On Tue, Jul 08, 2003 at 08:39:55AM +0300, Oron Peled wrote:
>
> So for your case you need two bits -- i.e: 255.255.255.252
> Now calculate the 2 correct host numbers as an exercise (and
> reply to the group, so others will benefit as well...)
>
A possible allocation of IPs with a 255.255.255.252 netmask might be:
network address: 192.168.0.8.
1st host: 192.168.0.9.
2nd host: 192.168.0.10.
broadcast address: 192.168.0.11.
Another similar example:
network address: 192.168.25.252.
1st host: 192.168.25.253.
2nd host: 192.168.25.254.
broadcast address: 192.168.25.255.
I still don't get something. Quoting section 7 of the IP Sub-Networking
Mini-Howto:
\bgein{quotation}
For the sake of this example, let us assume that you have decided to
subnetwork you C class IP network number 192.168.1.0 into 4 subnets
(each of 62 usable interface/host IP numbers). However, two of these
subnets are being combined into a larger single network, giving three
physical networks.
These are :-
______________________________________________________________________
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}
This illustrates an important point with subnetworks - the usable
addresses are determined by the SMALLEST subnetwork in that address
space.
7.1. The routing tables
Let us assume that a computer running Linux is acting as a router for
this network. It will have three network interfaces to the local LANs
and possibly a fourth interface to the Internet (which would be its
default route.
Let us assume that the Linux computer uses the lowest available IP
address in each subnetwork on its interface to that network. It would
configure its network interfaces as
______________________________________________________________________
Interface IP Address Netmask
eth0 192.168.1.1 255.255.255.192
eth1 192.168.1.65 255.255.255.192
eth2 192.168.1.129 255.255.255.128
______________________________________________________________________
The routing it would establish would be
______________________________________________________________________
Destination Gateway Genmask Iface
192.168.1.0 0.0.0.0 255.255.255.192 eth0
192.168.1.64 0.0.0.0 255.255.255.192 eth1
192.168.1.128 0.0.0.0 255.255.255.128 eth2
______________________________________________________________________
On each of the subnetworks, the hosts would be configured with their
own IP number and net mask (appropriate for the particular network).
Each host would declare the Linux PC as its gateway/router, specifying
the Linux PCs IP address for its interface on to that particular
network.
Robert Hart Melbourne, Australia March 1997.
\end{quotation}
I will repeat the question in case you haven't noticed that I have
\begin{interruptRequest} in the middle:
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?
--
Shaul Karl, [EMAIL PROTECTED] e t
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