关于调度作业分析 [英] Regarding scheduling jobs analysis

问题描述

我读调度算法分析的主题：

I am reading topic on scheduling algorithm analysis:

假设我们有四个工作和相关的运行时间显示   下面。一个可能的时间表J1，J2，J3和J4由于J1饰面   在图15（时间单位），J2在23，j3中在26和J4在36，平均   完成时间是25一个更好的时间表，这将产生一个平均值   17.75完成时间，为J3，J2，J4和J1。

suppose we have the four jobs and associated running times shown below. One possible schedule j1, j2, j3, and j4 Because j1 finishes in 15 (time units), j2 in 23, j3 in 26, and j4 in 36, the average completion time is 25. A better schedule, which yields a mean completion time of 17.75, is j3, j2, j4 and j1.

工作时间

J1 15

J2 8

J3 3

J4 10

我的问题是如何笔者计算出的平均完成时间，即，我们如何在上面的文字得到25和17.75？

My question how author calculated average completion time i.e., how we got 25 and 17.75 in above text?

谢谢！

推荐答案

该作业的运行顺序（一次）。

The jobs run serially (one at a time).

因此​​，订货 J1-J2-J3-J4 ， J1 结束在 15 ， J2 结束在 15 + 8 = 23 ， J3 结束在 15 + 8 + 3 = 26 和 J4 结束在 15 + 8 + 3 + 10 = 36 。然后，他们平均 15 ， 23 ， 26 和 36 获得 25 使用标准总和/计数公式：

Hence with order j1-j2-j3-j4, j1 finishes at 15, j2 finishes at 15+8=23, j3 finishes at 15+8+3=26 and j4 finishes at 15+8+3+10=36. They then average 15, 23, 26 and 36 to get 25 using your standard sum/count formula:

  (15 + 23 + 26 + 36) / 4
=         100         / 4
=                25

在换句话说，完成时间，他们正在谈论的是一份工作，从什么时候开始没有多久了，但它从第一的作业开始（即一个点了多久在时间而非持续时间）。我不知道怎么的有用的这样一个指标是，但是这是他们在做什么，基于这些数字。

In other words, the completion time they're talking about is not how long a job took from when it started but how long it took from the start of the first job (ie, a point in time rather than a duration). I'm not sure how useful such a metric is, but that's what they're doing, based on the figures.

使用顺序 J3-J2-J4-J1 ， J3 结束在 3 ， J2 结束在 3 + 8 = 11 ， J4 结束在 3 + 8 + 10 = 21 和 J1 结束在 3 + 8 + 10 + 15 = 36 。这（平均 3 ， 11 ， 21 和 36 ）的 17.75 。

With order j3-j2-j4-j1, j3 finishes at 3, j2 finishes at 3+8=11, j4 finishes at 3+8+10=21 and j1 finishes at 3+8+10+15=36. The average of that (3, 11, 21 and 36) is 17.75.

对于一般完成时间的最佳（最低）的解决方案是做工作，以增加其持续时间。

The optimum (lowest) solution for the average finishing time is to do the jobs in order of increasing duration.

这是因为在一套四的最后一个作业将始终结束的同时，不论顺序（ 36 在这种情况下）。

That's because the last job in a set of four will always finish at the same time, regardless of order (36 in this case).

因此​​，为了降低平均终点，其他三个作业的精加工点应尽可能地低。

So, in order to reduce the average finishing point, the finishing points of the other three jobs should be as low as possible.

和，适用于四份工作相同的规则也适用于三个作业（一次运行时间最长的 J1 取出的组合）。然后两份工作，一旦你删除 J4 。

And, the same rule that applies to four jobs also applies to three jobs (once the longest-running j1 is taken out of the mix). Then two jobs, once you've removed j4.

一旦你删除三个作业，只有一个站立的是你应该选择（当然）之一。

Once you've removed three jobs, the only one standing is the one you should choose (of course).

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