GATE CSE 2011 | Question: 20, UGCNET-June2013-II: 48

Question

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?

• A. $21$ ns
• B. $30$ ns
• C. $23$ ns
• D. $35$ ns

Comments

@Arjun sir does the page service time include memory access time or not

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Doesn’t avg. memory access time same as EMAT

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References : Galvin 

Note : This Question is inspired from Galvin book examples 

Formula from Galvin book

Example of Galvin book 

Answer 1

Open slides $12-13$ to check : 

http://web.cs.ucla.edu/~ani/classes/cs111.08w/Notes/Lecture%2016.pdf$$\begin{align*} \text{EMAT} &= \frac{1}{10^6} \times 10\ \text{ms} + \left(1-\frac{1}{10^6} \right ) \times 20\ \text{ns} \\ &= 29.99998\ \text{ns} \\ &\approx 30\ \text{ns} \end{align*}$$Answer = option B

Comments

the link is broken now

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@himanshud2611
For a page fault miss there should ideally be 2 memory accesses
one for accessing the page table from MM(main memory) and second for accessing the content from MM
@Ritik gupta, I also have same doubt...if you have cleared your doubt..kindly clear mine tooo
Thanks in Adv:)

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i think options given is wrong because :
when no page fault : page table access + main memory access = 40ns.

if page fault : page table access + page fault service time = 20ns + 10 ms = approx 10ms.

then : R = (1/10^6)

40ns * (1-R)  + 10ms (R)  = approx 50ns.

Answer 2

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.

Comments

 @ neha pawar u r wrong

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

so here

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@neha pawar

any valid source of this formula: "Effective memory access time = Memory access time + page fault rate *page fault service time" ?

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Actual formula is this:

Effective Access Time (EAT) EAT = (1 – p) x memory access + p (page fault overhead + swap page out + swap page in + restart overhead )

Source: http://web.cs.ucla.edu/~ani/classes/cs111.08w/Notes/Lecture%2016.pdf (page 12 - 13 as already mentioned in above answer).

Answer 3

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.

Comments

"EMAT = (Page service) P + M"

From where u wrote this formula ? It is working in this question only ?

or it is some standard ? @Prasanna

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EMAT= P*(PS + M) + (1-P)*(M)

   Here P=Miss Rate , 1-P= Hit Rate ,PS=Page Fault Service Time, M=Main Memory Access Time

We can derive as 

P*(PS+M)+(1-P)*(M)

P*PS+P*M+M-P*M

Hence EMAT =P*PS+M

 

Answer 4

Here the explanation

Comments

http://www4.comp.polyu.edu.hk/~csajaykr/myhome/teaching/eel602/majors.htm#:~:text=It%20takes%208%20milliseconds%20to,access%20time%20is%20100%20nanoseconds.

Refer 1^st Question in the link. They have used  

EMAT=(1-p)Memory Access + p (Memory Access+Page Service Time) 

p= Page fault Rate

Answer 5

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

now solving it when we convert 10ms in ns it wil be 10*10^6 ns which is huge incomparison to 20ns so neglect 20ns 

and also when are multiplying (1−1/10^6)×20 ns it is approx 20ns again 

Now EMAT = 10*10^6ns * 1/10^6 + 20ns 

So EMAT = 10ns + 20ns = 30ns approx

So option B is right answer 30ns

Answer 6

Ans (30)

Refer this,