An application generates 4200 small random I/Os at peak workloads with a read/write ratio of 2:1. What is the disk load at peak activity for a RAID 6 configuration?

Question

Options

• A2800
• B5600
• C8400
• D11200

• A
  19% (5)
• B
  4% (1)
• C
  7% (2)
• D
  70% (19)

Explanation

D (11,200) is correct because RAID 6 carries a write penalty of 6 - each logical write generates 6 physical disk I/Os (read old data + read both parity blocks + write new data + write both new parity blocks). With a 2:1 read/write ratio across 4,200 I/Os, there are 2,800 reads and 1,400 writes. The total disk load is 2,800 + (1,400 × 6) = 11,200.

A (2,800) only accounts for the reads and ignores writes entirely. B (5,600) represents the raw write count (1,400) multiplied by RAID 5's write penalty of 4, dropping the reads - a double mistake. C (8,400) is the correct answer for RAID 5, not RAID 6 - it applies a 4× write penalty: 2,800 + (1,400 × 4) = 8,400; this is the most dangerous distractor.

Memory tip: Think of the parity count: RAID 5 has 1 parity → 4 I/Os per write; RAID 6 has 2 parities → 6 I/Os per write. Each extra parity stripe adds 2 more I/Os (one read + one write). If you see "RAID 6 write penalty," immediately think ×6.

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