Question

A cache memory that has a hit rate of 0.8 has an access latency of 10 ns and a miss penalty of 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 \(\underline{\hspace{1cm}}\).

Hint:

To maintain the same or better average memory access time after optimization, adjust the hit rate to compensate for the increase in hit time while keeping the miss penalty constant.

Answer 1

Correct Answer: 0.85

The average memory access time (AMAT) before the optimization is given by: \[ AMAT_{\text{before}} = (H \times \text{Hit Time}) + (1 - H) \times \text{Miss Penalty}, \] where \( H = 0.8 \) is the initial hit rate, the hit time is 10 ns, and the miss penalty is 100 ns. Therefore: \[ AMAT_{\text{before}} = (0.8 \times 10) + (0.2 \times 100) = 8 + 20 = 28 \, \text{ns}. \] After the optimization, the hit time becomes 15 ns and the miss penalty remains 100 ns. Let the new hit rate be \( H' \). The AMAT after optimization is: \[ AMAT_{\text{after}} = (H' \times 15) + (1 - H') \times 100. \] To ensure that the average memory access time does not increase: \[ AMAT_{\text{after}} \leq AMAT_{\text{before}}. \] Substituting the values: \[ (15H' + 100 - 100H') \leq 28, \] \[ 15H' + 100 - 100H' \leq 28, \] \[ -85H' \leq -72, \] \[ H' \geq \frac{72}{85} \approx 0.85. \] Thus, the minimum hit rate needed is: \[ \boxed{0.85}. \]