Re: [Make-wifi-fast] [RFC/RFT] mac80211: Switch to a virtual time-based airtime scheduler

[Yibo Zhao <yiboz@codeaurora.org>]

On 2019-04-11 19:24, Toke Høiland-Jørgensen wrote:
> Yibo Zhao <yiboz@codeaurora.org> writes:
> 
>> On 2019-04-10 18:40, Toke Høiland-Jørgensen wrote:
>>> Yibo Zhao <yiboz@codeaurora.org> writes:
>>> 
>>>> On 2019-04-10 04:41, Toke Høiland-Jørgensen wrote:
>>>>> Yibo Zhao <yiboz@codeaurora.org> writes:
>>>>> 
>>>>>> On 2019-04-04 16:31, Toke Høiland-Jørgensen wrote:
>>>>>>> Yibo Zhao <yiboz@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.
> 
> Yeah, in many cases we will end up throttling all but the first (couple
> of) node(s). This is by design; otherwise we can't ensure fairness. As
> long as we are making forward progress that is fine, though...
All right, I see your point. Fairness is our first priority. :)
> 
> -Toke

-- 
Yibo