[Gluster-users] Need some clarifications about the disperse feature

Atin Mukherjee amukherj at redhat.com
Tue Nov 25 15:02:56 UTC 2014


Xavi will be the better person to clear all your doubts on this feature,
however as per my understanding please see the response inline.

~Atin

On 11/25/2014 07:11 PM, Ayelet Shemesh wrote:
> Hello Gluster experts,
> 
> I have been using gluster for a small cluster for a few years now and I
> have a question regarding the new disperse feature, which is for me a much
> anticipated addition.
> 
> *Suppose* I create a volume with a disperse set of 3, redundancy 1 (let's
> call them A1, A2, A3) and then I add 3 more bricks to that volume (we'll
> call them B1, B2, B3).
> 
> *First question* - which of the bricks will be the one carrying the
> redundancy data?
The current implementation is *non systematic* which means we don't have
any dedicated parity/redundancy brick.
> 
> *Second question* - If I have machines with faster disk - should I assign
> them to the data or the redundancy bricks? What should I expect the load to
> be on the redundancy machine in heavy read scenarios and in heavy write
> scenarios?
As mentioned above, this configuration is not possible for non
systematic implementation.
> 
> *Third question* - *does this require reading the entire data* of A1, A2
> and A3 by initiating a heal or another operation?
If the configuration is 2+1 as you mentioned, you can recover the whole
set of data from any two of three bricks, the algorithm provides the
intelligence of constructing the chunk of data which resides in a brick
which might be down for this configuration.
> 
> *4th question* (and most important for me) - I saw in the list that it is
> now a Distributed-Dispersed volume. I understand I can now lose, for
> example bricks A1 and B1 and still have my entire data intact. Is this also
> correct for bricks from the same set, for example A1 and A2?
> Or to put it in a more generic way - *does this create the exact same
> dispersed volume as if I created it originally with A1, A2 A3 B1 B2 B3 and
> a redundancy of 2?*
No, if you see the volume info with this configuration it will show you
2 X (2+1) which means on every set the quorum is two i.e. you need to
have atleast two bricks running.
> 
> 
> Many thanks for your work and for your help on this list,
> Ayelet
> 
> 
> 
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