An administrator is tasked to create a new storage policy for an eight-node VMware vSAN Original Storage Architecture (OSA) cluster. The following specifications have been provided: - Tolerate up to…

Question

Options

• A1 failure - RAID-1
• B2 failures - RAID-5
• C3 failures - RAID-1
• D1 failure - RAID-5

• A
  12% (3)
• B
  77% (20)
• C
  4% (1)
• D
  8% (2)

Explanation

Option B is correct because the requirement to tolerate two host failures mandates an FTT (Failures to Tolerate) value of 2, and RAID-5/6 erasure coding delivers maximum usable capacity by using only ~1.5x storage overhead compared to RAID-1 mirroring's 3x overhead for the same FTT level. An 8-node cluster exceeds the 6-node minimum required for FTT=2 erasure coding (RAID-6 mode), making this combination fully supported.

Why the distractors fail:

• A (1 failure - RAID-1): FTT=1 only protects against a single host failure, violating the two-failure requirement.
• C (3 failures - RAID-1): Over-protects beyond the stated requirement and RAID-1 with FTT=3 demands 7 nodes while consuming 4x storage overhead - the worst possible capacity efficiency.
• D (1 failure - RAID-5): Again, FTT=1 fails the two-failure tolerance requirement, even though erasure coding is the right type of policy.

Memory tip: Think "FTT must match the spec exactly, then pick erasure coding (RAID-5/6) over mirroring (RAID-1) whenever capacity efficiency is the goal" - RAID-1 always mirrors full copies, while erasure coding spreads parity across fewer drives.

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