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Demand Paging

A demand paging system, with page table held in the register, takes 5 ms to service a page fault if an empty page is available, or if the page to be replaced is not dirty. It takes 15 ms if the replaced page is dirty. Memory access time is 1μs. Assume we want an effective access time of 2μs and that the page to be replaced is dirty 60% of the time. The approximate maximum acceptable page fault rate to meet this time requirement will be ____________ % (correct to two decimal places).

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Gahan Saraiya @gahan
1 Feb 2018 09:30 pm

\(Effective\ Memory\ Access\ time = {EAT} = {2 \mu s}\)

\(Page\ fault\ Rate = p\)

\(Dirty\ Page\ Rate = PR = 60 \%\)

\( Memory\ Access\ time = MAT = 1 \mu s\)

\(Servicing\ time\ for\ dirty\ page = S\)

\(Page\ fault\ service\ time = P_t\)

\(EAT = {(1-p) \times MA + p \times (PR \times S_d+ (1 - PR) \times P_t)}\)

\(2\mu s = {(1-p) \times 1 \mu s + p \times (0.6 \times 15000 ms+ (1 - 0.6) \times 5000ms)}\)

 

\(2\mu s = {(1-p) \times 1 \mu s + p \times (0.6 \times 15000\mu s + 0.4 \times 5000\mu s)}\)

\(2 = {1-p + p \times (9000 + 2000)}\)

 

\(1 = -p + 11000p\)

 

\(p = {1 \over 10999}\)

 

Hence p is 0.009091735 %

As this calculation I got this answer but in virtualgate OS section it's answer is considered as 0.01% because (1-p) is completely igonered.