Message-ID: <51B1A04B.7030003@yandex-team.ru>
Date: Fri, 07 Jun 2013 12:56:43 +0400
From: Roman Gushchin <klamm@yandex-team.ru>
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To: cl@linux-foundation.org, penberg@kernel.org, mpm@selenic.com,
        yanmin.zhang@intel.com
CC: linux-mm@kvack.org, linux-kernel@vger.kernel.org
Subject: slub: slab order on multi-processor machines
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Hi!

While investigating some compaction-related problems, I noticed, that many (even most)
kernel objects are allocated on slabs with order 2 or 3.

This behavior was introduced by commit 9b2cd506e "slub: Calculate min_objects based on
number of processors." by Christoph Lameter.
As I understand, the idea was to make kernel allocations cheaper by reducing the total
number of page allocations (allocating 1 page with order 3 is cheaper than allocating
8 1-ordered pages).

I'm sure, it's true for recently rebooted machine with a lot of free non-fragmented memory.
But is it also true for heavy-loaded machine with fragmented memory?
Are we sure, that it's cheaper to run compaction and allocate order 3 page than to use
small 1-pages slabs?
Do I miss something?

Disabling this behavior dramatically reduces the number of 2- and 3-ordered allocations.
Compaction is performed significantly rarer. This is especially noticeable on machines
with intensive disk i/o. I do not see any performance degradation. But I'm not sure,
that I'm not missing something.

Any comments and/or ideas are welcomed.

Thanks!

Regards,
Roman
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