linux-kernel - Re: [PATCH 5/5] usb_debug: EXPERIMENTAL

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Message-Id: <200905062201.42965.oliver@neukum.org>
Date:	Wed, 6 May 2009 22:01:42 +0200
From:	Oliver Neukum <oliver@...kum.org>
To:	Alan Stern <stern@...land.harvard.edu>
Cc:	Alan Cox <alan@...rguk.ukuu.org.uk>,
	Jason Wessel <jason.wessel@...driver.com>, greg@...ah.com,
	linux-usb@...r.kernel.org, linux-kernel@...r.kernel.org
Subject: Re: [PATCH 5/5] usb_debug: EXPERIMENTAL - poll hcd device to force writes

Am Mittwoch, 6. Mai 2009 21:24:56 schrieb Alan Stern:
> On Wed, 6 May 2009, Oliver Neukum wrote:

> A little thought yielded the following algorithm.  It assumes there is
> a fixed set of URBs allocated, unlike what you have done.  Does it make

No, it does not ;-) Your approach is more general than you think.
The only important constraint is that the number of URBs in flight
be limited. It doesn't matter when they are allocated.

> sense to take this approach?

Yes.

> Let N be the total number of URBs allocated, each capable of holding up
> to B bytes.  Let NIF be the number of URBs in flight at any time, so
> the number of available URBs is N - NIF.  The number of available bytes
> might be < (N - NIF)*B because the next URB might be partially full.
>
> P is an adjustable parameter of the algorithm.  For simplicity you can
> take P = 1, but increasing P (any value below N is okay) would yield
> reduced latency at the cost of more partially-filled URB submissions
> (so possibly reduced throughput).
>
>     Write routine:
> 	Copy bytes into the available URB buffers, submitting URBs as
> 	they get filled.  At the end, if the next URB is partially full
> 	then submit it only if NIF < P.

I did so. In principle. I am leaving the iteration to the tty layer.

>     Completion routine:
> 	If the next URB to send is partially filled, submit it.

Much easier if P = N - 1

>     write_room routine:
> 	Return the actual number of bytes remaining in the available
> 	URBs, but no more than (N-P)*B.

Yes.

> How does that sound?  Converting \n to \r\n will add some complication
> but not too much.
>
> Allocating URBs on the fly adds a lot of complication.  There has to be
> a minimum number of pre-allocated URBs; otherwise write_room could

Why? You can always calculate with the number of URBs you'd
allocate as a maximum.

> never return a positive value.  If you allocate additional URBs
> later on, when would you free them?

Like you free all URBs, on completion.

	Regards
		Oliver

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