<div dir="ltr"><div><div>Thank you Xavi, it's very helpful (also to Atin).<br><br></div><div>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.<br><br></div><div>Ayelet</div><div><br></div></div><div><br><br><div><div><div class="gmail_extra"><br><div class="gmail_quote">On Tue, Nov 25, 2014 at 5:19 PM, Xavier Hernandez <span dir="ltr"><<a href="mailto:xhernandez@datalab.es" target="_blank">xhernandez@datalab.es</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">Hi Ayelet,<span class=""><br>
<br>
On 11/25/2014 02:41 PM, Ayelet Shemesh wrote:<br>
</span><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><span class="">
Hello Gluster experts,<br>
<br>
I have been using gluster for a small cluster for a few years now and I<br>
have a question regarding the new disperse feature, which is for me a<br>
much anticipated addition.<br>
<br></span>
*Suppose* I create a volume with a disperse set of 3, redundancy 1<span class=""><br>
(let's call them A1, A2, A3) and then I add 3 more bricks to that volume<br>
(we'll call them B1, B2, B3).<br>
<br></span>
*First question* - which of the bricks will be the one carrying the<br>
redundancy data?<br>
</blockquote>
<br>
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.<br>
<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<br>
*Second question* - If I have machines with faster disk - should I<span class=""><br>
assign them to the data or the redundancy bricks? What should I expect<br>
the load to be on the redundancy machine in heavy read scenarios and in<br>
heavy write scenarios?<br>
</span></blockquote>
<br>
As I said, there isn't a dedicated redundancy brick, so there's no benefit in assigning the fast disk to a specific brick.<br>
<br>
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.<br>
<br>
Write requests are processed by all bricks, so the load is the same on all of them.<br>
<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<br>
*Third question* - _does this require reading the entire data_ of A1, A2<span class=""><br>
and A3 by initiating a heal or another operation?<br>
<br>
</span></blockquote>
<br>
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.<br>
<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
*4th question* (and most important for me) - I saw in the list that it<span class=""><br>
is now a Distributed-Dispersed volume. I understand I can now lose, for<br>
example bricks A1 and B1 and still have my entire data intact.<br>
</span></blockquote>
<br>
Correct<span class=""><br>
<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
Is this also correct for bricks from the same set, for example A1 and A2?<br>
</blockquote>
<br></span>
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.<br>
<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
Or to put it in a more generic way - _does this create the exact same<span class=""><br>
dispersed volume as if I created it originally with A1, A2 A3 B1 B2 B3<br>
and a redundancy of 2?<br>
</span></blockquote>
<br>
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.<br>
<br>
Xavi<br>
</blockquote></div><br></div></div></div></div></div>