[Make-wifi-fast] [RFC/RFT] mac80211: Switch to a virtual time-based airtime scheduler
Yibo Zhao
yiboz at codeaurora.org
Wed Apr 10 23:12:33 EDT 2019
On 2019-04-10 18:40, Toke Høiland-Jørgensen wrote:
> Yibo Zhao <yiboz at codeaurora.org> writes:
>
>> On 2019-04-10 04:41, Toke Høiland-Jørgensen wrote:
>>> Yibo Zhao <yiboz at codeaurora.org> writes:
>>>
>>>> On 2019-04-04 16:31, Toke Høiland-Jørgensen wrote:
>>>>> Yibo Zhao <yiboz at codeaurora.org> writes:
>>>>>
>>>>>> On 2019-02-16 01:05, Toke Høiland-Jørgensen wrote:
>>>>>>> This switches the airtime scheduler in mac80211 to use a virtual
>>>>>>> time-based
>>>>>>> scheduler instead of the round-robin scheduler used before. This
>>>>>>> has
>>>>>>> a
>>>>>>> couple of advantages:
>>>>>>>
>>>>>>> - No need to sync up the round-robin scheduler in
>>>>>>> firmware/hardware
>>>>>>> with
>>>>>>> the round-robin airtime scheduler.
>>>>>>>
>>>>>>> - If several stations are eligible for transmission we can
>>>>>>> schedule
>>>>>>> both of
>>>>>>> them; no need to hard-block the scheduling rotation until the
>>>>>>> head
>>>>>>> of
>>>>>>> the
>>>>>>> queue has used up its quantum.
>>>>>>>
>>>>>>> - The check of whether a station is eligible for transmission
>>>>>>> becomes
>>>>>>> simpler (in ieee80211_txq_may_transmit()).
>>>>>>>
>>>>>>> The drawback is that scheduling becomes slightly more expensive,
>>>>>>> as
>>>>>>> we
>>>>>>> need
>>>>>>> to maintain an rbtree of TXQs sorted by virtual time. This means
>>>>>>> that
>>>>>>> ieee80211_register_airtime() becomes O(logN) in the number of
>>>>>>> currently
>>>>>>> scheduled TXQs. However, hopefully this number rarely grows too
>>>>>>> big
>>>>>>> (it's
>>>>>>> only TXQs currently backlogged, not all associated stations), so
>>>>>>> it
>>>>>>> shouldn't be too big of an issue.
>>>>>>>
>>>>>>> @@ -1831,18 +1830,32 @@ void
>>>>>>> ieee80211_sta_register_airtime(struct
>>>>>>> ieee80211_sta *pubsta, u8 tid,
>>>>>>> {
>>>>>>> struct sta_info *sta = container_of(pubsta, struct sta_info,
>>>>>>> sta);
>>>>>>> struct ieee80211_local *local = sta->sdata->local;
>>>>>>> + struct ieee80211_txq *txq = sta->sta.txq[tid];
>>>>>>> u8 ac = ieee80211_ac_from_tid(tid);
>>>>>>> - u32 airtime = 0;
>>>>>>> + u64 airtime = 0, weight_sum;
>>>>>>> +
>>>>>>> + if (!txq)
>>>>>>> + return;
>>>>>>>
>>>>>>> if (sta->local->airtime_flags & AIRTIME_USE_TX)
>>>>>>> airtime += tx_airtime;
>>>>>>> if (sta->local->airtime_flags & AIRTIME_USE_RX)
>>>>>>> airtime += rx_airtime;
>>>>>>>
>>>>>>> + /* Weights scale so the unit weight is 256 */
>>>>>>> + airtime <<= 8;
>>>>>>> +
>>>>>>> spin_lock_bh(&local->active_txq_lock[ac]);
>>>>>>> +
>>>>>>> sta->airtime[ac].tx_airtime += tx_airtime;
>>>>>>> sta->airtime[ac].rx_airtime += rx_airtime;
>>>>>>> - sta->airtime[ac].deficit -= airtime;
>>>>>>> +
>>>>>>> + weight_sum = local->airtime_weight_sum[ac] ?:
>>>>>>> sta->airtime_weight;
>>>>>>> +
>>>>>>> + local->airtime_v_t[ac] += airtime / weight_sum;
>>>>>> Hi Toke,
>>>>>>
>>>>>> Please ignore the previous two broken emails regarding this new
>>>>>> proposal
>>>>>> from me.
>>>>>>
>>>>>> It looks like local->airtime_v_t acts like a Tx criteria. Only the
>>>>>> stations with less airtime than that are valid for Tx. That means
>>>>>> there
>>>>>> are situations, like 50 clients, that some of the stations can be
>>>>>> used
>>>>>> to Tx when putting next_txq in the loop. Am I right?
>>>>>
>>>>> I'm not sure what you mean here. Are you referring to the case
>>>>> where
>>>>> new
>>>>> stations appear with a very low (zero) airtime_v_t? That is handled
>>>>> when
>>>>> the station is enqueued.
>>>> Hi Toke,
>>>>
>>>> Sorry for the confusion. I am not referring to the case that you
>>>> mentioned though it can be solved by your subtle design, max(local
>>>> vt,
>>>> sta vt). :-)
>>>>
>>>> Actually, my concern is situation about putting next_txq in the
>>>> loop.
>>>> Let me explain a little more and see below.
>>>>
>>>>> @@ -3640,126 +3638,191 @@ EXPORT_SYMBOL(ieee80211_tx_dequeue);
>>>>> struct ieee80211_txq *ieee80211_next_txq(struct ieee80211_hw *hw,
>>>>> u8
>>>>> ac)
>>>>> {
>>>>> struct ieee80211_local *local = hw_to_local(hw);
>>>>> + struct rb_node *node = local->schedule_pos[ac];
>>>>> struct txq_info *txqi = NULL;
>>>>> + bool first = false;
>>>>>
>>>>> lockdep_assert_held(&local->active_txq_lock[ac]);
>>>>>
>>>>> - begin:
>>>>> - txqi = list_first_entry_or_null(&local->active_txqs[ac],
>>>>> - struct txq_info,
>>>>> - schedule_order);
>>>>> - if (!txqi)
>>>>> + if (!node) {
>>>>> + node = rb_first_cached(&local->active_txqs[ac]);
>>>>> + first = true;
>>>>> + } else
>>>>> + node = rb_next(node);
>>>>
>>>> Consider below piece of code from ath10k_mac_schedule_txq:
>>>>
>>>> ieee80211_txq_schedule_start(hw, ac);
>>>> while ((txq = ieee80211_next_txq(hw, ac))) {
>>>> while (ath10k_mac_tx_can_push(hw, txq)) {
>>>> ret = ath10k_mac_tx_push_txq(hw, txq);
>>>> if (ret < 0)
>>>> break;
>>>> }
>>>> ieee80211_return_txq(hw, txq);
>>>> ath10k_htt_tx_txq_update(hw, txq);
>>>> if (ret == -EBUSY)
>>>> break;
>>>> }
>>>> ieee80211_txq_schedule_end(hw, ac);
>>>>
>>>> If my understanding is right, local->schedule_pos is used to record
>>>> the
>>>> last scheduled node and used for traversal rbtree for valid txq.
>>>> There
>>>> is chance that an empty txq is feeded to return_txq and got removed
>>>> from
>>>> rbtree. The empty txq will always be the rb_first node. Then in the
>>>> following next_txq, local->schedule_pos becomes meaningless since
>>>> its
>>>> rb_next will return NULL and the loop break. Only rb_first get
>>>> dequeued
>>>> during this loop.
>>>>
>>>> if (!node || RB_EMPTY_NODE(node)) {
>>>> node = rb_first_cached(&local->active_txqs[ac]);
>>>> first = true;
>>>> } else
>>>> node = rb_next(node);
>>>
>>> Ah, I see what you mean. Yes, that would indeed be a problem - nice
>>> catch! :)
>>>
>>>> How about this? The nodes on the rbtree will be dequeued and removed
>>>> from rbtree one by one until HW is busy. Please note local vt and
>>>> sta
>>>> vt will not be updated since txq lock is held during this time.
>>>
>>> Insertion and removal from the rbtree are relatively expensive, so
>>> I'd
>>> rather not do that for every txq. I think a better way to solve this
>>> is to just defer the actual removal from the tree until
>>> ieee80211_txq_schedule_end()... Will fix that when I submit this
>>> again.
>>
>> Do you mean we keep the empty txqs in the rbtree until loop finishes
>> and
>> remove them in ieee80211_txq_schedule_end(may be put return_txq in
>> it)?
>> If it is the case, I suppose a list is needed to store the empty txqs
>> so
>> as to dequeue them in ieee80211_txq_schedule_end.
>
> Yeah, return_txq() would just put "to be removed" TXQs on a list, and
> schedule_end() would do the actual removal (after checking whether a
> new
> packet showed up in the meantime).
SGTM
>
>> And one more thing,
>>
>>> + if (sta->airtime[ac].v_t > local->airtime_v_t[ac]) {
>>> + if (first)
>>> + local->airtime_v_t[ac] =
>>> sta->airtime[ac].v_t;
>>> + else
>>> + return NULL;
>>
>> As local->airtime_v_t will not be updated during loop, we don't need
>> to
>> return NULL.
>
> Yes we do; this is actually the break condition. I.e., stations whose
> virtual time are higher than the global time (in local->airtime_v_t)
> are
> not allowed to transmit. And since we are traversing them in order,
> when
> we find the first such station, we are done and can break out of the
> scheduling loop entirely (which is what we do by returning NULL). The
> other branch in the inner if() is just for the case where no stations
> are currently eligible to transmit according to this rule; here we
> don't
> want to stall, so we advance the global timer so the first station
> becomes eligible...
Yes，the inner if() make sure first node always get scheduled no matter
its vt.
To detail my concern, let's assume only two nodes in the tree and empty
nodes will be in tree until schedule_end(). In the loop and in case hw
is not busy, ath10k will drain every node next_txq returned before
asking for another txq again. Then as we are traversing to next rb node,
it is highly possible the second node is not allowed to transmit since
the global time has not been updated yet as the active txq lock is held.
At this time, only second node on the tree has data and hw is capable of
sending more data. I don't think the second node is not valid for
transmission in this situation.
With more nodes in the tree in this situation, I think same thing
happens that all nodes except the first node are not allowed to transmit
since none of their vts are less than the global time which is not
updated in time. The loop breaks when we are checking the second node.
>
> -Toke
--
Yibo
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