Re: [RFC PATCH v3 00/16] Core scheduling v3

[Aubrey Li]

On Thu, Sep 26, 2019 at 1:24 AM Tim Chen <tim.c.chen@xxxxxxxxxxxxxxx> wrote:
>
> On 9/24/19 7:40 PM, Aubrey Li wrote:
> > On Sat, Sep 7, 2019 at 2:30 AM Tim Chen <tim.c.chen@xxxxxxxxxxxxxxx> wrote:
> >> +static inline s64 core_sched_imbalance_delta(int src_cpu, int dst_cpu,
> >> +                       int src_sibling, int dst_sibling,
> >> +                       struct task_group *tg, u64 task_load)
> >> +{
> >> +       struct sched_entity *se, *se_sibling, *dst_se, *dst_se_sibling;
> >> +       s64 excess, deficit, old_mismatch, new_mismatch;
> >> +
> >> +       if (src_cpu == dst_cpu)
> >> +               return -1;
> >> +
> >> +       /* XXX SMT4 will require additional logic */
> >> +
> >> +       se = tg->se[src_cpu];
> >> +       se_sibling = tg->se[src_sibling];
> >> +
> >> +       excess = se->avg.load_avg - se_sibling->avg.load_avg;
> >> +       if (src_sibling == dst_cpu) {
> >> +               old_mismatch = abs(excess);
> >> +               new_mismatch = abs(excess - 2*task_load);
> >> +               return old_mismatch - new_mismatch;
> >> +       }
> >> +
> >> +       dst_se = tg->se[dst_cpu];
> >> +       dst_se_sibling = tg->se[dst_sibling];
> >> +       deficit = dst_se->avg.load_avg - dst_se_sibling->avg.load_avg;
> >> +
> >> +       old_mismatch = abs(excess) + abs(deficit);
> >> +       new_mismatch = abs(excess - (s64) task_load) +
> >> +                      abs(deficit + (s64) task_load);
> >
> > If I understood correctly, these formulas made an assumption that the task
> > being moved to the destination is matched the destination's core cookie.
>
> That's not the case.  We do not need to match the destination's core cookie,

I actually meant destination core's core cookie.

> as that may change after context switches. It needs to reduce the load mismatch
> with the destination CPU's sibling for that cgroup.

So the new_mismatch is not always true, especially when there are more
cgroups and
more core cookies on the system.

Thanks,
-Aubrey