Minimizing Makespan Scheduling on a Single Machine with General Positional Deterioration Effects

Abstract

This work studies single-machine scheduling with general position-dependent deterioration, where job processing times are general non-decreasing functions dependent on their positions in a sequence. The goal is to find a job sequence such that makespan is minimized. The problem can be extended to deal with green scheduling environment where processing time increases due to additional carbon-reduction procedure. Under some optimal properties, we prove that the problem is solved by the largest processing time (denoted by LPT) first rule.

1. Introduction

Scheduling theory, as a core discipline in operations research, focuses on optimizing the temporal allocation of jobs to machines under specific constraints. Originating during World War II with the development of operations research, this field has evolved to address complex real-world problems ranging from manufacturing systems to healthcare management. Modern scheduling research encompasses two fundamental aspects: job sequencing and resource–time coordination. The terminology distinction between “Scheduling” and “Sequencing” is crucial (see Baker [1], Pinedo [2], Biskup [3], and Zhang et al. [4]), where scheduling involves comprehensive temporal planning while Sequencing focuses purely on order optimization. This theoretical framework has been significantly advanced through cross-disciplinary contributions from mathematics, computer science, and economics (see Potts and Strusevich [5], Azzouz et al. [6], Khatami et al. [7], Li and Goossens [8]).

In many real manufacturing processes and scheduling, one of the assumptions is that the job processing times have deterioration effects, i.e., the of jobs processing times are non-decreasing (or an increasing) functions of their starting times or positions in a sequence. This type of scenario is widely present in the context of green manufacturing where extra decarbonation procedure is introduced. Generally, there are two deterioration effects; one is the time-dependent deterioration (denoted by $\widetilde{TDD}$, see Agnetis et al. [9], Strusevich and Rustogi [10], Gawiejnowicz [11], Huang [12], Pei et al. [13]). Recently, Sun et al. [14] considered scheduling with the $\widetilde{TDD}$ and a maintenance activity. Jiang et al. [15] and Wang et al. [16] studied resource allocation scheduling problems with the $\widetilde{TDD}$. Miao et al. [17] studied total (weighted) completion time minimization scheduling with the step-$\widetilde{TDD}$. Wu et al. [18] and Cheng et al. [19] considered scheduling problems with the step-improving processing times, i.e., the step-$\widetilde{TDD}$. For minimizing the linear sum of total weighted completion time and total tardiness, Wu et al. [18] proposed a branch-and-bound algorithm and some heuristics. To minimize the total weighted completion time, Cheng et al. [19] proposed a pseudo-polynomial algorithm. Zhang et al. [20] considered due window assignment scheduling with the $\widetilde{TDD}$. Gkikas et al. [21] studied scheduling with the $\widetilde{TDD}$, where the deteriorating rates are identical. For the total weighted completion time minimization, they presented some new bounds. Lu et al. [22] addressed delivery times scheduling with the $\widetilde{TDD}$. Zhang et al. [23] considered two-agent problems with the $\widetilde{TDD}$. Lv and Wang [24] studied no-idle flow shop problems with the $\widetilde{TDD}$. Under the common due date, they proved that some special problems are polynomially solvable. Yin and Gao [25] conducted group scheduling with the $\widetilde{TDD}$ and learning effects. Li et al. [26] considered the flexible flow shop problem with the step-$\widetilde{TDD}$. For the makespan minimization, they proposed a discrete artificial bee colony algorithm. Qiu and Wang [27] studied mixed due-windows scheduling with the $\widetilde{TDD}$. Choi et al. [28] addressed scheduling with the step-$\widetilde{TDD}$ and multiple critical dates. For the NP-hard problem of minimizing total weighted completion time, they presented some solution algorithms. Sun et al. [29]) considered the flow shop problem with the shortening-$\widetilde{TDD}$. For the makespan minimization, they proposed some algorithms and a mathematical programming method.

In addition, there are situations in which the job processing times are non-increasing functions of their positions in a sequence; this is the positional learning effects (denoted by $\widetilde{PLE}$, see Vitaly and Strusevich [30], Jiang et al. [31], Sun et al. [32], Zhao [33], Liu and Wang [34], Bai et al. [35], Zhang et al. [36], Saavedra-Nieves et al. [37], Zhang et al. [38], Lv and Wang [39]). In the case of the opposite effects, this is the positional deterioration effects (denoted by $\widetilde{PDE}$, see Cohen and Shapira [40], Gerstl and Mosheiov [41], Hu et al. [42]). The real-world applications of $\widetilde{PDE}$ often appear in steel productions, logistics and manufacturing. For example, in a steel mill, the ingot needs to be preheated to a certain temperature before processing. If the ingot is waiting in a buffer, then the ingot needs to be reheated for rolling (Gawiejnowicz [11], Bachman and Janiak [43], Mosheiov [44]). Bachman and Janiak [43] considered the linear $\widetilde{PDE}$, i.e., if job $JO{B}_j$ is placed in the rth position, the actual processing time of $JO{B}_j$ is

$${\widehat{Pro}}_{jr}^a=A_j+B_{j}r,$$

where $A_j\ge 0$ (resp. $B_j\ge 0$) is the basic processing time (resp. deterioration factor) of $JO{B}_j$. Mosheiov [44] studied the polynomial $\widetilde{PDE}$:

$${\widehat{Pro}}_{jr}^a=A_j{r}^{\tilde{a}},$$

where $\tilde{a}\ge 0$ is the deterioration (aging) index. Gordon et al. [45] considered the exponential $\widetilde{PDE}$:

$${\widehat{Pro}}_{jr}^a=A_j{b}^{r-1},$$

where $b\ge 1$ ($0<b\le 1$) is the deterioration (learning) index. Strusevich and Rustogi [10] considered the $\widetilde{PDE}$ model:

$${\widehat{Pro}}_{jr}^a=A_{j}h\left(r\right),$$

where $h\left(r\right)$ is non-increasing (non-decreasing) order on r. Wang et al. [46] considered the cumulative $\widetilde{PDE}$:

$${\widehat{Pro}}_{jr}^a=A_j{\left(1+\sum _{l=1}^{r-1}{A}_{<l>}\right)}^{\alpha 1},$$

where $\alpha 1\ge 0$ is the deterioration index, and $<l>$ denotes some job scheduled in lth position. Lee et al. [47] considered the $\widetilde{PDE}$ model:

$${\widehat{Pro}}_{jr}^a=A_j{\left(1+{\displaystyle \frac{{\sum }_{l=1}^{r-1}{A}_{<l>}}{{\sum }_{l=1}^n{A}_l}}\right)}^{\alpha 2},$$

where $0\le \alpha 2\le 1$ is a deterioration index. Lai et al. [48] considered the $\widetilde{PDE}$ model:

$${\widehat{Pro}}_{jr}^a=A_j{\left(z\left(r\right)+\sum _{l=1}^{r-1}{\Theta }_{l}log{A}_{<l>}\right)}^{\alpha 3},$$

where $\alpha 3\ge 1$ is the deterioration index, $z\left(r\right)\ge 1$ is a non-increasing function on position r, ${\Theta }_l$ is the positional-weight of lth position and $0\le {\Theta }_n\le {\Theta }_{n-1}\le \dots \le {\Theta }_2\le {\Theta }_1$. Huang and Wang [49] considered the $\widetilde{PDE}$ model:

$${\widehat{Pro}}_{jr}^a=A_j{\left(1+\sum _{l=1}^{r-1}{\Theta }_l{A}_{<l>}\right)}^{\alpha 4},$$

where $\alpha 4\ge 1$ is the deterioration index. Miao et al. [50] considered the $\widetilde{PDE}$ models:

$${\widehat{Pro}}_{jr}^a=A_j\left[R+(1-R){\left(1+{\displaystyle \frac{{\sum }_{l=1}^{r-1}ln{A}_{<l>}}{{\sum }_{l=1}^n{A}_l}}\right)}^{\alpha 5}\right],$$

and

$${\widehat{Pro}}_{jr}^a=A_j\left[R+(1-R){\left(1+\sum _{l=1}^{r-1}{\Theta }_l{A}_{<l>}\right)}^{\alpha 6}\right],$$

where $0<R<1$, $0\le \alpha 5\le 1$, $\alpha 6\ge 0$ is the deterioration index, and $A_j\ge e$ ($ln{A}_j\ge 0$).

In this paper, we extend the above models (including the references Strusevich and Rustogi [10], Bachman and Janiak [43], Mosheiov [44], Gordon et al. [45], Wang et al. [46], Lee et al. [47], Lai et al. [48], Huang and Wang [49], and Miao et al. [50]) to a more general $\widetilde{PDE}$ model.

Table 1. Comparative summary of key $\widetilde{PDE}$ models and their characteristics.

Reference	Model Formulation	Limitations
Bachman & Janiak [43]	${\widehat{Pro}}_{jr}^a=A_j+B_{j}r$	Fixed linear structure, no constraint handling
Mosheiov [44]	${\widehat{Pro}}_{jr}^a=A_j{r}^{\tilde{a}}$	Monotonous deterioration pattern
Gordon et al. [45]	${\widehat{Pro}}_{jr}^a=A_j{b}^{r-1}$	Limited to geometric progression
Wang et al. [46]	${\widehat{Pro}}_{jr}^a=A_j{(1+\sum A_{<l>})}^{\alpha 1}$	No position-dependent weights
Lee et al. [47]	${\widehat{Pro}}_{jr}^a=A_j{(1+{\displaystyle \frac{\sum A_{<l>}}{\sum A_l}})}^{\alpha 2}$	Requires total sum constraint
Lai et al. [48]	${\widehat{Pro}}_{jr}^a=A_j{(z\left(r\right)+\sum {\Theta }_{l}log{A}_{<l>})}^{\alpha 3}$	Complex parameter tuning
Huang & Wang [49]	${\widehat{Pro}}_{jr}^a=A_j{(1+\sum {\Theta }_l{A}_{<l>})}^{\alpha 4}$	No adaptive constraints
Miao et al. [50]	${\widehat{Pro}}_{jr}^a=A_j[R+(1-R)(·)]$	Limited functional combinations
Proposed	${\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)min\left\{f(·)h\left(r\right),\Xi \right\}\right)$	Unified framework with adaptive constraints

provides a comparative analysis of these models in terms of formulation characteristics and limitations, which motivates the need for our generalized $\widetilde{PDE}$ framework. The aim is to minimize the makespan. It is shown that the problem is polynomially solvable. The paper is organized as follows: Section 2 presents the notations and model. Section 3 proves that the problem is polynomially solvable. Section 4 concludes the paper.

2. Formulation

We assume that n jobs $JO{B}_1,JO{B}_2,\cdots ,JO{B}_n$ are to be processed on a single machine, and all jobs are available at time 0. Let $A_j$ be the basic processing time of $JO{B}_j$. In this article, we define a general positional deterioration (denoted by $\widetilde{GPDE}$), i.e., if job $JO{B}_j$ is placed in the rth position, the actual processing time of $JO{B}_j$ is

$${\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)min\left\{f\left(\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)h\left(r\right),\Xi \right\}\right),$$

(1)

$${\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)min\left\{f\left(z\left(r\right)+\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right),\Xi \right\}\right),$$

(2)

where the following is true:

• $f:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is a non-decreasing concave function satisfying $f^{\prime }\left(x\right)\ge 0$ and $f^{″}\left(x\right)\le 0$. This converts accumulated deterioration effects into time-increasing factors. Typical examples include $f\left(x\right)=ln(1+x)$ or saturation functions like $f\left(x\right)={\displaystyle \frac{x}{K+x}}$, where $K>0$ is a saturation constant [45,46].
• $g:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is a concave transformation of basic processing times with $g^{\prime }\left(x\right)\ge 0$, $g^{″}\left(x\right)\le 0$, modeling diminishing marginal effects of accumulated workloads [48].
• $h\left(r\right):{\mathbb{N}}^{+}\to {\mathbb{R}}^{+}$ is a non-decreasing positional weighting function with $h\left(1\right)=1$, reflecting progressive deterioration through job positions [44].
• $z\left(r\right):{\mathbb{N}}^{+}\to {\mathbb{R}}^{+}$ is a non-decreasing baseline function with $z\left(1\right)=1$, representing intrinsic positional complexity. Common forms include $z\left(r\right)=r^{\alpha }$ ($\alpha \ge 0$) or $z\left(r\right)=log(1+r)$ [49].
• $0\le R\le 1$ controls the baseline processing proportion [50].
• ${\Theta }_l$ denotes positional weights satisfying $0\le {\Theta }_n\le {\Theta }_{n-1}\le \cdots \le {\Theta }_1$, emphasizing earlier positions’ contribution to deterioration [49].
• $\Xi$ denotes truncated deterioration parameter and can be determined in practice, which emphasizing the upper limit of the positional deteriorating effect.

The objective is minimum the makespan, i.e., maximal completion time: $C_{max}=max\{C_j,j=1,\dots ,n\}$, where $C_j$ denotes the completion time of $JO{B}_j$, this problem studied here is expressed as

$$1\left|{\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)min\left\{f\left(\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)h\left(r\right),\Xi \right\}\right)\right|C_{max}$$

(3)

and

$$1\left|{\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)min\left\{f\left(z\left(r\right)+\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right),\Xi \right\}\right)\right|C_{max}.$$

(4)

The papers related to $C_{max}$ with $\widetilde{PDE}$ and/or $\widetilde{PLE}$ are presented by

Table 2. $C_{max}$ scheduling with $\widetilde{PDE}$ and/or $\widetilde{PLE}$.

Problem	Complexity	Paper
$1\left|{\widehat{Pro}}_{jr}^a=A_j+B_{j}r\right|C_{max}$	$O(nlogn)$, LPT ($B_j\downarrow$)	Bachman and Janiak [43]
$1\left|{\widehat{Pro}}_{jr}^a=A_j{r}^{\tilde{a}},\tilde{a}\ge 0\right|C_{max}$	$O(nlogn)$, LPT ($A_j\downarrow$)	Mosheiov [44]
$1\left|{\widehat{Pro}}_{jr}^a=A_j{b}^{r-1},0<b\le 1\right|C_{max}$	$O(nlogn)$, SPT ($A_j\uparrow$)	Gordon  et al. [45]
$1\left|{\widehat{Pro}}_{jr}^a=A_j{b}^{r-1},b\ge 1\right|C_{max}$	$O(nlogn)$, LPT ($A_j\downarrow$)	Gordon et al. [45]
$1\left|{\widehat{Pro}}_{jr}^a=A_{j}h\left(r\right),h\left(r\right)\downarrow \right|C_{max}$	$O(nlogn)$, SPT ($A_j\uparrow$)	Strusevich and Rustogi [10]
$1\left|{\widehat{Pro}}_{jr}^a=A_{j}h\left(r\right),h\left(r\right)\uparrow \right|C_{max}$	$O(nlogn)$, LPT ($A_j\downarrow$)	Strusevich and Rustogi [10]
$1\left|{\widehat{Pro}}_{jr}^a=A_j{\left(1+{\sum }_{l=1}^{r-1}{A}_{<l>}\right)}^{\alpha 1},\alpha 1\ge 1\right|C_{max}$	$O(nlogn)$, SPT ($A_j\uparrow$)	Wang et al. [46]
$1\left|{\widehat{Pro}}_{jr}^a=A_j{\left(1+{\sum }_{l=1}^{r-1}{A}_{<l>}\right)}^{\alpha 1},0\le \alpha 1\le 1\right|C_{max}$	$O(nlogn)$, LPT ($A_j\downarrow$)	Wang et al. [46]
$1\left|{\widehat{Pro}}_{jr}^a=A_j{\left(1+{\displaystyle \frac{{\sum }_{l=1}^{r-1}{A}_{<l>}}{{\sum }_{l=1}^n{A}_l}}\right)}^{\alpha 2},0\le \alpha 2\le 1\right|C_{max}$	$O(nlogn)$, LPT ($A_j\downarrow$)	Lee et al. [47]
$1\left|{\widehat{Pro}}_{jr}^a=A_j{\left(z\left(r\right)+{\sum }_{l=1}^{r-1}{\Theta }_{l}log{A}_{<l>}\right)}^{\alpha 3},\alpha 3\ge 1,0\le {\Theta }_n\le {\Theta }_{n-1}\le \cdots \le {\Theta }_1\right|C_{max}$	$O(nlogn)$, SPT ($A_j\uparrow$)	Lai et al. [48]
$1\left|{\widehat{Pro}}_{jr}^a=A_j{\left(1+{\sum }_{l=1}^{r-1}{\Theta }_l{A}_{<l>}\right)}^{\alpha 4},\alpha 4\ge 1,0\le {\Theta }_n\le {\Theta }_{n-1}\le \dots \le {\Theta }_2\le {\Theta }_1\right|C_{max}$	$O(nlogn)$, SPT ($A_j\uparrow$)	Huang and Wang [49]
$1\left|{\widehat{Pro}}_{jr}^a=A_j{\left(1+{\sum }_{l=1}^{r-1}{\Theta }_l{A}_{<l>}\right)}^{\alpha 4},0\le \alpha 4\le 1,0\le {\Theta }_1\le {\Theta }_2\le \dots \le {\Theta }_{n-1}\le {\Theta }_n\right|C_{max}$	$O(nlogn)$, LPT ($A_j\downarrow$)	Miao et al. [50]
$1\left|{\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R){\left(1+{\displaystyle \frac{{\sum }_{l=1}^{r-1}ln{A}_{<l>}}{{\sum }_{l=1}^n{A}_l}}\right)}^{\alpha 5}\right),0\le \alpha 5\le 1\right|C_{max}$	$O(nlogn)$, LPT ($A_j\downarrow$)	Miao et al. [50]
$1\left|{\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R){\left(1+{\sum }_{l=1}^{r-1}{\Theta }_l{A}_{<l>}\right)}^{\alpha 6}\right),0\le \alpha 6\le 1,0\le {\Theta }_1\le {\Theta }_2\le \dots \le {\Theta }_{n-1}\le {\Theta }_n\right|C_{max}$	$O(nlogn)$, LPT ($A_j\downarrow$)	Miao et al. [50]
$1\left|{\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R){\left(1+{\sum }_{l=1}^{r-1}{\Theta }_l{A}_{<l>}\right)}^{\alpha 6}\right),\alpha 6\ge 1,0\le {\Theta }_n\le {\Theta }_{n-1}\le \dots \le {\Theta }_2\le {\Theta }_1\right|C_{max}$	$O(nlogn)$, SPT ($A_j\downarrow$)	Miao et al. [50]
$1\left|{\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)min\left\{f\left({\sum }_{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)h\left(r\right),\Xi \right\}\right)\right|C_{max}$	$O(nlogn)$, LPT ($A_j\downarrow$)	this paper
$1\left|{\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)min\left\{f\left(z\left(r\right)+{\sum }_{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right),\Xi \right\}\right)\right|C_{max}$	$O(nlogn)$, LPT ($A_j\downarrow$)	this paper
$1\left|{\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)max\left\{f\left({\sum }_{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)h\left(r\right),\Xi \right\}\right)\right|C_{max}$	$O(nlogn)$, LPT ($A_j\downarrow$)	this paper
$1\left|{\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)max\left\{f\left(z\left(r\right)+{\sum }_{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right),\Xi \right\}\right)\right|C_{max}$	$O(nlogn)$, LPT ($A_j\downarrow$)	this paper

$y\uparrow$ (resp. $y\downarrow$) represents non-decreasing (resp. non-increasing) order of some parameter y.

.

3. Main Result

Let $JO{B}_{\left[j\right]}$ be some job scheduled in jth position, for a given schedule

$$\varpi =\{JO{B}_{\left[1\right]},JO{B}_{\left[2\right]},\cdots ,JO{B}_{\left[n\right]}\}$$

of the models

$${\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)min\left\{f\left(\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)h\left(r\right),\Xi \right\}\right)$$

and

$${\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)min\left\{f\left(z\left(r\right)+\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right),\Xi \right\}\right),$$

we have

$$\begin{array}{cc}\hfill C_{max}\left(\varpi \right)& =\sum _{j=1}^n{A}_{\left[j\right]}\left(R+(1-R)min\left\{f\left(\sum _{l=1}^{j-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)h\left(j\right),\Xi \right\}\right)\hfill \end{array}$$

(5)

and

$$\begin{array}{cc}\hfill C_{max}\left(\varpi \right)& =\sum _{j=1}^n{A}_{\left[j\right]}\left(R+(1-R)min\left\{f\left(z\left(j\right)+\sum _{l=1}^{j-1}{\Theta }_{l}g\left(A_{<l>}\right)\right),\Xi \right\}\right).\hfill \end{array}$$

(6)

Lemma 1.

$W\left(x\right)={\displaystyle \frac{f(Z+{\Theta }_{r}g\left(x\right))h(r+1)-f\left(Z\right)h\left(r\right)}{x}}$ is a non-increasing function on $x\ge 0$, where $f^{\prime }\left(x\right)\ge 0$, $f^{″}\left(x\right)\le 0$, $g^{″}\left(x\right)\le 0$, and $h\left(r\right)\ge 1$ is a non-decreasing function.

Proof. 

We first compute the derivative of $W\left(x\right)$. For $x>0$,

$$W^{\prime }\left(x\right)={\displaystyle \frac{{\Theta }_r{f}^{\prime }\left(Z+{\Theta }_{r}g\left(x\right)\right)h(r+1)g^{\prime }\left(x\right)x-\left[f\left(Z+{\Theta }_{r}g\left(x\right)\right)h(r+1)-f\left(Z\right)h\left(r\right)\right]}{x^2}}.$$

Define the numerator as:

$$V\left(x\right)={\Theta }_r{f}^{\prime }\left(Z+{\Theta }_{r}g\left(x\right)\right)h(r+1)g^{\prime }\left(x\right)x-f\left(Z+{\Theta }_{r}g\left(x\right)\right)h(r+1)+f\left(Z\right)h\left(r\right).$$

We analyze $V\left(x\right)$ by computing its derivative:

$$\begin{array}{cc}\hfill V^{\prime }\left(x\right)& ={\Theta }_r^{2}xh(r+1)f^{″}\left(Z+{\Theta }_{r}g\left(x\right)\right){\left(g^{\prime }\left(x\right)\right)}^2+{\Theta }_{r}xh(r+1)f^{\prime }\left(Z+{\Theta }_{r}g\left(x\right)\right)g^{″}\left(x\right)\hfill \\ \hfill & ={\Theta }_{r}xh(r+1)\left[\underset{\le 0}{\underbrace{{\Theta }_r{f}^{″}\left(Z+{\Theta }_{r}g\left(x\right)\right){\left(g^{\prime }\left(x\right)\right)}^2}}+\underset{\le 0}{\underbrace{f^{\prime }\left(Z+{\Theta }_{r}g\left(x\right)\right)g^{″}\left(x\right)}}\right].\hfill \end{array}$$

Since $f^{″}\left(x\right)\le 0$, $g^{″}\left(x\right)\le 0$, and ${\Theta }_r,h(r+1)>0$, all terms in $V^{\prime }\left(x\right)$ are non-positive. Thus, $V^{\prime }\left(x\right)\le 0$, implying $V\left(x\right)$ is non-increasing on $x\ge 0$.

Next, we evaluate the limit as $x\to 0^{+}$:

$$\underset{x\to 0^{+}}{lim}V\left(x\right)=-f(Z+{\Theta }_{r}g\left(0\right))h(r+1)+f\left(Z\right)h\left(r\right).$$

Given $h(r+1)>h\left(r\right)$ and f is non-decreasing, if ${\Theta }_{r}g\left(0\right)\ge 0$, then $f(Z+{\Theta }_{r}g\left(0\right))\ge f\left(Z\right)$. Thus,

$$\begin{array}{cc}\hfill -f(Z+{\Theta }_{r}g\left(0\right))h(r+1)+f\left(Z\right)h\left(r\right)& \le -f\left(Z\right)h(r+1)+f\left(Z\right)h\left(r\right)\hfill \\ \hfill & =f\left(Z\right)\left(h\left(r\right)-h(r+1)\right)\le 0.\hfill \end{array}$$

Since $V\left(x\right)$ is non-increasing and $V\left(0^{+}\right)\le 0$, it follows that $V\left(x\right)\le 0$ for all $x\ge 0$. Therefore, $W^{\prime }\left(x\right)=V\left(x\right)/x^2\le 0$, proving that $W\left(x\right)$ is non-increasing on $x\ge 0$. □

Lemma 2.

$Q\left(x\right)={\displaystyle \frac{f[G+Z(r+1)+{\Theta }_{r}g\left(x\right)]-f[G+Z\left(r\right)]}{x}}$ is a non-increasing function on $x\ge 0$, where where $f^{\prime }\left(x\right)\ge 0$, $f^{″}\left(x\right)\le 0$, and $g^{″}\left(x\right)\le 0$.

Proof. 

Similar to the proof of Lemma 1. □

Theorem 1.

For $1\left|{\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)min\left\{f\left({\sum }_{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)h\left(r\right),\Xi \right\}\right)\right|C_{max}$, if $f^{\prime }\left(x\right)$$={\displaystyle \frac{df\left(x\right)}{dx}}\ge 0,f^{″}\left(x\right)={\displaystyle \frac{d^{2}f\left(x\right)}{d{x}^2}}\le 0$, $g^{\prime }\left(x\right)={\displaystyle \frac{dg\left(x\right)}{dx}}\ge 0$, $g^{″}\left(x\right)={\displaystyle \frac{d^{2}g\left(x\right)}{d{x}^2}}\le 0$, $h\left(r\right)\ge 1$ is a non-decreasing function, and $0\le {\Theta }_1\le {\Theta }_2\le \dots \le {\Theta }_{n-1}\le {\Theta }_n$, the optimal schedule is obtained in $O(nlogn)$ time by the LPT (largest processing time) rule, i.e., by renumbering the jobs in non-increasing order of $A_j$.

Proof. 

Consider two adjacent job sequences ${\varpi }^{\prime }=({\vartheta }_1,J_i,J_j,{\vartheta }_2)$ and ${\varpi }^{″}=({\vartheta }_1,J_j,J_i,{\vartheta }_2)$ where $A_i\le A_j$. Let H be the completion time of the last job in ${\vartheta }_1$ containing $r-1$ jobs. Define the cumulative learning effect as $Z_r:={\sum }_{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)$. The completion times are

$$C_i\left({\varpi }^{\prime }\right)=H+A_i\left[R+(1-R)min\left\{f\left(Z_r\right)h\left(r\right),\Xi \right\}\right]$$

$$C_j\left({\varpi }^{\prime }\right)=C_i\left({\varpi }^{\prime }\right)+A_j\left[R+(1-R)min\left\{f(Z_r+{\Theta }_{r}g\left(A_i\right))h(r+1),\Xi \right\}\right]$$

(7)

$$C_j\left({\varpi }^{″}\right)=H+A_j\left[R+(1-R)min\left\{f\left(Z_r\right)h\left(r\right),\Xi \right\}\right]$$

$$C_i\left({\varpi }^{″}\right)=C_j\left({\varpi }^{″}\right)+A_i\left[R+(1-R)min\left\{f(Z_r+{\Theta }_{r}g\left(A_j\right))h(r+1),\Xi \right\}\right]$$

(8)

The makespan difference is analyzed through four exhaustive cases:

Case 1: if $f(Z_r+{\Theta }_{r}g\left(A_j\right))h(r+1)\le \Xi$, hence $f\left(Z_r\right)h\left(r\right)\le f(Z_r+{\Theta }_{r}g\left(A_i\right))h(r+1)\le f(Z_r+{\Theta }_{r}g\left(A_j\right))h(r+1)\le \Xi$, by Lemma 1, Equations (7) and (8), it obtains

$$\begin{array}{cc}\hfill \Delta & =C_i\left({\varpi }^{″}\right)-C_j\left({\varpi }^{\prime }\right)\hfill \\ & =(1-R)(A_j-A_i)min\left\{f\left(\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)h\left(r\right),\Xi \right\}\hfill \\ & +(1-R)A_{i}min\left\{f\left(\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)+{\Theta }_{r}g\left(A_j\right)\right)h(r+1),\Xi \right\}\hfill \\ & -(1-R)A_{j}min\left\{f\left(\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)+{\Theta }_{r}g\left(A_i\right)\right)h(r+1),\Xi \right\}\hfill \\ \hfill & =(1-R)[A_{j}f\left(Z_r\right)h\left(r\right)-A_{i}f\left(Z_r\right)h\left(r\right)\hfill \\ \hfill & +A_{i}f(Z_r+{\Theta }_{r}g\left(A_j\right))h(r+1)-A_{j}f(Z_r+{\Theta }_{r}g\left(A_i\right))h(r+1)]\hfill \\ \hfill & =(1-R)A_i{A}_j\left[{\displaystyle \frac{f(Z_r+{\Theta }_{r}g\left(A_j\right))h(r+1)-f\left(Z_r\right)h\left(r\right)}{A_j}}\right.\hfill \\ \hfill & \left.-{\displaystyle \frac{f(Z_r+{\Theta }_{r}g\left(A_i\right))h(r+1)-f\left(Z_r\right)h\left(r\right)}{A_i}}\right]\hfill \\ \hfill & \le 0\hfill \end{array}$$

by the concave function’s subadditivity:

$${\displaystyle \frac{f(x+{\Delta }_2)-f\left(x\right)}{{\Delta }_2}}\le {\displaystyle \frac{f(x+{\Delta }_1)-f\left(x\right)}{{\Delta }_1}}$$

for ${\Delta }_2\ge {\Delta }_1$.

Case 2: if $f\left(Z_r\right)h\left(r\right)\le f(Z_r+{\Theta }_{r}g\left(A_i\right))h(r+1)\le \Xi \le f(Z_r+{\Theta }_{r}g\left(A_j\right))h(r+1)$, we have

$$\begin{array}{cc}\hfill \Delta & =(1-R)\left[(A_j-A_i)f\left(Z_r\right)h\left(r\right)+A_i\Xi -A_{j}f(Z_r+{\Theta }_{r}g\left(A_i\right))h(r+1)\right]\hfill \\ \hfill & \le (1-R)\left[(A_j-A_i)f\left(Z_r\right)h\left(r\right)+A_i\Xi -A_j\Xi \right](\sin \mathrm{ce}f(Z_r+{\Theta }_{r}g\left(A_i\right))h(r+1)\le \Xi )\hfill \\ \hfill & =(1-R)\left[A_j(f\left(Z_r\right)h\left(r\right)-\Xi )-A_i(f\left(Z_r\right)h\left(r\right)-\Xi )\right]\hfill \\ \hfill & =(1-R)(A_j-A_i)(f\left(Z_r\right)h\left(r\right)-\Xi )\le 0(\mathrm{by}{A}_j\ge A_i\mathrm{and}f\left(Z_r\right)h\left(r\right)\le \Xi )\hfill \end{array}$$

Case 3: if $f\left(Z_r\right)h\left(r\right)\le \Xi \le f(Z_r+{\Theta }_{r}g\left(A_i\right))h(r+1)\le f(Z_r+{\Theta }_{r}g\left(A_j\right))h(r+1)$, we have

$$\begin{array}{cc}\hfill \Delta & =(1-R)\left[(A_j-A_i)f\left(Z_r\right)h\left(r\right)+A_i\Xi -A_j\Xi \right]\hfill \\ \hfill & =(1-R)\left[A_j(f\left(Z_r\right)h\left(r\right)-\Xi )-A_i(f\left(Z_r\right)h\left(r\right)-\Xi )\right]\hfill \\ \hfill & =(1-R)(A_j-A_i)(f\left(Z_r\right)h\left(r\right)-\Xi )\le 0(\mathrm{same}\mathrm{reasoning}\mathrm{as}\mathrm{Case}2)\hfill \end{array}$$

Case 4: if $\Xi \le f\left(Z_r\right)h\left(r\right)\le f(Z_r+{\Theta }_{r}g\left(A_i\right))h(r+1)\le f(Z_r+{\Theta }_{r}g\left(A_j\right))h(r+1)$, we have

$$\begin{array}{cc}\hfill \Delta & =(1-R)\Xi [(A_j-A_i)+(A_i-A_j)]=0(\mathrm{both}\mathrm{schedules}\mathrm{hit}\mathrm{the}\mathrm{upper}\mathrm{bound}\Xi )\hfill \end{array}$$

From $A_i\le A_j$, $f^{\prime }\left(x\right)={\displaystyle \frac{df\left(x\right)}{dx}}\ge 0$, $g\left(A_i\right)\le g\left(A_j\right)$ ($g^{\prime }\left(x\right)={\displaystyle \frac{dg\left(x\right)}{dx}}\ge 0$), ${\Theta }_r\le {\Theta }_{r+1}$, and ${\Theta }_{r}g\left(A_i\right)+{\Theta }_{r+1}g\left(A_j\right)\ge {\Theta }_{r}g\left(A_j\right)+{\Theta }_{r+1}g\left(A_i\right)$. The inductive step for subsequent jobs $J_q$ follows from:

$$\begin{array}{cc}\hfill & \underset{\mathrm{Original}\mathrm{sequence}}{\underbrace{{\Theta }_{r}g\left(A_j\right)+{\Theta }_{r+1}g\left(A_i\right)}}-\underset{\mathrm{Swapped}\mathrm{sequence}}{\underbrace{{\Theta }_{r}g\left(A_i\right)+{\Theta }_{r+1}g\left(A_j\right)}}\hfill \\ \hfill & =({\Theta }_r-{\Theta }_{r+1})(g\left(A_j\right)-g\left(A_i\right))\ge 0\hfill \\ \hfill \Rightarrow & f\left(Z_r+{\Theta }_{r}g\left(A_j\right)+{\Theta }_{r+1}g\left(A_i\right)\right)\ge f\left(Z_r+{\Theta }_{r}g\left(A_i\right)+{\Theta }_{r+1}g\left(A_j\right)\right)\hfill \end{array}$$

In addition, from

$$\begin{array}{cc}\hfill C_q\left({\varpi }^{\prime }\right)& =C_j\left({\varpi }^{\prime }\right)+A_q(R+(1-R)\times \hfill \\ \hfill & min\left\{f\left(\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)+{\Theta }_{r}g\left(A_i\right)+{\Theta }_{r+1}g\left(A_j\right)\right)h(r+2),\Xi \right\}),\hfill \end{array}$$

(9)

and

$$\begin{array}{cc}\hfill C_q\left({\varpi }^{″}\right)& =C_i\left({\varpi }^{″}\right)+A_q(R+(1-R)\times \hfill \\ \hfill & min\left\{f\left(\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)+{\Theta }_{r}g\left(A_j\right)+{\Theta }_{r+1}g\left(A_i\right)\right)h(r+2),\Xi \right\})\hfill \end{array}$$

(10)

we have $C_q\left({\varpi }^{\prime }\right)\ge C_q\left({\varpi }^{″}\right)$. Similarly, we have $C_{\left[n\right]}\left({\varpi }^{\prime }\right)\ge C_{\left[n\right]}\left({\varpi }^{″}\right)$, it implies $C_{max}\left({\varpi }^{\prime }\right)\ge C_{max}\left({\varpi }^{″}\right)$. By mathematical induction, the LPT ordering maintains optimality throughout all positions. The $O(nlogn)$ complexity comes from the initial sorting operation; hence, the result holds. □

Theorem 2.

For $1\left|{\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)min\left\{f\left(z\left(r\right)+{\sum }_{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right),\Xi \right\}\right)\right|C_{max}$, if $f^{\prime }\left(x\right)={\displaystyle \frac{df\left(x\right)}{dx}}\ge 0,f^{″}\left(x\right)={\displaystyle \frac{d^{2}f\left(x\right)}{d{x}^2}}\le 0$, $g^{\prime }\left(x\right)={\displaystyle \frac{dg\left(x\right)}{dx}}\ge 0$, $g^{″}\left(x\right)={\displaystyle \frac{d^{2}g\left(x\right)}{d{x}^2}}\le 0$, $z\left(r\right)\ge 1$ is a non-decreasing function, and $0\le {\Theta }_1\le {\Theta }_2\le \dots \le {\Theta }_{n-1}\le {\Theta }_n$, the optimal schedule is obtained in $O(nlogn)$ time by the LPT rule of $A_j$.

Proof. 

Similar to Theorem 1, by Lemma 2, the result holds. □

Example 1.

Consider the model ${\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)min\left\{f\left({\sum }_{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)h\left(r\right),\Xi \right\}\right)$$=A_j\left(R+(1-R)min\left\{{\left(1+{\sum }_{l=1}^{r-1}{\Theta }_{l}ln{A}_{<l>}\right)}^{\alpha }{r}^{\tilde{a}},\Xi \right\}\right)$, i.e., $f\left(G\right)={(1+G)}^{\alpha }$ ($0\le \alpha \le 1$, $f{\left(G\right)}^{\prime }=\alpha {(1+G)}^{\alpha -1}\ge 0$, $f{\left(G\right)}^{″}=\alpha (\alpha -1){(1+G)}^{\alpha -2}\le 0$), $g\left(x\right)=lnx$ ($g{\left(x\right)}^{\prime }={\displaystyle \frac{1}{x}}\ge 0$, $g{\left(x\right)}^{″}=-{\displaystyle \frac{1}{x^2}}\le 0$), $h\left(r\right)=r^{\tilde{a}}$ ($\tilde{a}\ge 0$), where $n=9$, $R=0.4$, $\Xi =3$, $\alpha =0.5$, $\tilde{a}=0.5$, ${\Theta }_1=0.02$, ${\Theta }_2=0.03$, ${\Theta }_3=0.04$, ${\Theta }_4=0.05$, ${\Theta }_5=0.06$, ${\Theta }_6=0.07$, ${\Theta }_7=0.08$, ${\Theta }_8=0.09$, ${\Theta }_9=0.1$, $A_1=16$, $A_2=19$, $A_3=8$, $A_4=9$, $A_5=10$, $A_6=11$, $A_7=28$, $A_8=17$, $A_9=6$.

LPT rule: By the LPT rule, the optimal sequence is

$${\varpi }^{*}=\left\{JO{B}_7\to JO{B}_2\to JO{B}_8\to JO{B}_1\to JO{B}_6\to JO{B}_5\to JO{B}_4\to JO{B}_3\to JO{B}_9\right\},$$

we have

$$\begin{array}{cc}\hfill & C_{max}\left({\varpi }^{*}\right)\hfill \\ \hfill =& 28+19\ast (0.4+0.6\ast min\{{\left(1+0.02\ast ln28\right)}^{0.5}\ast 2^{0.5},3\})\hfill \\ & +17\ast (0.4+0.6\ast min\{{\left(1+0.02\ast ln28+0.03\ast ln19\right)}^{0.5}\ast 3^{0.5},3\})\hfill \\ & +16\ast (0.4+0.6\ast min\{{\left(1+0.02\ast ln28+0.03\ast ln19+0.04\ast ln17\right)}^{0.5}\ast 4^{0.5},3\})\hfill \\ \hfill & +11\ast (0.4+0.6\ast min\{(1+0.02\ast ln28+0.03\ast ln19+0.04\ast ln17\hfill \\ & {+0.05\ast ln16)}^{0.5}\ast 5^{0.5},3\left\}\right)\hfill \\ \hfill & +10\ast (0.4+0.6\ast min\{(1+0.02\ast ln28+0.03\ast ln19+0.04\ast ln17\hfill \\ & {+0.05\ast ln16+0.06\ast ln11)}^{0.5}\ast 6^{0.5},3\left\}\right)\hfill \\ \hfill & +9\ast (0.4+0.6\ast min\{(1+0.02\ast ln28+0.03\ast ln19+0.04\ast ln17\hfill \\ & {+0.05\ast ln16+0.06\ast ln11+0.07\ast ln10)}^{0.5}\ast 7^{0.5},3\left\}\right)\hfill \\ \hfill & +8\ast (0.4+0.6\ast min\{(1+0.02\ast ln28+0.03\ast ln19+0.04\ast ln17\hfill \\ & {+0.05\ast ln16+0.06\ast ln11+0.07\ast ln10+0.08\ast ln9)}^{0.5}\ast 8^{0.5},3\left\}\right)\hfill \\ \hfill & +6\ast (0.4+0.6\ast min\{(1+0.02\ast ln28+0.03\ast ln19+0.04\ast ln17\hfill \\ & {+0.05\ast ln16+0.06\ast ln11+0.07\ast ln10+0.08\ast ln9+0.09\ast ln8)}^{0.5}\ast 9^{0.5},3\left\}\right)\hfill \\ \hfill =& 200.565226.\hfill \end{array}$$

SPT rule: If the rule is SPT, i.e.,

$$\varpi =\left\{JO{B}_9\to JO{B}_3\to JO{B}_4\to JO{B}_5\to JO{B}_6\to JO{B}_1\to JO{B}_8\to JO{B}_2\to JO{B}_7\right\},$$

we have

$$\begin{array}{cc}\hfill C_{max}\left(SPT\right)=& 6+8\ast (0.4+0.6\ast min\{{\left(1+0.02\ast ln6\right)}^{0.5}\ast 2^{0.5},3\})\hfill \\ & +9\ast (0.4+0.6\ast min\{{\left(1+0.02\ast ln6+0.03\ast ln8\right)}^{0.5}\ast 3^{0.5},3\})\hfill \\ & +10\ast (0.4+0.6\ast min\{{\left(1+0.02\ast ln6+0.03\ast ln8+0.04\ast ln9\right)}^{0.5}\ast 4^{0.5},3\})\hfill \\ \hfill & +11\ast (0.4+0.6\ast min\{(1+0.02\ast ln6+0.03\ast ln8+0.04\ast ln9\hfill \\ & {+0.05\ast ln10)}^{0.5}\ast 5^{0.5},3\left\}\right)\hfill \\ \hfill & +16\ast (0.4+0.6\ast min\{(1+0.02\ast ln6+0.03\ast ln8+0.04\ast ln9\hfill \\ & {+0.05\ast ln10+0.06\ast ln11)}^{0.5}\ast 6^{0.5},3\left\}\right)\hfill \\ \hfill & +17\ast (0.4+0.6\ast min\{(1+0.02\ast ln6+0.03\ast ln8+0.04\ast ln9\hfill \\ & {+0.05\ast ln10+0.06\ast ln11+0.07\ast ln16)}^{0.5}\ast 7^{0.5},3\left\}\right)\hfill \\ \hfill & +19\ast (0.4+0.6\ast min\{(1+0.02\ast ln6+0.03\ast ln8+0.04\ast ln9\hfill \\ & {+0.05\ast ln10+0.06\ast ln11+0.07\ast ln16+0.08\ast ln17)}^{0.5}\ast 8^{0.5},3\left\}\right)\hfill \\ \hfill & +28\ast (0.4+0.6\ast min\{(1+0.02\ast ln6+0.03\ast ln8+0.04\ast ln9\hfill \\ & +0.05\ast ln10+0.06\ast ln11+0.07\ast ln16+0.08\ast ln17\hfill \\ & {+0.09\ast ln19)}^{0.5}\ast 9^{0.5},3\left\}\right)\hfill \\ \hfill =& 243.282365.\hfill \end{array}$$

FCFS rule (Fist Come First Service): If the rule is FCFS, i.e.,

$$\varpi =\left\{JO{B}_1\to JO{B}_2\to JO{B}_3\to JO{B}_4\to JO{B}_5\to JO{B}_6\to JO{B}_7\to JO{B}_8\to JO{B}_9\right\},$$

we have

$$\begin{array}{cc}\hfill & C_{max}\left(\mathrm{FCFS}\right)\hfill \\ \hfill =& 16+19\ast \left(0.4+0.6\ast min\left\{{\left(1+0.02\ast ln16\right)}^{0.5}\ast 2^{0.5},3\right\}\right)\hfill \\ \hfill & +8\ast \left(0.4+0.6\ast min\left\{{\left(1+0.02\ast ln16+0.03\ast ln19\right)}^{0.5}\ast 3^{0.5},3\right\}\right)\hfill \\ \hfill & +9\ast \left(0.4+0.6\ast min\left\{\left(1+0.02\ast ln16+0.03\ast ln19\right.\right.\right.\hfill \\ \hfill & \left.\left.{\left.+0.04\ast ln8\right)}^{0.5}\ast 4^{0.5},3\right\}\right)\hfill \\ \hfill & +10\ast \left(0.4+0.6\ast min\left\{\left(1+0.02\ast ln16+0.03\ast ln19+0.04\ast ln8\right.\right.\right.\hfill \\ \hfill & \left.\left.{\left.+0.05\ast ln9\right)}^{0.5}\ast 5^{0.5},3\right\}\right)\hfill \\ \hfill & +11\ast \left(0.4+0.6\ast min\left\{\left(1+0.02\ast ln16+0.03\ast ln19+0.04\ast ln8\right.\right.\right.\hfill \\ \hfill & \left.\left.{\left.+0.05\ast ln9+0.06\ast ln10\right)}^{0.5}\ast 6^{0.5},3\right\}\right)\hfill \\ \hfill & +28\ast \left(0.4+0.6\ast min\left\{\left(1+0.02\ast ln16+0.03\ast ln19+0.04\ast ln8\right.\right.\right.\hfill \\ \hfill & \left.\left.{\left.+0.05\ast ln9+0.06\ast ln10+0.07\ast ln11\right)}^{0.5}\ast 7^{0.5},3\right\}\right)\hfill \\ \hfill & +17\ast \left(0.4+0.6\ast min\left\{\left(1+0.02\ast ln16+0.03\ast ln19+0.04\ast ln8\right.\right.\right.\hfill \\ \hfill & \left.\left.{\left.+0.05\ast ln9+0.06\ast ln10+0.07\ast ln11+0.08\ast ln28\right)}^{0.5}\ast 8^{0.5},3\right\}\right)\hfill \\ \hfill & +6\ast \left(0.4+0.6\ast min\left\{\left(1+0.02\ast ln16+0.03\ast ln19+0.04\ast ln8\right.\right.\right.\hfill \\ \hfill & \left.\left.{\left.+0.05\ast ln9+0.06\ast ln10+0.07\ast ln11+0.08\ast ln28+0.09\ast ln17\right)}^{0.5}\ast 9^{0.5},3\right\}\right)\hfill \\ \hfill =& 223.548165.\hfill \end{array}$$

NEH rule (Nawaz–Enscore–Ham rule, Nawaz et al. [51]): If the rule is NEH, i.e.,

$${\varpi }^{*}=\left\{JO{B}_7\to JO{B}_2\to JO{B}_8\to JO{B}_1\to JO{B}_6\to JO{B}_5\to JO{B}_4\to JO{B}_3\to JO{B}_9\right\},$$

we have

$$\begin{array}{cc}\hfill & C_{max}\left(\mathrm{NEH}\right)\hfill \\ \hfill =& 28+19\ast (0.4+0.6\ast min\{{(1+0.02\ast ln28)}^{0.5}\ast 2^{0.5},3\})\hfill \\ \hfill & +17\ast (0.4+0.6\ast min\{{(1+0.02\ast ln28+0.03\ast ln19)}^{0.5}\ast 3^{0.5},3\})\hfill \\ \hfill & +16\ast (0.4+0.6\ast min\{{(1+0.02\ast ln28+0.03\ast ln19+0.04\ast ln17)}^{0.5}\ast 4^{0.5},3\})\hfill \\ \hfill & +11\ast (0.4+0.6\ast min\{(1+0.02\ast ln28+0.03\ast ln19+0.04\ast ln17\hfill \\ \hfill & {+0.05\ast ln16)}^{0.5}\ast 5^{0.5},3\left\}\right)\hfill \\ \hfill & +10\ast (0.4+0.6\ast min\{(1+0.02\ast ln28+0.03\ast ln19+0.04\ast ln17\hfill \\ \hfill & {+0.05\ast ln16+0.06\ast ln11)}^{0.5}\ast 6^{0.5},3\left\}\right)\hfill \\ \hfill & +9\ast (0.4+0.6\ast min\{(1+0.02\ast ln28+0.03\ast ln19+0.04\ast ln17\hfill \\ \hfill & {+0.05\ast ln16+0.06\ast ln11+0.07\ast ln10)}^{0.5}\ast 7^{0.5},3\left\}\right)\hfill \\ \hfill & +8\ast (0.4+0.6\ast min\{(1+0.02\ast ln28+0.03\ast ln19+0.04\ast ln17\hfill \\ \hfill & {+0.05\ast ln16+0.06\ast ln11+0.07\ast ln10+0.08\ast ln9)}^{0.5}\ast 8^{0.5},3\left\}\right)\hfill \\ \hfill & +6\ast (0.4+0.6\ast min\{(1+0.02\ast ln28+0.03\ast ln19+0.04\ast ln17\hfill \\ \hfill & {+0.05\ast ln16+0.06\ast ln11+0.07\ast ln10+0.08\ast ln9+0.09\ast ln8)}^{0.5}\ast 9^{0.5},3\left\}\right)\hfill \\ \hfill =& 200.565226.\hfill \end{array}$$

Example 2.

Consider the model ${\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)min\left\{f\left({\sum }_{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)h\left(r\right),\Xi \right\}\right)$$=A_j\left(R+(1-R)min\left\{{\left(1+{\sum }_{l=1}^{r-1}{\Theta }_{l}ln{A}_{<l>}\right)}^{\alpha }{r}^{\tilde{a}},\Xi \right\}\right)$, i.e., $f\left(G\right)={(1+G)}^{\alpha }$ ($0\le \alpha \le 1$, $f{\left(G\right)}^{\prime }=\alpha {(1+G)}^{\alpha -1}\ge 0$, $f{\left(G\right)}^{″}=\alpha (\alpha -1){(1+G)}^{\alpha -2}\le 0$), $g\left(x\right)=lnx$ ($g{\left(x\right)}^{\prime }={\displaystyle \frac{1}{x}}\ge 0$, $g{\left(x\right)}^{″}=-{\displaystyle \frac{1}{x^2}}\le 0$), $h\left(r\right)=r^{\tilde{a}}$ ($\tilde{a}\ge 0$), where $n=8$, $R=0.4$, $\Xi =3$, $\alpha =0.5$, $\tilde{a}=0.5$, ${\Theta }_1=0.02$, ${\Theta }_2=0.04$, ${\Theta }_3=0.06$, ${\Theta }_4=0.08$, ${\Theta }_5=0.10$, ${\Theta }_6=0.12$, ${\Theta }_7=0.14$, ${\Theta }_8=0.16$, $A_1=15$, $A_2=18$, $A_3=7$, $A_4=10$, $A_5=12$, $A_6=20$, $A_7=25$, $A_8=5$.

LPT rule: By the LPT rule, the optimal sequence is

$${\varpi }^{*}=\left\{JO{B}_7\to JO{B}_6\to JO{B}_2\to JO{B}_1\to JO{B}_5\to JO{B}_4\to JO{B}_3\to JO{B}_8\right\},$$

we have

$$\begin{array}{cc}\hfill & C_{max}\left(\mathrm{LPT}\right)\hfill \\ \hfill =& 25+20\ast (0.4+0.6\ast min\{{(1+0.02\ast ln25)}^{0.5}\ast 2^{0.5},3\})\hfill \\ \hfill & +18\ast (0.4+0.6\ast min\{{(1+0.02\ast ln25+0.04\ast ln20)}^{0.5}\ast 3^{0.5},3\})\hfill \\ \hfill & +15\ast (0.4+0.6\ast min\{{(1+0.02\ast ln25+0.04\ast ln20+0.06\ast ln18)}^{0.5}\ast 4^{0.5},3\})\hfill \\ \hfill & +12\ast (0.4+0.6\ast min\{(1+0.02\ast ln25+0.04\ast ln20+0.06\ast ln18\hfill \\ \hfill & {+0.08\ast ln15)}^{0.5}\ast 5^{0.5},3\left\}\right)\hfill \\ \hfill & +10\ast (0.4+0.6\ast min\{(1+0.02\ast ln25+0.04\ast ln20+0.06\ast ln18\hfill \\ \hfill & {+0.08\ast ln15+0.10\ast ln12)}^{0.5}\ast 6^{0.5},3\left\}\right)\hfill \\ \hfill & +7\ast (0.4+0.6\ast min\{(1+0.02\ast ln25+0.04\ast ln20+0.06\ast ln18\hfill \\ \hfill & {+0.08\ast ln15+0.10\ast ln12+0.12\ast ln10)}^{0.5}\ast 7^{0.5},3\left\}\right)\hfill \\ \hfill & +5\ast (0.4+0.6\ast min\{(1+0.02\ast ln25+0.04\ast ln20+0.06\ast ln18\hfill \\ \hfill & {+0.08\ast ln15+0.10\ast ln12+0.12\ast ln10+0.14\ast ln7)}^{0.5}\ast 8^{0.5},3\left\}\right)\hfill \\ \hfill =& 177.566000.\hfill \end{array}$$

SPT rule: By the SPT rule, the optimal sequence is

$${\varpi }^{*}=\left\{JO{B}_8\to JO{B}_3\to JO{B}_4\to JO{B}_5\to JO{B}_1\to JO{B}_2\to JO{B}_6\to JO{B}_7\right\},$$

we have

$$\begin{array}{cc}\hfill & C_{max}\left(\mathrm{SPT}\right)\hfill \\ \hfill =& 5+7\ast (0.4+0.6\ast min\{{(1+0.02\ast ln5)}^{0.5}\ast 2^{0.5},3\})\hfill \\ \hfill & +10\ast (0.4+0.6\ast min\{{(1+0.02\ast ln5+0.04\ast ln7)}^{0.5}\ast 3^{0.5},3\})\hfill \\ \hfill & +12\ast (0.4+0.6\ast min\{{(1+0.02\ast ln5+0.04\ast ln7+0.06\ast ln10)}^{0.5}\ast 4^{0.5},3\})\hfill \\ \hfill & +15\ast (0.4+0.6\ast min\{(1+0.02\ast ln5+0.04\ast ln7+0.06\ast ln10\hfill \\ \hfill & {+0.08\ast ln12)}^{0.5}\ast 5^{0.5},3\left\}\right)\hfill \\ \hfill & +18\ast (0.4+0.6\ast min\{(1+0.02\ast ln5+0.04\ast ln7+0.06\ast ln10\hfill \\ \hfill & {+0.08\ast ln12+0.10\ast ln15)}^{0.5}\ast 6^{0.5},3\left\}\right)\hfill \\ \hfill & +20\ast (0.4+0.6\ast min\{(1+0.02\ast ln5+0.04\ast ln7+0.06\ast ln10\hfill \\ \hfill & {+0.08\ast ln12+0.10\ast ln15+0.12\ast ln18)}^{0.5}\ast 7^{0.5},3\left\}\right)\hfill \\ \hfill & +25\ast (0.4+0.6\ast min\{(1+0.02\ast ln5+0.04\ast ln7+0.06\ast ln10\hfill \\ \hfill & {+0.08\ast ln12+0.10\ast ln15+0.12\ast ln18+0.14\ast ln20)}^{0.5}\ast 8^{0.5},3\left\}\right)\hfill \\ \hfill =& 202.716359.\hfill \end{array}$$

FCFS rule: If the rule is FCFS, i.e.,

$${\varpi }^{*}=\left\{JO{B}_1\to JO{B}_2\to JO{B}_3\to JO{B}_4\to JO{B}_5\to JO{B}_6\to JO{B}_7\to JO{B}_8\right\},$$

we have

$$\begin{array}{cc}\hfill & C_{max}\left(\mathrm{FCFS}\right)\hfill \\ \hfill =& 15+18\ast (0.4+0.6\ast min\{{(1+0.02\ast ln15)}^{0.5}\ast 2^{0.5},3\})\hfill \\ \hfill & +7\ast (0.4+0.6\ast min\{{(1+0.02\ast ln15+0.04\ast ln18)}^{0.5}\ast 3^{0.5},3\})\hfill \\ \hfill & +10\ast (0.4+0.6\ast min\{{(1+0.02\ast ln15+0.04\ast ln18+0.06\ast ln7)}^{0.5}\ast 4^{0.5},3\})\hfill \\ \hfill & +12\ast (0.4+0.6\ast min\{(1+0.02\ast ln15+0.04\ast ln18+0.06\ast ln7\hfill \\ \hfill & {+0.08\ast ln10)}^{0.5}\ast 5^{0.5},3\left\}\right)\hfill \\ \hfill & +20\ast (0.4+0.6\ast min\{(1+0.02\ast ln15+0.04\ast ln18+0.06\ast ln7\hfill \\ \hfill & {+0.08\ast ln10+0.10\ast ln12)}^{0.5}\ast 6^{0.5},3\left\}\right)\hfill \\ \hfill & +25\ast (0.4+0.6\ast min\{(1+0.02\ast ln15+0.04\ast ln18+0.06\ast ln7\hfill \\ \hfill & {+0.08\ast ln10+0.10\ast ln12+0.12\ast ln20)}^{0.5}\ast 7^{0.5},3\left\}\right)\hfill \\ \hfill & +5\ast (0.4+0.6\ast min\{(1+0.02\ast ln15+0.04\ast ln18+0.06\ast ln7\hfill \\ \hfill & {+0.08\ast ln10+0.10\ast ln12+0.12\ast ln20+0.14\ast ln25)}^{0.5}\ast 8^{0.5},3\left\}\right)\hfill \\ \hfill =& 201.379546.\hfill \end{array}$$

NEH rule: If the rule is NEH, i.e.,

$${\varpi }^{*}=\left\{JO{B}_7\to JO{B}_6\to JO{B}_2\to JO{B}_1\to JO{B}_5\to JO{B}_4\to JO{B}_3\to JO{B}_8\right\},$$

we have

$$\begin{array}{cc}\hfill & C_{max}\left(\mathrm{NEH}\right)\hfill \\ \hfill =& 25+20\ast (0.4+0.6\ast min\{{(1+0.02\ast ln25)}^{0.5}\ast 2^{0.5},3\})\hfill \\ \hfill & +18\ast (0.4+0.6\ast min\{{(1+0.02\ast ln25+0.04\ast ln20)}^{0.5}\ast 3^{0.5},3\})\hfill \\ \hfill & +15\ast (0.4+0.6\ast min\{{(1+0.02\ast ln25+0.04\ast ln20+0.06\ast ln18)}^{0.5}\ast 4^{0.5},3\})\hfill \\ \hfill & +12\ast (0.4+0.6\ast min\{(1+0.02\ast ln25+0.04\ast ln20+0.06\ast ln18\hfill \\ \hfill & {+0.08\ast ln15)}^{0.5}\ast 5^{0.5},3\left\}\right)\hfill \\ \hfill & +10\ast (0.4+0.6\ast min\{(1+0.02\ast ln25+0.04\ast ln20+0.06\ast ln18\hfill \\ \hfill & {+0.08\ast ln15+0.10\ast ln12)}^{0.5}\ast 6^{0.5},3\left\}\right)\hfill \\ \hfill & +7\ast (0.4+0.6\ast min\{(1+0.02\ast ln25+0.04\ast ln20+0.06\ast ln18\hfill \\ \hfill & {+0.08\ast ln15+0.10\ast ln12+0.12\ast ln10)}^{0.5}\ast 7^{0.5},3\left\}\right)\hfill \\ \hfill & +5\ast (0.4+0.6\ast min\{(1+0.02\ast ln25+0.04\ast ln20+0.06\ast ln18\hfill \\ \hfill & {+0.08\ast ln15+0.10\ast ln12+0.12\ast ln10+0.14\ast ln7)}^{0.5}\ast 8^{0.5},3\left\}\right)\hfill \\ \hfill =& 177.566000.\hfill \end{array}$$

Remark 1.

If ${\widehat{Pro}}_{jr}^a=A_j{\left(1+{\sum }_{l=1}^{r-1}{\Theta }_{l}ln{A}_{<l>}\right)}^{\alpha }{r}^{\tilde{a}}$, where $n=4$, $R=0$, $\Xi =100$, α = 1.5, $\tilde{a}=0.5$, ${\Theta }_1=0.9$, ${\Theta }_2=0.7$, ${\Theta }_3=0.5$, ${\Theta }_4=0.3$, $A_1=16$, $A_2=19$, $A_3=28$, $A_4=79$ (i.e., $f\left(G\right)={(1+G)}^{\alpha }$, $f{\left(G\right)}^{\prime }=\alpha {(1+G)}^{\alpha -1}\ge 0$, $f{\left(G\right)}^{″}=\alpha (\alpha -1){(1+G)}^{\alpha -2}\ge 0$), the SPT rule is not an optimal sequence,

$$\begin{array}{cc}\hfill C_{max}\left(SPT\right)=& 16+19\ast {\left(1+0.9ln16\right)}^{1.5}\ast 2^{0.5}+28\ast {\left(1+0.9ln16+0.7ln19\right)}^{1.5}\ast 3^{0.5}\hfill \\ & +79\ast {\left(1+0.9ln16+0.7ln19+0.5ln28\right)}^{1.5}\ast 4^{0.5}\hfill \\ \hfill =& 3893.6405,\hfill \end{array}$$

$$\begin{array}{cc}\hfill C_{max}\left(LPT\right)=& 79+28\ast {\left(1+0.9ln79\right)}^{1.5}\ast 2^{0.5}+19\ast {\left(1+0.9ln79+0.7ln28\right)}^{1.5}\ast 3^{0.5}\hfill \\ & +16\ast {\left(1+0.9ln79+0.7ln28+0.5ln19\right)}^{1.5}\ast 4^{0.5}\hfill \\ \hfill =& 1983.6524.\hfill \end{array}$$

If ${\widehat{Pro}}_{jr}^a=A_{j}f\left({\sum }_{l=1}^{r-1}{\Theta }_l{A}_{<l>}\right)h\left(r\right)=A_j{\left(1+{\sum }_{l=1}^{r-1}{\Theta }_l{A}_{<l>}\right)}^{\alpha }{r}^{\tilde{a}}$, where $n=4$, $R=0$, $\Xi =3000$, α = 1.5, $\tilde{a}=0.5$, ${\Theta }_1=0.9$, ${\Theta }_2=0.7$, ${\Theta }_3=0.5$, ${\Theta }_4=0.3$, $A_1=16$, $A_2=19$, $A_3=28$, $A_4=79$ (i.e., $f\left(G\right)={(1+G)}^{\alpha }$, $f{\left(G\right)}^{\prime }=\alpha {(1+G)}^{\alpha -1}\ge 0$, $f{\left(G\right)}^{″}=\alpha (\alpha -1){(1+G)}^{\alpha -2}\ge 0$, $g\left(x\right)=x$), the LPT rule is not an optimal sequence,

$$\begin{array}{cc}\hfill C_{max}\left(LPT\right)=& 79+28\ast {\left(1+0.9\ast 79\right)}^{1.5}\ast 2^{0.5}+19\ast {\left(1+0.9\ast 79+0.7\ast 28\right)}^{1.5}\ast 3^{0.5}\hfill \\ & +16\ast {\left(1+0.9\ast 79+0.7\ast 28+0.5\ast 19\right)}^{1.5}\ast 4^{0.5}\hfill \\ \hfill =& 85797.1717,\hfill \end{array}$$

$$\begin{array}{cc}\hfill C_{max}\left(SPT\right)=& 16+19\ast {\left(1+0.9\ast 16\right)}^{1.5}\ast 2^{0.5}+28\ast {\left(1+0.9\ast 16+0.7\ast 19\right)}^{1.5}\ast 3^{0.5}\hfill \\ & +79\ast {\left(1+0.9\ast 16+0.7\ast 19+0.5\ast 28\right)}^{1.5}\ast 4^{0.5}\hfill \\ \hfill =& 53182.2930.\hfill \end{array}$$

Remark 2.

Obviously, for the models

$${\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)max\left\{f\left(\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)h\left(r\right),\Xi \right\}\right)$$

and

$${\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)max\left\{f\left(z\left(r\right)+\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right),\Xi \right\}\right),$$

we have the following results:

Theorem 3.

For $1\left|{\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)max\left\{f\left({\sum }_{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)h\left(r\right),\Xi \right\}\right)\right|C_{max}$, if $f^{\prime }\left(x\right)$$={\displaystyle \frac{df\left(x\right)}{dx}}\ge 0,f^{″}\left(x\right)={\displaystyle \frac{d^{2}f\left(x\right)}{d{x}^2}}\le 0$, $g^{\prime }\left(x\right)={\displaystyle \frac{dg\left(x\right)}{dx}}\ge 0$, $g^{″}\left(x\right)={\displaystyle \frac{d^{2}g\left(x\right)}{d{x}^2}}\le 0$, $h\left(r\right)\ge 1$ is a non-decreasing function, and $0\le {\Theta }_1\le {\Theta }_2\le \dots \le {\Theta }_{n-1}\le {\Theta }_n$, the optimal schedule is obtained in $O(nlogn)$ time by the LPT rule of $A_j$.

Theorem 4.

For $1\left|{\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)max\left\{f\left(z\left(r\right)+{\sum }_{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right),\Xi \right\}\right)\right|C_{max}$, if $f^{\prime }\left(x\right)={\displaystyle \frac{df\left(x\right)}{dx}}\ge 0,f^{″}\left(x\right)={\displaystyle \frac{d^{2}f\left(x\right)}{d{x}^2}}\le 0$, $g^{\prime }\left(x\right)={\displaystyle \frac{dg\left(x\right)}{dx}}\ge 0$, $g^{″}\left(x\right)={\displaystyle \frac{d^{2}g\left(x\right)}{d{x}^2}}\le 0$, $z\left(r\right)\ge 1$ is a non-decreasing function, and $0\le {\Theta }_1\le {\Theta }_2\le \dots \le {\Theta }_{n-1}\le {\Theta }_n$, the optimal schedule is obtained in $O(nlogn)$ time by the LPT rule of $A_j$.

Remark 3.

Obviously, for the models

$${\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)min\left\{\beta f\left(\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)+(1-\beta )h\left(r\right),\Xi \right\}\right)$$

and

$${\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)max\left\{\beta f\left(\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)+(1-\beta )h\left(r\right),\Xi \right\}\right)$$

where $0\le \beta \le 1$ is a given constant, we have the following results:

Theorem 5.

For

$$1\left|{\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)min\left\{\beta f\left(\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)+(1-\beta )h\left(r\right),\Xi \right\}\right)\right|C_{max},$$

if $f^{\prime }\left(x\right)$$={\displaystyle \frac{df\left(x\right)}{dx}}\ge 0,f^{″}\left(x\right)={\displaystyle \frac{d^{2}f\left(x\right)}{d{x}^2}}\le 0$, $g^{\prime }\left(x\right)={\displaystyle \frac{dg\left(x\right)}{dx}}\ge 0$, $g^{″}\left(x\right)={\displaystyle \frac{d^{2}g\left(x\right)}{d{x}^2}}\le 0$, $h\left(r\right)\ge 1$ is a non-decreasing function, and $0\le {\Theta }_1\le {\Theta }_2\le \dots \le {\Theta }_{n-1}\le {\Theta }_n$, the optimal schedule is obtained in $O(nlogn)$ time by the LPT rule of $A_j$.

Theorem 6.

For

$$1\left|{\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)max\left\{\beta f\left(\sum _{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)+(1-\beta )h\left(r\right),\Xi \right\}\right)\right|C_{max},$$

if $f^{\prime }\left(x\right)={\displaystyle \frac{df\left(x\right)}{dx}}\ge 0,f^{″}\left(x\right)={\displaystyle \frac{d^{2}f\left(x\right)}{d{x}^2}}\le 0$, $g^{\prime }\left(x\right)={\displaystyle \frac{dg\left(x\right)}{dx}}\ge 0$, $g^{″}\left(x\right)={\displaystyle \frac{d^{2}g\left(x\right)}{d{x}^2}}\le 0$, $z\left(r\right)\ge 1$ is a non-decreasing function, and $0\le {\Theta }_1\le {\Theta }_2\le \dots \le {\Theta }_{n-1}\le {\Theta }_n$, the optimal schedule is obtained in $O(nlogn)$ time by the LPT rule of $A_j$.

Remark 4.

This paper proves that the Largest Processing Time (LPT) rule is always optimal, it looks like this is the natural results of the following rule: The term ${\sum }_{j=1}^n{x}_j{y}_j$ is minimized if two sequences of $(x_1,x_2,\dots ,x_n)$ and $(y_1,y_2,\dots ,y_n)$ are ordered in the reverse way (Hardy et al. [52], $f\left(z\left(r\right)+{\sum }_{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)$ is a non-decreasing order on position r). However, some special functions do not satisfy this property (e.g., the SPT rule is an optimal schedule for $1\left|{\widehat{Pro}}_{jr}^a=A_j{\left(z\left(r\right)+{\sum }_{l=1}^{r-1}{\Theta }_{l}log{A}_{<l>}\right)}^{\alpha 3},\alpha 3\ge 1,0\le {\Theta }_n\le {\Theta }_{n-1}\le \cdots \le {\Theta }_1\right|$$C_{max}$, Lai et al. [48]).

4. Conclusions

In this paper, the single-machine makespan minimization with general positional-dependent deterioration has been considered. It was proved that $1\left|X\right|C_{max}$ is polynomially solvable, where $X\in \{{\widehat{Pro}}_{jr}^a=A_j\left(R+(1-R)Y\left\{f\left({\sum }_{l=1}^{r-1}{\Theta }_{l}g\left(A_{<l>}\right)\right)h\left(r\right),\Xi \right\}\right)$, ${\widehat{Pro}}_{jr}^a=A_{j}R+(1-R)Y\left\{f\right(z\left(r\right)+{\sum }_{l=1}^{r-1}{\Theta }_{l}g$ $\left(A_{<l>}\right)),\Xi \}\left)\right\},Y\in \{min,max\}$. However, there are some limitations for the $1\left|X\right|C_{max}$, for instance, in this problem, $f^{\prime }\left(x\right)={\displaystyle \frac{df\left(x\right)}{dx}}\ge 0,f^{″}\left(x\right)={\displaystyle \frac{d^{2}f\left(x\right)}{d{x}^2}}\le 0$, $g^{\prime }\left(x\right)={\displaystyle \frac{dg\left(x\right)}{dx}}\ge 0$, $g^{″}\left(x\right)={\displaystyle \frac{d^{2}g\left(x\right)}{d{x}^2}}\le 0$ and $h\left(r\right)\ge 1$ ($z\left(r\right)\ge 1$). A possible extension of the problem could be to study the condition $f^{″}\left(x\right)={\displaystyle \frac{d^{2}f\left(x\right)}{d{x}^2}}\ge 0$. Furthermore, other research may consider the sensitivity analysis to examine how different job distributions or deterioration parameter variations affect scheduling performance, deal with $\widetilde{PDE}$ scheduling problems with positional weights (see Sun et al. [53], Wang et al. [54]), study $\widetilde{PDE}$ scheduling problems with due date (window) assignments (see Qian et al. [55], Qian and Guo [56], Geng et al. [57], Lv and Wang [58]) or delve into $\widetilde{PDE}$ scheduling problems with job rejection (see Mor et al. [59], Shabtay and Oron [60], Mor and Shapira [61], Chen and Li [62]).