GATE CSE 2001 | Question: 20

Question

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.

• A. 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?
• B. 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?

Comments

@Akash Kanase @Bikram

why we are not trvl only given track 

like 25 -- 32--45--99--10--7--5

please clear 

---

@mohan123

We are moving on all the given tracks:

$25\overset{7}{\rightarrow}32\overset{13}{\rightarrow}45\overset{54}{\rightarrow}99\overset{89}{\rightarrow}10\overset{3}{\rightarrow}7\overset{2}{\rightarrow}5=168$

---

In the previous year exam paper this question is an mcq without any options @Counsellor

link

---

For anyone stuck for B

Do you seek random tracks in SCAN/Elevator algorithm?

Answer -No ( that is why average seek not taken ), all tracks are accessed one by one from initial point to one end .

Here initial point is track 0 , As static list of request is served so only one seek from one end to another end is enough.

We took average rotational latency as the starting point of our required block may be anywhere on track.

We did not took transfer time in this question because we don't know how much sectors we have to serve in our current track.

 

---

Rotational latency is included because even if the head reaches the correct track, the requested sector may not be immediately under the read/write head. So in worst case we will have to do this exercise for each track.

---

@Sun Wukong Thanks for the clarification about transfer time. it is a very useful comment.

---

https://gateoverflow.in/761/gate-cse-2001-question-20?show=149955#c149955
Why avg rotational latency over 1 revol time

Answer 1

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, average 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 $= 99  \times 0.2 + 10 \times 100  = 1019.8 \ msec = 1.019 \ sec$

Comments

@camelCase.ai

As we don't know the amount of data in each sector, we cannot calculate time spent in data transfer

---

@Krish_Vg

Part A asks only for the total seek time. Not the total time spent servicing the requests (like Part B did) 

---

data transfer time will be also added because time ti take transfer equal to ek sector ko transfer me kitna time laga to rpm given hai and no of sector in track given hai to 20 mili sec data  transfer rate nikalaayega

 

---

@KESHAV_M

Yes bro, we can calculate the data transfer time for 1 sector.

Time to complete 1 rotation = 20 msec

Number of sectors = 100

Data transfer time for 1 sector = 20msec/100 = 0.2 msec

But, the question doesn't mention that each disk request involves only 1 sector.

So, we cannot tell how many sectors, each disk request entails

Answer 2

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

Comments

You missed rotational latency

---

Akash Kanase

why rotational latency is needed here?

---

Because the question is asking for worst case.

Answer 3

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...