Re: [PATCH -next v7 2/3] block, bfq: refactor the counting of 'num_groups_with_pending_reqs'

[Yu Kuai <yukuai3@huawei.com>]

在 2022/05/31 18:01, Jan Kara 写道:
> On Tue 31-05-22 17:33:25, Yu Kuai wrote:
>> 在 2022/05/31 17:19, Paolo Valente 写道:
>>>> Il giorno 31 mag 2022, alle ore 11:06, Yu Kuai <yukuai3@huawei.com> ha scritto:
>>>>
>>>> 在 2022/05/31 16:36, Paolo VALENTE 写道:
>>>>>> Il giorno 30 mag 2022, alle ore 10:40, Yu Kuai <yukuai3@huawei.com> ha scritto:
>>>>>>
>>>>>> 在 2022/05/30 16:34, Yu Kuai 写道:
>>>>>>> 在 2022/05/30 16:10, Paolo Valente 写道:
>>>>>>>>
>>>>>>>>
>>>>>>>>> Il giorno 28 mag 2022, alle ore 11:50, Yu Kuai <yukuai3@huawei.com> ha scritto:
>>>>>>>>>
>>>>>>>>> Currently, bfq can't handle sync io concurrently as long as they
>>>>>>>>> are not issued from root group. This is because
>>>>>>>>> 'bfqd->num_groups_with_pending_reqs > 0' is always true in
>>>>>>>>> bfq_asymmetric_scenario().
>>>>>>>>>
>>>>>>>>> The way that bfqg is counted into 'num_groups_with_pending_reqs':
>>>>>>>>>
>>>>>>>>> Before this patch:
>>>>>>>>> 1) root group will never be counted.
>>>>>>>>> 2) Count if bfqg or it's child bfqgs have pending requests.
>>>>>>>>> 3) Don't count if bfqg and it's child bfqgs complete all the requests.
>>>>>>>>>
>>>>>>>>> After this patch:
>>>>>>>>> 1) root group is counted.
>>>>>>>>> 2) Count if bfqg have at least one bfqq that is marked busy.
>>>>>>>>> 3) Don't count if bfqg doesn't have any busy bfqqs.
>>>>>>>>
>>>>>>>> Unfortunately, I see a last problem here. I see a double change:
>>>>>>>> (1) a bfqg is now counted only as a function of the state of its child
>>>>>>>>        queues, and not of also its child bfqgs
>>>>>>>> (2) the state considered for counting a bfqg moves from having pending
>>>>>>>>        requests to having busy queues
>>>>>>>>
>>>>>>>> I'm ok with with (1), which is a good catch (you are lady explained
>>>>>>>> the idea to me some time ago IIRC).
>>>>>>>>
>>>>>>>> Yet I fear that (2) is not ok.  A bfqq can become non busy even if it
>>>>>>>> still has in-flight I/O, i.e.  I/O being served in the drive.  The
>>>>>>>> weight of such a bfqq must still be considered in the weights_tree,
>>>>>>>> and the group containing such a queue must still be counted when
>>>>>>>> checking whether the scenario is asymmetric.  Otherwise service
>>>>>>>> guarantees are broken.  The reason is that, if a scenario is deemed as
>>>>>>>> symmetric because in-flight I/O is not taken into account, then idling
>>>>>>>> will not be performed to protect some bfqq, and in-flight I/O may
>>>>>>>> steal bandwidth to that bfqq in an uncontrolled way.
>>>>>>> Hi, Paolo
>>>>>>> Thanks for your explanation.
>>>>>>> My orginal thoughts was using weights_tree insertion/removal, however,
>>>>>>> Jan convinced me that using bfq_add/del_bfqq_busy() is ok.
>>>>>>>   From what I see, when bfqq dispatch the last request,
>>>>>>> bfq_del_bfqq_busy() will not be called from __bfq_bfqq_expire() if
>>>>>>> idling is needed, and it will delayed to when such bfqq get scheduled as
>>>>>>> in-service queue again. Which means the weight of such bfqq should still
>>>>>>> be considered in the weights_tree.
>>>>>>> I also run some tests on null_blk with "irqmode=2
>>>>>>> completion_nsec=100000000(100ms) hw_queue_depth=1", and tests show
>>>>>>> that service guarantees are still preserved on slow device.
>>>>>>> Do you this is strong enough to cover your concern?
>>>>> Unfortunately it is not.  Your very argument is what made be believe
>>>>> that considering busy queues was enough, in the first place.  But, as
>>>>> I found out, the problem is caused by the queues that do not enjoy
>>>>> idling.  With your patch (as well as in my initial version) they are
>>>>> not counted when they remain without requests queued.  And this makes
>>>>> asymmetric scenarios be considered erroneously as symmetric.  The
>>>>> consequence is that idling gets switched off when it had to be kept
>>>>> on, and control on bandwidth is lost for the victim in-service queues.
>>>>
>>>> Hi，Paolo
>>>>
>>>> Thanks for your explanation, are you thinking that if bfqq doesn't enjoy
>>>> idling, then such bfqq will clear busy after dispatching the last
>>>> request?
>>>>
>>>> Please kindly correct me if I'm wrong in the following process:
>>>>
>>>> If there are more than one bfqg that is activatied, then bfqqs that are
>>>> not enjoying idle are still left busy after dispatching the last
>>>> request.
>>>>
>>>> details in __bfq_bfqq_expire:
>>>>
>>>>          if (RB_EMPTY_ROOT(&bfqq->sort_list) &&
>>>>          ┊   !(reason == BFQQE_PREEMPTED &&
>>>>          ┊     idling_needed_for_service_guarantees(bfqd, bfqq))) {
>>>> -> idling_needed_for_service_guarantees will always return true,
>>>
>>> It returns true only is the scenario is symmetric.  Not counting bfqqs
>>> with in-flight requests makes an asymmetric scenario be considered
>>> wrongly symmetric.  See function bfq_asymmetric_scenario().
>>
>> Hi, Paolo
>>
>> Do you mean this gap?
>>
>> 1. io1 is issued from bfqq1(from bfqg1)
>> 2. bfqq1 dispatched this io, it's busy is cleared
>> 3. *before io1 is completed*, io2 is issued from bfqq2(bfqg2)
> 
> Yes. So as far as I understand Paolo is concerned about this scenario.
> 
>> 4. with this patchset, while dispatching io2 from bfqq2, the scenario
>> should be symmetric while it's considered wrongly asymmetric.
> 
> But with this patchset, we will consider this scenario symmetric because at
> any point in time there is only one busy bfqq. Before, we considered this
> scenario asymmetric because two different bfq groups have bfqq in their
> weights_tree. So before this patchset
> idling_needed_for_service_guarantees() returned true, after this patchset
> the function returns false so we won't idle anymore and Paolo argues that
> bfqq1 does not get adequate protection from bfqq2 as a result.
> 
> I agree with Paolo this seems possible. The fix is relatively simple though
> - instead of changing how weights_tree is used for weight raised queues as
> you did originally, I'd move the accounting of groups with pending requests
> to bfq_add/del_bfqq_busy() and bfq_completed_request().
> 
> 								Honza

Thanks for your explanation, I'll send a new version.

Kuai