Re: [RFC v5 8/9] sched/deadline: base GRUB reclaiming on the inactive utilization

[Peter Zijlstra]

On Mon, Mar 27, 2017 at 04:56:51PM +0200, Luca Abeni wrote:

> > > +u64 grub_reclaim(u64 delta, struct rq *rq, u64 u)
> > >  {
> > > +	u64 u_act;
> > > +
> > > +	if (rq->dl.this_bw - rq->dl.running_bw > (1 << 20) - u)
> > > +		u_act = u;
> > > +	else
> > > +		u_act = (1 << 20) - rq->dl.this_bw +
> > > rq->dl.running_bw; +
> > > +	return (delta * u_act) >> 20;  
> > 
> > But that's not what is done here I think, something like this instead:
> > 
> > 	Uinact = Utot - Uact
> > 
> > 		-t_u dt ; Uinact > (1 - t_u)
> > 	dq = {
> > 		-(1 - Uinact) dt
> > 
> > 
> > And nowhere do we have an explanation for that.
> 
> Sorry about this confusion... The accounting should be
> 	dq = -(1 - Uinact)dt
> but if (1 - Uinact) is too large (larger than the task's utilization)
> then we use the task's utilization instead (otherwise, we end up
> reclaiming other runqueues' time). I realized that this check was
> needed after writing the comments, and I forgot to update the comments
> when I fixed the code :(
> 
> > Now, I suspect we can write that like: dq = -max{ t_u, (1 - Uinact) }
> > dt, which would suggest this is a sanity check on Utot, which I
> > suspect can be over 1. Is this what is happening?
> 
> Right... I'll fix the code and comments according to your suggestion.

But doesn't that suggest there is now another corner case where we
'always' select t_u because of Utot overload?

My intuition suggests we'd reclaim insufficient time in that case, but
I've not thought much about it.

I feel we want a few words explaining the trade-offs made here and the
corner cases explored.

Does that make sense?