Made easy test series

Question

A demand paging system has page fault service time as 125 time units if page is not dirty and 400 times units of page fault service time if it is a dirty page. Memory access time is 10 time units. The probability of a page fault is 0.3. In case of page fault, the probability of page being dirty is P. It is observed that average access time is 50 time units. Then, the value of P is ______? [upto four decimal places]

Comments

I am getting 0.0303.....but the given ans is 0.0667

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Effective access time = (1-p) (memory access time) + p[ (q * required time) + ((1-q) * required time)]

here p = page fault probability and q= probability of dirty page.

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take a look at it....

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@akash
Emat=p(pfst+m)+(1-p)(m)
This would give us p(pfst)+m
why we are not considering it like this

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Page fault service time also include memory access time so insteed of p(pfst+m) we use p (pfst).

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@Anu007 any references.....as many questions consider it and many don't..

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Reffer galvin.

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I

In this memory access time is not taken that is considered in pfst itself

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can someone explain dirty bit concept in this question

Answer 1

equation will be

50 = .7*10 + .3*125*(1-p) + .3*400*p

gives 0.0667

Comments

t = (page fault dosenot occure * mem access time) + page fault occure((probability of page being dirty* service time when page dirty) + (probability of page not being dirty * service time when page not dirty))

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@sunaina

some solutions use this way like u said and some solutions like akash said

even i am confused when to use which one :(

can someone clear out the difference betwenen both approaches

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actually memory acess time is in nano seconds and page fault service time is in miliseconds .memory acess time is compared negligible(though it shouldn't)  so it some authors ignore memory acess time during page fault

Answer 2

Refer this 

https://gateoverflow.in/177643/made-easy-test-series