Assume a demand-paged memory system with an average memory-access time of 100 ns

Question

Assume a demand-paged memory system with an average memory-access time of 100 ns (including the time required to look up a frame number in the page table). It takes 8 ms to service a page fault if an empty frame is available or if the replaced page is not modified, or 16 ms if the replaced page must be written back to disk, which is the case 10% of the time. What is the effective access time (in nanoseconds) if the page fault rate is 0.000 01? Show your work. Give the formula for calculating the effective access time (E.A.T.), then calculate the E.A.T. for this case.

Explanation / Answer

See, Let p be the page fault occurance probability, so the probab that page is found in Main Memory becomes(1-p).

When a page is not found in Main Memory it is fetched from the disk => page fault service. (page fault service time.)

Thus,
Effective Memory Access Time = (pagefault occurance probability) * (page fault service time) + ( no page fault Probability) * (Memory access time)
=> E.A.T. = p * (page fault service time) + (1 - p) * (Memory access time)

Now you could subsitute values for finding the answer.

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