Official PYQ · 2015Set 2 · Q24Memory Hierarchy: Cache, Main Memory and Secondary StorageNAT1 mark

GATE CS 2015 Set 2 Q24 — Memory Hierarchy: Cache, Main Memory and Secondary Storage

Assume that for a certain processor, a read request takes

50\:\text{nanoseconds}

on a cache miss and

5\:\text{nanoseconds}

on a cache hit. Suppose while running a program, it was observed that

80\%

of the processor's read requests result in a cache hit. The average read access time in nanoseconds is ______.

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Answer: 14

To find the average read access time ( $T_{\text{avg}}$ ), we use the weighted average of the cache hit and cache miss times based on their respective probabilities.

Given data:

Cache hit time ( $T_{\text{hit}}$ ) = $5\text{ ns}$
Cache miss time ( $T_{\text{miss}}$ ) = $50\text{ ns}$
Hit ratio ( $H$ ) = $80\% = 0.8$
Miss ratio ( $M$ ) = $1 - H = 20\% = 0.2$

Formula:

T_{\text{avg}} = (H \times T_{\text{hit}}) + (M \times T_{\text{miss}})

Calculation:

T_{\text{avg}} = (0.8 \times 5\text{ ns}) + (0.2 \times 50\text{ ns})

T_{\text{avg}} = 4\text{ ns} + 10\text{ ns} = 14\text{ ns}

Thus, the average read access time is 14 nanoseconds.

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Published 10 Oct 2026

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