#Applied Course

Question

Consider a demand-paging system with a paging disk that has an 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. The effective memory access time is _____ micro-seconds (Round off to 1 decimal place)

Answer 1

Average access and transfer time = 20 ms = 20000 µs

Address Translation time = 1 µs

% of access in the associated memory = 80% = 0.8

% of page faults = 2% = 0.02

[We have to find the effective access time.]

EMAT(Effective Memory Access Time) = Time taken by 80% accesses in the memory + Time taken by 18% accesses not in the memory + Time taken by 2% accesses not in the memory causing page fault 

 = 0.8(1µs) + 0.18(2µs) + 0.02(20000µs + 2µs)

 = 0.8µs + 0.36µs + 400.02µs = 401.2µs

Hence Answer : 401.2µs