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# Transfer Rate and Hard Disks

Posted on 2011-04-28
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I'm not looking for direct answers.

I'm focusing on part c. Does this show you anything about how to find the transfer rate in this situation?
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Question by:JCW2

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Accepted Solution

Since the track size is not specified, assume that the entire file is on the single track, or assume CLV drive, and should be fairly simple. Especially considering this looks like a textbook excerpt.

So record size R is 9+20+20+1+10+35+12+9+4+4+1=125 bytes.

Now we have 19.2 records per block. Unspanned blocking means 19 records are stored, using 2375 bytes. Wasted space 2400 - 2375 = 25 bytes per block

30 000 records are using 1579 blocks at 19 records per block.

Then, 1 msec to read 2400 bytes block. This means transfer rate TR = 2400 bytes/msec. Also gaps would require 600/2400 = 0.25 msec each.

However, this looks not the answer they want in a book because rotational delay, seek time, and file size are still unused. Probably, their definition of the bulk transfer rate (BTR) is required to properly answer the last part (C).

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