126 124 (see note)
______________________________________________________________________
Note: the reason the last network has only 124 usable network
addresses (not 126 as would be expected from the network mask) is that
it is really a 'super net' of two subnetworks. Hosts on the other two
networks will interpret 192.168.1.192 as the network address of the
'non-existent' subnetwork. Similarly, they will interpret
192.168.1.191 as the broadcast address of the 'non-existent'
subnetwork.
So, if you use 192.168.1.191 or 192 as host addresses on the third
network, then machines on the two smaller networks will not be able to
communicate with them.
******--------------------------
Now the above Howto is saying something different that .191 and .192 cannot
be used in the third subnet. Why is this information different ? Can anyone
explain.
Irfan Akber
----------
> From: Kris Carlier <[EMAIL PROTECTED]>
> To: Irfan Akber <[EMAIL PROTECTED]>
> Cc: [EMAIL PROTECTED]
> Subject: Re: Subnetting
> Date: Friday, January 01, 1999 1:35 PM
>
> Irfan,
>
> > 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
> This is where 191 and 192 will be, and yes they are valid host IP's.
>
> For the last two lines of course you'd need to set a gw statement, the
> first one would be local I presume ?
> Because these three statements of course define the complete C-class
> subnet.
>
> kr=
>
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