Re: how to multiplex ocr and voting disk

"Fabrizio Magni" <fabrizio.magni_at_mycontinent.com> wrote in message news:43cd12ef$0$1089$4fafbaef_at_reader1.news.tin.it...
> Jonathan Lewis wrote:
>>
>> Got my own answer:
>> it has to be an odd number of copies.
>>
>> So I guess no single instance can take
>> charge unless it has exclusively acquired
>> more than half the known copies.
>>
>
> Some days ago I wished to give the same answer...
> ...but then I changed my mind.
>
> I couldn't convince myself that the odd number of voting disks is enough.
>
> I concluded I don't know how the voting mechanism in RAC is truly working
> (and I couldn't find documentation on the topic).
>
> Maybe it is the flu or I'm naturally dumb but what happens if every voting
> disk is owned by a different node or no node owns enough disks?
>
> Being at home I have no chance to play with a test system so if someone
> can post the list of file descriptors involving every voting disk grouped
> by process (on any node)...
>

I don't know how the voting mechanism in RAC works either, I was just guessing. But if the requirement is to have at least half of the known devices then your example with every node having one device each means no node has enough to take control. It does mean that the system is dead, but at least it isn't in state where conflicting subsets of instances can damage the data.

-- 
Regards

Jonathan Lewis

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