<p dir="ltr">Il 09 ago 2016 19:57, "Ashish Pandey" <<a href="mailto:aspandey@redhat.com">aspandey@redhat.com</a>> ha scritto:<br>
> Yes, redundant data spread across multiple servers. In my example I mentioned 6 different nodes each with one brick.<br>
> Point is that for 4+2 you can loose any 2 bricks. It could be because of node failure or brick failure.<br>
> 1 - 6 bricks on 6 different nodes - any 2 nodes may go down - EC win <br>
><br>
> However if you have only 2 nodes and 3 bricks on each nodes, then yes in this case even if one node goes down, ec will fail because that will cause 3 bricks down.<br>
> In this case replica 3 would win.</p>
<p dir="ltr">6 nodes with 1 brick each is a surreal case.<br>
A much common case is multiple nodes with multiple bricks, something like 9 nodes with 12 bricks each. (In example, a 2U supermicro server with 12 disks)</p>
<p dir="ltr">In this case, EC replicas could be placed on a single server.</p>
<p dir="ltr">And with 9*12 bricks you still have 2 single disks (or one server if both are placed on the same hardware) as failure domains.<br>
Yes, you'll get 9*(12-2) usable bricks and not (9*12)/3 but you risk data loss for sure.</p>
<p dir="ltr">Just a question: with EC which is the right calc method between these 3:</p>
<p dir="ltr">a) (#servers*#bricks)-#replicas</p>
<p dir="ltr">Or</p>
<p dir="ltr">b) #servers*(#bricks - #replicas)</p>
<p dir="ltr">Or</p>
<p dir="ltr">c) (#servers-#replicas)*#bricks</p>
<p dir="ltr">In case A I'll use 2 disks as replica for the whole volume (exactly like a raid6)</p>
<p dir="ltr">In case B I'll use 2 disks from each server as replica</p>
<p dir="ltr">in case C I'll use 2 whole servers as replica (this is the most secure as i can loose 2 whole servers)</p>