A system administrator wants to limit I/O for the following LUNs: - LUN A: Size 100 GB, Limit:500 IOPS - LUN B: Size 500 GB, Limit: 1000 1OPS - LUN C: Size 500 GB, Limit: 500 1OPS - LUN D: Size 1000…

Question

Options

• AOne shared absolute limit policy
• BTwo absolute limit policies
• COne absolute policy and one density based limit policy
• DTwo density based limit policies

• A
  6% (2)
• B
  55% (17)
• C
  10% (3)
• D
  29% (9)

Explanation

Option B is correct because the four LUNs can be grouped by their absolute IOPS limits into exactly two groups: a 500 IOPS policy covering LUN A and LUN C, and a 1000 IOPS policy covering LUN B and LUN D. An absolute limit policy caps I/O at a fixed value regardless of LUN size, making this the simplest and most efficient solution here.

Why the distractors fail:

• A (One shared absolute policy) is wrong because the LUNs have two distinct IOPS ceilings (500 and 1000), so no single absolute limit policy can satisfy all four requirements simultaneously.
• C (One absolute + one density-based) is unnecessary complexity - mixing policy types is only warranted when the limits scale with LUN size, which isn't required here.
• D (Two density-based policies) fails because density-based limits are expressed as IOPS-per-GB. Computing the densities (LUN A = 5 IOPS/GB, B = 2 IOPS/GB, C = 1 IOPS/GB, D = 1 IOPS/GB) reveals three distinct densities, so you'd need at least three policies - not two.

Memory tip: Ask yourself "do the limits scale with size?" If yes → density-based. If no → absolute. Then count how many unique limit values exist - that tells you the minimum number of policies needed. Here, two unique absolute values (500 and 1000 IOPS) → two absolute policies.

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