Solutions needed a,b,c and d Consider a disk with the following characteristics
ID: 450845 • Letter: S
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Solutions needed a,b,c and d
Consider a disk with the following characteristics (these are not parameters of any particular disk unit): block size B = 512 bytes; interblock gap size G - 128 bytes; number of blocks per track = 20; number of tracks per surface - 400. A disk pack consists of 15 double-sided disks. What is the total capacity of a track, and what is its useful capacity (excluding interblock gaps)? How many cylinders are there? What are the total capacity and the useful capacity of a cylinder? What are the total capacity and the useful capacity of a disk pack? Suppose that the disk drive rotates the disk pack at a speed of 2,400 rpm (revolutions per minute); what are the transfer rate (tr) in bytes/msec and the block transfer time (btt) in msec? What is the average rotational delay (rd) in msec? What is the bulk transfer rate? (See Appendix B.)Explanation / Answer
a) What is the total capacity of a track and what is its useful capacity (excluding interblock gaps)?
Total capacity of a track = no of blocks per track * (block size + interblock gap size)
= 20 * (512+128) = 12,800 bytes
Useful capacity = no of blocks * block size
= 20 * 512 = 10240 bytes
b) How many cylinders are there?
No of cylinders = No of tracks = 400
c) What are the total capacity and the useful capacity of a cylinder?
Total cylinder capacity = Total capacity of the track * no of disks in pack
no of disks in pack = 15 double sided disks = 15* 2= 30
Total cylinder capacity = 20 * (512+128) * 15*2
= 384000 bytes
USeful cylinder capacity = 20 * 512 * 15*2
= 307200bytes
d) What are the total capacity and the useful capacity of a disk pack?
Total capacity of the disk pack = 15*2*20*400*(512+128)
= 15,36,00,000 bytes
Useful capacity of the disk pack = 15*2*20*400*512
= 12,28,80,000bytes
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