[Openstack] [SWIFT] What happens when the shard count on available disk is less than the total shard count?

Pete Zaitcev zaitcev at redhat.com
Fri Feb 28 19:20:30 UTC 2014


On Fri, 28 Feb 2014 09:10:06 -0800
Stephen Wood <smwood4 at gmail.com> wrote:

> However I realize that the shard count is completely different now.

What is a "shard count"? Do you have a document that uses such
terminology?

> I
> originally used a partition value of 15 but this now seems much to high for
> 4 servers with only disk each.

So what? As long as there are no ill effects, it's all good.
Meaning if you have enough RAM to keep your ring once it's loaded,
then no problem, isn't it? It's not like your A+C servers magically
shrunk when you swapped the winchesters for SSDs, right?

> Can I dynamically
> adjust the partition values after the swift ring has been created?

No, you can't.

>  Or
> should I just take the disks on my 4 SSD hosts and put their weight as 2^15
> / 4 so the overall shard count stays the same?

I am failing to make sense of the above sentence. Weight only matters
for builder scattering partitions at devices relative to each other.
So, if one replaces rotating media with SSDs, but keeps the cluster
running, the number of parititions stays the same, right? At that point
weights can be redefined at, say, 100, or any other number, without
any effect on total or per-device number of partitions.

I think we need to circle back to the definition of the mysterious
"shard count" before we can get to the bottom of this.

-- Pete




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