Re: [PATCH 06/10] sched/fair: Add avg_vruntime

[Peter Zijlstra <peterz@infradead.org>]

On Fri, Mar 24, 2023 at 03:12:19PM +0800, Chen Yu wrote:

> > Or worse, SCHED_IDLE, where weight is 2 (IIRC) or cgroups, then vtime
> > advances at 512 times realtime. Now, the tick puts a limit on how long
> > we'll overshoot these super low weight entities, for HZ=1000 we still
> > only get 0.5s of vtime for weight=2.
> >
> > That would be only 30 bits used, except we use double FIXEDPOINT_SHIFT
> > on 64bit, so we'll end up at 40-ish.
> > 
> > That should give us enough room to carry an average of deltas around
> > min_vruntime.
> > 
> I'm trying to digest how ticks could prevent the overflow.

They don't prevent overflow per se, but they do limit on how far
vruntime can advance ahead of the pack.

> In update_curr() -> update_min_vruntime(cfs_rq), the cfs_rq->min_vruntime
> is set to
> max (cfs_rq->min_vruntime, min(curr->vruntime, leftmost(se->vruntime)))
> so, although curr->vruntime increase by 0.5 seconds in each tick,
> the leftmost(se->vruntime) could still be very small and unchanged,
> thus the delta between v_i and cfs_rq->min_vruntime is still large.

Well, since the basic task selection rule is: pick leftmost, the idea is
that leftmost and hence min_vruntime advances. The only problem is that
placement can place new entities left of min_vruntime and then it stalls
for a bit. But then rightmost tasks shouldn't get more runtime and the
whole situation should be 'stable'-ish.

> Instead sysctl_sched_latency could decide how far it is between the
> se.vruntime and the cfs_rq.min_vruntime, by calculating the vruntime
> delta between task1 and task2:
> 
>     sched_vslice(task1) = (NICE0_LOAD/se1.weight)  * (w1/Sum wi * sysctl_sched_latency)
>     sched_vslice(task2) = (NICE0_LOAD/se2.weight)  * (w2/Sum wi * sysctl_sched_latency)

Yes, vslice is obviously involved, but low weight tasks are the ones
that tend to shoot away and are tick limited.

> Besides in patch 10, entity_eligible() checks
> \Sum (v_i - v)*w_i >= (v_i - v)*(\Sum w_i)
> and the \Sum w_i could become large if there are many runnable tasks and
> bring overflow?

Indeed; I'll check there too. I think I'll make it do the division on
32bit and use 64x64->128 on 64bit.

Let me have a play..