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Let the page fault service time be $10$ milliseconds(ms) in a computer with average memory access time being $20$ nanoseconds (ns). If one page fault is generated every $10^6$ memory accesses, what is the effective access time for memory?

1. $21$ ns
2. $30$ ns
3. $23$ ns
4. $35$ ns
edited | 7.2k views
+1

http://web.cs.ucla.edu/~ani/classes/cs111.08w/Notes/Lecture%2016.pdf

In this last step is restart the instruction so page fault does not include memory access time.

http://stackoverflow.com/questions/40027359/what-is-page-fault-service-time

Page Fault service time = TLB time + page table time + page fetch from disk

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Why the page table access time is not included here in any of the answers? Does Page Fault Service time include it or  has it been ignored just  because it is negligible?

open slide 12-13 to check :

http://web.cs.ucla.edu/~ani/classes/cs111.08w/Notes/Lecture%2016.pdf

\begin{align*} \text{EMAT} &= \frac{1}{10^6} \times 10\ ms + \left(1-\frac{1}{10^6} \right ) \times 20\ ns \\ &= 29.99998\ ns \\ &\approx 30\ ns \end{align*}

answer = option B

answered by Boss (31.1k points)
selected
+15
That's correct one, bcoz page fault service time includes memory access time.
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can u please tell me how to download the notes from the link u have suggested ( not only that which u have sent but also other material)

very happy wiht ur faster reply
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>That's correct one, bcoz page fault service time includes memory access time.

NO. Please correct me if i am wrong.

Refer - https://en.wikipedia.org/wiki/Page_fault and

http://stackoverflow.com/questions/40027359/what-is-page-fault-service-time

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Why page table access time is not included?
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page table is accessed from memory only when TLB misses.
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in page fault service time we have to add average memory access time,but it is very less compare to pagefault service time .thts why just neglect it
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Actually EMAT = PFrate (PF servicetime + MAT) + (1-PFrate) MAT

But in this question why that first MAT is ignored because it is negligible compared to PF service time.
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I would choose to write $20+10$ because

with average memory access time being 20nanoseconds (ns)

That means $20$ is average memory access. Average in the sense it's overall. Right?

+1

I think

EMAT$=$page fault rate$\times$(Service time or page fault time+memory access time)$+$hit rate$\times$(memory access time)

Here let suppose $p=$page fault rate$,$ps$=$Service time or page fault time$,$ma$=$memory access time

Now $EMAT=p\times(ps+ma)+(1-p)\times(ma)$

$EMAT=p\times(ps)+p\times(ma)+ma-p\times(ma)$

$EMAT=p\times(ps)+ma$

Now put the value $p=\frac{1}{10^{6}},ps=10msec,ma=20nsec$

$EMAT=\frac{1}{10^{6}}\times(10\times10^{-3}sec)+20\times10^{-9}sec$

$EMAT={10^{-6}}\times(10\times10^{-3}sec)+20\times10^{-9}sec$

$EMAT=10\times10^{-9}sec+20\times10^{-9}sec$

$EMAT=(10+20)\times10^{-9}sec$

$EMAT=30\times10^{-9}sec$

$EMAT=30sec$

+1

you are correct but in many questions

EMAT =page fault (page fault service time) + not page fault(mm access time).

some people say page fault service time already includes mm access time.

i am confused about where to use which formula.

If it's numerical type then more or less answer will be same if mm access time is very small compared to page fault service time.

+1
i too have the same confusion
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yes because memory acess time is in nana seconds but page fault service time is in mili seconds

Effective memory access time = Memory access time + page fault rate *page fault service time

so here

$\text{EMAT} = 20 ns + \frac{ 1}{10^6} \times 10 \times 10^6 ns$

$= 20 + 10 =30 ns$

Ans is B.

answered by Active (4.1k points)
0

@ neha pawar u r wrong

Effective memory access time = Memory access time + page fault rate *page fault service time

so here

Page Service = 10 ms

P = 1 / 10^6

M = 20ns

EMAT = (Page service) P + M

EMAT = (10 ms) * (1 / 10^6) + 20 ns

=10ns + 20ns

so, EMAT = 30ns

Ans) Option B.

answered by Active (4.9k points)

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