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

[Sebastian Andrzej Siewior]

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.

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