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How to Calculate access time of Disk.....

Posted on 2013-06-08
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Seek Time on a hard disk is 60 ms.It rotates at the rate of 12000 rpm.Each tract has 300 sectors.Calculate the access time of disk?
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Question by:sarang8180
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by:MacroShadow
ID: 39232534
average seek time + average rotational delay + transfer time + controller overhead + queuing delay

See: http://www.csc.villanova.edu/~japaridz/8400/sld012.htm
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by:sarang8180
ID: 39232542
Calculation is: Access time=Seek time+latency time

    =60 ms+(300/2)/12000*100

how 100 they have put here?Each time i have to divide 2 from number of sectors ie. 300
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by:nobus
ID: 39232589
where does that calculation come from??
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by:sarang8180
ID: 39232597
from my book..Another example..

Seek time of dsk is 20ms.It rotates at the rate of 6000 r.pm.Each track on this disk has 200 sectors. Calculate access time of disk.

Solution: (200/2 /6000*200)*60

     now please tell how above calculation formed?....
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by:sarang8180
ID: 39232599
book name is dotcombooks4u.com
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by:nobus
ID: 39232640
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by:garycase
ID: 39232818
The statement that "... Calculation is: Access time=Seek time+latency time " is correct, but the formula does not match that.

Average latency time is simply 1/2 the time it takes to rotate the platter ... i.e. on average you will get to the data quicker sometimes, a bit longer others, but will average 1/2 of the rotation.

The number of sectors/cylinder is irrelevant (and is in fact DIFFERENT on different sections of the disk).

So ... for the first example in your question ...

Average seek time = 60 ms

Disk rotates at 12,000 rpm, so that's 12000/60 = 200 rotations/second.   So it takes 1/200 th of second to rotate once, which is 5ms.     1/2 of that is 2.5ms, so the average latency is 2.5ms.

So the average access time is 60 + 2.5 = 62.5ms


In your 2nd example, the seek time is 20ms;  the average latency for a 6000rpm disk will be 5ms (since it takes 10ms to rotate once);  so the average access = 20ms + 5ms = 25ms
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by:sarang8180
ID: 39232980
why have you divided avg. latency by 2?
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garycase earned 500 total points
ID: 39233355
I didn't divide the average latency by 2.   I divided the time it takes for one rotation of the platter by 2 ==> that's how you compute the average latency.

Remember, the disk is rotating.   To access data there are fundamentally two things that have to happen [there are actually others, but they take so little time it's not worth considering]:  (a)  you have to move the heads to the right cylinder ("seek");  and then you have to wait until the platter rotates so the sector you want is under the heads so it can be read/written.

The seek time obviously varies depending on where the cylinder you want is relative to where the head is right now; but the published seek times are the average, so that's what you generally use.   For more detailed calculations, you can check the manufacturer's site and get data for track-track seek; head settling time; max stroke time (time to move from the innermost cylinder to the outermost cylinder), etc. ==> but for your purposes just using the average seek time is fine.

The "latency" refers to how long you're waiting for the platter to rotate to the sector you want.    On average, this will be 1/2 of the time it takes to rotate.   It could be as little as zero ... i.e. the sector you want might just happen to be where it's at when the seek is completed;  or it could be a full rotation ... if you "just missed" the sector when the seek finished.   But statistically, it will be 1/2 of the rotation time.
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