Sequencing to Minimize Weighted Tardiness

Happy New Year, everyone!  I just checked ahead in the appendix, and at my current pace of about 1 problem a week, we should get through appendices A5 (Sequencing and Scheduling), A6 (Mathematical Programming), and possibly A7 (Algebra and Number Theory) this year.  The appendix goes through A12, so it looks like I have plenty of work ahead of me.

The problem: Sequencing to Minimize Weighted Tardiness.  This is problem SS5 in the appendix.

The description: Given a set T of tasks, each task t  has a length l(t), a weight w(t) (which is really a penalty for finishing a task late) and a deadline d(t), and a positive integer K.  Is there a one-processor schedule for all tasks in T where each task is penalized by w(t) for each time unit it goes beyond its deadline, such that the total penalty is K or less?

Example: Suppose I have 3 tasks:

Task Length Deadine Weight
1 3 3 100
2 1 4 3
3 2 4 1

If task 1 is going to not miss its deadline, it needs to start right away.  Since the weight penalty for missing the deadline is pretty disastrous, we should do that.

At time 3, when task 1 is done, we have a choice: Schedule task 2, and have task 2 finish 2 time units late, or schedule task 3, have it finish just 1 time unit late and schedule task 2 afterward.  With these weights, scheduling task 2 first gives a penalty of 2 (task 3 finishes 2 time units late, for a penalty of 1, and 1×2=2).  The other plan would have a penalty of 7 (Task 3 finishes 1 unit late, and task 2 finishes 2 units late).

If the weight of task 2 was 1, and the weight of task 3 was 5, then scheduling task 2 first gives a penalty of 10, but scheduling task 3 first gives a penalty of 6.

Reduction: G&J reference a paper by Lawler, which itself references G&J.  Weird.   Anyway, they reduce from 3-Partition.  Given a 3-Partition instance (a set of 3n elements, and a bound B.  Each element is between B/4 and B/2, and the sum of all elements is m*B), we create 4n jobs.  3n of them will be “A” jobs, each one corresponding to an element in the 3-Partition instance.  We will also have n “X” jobs.

Each “A” job Ai corresponds to an element in the 3-partition instance ai.  These jobs have weight and length B+ai and deadline 0.

The X jobs all have processing time 16B2+n(n+1)/2 + 1.  Call this number L.  The X jobs have weight (L+4B)*(4B)*n(n+1)/2 + 1.  Call this number W. Each X job Xi has deadline iL+(i-1)*4B.    Notice that these X jobs don’t depend on the values in the partition instance at all (they do depend on the size).

Our K value is W-1.

If the original set has a partition, then we can make a schedule of: X1 then the A jobs corresponding to a 3 element partition, then X2, then another set of 3 A jobs corresponding to another 3 element partition, and so on down.  Since we know the  3 elements in a partition sum to exactly B, we know that the processing time of each set of 3 A jobs is exactly 4B.  So each X job will finish at time i*L+(i-1)*4B, and not be tardy.  The A jobs all have deadline 0 and will all be tardy and will pay penalties related to their completion times, but that will sum to less than K.

In the other direction, suppose that we have a legal schedule.  Since all of the X jobs have weight > K, they all have to be scheduled before their deadlines.  Then we can talk about the weights and completion times that the A jobs need to have to fit around these X jobs.  It’s a lot of algebra to make it all work out, but the math isn’t too hard.

Difficulty: 7.  While the algebra is followable, the idea that you’d need to come up with these numbers in the first place is a hard one- I know they’re picked because they “work”, but I feel like there must be an easier way.

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