linux-kernel - Re: [RFC PATCH] Add /proc/<pid>/numa

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Message-Id: <33f96879-351f-674a-ca23-43f233f4eb1d@linux.vnet.ibm.com>
Date:   Thu, 3 May 2018 14:16:02 +0530
From:   Anshuman Khandual <khandual@...ux.vnet.ibm.com>
To:     Dave Hansen <dave.hansen@...el.com>,
        Andrew Morton <akpm@...ux-foundation.org>,
        Prakash Sangappa <prakash.sangappa@...cle.com>
Cc:     linux-kernel@...r.kernel.org, linux-mm@...ck.org,
        linux-api@...r.kernel.org, mhocko@...e.com,
        kirill.shutemov@...ux.intel.com, n-horiguchi@...jp.nec.com,
        drepper@...il.com, rientjes@...gle.com,
        Naoya Horiguchi <nao.horiguchi@...il.com>
Subject: Re: [RFC PATCH] Add /proc/<pid>/numa_vamaps for numa node information

On 05/03/2018 03:58 AM, Dave Hansen wrote:
> On 05/02/2018 02:33 PM, Andrew Morton wrote:
>> On Tue,  1 May 2018 22:58:06 -0700 Prakash Sangappa <prakash.sangappa@...cle.com> wrote:
>>> For analysis purpose it is useful to have numa node information
>>> corresponding mapped address ranges of the process. Currently
>>> /proc/<pid>/numa_maps provides list of numa nodes from where pages are
>>> allocated per VMA of the process. This is not useful if an user needs to
>>> determine which numa node the mapped pages are allocated from for a
>>> particular address range. It would have helped if the numa node information
>>> presented in /proc/<pid>/numa_maps was broken down by VA ranges showing the
>>> exact numa node from where the pages have been allocated.
> 
> I'm finding myself a little lost in figuring out what this does.  Today,
> numa_maps might us that a 3-page VMA has 1 page from Node 0 and 2 pages
> from Node 1.  We group *entirely* by VMA:
> 
> 1000-4000 N0=1 N1=2
> 
> We don't want that.  We want to tell exactly where each node's memory is
> despite if they are in the same VMA, like this:
> 
> 1000-2000 N1=1
> 2000-3000 N0=1
> 3000-4000 N1=1

I am kind of wondering on a big memory system how many lines of output
we might have for a large (consuming lets say 80 % of system RAM) VMA
in interleave policy. Is not that a problem ?