Date: Wed, 28 Nov 2012 22:00:45 +0100 From: "Ronald Klop" <ronald-freebsd8@klop.yi.org> To: freebsd-arm@freebsd.org, "Matthias Reyelt" <Matthias.Reyelt@brunel.de> Subject: Re: Improve system latency during harddisk access Message-ID: <op.woh4njgy8527sy@212-182-167-131.ip.telfort.nl> In-Reply-To: <201211280842.51271.Matthias.Reyelt@brunel.de> References: <201211280842.51271.Matthias.Reyelt@brunel.de>
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On Wed, 28 Nov 2012 08:42:51 +0100, Matthias Reyelt
<Matthias.Reyelt@brunel.de> wrote:
> Hi,
>
> We have a cyclic test application running at a 50ms cycle running on a
> Marvell OpenRD with 9.0 with an attached sata harddisk.
>
> We have observed that the system is sometimes blocked up to about 18ms by
> accessing the harddisk. We have checked the ULE and 4BSD scheduler and
> preemption, but without any significant difference.
>
> I actually do not really believe that this is somewhat ARM specific.
> Generally, harddisk I/O is critical for latency, but I've cross checked
> with a
> Linux installed OpenRD and had less impact of harddisk I/O.
>
> Do you have any idea how to minimize the influence of harddisk I/O under
> FreeBSD?
>
> Regards
> Matthias
I think your question is to open to give a good answer. Plus, I don't
think 18ms for harddisk IO is very much. My (not so new) desktop PC gives:
[root@sjakie ~]# diskinfo -t ada0
ada0
512 # sectorsize
320072933376 # mediasize in bytes (298G)
625142448 # mediasize in sectors
0 # stripesize
0 # stripeoffset
620181 # Cylinders according to firmware.
16 # Heads according to firmware.
63 # Sectors according to firmware.
GEA354RC3B5S6L # Disk ident.
Seek times:
Full stroke: 250 iter in 7.440554 sec = 29.762 msec
Half stroke: 250 iter in 5.806944 sec = 23.228 msec
Quarter stroke: 500 iter in 9.120537 sec = 18.241 msec
Short forward: 400 iter in 2.397724 sec = 5.994 msec
Short backward: 400 iter in 1.500501 sec = 3.751 msec
Seq outer: 2048 iter in 0.208478 sec = 0.102 msec
Seq inner: 2048 iter in 0.143242 sec = 0.070 msec
Transfer rates:
outside: 102400 kbytes in 1.578200 sec = 64884
kbytes/sec
middle: 102400 kbytes in 1.726028 sec = 59327
kbytes/sec
inside: 102400 kbytes in 2.321649 sec = 44107
kbytes/sec
So if your application has to do a little bit of seeking on the disk it
has to wait.
As there is no information about what your program actually does I can't
say anything else about it.
Ronald.
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