then time required to read 500KB of data in milliseconds is _______.

Question

A hard disk needs 50ms for placing the head on the right track. One track rotational speed is 5ms. There are 63 sectors on each track, each sector storing 512 bytes and data is supposed to be distributed in the best possible way. If transfer time is 0.5 ms/sector then time required to read 500KB of data in milliseconds is _______.

Comments

53 s ?

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no..answer is 1379.36 ms

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A single track can accomdate only 31.5 kb of data. As the file size is 500 kb we ll require 15 tracks and 55 sectors. The question also includes the statement to assume that data is distributed in best possible way. So must be in adjacent tracks. But still I think seek time(even if its adjacent rw head has to be moved to it) and rotational latencies for each of the 16 tracks has to be taken into account.

So I think the answer is = 16*seek time + 15.87*rotational latency + transfer time

bandwidth = 0.5 ms / sector = 1000KBPS and hence transfer time = 500ms

so ans is = 1379.36ms

Answer 1

First check no. of sectors neded to store data = 500KB/512B = 500 * 1024 / 512 = 1000 sectors

These sector will be stored in 1000/63 = 15.87 tracks

This mean that we need total 16 tracks to store this amount of data.

Thus total seek time is 50*16 = 800ms

Total rotational time = 15.87 * 5 = 79.36ms (note here we use exact 15.87 rather than 16 because rotational time is dependent on sectors visited that is 1000/63)

Now transfer time = (0.5 / 512) * 500 * 1024 = 500ms

Total time required = 800 + 500 + 79.36 = 1379.36ms

Comments

what is adjacent seek time?

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Time required to visit one track to its next track is adjacent seek time.

Average seek time is basically calculated from adjacent seek time based on the probability distribution. Simply it is expected no. of head movements in one disk access * adjacent seek time

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@Mandeep. We should be ignoring the adjacent seek time. I am doubtful about this

Answer 2

A single track can accomdate only 31.5 kb of data. As the file size is 500 kb we ll require 15 tracks and 55 sectors. The question also includes the statement to assume that data is distributed in best possible way. So must be in adjacent tracks. But still I think seek time(even if its adjacent rw head has to be moved to it) and rotational latencies for each of the 16 tracks has to be taken into account.

So I think the answer is = 16*seek time + 15.87*rotational latency + transfer time

bandwidth = 0.5 ms / sector = 1000KBPS and hence transfer time = 500ms

so ans is = 1379.36ms