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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]
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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..

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Harikesh Kumar asked Feb 3, 2018
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There is any difference or same these two statement?1. Page fault service time is 10ms.2. The Time to service a page fault is on average 10ms.