Study on Single-Machine Common/Slack Due-Window Assignment Scheduling with Delivery Times, Variable Processing Times and Outsourcing

Abstract

Single-machine due-window assignment scheduling with delivery times and variable processing times is investigated, where the variable processing time of a job means that the processing time is a function of its position in a sequence and its resource allocation. Currently, there are multiple optimization objectives for the due-window assignment problem, and there is a small amount of research on optimization problems where the window starting time, the rejected cost and the optimal scheduling are jointly required. The goal of this paper is to minimize the weighed sum of scheduling cost, resource consumption cost and outsourcing measure under the optional job outsourcing (rejection). Under two resource allocation models (i.e., linear and convex resource allocation models), the scheduling cost is the weighted sum of the number of early–tardy jobs, earliness–tardiness penalties and due-window starting time and size, where the weights are positional-dependent. The main contributions of this paper include the study and data simulation of single-machine scheduling with learning effects, delivery times and outsourcing cost. For the weighed sum of scheduling cost, resource consumption cost and outsourcing measure, we prove the polynomial solvability of the problem. Under the common and slack due-window assignments, through the theoretical analysis of the optimal solution, we reveal that four problems can be solved in $O\left(n^6\right)$ time, where n is the number of jobs.

1. Introduction

In many real-world producing and scheduling problems, the due-window assignment is a growing concern, especially for the earliness and tardiness penalty problems (Tian [1]). Under the common/slack due-window (CONDW/SLKDW), if a job is earlier (later) than the due-window, it will incur an early (delayed) penalty (Xu et al. [2], Shabtay et al. [3] and Zhang et al. [4]). For any enterprise in production, it is necessary to specify the delivery time period of the product, which is the due window. Qian and Zhan [5] and Mao et al. [6] thought over the single-machine scheduling problems with a due-window and delivery time, where the delivery time is the past-sequence-dependent delivery time (denoted by $q_{psd}$). Mao et al. [7] considered the single-machine scheduling with deteriorating jobs and $q_{psd}$. They proved that the makespan minimization can be solved in polynomial time. Ren et al. [8] explored the single-machine scheduling with learning effects and exponential $q_{psd}$. Moreover, minimizing the total cost is an important optimization objective in the problem of due-window assignment in single-machine scheduling. The total cost is a weighted sum function of the earliness and tardiness, the start of due window and the size of due window, where the weights depend only on their positions in the sequence (i.e., position-dependent weights, Lin [9]). Wang et al. [10] considered single-machine position-dependent weights scheduling problem with $q_{psd}$ and truncated sum-of-processing-times-based learning effect. Recently, under common and slack due-window assignments, Wang et al. [11] examined single-machine problems with due-window assignment and positional-dependent weights. They demonstrated that some versions of common and slack due-window assignments are polynomially solvable. Nevertheless, in real-world production processes, job processing times need to be controlled by allocating additional resources, which is the scheduling with resource allocation (see Lu et al. [12] and Wang et al. [13]). Zhao et al. [14] and Sun et al. [15] considered the resource allocation scheduling with slack due-window assignment.

In addition to the due-window assignments, there are more factors to consider in actual scheduling problems, namely the cost of job rejection and/or learning effect. In the classical scheduling models, all jobs must be processed; nevertheless, in actual production, due to the requirements of processing capacity and profitability, vendors need to reject some jobs that take longer to process or are less profitable, thus incurring rejection costs. The learning effect is due to the accumulation of experience, which shortens the processing times of jobs (Lv and Wang [16] and Zhang et al. [17]). Li and Chen [18] conducted single-machine scheduling problems with job rejection and a deteriorating maintenance activity. Mor [19] considered single-machine scheduling problems with job rejection and deteriorating effects. Geng et al. [20] studied proportionate flow shop scheduling with job rejection. Under common due-date assignment, they proved that some problems are polynomially solvable. Toksari and Atalay [21] and Liu et al. [22] considered single-machine scheduling problems with job rejection and learning effects. Atsmony and Mosheiov [23] studied single-machine scheduling with job rejection. For the maximum earliness/tardiness cost minimization under an upper bound on the total permitted rejection cost, they proposed a pseudo-polynomial dynamic programming algorithm and a heuristic algorithm. The learning effect is specifically seen in the fact that workers repeat their work many times and their skills become more proficient (Sun et al. [24], Zhao  [25] and Chen et al. [26]). Li et al. [27] considered the flow shop scheduling with truncated learning effects. For the makespan objective, they proposed the branch-and-bound and heuristic algorithms.

Currently, in the practical application of due-window assignment scheduling problems, there are multiple optimization objectives and, therefore, some limitations. Recently, Wang et al. [28] considered the single-machine resource allocation scheduling with the CONDW assignment. Under the linear and convex resource allocations, they showed that four versions of general earliness–tardiness cost and resource consumption cost are polynomially solvable. The goal of this paper is to further extend the results of Wang et al. [28] by combining delivery times and optional outsourcing (job rejection), i.e., under CONDW/SLKDW assignment; the method of outsourcing enables the operations manager to improve the overall performance; obviously, the outsourced jobs have outsourcing costs (see Fang et al. [29] and Freud and Mosheiov [30]). The main contributions of this paper are as follows: (i) single-machine scheduling with learning effects, delivery times and outsourcing costs is studied and simulated with data; (ii) for a given schedule, the optimal solution properties are given; (iii) for the weighed sum of scheduling cost, resource consumption cost and outsourcing measure, we prove that the problem can be solved in polynomial time. The paper is organized as follows: Section 2 provide a description of the problem. In Section 3, we prove that four problems are polynomially solvable. Section 4 conducts the experiment. In Section 5, we summarize the conclusions.

2. Problem Statement

In this paper, the mathematical notations used are listed in Table 1. There are n independent and non-preemptive jobs $J=\{J_1,J_2,\cdots ,J_n\}$ are processed on a single machine, all jobs are available at time zero and the machine processes at most one job at the same time. First, we determine whether the job is processed or not, and classify the jobs into an acceptable set $AS$ and unacceptable (rejected) set $\overline{AS}$.

Table 1. Mathematical notations.

Notation	Meaning
n	The number of jobs
$AS$	The acceptable set of jobs
$\overline{AS}$	The unacceptable (rejected) set of jobs
$n_a$	the number of jobs in acceptable set
$J_j$	The jth job $J_j$
$J_{\left[j\right]}$	The job scheduled in the jth position
$p_j$	The normal processing time of job $J_j$
$p_{\left[j\right]}$	The normal processing time of the job scheduled in the jth position $J_j$
$p_j^A$	The actual processing time of job $J_j$
$a_j$	The position-dependent learning index
$b_j$	The positive compression rate
$u_j\left(u_{\left[j\right]}\right)$	The resource allocation to job $J_j\left(J_{\left[j\right]}\right)$
$q_{\left[j\right]}$	The past-sequence-dependent delivery time of job $J_{\left[j\right]}$
$C_j\left(C_{\left[j\right]}\right)$	The completion time of job $J_j\left(J_{\left[j\right]}\right)$
$G_j\left(H_j\right)$	The number of early (tardy) jobs
${\sigma }_j({\gamma }_j,{\alpha }_j,{\beta }_j)$	The positional-dependent weights
${\upsilon }_j$	The unit cost associated with $u_j$
$e_j$	The rejected cost of job $J_j$

Under linear resource allocation, the actual processing time of job $J_j$ in position r can be expressed as

$$p_j^A=p_j{r}^{a_j}-b_j{u}_j,j,r=1,2,\cdots ,n_a,$$

(1)

where $a_j\le 0$ is the position-dependent learning index, $0\le u_j\le \overline{u_j}\le {\displaystyle \frac{p_j{\left(n_a\right)}^{a_j}}{b_j}}$ and $\overline{u_j}$ is the upper bound of $u_j$. Under convex resource allocation, $p_j^A$ can be expressed as

$$p_j^A={\left({\displaystyle \frac{p_j{r}^{a_j}}{u_j}}\right)}^m,j,r=1,2,\cdots ,n_a,$$

(2)

where $m>0$ is a constant. Let $\left[d_1^j,d_2^j\right]$$\left(d_1^j<d_2^j\right)$ be the due-window of job $J_j$, where $d_1^j$ (resp. $d_2^j$) is the starting (resp. finishing) time, and $D_j=d_2^j-d_1^j$ is the due-window size of job $J_j$. Under the common due-window (CONDW) assignment, it is assumed that $d_1^j=d_1,d_2^j=d_2$$(d_1<d_2)$ and $D_j=D=d_2-d_1$ is the size of the common due-window. Under slack due-window (SLKDW) assignment, it is assumed that $d_1^j=p_j^A+q^{\prime },d_2^j=p_j^A+q^{″}$$(q^{\prime }<q^{″})$ and $D_j=D=q^{″}-q^{\prime }$ is the size of the slack due-window. Let $\left[j\right]$ be some job placed in the jth position, as in Qian and Zhan [5], the past-sequence-dependent delivery time (denoted by $q_{psd}$) of job $J_{\left[j\right]}$ is

$${\displaystyle q_{\left[j\right]}=X{W}_{\left[j\right]}=X\sum _{l=1}^{j-1}{p}_{\left[l\right]}^A,}$$

(3)

where the delivery rate is $X\left(X>0\right)$; then, the completion time of job $J_j$ is

$${\displaystyle C_{\left[j\right]}=\sum _{l=1}^j{p}_{\left[l\right]}^A+q_{\left[j\right]}.}$$

(4)

Let the number of early job $G_j$ and tardy job $H_j$ be

$$G_j=\left\{\begin{array}{cc}1,\hfill & if{d}_1^j>C_j,\hfill \\ 0,\hfill & otherwise,\hfill \end{array}\right.$$

(5)

and

$$H_j=\left\{\begin{array}{cc}1,\hfill & if{d}_2^j<C_j,\hfill \\ 0,\hfill & otherwise.\hfill \end{array}\right.$$

(6)

Let the earliness and tardiness of job $J_j$ be $E_j=\max \left\{0,d_1^j-C_j\right\}$ and $T_j=\max \left\{0,C_j-d_2^j\right\}$, the goal of this paper is to determine the optimal schedule ${\pi }^{\ast }$, $d_1$ ($q^{\prime }$), $d_2$ ($q^{″}$) (such D can be obtained) and resource allocation $u_j$ for the jobs in the acceptable set. This paper is to minimize the weighed sum of scheduling cost (${\displaystyle \sum _{j=1}^{n_a}{S}_j}$), resource consumption cost (${\displaystyle \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}}$) and outsourcing measure (${\displaystyle \sum _{j\in \overline{AS}}{e}_j}$), i.e., ${\displaystyle \sum _{j=1}^{n_a}{S}_j+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j}$, where $S_j={\sigma }_j{G}_{\left[j\right]}+{\gamma }_j{H}_{\left[j\right]}+{\alpha }_j{E}_{\left[j\right]}+{\beta }_j{T}_{\left[j\right]}+\rho d_1/q^{\prime }+\eta D$, ${\sigma }_j$, ${\gamma }_j$, ${\alpha }_j$ and ${\beta }_j$ are positional-dependent weights (Wang et al. [31] and Qian et al. [32]); $\rho$ and $\eta$ are the unit time weight for the due-window starting time $d_1/q^{\prime }$ and size D; $\lambda$ is the weight of resource consumption cost; $\epsilon \ge 0$ is the weight of unacceptable set of jobs (where $e_j$ is the rejected cost of job $J_j$). If the above parameters are given (i.e., ${\sigma }_j$, ${\gamma }_j$, ${\alpha }_j$, ${\beta }_j$, $p_j$, $b_j$, ${\overline{u}}_j$, $a_j$, $e_j$, $\rho$, $\eta$, $\lambda$, $\epsilon$, X are the assumptions made), the problems can be denoted by

$${\displaystyle P1/P2:1\left|p_j^A=p_j{r}^{a_j}-b_j{u}_j,q_{psd},CONDW/SLKDW\right|\sum _{j=1}^{n_a}{S}_j+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j}$$

(7)

and

$${\displaystyle P3/P4:1\left|p_j^A={\left(\frac{p_j{r}^{a_j}}{u_j}\right)}^m,q_{psd},CONDW/SLKDW\right|\sum _{j=1}^{n_a}{S}_j+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j.}$$

(8)

Table 2 lists the relevant models studied.

Table 2. Model studied.

Reference	Scheduling Problem	Time Complexity
Qian and Zhan [5]	$1\left|CON/SLK,DES,q_{psd}\right|{\sum }_{j=1}^n\alpha E_j+\beta T_j+\rho d_k^1+\eta D$	$O\left(n\log n\right)$
Wang et al. [11]	$1\left|CONDW\right|{\sum }_{j=1}^n({\alpha }_j{U}_{\left[j\right]}+{\beta }_j{V}_{\left[j\right]}+{\eta }_j{E}_{\left[j\right]}+{\delta }_j{T}_{\left[j\right]}+\theta d+\vartheta D)$	$O\left(n^5\right)$
	$1\left|CONDW\right|{\sum }_{j=1}^n({\alpha }_j{U}_{\left[j\right]}+{\beta }_j{V}_{\left[j\right]}+{\eta }_j{E}_{\left[j\right]}+{\delta }_j{T}_{\left[j\right]}+\theta d$.	$O\left(n^4\right)$
Wang et al. [28]	$1\left|p_j^A=p_j-b_j{u}_j,CONDW\right|{\sum }_{j=1}^n({\alpha }_j{U}_{\left[j\right]}+{\beta }_j{V}_{\left[j\right]}+{\eta }_j{E}_{\left[j\right]}+{\delta }_j{T}_{\left[j\right]}+\theta d_1+\lambda D)+$$\rho {\sum }_{j=1}^n{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}$	$O\left(n^5\right)$
	$1\left|p_j^A={\left(\frac{p_j}{u_j}\right)}^k,CONDW\right|{\sum }_{j=1}^n({\alpha }_j{U}_{\left[j\right]}+{\beta }_j{V}_{\left[j\right]}+{\eta }_j{E}_{\left[j\right]}+{\delta }_j{T}_{\left[j\right]}+\theta d_1+\lambda D)+$$\rho {\sum }_{j=1}^n{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}$	$O\left(n^5\right)$
This article	${\displaystyle 1\left|p_j^A=p_j{r}^{a_j}-b_j{u}_j,q_{psd},CONDW\right|\sum _{j=1}^{n_a}{S}_j+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j}$	$O\left(n^6\right)$
	${\displaystyle 1\left|p_j^A=p_j{r}^{a_j}-b_j{u}_j,q_{psd},SLKDW\right|\sum _{j=1}^{n_a}{S}_j+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j}$	$O\left(n^6\right)$
	${\displaystyle 1\left|p_j^A={\left(\frac{p_j{r}^{a_j}}{u_j}\right)}^m,q_{psd},CONDW\right|\sum _{j=1}^{n_a}{S}_j+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j}$	$O\left(n^6\right)$
	${\displaystyle 1\left|p_j^A={\left(\frac{p_j{r}^{a_j}}{u_j}\right)}^m,q_{psd},SLKDW\right|\sum _{j=1}^{n_a}{S}_j+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j}$	$O\left(n^6\right)$

DES denotes deterioration effects, CON (resp. SLK) denotes common (resp. slack) due-date assignment.

3. Method

For the P1–P4, there exists an optimal schedule ${\pi }^{\ast }$ such that all jobs are not idle during processing and processing starts at moment 0 from Wang et al. [11].

Lemma 1.

In the case that $n_a$ is given for the P1 and P3 of the CONDW assignment, there exists an optimal schedule π such that the optimal $d_1$ (resp. $d_2$) is equal to $C_{\left[k\right]}$ (resp. $C_{\left[w\right]}$), i.e., $d_1=C_{\left[k\right]}$, where

$${\displaystyle \sum _{j=1}^{k-1}{\alpha }_j\le n_a\left(\eta -\rho \right),}$$

(9)

$d_2=C_{\left[w\right]}$, where

$${\displaystyle \sum _{j=w+1}^{n_a}{\beta }_j\le n_a\eta ,}$$

(10)

$1\le k\le w\le n_a.$

Proof. 

It is similar to the proof of Wang et al. [11] and Wang et al. [28]:

(1) Let $C_{[k-1]}<d_1<C_{\left[k\right]}$ and $d_2=C_{\left[w\right]}$, we have

$$\begin{array}{cc}\hfill F& {\displaystyle =\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{k-1}{\alpha }_j(d_1-C_{\left[j\right]})+\sum _{j=w+1}^{n_a}{\beta }_j(C_{\left[j\right]}-d_2)}\hfill \\ \hfill & {\displaystyle +n\rho d_1+n\eta (d_2-d_1)+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j.}\hfill \end{array}$$

(11)

When $d_1=C_{[k-1]}$ and $d_2=C_{\left[w\right]}$, we have

$$\begin{array}{cc}\hfill F_1& {\displaystyle =\sum _{j=1}^{k-2}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{k-1}{\alpha }_j(C_{[k-1]}-C_{\left[j\right]})+\sum _{j=w+1}^{n_a}{\beta }_j(C_{\left[j\right]}-d_2)}\hfill \\ \hfill & {\displaystyle +n\rho C_{[k-1]}+n\eta (d_2-C_{[k-1]})+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j.}\hfill \end{array}$$

(12)

When $d_1=C_{\left[k\right]}$ and $d_2=C_{\left[w\right]}$, we have

$$\begin{array}{cc}\hfill F_2& {\displaystyle =\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{k-1}{\alpha }_j(C_{\left[k\right]}-C_{\left[j\right]})+\sum _{j=w+1}^{n_a}{\beta }_j(C_{\left[j\right]}-d_2)}\hfill \\ \hfill & {\displaystyle +n\rho C_{\left[k\right]}+n\eta (d_2-C_{\left[k\right]})+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j.}\hfill \end{array}$$

(13)

$$\begin{array}{c}\hfill F-F_1={\sigma }_{k-1}+\left[{\displaystyle \sum _{j=1}^{k-1}{\alpha }_j+n(\rho -\eta )}\right](d_1-C_{[k-1]})\end{array}$$

(14)

and

$$\begin{array}{c}\hfill F-F_2=\left[{\displaystyle \sum _{j=1}^{k-1}{\alpha }_j+n(\rho -\eta )}\right](d_1-C_{\left[k\right]}).\end{array}$$

(15)

If ${\displaystyle \sum _{j=1}^{k-1}{\alpha }_j+n(\rho -\eta )\ge 0}$, then $F-F_1\ge 0$; if ${\displaystyle \sum _{j=1}^{k-1}{\alpha }_j+n(\rho -\eta )\le 0}$; thus, $F-F_2\ge 0$. Hence, $d_1$ is equal to the completion time of some job, i.e., $d_1=C_k$. Let $d_1=C_{\left[k\right]}$ and $C_{[w-1]}<d_2<C_{\left[w\right]}$, similarly, $d_2$ is equal to the completion time of some job, i.e., $d_2=C_{\left[w\right]}$.

(2) Let $d_1=C_{\left[k\right]}$ and $d_2=C_{\left[w\right]}$, we have

$$\begin{array}{cc}\hfill F& {\displaystyle =\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{k-1}{\alpha }_j(C_{\left[k\right]}-C_{\left[j\right]})+\sum _{j=w+1}^{n_a}{\beta }_j(C_{\left[j\right]}-d_2)}\hfill \\ \hfill & {\displaystyle +n\rho C_{\left[k\right]}+n\eta (d_2-C_{\left[k\right]})+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j.}\hfill \end{array}$$

(16)

When $d_1$ reduces x (i.e., $d_1=C_{\left[k\right]}-x$) and $d_2=C_{\left[w\right]}$, we have

$$\begin{array}{cc}\hfill F_1& {\displaystyle =\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{k-1}{\alpha }_j(C_{\left[k\right]}-x-C_{\left[j\right]})+\sum _{j=w+1}^{n_a}{\beta }_j(C_{\left[j\right]}-d_2)}\hfill \\ \hfill & {\displaystyle +n\rho (C_{\left[k\right]}-x)+n\eta (d_2-C_{\left[k\right]}+x)+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j.}\hfill \end{array}$$

(17)

When $d_1$ induces x (i.e., $d_1=C_{\left[k\right]}+x$) and $d_2=C_{\left[w\right]}$, we have

$$\begin{array}{cc}\hfill F_2& {\displaystyle =\sum _{j=1}^k{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^k{\alpha }_j(C_{\left[k\right]}+x-C_{\left[j\right]})+\sum _{j=w+1}^{n_a}{\beta }_j(C_{\left[j\right]}-d_2)}\hfill \\ \hfill & {\displaystyle +n\rho (C_{\left[k\right]}+x)+n\eta (d_2-C_{\left[k\right]}-x)+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j.}\hfill \end{array}$$

(18)

By using the classical small perturbation technique, we have

$$\begin{array}{c}\hfill {\displaystyle F-F_1=x(\sum _{j=1}^{k-1}{\alpha }_j+n\rho -n\eta )\le 0}\end{array}$$

(19)

and

$$\begin{array}{c}\hfill {\displaystyle F-F_2=-{\sigma }_{\left[k\right]}-x(\sum _{j=1}^k{\alpha }_j+n\rho -n\eta )\le 0.}\end{array}$$

(20)

Accordingly, ${\displaystyle \sum _{j=1}^{k-1}{\alpha }_j\le n(\eta -\rho )\le \sum _{j=1}^k{\alpha }_j+\frac{{\sigma }_k}{x}}$. Since it is impossible to determine the value of ${\displaystyle \frac{{\sigma }_k}{x}}$, and $n_a$ is given, it can only determine that ${\displaystyle \sum _{j=1}^{k-1}{\alpha }_j\le n_a(\eta -\rho ).}$

Similarly, following the above steps, we can get ${\displaystyle \sum _{j=w+1}^{n_a}{\beta }_j\le n\eta \le \sum _{j=w}^{n_a}{\beta }_j+\frac{{\gamma }_w}{x}}$. Since it is impossible to determine the value of ${\displaystyle \frac{{\gamma }_w}{x}}$, and $n_a$ is given, it can only determine that ${\displaystyle \sum _{j=w+1}^{n_a}{\beta }_j\le n\eta }$.    □

Lemma 2.

In the case that $n_a$ is given for the P2 and P4 of the SLKDW assignment,, there exists an optimal schedule π such that the optimal $q^{\prime }$ (resp. $q^{″}$) is equal to $C_{\left[k-1\right]}$ (resp. $C_{\left[w-1\right]}$), i.e., $q^{\prime }=C_{\left[k-1\right]}$, $q^{″}=C_{\left[w-1\right]}$ where k (resp. w) is given by Equation (9) (resp. Equation (10)), and $1\le k\le w\le n_a$.

Proof. 

It is similar to the proof of Lemma 1 (see Wang et al. [11] and Wang et al. [28]).    □

3.1. Problem P1

For a given schedule of the acceptable job set, from Lemma 1, $d_1$, $d_2$ and D can be computed as follows: ${\displaystyle d_1=C_{\left[k\right]}=\sum _{l=1}^k{p}_{\left[l\right]}^A+X\sum _{l=1}^{k-1}{p}_{\left[l\right]}^A}$, ${\displaystyle d_2=C_{\left[w\right]}=\sum _{l=1}^w{p}_{\left[l\right]}^A+X\sum _{l=1}^{w-1}{p}_{\left[l\right]}^A}$ and ${\displaystyle D=C_{\left[w\right]}-C_{\left[k\right]}=\sum _{l=k+1}^w{p}_{\left[l\right]}^A+X\sum _{l=k}^{w-1}{p}_{\left[l\right]}^A}$. Thus

$$\begin{array}{cc}\hfill F& {\displaystyle =\sum _{j=1}^{n_a}\left({\sigma }_j{G}_{\left[j\right]}+{\gamma }_j{H}_{\left[j\right]}+{\alpha }_j{E}_{\left[j\right]}+{\beta }_j{T}_{\left[j\right]}+\rho d_1+\eta D\right)+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j}\hfill \\ \hfill & {\displaystyle =\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{k-1}{\alpha }_j{E}_{\left[j\right]}+\sum _{j=w+1}^{n_a}{\beta }_j{T}_{\left[j\right]}+n\rho d_1+n\eta D+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j}\hfill \\ \hfill & {\displaystyle =\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{k-1}{\alpha }_j\left(d_1-C_{\left[j\right]}\right)+\sum _{j=w+1}^{n_a}{\beta }_j\left(C_{\left[j\right]}-d_2\right)+n\rho d_1+n\eta D}\hfill \\ \hfill & {\displaystyle +\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j}\hfill \\ \hfill & {\displaystyle =\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{k-1}{\alpha }_j\left({\displaystyle \sum _{l=1}^k{p}_{\left[l\right]}^A+X\sum _{l=1}^{k-1}{p}_{\left[l\right]}^A-\sum _{l=1}^j{p}_{\left[l\right]}^A-X\sum _{l=1}^{j-1}{p}_{\left[l\right]}^A}\right)}\hfill \\ \hfill & {\displaystyle +\sum _{j=w+1}^{n_a}{\beta }_j\left({\displaystyle \sum _{l=1}^j{p}_{\left[j\right]}^A+X\sum _{l=1}^{j-1}{p}_{\left[l\right]}^A-\sum _{l=1}^w{p}_{\left[l\right]}^A-X\sum _{l=1}^{w-1}{p}_{\left[l\right]}^A}\right)+n\rho \left({\displaystyle \sum _{l=1}^k{p}_{\left[l\right]}^A+X\sum _{l=1}^{k-1}{p}_{\left[l\right]}^A}\right)}\hfill \\ \hfill & {\displaystyle +n\eta \left({\displaystyle \sum _{l=k+1}^w{p}_{\left[l\right]}^A+X\sum _{l=k}^{w-1}{p}_{\left[l\right]}^A}\right)+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j}\hfill \\ \hfill & {\displaystyle =\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{n_a}{\phi }_j{p}_{\left[j\right]}^A+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j,}\hfill \end{array}$$

(21)

where

$${\phi }_j=\left\{\begin{array}{cc}\left(X+1\right)\left({\displaystyle \sum _{l=1}^j{\alpha }_l+n\rho }\right),\hfill & j=1,2,\cdots ,k-1,\hfill \\ {\alpha }_{j-1}+n\rho +Xn\eta ,\hfill & j=k,\hfill \\ \left(X+1\right)n\eta ,\hfill & j=k+1,k+2,\cdots ,w-1,\hfill \\ n\eta +X{\beta }_{j+1},\hfill & j=w,\hfill \\ {\displaystyle \left(X+1\right)\sum _{l=j}^{n_a}{\beta }_l,}\hfill & j=w+1,w+2,\cdots ,n_a-1,\hfill \\ {\beta }_j,\hfill & j=n_a.\hfill \end{array}\right.$$

(22)

When $p_{\left[j\right]}^A=p_{\left[j\right]}{j}^{a_{\left[j\right]}}-b_{\left[j\right]}{u}_{\left[j\right]}$, we have

$$\begin{array}{cc}\hfill F& {\displaystyle =\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{n_a}{\phi }_j\left(p_{\left[j\right]}{j}^{a_{\left[j\right]}}-b_{\left[j\right]}{u}_{\left[j\right]}\right)+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j}\hfill \\ \hfill & {\displaystyle =\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{n_a}{\phi }_j{p}_{\left[j\right]}{j}^{a_{\left[j\right]}}+\sum _{j=1}^{n_a}\left(\lambda {\upsilon }_{\left[j\right]}-b_{\left[j\right]}{\phi }_j\right)u_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j.}\hfill \end{array}$$

(23)

Lemma 3.

For problem P1, the optimal resource allocation of the jth ($j=1,2,\cdots ,n_a$) position is

$$u_{\left[j\right]}^{\ast }=\left\{\begin{array}{cc}{\overline{u}}_{\left[j\right]},\hfill & \lambda {\upsilon }_{\left[j\right]}-b_{\left[j\right]}{\phi }_j<0,\hfill \\ u_{\left[j\right]}\in \left[0,{\overline{u}}_{\left[j\right]}\right],\hfill & \lambda {\upsilon }_{\left[j\right]}-b_{\left[j\right]}{\phi }_j=0,\hfill \\ 0,\hfill & \lambda {\upsilon }_{\left[j\right]}-b_{\left[j\right]}{\phi }_j>0.\hfill \end{array}\right.$$

(24)

Proof. 

We take the partial derivation of Equation (23); it follows that ${\displaystyle \frac{\partial F\left(\pi ,u_{\left[j\right]}\right)}{\partial u_{\left[j\right]}}}=\lambda {\upsilon }_{\left[j\right]}-b_{\left[j\right]}{\phi }_{\left[j\right]},j=1,2,\cdots ,n_a$. When $\lambda {\upsilon }_{\left[j\right]}-b_{\left[j\right]}{\phi }_{\left[j\right]}<0$, the optimal resource allocation is $u_{\left[j\right]}^{\ast }={\overline{u}}_{\left[j\right]}$; when $\lambda {\upsilon }_{\left[j\right]}-b_{\left[j\right]}{\phi }_{\left[j\right]}=0$, the optimal resource allocation is $u_{\left[j\right]}^{\ast }=u_j\in \left[0,{\overline{u}}_{\left[j\right]}\right]$; when $\lambda {\upsilon }_{\left[j\right]}-b_{\left[j\right]}{\phi }_{\left[j\right]}>0$, the optimal resource allocation is $u_{\left[j\right]}^{\ast }=0$.    □

From the distribution of resources requested and substituted into Equation (23), if k and w are given, let

$$x_{jr}=\left\{\begin{array}{cc}1,\hfill & ifjob{J}_{j}isplacedintherthposition,\hfill \\ 0,\hfill & othersie,\hfill \end{array}\right.$$

(25)

then, the optimal job schedule ${\pi }^{\ast }$ of the P1 can be transformed into the following assignment problem (denoted by $\widetilde{\tilde{AP}}$):

$${\displaystyle \min F=\sum _{j=1}^n\sum _{r=1}^n{\theta }_{jr}{x}_{jr}}$$

(26)

$$s.t.\left\{\begin{array}{cc}{\displaystyle \sum _{j=1}^n{x}_{jr}=1,}\hfill & r=1,2,\cdots ,n,\hfill \\ {\displaystyle \sum _{r=1}^n{x}_{jr}=1,}\hfill & j=1,2,\cdots ,n,\hfill \\ x_{jr}=0or1,\hfill \end{array}\right.$$

(27)

included among these is

$${\theta }_{jr}=\left\{\begin{array}{cc}{\mu }_r+{\phi }_r{p}_j{r}^{a_j},\hfill & \lambda {\upsilon }_j-b_j{\phi }_r\ge 0,1\le r\le n_a,\hfill \\ {\mu }_r+{\phi }_r{p}_j{r}^{a_j}+\left(\lambda {\upsilon }_j-b_j{\phi }_r\right){\overline{u}}_j,\hfill & \lambda {\upsilon }_j-b_j{\phi }_r<0,1\le r\le n_a,\hfill \\ \epsilon e_j,\hfill & n_a+1\le r\le n,\hfill \end{array}\right.$$

(28)

where ${\phi }_r$ is given from Equation (22) and

$${\mu }_r=\left\{\begin{array}{cc}{\sigma }_r,\hfill & r=1,2,\cdots ,k-1,\hfill \\ 0,\hfill & r=k,k+1,\cdots ,w,\hfill \\ {\gamma }_r,\hfill & r=w+1,w+2,\cdots ,n_a.\hfill \end{array}\right.$$

(29)

From the above analysis, the algorithm for solving P1 can be given as follows (Algorithm 1):

Algorithm 1 P1

Input: ${\sigma }_j$, ${\gamma }_j$, ${\alpha }_j$, ${\beta }_j$, $p_j$, $b_j$, ${\overline{u}}_j$, $a_j$, $e_j$, $\rho$, $\eta$, $\lambda$, $\epsilon$, X Output: The optimal schedule ${\pi }^{\ast }$, $d_1^{\ast }$, $D^{\ast }$, $F^{\ast }$, $u_j^{\ast }$ Step 1. Compute ${\displaystyle F=\sum _{j=1}^n{e}_j}$ when $n_a=0$. Step 2. Compute the ranges of k and w from Lemma 1. Step 3. When $1\le n_a\le n$, for every pair of k and w $\left(k,w=1,2,\cdots ,n_a,k\le w\right)$, calculate ${\theta }_{jr}$ (from Equations (28) and (29)) to solve $\widetilde{\tilde{AP}}$ (26)–(27). Step 4. Compute the value $F\left(k,w\right)$ and a suboptimal schedule $\pi \left(k,w\right)$ from each pair of k and w. Step 5. Compute the global optimal schedule ${\pi }^{\ast }$ which is the one with the minimum value $F^{\ast }=\min \left\{F\left(k,w\right),\mathrm{where}1\le k\le w\le n_a\le n\right\}$. Step 6. By Lemma 3 (i.e., Equation (24)), compute the optimal resource allocation $u_j^{\ast }$. Step 7. Compute $d_1^{\ast }=C_{\left[k\right]}$ and $D^{\ast }=C_{\left[w\right]}-C_{\left[k\right]}$.

Theorem 1.

The P1 can be solved in $O\left(n^6\right)$ time by Algorithm 1.

Proof. 

For each given value of $n_a$$(n_a=0,1,\cdots ,n)$, Algorithm 1 needs to recursively k and w from 1 to $n_a$, which takes a total of $O\left(n^2\right)$ time. At the same time, it takes $O\left(n\right)$ time to calculate $F\left(k,w\right)$. It takes $O\left(n^3\right)$ time to solve $\widetilde{\tilde{AP}}$. Thus, the P1 can be solved by Algorithm 1 in $O\left(n^6\right)$ time.    □

3.2. Problem P2

For any given schedule, from Lemma 2, $q^{\prime }$, $q^{″}$ and D can be computed as follows: ${\displaystyle q^{\prime }=C_{[k-1]}=\sum _{l=1}^{k-1}{p}_{\left[l\right]}^A+X\sum _{l=1}^{k-2}{p}_{\left[l\right]}^A}$, ${\displaystyle q″=C_{[w-1]}=\sum _{l=1}^{w-1}{p}_{\left[l\right]}^A+X\sum _{l=1}^{w-2}{p}_{\left[l\right]}^A}$ and ${\displaystyle D=q″-q^{\prime }=\sum _{l=k}^{w-1}{p}_{\left[l\right]}^A+X\sum _{l=k-1}^{w-2}{p}_{\left[l\right]}^A}$, we have

$$\begin{array}{cc}\hfill F& {\displaystyle =\sum _{j=1}^{n_a}\left({\sigma }_j{G}_{\left[j\right]}+{\gamma }_j{H}_{\left[j\right]}+{\alpha }_j{E}_{\left[j\right]}+{\beta }_j{T}_{\left[j\right]}+\rho q^{\prime }+\eta D\right)+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j}\hfill \\ \hfill & {\displaystyle =\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{k-1}{\alpha }_j{E}_{\left[j\right]}+\sum _{j=w+1}^{n_a}{\beta }_j{T}_{\left[j\right]}+n\rho q^{\prime }+n\eta D+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j}\hfill \\ \hfill & {\displaystyle =\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{k-1}{\alpha }_j\left(d_1-C_{\left[j\right]}\right)+\sum _{j=w+1}^{n_a}{\beta }_j\left(C_{\left[j\right]}-d_2\right)+n\rho q^{\prime }+n\eta D}\hfill \\ \hfill & {\displaystyle +\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j}\hfill \\ \hfill & {\displaystyle =\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{k-1}{\alpha }_j\left({\displaystyle p_j^A+\sum _{l=1}^{k-1}{p}_{\left[l\right]}^A+X\sum _{l=1}^{k-2}{p}_{\left[l\right]}^A-\sum _{l=1}^j{p}_{\left[l\right]}^A-X\sum _{l=1}^{j-1}{p}_{\left[l\right]}^A}\right)}\hfill \\ \hfill & {\displaystyle +\sum _{j=w+1}^{n_a}{\beta }_j\left({\displaystyle \sum _{l=1}^j{p}_{\left[j\right]}^A+X\sum _{l=1}^{j-1}{p}_{\left[l\right]}^A-p_{\left[j\right]}^A-\sum _{l=1}^{w-1}{p}_{\left[l\right]}^A-X\sum _{l=1}^{w-2}{p}_{\left[l\right]}^A}\right)+n\rho \left({\displaystyle \sum _{l=1}^{k-1}{p}_{\left[l\right]}^A+X\sum _{l=1}^{k-2}{p}_{\left[l\right]}^A}\right)}\hfill \\ \hfill & {\displaystyle +n\eta (\sum _{l=k}^{w-1}{p}_{\left[l\right]}^A+X\sum _{l=k-1}^{w-2}{p}_{\left[l\right]}^A)+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j}\hfill \\ \hfill & {\displaystyle =\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{n_a}{\varphi }_j{p}_{\left[j\right]}^A+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j,}\hfill \end{array}$$

(30)

where

$${\varphi }_j=\left\{\begin{array}{cc}\left(X+1\right)\left({\displaystyle \sum _{l=1}^j{\alpha }_l+n\rho }\right),\hfill & j=1,2,\cdots ,k-2,\hfill \\ {\alpha }_j+n\rho +Xn\eta ,\hfill & j=k-1,\hfill \\ \left(X+1\right)n\eta ,\hfill & j=k,k+1,\cdots ,w-2,\hfill \\ n\eta +X{\beta }_{j+2},\hfill & j=w-1,\hfill \\ {\displaystyle \left(X+1\right)\sum _{l=j}^{n_a}{\beta }_l,}\hfill & j=w,w+1,\cdots ,n_a-1,\hfill \\ {\beta }_j,\hfill & j=n_a.\hfill \end{array}\right.$$

(31)

When $p_{\left[j\right]}^A=p_{\left[j\right]}{j}^{a_{\left[j\right]}}-b_{\left[j\right]}{u}_{\left[j\right]}$, we have

$$\begin{array}{cc}\hfill F& {\displaystyle =\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{n_a}{\varphi }_j\left(p_{\left[j\right]}{j}^{a_{\left[j\right]}}-b_{\left[j\right]}{u}_{\left[j\right]}\right)+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j}\hfill \\ \hfill & {\displaystyle =\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{n_a}{\varphi }_j{p}_{\left[j\right]}{j}^{a_{\left[j\right]}}+\sum _{j=1}^{n_a}\left(\lambda {\upsilon }_{\left[j\right]}-b_{\left[j\right]}{\varphi }_j\right)u_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j.}\hfill \end{array}$$

(32)

Lemma 4.

For the P2, the optimal resource allocation of the jth ($j=1,2,\cdots ,n_a$) position is

$$u_{\left[j\right]}^{\ast }=\left\{\begin{array}{cc}{\overline{u}}_{\left[j\right]},\hfill & \lambda {\upsilon }_{\left[j\right]}-b_{\left[j\right]}{\varphi }_j<0,\hfill \\ u_{\left[j\right]}\in \left[0,{\overline{u}}_{\left[j\right]}\right],\hfill & \lambda {\upsilon }_{\left[j\right]}-b_{\left[j\right]}{\varphi }_j=0,\hfill \\ 0,\hfill & \lambda {\upsilon }_{\left[j\right]}-b_{\left[j\right]}{\varphi }_j>0.\hfill \end{array}\right.$$

(33)

Proof. 

It is similar to the proof of the Lemma 3.    □

From the distribution of resources requested and substituted into Equation (32), if k and w are given, the optimal job schedule ${\pi }^{\ast }$ of the P2 can be transformed into the following $\widetilde{\tilde{AP}}$:

$${\displaystyle \min F=\sum _{j=1}^n\sum _{r=1}^n{\theta }_{jr}{x}_{jr}}$$

(34)

$$s.t.\left\{\begin{array}{cc}{\displaystyle \sum _{j=1}^n{x}_{jr}=1,}\hfill & r=1,2,\cdots ,n,\hfill \\ {\displaystyle \sum _{r=1}^n{x}_{jr}=1,}\hfill & j=1,2,\cdots ,n,\hfill \\ x_{jr}=0or1,\hfill \end{array}\right.$$

(35)

included among these is

$${\theta }_{jr}=\left\{\begin{array}{cc}{\mu }_r+{\varphi }_r{p}_j{r}^{a_j},\hfill & \lambda {\upsilon }_j-b_j{\varphi }_r\ge 0,1\le r\le n_a,\hfill \\ {\mu }_r+{\varphi }_r{p}_j{r}^{a_j}+\left(\lambda {\upsilon }_j-b_j{\varphi }_r\right){\overline{u}}_j,\hfill & \lambda {\upsilon }_j-b_j{\varphi }_r<0,1\le r\le n_a,\hfill \\ \epsilon e_j,\hfill & n_a+1\le r\le n,\hfill \end{array}\right.$$

(36)

where ${\varphi }_r$ is given from Equation (31) and ${\mu }_r$ is given from Equation (29).

From the above analysis, the algorithm for solving P2 can be given as follows (Algorithm 2):

Algorithm 2 P2

Input: ${\sigma }_j$, ${\gamma }_j$, ${\alpha }_j$, ${\beta }_j$, $p_j$, $b_j$, ${\overline{u}}_j$, $a_j$, $e_j$, $n_a$, $\rho$, $\eta$, $\lambda$, $\epsilon$, X Output: The optimal schedule ${\pi }^{\ast }$, ${q^{\prime }}^{\ast }$, $D^{\ast }$, $F^{\ast }$, $u_j^{\ast }$ Step 1. Compute ${\displaystyle F=\sum _{j=1}^n{e}_j}$ when $n_a=0$. Step 2. Compute the ranges of k and w from Lemma 2. Step 3. When $1\le n_a\le n$, for every pair of k and w $\left(k,w=1,2,\cdots ,n_a,k\le w\right)$, calculate ${\theta }_{jr}$ (from Equation (36) to solve $\widetilde{\tilde{AP}}$ (34)–(35). Step 4. Compute the value $F\left(k,w\right)$ and a suboptimal schedule $\pi \left(k,w\right)$ from each pair of k and w. Step 5. Compute the global optimal schedule ${\pi }^{\ast }$ which is the one with the minimum value $F^{\ast }=\min \left\{F\left(k,w\right),\mathrm{where}1\le k\le w\le n_a\le n\right\}$. Step 6. By Lemma 4 (i.e., Equation (33)), compute the optimal resource allocation $u_j^{\ast }$. Step 7. Compute ${q^{\prime }}^{\ast }=C_{[k-1]}$ and $D^{\ast }=C_{[w-1]}-C_{[k-1]}$.

Theorem 2.

The P2 can be solved in $O\left(n^6\right)$ time by Algorithm 2.

Proof. 

It is similar to the proof of the Theorem 1.    □

The corresponding explanation for Algorithms 1 and 2 can be seen in the following flowchart (see Figure 1).

Figure 1. The flowchart of Algorithms 1 and 2.

3.3. Problem P3

According to Section 3.1, for the P3, when $p_j^A={\left({\displaystyle \frac{p_j{r}^{a_j}}{u_j}}\right)}^m$, we have

$${\displaystyle F=\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{n_a}{\phi }_j{\left(\frac{p_{\left[j\right]}{j}^{a_{\left[j\right]}}}{u_{\left[j\right]}}\right)}^m+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j.}$$

(37)

Lemma 5.

The optimal resource allocation of the P3 is

$$u_{\left[j\right]}^{\ast }={\displaystyle \frac{{\left(m{\phi }_j\right)}^{\frac{1}{m+1}}{\left(p_{\left[j\right]}{j}^{a_{\left[j\right]}}\right)}^{\frac{m}{m+1}}}{{\left(\lambda {\upsilon }_{\left[j\right]}\right)}^{\frac{m}{m+1}}}},j=1,2,\cdots ,n_a.$$

(38)

Proof. 

We take the partial derivation of Equation (37) and, making it equal to 0, it follows that

$$\lambda {\upsilon }_{\left[j\right]}-m{\phi }_j{\displaystyle \frac{{\left(p_{\left[j\right]}{j}^{a_{\left[j\right]}}\right)}^m}{{\left(u_{\left[j\right]}\right)}^{m+1}}}=0,$$

(39)

and the result of Equation (38) can be obtained.    □

Substituting Equation (38) into Equation (37) yields

$${\displaystyle F=\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=k+1}^{n_a}{\gamma }_j+\left(m^{\frac{-m}{m+1}}+m^{\frac{1}{m+1}}\right){\lambda }^{\frac{m}{m+1}}\sum _{j=1}^{n_a}{{\phi }_j}^{\frac{1}{m+1}}{\left({\upsilon }_{\left[j\right]}{p}_{\left[j\right]}{j}^{a_{\left[j\right]}}\right)}^{\frac{m}{m+1}}+\epsilon \sum _{j\in \overline{AS}}{e}_j.}$$

(40)

In the same way, if k and w are given, from Equation (40), the optimal schedule $\pi$ of the P3 can be transformed into the following $\widetilde{\tilde{AP}}$:

$${\displaystyle \min F=\sum _{j=1}^n\sum _{r=1}^n{\theta }_{jr}{x}_{jr}}$$

(41)

$$s.t.\left\{\begin{array}{cc}{\displaystyle \sum _{j=1}^n{x}_{jr}=1,}\hfill & r=1,2,\cdots ,n,\hfill \\ {\displaystyle \sum _{r=1}^n{x}_{jr}=1,}\hfill & j=1,2,\cdots ,n,\hfill \\ x_{jr}=0or1,\hfill \end{array}\right.$$

(42)

where

$${\theta }_{jr}=\left\{\begin{array}{cc}\left(m^{\frac{-m}{m+1}}+m^{\frac{1}{m+1}}\right){\lambda }^{\frac{m}{m+1}}{{\phi }_r}^{\frac{1}{m+1}}{\left({\upsilon }_j{p}_j{r}^{a_j}\right)}^{\frac{m}{m+1}}+{\sigma }_r,\hfill & r=1,2,\cdots ,k-1,\hfill \\ \left(m^{\frac{-m}{m+1}}+m^{\frac{1}{m+1}}\right){\lambda }^{\frac{m}{m+1}}{{\phi }_r}^{\frac{1}{m+1}}{\left({\upsilon }_j{p}_j{r}^{a_j}\right)}^{\frac{m}{m+1}},\hfill & r=k,k+1,\cdots ,w,\hfill \\ \left(m^{\frac{-m}{m+1}}+m^{\frac{1}{m+1}}\right){\lambda }^{\frac{m}{m+1}}{{\phi }_r}^{\frac{1}{m+1}}{\left({\upsilon }_j{p}_j{r}^{a_j}\right)}^{\frac{m}{m+1}}+{\gamma }_r\hfill & r=w+1,w+2,\cdots ,n_a,\hfill \\ \epsilon e_j,\hfill & n_a+1\le r\le n,\hfill \end{array}\right.$$

(43)

and ${\phi }_r$ is given from Equation (22). From the above analysis, the algorithm for solving P3 can be given as follows (Algorithm 3):

Algorithm 3 P3

Input: ${\sigma }_j$, ${\gamma }_j$, ${\alpha }_j$, ${\beta }_j$, $p_j$,$b_j$, $q_j$, $e_j$, $n_a$, $\rho$, $\eta$, $\lambda$, $\epsilon$ Output: The optimal schedule ${\pi }^{\ast }$, $d_1^{\ast }$, $D^{\ast }$, $F^{\ast }$, $u_j^{\ast }$ Step 1. Compute ${\displaystyle F=\sum _{j=1}^n{e}_j}$ when $n_a=0$. Step 2. Compute the ranges of k and w from Lemma 1. Step 3. When $1\le n_a\le n$, for every pair of k and w$\left(k=1,2,\cdots ,n,w=1,2,\cdots ,n,k\le w\right)$,   compute ${\theta }_{jr}$ (from Equation (43)) to solve $\widetilde{\tilde{AP}}$ (41)–(42). Step 4. Compute the value $F\left(k,w\right)$ and a suboptimal schedule $\pi \left(k,w\right)$ from each pair of k and w. Step 5. Compute the global optimal schedule ${\pi }^{\ast }$ which is the one with the minimum value  $F^{\ast }=\min \left\{F\left(k,w\right),\mathrm{where}1\le k\le w\le n_a\le n\right\}$. Step 6. By Lemma 5 (i.e., Equation (38)), compute the optimal resource allocation $u_j^{\ast }$. Step 7. Compute $d_1^{\ast }=C_{\left[k\right]}$ and $D^{\ast }=C_{\left[w\right]}-C_{\left[k\right]}$.

Theorem 3.

The P3 can be solved in $O\left(n^6\right)$ time by Algorithm 3.

Proof. 

It is similar to the proof of Theorem 1.    □

3.4. Problem P4

According to Section 3.2, for the P4, where $p_j^A={\left(\frac{p_j{r}^{a_j}}{u_j}\right)}^m$, we have

$${\displaystyle F=\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\sum _{j=1}^{n_a}{\varphi }_j{\left(\frac{p_{\left[j\right]}{j}^{a_{\left[j\right]}}}{u_{\left[j\right]}}\right)}^m+\lambda \sum _{j=1}^{n_a}{\upsilon }_{\left[j\right]}{u}_{\left[j\right]}+\epsilon \sum _{j\in \overline{AS}}{e}_j.}$$

(44)

Lemma 6.

The optimal resource allocation of the P4 is

$$u_{\left[j\right]}^{\ast }={\displaystyle \frac{{\left(m{\varphi }_j\right)}^{\frac{1}{m+1}}{\left(p_{\left[j\right]}{j}^{a_{\left[j\right]}}\right)}^{\frac{m}{m+1}}}{{\left(\lambda {\upsilon }_{\left[j\right]}\right)}^{\frac{m}{m+1}}}},j=1,2,\cdots ,n_a.$$

(45)

Proof. 

It is similar to the proof of the Lemma 5.    □

Substituting Equation (45) into Equation (44) yields

$${\displaystyle F=\sum _{j=1}^{k-1}{\sigma }_j+\sum _{j=w+1}^{n_a}{\gamma }_j+\left(m^{\frac{-m}{m+1}}+m^{\frac{1}{m+1}}\right){\lambda }^{\frac{m}{m+1}}\sum _{j=1}^{n_a}{{\varphi }_j}^{\frac{1}{m+1}}{\left({\upsilon }_{\left[j\right]}{p}_{\left[j\right]}{j}^{a_{\left[j\right]}}\right)}^{\frac{m}{m+1}}+\epsilon \sum _{j\in \overline{AS}}{e}_j.}$$

(46)

In the same way, if k and w are given, from Equation (46), the optimal schedule $\pi$ of the P4 can be transformed into the following $\widetilde{\tilde{AP}}$:

$${\displaystyle \min F=\sum _{j=1}^n\sum _{r=1}^n{\theta }_{jr}{x}_{jr}}$$

(47)

$$s.t.\left\{\begin{array}{cc}{\displaystyle \sum _{j=1}^n{x}_{jr}=1,}\hfill & r=1,2,\cdots ,n,\hfill \\ {\displaystyle \sum _{r=1}^n{x}_{jr}=1,}\hfill & j=1,2,\cdots ,n,\hfill \\ x_{jr}=0or1,\hfill \end{array}\right.$$

(48)

where

$${\theta }_{jr}=\left\{\begin{array}{cc}\left(m^{\frac{-m}{m+1}}+m^{\frac{1}{m+1}}\right){\lambda }^{\frac{m}{m+1}}{{\varphi }_r}^{\frac{1}{m+1}}{\left({\upsilon }_j{p}_j{r}^{a_j}\right)}^{\frac{m}{m+1}}+{\sigma }_r,\hfill & r=1,2,\cdots ,k-1,\hfill \\ \left(m^{\frac{-m}{m+1}}+m^{\frac{1}{m+1}}\right){\lambda }^{\frac{m}{m+1}}{{\varphi }_r}^{\frac{1}{m+1}}{\left({\upsilon }_j{p}_j{r}^{a_j}\right)}^{\frac{m}{m+1}},\hfill & r=k,k+1,\cdots ,w,\hfill \\ \left(m^{\frac{-m}{m+1}}+m^{\frac{1}{m+1}}\right){\lambda }^{\frac{m}{m+1}}{{\varphi }_r}^{\frac{1}{m+1}}{\left({\upsilon }_j{p}_j{r}^{a_j}\right)}^{\frac{m}{m+1}}+{\gamma }_r\hfill & r=w+1,w+2,\cdots ,n_a,\hfill \\ \epsilon e_j,\hfill & n_a+1\le r\le n,\hfill \end{array}\right.$$

(49)

and ${\varphi }_r$ is given from Equation (31). From the above analysis, the algorithm for solving P4 can be given as follows (Algorithm 4):

Algorithm 4 P4

Input: ${\sigma }_j$, ${\gamma }_j$, ${\alpha }_j$, ${\beta }_j$, $p_j$,$b_j$, $,q_j$, $e_j$, $n_a$, $\rho$, $\eta$, $\lambda$, $\epsilon$ Output: The optimal schedule ${\pi }^{\ast }$, ${q^{\prime }}^{\ast }$, $D^{\ast }$, $F^{\ast }$, $u_j^{\ast }$ Step 1. Compute ${\displaystyle F=\sum _{j=1}^n{e}_j}$ when $n_a=0$. Step 2. Compute the ranges of k and w from Lemma 2. Step 3. When $1\le n_a\le n$, for every pair of k and w $\left(k=1,2,\cdots ,n,w=1,2,\cdots ,n,k\le w\right)$,   compute ${\theta }_{jr}$ (from Equation (49)) to solve $\widetilde{\tilde{AP}}$ (47)–(48). Step 4. Compute the value $F\left(k,w\right)$ and a suboptimal schedule $\pi \left(k,w\right)$ from each pair of k and w. Step 5. Compute the global optimal schedule ${\pi }^{\ast }$ which is the one with the minimum value  $F^{\ast }=\min \left\{F\left(k,w\right),\mathrm{where}1\le k\le w\le n_a\le n\right\}$. Step 6. By Lemma 6 (i.e., Equation (45)), compute the optimal resource allocation $u_j^{\ast }$. Step 7. Compute ${q^{\prime }}^{\ast }=C_{[k-1]}$ and $D^{\ast }=C_{[w-1]}-C_{[k-1]}$.

Theorem 4.

The P4 can be solved in $O\left(n^6\right)$ time by Algorithm 4.

Proof. 

It is similar to the proof of Theorem 1. □

4. Experiment

4.1. An Example

We only consider the CONDW assignment, where $n=5$, $\rho =3$, $\eta =4$. $m=2$, $X=1$, $\epsilon =1$ and $\lambda =5$; the parameters of jobs setting are shown in Table 3 below and the parameters of position-dependent weights setting are shown in Table 4 below.

Table 3. Values of job-dependent parameters.

	$\mathit{j}=1$	$\mathit{j}=2$	$\mathit{j}=3$	$\mathit{j}=4$	$\mathit{j}=5$
$p_j$	20	26	18	30	33
${\upsilon }_j$	15	18	13	14	19
$b_j$	2	4	5	6	8
${\overline{u}}_j$	5	3	1	1	2
$a_j$	−0.1	−0.1	−0.2	−0.3	−0.4
$e_j$	120	240	450	600	890

Table 4. Values of position-dependent weights.

	$\mathit{j}=1$	$\mathit{j}=2$	$\mathit{j}=3$	$\mathit{j}=4$	$\mathit{j}=5$
${\sigma }_j$	10	14	17	18	22
${\gamma }_j$	15	12	8	10	17
${\alpha }_j$	3	1	2	6	11
${\beta }_j$	7	4	9	8	10

In relation to problem P1, $1\le k\le w\le n_a$ and Table 5 shows all the results.

Table 5. Results of P1.

${\mathit{n}}_{\mathit{a}}$	k	w	$\mathit{\pi }\left(\mathit{k},\mathit{w}\right)$	$\mathit{F}\left(\mathit{k},\mathit{w}\right)$
0	-	-	-	2300
1	1	1	$J_5$	1804
2	1	2	$J_3,J_5$	1307.04
3	1	2	$J_3,J_5,J_4$	1279.39
	1	3	$J_3,J_5,J_4$	1298.41
	2	3	$J_3,J_5,J_4$	1347.41
4	1	3	$J_3,J_5,J_2,J_4$	1662.21
	1	4	$J_2,J_3,J_5,J_4$	1705.26
	2	3	$J_3,J_2,J_5,J_4$	1689.20
	2	4	$J_3,J_2,J_5,J_4$	1732.59
	3	4	$J_3,J_2,J_5,J_4$	1741.46
5	1	3	$J_1,J_3,J_5,J_2,J_4$	2242.62
	1	4	$J_1,J_2,J_3,J_5,J_4$	2314.52
	1	5	$J_1,J_2,J_3,J_5,J_4$	2396.58
	2	3	$J_3,J_2,J_5,J_1,J_4$	2251.28
	2	4	$J_1,J_3,J_2,J_5,J_4$	2323.18
	2	5	$J_1,J_2,J_3,J_5,J_4$	2402.06
	3	4	$J_3,J_1,J_2,J_5,J_4$	2316.12
	3	5	$J_2,J_1,J_3,J_5,J_4$	2374.71

From Table 5, the optimal solution is $n_a=3$, $k=1$, $w=2$ and $F^{\ast }=1279.39$; the optimal schedule is $J_3\to J_5\to J_4$; the optimal resources are $u_1^{\ast }=0$, $u_2^{\ast }=0$ and $u_3^{\ast }=1$; $d_1^{\ast }=C_{\left[1\right]}=13$ and $D^{\ast }=C_{\left[2\right]}-C_{\left[1\right]}=22.0093$. In addition, for the optimal solution, the values ${\phi }_r$ (from Equation (12)) are ${\phi }_1=21$, ${\phi }_2=21$ and ${\phi }_3=9$ and the values for ${\theta }_{jr}$ are presented in Table 6 (see Equation (17)).

Table 6. Values ${\theta }_{jr}$ for $n_a=3$, $k=1$ and $w=2$ (bold numbers are the optimal solution).

	$\mathit{r}=1$	$\mathit{r}=2$	$\mathit{r}=3$	$\mathit{r}=4$	$\mathit{r}=5$
$J_1$	420	391.8739	169.2725	120	120
$J_2$	546	509.436	217.6543	240	240
$J_3$	338	289.0681	138.0441	450	450
$J_4$	574	455.719	202.1902	600	600
$J_5$	560	379.1958	199.385	890	890

Meanwhile, for the problem P3, by Algorithm 3, the results are given by Table 7.

Table 7. Results of P3.

${\mathit{n}}_{\mathit{a}}$	k	w	$\mathit{\pi }\left(\mathit{k},\mathit{w}\right)$	$\mathit{F}\left(\mathit{k},\mathit{w}\right)$
0	-	-	-	2300
1	1	1	$J_5$	1527.93
2	1	2	$J_4,J_5$	978.19
3	1	2	$J_3,J_4,J_5$	653.46
	1	3	$J_3,J_4,J_5$	650.30
	2	3	$J_3,J_4,J_5$	664.58
4	1	3	$J_2,J_3,J_4,J_5$	549.09
	1	4	$J_2,J_3,J_4,J_5$	549.86
	2	3	$J_2,J_3,J_4,J_5$	561.04
	2	4	$J_2,J_3,J_4,J_5$	561.80
	3	4	$J_2,J_3,J_4,J_5$	573.09
5	1	3	$J_1,J_3,J_2,J_4,J_5$	585.98
	1	4	$J_2,J_1,J_3,J_4,J_5$	583.36
	1	5	$J_2,J_1,J_3,J_4,J_5$	577.24
	2	3	$J_1,J_2,J_3,J_4,J_5$	595.24
	2	4	$J_1,J_2,J_3,J_4,J_5$	592.49
	2	5	$J_1,J_2,J_3,J_4,J_5$	586.76
	3	4	$J_2,J_1,J_3,J_4,J_5$	602.76
	3	5	$J_2,J_1,J_3,J_4,J_5$	597.02

From Table 7, the optimal solution is $n_a=4$, $k=1$, $w=3$ and $F^{\ast }=549.09$; the optimal schedule is $J_2\to J_3\to J_4\to J_5$; the optimal resources are $u_1^{\ast }=1.585$, $u_2^{\ast }=1.748$, $u_3^{\ast }=1.5494$ and $u_4^{\ast }=1.3518$; $d_1^{\ast }=C_{\left[1\right]}=14$ and $D^{\ast }=C_{\left[3\right]}-C_{\left[1\right]}=50.92$.

4.2. Numerical Study

We only consider the CONDW assignment, for Algorithms 1 and 3; the program was coded and run on Pycharm and the testing computer was a desktop machine (HUAWEI, Shenzhen, China) with CPU 12th Gen Intel(R) Core(TM) i5-12400 2.50 GHz, 16.00 GB RAM and a Windows 11 operating system. The coefficients of instances are listed below:

(1)

$n=30,35,40,45,50,55,60,65,70,75,80,85,90,95,100,105,110,115,120,125,130,135,140,145$ and 150;

(2)

$\eta =16,\rho =15,\lambda =25,X=1,m=2$ and $\epsilon =1$;

(3)

$p_j,v_j$ ($j=1,2,\dots ,n$) were generated using the uniform distribution over [1, 100];

(4)

$b_j,{\overline{u}}_j$ ($j=1,2,\dots ,n$) were generated using the uniform distribution over [1, 10] (such that $p_j^A>0$);

(5)

$a_j$ was generated using the uniform distribution over [−0.9, −0.1];

(6)

$e_j$ was generated using the uniform distribution over [1, 1000];

(7)

${\sigma }_j,{\gamma }_j,{\alpha }_j,{\beta }_j$ ($j=1,2,\dots ,n$) were generated using the uniform distribution over [1, 30].

For each experimental condition, 20 replications were generated randomly, and the minimum (min), mean and maximum (max) CPU times (milliseconds, denoted by ms) are given in Table 8. From Table 8, it is found that Algorithms 1 and 3 are effective as n increases from 30 to 150, and the maximum CPU time of Algorithm 3 is 1,087,899.14 ms.

Table 8. CPU times of algorithms.

	Algorithm 1			Algorithm 3
Jobs ($\mathit{n}$)	Min (ms)	Mean (ms)	Max (ms)	Min (ms)	Mean (ms)	Max (ms)
30	142.85	235.43	409.39	157.18	255.57	445.18
35	269.01	426.10	910.82	294.11	458.24	974.10
40	521.64	895.93	1449.04	527.10	986.09	1597.00
45	1128.91	1654.00	2948.35	1219.00	1775.63	3166.93
50	1269.78	2687.22	3837.23	1396.20	2913.98	4138.61
55	2488.75	4429.17	6321.08	2655.95	4740.20	7015.84
60	4049.52	7007.70	11,169.58	4263.45	7591.72	11,818.76
65	6910.82	11,037.27	15,692.68	7489.74	11,733.91	16,679.26
70	7044.36	13,262.64	24,045.75	8041.63	14,145.77	24,259.30
75	12,064.63	18,053.98	26,240.19	12,696.11	19,176.10	28,541.88
80	17,233.22	25,217.32	33,648.05	18,111.85	26,875.97	36,438.14
85	22,911.99	36,814.51	48,270.63	24,128.63	39,104.94	52,053.80
90	35,334.52	47,669.83	68,108.47	36,176.71	50,825.94	73,510.58
95	40,514.55	65,156.01	92,957.04	44,534.29	69,221.00	95,678.19
100	50,647.46	80,251.27	116,104.41	53,037.41	85,215.23	122,046.86
105	69,987.21	100,446.61	141,959.81	75,166.38	107420.80	150,051.81
110	84,879.59	124,397.83	181,174.65	91,031.78	132,078.97	192,292.14
115	112,121.22	181,991.24	267,473.79	118,355.51	192,687.26	282,627.60
120	137,110.84	178,208.74	217,306.21	143,764.50	188815.01	243,220.70
125	187,556.82	250,157.16	330,993.79	197,813.28	263,503.63	353,033.11
130	229,027.22	300,983.57	479,574.26	243,667.19	319,436.17	513,703.90
135	306,409.13	380,366.81	503,867.34	324,645.02	402,716.59	534,332.42
140	301,224.31	461,883.33	612,613.20	318,772.65	487,794.09	648,533.54
145	351,303.26	537,853.02	715,481.80	368,110.65	573,954.21	769,660.68
150	425,236.54	661,529.77	930,615.10	446,659.95	706,957.81	1,087,899.14

5. Conclusions

This paper studied resource allocation single-machine scheduling problems with delivery times and due-window assignments under the premise that jobs have learning effects and a rejection cost. Under common and slack due-window assignments, we proved that the problems presented in this paper can be solved in $O\left(n^6\right)$ time, i.e., these problems are polynomially solvable. The managerial implications of our approach are as follows: the single-machine scheduling with delivery times, variable processing times, outsourcing cost and due-window allocation combine to affect the order in which jobs are processed and, thus, production decisions. We conducted computational experiments on randomly generated data. As can be seen in Table 8, our approach is effective and helps to improve resource utilization, reduce resource consumption costs, reduce inventory backlogs and, thus, reduce overheads such as warehousing costs in practical applications.

However, the current algorithms have some limitations because there are many dynamic factors in scheduling in real production, and the existing algorithms may perform well under specific conditions, but may need to be adjusted differently under different production environments and scheduling goals. In future research, the scheduling problem presented in this paper can be further generalized to existing algorithms by taking the deterioration effects (i.e., deteriorating jobs) into account (see Sun et al. [33], Lv et al. [34], Miao et al. [35] and Lu et al. [36]), or applied to flow shops (see Rossit et al. [37] and Panwalkar and Koulamas [38]).