41 votes 41 votes 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? $21$ ns $30$ ns $23$ ns $35$ ns Operating System gatecse-2011 operating-system virtual-memory normal ugcnetcse-june2013-paper2 + – go_editor asked Sep 29, 2014 edited Jun 23, 2018 by Pooja Khatri go_editor 26.5k views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Chhotu commented Dec 5, 2017 reply Follow Share 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 https://en.wikipedia.org/wiki/Page_fault 2 votes 2 votes Saurabh666 commented Dec 28, 2017 i edited by Saurabh666 Dec 28, 2017 reply Follow Share 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? 1 votes 1 votes Minal Patil commented Sep 16, 2020 reply Follow Share What is 1 page fault for every 10^6 memory access ,could you please tell? 0 votes 0 votes JAINchiNMay commented Jul 19, 2022 reply Follow Share @Arjun sir does the page service time include memory access time or not 0 votes 0 votes N3314nch41 commented Jun 16, 2023 reply Follow Share Doesn’t avg. memory access time same as EMAT 0 votes 0 votes Please log in or register to add a comment.
Best answer 72 votes 72 votes 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 amarVashishth answered Oct 23, 2015 edited Jun 21, 2021 by Lakshman Bhaiya amarVashishth comment Share Follow See all 16 Comments See all 16 16 Comments reply Show 13 previous comments shivam001 commented Jun 9, 2020 reply Follow Share @amarVashishth whenever access time given , it already include hit rate . so, you need not to use hit time again so, EMAT = (20*$10^{-9}$) + ($\frac{1}{10^{6}}$*10*${10^{-3}}$) EMAT = 20+10 = 30ns 0 votes 0 votes Ritik gupta commented Jul 22, 2020 reply Follow Share why we are not multiplying the MAT with 2 as the memory access is done two times? EMAT=P*S+2*MAT EMAT=(1/10^6)*10ms+(1-1/10^6)*2*20 EMAT=50ns but there is no option for that why? 4 votes 4 votes himanshud2611 commented Oct 4, 2023 reply Follow Share @Ritik gupta that is case when TLB comes into the picture. 0 votes 0 votes Please log in or register to add a comment.
25 votes 25 votes 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. neha pawar answered Oct 28, 2014 neha pawar comment Share Follow See all 3 Comments See all 3 3 Comments reply suvasish pal commented Sep 22, 2017 reply Follow Share @ neha pawar u r wrong Effective memory access time = Memory access time + page fault rate *page fault service time so here 0 votes 0 votes Kuljeet Shan commented Apr 19, 2019 reply Follow Share @neha pawar any valid source of this formula: "Effective memory access time = Memory access time + page fault rate *page fault service time" ? 0 votes 0 votes Kuljeet Shan commented Apr 19, 2019 reply Follow Share 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). 2 votes 2 votes Please log in or register to add a comment.
5 votes 5 votes 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. Prasanna answered Nov 12, 2015 Prasanna comment Share Follow See all 2 Comments See all 2 2 Comments reply Kuljeet Shan commented Apr 19, 2019 reply Follow Share "EMAT = (Page service) P + M" From where u wrote this formula ? It is working in this question only ? or it is some standard ? @Prasanna 0 votes 0 votes Pathki Shivamsh commented Aug 16, 2020 reply Follow Share 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 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes Here the explanation shruti gupta1 answered May 23, 2019 shruti gupta1 comment Share Follow See 1 comment See all 1 1 comment reply Jayprakash Ray commented Oct 11, 2020 reply Follow Share http://www4.comp.polyu.edu.hk/~csajaykr/myhome/teaching/eel602/majors.htm#:~:text=It%20takes%208%20milliseconds%20to,access%20time%20is%20100%20nanoseconds. Refer 1st Question in the link. They have used EMAT=(1-p)Memory Access + p (Memory Access+Page Service Time) p= Page fault Rate 0 votes 0 votes Please log in or register to add a comment.