Consider a disk with the following characteristics (these are not parameters of

Question

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.

a. What is the total capacity of a track, and what is its useful capacity
(excluding interblock gaps)?
b. How many cylinders are there?
c. What are the total capacity and the useful capacity of a cylinder?
d. What are the total capacity and the useful capacity of a disk pack?
e. 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? (The bulk transfer rate is the rate of transferring "useful" bytes of data, which exclude interlock gap bytes.
f. Suppose that the average seek time is 30 msec. How much time does it
take (on the average) in msec to locate and transfer a single block, given
its block address?
g. Calculate the average time it would take to transfer 20 random blocks,
and compare this with the time it would take to transfer 20 consecutive
blocks using double buffering to save seek time and rotational delay.

Explanation / Answer

Block size B = 512 bytes ;

Interblock Gap Size G = 128 bytes ;

Number of Blocks per track = 20 ;

Number of tracks per surface = 400 ;

15 double sided - disks in a disk pack ;

(a) Total capacity of a track = Number of blocks per track x (block size + interblock size) = 20 x ( 512 + 128 ) = 12800 bytes = 12800/1024 KBytes = 12.5 KB.

Useful capacity = 20 x 512 / 1024 KB = 10 KB

(b) No. of cylinders = Number of tracks (assume a bunch of disks stacked one on top of the other that will form 400 cylinders corresponding to the 400 tracks on each disk.) = 400 cylinders.

(c) For 1 track the total capacity is 12.5 KB and useful capacity is 10 KB, therefore, for 400 tracks (per cylinder) total capacity = 12.5 x 400 = 5000 KB = 5000/1024 = 4.8828125 MB

Useful capacity = 10 x 400 = 4000 KB = 4000/1024 = 3.90625 MB

(d) For disk pack containing 15 disks, total capacity = 5000 x 15 = 75000 KB = 73.2421875 MB

Useful capacity = 4000 x 15 = 60000 KB = 58.59375 MB

(e) No of revolutions per minute = 2400, No of revolutions per msec = 2400 / ( 60 x 1000 )

Therefore, transfer rate = Total capacity of a track x No of revolutions per msec = 12800 x 2400 / 60000 = 512 bytes/msec

Block transfer time = Block size / transfer rate = 512 bytes / 512 bytes/msec = 1 msec

Average rotational delay = one rotation time / 2 = 60000/(2x2400) = 25/2 = 12.5 msec

(f) Avg Time to transfer a block = Avg rotational delay + Avg seek time + Block transfer time = 12.5 + 30 + 1 = 43.5 msec

(e) To transfer 20 random blocks = 20 * Avg Time to transfer a block = 20 * 43.5 = 870 msec

For 20 consecutive blocks = 12.5 + 30 + (1 * 20) = 62.5 msec

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.