> Il giorno 9 dic 2022, alle ore 02:46, Damien Le Moal <damien.lemoal@xxxxxxxxxxxxxxxxxx> ha scritto:
>
> On 12/8/22 19:43, Paolo Valente wrote:
>> From: Davide Zini <davidezini2@xxxxxxxxx>
>>
>> The main service scheme of BFQ for sync I/O is serving one sync
>> bfq_queue at a time, for a while. In particular, BFQ enforces this
>> scheme when it deems the latter necessary to boost throughput or
>> to preserve service guarantees. Unfortunately, when BFQ enforces
>> this policy, only one actuator at a time gets served for a while,
>> because each bfq_queue contains I/O only for one actuator. The
>> other actuators may remain underutilized.
>>
>> Actually, BFQ may serve (inject) extra I/O, taken from other
>> bfq_queues, in parallel with that of the in-service queue. This
>> injection mechanism may provide the ground for dealing also with
>> the above actuator-underutilization problem. Yet BFQ does not take
>> the actuator load into account when choosing which queue to pick
>> extra I/O from. In addition, BFQ may happen to inject extra I/O
>> only when the in-service queue is temporarily empty.
>>
>> In view of these facts, this commit extends the
>> injection mechanism in such a way that the latter:
>> (1) takes into account also the actuator load;
>> (2) checks such a load on each dispatch, and injects I/O for an
>> underutilized actuator, if there is one and there is I/O for it.
>>
>> To perform the check in (2), this commit introduces a load
>> threshold, currently set to 4. A linear scan of each actuator is
>> performed, until an actuator is found for which the following two
>> conditions hold: the load of the actuator is below the threshold,
>> and there is at least one non-in-service queue that contains I/O
>> for that actuator. If such a pair (actuator, queue) is found, then
>> the head request of that queue is returned for dispatch, instead
>> of the head request of the in-service queue.
>>
>> We have set the threshold, empirically, to the minimum possible
>> value for which an actuator is fully utilized, or close to be
>> fully utilized. By doing so, injected I/O 'steals' as few
>> drive-queue slots as possibile to the in-service queue. This
>> reduces as much as possible the probability that the service of
>> I/O from the in-service bfq_queue gets delayed because of slot
>> exhaustion, i.e., because all the slots of the drive queue are
>> filled with I/O injected from other queues (NCQ provides for 32
>> slots).
>>
>> This new mechanism also counters actuator underutilization in the
>> case of asymmetric configurations of bfq_queues. Namely if there
>> are few bfq_queues containing I/O for some actuators and many
>> bfq_queues containing I/O for other actuators. Or if the
>> bfq_queues containing I/O for some actuators have lower weights
>> than the other bfq_queues.
>>
>> Reviewed-by: Damien Le Moal <damien.lemoal@xxxxxxxxxxxxxxxxxx>
>> Signed-off-by: Paolo Valente <paolo.valente@xxxxxxxxxx>
>> Signed-off-by: Davide Zini <davidezini2@xxxxxxxxx>
>
> [...]
>
>> @@ -4792,22 +4799,69 @@ bfq_choose_bfqq_for_injection(struct bfq_data *bfqd)
>> else
>> limit = in_serv_bfqq->inject_limit;
>>
>> - if (bfqd->rq_in_driver < limit) {
>> + if (bfqd->tot_rq_in_driver < limit) {
>> bfqd->rqs_injected = true;
>> return bfqq;
>> }
>> }
>> + }
>> +
>> + return NULL;
>> +}
>> +
>> +static struct bfq_queue *
>> +bfq_find_active_bfqq_for_actuator(struct bfq_data *bfqd, int idx)
>> +{
>> + struct bfq_queue *bfqq = NULL;
>
> I do not think that you need the NULL initialization here.
>
>> +
>> + if (bfqd->in_service_queue &&
>> + bfqd->in_service_queue->actuator_idx == idx)
>> + return bfqd->in_service_queue;
>> +
>> + list_for_each_entry(bfqq, &bfqd->active_list[idx], bfqq_list) {
>> + if (!RB_EMPTY_ROOT(&bfqq->sort_list) &&
>> + bfq_serv_to_charge(bfqq->next_rq, bfqq) <=
>> + bfq_bfqq_budget_left(bfqq)) {
>> + return bfqq;
>> + }
>> + }
>>
>> return NULL;
>> }
>
> Otherwise looks OK.
>
> Reviewed-by: Damien Le Moal <damien.lemoal@xxxxxxxxxxxxxxxxxx>
Hi Damien,
I've applied your recommendations in this and the other replies of
yours. I'm about to send a V10.
Thank you very much for checking this code and helping me improve it,
Paolo
>
> --
> Damien Le Moal
> Western Digital Research
