Question 9394 – Cache
November 12, 2023Question 10934 – Cache
November 12, 2023Question 16133 – Cache
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 _____________.
Correct Answer: A
Question 16 Explanation:
The initial values are:
h1 = 0.8
t1 = 10 ns
t2 = 100ns
After the optimization the average memory access time remains the same,
Let the new hit rate be h2. The new cache access time is 15ns.
Applying the average access time formula h1t1+(1-h1)(t1+t2)
0.8*10+0.2*(10+100) = h2*15+(1-h2)(15+100)
30 = 15 + (1-h2)*100
1-h2 = 0.15
h2 = 0.85
h1 = 0.8
t1 = 10 ns
t2 = 100ns
After the optimization the average memory access time remains the same,
Let the new hit rate be h2. The new cache access time is 15ns.
Applying the average access time formula h1t1+(1-h1)(t1+t2)
0.8*10+0.2*(10+100) = h2*15+(1-h2)(15+100)
30 = 15 + (1-h2)*100
1-h2 = 0.15
h2 = 0.85
0.85
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