Re: [HACKERS] Partition-wise aggregation/grouping

On 10 October 2017 at 17:57, Ashutosh Bapat
<ashutosh.bapat@enterprisedb.com> wrote:
> Append node just returns the result of ExecProcNode(). Charging
> cpu_tuple_cost may make it too expensive. In other places where we
> charge cpu_tuple_cost there's some processing done to the tuple like
> ExecStoreTuple() in SeqNext(). May be we need some other measure for
> Append's processing of the tuple.

I don't think there's any need to invent any new GUC. You could just
divide cpu_tuple_cost by something.

I did a quick benchmark on my laptop to see how much Append really
costs, and with the standard costs the actual cost seems to be about
cpu_tuple_cost / 2.4. So probably cpu_tuple_cost / 2 might be
realistic. create_set_projection_path() does something similar and
brincostestimate() does some similar magic and applies 0.1 *
cpu_operator_cost to the total cost.

# create table p (a int, b int);
# create table p1 () inherits (p);
# insert into p1 select generate_series(1,1000000);
# vacuum analyze p1;
# \q
$ echo "select count(*) from p1;" > p1.sql
$ echo "select count(*) from p;" > p.sql
$ pgbench -T 60 -f p1.sql -n

latency average = 58.567 ms

$ pgbench -T 60 -f p.sql -n
latency average = 72.984 ms

$ psql
psql (11devel)
Type "help" for help.

# -- check the cost of the plan.
# explain select count(*) from p1;
                            QUERY PLAN
------------------------------------------------------------------
 Aggregate  (cost=16925.00..16925.01 rows=1 width=8)
   ->  Seq Scan on p1  (cost=0.00..14425.00 rows=1000000 width=0)
(2 rows)

# -- selecting from the parent is the same due to zero Append cost.
# explain select count(*) from p;
                               QUERY PLAN
------------------------------------------------------------------------
 Aggregate  (cost=16925.00..16925.01 rows=1 width=8)
   ->  Append  (cost=0.00..14425.00 rows=1000001 width=0)
         ->  Seq Scan on p  (cost=0.00..0.00 rows=1 width=0)
         ->  Seq Scan on p1  (cost=0.00..14425.00 rows=1000000 width=0)
(4 rows)

# -- extrapolate the additional time taken for the Append scan and
work out what the planner
# -- should add to the plan's cost, then divide by the number of rows
in p1 to work out the
# -- tuple cost of pulling a row through the append.
# select (16925.01 * (72.984 / 58.567) - 16925.01)  / 1000000;
        ?column?
------------------------
 0.00416630302337493743
(1 row)

# show cpu_tuple_cost;
 cpu_tuple_cost
----------------
 0.01
(1 row)

# -- How does that compare to the cpu_tuple_cost?
# select current_Setting('cpu_tuple_cost')::float8 / 0.00416630302337493743;
    ?column?
----------------
 2.400209476818
(1 row)

Maybe it's worth trying with different row counts to see if the
additional cost is consistent, but it's probably not worth being too
critical here.

--
 David Rowley                   http://www.2ndQuadrant.com/
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