Re: [PATCH 1/3] LoongArch: Reduce min_delta for the arch clockevent device

[Huacai Chen]

On Thu, Nov 14, 2024 at 9:27 PM Sebastian Andrzej Siewior
<bigeasy@xxxxxxxxxxxxx> wrote:
>
> On 2024-11-14 19:46:39 [+0800], Huacai Chen wrote:
> > Hi, Sebastian,
> Hi,
>
> > On Thu, Nov 14, 2024 at 6:21 PM Sebastian Andrzej Siewior
> > <bigeasy@xxxxxxxxxxxxx> wrote:
> > >
> > > On 2024-11-08 17:15:43 [+0800], Huacai Chen wrote:
> > > > Now the min_delta is 0x600 (1536) for LoongArch's constant clockevent
> > > > device. For a 100MHz hardware timer this means ~15us. This is a little
> > > > big, especially for PREEMPT_RT enabled kernels. So reduce it to 1000
> > > > (we don't want too small values to affect performance).
> > >
> > > So this reduces it to 10us. Is anything lower than that bad performance
> > > wise?
> > Maybe I misunderstood the meaning of min_delta, but if I'm correct,
> > small min_delta may cause more timers to be triggered, because timers
> > are aligned by the granularity (min_delta). So I think min_delta
> > affects performance.
>
> They are not aligned. Well they get aligned due to the consequences.
Then I still think it affects performance (and power
consumption).Because it is different to fire a timer every 1us and
fire 10 timers together at the end of 10us.

Huacai

>
> In one-shot mode you program the device for the next timer to expire. It
> computes the delta between expire-time and now. This delta is then
> clamped between min & max delta. See clockevents_program_event().
>
> This means if your timer is supposed to expire in 5us (from now) but
> your min delta is set to 15us then the timer device will be programmed
> to 15us from now. This is 10us after the expire time of your first
> timer. Once the timer devices fires, it will expire all hrtimers which
> expired at this point. This includes that timer, that should have fired
> 10us ago, plus everything else following in the 10us window.
>
> > Huacai
>
> Sebastian