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

Xavier Hernandez xhernandez at datalab.es
Wed Nov 26 09:50:57 UTC 2014

Hi Ayelet,

I attached a spread sheet with some test data. It contains 3 sets of tests:

* Tests with 0-byte files (touch)
* Tests with 1-byte files
* Tests with 1 MB files

For each set, a creation (write), a recursive ls, a read and a rm is 
done over 11100 files distributed in 111 directories by 10 concurrent 

Each individual test is performed twice: one run with all caches 
(gluster and filesystem) cleared, and another run just after finishing 
the first one (that should have caches full). The only exception is rm, 
that cannot be done twice (the second test takes almost no time because 
there's nothing to delete).

x1 means a single brick volume.
x3 means a distributed volume with 3 bricks.
x1_r2 means a replica-2 volume.
x1_r3 means a replica-3 volume.
x1_d3_1 means a dispersed 3:1 volume.
x1_d5_1 means a dispersed 5:1 volume.
x1_d6_2 means a dispersed 6:2 volume.

These tests have been made in a virtualized environment, so I'm not very 
sure if the comparison is reliable or not. You should test it in your 
environment with your workload and see how it works. If possible, I 
would also be very interested to know your results :)

I think there's some room to improve read performance, that probably is 
the worst part, but this will be done after being sure it's really stable.


On 11/26/2014 09:35 AM, Ayelet Shemesh wrote:
> Thank you Xavi, it's very helpful (also to Atin).
> Have you had any benchmarks of how much penalty in performance I should
> expect for an intense reading using this feature? Naturaly I will test
> in my specific environment, just want to know if there are any
> benchmarks I can see for now.
> Ayelet
> On Tue, Nov 25, 2014 at 5:19 PM, Xavier Hernandez <xhernandez at datalab.es
> <mailto:xhernandez at datalab.es>> wrote:
>     Hi Ayelet,
>     On 11/25/2014 02:41 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?
>     In current implementation, there's no difference between data and
>     redundancy. All bricks behave exactly equal and there isn't anyone
>     more important than another. In a configuration with 3 bricks and
>     redundancy 1, you can lose any brick and everything will continue
>     working normally.
>         *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 I said, there isn't a dedicated redundancy brick, so there's no
>     benefit in assigning the fast disk to a specific brick.
>     Read requests only need to be processed on N - R bricks (N = total
>     number of bricks, R = redundancy). This means that in your
>     configuration, each read will be sent to 2 bricks. If all bricks are
>     alive and healthy, the disperse translator balances these reads
>     among all nodes, giving 2/3 of the load to each brick.
>     Write requests are processed by all bricks, so the load is the same
>     on all of them.
>         *Third question* - _does this require reading the entire data_
>         of A1, A2
>         and A3 by initiating a heal or another operation?
>     Healing operations are on file basis. If only some files of A3 have
>     been damaged, it will only read the corresponding data from A1 and
>     A2, but not the entire contents of A1 and A2. To heal a file, all
>     file contents are read.
>         *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.
>     Correct
>         Is this also correct for bricks from the same set, for example
>         A1 and A2?
>     No, each disperse set is independent and have the same redundancy.
>     It's equivalent to a distributed replicated: if you lose both bricks
>     of the same replica set, you will lose access to the data stored in
>     that replica set.
>         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. These are two different configurations. Both have the same
>     effective capacity, but the probability of failure in the second
>     case is several times lower than the first one (you can lose *any*
>     two bricks without losing access to the data). However it's more
>     expensive to grow the volume because you will need to add 6 new
>     bricks at the same time, while with the first case you only need to
>     add 3.
>     Xavi
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