Re: [RFC PATCH v2 2/8] cpuidle: record the overhead of idle entry

Rafael J. Wysocki
2017-10-17T00:15:16Z

On Monday, October 16, 2017 5:11:57 AM CEST Li, Aubrey wrote:
> On 2017/10/14 8:35, Rafael J. Wysocki wrote:
> > On Saturday, September 30, 2017 9:20:28 AM CEST Aubrey Li wrote:
> >> Record the overhead of idle entry in micro-second
> >>
> >
> > What is this needed for?
>
> We need to figure out how long of a idle is a short idle and recording
> the overhead is for this purpose. The short idle threshold is based
> on this overhead.

I don't really understand this statement.

Pretent I'm not familiar with this stuff and try to explain it to me. :-)

> >
> >> +void cpuidle_entry_end(void)
> >> +{
> >> + struct cpuidle_device *dev = cpuidle_get_device();
> >> + u64 overhead;
> >> + s64 diff;
> >> +
> >> + if (dev) {
> >> + dev->idle_stat.entry_end = local_clock();
> >> + overhead = div_u64(dev->idle_stat.entry_end -
> >> + dev->idle_stat.entry_start, NSEC_PER_USEC);
> >
> > Is the conversion really necessary?
> >
> > If so, then why?
>
> We can choose nano-second and micro-second. Given that workload results
> in the short idle pattern, I think micro-second is good enough for the
> real workload.
>
> Another reason is that prediction from idle governor is micro-second, so
> I convert it for comparing purpose.
> >
> > And if there is a good reason, what about using right shift to do
> > an approximate conversion to avoid the extra division here?
>
> Sure >> 10 works for me as I don't think here precision is a big deal.
>
> >
> >> + diff = overhead - dev->idle_stat.overhead;
> >> + dev->idle_stat.overhead += diff >> 3;
> >
> > Can you please explain what happens in the two lines above?
>
> Online average computing algorithm, stolen from update_avg() @ kernel/sched/core.c.

OK

Maybe care to add a comment to that effect?

Thanks,
Rafael