[ee122] [EE122] Question on IP Fragmentation
kedar at berkeley.edu
Thu Oct 11 21:06:53 PDT 2007
From HW1 solutions. Chapter 4 P16:
"Note #1: the slides presented in lecture to illustrate fragmentation
are wrong, for which Prof.
Paxson offers his apologies :-(."
Also, the grouping of data will have to be slightly different than what
you stated below. Since 1476 data bytes is not a multiple of 8, that's
not a valid fragment; you'd have to use either 1472 or 1480 bytes (and
therefore an offset of 184 or 185, respectively).
On 10/11/2007 7:54 PM, Jonathan D. Ellithorpe wrote:
> On slides 21, 22, and 23 of lecture 4, I noticed something confusing
> about the length field of fragment 1 (slide 21), and the offset of
> fragment 2 (slide 22). Shouldn't the offset of fragment 2 denote the
> location of where data should be concatenated in memory on the receiver?
> If so, I'm not sure why the length of frag 1 is 1496 and the offset of
> frag 2 is 1496. It would seem to me that, if the IP header were 20
> bytes, then the offset of frag 2 should be 1476, because the data of
> frag 1 was 1476 bytes long, (bytes 0 to 1475 in the receiving host's
> memory). Thus, frag 2's offset should be 1476, to denote where frag 2's
> data should start.
> Thanks for any help I can get on understanding this!
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