On Sat, Jun 12, 2010 at 12:15 AM, Stefan Kaltenbrunner
<stefan@kaltenbrunner.cc> wrote:
> hmm ok - but assuming sync rep we would end up with something like the
> following(hypotetically assuming each operation takes 1 time unit):
>
> originally:
>
> write 1
> sync 1
> network 1
> write 1
> sync 1
>
> total: 5
>
> whereas in the new case we would basically have the write+sync compete with
> network+write+sync in parallel(total 3 units) and we would only have to wait
> for the slower of those two sets of operations instead of the total time of
> both or am I missing something.
Yeah, this is what I'd like to say. Thanks!
Regards,
--
Fujii Masao
NIPPON TELEGRAPH AND TELEPHONE CORPORATION
NTT Open Source Software Center
