| Message ID | 20220528095020.186970-3-yukuai3@huawei.com (mailing list archive) |
|---|---|
| State | New, archived |
| Headers | show |
| Series | support concurrent sync io for bfq on a specail occasion | expand |
> 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. I verified this also experimentally a few years ago, when I added this weights_tree stuff. That's the rationale behind the part of bfq_weights_tree_remove that this patch eliminates. IOW, for a bfqq and its parent bfqg to be out of the count for symmetry, all bfqq's requests must also be completed. Thanks, Paolo > > The main reason to use busy state of bfqq instead of 'pending requests' > is that bfqq can stay busy after dispatching the last request if idling > is needed for service guarantees. > > With this change, the occasion that only one group is activated can be > detected, and next patch will support concurrent sync io in the > occasion. > > This patch also rename 'num_groups_with_pending_reqs' to > 'num_groups_with_busy_queues'. > > Signed-off-by: Yu Kuai <yukuai3@huawei.com> > Reviewed-by: Jan Kara <jack@suse.cz> > --- > block/bfq-iosched.c | 46 ++----------------------------------- > block/bfq-iosched.h | 55 ++++++--------------------------------------- > block/bfq-wf2q.c | 19 ++++------------ > 3 files changed, 13 insertions(+), 107 deletions(-) > > diff --git a/block/bfq-iosched.c b/block/bfq-iosched.c > index 0d46cb728bbf..eb1da1bd5eb4 100644 > --- a/block/bfq-iosched.c > +++ b/block/bfq-iosched.c > @@ -852,7 +852,7 @@ static bool bfq_asymmetric_scenario(struct bfq_data *bfqd, > > return varied_queue_weights || multiple_classes_busy > #ifdef CONFIG_BFQ_GROUP_IOSCHED > - || bfqd->num_groups_with_pending_reqs > 0 > + || bfqd->num_groups_with_busy_queues > 0 > #endif > ; > } > @@ -970,48 +970,6 @@ void __bfq_weights_tree_remove(struct bfq_data *bfqd, > void bfq_weights_tree_remove(struct bfq_data *bfqd, > struct bfq_queue *bfqq) > { > - struct bfq_entity *entity = bfqq->entity.parent; > - > - for_each_entity(entity) { > - struct bfq_sched_data *sd = entity->my_sched_data; > - > - if (sd->next_in_service || sd->in_service_entity) { > - /* > - * entity is still active, because either > - * next_in_service or in_service_entity is not > - * NULL (see the comments on the definition of > - * next_in_service for details on why > - * in_service_entity must be checked too). > - * > - * As a consequence, its parent entities are > - * active as well, and thus this loop must > - * stop here. > - */ > - break; > - } > - > - /* > - * The decrement of num_groups_with_pending_reqs is > - * not performed immediately upon the deactivation of > - * entity, but it is delayed to when it also happens > - * that the first leaf descendant bfqq of entity gets > - * all its pending requests completed. The following > - * instructions perform this delayed decrement, if > - * needed. See the comments on > - * num_groups_with_pending_reqs for details. > - */ > - if (entity->in_groups_with_pending_reqs) { > - entity->in_groups_with_pending_reqs = false; > - bfqd->num_groups_with_pending_reqs--; > - } > - } > - > - /* > - * Next function is invoked last, because it causes bfqq to be > - * freed if the following holds: bfqq is not in service and > - * has no dispatched request. DO NOT use bfqq after the next > - * function invocation. > - */ > __bfq_weights_tree_remove(bfqd, bfqq, > &bfqd->queue_weights_tree); > } > @@ -7118,7 +7076,7 @@ static int bfq_init_queue(struct request_queue *q, struct elevator_type *e) > bfqd->idle_slice_timer.function = bfq_idle_slice_timer; > > bfqd->queue_weights_tree = RB_ROOT_CACHED; > - bfqd->num_groups_with_pending_reqs = 0; > + bfqd->num_groups_with_busy_queues = 0; > > INIT_LIST_HEAD(&bfqd->active_list); > INIT_LIST_HEAD(&bfqd->idle_list); > diff --git a/block/bfq-iosched.h b/block/bfq-iosched.h > index d92adbdd70ee..6c6cd984d769 100644 > --- a/block/bfq-iosched.h > +++ b/block/bfq-iosched.h > @@ -197,9 +197,6 @@ struct bfq_entity { > /* flag, set to request a weight, ioprio or ioprio_class change */ > int prio_changed; > > - /* flag, set if the entity is counted in groups_with_pending_reqs */ > - bool in_groups_with_pending_reqs; > - > /* last child queue of entity created (for non-leaf entities) */ > struct bfq_queue *last_bfqq_created; > }; > @@ -496,52 +493,14 @@ struct bfq_data { > struct rb_root_cached queue_weights_tree; > > /* > - * Number of groups with at least one descendant process that > - * has at least one request waiting for completion. Note that > - * this accounts for also requests already dispatched, but not > - * yet completed. Therefore this number of groups may differ > - * (be larger) than the number of active groups, as a group is > - * considered active only if its corresponding entity has > - * descendant queues with at least one request queued. This > - * number is used to decide whether a scenario is symmetric. > - * For a detailed explanation see comments on the computation > - * of the variable asymmetric_scenario in the function > - * bfq_better_to_idle(). > - * > - * However, it is hard to compute this number exactly, for > - * groups with multiple descendant processes. Consider a group > - * that is inactive, i.e., that has no descendant process with > - * pending I/O inside BFQ queues. Then suppose that > - * num_groups_with_pending_reqs is still accounting for this > - * group, because the group has descendant processes with some > - * I/O request still in flight. num_groups_with_pending_reqs > - * should be decremented when the in-flight request of the > - * last descendant process is finally completed (assuming that > - * nothing else has changed for the group in the meantime, in > - * terms of composition of the group and active/inactive state of child > - * groups and processes). To accomplish this, an additional > - * pending-request counter must be added to entities, and must > - * be updated correctly. To avoid this additional field and operations, > - * we resort to the following tradeoff between simplicity and > - * accuracy: for an inactive group that is still counted in > - * num_groups_with_pending_reqs, we decrement > - * num_groups_with_pending_reqs when the first descendant > - * process of the group remains with no request waiting for > - * completion. > - * > - * Even this simpler decrement strategy requires a little > - * carefulness: to avoid multiple decrements, we flag a group, > - * more precisely an entity representing a group, as still > - * counted in num_groups_with_pending_reqs when it becomes > - * inactive. Then, when the first descendant queue of the > - * entity remains with no request waiting for completion, > - * num_groups_with_pending_reqs is decremented, and this flag > - * is reset. After this flag is reset for the entity, > - * num_groups_with_pending_reqs won't be decremented any > - * longer in case a new descendant queue of the entity remains > - * with no request waiting for completion. > + * Number of groups with at least one bfqq that is marked busy, > + * and this number is used to decide whether a scenario is symmetric. > + * Note that bfqq is busy doesn't mean that the bfqq contains requests. > + * If idling is needed for service guarantees, bfqq will stay busy > + * after dispatching the last request, see details in > + * __bfq_bfqq_expire(). > */ > - unsigned int num_groups_with_pending_reqs; > + unsigned int num_groups_with_busy_queues; > > /* > * Per-class (RT, BE, IDLE) number of bfq_queues containing > diff --git a/block/bfq-wf2q.c b/block/bfq-wf2q.c > index b97e33688335..48ca7922035c 100644 > --- a/block/bfq-wf2q.c > +++ b/block/bfq-wf2q.c > @@ -221,13 +221,15 @@ static bool bfq_no_longer_next_in_service(struct bfq_entity *entity) > static void bfq_inc_busy_queues(struct bfq_queue *bfqq) > { > bfqq->bfqd->busy_queues[bfqq->ioprio_class - 1]++; > - bfqq_group(bfqq)->busy_queues++; > + if (!(bfqq_group(bfqq)->busy_queues++)) > + bfqq->bfqd->num_groups_with_busy_queues++; > } > > static void bfq_dec_busy_queues(struct bfq_queue *bfqq) > { > bfqq->bfqd->busy_queues[bfqq->ioprio_class - 1]--; > - bfqq_group(bfqq)->busy_queues--; > + if (!(--bfqq_group(bfqq)->busy_queues)) > + bfqq->bfqd->num_groups_with_busy_queues--; > } > > #else /* CONFIG_BFQ_GROUP_IOSCHED */ > @@ -1006,19 +1008,6 @@ static void __bfq_activate_entity(struct bfq_entity *entity, > entity->on_st_or_in_serv = true; > } > > -#ifdef CONFIG_BFQ_GROUP_IOSCHED > - if (!bfq_entity_to_bfqq(entity)) { /* bfq_group */ > - struct bfq_group *bfqg = > - container_of(entity, struct bfq_group, entity); > - struct bfq_data *bfqd = bfqg->bfqd; > - > - if (!entity->in_groups_with_pending_reqs) { > - entity->in_groups_with_pending_reqs = true; > - bfqd->num_groups_with_pending_reqs++; > - } > - } > -#endif > - > bfq_update_fin_time_enqueue(entity, st, backshifted); > } > > -- > 2.31.1 >
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? Thanks, Kuai > > I verified this also experimentally a few years ago, when I added this > weights_tree stuff. That's the rationale behind the part of > bfq_weights_tree_remove that this patch eliminates. IOW, > for a bfqq and its parent bfqg to be out of the count for symmetry, > all bfqq's requests must also be completed. > > Thanks, > Paolo > >> >> The main reason to use busy state of bfqq instead of 'pending requests' >> is that bfqq can stay busy after dispatching the last request if idling >> is needed for service guarantees. >> >> With this change, the occasion that only one group is activated can be >> detected, and next patch will support concurrent sync io in the >> occasion. >> >> This patch also rename 'num_groups_with_pending_reqs' to >> 'num_groups_with_busy_queues'. >> >> Signed-off-by: Yu Kuai <yukuai3@huawei.com> >> Reviewed-by: Jan Kara <jack@suse.cz> >> --- >> block/bfq-iosched.c | 46 ++----------------------------------- >> block/bfq-iosched.h | 55 ++++++--------------------------------------- >> block/bfq-wf2q.c | 19 ++++------------ >> 3 files changed, 13 insertions(+), 107 deletions(-) >> >> diff --git a/block/bfq-iosched.c b/block/bfq-iosched.c >> index 0d46cb728bbf..eb1da1bd5eb4 100644 >> --- a/block/bfq-iosched.c >> +++ b/block/bfq-iosched.c >> @@ -852,7 +852,7 @@ static bool bfq_asymmetric_scenario(struct bfq_data *bfqd, >> >> return varied_queue_weights || multiple_classes_busy >> #ifdef CONFIG_BFQ_GROUP_IOSCHED >> - || bfqd->num_groups_with_pending_reqs > 0 >> + || bfqd->num_groups_with_busy_queues > 0 >> #endif >> ; >> } >> @@ -970,48 +970,6 @@ void __bfq_weights_tree_remove(struct bfq_data *bfqd, >> void bfq_weights_tree_remove(struct bfq_data *bfqd, >> struct bfq_queue *bfqq) >> { >> - struct bfq_entity *entity = bfqq->entity.parent; >> - >> - for_each_entity(entity) { >> - struct bfq_sched_data *sd = entity->my_sched_data; >> - >> - if (sd->next_in_service || sd->in_service_entity) { >> - /* >> - * entity is still active, because either >> - * next_in_service or in_service_entity is not >> - * NULL (see the comments on the definition of >> - * next_in_service for details on why >> - * in_service_entity must be checked too). >> - * >> - * As a consequence, its parent entities are >> - * active as well, and thus this loop must >> - * stop here. >> - */ >> - break; >> - } >> - >> - /* >> - * The decrement of num_groups_with_pending_reqs is >> - * not performed immediately upon the deactivation of >> - * entity, but it is delayed to when it also happens >> - * that the first leaf descendant bfqq of entity gets >> - * all its pending requests completed. The following >> - * instructions perform this delayed decrement, if >> - * needed. See the comments on >> - * num_groups_with_pending_reqs for details. >> - */ >> - if (entity->in_groups_with_pending_reqs) { >> - entity->in_groups_with_pending_reqs = false; >> - bfqd->num_groups_with_pending_reqs--; >> - } >> - } >> - >> - /* >> - * Next function is invoked last, because it causes bfqq to be >> - * freed if the following holds: bfqq is not in service and >> - * has no dispatched request. DO NOT use bfqq after the next >> - * function invocation. >> - */ >> __bfq_weights_tree_remove(bfqd, bfqq, >> &bfqd->queue_weights_tree); >> } >> @@ -7118,7 +7076,7 @@ static int bfq_init_queue(struct request_queue *q, struct elevator_type *e) >> bfqd->idle_slice_timer.function = bfq_idle_slice_timer; >> >> bfqd->queue_weights_tree = RB_ROOT_CACHED; >> - bfqd->num_groups_with_pending_reqs = 0; >> + bfqd->num_groups_with_busy_queues = 0; >> >> INIT_LIST_HEAD(&bfqd->active_list); >> INIT_LIST_HEAD(&bfqd->idle_list); >> diff --git a/block/bfq-iosched.h b/block/bfq-iosched.h >> index d92adbdd70ee..6c6cd984d769 100644 >> --- a/block/bfq-iosched.h >> +++ b/block/bfq-iosched.h >> @@ -197,9 +197,6 @@ struct bfq_entity { >> /* flag, set to request a weight, ioprio or ioprio_class change */ >> int prio_changed; >> >> - /* flag, set if the entity is counted in groups_with_pending_reqs */ >> - bool in_groups_with_pending_reqs; >> - >> /* last child queue of entity created (for non-leaf entities) */ >> struct bfq_queue *last_bfqq_created; >> }; >> @@ -496,52 +493,14 @@ struct bfq_data { >> struct rb_root_cached queue_weights_tree; >> >> /* >> - * Number of groups with at least one descendant process that >> - * has at least one request waiting for completion. Note that >> - * this accounts for also requests already dispatched, but not >> - * yet completed. Therefore this number of groups may differ >> - * (be larger) than the number of active groups, as a group is >> - * considered active only if its corresponding entity has >> - * descendant queues with at least one request queued. This >> - * number is used to decide whether a scenario is symmetric. >> - * For a detailed explanation see comments on the computation >> - * of the variable asymmetric_scenario in the function >> - * bfq_better_to_idle(). >> - * >> - * However, it is hard to compute this number exactly, for >> - * groups with multiple descendant processes. Consider a group >> - * that is inactive, i.e., that has no descendant process with >> - * pending I/O inside BFQ queues. Then suppose that >> - * num_groups_with_pending_reqs is still accounting for this >> - * group, because the group has descendant processes with some >> - * I/O request still in flight. num_groups_with_pending_reqs >> - * should be decremented when the in-flight request of the >> - * last descendant process is finally completed (assuming that >> - * nothing else has changed for the group in the meantime, in >> - * terms of composition of the group and active/inactive state of child >> - * groups and processes). To accomplish this, an additional >> - * pending-request counter must be added to entities, and must >> - * be updated correctly. To avoid this additional field and operations, >> - * we resort to the following tradeoff between simplicity and >> - * accuracy: for an inactive group that is still counted in >> - * num_groups_with_pending_reqs, we decrement >> - * num_groups_with_pending_reqs when the first descendant >> - * process of the group remains with no request waiting for >> - * completion. >> - * >> - * Even this simpler decrement strategy requires a little >> - * carefulness: to avoid multiple decrements, we flag a group, >> - * more precisely an entity representing a group, as still >> - * counted in num_groups_with_pending_reqs when it becomes >> - * inactive. Then, when the first descendant queue of the >> - * entity remains with no request waiting for completion, >> - * num_groups_with_pending_reqs is decremented, and this flag >> - * is reset. After this flag is reset for the entity, >> - * num_groups_with_pending_reqs won't be decremented any >> - * longer in case a new descendant queue of the entity remains >> - * with no request waiting for completion. >> + * Number of groups with at least one bfqq that is marked busy, >> + * and this number is used to decide whether a scenario is symmetric. >> + * Note that bfqq is busy doesn't mean that the bfqq contains requests. >> + * If idling is needed for service guarantees, bfqq will stay busy >> + * after dispatching the last request, see details in >> + * __bfq_bfqq_expire(). >> */ >> - unsigned int num_groups_with_pending_reqs; >> + unsigned int num_groups_with_busy_queues; >> >> /* >> * Per-class (RT, BE, IDLE) number of bfq_queues containing >> diff --git a/block/bfq-wf2q.c b/block/bfq-wf2q.c >> index b97e33688335..48ca7922035c 100644 >> --- a/block/bfq-wf2q.c >> +++ b/block/bfq-wf2q.c >> @@ -221,13 +221,15 @@ static bool bfq_no_longer_next_in_service(struct bfq_entity *entity) >> static void bfq_inc_busy_queues(struct bfq_queue *bfqq) >> { >> bfqq->bfqd->busy_queues[bfqq->ioprio_class - 1]++; >> - bfqq_group(bfqq)->busy_queues++; >> + if (!(bfqq_group(bfqq)->busy_queues++)) >> + bfqq->bfqd->num_groups_with_busy_queues++; >> } >> >> static void bfq_dec_busy_queues(struct bfq_queue *bfqq) >> { >> bfqq->bfqd->busy_queues[bfqq->ioprio_class - 1]--; >> - bfqq_group(bfqq)->busy_queues--; >> + if (!(--bfqq_group(bfqq)->busy_queues)) >> + bfqq->bfqd->num_groups_with_busy_queues--; >> } >> >> #else /* CONFIG_BFQ_GROUP_IOSCHED */ >> @@ -1006,19 +1008,6 @@ static void __bfq_activate_entity(struct bfq_entity *entity, >> entity->on_st_or_in_serv = true; >> } >> >> -#ifdef CONFIG_BFQ_GROUP_IOSCHED >> - if (!bfq_entity_to_bfqq(entity)) { /* bfq_group */ >> - struct bfq_group *bfqg = >> - container_of(entity, struct bfq_group, entity); >> - struct bfq_data *bfqd = bfqg->bfqd; >> - >> - if (!entity->in_groups_with_pending_reqs) { >> - entity->in_groups_with_pending_reqs = true; >> - bfqd->num_groups_with_pending_reqs++; >> - } >> - } >> -#endif >> - >> bfq_update_fin_time_enqueue(entity, st, backshifted); >> } >> >> -- >> 2.31.1 >> > > . >
> 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. Thanks, Paolo >> Thanks, >> Kuai >>> >>> I verified this also experimentally a few years ago, when I added this >>> weights_tree stuff. That's the rationale behind the part of >>> bfq_weights_tree_remove that this patch eliminates. IOW, >>> for a bfqq and its parent bfqg to be out of the count for symmetry, >>> all bfqq's requests must also be completed. >>> >>> Thanks, >>> Paolo > > I forgot to cc Jan for this patchset... This is a reply for Jan.
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, bfqq(whether or not enjoy idling) will stay busy. if (bfqq->dispatched == 0) /* ┊* Overloading budget_timeout field to store ┊* the time at which the queue remains with no ┊* backlog and no outstanding request; used by ┊* the weight-raising mechanism. ┊*/ bfqq->budget_timeout = jiffies; bfq_del_bfqq_busy(bfqd, bfqq, true); Thanks, Kuai
> 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(). Paolo > bfqq(whether or not enjoy idling) will stay busy. > if (bfqq->dispatched == 0) > /* > ┊* Overloading budget_timeout field to store > ┊* the time at which the queue remains with no > ┊* backlog and no outstanding request; used by > ┊* the weight-raising mechanism. > ┊*/ > bfqq->budget_timeout = jiffies; > > bfq_del_bfqq_busy(bfqd, bfqq, true); > > Thanks, > Kuai
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, Yes, with this patchset, If there are more than one bfqg that is activatied(contain busy bfqq), bfq_asymmetric_scenario() will return true: bfq_asymmetric_scenario() return varied_queue_weights || multiple_classes_busy #ifdef CONFIG_BFQ_GROUP_IOSCHED || bfqd->num_groups_with_busy_queues > 1 #endif From what I see, bfqd->num_groups_with_busy_queues > 1 is always true... > > Paolo > >> bfqq(whether or not enjoy idling) will stay busy. >> if (bfqq->dispatched == 0) >> /* >> ┊* Overloading budget_timeout field to store >> ┊* the time at which the queue remains with no >> ┊* backlog and no outstanding request; used by >> ┊* the weight-raising mechanism. >> ┊*/ >> bfqq->budget_timeout = jiffies; >> >> bfq_del_bfqq_busy(bfqd, bfqq, true); >> >> Thanks, >> Kuai > > . >
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) 4. with this patchset, while dispatching io2 from bfqq2, the scenario should be symmetric while it's considered wrongly asymmetric. > > Paolo > >> bfqq(whether or not enjoy idling) will stay busy. >> if (bfqq->dispatched == 0) >> /* >> ┊* Overloading budget_timeout field to store >> ┊* the time at which the queue remains with no >> ┊* backlog and no outstanding request; used by >> ┊* the weight-raising mechanism. >> ┊*/ >> bfqq->budget_timeout = jiffies; >> >> bfq_del_bfqq_busy(bfqd, bfqq, true); >> >> Thanks, >> Kuai > > . >
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
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
> Il giorno 31 mag 2022, alle ore 12:01, Jan Kara <jack@suse.cz> ha scritto: > > 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(). > Why don't we use simply the existing logic? I mean, as for the changes made by this patch, we could simply turn the loop: void bfq_weights_tree_remove(struct bfq_data *bfqd, struct bfq_queue *bfqq) { ... for_each_entity(entity) { struct bfq_sched_data *sd = entity->my_sched_data; ... if (entity->in_groups_with_pending_reqs) { entity->in_groups_with_pending_reqs = false; bfqd->num_groups_with_pending_reqs--; } } ... } into a single: bfqd->num_groups_with_pending_reqs--; so that only the parent group is concerned. Thanks, Paolo > Honza > -- > Jan Kara <jack@suse.com> > SUSE Labs, CR
在 2022/05/31 20:57, Paolo Valente 写道: > > >> Il giorno 31 mag 2022, alle ore 12:01, Jan Kara <jack@suse.cz> ha scritto: >> >> 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(). >> > > Why don't we use simply the existing logic? I mean, as for the changes made by this patch, we could simply turn the loop: > > void bfq_weights_tree_remove(struct bfq_data *bfqd, > struct bfq_queue *bfqq) > { > ... > for_each_entity(entity) { > struct bfq_sched_data *sd = entity->my_sched_data; > > ... > if (entity->in_groups_with_pending_reqs) { > entity->in_groups_with_pending_reqs = false; > bfqd->num_groups_with_pending_reqs--; > } > } > ... > } > > into a single: > > bfqd->num_groups_with_pending_reqs--; > > so that only the parent group is concerned. It's ok to decrease it here, however, we need another place to increase it in order to count root group... And bfq_weights_tree_add is not good because it bypass wr queues. Thanks, Kuai
在 2022/05/31 20:57, Paolo Valente 写道: > > >> Il giorno 31 mag 2022, alle ore 12:01, Jan Kara <jack@suse.cz> ha scritto: >> >> 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(). >> > > Why don't we use simply the existing logic? I mean, as for the changes made by this patch, we could simply turn the loop: > > void bfq_weights_tree_remove(struct bfq_data *bfqd, > struct bfq_queue *bfqq) > { > ... > for_each_entity(entity) { > struct bfq_sched_data *sd = entity->my_sched_data; > > ... > if (entity->in_groups_with_pending_reqs) { > entity->in_groups_with_pending_reqs = false; > bfqd->num_groups_with_pending_reqs--; > } > } > ... > } > > into a single: > > bfqd->num_groups_with_pending_reqs--; > > so that only the parent group is concerned. > > Thanks, > Paolo Hi, Paolo Can you please take a look if this patchset(v9) delivered to you? There are still some problems with out mail server... https://patchwork.kernel.org/project/linux-block/cover/20220601114340.949953-1-yukuai3@huawei.com/ Thanks, Kuai
diff --git a/block/bfq-iosched.c b/block/bfq-iosched.c index 0d46cb728bbf..eb1da1bd5eb4 100644 --- a/block/bfq-iosched.c +++ b/block/bfq-iosched.c @@ -852,7 +852,7 @@ static bool bfq_asymmetric_scenario(struct bfq_data *bfqd, return varied_queue_weights || multiple_classes_busy #ifdef CONFIG_BFQ_GROUP_IOSCHED - || bfqd->num_groups_with_pending_reqs > 0 + || bfqd->num_groups_with_busy_queues > 0 #endif ; } @@ -970,48 +970,6 @@ void __bfq_weights_tree_remove(struct bfq_data *bfqd, void bfq_weights_tree_remove(struct bfq_data *bfqd, struct bfq_queue *bfqq) { - struct bfq_entity *entity = bfqq->entity.parent; - - for_each_entity(entity) { - struct bfq_sched_data *sd = entity->my_sched_data; - - if (sd->next_in_service || sd->in_service_entity) { - /* - * entity is still active, because either - * next_in_service or in_service_entity is not - * NULL (see the comments on the definition of - * next_in_service for details on why - * in_service_entity must be checked too). - * - * As a consequence, its parent entities are - * active as well, and thus this loop must - * stop here. - */ - break; - } - - /* - * The decrement of num_groups_with_pending_reqs is - * not performed immediately upon the deactivation of - * entity, but it is delayed to when it also happens - * that the first leaf descendant bfqq of entity gets - * all its pending requests completed. The following - * instructions perform this delayed decrement, if - * needed. See the comments on - * num_groups_with_pending_reqs for details. - */ - if (entity->in_groups_with_pending_reqs) { - entity->in_groups_with_pending_reqs = false; - bfqd->num_groups_with_pending_reqs--; - } - } - - /* - * Next function is invoked last, because it causes bfqq to be - * freed if the following holds: bfqq is not in service and - * has no dispatched request. DO NOT use bfqq after the next - * function invocation. - */ __bfq_weights_tree_remove(bfqd, bfqq, &bfqd->queue_weights_tree); } @@ -7118,7 +7076,7 @@ static int bfq_init_queue(struct request_queue *q, struct elevator_type *e) bfqd->idle_slice_timer.function = bfq_idle_slice_timer; bfqd->queue_weights_tree = RB_ROOT_CACHED; - bfqd->num_groups_with_pending_reqs = 0; + bfqd->num_groups_with_busy_queues = 0; INIT_LIST_HEAD(&bfqd->active_list); INIT_LIST_HEAD(&bfqd->idle_list); diff --git a/block/bfq-iosched.h b/block/bfq-iosched.h index d92adbdd70ee..6c6cd984d769 100644 --- a/block/bfq-iosched.h +++ b/block/bfq-iosched.h @@ -197,9 +197,6 @@ struct bfq_entity { /* flag, set to request a weight, ioprio or ioprio_class change */ int prio_changed; - /* flag, set if the entity is counted in groups_with_pending_reqs */ - bool in_groups_with_pending_reqs; - /* last child queue of entity created (for non-leaf entities) */ struct bfq_queue *last_bfqq_created; }; @@ -496,52 +493,14 @@ struct bfq_data { struct rb_root_cached queue_weights_tree; /* - * Number of groups with at least one descendant process that - * has at least one request waiting for completion. Note that - * this accounts for also requests already dispatched, but not - * yet completed. Therefore this number of groups may differ - * (be larger) than the number of active groups, as a group is - * considered active only if its corresponding entity has - * descendant queues with at least one request queued. This - * number is used to decide whether a scenario is symmetric. - * For a detailed explanation see comments on the computation - * of the variable asymmetric_scenario in the function - * bfq_better_to_idle(). - * - * However, it is hard to compute this number exactly, for - * groups with multiple descendant processes. Consider a group - * that is inactive, i.e., that has no descendant process with - * pending I/O inside BFQ queues. Then suppose that - * num_groups_with_pending_reqs is still accounting for this - * group, because the group has descendant processes with some - * I/O request still in flight. num_groups_with_pending_reqs - * should be decremented when the in-flight request of the - * last descendant process is finally completed (assuming that - * nothing else has changed for the group in the meantime, in - * terms of composition of the group and active/inactive state of child - * groups and processes). To accomplish this, an additional - * pending-request counter must be added to entities, and must - * be updated correctly. To avoid this additional field and operations, - * we resort to the following tradeoff between simplicity and - * accuracy: for an inactive group that is still counted in - * num_groups_with_pending_reqs, we decrement - * num_groups_with_pending_reqs when the first descendant - * process of the group remains with no request waiting for - * completion. - * - * Even this simpler decrement strategy requires a little - * carefulness: to avoid multiple decrements, we flag a group, - * more precisely an entity representing a group, as still - * counted in num_groups_with_pending_reqs when it becomes - * inactive. Then, when the first descendant queue of the - * entity remains with no request waiting for completion, - * num_groups_with_pending_reqs is decremented, and this flag - * is reset. After this flag is reset for the entity, - * num_groups_with_pending_reqs won't be decremented any - * longer in case a new descendant queue of the entity remains - * with no request waiting for completion. + * Number of groups with at least one bfqq that is marked busy, + * and this number is used to decide whether a scenario is symmetric. + * Note that bfqq is busy doesn't mean that the bfqq contains requests. + * If idling is needed for service guarantees, bfqq will stay busy + * after dispatching the last request, see details in + * __bfq_bfqq_expire(). */ - unsigned int num_groups_with_pending_reqs; + unsigned int num_groups_with_busy_queues; /* * Per-class (RT, BE, IDLE) number of bfq_queues containing diff --git a/block/bfq-wf2q.c b/block/bfq-wf2q.c index b97e33688335..48ca7922035c 100644 --- a/block/bfq-wf2q.c +++ b/block/bfq-wf2q.c @@ -221,13 +221,15 @@ static bool bfq_no_longer_next_in_service(struct bfq_entity *entity) static void bfq_inc_busy_queues(struct bfq_queue *bfqq) { bfqq->bfqd->busy_queues[bfqq->ioprio_class - 1]++; - bfqq_group(bfqq)->busy_queues++; + if (!(bfqq_group(bfqq)->busy_queues++)) + bfqq->bfqd->num_groups_with_busy_queues++; } static void bfq_dec_busy_queues(struct bfq_queue *bfqq) { bfqq->bfqd->busy_queues[bfqq->ioprio_class - 1]--; - bfqq_group(bfqq)->busy_queues--; + if (!(--bfqq_group(bfqq)->busy_queues)) + bfqq->bfqd->num_groups_with_busy_queues--; } #else /* CONFIG_BFQ_GROUP_IOSCHED */ @@ -1006,19 +1008,6 @@ static void __bfq_activate_entity(struct bfq_entity *entity, entity->on_st_or_in_serv = true; } -#ifdef CONFIG_BFQ_GROUP_IOSCHED - if (!bfq_entity_to_bfqq(entity)) { /* bfq_group */ - struct bfq_group *bfqg = - container_of(entity, struct bfq_group, entity); - struct bfq_data *bfqd = bfqg->bfqd; - - if (!entity->in_groups_with_pending_reqs) { - entity->in_groups_with_pending_reqs = true; - bfqd->num_groups_with_pending_reqs++; - } - } -#endif - bfq_update_fin_time_enqueue(entity, st, backshifted); }