linux-kernel - Re: [RFC PATCH v3 1/3] sched: schedule balance map foundation

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Date:	Fri, 22 Feb 2013 10:53:12 +0800
From:	Michael Wang <wangyun@...ux.vnet.ibm.com>
To:	Peter Zijlstra <a.p.zijlstra@...llo.nl>
CC:	LKML <linux-kernel@...r.kernel.org>,
	Ingo Molnar <mingo@...nel.org>, Paul Turner <pjt@...gle.com>,
	Mike Galbraith <efault@....de>,
	Andrew Morton <akpm@...ux-foundation.org>, alex.shi@...el.com,
	Ram Pai <linuxram@...ibm.com>,
	"Nikunj A. Dadhania" <nikunj@...ux.vnet.ibm.com>,
	Namhyung Kim <namhyung@...nel.org>
Subject: Re: [RFC PATCH v3 1/3] sched: schedule balance map foundation

On 02/21/2013 07:37 PM, Peter Zijlstra wrote:
> On Thu, 2013-02-21 at 12:58 +0800, Michael Wang wrote:
>>
>> You are right, it cost space in order to accelerate the system, I've
>> calculated the cost once before (I'm really not good at this, please
>> let
>> me know if I make any silly calculation...), 
> 
> The exact size isn't that important, but its trivial to see its quadric.
> You have a NR_CPUS array per-cpu, thus O(n^2).
> 
> ( side note; invest in getting good at complexity analysis -- or at
> least competent, its the single most important aspect of programming. )
> 
> ...
> 
>> And the final cost is 3000 int and 1030000 pointer, and some padding,
>> but won't bigger than 10M, not a big deal for a system with 1000 cpu
>> too.
> 
> Maybe, but quadric stuff should be frowned upon at all times, these
> things tend to explode when you least expect it.
> 
> For instance, IIRC the biggest single image system SGI booted had 16k
> cpus in there, that ends up at something like 14+14+3=31 aka as 2G of
> storage just for your lookup -- that seems somewhat preposterous.

Honestly, if I'm a admin who own 16k cpus system (I could not even image
how many memory it could have...), I really prefer to exchange 2G memory
to gain some performance.

I see your point here, the cost of space will grow exponentially, but
the memory of system will also grow, and according to my understanding ,
it's faster.

Regards,
Michael Wang

> 
> The domain levels are roughly O(log n) related to the total cpus, so
> what you're doing is replacing an O(log n) lookup with an O(1) lookup,
> but at the cost of O(n^2) storage. 
> 
> 
> 
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