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Date:   Tue, 09 Apr 2019 21:25:34 +0800
From:   Yibo Zhao <yiboz@codeaurora.org>
To:     =?UTF-8?Q?Toke_H=C3=B8iland-J=C3=B8rgensen?= <toke@redhat.com>
Cc:     make-wifi-fast@lists.bufferbloat.net,
        linux-wireless@vger.kernel.org, Felix Fietkau <nbd@nbd.name>,
        Rajkumar Manoharan <rmanohar@codeaurora.org>,
        Kan Yan <kyan@google.com>, linux-wireless-owner@vger.kernel.org
Subject: Re: [RFC/RFT] mac80211: Switch to a virtual time-based airtime
 scheduler
In-Reply-To: <87ftqy41ea.fsf@toke.dk>
References: <20190215170512.31512-1-toke@redhat.com>
 <753b328855b85f960ceaf974194a7506@codeaurora.org> <87ftqy41ea.fsf@toke.dk>
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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);

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.


> 
>>> +	sta->airtime[ac].v_t += airtime / sta->airtime_weight;
>> Another question. Any plan for taking v_t overflow situation into
>> consideration? u64 might be enough for low throughput products but not
>> sure for high end products. Something like below for reference:
> 
> The unit for the variable is time, not bytes, so it is unaffected by
> throughput. 2**64 microseconds is 584554 *years* according to my
> 'units' binary, so don't think we have to worry too much about this
> overflowing ;)

Cool, Great! Then no problem with that now.

> 
> -Toke

-- 
Yibo