Question

I am reading topic on scheduling algorithm analysis:

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.

Job Time

---

j1 15

j2 8

j3 3

j4 10

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

Thanks!

Answer 1

The jobs run serially (one at a time).

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

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.

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.

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

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).

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

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.

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