GATE CSE 2000 | Question: 2.22

Question

Suppose the time to service a page fault is on the average $10$ milliseconds, while a memory access takes $1$ microsecond. Then a $99.99\%$ hit ratio results in average memory access time of

• A. $1.9999$ milliseconds
• B. $1$ millisecond
• C. $9.999$ microseconds
• D. $1.9999$ microseconds 

Comments

@꧁༒☬ĿọŗԀ 🆂🅷🅸🆅🅰☬༒꧂ Page fault time is not same page fault service time. Page fault service time includes memory access time while page fault time doesn't.

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@https_guru i m not able to visulaize this what do u mean  be PAge fault time ? service time ik that we've to load it from the disk 

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@꧁༒☬ĿọŗԀ 🆂🅷🅸🆅🅰☬༒꧂ 

Page fault time = Time to load faulted page into main memory

Page fault service time = Page fault time + Time to access data from MM

Answer 1

Since nothing is told about page tables, we can assume page table access time is included in memory access time.

So, average memory access time

$= .9999 \times 1 + 0.0001 \times 10,000$
$= 0.9999 + 1$
$= 1.9999$ microseconds

Correct Answer: $D$

Comments

why 1000*10?

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They have converted everything in terms of microseconds. 1 microsecond = 10^3 millisec. And it's given in question that s=10 millisec so if we convert it to microseconds, it's gonna be 10 *10^3 microseconds.

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@Arjun sir but tlb contains some mapping translation after refering from here when we go to memory then there is chnce tht we may get page fault then ???

Answer 2

p=miss rate, 1−p=hit rate, ma=memory access time, ps=page fault time (or) service time

EMAT = p(ps+ma)+(1−p)×ma = p(ps)+p(ma)+ma−p(ma) = p(ps)+ma

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Given Values:  p = $\frac{0.001}{100}$ = $10^{-4}$,  ps = 10×$10^{-3}$ sec,  ma=$10^{-6}$ sec

EMAT = $10^{-4}$x(10×$10^{-3}$)+$10^{-6}$ sec

EMAT = $10^{-4}$×($10^{-2}$)+$10^{-6}$ sec

EMAT = $10^{-6}$+$10^{-6}$ sec

EMAT = 1+1 μsec

EMAT = 2 μsec
 

So correct answer is D

Comments

 Please correct if i am mistaken ,

 

Page Fault service time itself includes accessing specific page from the process and then loading in the main memory and updating the page table , so explicitly we should not consider memory access time though in this case we are almost close to the right answer. The right formula

p-miss rate
EMAT = p(ps)+(1−p)×ma 

PS- If its memory hit , we check the page table ,get access to the frame no and go to the respective frame in main memory ,since page table accesss time is not given in ques ,it is assumed "Ma" ( memory access time ) includes both the access time

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Yeah, you are right. In galvin also they have said the same thing and in the question below also they are using the same concept.
https://gateoverflow.in/3500/gate2007-it-58

Answer 3

CAN ANYONE TELL ME WHERE AM I WRONG

Step 1: memory read to access the page table (1 micro second)

case 1) frame number found -> second memory access to access the desired page from memory (1 microsecond)

case 2) frame number not found -> page fault -> service the page fault (10000 microseconds)

Step 2:   (.9999*(first memory access time + second memory access time)) +

             (.0001*(first memory access time + time to service page fault))

             = (.9999*2)+(.0001*10001)

             =2.9999 micro seconds

Comments

You are not wrong. But question doesn't say anything about page tables so we can assume nearly 100% TLB hit or that page table access time is also counted as part of memory access time given. (If 2 level paging is used we will need 2 page table accesses, so we can't assume a memory access for a page table access every time)

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Thanks !!!

Answer 4

option d

Comments

thanks bro , I had same thought during this question but little bit unit mistake....

Answer 5

Average memory access time = (P*t1) + [(1-P)t2]
= (0.9999*1) + [(1-0.9999) *10000]
= (0.9999) + (0.0001 * 10000)
= 0.9999 + 1
= 1.9999 microseconds