Improved algorithms for single machine scheduling with release dates and rejections

Abstract

We consider bi-criteria scheduling problems on a single machine with release dates and rejections and both the makespan and the total rejection cost have to be minimized. We consider three scenarios: (1) minimize the sum of the two objectives: makespan and total rejection cost, (2) minimize the makespan subject to a bound on the total rejection cost and (3) minimize the total rejection cost subject to a bound on the makespan. We summarize the results obtained in the literature and provide for several cases improved approximation algorithms and FPTASs.

Appendices

Appendix 1

This appendix describes a heuristic developed by Csirik et al. (1991) and its worst case behaviour. The heuristic is used in the body of the paper.

Let $\overline{W}_0 := \sum _{j=1}^n w_j - W_0$. It is obvious that $\sum _{j \in A} w_j \ge \overline{W}_0$ if and only if $\sum _{j \in R} w_j \le W_0$. Let $p_j /w_j$ be defined as the relative processing time of job j.

Algorithm H

Step 1 Sort the jobs in nondecreasing order of their relative processing times. From here on, assume that

$$\begin{aligned} p_1 /w_1 \le p_2/w_2 \le \cdots \le p_n/w_n. \end{aligned}$$

Step 2 (a) Let $k_1$ be the index for which

$$\begin{aligned} \sum _{i=1}^{k_1} w_i < \overline{W}_0 \le \sum _{i=1}^{k_1+1} w_i. \end{aligned}$$

The sublist $(1, 2, \ldots , k_1+1)$ is a candidate for the solution given by Algorithm H. Let

$$\begin{aligned} S_1 = (1, 2, \ldots , {k_1}), \end{aligned}$$

then this candidate can be written as

$$\begin{aligned} S_1 \cup \{ {k_1+1} \}. \end{aligned}$$

(b) Let $k_1 +2, k_1 + 3, \ldots , k_2 - 1$ be a (possibly empty) series of jobs so that for all corresponding jobs (i.e., $j \in \{k_1 +2, \ldots k_2 -1 \}$) the following holds:

$$\begin{aligned} \sum _{i=1}^{k_1} w_i + w_j \ge \overline{W}_0. \end{aligned}$$

Let

$$\begin{aligned} B_1 = ({k_1+1}, \ldots , {k_2 -1}), \end{aligned}$$

then all $S_1 \cup \{ j\}$, $j \in \{k_1 + 2, \ldots , k_2 -1 \}$, are also candidate solutions.

(c) Let $k_2$ be the first next index for which

$$\begin{aligned} \sum _{i=1}^{k_1} w_i + w_{k_2} < \overline{W}_0, \end{aligned}$$

and let $k_3 \ge k_2$ be the index for which

$$\begin{aligned} \sum _{i=1}^{k_1} w_i +\sum _{i=k_2}^{k_3} w_i < \overline{W}_0 \le \sum _{i=1}^{k_1} w_i +\sum _{i=k_2}^{k_3+1} w_i. \end{aligned}$$

Set

$$\begin{aligned} S_2 = ({k_2}, {k_2+1}, \ldots , {k_3}). \end{aligned}$$

Then, $S_1 \cup S_2 \cup \{{k_3+1} \}$ is also a candidate solution.

Now iterate from (b): in the first iteration use $k_3$ instead of $k_1$ and $k_4$ instead of $k_2$; in the i-th iteration, use $k_{2i+1}$ and $k_{2i+2}$. Repeat this until the end of the list. The solution generated by Algorithm H is the minimum cost candidate.

By applying Algorithm H to problem instance $\mathcal{J}$, we subdivide $\mathcal{J}$ into a sequence of alternating sublists as follows:

$$\begin{aligned}&\overbrace{1, 2, \ldots , k_1}^{S_1},\ \overbrace{k_1+1, \ldots , k_2-1}^{B_1}, \ \overbrace{k_2, \ldots , k_3}^{S_2}, \ \overbrace{k_3+1, \ldots , k_4-1}^{B_2}, \ \ldots ,\\&\qquad \qquad \ldots , \overbrace{k_{2\ell -1}+1,\ldots , k_{2\ell }-1}^{B_{\ell }}, \ \overbrace{k_{2\ell }, \ldots , k_{2\ell +1}}^{S_{\ell +1}}, \ \ldots ,\\&\qquad \qquad \qquad \ldots , \overbrace{k_{2m-1}+1,\ldots , k_{2m}-1}^{B_{m}}, \ \overbrace{k_{2m}, \ldots , n}^{S_{m+1}} \end{aligned}$$

where the last set is possibly empty, in which case $k_{2m}-1 = n$. We call the elements in the S-lists small and in the B-lists big.

It is easy to see that Steps 1 and 2 have running times of $O(n \log n)$ and O(n), respectively. Let $\sigma _1$ be the schedule obtained by algorithm H and let $\pi ^*$ be the optimal schedule. Then, it is proved by Csirik et al. (1991) that

$$\begin{aligned} C_{\max }(\sigma _1) \le 2 \cdot C_{\max }(\pi ^*). \end{aligned}$$

Appendix 2: Dynamic programming $DP^{*}$

Sort the jobs such that $r_1\le \ldots \le r_n$. Let $f_j(t)$ be the minimum value of the total rejection penalty when the jobs under consideration are jobs $1, \ldots , j$ and the makespan of the accepted jobs among $1, \ldots , j$ is exactly t. Now, we consider any optimal schedule for the jobs $1, \ldots , j$ in which the makespan of the accepted jobs among $1, \ldots , j$ is exactly t.

The initialization is

$$\begin{aligned} f_1(t) = \left\{ \begin{array}{ll} w_1, &{} \quad \text {if}\;t = 0;\\ 0, &{} \quad \text {if}\;t = r_1 + p_1; \\ + \infty , &{} \quad \text {otherwise}. \\ \end{array} \right. \end{aligned}$$

The recursion relation for $j = 1, \ldots , n$ is

$$\begin{aligned} f_j(t) = \left\{ \begin{array}{l l} f_{j-1}(t) + w_j, &{} \quad \text {if}\;t < r_j + p_j;\\ \min \left\{ f_{j-1}(t) + w_j,~~ \min \{f_{j-1}(t'): 0 \le t' \le r_j \} \right\} , &{} \quad \text {if}\;t = r_j + p_j; \\ \min \left\{ f_{j-1}(t) + w_j,~~ f_{j-1}(t - p_j)\right\} , &{} \quad \text {if}\;t > r_j + p_j; \\ \end{array} \right. \end{aligned}$$

Hence the optimal objective function value is

$$\begin{aligned} \min \left\{ t : 0 \le t \le r_n + \sum _{j=1}^{n}p_{j} \text { and } f_n(t) \le W_0 \right\} . \end{aligned}$$