[openstack-dev] Scheduler proposal
alawson at aqorn.com
Sun Oct 11 15:06:01 UTC 2015
I have a quick question: how is Amazon doing this? When choosing a next
path forward that reliably scales, would be interesting to know how this is
already being done.
On Oct 9, 2015 10:12 AM, "Zane Bitter" <zbitter at redhat.com> wrote:
> On 08/10/15 21:32, Ian Wells wrote:
>> > 2. if many hosts suit the 5 VMs then this is *very* unlucky,because
>> we should be choosing a host at random from the set of
>> suitable hosts and that's a huge coincidence - so this is a tiny
>> corner case that we shouldn't be designing around
>> Here is where we differ in our understanding. With the current
>> system of filters and weighers, 5 schedulers getting requests for
>> identical VMs and having identical information are *expected* to
>> select the same host. It is not a tiny corner case; it is the most
>> likely result for the current system design. By catching this
>> situation early (in the scheduling process) we can avoid multiple
>> RPC round-trips to handle the fail/retry mechanism.
>> And so maybe this would be a different fix - choose, at random, one of
>> the hosts above a weighting threshold, not choose the top host every
>> time? Technically, any host passing the filter is adequate to the task
>> from the perspective of an API user (and they can't prove if they got
>> the highest weighting or not), so if we assume weighting an operator
>> preference, and just weaken it slightly, we'd have a few more options.
> The optimal way to do this would be a weighted random selection, where the
> probability of any given host being selected is proportional to its
> weighting. (Obviously this is limited by the accuracy of the weighting
> function in expressing your actual preferences - and it's at least
> conceivable that this could vary with the number of schedulers running.)
> In fact, the choice of the name 'weighting' would normally imply that it's
> done this way; hearing that the 'weighting' is actually used as a 'score'
> with the highest one always winning is quite surprising.
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