A storage system is configured with erasure coding, divided up into 10 data segments and...

As with any erasure-coded configuration, the maximum number of simultaneous drive (segment) failures the system can withstand without data loss is equal to the number of coding (parity) segments configured, since the mathematical reconstruction process requires at least as many surviving segments as there are original data segments, meaning it can tolerate losing exactly as many segments as were added for redundancy. In this configuration, described as 10 data segments plus 2 coding segments (a 10+2 scheme), there are exactly 2 coding segments, which directly sets the maximum tolerable simultaneous drive failure count at 2. If a third drive were to fail at the same time, only 9 of the original 10 data segments' worth of reconstructable information would remain available, which is insufficient to fully reconstruct the original data, resulting in permanent data loss. This configuration trades some fault tolerance for storage efficiency compared to the earlier 10+4 example, since fewer coding segments consume less overhead capacity but provide a correspondingly lower resilience ceiling, a classic and deliberate tradeoff administrators make when balancing storage efficiency against desired fault tolerance levels in large-scale, erasure-coded storage systems. Options B, C, and D all overstate the tolerance beyond what the 2 configured coding segments can mathematically support. The correct maximum is 2, directly matching the number of coding segments.

Reference topic: Fault Tolerance Techniques - Erasure Coding Fault Tolerance Calculation (10+2 Configuration).