In a memory subsystem it takes 43 nanoseconds to access the main memory and 27 m

Question

In a memory subsystem it takes 43 nanoseconds to access the main memory and 27 milliseconds to access a hard disk drive which contains the page file. What is the Effective Access Time (EAT) of a memory access if the page fault rate is: 100% 4.13% 2.03% 0% A new design of the memory subsystem described in Question 1 above, implements a Translation Look-aside Buffer (TLB). Assuming that the TLB is stored in the cache with an access time of 3 nanoseconds, what is the EAT of a memory access if the page fault rate is 3.26%?

Explanation / Answer

ANSWER:

EAT = [(page fault rate * hard drive access time) + ((100 - page fault rate) * memory access time)]/100

For the first case,(a) EAT = 27ms

(b) EAT = 0.0413*27m + 0.9587*43n = 1.1151m + 0.041u ~ 1.1151ms

Similarly, (c) and (d) could be found out.

2.EAT = TLB access time + [(page fault rate * (hard drive access time) + ((100 - page fault rate) * memory access time)]/100

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