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 10.8k 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.