ITPEC FE Subject A April 2025 Question 10
The table below shows the access times for the cache memory and main memory in systems A and B. Which of the following is the approximate hit ratio of the cache memory, when the effective access time and the cache memory hit ratio are the same for systems A and B?
Explanation
0.83 — write each system's effective access time with the same hit ratio h, set them equal and solve.
Effective access time = h × cache time + (1 − h) × main memory time.
- •System A: 14h + 80(1 − h) = 80 − 66h
- •System B: 12h + 90(1 − h) = 90 − 78h
- •Equal: 80 − 66h = 90 − 78h, so 12h = 10 and h = 10 ÷ 12 ≈ 0.83
Check: both systems then take about 25 ns per access.
Answer (c)
Why not others:
- •(a) 0.17 — the miss ratio 1 − h = 2 ÷ 12, not the hit ratio
- •(b) 0.20 — uses a model where every access pays the cache time and a miss adds the main memory time, 14 + 80m = 12 + 90m, which gives a miss ratio m = 0.2; it is neither the question's formula nor a hit ratio
- •(d) 1.00 — if every access hit the cache, the times would be 14 ns and 12 ns, which are not equal
Key rule: Effective access time is the average of the cache and main memory times weighted by the hit and miss ratios; write one equation per system and solve for the shared unknown.
Computer Component · AI-written, may contain errors
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