Question

Consider a disk with a sector size of 512 bytes, 2000 tracks per surface,50 sectors per track five double sided platters,and average seek time of 10 msec.What is the capacity of a track in byteswhat is the capacity of each surfacewhat is the capacity of the diskhow many cylinders does the disk haveif the disk platters rotate at 5400 rpm(revolutions per minute),what is the maximum rotational delay

Answer 1

51200 is not a valid block size in this case because block size cannot exceed the size of a track, which is 25600 bytes. Consider a disk with a sector size of 512 bytes, 2000 tracks per surface, 50 sectors per track, five double-sided platters, and average seek time of 10 msec

Answer 2

Capacity for 1 platter = Number of track/surface * Number of sector/track * sector size

=2*5*2000*50*512 bytes

=500000Kbytes

• The length of time it takes for the desired sector of a disc drive to rotate under the read-write heads is known as rotational latency.
• The average rotational latency of magnetic media-based drives is commonly calculated using the empirical relation that the average latency in milliseconds for such a drive is equal to one-half the rotational period.
• Since both a drum and a fixed-head disc have heads that are fixed over the data regions, the average access time for each is a half-revolution, while the maximum access time is a complete revolution. Drum access times typically range from 5 to 10 ms.
• Spin rates for hard drives have been designed to range from 1200 RPM to 15K RPM. However, the most typical RPM range today for desktop and laptop PCs is between 5400 and 7200 RPM.

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