linux-kernel - Re: [PATCH v0] sched: change how run-queue is selected for RT task

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Date:	Tue, 24 May 2011 09:31:12 -0400
From:	Steven Rostedt <rostedt@...dmis.org>
To:	Hillf Danton <dhillf@...il.com>
Cc:	LKML <linux-kernel@...r.kernel.org>, Ingo Molnar <mingo@...e.hu>,
	Peter Zijlstra <peterz@...radead.org>,
	Mike Galbraith <efault@....de>,
	Yong Zhang <yong.zhang0@...il.com>
Subject: Re: [PATCH v0] sched: change how run-queue is selected for RT task

On Sat, 2011-05-21 at 23:28 +0800, Hillf Danton wrote:
> When selecting run-queue for a given RT task, we have to take a few
> factors, such as task priority and CPU cache affinity, into
> consideration. In this work, a simpler method is proposed, which is
> focusing on the relation between the current run-queue of the given
> task and the given run-queue.
> 
> If the current run-queue of task is the given run-queue, the run-queue
> of task keeps unchanged, so the CPU cache affinities of both task and
> the current task of run-queue remain unchanged. Then there are at
> least two tasks competing one CPU, and in the worst case that both
> competitors are RT tasks the victim will be selected and processed by
> pusher later.
> 
> On other hand, if the current run-queue of task is different from the
> given run-queue, task is simply delivered to its current run-queue,
> since pusher is always willing to do hard works.
> 
> In summary, the burden of RT task is always processed first by the
> pusher of its current run-queue.

Why? Why should we preempt a high prio task to make it push off a task
that has just woken up on its CPU? If we know that we are about to
preempt a high prio RT task, why interrupt it, when we could simply make
this task wake up on another CPU?

-- Steve

> 
> Signed-off-by: Hillf Danton <dhillf@...il.com>
> ---
> 
> --- a/kernel/sched_rt.c	2011-04-27 11:48:50.000000000 +0800
> +++ b/kernel/sched_rt.c	2011-05-21 22:19:52.000000000 +0800
> @@ -998,14 +998,12 @@ select_task_rq_rt(struct rq *rq, struct
>  	 *
>  	 * For equal prio tasks, we just let the scheduler sort it out.
>  	 */
> -	if (unlikely(rt_task(rq->curr)) &&
> -	    (rq->curr->rt.nr_cpus_allowed < 2 ||
> -	     rq->curr->prio < p->prio) &&
> -	    (p->rt.nr_cpus_allowed > 1)) {
> -		int cpu = find_lowest_rq(p);
> 
> -		return (cpu == -1) ? task_cpu(p) : cpu;
> -	}
> +	if (task_cpu(p) == rq->cpu)
> +		return rq->cpu;
> +
> +	if (likely(!rt_task(rq->curr)))
> +		return rq->cpu;
> 
>  	/*
>  	 * Otherwise, just let it ride on the affined RQ and the


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