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Consider a disk pack with a seek time of $4$ milliseconds and rotational speed of $10000$ rotations per minute (RPM). It has $600$ sectors per track and each sector can store $512$ bytes of data. Consider a file stored in the disk. The file contains $2000$ sectors. Assume that every sector access necessitates a seek, and the average rotational latency for accessing each sector is half of the time for one complete rotation. The total time (in milliseconds) needed to read the entire file is__________________
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1 votes

Seek time $ =4ms$ 
In 60 seconds 1000 rotations.

Rotation Time$ = (60 / 10000) = 6 ms $  (Time to read one Track)
Rotational latency$ = (1 / 2 ) * 6ms = 3ms $

Transfer time = No. of tracks * time to read one track
          $= $(Sectors in file/Sectors in one track) * (Time to read one track)
          $= (2000/600)*6
          =  20$ ms
total time needed to read the entire file is: 
$= 2000*(4+3) + 20 
= 8000 + 6000 + 20 = 14020$ ms

So, correct answer is 14020.

0 votes
0 votes

In the question, we have to Find the time required to read / transfer the data (2000 sector /file ) data from disk 

So we have to find the time required to read /transfer for 1 sector data than we find 2000(time required for one sector).

so time for one sector is Total time =2000  (seek time +avg. rotational latency + data transfer time)
–4 votes
–4 votes
SeekTime = 4ms
OneRotationalLatency = (60/10000) = 0.006s = 6ms
AverageRotationalLatency = 0.5 * 6 = 3ms

1 TrackAccessTime = 1 Rotation
4 TrackAccessTime = 4 * OneRotationalLatency  = 24ms

600 SectorsAccessTime = 1 Rotation
600 SectorsAccessTime = 6ms
1 SectorAccessTime = 6/600 = 0.01ms = 10 ms

FullTracksToBeAccessed = floor(2000/600) = 3
ExtraSectorsToBeAccessed = 2000 - (600*3) = 200

TotalTime =

(4 * TrackAccessTime) + (200 * SectorAccessTime)

= (4 * 24) + (10*200) = 2096ms
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