为什么平均磁盘寻道时间的全部三分之一？ [英] Why is average disk seek time one-third of the full seek time?

问题描述

考虑到磁盘性能，我已经阅读了很多书籍和论文，平均搜索时间大约是完整搜索时间的三分之一，但没有人真正提供任何解释。这是从哪里来的？

解决方案

平均值是使用微积分计算的。
我们用平均值计算的基本公式来计算平均值。

平均寻道时间=（所有可能寻道时间的总和）/时间）

假设磁盘有N个磁道，所以这些磁道的编号从1 ... N
磁头在任何位置时间点可以是从0到N（包含）的任何值。
让我们说，磁头的初始位置是在轨道'x'，磁头的最终位置在轨道'y'，所以x可以从0到N变化，而且y可以

与我们定义的平均寻道时间类似，我们可以说，

平均寻道距离=（所有可能寻道距离之和）/（可能寻道距离的总数）

根据x和y的定义，
总数没有。 （x = 0，N）SIGMA（y = 0，N）| x-y |
= INTEGRAL（x = 0，N）INTEGRAL（y = 0，N）| x-y |为了解决这个问题，使用将表达式的模数分解为y = 0到x和y = x到N的技巧。然后求解x = 0 （N ^ 3）/ 3。

平均搜寻距离= （N ^ 3）/ 3 * N * N = N / 3

平均查找时间=平均查找距离/查找率

如果从位置0到轨道N的寻道时间取't'秒，则寻道率= N / t因此，avg寻道时间= （N / 3）/（N / t）= t / 3

参考：

http://pages.cs.wisc.edu/~remzi/OSFEP/file-disks.pdf
Page-9给出了非常好的答案。

I have read in many books and papers, considering disk performance, that the average seek time is roughly one-third of the full seek time, but no one really offers any explanation about that. Where does this come from?

解决方案

The average is calculated mathematically using calculus. We use the very basic formula for calculation of average.

Average seek time = (Sum of all possible seek times)/(Total no. of possible seek times)

The disk is assumed to have N number of tracks, so that these are numbered from 1...N The position of the head at any point of time can be anything from 0 to N (inclusive). Let us say that the initial position of the disk head is at track 'x' and the final position of the disk head is at track 'y' , so that x can vary from 0 to N and also, y can vary from 0 to N.

On similar lines as we defined average seek time, we can say that,

Average seek distance = (Sum of all possible seek distances)/(total no. of possible seek distances)

By definition of x and y, Total no. of possible seek distances = N*N and Sum of all possible seek distances = SIGMA(x=0,N) SIGMA(y=0,N) |x-y| = INTEGRAL(x=0,N)INTEGRAL(y=0,N) |x-y| dy dx

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

This comes out to be (N^3)/3.

Avg seek distance = (N^3)/3*N*N = N/3

Average seek time = Avg seek distance / seek rate

If the seek time for the from position 0 to track N takes 't' seconds then seek rate = N/t

Therefore, avg seek time = (N/3)/(N/t) = t/3

Reference:

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

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