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Date:      Wed, 24 Oct 2001 01:06:17 +0100
From:      j mckitrick <jcm@FreeBSD-uk.eu.org>
To:        cjclark@alum.mit.edu
Cc:        freebsd-questions@FreeBSD.ORG
Subject:   Re: question about mtu and fragment offset
Message-ID:  <20011024010617.A50480@dogma.freebsd-uk.eu.org>
In-Reply-To: <20011022225420.E364@blossom.cjclark.org>; from cristjc@earthlink.net on Mon, Oct 22, 2001 at 10:54:21PM -0700
References:  <20011019134824.A9949@dogma.freebsd-uk.eu.org> <20011022225420.E364@blossom.cjclark.org>

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On Mon, Oct 22, 2001 at 10:54:21PM -0700, Crist J. Clark wrote:
| On Fri, Oct 19, 2001 at 01:48:24PM +0100, j mckitrick wrote:
| > 
| > This is for a class project  :-)
| > 
| > If the mtu of a gateway is 1500, does that mean all fragments larger
| > than this will be broken into packets of 1500 (20 header, 1480 data) or
| > packets of 1496 (20 header, 1476 data) since the fragment offset must be
| > even multiples of 8?
| 
| 1500. 1480 is a multiple of 8 last time I checked.

Okay, I took my time and thought about it to make sure I phrased my
answer correctly.  As I understand it, the fragment offset equals the
length of the entire datagram, not just the data.  So, if a gateway has
an mtu of 1500, since the offset must be a multiple of 8, and 1500/8 is
not even, then the total datagram length could be no more than 1496.  If
the header is 20 bytes, that leaves 1476 for data.  Even though 1480 is
a multiple of 8, 1500 is not.  So 1480 + 20 header bytes would still be
a non-integer number of octets.  Am I wrong?  Or what have I
misunderstood?



jm
-- 
My other computer is your windows box.

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