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I have read that EMAT when the page fault occurs is =p(PS+MA)+(1-p)(MA)

p is the probability of page fault

i-p is the probability that page is present in Main Memory

PS is the Page fault service time

MA is the memory access time

but according to me first it should check that page is Present in main memory by Page table entry which is present in MM which is going to take one Memory access

So,EMAT should be equals=p(MA+PS+MA)+(1-p)(MA+MA).

why it is not so??
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Actually it is depends on implementation that we are using.(see Galvin book 9E PAGE NO 374  for better explanation )

 However ;

EMAT = p * (page table access + page fault service ) + (1-p) (page table access + access frame or byte in memory)

Here we are not using memory access two times in page fault occuring. Bcoz in this scenario page  table says the page that you are refered not in memory so why you put memory access two times okk.
After accessing the page table if page is present in memory then one memory access is also required to access the word.So total two memory accesses should required.
Ya sure ... i also saying this things in above. But you use only two memory access when page is present in memory that is page hit  . Occuring of page fault we are not going to be use two memory access.

In case of Page Fault,First it access the memory for page table and found that required page is not present in memory by present or absent bit and do page service.After Page Service ,page is loaded into Memory  ,So it again take one memory access to access the word or required page.So total Two Memory access Should required in case of Page Fault.

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