Problem 2 Consider a disk with the following specifications: sector size = 1024,
ID: 3728894 • Letter: P
Question
Problem 2
Consider a disk with the following specifications: sector size = 1024, 4000 tracks per surface, 100 sectors per track, 10 double-side platters, average seek time of 10 msec, and the disk platters rotate 7,200 rpm (revolutions per minute).
What is the capacity of a track in KBs (an KB = 1024 bytes)?
What is the capacity of each platter surface?
What is the capacity of the disk?
How many cylinders does the disk have?
What is the maximum rotational delay?
If an entire track of data can be transfer per revolution, what is the transfer rate?
Explanation / Answer
Given that sector size = 1024, 4000 tracks per surface, 100 sectors per track, 10 double-side platters, average seek time of 10 msec, and the disk platters rotate 7,200 rpm (revolutions per minute).
Assumption: sector size = 1024 bytes
What is the capacity of a track in KBs (an KB = 1024 bytes)?
Ans: capacity of a track in bytes = bytes/ track = (bytes/sector )* (sectors/track )= 1024 * 100
= 102400 (bytes)
So capacity of a track in KBs = 100 KB
What is the capacity of each platter surface?
Ans: capacity of each platter surface = bytes/surface =( bytes/track )*( tracks/surface)
= 1024 *100 * 4000 = 400000 KB
What is the capacity of the disk?
Ans: capacity of disk = bytes/disk = (bytes/surface )*( surfaces/disk) = 400000 *1024 * 10 *2 = 8000000 KB
Explanation: surface number of the disk = 10*2 as 10 double-side platters present in the given disk;
How many cylinders does the disk have?
Ans: The number of cylinders in the disk is same as the number of tracks on each surface, here it is 4000; so the answer is 4000.
What is the maximum rotational delay?
Ans: Given that the disk platters rotate at 7,200 rpm, the time required for a rotation is 1/7200 m or 60/7200 sec or 1/120 sec; this is the maximum rotational delay. So the answer is :
0.0083 sec is the maximum rotational delay .
If an entire track of data can be transfer per revolution, what is the transfer rate?
Ans: The capacity of a track is 100KB. Now one track of data can be transferred per revolution, one rotation takes 0.0083 sec so the data transfer rate is = 100/0.0083 KB/sec = 12000 KB/sec
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