From owner-freebsd-questions Tue Oct 23 17: 6:28 2001 Delivered-To: freebsd-questions@freebsd.org Received: from probity.mcc.ac.uk (probity.mcc.ac.uk [130.88.200.94]) by hub.freebsd.org (Postfix) with ESMTP id 4DA6037B403 for ; Tue, 23 Oct 2001 17:06:23 -0700 (PDT) Received: from dogma.freebsd-uk.eu.org ([130.88.200.97] helo=dogma) by probity.mcc.ac.uk with esmtp (Exim 2.05 #7) id 15wBYo-000Ars-00; Wed, 24 Oct 2001 01:06:18 +0100 Received: (from jcm@localhost) by dogma (8.11.4/8.11.1) id f9O06IT50511; Wed, 24 Oct 2001 01:06:18 +0100 (BST) (envelope-from jcm) Date: Wed, 24 Oct 2001 01:06:17 +0100 From: j mckitrick 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> References: <20011019134824.A9949@dogma.freebsd-uk.eu.org> <20011022225420.E364@blossom.cjclark.org> Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii X-Mailer: Mutt 1.0.1i In-Reply-To: <20011022225420.E364@blossom.cjclark.org>; from cristjc@earthlink.net on Mon, Oct 22, 2001 at 10:54:21PM -0700 Sender: owner-freebsd-questions@FreeBSD.ORG Precedence: bulk List-ID: List-Archive: (Web Archive) List-Help: (List Instructions) List-Subscribe: List-Unsubscribe: X-Loop: FreeBSD.ORG 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. To Unsubscribe: send mail to majordomo@FreeBSD.org with "unsubscribe freebsd-questions" in the body of the message