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SubjectRe: [PATCH 1/2] cpuidle : auto-promotion for cpuidle states
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On 03/22/2019 06:56 PM, Daniel Lezcano wrote:
> On 22/03/2019 10:45, Rafael J. Wysocki wrote:
>> On Fri, Mar 22, 2019 at 8:31 AM Abhishek Goel
>> <huntbag@linux.vnet.ibm.com> wrote:
>>> Currently, the cpuidle governors (menu /ladder) determine what idle state
>>> an idling CPU should enter into based on heuristics that depend on the
>>> idle history on that CPU. Given that no predictive heuristic is perfect,
>>> there are cases where the governor predicts a shallow idle state, hoping
>>> that the CPU will be busy soon. However, if no new workload is scheduled
>>> on that CPU in the near future, the CPU will end up in the shallow state.
>>>
>>> In case of POWER, this is problematic, when the predicted state in the
>>> aforementioned scenario is a lite stop state, as such lite states will
>>> inhibit SMT folding, thereby depriving the other threads in the core from
>>> using the core resources.
>>>
>>> To address this, such lite states need to be autopromoted. The cpuidle-
>>> core can queue timer to correspond with the residency value of the next
>>> available state. Thus leading to auto-promotion to a deeper idle state as
>>> soon as possible.
>> Isn't the tick stopping avoidance sufficient for that?
> I was about to ask the same :)
>
>
>
>
Thanks for the review.
I performed experiments for three scenarios to collect some data.

case 1 :
Without this patch and without tick retained, i.e. in a upstream kernel,
It would spend more than even a second to get out of stop0_lite.

case 2 : With tick retained(as suggested) -

Generally, we have a sched tick at 4ms(CONF_HZ = 250). Ideally I expected
it to take 8 sched tick to get out of stop0_lite. Experimentally,
observation was

===================================
min            max            99percentile
4ms            12ms          4ms
===================================
*ms = milliseconds

It would take atleast one sched tick to get out of stop0_lite.

case 2 :  With this patch (not stopping tick, but explicitly queuing a
timer)

min            max              99.5percentile
===============================
144us       192us              144us
===============================
*us = microseconds

In this patch, we queue a timer just before entering into a stop0_lite
state. The timer fires at (residency of next available state + exit
latency of next available state * 2).
Let's say if next state(stop0) is available which has residency of 20us, it
should get out in as low as (20+2*2)*8 [Based on the forumla (residency +
2xlatency)*history length] microseconds = 192us. Ideally we would expect 8
iterations, it was observed to get out in 6-7 iterations.
Even if let's say stop2 is next available state(stop0 and stop1 both are
unavailable), it would take (100+2*10)*8 = 960us to get into stop2.

So, We are able to get out of stop0_lite generally in 150us(with this
patch) as
compared to 4ms(with tick retained). As stated earlier, we do not want
to get
stuck into stop0_lite as it inhibits SMT folding for other sibling
threads, depriving
them of core resources. Current patch is using auto-promotion only for
stop0_lite,
as it gives performance benefit(primary reason) along with lowering down
power
consumption. We may extend this model for other states in future.

--Abhishek

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