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Deteriorating jobs scheduling on a single machine with release dates, rejection and a fixed non-availability interval

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Abstract

This paper studies single machine scheduling with a fixed non-availability interval. The processing time of a job is a linear increasing function of its starting time, and each job has a release date. A job is either rejected by paying a penalty cost or accepted and processed on the machine. The objective is to minimize the makespan of the accepted jobs and the total rejection penalties of the rejected jobs. We present a fully polynomial-time approximation scheme for the problem. We also show that the special case without non-availability interval can be solved using the same method with a lower order.

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Acknowledgments

We are very grateful to thank the referees for careful reading and valuable comments which led to improvements of our original manuscript.

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Correspondence to Wei-Xuan Li.

Appendix

Appendix

This appendix proves two important results which corresponding to (3) (5) and (4) in the proof of Theorem 1, respectively.

Result 1 \(\left| {P_{j+1}^k (x^{*})-P_{j+1}^k (\tilde{x}^{(a_1 a_2 bc)})} \right| \le \delta _1 P_{j+1}^k (x^{*})\) for \(x_{j+1} =k, \,k=1,2.\)

$$\begin{aligned}&\left| {P_{j+1}^k (x^{*})-P_{j+1}^k \left( \tilde{x}^{(a_1 a_2 bc)}\right) } \right| \\&=\left| (1+bp_{j+1} )\max \left\{ P_j^k (x^{*}),r_{j+1} \right\} +ap_{j+1}\right. \\&\left. \quad -(1+bp_{j+1} )\max \left\{ P_j^k \left( x^{(a_1 a_2 bc)}\right) ,r_{j+1} \right\} -ap_{j+1} \right| \\&=\left| {(1+bp_{j+1} )\left( {\max \left\{ P_j^k (x^{*}),r_{j+1} \right\} } \right. \left. {-\max \left\{ P_j^k \left( x^{(a_1 a_2 bc)}\right) ,r_{j+1} \right\} } \right) } \right| \\&=(1+bp_{j+1} )\left| {\max \left\{ P_j^k (x^{*}),r_{j+1} \right\} -\max \left\{ P_j^k \left( x^{(a_1 a_2 bc)}\right) ,r_{j+1} \right\} } \right| \end{aligned}$$

It can be divided into the following four cases, and we show that Result 1 holds for each of them.

Case 1.1 \(r_{j+1} \ge P_j^k (x^{*})\) and \(r_{j+1} \ge P_j^k (x^{(a_1 a_2 bc)})\)

$$\begin{aligned} \left| {P_{j+1}^k (x^{*})-P_{j+1}^k \left( \tilde{x}^{(a_1 a_2 bc)}\right) } \right|&= (1+bp_{j+1} )\left| {r_{j+1} -r_{j+1} } \right| \le \delta P_{j+1}^k (x^{*})\\&=\delta _1 P_{j+1}^k (x^{*}). \end{aligned}$$

Case 1.2 \(P_j^k (x^{(a_1 a_2 bc)})>r_{j+1} \ge P_j^k (x^{*})\)

$$\begin{aligned} \left| {P_{j+1}^k (x^{*})-P_{j+1}^k \left( \tilde{x}^{(a_1 a_2 bc)}\right) } \right|&= (1+bp_{j+1} )\left| {r_{j+1} -P_j^k \left( x^{(a_1 a_2 bc)}\right) } \right| \\&= (1+bp_{j+1} )(P_j^k \left( x^{(a_1 a_2 bc)}\right) -r_{j+1} )\\&\le (1+bp_{j+1} )(P_j^k \left( x^{(a_1 a_2 bc)}\right) -P_j^k (x^{*}))\\&= (1+bp_{j+1} )\left| {P_j^k \left( x^{(a_1 a_2 bc)}\right) -P_j^k (x^{*})} \right| \\&\le (1+bp_{j+1} )\delta P_j^k (x^{*})\\&\le \delta (1+bp_{j+1} )\max \left\{ P_j^k (x^{*}),r_{j+1} \right\} \\&\le \delta _1 P_{j+1}^k (x^{*}) \end{aligned}$$

Case 1.3 \(P_j^k (x^{*})>r_{j+1} \ge P_j^k (x^{(a_1 a_2 bc)})\)

$$\begin{aligned} \left| {P_{j+1}^k (x^{*})-P_{j+1}^k \left( \tilde{x}^{(a_1 a_2 bc)}\right) } \right|&= (1+bp_{j+1} )\left| {P_j^k (x^{*})-r_{j+1} } \right| \\&= (1+bp_{j+1} )\left( P_j^k (x^{*})-r_{j+1} \right) \\&\le (1+bp_{j+1} )\left( P_j^k (x^{*})-P_j^k \left( x^{(a_1 a_2 bc)}\right) \right) \\&= (1+bp_{j+1} )\left| {P_j^k (x^{*})-P_j^k \left( x^{(a_1 a_2 bc)}\right) } \right| \\&\le (1+bp_{j+1} )\delta P_j^k (x^{*})\\&= \delta (1+bp_{j+1} )\max \left\{ P_j^k (x^{*}),r_{j+1} \right\} \\&\le \delta _1 P_{j+1}^k (x^{*}) \end{aligned}$$

Case 1.4 \(r_{j+1} <P_j^k (x^{*})\) and \(r_{j+1} <P_j^k (x^{(a_1 a_2 bc)})\)

$$\begin{aligned} \left| {P_{j+1}^k (x^{*})-P_{j+1}^k \left( \tilde{x}^{(a_1 a_2 bc)}\right) } \right|&\le (1+bp_{j+1} )\left| {P_j^k (x^{*})-P_j^k (x^{(a_1 a_2 bc)})} \right| \\&\le (1+bp_{j+1} )\delta P_j^k (x^{*})\\&= \delta (1+bp_{j+1} )\max \left\{ P_j^k (x^{*}),r_{j+1} \right\} \\&\le \delta _1 P_{j+1}^k (x^{*}) \end{aligned}$$

Based on the above analysis, obviously we have

$$\begin{aligned} \left| {P_{j+1}^k (x^{*})-P_{j+1}^k \left( \tilde{x}^{(a_1 a_2 bc)}\right) } \right| \le \delta _1 P_{j+1}^k (x^{*}),\hbox { for}\quad k=1,2. \end{aligned}$$

This completes the proof of Result 1.

Result 2 \(\left| {f_{j+1} (x^{*})-f_{j+1} (\tilde{x}^{(a_1 a_2 bc)})} \right| \le \delta _1 f_{j+1} (x^{*})\), for \(x_{j+1} =1\).

Since \(\left| {f_j (x^{*})-f_j (x^{(a_1 a_2 bc)})} \right| \le \delta f_j (x^{*})\), then it implies the following four cases:

Case 2.1 \(P_j^2 (x^{*})=T_2,\, P_j^2 (x^{(a_1 a_2 bc)})=T_2 \)

$$\begin{aligned} \left| {f_j (x^{*})-f_j \left( x^{(a_1 a_2 bc)}\right) } \right|&= \left| P_j^1 (x^{*})+W_j (x^{*})-P_j^1 \left( x^{(a_1 a_2 bc)}\right) \right. \\&\left. -W_j (x^{(a_1 a_2 bc)}) \right| \le \delta f_j (x^{*}) \end{aligned}$$

Case 2.2 \(P_j^2 (x^{*})>T_2,\, P_j^2 (x^{(a_1 a_2 bc)})>T_2 \)

$$\begin{aligned} \left| {f_j (x^{*})-f_j \left( x^{(a_1 a_2 bc)}\right) } \right|&= \left| P_j^2 (x^{*})+W_j (x^{*})-P_j^2 \left( x^{(a_1 a_2 bc)}\right) \right. \\&\left. -W_j (x^{(a_1 a_2 bc)}) \right| \le \delta f_j (x^{*}) \end{aligned}$$

Case 2.3 \(P_j^2 (x^{*})>T_2,\, P_j^2 (x^{(a_1 a_2 bc)})=T_2 \)

$$\begin{aligned} \left| {f_j (x^{*})-f_j \left( x^{(a_1 a_2 bc)}\right) } \right|&= \left| P_j^2 (x^{*})+W_j (x^{*})-P_j^1 \left( x^{(a_1 a_2 bc)}\right) \right. \\&\left. -W_j (x^{(a_1 a_2 bc)}) \right| \le \delta f_j (x^{*}) \end{aligned}$$

Case 2.4 \(P_j^2 (x^{*})=T_2,\, P_j^2 (x^{(a_1 a_2 bc)})>T_2 \)

$$\begin{aligned} \left| {f_j (x^{*})-f_j \left( x^{(a_1 a_2 bc)}\right) } \right|&= \left| P_j^1 (x^{*})+W_j (x^{*})-P_j^2 \left( x^{(a_1 a_2 bc)}\right) \right. \\&\left. -W_j (x^{(a_1 a_2 bc)}) \right| \le \delta f_j (x^{*}) \end{aligned}$$

In the following we show that Result 2 holds for each of the above four cases. We only consider Case 2.1 and Case 2.3 since Case 2.2 and Case 2.4 can be similarly proved.

For Case 2.1,

$$\begin{aligned} \left| {f_{j+1} (x^{*})-f_{j+1} \left( \tilde{x}^{(a_1 a_2 bc)}\right) } \right|&= \left| P_{j+1}^1 (x^{*})+W_{j+1} (x^{*})-P_{j+1}^1 \left( \tilde{x}^{(a_1 a_2 bc)}\right) \right. \\&\left. -W_{j+1} \left( \tilde{x}^{(a_1 a_2 bc)}\right) \right| \\&\le \delta _1 P_{j+1}^1 (x^{*})+\delta _1 W_{j+1} (x^{*})\\&= \delta _1 f_{j+1} (x^{*}) \end{aligned}$$

For Case 2.3,

$$\begin{aligned} \left| {f_j (x^{*})-f_j \left( x^{(a_1 a_2 bc)}\right) } \right| =\left| {P_j^2 (x^{*})+W_j (x^{*})-P_j^1 \left( x^{(a_1 a_2 bc)}\right) -W_j (x^{(a_1 a_2 bc)})} \right| \end{aligned}$$

If \(P_j^2 (x^{*})+W_j (x^{*})-P_j^1 (x^{(a_1 a_2 bc)})-W_j (x^{(a_1 a_2 bc)})\ge 0\), then

$$\begin{aligned} \left| {f_j (x^{*})-f_j \left( x^{(a_1 a_2 bc)}\right) } \right|&= P_j^2 (x^{*})-P_j^1 (x^{(a_1 a_2 bc)})+W_j (x^{*})\nonumber \\&-W_j \left( x^{(a_1 a_2 bc)}\right) \le \delta f_j (x^{*}) \end{aligned}$$
(12)

Otherwise,

$$\begin{aligned} \left| {f_j (x^{*})-f_j \left( x^{(a_1 a_2 bc)}\right) } \right|&= P_j^1 (x^{(a_1 a_2 bc)})-P_j^2 (x^{*})+W_j \left( x^{(a_1 a_2 bc)}\right) \nonumber \\&-W_j (x^{*})\le \delta f_j (x^{*}) \end{aligned}$$
(13)

Either (12) or (13) is hold. Hence, we consider (12) and (13) in the following, respectively.

  • If (12) is established, then we have

    $$\begin{aligned} \left| {f_{j+1} (x^{*})-f_{j+1} \left( \tilde{x}^{(a_1 a_2 bc)}\right) } \right|&= P_{j+1}^2 (x^{*})-P_{j+1}^1 \left( \tilde{x}^{(a_1 a_2 bc)}\right) +W_{j+1} (x^{*})\nonumber \\&-W_{j+1} \left( \tilde{x}^{(a_1 a_2 bc)}\right) \end{aligned}$$
    (14)

or

$$\begin{aligned} \left| {f_{j+1} (x^{*})-f_{j+1} \left( \tilde{x}^{(a_1 a_2 bc)}\right) } \right|&= P_{j+1}^1 \left( \tilde{x}^{(a_1 a_2 bc)}\right) -P_{j+1}^2 (x^{*})\nonumber \\&+W_{j+1} \left( \tilde{x}^{(a_1 a_2 bc)}\right) -W_{j+1} (x^{*}) \end{aligned}$$
(15)

Consider (14), we have

$$\begin{aligned} \left| {f_{j+1} (x^{*})-f_{j+1} \left( \tilde{x}^{(a_1 a_2 bc)}\right) } \right|&= P_{j+1}^2 (x^{*})-P_{j+1}^1 \left( \tilde{x}^{(a_1 a_2 bc)}\right) +W_{j+1} (x^{*})\\&-W_{j+1} \left( \tilde{x}^{(a_1 a_2 bc)}\right) \\&= P_j^2 (x^{*})-P_{j+1}^1 \left( \tilde{x}^{(a_1 a_2 bc)}\right) +W_j (x^{*})\\&-W_j \left( x^{(a_1 a_2 bc)}\right) \\&\le P_j^2 (x^{*})-P_j^1 \left( x^{(a_1 a_2 bc)}\right) +W_j (x^{*})\\&-W_j \left( x^{(a_1 a_2 bc)}\right) \\&\le \delta f_j (x^{*})\le \delta _1 f_{j+1} (x^{*})\, (\hbox {obtained from }(12)). \end{aligned}$$

Consider (15), we have

$$\begin{aligned} \left| {f_{j+1} (x^{*})-f_{j+1} \left( \tilde{x}^{(a_1 a_2 bc)}\right) } \right|&= P_{j+1}^1 \left( \tilde{x}^{(a_1 a_2 bc)}\right) -P_{j+1}^2 (x^{*})+W_{j+1} \left( \tilde{x}^{(a_1 a_2 bc)}\right) \nonumber \\&-W_{j+1} (x^{*})\nonumber \\&\le P_{j+1}^1 \left( \tilde{x}^{(a_1 a_2 bc)}\right) -P_{j+1}^1 (x^{*})+W_{j+1} \left( \tilde{x}^{(a_1 a_2 bc)}\right) \nonumber \\&-W_{j+1} (x^{*})\nonumber \\&\le \delta _1 P_{j+1}^1 (x^{*})+\delta _1 W_{j+1} (x^{*})\nonumber \\&\le \delta _1 P_{j+1}^2 (x^{*})+\delta _1 W_{j+1} (x^{*})\nonumber \\&= \delta _1 f_{j+1} (x^{*}) \end{aligned}$$
(16)
  • While if (13) is established, then (15) must follow, and (14) is impossible.

Consider (15), we have

$$\begin{aligned} \left| {f_{j+1} (x^{*})-f_{j+1} (\tilde{x}^{(a_1 a_2 bc)})} \right| \le \delta _1 f_{j+1} (x^{*})\, (\hbox {obtained from }(16)). \end{aligned}$$

Consequently, for both (12) and (13) in Case 2.3, we can obtain

$$\begin{aligned} \left| {f_{j+1} (x^{*})-f_{j+1} (\tilde{x}^{(a_1 a_2 bc)})} \right| \le \delta _1 f_{j+1} (x^{*}),\quad \hbox { for }x_{j+1} =1. \end{aligned}$$

Based on the analysis above, for each of the four cases we have

$$\begin{aligned} \left| {f_{j+1} (x^{*})-f_{j+1} (\tilde{x}^{(a_1 a_2 bc)})} \right| \le \delta _1 f_{j+1} (x^{*}),\quad \hbox { for }x_{j+1} =1. \end{aligned}$$

This completes the proof of Result 2.

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Li, WX., Zhao, CL. Deteriorating jobs scheduling on a single machine with release dates, rejection and a fixed non-availability interval. J. Appl. Math. Comput. 48, 585–605 (2015). https://doi.org/10.1007/s12190-014-0820-3

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