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Article

Single-Machine Maintenance Activity Scheduling with Convex Resource Constraints and Learning Effects

1
School of Science, Shenyang Aerospace University, Shenyang 110136, China
2
College of Electronic and Information Engineering, Shenyang Aerospace University, Shenyang 110136, China
*
Author to whom correspondence should be addressed.
Mathematics 2023, 11(16), 3536; https://doi.org/10.3390/math11163536
Submission received: 1 July 2023 / Revised: 7 August 2023 / Accepted: 11 August 2023 / Published: 16 August 2023
(This article belongs to the Special Issue Systems Engineering, Control, and Automation)

Abstract

:
In this paper, the single-machine scheduling problems under the common and slack due date assignments are studied, where the actual processing time of the job needs to consider some factors, such as convex resource allocation, maintenance activity, and learning effects. The goal of this study is to find the optimal sequence, maintenance activity location, resource allocation and common due date (flow allowance). The objective function is (1) to minimize the sum of scheduling cost (including the weighted sum of earliness, tardiness and common due date (flow allowance), where the weights are position-dependent weights) and resource consumption cost, and (2) to minimize the scheduling cost under the resource consumption cost which is bounded. We prove that these problems can be solved in polynomial time.

1. Introduction

In a realistic scheduling system, the job-processing time is often affected by many practical settings, such as learning effects, resource allocation, and maintenance activities. Learning effects appear such that, for example, workers continue to process the same jobs, experience increases, and the actual processing time of the work is gradually shortened (see Mosheiov [1]; Cheng et al. [2]; Wu et al. [3]; Azzouz et al. [4]; Sun et al. [5]; Zhao [6]; and Wang et al. [7]). The manager assigns a certain amount of additional resources to the job to reduce and control the processing time of the artifact. Common limited resources are the financial budget, energy, fuel, or manpower (see Vixkson [8]; Shabtay and Kasoi [9]; Wang and Cheng [10]; Shabtay and Steiner [11]; Zhang et al. [12]; Wang et al. [13]). In production, due to machine failure or the processing time being too long, wear will inevitably occur and reduce the working rate, so maintenance activities can be carried out to reduce the job processing time (see Lee and Leon [14]; Wang and Wang [15]; Mosheiov and Sidney [16]; Bai et al. [17]; Yin et al. [18]; and Strusevich and Rustogi [19]).
In order to strengthen global business competition and improve customer satisfaction, new production technologies such as “just-in-time” (JIT) production are adopted. The emergence of the concept of JIT production and the sequencing problems have attracted widespread attention; in a just-in-time system, jobs (work pieces) can neither be completed early nor late, otherwise penalty costs will be incurred (see Parwalker et al. [20], and Cheng et al. [21]). In addition, in actual production, we found that task-processing rates can be affected by several factors simultaneously. Therefore, in previous studies, Ji et al. [22] considered a single machine due date assignment scheduling problem with job-dependent aging effects and a deteriorating maintenance activity, where due dates are assigned using the SLK due date determination method. He et al. [23] studied the single-machine sequencing problem of resource allocation with general truncated learning effects. A convex resource allocation model under the condition of finite resource consumption cost is proposed, and various optimal algorithms are given for different cases of the problem. Liu and Jiang [24] delved into due date assignment scheduling problems with learning effects and resource allocation. Under common due date assignment and slack due date assignment rules, a bi-criteria analysis is provided. Zhao et al. [25] examined a single machine scheduling problem with slack due date assignment in which the actual processing time of a job is determined by its position in a sequence, its resource allocation function, and a rate-modifying activity simultaneously.
In previous studies, two different resource allocation functions were usually employed. One was a linear function setting for the amount of resources and actual processing time associated with each job (Janiak and Kovalyov [26]), and the other was a convex function setting for the amount of resources assigned to each job (Monma et al. [27]). In general, there are two ways to model learning effects: one is a location-dependent learning effect (Biskup [28]), and the other is a time-dependent (sum-of-processing time) learning effect (Azzouz et al. [4]). Wang et al. [29] considered location-dependent learning effects and convex resource allocation to build the model p j r A ( u j ) = θ j r α u j η , where η > 0 is a given constant, α 0 is the learning factor, θ j is the normal processing time of job J j , and  u j is the resource allocated to the job J j . Zhu et al. [30] addressed the job processing time considering the rate modification activity, learning effect and convex resource allocation, described as follows: when the rate modification activity is not carried out, the job-processing time is p j r A ( u j ) = θ j r α u j η ; otherwise, it is p j r A ( u j ) = a j θ j r α u j η , where a j is the modifying rate. For a comparison with other similar papers (see Table 1; the related symbols are given later), this article extends the results of Wang and Wang [15], Bai et al. [17], Ji et al. [22], Zhao et al. [25], Wang et al. [29], and Zhu et al. [30] by scrutinizing a more general scheduling model.
This paper’s contributions and novelties are as follows:
  • Single-machine maintenance scheduling with convex resource constraint and learning effect is modeled and studied;
  • Four algorithms are provided for the following two objective functions: (1) minimize the sum of scheduling cost (including the weighted sum of earliness, tardiness and common due date (flow allowance), where the weight is the position-dependent weight) and resource consumption cost; and (2) the resource consumption cost has an upper bound, minimizing the dispatch cost.
  • It is shown that these problems can be solved in polynomial time, and the effectiveness of the algorithms is presented by numerical study.
The paper is organized as follows. Section 2 introduces the model. Section 3 describes the optimal properties. Section 4 performs an optimal analysis of the objective function and proves that it can be solved in polynomial time. In Section 5, an example is calculated, and numerical experiments are carried out to verify the effectiveness of the algorithm. Section 6 concludes this paper.

2. Problem Description

In this article, the use of symbols is listed in Table 2, and the problem can be stated as follows: there are n independent and non-preemptive jobs J = J 1 , J 2 , , J n to be processed on a single machine. Each job J j is available for processing at time zero. The machine can handle one job at a time. In addition, for any sequence ξ = ( J [ 1 ] , J [ 2 ] , , J [ n ] ) , there is a maintenance activity whose duration time is t and no jobs are processed during this execution.
In this paper, the model considered is as follows
P j r A = ( θ j ( r ) α u j ) η , r l ( β j θ j ( r ) α u j ) η , r > l
where j = 1 , 2 , , n , η > 0 is a constant, α 0 is the non-negative learning index, β j is the modifying rate (it satisfies 0 < β j 1 ) and l denotes the position of the job preceding the maintenance activity (i.e., position l + 1 is the first position after the maintenance activity). In addition, in this article, we discuss two due date assignment models, including the common (CON) due date and slack (SLK) due date assignment. For the CON, d j = d o p t , where d o p t is a decision variable. For the SLK, d j = p j A + q o p t , where the common flow allowance q o p t is a decision variable. Given that the maintenance activity time is a fixed value, our goal is to determine the optimal sequence of jobs ξ * , optimal amount of resource allocation u * , optimal due date ( d o p t or q o p t ) and maintenance location l * . The first problem of this paper is to minimize
Z ( u , ξ , d , l ) = j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ d o p t ) + j = 1 n v j u j ,
using three-field representation, the first problem ( P 1 ) can be expressed as
1 | M A L E , C R E | j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ d o p t ) + j = 1 n v j u j ,
where M A L E means “a maintenance activity and learning effect”, and C R E means “convex resource allocation”.
The second problem is to minimize
Z ( u , ξ , q , l ) = j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ q o p t ) + j = 1 n v j u j ,
using three-field representation, and the second problem ( P 2 ) can be expressed as
1 | M A L E , C R E | j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ q o p t ) + j = 1 n v j u j .
The third problem is to minimize j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ d o p t ) under a common due date, and the cost of the resource consumption cannot exceed a ceiling, i.e.,  j = 1 n v j u j U , and this problem (denoted by P 3 ) is
1 | M A L E , C R E , j = 1 n v j u j U | j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ d o p t ) ,
where U > 0 is an upper bound on j = 1 n v j u j . The last problem is to consider the slack due date assignment, i.e., the fourth problem (denoted by P4) is
1 | M A L E , C R E , j = 1 n v j u j U | j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ q o p t ) .

3. Main Properties

In this section, we show some main properties of the problems. Based on the above notations and allowing to perform a maintenance activity in position l,  the maintenance time is a fixed constant t, and the completion time of each job j ( j = 1 , 2 , , n ) can be presented in the following:
                      C [ j ] = C [ j 1 ] + θ [ j ] ( j ) α u [ j ] , j = 1 , 2 , . . . , l ,
                      C [ j ] = C [ j 1 ] + t + β [ j ] θ [ j ] ( j ) α u [ j ] , j = l + 1 ,
                      C [ j ] = C [ j 1 ] + β [ j ] θ [ j ] ( j ) α u [ j ] , j = l + 2 , . . . , n .
Under the optimal schedule, the first job starts processing at time 0 and there is no idle time between jobs.
Lemma 1.
For any given sequence ξ = J [ 1 ] , J [ 2 ] , , J [ n ] , the optimal value of d o p t is determined by the completion time of the k-th job, i.e.,  d o p t = C [ k ] . The optimal value of the common flow allowance q o p t is decided by the start time of the k-th job, i.e.,  q o p t = S [ k ] .
Proof. 
Now, we prove d o p t = C [ k ] . Consider a schedule ξ = J [ 1 ] , J [ 2 ] , , J [ n ] , d o p t with C [ k ] < d o p t < C [ k + 1 ] and let Z be the corresponding objective value. Define x = d o p t C [ k ] and y = C [ k + 1 ] d o p t . Let Z and Z be the objective values for d o p t = C [ k ] and d o p t = C [ k + 1 ] . Then
Z = Z x j = 1 k + 1 δ j + w j + γ + x j = k + 2 n δ j + w j + γ
= Z + x [ j = k + 2 n δ j + w j + γ j = 1 k + 1 δ j + w j + γ ]
Z = Z + y j = 1 k + 1 δ j + w j + γ y j = k + 2 n δ j + w j + γ
= Z y [ j = k + 2 n δ j + w j + γ j = 1 k + 1 δ j + w j + γ ]
Thus, Z Z if j = k + 2 n δ j + w j + γ j = 1 k + 1 δ j + w j + γ and Z Z otherwise. This implies that an optimal solution exists in which d o p t is equal to the completion time of some job.
The proof for q o p t = S [ k ] is similar to that for d o p t = C [ k ] . □
Lemma 2.
In the optimal sequence, d o p t = C [ k ] , q o p t = S [ k ] , where k satisfies
               j = 1 k 1 δ j j = k n ω j + n γ 0 and j = 1 k δ j j = k + 1 n ω j + n γ 0 .
Proof. 
From Lemma 1, in the CON, d o p t = C [ k ] , through the small disturbance technique, we move d o p t = C [ k ] .
(1) If d o p t = C [ k ] , the total cost is
Z = j = 1 k 1 δ j ( C [ k ] C [ j ] ) + j = k + 1 n ω j ( C [ j ] C [ k ] ) + n γ C [ k ] + j = 1 n v j u j .
(2) If d o p t = C ξ ( k ) x , the total cost is
Z 1 = j = 1 k 1 δ j ( C [ k ] x C [ j ] ) + j = k n ω j ( C [ j ] + x C [ k ] ) + n γ ( C [ k ] x ) + j = 1 n v j u j .
(3) If d o p t = C ξ ( k ) + y , the total cost is
Z 2 = j = 1 k δ j ( C [ k ] + y C [ j ] ) + j = k + 1 n ω j ( C [ j ] y C [ k ] ) + n γ ( C [ k ] + y ) + j = 1 n v j u j .
We have
Z Z 1 = j = 1 k 1 δ j x j = k n ω j x + n γ x 0 ,
Z Z 2 = j = 1 k δ j y + j = k + 1 n ω j y n γ y 0 .
So k satisfies both
               ( j = 1 k 1 δ j j = k n ω j + n γ ) 0 and ( j = 1 k δ j j = k + 1 n ω j + n γ ) 0 . □
Remark 1.
For a given schedule, if k satisfies both the above inequalities, the optimal common due date can be determined by Lemma 2. But k may not meet both of the above inequalities, so we need to set d o p t = q o p t = 0 , and this term can be minimized by the HLP rule (see Hardy et al. [31]).

4. Optimal Analysis

In this section, we perform an optimal analysis of single-machine maintenance activity scheduling problems with convex resource constraints and learning effects. In each case of the above problems in Section 2, the decision consists of four parts: optimal sequence of jobs ξ * , optimal amount of resource allocation u * , optimal due date ( d o p t or q o p t ) and maintenance location l * .

4.1. Results of P 1

In this section, we provide an optimal solution to the P1 problem. For the 1 | M A L E , C R E | j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ d o p t ) + j = 1 n v j u j , its objective function can be expressed as follows.
If k l , where d o p t = C [ k ] = j = 1 k P [ j ] A , we have
Z ( u , ξ , d o p t , l ) = n γ j = 1 k P [ j ] A + j = 1 k 1 δ j ( C [ k ] C [ j ] ) + j = k + 1 n ω j ( C [ j ] C [ k ] ) + j = 1 n v [ j ] u [ j ] = j = 1 k P [ j ] A ( n γ + m = 1 j 1 δ m ) + j = k + 1 l P [ j ] A ( m = j l ω m + m = l + 1 n ω m ) + j = l + 1 n P [ j ] A ( m = j n ω m ) + j = 1 n v [ j ] u [ j ] + j = l + 1 n ω j t = j = 1 l λ j θ [ j ] ( j ) α u [ j ] η + j = l + 1 n λ j β [ j ] θ [ j ] ( j ) α u [ j ] η + j = 1 n v [ j ] u [ j ] + j = l + 1 n ω j t ,
 where
λ j = n γ + m = 1 j 1 δ m , j = 1 , 2 , , k , m = j l ω m + m = l + 1 n ω m , j = k + 1 , k + 2 , , l , m = j n ω m , j = l + 1 , l + 2 , , n .
If k > l , where d o p t = C [ k ] = j = 1 l P [ j ] A + t + j = l + 1 k P [ j ] A , we have
Z ( u , ξ , d o p t , l ) = n γ ( j = 1 l P [ j ] A + t + j = l + 1 k P [ j ] A ) + j = 1 k δ j ( C [ k ] C [ j ] ) + j = k + 1 n ω j ( C [ j ] C [ k ] ) + j = 1 n v [ j ] u [ j ] = j = 1 l P [ j ] A ( n γ + m = 1 j 1 δ m ) + j = l + 1 k P [ j ] A ( n γ + m = 1 l δ m + m = l + 1 j 1 δ m ) + j = k + 1 n P [ j ] A m = j n ω m + ( n γ + j = 1 l δ j ) t + j = 1 n v [ j ] u [ j ] = j = 1 l ψ j θ [ j ] ( j ) α u [ j ] η + j = l + 1 n ψ j β [ j ] θ [ j ] ( j ) α u [ j ] η + j = 1 n v [ j ] u [ j ] + ( n γ + j = 1 l δ j ) t ,
where
ψ j = n γ + m = 1 j 1 δ m , j = 1 , 2 , , l , n γ + m = 1 l δ m + m = l + 1 j 1 δ m , j = l + 1 , l + 2 , , k , m = j n ω m , j = k + 1 , k + 2 , , n .
Lemma 3.
For a given sequence ξ, the optimal resources u [ j ] * allocation of P1 is as follows.
If k l ,
u [ j ] * = η λ j v [ j ] 1 η + 1 θ [ j ] ( j ) α η η + 1 , j = 1 , 2 , , l , η λ j v [ j ] 1 η + 1 ( β [ j ] θ [ j ] ( j ) α ) η η + 1 , j = l + 1 , l + 2 , , n .
If k > l ,
u [ j ] * = η ψ j v [ j ] 1 η + 1 ( θ [ j ] ( j ) α ) η η + 1 , j = 1 , 2 , , l , η ψ j v [ j ] 1 η + 1 ( β [ j ] θ [ j ] ( j ) α ) η η + 1 , j = l + 1 , l + 1 , , n .
Proof. 
From (8) and (10), the objective function is a convex function of u [ j ] . For the case of k l , deriving (8) with respect to u [ j ] , we have   Z u [ j ] = λ j θ [ j ] ( j ) α u [ j ] η + v [ j ] u [ j ] u [ j ] = v [ j ] η λ j ( θ [ j ] ( j ) α ) η 1 u [ j ] η + 1 , j = 1 , 2 , , l .
Let Z u [ j ] = 0 , we have u [ j ] * = η λ j v [ j ] 1 η + 1 ( θ [ j ] ( j ) α ) η η + 1 .
Similarly, if  k > l , the result (13) can be obtained. □
Bringing optimal u * (i.e., Equations (12) and (13)) into the objective function (i.e., Equations (8) and (10)) yields
Z [ l ] = ( η η η + 1 + η 1 η + 1 ) Z 1 [ l ] + f 1 ( t )
 where
f 1 ( t ) = j = l + 1 n ω j t , k l , ( n γ + j = 1 l δ j ) t , k > l .
Z 1 [ l ] = j = 1 l v [ j ] θ [ j ] ( j ) α η η + 1 μ j 1 η + 1 + j = l + 1 n v [ j ] β [ j ] θ [ j ] ( j ) α η η + 1 μ j 1 η + 1
μ j = λ j , k l , ψ j , k > l .
It is obvious that f 1 ( t ) is a constant (if the location l of the maintenance activity location is deterministic), and  Z [ l ] is only related to the the location l so that we find the minimal value of Z [ l ] is actually the minimal value of Z 1 [ l ] . Let
χ j r = v j θ j η η + 1 μ r ( r ) α η 1 η + 1 , 1 r l v j β j θ j η η + 1 μ r ( r ) α η 1 η + 1 , l + 1 r n
Z 1 [ l ] can be minimized by solving the next assignment problem:
M i n Z 1 [ l ] = j = 1 n r = 1 n χ j r M j r
s . t j = 1 n M j r = 1 r = 1 , 2 , , n
r = 1 n M j r = 1 j = 1 , 2 , , n
M j r = 0 o r 1 1 j , r n
where M j r is a 0 or 1 variable, if the job J j is at location r, M j r = 1 , otherwise M j r = 0 .
Based on the above analysis, it is obtained that the problem P 1 can be solved by the next Algorithm 1.
Algorithm 1: Solution for problem P 1 .
Initialization: Let Z = , ξ * = 0 , d o p t * = 0 , u * = 0 and l * = 0 .
Step 1. For l = 1 n
Step 2. If k l , then
          obtain the minimum value Z 1 l and the schedule ξ by using (14)–(19);
            If Z 1 l < Z , then
            let Z = Z 1 l , l * = l , u * = u , d o p t * = d o p t and ξ * = ξ ;
        If k > l , then
          obtain the minimum value Z 1 l and the schedule ξ by using (14)–(19);
            If Z 1 l < Z , then
            let Z = Z 1 l , l * = l , u * = u , d o p t * = d o p t and ξ * = ξ .
Step 3. Choose the minimum value Z * = min { Z 1 [ l ] , l = 1 , 2 , , n } , and obtain the
      corresponding schedule ξ * , d o p t * , u * and l * .
Theorem 1.
The problem P 1 can be solved by Algorithm 1 in O ( n 4 ) time.
Proof. 
The correctness of Algorithm 1 follows the above analysis. Steps 1 need O ( n ) time; for each maintenance position l, the complexity of the assignment problem is O ( n 3 ) . Hence, Algorithm 1 can be solved in O ( n 4 ) time. □

4.2. Results of P 2

For the 1 | M A L E , C R E | j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ q o p t ) + j = 1 n v j u j , its objective function (4) can be expressed as follows.
If k 1 l , where q o p t = S [ k ] = j = 1 k 1 P [ j ] A ,
Z ( u , ξ , q o p t , l ) = n γ j = 1 k 1 P [ j ] A + j = 1 k 1 δ j ( C [ k 1 ] C [ j ] ) + j = k n ω j ( C [ j ] C [ k 1 ] ) + j = 1 n v [ j ] u [ j ] = j = 1 k 1 P [ j ] A ( n γ + m = 1 j 1 δ m ) + j = k l P [ j ] A ( m = l + 1 n ω m + m = j l ω m ) + j = l + 1 n P [ j ] A m = j n ω m + j = 1 n v [ j ] u [ j ] + j = l + 1 n ω j t = j = 1 l λ j θ [ j ] ( j ) α u [ j ] η + j = l + 1 n λ j β [ j ] θ [ j ] ( j ) α u [ j ] η + j = 1 n v [ j ] u [ j ] + j = l + 1 n ω j t ,
where
λ j = n γ + m = 1 j 1 δ m , j = 1 , 2 , , k 1 , m = j l ω m + m = l + 1 n ω m , j = k , k + 1 , , l , m = j n ω m , j = l + 1 , l + 2 , , n .
If k 1 > l , where q o p t = S [ k ] = j = 1 l P [ j ] A + t + j = l + 1 k 1 P [ j ] A
Z ( u , ξ , q o p t , l ) = n γ ( j = 1 l P [ j ] A + t + j = l + 1 k 1 P [ j ] A ) + j = 1 k 1 δ j ( C [ k 1 ] C [ j ] ) + j = k n ω j ( C [ j ] C [ k 1 ] ) + j = 1 n v [ j ] u [ j ] = j = 1 l P [ j ] A ( n γ + m = 1 j 1 δ m ) + j = l + 1 k 1 P [ j ] A ( n γ + m = 1 l δ m + m = l + 1 j 1 δ m ) + j = k n P [ j ] A m = j n ω m + ( n γ + m = 1 l δ m ) t + j = 1 n v [ j ] u [ j ] = j = 1 l ψ j θ [ j ] ( j ) α u [ j ] η + j = l + 1 n ψ j β [ j ] θ [ j ] ( j ) α u [ j ] η + j = 1 n v [ j ] u [ j ] + ( n γ + m = 1 l δ m ) t ,
 where
ψ j = n γ + m = 1 j 1 δ m , j = 1 , 2 , , l , n γ + m = 1 l δ m + m = l + 1 j 1 δ m , j = l + 1 , l + 2 , , k 1 , m = j n ω m , j = k , k + 1 , , n .
Lemma 4.
The optimal resource u [ j ] * allocation of the problem P 2 is as follows.
If k 1 l
u [ j ] * = η λ j v [ j ] 1 η + 1 ( θ [ j ] ( j ) α ) η η + 1 , j = 1 , 2 , , l , η λ j v [ j ] 1 η + 1 ( β [ j ] θ [ j ] ( j ) α ) η η + 1 , j = l + 1 , l + 2 , , n .
If k 1 > l
u [ j ] * = η ψ j v [ j ] 1 η + 1 ( θ [ j ] ( j ) α ) η η + 1 , j = 1 , 2 , , l , η ψ j v [ j ] 1 η + 1 ( β [ j ] θ [ j ] ( j ) α ) η η + 1 , j = l + 1 , l + 1 , , n .
Proof. 
The proof process is similar to that of Lemma 3. Taking a partial derivative of Equation (20) with respect to u [ j ] and making it equal to 0, result Equation (25) can be obtained. □
Substituting Equations (24) and (25) into Equations (20) and (22), we have
Z [ l ] = ( η η η + 1 + η 1 η + 1 ) Z 2 [ l ] + f 2 ( t )
where
f 2 ( t ) = j = l + 1 n ω j t , k 1 l , ( n γ + j = 1 l δ j ) t , k 1 > l .
Z 2 [ l ] = j = 1 l v [ j ] θ [ j ] ( j ) α η η + 1 μ j 1 η + 1 + j = l + 1 n v [ j ] β [ j ] θ [ j ] ( j ) α η η + 1 μ j 1 η + 1
μ j = λ j , k 1 l , ψ j , k 1 > l ,
and it is obvious that f 2 ( t ) is a constant (if the location l of the maintenance activity location is deterministic), and  Z [ l ] is only related to the location l, so that we find the minimal value of Z [ l ] is actually the minimal value of Z 2 [ l ] . Let
χ j r = v j θ j η η + 1 μ r ( r ) α η 1 η + 1 , 1 r l v j β j θ j η η + 1 μ r ( r ) α η 1 η + 1 , l + 1 r n
and Z 2 [ l ] can be minimized by solving the next assignment problem:
M i n Z 2 [ l ] = j = 1 n r = 1 n χ j r M j r
s . t j = 1 n M j r = 1 r = 1 , 2 , , n
r = 1 n M j r = 1 j = 1 , 2 , , n
M j r = 0 o r 1 1 j , r n
where M j r is a 0 or 1 variable, if the job J j is at location r, M j r = 1 , otherwise M j r = 0 .
Based on the above analysis, the problem P 2 can be solved by the next Algorithm 2:
Algorithm 2: Solution for problem P 2 .
Initialization: Let Z = , ξ * = 0 , q o p t * = 0 , u * = 0 and l * = 0 .
Step 1. For l = 1 n
Step 2. If k 1 l , then
          obtain the minimum value Z 2 l and the schedule ξ by using (26)–(30);
            If Z 2 l < Z , then
            let Z = Z 2 l , l * = l , u * = u , q o p t * = q o p t and ξ * = ξ ;
        If k 1 > l , then
          obtain the minimum value Z 2 l and the schedule ξ by using (26)–(30);
            If Z 2 l < Z , then
            let Z = Z 2 l , l * = l , u * = u , q o p t * = q o p t and ξ * = ξ .
Step 3. Choose the minimum value Z * = min { Z 2 [ l ] , l = 1 , 2 , , n } , and obtain the
      corresponding schedule ξ * , q o p t * , u * and l * .
Theorem 2.
The problem P 2 can be solved by Algorithm 2 in O ( n 4 ) time.

4.3. Results of P 3

In this subsection, we consider the problem P 3 , i.e.,
1 | M A L E , C R E , j = 1 n v j u j U | j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ d o p t ) .
If k l , similarly, we have
j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ d o p t ) = j = 1 l κ j θ [ j ] ( j ) α u [ j ] η + j = l + 1 n κ j β [ j ] θ [ j ] ( j ) α u [ j ] η + j = l + 1 n ω j t .
If k > l , similarly, we have
j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ d o p t ) = j = 1 l σ j θ [ j ] ( j ) α u [ j ] η + j = l + 1 n σ j β [ j ] θ [ j ] ( j ) α u [ j ] η + ( n γ + j = 1 l δ j ) t ,
where
κ j = n γ + m = 1 j 1 δ m , j = 1 , 2 , , k , m = l + 1 n ω m + m = j l ω m , j = k + 1 , k + 2 , , l , m = j n ω m , j = l + 1 , l + 2 , , n ,
σ j = n γ + m = 1 j 1 δ m , j = 1 , 2 , , l , n γ + m = 1 l δ m + m = l + 1 j 1 δ m , j = l + 1 , l + 2 , , k , m = j n ω m , j = k + 1 , k + 2 , , n .
Lemma 5.
For a given sequence ξ, the corresponding optimal resources u [ j ] * allocation for P3 is as follows.
If k l
u [ j ] * = U 1 v [ j ] 1 η + 1 κ j 1 η + 1 θ [ j ] ( j ) α + β [ j ] θ [ j ] ( j ) α η η + 1 j = 1 l κ j 1 η + 1 v [ j ] θ [ j ] ( j ) α η η + 1 + j = l + 1 n κ j 1 η + 1 v [ j ] β [ j ] θ [ j ] ( j ) α η η + 1 , j = 1 , 2 , , n .
If k > l
u [ j ] * = U 1 v [ j ] 1 η + 1 σ j 1 η + 1 θ [ j ] ( j ) α + β [ j ] θ [ j ] ( j ) α η η + 1 j = 1 l σ j 1 η + 1 v [ j ] θ [ j ] ( j ) α η η + 1 + j = l + 1 n σ j 1 η + 1 v [ j ] β [ j ] θ [ j ] ( j ) α η η + 1 , j = 1 , 2 , , n .
Proof. 
Obviously, when j = 1 n v j u j = U , j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ d o p t ) is the smallest. Thus, this problem becomes a conditional extreme problem of minimizing j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ d o p t ) subject to j = 1 n v j u j = U . For this problem, we use the Lagrange multiplier method to solve it, and the Lagrange function is
L ( u , π , l , ϕ ) = j = 1 l κ j θ [ j ] ( j ) α u [ j ] η + j = l + 1 n κ j β [ j ] θ [ j ] ( j ) α u [ j ] η + j = l + 1 n ω j t + ϕ j = 1 n v [ j ] u [ j ] U
where ϕ ( 0 ϕ ) is the Lagrange multiplier.
Deriving (37) with respect to f u [ j ] and ϕ , we have
L u = ϕ v [ j ] η κ j θ [ j ] ( j ) α η u [ j ] η + 1 η κ j β [ j ] θ [ j ] ( j ) α η u [ j ] η + 1 = 0 , j = 1 , 2 , . . . , n .
L ϕ = j = 1 n v [ j ] u [ j ] U = 0 , j = 1 , 2 , , n .
From (38), we have
u [ j ] * = η κ j ϕ v [ j ] 1 η + 1 × θ [ j ] ( j ) α + β [ j ] θ [ j ] ( j ) α η η + 1 , j = 1 , 2 , , n .
From (39) and (40), we have
ϕ 1 η + 1 = j = 1 n ( η κ j ) 1 η + 1 ( θ [ j ] ( j ) α + β [ j ] θ [ j ] ( j ) α ) v [ j ] η η + 1 U , j = 1 , 2 , , n .
From (40) and (41), we have
u [ j ] * = U 1 v [ j ] 1 η + 1 κ j 1 η + 1 θ [ j ] ( j ) α + β [ j ] θ [ j ] ( j ) α η η + 1 j = 1 l κ j 1 η + 1 v [ j ] θ [ j ] ( j ) α η η + 1 + j = l + 1 n κ j 1 η + 1 v [ j ] β [ j ] θ [ j ] ( j ) α η η + 1 , j = 1 , 2 , , n .
   □
Bringing optimal u * (i.e., Equations (35) and (36)) into the objective function (i.e., Equations (31) and (32)) yields
Z ˙ [ l ] = U η Z ˙ 3 η + 1 [ l ] + f 1 ( t )
where f 1 ( t ) is given by Equation (15) (is a constant), and 
Z ˙ 3 [ l ] = j = 1 l μ j ˙ 1 η + 1 v [ j ] θ [ j ] ( j ) α η η + 1 + j = l + 1 n μ j ˙ 1 η + 1 v [ j ] β [ j ] θ [ j ] ( j ) α η η + 1
μ j ˙ = κ j , k l , σ j , k > l ,
it is obvious that the minimal value of Z ˙ [ l ] is actually the minimal value of Z ˙ 3 [ l ] . Let
B ˙ j r = v j θ j η η + 1 μ r ˙ ( r ) α η 1 η + 1 , 1 r l v j β j θ j η η + 1 μ r ˙ ( r ) α η 1 η + 1 , l + 1 r n
Z ˙ 3 [ l ] can be minimized by solving the next assignment problem:
M i n Z ˙ 3 [ l ] = j = 1 n r = 1 n B ˙ j r N ˙ j r
s . t j = 1 n N ˙ j r = 1 r = 1 , 2 , , n
r = 1 n N ˙ j r = 1 j = 1 , 2 , , n
N ˙ j r = 0 o r 1 1 j , r n
where N ˙ j r is a 0 or 1 variable, and if the job J j is at location r, N ˙ j r = 1 ; otherwise, N ˙ j r = 0 .
Based on the above analysis, the problem P 3 can be solved by the next Algorithm 3:
Algorithm 3: Solution for problem P 3 .
Initialization: Let Z ˙ = , ξ * = 0 , d o p t * = 0 , u * = 0 and l * = 0 .
Step 1. For l = 1 n
Step 2. If k l , then
          obtain the minimum value Z ˙ 3 [ l ] and the schedule ξ by using (43)–(46);
            If Z ˙ 3 [ l ] < Z ˙ , then
            let Z ˙ = Z ˙ 3 [ l ] , l * = l , u * = u , d o p t * = d o p t and ξ * = ξ ;
        If k > l , then
          obtain the minimum value Z ˙ 3 [ l ] and the schedule ξ by using (43)–(46);
            If Z ˙ 3 [ l ] < Z ˙ , then
            let Z ˙ = Z ˙ 3 [ l ] , l * = l , u * = u , d o p t * = d o p t and ξ * = ξ .
Step 3. Choose the minimum value Z * ˙ = min { Z ˙ 3 [ l ] , l = 1 , 2 , , n } , and obtain the
       corresponding schedule ξ * , d o p t * , u * and l * .
Theorem 3.
The problem P 3 can be solved by Algorithm 3 in O ( n 4 ) time.

4.4. Results of P 4

Similar to Section 4.3, the  1 | M A L E , C R E , j = 1 n v j u j U | j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ q o p t ) problem can be expressed as follows.
If k 1 l
j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ q o p t ) = j = 1 l κ j θ [ j ] ( j ) α u [ j ] η + j = l + 1 n κ j β j θ [ j ] ( j ) α u [ j ] η + j = l + 1 n ω j t .
If k 1 > l
j = 1 n ( δ j E [ j ] + ω j T [ j ] + γ q o p t ) = j = 1 l σ j θ [ j ] ( j ) α u [ j ] η + j = l + 1 n σ j β j θ [ j ] ( j ) α u [ j ] η + j = 1 l δ j t .
where
κ j = n γ + m = 1 j 1 δ m , j = 1 , 2 , , k 1 , m = j l ω m + m = l + 1 n ω m , j = k , k + 1 , , l , m = j n ω m , j = l + 1 , l + 2 , , n .
σ j = n γ + m = 1 j 1 δ m , j = 1 , 2 , , l , n γ + m = 1 l δ m + m = l + 1 j 1 δ m , j = l + 1 , l + 2 , , k 1 , m = j n ω m , j = k , k + 1 , , n .
Lemma 6.
The optimal resource allocation of the problem P 4 is as follows.
If k 1 l
u [ j ] * = U 1 v [ j ] 1 η + 1 κ j 1 η + 1 θ [ j ] ( j ) α + β [ j ] θ [ j ] ( j ) α η η + 1 j = 1 l κ j 1 η + 1 v [ j ] θ [ j ] ( j ) α η η + 1 + j = l + 1 n κ j 1 η + 1 v [ j ] β [ j ] θ [ j ] ( j ) α η η + 1 , j = 1 , 2 , , n .
If k 1 > l
u [ j ] * = U 1 v [ j ] 1 η + 1 σ j 1 η + 1 θ [ j ] ( j ) α + β [ j ] θ [ j ] ( j ) α η η + 1 j = 1 l σ j 1 η + 1 v [ j ] θ [ j ] ( j ) α η η + 1 + j = l + 1 n σ j 1 η + 1 v [ j ] β [ j ] θ [ j ] ( j ) α η η + 1 , j = 1 , 2 , , n .
Proof. 
Similar to the proof of Lemma 5. □
Bringing optimal u * (i.e., Equations (51) and (52)) into the objective function (i.e., Equations (47) and (48)) yields
Z ¨ [ l ] = U η Z ¨ 4 η + 1 [ l ] + f 2 ( t )
where f 2 ( t ) is given by Equation (27) (is a constant), and 
Z ¨ 4 [ l ] = j = 1 l μ j ¨ 1 η + 1 v [ j ] θ [ j ] ( j ) α η η + 1 + j = l + 1 n μ j ¨ 1 η + 1 v [ j ] β [ j ] θ [ j ] ( j ) α η η + 1
μ j ¨ = κ j , k 1 l , σ j , k 1 > l .
Let
B ¨ j r = v j θ j η η + 1 μ r ¨ ( r ) α η 1 η + 1 , 1 r l v j β j θ j η η + 1 μ r ¨ ( r ) α η 1 η + 1 , l + 1 r n ,
and Z ¨ 4 [ l ] can be minimized by solving the next assignment problem:
M i n Z ¨ 4 [ l ] = j = 1 n r = 1 n B ¨ j r N ¨ j r
s . t j = 1 n N ¨ j r = 1 r = 1 , 2 , , n
r = 1 n N ¨ j r = 1 j = 1 , 2 , , n
N ¨ j r = 0 o r 1 1 j , r n
where N ¨ j r is a 0 or 1 variable, and if the job J j is at location r, N ¨ j r = 1 ; otherwise, N ¨ j r = 0 .
The problem P 4 can be solved by the following Algorithm 4:
Algorithm 4: Solution for problem P 4 .
Initialization: Let Z ¨ = , ξ * = 0 , q o p t * = 0 , u * = 0 and l * = 0 .
Step 1. For l = 1 n
Step 2. If k 1 l , then
          obtain the minimum value Z ¨ 4 [ l ] and the schedule ξ by using (52)–(56);
            If Z ¨ 4 [ l ] < Z ¨ , then
            let Z ¨ = Z ¨ 4 [ l ] , l * = l , u * = u , q o p t * = q o p t and ξ * = ξ ;
        If k 1 > l , then
          obtain the minimum value Z ¨ 4 [ l ] and the schedule ξ by using (52)–(56);
            If Z ¨ 4 [ l ] < Z ¨ , then
            let Z ¨ = Z ¨ 4 [ l ] , l * = l , u * = u , q o p t * = q o p t and ξ * = ξ .
Step 3. Choose the minimum value Z ¨ * = min { Z ¨ 4 [ l ] , l = 1 , 2 , , n } , and obtain the
        corresponding schedule ξ * , q o p t * , u * and l * .
Theorem 4.
The problem P 4 can be solved by Algorithm 4 in O ( n 4 ) time.

5. An Example and Numerical Study

5.1. An Example

Since algorithms of P 1 and P 2 (resp. P 3 and P 4 ) are similar, we only consider the calculation steps of P 1 (resp. P 3 ).
Let n = 6 , t = 3 , α = 0.2 , η = 1 , γ = 4 , U = 100 , and the remaining data are given in Table 3.
Solution: According to Lemma 2, we can obtain k = 2 .
Case 1. First, we consider the problem P 1 .
When l = 3 , from Equations (17) and (18), we can obtain
χ j r = v j θ j 1 2 ( n γ + m = 1 r 1 δ m ) ( r ) 0.2 1 2 r = 1 , 2 v j θ j 1 2 m = r n ω m ( r ) 0.2 1 2 r = 3 v j β j θ j 1 2 m = r n ω m ( r ) 0.2 1 2 r = 4 , 5 , 6
and it is easy to see that
χ 11 = v 1 θ 1 1 2 ( n γ ) ( 1 ) 0.2 1 2 = ( 22 × 9 ) 1 2 × ( 24 ) 1 2 = 68.93475 ;
χ 12 = v 1 θ 1 1 2 ( n γ + δ 1 ) ( 2 ) 0.2 1 2 = ( 22 × 9 ) 1 2 × [ ( 24 + 7 ) × 2 0.2 ] 1 2 = 73.09883 ;
χ 13 = v 1 θ 1 1 2 ( ω 3 + ω 4 + ω 5 + ω 6 ) ( 3 ) 0.2 1 2 = ( 22 × 9 ) 1 2 × [ ( 2 + 12 + 6 + 9 )
× 3 0.2 ] 1 2 = 67.89213 ;
χ 14 = v 1 β 1 θ 1 1 2 ( ω 4 + ω 5 + ω 6 ) ( 4 ) 0.2 1 2 = ( 22 × 0.4 × 9 ) 1 2 × [ ( 12 + 6 + 9 ) × 4 0.2 ] 1 2
= 40.25672 ;
χ 15 = v 1 β 1 θ 1 1 2 ( ω 5 + ω 6 ) ( 5 ) 0.2 1 2 = ( 22 × 0.4 × 9 ) 1 2 × [ ( 6 + 9 ) × 5 0.2 ] 1 2 = 29.34345;
χ 16 = v 1 β 1 θ 1 1 2 ω 6 ( 6 ) 0.2 1 2 = ( 22 × 0.4 × 9 ) 1 2 × [ 9 × 6 0.2 ] 1 2 = 22.31869 .
Similarly, the rest of the values of χ j r are given in Table 4.
We can obtain the optimal schedule ξ = [ J 3 , J 4 , J 1 , J 6 , J 5 , J 2 ] by solving the assignment problem (19), and Z 1 [ 3 ] = 67.89213 + 18.9772 + 37.94733 + 28.59419 + 18.29885 + 38.17462 = 209.88434 , u = ( 2.5298 , 4.18827 , 3.086 , 3.3035 , 2.61412 , 0.65757 ) , d o p t = p 3 + p 4 = 1.58115 + 2.7004 = 4.28155 , f 1 ( t ) = t ( ω 4 + ω 5 + ω 6 ) = 3 × ( 12 + 6 + 9 ) = 81 , so Z [ 3 ] = ( η η η + 1 + η 1 η + 1 ) Z 1 [ 3 ] + f 1 ( t ) = 2 × 209.88434 + 81 = 500.76868 .
l = 1 : Similarly, we can easily find the values of χ j r from Equations (17) and (18), and solve the assignment problem (19) to obtain the optimal schedule ξ = [ J 5 , J 2 , J 3 , J 6 , J 4 , J 1 ] , and Z 1 [ 1 ] = 169.78529 , u = ( 8.68496 , 1.36212 , 1.93 , 3.3035 , 1.8797 , 1.0145 ) , d o p t = p 5 + t + p 2 = 2.53311 + 3 + 0.70258 = 6.23569 , f 1 ( t ) = t ( n γ + δ 1 ) = 3 × ( 24 + 7 ) = 93 , so Z [ 1 ] = ( η η η + 1 + η 1 η + 1 ) Z 1 [ 1 ] + f 1 ( t ) = 2 × 169.78529 + 93 = 432.57051 .
l = 2 : With the above calculation, we can obtain the optimal schedule ξ = [ J 5 , J 4 , J 1 , J 6 , J 3 , J 2 ] , and Z 1 [ 1 ] = 179.81382 , u = ( 8.68496 , 4.18827 , 1.95176 , 3.3035 , 1.31889 , 0.65757 ) , d o p t = p 5 + p 4 = 2.53311 + 2.7004 = 5.23351 , f 1 ( t ) = t ( ω 3 + ω 4 + ω 5 + ω 6 ) = 3 × ( 2 + 12 + 6 + 9 ) = 87 , so Z [ 2 ] = ( η η η + 1 + η 1 η + 1 ) Z 1 [ 2 ] + f 1 ( t ) = 2 × 179.81382 + 87 = 446.62764 .
l = 4 : As above, we can obtain the optimal schedule ξ = [ J 4 , J 5 , J 1 , J 6 , J 3 , J 2 ] , and Z 1 [ 4 ] = 242.07549, u = ( 3.94968 , 9.20959 , 3.086 , 3.69343 , 1.31889 , 0.65757 ) , d o p t = p 4 + p 5 = 3.29141 + 2.07827 = 5.36968 , f 1 ( t ) = t ( ω 5 + ω 6 ) = 3 × ( 6 + 9 ) = 45 , so Z [ 4 ] = ( η η η + 1 + η 1 η + 1 ) Z 1 [ 4 ] + f 1 ( t ) = 2 × 242.07549 + 45 = 529.15098 .
l = 5 : We can obtain the optimal schedule ξ = [ J 4 , J 6 , J 1 , J 5 , J 3 , J 2 ] , and Z 1 [ 5 ] = 263.33595 , u = ( 3.94968 , 3.9395 , 3.086 , 8.0193 , 1.70268 , 0.65757 ) , d o p t = p 4 + p 6 = 3.29141 + 1.32504 = 4.61645 , f 1 ( t ) = t ( ω 6 ) = 3 × ( 9 ) = 27 , so Z [ 5 ] = ( η η η + 1 + η 1 η + 1 ) Z 1 [ 5 ] + f 1 ( t ) = 2 × 263.33595 + 27 = 553.6719 .
l = 6 : We can obtain the optimal schedule ξ = [ J 4 , J 5 , J 1 , J 6 , J 3 , J 2 ] , and Z 1 [ 6 ] = 283.82959 , u = ( 3.94968 , 9.20959 , 3.086 , 2.69217 , 1.70268 , 2.07942 ) , d o p t = p 4 + p 5 = 3.29141 + 2.07827 = 5.36968 , f 1 ( t ) = 0 , so Z [ 6 ] = ( η η η + 1 + η 1 η + 1 ) Z 1 [ 46 ] + f 1 ( t ) = 2 × 283.82959 = 567.65918 .
After the above calculation, the results of Case 1 are shown in Table 5.
We can see that the optimal solution for P 1 is l = 1 , ξ * = [ J 5 , J 2 , J 3 , J 6 , J 4 , J 1 ] , d o p t * = 6.23569 and u * = ( 8.68496 , 1.36212 , 1.93 , 3.3035 , 1.8797 , 1.0145 ) .
Case 2. Next, we compute the problem P 3 .
When l = 1 , from Equations (44) and (45), we can obtain
B ˙ j r = v j θ j 1 2 ( n γ + m = 1 r 1 δ m ) ( r ) 0.2 1 2 r = 1 v j β j θ j 1 2 ( n γ + m = 1 r 1 δ m ) ( r ) 0.2 1 2 r = 2 v j β j θ j 1 2 m = r n ω m ( r ) 0.2 1 2 r = 3 , 4 , 5 , 6
and it is easy to see that
B ˙ 11 = v 1 θ 1 1 2 ( n γ ) ( 1 ) 0.2 1 2 = ( 22 × 9 ) 1 2 × ( 24 ) = 337.70993 ;
B ˙ 12 = v 1 β 1 θ 1 1 2 ( n γ + δ 1 ) ( 2 ) 0.2 1 2 = ( 22 × 0.4 × 9 ) 1 2 × [ ( 24 + 7 ) × 2 0.2 ] 1 2
= 257.407552 ;
B ˙ 13 = v 1 β 1 θ 1 1 2 ( ω 3 + ω 4 + ω 5 + ω 6 ) ( 3 ) 0.2 1 2 = ( 22 × 0.4 × 9 ) 1 2 × [ ( 2 + 12 + 6 + 9 )
× 3 0.2 ] 1 2 = 231.23228 ;
B ˙ 14 = v 1 β 1 θ 1 1 2 ( ω 4 + ω 5 + ω 6 ) ( 4 ) 0.2 1 2 = ( 22 × 0.4 × 9 ) 1 2 × [ ( 12 + 6 + 9 ) × 4 0.2 ] 1 2
= 209.180094 ;
B ˙ 15 = v 1 β 1 θ 1 1 2 ( ω 5 + ω 6 ) ( 5 ) 0.2 1 2 = ( 22 × 0.4 × 9 ) 1 2 × [ ( 6 + 9 ) × 5 0.2 ] 1 2
= 113.64671 ;
B ˙ 16 = v 1 β 1 θ 1 1 2 ω 6 ( 6 ) 0.2 1 2 = ( 22 × 0.4 × 9 ) 1 2 × [ 9 × 6 0.2 ] 1 2 = 66.95607 .
Similarly, the rest of the values of B ˙ j r are given in Table 6,
We can obtain the optimal schedule ξ = [ J 5 , J 2 , J 3 , J 6 , J 4 , J 1 ] by solving the assignment problem (46), and Z ˙ 3 [ 1 ] = 169.78528 , u = ( 5.60353 , 2.65995 , 1.22568 , 2.92227 , 1.91124 , 1.11076 ) , d o p t = p 5 + p 2 = 4.33963 , f 1 ( t ) = t ( n γ + δ 1 ) = 3 × ( 24 + 7 ) = 93 , so Z ˙ [ 1 ] = U η Z ˙ 3 η + 1 [ 1 ] + f 1 ( t ) = 100 1 × 169 . 78528 2 + 93 = 381.27041 .
l = 2 : Similarly, we can easily find the values of B ˙ j r from Equations (44) and (45), and solve the assignment problem (46) to obtain the optimal schedule ξ = [ J 5 , J 4 , J 1 , J 6 , J 3 , J 2 ] , and Z ˙ 3 [ 2 ] = 179.81382 , u = ( 5.29097 , 2.85518 , 2.202719 , 2.75966 , 1.19380 , 1.20519 ) , d o p t = p 5 + p 4 = 8.7116 , f 1 ( t ) = t ( ω 3 + ω 4 + ω 5 + ω 6 ) = 3 × ( 2 + 12 + 6 + 9 ) = 87 , so Z ˙ [ 2 ] = U η Z ˙ 3 η + 1 [ 2 ] + f 1 ( t ) = 100 1 × 179 . 81382 2 + 87 = 404.331 .
l = 3 : We can obtain the optimal schedule ξ = [ J 3 , J 4 , J 1 , J 6 , J 5 , J 2 ] , and Z ˙ 3 [ 3 ] = 209.88434 , u = ( 3.02238 , 2.44322 , 1.73675 , 2.36428 , 3.04078 , 1.03252 ) , d o p t = p 3 + p 4 = 6.64431 , f 1 ( t ) = t ( ω 4 + ω 5 + ω 6 ) = 3 × ( 12 + 6 + 9 ) = 81 , so Z ˙ [ 3 ] = U η Z ˙ 3 η + 1 [ 3 ] + f 1 ( t ) = 100 1 × 209 . 88434 2 + 81 = 521.54136 .
l = 4 : As given above, we can obtain the optimal schedule ξ = [ J 4 , J 5 , J 1 , J 6 , J 3 , J 2 ] , and Z ˙ 3 [ 4 ]   = 242.07549 , u = ( 1.99828 , 4.16622 , 1.5058 , 2.04988 , 0.88676 , 0.89522 ) , d o p t = p 4 + p 5 = 11.78616 , f 1 ( t ) = t ( ω 5 + ω 6 ) = 3 × ( 6 + 9 ) = 45 , so Z ˙ [ 4 ] = U η Z ˙ 3 η + 1 [ 4 ] + f 1 ( t ) = 100 1 × 242 . 07549 2 + 45 = 631.00543 .
l = 5 : We can obtain the optimal schedule ξ = [ J 4 , J 6 , J 1 , J 5 , J 3 , J 2 ] , and Z ˙ 3 [ 5 ] = 263.33595 , u = ( 1.83695 , 2.16033 , 1.38423 , 3.34065 , 0.81517 , 0.82294 ) , d o p t = p 4 + p 6 = 9.85430 , f 1 ( t ) = t ( ω 6 ) = 3 × ( 9 ) = 27 , so Z ˙ [ 5 ] = U η Z ˙ 3 η + 1 [ 5 ] + f 1 ( t ) = 100 1 × 263 . 33595 2 + 27 = 720.45823 .
l = 6 : We can obtain the optimal schedule ξ = [ J 4 , J 5 , J 1 , J 6 , J 3 , J 2 ] , and Z ˙ 3 [ 6 ] = 283.82959 , u = ( 1.70432 , 3.55333 , 1.2843 , 1.74832 , 0.75632 , 0.76352 ) , d o p t = p 4 + p 5 = 13.81905, f 1 ( t ) = 0 , so Z ˙ [ 6 ] = U η Z ˙ 3 η + 1 [ 6 ] + f 1 ( t ) = 100 1 × 283 . 82959 2 = 805.59236 .
After the above calculation, the results of Case 2 are shown in Table 7.
We can see that the optimal solution for P 3 is l = 1 , ξ * = [ J 5 , J 2 , J 3 , J 6 , J 4 , J 1 ] , d o p t * = 4.33963 and u * = ( 5.60353 , 2.65995 , 1.22568 , 2.92227 , 1.91124 , 1.11076 ) .

5.2. Numerical Study

To test the validity of Algorithms 1–4, we randomly generated the instances. We implemented all the algorithms in Java language on JetBrains 2023, and coded on a PC with Intel(R) Core(TM) i5-10500 CPU @ 3.10 GHz, 8.00 GB of RAM on a Windows 10 OS. The features of the examples are listed below:
(1)
n = 35 , 45 , 55 , 65 , 75 , 85 , 95 , 105 , 115 , 125 , 135 , t = 10 and α = 0.3 ;
(2)
η = 2 , γ = 12 and U = 500 ;
(3)
θ j ( j = 1 , 2 , , n ) is drawn from a discrete uniform distribution in [1, 100] (i.e., θ j [ 1 , 100 ] );
(4)
β j ( j = 1 , 2 , , n ) ∼ [0.5, 1];
(5)
δ j , ω j ( j = 1 , 2 , , n ) ∼ [1, 40];
(6)
v j ( j = 1 , 2 , , n ) ∼ [1, 50].
For each n, 20 instances are generated randomly. The computational tests for Algorithms 1–4 are given as follows. The average (mean) and maximum (max) CPU times (milliseconds (ms)) are shown in Table 8. From Table 8, we can see that Algorithms 1–4 are effective, and their CPU times increase moderately as n increases from 35 to 135, and the maximum CPU time is 139,834.75 ms for n = 135 .

6. Conclusions

In previous studies, the influence of learning effects and resource allocation factors in single-machine scheduling was considered. In this paper, we extend this setting by allowing the execution of a maintenance activity. It is revealed that these four problems can be solved in O ( n 4 ) time (see Table 1). In future work, we will incorporate more realistic settings, such as multiple maintenance activities, or multiple machine (e.g., flow shop) conditions.

Author Contributions

Investigation, L.-Y.W.; Writing—original draft, Z.-J.W.; Writing—review and editing, Z.-J.W., L.-Y.W., L.Z., J.-B.W. and E.W. All authors have read and agreed to the published version of the manuscript.

Funding

This Work was supported by the Science Research Foundation of Educational Department of Liaoning Province (LJKMZ20220527). Ershen Wang was also supported by the SongShan Laboratory Foundation (YYJC062022017), the Open Fund of Key Laboratory of Flight Techniques and Flight Safety CAAC (FZ2021KF15, FZ2021ZZ06), the Applied Basic Research Programs of Liaoning Province (2022020502-JH2/1013), and the Special Funds program of Shenyang Science and technology (22-322-3-34).

Data Availability Statement

The data used to support the findings of this study are available from the corresponding author upon request.

Conflicts of Interest

The authors declare no conflict of interest.

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Table 1. Models studied.
Table 1. Models studied.
ReferencesScheduling ProblemTime Complexity
Wang et al. [29] 1 P j r A ( u j ) = θ j r α u j η δ 1 C max + δ 2 T C + δ 3 T A D C + δ 4 j = 1 n ˇ v j u j O ( n log ( n ) )
1 P j r A ( u j ) = θ j r α u j η δ 1 C max + δ 2 T W + δ 3 T A D W + δ 4 j = 1 n ˇ v j u j O ( n log ( n ) )
Zhu et al. [30] 1 M A L E , R E δ 1 C max + δ 2 T C + δ 3 T A D C + δ 4 j = 1 n ˇ v j u j O ( n 2 log ( n ) )
1 M A L E , R E δ 1 C max + δ 2 T W + δ 3 T A D W + δ 4 j = 1 n ˇ v j u j O ( n 2 log ( n ) )
Wang and Wang [15] 1 M A , S L K , P j = ( θ j , β j θ j ) j = 1 n δ E j + ω T j + γ q o p t O ( n 4 )
Bai et al. [17] 1 M A D E , P j = ( θ j + b s j , β j θ j + b s j ) j = 1 n δ E j + ω T j + γ q o p t O ( n 4 )
Ji et al. [22] 1 M A D E , P j r A = θ j ( r l ) a j j = 1 n δ E j + ω T j + γ q o p t O ( n 4 )
Zhao et al. [25] 1 M A D E , C R E , j = 1 n v [ j ] u [ j ] U j = 1 n δ E j + ω T j + γ q o p t O ( n 4 )
1 M A D E , C R E , j = 1 n δ E j + ω T j + γ q o p t V j = 1 n v [ j ] u [ j ] O ( n 4 )
This article 1 M A L E , C R E j = 1 n δ j E [ j ] + ω j T [ j ] + γ d o p t + j = 1 n v [ j ] u [ j ] O ( n 4 )
1 M A L E , C R E j = 1 n δ j E [ j ] + ω j T [ j ] + γ q o p t + j = 1 n v [ j ] u [ j ] O ( n 4 )
1 M A L E , C R E , j = 1 n v [ j ] u [ j ] U j = 1 n δ j E [ j ] + ω j T [ j ] + γ d o p t O ( n 4 )
1 M A L E , C R E , j = 1 n v [ j ] u [ j ] U j = 1 n δ j E [ j ] + ω j T [ j ] + γ q o p t O ( n 4 )
M A means “maintenance activity”; M A D E means “maintenance activity and aging effect”; R E means “resource allocation”; b means common deterioration rate of all jobs; a j means aging factor of job J j ; s j means starting time of job J j ; δ , ω , δ 1 , δ 2 , δ 3 and δ 4 are given constants.
Table 2. Notations.
Table 2. Notations.
NotationMeaning
nthe number of jobs
J j the j-th job
J [ j ] the job scheduled in the j-th position
θ j (resp. θ j )the normal processing time of job J [ j ] (resp. J j )
p j A ( p j A )the actual processing time of job J [ j ] (resp. J j )
β j the modifying rate of job J j
p j r A the actual processing time of job J j in position r
α the learning factor
tthe maintenance duration
lthe location of the maintenance activity
u [ j ] (resp. u j )the resource allocated to job J [ j ] (resp. J j )
C j (resp. C j )the completion time of job J [ j ] (resp. J j )
S j (resp. S j )the start time of job J [ j ] (resp. J j )
E j (resp. E j ) ( = max 0 , d j C j ) the earliness of job J [ j ] (resp. J j )
T j (resp. T j ) ( = max 0 , C j d j ) the tardiness of job J [ j ] (resp. J j )
v j (resp. v [ j ] )the cost when allocating unit resource to job J [ j ] (resp. J j )
δ j , ω j the position-dependent (but job-independent) weight (cost) of the j-th job
γ ( η , U )the given constant
Table 3. The remaining data.
Table 3. The remaining data.
J j J 1 J 2 J 3 J 4 J 5 J 6
θ j 911413226
β j 0.4 0.1 0.6 0.5 0.2 0.8
δ j 7285410
ω j 5421269
v j 2216152079
Table 4. The values of χ j r .
Table 4. The values of χ j r .
jr123456
1 68.93475 73.09883 67.89213 40.25672 29.34345 22.31869
2 64.9923 68.91824 64.00931 18.9772 13.83263 10.52113
3 37 . 94733 40.23958 37.37339 27.14108 19.78335 15.04725
4 78.99367 83.76537 77.79892 51.57599 37.59415 28 . 59419
5 60.79474 64.46711 59.87523 25.10448 18 . 29885 13.9184
636 38 . 17462 35.45551 29.73156 21.67157 16.48345
The bold values are the optimal solution.
Table 5. The results of Case 1.
Table 5. The results of Case 1.
l d opt u Z ([l]) ξ
16.23569 ( 8.68496 , 1.36212 , 1.93 , 3.3035 , 1.8797 , 1.0145 ) 432.57051 [ J 5 , J 2 , J 3 , J 6 , J 4 , J 1 ]
2 5.23351 ( 8.68496 , 4.18827 , 1.95176 , 3.3035 , 1.31889 , 0.65757 ) 446.62764 [ J 5 , J 4 , J 1 , J 6 , J 3 , J 2 ]
3 4.28155 ( 2.5298 , 4.18827 , 3.086 , 3.3035 , 2.61412 , 0.65757 ) 500.76868 [ J 3 , J 4 , J 1 , J 6 , J 5 , J 2 ]
4 5.36968 ( 3.94968 , 9.20959 , 3.086 , 3.69343 , 1.31889 , 0.65757 ) 529.15098 [ J 4 , J 5 , J 1 , J 6 , J 3 , J 2 ]
5 4.61645 ( 3.94968 , 3.9395 , 3.086 , 8.0193 , 1.70268 , 0.65757 ) 553.6719 [ J 4 , J 6 , J 1 , J 5 , J 3 , J 2 ]
6 5.36968 ( 3.94968 , 9.20959 , 3.086 , 2.69217 , 1.70268 , 2.07942 ) 567.65918 [ J 4 , J 5 , J 1 , J 6 , J 3 , J 2 ]
Table 6. The values of B ˙ j r .
Table 6. The values of B ˙ j r .
jr123456
1 68.93475 46.23176 42.93875 40.25672 29 . 34345 22.31869
2 64.99230 21 . 79386 20.24152 18.9772 13.83264 10.52113
3 37.94733 31.16945 28 . 94930 27.14108 19.78335 15.04726
4 78.99367 59.23106 55.01214 51.57599 37.59415 28 . 59419
5 60.79474 28.83057 26.77702 25 . 10448 18.29886 13.91815
6 36 34.14442 31.7124 29.73156 21.67158 16.48345
The bold values are the optimal solution.
Table 7. The results of Case 2.
Table 7. The results of Case 2.
l d opt u Z ˙ [ l ] ξ
1 4.33963 ( 5.60353 , 2.65995 , 1.22568 , 2.92227 , 1.91124 , 1.11076 ) 381.27041 [ J 5 , J 2 , J 3 , J 6 , J 4 , J 1 ]
2 8.71116 ( 5.29097 , 2.85518 , 2.202719 , 2.75966 , 1.19380 , 1.20519 ) 404.331 [ J 5 , J 4 , J 1 , J 6 , J 3 , J 2 ]
3 6.64431 ( 3.02238 , 2.44322 , 1.73675 , 2.36428 , 3.04078 , 1.03252 ) 521.54136 [ J 3 , J 4 , J 1 , J 6 , J 5 , J 2 ]
4 11.78616 ( 1.99828 , 4.16622 , 1.5058 , 2.04988 , 0.88676 , 0.89522 ) 631.00543 [ J 4 , J 5 , J 1 , J 6 , J 3 , J 2 ]
5 9.85430 ( 1.83695 , 2.16033 , 1.38423 , 3.34065 , 0.81517 , 0.82294 ) 720.45823 [ J 4 , J 6 , J 1 , J 5 , J 3 , J 2 ]
6 13.81905 ( 1.70432 , 3.55333 , 1.2843 , 1.74832 , 0.75632 , 0.76352 ) 805.59236 [ J 4 , J 5 , J 1 , J 6 , J 3 , J 2 ]
Table 8. CPU times (ms) of algorithms.
Table 8. CPU times (ms) of algorithms.
Algorithm 1Algorithm 2Algorithm 3Algorithm 4
nmeanmaxmeanmaxmeanmaxmeanmax
35367.09390.12408.67442.16302.86352.29322.18349.82
45896.46925.061250.461346.82759.28829.51762.24837.24
551864.521876.592386.352587.351562.371582.652118.322297.14
653624.213703.294103.544346.542993.523121.573314.173504.02
756735.586898.347827.508786.355615.585754.396849.916928.42
8513,683.4513,968.3215,072.5316,855.5011,394.3511,528.3913,864.4714,941.57
9523,877.0324,120.5726,147.3527,845.3621,572.7023,489.8624,478.5026,116.25
10534,453.8534,635.5840,527.3242,965.5028,712.1528,892.2734,785.2337,123.85
11553,679.5454,008.9461,296.8663,085.4544,731.6245,021.2352,246.2154,004.32
12577,623.5377,985.7289,614.3695,238.6264,685.8365,023.2176,372.2381,823.32
135114,304.61114,892.56136,835.45139,834.7595,153.3095,802.52117,325.23120,115.84
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Wei, Z.-J.; Wang, L.-Y.; Zhang, L.; Wang, J.-B.; Wang, E. Single-Machine Maintenance Activity Scheduling with Convex Resource Constraints and Learning Effects. Mathematics 2023, 11, 3536. https://doi.org/10.3390/math11163536

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Wei Z-J, Wang L-Y, Zhang L, Wang J-B, Wang E. Single-Machine Maintenance Activity Scheduling with Convex Resource Constraints and Learning Effects. Mathematics. 2023; 11(16):3536. https://doi.org/10.3390/math11163536

Chicago/Turabian Style

Wei, Zong-Jun, Li-Yan Wang, Lei Zhang, Ji-Bo Wang, and Ershen Wang. 2023. "Single-Machine Maintenance Activity Scheduling with Convex Resource Constraints and Learning Effects" Mathematics 11, no. 16: 3536. https://doi.org/10.3390/math11163536

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