At Thu, 16 Mar 2017 17:16:32 -0400, Corey Huinker <corey.huinker@gmail.com> wrote in
<CADkLM=cBZEX9L9HnhJYrtfiAN5Ebdu=xbvM_poWVGBR7yN3gVw@mail.gmail.com>
> On Thu, Mar 16, 2017 at 4:17 PM, Tom Lane <tgl@sss.pgh.pa.us> wrote:
>
> > Corey Huinker <corey.huinker@gmail.com> writes:
> > > I reworked the test such that all of the foreign tables inherit from the
> > > same parent table, and if you query that you do get async execution. But
> > It
> > > doesn't work when just stringing together those foreign tables with UNION
> > > ALLs.
> >
> > > I don't know how to proceed with this review if that was a goal of the
> > > patch.
> >
> > Whether it was a goal or not, I'd say there is something either broken
> > or incorrectly implemented if you don't see that. The planner (and
> > therefore also the executor) generally treats inheritance the same as
> > simple UNION ALL. If that's not the case here, I'd want to know why.
> >
> > regards, tom lane
> >
>
> Updated commitfest entry to "Returned With Feedback".
Sorry for the absense. For information, I'll continue to write
some more.
At Tue, 14 Mar 2017 10:28:36 +0900, Amit Langote <Langote_Amit_f8@lab.ntt.co.jp> wrote in
<e7dc8128-f32b-ff9a-870e-f1117b8e4fa6@lab.ntt.co.jp>
> async
> t0: 3806.31 ( 4.49) 0.4% faster (should be error)
> pl: 1629.17 ( 0.29) 1.3% slower
> pf0: 6447.07 ( 25.19) 18.7% faster
> pf1: 1876.80 ( 47.13) 76.6% faster
> </quote>
>
> IIUC, pf0 and pf1 is the same test case (all 4 foreign tables target the
> same server) measured with different implementations of the patch.
pf0 is measured on a partitioned(sharded) tables on one foreign
server, that is, sharing a connection. pf1 is in contrast sharded
tables that have dedicate server (or connection). The parent
server is async-patched and the child server is not patched.
Async-capable plan is generated in planner. An Append contains at
least one async-capable child becomes async-aware Append. So the
async feature should be effective also for the UNION ALL case.
The following will work faster than unpatched version.I
SELECT sum(a) FROM (SELECT a FROM ft10 UNION ALL SELECT a FROM ft20 UNION ALL SELECT a FROM ft30 UNION ALL SELECT a
FROMft40) as ft;
I'll measure the performance for the case next week.
regards,
--
Kyotaro Horiguchi
NTT Open Source Software Center
