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On Thu, Apr 25, 2002 at 12:17:16PM +0300, Tero Kivinen wrote:
>=20
> That is going to be 8000 fragments. The ip_input will find the struct
> ipq *fp to match that packet. The struct ipq *fp will have list of all
> those fragments belonging to same packet as a list and it will go
> through that list twice in the ip_reass function (one to find out
> where to put the fragment, and second time to see if the packet is
> complete).
>=20
> This means that for each of those 56 byte (48 bytes of ip header
> (header + some options) and 8 bytes of actual data) * 8000 fragment
> packets, we do 2 * 8000 * 8000 / 2 list operations. This means that we
> do 64000000 list operations for each 448000 bytes of data. If each

I wonder: shouldn't that per-packet list be a per-packet balanced tree of
some sort, such that the case above would be some constant times
8000 log 8000?

Regards,
	Ignatios

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