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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.

  1. 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?
  2. 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?
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3 Answers

Best answer
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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 $= 99  \times 0.2 + 10 \times 100  = 1019.8 \ msec = 1.019 \ sec$

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–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
–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...
Answer:

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