Discussion
An erasure coding scheme stores 10 data fragments and 4 parity fragments. What storage overhead does that represent?
Answer: D
Correct answer: 40 per cent
\[ \frac{4\ \mathrm{parity}}{10\ \mathrm{data}} = 0.40 = 40\%, \] so 14 fragments are stored for 10 of data. That is the attraction of erasure coding: triple replication costs 200 per cent overhead for the loss of two copies, while this scheme tolerates the loss of any four fragments for 40 per cent. The cost is computation on every read that must reconstruct, and higher latency when fragments are missing.
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