Discussion
The access time of a word in 4 MB main memory is 100 ms. The access time of a word in a 32 kb data cache memory is 10 ns. The average data cache bit ratio is 0.95. The efficiency of memory access time is
Answer: B
The answer is 14.5 ns, the average memory access time.
With a cache hit ratio \(h\), a cache access time \(T_{c}\) and a main memory access time \(T_{m}\), the average access time is \(hT_{c} + (1 - h)T_{m}\). The paper prints the main memory time as 100 ms and calls the hit ratio a 'bit ratio'; all the options are in nanoseconds, and the keyed value is obtained with a main memory time of 100 ns:
\[ \begin{aligned} t &= 0.95 \times 10 + 0.05 \times 100 \\ &= 9.5 + 5 = 14.5\ \text{ns} \end{aligned} \]
9.5 ns counts only the cache hits and forgets the misses, and 95 ns and 20 ns do not follow from the formula.
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