[tahoe-dev] How many servers can fail?

Terrell Russell terrellrussell at gmail.com
Wed Oct 26 15:14:49 UTC 2011


R could be for Reconstruction

but also

Recovery





On 10/26/11 10:00 AM, Greg Troxel wrote:
> 
> Shawn Willden <shawn at willden.org> writes:
> 
>> I think it would simplify things greatly to further constrain share
>> placement so that each server gets no more than one share, so that N,
>> H and K all refer to servers.  I realize that there are some
>> interesting things that can be achieved by setting N to be a multiple
>> of the number of servers available, but in practice I don't think they
>> add enough value to offset the conceptual complexity.
> 
> In a grid with a lot of servers S >> N, I can see your point.
> 
> I have a private grid that had 3 servers, and now has 4.  But I'm using
> 3/10 encoding, so that I can gradually add servers without having to
> torque the world around.
> 
> I realize this is tricky business, but I've come to think that H isn't
> the right concept.  What I really care about is knowing how many servers
> can fail without causing me to lose files.  If I had 5 servers but was
> using 3/10 (to enable migration to larger-N grids), then optimal share
> placement would be 5x2, and I could lose 3/5 and reconstruct with any 2.
> But with 4x1 and 1x5, I could lose at most 2/5 while being sure of
> reconstruction.  Both situations have H=5.
> 
> It would be nice instead to have a way to express: place shares so that
> I am guaranteed to be able to reconstruct if any R (for reconstruction)
> servers which have a share have not failed.
> 
> For S >> N, then perhaps R = N - H.
> 
> For S << N, then H ensures that H servers have 1 share, whereas R will tend
> to ensure that that shares are spread evenly.
> 
> As an example, consider 3/10 encoding with R=2 and S=5.
> 
> Placing 2 shares each on 5 servers meets that goal, and placing 1 on 1,
> 3 on 1, and 2 on the last 3 also does.  But any less good distribution
> does not.  Here, there are many distributions which meet the same H
> (even if H=S=5!) but only some disitrbutions meet R=2.
> 
> 
> So, if it turns out I'm not confused, my proposal is to replace H with
> R.
> 
> 
> 
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