In a typical erasure coded storage system, a set of n disks is divided into m disks to hold data and k disks to hold coding information, where n, m, and k are integers. The coding information is calculated from the data. If up to k of the n disks fail, their contents can be recomputed from the surviving disks. In this example k=4, so the correct answer is B.

B. 4 "Provides space-optimal data redundancy to protect data loss against multiple drive failures – A set of n disks is divided into m disks to hold data and k disks to hold coding information – Coding information is calculated from data The figure illustrates an example of dividing a data into nine data segments (m = 9) and three coding fragments (k = 3). The maximum number of drive failure supported in this example is three."

The figure illustrates an example of dividing a data into nine data segments (m = 9) and three coding fragments (k = 3). The maximum number of drive failure supported in this example is three. Erasure coding offers higher fault tolerance (tolerates k faults) than replication with less storage cost.

"Provides space-optimal data redundancy to protect data loss against multiple drive failures – A set of n disks is divided into m disks to hold data and k disks to hold coding information – Coding information is calculated from data The figure illustrates an example of dividing a data into nine data segments (m = 9) and three coding fragments (k = 3). The maximum number of drive failure supported in this example is three."

B 4 , not A

The answer is 3. 12/4=3

Can tolerate up to K disk (here k=4)

Correct answer is B https://searchstorage.techtarget.com/definition/erasure-coding

correct A 12/4= 3

I think the correct is A. Because the data segments is 12 and the coding segments is 4 then: M/K=9/4= 3