GATE | Gate IT 2007 | Question 56

A demand paging system takes 100 time units to service a page fault and 300 time units to replace a dirty page. Memory access time is 1 time unit. The probability of a page fault is p. In case of a page fault, the probability of page being dirty is also p. It is observed that the average access time is 3 time units. Then the value of p is
(A) 0.194
(B) 0.233
(C) 0.514
(D) 0.981


Answer: (A)

Explanation: Page fault service time =100
page fault service time =300 (if there is dirty page)

probability of page fault =p

probability of being dirty is =p

so total page fault service time(ps) = p(300)+ (1-p)(100) = 200p+100

now given EAT =3 so
EAT = p(page fault service time +memory access time)+(1-p) memory access time

3= p*page fault service time + memory access time
3 =p(200p+100)+1= 200p²+100p+1
p=0.0194


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