Re: [RFC 1/1] sched/fair: Consider asymmetric scheduler groups in load balancer
From: Tobias Huschle
Date: Fri Jul 07 2023 - 11:59:29 EST
On 2023-07-07 16:33, Shrikanth Hegde wrote:
On 7/7/23 1:14 PM, Tobias Huschle wrote:
On 2023-07-05 09:52, Vincent Guittot wrote:
Le lundi 05 juin 2023 à 10:07:16 (+0200), Tobias Huschle a écrit :
On 2023-05-16 15:36, Vincent Guittot wrote:
> On Mon, 15 May 2023 at 13:46, Tobias Huschle <huschle@xxxxxxxxxxxxx>
> wrote:
> >
> > The current load balancer implementation implies that scheduler
> > groups,
> > within the same domain, all host the same number of CPUs. This is
> > reflected in the condition, that a scheduler group, which is load
> > balancing and classified as having spare capacity, should pull work
> > from the busiest group, if the local group runs less processes than
> > the busiest one. This implies that these two groups should run the
> > same number of processes, which is problematic if the groups are not
> > of the same size.
> >
> > The assumption that scheduler groups within the same scheduler
domain
> > host the same number of CPUs appears to be true for non-s390
> > architectures. Nevertheless, s390 can have scheduler groups of
unequal
> > size.
> >
> > This introduces a performance degredation in the following scenario:
> >
> > Consider a system with 8 CPUs, 6 CPUs are located on one CPU socket,
> > the remaining 2 are located on another socket:
> >
> > Socket -----1----- -2-
> > CPU 1 2 3 4 5 6 7 8
> >
> > Placing some workload ( x = one task ) yields the following
> > scenarios:
> >
> > The first 5 tasks are distributed evenly across the two groups.
> >
> > Socket -----1----- -2-
> > CPU 1 2 3 4 5 6 7 8
> > x x x x x
> >
> > Adding a 6th task yields the following distribution:
> >
> > Socket -----1----- -2-
> > CPU 1 2 3 4 5 6 7 8
> > SMT1 x x x x x
> > SMT2 x
>
> Your description is a bit confusing for me. What you name CPU above
> should be named Core, doesn' it ?
>
> Could you share with us your scheduler topology ?
>
You are correct, it should say core instead of CPU.
One actual configuration from one of my test machines (lscpu -e):
[...]
So, 6 cores / 12 CPUs in one group 2 cores / 4 CPUs in the other.
Thaks for the details
If I run stress-ng with 8 cpu stressors on the original code I get a
distribution
like this:
00 01 02 03 04 05 06 07 08 09 10 11 || 12 13 14 15
x x x x x x x x
Which means that the two cores in the smaller group are running into
SMT
while two
cores in the larger group are still idle. This is caused by the
prefer_sibling path
which really wants to see both groups run the same number of tasks.
yes and it considers that there are the same number of CPUs per group
> >
> > The task is added to the 2nd scheduler group, as the scheduler
has the
> > assumption that scheduler groups are of the same size, so they
should
> > also host the same number of tasks. This makes CPU 7 run into SMT
> > thread, which comes with a performance penalty. This means, that in
> > the window of 6-8 tasks, load balancing is done suboptimally,
because
> > SMT is used although there is no reason to do so as fully idle CPUs
> > are still available.
> >
> > Taking the weight of the scheduler groups into account, ensures that
> > a load balancing CPU within a smaller group will not try to pull
tasks
> > from a bigger group while the bigger group still has idle CPUs
> > available.
> >
> > Signed-off-by: Tobias Huschle <huschle@xxxxxxxxxxxxx>
> > ---
> > kernel/sched/fair.c | 3 ++-
> > 1 file changed, 2 insertions(+), 1 deletion(-)
> >
> > diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
> > index 48b6f0ca13ac..b1307d7e4065 100644
> > --- a/kernel/sched/fair.c
> > +++ b/kernel/sched/fair.c
> > @@ -10426,7 +10426,8 @@ static struct sched_group
> > *find_busiest_group(struct lb_env *env)
> > * group's child domain.
> > */
> > if (sds.prefer_sibling && local->group_type ==
> > group_has_spare &&
> > - busiest->sum_nr_running > local->sum_nr_running + 1)
> > + busiest->sum_nr_running * local->group_weight >
> > + local->sum_nr_running *
> > busiest->group_weight + 1)
what you want to test here is that moving 1 task from busiest to
local
group
would help and balance the ratio of tasks per cpu
(busiest->sum_nr_running - 1) / busiest->group_weight >
(local->sum_nr_running + 1) / local->group_weight
which can be develop into
busiest->sum_nr_running * local->group_weight >=
local->sum_nr_running
* busiest->group_weight + busiest->group_weight + local->group_weight
and you also have to change how we calculate the imbalance which just
provide the half of the diff of nr_running
by something like
(busiest->sum_nr_running * local->group_weight) -
(local->sum_nr_running * busiest->group_weight) /
(busiest->group_weight + local->group_weight)
Ahh right, I had a look at the imbalance part now and your suggestion
works
pretty well. Just had to make some minor adjustments so far.
Nice side effect is, that this allows the load balancer to behave
exactly the
same as before in the cases where local->group_weight ==
busiest->group_weight.
The behavior only changes for the case where the groups are not of
equal
size.
Not sure if it has been figured/discussed out already, pointing one
possible scenario.
Taking the formulas:
busiest->sum_nr_running * local->group_weight >= local->sum_nr_running
* busiest->group_weight + busiest->group_weight + local->group_weight
and calulate_imbalance:
(busiest->sum_nr_running * local->group_weight) -
(local->sum_nr_running * busiest->group_weight) /
(busiest->group_weight + local->group_weight)
I was considering to just use the imbalance as an indicator whether we
should
balance or not, i.e. check if the second formula yields a value greater
than 0.
Will have to play around with that a bit though.
First lets say imbalance was like this. same example as before.
sched_group in [busy_cpus/idle_cpus/group_weight]
[3/9/12] - local group.
[3/1/4] - busy group.
3*12 >= 3*4+12+4 --> true and imbalance would be (3*12-3*4)/(12+4) --
24/16 >> 1 -- lets say 1.
we will balance, good.
[4/8/12]
[2/2/4]
There will not be further load balances. good.
a new process comes, it would be scheduled on [4/8/120 sched group as
that would be idlest.
[5/7/12]
[2/2/4]
Process running on [2/2/4] exits. then in this scenario do you expect
the balance to happen again? Since balancing would result into optimal
performance.
[5/7/12] - busy_group
[1/3/4] - local group
5*4 >= 1*12+12+4 --> will not balance.
[5/7/12] - local group
[1/3/4] - busy group
1*12 >= 5*4 + 12 + 4 --> will not balance.
Is this scenario needs to be handled as well?
So, from an SMT standpoint, we would not need to balance here, both
groups
should not run into SMT. Now, would it be beneficial to balance anyway?
Now we have:
[5/7/12] -> 42% busy
[1/3/4] -> 25% busy
If we would now balance and move one task around we would get either
[6/6/12] -> 50% busy
[0/4/4] -> 0% busy
or
[4/8/12] -> 33% busy
[2/2/4] -> 50% busy
The first case does probably not make that much sense (unless we have
workload
which would benefit from maybe cache locality) and we want everything to
run
in one group.
The second case brings the smaller group right onto the edge of using
SMT, while
also creating the possibility (depending on the algorithm we would use),
that
now the larger group will attempt to pull work from the smaller group
again,
ending up in a back and forth between the two. This is obviously also
true for
the first variant.
Could you maybe elaborate on what you meant by optimal performance?
I will figure out a solution and send a patch soon which incorporates
these
adjustments plus a more detailed description.
Thanks for the feedback.
>
> This is the prefer_sibling path. Could it be that you should disable
> prefer_siling between your sockets for such topology ? the default
> path compares the number of idle CPUs when groups has spare capacity
>
>
If I disable prefer_sibling (I played around with it a bit), I run
into the
problem,
that the tasks are distributed s.t. each group has the same amount
of
idle
CPUs, which
yields distributions similar to this:
00 01 02 03 04 05 06 07 08 09 10 11 || 12 13 14 15
x x x x x x x x
Now both groups have 4 idle CPUs which fulfills the criteria imposed
by the
load balancer,
but the larger group is now running SMT while the smaller one is
just
idle.
So, in this asymmetric setup, both criteria seem to not work in an
optimal
way. Going for
the same number of idle CPUs or alternatively for the same number of
running
processes
both cause a sub-optimal distribution of tasks, leading to
unnecessary SMT.
there is the same behavior and assumption here too
It seems also to be possible to address the regular load balancing
path by
aiming to have the
same unused capacity between groups instead of the same number of
idle CPUs.
This seems to
have been considered in the past, but the choice went in favor of
the
number
of idle CPUs.
unused capacity doesn't give the instantaneous state so a group can
be
idle but without
unused capacity
Since this decision was actively taken already, I focused on the
prefer_sibling path.
The question now would be how to address this properly (or if I'm
missing
something here).
As mentioned in the cover letter, this was the most simplistic and
least
invasive approach
I could find, others might be more sophisticated but also have some
side-effects.
I have a bit of a hard time leaving this one as-is, as it just
introduces
two additional
multiplications with no effect for most architectures. Maybe an
architectures specific
inline function that the compiler can optimize away if not needed?
> > goto force_balance;
> >
> > if (busiest->group_type != group_overloaded) {
> > --
> > 2.34.1
> >