Virtual Gate Test Series: Operating System - Demand Paging

Question

A demand paging system,with page table held in integer,takes $5ms$ to service a page fault if an empty page is available,or if the page to be replaced  is not dirty.It takes $15ms$ if the replaced page is dirty. Memory access time is $1\mu s.$ Assume we want an effective access time of $2\mu s$ and that the page to be  replaced is dirty $60\%$ of the time. The approximate maximum acceptable  page fault rate to meet this time requirement will be______$\%?$
$\text{(Correct to two decimal places).}$

Comments

I don't know your doubt. But the solution looks fine. They have neglected the (1-p) term. Also, when your page fault rate come over 90 percent, then it means you have done something horribly wrong.

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What is correct?..can we ignore 1-p term here n why??

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You can ignore the (1-p) term here, since p will be very small. Consider the the equation without ignoring the (1-p) term, we can have $p=\frac{1}{10999}$ and this turn out to be $p\approx 0.0001$

Thus, $1 - p = 0.9999$

Since, the answer is required up to two decimal places, we can safely ignore the (1-p) term. After ignoring the term, they solved it and got $p\approx 0.01$

Answer 1

2 = (1 - p)*1 + p * [ (0.4 * 5000 ) + ( 0.6 * 15000 )]

2 = 1 - p + p [ 2000 + 9000 ]

2 -1 = -p + p * 11000

1 = p [11000 - 1 ]

1 = p * 10999

p = ( 1 / 10999)  * 100 %

p = 0.00909

Therefore p  ≈ 0.01

Comments

thanx bro

Answer 2

2*10^-6=.6((1-p)*10^-6+p*15*10^-3)+.4((1-p)*10^-6 +p*5*10^-3) this is equation for the given problem.here i converts all the units into second.for understanding purpose first we for solve 60% dirty then for remaining pages then combinly it will give effective memory access time. after solving equation p=1/10999 = 0.0000909 = 0.00909 %

Comments

answer given by @aashish s is also correct . he follows the standerd way of solving memory question. but i want to show u how it can be understood easily.