On Thu, Oct 17, 2019 at 03:23:04PM +0100, Douglas Raillard wrote: > On 10/17/19 10:50 AM, Peter Zijlstra wrote:
> > I'm still thinking about the exact means you're using to raise C; that > > is, the 'util - util_est' as cost_margin. It hurts my brain still. > > util_est is currently the best approximation of the actual portion of the CPU > the task needs: > 1) for periodic tasks, it's not too far from the duty cycle, and is always > higher > > 2) for aperiodic tasks, it (indirectly) takes into account the total time it > took > to complete the previous activation, so the signal is not 100% composed of > logical signals > only relevant for periodic tasks (although it's a big part of it). > > 3) Point 1) and 2) together allows util_est to adapt to periodic tasks that > changes > their duty cycle over time, without needing a very long history (the last > task period > is sufficient). > > For periodic tasks, the distance between instantaneous util_avg and the > actual task > duty cycle indicates somehow what is our best guess of the (potential) change > in the task > duty cycle. > > util_est is the threshold (assuming util_avg increasing) for util_avg after > which we know > for sure that even if the task stopped right now, its duty cycle would be > higher than > during the previous period. > This means for a given task and with (util >= util_est): > > 1) util - util_est == 0 means the task duty cycle will be equal to the one > during > during the previous activation, if the tasks stopped executing right now. > > 2) util - util_est > 0 means the task duty cycle will be higher to the one > during > during the previous activation, if the tasks stopped executing right now. So far I can follow, 2) is indeed a fairly sane indication that utilization is growing. > Using the difference (util - util_est) will therefore give these properties > to the boost signal: > * no boost will be applied as long as the task has a constant or decreasing > duty cycle. > > * when we can detect that the duty cycle increases, we temporarily increase > the frequency. > We start with a slight increase, and the longer we wait for the current > period to finish, > the more we boost, since the more likely it is that the task has a much > larger duty cycle > than anticipated. More specifically, the evaluation of "how much more" is > done the exact > same way as it is done for PELT, since the dynamic of the boost is > "inherited" from PELT. Right, because as long it keeps running, util_est will not be changed, so the difference will continue to increase. What I don't see is how that that difference makes sense as input to: cost(x) : (1 + x) * cost_j I suppose that limits the additional OPP to twice the previously selected cost / efficiency (see the confusion from that other email). But given that efficency drops (or costs rise) for higher OPPs that still doesn't really make sense.. > Now if the task is aperiodic, the boost will allow reaching the highest > frequency faster, > which may or may not be desired. Ultimately, it's not more or less wrong than > just picking > the freq based on util_est alone, since util_est is already somewhat > meaningless for aperiodic > tasks. It just allows reaching the max freq at some point without waiting for > too long, which is > all what we can do without more info on the task. > > When applying these boosting rules on the runqueue util signals, we are able > to detect if at least one > task needs boosting according to these rules. That only holds as long as the > history we look at is > the result of a stable set of tasks, i.e. no tasks added or removed from the > rq. So while I agree that 2) is a reasonable signal to work from, everything that comes after is still much confusing me.
