Test by Bikram | Operating Systems | Test 2 | Question: 13

Question

The virtual memory system uses the demand paging for its implementation. The probability of getting page faults is $0.25$, the normal memory access time is $200$ nanoseconds. If it takes $2$ millseconds to service a page fault, then what is effective memory access time?

• A. $500000$ ns
• B. $500075$ ns
• C. $500150$ ns
• D. $500250$ ns

Answer 1

E.M.A.T. = (1 - p) * ma + p * page fault service time     ,   where p = page fault rate
EMAT = [ 0.75 * 200  + 0.25 * 2000000 ]ns =  500150 n.s

Comments

@ hem chandra joshi 

yes,  we consider memory access time .

in this formula Effective Memory Access Time 

= (1 - p) * ma + p * page fault service time  

we use memory access time as ma .

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Sir,

We should use hierarchical  approach

as page fault service time to move page from disk to memory and update page table,

after that again memory reference takes place.

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For those who are confused about which formula to use.

https://gateoverflow.in/41917/confusion-in-effective-memory-access