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Message-ID: <46266E5F.4060205@bigpond.net.au>
Date: Thu, 19 Apr 2007 05:15:43 +1000
From: Peter Williams <pwil3058@...pond.net.au>
To: William Lee Irwin III <wli@...omorphy.com>,
Ingo Molnar <mingo@...e.hu>
CC: linux-kernel@...r.kernel.org,
Linus Torvalds <torvalds@...ux-foundation.org>,
Andrew Morton <akpm@...ux-foundation.org>,
Con Kolivas <kernel@...ivas.org>,
Nick Piggin <npiggin@...e.de>, Mike Galbraith <efault@....de>,
Arjan van de Ven <arjan@...radead.org>,
Thomas Gleixner <tglx@...utronix.de>, caglar@...dus.org.tr,
Willy Tarreau <w@....eu>,
Gene Heskett <gene.heskett@...il.com>,
Dmitry Adamushko <dmitry.adamushko@...il.com>
Subject: Re: [patch] CFS (Completely Fair Scheduler), v2
Peter Williams wrote:
> William Lee Irwin III wrote:
>> Ingo Molnar wrote:
>>>> this is the second release of the CFS (Completely Fair Scheduler)
>>>> patchset, against v2.6.21-rc7:
>>>> http://redhat.com/~mingo/cfs-scheduler/sched-cfs-v2.patch
>>>> i'd like to thank everyone for the tremendous amount of feedback and
>>>> testing the v1 patch got - i could hardly keep up with just reading
>>>> the mails! Some of the stuff people addressed i couldnt implement
>>>> yet, i mostly concentrated on bugs, regressions and debuggability.
>>
>> On Tue, Apr 17, 2007 at 04:46:57PM +1000, Peter Williams wrote:
>>> Have you considered using rq->raw_weighted_load instead of
>>> rq->nr_running in calculating fair_clock? This would take the nice
>>> value (or RT priority) of the other tasks into account when
>>> determining what's fair.
>>
>> I suspect you mean (curr->load_weight*delta_exec)/rq->raw_weighted_load
>> in update_curr().
>
> Or something like that, yes. :-)
Actually, this formula can't be used for the migration thread itself as
its load_weight isn't an accurate reflection of its static priority.
But as the migration thread is a real time task this probably isn't an
issue, right?
If this assumption is correct (i.e. curr is never a real time task) then
my earlier caveat re division by zero being possible is invalid because
the migration task will never be the only task on the runqueue when this
code is called.
I'm also assuming here that (because of its name) curr is already on the
runqueue when this code is called. If it isn't the divisor in the above
expression should be (rq->raw_weighted_load + curr->load_weight). This
would also preclude the possibility of divide by zero.
Peter
--
Peter Williams pwil3058@...pond.net.au
"Learning, n. The kind of ignorance distinguishing the studious."
-- Ambrose Bierce
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