Probability of a Page being Dirty

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

0.0303 ??

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Plz check my answer,..

Answer 1

Page fault rate = 0.3

Hence page hit rate = 0.7

Memory access time = 10 time units

Page fault service time  = 125 time units

Let probability of page being dirty = p

Given effective access time  =  50 ns

Thus E.M.A.T  = 0.7 * [Memory access time] + 0.3*[p * (400 PFST) + (1-p) * PFST]

 ==> 0.7 * 10 + 0.3 *[400*125*p + 125 - 125p]   =   50

 ==> 7 + 0.3 * [50000p - 125p + 125]  =  50

 ==> 0.3 * [49875p + 125]  = 43

 ==> p   =   0.0004 [correct upto 4 decimal places]

NOTE : I have used the expression given in Galvin for effective time calculation..and rest 400 PFST according to the question in case the page is dirty..

Comments

Won't there be 2 memory access one for page table and the other for memory?

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Then how we know the TLB hit rate? Unless that is given, we can assume memory access time includes page table access time also. Seeing standard book examples make this clear.

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Why are we having 

400*125*p + 125 - 125p

In case of dirty it takes ,then PFST=400 else it is 125.It does not say in case of dirty it is 400 times the normal PFST

Answer 2

Please check the answer and apprach to solve the question: