Filer upstart Qumulo rubs out mirroring, slots in erasure coding

Re: Reed-Solomon erasure coding offers +4 and +5 protection with +2 overhead

Terrible phrasing. An RS code is characterised directly as a ratio, m/n >= 1 where m is the number of "shadows" or "shares" (I forget the usual nomenclature) produced and stored, and n is the number of such "shadows" needed to decode the data.

I assume that +2 overhead means that m/n = 3. That is, the encoded data takes up 3 times as much space, which is an overhead of 200%. Rewrite this as m=3n (eq. a).

Usually, when talking about erasure codes, people talk about how many "erasures" (ie, missing shadows) the scheme can tolerate. So I assume that "+x protection" is simply counting the number of such erasures. This gives two more equations:

• "+4 protection": m - n = 4 (eq. b)
• "+5 protection": m - n = 5 (eq. c)

Substitute equation a into b and you get an answer n = 2, m = 6. This is like saying the data is spread over 6 disks and data can be recovered if up to 4 of those disks fail.

Substituting a into c gives a nonsensical solution of n = 2.5. This is nonsense because m and n have to be integers. I assume that what's implied here is that 5 erasures are tolerated (eq. c) and that the overhead is no more than 200% (m/n < 3). This has an infinite number of solutions. Looking at the sequence 6/1, 7/2, 8/3, 9/4, 10/5, ... (ie, all the schemes that tolerate exactly 5 erasures), the first one that satisfies m/n < 3 is 8/3, so I'm going to to call that the answer: there are 8 shadows, any three of which can reconstruct the data, for an overhead of 166.6%.