On Sat, Feb 14, 2026 at 08:51:53PM +0100, Simon Schippers wrote:
> On 2/14/26 19:18, Michael S. Tsirkin wrote:
> > On Sat, Feb 14, 2026 at 06:13:14PM +0100, Simon Schippers wrote:
> > 
> > ...
> > 
> >> Patched: Waking on __ptr_ring_produce_created_space() is too early. The
> >>          stop/wake cycle occurs too frequently which slows down
> >>          performance as can be seen for TAP.
> >>
> >> Wake on empty variant: Waking on __ptr_ring_empty() is (slightly) too
> >>                        late. The consumer starves because the producer
> >>                        first has to produce packets again. This slows
> >>                        down performance aswell as can be seen for TAP
> >>                   and TAP+vhost-net (both down ~30-40Kpps).
> >>
> >> I think something inbetween should be used.
> >> The wake should be done as late as possible to have as few
> >> NET_TX_SOFTIRQs as possible but early enough that there are still
> >> consumable packets remaining to not starve the consumer.
> >>
> >> However, I can not think of a proper way to implement this right now.
> >>
> >> Thanks!
> > 
> > What is the difficulty?
> 
> There is no way to tell how many entries are currently in the ring.
> 
> > 
> > Your patches check __ptr_ring_consume_created_space(..., 1).
> 
> Yes, and this returns if either 0 space or a batch size space was
> created.
> (In the current implementation it would be false or true, but as
> discussed earlier this can be changed.)
> 
> > 
> > How about __ptr_ring_consume_created_space(..., 8) then? 16?
> > 
> 
> This would return how much space the last 8/16 consume operations
> created. But in tap_ring_consume() we only consume a single entry.
> 
> Maybe we could avoid __ptr_ring_consume_created_space with this:
> 1. Wait for the queue to stop with netif_tx_queue_stopped()
> 2. Then count the numbers of consumes we did after the queue stopped
> 3. Wake the queue if count >= threshold with threshold >= ring->batch
> 
> I would say that such a threshold could be something like ring->size/2.

OK


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