sarang8180
asked on
How to Calculate access time of Disk.....
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?
ASKER
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
=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
where does that calculation come from??
ASKER
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?....
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?....
ASKER
book name is dotcombooks4u.com
here all you need : http://en.wikipedia.org/wiki/Access_time
exam info : http://www.google.be/url?sa=t&rct=j&q=disk%20access%20time%20formula&source=web&cd=1&sqi=2&ved=0CCwQFjAA&url=http%3A%2F%2Fwww.cs.ucla.edu%2Fclasses%2Fwinter03%2Fcs143%2Fl1%2Fhandouts%2Fexam_notes.doc&ei=0zG0UaPnFMu10QW-zYGAAQ&usg=AFQjCNHwWWrbjNpg9O_pufJhnXa3ZzpsMQ
the pdf they reference to is a nice explanation : http://stackoverflow.com/questions/9828736/why-is-average-disk-seek-time-one-third-of-the-full-access-time
exam info : http://www.google.be/url?sa=t&rct=j&q=disk%20access%20time%20formula&source=web&cd=1&sqi=2&ved=0CCwQFjAA&url=http%3A%2F%2Fwww.cs.ucla.edu%2Fclasses%2Fwinter03%2Fcs143%2Fl1%2Fhandouts%2Fexam_notes.doc&ei=0zG0UaPnFMu10QW-zYGAAQ&usg=AFQjCNHwWWrbjNpg9O_pufJhnXa3ZzpsMQ
the pdf they reference to is a nice explanation : http://stackoverflow.com/questions/9828736/why-is-average-disk-seek-time-one-third-of-the-full-access-time
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
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
ASKER
why have you divided avg. latency by 2?
ASKER CERTIFIED SOLUTION
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See: http://www.csc.villanova.edu/~japaridz/8400/sld012.htm