Improve system latency during harddisk access

[Ronald Klop]

On Wed, 28 Nov 2012 08:42:51 +0100, Matthias Reyelt  
<Matthias.Reyelt at 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 at 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.