Re: [PATCH 1/4] sched/eevdf: Fix vruntime adjustment on reweight

[Tianchen Ding <dtcccc@linux.alibaba.com>]

On 2024/3/1 16:30, Abel Wu wrote:
> On 3/1/24 2:41 PM, Tianchen Ding Wrote:
>> On 2024/2/29 22:25, Abel Wu wrote:
>>> Good catch. And to the best of my knowledge, the answer is YES. The
>>> above Equation in the paper, which is Eq. (20), is based on the
>>> assumption that:
>>>
>>>      "once client 3 leaves, the remaining two clients will
>>>       proportionally support the eventual loss or gain in the
>>>       service time"  -- Page 10
>>>
>>>      "by updating the virtual time according to Eq. (18,19) we
>>>       ensure that the sum over the lags of all active clients
>>>       is always zero"  -- Page 11
>>>
>>> But in Peter's implementation, it is the competitors in the new group
>>> that client 3 later joins in who actually support the effect. So when
>>> client 3 leaves competition with !0-lag in Linux, the rq's sum(lag_i)
>>> is no longer zero.
>>>
>>
>> I've different opinions. According to the comments above avg_vruntime_add(), V
>> is calculated exactly to satisfy sum(lag_i)=0. This is guaranteed by math.
> 
> Yes, you are right. I mixed another fairness issue with this. What I
> was thinking is that considering multiple competition groups (e.g.
> runqueues), the latency bound could be violated, that is someone could
> starve a bit. Say one entity even with positive lag could become less
> competitive if migrated to a higher competitive group.
> 
> Staring at Eq. (20) again, what if we do a fake reweight? I mean let
> the client leave and rejoin at the same time without changing weight?
> IMHO it should have no effects, but according to Eq. (20) the V will
> change to:
> 
>      V' = V + lag(j)/(W - w_j) - lag(j)/W != V
> 
> Have I missed anything?
> 

Good point! I've not ever noticed this conflict.

I tried to modify reweight_entity() to run dequeue_entity() -> adjust se->vlag ->
enqueue_entity(). And I found V do not changed.

The difference is, when doing enqueue_entity(), Peter enlarges the lag in place_entity().
Because after enqueue, the lag will evaporate.
In order to keep the same lag after enqueue, during place_entity(),
the new lag(t) will be enlarged with (W+w_i)/W.

So the Eq. (20) should be:


     V' = V + lag(j)/(W - w_j) - lag'(j)/(W - w_j + w'_j)

     lag'(j) = lag(j) * (W - w_j + w'_j)/(W - w_j)

So we can get

     V' = V + lag(j)/(W - w_j) - lag(j) * (W - w_j + w'_j)/(W - w_j)/(W - w_j + w'_j) = V

So COROLLARY #2 is correct.