> ______________________________________________________________________
> 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
> 192.168.1.128 192.168.1.255 255.255.255.128 124 (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.
>
>
> \begin{interruptRequest}
>
> How does the 2 smaller networks know that 192.168.1.191 and 192 were
> initially a broadcast and network addresses? Would they treat any one of
> 192.168.*.19[12] in the same way?
>
> \end{interruptRequest}
>
>
First of all, just to make sure we're on solid ground - you need to
realize that once you subnet, the "lowest common denominator" is used,
that is, as in the example, if you have 8 bits of hosts, and decided to
subnet that to 11000000, you now have 4 subnetworks. Essentially what
they did in the example was a two stage operation:
A) create 4 different subnetworks -> subnetmask(in our case) is 26 bits
B) supernet between two of the newly created subnetworks -> subnetmask
is 25 bits on two of the subnetworks.
I'm stating this because your question was defocused. the 2 smaller
networks know that 192.168.1.191 and 192 are "special" addresses,
because they still are. Lets assume that our networks are A,B,CD.
When the tcp-ip stack in a host in A initializes, it ANDs its own IP
address against the subnetmask to determine its home network number.
Then, every outgoing packet goes through the same process. if the
resulting bit string is different than that host's own result, the
packet is known to be out of this host's network, and the ARP request is
made to the default gateway, for forwarding. What will happen with 191
is that you will be requesting the host on subnet A to send a broadcast
packet to network C(again, by definition, supernetting 2 smaller
networks out of 4 is completely transparent to hosts in the 2 small
networks). And with 192, you'll be requesting the host on subnet A to
send a packet to the network address of network D. Both of these cases
are obviously broken.
The funny thing is that the so called CD network does not really exist.
Supernetting is actually a sort of a hack. you simply remove 1 bit on
all the hosts on networks C and D, and it "just works".. You couldn't
have supernetted it AC,D,B, for example. You should play a bit with the
binary versions of the IP addresses to get a hang of this..
To summon it up: from the two other networks' point of view, you're
trying to send packets to broadcast and network addresses. And they know
it because they're supposed to know it. Their subnet masks tells them
that. Hosts in A would say something like "Hey.. if hosts in C and D
want to ignore proper subnet mask rules, it's their own lookout. I still
know 191 and 192 are special addresses, and that's how I'll treat them"
I hope this helps..
Amir.
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