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 Show 2 previous comments 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 Sachin Mittal 1 commented Jan 29, 2017 reply Follow Share That's correct one, bcoz page fault service time includes memory access time. 36 votes 36 votes dattasai commented Apr 4, 2017 reply Follow Share 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 0 votes 0 votes Chhotu commented Apr 23, 2017 reply Follow Share >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 1 votes 1 votes rahul sharma 5 commented Aug 29, 2017 reply Follow Share Why page table access time is not included? 3 votes 3 votes set2018 commented Aug 31, 2017 reply Follow Share page table is accessed from memory only when TLB misses. 1 votes 1 votes set2018 commented Aug 31, 2017 reply Follow Share 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 0 votes 0 votes Ashwin Kulkarni commented Dec 5, 2017 reply Follow Share 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. 6 votes 6 votes 2019_Aspirant commented Dec 1, 2018 reply Follow Share 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? 0 votes 0 votes Lakshman Bhaiya commented Dec 25, 2018 i edited by Lakshman Bhaiya Jan 3, 2021 reply Follow Share 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 $p=\frac{1}{10^{6}},ps=10msec,ma=20nsec$ $EMAT=\frac{1}{10^{6}}\times(10\times10^{-3}sec)+20\times10^{-9}sec$ $EMAT={10^{-6}}\times(10\times10^{-3}sec)+20\times10^{-9}sec$ $EMAT=10\times10^{-9}sec+20\times10^{-9}sec$ $EMAT=(10+20)\times10^{-9}sec$ $EMAT=30\times10^{-9}sec$ $EMAT=30nsec$ 12 votes 12 votes Rishav Kumar Singh commented Dec 26, 2018 reply Follow Share @Lakshman Patel RJIT 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. 4 votes 4 votes RAVIRAJ03 commented Jan 9, 2019 reply Follow Share i too have the same confusion 1 votes 1 votes G Phalkey commented Jan 19, 2019 reply Follow Share yes because memory acess time is in nana seconds but page fault service time is in mili seconds 0 votes 0 votes val_pro20 commented Nov 16, 2019 reply Follow Share given that effective memory access time without page fault = 20 ns. Now, substituting values in the above formula, we get- Effective access time with page fault = 10 ^-6 x { 20 ns + 10 ms } + ( 1 – 10^-6 ) x { 20 ns } = 10^-6 x 10 ms + 20 ns = 10^-5 ms + 20 ns = 10 ns + 20 ns = 30 ns 0 votes 0 votes 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.