On Tue, 8 Jul 2003 04:01:25 +0300
Shaul Karl <[EMAIL PROTECTED]> wrote:
> Am I right that in theory one can construct a 2 interfaces Ethernet
> network with a netmask of 255.255.255.254?
No. You gave only 1 bit for the "host" part. This is not enough as
in any subnet there are two addresses that are reserved:
A host number "all zero" -- represent the network address
A host number "all ones" -- represent the broadcast address
E.g: With a netmask of 255.255.255.224 (three most significant bits of 1)
you have 5 "host" bits. However, in each of the subnet you may
have only 2^{5} - 2 == 30 host addresses.
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...)
> Will this setup work under Linux? Should ifconfig for such a setup be
> run with the pointopoint (point-to-point) flag?
You do not need any point to point flag (assuming your connection is
ethernet). If you just cable them back to back, than you need crossed
cable (to save a hub/switch).
> Are there reasons to skip such an Ethernet network and keep the
> smallest network with a minimum of 4 hosts and a netmask of
> 255.255.255.252?
Yes, now you know them...
Hope it helps.
--
Oron Peled Voice/Fax: +972-4-8228492
[EMAIL PROTECTED] http://www.actcom.co.il/~oron
"Your fair use of this book is restricted"
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