Consider a computer system that has cache memory, main memory (RAM) and disk and the operating system uses virtual memory. It takes 2 nsec to access a word from cache, 10 nsec to access a word from the RAM, and 10 ms to access a word from the disk. If the cache hit rate is 95% and main memory hit rate (after a cache miss) is 99%, what is the average time to access a word?
If we use 10,000 words as a sample the system would hit 9,500 words in cache, 495 words in ram and 5 words in disk. 19,000 words * 2 nsec = 19,000 nsec. 500 words (from ram and disk) * 10 nsecs = 5,000 nsecs. 19,000 + 5,000 = 24,000 nsecs total divided by 10,000 words would equal an average seek time of 2.4 nsecs.