平成30年度春期 ITパスポート試験問77

This question's explanation and your answers so far.

問77

頻出度★★☆

4台の HDD を使い, 障害に備えるために, 1台分の容量をパリティ情報の記録に使用する RAID5 を構成する。1台の HDD の容量が500G バイトのとき, 実効データ容量はおよそ何バイトか。

出典　平成30年度春期 ITパスポート試験問77

解説

問 77

Incorrectア
500 GB is merely the capacity of a single HDD. In RAID5, 3 of the 4 drives are usable for data, so counting only one drive's worth underestimates the capacity and is wrong.
Incorrectイ
1 TB corresponds to 2 drives' worth (2 × 500 GB). This comes from mistakenly assuming parity uses 2 drives, but RAID5's parity consumes only one drive's worth in total, so this value is too small and wrong.
ウ
正解 RAID5 spreads data and parity (redundant information used for error correction) across multiple HDDs, so that even if any one drive fails, the original data can be recovered from the remaining drives. Because the parity as a whole consumes exactly one HDD's worth of capacity, the effective data capacity = (number of HDDs − 1) × capacity of one drive. In a 4-drive setup, (4 − 1) = 3 drives' worth is available for data. With the 500 GB per drive assumed in this question, 3 × 500 GB = 1500 GB = about 1.5 TB, so ウ is correct.
Incorrectエ
2 TB is the combined capacity of all 4 drives (4 × 500 GB = 2000 GB); it forgets to subtract the one drive's worth lost to parity. In RAID5, one drive's worth is used for redundant information and cannot hold actual data, so this value is too large and wrong.

ここがポイント

RAID5 effective data capacity = (number of HDDs − 1) × capacity of one drive. Because parity is spread across all disks, one drive's worth of capacity is always consumed by redundant information in total.
Thanks to parity, RAID5 tolerates the failure of one HDD and can recover the data from the remaining disks.

用語

解答問77ウ