Consider a magnetic disk drive with 1 surfaces, 500 tracks per surface, and 32 s

Question

Consider a magnetic disk drive with 1 surfaces, 500 tracks per surface, and 32 sectors per track. Sector size is 512B. The average seek time is 10 ms, the track-to-track access time is 1.5 ms, and the drive rotates at 3600 rpm.

Assume a file of 10 KB stored as the next two configuration:

Sequentially from sector 0 to sector 3 from track 50.

Fragmented as follows:

2 KB      in Sector-1    in track-6

2 KB      in Sector-6    in track-12

2 KB      in sector-9    in track-30

2 KB      in sector-2    in track-33

2 KB      in sector-30   in track 2

Sectors should be accessed sequentially as they represent the 2 KB block from the required file. Calculate the total time required to access the same file for each of the two configurations A and B?

Explanation / Answer

Answer:

Given:

No of surfaces             =1

No of tracks per surface=500

Sectors for track          =32

Sector size                   =512 B

Average Seek time      =10ms

Track-to-track access time= 1.5 ms

Spindle speed              =3600 rpm

File size                       = 10 KB

From this we can calculate:

Average rotational delay=3 ms

Configuration A:

In this File is read sequentially from sector 0 to sector3 that starts with tack50

Average transfer time for1 track = 60/3600

    = 16.6 ms

Average total time per track=Average seek time + Average rotational delay + average transfer time

= 10 + 3 + 16.6 ms

= 29.6 ms

Total time to access file = 29.6 * 5 ms

=148 ms

Time taken to read 10KB file sequentially by configuration A would take 148 ms.

Configuration B:

In this case file is fragmented in different sectors in different tracks i.e. File is read in random manner.

Transfer time for 1 sector       = revolution time/no of sectors per track

= (1/3600) min/32

= (1/3600 *60)/32

= 60/ (3.6*32) ms

= 0.52 ms

Average time to read one sector=average seek time + average rotational delay + average transfer time

= 10+3+0.52 ms

= 13.52 ms

Average time to read the file = no of sectors to read * average time to read one sector

= 5 * 13.52 ms

= 67.6 ms

Time taken to read 10KB file randomly by configuration B would take 67.6 ms.

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