Question

A demand paging system takes 100 times 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 access time is 3 times units. Then the value of P is

• A. 0.233
• B. 0.981
• C. 0.194
• D. 0.514

Answer 1

The correct option is C 0.194

Average Access Time $=(1-p)\left(T_{mem}\right)+p\left[p\right(S.T.dirty)+(1-p)$(S.T.not dirty))]
$3=(1-p)\left(1\right)+p(p\times 300)+(1-p)\times 100$
$3=1-p+p^{2}300p100-p^{2}100$
$3=1+99p+200p^2$
$200p^2+99p-2=0$
$p=\frac{b\pm \sqrt{b^2-4ac}}{2a}$
$p\cong 0.0194$