Re: On disable_cost

From: Andres Freund <andres(at)anarazel(dot)de>
To: Robert Haas <robertmhaas(at)gmail(dot)com>
Cc: Thomas Munro <thomas(dot)munro(at)gmail(dot)com>, Zhenghua Lyu <zlv(at)pivotal(dot)io>, PostgreSQL Hackers <pgsql-hackers(at)lists(dot)postgresql(dot)org>
Subject: Re: On disable_cost
Date: 2019-11-01 17:34:10
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On 2019-11-01 12:56:30 -0400, Robert Haas wrote:
> On Fri, Nov 1, 2019 at 12:43 PM Andres Freund <andres(at)anarazel(dot)de> wrote:
> > As a first step I'd be inclined to "just" adjust disable_cost up to
> > something like 1.0e12. Unfortunately much higher and and we're getting
> > into the area where the loss of precision starts to be significant
> > enough that I'm not sure that we're always careful enough to perform
> > math in the right order (e.g. 1.0e16 + 1 being 1.0e16, and 1e+20 + 1000
> > being 1e+20). I've seen queries with costs above 1e10 where that costing
> > wasn't insane.
> We've done that before and we can do it again. But we're going to need
> to have something better eventually, I think, not just keep kicking
> the can down the road.

Yea, that's why I continued on to describe what we should do afterwards

> Yet another approach would be to divide the cost into two parts, a
> "cost" component and a "violations" component. If two paths are
> compared, the one with fewer violations always wins; if it's a tie,
> they compare on cost. A path's violation count is the total of its
> children, plus one for itself if it does something that's disabled.
> This would be more principled than the current approach, but maybe
> it's too costly.

Namely go for something like this. I think we probably get away with the
additional comparison, especially if we were to store the violations as
an integer and did it like if (unlikely(path1->nviolations !=
path2->nviolations)) or such - that ought to be very well predicted in
nearly all cases.

I wonder how much we'd need to reformulate
compare_path_costs/compare_path_costs_fuzzily to allow the compiler to
auto-vectorize. Might not be worth caring...


Andres Freund

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