On Mon, Feb 22, 2021 at 12:23:04PM +0000, Quentin Perret wrote: > On Monday 22 Feb 2021 at 11:36:03 (+0000), Vincent Donnefort wrote: > > Here's with real life numbers. > > > > The task: util_avg=3 (1) util_est=11 (2) > > > > pd0 (CPU-0, CPU-1, CPU-2) > > > > cpu_util_next(CPU-0, NULL): 7 > > cpu_util_next(CPU-1, NULL): 3 > > cpu_util_next(CPU-2, NULL): 0 <- Most capacity, try to place task here. > > > > cpu_util_next(CPU-2, task): 0 + 11 (2) > > > > > > pd1 (CPU-3): > > > > cpu_util_next(CPU-3, NULL): 77 > > > > cpu_util_next(CPU-3, task): 77 + 3 (1) > > > > > > On pd0, the task contribution is 11. On pd1, it is 3. > > Yes but that accurately reflects what the task's impact on frequency > selection of those CPUs if it was enqueued there, right? > > This is an important property we should aim to keep, the frequency > prediction needs to be in sync with the actual frequency request, or > the energy estimate will be off.
You mean that it could lead to a wrong frequency estimation when doing freq = map_util_freq() in em_cpu_energy()? But in any case, the computed energy, being the product of sum_util with the OPP's cost, it is directly affected by this util_avg/util_est difference. In the case where the task placement doesn't change the OPP, which is often the case, we can simplify the comparison and end-up with the following: delta_energy(CPU-3): OPP3 cost * (cpu_util_avg + task_util_avg - cpu_util_avg) delta_energy(CPU-2): OPP2 cost * (cpu_util_est + task_util_est - cpu_util_est) => OPP3 cost * task_util_avg < task_util_est * OPP2 cost With the same example I described previously, if you add the scaled OPP cost of 0.76 for CPU-3 and 0.65 for CPU-2 (real life OPP scaled costs), we have: 2.3 (CPU-3) < 7.15 (CPU-2) The task is placed on CPU-3, while it would have been much more efficient to use CPU-2. > > > When computing the energy > > deltas, pd0's is likely to be higher than pd1's, only because the task > > contribution is higher for one comparison than the other. > > You mean the contribution to sum_util right? I think I see what you mean > but I'm still not sure if this really is an issue. This is how util_est > works, and the EM stuff is just consistent with that. > > The issue you describe can only happen (I think) when a rq's util_avg is > larger than its util-est emwa by some margin (that has to do with the > ewma-util_avg delta for the task?). But that means the ewma is not to be > trusted to begin with, so ... cfs_rq->avg.util_est.ewma is not used. cpu_util() will only return the max between ue.enqueued and util_avg.
