Messages in this thread | | | Subject | Re: [RFCv5 PATCH 32/46] sched: Energy-aware wake-up task placement | From | Steve Muckle <> | Date | Sun, 20 Sep 2015 11:39:16 -0700 |
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On 09/18/2015 03:34 AM, Dietmar Eggemann wrote: >> Here should consider scenario for two groups have same capacity? >> This will benefit for the case LITTLE.LITTLE. So the code will be >> looks like below: >> >> int target_sg_cpu = INT_MAX; >> >> if (capacity_of(max_cap_cpu) <= target_max_cap && >> task_fits_capacity(p, max_cap_cpu)) { >> >> if ((capacity_of(max_cap_cpu) == target_max_cap) && >> (target_sg_cpu < max_cap_cpu)) >> continue; >> >> target_sg_cpu = max_cap_cpu; >> sg_target = sg; >> target_max_cap = capacity_of(max_cap_cpu); >> } >> > > It's true that on your SMP system the target sched_group 'sg_target' > depends only on 'task_cpu(p)' because this determines sched_domain 'sd' > (and so the order of sched_groups for the iteration). > > So the current do-while loop to select 'sg_target' for an SMP system > makes little sense. > > But why should we favour the first sched_group (cluster) (the one w/ the > lower max_cap_cpu number) in this situation?
Running the originally proposed code on a system with two identical clusters, it looks like we'll always end up doing an energy-aware search in the task's prev_cpu cluster (sched_group). If you had small tasks scattered across both clusters, energy_aware_wake_cpu() would not consider condensing them on a single cluster. Leo was this the issue you were seeing?
However I think there may be negative side effects with the proposed policy above as well - won't this cause us to pack the first cluster until it's 100% full (running at fmax) before using the second cluster? That would also be bad for power.
thanks, Steve
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