MULTITASKING SCHEDULING PROBLEMS WITH A COMMON DUE-WINDOW

We study multitasking scheduling and due-window assignment problems in a single machine, which can be found in various application domains. In multitasking environment, unfinished job always interrupts in-processing job. In common due window assignment, the aim is to find optimal due window to minimise the value of the earliness and tardiness penalty. In this paper, we study two problems, where the objective of the first problem is minimise the earliness, tardiness, due-window starting time, and due-window size costs, the objective of the second problem is minmax common due-date with completion time penalty, then we obtain some analytical properties and provide polynomial time solutions. Finally, the experimental results show that the proposed methods are effective. Mathematics Subject Classification. 90B35. Received June 8, 2020. Accepted May 9, 2021.


Introduction
Multitasking scheduling has attracted much attention in recently years. The multitasking phenomenon can be frequently observed in our daily lives. One of the examples is human multitasking in which a continuing schedule is interrupted by many routine activities such as sending e-mail or making end-of-day file backup in the workplace today. A possible reason is why people multitask are interrupted by clients or colleagues, another possible reason is due to a personality trait because working the same things for a long time is ineffective [15,16].
At the same time, commom due window assignment has remained an important topic in scheduling research. In business, managers usually need to arrange a common due-window. In such a case, a batch of work piece are ordered by the same customer, because of the different completion time of each work piece, the cost of batch shipment is relatively high. In order to save transportation costs, it is usually uniform shipment after all the work pieces are processed, so managers should make a common due date which is a predetermined time point. But the customers may have some amount of inventory when customers order a batch of work piece. Furthermore, there may be some force majeure factors during the long-distance transportation of goods, most customers can tolerant the common due date. Moreover, Just-in-time scheduling can improve customer satisfaction, but the cost is higher. So we relax the assumption of a common due date to a common due window.
To the best of our knowledge, We are the first to study the multitasking scheduling with common due-window. Our contributions mainly include: (1) We consider the multitasking scheduling with common due-window to minimize two objectives, and give some analytical results. (2) We propose two polynomial-time algorithms for the two problems.
(3) We do some numerical experiments to demonstrate the efficiency of the algorithm, and give some future issues.
Owing to the practical significance of common due window scheduling in the multitasking environment, there is a need to study the problem under consideration in this research. We organise the rest of the paper as follows: In Section 2, we introduce some literature about multitasking and common due-window. In Section 3, we give some definitions, then we introduce new multitasking model and give analytical results. In Section 4, We propose a polynomial-time algorithm for the two problems, and give some numerical experiments to demonstrate the efficiency of the algorithm. Finally, we conclude the paper and suggest topics for future research in Section 5.

Literature interview
Research on multitasking scheduling was in initiated by Hall et al. [3], who considered interruption and switching factors, in which a job was interrupted by other unfinished jobs, then they developed some new solution algorithms to solve the classical scheduling model (i.e. the total completion time). Hall et al. [4] considered two scheduling system, which are alternate period processing and shared processing. Zhu et al. [24] considered a rate-modifying activity (RMA) into multitasking scheduling to minimize the maskspan, total completion time, maximum lateness, and due-date assignment related cost. Zhu et al. [25] then extended their work to consider multiple rate-modifying activities to minimise the other objectives about the total completion time, the total waiting time, the total absolute differences in completion time and the total absolute differences in completion time. Liu et al. [12] considered multitasking scheduling on a single machine with the common due-date assignment to minimize the weighted earliness, tardiness, and due date assignment cost, and developed the DDA algorithm to solve the problem. Ji et al. [7] studied parallel-machine scheduling with machine-dependent duewindow assignment to minimize the total cost that comprises the earliness, tardiness, and due-window related costs. Xiong et al. [19] considered the multitasking scheduling problem on unrelated parallel machines to minimize the total weighted completion time and propose an exact branch-and-price algorithm. Li et al. [8] addressed several two-agent scheduling problems in the presence of multitasking, polynomial and pseudo-polynomial time algorithms are proposed to solve the setting, involving various combinations of cost functions. Wang et al. [17] considered a multitasking scheduling model with multiple agents and ascertained the computational complexity status of each of the problems. Yang et al. [21] addressed a two-agent scheduling problem with due date assignment under multitasking environment, they showed that the problem is NP-hard and devised a pseudopolynomial time dynamic programming algorithm. Wang et al. [18] investigated a multitasking scheduling and due date assignment problem with position-dependent deterioration effect and efficiency promotion, then they designed an efficient polynomial time algorithm for several scheduling criteria including the makespan, the total completion time, and two due date-related criteria.
For common due window assignment problems, Liman et al. [9] considered a single machine static and deterministic scheduling problem, the objective is to find the optimal size and location the window and an optimal sequence to minimise earliness, tardiness, window size and window location, they proposed an O(n log n) algorithm to solve the problem. Gur mosheiov et al. [13] addressed a common due-window assignment problem on parallel identical machines. Adam Janiak et al. [5] studied problems of scheduling n unit-time jobs on m identical parallel machines to minimize a weighted sum or maximum of costs associated with job earliness, job tardiness and due window location and size, establish properties of optimal solutions of these min-sum and min-max problems and reduce them to min-sum or min-max assignment problems solvable in O(n 5 /m 2 ) and O(n 4.5 log 0.5 n/m 2 ) time. Yin et al. [22] addressed a batch delivery single-machine scheduling problem in which jobs have an assignable common due window and showed proposed problem can be optimally solved in O(n 8 ) time by a dynamic programming algorithm. Yin et al. [22] consider single-machine batch delivery scheduling with an assignable common due date and controllable processing times and provided an O(n 5 ) dynamic programming algorithm to find the optimal job sequence. Enrique Gerstl et al. [2] studied both cases of a non-restrictive and a restrictive due-window, for both settings they introduce algorithms requiring O(2 m n 3 ) time and O(3 m n 3 ) time, respectively, where n is the number of jobs and m is the number of uniform machines. Recently, due window assignment problems have been studied, among others, by Yang et al. [20], Ji et al. [6], Liu et al. [10], Liu et al. [12], etc.

Problem statement
We assume that n jobs J = {J 1 , J 2 , . . . , J n } are processed on a single machine, and the start time is zero, each job J j has a normal processing time p j , and the machine is working all the time. There is no idle time between any two adjacent jobs. All jobs have common due window, the due-window starting time is d 1 , the due-window finishing time is d 2 , D is equal to d 2 −d 1 , which is defined the window size. Jobs completed in due-window incurs no penalties, other jobs incurs either earliness or tardiness penalties. If the job is early job, Only one job can be processed at a certain time. This job is called the primary job, the interrupted and unfinished jobs are called the waiting jobs of the primary job. Every waiting job must interrupt the primary job one time, which we called as the interruption time. The interruption time caused by job i during the processing of j is given by function The above studies all assume that the interruption function g k (p k ) is Dp k , where 0 < D < 1, in fact, the interrupted time of waiting jobs is uncertain. But for the convenience of calculation, it is also considered as a function of the remaining time of waiting job in this paper. The time during finished time of the former job and start time of the next job is referred to as the switching time, which function is defined as f (|S j |). We assume that the switching time during every two job is same, if the number of the waiting jobs of primary job is k, the f (|S j |) is equal to kδ. In other words, the remaining processing time of the waiting job will decrease because a part of the processing time is completed in the interruptions of previous jobs. When job j is the primary job, we denote S j as all waiting jobs. Each job can be arranged as primary job a time, the multitasking function is defined as f (|S j |) + i∈Sj g i (p i ), which expresses the total amount of interruption during the processing of job j.
As in Hall et al. [3], we define the remaining processing time p i of a waiting job i as a function h i (·) := {1, 2, . . . , n} → R. Specifically, with respect to interruption function g i (p i ), the remaining processing time of job i after it has interrupted l primary jobs is given by h i (l), where 1 ≤ l ≤ n. That is, h i (l) is the remaining processing time of job i when it is considered at the lth position in a schedule.
Observation 1. C max is a constant in an optimal schedule.
Proof. For any scheduling, the total completion time is divided into three parts. The first part is the working time of the main job, the second part is the working time of the waiting job, and the third part is the switching time between the waiting jobs. The main job is a interrupted job, its working time is low p i , but the rest of p i is accomplished on front of itself. So the working time is always equal to p i . What's more, the number of jobs switching is always same for any schedule, in this paper, we consider the switching function is a index 1 function of the number of jobs, so the switching time is equal. The completion time of the last job is equal to the processing time and the switching time of all jobs, so it is same.

Multitasking scheduling problem
We study two problems, the first objective is minimise (αE j + βT j + γd 1 + δD), every early job has (αE j + γd 1 + δD) cost, every late job has (αT j + γd 1 + δD) cost, in order to minimise (αE j + βT j + γd 1 + δD) cost, we hope to minimise the highest cost among all jobs, so another is to minimized max 1≤j≤n {max{αE j + γd 1 + δD, βT j + γd 1 + δD}}, which α is the unit penalty for earliness, β is the unit penalty for tardiness, γ is the unit penalty for the due-window starting time, δ is the unit penalty for the window size.

Common due-window problem
A previous study on common due-window assignment problem is given by Gur Mosheiov, and Assaf Sarig. They consider the problem 1 (αE j +βT j +γd 1 +δD). We study the problem 1 | mt | (αE j +βT j +γd 1 +δD), if the interruption function g(·) ≡ 0 and the switching function f (·) ≡ 0 for all jobs in multitasking environment, which is reduced to 1 (αE j + βT j + γd 1 + δD). So we have some similar results in this case.
an optimal schedule exists in which the due window starts at time zero.
Proof. Given any schedule σ and d 1 ≥ 0, we shift d 1 , units of time to the left, the value of the first term αE j decreases αk , where k denotes the numbers of early jobs, the value of the second term βT j keeps the same with unchanged, the value of the third term γd 1 decreases γn , while the value of the fourth term δD increases δn . Then the change in the total cost is given by: Z = (δ − γ)n -αk ≤ 0. Therefore, implying the optimal d 1 = 0.
Proof. (i) β < γ < δ, suppose that the d 1 > 0, we shift d 2 , units of time to the left, the value of the first term αE j is unchanged, the value of the second term βT j increases βk , where k denotes the numbers of tardy jobs, the value of the third term γd 1 is unchanged, while the value of the fourth term δD decreases δn . Then the change in the total cost is given by: Z = (βk −δn) < (βk −δk) = (β −δ)k < 0. A further shift of d 2 to coincide with d 1 can reduce the cost. The problem 1 | mt | (αE j +βT j +γd 1 +δD) reduces to the problem the 1 | mt | (αE j + βT j + δd 1 ). (ii) β < δ < γ, proposition 1 proposed that the optimal d 1 = d 2 = 0.
So if β < min{δ, γ}, an optimal schedule exists in which the due-window is reduced to a common due-date that starts at time zero.
Lemma 4.4. For the problem 1 | mt | (αE j + βT j + γd 1 + δD), an optimal schedule exists in which the due-window starting time and the due-window completion time coincide with the completion time of a job in a schedule.
, for some k, h ∈ [1, n], we first discuss the d 1 , the following two cases are discussed: (i) We denote 1 = C [k+1] − d 1 increasing the d 1 by 1 , the value of the first term αE j increases by αk 1 , the value of the second term βT j is unchanged, the value of the third term γd 1 increases by γn 1 , the value of the third term δD decreases by γn 1 , then the total increase of the objective function value is [αk + (γ − δ)n] 1 . (ii) We denote 2 = d 1 − C [k] decreasing the d 1 by 2 ,the value of the first term αE j decreases by αk 2 , the value of the second term βT j is unchanged, the value of the third term γd 1 decreases by γn 2 , the value of the third term δD increases by γn 2 , then the total decrease of the objective function value is [αk + (γ − δ)n] 2 .
Notice that in both case, the objective function can express that G , The proof about d 2 is similar to above about d 1 . Therefore, an optimal schedule exists such that both d 1 and d 2 coincide with job completion times.
Hence, the problem can be formulated as following assignment problem: Step 1. Compute d 1 and d 2 at the competition time of the kth job and hth job, where k = n(δ − γ)/α , h = n(β − δ)/β Step 2. For i from 1 to n do Step 2.1. h i (1) = p i .
Step 2.2. For l from 1 to n − 1 do Step 3. Compute all B jr for j, r = 1, . . . , n Step 4. Solve the assignment to determine the local optimal and the total cost Step 5. Schedule the jobs with multitasking in order of σ Property 4.6. For the problem 1 | mt | (αE j + βT j + γd 1 + δD), an optimal schedule σ can be obtained in O(n 3 ) time.

Proof.
Step 1 of algorithm 1 can be solved with constant time.

Minmax common due-window problem
In this section, the aim is to find the optimal sequence and the start time and size of the due-window that minimize the maximum cost. To do this, P2 is formulated as a linear programming (LP) problem. minZ Z ≥ αE j + γd 1 + δD, j = 1, 2, . . . , n, Z ≥ βT j + γd 1 + δD, j = 1, 2, . . . , n, d 1 , D, Z ≥ 0, j = 1, 2, . . . , n, Property 4.8. There are four different cases with specific cost functions: Case 1. β < γ, δ ≥ β, the due window is reduced to a due date at time zero, implying that all jobs are tardy.
Case 3. β ≥ γ, δ > γ and δ ≥ β α+γ α+β , the start time of the due window coincides with the completion time of the first job and all jobs are on time.
So we can reformulate the objective functions as follows.
Step 1. Compute d 1 and d 2 at the competition time of the kth job and hth job, where k = n(δ − γ)/α , h = n(β − δ)/β Step 2. For i from 1 to n do Step 2.1. h i (1) = p i .
Step 2.2. For l from 1 to n − 1 do Step 3. Compute all C jr by for j, r = 1, . . . , n Step 4. Solve the assignment to determine the local optimal and the total cost Step 5. Schedule the jobs with multitasking in order of σ Property 4.9. For the problem 1 | mt | max{max{αE j + γd 1 + δD, βT j + γd 1 + δD}}, an optimal schedule σ can be obtained in O(n 3 ) time.

Proof.
Step 1 of algorithm 2 can be solved with constant time.
Step 2 of algorithm 2, as the processing step by the remaining processing times of all jobs after they have interrupted j primary jobs, for j = 1, . . . , n, are computed, required O(n 2 ) times, step 3 of algorithm 2 can be obtained in O(n 2 ) times. The classic assignment problem can be solved with O(n 3 ) times. Thus the above theorem holds. We solve the 8 × 8 job-position processing times problem given in Table 1 with four different combinations of the cost parameters, depicting one of the four sub-cases discussed in Property 4.8. The results are shown in Tables 3-7.

Conclusion
We study the method of common due-window assignment and multitasking scheduling problem. We have proposed two polynomial algorithms to solve the single-machine common due window problem to minimise a cost function based on earliness, tardiness, window size, and due-window starting times, we illustrate the algorithm with a numerical example. We then show that the solutions can be extended to the problem with minmax common due-window assignment.