[libvirt-users] starting a domain only when you have enough resources

[Andrei Perietanu]

On Tue, Jan 26, 2016 at 12:39 PM, Michal Privoznik <mprivozn at redhat.com>
wrote:

> On 26.01.2016 12:30, Andrei Perietanu wrote:
> > Hi all,
> >
> > I am running KVM on a 3.18 kernel. The system runs and Atom processor
> with
> > 2Gb RAM.
> >
> > Using KVM you obviously can over allocate your resources: say you have 4
> > guests each configured with 1GB ram. Running all four at the same time,
> > depending on the workload, can crash the system - I get a kernel trace
> when
> > this happens.
> >
> > But let's consider a simpler case: one guest with 1.5 Gb RAM, ubuntu
> 14.03.
> > During the installation the system will again crash.
> > The memory statistics (top or proc/meminfo) will show that the FreeMemory
> > goes down to 12Mb when this happens - which kind of makes sense
> considering
> > the host will require some RAM to run.
> >
> > But the question is: does libvirt offer any way to prevent this from
> > happening?
> >
> > Some way of not allowing the user to start a guest unless you have enough
> > free memory. I know how much ram each guest has configured but that is
> not
> > enough. I need to know how much the system has available, and just
> reading
> > the free memory statistic does not help much since that is only a
> snapshot
> > - when running a guest you can have 1gb free now, and 10 mb free 2 min
> > later.
> >
> > Any ideas?
>
> There is one option I see, use -mem-prealloc. Either you can passthrough
> it onto qemu commandline [1] or use locked memoryBacking [2]. I advocate
> for the latter though. Not only it will allocate all the memory at qemu
> startup it will also lock it so it won't get swapped off.
>
> Michal
>
> 1: http://libvirt.org/drvqemu.html#qemucommand
> 2: http://libvirt.org/formatdomain.html#elementsMemoryBacking
>

I tried memoryBacking: I added this to the domain xml:
<memoryBacking>
  <locked/>
</memoryBacking>

And got an error when attempting to start the vm: memory locking no
supported by QEMU binary.

Aside from that I don't really understand how this helps solve my issue -
if you don't mind going a bit into details I'd appreciate it.

Thanks!
Andrei

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