Re: [PATCH] Use extents for recording what swap is allocated.

["Rafael J. Wysocki" <rjw@sisk.pl>]

On Wednesday, 25 October 2006 12:05, Nigel Cunningham wrote:
> Hi.
> 
> On Wed, 2006-10-25 at 11:10 +0200, Pavel Machek wrote:
> > Hi!
> > 
> > > > > > And now, can you do same computation assuming the swap allocator goes
> > > > > > completely crazy, and free space is in 1-page chunks?
> > > > > 
> > > > > The worst case is 3 * sizeof(unsigned long) *
> > > > > number_of_swap_extents_allocated bytes.
> > > > 
> > > > Okay, so if we got 4GB of swap space, thats 1MB swap pages, worst case
> > > > is you have one extent per page, on x86-64 that's 24MB. +kmalloc
> > > > overhead, I assume?
> > > 
> > > Sounds right.
> > 
> > Ok, 24-50MB per 4GB of swap space is not _that_ bad...
> 
> Other way round: 12MB for x86, 24 for x86_64 is the worst case.
> Actually, come to think of it, that would be for 8GB of swap space. The
> worst case would require every page of swap to be alternately free and
> allocated, so for 4GB you'd only have 2GB of swap allocated = 1/2 the
> number of extents and half the memory requirements.
> 
> > > > And you do linear walks over those extents, leading to O(n^2)
> > > > algorithm, no? That has bitten us before...
> > > 
> > > We start from where we last added an extent on the chain by default.
> > 
> > ...but linear search through 24MB _is_ going to hurt.
> 
> If we did one, yes it would. But this will be O(1) since we start at the
> last value, and get_swap_page goes through the space sequentially.

No, it doesn't, AFAICT, but I don't think it matters.

The question is whether there is a chance we'll get anywhere close to the
worst case and I think the answer is 'no' due to the way in which the swap
allocation works.

Greetings,
Rafael


-- 
You never change things by fighting the existing reality.
		R. Buckminster Fuller
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