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Article

Optimal Different Due-Date Assignment Scheduling with Group Technology and Resource Allocation

Business School, Northwest Normal University, Lanzhou 730070, China
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Author to whom correspondence should be addressed.
Mathematics 2024, 12(3), 436; https://doi.org/10.3390/math12030436
Submission received: 27 December 2023 / Revised: 25 January 2024 / Accepted: 26 January 2024 / Published: 29 January 2024
(This article belongs to the Special Issue Advances in Scheduling Optimization and Computational Intelligence)

Abstract

:
In this paper, we consider different due-date assignment scheduling with group technology and resource allocation on a single machine, where the due date of each job may be different. Under constant processing times, the objective function is to minimize the scheduling cost (i.e., the weighted sum of earliness, tardiness, and due-date assignment cost, where the weights are position dependent). Under some optimal properties, we prove that this problem can be solved in  O ( ζ log ζ )  time, where  ζ  is the number of jobs. The problem is also extended to cases which include linear and convex functions of the quantity of resource allocation. The objective function is minimizing the sum of the scheduling cost and the resource-consumption cost. For the special case of linear and convex functions, we show that the problem is polynomially solvable in  O ( ζ 3 )  time.

1. Introduction

Scheduling problems and models with resource allocation (also called controllable processing times) have garnered extensive attention (see Shabtay and Steiner [1], Wang and Wang [2,3], Strusevich and Rustogi [4], Yedidsiona and Shabtay [5], Sun et al. [6]). In 2021, Mor et al. [7] considered single-machine scheduling with resource allocation. For a large set of scheduling problems, they proposed heuristic algorithms. Lu et al. [8] studied single-machine scheduling with learning effects and resource allocation. Lv and Wang [9] revisited no-wait flow-shop scheduling with learning effects and resource allocation. They showed that four versions of the problem can be solved in polynomial time. In 2023, Shioura et al. [10] delved into a parallel-machine problem with resource allocation. Wang and Wang [11] investigated single-machine resource-allocation scheduling with a time-dependent learning effect. Under the convex resource function, for three versions of the scheduling cost and total-resource-consumption cost, they demonstrated that some special cases of the problem can be solved in polynomial time. Wang et al. [12] investigated single-machine resource allocation scheduling with a deterioration effect. Under position-dependent workloads, they showed that four versions of scheduling cost and resource-consumption cost are polynomially solvable. Zhang et al. [13] considered two-agent single-machine scheduling with deteriorating jobs. Under slack due-date assignment and position-dependent workloads, they proved that the problem of minimizing the maximum value of the earliness–tardiness cost can be solved in polynomial time. Pang and Meng [14] studied robust project scheduling with resource allocation. For maximizing the use of the precedence relation, they proposed a heuristic algorithm. Prabhu and Rajesh [15] considered advanced dynamic scheduling with resource allocation. Wang et al. [16] revisited single-machine resource-allocation scheduling with deteriorating jobs. For two versions of total weighted completion time and total resource-consumption cost, they proved that the problem is NP-hard, and they proposed branch-and-bound and heuristic algorithms to solve the problem.
In addition, Webster and Baker [17], Wang and Wang [18], Huang [19], Bajwa et al. [20], and Liu et al. [21] addressed the study of scheduling with group technology (denoted by  G T ^ ). Shabtay et al. [22] considered single-machine scheduling with  G T ^  and resource allocations. Under the common due-date assignment, they proved that a special case of the problem can be solved in polynomial time. Wang et al. [23] and Wang and Liang [24] studied single-machine scheduling with  G T ^ , resource allocations and deteriorating jobs concurrently. For some special cases Wang et al. [23] proved that the problem can be solved in polynomial time. For general case of the problem, Wang and Liang [24] proposed solution algorithms.
In 2023, Yan et al. [25] considered single-machine  G T ^  scheduling with learning effects and resource allocation. For total-completion-time minimization subject to limited resource availability, they proposed some heuristic algorithms and a branch-and-bound algorithm. Chen et al. [26] focused on single-machine  G T ^  scheduling with resource allocation and due-date assignment. Under different due-date assignments, they proved that some special cases of the weighted sum of the scheduling cost (including earliness, tardiness, and the due-date assignment cost) and resource-consumption-cost minimization can be solved in polynomial time. Liu and Wang [27] studied single-machine  G T ^  scheduling with resource allocation. Under common and slack due-date assignments, the objective was to minimize the weighted sum of earliness, tardiness, due-date assignment cost, and resource-consumption cost. For linear and convex resource functions, Liu and Wang [27] proved that two special cases of the problem are polynomially solvable.
To the best of our knowledge, very few articles (see Chen et al. [26]) have studied  G T ^  scheduling with resource allocation and different due-date assignments. We will continue the work of Liu and Wang [27] but consider different due-date assignments. First, for constant processing times, the objective is to minimize the weighted sum of earliness, tardiness, and due-date assignments cost (i.e., the scheduling cost, and the weights are position dependent (Lv and Wang [28], Wu et al. [29], and Pan et al. [30])). Second (resp. third), for the linear (convex) resource function, the objective is to minimize the sum of the scheduling cost and the consumption cost.
The rest of this paper is structured as follows: In Section 2, the model is presented. In Section 3, for the general case of constant processing times, we show that the problem is polynomially solvable. In Section 4 (resp. 5), for a special case of the linear (resp. convex) resource function, we prove that the problem is polynomially solvable. In Section 6, numerical examples are presented, and we conclude the paper in Section 7.

2. Problem Definition

Assume that there are  ζ  jobs  { j 1 , j 2 , , j ζ }  to be processed on a single machine. Under the  G T ^ , based on the similarities of the processing process, all the jobs are divided into  ψ  groups  { g 1 , g 2 , , g ψ }  in advance. All the jobs and the machine are available at time zero. Let  o h  be the number of jobs for group  g h  (i.e.,  o 1 + o 2 + + o ψ = ζ ) and  s h  be the setup time of group  g h  (i.e., the time,  s h , required to process the group  g h ). Let  j h i  be the ith job in  g h h = 1 , 2 , , ψ ; i = 1 , 2 , , o h . Due to the linear resource-consumption function, the actual processing time of  j h i  is given by
p h i = a h i b h i u h i , 0 u h i u ¯ h i < a h i b h i ,
where  a h i  (respective  b h i ) is the basic processing time (the respective compression rate) of job  j h i u h i  is the amount of nonrenewable resource allocated to  j h i , and  u ¯ h i  is the upper bound on the resource allocated to job  j h i .
For some resource-allocation problems in economic or physical systems, however, the linear resource-consumption function fails to reflect the law of diminishing marginal returns (Monma et al. [31]); hence, for the convex resource-consumption function,
p h i = a h i + ϑ h i u h i κ ,
where  ϑ h i  is the workload of  j h i  and  κ > 0  is a constant.
Let  E h i = max { 0 , d h i C h i }  (respectively,  T h i = max { 0 , C h i d h i }  be the earliness (respectively, the tardiness) of job  j h i , where  C h i  (respectively,  d h i ) is the completion time (respectively, the due date) of  j h i . Our  main question is to determine the group schedule  π , the internal job schedule  π h  within  g h , the due dates  d h i , and the resource allocations  u h i  ( h = 1 , 2 , , ψ ; i = 1 , 2 , , o h ), to minimize
h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i ,
where  α h i  (respectively,  β i h γ h i ) is the position-dependent weight for earliness (tardiness, due-date assignment cost), i.e.,  α h i β i h , and  γ h i  are not related to job  j h i  but to position i in group  g h v h i  is the cost of allocating one unit of the resource to job  j h i , and  [ i ]  is a scheduled job (or group) in the ith position. Following the notation in Pinedo [32], this problem is denoted as
1 p h i = a h i b h i u h i , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i
and
1 p h i = a h i + ϑ h i u h i κ , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i ,
where  Θ D I F  denotes the different due-date assignments. The comparison to  G T ^ , resource allocation, and due-date assignment is given in Table 1.

3. Constant Processing Times

In this section, we consider the problem with constant processing times, i.e., the problem  1 p h i , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) . Let  S h  be the starting time of  g h  for a given job schedule  π h . The completion time of  j h i  is
C [ h ] [ i ] = S [ h ] + s [ h ] + l = 1 i p [ h ] [ l ] = l = 1 h s [ l ] + f = 1 h 1 l = 1 o f p [ f ] [ l ] + l = 1 i p [ h ] [ l ] .
Lemma 1. 
Under group  g h  ( h = 1 , 2 , , ψ ), for the given resource allocation and job schedule  π h  there exists an optimal solution, such that  0 d h [ i ] C h [ i ]  ( i = 1 , 2 , , o h ).
Proof. 
If  d h [ i ] > C h [ i ] , for the job  j h [ i ]  the objective cost is
O C ˜ h [ i ] ( d h [ i ] > C h [ i ] ) = α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] = α h i ( d h [ i ] C h [ i ] ) + γ h i d h [ i ] .
Move  d h [ i ]  to  C h [ i ] ; then, the objective cost is
O C ˜ h [ i ] ( d h [ i ] = C h [ i ] ) = γ h i C h [ i ] < O C ˜ h [ i ] ( d h [ i ] > C h [ i ] ) .
Hence,  0 d h [ i ] C h [ i ] .    □
Lemma 2. 
Under group  g h  ( h = 1 , 2 , , ψ ), for the given resource allocation and job schedule  π h  there exists an optimal solution, such that
d h [ i ] = 0 , i f γ h i β h i 0 , C h [ i ] , i f γ h i β h i < 0 .
Proof. 
From Lemma 1, for the job  j h [ i ]  the objective cost is
O C ˜ h [ i ] = β h i ( C h [ i ] d h [ i ] ) + γ h i d h [ i ] = β h i C h [ i ] + ( γ h i β h i ) d h [ i ] .
Evidently, if  γ h i β h i 0 , then  d h [ i ] = 0 ; if  γ h i β h i < 0 , then  d h [ i ] = C h [ i ] . This completes the proof.    □
From Lemma 2, for group  g h  ( h = 1 , 2 , , ψ ), if  γ h i β h i 0 , we have
O C ˜ h = i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) = i = 1 o h β h i C h [ i ] .
Similarly, if  γ h i β h i < 0 , we have
O C ˜ h = i = 1 o h γ h i C h [ i ] .
Hence, for group  g h , we have
O C ˜ h = i = 1 o h ϱ h i C h [ i ] ,
where
ϱ h i = min { β h i , γ h i } .
Let  t h  be the starting time of group  g h . From Equation (13), the objective cost of group  g h  is
O C ˜ h = i = 1 o h ϱ h i C h i = i = 1 o h ϱ h i t h + s h + l = 1 i p h [ l ] = i = 1 o h ϱ h i l = 1 i p h [ l ] + i = 1 o h ϱ h i t h + s h = i = 1 o h l = i o h ϱ h l p h [ i ] + i = 1 o h ϱ h i t h + s h .
Lemma 3. 
Given the group schedule π, the optimal job schedule  π h * ( π )  within group  g h ( π )  can be obtained by the smallest-processing-time-(SPT) first rule, i.e., the non-decreasing order of  p h i .
Proof. 
Given the group schedule  π , from Equation (15),  h = 1 ψ l = h ψ i = 1 o l ϱ [ l ] i s [ h ]  and  l = h + 1 ψ i = 1 o l ϱ [ l ] i  are constants. As  l = i o h ϱ h l  is monotonically decreasing in i (i.e.,  l = 1 o h ϱ h l l = 2 o h ϱ h l ϱ h o h 1 + ϱ h o h ϱ h o h ), by the HLP rule (see Hardy et al. [33]) the lemma follows.    □
Lemma 4. 
For the problem  1 p h i , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) , the optimal group schedule is arranged in the non-descending order of  s h + l = 1 o h p h l l = 1 o h ϱ h l h = 1 , 2 , , ψ .
Proof. 
Let groups  g f  and  g h  be two adjacent group pairs, such that  π = ( , g f , g h , )  and  π = ( , g h , g f , ) . Let S be the starting time of group  g f  in  π . From Equation (15) we have
O C ˜ f ( π ) + O C ˜ h ( π ) O C ˜ h ( π ) O C ˜ f ( π ) = i = 1 o f ϱ f i S + s f + i = 1 o h ϱ h i S + s f + l = 1 o f p f l + s h i = 1 o h ϱ f i S + s h i = 1 o f ϱ f i S + s h + l = 1 o h p h l + s f = i = 1 o h ϱ h i s f + l = 1 o f p f l i = 1 o f ϱ f i s h + l = 1 o h p h l = i = 1 o f ϱ f i i = 1 o h ϱ h i s f + l = 1 o f p f l l = 1 o f ϱ f l s h + l = 1 o h p h l l = 1 o h ϱ h l 0 s f + l = 1 o f p f l l = 1 o f ϱ f l s h + l = 1 o h p h l l = 1 o h ϱ h l ,
and the lemma follows.    □
From Lemmas 1–4, the following Algorithm 1 is proposed, to solve the problem  1 p h i , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) .
Theorem 1. 
The problem  1 p h i , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] )  can be solved by Algorithm 1 in  O ( ζ log ζ )  time.
Algorithm 1 Constant processing times
Step 1. For each group  g h  ( h = 1 , 2 , , ψ ), by Lemma 3 arrange the jobs by the SPT rule of  p h i , to find the internal job schedule  π h * .
Step 2. By Lemma 4, the optimal group schedule  π *  is obtained by the non-descending order of  s h + l = 1 o h p h l l = 1 o h ϱ h l .
Step 3. By Lemma 2, calculate  d h i .
Proof. 
For each group  g h , Step 1 needs  O ( o h log o h )  time; hence, the total time of Step 1 is  O ( h = 1 ψ o h log o h ) O ( ζ log ζ ) ; Step 2 needs  O ( ψ log ψ ) O ( ζ log ζ )  time; Step 3 needs  O ( ζ )  time. Hence, the complexity of Algorithm 1 is  O ( ζ log ζ ) .    □

4. Linear Resource Problem

From Section 3, if  p h i = a h i b h i u h i  the objective cost is
O C ˜ = h = 1 ψ i = 1 o [ h ] ϱ [ h ] i C [ h ] [ i ] + h = 1 ψ i = 1 o [ h ] v [ h ] [ i ] u [ h ] [ i ] = h = 1 ψ i = 1 o [ h ] ϱ [ h ] i l = 1 h s [ l ] + f = 1 h 1 l = 1 o [ f ] p [ f ] [ l ] + l = 1 i p [ h ] [ l ] + h = 1 ψ i = 1 o [ h ] v [ h ] [ i ] u [ h ] [ i ] = h = 1 ψ i = 1 o [ h ] l = i o [ h ] ϱ [ h ] l + l = h + 1 ψ B [ l ] a [ h ] [ i ] b [ h ] [ i ] u [ h ] [ i ] + h = 1 ψ B [ h ] l = 1 h s [ l ] + h = 1 ψ i = 1 o [ h ] v [ h ] [ i ] u [ h ] [ i ] = h = 1 ψ i = 1 o [ h ] a [ h ] [ i ] A [ h ] i + h = 1 ψ B [ h ] l = 1 h s [ l ] + h = 1 ψ i = 1 o [ h ] v [ h ] [ i ] b [ h ] [ i ] A [ h ] i u [ h ] [ i ]
where  A [ h ] i = l = i o [ h ] ϱ [ h ] l + l = h + 1 ψ B [ l ]  and  B [ h ] = i = 1 o [ h ] ϱ [ h ] i .
Lemma 5. 
For the given group schedule and job schedule within each group of the problem  1 p h i = a h i b h i u h i , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i , the optimal resource allocation is
u [ h ] [ i ] * = u ¯ [ h ] [ i ] , if v [ h ] [ i ] b [ h ] [ i ] A [ h ] i < 0 , a n y v a l u e i n 0 , u ¯ [ h ] [ i ] , if v [ h ] [ i ] b [ h ] [ i ] A [ h ] i = 0 , 0 , if v [ h ] [ i ] b [ h ] [ i ] A [ h ] i > 0 .
Proof. 
In Equation (16), for the given group schedule and job schedule within each group, the terms  h = 1 ψ i = 1 o h a [ h ] [ i ] A [ h ] i  and  h = 1 ψ B [ h ] s [ h ]  are constant. Evidently, if  v [ h ] [ i ] b [ h ] [ i ] A [ h ] i < 0 , the optimal resource allocation should be its upper bound  u ¯ [ h ] [ i ] ; similarly, the results can be obtained.    □
To determine the optimal internal job schedule within each group, we only study a special case, i.e.,  ϱ h i = ϱ h  ( h = 1 , 2 , , ψ ; i = 1 , 2 , , o h ). For the given group schedule  π , we will prove that the optimal internal job schedule within each group can be determined by an assignment problem.
Lemma 6. 
For the problem  1 p h i = a h i b h i u h i , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i , if  ϱ h i = ϱ h  ( h = 1 , 2 , , ψ ; i = 1 , 2 , , o h ), given the group schedule π, the optimal job schedule  π h * ( π )  within group  g h ( π )  can be obtained by an assignment problem.
Proof. 
For a given  g [ h ] , let  x [ h ] i r = 1  if  j [ h ] i  is assigned to the rth position; otherwise,  x [ h ] i r = 0 i , r = 1 , 2 , , o [ h ] . When the group schedule  π  is given, if  ϱ h i = ϱ h  ( h = 1 , 2 , , ψ ; i = 1 , 2 , , o h ), from Equation (16) let
F [ h ] i r = v [ h ] i u ¯ [ h ] i + ( a [ h ] i b [ h ] i u ¯ [ h ] i ) ( o [ h ] r + 1 ) ϱ [ h ] + l = h + 1 ψ B [ l ] , if v [ h ] i b [ h ] i ( o [ h ] r + 1 ) ϱ [ h ] + l = h + 1 ψ B [ l ] 0 ; a [ i ] j ( o [ h ] r + 1 ) ϱ [ h ] + l = h + 1 ψ B [ l ] , if v [ h ] i b [ h ] i ( o [ h ] r + 1 ) ϱ [ h ] + l = h + 1 ψ B [ l ] > 0 .
According to above analysis, the optimal job schedule  π h * ( π )  is obtained by the following assignment problem:
Minimize i = 1 o [ h ] r = 1 o [ h ] F [ h ] i r x [ h ] i r
s . t .
i = 1 o [ h ] x [ h ] i r = 1 , r = 1 , 2 , , o [ h ] ,
r = 1 o [ h ] x [ h ] i r = 1 , i = 1 , , [ i ] ,
x j h [ i ] { 0 , 1 } , j , h = 1 , 2 , , o [ h ] .
   □
To determine the optimal group schedule, we study a special case, i.e.,  o h = o ¯  ( ψ o ¯ = ζ ) and  ϱ h i = ϱ ¯  ( h = 1 , 2 , , ψ ; i = 1 , 2 , , o ¯ ).
Lemma 7. 
For the problem  1 p h i = a h i b h i u h i , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i , if  o h = o ¯  and  ϱ h i = ϱ ¯  ( h = 1 , 2 , , ψ ; i = 1 , 2 , , o h , the optimal group schedule  π *  is obtained by an assignment problem.
Proof. 
If  o h = o ¯  and  ϱ h i = ϱ ¯  ( h = 1 , 2 , , ψ ; i = 1 , 2 , , o h ), we have  B h = o ¯ ϱ ¯ . From Equation (16), objective function (3) is determined by the resource allocation, the group schedule, and the job schedule. The optimal resource allocation is obtained by Lemma 5. The optimal job schedule  π h *  is obtained by Lemma 6. Let  x h r  be a 0–1 binary variable. If group  g h  is assigned to the rth position,  x h r = 1 ; otherwise,  x h r = 0 r , h = 1 , 2 , , ψ . From Equation (16), if group  g h  is assigned to the rth position, we have  i = 1 o [ r ] a [ r ] [ i ] A [ r ] i = i = 1 o ¯ a [ r ] [ i ] l = i o ¯ ϱ [ r ] l + l = r + 1 ψ B [ l ] = i = 1 o ¯ a h [ i ] ( o ¯ i + 1 ) ϱ ¯ + ( ψ r ) o ¯ ϱ ¯ h = 1 ψ B [ h ] l = 1 h s [ l ] = r = 1 ψ o ¯ ϱ ¯ ( ψ r + 1 ) s [ r ] .
Let
G h r = i = 1 o ¯ a h [ i ] ( o ¯ i + 1 ) ϱ ¯ + ( ψ r ) o ¯ ϱ ¯ + o ¯ ϱ ¯ ( ψ r + 1 ) s h + i = 1 o ¯ v h [ i ] b h [ i ] ( o ¯ i + 1 ) ϱ ¯ + ( ψ r ) o ¯ ϱ ¯ u ¯ h [ i ] , if v h [ i ] b h [ i ] ( o ¯ i + 1 ) ϱ ¯ + ( ψ r ) o ¯ ϱ ¯ 0 ; i = 1 o ¯ a h [ i ] ( o ¯ i + 1 ) ϱ ¯ + ( ψ r ) o ¯ ϱ ¯ + o ¯ ϱ ¯ ( ψ r + 1 ) s h , if v h [ i ] b h [ i ] ( o ¯ i + 1 ) ϱ ¯ + ( ψ r ) o ¯ ϱ ¯ > 0 .
Similarly, the optimal group schedule  π *  can be obtained by the following assignment problem:
Minimize h = 1 ψ r = 1 ψ G h r x h r
s . t .
h = 1 ψ x h r = 1 , r = 1 , 2 , , ψ ,
r = 1 ψ x h r = 1 , h = 1 , 2 , , ψ ,
x h r { 0 , 1 } , i , r = 1 , 2 , , ψ .
   □
Similarly, if  o h = o ¯  and  ϱ h i = ϱ ¯  ( h = 1 , 2 , , ψ ; i = 1 , 2 , , o ¯ ), the following algorithm is given, to solve
1 p h i = a h i b h i u h i , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i .
Theorem 2. 
If  o h = o ¯  and  ϱ h i = ϱ ¯  ( h = 1 , 2 , , ψ ; i = 1 , 2 , , o ¯ ), Algorithm 2 solves
1 p h i = a h i b h i u h i , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i
in  O ( ζ 3 )  time.
Algorithm 2 Linear resource problem
Step 1. According to Lemma 6, solve the assignment problem (19)–(22) to obtain an optimal job schedule  π h *  within each group  g h .
Step 2. According to Lemma 7, solve the assignment problem (24)–(27) to obtain an optimal group schedule  π * .
Step 3. Calculate the optimal resource allocation by Lemma 5.
Step 4. By Lemma 2, calculate  d h i  ( h = 1 , 2 , , ψ ; i = 1 , 2 , , o ¯ ).
Proof. 
Similarly to Theorem 1, Steps 1 and 2, respectively, need  O ( ζ 3 )  time; hence, the complexity of Algorithm 2 is  O ( ζ 3 ) .    □

5. Convex Resource Problem

Similarly, if  p h i = a h i + ϑ h i u h i κ , the objective cost is
O C ˜ = h = 1 ψ i = 1 o [ h ] ϱ [ h ] i C [ h ] [ i ] + h = 1 ψ i = 1 o [ h ] v [ h ] [ i ] u [ h ] [ i ] = h = 1 ψ i = 1 o [ h ] l = i o [ h ] ϱ [ h ] l + l = h + 1 ψ B [ l ] a [ h ] [ i ] + ϑ [ h ] [ i ] u [ h ] [ i ] κ + h = 1 ψ B [ h ] l = 1 h s [ l ] + h = 1 ψ i = 1 o [ h ] v [ h ] [ i ] u [ h ] [ i ] ,
where  A [ h ] i = l = i o [ h ] ϱ [ h ] l + l = h + 1 ψ B [ l ]  and  B [ h ] = i = 1 o [ h ] ϱ [ h ] i .
Lemma 8. 
Under the given group schedule and job schedule in each group, the optimal resource allocation of
1 p h i = a h i + ϑ h i u h i κ , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i
is
u [ h ] [ i ] * = κ l = i o [ h ] ϱ [ h ] l + l = h + 1 ψ B [ l ] v [ h ] [ i ] 1 κ + 1 × ϑ [ h ] [ i ] κ κ + 1 .
Proof. 
From Equation (28),  O C ˜  is a convex function of  u [ h ] [ i ] ; hence, let
O C ˜ u [ h ] [ i ] = v [ h ] [ i ] κ l = i o [ h ] ϱ [ h ] l + l = h + 1 ψ B [ l ] ϑ [ h ] [ i ] κ u [ h ] [ i ] κ + 1 = 0 ,
and we have  u [ h ] [ i ] * = κ l = i o [ h ] ϱ [ h ] l + l = h + 1 ψ B [ l ] v [ h ] [ i ] 1 κ + 1 × ϑ [ h ] [ i ] κ κ + 1 .     □
By substituting Equation (29) into Equation (28), we have
O C ˜ ( u [ h ] [ i ] * ) = h = 1 ψ i = 1 o [ h ] a [ h ] [ i ] l = i o [ h ] ϱ [ h ] l + l = h + 1 ψ B [ l ] + κ κ κ + 1 + κ 1 κ + 1 h = 1 ψ i = 1 o [ h ] l = i o [ h ] ϱ [ h ] l + l = h + 1 ψ B [ l ] 1 κ + 1 v [ h ] [ i ] ϑ [ h ] [ i ] κ κ + 1 + h = 1 ψ B [ h ] l = 1 h s [ l ] .
To find the optimal job schedule within each group, we consider a special case, i.e., for each group  g h a h i a h j v h i ϑ h i v h j ϑ h j .
Lemma 9. 
Given the group schedule π, for each group  g h , if  a h i a h j v h i ϑ h i v h j ϑ h j  the optimal job schedule  π h * ( π )  within group  g h ( π )  can be obtained by the non-decreasing order of  a h i  (or  v h i ϑ h i ).
Proof. 
Given the group schedule  π , from Equation (30 h = 1 ψ B [ h ] l = 1 h s [ l ]  and  l = h + 1 ψ B [ l ]  are constants. Similar to Lemma 3, for each group  g h l = i o h ϱ h l  is monotonically decreasing in i if  a h i a h j v h i ϑ h i v h j ϑ h j . By the HLP rule, the term  i = 1 o [ h ] a [ h ] [ i ] l = i o [ h ] ϱ [ h ] l + l = h + 1 ψ B [ l ]  is minimized by the non-decreasing order of  a h i , the term  i = 1 o [ h ] l = i o [ h ] ϱ [ h ] l + l = h + 1 ψ B [ l ] 1 κ + 1 v [ h ] [ i ] ϑ [ h ] [ i ] κ κ + 1  is minimized by the non-decreasing order of  v h i ψ h i , and the lemma follows.    □
To find the optimal group schedule, we study a special case, i.e.,  o h = o ¯  ( ψ o ¯ = ζ ) and  ϱ h i = ϱ ¯  ( h = 1 , 2 , , ψ ; i = 1 , 2 , , o ¯ ).
Lemma 10. 
For the problem
1 p h i = a h i + ϑ h i u h i κ , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i ,
if  o h = o ¯  and  ϱ h i = ϱ ¯  ( h = 1 , 2 , , ψ ; i = 1 , 2 , , o h , the optimal group schedule  π *  is obtained by an assignment problem.
Proof. 
Similar to Lemma 7, from Equation (28), if group  g h  is assigned to the rth position, we have
i = 1 o [ h ] a [ h ] [ i ] l = i o [ h ] ϱ [ h ] l + l = h + 1 ψ B [ l ] = i = 1 o ¯ a h [ i ] ( o ¯ i + 1 ) ϱ ¯ + ( ψ r ) o ¯ ϱ ¯ ,
i = 1 o [ h ] l = i o [ h ] ϱ [ h ] l + l = h + 1 ψ B [ l ] 1 κ + 1 v [ h ] [ i ] ϑ [ h ] [ i ] κ κ + 1 = i = 1 o ¯ v h [ i ] ϑ h [ i ] κ κ + 1 ( o ¯ i + 1 ) ϱ ¯ + ( ψ r ) o ¯ ϱ ¯ 1 κ + 1 ,
and
h = 1 ψ B [ h ] l = 1 h s [ l ] = r = 1 ψ o ¯ ϱ ¯ ( ψ r + 1 ) s [ r ] .
Let
G h r = + o ¯ ϱ ¯ ( ψ r + 1 ) s h + i = 1 o ¯ a h [ i ] ( o ¯ i + 1 ) ϱ ¯ + ( ψ r ) o ¯ ϱ ¯ + κ κ κ + 1 + κ 1 κ + 1 i = 1 o ¯ v h [ i ] ϑ h [ i ] κ κ + 1 ( o ¯ i + 1 ) ϱ ¯ + ( ψ r ) o ¯ ϱ ¯ 1 κ + 1 .
The optimal group schedule  π *  is obtained by an assignment problem (24)–(27), where  G h r  is given by Equation (31).    □
Similarly, if  o h = o ¯ ϱ h i = ϱ ¯  and for each group  g h a h i a h j v h i ϑ h i v h j ϑ h j , the following algorithm is given to solve
1 p h i = a h i + ϑ h i u h i κ , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i .
Theorem 3. 
If  o h = o ¯ ϱ h i = ϱ ¯  and for each group  g h a h i a h j v h i ϑ h i v h j ϑ h j , Algorithm 3 solves
1 p h i = a h i + ϑ h i u h i κ , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i
in  O ( ζ 3 )  time.
Algorithm 3 Convex resource problem
Step 1. According to Lemma 9, an optimal job schedule  π h *  within each group  g h  is obtained.
Step 2. According to Lemma 10, solve the assignment problem (24)–(27) to obtain an optimal group schedule  π * , where  G h r  is given by Equation (31).
Step 3. Calculate the optimal resource allocation by Lemma 8.
Step 4. By Lemma 2, calculate  d h i  ( h = 1 , 2 , , ψ ; i = 1 , 2 , , o ¯ ).
Proof. 
Similarly to Theorem 2. □

6. Examples

Example 1. 
For the problem with constant processing times, we assume that  ζ = 12 ψ = 3 s 1 = 3 , s 2 = 5 , s 3 = 9 ; the job parameters are given in Table 2, the position-dependent weights (i.e.,  α h i β h i , and  γ h i ) are given in Table 3.
For  1 p h i , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) , from Algorithm 1 the optimal job schedule of group  g 1  is  π 1 * : { j 1 3 j 1 2 j 1 1 } ; similarly,  π 2 * : { j 2 2 j 2 4 j 2 5 j 2 1 j 2 3 } π 3 * : { j 3 4 j 3 1 j 3 2 J 3 3 } . In addition, for group  g 1 s 1   +   l = 1 o 1 p 1 l l = 1 o 1 ϱ 1 l = 3   +   16   +   13   +   3 1   +   4   +   12 = 35 17 , for group  g 2 s 2   +   l = 1 o 2 p 2 l l = 1 o 2 ϱ 2 l = 5   +   16   +   7   +   18   +   12   +   15 3   +   1   +   15   +   4   +   16 = 73 39 , and for group  g 3 s 3   +   l = 1 o 3 p 3 l l = 1 o 3 ϱ 3 l = 9   +   9   +   15   +   20   +   6 17   +   7   +   2   +   12 = 59 38 ; hence, by the non-descending order of  s h   +   l = 1 o h p h l l = 1 o h ϱ h l  the optimal group schedule is  π * : { g 3 g 2 g 1 } . For the schedule  π * : { j 3 4 j 3 1 j 3 2 J 3 3 j 2 2 j 2 4 j 2 5 j 2 1 j 2 3 j 1 3 j 1 2 j 1 1 } , the completion times are  C 3 4 = 9   +   6   =   15 C 3 1   =   15   +   9   =   24 C 3 2   =   24   +   15   =   39 C 3 3   =   39   +   20   =   59 C 2 2   =   59   +   5   +   7   =   71 C 2 4   =   71   +   12   =   83 C 2 5   =   83   +   15   =   98 C 2 1   =   98   +   16   =   114 ,   C 2 3   =   114   +   18   =   132 C 1 3   =   132   +   3   +   3   =   138 C 1 2   =   138   +   13   =   151 C 1 1   =   151   +   16   =   167 . By Lemma 2, for  γ 3 1 = 18 > β 3 1   =   17  (position-dependent weight); hence,  d 3 4 = 0 . Similarly, the due dates are  d 3 1 = 24 d 3 2 = 39 d 3 3 = 0 d 2 2 = 0 d 2 4 = 0 d 2 5 = 0 d 2 1 = 114 d 2 3 = 0 d 1 3 = 138 ,   d 1 2 = 151 ,   d 1 1 = 0  and the objective cost is  h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) = 8289 . The flowchart of Example 1 can be found in Figure 1.
Example 2. 
For the linear resource problem, we assume that  ζ = 12 ψ = 3 o h = o ¯ = 4 s 1 = 3 , s 2 = 5 , s 3 = 9 ; the job parameters are given in Table 4. The position-dependent weights (i.e.,  α h i β h i , and  γ h i ) are given in Table 5 ( ϱ h i = ϱ ¯ = 1 ).
For  1 p h i = a h i b h i u h i , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) , from Algorithm 2, if group  g 1  is scheduled in 1st position, the values of  F [ h ] i r  are given in Table 6. From the assignment problem (19)–(22) the optimal job schedule of group  g 1  is  π 1 1 * : { j 1 2 j 1 1 j 1 3 j 1 4 } . Similarly, if group  g 1  is scheduled in 2nd position, the optimal job schedule of group  g 1  is  π 1 2 * : { j 1 2 j 1 1 j 1 3 j 1 4 } . If group  g 1  is scheduled in 3rd position, the optimal job schedule of group  g 1  is  π 1 3 * : { j 1 2 j 1 3 j 1 1 j 1 4 } .
Similarly, if group  g 2  is scheduled in 1st position, the optimal job schedule of group  g 2  is  π 2 1 * : { j 2 1 j 2 2 j 2 3 j 2 4 } . If group  g 2  is scheduled in 2nd position, the optimal job schedule of group  g 2  is  π 2 2 * : { j 2 1 j 2 2 j 2 3 j 2 4 } . If group  g 2  is scheduled in 3rd position, the optimal job schedule of group  g 2  is  π 2 3 * : { j 2 1 j 2 2 j 2 4 j 2 3 } .
Similarly, if group  g 3  is scheduled in 1st position, the optimal job schedule of group  g 3  is  π 3 1 * : { j 3 2 j 3 1 j 3 4 j 3 3 } . If group  g 3  is scheduled in 2nd position, the optimal job schedule of group  g 3  is  π 3 2 * : { j 3 3 j 3 2 j 3 1 j 3 4 } . If group  g 3  is scheduled in 3rd position, the optimal job schedule of group  g 3  is  π 3 3 * : { j 3 2 j 3 4 j 3 1 j 3 3 } .
Stemming from Lemma 7, the values of  G h r  are given in Table 7. From Table 7, we have  π * : { g 1 g 2 g 3 } .
According to Lemma 6, for the schedule  π * : { j 1 2 j 1 1 j 1 3 j 1 4 j 2 1 j 2 2 j 2 3 j 2 4 j 3 2 j 3 4 j 3 1 J 3 3 }  the optimal resource allocations are  u 1 2 * = u ¯ 1 2 = 4 ,   u 1 1 * = 7 ,   u 1 3 * = 5 ,   u 1 4 * = 10 u 2 1 * = 4 ,   u 2 2 * = 3 ,   u 2 3 * = 4 ,   u 2 4 * = 2 u 3 2 * = 2 ,   u 3 4 * = 3 ,   u 3 1 * = 0 u 3 3 * = 0 . The actual processing times are  p 1 2 = a 1 2 b 1 2 u ¯ 1 2 = 13 3 × 4 = 1 ,   p 1 1 = 2 ,   p 1 3 = 4 ,   p 1 4 = 6 p 2 1 = 1 ,   p 2 2 = 3 ,   p 2 3 = 4 ,   p 2 4 = 5 p 3 2 = 5 ,   p 3 4 = 3 ,   p 3 1 = 19 ,   p 3 3 = 20 . The completion times are  C 1 2 = 3 + 1 = 4 ,   C 1 1 = 4 + 2 = 6 ,   C 1 3 = 6 + 4 = 10 ,   C 1 4 = 10 + 6 = 16 C 2 1 = 16 + 5 + 1 = 22 ,   C 2 2 = 22 + 3 = 25 ,   C 2 3 = 25 + 4 = 29 ,   C 2 4 = 29 + 5 = 34 C 3 2 = 34 + 9 + 5 = 48 ,   C 3 4 = 48 + 3 = 51 ,   C 3 1 = 51 + 19 = 70 ,   C 3 3 = 70 + 20 = 90 . Similarly to Example 1, the optimal due dates are  d 1 2 = C 1 2 = 4 ,   d 1 1 = 0 ,   d 1 3 = 0 ,   d 1 4 = 16 d 2 1 = 0 ,   d 2 2 = 0 ,   d 2 3 = 0 ,   d 2 4 = 34 d 3 2 = 48 ,   d 3 4 = 0 ,   d 3 1 = 70 ,   d 3 3 = 0  and the objective cost is  h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i = 609 .
Example 3. 
For the convex resource problem, we assume that  ζ = 12 ψ = 3 o h = o ¯ = 4 κ = 1 s 1 = 3 ,   s 2 = 5 ,   s 3 = 9 ; the job parameters are given in Table 8 and the position-dependent weights are the same as Table 5.
For  1 p h i = a h i + ϑ h i u h i κ , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i , from Lemma 9 the optimal job schedule of group  g 1  is  π 1 * : { j 1 2 j 1 3 j 1 1 j 1 4 } ; similarly,  π 2 * : { j 2 3 j 2 4 j 2 1 j 2 2 } π 3 * : { j 3 2 j 3 1 j 3 3 J 3 4 } . Stemming from Lemma 10, the values of  G h r  are given in Table 9. From Table 9, we have  π * : { g 1 g 2 g 3 } .
According to Lemma 8, for the schedule  π * : { j 1 2 j 1 3 j 1 1 j 1 4 j 2 3 j 2 4 j 2 1 j 2 2 j 3 2 j 3 1 j 3 3 J 3 4 }  the optimal resource allocations are  u 1 2 * = 12 2 1 2 × 13 1 2 = 8.831761 ,   u 1 3 * = 5.744563 ,   u 1 1 * = 5.477226 ,   u 1 4 * = 4.242641 u 2 3 * = 1.414214 ,   u 2 4 * = 2.645751 ,   u 2 1 * = 1.851640 ,   u 2 2 * = 3.872983 u 3 2 * = 2.000000 ,   u 3 1 * = 1.000000 ,   u 3 3 * = 2.000000 u 3 4 * = 0.7905694 . The actual processing times are  p 1 2 = a 1 2 + ϑ 1 2 u 1 2 κ = 13 + 13 8.831761 = 14.47196 ,   p 1 3 = 15.56670 ,   p 1 1 = 18.19089 ,   p 1 4 = 22.35702 p 2 3 = 13.41421 ,   p 2 4 = 16.88982 ,   p 2 1 = 19.16025 ,   p 2 2 = 21.87298 p 3 2 = 17.50000 ,   p 3 1 = 22.00000 ,   p 3 3 = 24.00000 ,   p 3 4 = 27.32456 . The completion times are  C 1 2 = 3 + 14.47196 = 17.47196 C 1 3 = 17.47196 + 15.56670 = 33.03866 C 1 1 = 33.03866 + 18.19089 = 51.22955 ,   C 1 4 = 51.22955 + 22.35702 = 73.58657 C 2 3 = 73.58657 + 5 + 13.41421 = 92.00078 C 2 4 = 92.00078 + 16.88982 = 108.8906 C 2 1 = 108.8906 + 19.16025 = 128.0508 C 2 2 = 128.0508 + 21.87298 = 149.9238 C 3 2 = 149.9238 + 9 + 17.50000 = 176.4238 C 3 1 = 176.4238 + 22.00000 = 198.4238 C 3 3 = 198.4238 + 24.00000 = 222.4238 C 3 4 = 222.4238 + 27.32456 = 249.7484 . The optimal due dates are  d 1 2 = C 1 2 = 17.47196 d 1 3 = 0 d 1 1 = 0 d 1 4 = 73.58657 d 2 3 = 0 d 2 4 = 0 d 2 1 = 0 d 2 2 = 149.9238 d 3 2 = 176.4238 d 3 1 = 0 d 3 3 = 222.4238 d 3 4 = 0  and the objective cost is  h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i = 1669.425 .
Remark 1. 
The observations of three examples were compared in the following table (i.e., Table 10).

7. Conclusions

Single-machine  G T ^  scheduling with resource allocation was studied. Under  Θ D I F  assignment, we showed that the problem with constant processing times is polynomially solvable. For linear and convex resource functions, we also proved that a special case of the problem is polynomially solvable.
Future work could consider  G T ^  problems with resource allocation and a maintenance activity (Atsmony et al. [34] and Phosavanh and Oron [35]), study the general version of  1 p h i , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i , where  p h i a h i b h i u h i , a h i + ϑ h i u h i κ , take into account  G T ^  problems with resource allocation and deteriorating jobs (Lv et al. [36] and Zhang et al. [37]), address the potential impact of uncertainties or disruptions in the scheduling process, which may be relevant in real-world manufacturing scenarios, or investigate how our model is applied to real data from the company.

Author Contributions

Methodology, X.W.; writing, review, and editing, X.W. and W.L. All authors have read and agreed to the published version of the manuscript.

Funding

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

Data Availability Statement

Data are contained within the article.

Acknowledgments

We are grateful to the three anonymous referees for their helpful comments on an earlier version of this article.

Conflicts of Interest

The authors declare there are no conflicts of interest.

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Figure 1. Flowchart of Example 1.
Figure 1. Flowchart of Example 1.
Mathematics 12 00436 g001
Table 1. Summary results of proportional job deterioration.
Table 1. Summary results of proportional job deterioration.
ProblemComplexityReference
  1 c o n s t a n t , Θ C O N , G T ^ h = 1 ψ i = 1 o h ( α E h i + β T h i + γ d h ) + θ C max + h = 1 ψ i = 1 o h v h i u h i   O ( ζ log ζ ) Shabtay et al. [22]
  1 p h i = a h i b h i u h i , Θ C O N , G T ^ , o h = o ¯ h = 1 ψ i = 1 o h ( α E h i + β T h i + γ d h ) + θ C max + h = 1 ψ i = 1 o h v h i u h i   O ( ζ 3 ) Shabtay et al. [22]
  1 p h i = ϑ h i u h i κ , Θ C O N , G T ^ , o h = o ¯ h = 1 ψ i = 1 o h ( α E h i + β T h i + γ d h ) + θ C max + h = 1 ψ i = 1 o h v h i u h i   O ( max { ζ log ζ , ψ ζ , ψ 3 } ) Shabtay et al. [22]
  1 p h i = a h i b h i u h i , Θ D I F , G T ^ , o h = o ¯ , ϱ ˜ h i = ϱ ¯ h = 1 ψ i = 1 o h ( α ˜ h i E h i + β ˜ h i T h i + γ ˜ h i d h i ) + h = 1 ψ i = 1 o h v h i u h i   O ( ζ 3 ) Chen et al. [26]
  1 p h i = ϑ h i u h i κ , Θ D I F , G T ^ , o h = o ¯ , ϱ ˜ h i = ϱ ¯ h = 1 ψ i = 1 o h ( α ˜ h i E h i + β ˜ h i T h i + γ ˜ h i d h i ) + h = 1 ψ i = 1 o h v h i u h i   O ( ζ log ζ ) Chen et al. [26]
  1 p h i = a h i b h i u h i , Θ C O N , G T ^ , o h = o ¯ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ d h ) + h = 1 ψ i = 1 o h v h i u h i   O ( ζ 3 ) Liu and Wang [27]
  1 p h i = a h i b h i u h i , Θ S L K , G T ^ , o h = o ¯ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ q h ) + h = 1 ψ i = 1 o h v h i u h i   O ( ζ 3 ) Liu and Wang [27]
  1 p h i = ϑ h i u h i κ , Θ C O N , G T ^ , o h = o ¯ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ d h ) + h = 1 ψ i = 1 o h v h i u h i   O ( ζ log ζ ) Liu and Wang [27]
  1 p h i = ϑ h i u h i κ , Θ S L K , G T ^ , o h = o ¯ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ q h ) + h = 1 ψ i = 1 o h v h i u h i   O ( ζ log ζ ) Liu and Wang [27]
  1 c o n s t a n t , Θ D I F , G T ^ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i   O ( ζ log ζ ) Theorem 1, this paper
  1 p h i = a h i b h i u h i , Θ D I F , G T ^ , o h = o ¯ , ϱ h i = ϱ ¯ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i   O ( ζ 3 ) Theorem 2, this paper
  1 p h i = a h i + ϑ h i u h i κ , Θ D I F , G T ^ , c o n s i s t e n c y , o h = o ¯ , ϱ h i = ϱ ¯ h = 1 ψ i = 1 o h ( α h i E h [ i ] + β h i T h [ i ] + γ h i d h [ i ] ) + h = 1 ψ i = 1 o h v h i u h i   O ( ζ log ζ ) Theorem 3, this paper
Given integers— α , β , γ , and  θ . The constant processing times are denoted by  c o n s t a n t . The common due date, i.e.,  d h i = d h  is denoted by  Θ C O N . The slack due date, i.e.,  d h i = d h i + q h  is denoted by  Θ S L K . The term  c o n s i s t e n c y  denotes  a h i a h j v h i ϑ h i v h j ϑ h j . The job-dependent weights of job  j h i  are  α ˜ h i β ˜ h i , and  γ ˜ h i , and  ϱ ˜ h i = min { β ˜ h i , γ ˜ h i } C max  denotes the makespan.
Table 2. Job parameters of constant processing times.
Table 2. Job parameters of constant processing times.
  g h   g 1   g 2   g 3
  j h i   j 1 1   j 1 2   j 1 3   j 2 1   j 2 2   j 2 3   j 2 4   j 2 5   j 3 1   j 3 2   j 3 3   j 3 4
  p h i 16133167181215915206
Table 3. Position-dependent weights of Example 1.
Table 3. Position-dependent weights of Example 1.
  g h   g 1   g 2   g 3
  α h i 4861010910139141611
  β h i 111412311514161792012
  γ h i 14206521419187217
  ϱ h i = min { β h i , γ h i } 14123115416177212
Table 4. Job parameters of Example 2.
Table 4. Job parameters of Example 2.
  g h   g 1   g 2   g 3
  j h i   j 1 1   j 1 2   j 1 3   j 1 4   j 2 1   j 2 2   j 2 3   j 2 4   j 3 1   j 3 2   j 3 3   j 3 4
  a h i 161314161718121519152021
  b h i 232145253526
  u ¯ h i 7451043426273
  v h i 572345857546
Table 5. Position-dependent weights of Example 2.
Table 5. Position-dependent weights of Example 2.
  g h   g 1   g 2   g 3
  α h i 4861291039141211
  β h i 1111311116171201
  γ h i 142015214117117
  ϱ h i = min { β h i , γ h i } 111111111111
Table 6. Values of  F [ h ] i r  if  g 1  is scheduled in 1st position.
Table 6. Values of  F [ h ] i r  if  g 1  is scheduled in 1st position.
  r = 1   r = 2   r = 3   r = 4
  j 1 1 59575553
  j 1 2 40393837
  j 1 3 58545046
  j 1 4 102969075
Table 7. G h r  values of Example 2.
Table 7. G h r  values of Example 2.
  r = 1   r = 2   r = 3
  g 1 267203106
  g 2 26319196
  g 3 361274151
Table 8. Job parameters of Example 3.
Table 8. Job parameters of Example 3.
  g h   g 1   g 2   g 3
  j h i   j 1 1   j 1 2   j 1 3   j 1 4   j 2 1   j 2 2   j 2 3   j 2 4   j 3 1   j 3 2   j 3 3   j 3 4
  a h i 161314201718121519152021
  ϑ h i 1213910415253585
  v h i 423575859548
Table 9. G h r  values of Example 3.
Table 9. G h r  values of Example 3.
  r = 1   r = 2   r = 3
  g 1 842.0386543.9038230.1341
  g 2 847.3070546.7377230.6076
  g 3 1028.833661.7768280.6491
Table 10. Results of Examples 1, 2, and 3.
Table 10. Results of Examples 1, 2, and 3.
ExampleOptimal Schedule
Example 1   π * : { j 3 4 j 3 1 j 3 2 J 3 3 j 2 2 j 2 4 j 2 5 j 2 1 j 2 3 j 1 3 j 1 2 j 1 1 }
Example 2   π * : { j 1 2 j 1 1 j 1 3 j 1 4 j 2 1 j 2 2 j 2 3 j 2 4 j 3 2 j 3 4 j 3 1 J 3 3 }
Examples 3   π * : { j 1 2 j 1 3 j 1 1 j 1 4 j 2 3 j 2 4 j 2 1 j 2 2 j 3 2 j 3 1 j 3 3 J 3 4 }
Exampleoptimal resource
Example 1not have
Example 2 u 1 2 * = 4 ,   u 1 1 * = 7 ,   u 1 3 * = 5 ,   u 1 4 * = 10 ;
u 2 1 * = 4 ,   u 2 2 * = 3 ,   u 2 3 * = 4 ,   u 2 4 * = 2 ;
u 3 2 * = 2 ,   u 3 4 * = 3 ,   u 3 1 * = 0 u 3 3 * = 0
Examples 3 u 1 2 * = 8.831761 ,   u 1 3 * = 5.744563 ,   u 1 1 * = 5.477226 ,   u 1 4 * = 4.242641 ;
u 2 3 * = 1.414214 ,   u 2 4 * = 2.645751 ,   u 2 1 * = 1.851640 ,   u 2 2 * = 3.872983 ;
u 3 2 * = 2.000000 ,   u 3 1 * = 1.000000 ,   u 3 3 * = 2.000000 u 3 4 * = 0.7905694
Exampleoptimal due date
Example 1 d 3 1 = 24 ,   d 3 2 = 39 ,   d 3 3 = 0 d 2 2 = 0 ,   d 2 4 = 0 ,   d 2 5 = 0 ,   d 2 1 = 114 ,   d 2 3 = 0 ;
  d 1 3 = 138 ,   d 1 2 = 151 ,   d 1 1 = 0
Example 2 d 1 2 = C 1 2 = 4 ,   d 1 1 = 0 ,   d 1 3 = 0 ,   d 1 4 = 16 d 2 1 = 0 ,   d 2 2 = 0 ,   d 2 3 = 0 ,   d 2 4 = 34 ;
  d 3 2 = 48 ,   d 3 4 = 0 ,   d 3 1 = 70 ,   d 3 3 = 0
Examples 3 d 1 2 = 17.47196 ,   d 1 3 = 0 ,   d 1 1 = 0 ,   d 1 4 = 73.58657 ;
d 2 3 = 0 ,   d 2 4 = 0 ,   d 2 1 = 0 ,   d 2 2 = 149.9238 ;
  d 3 2 = 176.4238 ,   d 3 1 = 0 ,   d 3 3 = 222.4238 ,   d 3 4 = 0
Exampleoptimal objective cost
Example 18289
Example 2609
Examples 31669.425
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Wang, X.; Liu, W. Optimal Different Due-Date Assignment Scheduling with Group Technology and Resource Allocation. Mathematics 2024, 12, 436. https://doi.org/10.3390/math12030436

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Wang X, Liu W. Optimal Different Due-Date Assignment Scheduling with Group Technology and Resource Allocation. Mathematics. 2024; 12(3):436. https://doi.org/10.3390/math12030436

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Wang, Xuyin, and Weiguo Liu. 2024. "Optimal Different Due-Date Assignment Scheduling with Group Technology and Resource Allocation" Mathematics 12, no. 3: 436. https://doi.org/10.3390/math12030436

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