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Research article

High-dimensional Lehmer problem on Beatty sequences

  • Let q be a positive integer. For each integer a with 1a<q and (a,q)=1, it is clear that there exists one and only one ˉa with 1ˉa<q such that aˉa1(q). Let k be any fixed integer with k2,0<δi1,i=1,2,,k. rn(δ1,δ2,,δk,α,β,c;q) denotes the number of all k-tuples with positive integer coordinates (x1,x2,,xk) such that 1xiδiq,(xi,q)=1,x1x2xkc(q), and x1,x2,,xk1Bα,β. In this paper, we consider the high-dimensional Lehmer problem related to Beatty sequences over incomplete intervals and give an asymptotic formula by the properties of Beatty sequences and the estimates for hyper Kloosterman sums.

    Citation: Xiaoqing Zhao, Yuan Yi. High-dimensional Lehmer problem on Beatty sequences[J]. AIMS Mathematics, 2023, 8(6): 13492-13502. doi: 10.3934/math.2023684

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  • Let q be a positive integer. For each integer a with 1a<q and (a,q)=1, it is clear that there exists one and only one ˉa with 1ˉa<q such that aˉa1(q). Let k be any fixed integer with k2,0<δi1,i=1,2,,k. rn(δ1,δ2,,δk,α,β,c;q) denotes the number of all k-tuples with positive integer coordinates (x1,x2,,xk) such that 1xiδiq,(xi,q)=1,x1x2xkc(q), and x1,x2,,xk1Bα,β. In this paper, we consider the high-dimensional Lehmer problem related to Beatty sequences over incomplete intervals and give an asymptotic formula by the properties of Beatty sequences and the estimates for hyper Kloosterman sums.



    Scheduling models with setup times are widely used in manufacture and operational processes (see Allahverdi et al. [1] and Allahverdi [2]). Koulamas and Kyparisis [3,4] and Biskup and Herrmann [5] investigated single-machine scheduling with past-sequence-dependent setup times (~psdst). They showed that several regular objective function minimizations remain polynomially solvable. Wang [6] and Wang and Li [7] examined single-machine problems with learning effects and ~psdst. Hsu et al. [8] studied unrelated parallel machine scheduling problems with learning effects and ~psdst. They proved that the total completion time minimization remains polynomially solvable. Cheng et al. [9] investigated scheduling problems with ~psdst and deterioration effects in a single machine. Huang et al. [10] and Wang and Wang [11] studied scheduling jobs with ~psdst, learning and deterioration effects. They showed that the single-machine makespan and the sum of the αth (α>0) power of job completion times minimizations remain polynomially solvable. Wang et al. [12] dealt with scheduling with ~psdst and deterioration effects. Under job rejection, they showed that the the sum of scheduling cost and rejection cost minimization can be solved in polynomial time.

    In the real production scheduling, the jobs often have due dates (see Gordon et al. [13,14] and the recent survey papers Rolim and Nagano [15], and Sterna [16]). Recently, Wang [17] and Wang et al. [18] studied single-machine scheduling problems with ~psdst and due-date assignment. Under common, slack and different due-date assignment methods, Wang [17] proved that the linear weighted sum of earliness-tardiness, number of early and delayed jobs, and due date penalty minimization can be solved in polynomial time. Under common and slack due date assignment methods, Wang et al. [18] showed that the weighted sum of earliness, tardiness and due date minimization can be solved in polynomial time, where the weights are position-dependent weights. The real application of the position-dependent weights can be found in production services and resource utilization (see Brucker [19], Liu et al. [20] and Jiang et al. [21]). Hence, it would be interesting to investigate due date assignment scheduling with ~psdst and position-dependent weights. The purpose of this article is to determine the optimal due dates and job sequence to minimize the weight sum of generalized earliness-tardiness penalties, where the weights are position-dependent weights. The contributions of this study are given as follows:

    We focus on the due date assignment single-machine scheduling problems with ~psdst and position-dependent weights;

    We provide an analysis for the non-regular objective function (including earliness, tardiness, number of early and delayed jobs, and due date cost);

    We derive the structural properties of the position-dependent weights and show that three due date assignments can be solved in polynomial time, respectively.

    The problem formulation is described in Section 2. Three due-date assignments are discussed in Section 3. An example is presented in Section 4. In Section 5, the conclusions are given.

    The symbols used throughout the article are introduced in Table 1.

    Table 1.  Symbols used in this article.
    Symbol Meaning
    N number of jobs
    Jl index of job
    pl processing time of Jl
    ~psdst past-sequence-dependent setup times
    sl setup time of ~psdst of Jl
    Cl completion time of Jl
    β a normalizing constant
    dl due date of Jl
    d common due date
    q common flow allowance
    [l] lth position in a sequence
    Ll=Cldl lateness of Jl
    Ul earliness indicator viable of Jl
    Vl tardiness indicator viable of job Jl
    ζl positional-dependent weight of lateness cost
    ηl (θl) positional-dependent weight of earliness (tardiness) indicator viable
    ϑl positional-dependent weight of due date cost
    ϱ sequence of all jobs
    ~con (~slk,~dif) common (slack, different) due date

     | Show Table
    DownLoad: CSV

    Suppose there are N independent jobs ˜V={J1,J2,,JN} need to be processed on a single-machine. The ~psdst setup time s[l] of job J[l] is s[l]=βl1j=1p[j], where β0 is a normalizing constant, s[1]=0, and βl1j=1p[j]+p[l] is the total processing requirement of job J[l]. Let Ll=Cldl denote the lateness of job Jl, Ul (Vl) be earliness (tardiness) indicator viable of job Jl, i.e., if Cl<dl, Ul=1, otherwise, Ul=0; if Cl>dl, Vl=1, otherwise, Vl=0.

    For the common (~con) due date assignment, dl=d (l=1,2,,N) and d is a decision variable. For the slack (~slk) due date assignment, dl=sl+pl+q and q is a decision variable. For the different due date (~dif) assignment, dl is a decision variable for l=1,2,,N. The target is to determine dl and a sequence ϱ such that is minimized.

    M=Nl=1(ζl|L[l]|+ηlU[l]+θlV[l]+ϑld[l]), (1)

    where ζl0, ηl0, ηl0 and δl0 are given positional-dependent weight constants. From Pinedo [22], the problem can be defined as:

    1|~psdst,H|Nl=1(ζl|L[l]|+ηlU[l]+θlV[l]+ϑld[l]), (2)

    where H{~con,~slk,~dif}. The literature review related to the scheduling problems with ~psdst and due date assignment is given in Table 2. For a given sequence ϱ=(J[1],J[2],,J[N]), from (Wang [17]), we have

    C[l]=lj=1(s[j]+p[j])=lj=1[1+β(lj)]p[j],l=1,2,,N. (3)
    Table 2.  Problems with ~psdst and due date assignment.
    Problem Complexity Reference
    1|~psdst,~con|Nl=1(˜αEl+˜δTl+˜ηlUl+˜θlVl+˜ϑd) O(N4) Wang [17]
    1|~psdst,~con|Nl=1(˜αEl+˜δTl+˜ϑd) O(NlogN) Wang [17]
    1|~psdst,~slk|Nl=1(˜αEl+˜δTl+˜ηlUl+˜θlVl+˜ϑq) O(N4) Wang [17]
    1|~psdst,~slk|Nl=1(˜αEl+˜δTl+˜ϑq) O(NlogN) Wang [17]
    1|~psdst,~dif|Nl=1(˜αEl+˜δTl+˜ηlUl+˜θlVl+˜ϑdj) O(NlogN) Wang [17]
    1|~psdst,~con|Nl=1ζl|L[l]|+˜ϑd O(NlogN) Wang et al. [18]
    1|~psdst,~slk|Nl=1ζl|L[l]|+˜ϑq O(NlogN) Wang et al. [18]
    1|~psdst,~con|Nl=1(ζl|L[l]|+ηlU[l]+θlV[l]+ϑld[l]) O(N4) This paper
    1|~psdst,~con|Nl=1(ζl|L[l]|+ϑld[l]) O(NlogN) This paper
    1|~psdst,~slk|Nl=1(ζl|L[l]|+ηlU[l]+θlV[l]+ϑld[l]) O(N4) This paper
    1|~psdst,~slk|Nl=1(ζl|L[l]|+ϑld[l]) O(NlogN) This paper
    1|~psdst,~dif|Nl=1(ζl|L[l]|+ηlU[l]+θlV[l]+ϑld[l]) O(NlogN) This paper

     | Show Table
    DownLoad: CSV

    where ˜α,˜δ,˜ϑ are given constants, ˜ηl (˜θl) is the earliness (tardiness) penalty of job Jl, El=max{0,dlCl} (Tl=max{0,Cldl}) is the earliness (tardiness) of job Jl.

    Lemma 1. For 1|~psdst,H|Nl=1(ζl|L[l]|+ηlU[l]+θlV[l]+ϑld[l]) (H{~con,~slk,~dif}), an optimal sequence exists such that the first job is processed at time zero and contains no machine idle time.

    Proof. The result is obvious (see Brucker [19] and Liu et al. [20]).

    Lemma 2. For any given sequence ϱ, the optimal d is equal to the completion time of some job, i.e., d=C[a], a=1,2,,N.

    Proof. For any given sequence ϱ=(J[1],J[2],,J[N]), suppose that d is not equal to the completion time of some job, i.e., C[a]<d<C[a+1], 0a<n, C[0]=0, we have

    M=al=1ζl(dC[l])+Nl=a+1ζl(C[l]d)+aj=1ηl+nj=a+1θl+Nl=1dϑl.

    (i) When d=C[a], we have

    M1=al=1ζl(C[a]C[l])+Nl=a+1ζl(C[l]C[a])+a1l=1ηl+nl=a+1θl+Nl=1C[a]ϑl.

    (ii) When d=C[a+1], we have

    M2=al=1ζl(C[a+1]C[l])+Nl=a+1ζl(C[l]C[a+1])+al=1ηl+nl=a+2θl+Nl=1C[a+1]ϑl,
    MM1=al=1ζl(dC[a])Nl=a+1ζl(dC[a])+ηa+Nl=1ϑl(dC[a])=(al=1ζlNl=a+1ζl+Nl=1ϑl)(dC[a])+ηa

    and

    MM2=al=1ζl(dC[a+1])Nl=a+1ζl(dC[a+1])+θa+1+Nl=1ϑl(dC[a+1])=(al=1ζlNl=a+1ζl+Nl=1ϑl)(dC[a+1])+θa+1.

    If al=1ζlNl=a+1ζl+Nl=1ϑl0 and C[a]<d<C[a+1], then MM10; If al=1ζlNl=a+1ζl+Nl=1ϑl0 and C[a]<d<C[a+1], then MM20. Therefore, d is the completion time of some job.

    Lemma 3. For any given sequence ϱ=(J[1],J[2],,J[N]), if θl=ϑl=0 (l=1,2,N), there exists an optimal common due date d=C[a], where a is determined by

    a1l=1ζlNl=aζl+Nl=1ϑl0 (4)

    and

    al=1ζlNl=a+1ζl+Nl=1ϑl0. (5)

    Proof. From Lemma 2, when d=C[a], we have

    M=a1l=1ζl(C[a]C[l])+Nl=a+1ζl(C[l]C[a])+Nl=1C[a]ϑl.

    (i) When d reduces ε (i.e., d=C[a]ε), we have

    M=a1l=1ζl(C[a]εC[l])+Nl=aζl(C[l]C[a]+ε)+Nl=1(C[a]ε)ϑl.

    (ii) When d increases ε (i.e., d=C[a]+ε), we have

    M

    Hence, we have

    M-M' = \varepsilon \left(\sum\limits_{l = 1}^{a-1}\zeta_l-\sum\limits_{l = a}^N\zeta_l+\sum\limits_{l = 1}^{N}\vartheta_l\right) \leq0
    M-M'' = -\varepsilon \left(\sum\limits_{l = 1}^{a}\zeta_l-\sum\limits_{l = a+1}^N\zeta_l+\sum\limits_{l = 1}^{N}\vartheta_l\right) \leq0,

    i.e., a is determined by \sum_{l = 1}^{a-1}\zeta_l-\sum_{l = a}^N\zeta_l+\sum_{l = 1}^{N}\vartheta_l\leq0 and \sum_{l = 1}^{a}\zeta_l-\sum_{l = a+1}^N\zeta_l+\sum_{l = 1}^{N}\vartheta_l\geq0 .

    From Lemma 2, if d = C_{[a]} , the objective function is:

    M = \sum\limits_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}+d\vartheta_l \right)\\ = \sum\limits_{l = 1}^{a-1}\zeta_l(C_{[a]}-C_{[l]} ) +\sum\limits_{l = a+1}^N\zeta_l (C_{[l]} -C_{[a]})+\sum\limits_{l = 1}^{a-1}\eta_l +\sum\limits_{l = a+1}^N \theta_l +\sum\limits_{l = 1}^{N}C_{[a]}\vartheta_l\\ = \sum\limits_{l = 1}^{a-1}\zeta_l\left\{\sum\limits_{j = 1}^{a}\left[1+\beta(a-j)\right]p_{[j]}-\sum\limits_{j = 1}^{l} \left[1+\beta(l-j)\right] p_{[j]}\right\} \\ +\sum\limits_{l = a+1}^N\zeta_l \left\{\sum\limits_{j = 1}^{l}\left[1+\beta(l-j)\right]p_{[j]}-\sum\limits_{j = 1}^{a}\left[1+\beta(a-j)\right] p_{[j]}\right\}\\+\sum\limits_{l = 1}^{a-1}\eta_l +\sum\limits_{l = a+1}^N \theta_l +\sum\limits_{l = 1}^{N}\vartheta_l\left\{\sum\limits_{j = 1}^{a}\left[1+\beta(a-j)\right]p_{[j]}\right\}\\ = \sum\limits_{l = 1}^N \Psi_l p_{[l]}+\sum\limits_{l = 1}^{a-1}\eta_l +\sum\limits_{l = a+1}^N \theta_l, (6)

    where

    \begin{align} \Psi_l = \left\{ \begin{array}{lllllllllllll} \beta(a-1)\zeta_1+\beta(a-2)\zeta_2+\beta(a-3)\zeta_3+\ldots+\beta\zeta_{a-1}\\ +\beta\zeta_{a+1}+2\beta\zeta_{a+2}+\ldots+\beta(N-a)\zeta_{N}+[1+\beta(a-1)]\sum\limits_{j = 1}^{N}\vartheta_j, &&{l = 1, }\\ (1+\beta(a-2))\zeta_1+\beta(a-2)\zeta_2+\beta(a-3)\zeta_3+\ldots+\beta\zeta_{a-1}\\ +\beta\zeta_{a+1}+2\beta\zeta_{a+2}+\ldots+\beta(N-a)\zeta_{N}+[1+\beta(a-2)]\sum\limits_{j = 1}^{N}\vartheta_j, &&{l = 2, }\\ (1+\beta(a-3))(\zeta_1+\zeta_2)+\beta(a-3)\zeta_3\ldots+\beta\zeta_{a-1}\\ +\beta\zeta_{a+1}+2\beta\zeta_{a+2}+\ldots+\beta(N-a)\zeta_{N}+[1+\beta(a-3)]\sum\limits_{j = 1}^{N}\vartheta_j, &&{l = 3, }\\ \ldots&&\ldots\\ (1+\beta)(\zeta_1+\zeta_2+\ldots+\zeta_{a-2})+\beta\zeta_{a-1}\\ +\beta\zeta_{a+1}+2\beta\zeta_{a+2}+\ldots+\beta(N-a)\zeta_{N}+(1+\beta)\sum\limits_{j = 1}^{N}\vartheta_j, &&{l = a-1, }\\ \zeta_1+\zeta_2+\ldots+\zeta_{a-1}\\+\beta\zeta_{a+1}+2\beta\zeta_{a+2}+\ldots+\beta(N-a)\zeta_{N}+\sum\limits_{j = 1}^{N}\vartheta_j, &&{l = a, }\\ \zeta_{a+1}+(1+\beta)\zeta_{a+2}+(1+2\beta)\zeta_{a+3}+\ldots+(1+\beta(N-a-1))\zeta_{N}, &&{l = a+1, }\\ \zeta_{a+2}+(1+\beta)\zeta_{a+3}+(1+2\beta)\zeta_{a+4}+\ldots+(1+\beta(N-a-2))\zeta_{N}, &&{l = a+2, }\\ \ldots&&\ldots\\ \zeta_{N-1}+(1+\beta)\zeta_{N}, &&{N-1, }\\ \zeta_{N}, &&{N.} \end{array}\right. \end{align} (7)

    Let x_{l, r} = 1 if J_l is placed in r th position, and x_{l, r} = 0 ; otherwise. From Eq (6), the optimalsequence of 1|{\widetilde{psdst}}, \widetilde{con}|\sum_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}+\vartheta_ld_{[l]}\right) can be formulatedasthe following assignmentproblem:

    \begin{align} \mbox{Min} \ \sum\limits_{l = 1}^N\sum\limits_{r = 1}^N \Theta_{l, r} x_{l, r} \end{align} (8)
    \begin{align} s.t.\left\{ \begin{array}{rcl} \sum\limits_{h = 1}^N x_{l, r} = 1, &&{r = 1, 2, ..., N, }\\ \sum\limits_{r = 1}^N x_{l, r} = 1, &&{l = 1, 2, ..., N, }\\ x_{l, r} = 0 \;\;\mbox{or}\;\; 1, \end{array}\right. \end{align} (9)

    where

    \begin{align} \Theta_{l, r} = \left\{ \begin{array}{lll} \Psi_rp_l+\eta_r , &&{r = 1, 2, ..., a-1, }\\ \Psi_rp_l, &&{r = a, }\\ \Psi_rp_l+\theta_r , &&{r = a+1, a+2, ..., N, } \end{array}\right. \end{align} (10)

    and \Psi_r is given by Eq (7).

    Based on the above analysis, to solve 1|{\widetilde{psdst}}, \widetilde{con}|\sum_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}+\vartheta_ld_{[l]}\right) , Algorithm 1 was summarized as follows:

    Algorithm 1
    Require: \beta , {p}_{l}, \zeta_l, \eta_l, \theta_l, \vartheta_l for 1\leq l\leq N .
    Ensure: An optimal sequence \varrho^* , optimal common due date d^* .
    Step 1. For each a ( a = 1, 2, \ldots, N ), calculate \Psi_r (see Eq (7)) and \Theta_{l, r} (see Eq (10)), to solve the assignment problem (8)–(10), a suboptimal sequence \varrho(a) and objective function value M(a) can be obtained.
    Step 2. The (global) optimal sequence (i.e., \varrho^* ) is the one with the minimum value
    M^* = \min\left\{M(a)|a = 1, 2, \ldots, N\right\}.
    Step 3. Set d^* = C_{[a]} .

     | Show Table
    DownLoad: CSV

    Theorem 1. The 1|{\widetilde{psdst}}, \widetilde{con}|\sum_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}+\vartheta_ld_{[l]}\right) can be solved by Algorithm 1, and time complexity was O(N^4) .

    Proof. The correctness of Algorithm 1 follows the above analysis. In Step 1, for each a , solving the assignment problem needs O(N^3) time; Steps 2 and 3 require O(N) time; a = 1, 2, \ldots, N . Therefore, the total time complexity was O(N^4) .

    Lemma 4. (Hardy et al. [23]). "The sum of products \sum_{l = 1}^N a_l b_{l} is minimized if sequence a_1, a_2, \ldots, a_N is ordered nondecreasingly and sequence b_1, b_2, \ldots, b_N is ordered nonincreasingly or vice versa."

    If \eta_l = \theta_l = 0 , a can be determined by Lemma 3 (see Eqs (4) and (5)), We

    \begin{align} M = \sum\limits_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\vartheta_ld_{[l]}\right) = \sum\limits_{l = 1}^N \Psi_l p_{[l]}, \end{align} (11)

    where \Omega_j is given by Eq (6).

    Equation (11) can be minimized by Lemma 4 in O(N\log N) time (i.e., a_l = \Psi_l, b_l = p_{l} ), hence, to solve 1|{\widetilde{psdst}}, \widetilde{con}|\sum_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\vartheta_ld_{[l]}\right) , the following algorithm was summarized as follows:

    Theorem 2. The 1|{\widetilde{psdst}}, \widetilde{con}|\sum_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\vartheta_ld_{[l]}\right) can be solved by Algorithm 2, and time complexity was O(N\log N) .

    Algorithm 2
    Require: \beta , {p}_{l}, \zeta_l, \vartheta_l for 1\leq l\leq N .
    Ensure: An optimal sequence \varrho^* , optimal common due date d^* .
    Step 1. Calculate a by Lemma 3 (see Eqs (4) and (5)).
    Step 2. By using Lemma 4 (let a_l = \Psi_l, b_l = p_{l} ) to determine the optimal job sequence (i.e., \varrho^* ), i.e., place the largest p_l at the smallest \Psi_l position, place the second largest p_l at the second smallest \Psi_l position, etc.
    Step 3. Set d^* = C_{[a]} .

     | Show Table
    DownLoad: CSV

    Similarly, we have

    Lemma 5. For any given sequence \varrho of 1|{\widetilde{psdst}}, \widetilde{slk}|\sum_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}+\vartheta_ld_{[l]}\right) , an optimal sequence exists in which

    1) {C}_{[l]}\leq {d}_{[l]} implies {C}_{[l-1]}\leq {d}_{[l-1]} and {C}_{[l]}\geq {d}_{[l]} implies {C}_{[l+1]}\geq {d}_{[l+1]} for all l ;

    2) the optimal q is equal to the completion time of some job, i.e., q = C_{[b-1]} , b = 1, 2, \ldots, N .

    Lemma 6. For any given sequence \varrho = (J_{[1]}, J_{[2]}, \ldots, J_{[N]}) , if \theta_l = \vartheta_l = 0 ( l = 1, 2, \ldots N ), there exists an optimal common due date q = C_{[b-1]} , where b is determined by

    \begin{align} \sum\limits_{l = 1}^{b-1}\zeta_l-\sum\limits_{l = b}^N\zeta_l+\sum\limits_{l = 1}^{N}\vartheta_l\leq0 \end{align} (12)

    and

    \begin{align} \sum\limits_{l = 1}^{b}\zeta_l-\sum\limits_{l = b+1}^N\zeta_l+\sum\limits_{l = 1}^{N}\vartheta_l\geq0. \end{align} (13)

    Proof. From Lemma 5, when q = C_{[b-1]} , we have

    M = \sum\limits_{l = 1}^{b-1}\zeta_l(s_{[b]}+p_{[b]}+C_{[b-1]}-C_{[l]}) +\sum\limits_{l = b+1}^N\zeta_l (C_{[l]} -s_{[b]}-p_{[b]}-C_{[b-1]})+\sum\limits_{l = 1}^{N}\vartheta_l(s_{[b]}+p_{[b]}+C_{[b-1]}).

    (i) When q reduces \varepsilon (i.e., q = C_{[b-1]}-\varepsilon ), we have

    M' = \sum\limits_{l = 1}^{b-1}\zeta_l(s_{[b]}+p_{[b]}+C_{[b-1]}-\varepsilon-C_{[l]} ) +\sum\limits_{l = b}^N\zeta_l (C_{[l]} -s_{[b]}-p_{[b]} -C_{[b-1]}+\varepsilon)+\sum\limits_{l = 1}^{N}(s_{[b]}+p_{[b]}+C_{[b-1]}-\varepsilon)\vartheta_l.

    (ii) When q increases \varepsilon (i.e., q = C_{[b-1]}+\varepsilon ), we have

    M'' = \sum\limits_{l = 1}^{b}\zeta_l(s_{[b]}+p_{[b]}+C_{[b-1]}+\varepsilon-C_{[l]}) +\sum\limits_{l = b+1}^N\zeta_l (C_{[l]} -s_{[b]}-p_{[b]}-C_{[b-1]}-\varepsilon)+\sum\limits_{l = 1}^{N}(s_{[b]}+p_{[b]}+C_{[b-1]}+\varepsilon)\vartheta_l.

    Hence, we have

    M-M' = \varepsilon \left(\sum\limits_{l = 1}^{b-1}\zeta_l-\sum\limits_{l = b}^N\zeta_l+\sum\limits_{l = 1}^{N}\vartheta_l\right) \leq0
    M-M'' = -\varepsilon \left(\sum\limits_{l = 1}^{b}\zeta_l-\sum\limits_{l = b+1}^N\zeta_l+\sum\limits_{l = 1}^{N}\vartheta_l\right) \leq0,

    i.e., b is determined by \sum_{l = 1}^{b-1}\zeta_l-\sum_{l = b}^N\zeta_l+\sum_{l = 1}^{N}\vartheta_l\leq0 and \sum_{l = 1}^{b}\zeta_l-\sum_{l = b+1}^N\zeta_l+\sum_{l = 1}^{N}\vartheta_l\geq0 .

    From Lemma 5, if q = C_{[b-1]} (i.e., d_{[l]} = s_{[l]}+p_{[l]}+C_{[b-1]} ), the objective function is:

    \begin{eqnarray*} M& = &\sum\limits_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}+\vartheta_ld_{[l]} \right)\\ & = &\sum\limits_{l = 1}^{b-1}\zeta_l(s_{[l]}+p_{[l]}+C_{[b-1]}-C_{[l]} ) +\sum\limits_{l = b+1}^N\zeta_l (C_{[l]} -s_{[l]}-p_{[l]}-C_{[b-1]})\\&&+\sum\limits_{l = 1}^{b-1}\eta_l +\sum\limits_{l = b+1}^N \theta_l +\sum\limits_{l = 1}^{N}(s_{[l]}+p_{[l]}+C_{[b-1]})\vartheta_l\\ & = &\sum\limits_{l = 1}^{b-1}\zeta_l(C_{[b-1]}-C_{[l-1]} ) +\sum\limits_{l = b+1}^N\zeta_l (C_{[l-1]} -C_{[b-1]})+\sum\limits_{l = 1}^{b-1}\eta_l +\sum\limits_{l = b+1}^N \theta_l \\&&+\sum\limits_{l = 1}^{N}(s_{[l]}+p_{[l]})\vartheta_l+\sum\limits_{l = 1}^{N}C_{[b-1]}\vartheta_l\\ & = &\sum\limits_{l = 1}^{b-1}\zeta_l\left\{\sum\limits_{j = 1}^{b-1}\left[1+\beta(b-1-j)\right]p_{[j]} -\sum\limits_{j = 1}^{l-1}\left[1+\beta(l-1-j)\right] p_{[j]}\right\} \\ &&+\sum\limits_{l = b+1}^N\zeta_l \left\{\sum\limits_{j = 1}^{l-1}\left[1+\beta(l-1-j)\right]p_{[j]}-\sum\limits_{j = 1}^{b-1} \left[1+\beta(b-1-j)\right] p_{[j]}\right\}\\&&+\sum\limits_{l = 1}^{b-1}\eta_l +\sum\limits_{l = b+1}^N \theta_l +\sum\limits_{l = 1}^{N}\left(\beta\sum\limits_{j = 1}^{l-1}p_{[j]}+p_{[l]}\right)\vartheta_l +\sum\limits_{l = 1}^{N}\vartheta_l\left\{\sum\limits_{j = 1}^{b-1}\left[1+\beta(b-1-j)\right]p_{[j]}\right\}\\ & = &\sum\limits_{l = 1}^N \Phi_l p_{[l]}+\sum\limits_{l = 1}^{b-1}\eta_l +\sum\limits_{l = b+1}^N \theta_l, \end{eqnarray*} (14)

    where

    \begin{align} \Phi_l = \left\{ \begin{array}{lllllllllllllll} (1+\beta(b-2))\zeta_1+\beta(b-2)\zeta_2+\beta(b-3)\zeta_3+\ldots+\beta\zeta_{b-1}\\ +\beta\zeta_{b+1}+2\beta\zeta_{b+2}+\ldots+\beta(N-b)\zeta_{N}+[1+\beta(b-2)]\sum\limits_{j = 1}^{N}\vartheta_j\\ +\vartheta_1+\beta\sum\limits_{j = 2}^{N}\vartheta_j, &&{l = 1, }\\ (1+\beta(b-3))(\zeta_1+\zeta_2)+\beta(b-3)\zeta_3+\beta(b-4)\zeta_4+\ldots+\beta\zeta_{b-1}\\ +\beta\zeta_{b+1}+2\beta\zeta_{b+2}+\ldots+\beta(N-b)\zeta_{N} +[1+\beta(b-3)]\sum\limits_{j = 1}^{N}\vartheta_j \\+\vartheta_2+\beta\sum\limits_{j = 3}^{N}\vartheta_j, &&{l = 2, }\\ (1+\beta(b-4))(\zeta_1+\zeta_2+\zeta_3)+\beta(b-4)\zeta_4+\ldots+\beta\zeta_{b-1}\\ +\beta\zeta_{b+1}+2\beta\zeta_{b+2}+\ldots+\beta(N-b)\zeta_{N}+[1+\beta(b-4)] \sum\limits_{j = 1}^{N}\vartheta_j\\+\vartheta_3+\beta\sum\limits_{j = 4}^{N}\vartheta_j, &&{l = 3, }\\ \ldots&&\ldots\\ (1+\beta)(\zeta_1+\zeta_2+\ldots+\zeta_{b-2})+\beta\zeta_{b-1}\\ +\beta\zeta_{b+1}+2\beta\zeta_{b+2}+\ldots+\beta(N-b)\zeta_{N} +(1+\beta)\sum\limits_{j = 1}^{N}\vartheta_j\\+\vartheta_{b-2}+\beta\sum\limits_{j = b-1}^{N}\vartheta_j, &&{l = b-2, }\\ \zeta_1+\zeta_2+\ldots+\zeta_{b-1}\\ +\beta\zeta_{b+1}+2\beta\zeta_{b+2}+\ldots+\beta(N-b)\zeta_{N} +\sum\limits_{j = 1}^{N}\vartheta_j+\vartheta_{b-1}+\beta\sum\limits_{j = b}^{N}\vartheta_j, &&{l = b-1, }\\ \zeta_{b+1}+(1+\beta)\zeta_{b+2}+(1+2\beta)\zeta_{b+3}+\ldots+(1+\beta(N-b-1))\zeta_{N}\\ +\vartheta_{b}+\beta\sum\limits_{j = b+1}^{N}\vartheta_j, &&{l = b, }\\ \zeta_{b+2}+(1+\beta)\zeta_{b+3}+(1+2\beta)\zeta_{b+4}+\ldots+(1+\beta(N-b-2))\zeta_{N} \\+\vartheta_{b+1}+\beta\sum\limits_{j = b+2}^{N}\vartheta_j, &&{l = b+1, }\\ \ldots&&\ldots\\ \zeta_{N}+\vartheta_{N-1}+\beta\vartheta_N, &&{N-1, }\\ \vartheta_N, &&{N.} \end{array}\right. \end{align} (15)

    Similarly, from Eq (14), the optimal sequence of 1|{\widetilde{psdst}}, \widetilde{slk}|\sum_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}+\vartheta_ld_{[l]}\right) can be obtained as follows:

    \begin{align} \mbox{Min} \ \sum\limits_{l = 1}^N\sum\limits_{r = 1}^N \Xi_{l, r} x_{l, r} \end{align} (16)
    \begin{align} s.t.\left\{ \begin{array}{rcl} \sum\limits_{h = 1}^N x_{l, r} = 1, &&{r = 1, 2, ..., N, }\\ \sum\limits_{r = 1}^N x_{l, r} = 1, &&{l = 1, 2, ..., N, }\\ x_{l, r} = 0 \;\;\mbox{or}\;\; 1, \end{array}\right. \end{align} (17)

    where

    \begin{align} \Xi_{l, r} = \left\{ \begin{array}{lll} \Phi_rp_l+\eta_r , &&{r = 1, 2, ..., b-1, }\\ \Phi_rp_l, &&{r = b, }\\ \Phi_rp_l+\theta_r , &&{r = b+1, b+2, ..., N, } \end{array}\right. \end{align} (18)

    and \Phi_r is given by (15).

    Similarly, to solve 1|{\widetilde{psdst}}, \widetilde{slk}|\sum_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}+\vartheta_ld_{[l]}\right) , the following algorithm can be proposed:

    Theorem 3. The 1|{\widetilde{psdst}}, \widetilde{slk}|\sum_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}+\vartheta_ld_{[l]}\right) can be solved by Algorithm 3, and time complexity was O(N^4) .

    Algorithm 3
    Require: \beta , {p}_{l}, \zeta_l, \eta_l, \theta_l, \vartheta_l for 1\leq l\leq N .
    Ensure: An optimal sequence \varrho^* , optimal common flow allowance q^* .
    Step 1. For each b ( b = 1, 2, \ldots, N ), calculate \Phi_r (see Eq (15)) and \Xi_{l, r} (see Eq (18)), to solve the assignment problem (16)–(18), a suboptimal sequence \varrho(b) and objective function value M(b) can be obtained.
    Step 2. The (global) optimal sequence (i.e., \varrho^* ) is the one with the minimum value
    M^* = \min\left\{M(b)|b = 1, 2, \ldots, N\right\}.
    Step 3. Set q^* = C_{[b-1]} .

     | Show Table
    DownLoad: CSV

    Similarly, if \eta_l = \theta_l = 0 , we have

    Theorem 4. The problem 1|{\widetilde{psdst}}, \widetilde{slk}|\sum_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\vartheta_ld_{[l]}\right) can be solved in O(N\log N) time.

    Lemma 7. For a given sequence \pi of 1|{\widetilde{psdst}}, \widetilde{dif}|\sum_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}+\vartheta_ld_{[l]}\right) , an optimal solution exists such that d_{[l]}\leq C_{[l]} .

    Proof. For a given sequence \varrho , the objective function for job J_{[l]} was:

    \begin{align} M_{[l]} = \zeta_l |C_{[l]}-d_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}+\vartheta_ld_{[l]}. \end{align} (19)

    If d_{[l]} > C_{[l]} (i.e., the job J_{[l]} is an early job), it follows that

    M_{[l]} = \zeta_l (d_{[l]}-C_{[l]})+\eta_l{U_{[l]}}+\vartheta_ld_{[l]}.

    Move d_{[l]} to the left such that d_{[l]} = C_{[l]} , we have

    M'_{[l]} = \vartheta_ld_{[l]} = \vartheta_lC_{[l]} < M_{[l]},

    therefore, d_{[l]}\leq C_{[l]} .

    Lemma 8. For a given sequence \varrho , if \vartheta_l\geq\zeta_l , d_{[l]} = 0 ; otherwise d_{[l]} = C_{[l]} ( l = 1, 2, \ldots, N ).

    Proof. For a given sequence \varrho , from Lemma 7, we have d_{[l]}\leq C_{[l]} and

    \begin{align} M_{[l]} = \zeta_l (C_{[l]}-d_{[l]})+\theta_l{V_{[l]}}+\vartheta_ld_{[l]} = \zeta_lC_{[l]}+\theta_l+(\vartheta_l-\zeta_l)d_{[l]}. \end{align} (20)

    From Eq (20), when \vartheta_l-\zeta_l\geq 0 , d_{[l]} was equal to 0; otherwise, then d_{[l]} was equal to C_{[l]} .

    From Lemma 8, if \vartheta_l\geq\zeta_l , we have d_{[l]} = 0 and

    \begin{align} M = \sum\limits_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}+\vartheta_ld_{[l]}\right) = \sum\limits_{l = 1}^{N}\zeta_l C_{[l]}+\sum\limits_{l = 1}^{N}\theta_l. \end{align} (21)

    If \vartheta_l < \zeta_l , we have d_{[l]} = C_{[l]} and

    \begin{align} M = \sum\limits_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}+\vartheta_ld_{[l]}\right) = \sum\limits_{l = 1}^{N}\vartheta_l C_{[l]}. \end{align} (22)

    From Eqs (21) and (22), minimizing \sum_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}+\vartheta_ld_{[l]}\right) is equal to minimizing the expression

    \begin{align} M = \sum\limits_{l = 1}^{N}\min\{\vartheta_l, \zeta_l\} C_{[l]} = \sum\limits_{l = 1}^{N}\min\{\vartheta_l, \zeta_l\}\sum\limits_{j = 1}^{l}\left[1+\beta(l-j)\right]p_{[j]} = \sum\limits_{l = 1}^N \Upsilon_l p_{[l]}, \end{align} (23)

    where

    \begin{align} \Upsilon_l = \left\{ \begin{array}{lllllllllllllll} \min\{\vartheta_1, \zeta_1\}+(1+\beta)\min\{\vartheta_2, \zeta_2\}+\ldots+(1+(N-1)\beta)\min\{\vartheta_N, \zeta_N\}, &&{l = 1, }\\ \min\{\vartheta_2, \zeta_2\}+(1+\beta)\min\{\vartheta_3, \zeta_3\}+\ldots+(1+(N-2)\beta)\min\{\vartheta_N, \zeta_N\}, &&{l = 2, }\\ \ldots&&\ldots\\ \min\{\vartheta_{N-1}, \zeta_{N-1}\}+(1+\beta)\min\{\vartheta_N, \zeta_N\}, &&{N-1, }\\ \min\{\vartheta_N, \zeta_N\}, &&{N, } \end{array}\right. \end{align} (24)

    i.e.,

    \begin{align} \Upsilon_l = \sum\limits_{j = l}^{N}[1+\beta(j-l)]\min\{\vartheta_j, \zeta_j\}, \ \ \ \ {l = 1, 2, \ldots, N.} \end{align} (24')

    Obviously, Eq (23) can be minimized by Lemma 4.

    Theorem 5. The 1|{\widetilde{psdst}}, \widetilde{dif}|\sum_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}+\vartheta_ld_{[l]}\right) can be solved by Algorithm 4, and time complexity was O(N\log N) .

    Algorithm 4
    Require: \beta , {p}_{l}, \zeta_l, \eta_l, \theta_l, \vartheta_l for 1\leq l\leq N .
    Ensure: An optimal sequence \varrho^* , optimal common due date d_l^* .
    Step 1. By using Lemma 4 (let a_l = \Upsilon_l, b_l = p_{l} ) to determine the optimal job sequence (i.e., \varrho^* ), i.e., place the largest p_l at the smallest \Upsilon_l position, place the second largest p_l at the second smallest \Upsilon_l position, etc.
    Step 2. If \vartheta_l\geq\zeta_l , d_{[l]}^* = 0 ; otherwise d_{[l]}^* = C_{[l]} ( l = 1, 2, \ldots, N ).

     | Show Table
    DownLoad: CSV

    We present an example to illustrate the calculation steps and results of the three due date assignments.

    Example 1. Consider a 6-job problem, where \beta = 1 , p_1 = 7 , p_2 = 9 , p_3 = 4 , p_4 = 6 , p_5 = 8 , p_6 = 5 , \zeta_l, \eta_l, \theta_l and \vartheta_l are given in Table 3.

    Table 3.  Values of \zeta_l, \eta_l, \theta_l and \vartheta_l.
    l = 1 l = 2 l = 3 l = 4 l = 5 l = 6
    \zeta_l 6 8 14 3 15 7
    \eta_l 8 4 9 10 12 5
    \theta_l 10 8 6 5 14 17
    \vartheta_l 12 16 7 13 8 9

     | Show Table
    DownLoad: CSV

    From Algorithm 1, For the \widetilde{con} assignment, if a = 1 , the values \Psi_1 = 205, \Psi_2 = 140, \Psi_3 = 93, \Psi_4 = 54, \Psi_5 = 29, \Psi_6 = 7, (see Eqs (7) or (7')) and \Theta_{l, r} (see Eq (10)) are given in Table 4. By the assignment problems (8)–(10), the sequence is \varrho(1) = (J_3, J_6, J_4, J_1, J_5, J_2) and M(1) = 2801 . Similarly, for a = 2, 3, 4, 5, 6 , the results are shown in Table 5. From Table 5, the optimal sequence is \varrho^* = (J_3, J_6, J_4, J_1, J_5, J_2) , M^* = 2801 and d^* = {C}_{[2]} = 14 .

    Table 4.  Values \Theta_{l, r} for a = 1.
    {r = 1} {r = 2} {r = 3} {r = 4} {r = 5} {r = 6}
    J_1 1435 988 657 383 217 66
    J_2 1845 1268 843 491 275 80
    J_3 820 568 378 221 130 45
    J_4 1230 848 564 329 188 59
    J_5 1640 1128 750 437 246 73
    J_6 1025 708 471 275 159 52

     | Show Table
    DownLoad: CSV
    Table 5.  Results for \widetilde{con}.
    a \varrho(a) M(a)
    1 (J_3, J_6, J_4, J_1, J_5, J_2) 2801
    2 (J_3, J_6, J_4, J_1, J_5, J_2) 3017
    3 (J_3, J_6, J_4, J_1, J_5, J_2) 3615
    4 (J_3, J_6, J_4, J_1, J_5, J_2) 5335
    5 (J_3, J_6, J_4, J_1, J_5, J_2) 7451
    6 (J_3, J_6, J_4, J_1, J_5, J_2) 11,382

     | Show Table
    DownLoad: CSV

    For the \widetilde{slk} assignment, the results are shown in Table 6. From Table 6, the optimal sequence is \varrho^* = (J_3, J_6, J_4, J_1, J_5, J_2) , M^* = 2832 and q^* = {C}_{[0]} = 0 .

    Table 6.  Results for \widetilde{slk}.
    b \varrho(b) M(b)
    1 (J_3, J_6, J_4, J_1, J_5, J_2) 2832
    2 (J_3, J_6, J_4, J_1, J_5, J_2) 2928
    3 (J_3, J_6, J_4, J_1, J_5, J_2) 3286
    4 (J_3, J_6, J_4, J_1, J_5, J_2) 4310
    5 (J_3, J_6, J_4, J_1, J_5, J_2) 5934
    6 (J_3, J_6, J_4, J_1, J_5, J_2) 9049

     | Show Table
    DownLoad: CSV

    For the \widetilde{dif} assignment, \Upsilon_1 = 137, \Upsilon_2 = 98, \Upsilon_3 = 65, \Upsilon_4 = 40, \Upsilon_5 = 22, \Upsilon_6 = 7 , the optimal sequence is \varrho^* = (J_3, J_6, J_4, J_1, J_5, J_2) , M^* = 1987 , d_3^* = 0 , d_6^* = 0 , d_4^* = {C}_{4} = 28 , d_1^* = 0 , d_5^* = {C}_{5} = 80 and d_2^* = 0 .

    Under \widetilde{con} , \widetilde{slk} and \widetilde{dif} assignments, the single-machine scheduling problem with \widetilde{psdst} and position-dependent weights had been addressed. The goal was to minimize the weighted sum of lateness, number of early and delayed jobs and due date cost. Here we showed that the problem remains polynomially solvable. If the due dates are given, from Brucker [19], the problem 1|{\widetilde{psdst}}|\sum_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}\right) is NP-dard. For future research, we suggest some interesting topics as follows:

    1) Considering the problem 1|{\widetilde{psdst}}|\sum_{l = 1}^{N}\left(\zeta_l |L_{[l]}|+\eta_l{U_{[l]}}+\theta_l{V_{[l]}}\right) ;

    2) Investigating the problem in a flow shop setting;

    3) Studying the group technology problem with learning effects (deterioration effects) and/or resource allocation (see Wang et al. [24], Huang [25] and Liu and Xiong [26]);

    4) Investigating scenario-dependent processing times (see Wu et al. [27] and Wu et al. [28]).

    This research was supported by the National Natural Science Regional Foundation of China (71861031 and 72061029).

    The authors declare that they have no conflicts of interest.



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