8+2 is recommended for 10 servers. From n+k servers it will be good to
choose n with a power of 2(4,8,16,vs)
You need to add 10 bricks if you want to extend the volume.
On Wed, Sep 21, 2016 at 4:12 PM, Gandalf Corvotempesta
<gandalf.corvotempesta at gmail.com> wrote:> 2016-09-21 14:42 GMT+02:00 Xavier Hernandez <xhernandez at
datalab.es>:
>> You *must* ensure that *all* bricks forming a single disperse set are
placed
>> in a different server. There are no 4 special fragments. All fragments
have
>> the same importance. The way to do that is ordering them when the
volume is
>> created:
>>
>> gluster volume create test disperse 16 redundancy 4
>> server{1..20}:/bricks/test1 server{1..20}:/bricks/test2
>> server{1..20}:/bricks/test3
>>
>> This way all 20 fragments from each disperse set will be placed in a
>> different server. However each server will have 3 bricks and no
fragment
>> from a single file will be stored in more than one brick of each
server.
>
> Now it's clear.
> So, at very minimum, EC is good starting from 7+3 (10 servers with 1
> brick each) because: 512*(7-3) = 2048
> Any smaller combinations would mean less redundancy (6+2) or not
> optimized stripe size like 512*(5-3)=1024
>
> Is this correct ? And what if I have to add some brick to the current
> servers or I have to add new servers?
> Can I add them freely or I have to follow some rules ?
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