linux-kernel - Re: [RFC PATCH 0/4] Enable >0 order folio memory compaction

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Date:   Wed, 20 Sep 2023 19:05:25 -0700
From:   John Hubbard <jhubbard@...dia.com>
To:     Luis Chamberlain <mcgrof@...nel.org>, Zi Yan <ziy@...dia.com>
CC:     <linux-mm@...ck.org>, <linux-kernel@...r.kernel.org>,
        Ryan Roberts <ryan.roberts@....com>,
        Andrew Morton <akpm@...ux-foundation.org>,
        "Matthew Wilcox (Oracle)" <willy@...radead.org>,
        David Hildenbrand <david@...hat.com>,
        "Yin, Fengwei" <fengwei.yin@...el.com>,
        Yu Zhao <yuzhao@...gle.com>, Vlastimil Babka <vbabka@...e.cz>,
        Johannes Weiner <hannes@...xchg.org>,
        Baolin Wang <baolin.wang@...ux.alibaba.com>,
        Kemeng Shi <shikemeng@...weicloud.com>,
        Mel Gorman <mgorman@...hsingularity.net>,
        "Rohan Puri" <rohan.puri15@...il.com>,
        Adam Manzanares <a.manzanares@...sung.com>
Subject: Re: [RFC PATCH 0/4] Enable >0 order folio memory compaction

On 9/20/23 18:16, Luis Chamberlain wrote:
> On Wed, Sep 20, 2023 at 05:55:51PM -0700, Luis Chamberlain wrote:
>> Are there other known recipes test help test this stuff?
> 
> You know, it got me wondering... since how memory fragmented a system
> might be by just running fstests, because, well, we already have
> that automated in kdevops and it also has LBS support for all the
> different large block sizes on 4k sector size. So if we just had a
> way to "measure" or "quantify" memory fragmentation with a score,
> we could just tally up how we did after 4 hours of testing for each
> block size with a set of memory on the guest / target node / cloud
> system.
> 
>    Luis

I thought about it, and here is one possible way to quantify
fragmentation with just a single number. Take this with some
skepticism because it is a first draft sort of thing:

a) Let BLOCKS be the number of 4KB pages (or more generally, then number
of smallest sized objects allowed) in the area.

b) Let FRAGS be the number of free *or* allocated chunks (no need to
consider the size of each, as that is automatically taken into
consideration).

Then:
       fragmentation percentage = (FRAGS / BLOCKS) * 100%

This has some nice properties. For one thing, it's easy to calculate.
For another, it can discern between these cases:

Assume a 12-page area:

Case 1) 6 pages allocated allocated unevenly:

1 page allocated | 1 page free | 1 page allocated | 5 pages free | 4 pages allocated

fragmentation = (5 FRAGS / 12 BLOCKS) * 100% = 41.7%

Case 2) 6 pages allocated evenly: every other page is allocated:

fragmentation = (12 FRAGS / 12 BLOCKS) * 100% = 100%



thanks,
-- 
John Hubbard
NVIDIA