[2026 AP Subject A #8] RAID, Availability, and Reliability

Revised / source-checked: 2026-10-20 Scope source: IPA Applied Information Technology Engineer Examination (Level 3) Syllabus Ver.7.2

Disclaimer: This is independent study material written by Motitory Labs. It is not approved, reviewed, endorsed by, or affiliated with IPA (Information-technology Promotion Agency, Japan). Exam rules, scope, and syllabus may change; always check the latest information on the official IPA site before the exam. All five questions are original questions; no past exam questions are used.

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Learning Index

1. Terminology (KO/JA/EN)
2. RAID 0/1/5/6/10 — capacity and fault-tolerance conditions
3. Capacity worked examples (4 TB disks)
4. RAID is not backup
5. Series and parallel reliability
6. MTBF, MTTR, and availability
7. Five original questions (with explanations)
8. Quick sheet · common traps · final self-check · review plan
9. Source and syllabus mapping

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1. Terminology

한국어	日本語	English	Meaning
신뢰도	信頼度	reliability	Probability of functioning without failure over a specified interval (mission time) under stated conditions
가용성(가동률)	可用性（稼働率）	availability	Fraction of time / probability that the system is in an operable state
평균 고장 간격	平均故障間隔 (MTBF)	mean time between failures	Mean operating time between failures
평균 수리 시간	平均修復時間 (MTTR)	mean time to repair/restore	Mean repair/restore time after a failure (exact definition can vary by context)
스트라이핑	ストライピング	striping	Distributing data across multiple disks
미러링	ミラーリング	mirroring	Writing identical data to two or more disks
패리티	パリティ	parity	Redundant information used to reconstruct lost data
리빌드	リビルド	rebuild	Reconstructing data onto a replacement disk
직렬 시스템	直列システム	series system	Works only if all components work
병렬 시스템	並列システム	(active) parallel system	Works if at least one component works
백업	バックアップ	backup	Separate copy independent of the original (with history/versions)

Note: Reliability (probability of surviving a given interval) and availability (steady-state fraction of up time) are different metrics. Exam questions often apply the same series/parallel formulas to “availability (稼働率)”, but do not confuse the concepts themselves.

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2. RAID 0/1/5/6/10

Assumptions: every disk has the same usable size S; filesystem, controller, hot-spare, and metadata overhead are ignored unless stated. If physical sizes differ, capacity is generally based on the smallest member. “Fault tolerance” below refers to disk failures only; RAID does not protect against controller failure or correlated disasters such as fire or flood.

Level	Min. disks	Usable capacity	Disk-failure tolerance
RAID 0 (striping)	2	N×S	0 — any single failure loses the array
RAID 1 (mirroring, standard 2-disk)	2	S	Either one of the two disks
RAID 5 (distributed parity)	3	(N−1)×S	Any one disk. A second failure before rebuild completes loses the array
RAID 6 (dual parity)	4	(N−2)×S	Any two disks. A third failure loses the array
RAID 10 (mirror + stripe)	4 (even)	(N/2)×S	Guaranteed: any one disk. May survive up to N/2 failures only if each mirror pair loses at most one disk. Both disks of the same pair failing loses the array

• RAID 1: this article uses the standard two-disk mirror. Configurations with more members vary by implementation; do not overgeneralize.
• RAID 10 survival depends on which disks fail, not just how many. Example: six disks as mirror pairs (D1,D2)(D3,D4)(D5,D6):
  • D1, D3, D5 fail → each pair still has one disk → survives (3 failures = N/2)
  • D1, D2 fail → one pair completely lost → array lost (only 2 failures)
  • So “RAID 10 tolerates N/2 failures” is wrong as a guarantee; only one arbitrary disk is guaranteed.

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3. Capacity Worked Examples

Example A — six equal 4 TB disks (S = 4 TB, N = 6)

Level	Calculation	Usable capacity
RAID 0	6 × 4	24 TB
RAID 5	(6 − 1) × 4	20 TB
RAID 6	(6 − 2) × 4	16 TB
RAID 10	(6 / 2) × 4	12 TB

Example B — RAID 1 (separate example): two equal 4 TB disks → usable capacity 4 TB. (A six-disk “RAID 1” is implementation-dependent and ambiguous, so it is not used here.)

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4. RAID Is Not Backup

RAID is an availability technique that keeps the service running through some disk failures. Because all members reflect the current state simultaneously, it does not help against:

• Accidental deletion or overwrite → applied to every member immediately
• Ransomware encryption or data corruption from software bugs → propagated across members
• Theft, fire, flood, or loss of the entire site
• Controller failure or operator error (wrong disk pulled, array re-initialized, etc.)

A backup requires a separate copy independent of the original, retained versions/history, and regular restore testing.

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5. Series and Parallel Reliability

Assumptions (for every calculation): each component reliability Ri refers to the same mission interval, and failures are statistically independent. With common-cause failures (e.g., a shared power supply) or correlated failures, the formulas below do not apply unchanged.

Series (all components must work)

R_series = R1 × R2 × … × Rn

Example: independent R1 = 0.98, R2 = 0.99 → 0.98 × 0.99 = 0.9702 (97.02%) → A series system is less reliable than its weakest component.

Active parallel (works if at least one works)

R_parallel = 1 − (1 − R1)(1 − R2)…(1 − Rn)

Complement logic: “system fails” = “every component fails”. Under independence, the probability that all fail is the product of (1 − Ri); subtracting from 1 gives the probability the system works.

Example: two independent components, each R = 0.90 → 1 − (0.10 × 0.10) = 1 − 0.01 = 0.99 (99%)

Standby configurations differ. When a spare is switched in on failure, the result depends on switching success (coverage), whether the standby can fail while idle, and similar assumptions. Do not apply the simple active-parallel formula unchanged.

These values are mission reliability probabilities, not the steady-state availability in the next section.

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6. MTBF, MTTR, and Availability

Assumptions: a repairable system alternating between up and down states, a steady-state approximation, and consistent units.

• MTBF: mean operating time between failures
• MTTR: mean repair/restore time (whether detection or waiting time is included can vary by context — check the problem’s definition)

A = MTBF / (MTBF + MTTR) Unavailability = MTTR / (MTBF + MTTR) = 1 − A

Example: MTBF = 990 h, MTTR = 10 h → A = 990 / 1000 = 0.99 (99%), expected unavailable fraction 0.01 (1%)

Expected annual downtime (expectation under a stationary model): 0.01 × 8,760 h = 87.6 h/year → This is an expected value, not a schedule or a guarantee that the system stops exactly 87.6 hours every year.

Unit trap: MTBF = 49.5 days, MTTR = 12 hours

• Wrong: 49.5 / (49.5 + 12) ≈ 0.8049 (days mixed with hours)
• Right: 49.5 days = 49.5 × 24 = 1,188 h → 1,188 / (1,188 + 12) = 1,188 / 1,200 = 0.99

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7. Original Questions (5)

All questions below are original questions written by Motitory Labs. They are not past exam questions. Choice labels follow the Japanese exam convention (ア, イ, ウ, エ = 1st–4th).

Question 1 — RAID capacity

RAID 6 is built from eight equal 2 TB disks. Ignoring overhead, what is the usable capacity?

• ア. 16 TB
• イ. 14 TB
• ウ. 12 TB
• エ. 8 TB

Answer: ウ (12 TB)

Derivation: RAID 6 = (N − 2) × S = (8 − 2) × 2 = 12 TB

• ア 16 TB: 8 × 2 — RAID 0 capacity (no parity)
• イ 14 TB: (8 − 1) × 2 — RAID 5 capacity (one disk of parity)
• エ 8 TB: (8 / 2) × 2 — RAID 10 capacity (half mirrored)

Question 2 — RAID 10 failure pattern

A RAID 10 array stripes across six disks grouped as mirror pairs (D1,D2), (D3,D4), (D5,D6). Which simultaneous two-disk failure loses the array? (Before any rebuild; no other failures.)

• ア. D1 and D3
• イ. D3 and D4
• ウ. D2 and D5
• エ. D4 and D6

Answer: イ (D3 and D4)

Derivation: D3 and D4 form the same mirror pair, so that pair’s data is completely lost and the whole stripe is lost.

• ア D1 & D3: different pairs → each pair keeps one disk → survives
• ウ D2 & D5: different pairs → survives
• エ D4 & D6: different pairs → survives

RAID 10 survival is determined by failure location (same pair or not), not by the count alone.

Question 3 — Series/parallel reliability

Components A and B (reliability 0.90 each) are in active parallel, and that block is in series with component C (reliability 0.95). All reliabilities refer to the same mission interval and failures are independent. What is the system reliability?

• ア. 0.7695
• イ. 0.8550
• ウ. 0.9995
• エ. 0.9405

Answer: エ (0.9405 = 94.05%)

Derivation:

1. A∥B: 1 − (1 − 0.90)(1 − 0.90) = 1 − 0.01 = 0.99
2. (A∥B) in series with C: 0.99 × 0.95 = 0.9405

• ア 0.7695: 0.90 × 0.90 × 0.95 — treats all three as series
• イ 0.8550: 0.90 × 0.95 — counts only one of the parallel components
• ウ 0.9995: 1 − (0.10 × 0.10 × 0.05) — treats all three as parallel

Question 4 — MTBF/MTTR with units

A repairable server has MTBF = 49.5 days and MTTR = 12 hours. Under the steady-state approximation, what is its availability?

• ア. 0.9900
• イ. 0.0100
• ウ. 0.8049
• エ. 0.9899

Answer: ア (0.99 = 99%)

Derivation: Convert units: 49.5 days × 24 = 1,188 h. A = 1,188 / (1,188 + 12) = 1,188 / 1,200 = 0.99

• イ 0.0100: MTTR / (MTBF + MTTR) — this is unavailability, not availability
• ウ 0.8049: 49.5 / (49.5 + 12) — unit error mixing days and hours
• エ 0.9899: 1 − MTTR/MTBF = 1 − 12/1,188 ≈ 0.98990 — approximation that drops MTTR from the denominator

Question 5 — RAID vs. backup, and assumptions

Which statement is most appropriate?

• ア. RAID 1 keeps the same data on two disks, so no separate backup is needed.
• イ. RAID 6 can keep operating through any two disk failures, but a mistakenly deleted file is deleted on all members, so a separate versioned backup with restore testing is still needed.
• ウ. Two servers of reliability 0.90 in parallel give system reliability 0.99 even if both share one power supply whose failure stops both.
• エ. An availability of 0.99 guarantees that the system stops exactly 87.6 hours every year.

Answer: イ

Derivation: RAID 6 maintains availability through two disk failures, but deletion and overwrite propagate to every member, so a backup with independent copies, history, and restore testing is needed separately.

• ア: A mirror faithfully replicates deletion, ransomware, and corruption; it is not a backup.
• ウ: A shared power supply is a common-cause failure that breaks the independence assumption; 0.99 holds only under independence.
• エ: 87.6 h/year is an expected value under a stationary model, not a schedule or guarantee.

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8. Quick Sheet

Item	Formula / key point
RAID 0	N×S, tolerates 0, min. 2
RAID 1 (2 disks)	S, either one disk
RAID 5	(N−1)S, any one disk, min. 3
RAID 6	(N−2)S, any two disks, min. 4
RAID 10	(N/2)S, guaranteed any one; same-pair double failure loses array; even N ≥ 4
Series	∏Ri (independent, same interval)
Active parallel	1 − ∏(1 − Ri) (independent, same interval)
Availability	MTBF / (MTBF + MTTR), consistent units
Unavailability	MTTR / (MTBF + MTTR)

Common Traps

1. Claiming RAID 10 “tolerates N/2 failures” — only one is guaranteed.
2. Ignoring a second failure during a RAID 5 rebuild.
3. Treating RAID (availability) as backup (historical copies).
4. Mixing MTBF in days with MTTR in hours.
5. Using 1 − MTTR/MTBF, or answering with unavailability instead of availability.
6. Applying the independent-parallel formula despite a common-cause failure.
7. Calculating standby configurations as if they were active parallel.
8. Confusing mission reliability with steady-state availability, or reading expected annual downtime as a guarantee.
9. Mixing 0.99 and 99% mid-calculation and getting a 100× error.

Final Self-Check

• I can state RAID 0/5/6/10 capacities for six 4 TB disks (24/20/16/12 TB) instantly.
• I can give a RAID 10 example where three failures survive and one where two failures lose the array.
• I can give at least three reasons RAID is not backup.
• I can explain 0.98 × 0.99 = 0.9702 and 1 − 0.1² = 0.99 along with their assumptions.
• I can explain MTBF 990 h, MTTR 10 h → 0.99 and 87.6 h/year expected downtime (an expectation).
• I can convert days/hours consistently before calculating.

Review Plan

• Day 1: Cover the quick sheet and reproduce the RAID table and the three formulas on blank paper.
• Day 3: Re-solve Questions 1–5 without the explanations and say aloud why each wrong choice is wrong.
• Day 7: Make your own variants by changing the numbers (e.g., eight-disk RAID 10, MTBF 30 days / MTTR 6 h).

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9. Source and Syllabus Mapping

• Official source: IPA 応用情報技術者試験（レベル3）シラバス Ver.7.2 https://www.ipa.go.jp/shiken/syllabus/omgdg50000005kq5-att/syllabus_ap_ver7_2.pdf (checked 2026-10-20)
• Mapping (大分類2：コンピュータシステム／中分類4：システム構成要素):
  • 1. システムの構成 —（2）システム構成 → Sections 2, 5 (reliability improvement through redundancy)
  • 1. システムの構成 —（6）ストレージ → Sections 2, 3, 4 (RAID0, RAID1, RAID10, RAID5, RAID6, striping, mirroring, parity)
  • 1. システムの構成 —（7）信頼性設計 → Section 5 (reliability block diagrams, parallel operation, standby switching)
  • 2. システムの評価指標 —（2）システムの信頼性特性と評価 — ① RASIS → Section 1 (Reliability vs. Availability)
  • 2. システムの評価指標 —（2）システムの信頼性特性と評価 — ② 信頼性指標と信頼性計算 → Sections 5, 6 (MTBF, MTTR, availability, series/parallel calculation)
• Out of scope this installment: RAID 2/3/4 details, chunk size, NAS/SAN, backup-site architecture, detailed RASIS, bathtub curve, storage performance benchmarking → later installments.
• Past questions used: none. All five questions are original.