David Lang <[email protected]> wrote: > if there is no resouce contention, they should be equal.
> In practice, since the network devices are more likely to run into
resource
> contention (think locking overhead between cores if nothing else), it can
> easily be faster to sort them at the destination.
The problem has been, as I understand it, is that many historic TCP receivers
think that receipt of packet X+n without having seen X, means that there is a
loss. This can be solved with appropriate tuning of n, and by how long to
wait, but this has usually required some uber-expert action.
It seems to me that it could also be learnt heuristically how many might be
out of order by observing how long it takes to see packet X.
(Perhaps the newer stacks do this... when it comes to latest TCPs algorithms,
I'm strictly in the gawking section)
--
] Never tell me the odds! | ipv6 mesh networks [
] Michael Richardson, Sandelman Software Works | IoT architect [
] [email protected] http://www.sandelman.ca/ | ruby on rails [
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