IPA IP 2024 Question 69
A question IPA Japan published in 令和6年度 ITパスポート試験 公開問題 問69, translated into English. ITPEC is based on the same exams.
In preparation for failures, RAID5 is configured by using four (4) HDDs, and the capacity of one (1) HDD is used for recording parity information. When the capacity of one (1) HDD is 1 Tbyte, approximately how many bytes is the effective data capacity?
Explanation
3 T — in RAID5 one drive’s worth of capacity holds parity, so four 1 Tbyte HDDs give three drives’ worth of data.
RAID5 stripes data across all the drives and spreads the parity blocks among them. The parity takes up the capacity of one drive in total, whatever the number of drives, which is what lets the array rebuild the data of any one failed drive.
- •Raw capacity: 4 × 1 = 4 Tbytes
- •Parity: 1 drive’s worth = 1 Tbyte
- •Effective data capacity: (4 − 1) × 1 = 3 Tbytes
Answer (b)
Why not others:
- •(a) 2 T — two drives’ worth set aside, as in RAID6 (double parity) or mirroring the four drives in pairs (RAID 1+0); the question says one drive’s worth
- •(c) 4 T — the raw total, forgetting the parity
- •(d) 5 T — more than the four drives hold together; no RAID level gives more than the raw capacity
Key rule: RAID5 effective capacity = (number of drives − 1) × capacity of one drive; RAID6 loses two drives, RAID1 half.
System Component · AI-written, may contain errors
This question comes from the past questions IPA Japan publishes, translated into English by ITPEC Practice. ITPEC Practice is an independent study tool and is not affiliated with IPA or ITPEC. See the original paper (PDF) or Report an issue.