Re: [PATCH 02/10] mm: vmscan: Obey proportional scanning requirements for kswapd

[Mel Gorman <mgorman@suse.de>]

On Thu, Mar 21, 2013 at 04:07:55PM +0100, Michal Hocko wrote:
> > > > diff --git a/mm/vmscan.c b/mm/vmscan.c
> > > > index 4835a7a..182ff15 100644
> > > > --- a/mm/vmscan.c
> > > > +++ b/mm/vmscan.c
> > > > @@ -1815,6 +1815,45 @@ out:
> > > >  	}
> > > >  }
> > > >  
> > > > +static void recalculate_scan_count(unsigned long nr_reclaimed,
> > > > +		unsigned long nr_to_reclaim,
> > > > +		unsigned long nr[NR_LRU_LISTS])
> > > > +{
> > > > +	enum lru_list l;
> > > > +
> > > > +	/*
> > > > +	 * For direct reclaim, reclaim the number of pages requested. Less
> > > > +	 * care is taken to ensure that scanning for each LRU is properly
> > > > +	 * proportional. This is unfortunate and is improper aging but
> > > > +	 * minimises the amount of time a process is stalled.
> > > > +	 */
> > > > +	if (!current_is_kswapd()) {
> > > > +		if (nr_reclaimed >= nr_to_reclaim) {
> > > > +			for_each_evictable_lru(l)
> > > > +				nr[l] = 0;
> > > > +		}
> > > > +		return;
> > > 
> > > Heh, this is nicely cryptically said what could be done in shrink_lruvec
> > > as
> > > 	if (!current_is_kswapd()) {
> > > 		if (nr_reclaimed >= nr_to_reclaim)
> > > 			break;
> > > 	}
> > > 
> > 
> > Pretty much. At one point during development, this function was more
> > complex and it evolved into this without me rechecking if splitting it
> > out still made sense.
> > 
> > > Besides that this is not memcg aware which I think it would break
> > > targeted reclaim which is kind of direct reclaim but it still would be
> > > good to stay proportional because it starts with DEF_PRIORITY.
> > > 
> > 
> > This does break memcg because it's a special sort of direct reclaim.
> > 
> > > I would suggest moving this back to shrink_lruvec and update the test as
> > > follows:
> > 
> > I also noticed that we check whether the scan counts need to be
> > normalised more than once
> 
> I didn't mind this because it "disqualified" at least one LRU every
> round which sounds reasonable to me because all LRUs would be scanned
> proportionally.

Once the scan count for one LRU is 0 then min will always be 0 and no
further adjustment is made. It's just redundant to check again.

> E.g. if swappiness is 0 then nr[anon] would be 0 and
> then the active/inactive aging would break? Or am I missing something?
> 

If swappiness is 0 and nr[anon] is zero then the number of pages to scan
from every other LRU will never be adjusted. I do not see how this would
affect active/inactive scanning but maybe I'm misunderstanding you.

-- 
Mel Gorman
SUSE Labs

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