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SubjectRe: [PATCH 6/9] Include idle and iowait fields in cpuacct
On 09/20/2011 10:05 AM, Peter Zijlstra wrote:
> On Tue, 2011-09-20 at 09:58 -0300, Glauber Costa wrote:
>> On 09/20/2011 09:58 AM, Peter Zijlstra wrote:
>>> On Tue, 2011-09-20 at 09:36 -0300, Glauber Costa wrote:
>>>> On 09/20/2011 06:21 AM, Peter Zijlstra wrote:
>>>>> On Wed, 2011-09-14 at 17:04 -0300, Glauber Costa wrote:
>>>>>> These are slightly different from the others though:
>>>>>> (note to reviewers: might be better to put those in a separate
>>>>>> array?)
>>>>>>
>>>>>> Since idle/iowait are a property of the system - by definition,
>>>>>> no process from any cgroup is running when the system is idle,
>>>>>> they are system wide. So what these fields really mean, are baselines
>>>>>> for when the cgroup was created. It allows the cgroup to start
>>>>>> counting idle/iowait from 0.
>>>>>
>>>>> Alternatively you can make iowait based on nr_uninterruptible per cgroup
>>>>> and count all ticks _this_ cgroup was idle.
>>>> You think?
>>>>
>>>> Humm,humm... maybe...
>>>> iowait can indeed be seen as a process group characteristic. I was
>>>> mainly concerned about overhead here, specially for the idle case:
>>>
>>> The overhead of accounting per cgroup nr_uninterruptible is the worst I
>>> think, that's in the sleep/wakeup paths.
>>>
>>>> If we are idle, there is no task context we can draw from, since the
>>>> task in the cpu is the idle task. So we end up having to touch all
>>>> cgroups... Or am I missing something?
>>>>
>>>> Sounds expensive.
>>>
>>> Count the total number of ticks on the cpu (I think we already have
>>> that) and subtract the number of ticks in this cgroup (I think we also
>>> already have that), which should yield: number of ticks not in this
>>> cgroup, aka number of ticks this cgroup was idle.
>> No , no... remember steal time.
>
> Of course I don't.. that's virt stuff, I repress that with all my might.
> But add or subtract steal ticks someplace and it doesn't come out right?

That's what I am here for...

But back to your answer:

>>> Count the total number of ticks on the cpu (I think we already have
>>> that) and subtract the number of ticks in this cgroup (I think we also
>>> already have that),

Not sure if we have ticks in this cgroup... anyway, it can be done. We
need a baseline for what was the tick situation when the cgroup started
anyway.



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