[sheepdog-users] Suggestion about redundance

[Andrew J. Hobbs]

No.  Stay with -c 3 or -c 2:1.  Each node will be a zone, so the 2 devices are part of that single zone.  UNLESS you were going to run two sheepdog servers per node, each exposing the disk, and specify the zones.

So I'd recommend (unless the disks are significantly mismatched in terms of speed or size) sheepdog in md mode giving three nodes and either -c3 or -c2:1.  If the drives are highly mismatched in terms of speed or storage, then consider running two sheepdog instances on each node with an explicit zone flag for each.  Problem here:  in the event of node failure, you're losing two at once.  So you gain no resiliency here over sheepdog md mode.  -c 2:2 needs at least 4 nodes to be a viable erasure encoding.

On 06/23/2014 11:12 AM, Valerio Pachera wrote:
Hi, I'm going to format a new cluster.
It has 3 nodes.
I'm not going to a 4th node very soon.
Each node has 2 devices.

I was considering to use -c 2:2 because:
it uses the same space of -c 2;
I can loose two hosts / devices in the same period.
(In my case 1 host).

What do you think?




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