On 06/22/2011 11:05 AM, Jiri Denemark wrote:
On Wed, Jun 22, 2011 at 16:47:18 +0100, Daniel P. Berrange wrote:
> If the QEMU process has been stopped (kill -STOP/gdb), or the
> QEMU process has live-locked itself, then we will never get a
> reply from the monitor. We should not wait forever in this
> case, but instead timeout after a reasonable amount of time.
>
> NB if the host has high CPU load, or a single monitor command
> intentionally takes a long time, then this will cause bogus
> failures. In the case of high CPU load, arguably the guest
> should have been migrated elsewhere, since you can't effectively
> manage guests on a host if QEMU is taking > 30 seconds to reply
> to simply commands. Since we use background migration, there
> should not be any commands which take significant time to
> execute any more
The thing I'm most concerned about is that is far too easy to get into such
situations especially since disk cache subsystem in Linux kernel is not the
best thing in the world. While I agree that running guests on a loaded host is
not very clever and guests should rather be migrated elsewhere, such situation
doesn't have to be intentional. In other words, in case of a malfunction of
some kind (some processes go crazy, network disruptions, ...) QEMU may require
more than a timeout seconds to respond and we will penalize an innocent QEMU
process because we won't be able to control it anymore even though the issues
get fixed.
Is there any way to measure time spent by the child process, rather than
just relying on wall-time elapsed? That is, when libvirt hits 30
seconds of wall time in waiting for a monitor, can it then check whether
the child process has accumulated any execution time (likely hung) vs.
no execution time (likely a starved system situation), and only give up
in the former case?
--
Eric Blake eblake(a)redhat.com +1-801-349-2682
Libvirt virtualization library
http://libvirt.org