ITPEC FE Morning October 2021 Question 12
In memory systems, when data is detected in the cache, it is known as a hit. When data is not detected in the cache, it is read from the main memory and is known as a miss. Which of the following is the approximate hit ratio for the specifications in the table below? Here, effective access time is calculated as the weighted average of hit time and miss time, and other overhead is ignored.
Explanation
0.81 — set the effective access time equal to the weighted average of the hit and miss times and solve for the hit ratio h.
- •250 = 100h + 900(1 − h)
- •250 = 900 − 800h
- •800h = 650, so h = 0.8125, about 0.81
Check: 0.81 × 100 + 0.19 × 900 = 81 + 171 = 252 ns, close to 250 ns.
Answer (d)
Why not others:
- •(a) 0.25 and (b) 0.35 — a hit ratio below 0.5 puts the average nearer the 900 ns memory time than the 100 ns cache time: they give 700 ns and 620 ns
- •(c) 0.73 — gives 73 + 243 = 316 ns, not 250
Key rule: Effective access time = h × tc + (1 − h) × tm; solve it as a linear equation in h, then substitute back to check.
Computer Component · AI-written, may contain errors
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