GATE CSE 2018 | Question: 10

Question

Consider a process executing on an operating system that uses demand paging. The average time for a memory access in the system is $M$ units if the corresponding memory page is available in memory, and $D$ units if the memory access causes a page fault. It has been experimentally measured that the average time taken for a memory access in the process is $X$ units.

Which one of the following is the correct expression for the page fault rate experienced by the process.

• A. $(D-M) / X-M)$
• B. $(X-M) / D-M)$
• C. $(D-X) / D-M)$
• D. $(X-M) / D-X)$

Comments

Here, Page Fault Service Time encompasses main memory access time!!

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That's always the case as per standard books.

Answer 1

Let $P$ be the page fault rate.

Average memory access time $= ( 1 -$ page fault rate$) \times$ memory access time when no page fault  $+$  Page fault rate  $\times$ Memory access time when page fault.

$X = (1-P) M + P \ D$

$X = M + P( D - M)$

$P = (X - M) / ( D - M)$

(B) is the answer.

Comments

i think D includes(page fault service time and memory acess).,they did n't explicitly mention about page fault service time.

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@Hemant Parihar

@chirudeepnamini

@Satbir

Why can't I write it as 

$\mathbf{EMAT = p(page\;fault\;service\;time)+(1-p)(Main \;Memory\; Access\; time)}$

With this formula I am getting $\mathbf{C}$ as the answer.

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Okay, got it then I have to consider $\mathbf{page\;fault}$ as the $\mathbf{1-p}$.

Answer 2

The average memory access time is m when no page fault and d when page fault

X = (1-p)m + p(d)

 

upon solving

 

$\frac{x-m}{d-m}$ is the answer

Answer 3

if not page fault rate = p

so page fault rate =1-p

then  X= P* M + (1-P)*D
  P=X-D/M-D
1-P = 1 – (X-D/M-D ) =M-X/M-D = X-M/D-M = page fault rate