From: | Anssi Kääriäinen <anssi(dot)kaariainen(at)thl(dot)fi> |
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To: | Robert Haas <robertmhaas(at)gmail(dot)com> |
Cc: | "pgsql-hackers(at)postgresql(dot)org" <pgsql-hackers(at)postgresql(dot)org> |
Subject: | Re: So, is COUNT(*) fast now? |
Date: | 2011-10-31 13:51:39 |
Message-ID: | 4EAEA7EB.2090501@thl.fi |
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Lists: | pgsql-hackers |
On 10/31/2011 02:44 PM, Robert Haas wrote:
> What I think you're probably measuring here (oprofile would tell us
> for sure) is that once the size of the table goes beyond about half a
> gigabyte, it will have more than one page in the visibility map. The
> index-only scan code keeps the most recently used visibility map page
> pinned to save on overhead, but if you're bouncing back and forth
> between data in the first ~500MB of the table and data in the last
> ~100MB, each switch will result in dropping the current pin and
> getting a new one, which figures to be fairly expensive. With the
> table is only a little over 500GB, you're probably only changing VM
> pages every couple of tuples, but with a 6GB table just about every
> tuple will switch to a new VM page.
>
> Now, maybe you're right and the CPU caches are the more significant
> effect. But I wouldn't like to bet on it without seeing how much the
> drop-and-get-new-pin operations are costing us.
>
Maybe I should have left the analysis part out of the post,
I don't know the internals, so my analysis is likely to be wrong.
Now that I think of it, claiming that the cache effect is 50%
of the runtime is likely a little wrong...
However the part about clustering being important is still correct.
According to the test, you can get 50% overhead because of
random access to the VM.
Stupid question, but why not keep the whole VM pinned?
- Anssi
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