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Weighted interval scheduling with m-machines ('Weighted interval scheduling with m-machines')

I encountered the problem of weighted interval scheduling on m identical machines (as discussed in the link above). My greedy solution is exactly the same as the answer provided by D.W. in that link. 1. sort the interval by profit-to-duration in decreasing order, 2. then select intervals in that order. If there is still an available machine left, select it; Otherwise, skip the currently scanned interval. The selection is done when all intervals are considered.

I found the greedy solution provided pretty good empirical performance, and I wish to derive a theoretical approximation factor for this greedy solution. From the empirical result, I guess the approximation factor is very close to 1. I read about how to prove the approximation factor for greedy solutions in most general cases, but I can not transfer that knowledge to my case here. I think it is necessary to derive certain characteristics out of my greedy solution (in a mathematical form), which I can't :(.

In short, I would like to know how to derive an approximation factor for my greedy solution.

EddieG
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  • I do understand this problem itself can be solved in O(n^2logn), but I still would like to know the approximation factor of my greedy solution :) – EddieG Sep 16 '23 at 08:30

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Unfortunately, the approximation guarantee of the algorithm is unbounded.

For example, Consider a single machine and only two intervals: interval $I_1$ with profit $1$ and length $1$; and interval $I_2$ with profit $w \gg 1$, and length $w+1$. Suppose the two intervals overlap. Then, the greedy algorithm gives a solution of profit $1$, i.e., interval $I_1$. Whereas, the optimal solution is of profit $w$, i.e., interval $I_2$.

Inuyasha Yagami
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  • Thank you very much for this clear example. Really appreciate it :). But if I combine another greedy principle (profit first sorting + profit-to-duration first sorting[the original one]) for the scheduling procedure, so that we can get two schedules for intervals of these two intervals. Under such a case, would the solution be bounded? – EddieG Sep 16 '23 at 10:33
  • By 'combining' I mean: conducting scheduling with these two principles and choosing the schedules with max sum of profit out of these two greedy solutions. – EddieG Sep 16 '23 at 10:34
  • @EddieG The approximation guarantee then would be $\Theta(n)$: the number of intervals. The algorithm can not do better than that. For example, suppose there are $n$ mutually non-overlapping intervals each of profit $w \gg 1$ and length $w + 1$. Each of these intervals further contain an interval of profit $1$ and length $1$. And, there is one large interval of length $(w+1) \cdot n$ and profit $w+1$ that overlaps all these intervals. The optimal solution is $w \cdot n$. However, the greedy solution is $\max(n, w+1)$. – Inuyasha Yagami Sep 16 '23 at 11:57
  • Thanks again for another clear example :). – EddieG Sep 16 '23 at 13:46
  • hello again pal :). I notice that there are several constraints for this problem under my application scenario: 1. the length and profit of an interval belong to $N^{+}$ (positive integers), and they are both $\geq$ 1; 2. number of intervals can be bounded by $the_number_of_machines \times the_max_end_time_of_intervals$ 3. max profit can be bounded by $\frac{c*(max_number_of_ machines)}{the_max_end_time_of_intervals - 1}$, where $c$ is a known constant. Based on these certain constraints, can we derive a better lower bound (maybe even parameterized)? – EddieG Sep 17 '23 at 02:28
  • @EddieG Hello! How about you post a separate question for that? I will check that. – Inuyasha Yagami Sep 17 '23 at 04:45
  • no problem pal. The link is provided in https://cs.stackexchange.com/questions/162090/weighted-interval-scheduling-on-k-identical-machines-approximation-factor. Let me know if there is anything that I do not state clearly :). – EddieG Sep 17 '23 at 05:35