From: Church, Chuck (cchurch@netcogov.com)
Date: Wed Sep 14 2005 - 16:26:58 GMT-3
The subnet mask defines what addresses are local to an interface. If
the network/subnet bits match the interface, it'll ARP for the
destination. If the bits are different, it's considered not on this
subnet, and the routing table is searched for the most specific match.
So if you network is 172.16.1.1/25 , that means the first 25 bits must
match to be local to this interface. 172.16.1.0 through 172.16.1.127
will all share the same first 25 bits. But at 128, the 25 bit jumps to
1, and no longer matches. So you can address your hosts with 172.16.1.x
and a /24 mask, but the router won't consider hosts with the last octet
of 128 through 255 as local, so they'll be dead in the water...
Chuck Church
Lead Design Engineer
CCIE #8776, MCNE, MCSE
Netco Government Services - Design & Implementation Team
1210 N. Parker Rd.
Greenville, SC 29609
Home office: 864-335-9473
Cell: 864-266-3978
cchurch@netcogov.com
PGP key: http://pgp.mit.edu:11371/pks/lookup?op=get&search=0x4371A48D
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
Anthony Sequeira
Sent: Wednesday, September 14, 2005 1:47 PM
To: Group Study
Subject: Subnet Mask Tricks - or - I Will Show You My Mask If You Show
Me Yours!
I was under the false assumption that subnet masks must be identical in
length for two systems to communicate. But I do not believe this to be
true.
Isn't it a fact that the addressing just has to work out so that the two
systems think they are on the same subnet?
Can someone help me with the binary math that you would do to figure
this
out?
For example, let's say my router interface is
172.16.1.1/25<http://172.16.1.1/25>and I want to use /24 for all of my
clients off of that interface. What is
the quickest way to calculate the usable range of addresses for these
client?
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