Consider a demand-paging system with a paging disk. The addresses are translated

Question

Consider a demand-paging system with a paging disk. The addresses are translated through a page table in main memory, with an access time of 2 us per memory access. Thus, each memory reference through the page table takes two accesses. To improve this time, we have added an associative memory (TLB). The TLB lookup consumes 50 ns.

It takes 20 ms to service a page fault if an empty page is available or the replaced page is not modified, and 50 ms if the replaced page is modified.

Assume that 70% of the accesses are in the associative memory, and that, of the remaining, 20% (or 6% of the total) cause page faults. Also assume that the page to be replaced is modified 60% of the time.

What is the effective memory access time?

What is the best case memory access time?

What is the worst case memory access time?

Explanation / Answer

We all know that the associative memory has the page in it, so just one access would do if found. Else, search in the page table. If found, it would take two accesses . If not found, need to access it from disk. So,

effective access time EAT = (0.7 x 0.002ms) + (0.24 x (0.002ms + 20ms)) + (0.06 x (0.002 + 20 + 50))

= (0.001) + (4.801) + (3.00)
= 7.80212

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