Date: Sat, 24 Jul 2010 23:23:07 +0300 From: Andriy Gapon <avg@freebsd.org> To: freebsd-hackers@freebsd.org Subject: pageout question Message-ID: <4C4B4BAB.3000005@freebsd.org>
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There is a good deal of comments in the vm_pageout.c code that imply that we use
a hysteresis approach to deal with low available pages condition.
Evidence 1:
/*
* v_free_target and v_cache_min control pageout hysteresis. Note
* that these are more a measure of the VM cache queue hysteresis
* then the VM free queue. Specifically, v_free_target is the
* high water mark (free+cache pages).
*
* v_free_reserved + v_cache_min (mostly means v_cache_min) is the
* low water mark, while v_free_min is the stop. [...]
Evidence 2:
/*
* [...] Do
* not clear vm_pages_needed until we reach our target,
* otherwise we may be woken up over and over again and
* waste a lot of cpu.
*/
In general, the hysteresis, the comments and the code make sense.
My doubt, though, is about the block of code that is right below the comment
quoted above:
if (vm_pages_needed && !vm_page_count_min()) {
if (!vm_paging_needed())
vm_pages_needed = 0;
wakeup(&cnt.v_free_count);
}
If I read this code correctly, pagedaemon would go idle as soon as
vm_paging_needed() becomes false. Given the defintion of vm_paging_needed as
(v_free_reserved + v_cache_min) > (v_free_count + v_cache_count)
this means that pagedaemon quits paging as soon as available page count is above
_low_ watermark. Which contradicts the comments and the general logic of the
code. (Paging also generally starts only when vm_paging_needed() is true).
I think that it also means that vm_pageout_scan is called with pass > 0 only in
very very low memory conditions, where the first pass (pass=0) wasn't able to
free up memory even above low watermark.
But my impression is that the intention was to keep pagedaemon working until
high watermark was reached (vm_paging_target() < 0). Plus, perhaps, some more
depending on vm_pageout_deficit.
And, yes, I see that vm_pageout_scan() has a goal of reaching vm_paging_target()
+ vm_pageout_deficit, but I am speaking of the case where it is unable to do so
on the first pass.
Do I misunderstand something?
Is there a flaw in the code or are the comments
outdated/misleading/not-descriptive-enough?
--
Andriy Gapon
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