A cache memory that has a hit rate of 0.8 has an access latency 10 ns and miss penalty 100 ns. An optimization is done on the cache to reduce the miss rate. However, the optimization results in an increase of cache access latency to 15 ns, whereas the miss penalty is not affected. The minimum hit rate ( rounded off to two decimal places) needed after the optimization such that it should not increase the average memory access time is …

Official answer: 0.85. Explanation: Average Memory Access Time (AMAT) is given by: AMAT = Cache Access Latency + (1 - h) × Miss Penalty Initial Configuration: - Hit rate (h1) = 0.8 - Access latency (T1) = 10 ns - Miss penalty (P) = 100 ns AMAT1 = 10 + (1 - 0.8) × 100 = 10 + 0.2 × 100 = 30 ns Optimized Configuration: - Access latency (T2) = 15 ns - Miss penalty (P) = 100 ns - New hit rate = h2 To ensure the average memory access time does not increase: AMAT2 ≤ AMAT1 15 + (1 - h2) × 100 ≤ 30 (1 - h2) × 100 ≤ 15 1 - h2 ≤ 0.15 h2 ≥ 0.85 Thus, the minimum hit rate required after optimization is 0.85.

Cache Memory and Hierarchy

GATE CSE 2022 · Q33

NAT1 markEasy

Time00:00

A cache memory that has a hit rate of $0.8$ has an access latency $10\text{ ns}$ and miss penalty $100\text{ ns}$ . An optimization is done on the cache to reduce the miss rate. However, the optimization results in an increase of cache access latency to $15\text{ ns}$ , whereas the miss penalty is not affected. The minimum hit rate ( $\textit{rounded off to two decimal places}$ ) needed after the optimization such that it should not increase the average memory access time is ________.

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Official answer: 0.85 (±0)

Cache Memory and Hierarchy

Average Memory Access Time ( $\text{AMAT}$ ) is given by:

\text{AMAT} = \text{Cache Access Latency} + (1 - h) \times \text{Miss Penalty}

Initial Configuration:

Hit rate ( $h_1$ ) = $0.8$
Access latency ( $T_1$ ) = $10\text{ ns}$
Miss penalty ( $P$ ) = $100\text{ ns}$

\text{AMAT}_1 = 10 + (1 - 0.8) \times 100 = 10 + 0.2 \times 100 = 30\text{ ns}

Optimized Configuration:

Access latency ( $T_2$ ) = $15\text{ ns}$
Miss penalty ( $P$ ) = $100\text{ ns}$
New hit rate = $h_2$

To ensure the average memory access time does not increase:

\text{AMAT}_2 \le \text{AMAT}_1

15 + (1 - h_2) \times 100 \le 30

(1 - h_2) \times 100 \le 15

1 - h_2 \le 0.15

h_2 \ge 0.85

Thus, the minimum hit rate required after optimization is 0.85.

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This Cache Memory and Hierarchy previous year question is Q33 of the GATE CSE 2022 paper. Find more GATE CSE PYQs on Cache Memory and Hierarchy in Computer Organization and Architecture, or browse all GATE CSE PYQs.

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Cache Memory and Hierarchy in GATE CSE

Cache Memory and Hierarchy has 47 GATE CSE PYQs across 23 papers from 2011 to 2026. 7 of them are 1-mark NAT questions like this one, and GATE CSE 2022 has 4.

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.