* Srivatsa Vaddagiri <[EMAIL PROTECTED]> wrote:
> On Mon, May 14, 2007 at 01:10:51PM +0200, Ingo Molnar wrote:
> > but let me give you some more CFS design background:
>
> Thanks for this excellent explanation. Things are much clearer now to
> me. I just want to clarify one thing below:
>
> > > 2. Preemption granularity - sysctl_sched_granularity
>
> [snip]
>
> > This granularity value does not depend on the number of tasks running.
>
> Hmm ..so does sysctl_sched_granularity represents granularity in
> real/wall-clock time scale then? AFAICS that doesnt seem to be the
> case.
there's only this small detail i mentioned:
> > ( small detail: the granularity value is currently dependent on the
> > nice level, making it easier for higher-prio tasks to preempt
> > lower-prio tasks. )
> __check_preempt_curr_fair() compares for the distance between the two
> task's (current and next-to-be-run task) fair_key values for deciding
> preemption point.
>
> Let's say that to begin with, at real time t0, both current task Tc
> and next task Tn's fair_key values are same, at value K. Tc will keep
> running until its fair_key value reaches atleast K + 2000000. The
> *real/wall-clock* time taken for Tc's fair_key value to reach K +
> 2000000 - is surely dependent on N, the number of tasks on the queue
> (more the load, more slowly the fair clock advances)?
well, it's somewhere in the [ granularity .. granularity*2 ] wall-clock
scale. Basically the slowest way it can reach it is 'half speed' (two
tasks running), the slowest way is 'near full speed' (lots of tasks
running).
> This is what I meant by my earlier remark: "If there a million cpu
> hungry tasks, then the (real/wall-clock) time taken to switch between
> two tasks is more compared to the case where just two cpu hungry tasks
> are running".
the current task is recalculated at scheduler tick time and put into the
tree at its new position. At a million tasks the fair-clock will advance
little (or not at all - which at these load levels is our smallest
problem anyway) so during a scheduling tick in kernel/sched_fair.c
update_curr() we will have a 'delta_mine' and 'delta_fair' of near zero
and a 'delta_exec' of ~1 million, so curr->wait_runtime will be
decreased at 'full speed': delta_exec-delta_mine, by almost a full tick.
So preemption will occur every sched_granularity (rounded up to the next
tick) points in time, in wall-clock time.
with 2 tasks running delta_exec-delta_mine is 0.5 million, so preemption
will occur in 2*sched_granularity (rounded up to the next timer tick)
wall-clock time.
Ingo
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