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Unformatted text preview: 82 Chapter 9 2. How many cylinders does the disk have? 3. Give examples of valid block sizes. Is 256 bytes a valid block size? 2048? 51200? 4. If the disk platters rotate at 5400 rpm (revolutions per minute), what is the maximum rotational delay? 5. If one track of data can be transferred per revolution, what is the transfer rate? Answer 9.5 1. bytes/track = bytes/sector × sectors/track = 512 × 50 = 25 K bytes/surface = bytes/track × tracks/surface = 25 K × 2000 = 50 , 000 K bytes/disk = bytes/surface × surfaces/disk = 50 , 000 K × 5 × 2 = 500 , 000 K 2. The number of cylinders is the same as the number of tracks on each platter, which is 2000. 3. The block size should be a multiple of the sector size. We can see that 256 is not a valid block size while 2048 is. 51200 is not a valid block size in this case because block size cannot exceed the size of a track, which is 25600 bytes. 4. If the disk platters rotate at 5400rpm, the time required for one complete rotation, which is the maximum rotational delay, is 1 5400 × 60 = 0 . 011 seconds . The average rotational delay is half of the rotation time, 0.006 seconds. 5. The capacity of a track is 25K bytes. Since one track of data can be transferred per revolution, the data transfer rate is 25 K . 011 = 2 , 250 Kbytes/second Exercise 9.6 Consider again the disk specifications from Exercise 9.5, and supposeConsider again the disk specifications from Exercise 9....
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- Fall '12
- Rotation, Database management system, Replacements, Seek time, buffer manager