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SubjectRe: [RFC] Defragmentation proposal: preventative maintenance and cleanup [LONG]

>> It becomes effectively useless. The probability of all 8 pages of a given
>> 8 page unit being free when only 1% of memory is free is (1/100)**8 =
>> 1/(10**16).

> Sorry to sound like a broken record, but apply the
> /proc/memareas patch and you can see this happening. After extensive
> activity, you see practically none of the free pages in order 0
> blocks. You might see only a small number (20 or 30 on a 64k
> machine) of (say) order 3 blocks, but if you run your stats
> you would have an expected value of well less than one, and the
> chance of having 20 or 30 would be vanishingly small.

Ooops, what I wrote was factually correct, but misleading.
What I meant was it looks like this:

Zone 4kB 8kB 16kB 32kB 64kB 128kB 256kB 512kB 1024kB 2048kB Tot Pages/kb
DMA 495 348 196 72 10 1 1 0 0 0 = 2807)
@frag 0% 18% 42% 70% 91% 97% 98% 100% 100% 100% = 11228kB
Normal 0 1579 1670 667 140 12 3 1 0 0 = 18118)
@frag 0% 0% 17% 54% 84% 96% 98% 99% 100% 100% = 72472kB

If your model was correct, you would see free pages
per order run like
N = a (K ^ (2^-o)); (for a>0, K>1, o=order)

This doesn't happen. Instead you get GOOD coalescence
at oder 0 (in the Normal zone they've ALL been coalesced),
and not bad at order 1 (see how many order 2's we have).

8 page unit is order 3 (32k). This system has 20% of
memory free at the point where I took the snap shot.
Probability would be, (1/5)^8 = 2^8 / 10^8 =
roughly p = 2.5 x 10^-6. In a system with 32000 pages
(128Kb), if you were right, I'd expect to see about
0.08 free pages at order 3. But here I see 750.

The chance of seeing more than 500 events of probability
p = 2.5 ^ (10^-6) across 32000 samples, is vanishingly
small. Yet it looks this way all the time.

Hence I conclude your model is wrong :-)

--
Alex Bligh
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