A computer uses an on-chip cache backed up by main memory (no off-chip cache). A

Question

A computer uses an on-chip cache backed up by main memory (no off-chip cache). Assume the on-chip cache has a hit time of .3 ns and main memory has an access time of 10 ns. The on-chip cache has a hit rate of 93.6%, and main memory has a hit rate of 100%.

a. What is the average memory access time for this machine?  

b. If we were to add an off-chip cache with an access time of 1 ns and a hit rate of 97.7%, what would the average memory access time become?

c. We add to part b virtual memory with a TLB with a hit time of .3 ns hit time and hit rate of 98.1%, a page table in memory with the same hit time as main memory (10 ns) and a 100% hit rate, and a main memory hit rate of 99.9969% with the swapping process taking 500,000 ns. What is the average memory access time for this machine now?

Explanation / Answer

a.) Average Memory access time for any machine with on-chip and main memory is :

(On-chip hit time ) + ((100 - accessRate of on-chip)/100)*(main-memory hit time)

So, the answer would be : 0.3 ns+ (0.064*10ns) = 0.94 ns.

b.) With an off-chip of access time 1 ns and 97.7% hit rate, the average memory access would become :

0.3ns + (0.064 * (1ns + 0.023*10ns)) = 0.3ns + (0.064 * 1.23ns) = 0.3 ns+ 0.07872ns = 0.37872 ns

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