Re: Observations in Parallel Append

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От Robert Haas
Тема Re: Observations in Parallel Append
Дата
Msg-id CA+TgmobUYcS4+_QPwsyUseDhBebqtmU_oL=rk3MKgBK-w52U9w@mail.gmail.com
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Ответ на Re: Observations in Parallel Append  (Amit Kapila <amit.kapila16@gmail.com>)
Ответы Re: Observations in Parallel Append  (Amit Kapila <amit.kapila16@gmail.com>)
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On Sun, Dec 24, 2017 at 8:37 PM, Amit Kapila <amit.kapila16@gmail.com> wrote:
> On Sun, Dec 24, 2017 at 12:06 PM, Robert Haas <robertmhaas@gmail.com> wrote:
>> On Fri, Dec 22, 2017 at 6:18 AM, Amit Kapila <amit.kapila16@gmail.com> wrote:
>>
>>> Also, don't we need to use parallel_divisor for partial paths instead
>>> of non-partial paths as those will be actually distributed among
>>> workers?
>>
>> Uh, that seems backwards to me.  We're trying to estimate the average
>> number of rows per worker.
>
> Okay, but is it appropriate to use the parallel_divisor?  The
> parallel_divisor means the contribution of all the workers (+
> leader_contribution) whereas for non-partial paths there will be
> always only the subset of workers which will operate on them.
> Consider a case with one non-partial subpath and five partial subpaths
> with six as parallel_divisor, now the current code will try to divide
> the rows of non-partial subpath with respect to six workers.  However,
> in reality, there will always be one worker which will execute that
> path.

That's true, of course, but if five processes each return 0 rows and
the sixth process returns 600 rows, the average number of rows per
process is 100, not anything else.

Here's one way to look at it.  Suppose there is a table with 1000
partitions.  If we do a Parallel Append over a Parallel Seq Scan per
partition, we will come up with a row estimate by summing the
estimated row count across all partitions and dividing by the
parallel_divisor.  This will give us some answer.  If we instead do a
Parallel Append over a Seq Scan per partition, we should really come
up with the *same* estimate.  The only way to do that is to also
divide by the parallel_divisor in this case.

-- 
Robert Haas
EnterpriseDB: http://www.enterprisedb.com
The Enterprise PostgreSQL Company


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