Research on Group Scheduling with General Logarithmic Deterioration Subject to Maximal Completion Time Cost

Abstract

Single-machine group scheduling with general logarithmic deterioration is investigated, where the actual job processing (resp. group setup) time is a non-decreasing function of the sum of the logarithmic job processing (resp. group setup) times of the jobs (resp. groups) already processed. Under some optimal properties, it is shown that the maximal completion time (i.e., makespan) cost is solved in polynomial time and the optimal algorithm is presented. In addition, an extension of the general weighted deterioration model is given.

1. Introduction

In many real production processes, group technology (denoted by $gro-tec$) is very important for reducing production costs and improving efficiency (Kuo and Yang [1], Lee and Wu [2], Kuo [3], He and Sun [4], Zhang et al. [5], Liu et al. [6], and Huang [7]). Under a single-machine setting, in 2021, Wang and Ye [8] considered stochastic $gro-tec$ scheduling with learning effects. For expected total completion time minimization, they proposed heuristic algorithms. In 2022, Qian and Zhan [9] studied $gro-tec$ scheduling with a learning (aging) effect. For the total completion time (makespan) minimization, they presented a polynomial time algorithm. In 2023, Liu and Wang [10] and Chen et al. [11] considered resource allocation scheduling with $gro-tec$. In 2024, Li et al. [12] addressed $gro-tec$ scheduling with resource allocation and a learning effect. For the non-regular cost of a common due date assignment, they proposed some heuristics. Wang and Liu [13,14] investigated $gro-tec$ scheduling with resource allocation. For the non-regular cost of different due dates, Wang and Liu [13] proved that a special case is polynomially solvable. For the general case, Wang and Liu [14] proposed some solution algorithms. Yin and Gao [15] studied $gro-tec$ scheduling with deterioration effects (learning effects) of group setup times (job processing times). Lv and Wang [16] considered $gro-tec$ scheduling with resource allocation. For a minmax criterion of slack due window assignment, they proposed some solution algorithms.

In addition, one of the assumptions in the scheduling is that the job processing (group setup) times have deterioration effects (Gawiejnowicz [17] and Strusevich and Rustogi [18]). Generally, there are two deterioration models; one is the time-dependent deterioration effect (denoted by $TDDE$, Shabtay and Mor [19], Sun et al. [20], Wang et al. [21], Zhang et al. [22], Lu et al. [23], and Qiu and Wang [24]) and the other is the position-dependent deterioration (resp. learning) effect (denoted by $DDDE$ (resp. $DDLE$)). For the $DDLE$, the processing times of jobs are a non-increasing function of their positions in a sequence (see Koulamas and Kyparisis [25], Zhao [26], Azzouz et al. [27], Jiang et al. [28], Liu and Wang [29], Gerstl and Mosheiov [30], Cohen and Shapira [31], and Zhang et al. [32]). For the $DDDE$, the processing time of jobs are a non-decreasing function of their positions in a sequence (see Gerstl and Mosheiov [30], Cohen and Shapira [31], Vitaly and Strusevich [33], Montoya-Torres et al. [34], Saavedra-Nieves et al. [35], and Hu et al. [36]). Real-world examples of the $DDDE$ can be found in steel production (see Liu et al. [37]), semiconductor production (see Sloan and Shanthikumar [38]), scheduling derusting operations (see Gawiejnowicz et al. [39]), and production systems that use cooling systems or cutting tools (see Bajestani et al. [40]). Mosheiov [41] considered the following model:

$$p_j^A=p_j{r}^{{\alpha }_{Mosh}},$$

where ${\alpha }_{Mosh}\ge 0$ (resp. ${\alpha }_{Mosh}\le 0$) is a deterioration (resp. learning, see Biskup [42], Mosheiov [43], and Biskup [44]) index. Gordon et al. [45] considered the following model:

$$p_j^A=p_j{\alpha }_{Gord}^{r-1},$$

where ${\alpha }_{Gord}\ge 1$ (resp. $0<{\alpha }_{Gord}\le 1$) is a deterioration (resp. learning) index and $p_j$ is the normal processing time of job $J_j$. Wang et al. [46] considered the following model: if job $J_j$ is placed in the lth position, the actual processing time of $J_j$ is

$$p_j^A=p_j{\left(1+\sum _{k=1}^{l-1}{p}_{\left[k\right]}\right)}^{{\alpha }_{Wang}},$$

where ${\alpha }_{Wang}\ge 0$ (resp. ${\alpha }_{Wang}\le 0$) is a deterioration (resp. learning, see Kuo and Yang [47]) index, and $\left[k\right]$ denotes some job scheduled in kth position. Lee et al. [48] considered the following model:

$$p_j^A=p_j{\left(1+{\displaystyle \frac{{\sum }_{k=1}^{l-1}{p}_{\left[k\right]}}{{\sum }_{k=1}^n{p}_k}}\right)}^{{\alpha }_{Lee}},$$

where $0\le {\alpha }_{Lee}\le 1$ is a deterioration index. Huang and Wang [49] considered the following model:

$$p_j^A=p_j{\left(1+\sum _{k=1}^{l-1}{\zeta }_k{p}_{\left[k\right]}\right)}^{{\alpha }_{Huang}},$$

where ${\alpha }_{Huang}\ge 1$ (resp. ${\alpha }_{Huang}\le 0$) is a deterioration (resp. learning, see Yang et al. [50]) index, ${\zeta }_k$ is the weight of kth position, and $0<{\zeta }_n\le {\zeta }_{n-1}\le \dots \le {\zeta }_2\le {\zeta }_1$.

In view of the importance of group technology, in this paper, we focus on another group scheduling model with logarithmic/weighted $DDDE$, that is, a job processing (group setup) time which cannot deteriorate quickly. The main contributions of this paper are as follows: (i) The single-machine $gro-tec$ scheduling with logarithmic/weighted $DDDE$ is modeled studied; (ii) for the maximal completion time (i.e., makespan) minimization, some optimal properties on group and job sequences are given; (iii) we show that the problem is polynomially solvable, and verify the performance through some examples. The rest of this paper is given as follows. In Section 2, we present the model. In Section 3, we show that the problem with logarithmic $DDDE$ is polynomially solvable. In Section 4, an extension with weighted $DDDE$ is given. In Section 5, we report the results of numberical tests. In Section 6, we conclude the paper.

2. Problem Formulation

Assume that n jobs are divided into m groups to be processed on one machine, and all jobs can be processed at time 0. Let the number of jobs in the i group (i.e., group $G_i$) be $n_i$, $n_1+n_2+\cdots +n_m=n$. Let $J_{ij}$ denote the jth job in $G_i$, $p_{ij}$ denotes the normal processing time of $J_{ij}$, and $s_i$ denotes the normal setup time of $G_i$. As in Lee et al. [48] and Cheng et al. [51], we define a general logarithmic deterioration model, i.e., if job $J_{ij}$ is placed in the lth position of $G_i$, the actual processing time of $J_{ij}$ is as follows:

$$p_{ij}^A=p_{ij}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{k=1}^{l-1}ln{p}_{i\left[k\right]}}{{\sum }_{k=1}^{n_i}{p}_{ik}}}\right)}^{a_i}\right],$$

(1)

where $0<M<1$ is a given constant, $0\le a_i\le 1$ is the job-deterioration index for $G_i$, and $p_{ij}\ge e$ (i.e., the “ln” is a natural logarithm, $ln{p}_{ij}\ge 1$). Similarly, if group $G_i$ is placed in the rth position, the actual setup time is as follows:

$$s_i^A=s_i\left[N+(1-N){\left(1+{\displaystyle \frac{{\sum }_{l=1}^{r-1}ln{s}_{\left[l\right]}}{{\sum }_{l=1}^m{s}_l}}\right)}^b\right],$$

(2)

where $0<N<1$ is a given constant, $0\le b\le 1$ is the group-deterioration index for the setup time and $s_i\ge e$ (i.e., $ln{s}_i\ge 1$). Let $C_{ij}$ be the completion time of $J_{ij}$. The goal of this paper is to minimize the maximal completion time (i.e., makespan, $C_{max}=max\{C_{ij},i=1,\dots ,m;j=1,\dots ,n_i\}$); this problem can be expressed as

$$1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{logarithmic}\right|C_{max},$$

(3)

where $DDD{E}_{logarithmic}$ denotes the logarithmic $DDDE$ model. The literature review related to the scheduling with $DDDE$ ($DDLE$) is given in

Table 1. Results of scheduling with $DDDE$ ($DDLE$).

Problem	Complexity	Paper
$1\left|p_j^A=p_j{r}^{{\alpha }_{Mosh}},{\alpha }_{Mosh}\ge 0\right|C_{max}$	$O(nlogn)$	Mosheiov [41]
$1\left|p_j^A=p_j{r}^{{\alpha }_{Mosh}},{\alpha }_{Mosh}\le 0\right|C_{max}$	$O(nlogn)$	Biskup [42]
$1\left|p_j^A=p_j{r}^{{\alpha }_{Mosh}},{\alpha }_{Mosh}\le 0\right|\sum \sum C_{ij}$	$O(nlogn)$	Biskup [42]
$1\left|p_j^A=p_j{\alpha }_{Gord}^{r-1},0<{\alpha }_{Gord}\le 1\right|C_{max}$	$O(nlogn)$	Gordon et al. [45]
$1\left|p_j^A=p_j{\alpha }_{Gord}^{r-1},{\alpha }_{Gord}\ge 1\right|C_{max}$	$O(nlogn)$	Gordon et al. [45]
$1\left|p_j^A=p_j{\left(1+{\sum }_{k=1}^{l-1}{p}_{\left[k\right]}\right)}^{{\alpha }_{Wang}},{\alpha }_{Wang}\ge 1\right|C_{max}$	$O(nlogn)$	Wang et al. [46]
$1\left|p_j^A=p_j{\left(1+{\sum }_{k=1}^{l-1}{p}_{\left[k\right]}\right)}^{{\alpha }_{Wang}},0\le {\alpha }_{Wang}\le 1\right|C_{max}$	$O(nlogn)$	Wang et al. [46]
$1\left|p_j^A=p_j{\left(1+{\sum }_{k=1}^{l-1}{p}_{\left[k\right]}\right)}^{{\alpha }_{Wang}},{\alpha }_{Wang}\le 0\right|C_{max}$	$O(nlogn)$	Kuo and Yang [47]
$1\left|p_j^A=p_j{\left(1+{\sum }_{k=1}^{l-1}{p}_{\left[k\right]}\right)}^{{\alpha }_{Wang}},{\alpha }_{Wang}\le 0\right|\sum \sum C_{ij}$	$O(nlogn)$	Kuo and Yang [47]
$1\left|p_j^A=p_j{\left(1+{\displaystyle \frac{{\sum }_{k=1}^{l-1}{p}_{\left[k\right]}}{{\sum }_{k=1}^n{p}_k}}\right)}^{{\alpha }_{Lee}},0\le {\alpha }_{Lee}\le 1\right|C_{max}$	$O(nlogn)$	Lee et al. [48]
$1\left|p_j^A=p_j{\left(1-{\displaystyle \frac{{\sum }_{k=1}^{l-1}{p}_{\left[k\right]}}{{\sum }_{k=1}^n{p}_k}}\right)}^{{\alpha }_{Lee}},{\alpha }_{Koul}\ge 1\right|C_{max}$	$O(nlogn)$	Koulamas and Kyparisis [25]
$1\left|p_j^A=p_j{\left(1-{\displaystyle \frac{{\sum }_{k=1}^{l-1}{p}_{\left[k\right]}}{{\sum }_{k=1}^n{p}_k}}\right)}^{{\alpha }_{Lee}},{\alpha }_{Koul}\ge 1\right|\sum \sum C_{ij}$	$O(nlogn)$	Koulamas and Kyparisis [25]
$1\left|p_j^A=p_j{\left(1+{\sum }_{k=1}^{l-1}{\zeta }_k{p}_{\left[k\right]}\right)}^{{\alpha }_{Huang}},{\alpha }_{Huang}\ge 1,{\zeta }_k\downarrow \right|C_{max}$	$O(nlogn)$	Huang and Wang [49]
$1\left|p_j^A=p_j{\left(1+{\sum }_{k=1}^{l-1}{\zeta }_k{p}_{\left[k\right]}\right)}^{{\alpha }_{Huang}},{\alpha }_{Huang}\ge 1,{\zeta }_k\downarrow \right|\sum \sum C_{ij}^{\vartheta }$	$O(nlogn)$	Huang and Wang [49]
$1\left|p_j^A=p_j{\left(1+{\sum }_{k=1}^{l-1}{\zeta }_k{p}_{\left[k\right]}\right)}^{{\alpha }_{Huang}}{r}^{{\alpha }_{Mosh}},{\alpha }_{Huang}\le 0,{\alpha }_{Mosh}\le 0\right|C_{max}$	$O(nlogn)$	Yang et al. [50]
$1\left|p_j^A=p_j{\left(1+{\sum }_{k=1}^{l-1}{\zeta }_k{p}_{\left[k\right]}\right)}^{{\alpha }_{Huang}}{r}^{{\alpha }_{Mosh}},{\alpha }_{Huang}\le 0,{\alpha }_{Mosh}\le 0\right|\sum \sum C_{ij}^{\vartheta }$	$O(nlogn)$	Yang et al. [50]
$1\left|p_j^A=p_j{\left(1+{\sum }_{k=1}^{l-1}ln{p}_{\left[k\right]}\right)}^{{\alpha }_{Cheng}},{\alpha }_{Cheng}\le 0\right|C_{max}$	$O(nlogn)$	Cheng et al. [51]
$1\left|p_j^A=p_j{\left(1+{\sum }_{k=1}^{l-1}ln{p}_{\left[k\right]}\right)}^{{\alpha }_{Cheng}},{\alpha }_{Cheng}\le 0\right|\sum \sum C_{ij}$	$O(nlogn)$	Cheng et al. [51]
$1\left|gro-tec,p_{ij}^A=p_{ij}{\left(1+{\sum }_{k=1}^{l-1}{p}_{i\left[k\right]}\right)}^{{\alpha }_{i,Kuo}},{\alpha }_{i,Kuo}\le 0,s_i^A=s_i\right|C_{max}$	$O(nlogn)$	Kuo and Yang [1]
$1\left|gro-tec,p_{ij}^A=p_{ij}{\left(1+{\sum }_{k=1}^{l-1}{p}_{i\left[k\right]}\right)}^{{\alpha }_{i,Kuo}},{\alpha }_{i,Kuo}\le 0,s_i^A=s_i\right|\sum \sum C_{ij}$	$O(nlogn)$	Kuo and Yang [1]
$1\left|gro-tec,p_{ij}^A=p_{ij}{l}^{{\alpha }_{1Lee}}{r}^{{\alpha }_{2Lee}},{\alpha }_{1Lee}\le 0,{\alpha }_{2Lee}\le 0,s_i^A=s_i{r}^{{\alpha }_{2Lee}}\right|C_{max}$	$O(nlogn)$	Lee and Wu [2]
$1\left|gro-tec,p_{ij}^A=p_{ij}{\left(1+{\sum }_{k=1}^{l-1}{p}_{i\left[k\right]}\right)}^{{\alpha }_{i,Kuo}},{\alpha }_{i,Kuo}\le 0,s_i^A=s_i{r}^{{\alpha }_{Kuo}},{\alpha }_{Kuo}\le 0\right|C_{max}$	$O(nlogn)$	Kuo [3]
$1\left|p_j^A=p_j{\left(1+{\sum }_{k=1}^{l-1}{\zeta }_k{p}_{\left[k\right]}\right)}^{{\alpha }_{Huang}},0\le {\alpha }_{Huang}\le 1,{\zeta }_k\uparrow \right|C_{max}$	$O(nlogn)$	this paper
$1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{logarithmic}\right|C_{max}$	$O(nlogn)$	this paper
$1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{weight}\right|C_{max}$	$O(nlogn)$	this paper

$\vartheta >0$ is a given constant, ${\zeta }_k\uparrow$ (resp. ${\zeta }_k\downarrow$) denotes the non-decreasing (resp. non-increasing) order of ${\zeta }_k$, ${\alpha }_{i,Kuo}$ (${\alpha }_{Kuo}$, and ${\alpha }_{1Lee}$, ${\alpha }_{2Lee}$) denotes the learning index.

.

3. Basic Result

Let $J_{\left[i\right]\left[j\right]}$ denote the job in the jth position of the ith group and $C_{\left[i\right]\left[j\right]}$ denote the completion time of $J_{\left[i\right]\left[j\right]}$; then, for a given sequence

$$\varrho =\{J_{\left[1\right]\left[1\right]},\cdots ,J_{\left[1\right]\left[n_1\right]},J_{\left[2\right]\left[1\right]},\cdots ,J_{\left[2\right]\left[n_2\right]},\cdots ,J_{\left[m\right]\left[1\right]},\cdots ,J_{\left[m\right]\left[n_m\right]}\},$$

by mathematical induction

$$C_{\left[1\right]\left[n_1\right]}=s_{\left[1\right]}+\sum _{j=1}^{n_{\left[1\right]}}{p}_{\left[1\right]\left[j\right]}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{k=1}^{j-1}ln{p}_{\left[1\right]\left[k\right]}}{{\sum }_{k=1}^{n_{\left[1\right]}}{p}_{\left[1\right]\left[k\right]}}}\right)}^{a_{\left[1\right]}}\right],$$

$$\begin{array}{cc}\hfill C_{\left[2\right]\left[n_2\right]}=& s_{\left[1\right]}+\sum _{j=1}^{n_{\left[1\right]}}{p}_{\left[1\right]\left[j\right]}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{k=1}^{j-1}ln{p}_{\left[1\right]\left[k\right]}}{{\sum }_{k=1}^{n_{\left[1\right]}}{p}_{\left[1\right]\left[k\right]}}}\right)}^{a_{\left[1\right]}}\right]\hfill \\ & +s_{\left[2\right]}\left[N+(1-N){\left(1+{\displaystyle \frac{ln{s}_{\left[1\right]}}{{\sum }_{k=1}^m{s}_{\left[k\right]}}}\right)}^b\right]\hfill \\ & +\sum _{j=1}^{n_{\left[2\right]}}{p}_{\left[2\right]\left[j\right]}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{k=1}^{j-1}ln{p}_{\left[2\right]\left[k\right]}}{{\sum }_{k=1}^{n_{\left[2\right]}}{p}_{\left[2\right]\left[k\right]}}}\right)}^{a_{\left[2\right]}}\right],\hfill \\ & ⋮\end{array}$$

$$\begin{array}{cc}\hfill C_{\left[m\right]\left[n_m\right]}& =\sum _{i=1}^m{s}_{\left[i\right]}\left[N+(1-N){\left(1+{\displaystyle \frac{{\sum }_{k=1}^{i-1}ln{s}_{\left[k\right]}}{{\sum }_{k=1}^m{s}_{\left[k\right]}}}\right)}^b\right]\hfill \\ & +\sum _{i=1}^m\sum _{j=1}^{n_{\left[i\right]}}{p}_{\left[i\right]\left[j\right]}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{k=1}^{j-1}ln{p}_{\left[i\right]\left[k\right]}}{{\sum }_{k=1}^{n_{\left[2\right]}}{p}_{\left[i\right]\left[k\right]}}}\right)}^{a_{\left[i\right]}}\right].\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill C_{max}& =\sum _{i=1}^m{s}_{\left[i\right]}\left[N+(1-N){\left(1+{\displaystyle \frac{{\sum }_{k=1}^{i-1}ln{s}_{\left[k\right]}}{{\sum }_{k=1}^m{s}_{\left[k\right]}}}\right)}^b\right]\hfill \\ & +\sum _{i=1}^m\sum _{j=1}^{n_{\left[i\right]}}{p}_{\left[i\right]\left[j\right]}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{k=1}^{j-1}ln{p}_{\left[i\right]\left[k\right]}}{{\sum }_{k=1}^{n_{\left[i\right]}}{p}_{\left[i\right]\left[k\right]}}}\right)}^{a_{\left[i\right]}}\right].\hfill \end{array}$$

(4)

Lemma 1.

$F\left(\eta \right)={(1+\eta )}^a-1-a\eta {(1+\eta )}^{a-1}\ge 0$ if $0\le a\le 1$, and $0\le \eta \le 1$.

Proof. 

Let $F\left(\eta \right)={(1+\eta )}^a-1-a\eta {(1+\eta )}^{a-1}$. Taking the first derivative of $F\left(\eta \right)$ to $\eta$, we have

$$F^{\prime }\left(\eta \right)=-a(a-1)\eta {(1+\eta )}^{a-2}\ge 0$$

for $0\le a\le 1$, $0\le \eta \le 1$. Thus, $F\left(\eta \right)$ is a non-decreasing function of $\eta$, and $F\left(\eta \right)\ge F\left(0\right)=0$.    □

Lemma 2.

$G\left(x\right)={(1+\eta x)}^a-1-a\eta {(1+\eta x)}^{a-1}\ge 0$ if $x\ge 1$, $0\le a\le 1$, and $0\le \eta \le 1$.

Proof. 

Let $G\left(x\right)={(1+\eta x)}^a-1-a\eta {(1+\eta x)}^{a-1}$; similarly, we have

$$G^{\prime }\left(x\right)=a\eta {(1+\eta x)}^{a-1}-a(a-1){\eta }^2{(1+\eta x)}^{a-2}\ge 0$$

for $0\le a\le 1$, $0\le \eta \le 1$. Thus, from Lemma 1,

$$G\left(x\right)\ge G\left(1\right)={(1+\eta )}^a-1-a\eta {(1+\eta )}^{a-1}\ge 0.$$

   □

Lemma 3.

$H\left(\varpi \right)=[1-{(1+\eta ln\varpi +\eta x)}^a]-\varpi [1-{(1+\eta x)}^a]\ge 0$ if $x\ge 1$, $0\le a\le 1$, $0\le \eta \le 1$, and $\varpi \ge 1$.

Proof. 

Let $H\left(\varpi \right)=[1-{(1+\eta ln\varpi +\eta x)}^a]-\varpi [1-{(1+\eta x)}^a]$; similarly, we have

$$H^{\prime }\left(\varpi \right)={(1+\eta x)}^a-1-a\eta {(1+\eta ln\varpi +\eta x)}^{a-1}/\varpi ,$$

$$H^{″}\left(\varpi \right)={\displaystyle \frac{a\eta {(1+\eta ln\varpi +\eta x)}^{a-1}-a(a-1){\eta }^2{(1+\eta ln\varpi +\eta x)}^{a-2}}{{\varpi }^2}}\ge 0,$$

for $0\le a\le 1$. Thus, from Lemma 2, $H^{\prime }\left(\varpi \right)\ge H^{\prime }\left(1\right)={(1+\eta x)}^a-1-a\eta {(1+\eta x)}^{a-1}\ge 0$. Hence, $H\left(\varpi \right)\ge H\left(1\right)=0$.    □

Lemma 4.

If $0\le a_i\le 1$, $0\le b\le 1$, for $1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{logarithmic}\right|C_{max}$; jobs within each group are arranged in non-increasing order of normal processing time, i.e., for $G_i$, the jobs are arranged in non-increasing order of $p_{ij}$ (the Largest Processing Time (LPT) rule).

Proof. 

Let ${\pi }_i^{\prime }=\{S_1,J_{ij}\to J_{ik}\to J_{iz},S_2\}$ and ${\pi }_i^{″}=\{S_1,J_{ik}\to J_{ij}\to J_{iz},S_2\}$, where → denotes the order relation, i.e., $x\to y$ denotes that x is scheduled in front of y in a sequence, $S_1$ and $S_2$ are partial job sequences, and $p_{ij}\le p_{ik}$. Let A be the completion time of last job in $S_1$ and there are $l-1$ jobs in $S_1$; then, we have

$$C_{ij}\left({\pi }_i^{\prime }\right)=A+p_{ij}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{i\left[\delta \right]}}{{\sum }_{\delta =1}^{n_i}{p}_{i\delta }}}\right)}^{a_i}\right],$$

$$\begin{array}{cc}\hfill C_{ik}\left({\pi }_i^{\prime }\right)& =A+p_{ij}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{i\left[\delta \right]}}{{\sum }_{\delta =1}^{n_i}{p}_{i\delta }}}\right)}^{a_i}\right]\hfill \\ & +p_{ik}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{i\left[\delta \right]}+ln{p}_{ij}}{{\sum }_{\delta =1}^{n_i}{p}_{i\delta }}}\right)}^{a_i}\right],\hfill \end{array}$$

(5)

$$C_{ik}\left({\pi }_i^{″}\right)=A+p_{ik}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{i\left[\delta \right]}}{{\sum }_{\delta =1}^{n_i}{p}_{i\delta }}}\right)}^{a_i}\right],$$

$$\begin{array}{cc}\hfill C_{ij}\left({\pi }_i^{″}\right)& =A+p_{ik}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{i\left[\delta \right]}}{{\sum }_{\delta =1}^{n_i}{p}_{i\delta }}}\right)}^{a_i}\right]\hfill \\ & +p_{ij}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{i\left[\delta \right]}+ln{p}_{ik}}{{\sum }_{\delta =1}^{n_i}{p}_{i\delta }}}\right)}^{a_i}\right].\hfill \end{array}$$

(6)

From (5) and (6), we have

$$\begin{array}{cc}\hfill C_{ik}\left({\pi }_i^{\prime }\right)-C_{ij}\left({\pi }_i^{″}\right)=& p_{ij}(1-M)\left[{\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{i\left[\delta \right]}}{{\sum }_{\delta =1}^{n_i}{p}_{i\delta }}}\right)}^{a_i}-{\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{i\left[\delta \right]}+ln{p}_{ik}}{{\sum }_{\delta =1}^{n_i}{p}_{i\delta }}}\right)}^{a_i}\right]\hfill \\ & -p_{ik}(1-M)\left[{\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{i\left[\delta \right]}}{{\sum }_{\delta =1}^{n_i}{p}_{i\delta }}}\right)}^{a_i}-{\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{i\left[\delta \right]}+ln{p}_{ij}}{{\sum }_{\delta =1}^{n_i}{p}_{i\delta }}}\right)}^{a_i}\right].\hfill \end{array}$$

Let ${\tilde{P}}^i={\sum }_{\delta =1}^{n_i}{p}_{i\delta }$, ${\tilde{P}}_l^i={\sum }_{\delta =1}^{l-1}ln{p}_{i\delta }$, $\varpi ={\displaystyle \frac{p_{ik}}{p_{ij}}}\ge 1$ ($\varpi \ge 1$), $\eta ={\displaystyle \frac{1}{{\tilde{P}}^i+{\tilde{P}}_l^i}}$ ($0\le \eta \le 1$), $x=ln{p}_{ij}$ ($x\ge 1$), then

$$\begin{array}{cc}\hfill C_{ik}\left({\pi }_i^{\prime }\right)-C_{ij}\left({\pi }_i^{″}\right)& =(p_{ij}-p_{ik})(1-M){\left(1+{\displaystyle \frac{{\tilde{P}}_l^i}{{\tilde{P}}^i}}\right)}^{a_i}+p_{ik}(1-M){\left(1+{\displaystyle \frac{{\tilde{P}}_l^i}{{\tilde{P}}^i}}+{\displaystyle \frac{ln{p}_{ij}}{{\tilde{P}}^i}}\right)}^{a_i}\hfill \\ & -p_{ij}(1-M){\left(1+{\displaystyle \frac{{\tilde{P}}_l^i}{{\tilde{P}}^i}}+{\displaystyle \frac{ln{p}_{ik}}{{\tilde{P}}^i}}\right)}^{a_i}\hfill \\ \hfill & =p_{ij}(1-M){\left(1+{\displaystyle \frac{{\tilde{P}}_l^i}{{\tilde{P}}^i}}\right)}^{a_i}\left[1-\varpi +\varpi {(1+\eta x)}^{a_i}-{(1+\eta ln\varpi +\eta x)}^{a_i}\right]\hfill \\ \hfill & =p_{ij}(1-M){\left(1+{\displaystyle \frac{{\tilde{P}}_l^i}{{\tilde{P}}^i}}\right)}^{a_i}\{[1-{(1+\eta ln\varpi +\eta x)}^{a_i}]-\varpi [1-{(1+\eta x)}^{a_i}]\}.\hfill \end{array}$$

Noting $0<M<1$, $x\ge 1$, $0\le a\le 1$, $0\le \eta \le 1$, $\varpi \ge 1$, from Lemma 3, it follows that

$$\begin{array}{cc}\hfill C_{ik}\left({\pi }_i^{\prime }\right)-C_{ij}\left({\pi }_i^{″}\right)& =p_{ij}(1-M){\left(1+{\displaystyle \frac{{\tilde{P}}_l^i}{{\tilde{P}}^i}}\right)}^{a_i}\{[1-{(1+\eta ln\varpi +\eta x)}^{a_i}]-\varpi [1-{(1+\eta x)}^{a_i}]\}\hfill \\ \hfill & \ge 0.\hfill \end{array}$$

In addition,

$$C_{iz}\left({\pi }_i^{\prime }\right)=C_{ik}\left({\pi }_i^{\prime }\right)+p_{iz}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{i\left[\delta \right]}+ln{p}_{ij}+ln{p}_{ik}}{{\sum }_{\delta =1}^{n_i}{p}_{i\delta }}}\right)}^{a_i}\right],$$

and

$$C_{iz}\left({\pi }_i^{″}\right)=C_{ij}\left({\pi }_i^{″}\right)+p_{iz}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{i\left[\delta \right]}+ln{p}_{ik}+ln{p}_{ij}}{{\sum }_{\delta =1}^{n_i}{p}_{i\delta }}}\right)}^{a_i}\right].$$

Obviously, $C_{iz}\left({\pi }_i^{\prime }\right)\ge C_{iz}\left({\pi }_i^{″}\right)$; this implies $C_{imax}\left({\pi }_i^{\prime }\right)\ge C_{imax}\left({\pi }_i^{″}\right)$, where $C_{imax}$ is the maximal completion time of group $G_i$.    □

Lemma 5.

If $0\le a_i\le 1$, $0\le b\le 1$, for $1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{logarithmic}\right|C_{max}$, the groups are listed in non-increasing order of normal setup time, i.e., groups are arranged in non-increasing order of $s_i$ (the LPT rule).

Proof. 

Let ${\varrho }^{\prime }=\{{\pi }^{\prime },G_i\to G_j\to G_h,{\pi }^{″}\}$ and ${\varrho }^{″}=\{{\pi }^{\prime },G_j\to G_i\to G_h,{\pi }^{″}\}$, where ${\pi }^{\prime }$ and ${\pi }^{″}$ are partial group sequences and $s_i\le s_j$. Let B be the starting time of $G_i$ (resp. $G_j$) in ${\varrho }^{\prime }$ (resp. ${\varrho }^{″}$) and there are $r-1$ groups in ${\pi }_1$; we have

$$\begin{array}{cc}\hfill C_i\left({\varrho }^{\prime }\right)& =B+s_i\left[N+(1-N){\left(1+{\displaystyle \frac{{\sum }_{\theta =1}^{r-1}ln{s}_{\left[\theta \right]}}{{\sum }_{\theta =1}^m{s}_{\theta }}}\right)}^b\right]\hfill \\ & +\sum _{l=1}^{n_i}{p}_{i\left[l\right]}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{i\left[\delta \right]}}{{\sum }_{\delta =1}^{n_i}{p}_{i\delta }}}\right)}^{a_i}\right],\hfill \end{array}$$

$$\begin{array}{cc}\hfill C_j\left({\varrho }^{\prime }\right)=& B+s_i\left[N+(1-N){\left(1+{\displaystyle \frac{{\sum }_{\theta =1}^{r-1}ln{s}_{\left[\theta \right]}}{{\sum }_{\theta =1}^m{s}_{\theta }}}\right)}^b\right]\hfill \\ & +\sum _{l=1}^{n_i}{p}_{i\left[l\right]}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{i\left[\delta \right]}}{{\sum }_{\delta =1}^{n_i}{p}_{i\delta }}}\right)}^{a_i}\right]\hfill \\ & +s_j\left[N+(1-N){\left(1+{\displaystyle \frac{{\sum }_{\theta =1}^{r-1}ln{s}_{\left[\theta \right]}+ln{s}_i}{{\sum }_{\theta =1}^m{s}_{\theta }}}\right)}^b\right]\hfill \\ & +\sum _{l=1}^{n_i}{p}_{j\left[l\right]}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{j\left[\delta \right]}}{{\sum }_{\delta =1}^{n_i}{p}_{j\delta }}}\right)}^{a_j}\right],\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill C_j\left({\varrho }^{″}\right)& =B+s_j\left[N+(1-N){\left(1+{\displaystyle \frac{{\sum }_{\theta =1}^{r-1}ln{s}_{\left[\theta \right]}}{{\sum }_{\theta =1}^m{s}_{\theta }}}\right)}^b\right]\hfill \\ & +\sum _{l=1}^{n_i}{p}_{j\left[l\right]}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{j\left[\delta \right]}}{{\sum }_{\delta =1}^{n_i}{p}_{j\delta }}}\right)}^{a_i}\right],\hfill \end{array}$$

$$\begin{array}{cc}\hfill C_i\left({\varrho }^{″}\right)=& B+s_j\left[N+(1-N){\left(1+{\displaystyle \frac{{\sum }_{\theta =1}^{r-1}ln{s}_{\left[\theta \right]}}{{\sum }_{\theta =1}^m{s}_{\theta }}}\right)}^b\right]\hfill \\ & +\sum _{l=1}^{n_i}{p}_{j\left[l\right]}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{j\left[\delta \right]}}{{\sum }_{\delta =1}^{n_i}{p}_{j\delta }}}\right)}^{a_j}\right]\hfill \\ & +s_i\left[N+(1-N){\left(1+{\displaystyle \frac{{\sum }_{\theta =1}^{r-1}ln{s}_{\left[\theta \right]}+ln{s}_j}{{\sum }_{\theta =1}^m{s}_{\theta }}}\right)}^b\right]\hfill \\ & +\sum _{l=1}^{n_i}{p}_{i\left[l\right]}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{\delta =1}^{l-1}ln{p}_{i\left[\delta \right]}}{{\sum }_{\delta =1}^{n_i}{p}_{i\delta }}}\right)}^{a_i}\right].\hfill \end{array}$$

(8)

From (7) and (8), we have

$$\begin{array}{cc}\hfill C_j\left({\varrho }^{\prime }\right)-C_i\left({\varrho }^{″}\right)=& s_i(1-N)\left[{\left(1+{\displaystyle \frac{{\sum }_{\theta =1}^{r-1}ln{s}_{\left[\theta \right]}}{{\sum }_{\theta =1}^m{s}_{\theta }}}\right)}^b-{\left(1+{\displaystyle \frac{{\sum }_{\theta =1}^{r-1}ln{s}_{\left[\theta \right]}+ln{s}_j}{{\sum }_{\theta =1}^m{s}_{\theta }}}\right)}^b\right]\hfill \\ & -s_j(1-N)\left[{\left(1+{\displaystyle \frac{{\sum }_{\theta =1}^{r-1}ln{s}_{\left[\theta \right]}}{{\sum }_{\theta =1}^m{s}_{\theta }}}\right)}^b-{\left(1+{\displaystyle \frac{{\sum }_{\theta =1}^{r-1}ln{s}_{\left[\theta \right]}+ln{s}_i}{{\sum }_{\theta =1}^m{s}_{\theta }}}\right)}^b\right].\hfill \end{array}$$

Let $S={\sum }_{\theta =1}^m{s}_{\theta }$, $S_r={\sum }_{\theta =1}^{r-1}{s}_{\left[\theta \right]}$, $\varpi ={\displaystyle \frac{s_j}{s_i}}$ ($\varpi \ge 1$), $\eta ={\displaystyle \frac{1}{S+S_r}}$ ($0\le \eta \le 1$), $x=ln{s}_i$ ($x\ge 1$); then, from Lemma 3, we have

$$\begin{array}{cc}\hfill C_j\left({\varrho }^{\prime }\right)-C_i\left({\varrho }^{″}\right)& =(s_i-s_j)(1-N){\left(1+{\displaystyle \frac{S_r}{S}}\right)}^b+s_j(1-N){\left(1+{\displaystyle \frac{S_r}{S}}+{\displaystyle \frac{ln{s}_i}{S}}\right)}^b\hfill \\ & -s_i(1-N){\left(1+{\displaystyle \frac{S_r}{S}}+{\displaystyle \frac{ln{s}_j}{S}}\right)}^b\hfill \\ \hfill & =s_i(1-N){\left(1+{\displaystyle \frac{S_r}{S}}\right)}^b\left[1-\varpi +\varpi {(1+\eta x)}^b-{(1+\eta ln\varpi +\eta x)}^b\right]\hfill \\ \hfill & =s_i(1-N){\left(1+{\displaystyle \frac{S_r}{S}}\right)}^b\{[1-{(1+\eta ln\varpi +\eta x)}^b]-\varpi [1-{(1+\eta x)}^b]\}\hfill \\ \hfill & \ge 0.\hfill \end{array}$$

In addition,

$$C_h\left({\varrho }^{\prime }\right)=C_j\left({\varrho }^{\prime }\right)+s_h\left[N+(1-N){\left(1+{\displaystyle \frac{{\sum }_{\theta =1}^{r-1}ln{s}_{\left[\theta \right]}+ln{s}_i+ln{s}_j}{{\sum }_{\theta =1}^m{s}_{\theta }}}\right)}^b\right],$$

and

$$C_h\left({\varrho }^{″}\right)=C_i\left({\varrho }^{″}\right)+s_h\left[N+(1-N){\left(1+{\displaystyle \frac{{\sum }_{\theta =1}^{r-1}ln{s}_{\left[\theta \right]}+ln{s}_j+ln{s}_i}{{\sum }_{\theta =1}^m{s}_{\theta }}}\right)}^b\right].$$

Obviously, $C_h\left({\varrho }^{\prime }\right)\ge C_h\left({\varrho }^{″}\right)$; this implies $C_{max}\left({\varrho }^{\prime }\right)\ge C_{max}\left({\varrho }^{″}\right)$.    □

Based on Lemmas 4 and 5, the following algorithm is proposed to solve

$$1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{logarithmic}\right|C_{max}.$$

Theorem 1.

If $0\le a_i\le 1$, $0\le b\le 1$, $1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{logarithmic}\right|C_{max}$ can be optimally solved by Algorithm 1 in $O(nlogn)$ time.

Algorithm 1 $1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{logarithmic}\right|C_{max}$.

Step 1. For $G_i$, jobs are arranged in non-increasing order of $p_{ij}$ (Lemma 4). Step 2. All groups are arranged in non-increasing order of $s_i$ (Lemma 5). Step 3. Calculate the value $C_{max}$ by (4).

Proof. 

The correctness of Algorithm 1 follows directly from Lemmas 4 and 5. For each group $G_i$, the simple sorting algorithm needs $n_{i}log{n}_i$ time; thus, Step 1 needs ${\sum }_{i=1}^{m}O(n_{i}log{n}_i)\le O(nlogn)$ time. Similarly, Step 2 needs $O(mlogm)\le O(nlogn)$ time. Step 3 needs $O\left(n\right)$ time. Thus, the total time of Algorithm 1 is $O(nlogn)$.    □

Example 1.

Consider $1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{logarithmic}\right|C_{max}$, where $n=24$, $m=4$, $n_1=n_2=n_3=n_4=6$, $M=0.5$, $N=0.5$, the deterioration index of group setup time is $b=0.25$, the deterioration index of jobs in each group is $a_1=0.11$, $a_2=0.63$, $a_3=0.48$, $a_4=0.79$, and the group- and job-related parameters are given in

Table 2. Parameters for Example 1.

${\mathit{G}}_{\mathit{i}}$	${\mathit{s}}_{\mathit{i}}$	${\mathit{p}}_{\mathit{i}1}$	${\mathit{p}}_{\mathit{i}2}$	${\mathit{p}}_{\mathit{i}3}$	${\mathit{p}}_{\mathit{i}4}$	${\mathit{p}}_{\mathit{i}5}$	${\mathit{p}}_{\mathit{i}6}$
$G_1$	38	88	91	35	75	73	82
$G_2$	72	92	29	67	27	59	88
$G_3$	86	91	49	85	73	53	74
$G_4$	82	78	90	79	21	34	64

.

By Algorithm 1, Example 1 can be solved as follows:

Step 1: Jobs within each group are arranged in non-increasing order of $p_{ij}$, i.e.,

• $G_1:{\pi }_1\ast =\{J_{12}\to J_{11}\to J_{16}\to J_{14}\to J_{15}\to J_{13}\}$,
  $G_2:{\pi }_2\ast =\{J_{21}\to J_{26}\to J_{23}\to J_{25}\to J_{22}\to J_{24}\}$,
  $G_3:{\pi }_3\ast =\{J_{31}\to J_{33}\to J_{36}\to J_{34}\to J_{35}\to J_{32}\}$,
  $G_4:{\pi }_4\ast =\{J_{42}\to J_{43}\to J_{41}\to J_{46}\to J_{45}\to J_{44}\}$.

Step 2: Arrange all groups in non-increasing order of $s_i$, i.e.,

• $\varrho \ast =\{G_3\to G_4\to G_2\to G_1\}$.

Step 3: By (4), the value of $C_{max}$ for the optimal sequence is as follows:

$$\begin{array}{cc}\hfill C_{max}(\varrho \ast )=& \sum _{i=1}^m{s}_{\left[i\right]}\left[N+(1-N){\left(1+{\displaystyle \frac{{\sum }_{k=1}^{i-1}ln{s}_{\left[k\right]}}{{\sum }_{k=1}^m{s}_{\left[k\right]}}}\right)}^b\right]\hfill \\ & +\sum _{i=1}^m\sum _{j=1}^{n_{\left[i\right]}}{p}_{\left[i\right]\left[j\right]}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{k=1}^{j-1}ln{p}_{\left[i\right]\left[k\right]}}{{\sum }_{k=1}^{n_{\left[i\right]}}{p}_{\left[i\right]\left[k\right]}}}\right)}^{a_{\left[i\right]}}\right]\hfill \\ \hfill =& 86+82\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln86}{86+82+72+38}}\right)}^{0.25}\right]\hfill \\ & +72\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln86+ln82}{86+82+72+38}}\right)}^{0.25}\right]\hfill \\ \hfill & +38\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln86+ln82+ln72}{86+82+72+38}}\right)}^{0.25}\right]\hfill \\ \hfill & +91+85\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln91}{91+85+74+73+53+49}}\right)}^{0.48}\right]\hfill \\ \hfill & +74\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln91+ln85}{91+85+74+73+53+49}}\right)}^{0.48}\right]\hfill \\ \hfill & +73\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln91+ln85+ln74}{91+85+74+73+53+49}}\right)}^{0.48}\right]\hfill \\ \hfill & +53\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln91+ln85+ln74+ln73}{91+85+74+73+53+49}}\right)}^{0.48}\right]\hfill \\ \hfill & +49\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln91+ln85+ln74+ln73+ln53}{91+85+74+73+53+49}}\right)}^{0.48}\right]\hfill \end{array}$$

$$\begin{array}{cc}\hfill & +90+79\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln90}{90+79+78+64+34+21}}\right)}^{0.79}\right]\hfill \\ & +78\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln90+ln79}{90+79+78+64+34+21}}\right)}^{0.79}\right]\hfill \\ \hfill & +64\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln90+ln79+ln78}{90+79+78+64+34+21}}\right)}^{0.79}\right]\hfill \\ \hfill & +34\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln90+ln79+ln78+ln64}{90+79+78+64+34+21}}\right)}^{0.79}\right]\hfill \\ \hfill & +21\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln90+ln79+ln78+ln64+ln34}{90+79+78+64+34+21}}\right)}^{0.79}\right]\hfill \end{array}$$

$$\begin{array}{cc}\hfill & +92+88\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln92}{92+88+67+59+29+27}}\right)}^{0.63}\right]\hfill \\ & +67\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln92+ln88}{92+88+67+59+29+27}}\right)}^{0.63}\right]\hfill \\ \hfill & +59\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln92+ln88+ln67}{92+88+67+59+29+27}}\right)}^{0.63}\right]\hfill \\ \hfill & +29\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln92+ln88+ln67+ln59}{92+88+67+59+29+27}}\right)}^{0.63}\right]\hfill \\ \hfill & +27\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln92+ln88+ln67+ln59+ln29}{92+88+67+59+29+27}}\right)}^{0.63}\right]\hfill \\ \hfill & +91+88\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln91}{91+88+82+75+73+35}}\right)}^{0.11}\right]\hfill \\ & +82\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln91+ln88}{91+88+82+75+73+35}}\right)}^{0.11}\right]\hfill \\ \hfill & +75\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln91+ln88+ln82}{91+88+82+75+73+35}}\right)}^{0.11}\right]\hfill \\ \hfill & +73\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln91+ln88+ln82+ln75}{91+88+82+75+73+35}}\right)}^{0.11}\right]\hfill \\ \hfill & +35\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{ln91+ln88+ln82+ln75+ln73}{91+88+82+75+73+35}}\right)}^{0.11}\right]\hfill \\ \hfill =& 1884.01556.\hfill \end{array}$$

Remark 1.

For Example 1, if jobs within each group are arranged in non-decreasing order of $p_{ij}$ (i.e., the Smallest Processing Time (SPT) rule), we have $G_1:{\pi }_1=\{J_{13}\to J_{15}\to J_{14}\to J_{16}\to J_{11}\to J_{12}\}$, $G_2:{\pi }_2=\{J_{24}\to J_{22}\to J_{25}\to J_{23}\to J_{26}\to J_{21}\}$, $G_3:{\pi }_3=\{J_{32}\to J_{35}\to J_{34}\to J_{36}\to J_{33}\to J_{31}\}$, $G_4:{\pi }_4=\{J_{44}\to J_{45}\to J_{46}\to J_{41}\to J_{42}\to J_{42}\}$. And when all groups are arranged by the SPT rule of $s_i$, i.e., $\varrho =\{G_1\to G_2\to G_4\to G_3\}$, we have $C_{max}\left(\varrho \right)=1887.64453.$

Remark 2.

If $a_i\ge 1$, $b\ge 1$, the optimal job sequence within each group cannot be obtained by the SPT or LPT rules of $p_{ij}$, and the optimal group sequence cannot be obtained by the SPT or LPT rules of $s_i$. For example, if $n=2$, $m=1$, $a_1=2$, $b=2$, $M=N=0$, $s_1=5, p_{11}=10, p_{12}=8$, we have $C_{max}\left(SPT\right)=5+8+10\ast {\left(1+{\displaystyle \frac{ln8}{10+8}}\right)}^2=25.44395>C_{max}\left(LPT\right)=5+10+8\ast {\left(1+{\displaystyle \frac{ln10}{10+8}}\right)}^2=25.17765$. If $n=2$, $m=1$, $a_1=21$, $b=2$, $s_1=5, p_{11}=10, p_{12}=8$, $C_{max}\left(SPT\right)=5+8+10\ast {\left(1+{\displaystyle \frac{ln8}{10+8}}\right)}^{21}=112.32570<C_{max}\left(LPT\right)=5+10+8\ast {\left(1+{\displaystyle \frac{ln10}{10+8}}\right)}^{21}=115.21795$.

Remark 3.

If $a_1=a_2=a_3=a_4=b=0$, $C_{max}=86+82+72+38+91+85+74+73+53+49+90+79+78+64+34+21+92+88+67+59+29+27+91+88+82+75+73+35=1875.$ Obviously, if $0\le a_i\le 1$, $0\le b\le 1$,

$$p_{ij}^A=p_{ij}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{k=1}^{l-1}{p}_{i\left[k\right]}}{{\sum }_{k=1}^{n_i}{p}_{ik}}}\right)}^{a_i}\right],$$

and

$$s_i^A=s_i\left[N+(1-N){\left(1+{\displaystyle \frac{{\sum }_{l=1}^{r-1}{s}_{\left[l\right]}}{{\sum }_{l=1}^m{s}_l}}\right)}^b\right],$$

the optimal solution of $1\left|gro-tec,p_{ij}^A,s_i^A,DDDE\right|C_{max}$ can be obtained by Algorithm 1, and

$$\begin{array}{cc}\hfill C_{max}=& \sum _{i=1}^m{s}_{\left[i\right]}\left[N+(1-N){\left(1+{\displaystyle \frac{{\sum }_{k=1}^{i-1}{s}_{\left[k\right]}}{{\sum }_{k=1}^m{s}_{\left[k\right]}}}\right)}^b\right]\hfill \\ & +\sum _{i=1}^m\sum _{j=1}^{n_{\left[i\right]}}{p}_{\left[i\right]\left[j\right]}\left[M+(1-M){\left(1+{\displaystyle \frac{{\sum }_{k=1}^{j-1}{p}_{\left[i\right]\left[k\right]}}{{\sum }_{k=1}^{n_{\left[i\right]}}{p}_{\left[i\right]\left[k\right]}}}\right)}^{a_{\left[i\right]}}\right]\hfill \\ \hfill =& 86+82\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{86}{86+82+72+38}}\right)}^{0.25}\right]\hfill \\ & +72\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{86+82}{86+82+72+38}}\right)}^{0.25}\right]\hfill \\ \hfill & +38\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{86+82+72}{86+82+72+38}}\right)}^{0.25}\right]\hfill \\ \hfill & +91+85\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{91}{91+85+74+73+53+49}}\right)}^{0.48}\right]\hfill \\ & +74\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{91+85}{91+85+74+73+53+49}}\right)}^{0.48}\right]\hfill \\ \hfill & +73\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{91+85+74}{91+85+74+73+53+49}}\right)}^{0.48}\right]\hfill \\ \hfill & +53\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{91+85+74+73}{91+85+74+73+53+49}}\right)}^{0.48}\right]\hfill \\ \hfill & +49\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{91+85+74+73+53}{91+85+74+73+53+49}}\right)}^{0.48}\right]\hfill \\ \hfill & +90+79\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{90}{90+79+78+64+34+21}}\right)}^{0.79}\right]\hfill \\ \hfill & +78\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{90+79}{90+79+78+64+34+21}}\right)}^{0.79}\right]\hfill \end{array}$$

$$\begin{array}{cc}\hfill & +64\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{90+79+78}{90+79+78+64+34+21}}\right)}^{0.79}\right]\hfill \\ \hfill & +34\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{90+79+78+64}{90+79+78+64+34+21}}\right)}^{0.79}\right]\hfill \\ \hfill & +21\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{90+79+78+64+34}{90+79+78+64+34+21}}\right)}^{0.79}\right]\hfill \end{array}$$

$$\begin{array}{cc}\hfill & +92+88\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{92}{92+88+67+59+29+27}}\right)}^{0.63}\right]\hfill \\ & +67\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{92+88}{92+88+67+59+29+27}}\right)}^{0.63}\right]\hfill \\ \hfill & +59\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{92+88+67}{92+88+67+59+29+27}}\right)}^{0.63}\right]\hfill \\ \hfill & +29\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{92+88+67+59}{92+88+67+59+29+27}}\right)}^{0.63}\right]\hfill \\ \hfill & +27\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{92+88+67+59+29}{92+88+67+59+29+27}}\right)}^{0.63}\right]\hfill \\ \hfill & +91+88\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{91}{91+88+82+75+73+35}}\right)}^{0.11}\right]\hfill \\ & +82\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{91+88}{91+88+82+75+73+35}}\right)}^{0.11}\right]\hfill \\ \hfill & +75\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{91+88+82}{91+88+82+75+73+35}}\right)}^{0.11}\right]\hfill \\ \hfill & +73\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{91+88+82+75}{91+88+82+75+73+35}}\right)}^{0.11}\right]\hfill \\ \hfill & +35\ast \left[0.5+0.5\ast {\left(1+{\displaystyle \frac{91+88+82+75+73}{91+88+82+75+73+35}}\right)}^{0.11}\right]\hfill \\ \hfill =& 2027.24376.\hfill \end{array}$$

4. Extension

Similar to Huang and Wang [49] and Section 3, the following general weighted deterioration model can be presented: If job $J_{ij}$ is placed in the lth position, the actual processing time is as follows:

$$p_{ij}^A=p_{ij}\left[M+(1-M){\left(1+\sum _{k=1}^{l-1}{\zeta }_{ik}{p}_{i\left[k\right]}\right)}^{{\alpha }_i}\right],$$

(9)

where ${\alpha }_i\ge 0$ is the deterioration index for $G_i$ and ${\zeta }_{ik}$ is the position-dependent weight of kth position in $G_i$. If $G_i$ is placed in the rth position, the actual setup time is as follows:

$$s_i^A=s_i\left[N+(1-N){\left(1+\sum _{l=1}^{r-1}{\tilde{\zeta }}_l{s}_{\left[l\right]}\right)}^{\beta }\right],$$

(10)

where $\beta \ge 0$ is the group-deterioration index and ${\tilde{\zeta }}_l$ is the position-dependent weight of lth group position. The problem can be denoted by

$$1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{weight}\right|C_{max},$$

where $DDD{E}_{weight}$ denotes the weighted-$DDDE$ models (9) and (10).

Similar to Huang and Wang [49] and Section 3, we have the following:

Lemma 6.

$I\left(\varpi \right)=[1-{(1+\varpi t)}^a]-\varpi [1-{(1+t)}^a]\ge 0$ if $t\ge 0$, $0\le a\le 1$, and $\varpi \ge 1$.

Lemma 7.

$I\left(\varpi \right)=[1-{(1+\varpi t)}^a]-\varpi [1-{(1+t)}^a]\le 0$ if $t\ge 0$, $a\ge 1$, and $\varpi \ge 1$.

Lemma 8.

If $0\le {\alpha }_i\le 1$ and ${\zeta }_{i{n}_i}\ge {\zeta }_{i,n_i-1}\ge \dots \ge {\zeta }_{i2}\ge {\zeta }_{i1}>0$, for

$$1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{weight}\right|C_{max},$$

and for $G_i$, the jobs are arranged in non-increasing order of $p_{ij}$.

Lemma 9.

If $0\le \beta \le 1$ and ${\tilde{\zeta }}_m\ge {\tilde{\zeta }}_{m-1}\ge \dots \ge {\tilde{\zeta }}_2\ge {\tilde{\zeta }}_1>0$, for

$$1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{weight}\right|C_{max},$$

the groups are arranged in non-increasing order of $s_i$.

Lemma 10.

If ${\alpha }_i\ge 1$ and $0<{\zeta }_{i{n}_i}\le {\zeta }_{i,n_i-1}\le \dots \le {\zeta }_{i2}\le {\zeta }_{i1}$, for

$$1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{weight}\right|C_{max},$$

and for $G_i$, the jobs are arranged in non-decreasing order of $p_{ij}$.

Lemma 11.

If $\beta \ge 1$ and $0<{\tilde{\zeta }}_m\le {\tilde{\zeta }}_{m-1}\le \dots \le {\tilde{\zeta }}_2\le {\tilde{\zeta }}_1$, for

$$1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{weight}\right|C_{max},$$

groups are arranged in non-decreasing order of $s_i$.

Based on Lemmas 8–11, the following algorithm is proposed to solve

$$1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{weight}\right|C_{max}.$$

Theorem 2.

$1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{weight}\right|C_{max}$ can be optimally solved by Algorithm 2 in $O(nlogn)$ time.

Algorithm 2 $1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{weight}\right|C_{max}$.

Step 1. If $0\le {\alpha }_i\le 1$ and ${\zeta }_{i{n}_i}\ge {\zeta }_{i,n_i-1}\ge \dots \ge {\zeta }_{i2}\ge {\zeta }_{i1}>0$, for $G_i$, the jobs are arranged in non-increasing order of $p_{ij}$ (Lemma 8). If ${\alpha }_i\ge 1$ and $0<{\zeta }_{i{n}_i}\le {\zeta }_{i,n_i-1}\le \dots \le {\zeta }_{i2}\le {\zeta }_{i1}$, for $G_i$, jobs are arranged in a non-decreasing order of $p_{ij}$ (Lemma 10). Step 2. If $0\le \beta \le 1$ and ${\tilde{\zeta }}_m\ge {\tilde{\zeta }}_{m-1}\ge \dots \ge {\tilde{\zeta }}_2\ge {\tilde{\zeta }}_1>0$, groups are arranged in a non-increasing order of $s_i$ (Lemma 9). If $\beta \ge 1$ and $0<{\tilde{\zeta }}_m\le {\tilde{\zeta }}_{m-1}\le \dots \le {\tilde{\zeta }}_2\le {\tilde{\zeta }}_1$, groups are arranged in a non-decreasing order of $s_i$ (Lemma 11). Step 3. Calculate $C_{max}$.

Remark 4.

From Lemma 8, if $0\le {\alpha }_{Huang}\le 1$ and ${\zeta }_n\ge {\zeta }_{m-1}\ge \dots \ge {\zeta }_2\ge {\zeta }_1>0$ (i.e., ${\zeta }_k\uparrow$), the problem $1\left|p_j^A=p_j{\left(1+{\sum }_{k=1}^{l-1}{\zeta }_k{p}_{\left[k\right]}\right)}^{{\alpha }_{Huang}},0\le {\alpha }_{Huang}\le 1,{\zeta }_k\uparrow \right|C_{max}$ can be solved by sequencing the jobs in non-increasing order of $p_j$.

Example 2.

Consider $1\left|gro-tec,p_{ij}^A,s_i^A,DDD{E}_{weight}\right|C_{max}$, where $n=15$, $m=3$, $n_1=n_2=n_3=5$, $M=0.5$, $N=0.5$, ${\alpha }_1=0.15$, ${\alpha }_2=0.65$, ${\alpha }_3=0.45$, $\beta =0.25$, ${\zeta }_{11}=0.1$, ${\zeta }_{12}=0.2$, ${\zeta }_{13}=0.3$, ${\zeta }_{14}=0.4$, ${\zeta }_{15}=0.5$, ${\zeta }_{21}=0.15$, ${\zeta }_{22}=0.25$, ${\zeta }_{23}=0.35$, ${\zeta }_{24}=0.45$, ${\zeta }_{25}=0.55$, ${\zeta }_{31}=0.11$, ${\zeta }_{32}=0.21$, ${\zeta }_{33}=0.36$, ${\zeta }_{34}=0.47$, ${\zeta }_{35}=0.51$, ${\tilde{\zeta }}_1=0.2$, ${\tilde{\zeta }}_2=0.3$, ${\tilde{\zeta }}_3=0.4$; the group- and job-related parameters are given in

Table 3. Parameters for Example 2.

${\mathit{G}}_{\mathit{i}}$	${\mathit{s}}_{\mathit{i}}$	${\mathit{p}}_{\mathit{i}1}$	${\mathit{p}}_{\mathit{i}2}$	${\mathit{p}}_{\mathit{i}3}$	${\mathit{p}}_{\mathit{i}4}$	${\mathit{p}}_{\mathit{i}5}$
$G_1$	8	12	13	15	25	17
$G_2$	7	9	12	27	26	21
$G_3$	10	11	19	8	7	5

.

Using Algorithm 2, Example 2 can be solved as follows:

Step 1: Jobs within each group are arranged in non-increasing order of $p_{ij}$, i.e.,

• $G_1:{\pi }_1\ast =\{J_{14}\to J_{15}\to J_{13}\to J_{12}\to J_{11}\}$,
• $G_2:{\pi }_2\ast =\{J_{23}\to J_{24}\to J_{25}\to J_{22}\to J_{21}\}$,
• $G_3:{\pi }_3\ast =\{J_{32}\to J_{31}\to J_{33}\to J_{34}\to J_{35}\}$.

Step 2: Arrange all groups in non-increasing order of $s_i$, i.e.,

• $\varrho \ast =\{G_3\to G_1\to G_2\}$.

Step 3: By (4), (9), and (10), we have

$$\begin{array}{cc}\hfill C_{max}(\varrho \ast )=& \sum _{i=1}^m{s}_{\left[i\right]}\left[N+(1-N){\left(1+\sum _{l=1}^{r-1}{\zeta }_l{s}_{\left[l\right]}\right)}^{\beta }\right]\hfill \\ & +\sum _{i=1}^m\sum _{j=1}^{n_{\left[i\right]}}{p}_{\left[i\right]\left[j\right]}\left[M+(1-M){\left(1+\sum _{k=1}^{l-1}{\zeta }_{ik}{p}_{\left[i\right]\left[k\right]}\right)}^{{\alpha }_{\left[i\right]}}\right]\hfill \\ \hfill =& 10+8\ast \left[0.5+0.5\ast {\left(1+0.2\ast 10\right)}^{0.25}\right]\hfill \\ & +7\ast \left[0.5+0.5\ast {\left(1+0.2\ast 10+0.3\ast 8\right)}^{0.25}\right]\hfill \\ \hfill & +19+11\ast \left[0.5+0.5\ast {\left(1+0.11\ast 19\right)}^{0.45}\right]\hfill \\ & +8\ast \left[0.5+0.5\ast {\left(1+0.11\ast 19+0.21\ast 11\right)}^{0.45}\right]\hfill \\ \hfill & +7\ast \left[0.5+0.5\ast {\left(1+0.11\ast 19+0.21\ast 11+0.36\ast 8\right)}^{0.45}\right]\hfill \\ \hfill & +5\ast \left[0.5+0.5\ast {\left(1+0.11\ast 19+0.21\ast 11+0.36\ast 8+0.47\ast 7\right)}^{0.45}\right]\hfill \\ \hfill & +25+17\ast \left[0.5+0.5\ast {\left(1+0.1\ast 25\right)}^{0.15}\right]\hfill \\ \hfill & +15\ast \left[0.5+0.5\ast {\left(1+0.1\ast 25+0.2\ast 17\right)}^{0.15}\right]\hfill \\ \hfill & +13\ast \left[0.5+0.5\ast {\left(1+0.1\ast 25+0.2\ast 17+0.3\ast 15\right)}^{0.15}\right]\hfill \\ \hfill & +12\ast \left[0.5+0.5\ast {\left(1+0.1\ast 25+0.2\ast 17+0.3\ast 15+0.4\ast 13\right)}^{0.15}\right]\hfill \\ \hfill & +27+26\ast \left[0.5+0.5\ast {\left(1+0.15\ast 27\right)}^{0.65}\right]\hfill \end{array}$$

$$\begin{array}{cc}\hfill & +21\ast \left[0.5+0.5\ast {\left(1+0.15\ast 27+0.25\ast 26\right)}^{0.65}\right]\hfill \\ \hfill & +12\ast \left[0.5+0.5\ast {\left(1+0.15\ast 27+0.25\ast 26+0.35\ast 21\right)}^{0.65}\right]\hfill \\ \hfill & +9\ast \left[0.5+0.5\ast {\left(1+0.15\ast 27+0.25\ast 26+0.35\ast 21+0.45\ast 12\right)}^{0.65}\right]\hfill \\ \hfill =& 415.24034.\hfill \end{array}$$

Remark 5.

Scheduling combined with $gro-tec$ and general logarithmic/weight deterioration can impact the group and job sequences and thus affect production processes and decisions. Our theoretical results and numerical analysis are conducted, we find that our theories and methods are very efficient. Hence, $gro-tec$ and general logarithmic/weight deterioration need to be taken into consideration to reduce costs and improve production efficiency.

5. Numerical Study

To evaluate the running time of Algorithms 1 and 2, the random instances were generated. The program was programmed in Visual Studio 2022 v17.1.0 using the C++ language and the testing computer was a desktop machine with a 12th Gen Intel(R) Core(TM) i5-12400 2.50 GHz CPU, 16.00GB RAM, and a Windows 11 operating system. The features of these instances were listed as follows:

(1)

Jobs: $n=200,400,600,800,1000,1200,1400,1600,1800$;

(2)

Groups: $m=7,19,23,35,46$;

(3)

$p_{ij}\in U[5,100],s_i\in U[5,100]$ (such that $ln{p}_{ij}>0$ and $ln{s}_i>0$, $i=1,2,\cdots ,m;j=1,2,\cdots ,n_i$);

(4)

$a_i\in U[0.1,0.9],b\in U[0.1,0.9]$, ${\alpha }_i\in U[0.1,0.9],\beta \in U[0.1,0.9]$ ($i=1,2,\cdots ,m$);

(5)

$M=N\in U[0.1,0.5]$, $M=N\in U[0.5,0.9]$, $M=N\in U[0.1,0.9]$;

(6)

${\tilde{\zeta }}_l={\zeta }_{ik}=1$ ($l=1,2,\cdots ,m;i=1,2,\cdots ,m;k=1,2,\cdots ,n_i$).

For each numerical study, 20 random replications were generated, where the minimum, average, and maximum (denoted by min, ave, and max) CPU times (the unit is milliseconds, denoted by ms) were given in

Table 4. $M=N\in U[0.1,0.5]$: CPU time (ms) of Algorithms 1 and 2.

		CPU of Algorithm 1			CPU of Algorithm 2
$\mathit{n}$	$\mathit{m}$	min	ave	max	min	ave	max
200	7	0.1774	0.2997	0.7437	0.2200	0.3867	1.0640
	19	0.2357	0.2976	0.4157	0.3529	0.4372	0.6122
	23	0.3163	0.3416	0.3691	0.5212	0.5437	0.5634
	35	0.4258	0.4617	0.4898	0.7274	0.7665	0.7931
	46	0.5438	0.7598	1.6726	0.9814	1.2339	1.5569
400	7	0.3684	0.4951	0.7386	0.6117	0.7653	0.9145
	19	0.4843	0.6571	0.8787	0.8341	1.0164	1.2412
	23	0.6028	0.7053	1.0639	1.1206	1.2842	1.7052
	35	0.7610	0.9271	1.4268	1.4281	1.6534	2.1336
	46	0.9153	1.1397	1.5983	1.7811	2.0814	3.0258
600	7	0.6473	0.7422	1.0708	1.2052	1.2830	1.5264
	19	0.7788	0.9706	1.5706	1.5413	1.7291	2.4133
	23	1.0146	1.1945	1.5703	1.9720	2.2973	4.1754
	35	1.1692	1.6870	2.6565	2.4915	2.9876	4.5148
	46	1.3560	1.6802	2.6189	2.7533	3.1205	4.0006
800	7	1.3822	1.7596	2.9700	2.9313	3.3962	4.2770
	19	1.8140	2.6647	4.7241	3.7607	4.6007	5.5952
	23	2.0440	2.7970	3.8305	4.3696	5.1033	6.4268
	35	2.2852	2.5880	3.9130	4.9293	5.2313	5.6840
	46	2.3976	2.7193	4.1999	5.2605	5.9354	7.6332
1000	7	3.9664	4.6102	5.9352	9.3543	9.9175	12.7181
	19	4.2255	4.6904	5.2505	9.8872	10.5733	11.4234
	23	4.5886	4.9357	6.0811	10.8033	11.4962	15.7895
	35	5.0376	5.5177	7.3171	11.6566	12.3367	14.8367
	46	5.3083	5.7599	6.4271	12.4811	13.0953	14.4685
1200	7	9.4020	11.6916	16.4927	14.0080	16.8438	21.0809
	19	9.6212	11.4964	14.7645	14.8975	16.2413	18.4022
	23	9.8770	11.4983	15.3408	15.6348	17.3383	21.5357
	35	10.8521	11.8056	14.7564	16.5247	17.6264	19.2839
	46	10.8900	11.7488	12.8379	17.8773	18.9050	21.0399
1400	7	18.3438	22.3562	33.2245	22.9130	27.1067	33.6057
	19	18.0093	21.1386	28.2760	24.3032	26.4216	32.3352
	23	19.0598	20.1206	24.6603	24.2044	26.1036	30.6468
	35	19.6248	20.5885	26.4030	25.5290	26.4424	29.5598
	46	20.2367	21.2529	25.1075	26.8124	27.5080	29.2252
1600	7	24.3825	28.5933	35.2687	27.8955	31.7924	35.7928
	19	25.3655	26.5179	28.1867	29.7942	31.0178	32.6665
	23	26.0762	27.5465	30.2032	31.5580	32.6388	35.8208
	35	26.9154	28.2394	31.8867	32.0396	33.4994	35.7076
	46	27.6596	29.1840	35.7480	33.6663	35.2880	37.8210
1800	7	32.1265	36.4415	47.7114	35.8050	40.1072	50.2326
	19	32.9307	34.9354	40.2402	37.1589	39.1894	43.9375
	23	32.9574	35.9799	42.5340	38.4970	40.7730	47.6961
	35	34.8087	36.5571	39.8731	40.6130	42.1835	45.2329
	46	35.7117	37.8375	41.6455	41.4760	43.6270	45.9044

,

Table 5. $M=N\in U[0.5,0.9]$: CPU time (ms) of Algorithms 1 and 2.

		CPU of Algorithm 1			CPU of Algorithm 2
$\mathit{n}$	$\mathit{m}$	min	ave	max	min	ave	max
200	7	0.2560	0.2938	0.3864	0.2082	0.2694	0.3556
	19	0.3681	0.4525	0.6267	0.3567	0.4219	0.5697
	23	0.5035	0.5471	0.8365	0.5155	0.5585	0.7156
	35	0.6510	0.6993	0.8260	0.7385	0.7973	0.9962
	46	0.7851	0.8552	1.1812	0.9350	1.0005	1.1266
400	7	1.2162	1.3856	1.8078	1.2118	1.2991	1.6526
	19	1.4383	1.6922	2.4148	1.5410	1.8255	2.4314
	23	1.7991	2.2955	4.8562	1.9864	2.2877	2.7729
	35	2.0593	2.5262	3.4860	2.4157	2.8369	4.6875
	46	2.2577	2.7155	3.9557	2.7797	3.2972	4.0864
600	7	1.8330	2.0841	2.8406	2.0888	2.4710	4.3851
	19	2.1347	2.5783	4.3220	2.5958	3.0260	4.4506
	23	2.4840	3.0206	3.7853	3.0453	3.7215	4.7296
	35	2.7277	3.0337	4.2711	3.4734	3.8319	4.6743
	46	3.0436	3.4667	5.2598	4.0126	4.5317	6.1291
800	7	3.4612	3.8030	4.8420	4.3228	5.0811	9.1788
	19	4.0267	4.6951	5.8460	5.2678	6.0208	7.7749
	23	4.1744	5.0696	7.1361	5.6216	6.4516	8.2103
	35	4.3251	5.0799	7.4577	5.9949	6.7402	7.7306
	46	4.7400	5.2664	6.7092	6.6353	7.1462	8.3423
1000	7	7.1940	8.3822	10.8126	7.7254	9.3606	13.4320
	19	7.9808	9.1992	11.7655	8.7158	10.3472	12.5973
	23	7.7566	8.2500	9.5571	8.7284	9.6805	11.4939
	35	8.2859	9.4391	12.4033	9.8956	11.2440	14.9506
	46	8.7764	9.4443	10.9769	10.7814	11.6629	14.1304
1200	7	19.7160	21.4685	24.9446	17.2613	19.8019	25.0173
	19	19.3804	21.0536	25.3733	17.1724	19.3152	22.8843
	23	19.6164	20.8600	22.4987	18.1426	19.2179	21.0724
	35	20.6578	22.1495	26.7464	19.3835	20.3367	22.5411
	46	21.3446	22.2309	23.2110	20.6118	21.6691	23.4645
1400	7	25.7501	28.5240	33.6090	22.1144	24.9513	30.0138
	19	26.1913	27.2424	28.3921	23.0511	24.6753	27.1149
	23	27.4405	28.6227	30.1684	24.7625	25.9299	29.5416
	35	27.8198	29.4367	31.5362	25.3246	26.8058	30.7956
	46	29.1443	30.2465	33.2877	26.7806	28.5316	31.1388
1600	7	27.2853	29.6356	33.5509	28.8794	32.4197	39.5174
	19	27.5957	29.4912	31.6098	29.6950	31.4061	35.1885
	23	28.9769	30.1650	31.6810	31.2172	32.6932	38.5413
	35	29.5302	30.8947	33.0866	33.2494	34.1973	35.4664
	46	30.2623	32.5103	39.9005	34.0117	35.7329	40.2877
1800	7	31.4170	35.6590	48.8139	36.2494	39.3941	51.7198
	19	32.5265	34.3566	39.0429	36.8608	38.7472	40.8807
	23	33.4835	35.2512	39.2264	39.2339	40.6845	42.2809
	35	35.0957	36.9180	40.9085	41.6383	43.3374	47.0185
	46	36.1107	37.2882	39.0944	42.7363	44.9120	49.8094

and

Table 6. $M=N\in U[0.1,0.9]$: CPU time (ms) of Algorithms 1 and 2.

		CPU of Algorithm 1			CPU of Algorithm 2
$\mathit{n}$	$\mathit{m}$	min	ave	max	min	ave	max
200	7	0.1644	0.2295	0.3673	0.2017	0.2988	0.6115
	19	0.2583	0.2810	0.3323	0.3488	0.3776	0.4798
	23	0.3501	0.3984	0.5243	0.5198	0.5879	0.8461
	35	0.4715	0.5201	0.6056	0.7477	0.8059	1.0859
	46	0.6150	0.9036	1.3041	1.0083	1.3354	1.6907
400	7	0.4787	0.5776	0.8025	0.6520	0.8205	1.1169
	19	0.5781	0.6488	0.9687	0.8273	0.9017	1.4398
	23	0.7159	0.7606	0.9290	1.0689	1.1596	1.5978
	35	0.8800	0.9908	1.3700	1.3657	1.5304	2.3361
	46	1.0122	1.1861	1.6928	1.6289	1.7787	2.2620
600	7	1.2740	1.4686	2.1558	2.0385	2.3889	3.7875
	19	1.5067	1.7785	2.6780	2.5963	2.9535	4.0530
	23	1.7398	2.1118	3.0119	2.9240	3.4155	4.4226
	35	2.0562	2.7242	3.7269	3.5862	4.5978	5.7118
	46	2.3481	3.1100	4.7084	4.2361	5.2430	6.1953
800	7	2.6546	3.4675	5.1269	4.4362	5.3880	7.2490
	19	3.0512	3.9569	4.9959	5.1926	6.0761	7.0845
	23	3.1419	3.6941	5.2662	5.6052	6.3130	8.4620
	35	3.4184	3.7762	4.7588	6.1878	6.6297	7.9533
	46	3.6231	3.9221	5.1038	6.6183	7.2700	8.9901
1000	7	6.0690	8.2522	12.4782	9.5774	11.8295	15.0412
	19	6.3727	7.2305	8.4022	10.2436	11.4977	14.2660
	23	6.4809	7.0419	7.9854	10.6934	11.5947	13.4705
	35	6.9904	7.4870	8.9789	11.7009	12.6167	14.1152
	46	7.3904	8.0823	11.6773	12.6301	13.7278	16.0588
1200	7	15.4206	17.5968	22.9041	17.6240	19.7965	27.7313
	19	15.6379	16.1995	17.0367	17.2107	18.3928	20.9206
	23	16.1442	16.9836	18.6367	18.0681	19.3553	20.5493
	35	17.3559	18.0029	19.1652	19.3204	20.3263	21.4951
	46	17.7533	18.5459	19.8889	21.1646	21.9092	23.8463
1400	7	22.6661	25.0604	32.0837	23.2173	26.1056	32.0133
	19	22.2896	23.4033	26.1901	23.4637	24.7351	26.6533
	23	22.9604	23.4402	23.8200	24.1449	25.9237	28.4372
	35	23.9459	25.4409	35.2886	26.0647	27.1852	30.2367
	46	24.9069	25.6918	26.8693	27.6540	28.3296	31.1208
1600	7	27.7084	29.5620	34.5762	29.0357	31.8625	38.9354
	19	27.4759	29.6487	40.7330	29.7080	30.9598	33.3532
	23	28.6234	30.0695	31.8289	31.3310	32.9337	34.8635
	35	29.2710	30.9799	33.2976	32.3556	34.2822	35.5770
	46	31.0965	32.1412	34.0411	34.2652	36.1895	38.8007
1800	7	35.8786	38.7141	45.0549	35.9549	39.3609	44.4885
	19	36.0342	37.8711	40.8724	37.4281	39.1624	41.6080
	23	36.8003	38.8389	41.5131	38.7891	40.4470	43.2618
	35	37.7501	39.9583	42.2122	40.6191	42.5458	44.7960
	46	39.4025	40.9845	43.6278	42.4824	43.9993	45.4539

. As seen from Table 4, Table 5 and Table 6, we can find that Algorithms 1 and 2 are effective and their CPU times increase moderately as n and m increases, and the maximum CPU times of Algorithms 1 and 2 in this experiment are only 48.8139 ms and 51.7198 ms, respectively, when the problem size is $n=1800$, $m=7$, and $M=N\in U[0.5,0.9]$.

6. Conclusions

In this article, the single-machine $gro-tec$ maximal completion time cost with general logarithmic/weight deterioration has been investigated. It was shown that $1|gro-tec,$$p_{ij}^A,s_i^A,DDD{E}_{logarithmic}|C_{max}$ ($Z\in \{DDD{E}_{logarithmic},DDD{E}_{weight}\}$) is polynomially solvable. Further research may consider $gro-tec$ scheduling in a flow shop setting (see Sun et al. [52], Lv and Wang [53], Geng et al. [54]), $DDDE$ scheduling with position-dependent weights (see Sun et al. [55], Wang et al. [56]), $DDDE$ scheduling with resource allocation (see Qian et al. [57], Bai et al. [58], Zhang et al. [59]), or $gro-tec$ scheduling with step-improving jobs (see Kim and Oron [60], Lim et al. [61], Wu et al. [62], Cheng et al. [63]).