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Mass storage

Assume a floppy disk has Rotation speed = 8.33 msec/rev, Arm movement time =0.5 msec fixed startup time + 0.05 msec for each track crossed. (The startup time is a constant no matter how far the arm moves.)
Number of surfaces =2 ( a double-sided floppy disk. A single read/write arm holds both read/write heads.)
Number of tracks per surface = 500, Number of sectors per track=20
Number of characters per sector =1,024
What is the worst-case on a single floppy disk?
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First, my question is what does 'each track crossed' mean?
This is what I did: 499*(0.5 msec+0.05 msec)*2+8.33 msec*2+1/20*8.33 msec*2 =566.393 msec. {*2 because it's two surfaces).
Now let's say if the disk has one read/write head per track instead of per surface, what's the worst case?
This is what I did: 499*(0.5 msec+0.05 msec)*2+8.33 msec*2+1/20*0 msec*2 = 565.56 msec. Im not sure about the transfer time because now it's per track, i put 0. Thanks a lot.

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Did you notice that this is the "New To Java" forum?

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yes, I noticed that, so what's your point? I'm a beginner, and it is a beginner CS question.

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Originally Posted by MIA6
...and it is a beginner CS question.
which has nothing to do with Java.

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