> Il giorno 5 feb 2021, alle ore 11:16, Paolo Valente <paolo.valente@xxxxxxxxxx> ha scritto:
>
>
>
>> Il giorno 3 feb 2021, alle ore 12:43, Jan Kara <jack@xxxxxxx> ha scritto:
>>
>> On Thu 28-01-21 18:54:05, Paolo Valente wrote:
>>>
>>>
>>>> Il giorno 26 gen 2021, alle ore 17:18, Jens Axboe <axboe@xxxxxxxxx> ha scritto:
>>>>
>>>> On 1/26/21 3:50 AM, Paolo Valente wrote:
>>>>> Consider a new I/O request that arrives for a bfq_queue bfqq. If, when
>>>>> this happens, the only active bfq_queues are bfqq and either its waker
>>>>> bfq_queue or one of its woken bfq_queues, then there is no point in
>>>>> queueing this new I/O request in bfqq for service. In fact, the
>>>>> in-service queue and bfqq agree on serving this new I/O request as
>>>>> soon as possible. So this commit puts this new I/O request directly
>>>>> into the dispatch list.
>>>>>
>>>>> Tested-by: Jan Kara <jack@xxxxxxx>
>>>>> Signed-off-by: Paolo Valente <paolo.valente@xxxxxxxxxx>
>>>>> ---
>>>>> block/bfq-iosched.c | 17 ++++++++++++++++-
>>>>> 1 file changed, 16 insertions(+), 1 deletion(-)
>>>>>
>>>>> diff --git a/block/bfq-iosched.c b/block/bfq-iosched.c
>>>>> index a83149407336..e5b83910fbe0 100644
>>>>> --- a/block/bfq-iosched.c
>>>>> +++ b/block/bfq-iosched.c
>>>>> @@ -5640,7 +5640,22 @@ static void bfq_insert_request(struct blk_mq_hw_ctx *hctx, struct request *rq,
>>>>>
>>>>> spin_lock_irq(&bfqd->lock);
>>>>> bfqq = bfq_init_rq(rq);
>>>>> - if (!bfqq || at_head || blk_rq_is_passthrough(rq)) {
>>>>> +
>>>>> + /*
>>>>> + * Additional case for putting rq directly into the dispatch
>>>>> + * queue: the only active bfq_queues are bfqq and either its
>>>>> + * waker bfq_queue or one of its woken bfq_queues. In this
>>>>> + * case, there is no point in queueing rq in bfqq for
>>>>> + * service. In fact, the in-service queue and bfqq agree on
>>>>> + * serving this new I/O request as soon as possible.
>>>>> + */
>>>>> + if (!bfqq ||
>>>>> + (bfqq != bfqd->in_service_queue &&
>>>>> + bfqd->in_service_queue != NULL &&
>>>>> + bfq_tot_busy_queues(bfqd) == 1 + bfq_bfqq_busy(bfqq) &&
>>>>> + (bfqq->waker_bfqq == bfqd->in_service_queue ||
>>>>> + bfqd->in_service_queue->waker_bfqq == bfqq)) ||
>>>>> + at_head || blk_rq_is_passthrough(rq)) {
>>>>> if (at_head)
>>>>> list_add(&rq->queuelist, &bfqd->dispatch);
>>>>> else
>>>>>
>>>>
>>>> This is unreadable... Just seems like you are piling heuristics in to
>>>> catch some case, and it's neither readable nor clean.
>>>>
>>>
>>> Yeah, these comments inappropriately assume that the reader knows the
>>> waker mechanism in depth. And they do not stress at all how important
>>> this improvement is.
>>>
>>> I'll do my best to improve these comments.
>>>
>>> To try to do a better job, let me also explain the matter early here.
>>> Maybe you or others can give me some early feedback (or just tell me
>>> to proceed).
>>>
>>> This change is one of the main improvements that boosted
>>> throughput in Jan's tests. Here is the rationale:
>>> - consider a bfq_queue, say Q1, detected as a waker of another
>>> bfq_queue, say Q2
>>> - by definition of a waker, Q1 blocks the I/O of Q2, i.e., some I/O of
>>> of Q1 needs to be completed for new I/O of Q1 to arrive. A notable
>> ^^ Q2?
>>
>
> Yes, thank you!
>
> (after this interaction, I'll fix and improve all this description,
> according to your comments)
>
>>> example is journald
>>> - so, Q1 and Q2 are in any respect two cooperating processes: if the
>>> service of Q1's I/O is delayed, Q2 can only suffer from it.
>>> Conversely, if Q2's I/O is delayed, the purpose of Q1 is just defeated.
>>
>> What do you exactly mean by this last sentence?
>
> By definition of waker, the purpose of Q1's I/O is doing what needs to
> be done, so that new Q2's I/O can finally be issued. Delaying Q2's I/O
> is the opposite of this goal.
>
>>
>>> - as a consequence if some I/O of Q1/Q2 arrives while Q2/Q1 is the
>>> only queue in service, there is absolutely no point in delaying the
>>> service of such an I/O. The only possible result is a throughput
>>> loss, detected by Jan's test
>>
>> If we are idling at that moment waiting for more IO from in service queue,
>> I agree.
>
> And I agree too, if the drive has no internal queueing, has no
> parallelism or pipeline, or is at least one order of magnitude slower
> than the CPU is processing I/O. In all other cases, serving the I/O
> of only one queue at a time means throwing away throughput. For
> example, on a consumer SSD, moving from one to two I/O threads served
> in parallel usually means doubling the throughput.
>
> So, the best thing to do, if all the above conditions are met, is to
> have this new I/O dispatched as soon as possible.
>
> The most efficient way to attain this goal is to just put the new I/O
> directly into the dispatch list.
>
>> But that doesn't seem to be part of your condition above?
>>
>>> - so, when the above condition holds, the most effective and efficient
>>> action is to put the new I/O directly in the dispatch list
>>> - as an additional restriction, Q1 and Q2 must be the only busy queues
>>> for this commit to put the I/O of Q2/Q1 in the dispatch list. This is
>>> necessary, because, if also other queues are waiting for service, then
>>> putting new I/O directly in the dispatch list may evidently cause a
>>> violation of service guarantees for the other queues
>>
>> This last restriction is not ideal for cases like jbd2 thread since it may
>> still lead to pointless idling but I understand that without some
>> restriction like this several waking threads could just starve other ones.
>
> Yeah, the goal here is to reduce a little bit false positives.
>
>> So I guess it's fine for now.
>>
>
> Yes, hopefully experience will lead us to even improvements or even
> better solutions.
>
Hi Jens,
on a separate thread, Jan told me that my last reply, and therefore
also this patch are ok for him. May I now proceed with a V2, in which
I'll report my extra comments? Or are there some other issues for you?
Thanks,
Paolo
> Thanks,
> Paolo
>
>> Honza
>> --
>> Jan Kara <jack@xxxxxxxx>
>> SUSE Labs, CR
