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From: "John W. Linville" <linville@tuxdriver.com>
To: Nathaniel Smith <njs@pobox.com>
Cc: bloat-devel@lists.bufferbloat.net, johannes@sipsolutions.net,
	linux-wireless@vger.kernel.org
Subject: Re: [RFC v2] mac80211: implement eBDP algorithm to fight bufferbloat
Date: Wed, 23 Feb 2011 17:28:43 -0500	[thread overview]
Message-ID: <20110223222842.GA20039@tuxdriver.com> (raw)
In-Reply-To: <AANLkTi=LnpFLzKbtaPxSjg9TRU_pY2oRi6tsFg5GkaGK@mail.gmail.com>

On Mon, Feb 21, 2011 at 03:26:32PM -0800, Nathaniel Smith wrote:
> On Mon, Feb 21, 2011 at 10:47 AM, John W. Linville
> <linville@tuxdriver.com> wrote:

> > I tried to see how your measurement would be useful, but I just don't
> > see how the number of frames ahead of me in the queue is relevant to
> > the measured link latency?  I mean, I realize that having more packets
> > ahead of me in the queue is likely to increase the latency for this
> > frame, but I don't understand why I should use that information to
> > discount the measured latency...?
> 
> It depends on which latency you want to measure. The way that I
> reasoned was, suppose that at some given time, the card is able to
> transmit 1 fragment every T nanoseconds. Then it can transmit n
> fragments in n*T nanoseconds, so if we want the queue depth to be 2
> ms, then we have
>   n * T = 2 * NSEC_PER_MSEC
>   n = 2 * NSEC_PER_MSEC / T
> 
> Which is the calculation that you're doing:
> 
> +                       sta->sdata->qdata[q].max_enqueued =
> +                               max_t(int, 2, 2 * NSEC_PER_MSEC / tserv_ns_avg);
> 
> But for this calculation to make sense, we need T to be the time it
> takes the card to transmit 1 fragment. In your patch, you're not
> measuring that. You're measuring the total time between when a packet
> is enqueued and when it is transmitted; if there were K packets in the
> queue ahead of it, then this is the time to send *all* of them --
> you're measuring (K+1)*T. That's why in my patch, I recorded the
> current size of the queue when each packet is enqueued, so I could
> compute T = total_time / (K+1).

Thanks for the math!  I think I see what you are saying now.  Since the
measured time is being used to determine the queue size, we need to
factor-in the length of the queue that resulted in that measurment.

Unfortunately, I'm not sure how to apply this with the technique I
am using for the timing measurements. :-(  I'll have to think about
this some more...

John
-- 
John W. Linville		Someday the world will need a hero, and you
linville@tuxdriver.com			might be all we have.  Be ready.

  reply	other threads:[~2011-02-23 22:30 UTC|newest]

Thread overview: 23+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
     [not found] <1297619803-2832-1-git-send-email-njs@pobox.com>
2011-02-17  1:49 ` [RFC] " John W. Linville
2011-02-17  3:31   ` Ben Greear
2011-02-17  4:26   ` Nathaniel Smith
2011-02-17  8:31   ` Johannes Berg
2011-02-18 21:21   ` [RFC v2] " John W. Linville
2011-02-19  3:44     ` Nathaniel Smith
2011-02-21 18:47       ` John W. Linville
2011-02-21 23:26         ` Nathaniel Smith
2011-02-23 22:28           ` John W. Linville [this message]
2011-02-25 18:21             ` Nathaniel Smith
2011-02-25 18:27               ` Nathaniel Smith
2011-02-20  0:37     ` Nathaniel Smith
2011-02-20  0:51       ` Jim Gettys
2011-02-20 15:24         ` Dave Täht
2011-02-21 18:52       ` John W. Linville
2011-02-21 15:28     ` Johannes Berg
2011-02-21 16:12       ` Jim Gettys
2011-02-21 19:15         ` John W. Linville
2011-02-21 19:06       ` John W. Linville
2011-02-21 19:29         ` [RFC v2] mac80211: implement eBDP algorithm to fight bufferbloat - AQM on hosts Jim Gettys
2011-02-21 20:26         ` [RFC v2] mac80211: implement eBDP algorithm to fight bufferbloat Tianji Li
2011-02-28 13:07         ` Johannes Berg
     [not found] <x1-oTZGm1A7eclvABnv1aK0z1Nc7iI@gwene.org>
2011-02-20  1:59 ` Dave Täht

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