Resource Allocation Scheduling with Position-Dependent Weights and Generalized Earliness–Tardiness Cost

Abstract

Under just-in-time production, this paper studies a single machine common due-window (denoted by CONW) assignment scheduling problem with position-dependent weights and resource allocations. A job’s actual processing time can be determined by the resource assigned to the job. A resource allocation model is divided into linear and convex resource allocations. Under the linear and convex resource allocation models, our goal is to find an optimal due-window location, job sequence and resource allocation. We prove that the weighted sum of scheduling cost (including general earliness–tardiness penalties with positional-dependent weights) and resource consumption cost minimization is polynomially solvable. In addition, under the convex resource allocation, we show that scheduling (resp. resource consumption) cost minimization is solvable in polynomial time subject to the resource consumption (resp. scheduling) cost being bounded.

1. Introduction

In scheduling models and problems, a due-window assignment has attracted growing attention, particularly in just-in-time (JIT) philosophy (Yin et al. [1]; Janiak et al. [2]; Liu [3]; Liu et al. [4]; Rolim and Nagano [5]; Qian and Zhan [6]). More recently, Wang et al. [7] studied a single machine production scheduling with the due-window assignment. The objective is to minimize the total cost, which is a weighted sum function of earliness, tardiness, due-window starting time, and due-window size, where the weights only depend on their position in a sequence (i.e., position-dependent weights, Wang et al. [8] and Wang et al. [9]). Sun et al. [10] and Qian and Han [11] investigated a proportionate flow shop with the CONW and slack due-window (denoted by SLKW) assignments. The objective is to minimize the generalized weighted sum of numbers of early and late jobs, earliness–tardiness penalties and due-window assignment cost, where the weights are the position-dependent weights. Yue and Zhou [12] studied single machine scheduling with the due-window assignment and stochastic processing times. Wang [13] considered single machine scheduling with past-sequence-dependent setup times. Under common, slack and different due-date assignments, he proved that a general earliness–tardiness cost minimization can be solved in polynomial time. Shabtay et al. [14] explored the single machine scheduling problem with common due-date/due-window assignments. They proved that a non-regular objective cost minimization can be solved in polynomial time under some conditions. They also showed that the problem is NP-hard when the length of the due-window is bounded. Jiang et al. [15] scrutinized the seru scheduling problem with a learning effect. Under multiple due-window assignments, they showed that some cases of the problem are polynomially solvable. Liu et al. [16] considered the single machine due-window assignment scheduling with past-sequence-dependent setup times. Under three due-window assignment models, the objective is to minimize a non-regular cost function (including the weighted sum of earliness–tardiness, number of early and tardy jobs, due-window assignment cost). They proved that this problem is polynomially solvable.

Recently, Wang et al. [17] explored single-machine problems with due-window assignment and positional-dependent weights. Under common and slack due-window assignments, the goal is to minimize the weighted sum of the number of early and late jobs, earliness–tardiness and due-window assignment costs, where the weights are positional-dependent weights. They demonstrated that some versions of common and slack due-window assignments are polynomially solvable. However, in real-world production processes, job processing times are complicated, which are controlled by allocating additional resources (e.g., fuel and gas, et al., see Shabtay and Steiner  [18], Liu et al. [19], Lu et al. [20], Wang et al. [21], Zhao [22], Lu et al. [23], and Yan et al. [24]). Hence, in this investigation, we extend the results of Wang et al. [17] by considering resource allocation that includes the one given in Wang et al. [17] as a special case. We summarize the contributions as follows: (1) We investigate single machine scheduling with resource allocation and CONW due-window assignment; (2) Under linear resource allocation, we provide an analysis for the weighted sum of scheduling cost (including generalized earliness–tardiness penalties with positional-dependent weights) and resource consumption costs; (3) Under convex resource allocation, we give a bicriteria analysis for scheduling costs and resource consumption costs; (4) We derive the structural properties of an optimal solution and demonstrate that the problem remains polynomially solvable. This article is organized as follows. Section 2 presents a description of the problem. In Section 3, we first give two lemmas, then we prove that four problems are polynomially solvable. In Section 4, a case study is given. In Section 5, conclusions are presented.

2. Problem Formulation

There are n independent non-preemptive jobs $J=\{J_1,J_2,\dots ,J_n\}$ that are to be processed on a single-machine, and all jobs are available at time zero. The machine can only process one job at a time. Let $p_j^A$ be the actual processing time of job $J_j$; under a linear resource allocation, we have

$$\begin{array}{c}{p}_j^A={\overline{p}}_j-b_j{u}_j,\hfill \end{array}$$

where ${\overline{p}}_j$ (resp. $b_j$) is the normal processing time (resp. compression rate) of job $J_j$, $u_j$ is the resource amount of job $J_j$, $0\le u_j\le {\overline{u}}_j<\frac{{\overline{p}}_j}{b_j}$ and ${\overline{u}}_j$ is the upper bound of $u_j$. For a convex resource allocation, we have

$$\begin{array}{c}{p}_j^A={\left(\frac{{\overline{p}}_j}{u_j}\right)}^k,\hfill \end{array}$$

where $k>0$ is a constant. In this paper, all the jobs are subject to a common due-window $[d_1,d_2]$ ($d_1\le d_2$), where $d_1$ is the due-window starting time and $d_2$ is the due-window finishing (completion) time; then, the due-window size is $D=d_2-d_1$. Let $C_j$ be the completion time of job $J_j$, the number of early ($U_j$) and tardy jobs ($V_j$) that are given as:

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

and

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

Let $E_j=max\{d_1-C_j,0\}$ (resp. $T_j=max\{C_j-d_2,0\}$) be the earliness (tardiness) of job $J_j$ ($j=1,2,\dots ,n$). Let $\left[j\right]$ be the job that is placed in the jth position; the goal is to determine the optimal job sequence $\pi$, $d_1$ and D (such as $d_2$) that minimizes the total cost. Formally, the  resource consumption (resp. scheduling) cost is ${\sum }_{j=1}^n{v}_j{u}_j$ (${\sum }_{j=1}^n({\alpha }_j{U}_{\left[j\right]}+{\beta }_j{V}_{\left[j\right]}+{\eta }_j{E}_{\left[j\right]}+{\delta }_j{T}_{\left[j\right]}+\theta d_1+\lambda D)$). For the linear (resp. convex) resource allocation, the first (second) problem (denoted by $P_1$ and $P_2$ respectively) is to  

$$Z=\sum _{j=1}^n\left({\alpha }_j{U}_{\left[j\right]}+{\beta }_j{V}_{\left[j\right]}+{\eta }_j{E}_{\left[j\right]}+{\delta }_j{T}_{\left[j\right]}+\theta d_1+\lambda D\right)+\rho \sum _{j=1}^n{v}_{\left[j\right]}{u}_{\left[j\right]},$$

(1)

where ${\alpha }_j$, ${\beta }_j$, ${\eta }_j$ and ${\delta }_j$ are positional-dependent weights (penalty factors) for the one-time penalties for earliness, tardiness and the unit time of earliness, tardiness, respectively; $\theta$ and $\lambda$ are the unit time weight (penalty) for the due-window starting time $d_1$ and size D, $\rho$ is the weight of resource costs and $v_j$ is the unit cost of processing job $J_j$. For the convex resource allocation, the third (resp. fourth) problem (denoted by $P_3$ and $P_4$, respectively) is to minimize scheduling cost ${\sum }_{j=1}^n\left({\alpha }_j{U}_{\left[j\right]}+{\beta }_j{V}_{\left[j\right]}+{\eta }_j{E}_{\left[j\right]}+{\delta }_j{T}_{\left[j\right]}+\theta d_1+\lambda D\right)$ (resource cost ${\sum }_{j=1}^n{v}_j{u}_j$) subject to ${\sum }_{j=1}^n{v}_j{u}_j$ $\le \widehat{U}$ (${\sum }_{j=1}^n\left({\alpha }_j{U}_{\left[j\right]}+{\beta }_j{V}_{\left[j\right]}+{\eta }_j{E}_{\left[j\right]}+{\delta }_j{T}_{\left[j\right]}+\theta d_1+\lambda D\right)\le \widehat{V}$), where $\widehat{U}$ ($\widehat{V}$) is a given constant.

Remark 1

(Liu et al. [16]). considered the single machine scheduling with past-sequence-dependent setup times, i.e., for the common due-window assignment; the objective function is to minimize ${\sum }_{j=1}^n\left({\tilde{\alpha }}_j{U}_j+{\tilde{\beta }}_j{V}_j+\tilde{\eta }{E}_{\left[j\right]}+\tilde{\delta }{T}_{\left[j\right]}+\theta d_1+\lambda D\right)$, where ${\tilde{\alpha }}_j$ (${\tilde{\beta }}_j$) is the weight of job $J_j$ (i.e., job-dependent weight), and $\tilde{\eta }$ and $\tilde{\delta }$ are given constants. They also addressed the slack and unrestricted due-window assignments.

3. Method

From Wang et al. [17], for the problems $P_1$–$P_4$, there exists an optimal sequence $\pi$ such that all jobs are processed at time 0, and there is no idle time during consecutive processing.

Lemma 1.

For the problems $P_1$-$P_4$, there exists an optimal sequence π such that

(1) The optimal $d_1$ is equal to $C_{\left[m\right]}$ (i.e., $d_1=C_{\left[m\right]}$) and the optimal $d_2$ is equal to $C_{\left[w\right]}$ (i.e., $d_2=C_{\left[w\right]}$), where $1\le m\le w\le n$;

(2) $d_1=C_{\left[m\right]}$, where ${\sum }_{j=1}^{m-1}{\eta }_j\le n(\lambda -\theta )$, $d_2=C_{\left[w\right]}$, where ${\sum }_{j=w+1}^n{\eta }_j\le n\lambda$.

Proof. 

Similar to the proof of Wang et al. [17].    □

3.1. Problem $P_1$

From Lemma 1, for any given sequence, $d_1$, $d_2$ and D can be calculated as follows: $d_1=C_{\left[m\right]}={\sum }_{l=1}^m{p}_{\left[l\right]}^A$, $d_2=C_{\left[w\right]}={\sum }_{l=1}^n{p}_{\left[l\right]}^A$, $D=C_{\left[w\right]}-C_{\left[m\right]}={\sum }_{l=m+1}^w{p}_{\left[l\right]}^A$. Hence, it follows that

$$\begin{array}{ccc}\hfill Z& =& \sum _{j=1}^n\left({\alpha }_j{U}_{\left[j\right]}+{\beta }_j{V}_{\left[j\right]}+{\eta }_j{E}_{\left[j\right]}+{\delta }_j{T}_{\left[j\right]}+\theta d_1+\lambda D\right)+\rho \sum _{j=1}^n{v}_{\left[j\right]}{u}_{\left[j\right]}\hfill \\ & =& \sum _{j=1}^{m-1}{\alpha }_j+\sum _{j=w+1}^n{\beta }_j+\sum _{j=1}^{m-1}{\eta }_j{E}_{\left[j\right]}+\sum _{j=w+1}^n{\delta }_j{T}_{\left[j\right]}+n\theta d_1+n\lambda D+\rho \sum _{j=1}^n{v}_{\left[j\right]}{u}_{\left[j\right]}\hfill \\ & =& \sum _{j=1}^{m-1}{\alpha }_j+\sum _{j=w+1}^n{\beta }_j+\sum _{j=1}^{m-1}{\eta }_j\left(d_1-C_{\left[j\right]}\right)+\sum _{j=w+1}^n{\delta }_j\left(C_{\left[j\right]}-d_2\right)+n\theta d_1+n\lambda D\hfill \\ & & +\rho \sum _{j=1}^n{v}_{\left[j\right]}{u}_{\left[j\right]}\hfill \\ & =& \sum _{j=1}^{m-1}{\alpha }_j+\sum _{j=w+1}^n{\beta }_j+\sum _{j=1}^{m-1}{\eta }_j\left(\sum _{l=1}^m{p}_{\left[l\right]}^A-\sum _{l=1}^j{p}_{\left[l\right]}^A\right)+\sum _{j=w+1}^n{\delta }_j\left(\sum _{l=1}^j{p}_{\left[l\right]}^A-\sum _{l=1}^w{p}_{\left[l\right]}^A\right)\hfill \\ & & +n\theta \sum _{l=1}^m{p}_{\left[l\right]}^A+n\lambda \sum _{l=m+1}^w{p}_{\left[l\right]}^A+\rho \sum _{j=1}^n{v}_{\left[j\right]}{u}_{\left[j\right]}\hfill \\ & =& \sum _{j=1}^{m-1}{\alpha }_j+\sum _{j=w+1}^n{\beta }_j+\sum _{j=1}^n{\tilde{\Phi }}_j{p}_{\left[j\right]}^A+\rho \sum _{j=1}^n{v}_{\left[j\right]}{u}_{\left[j\right]},\hfill \end{array}$$

(2)

where

$${\tilde{\mathsf{\Phi }}}_j=\left\{\begin{array}{ccc}\sum _{l=1}^{j-1}{\eta }_l+n\theta ,\hfill & & j=1,2,\dots ,m,\hfill \\ n\lambda ,\hfill & & j=m+1,m+2,\dots ,w\hfill \\ \sum _{l=j}^n{\delta }_l,\hfill & & j=w+1,w+2,\dots ,n.\hfill \end{array}\right.$$

(3)

When $p_j^A={\overline{p}}_j-b_j{u}_j$, we have

$$\begin{array}{c}Z=\sum _{j=1}^{m-1}{\alpha }_j+\sum _{j=w+1}^n{\beta }_j+\sum _{j=1}^n{\tilde{\mathsf{\Phi }}}_j({\overline{p}}_{\left[j\right]}-b_{\left[j\right]}{u}_{\left[j\right]})+\rho \sum _{j=1}^n{v}_{\left[j\right]}{u}_{\left[j\right]}\hfill \\ =\sum _{j=1}^{m-1}{\alpha }_j+\sum _{j=w+1}^n{\beta }_j+\sum _{j=1}^n{\tilde{\mathsf{\Phi }}}_j{\overline{p}}_{\left[j\right]}+\sum _{j=1}^n\left(\rho v_{\left[j\right]}-b_{\left[j\right]}{\tilde{\mathsf{\Phi }}}_j\right)u_{\left[j\right]}\hfill \end{array}$$

(4)

Let $u_{\left[j\right]}^{*}(1\le j\le n)$ be the optimal resource allocation of the jth position; we have

$$u_{\left[j\right]}^{*}=\left\{\begin{array}{ccc}{\overline{u}}_{\left[j\right]},\hfill & & \rho v_{\left[j\right]}-b_{\left[j\right]}{\tilde{\mathsf{\Phi }}}_j<0\hfill \\ u_{\left[j\right]}\in \left[0,{\overline{u}}_{\left[j\right]}\right],\hfill & & \rho v_{\left[j\right]}-b_{\left[j\right]}{\tilde{\mathsf{\Phi }}}_j=0\hfill \\ 0,\hfill & & \rho v_{\left[j\right]}-b_{\left[j\right]}{\tilde{\mathsf{\Phi }}}_j>0\hfill \end{array}\right.$$

(5)

From (2) and (3), if m and w are given, let $x_{jr}=1$; if job $J_j$ is placed in the rth position and $x_{jr}=0$ otherwise, then the optimal job sequence $\pi$ of the problem $P_1$ can be formulated as the following assignment problem:

$$minZ=\sum _{j=1}^n\sum _{r=1}^n{\tau }_{jr}{x}_{jr}$$

(6)

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

(7)

in which

$${\tau }_{jr}=\left\{\begin{array}{ccc}{\varpi }_r+{\chi }_r{\overline{p}}_{\left[j\right]},\hfill & & \rho v_{\left[j\right]}-b_{\left[j\right]}{\chi }_r\ge 0\hfill \\ {\varpi }_r+{\chi }_r{\overline{p}}_{\left[j\right]}+\left(\rho v_{\left[j\right]}-b_{\left[j\right]}{\chi }_r\right){\overline{u}}_{\left[j\right]},\hfill & & \rho v_{\left[j\right]}-b_{\left[j\right]}{\chi }_r<0\hfill \end{array}\right.$$

(8)

$${\chi }_r=\left\{\begin{array}{ccc}\sum _{l=1}^{r-1}{\eta }_l+n\theta ,\hfill & & r=1,2,\dots ,m\hfill \\ n\lambda ,\hfill & & r=m+1,m+2,\dots ,w\hfill \\ \sum _{l=r}^n{\delta }_l,\hfill & & r=w+1,w+2,\dots ,n\hfill \end{array}\right.$$

(9)

and

$${\varpi }_r=\left\{\begin{array}{ccc}{\alpha }_r,\hfill & & r=1,2,\dots ,m-1\hfill \\ 0,\hfill & & r=m,m+1,\dots ,w\hfill \\ {\beta }_r,\hfill & & r=w+1,w+2,\dots ,n.\hfill \end{array}\right.$$

(10)

Based on the above analysis, the problem $P_1$ can be solved by following Algorithm 1:

Algorithm 1: The problem $P_1$.

Input: ${\overline{p}}_j$, $b_j$, ${\overline{u}}_j$, $v_j$, ${\alpha }_j$, ${\beta }_j$, ${\eta }_j$, ${\delta }_j$ ($j=1,2,\dots ,n$), $n,\theta ,\lambda ,\rho$ Output: The optimal sequence ${\pi }^{*}$, $Z^{*}$, $u_j^{*}$, ${d_1}^{*}$, $D^{*}$ Step 1. From Lemma 1, calculate the range of m and w. Step 2. For each pair of m and w ($m=1,2,\dots ,n$, $w=1,2,\dots ,n$, $m\le w$), calculate ${\tau }_{jr}$ (see (8)–(10)), to solve the assignment problems (6) and (7). Step 3. For each pair of m and w, a suboptimal sequence $\pi \left(m,w\right)$ and value $Z(m,w)$ can be obtained. Step 4. The global optimal sequence ${\pi }^{*}$ is the one with the minimum value $Z^{*}=min\{Z(m,w),$ where $1\le m\le w\le n\}$. Step 5. Calculate $u_j^{*}$ by using Equation (5). Step 6. Calculate $d_1^{*}=C_{\left[m\right]}$ and $D^{*}=C_{\left[w\right]}-C_{\left[m\right]}$.

Theorem 1.

The problem $P_1$ can be solved by Algorithm 1 in $O\left(n^5\right)$ time.

Proof. 

The correctness of Algorithm 1 follows the above analysis. Steps 1 and 3–6 need $O\left(n\right)$ time; in Step 2, for each pair of m and w, solving the assignment problem requires $O(n^3$) time; m and w are less than n. Hence, the time complexity of Algorithm 1 is $O\left(n^5\right)$.    □

3.2. Problem $P_2$

Similar to Section 3.1, for the problem $P_2$, under $p_j^A={\left(\frac{{\overline{p}}_j}{u_j}\right)}^k$, we have

$$Z=\sum _{j=1}^{m-1}{\alpha }_j+\sum _{j=w+1}^n{\beta }_j+\sum _{j=1}^n{\tilde{\mathsf{\Phi }}}_j{\left(\frac{{\overline{p}}_{\left[j\right]}}{u_{\left[j\right]}}\right)}^k+\rho \sum _{j=1}^n{v}_{\left[j\right]}{u}_{\left[j\right]}.$$

(11)

Lemma 2.

The optimal resource allocation of the problem $P_2$ is:

$$u_{\left[j\right]}^{*}=\frac{{\left(k{\tilde{\mathsf{\Phi }}}_j\right)}^{\frac{1}{k+1}}{\left({\overline{p}}_{\left[j\right]}\right)}^{\frac{k}{k+1}}}{{\left(\rho v_{\left[j\right]}\right)}^{\frac{k}{k+1}}},j=1,2,\cdots n.$$

(12)

Proof. 

Taking a partial derivative of (11) with respect to $u_{\left[j\right]}$ and making it equal to 0, we have

$$\rho v_{\left[j\right]}-k{\tilde{\mathsf{\Phi }}}_j\frac{{\left({\overline{p}}_{\left[j\right]}\right)}^k}{{\left(u_{\left[j\right]}\right)}^{k+1}}=0,$$

(13)

and the result (12) can be obtained.    □

Substituting Equation (12) into Equation (11), we have

$$Z=\left(k^{-\frac{k}{k+1}}+k^{\frac{1}{k+1}}\right){\rho }^{\frac{k}{k+1}}\sum _{j=1}^n{\left(v_{\left[j\right]}{\overline{p}}_{\left[j\right]}\right)}^{\frac{k}{k+1}}{\tilde{\mathsf{\Phi }}}_j^{\frac{1}{k+1}}+\sum _{j=1}^{m-1}{\alpha }_j+\sum _{j=w+1}^n{\beta }_j.$$

(14)

Similarly, from Equation (14), if m and w are given, then the optimal job sequence $\pi$ of the problem $P_2$ can be formulated as the following assignment problem:

$$minZ=\sum _{j=1}^n\sum _{r=1}^n{\tau }_{jr}{x}_{jr}$$

(15)

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

(16)

in which

$${\tau }_{jr}=\left\{\begin{array}{ccc}\left(k^{-\frac{k}{k+1}}+k^{\frac{1}{k+1}}\right){\rho }^{\frac{k}{k+1}}{\chi }_r^{\frac{1}{k+1}}{\left(v_j{\overline{p}}_j\right)}^{\frac{k}{k+1}}+{\alpha }_r,\hfill & & r=1,2\cdots m-1\hfill \\ \left(k^{-\frac{k}{k+1}}+k^{\frac{1}{k+1}}\right){\rho }^{\frac{k}{k+1}}{\chi }_r^{\frac{1}{k+1}}{\left(v_j{\overline{p}}_j\right)}^{\frac{k}{k+1}},\hfill & & r=m,m+1,\dots ,w,\hfill \\ \left(k^{-\frac{k}{k+1}}+k^{\frac{1}{k+1}}\right){\rho }^{\frac{k}{k+1}}{\chi }_r^{\frac{1}{k+1}}{\left(v_j{\overline{p}}_j\right)}^{\frac{k}{k+1}}+{\beta }_r,\hfill & & r=w+1,w+2,\dots ,n,\hfill \end{array}\right.$$

(17)

and ${\chi }_r$ is given by Equation (10).

Based on the above analysis, the problem $P_2$ can be solved by following Algorithm 2:

Algorithm 2: The problem $P_2$.

Input: ${\overline{p}}_j$, $v_j$, ${\alpha }_j$, ${\beta }_j$, ${\eta }_j$, ${\delta }_j$ ($j=1,2,\dots ,n$), $n,k,\theta ,\lambda ,\rho$ Output: The optimal sequence ${\pi }^{*}$, $Z^{*}$, $u_j^{*}$, ${d_1}^{*}$, $D^{*}$ Step 1. From Lemma 1, calculate the range of m and w. Step 2. For each pair of m and w ($m=1,2,\dots ,n$, $w=1,2,\dots ,n$, $m\le w$), calculate ${\tau }_{jr}$ (see (17)), to solve the assignment problems (15) and (16). Step 3. For each pair of m and w, a suboptimal sequence $\pi \left(m,w\right)$ and value $Z(m,w)$ (see (15) and (16) ) can be obtained. Step 4. The global optimal sequence ${\pi }^{*}$ is the one with the minimum value $Z^{*}=min\{Z(m,w)$, where $1\le m\le w\le n\}$. Step 5. Calculate $u_j^{*}$ by using Equation (12). Step 6. Calculate $d_1^{*}=C_{\left[m\right]}$ and $D^{*}=C_{\left[w\right]}-C_{\left[m\right]}$.

Theorem 2.

The problem $P_2$ can be solved by Algorithm 2 in $O\left(n^5\right)$ time.

3.3. Problem $P_3$

In this subsection, we consider the problem $P_3$, i.e.,

$$minZ=\sum _{j=1}^{m-1}{\alpha }_j+\sum _{j=w+1}^n{\beta }_j+\sum _{j=1}^n{\tilde{\mathsf{\Phi }}}_j{p}_{\left[j\right]}^A=\sum _{j=1}^{m-1}{\alpha }_j+\sum _{j=w+1}^n{\beta }_j+\sum _{j=1}^n{\tilde{\mathsf{\Phi }}}_j{\left(\frac{{\overline{p}}_{\left[j\right]}}{u_{\left[j\right]}}\right)}^k$$

(18)

subject to ${\sum }_{j=1}^n{v}_{\left[j\right]}{u}_{\left[j\right]}\le \widehat{U}$.

Lemma 3.

The optimal resource allocation of the problem $P_3$ is:

$$u_{\left[j\right]}^{*}=\frac{{\left({\tilde{\mathsf{\Phi }}}_j{\left({\overline{p}}_{\left[j\right]}\right)}^k\right)}^{\frac{1}{k+1}}}{{\left(v_{\left[j\right]}\right)}^{\frac{1}{k+1}}{\sum }_{j=1}^n{\left(v_{\left[j\right]}{\overline{p}}_{\left[j\right]}\right)}^{\frac{k}{k+1}}{\left({\tilde{\mathsf{\Phi }}}_j\right)}^{\frac{1}{k+1}}}\times \widehat{U},j=1,2,\cdots ,n$$

(19)

Proof. 

For any given job sequence, by the Lagrange multiplier method, we have

$$\begin{array}{c}L\left(u_{\left[j\right]},\mu \right)=\sum _{j=1}^{m-1}{\alpha }_j+\sum _{j=w+1}^n{\beta }_j+\sum _{j=1}^n{\tilde{\mathsf{\Phi }}}_j{\left(\frac{{\overline{p}}_j}{u_j}\right)}^k+\mu \left(\sum _{j=1}^n{v}_{\left[j\right]}{u}_{\left[j\right]}-\widehat{U}\right),\hfill \end{array}$$

(20)

where $\mu$ is the Lagrangian multiplier, $\mu \ge 0$. Taking the partial derivative of $u_{\left[j\right]}$ and $\mu$, we have

$$\frac{\partial L(u,\mu )}{\partial u_{\left[j\right]}}=\mu v_{\left[j\right]}-k{\tilde{\mathsf{\Phi }}}_j\frac{{\left({\overline{p}}_{\left[j\right]}\right)}^k}{{\left(u_{\left[j\right]}\right)}^{k+1}}=0$$

(21)

$$u_{\left[j\right]}=\frac{{\left(k{\tilde{\mathsf{\Phi }}}_j{\left({\overline{p}}_{\left[j\right]}\right)}^k\right)}^{1/k+1}}{{\left(\mu v_{\left[j\right]}\right)}^{1/k+1}}$$

(22)

$$\frac{\partial L(u,\mu )}{\partial \mu }=\sum _{j=1}^n{v}_{\left[j\right]}{u}_{\left[j\right]}-\widehat{U}=0$$

(23)

$${\mu }^{1/k+1}=\frac{{\sum }_{j=1}^n{\left(k{\tilde{\mathsf{\Phi }}}_j\right)}^{1/k+1}{\left(v_{\left[j\right]}{\overline{p}}_{\left[j\right]}\right)}^{k/k+1}}{\widehat{U}}.$$

(24)

Substituting Equation (24) into Equation (22), Equation (19) can be obtained.    □

Putting $u_{\left[j\right]}^{*}$ (i.e., Equation (19)) in Equation (18), we have

$$\begin{array}{c}Z=\frac{{\left({\sum }_{j=1}^n{\left(v_{\left[j\right]}{\overline{p}}_{\left[j\right]}\right)}^{\frac{k}{k+1}}{\tilde{\mathsf{\Phi }}}_j^{\frac{1}{k+1}}\right)}^{k+1}}{{\widehat{U}}^k}+\sum _{j=1}^{m-1}{\alpha }_j+\sum _{j=w+1}^n{\beta }_j.\hfill \end{array}$$

(25)

If m and w are given, then ${\sum }_{j=1}^{m-1}{\alpha }_j+{\sum }_{j=w+1}^n{\beta }_j$ is a constant number. Hence, it can be obtained that solving problem $P_3$ is equal to minimizing ${\sum }_{j=1}^n{\left(v_{\left[j\right]}{\overline{p}}_{\left[j\right]}\right)}^{\frac{k}{k+1}}{\tilde{\mathsf{\Phi }}}_j^{\frac{1}{k+1}}$ and this term can be minimized by the HLP rule (see Hardy et al. [25]).

Based on the above analysis, the problem $P_3$ can be solved by following Algorithm 3:

Algorithm 3: The problem $P_3$.

Input: ${\overline{p}}_j$, $v_j$, ${\alpha }_j$, ${\beta }_j$, ${\eta }_j$, ${\delta }_j$ ($j=1,2,\dots ,n$), $n,k,\theta ,\lambda ,\widehat{U}$ Output: The optimal sequence ${\pi }^{*}$, $Z^{*}$, $u_j^{*}$, ${d_1}^{*}$, $D^{*}$ Step 1. From Lemma 1, calculate the range of m and w. Step 2. For each pair of m and w, a suboptimal sequence $\pi \left(m,w\right)$ and value $Z(m,w)$ (see Equation (25)) can be obtained by the HLP rule, i.e.,  by matching the largest ${\left(v_j{\overline{p}}_j\right)}^{\frac{k}{k+1}}$ with the smallest ${\tilde{\mathsf{\Phi }}}_j^{\frac{1}{k+1}}$, the second largest ${\left(v_j{\overline{p}}_j\right)}^{\frac{k}{k+1}}$ matches the second smallest ${\tilde{\mathsf{\Phi }}}_j^{\frac{1}{k+1}}$, and so on. Step 3. The global optimal sequence ${\pi }^{*}$ is the one with the minimum value $Z^{*}=min\{Z(m,w),$ where $1\le m\le w\le n\}$. Step 4. Calculate $u_j^{*}$ by using Equation (19). Step 5. Calculate $d_1^{*}=C_{\left[m\right]}$ and $D^{*}=C_{\left[w\right]}-C_{\left[m\right]}$.

Theorem 3.

The problem $P_3$ can be solved by Algorithm 3 in $O\left(n^5\right)$ time.

3.4. Problem $P_4$

Similar to Section 3.3, we have:

Lemma 4.

The optimal resource allocation of the problem $P_4$ is:

$$u_{\left[j\right]}^{*}=\frac{{\tilde{\mathsf{\Phi }}}_j^{\frac{1}{k+1}}{\left({\overline{p}}_{\left[j\right]}\right)}^{\frac{k}{k+1}}{\left({\sum }_{j=1}^n{\left({\tilde{\mathsf{\Phi }}}_j\right)}^{\frac{1}{k+1}}{\left(v_{\left[j\right]}{\overline{p}}_{\left[j\right]}\right)}^{\frac{k}{k+1}}\right)}^{\frac{1}{k}}}{{\left(v_{\left[j\right]}\right)}^{\frac{1}{k+1}}{\left(\widehat{V}-{\sum }_{j=1}^{m-1}{\alpha }_j-{\sum }_{j=w+1}^n{\beta }_j\right)}^{\frac{1}{k}}},j=1,2,\cdots ,n$$

(26)

Proof. 

For any given job sequence, we have:

$$\begin{array}{c}L\left(u_{\left[j\right]},\mu \right)=\sum _{j=1}^n{v}_{\left[j\right]}{u}_{\left[j\right]}+\mu \left(\sum _{j=1}^{m-1}{\alpha }_j+\sum _{j=w+1}^n{\beta }_j+\sum _{j=1}^n{\tilde{\mathsf{\Phi }}}_j{\left(\frac{{\overline{p}}_{\left[j\right]}}{u_{\left[j\right]}}\right)}^k-\widehat{V}\right),\hfill \end{array}$$

(27)

where $\mu$ is the Lagrangian multiplier, $\mu \ge 0$, we have:

$$\frac{\partial L(u_{\left[j\right]},\mu )}{\partial u_{\left[j\right]}}=v_{\left[j\right]}-\mu k{\tilde{\mathsf{\Phi }}}_j\frac{{\left({\overline{p}}_{\left[j\right]}\right)}^k}{{\left(u_{\left[j\right]}\right)}^{k+1}}=0$$

(28)

$$u_{\left[j\right]}=\frac{{\left(\mu k{\tilde{\mathsf{\Phi }}}_j{\left({\overline{p}}_{\left[j\right]}\right)}^k\right)}^{1/k+1}}{{\left(v_{\left[j\right]}\right)}^{\frac{1}{k+1}}}$$

(29)

$$\frac{\partial L(u_{\left[j\right]},\mu )}{\partial \mu }=\sum _{j=1}^{m-1}{\alpha }_j+\sum _{j=w+1}^n{\beta }_j+\sum _{j=1}^n{\tilde{\mathsf{\Phi }}}_j{\left(\frac{{\overline{p}}_{\left[j\right]}}{u_{\left[j\right]}}\right)}^k-\widehat{V}=0$$

(30)

$${\left(k\mu \right)}^{\frac{1}{k+1}}=\frac{{\left({\sum }_{j=1}^n{\left({\tilde{\mathsf{\Phi }}}_j\right)}^{\frac{1}{k+1}}{\left(v_{\left[j\right]}{\overline{p}}_{\left[j\right]}\right)}^{\frac{k}{k+1}}\right)}^{\frac{1}{k}}}{{\left(\widehat{V}-{\sum }_{j=1}^{m-1}{\alpha }_j-{\sum }_{j=w+1}^n{\beta }_j\right)}^{\frac{1}{k}}}.$$

(31)

Substituting Equation (31) into Equation (29), Equation (26) can be obtained.    □

Substituting Equation (26) into ${\sum }_{j=1}^n{v}_j{u}_j$, we have:

$$\begin{array}{ccc}\hfill Z& =& \sum _{j=1}^n{v}_{\left[j\right]}{u}_{\left[j\right]}\hfill \\ & =& \sum _{j=1}^n{v}_{\left[j\right]}\left[\frac{{\tilde{\mathsf{\Phi }}}_j^{\frac{1}{k+1}}{\left({\overline{p}}_{\left[j\right]}\right)}^{\frac{k}{k+1}}{\left({\sum }_{j=1}^n{\left({\tilde{\mathsf{\Phi }}}_j\right)}^{\frac{1}{k+1}}{\left(v_{\left[j\right]}{\overline{p}}_{\left[j\right]}\right)}^{\frac{k}{k+1}}\right)}^{\frac{1}{k}}}{{\left(v_{\left[j\right]}\right)}^{\frac{1}{k+1}}{\left(V-{\sum }_{j=1}^{m-1}{\alpha }_j-{\sum }_{j=w+1}^n{\beta }_j\right)}^{\frac{1}{k}}}\right]\hfill \\ & =& \frac{{\left({\sum }_{j=1}^n{\left({\tilde{\mathsf{\Phi }}}_j\right)}^{\frac{1}{k+1}}{\left(v_{\left[j\right]}{\overline{p}}_{\left[j\right]}\right)}^{\frac{k}{k+1}}\right)}^{\frac{1+k}{k}}}{{\left(\widehat{V}-{\sum }_{j=1}^{m-1}{\alpha }_j-{\sum }_{j=w+1}^n{\beta }_j\right)}^{\frac{1}{k}}}.\left(32\right)\hfill \end{array}$$

Similar to Section 3.3, the problem $P_4$ can be solved by following Algorithm 4:

Algorithm 4: The problem $P_4$.

Input: ${\overline{p}}_j$, $v_j$, ${\alpha }_j$, ${\beta }_j$, ${\eta }_j$, ${\delta }_j$ ($j=1,2,\dots ,n$), $n,k,\theta ,\lambda ,\widehat{V}$ Output: The optimal sequence ${\pi }^{*}$, $Z^{*}$, $u_j^{*}$, ${d_1}^{*}$, $D^{*}$ Step 1. From Lemma 1, calculate the range of m and w. Step 2. For each pair of m and w, a suboptimal sequence $\pi \left(m,w\right)$ and value $Z(m,w)$ (see Equation (32)) can be obtained by the HLP rule, i.e.,  by matching the largest ${\left(v_j{\overline{p}}_j\right)}^{\frac{k}{k+1}}$ with the smallest ${\tilde{\mathsf{\Phi }}}_j^{\frac{1}{k+1}}$, the second largest ${\left(v_j{\overline{p}}_j\right)}^{\frac{k}{k+1}}$ matches the second smallest ${\tilde{\mathsf{\Phi }}}_j^{\frac{1}{k+1}}$, and so on. Step 3. The global optimal sequence ${\pi }^{*}$ is the one with the minimum value $Z^{*}=min\{Z(m,w),$ where $1\le m\le w\le n\}$. Step 4. Calculate $u_j^{*}$ by using Equation (26). Step 5. Calculate $d_1^{*}=C_{\left[m\right]}$ and $D^{*}=C_{\left[w\right]}-C_{\left[m\right]}$.

Theorem 4.

The problem $P_4$ can be solved by Algorithm 4 in $O\left(n^5\right)$ time.

4. A Case Study

In this section, we present a case study to illustrate how the proposed algorithms works.

Example 1.

Consider a five-job problem, where $k=2,\theta =3,\lambda =4,\rho =6$, $\widehat{U}=100,\widehat{V}=1000$; the parameters of job $J_j$ ($j=1,2,\dots ,5$) are given in

Table 1. Values of job-dependent parameters.

	$\mathit{j}=1$	$\mathit{j}=2$	$\mathit{j}=3$	$\mathit{j}=4$	$\mathit{j}=5$
${\overline{p}}_j$	13	12	14	15	17
$b_j$	2	2	3	1	1
${\overline{u}}_j$	5	4	3	11	7
$v_j$	3	5	4	1	6

. The parameters of position-dependent weights (i.e., ${\alpha }_j,{\beta }_j,{\eta }_j$ and ${\delta }_j$, where $j=1,2,\dots ,5$) are given in

Table 2. Values of position-dependent weights.

	$\mathit{j}=1$	$\mathit{j}=2$	$\mathit{j}=3$	$\mathit{j}=4$	$\mathit{j}=5$
${\alpha }_j$	4	8	7	6	5
${\beta }_j$	8	7	3	2	6
${\eta }_j$	2	4	6	10	7
${\delta }_j$	2	5	4	3	6

.

For the problem $P_1$, from Algorithm 1, $m=1,2$, $w=3,4,5$, and all results are shown in

Table 3. Results of problem $P_1$ (bold numbers are the optimal solution).

m	w	$\mathit{\pi }(\mathit{m},\mathit{w})$	$\mathit{Z}(\mathit{m},\mathit{w})$
1	3	$(J_3,J_4,J_1,J_2,J_5)$	661
1	4	$(J_3,J_2,J_1,J_4,J_5)$	751
1	5	$(J_5,J_1,J_2,J_3,J_4)$	923
2	3	$({\mathit{J}}_{\mathbf{3}},{\mathit{J}}_{\mathbf{4}},{\mathit{J}}_{\mathbf{1}},{\mathit{J}}_{\mathbf{2}},{\mathit{J}}_{\mathbf{5}})$	653
2	4	$(J_3,J_4,J_2,J_1,J_5)$	743
2	5	$(J_5,J_3,J_1,J_4,J_2)$	912

. From Table 3, the optimal sequence ${\pi }^{*}(2,3)$ is $J_3\to J_4\to J_1\to J_2\to J_5$, $u_3^{*}=3$, $u_4^{*}=11$, $u_1^{*}=5$, $u_2^{*}=4$, $u_5^{*}=0$, $Z^{*}(2,3)=653$, $d_1^{*}=C_{\left[2\right]}=9$, and $D^{*}=C_{\left[3\right]}-C_{\left[2\right]}=3$. When $m=2$ and $w=3$, the values ${\tilde{\mathsf{\Phi }}}_1=15$, ${\tilde{\mathsf{\Phi }}}_2=17$, ${\tilde{\mathsf{\Phi }}}_3=20$, ${\tilde{\mathsf{\Phi }}}_4=9$, ${\tilde{\mathsf{\Phi }}}_5=6$ (see Equation (3)), ${\tau }_{jr}$ (see Equation (8)) are given in

Table 4. Values ${\tau }_{jr}$ for $m=2$ and $w=3$ (bold numbers are the optimal solution).

	$\mathit{r}=1$	$\mathit{r}=2$	$\mathit{r}=3$	$\mathit{r}=4$	$\mathit{r}=5$
$J_1$	139	141	150	119	84
$J_2$	184	188	200	110	78
$J_3$	151	157	172	119	90
$J_4$	130	134	146	104	96
$J_5$	259	289	340	155	108

, and the optimal solution of the assignment problem is given by Table 4.

Similarly, for the problem $P_2$ from Algorithm 2, the results are shown in

Table 5. Results of problem $P_2$ (bold numbers are the optimal solution).

m	w	$\mathit{\pi }(\mathit{m},\mathit{w})$	$\mathit{Z}(\mathit{m},\mathit{w})$
1	3	$(J_3,J_4,J_1,J_2,J_5)$	$977.570$
1	4	$(J_2,J_4,J_1,J_3,J_5)$	$1035.169$
1	5	$(J_5,J_1,J_3,J_4,J_2)$	$1141.332$
2	3	$({\mathit{J}}_{\mathbf{3}},{\mathit{J}}_{\mathbf{1}},{\mathit{J}}_{\mathbf{4}},{\mathit{J}}_{\mathbf{2}},{\mathit{J}}_{\mathbf{5}})$	971.297
2	4	$(J_2,J_3,J_1,J_4,J_5)$	$1026.095$
2	5	$(J_5,J_2,J_1,J_4,J_3)$	$1131.643$

. From Table 5, the optimal sequence ${\pi }^{*}(2,3)$ is $J_3\to J_1\to J_4\to J_2\to J_5$, $u_3^{*}=2.169$, $u_1^{*}=2.608$, $u_4^{*}=6.230$, $u_2^{*}=1.423$, $u_5^{*}=1.388$, $Z^{*}(2,3)=\mathbf{971.297}$, $d_1^{*}=C_{\left[2\right]}=66.505$ and $D^{*}=C_{\left[3\right]}-C_{\left[2\right]}=5.797$.

Similarly, for problem $P_3$, from Algorithm 3, the results are shown in

Table 6. Results of problem $P_3$ (bold numbers are the optimal solution).

m	w	$\mathit{\pi }(\mathit{m},\mathit{w})$	$\mathit{Z}(\mathit{m},\mathit{w})$
1	3	$(J_3,J_4,J_1,J_2,J_5)$	$383.086$
1	4	$(J_2,J_4,J_1,J_3,J_5)$	$454.595$
1	5	$(J_5,J_4,J_1,J_3,J_2)$	$611.830$
2	3	$({\mathit{J}}_{\mathbf{3}},{\mathit{J}}_{\mathbf{1}},{\mathit{J}}_{\mathbf{4}},{\mathit{J}}_{\mathbf{2}},{\mathit{J}}_{\mathbf{5}})$	375.290
2	4	$(J_2,J_3,J_4,J_1,J_5)$	$441.715$
2	5	$(J_5,J_2,J_4,J_1,J_3)$	$594.076$

. From Table 6, the optimal sequence ${\pi }^{*}(2,3)$ is $J_3\to J_1\to J_4\to J_2\to J_5$, $u_3^{*}=5.871$, $u_1^{*}=6.412$, $u_4^{*}=10.740$, $u_2^{*}=4.148$, $u_5^{*}=4.301$, $Z^{*}(2,3)=\mathbf{375.290}$, $d_1^{*}=C_{\left[2\right]}=9.798$, and $D^{*}=C_{\left[3\right]}-C_{\left[2\right]}=1.951$.

Similarly, for problem $P_4$, from Algorithm 4, the results are shown in

Table 7. Results of problem $P_4$ (bold numbers are the optimal solution).

m	w	$\mathit{\pi }(\mathit{m},\mathit{w})$	$\mathit{Z}(\mathit{m},\mathit{w})$
1	3	$(J_3,J_4,J_1,J_2,J_5)$	$61.491$
1	4	$(J_2,J_4,J_1,J_3,J_5)$	$67.180$
1	5	$(J_5,J_4,J_1,J_3,J_2)$	$78.220$
2	3	$({\mathit{J}}_{\mathbf{3}},{\mathit{J}}_{\mathbf{1}},{\mathit{J}}_{\mathbf{4}},{\mathit{J}}_{\mathbf{2}},{\mathit{J}}_{\mathbf{5}})$	60.516
2	4	$(J_2,J_3,J_4,J_1,J_5)$	$66.036$
2	5	$(J_5,J_2,J_4,J_1,J_3)$	$76.971$

. From Table 7, the optimal sequence is ${\pi }^{*}(2,3)$ = $J_3\to J_1\to J_4\to J_2\to J_5$, $u_3^{*}=3.563$, $u_1^{*}=3.892$, $u_4^{*}=6.519$, $u_2^{*}=2.518$, $u_5^{*}=2.611$, $Z^{*}(2,3)=\mathbf{60.516}$, $d_1^{*}=C_{\left[2\right]}=26.596$, and $D^{*}=C_{\left[3\right]}-C_{\left[2\right]}=5.295$.

5. Conclusions

This article explored resource allocation scheduling with general position-dependent weights and common due-window assignment. The resource allocation can be divided into linear and convex resource allocations. Under some combination of general earliness–tardiness cost and resource consumption cost, we showed that four versions of the problem are polynomially solvable. Future research and investigation will consider the scheduling under flow shop settings, or extending our model from resource allocation to general deterioration effects.