40 votes 40 votes Consider a disk with the $100$ tracks numbered from $0$ to $99$ rotating at $3000$ rpm. The number of sectors per track is $100$ and the time to move the head between two successive tracks is $0.2$ millisecond. Consider a set of disk requests to read data from tracks $32, 7, 45, 5$ and $10$. Assuming that the elevator algorithm is used to schedule disk requests, and the head is initially at track $25$ moving up (towards larger track numbers), what is the total seek time for servicing the requests? Consider an initial set of $100$ arbitrary disk requests and assume that no new disk requests arrive while servicing these requests. If the head is initially at track $0$ and the elevator algorithm is used to schedule disk requests, what is the worse case time to complete all the requests? Operating System gatecse-2001 operating-system disk normal descriptive + – Kathleen asked Sep 14, 2014 • edited Jun 21, 2018 by kenzou Kathleen 11.0k views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply mohan123 commented Oct 19, 2019 reply Follow Share @Akash Kanase @Bikram why we are not trvl only given track like 25 -- 32--45--99--10--7--5 please clear 0 votes 0 votes KUSHAGRA गुप्ता commented Nov 30, 2019 reply Follow Share @mohan123 We are moving on all the given tracks: $25\overset{7}{\rightarrow}32\overset{13}{\rightarrow}45\overset{54}{\rightarrow}99\overset{89}{\rightarrow}10\overset{3}{\rightarrow}7\overset{2}{\rightarrow}5=168$ 1 votes 1 votes JAINchiNMay commented Jul 20, 2022 reply Follow Share In the previous year exam paper this question is an mcq without any options @Counsellorlink 1 votes 1 votes Please log in or register to add a comment.
Best answer 42 votes 42 votes Answer for (A): We are using SCAN - Elevator algorithms. We will need to go from $25\rightarrow 99 \rightarrow 5$. (As we will move up all the way to $99$,servicing all request, then come back to $5$.) So, total seeks $= 74+94= 168$ Total time $= 168 \times 0.2 = 33.60000$ Answer for (B): We need to consider rotational latency too $\rightarrow $ $3000$ rpm I.e. $50$ rps $1 \ r = 1000 /50 \ msec = 20 \ msec$ So, rotational latency $= 20/2 = 10 \ msec$ per access. In worst case we need to go from tracks $0-99.$ I.e. $99$ seeks Total time $= 99 \times 0.2 + 10 \times 100 = 1019.8 \ msec = 1.019 \ sec$ Akash Kanase answered Nov 22, 2015 • edited Jun 26, 2018 by Milicevic3306 Akash Kanase comment Share Follow See all 52 Comments See all 52 52 Comments reply Show 49 previous comments samarpita commented Nov 23, 2022 i edited by samarpita Nov 24, 2022 reply Follow Share @Arjun @Deepak Poonia @Sachin Mittal 1 sirif FCFS has been asked:Consider a set of disk requests to read data from tracks 1,99,1,99,1,99,……….1,99.So, total seek time will be (50 * 99 * 0.2)ms Total avg latency is (10 ms * 100) So total time will be (seek time + avg latency) = 1990 ms 0 votes 0 votes Deepak Poonia commented Nov 23, 2022 reply Follow Share @samarpita Elevator algorithm is used, Not FCFS. Using Elevator algorithm, you can service ALL requrest in a Single Traversal from Track 0 to 99. 0 votes 0 votes samarpita commented Nov 24, 2022 reply Follow Share @Deepak Poonia yes sir...i don’t know what i was thinking on that time. 0 votes 0 votes Please log in or register to add a comment.
–2 votes –2 votes a) 5,7,10,32,45 end track: 99 current:25 total seeks=(99-25)+(99-5)=74+94=168 total seek time = 0.2 * 168 ms b) from 0 to 99 worst case time = 0.2 * 99 ms jayendra answered Jan 2, 2015 jayendra comment Share Follow See all 2 Comments See all 2 2 Comments reply Akash Kanase commented Nov 22, 2015 reply Follow Share You missed rotational latency 1 votes 1 votes Pranavpurkar commented Aug 15, 2022 reply Follow Share Akash Kanase why rotational latency is needed here? 0 votes 0 votes Please log in or register to add a comment.
–3 votes –3 votes B. Question is about how much time required to satisfied 100 requests... 1 request time is = Tseek +TrotLatency + data transfer time Datatransfer time = 0(as not given any info) 99 seeks and 100 rotations required in worst case... So 99 * seek time + 100*rotational latency... GateRank1 answered Dec 22, 2015 GateRank1 comment Share Follow See all 0 reply Please log in or register to add a comment.