Why is the average disk seek time one third of the total seek time?

I have read in many books and docs with disk performance that the average seek time is about one third of the total seek time, but no one really gives any explanation on this. Where does it come from?


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Average is calculated mathematically using calculus. We use the most basic formula to calculate the average.

Average search time = (Sum of all possible search times) / (Total number of possible search times)

It is assumed that there are N number of tracks on the disc, so they are numbered from 1 ... N The position of the head at any time can be any from 0 to N (inclusive). Let's say the start position of the disc head is on track 'x' and the end position of the head of the disc is on track 'y', so x can range from 0 to N, and also y can range from 0 to N.

Similar to how we determined the average seek time, we can say that

Average search distance = (Sum of all possible search distances) / (total number of possible search distances)

By the definition of x and y, there is no total. possible search distances = N * N and Sum of all possible search distances = SIGMA (x = 0, N) SIGMA (y = 0, N) | xy | = INTEGRAL (x = 0, N) INTEGRAL (y = 0, N) | xy | dy dx

To solve this problem, use the modulus technique of splitting expression for y = 0 for x and for y = x in N. Then solve for x = 0 to N.

This means that (N ^ 3) / 3.

Average search distance = (N ^ 3) / 3 * N * N = N / 3

Average search time = Average search time / search speed

If the search time for a position from 0 to track N takes seconds, then the search speed = N / t

Hence avg seek time = (N / 3) / (N / t) = t / 3


http://pages.cs.wisc.edu/~remzi/OSFEP/file-disks.pdf Page-9 gives a very good answer to this question.



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