Re: [RFC v5 5/9] sched/deadline: do not reclaim the whole CPU bandwidth

[luca abeni <luca.abeni@santannapisa.it>]

Hi Peter,

On Fri, 24 Mar 2017 15:00:15 +0100
Peter Zijlstra <peterz@infradead.org> wrote:

> On Fri, Mar 24, 2017 at 04:52:58AM +0100, luca abeni wrote:
> 
> > diff --git a/kernel/sched/core.c b/kernel/sched/core.c
> > index 20c62e7..efa88eb 100644
> > --- a/kernel/sched/core.c
> > +++ b/kernel/sched/core.c
> > @@ -6716,6 +6716,12 @@ static void sched_dl_do_global(void)
> >  		raw_spin_unlock_irqrestore(&dl_b->lock, flags);
> >  
> >  		rcu_read_unlock_sched();
> > +		if (dl_b->bw == -1)
> > +			cpu_rq(cpu)->dl.deadline_bw_inv = 1 << 8;
> > +		else
> > +			cpu_rq(cpu)->dl.deadline_bw_inv =
> > +				to_ratio(global_rt_runtime(),
> > +					 global_rt_period()) >>
> > 12;  
> 
> Coding style requires braces here (on both legs of the condition)..

Sorry about this; checkpatch did not complain and I did not check the
coding rules. I'll add the braces.


> Also, I find deadline_bw_inv an awkward name; would something like
> bw_ratio or so be more accurate?

I am not good at finding names :)
(I used "deadline_bw_inv" because it represents the inverse of the
deadline tasks bandwidth")
I'll change the name in bw_ratio or something better (suggestions?) 


> > +	if (global_rt_runtime() == RUNTIME_INF)
> > +		dl_rq->deadline_bw_inv = 1 << 8;
> > +	else
> > +		dl_rq->deadline_bw_inv =
> > +			to_ratio(global_rt_runtime(),
> > global_rt_period()) >> 12;  
> 
> That's almost the same code; do we want a helper function?

OK, I'll look at this.


> >  u64 grub_reclaim(u64 delta, struct rq *rq)
> >  {
> > +	return (delta * rq->dl.running_bw *
> > rq->dl.deadline_bw_inv) >> 20 >> 8; }  
> 
> At which point we might want a note about how this doesn't overflow I
> suppose.

I'll add it on Monday.


> 
> Also:
> 
> 	delta *= rq->dl.running_bw;
> 	delta *= rq->dl.bw_ratio;
> 	delta >>= 20 + 8;
> 
> 	return delta;
> 
> Might be more readable ?
> 
> Alternatively:
> 
> 	delta = (delta * rq->dl.running_bw) >> 8;
> 	delta = (delta * rq->dl.bw_ratio) >> 20;
> 
> 	return delta;
> 
> But I doubt we care about those extra 8 bit of space; delta should not
> be over 36 bits (~64 seconds) anyway I suppose.

I think the version with all the shifts after the multiplications is
more precise, right?


			Thanks,
				Luca