Chapter 10 — Quiz

Fourteen questions: ten multiple choice, two true/false, two short answer. Answers and a scoring band are at the bottom. Cover them until you have committed to each response — guessing teaches nothing.

Multiple choice

Q1. What is the minimum number of disks required to build a RAID 6 array? - A) 2 - B) 3 - C) 4 - D) 6

Q2. A four-disk RAID 5 can survive how many simultaneous member failures before data is unrecoverable from parity? - A) 0 - B) 1 - C) 2 - D) 3

Q3. Which parity layout is the Linux mdadm default — the one you will encounter most often, and the highest-performing for sequential reads? - A) Left-asymmetric - B) Left-symmetric - C) Right-asymmetric - D) Right-symmetric

Q4. A Linux software-RAID member can be identified by a superblock with which magic value? - A) 0x55AA - B) 0xEF53 - C) 0xA92B4EFC - D) 0xDE11DE11

Q5. The single most common RAID disaster that lands arrays on a recovery bench is: - A) A power surge that destroys all members at once - B) A forced rebuild of a degraded array that triggers a second failure - C) A firmware bug that silently corrupts parity - D) A user deleting files from the live volume

Q6. In RAID terminology as used in this chapter, a strip (or chunk) is: - A) The full horizontal row of blocks across all disks at the same offset - B) The contiguous block written to one disk before the controller moves to the next disk - C) The reserved metadata area at the start of each member - D) The parity block for a given stripe

Q7. In a standard RAID 5, the parity strip P for a stripe is computed as: - A) The arithmetic sum of the data strips, modulo 256 - B) The bitwise XOR of the data strips in that stripe - C) A Reed-Solomon syndrome over GF(2⁸) - D) An MD5 hash of the data strips

Q8. RAID 6's second syndrome, Q, differs from P because Q is: - A) Also a plain XOR, just stored on a different disk - B) A Reed-Solomon syndrome computed over the Galois field GF(2⁸) - C) A copy of P written for redundancy - D) The two's-complement of P

Q9. You suspect one member of a RAID 5 is stale (it dropped out earlier while the array kept running). The clearest single indicator in the metadata is: - A) A larger Avail Dev Size than its peers - B) A different Array UUID - C) A lower Events (sequence) counter than the other members - D) A Checksum line reading "correct"

Q10. A RAID 6 array of six 4 TB drives provides how much usable capacity? - A) 8 TB - B) 12 TB - C) 16 TB - D) 24 TB

True / False

Q11. A single member of a RAID 1 mirror is, by itself, a complete standalone copy of the volume that can be imaged and analyzed like an ordinary single disk. (True / False)

Q12. If two disks fail simultaneously in a RAID 5, the array can always be fully reconstructed from the surviving parity. (True / False)

Short answer

Q13. In two or three sentences, explain why you image every member of a failed RAID 5 — including the disk the controller marked "failed" — before attempting any reconstruction.

Q14. A reconstructed array mounts successfully and its boot sector and root directory look correct, but files deep in subdirectories open as corruption. Name the most likely cause and the verification step that catches this failure mode.

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Answer key

Q1 — C. RAID 6 needs a minimum of four disks: it stores two independent parity syndromes (P and Q), so usable capacity is N−2, and you need N ≥ 4 to have any usable data capacity worth the redundancy.

Q2 — B. RAID 5 has exactly one parity strip per stripe, so it survives exactly one missing strip (data or parity) per stripe — one disk. A second failure leaves each stripe missing two strips, which a single parity equation cannot solve.

Q3 — B. Left-symmetric. Data restarts on the disk immediately after the parity strip and wraps, keeping sequential reads moving forward across disks; it is the mdadm default and the layout you will see most.

Q4 — C. 0xA92B4EFC is the mdadm superblock magic. 0x55AA is the boot-sector signature, 0xEF53 is the ext4 superblock magic, and 0xDE11DE11 is the SNIA DDF anchor used by hardware controllers.

Q5 — B. A forced rebuild of a degraded array stresses aging, same-batch disks with a full-surface read just when redundancy is already zero; a second URE surfaces and converts a recoverable single failure into an unrecoverable double failure.

Q6 — B. A strip (chunk) is the per-disk contiguous block written before moving to the next disk. The full horizontal row (option A) is the stripe — conflating the two is the "stripe size" vocabulary trap.

Q7 — B. P is the bitwise XOR of the data strips. XOR is its own inverse, so any one missing strip is recovered by XOR-ing all the others.

Q8 — B. Q is a Reed-Solomon syndrome over GF(2⁸) (g = 0x02, reduction polynomial 0x11D). Because it is a finite-field computation, you cannot hand-compute it the way you can XOR P, and dual-failure recovery requires solving the P and Q equations together.

Q9 — C. A stale member shows a lower Events/sequence counter — a trace of when it dropped out. (Absent metadata, a parity mismatch isolated to one disk's strips is the empirical tell.)

Q10 — C. RAID 6 capacity is (N−2) × disk size = (6 − 2) × 4 TB = 16 TB.

Q11 — True. Every block is written to both members, so one mirror member is a full copy. The caution: if the mirrors diverged (degraded operation), image both and use the freshest — but a single member is still a complete volume.

Q12 — False. Two missing strips per stripe cannot be solved by one parity equation. The only paths are physically resurrecting one of the two failed disks to image it (reducing to a single failure) or accepting partial recovery.

Q13 — Model answer. RAID 5 parity covers one missing strip per stripe, so a disk the controller "failed" is usually still mostly readable and remains useful — its readable strips reduce how much you must reconstruct, and parity covers its gaps. Imaging every member first means each reconstruction attempt is a no-risk hypothesis test on copies, and a controller's "double failure" frequently becomes a near-total recovery on the bench because the two members' bad sectors rarely overlap at the same stripe offset.

Q14 — Model answer. The most likely cause is a wrong-but-close geometry — an incorrect strip size, disk order, or parity layout — that aligns the first stripe (boot sector, root directory) but scatters data once the parity rotation advances. The verification step that catches it is to mount read-only, run a non-writing consistency check, and open several large files of different types from deep in the directory tree; a correct geometry opens them cleanly, a near-miss corrupts past the first rotation.

Scoring: 13–14 correct — you can take a RAID job to the bench unsupervised. 11–12 — solid; revisit parity rotation and the dual-parity math. 8–10 — reread "How each RAID level fails" and "Parity rotation" in index.md and redo the parity exercises. Below 8 — work the whole chapter again before attempting a reconstruction lab; the cost of a wrong move here is measured in clients' irreplaceable data.