Disks

Question

Consider a magnetic disk containing 400 sectors per track. An application needs to read 100 randomly scattered sectors for loading the libraries at start up. Reading each sector requires exactly one disk access. The average seek time of the disk to a random location is given as 10ms. Rotational speed of the disk is 6000 rpm. How long does it take to load libraries at start up to run the application?

Comments

Here sequentially implies that their sector no is in sequence but the size of the file is more than no of sectors on a particular track.

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ok  :)

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yes, i mean to s ay what saxena said.

Answer 1

We have 6000 rpm

     6000 rounds -> 60 s

=> 1/2 round -> 60/(6000*2) s

= 5 ms   <- This is the average rotational delay.

Now, we have 400 sectors per track.

     6000 * 400 sectors -> 60 s

=> 1 sector -> 60/(6000 * 400) s

= 1/40000 s

= 0.025 ms

Total time to access a random sector = avg seek time + avg rotational delay + time to read from 1 sector

= 10 + 5 + 0.025 = 15.025 ms

We have to read 100 randomly located sectors,

so total time = 100 * 15.025 ms

= 1502.5 ms

Since the sectors are random, we have taken the average time to access 1 sector and multiplied it by the number of sectors to read.

Answer 2

seek time = 10 ms

average rotational latency = (60*1000)/(6000*2)ms = 5 ms

Since,there are 400 sector per track and the disk cover one track in one rotation.

So,time for reading 400 sector is 10 ms i.e. rotational latency

Hence,time for reading one sector = 10/400 = 0,025 ms

Access time for one sector = 10 + 5 + 0.025 ms = 15.025 ms

Access time for 100 sectors = 100 * 15.025 ms = 1502.5 ms i.e. 1.5 sec(approx