Calculate usable RAID capacity, parity or mirror overhead and how many drive failures RAID 0, 1, 5, 6 and 10 survive, in TB and the TiB Windows shows.
업데이트 검증한 예제: 6
RAID 5 on 4 drives of 8 TB gives 24 TB usable (21.83 TiB in the binary units Windows shows), 75% of the raw space, and always survives 1 failed drive.
Drive labels use 1 TB = 10¹² bytes. Windows divides by 2⁴⁰ but still prints “TB”, so a new drive looks about 9% smaller there; macOS reports decimal TB.
One drive's worth of parity is spread across all drives. Every member counts as the smallest drive in the set.
1 drive
Usable capacity is the drive size times the number of drives' worth of space that holds data: n for RAID 0 striping, 1 for a RAID 1 mirror, n − 1 for RAID 5 single parity, n − 2 for RAID 6 double parity and n ÷ 2 for RAID 10 striped mirrors. The rest of the raw space holds copies or parity, which is what lets the array keep running after a drive fails.
People use it to size a NAS or server before buying drives. The default, four 8 TB drives in RAID 5, gives 24 TB usable out of 32 TB raw, 75% efficiency, and survives one failed drive. In Windows the same space appears as 21.83 TB, because Windows divides by 2^40 (the binary TiB) but prints "TB".
The results assume identical drives; with mixed sizes, each drive counts as the smallest one. File-system metadata, hot spares and reserved space are not subtracted, so the room left for files is slightly lower.
검증 출처: Python 3.8 decimal: (4 − 1) × 8 = 24 TB; 24e12 / 2**40 = 21.8279 TiB; 8e12 / 2**40 = 7.2760 TiB
검증 출처: Python 3.8: (6 − 2) × 4 = 16; 16/24 = 66.67%
검증 출처: Python 3.8: 4 / 2 × 2 = 4 TB; with 2 mirror pairs one failure per pair is survivable, but two in the same pair are not
검증 출처: Definition: every member holds a full copy
RAID 5 gives (n − 1) × drive size, because one drive's worth of space holds parity. Four 8 TB drives give 24 TB (75%) and eight give 56 TB (87.5%). It needs at least 3 drives and survives exactly one failure. Efficiency rises as you add drives, but so does the amount of data a rebuild must read without hitting a second failure.
RAID 6 stores two independent parities (P and Q) instead of one, so it survives any 2 failed drives at the cost of a second drive's capacity. With six 4 TB drives, RAID 5 gives 20 TB and RAID 6 gives 16 TB. A small random write costs 4 disk I/Os on RAID 5 and 6 on RAID 6, so RAID 6 writes are slower.
RAID 10 is faster and rebuilds more safely, while RAID 5 stores more. RAID 10 always keeps 50% of raw space, so 4 × 8 TB gives 16 TB against 24 TB for RAID 5. A small random write costs 2 disk I/Os instead of 4, and a rebuild copies one mirror partner instead of reading every drive. RAID 10 survives 1 failure for certain and up to half its drives if each is in a different pair.
Drive makers use decimal units, 1 TB = 10^12 bytes, while Windows divides by 2^40 = 1,099,511,627,776 bytes and still labels the result TB. 8 × 10^12 ÷ 2^40 = 7.28, so no space is missing; the gap is about 9% at terabyte scale. IEC 80000-13 names the binary unit the tebibyte (TiB). Macs have reported decimal units since OS X Snow Leopard.
No. RAID keeps an array running through drive failures, but deletions, ransomware encryption and file-system corruption reach every drive at once, and RAID 0 has no redundancy at all. Keep separate backups: CISA recommends the 3-2-1 rule, which means 3 copies of important files, on 2 different media types, with 1 copy stored off-site.
정확도는 입력값과 계산 방법의 가정에 따라 달라집니다. 십진 연산은 유효숫자 50자리를 사용하지만, 추정값·수치해석 방법·원본 데이터의 정밀도는 더 낮을 수 있습니다. 표시값을 반올림해도 이러한 한계는 사라지지 않습니다. 독립적인 출처의 풀이와 대조한 계산 예시: 6. 예를 들어 “RAID 5 with 4 × 8 TB”은 Python 3.8 decimal: (4 − 1) × 8 = 24 TB; 24e12 / 2**40 = 21.8279 TiB; 8e12 / 2**40 = 7.2760 TiB와 대조해 확인합니다.
SNIA Common RAID Disk Data Format (DDF) Specification, rev. 2.0 — RAID level definitions; Patterson, Gibson & Katz (1988), A Case for Redundant Arrays of Inexpensive Disks (RAID), SIGMOD; IEC 80000-13:2008 — binary prefixes (1 TiB = 2⁴⁰ bytes).
이 계산기에는 독립적인 출처에서 답을 얻은 계산 예제가 6개 있습니다. 테스트 모음에서 실행되며 여기에서도 실행할 수 있습니다.
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