Consider a magnetic disk with the following parameters; 6 recording surfaces (la

Question

Consider a magnetic disk with the following parameters; 6 recording surfaces (labeled 0-5), 80 tracks per surface (labeled 0-79), 16 sectors per track (labeled 0-15); rotation time is 25 ms, and a seek between adjacent tracks takes 20 msec. Head switching time from surface to surface is minimal and can be ignored. Assume that Sector 0 in all tracks within a surface are aligned; further assume that Sector 0 of Track 0 in all surfaces align. We are interested in reading the entire disk and, obviously, in minimizing the overall time. Assume that the head is initially positioned on the outer track (Track 0), just near the beginning of Sector 0. Within each track, we would like to read all sectors in sequential order, starting from Sector 0. State precisely the order in which the sectors are to be read, in terms of surface, track and sector numbers. How long does it take to read the entire disk? Take into account the rotational time, head movement time and possible wait time for sector alignment, because we want to read starting from sector 0 in each track. You can ignore head switching time.

Explanation / Answer

Answer :

(a)
Here we have 6 recording surfaces(0-5), 0 tracks per surface(0-79) and 16 sectors per track(0-15).
Our head is initially positioned on the outer track (track 0),just near the begining of the
Sector 0.

As you can see inthe figure how tracks and sectors are distributed over the disk.
Now for reading operation head start with sector 0 which is Track 0 so sequence is as follows :
Surface 0 :Track 0 : Sector 0 ,Sector 1, Sector 2....Sector 15
Surface 0 : Track 1 : Sector 0 ,Sector 1, Sector 2....Sector 15
.....       ...       ....
Surfce 0 : Track 79 : Sector 0 ,Sector 1, Sector 2....Sector 15
Surface 1 : Track 0 : Sector 0 ,Sector 1, Sector 2....Sector 15
.....       ....       ....
Surface 1 : Track 79 : Sector 0 ,Sector 1, Sector 2....Sector 15
.....       ....       .....
Surface 5 : Track 79 :Sector 0 ,Sector 1, Sector 2....Sector 15

(b)
Now to find time to read entire disk.
We have rotation time = 25ms
Seek time = 20msec
Head switching time from surface to surface is minimal hence can be ignored.
Here we are ignoring transfer time as we have assumed that sectors are aligned and there is miniamal delay from transferring from surface to surface.
So average sector read time = rotation time + seek time
                     = 25 + 20 = 45 ms
now we have 16 sectors per track so average read time for per track,
T (Track)= 45 * 16 = 720 ms
And we have 80 sectors per surface,
T(Surface) = 720* 80 = 57600 ms
Finally we have 6 recording surfaces,
T(total) = 57600 * 6 = 345600 ms

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