Will m=6 cause huge CPU usage?

Best,

Feng

On Fri, May 3, 2019 at 11:57 AM Ashley Merrick <singap...@amerrick.co.uk> wrote:
>
> I may be wrong, but your correct with your m=6 statement.
>
> Your need atleast K amount of shards available. If you had k=8 and m=2 
> equally across 2 rooms (5 each), a faidlure in either room would cause an 
> outrage.
>
> With M=6 your atleast getting better disk space availability than 3 
> replication. But not sure if you may end up with some form of split brain if 
> just was a network issue between both sides and each side was still online 
> and working independently. As both would technically have enough shards to 
> continue to operate.
>
> On Fri, 3 May 2019, 11:46 PM Robert Sander, <r.san...@heinlein-support.de> 
> wrote:
>>
>> Hi,
>>
>> I would be glad if anybody could give me a tip for an erasure code
>> profile and an associated crush ruleset.
>>
>> The cluster spans 2 rooms with each room containing 6 hosts and each
>> host has 12 to 16 OSDs.
>>
>> The failure domain would be the room level, i.e. data should survive if
>> one of the rooms has a power loss.
>>
>> Is that even possible with erasure coding?
>> I am only coming up with profiles where m=6, but that seems to be a
>> little overkill.
>>
>> Regards
>> --
>> Robert Sander
>> Heinlein Support GmbH
>> Schwedter Str. 8/9b, 10119 Berlin
>>
>> https://www.heinlein-support.de
>>
>> Tel: 030 / 405051-43
>> Fax: 030 / 405051-19
>>
>> Amtsgericht Berlin-Charlottenburg - HRB 93818 B
>> Geschäftsführer: Peer Heinlein - Sitz: Berlin
>>
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