問69
頻出度★★☆
障害に備えるために，4台のHDDを使い，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1Tバイトのとき，実効データ容量はおよそ何バイトか。
令和6年度 ITパスポート試験 問69
テクノロジ系システム構成要素 › システムの構成
This question's explanation and your answers so far.
解説
問 69
Incorrectア
2T results from subtracting parity equivalent to two disks (N − 2 = 2), which is closer to the idea of RAID6. In RAID5 the parity uses one disk in total, so only one disk is subtracted. The correct value is 3T.
- イ
正解 RAID5 distributes data and parity (redundant information used for error correction) across multiple HDDs, and even if one disk fails the original data can be recovered from the parity. When built with N disks, the effective data capacity is (N − 1) disks, after subtracting the equivalent of one disk used for parity. Here there are 4 disks of 1T each, so the effective data capacity is (4 − 1) × 1T = 3T. Therefore イ is correct.
Incorrectウ
4T is the capacity when all four disks are used for data, with no parity subtracted at all. In RAID5 the parity consumes the equivalent of one disk, so the effective capacity is smaller than 4T.
Incorrectエ
5T even exceeds the total capacity of the four disks that make up the array (4T), which is physically impossible. It comes from confusing the relationship between the number of disks and the parity, while the correct value is 3T.
ここがポイント
- The effective data capacity of RAID5 is calculated as (number of HDDs − 1) × (capacity per disk). Parity consumes the equivalent of one disk in total.
- RAID5 uses parity equivalent to one disk and can recover from a single disk failure. Using parity equivalent to two disks to withstand two disk failures is RAID6.
用語