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Subject: Re: [PATCH v2] proc:  speedup /proc/stat handling
From: Eric Dumazet <eric.dumazet@gmail.com>
To: =?ISO-8859-1?Q?J=F6rg-Volker?= Peetz <jvpeetz@web.de>
Cc: linux-kernel@vger.kernel.org,
        KAMEZAWA Hiroyuki <kamezawa.hiroyu@jp.fujitsu.com>,
        Andrew Morton <akpm@linux-foundation.org>,
        Glauber Costa <glommer@parallels.com>,
        Peter Zijlstra <a.p.zijlstra@chello.nl>, Ingo Molnar <mingo@elte.hu>,
        Russell King - ARM Linux <linux@arm.linux.org.uk>,
        Paul Tuner <pjt@google.com>
Date: Mon, 30 Jan 2012 10:25:55 +0100
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Le lundi 30 janvier 2012 à 09:06 +0100, Jörg-Volker Peetz a écrit :
> Eric Dumazet wrote, on 01/25/12 01:26:
> > Le mercredi 25 janvier 2012 à 09:18 +0900, KAMEZAWA Hiroyuki a écrit :
> > 
> >> BTW, what is the reason of this change ?
> >>
> >>> -	unsigned size = 4096 * (1 + num_possible_cpus() / 32);
> >>> +	unsigned size = 1024 + 128 * num_possible_cpus();
> >>
> >> I think size of buffer is affected by the number of online cpus.
> >> (Maybe 128 is enough but please add comment why 128 ?)
> >>
> > 
> > There is no change, as 4096/32 is 128 bytes per cpu.
> > 
> 
> Wrong math, only num_possible_cpus() is divided by 32. Thus,
> 
> -	unsigned size = 4096 * (1 + num_possible_cpus() / 32);
> +	unsigned size = 4096 + 128 * num_possible_cpus();
> 
> <snip>


It is good math, once you take the time to think a bit about it.

The original question was about the 128 * num_possible_cpus()

4096/32 is 128 as I said. 

The 4096 -> 1024 is just taking into account fact that once you do the
correct computations, you dont need initial 4096 value, and 1024 is more
than enough.

Example on a dual core machine : 

# dmesg|grep nr_irq
[    0.000000] nr_irqs_gsi: 40
[    0.000000] NR_IRQS:2304 nr_irqs:712 16

size = 1024 + 2*128 + 2*712 = 2704 bytes (rounded to 4096 by kmalloc())

# wc -c /proc/stat
1767 /proc/stat

Problem with original math was that for a machine with 16 cpus or a
machine with 1 cpu, we ended with the same 4096 value. That was a real
problem.

If we instead use "unsigned size = 4096 + 128 * num_possible_cpus();" as
you suggest, we would always allocate 2 pages of memory, this is not
needed at all for typical 1/2/4 way machines.