Demand Paging

Question

Consider a demand-paging system with a paging disk that has  average access and transfer time of 20 milliseconds. Addresses are translated through a page table in main memory, with an access time of 1 microsecond per memory access. Thus, each memory reference through the page table takes two accesses. To improve this time, we have added an associative memory that reduces access time to one memory reference if the page-table entry is in the associative memory. Assume that 80 percent of the accesses are in the associative memory and that, of those remaining, 10 percent (or 2 percent of the total) cause page faults. What is the effective memory access time in milliseconds?

Answer 1

EMAT= 80% of Reference from associative memory + 18% form page table + 2% from page fault

         =0.8(1$\mu$s) + 0.18(2$\mu$s) + 0.02(20000$\mu$s + 2ns )

         =401.2 $\mu$ sec or 0.4ms

Comments

yes 0.481 millisec is correct , it is equivalent to 401.2 microsecs

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How do you calculate the page fault service time ?

Please explain with some authentic source if possible.

Thanks

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2 memory access is required for block transfer which needed 2microsecond extra and 20000 microsecond for page fault service which only include transfer time.