Re: controlling ACPI IRQ routing

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Len Brown wrote:
On Wednesday 02 January 2008 09:11, Dominique Michel wrote:
Le Sat, 29 Dec 2007 10:40:20 -0800,
Lee Howard <faxguy@xxxxxxxxxxxxxxxx> a écrit :


# cat /proc/interrupts
CPU0 0: 1051464247 IO-APIC-edge timer
  1:          8   IO-APIC-edge      i8042
  8:          0   IO-APIC-edge      rtc
  9:          0   IO-APIC-fasteoi   acpi
 12:        104   IO-APIC-edge      i8042
 14:    9414304   IO-APIC-edge      ide0
 16: 1051172722   IO-APIC-fasteoi   wct4xxp
 19:          1   IO-APIC-fasteoi   eth1
 21:  158008518   IO-APIC-fasteoi   eth0
 22:    6974044   IO-APIC-fasteoi   libata
 23:    7071112   IO-APIC-fasteoi   libata
NMI:          0
LOC: 1051371544
ERR:          0


Based on that, you may want to use rtirq. It is a boot script (need schedutils
as dependency) with associated config scripts. http://alsa.opensrc.org/Rtirq

In /etc/conf.d/rtirq, you will find:

# IRQ thread service names
# (space separated list, from higher to lower priority).
RTIRQ_NAME_LIST="rtc snd usb i8042"

The most important things is that the devices listed here doesn't have any
shared IRQ with some other device and that the rtc remain the first listed
device (the one with the higer priority), or the system will hang soon or later.

This setup is for audio workstation but is very easy to adapt to any kind of
work.


I've got no experience with the thread priority thing above.
However, the better way to handle quality of service contention
is to not do the things you don't care about.

Does the system really need HZ=1000?  you could slow that down to 100,
or run CONFIG_NO_HZ to reduce it even more...
Also, on a UP, the LAPIC timer interrupt is redundant --
you could build your kernel w/o APIC support or boot with nolapic.

This would get rid of the only interrupts which are frequent
enough to be competing with the device, clock ticks.

If I boot with nolapic then XT-PIC is used, and the wcfxo device ends up sharing an IRQ with libata which results in worse performance than with it the way that it is.

Thanks,

Lee.
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