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Scheduling Piecewise Linear Deteriorating Jobs to Minimize Makespan in a Two-Machine Flowshop

Abstract

A job processed later may consume more time in many realistic machine scheduling situations than the same job when processed earlier. This phenomenon is known as the deterioration of jobs. In this paper, two-machine flowshop scheduling problem under piecewise linear deteriorating jobs is investigated where the objective is to minimize the makespan. Deteriorating jobs may increase the makespan and it is important to obtain the optimal or near-optimal solution. Tab;At first, a Mixed Integer Non-Linear Programming (MINLP) model is presented for the problem and linearization of it is performed. After that, a branch and bound algorithm with dominance rules and lower bounds is established to solve the problem optimally. Several simple heuristics are also proposed to derive near-optimal solutions. The computational experiments are conducted through randomly generated examples to evaluate the performance of the proposed algorithms. The results demonstrate that the branch and bound algorithm can solve most medium-sized problems within a reasonable time. The heuristic algorithm is relatively accurate, with an average error percentage of less than 1.6%.

Introduction

In the classical scheduling problems, the job processing times are assumed to be constant and known during the processing [1]. A job processed later needs more time to be processed in some production environments than the same job when processed earlier [2]. This phenomenon is called the deteriorating effects and the actual processing time of the job depends on its starting time or its position in the sequence [3]. This situation appears in many real-world applications, e.g., fire-fighting, maintenance planning, medical procedures, steel production, and lathing process [4, 5]. The time and effort to cease fire will increase if there exists a delay in starting the fire-fighting operations. Any delay in medical procedures is penalized by incurring additional time for accomplishment. In steel production, the ingots are needed to be re-heated up to the required temperature before the hot-roll operation if their temperatures while waiting drop below a threshold level [6].

There is an interest in the literature to study the scheduling problems with deteriorating jobs, formulated numerous models with various criteria and solved by branch and bound (B&B), heuristic, and metaheuristic algorithms [4,5,6,7,8,9,10,11]. Flowshop scheduling problems have received increasing attention in recent years by various assumptions [12,13,14,15,16,17].

Generally, three types of function describe the deteriorating jobs: linear, piecewise linear, and non-linear [8]. Most of the previous works assumed a linear deterioration function. The actual processing time of ith job (Pi) in the linear deterioration is an increasing linear function of its start time (S); the deterioration rates of jobs (bi) may be similar or different [4]. In different deterioration rates, the normal processing times (ai) might be zero or positive. Therefore, the linear deterioration functions are presented as follows:\({P}_{i}={a}_{i}+{b}_{i}S, {P}_{i}={a}_{i}+bS, {P}_{i}={b}_{i}S\). Some authors assumed that the deterioration function is piecewise linear where the actual processing time of each job is a non-decreasing function of its start time based on two or more constant or linear criteria; it could be found in Moslehi and Jafari [6] and Kubiak and Velde [18]. The actual processing time of the job in the non-linear deterioration is an increasing non-linear function of its position in the schedule; it was used by a few authors [19]. Below, we provide a critical review of the literature regarding the linear and piecewise linear deteriorating job scheduling problems on a single machine, flowshop and parallel machines. We employ the three-field notation \(\alpha \left|\beta \right|\gamma\) where \(\alpha\) denotes the machine environment, e.g., single machine and flowshop, \(\beta\) represents the problem specification such as deteriorating effect, maintenance activity and ready time, and \(\gamma\) demonstrates the objective function, for example, makespan, the number of tardy jobs and total completion time and [20].

Woo et al. [21] addressed a linear deteriorating job scheduling problem (LDJSP) on a single machine. The research objective is to minimize the makespan and determine the number and positions of maintenance activities. They introduced a mixed integer linear programming (MILP) model to solve the problem. Jafari and Moslehi [4] proved that problem \(1\left|{P}_{i}={a}_{i}+bS\left|\sum {U}_{i}\right.\right.\) is NP-hard; hence, a B&B and heuristic procedure was proposed. Lee et al. [7] presented heuristic and B&B algorithms for LDJSP with release times, i.e., \(1\left|{r}_{i},{P}_{i}={a}_{i}+bS\left|{C}_{max}\right.\right.\), to minimize makespan on a single machine where the proposed algorithms were able to solve the problems up to 28 jobs. Mosheiov [22] presented the optimal solution for the simple linear deteriorating job scheduling problem on a single machine where \({P}_{i}={b}_{i}S\) using the simple rules for performance criteria Cmax, \(\sum {C}_{i}\),\(\sum {w}_{i}{C}_{i}\), Tmax, Lmax, and \(\sum {U}_{i}\). Some researches supposed that the setup time of each job is not constant, and similar to the processing time, it is a simple linear function of start time [23]. Lee et al. [24] proposed a B&B algorithm for the problem \(1\left|{P}_{i}={b}_{i}S,{S}_{i}={b}_{i}^{3}S\left|\sum {U}_{i}\right.\right.\), which could solve the samples up to 1000 jobs in a reasonable time. Kong et al. [25] addressed the problem \(1\left|{P}_{i}={b}_{i}S,{S}_{i}={b}_{i}^{^{\prime}}S, rej\left|{C}_{max}\right.\right.\) and developed a dynamic programming procedure to solve it. Cheng et al. [26] and Lee and Lu [2] considered the problems \(1\left|{P}_{i}={b}_{i}S,{S}_{i}={b}_{i}^{3}S\left|{T}_{max}\right.\right.\) and \(1\left|{P}_{i}={b}_{i}S,{S}_{i}={b}_{i}^{^{\prime}}S\left|\sum {w}_{i}{U}_{i}\right.\right.\) and presented a B&B algorithm to solve the problem, respectively.

Some authors introduced and investigated the piecewise linear deteriorating job scheduling problem (PLDJSP). Chung and Kim [27] considered PLDJSP on a single machine where the objective is to determine the number and positions of rate-modifying activities and minimize the makespan. They derived a MILP model to solve the problem optimally and proposed a hybrid genetic algorithm to obtain a near-optimal solution. Layegh et al. [28] addressed minimizing total weighted completion time on a single machine under piecewise linear deterioration. They proposed the memetic algorithm using dominance properties that the average percentage error of the algorithm from optimal solutions is about 2%. Cheng et al. [29] considered the problem \(1\left|{P}_{i}={a}_{i} or {a}_{i}+{b}_{i}\left|\sum {C}_{i}\right.\right.\) and showed it is NP-hard. They developed a dynamic programming algorithm that solved the problem in pseudo-polynomial time. Kubiak and Velde [18] studied the problem \(1\left|{P}_{i}\left|{C}_{max}\right.\right.\) where Pi is considered as the following non-decreasing three-part function where y1 and y2 are the input variables.

$$P_i=\left\{\begin{array}{cc}a_i&if\;S\;\leq\;y_1\\a_{ji}+b_i(S-y_1)&if\;y_1\;<\;S<y_2\\a_i+b_1(y_2-y_1)&if\;S\geq\;y_2\end{array}\right.$$
(1)

They showed that the particular case y2 = ∞ and y1 \(>\) 0 is NP-hard and proposed a binary B&B algorithm. Moslehi and Jafari [6] surveyed the PLDJSP, where the deterioration function was similar to Eq. (1) and the objective function was to minimize the number of tardy jobs. They proved that the problem \(1\left|{P}_{i}={a}_{i}+{b}_{i}\left(S-{y}_{1}\right), {y}_{1}>0,{y}_{2}>{y}_{1}\left|\sum {U}_{i}\right.\right.\) is NP-hard and developed a B&B procedure and a heuristic algorithm.

Ng et al. [10] proposed a B&B algorithm for the problem \(F2\left|{P}_{i}={a}_{i}+{b}_{i}S\left|\sum {C}_{i}\right.\right.\), which handled the problems up to 15 jobs. Also, they used a heuristic algorithm in the proposed B&B algorithm as an upper bound. Wu and Lee [8] presented a B&B and several heuristics for the problem \(F2\left|{P}_{i}={a}_{i}+bs\right|\overline{F }\). The research was continued by Lee et al. [12] where they developed a B&B and several heuristics to minimize the makespan. Wang et al. [15] showed that the problem \(F2\left|{P}_{i}={b}_{i}S\left|\sum {C}_{i}\right.\right.\) is NP-hard; they suggested a B&B algorithm that could handle up to 14 jobs. Yang and Wang [13] developed a B&B and heuristic algorithm for the problem \(F2\left|{P}_{i}={b}_{i}S\left|\sum {w}_{i}{C}_{i}\right.\right.\). Jafari et al. [30] studied the problem \(F2\left|{P}_{i}={a}_{i}+{b}_{i}s\right|{C}_{max}\) and proposed a B&B and several heuristic algorithms to solve it. Wang and Wang [9] studied the problem \(F3\left|{P}_{i}={a}_{i}+bs\right|{C}_{max}\). They derived several dominance properties, some lower bounds, and two heuristic algorithms applied in a proposed B&B algorithm to earn an optimal solution. Jafari et al. [31] showed that the dominance properties in Wang and Wang [9] are incorrect. Wang et al. [14] used a B&B and heuristic algorithm to obtain the optimal solution for the problem \(F3\left|{P}_{i}={b}_{i}s\right|{C}_{max}\) with 14 jobs. Jafari et al. [32] considered the problem and proved that the B&B algorithm in Wang et al. [14] is incorrect.

Mosheiov et al. [33] investigated an m-machine flowshop scheduling problem under linear deterioration where the normal processing time of each job on m machines is equal. They proved that the problem is optimally solved by sequencing the jobs in non-decreasing order \({a}_{i}{{b}_{i}}\). Lee et al. [34] considered the problem \(Fm\left|{P}_{i}={a}_{i}+{b}_{i}S\left|\sum {T}_{i}\right.\right.\) and developed a B&B algorithm and two metaheuristic approaches to solve it.

Ouazene and Yalaoui [35] considered the problem \(P\left|{P}_{i}={b}_{i}S\left|{C}_{max}\right.\right.\) and presented heuristic algorithms to solve it. Cheng et al. [36] proved that the problem \(Pm\left|{P}_{i}={b}_{i}S\left|{C}_{max}\right.\right.\) is NP-hard and proposed a heuristic algorithm. Mor and Mosheiov [3] addressed the problem \(Pm\left|{P}_{i}=1+{b}_{i}S\text{\hspace{0.17em}}\left|\sum {C}_{max}\right.\right.\) and proposed a heuristic algorithm to find the near-optimal solution. The result showed that the algorithm performs very well. Woo and Kim [37] studied PLDJSP with rate-modifying activities on the parallel machine. They presented a MILP model and metaheuristic algorithms such as simulated annealing and genetic algorithm to solve the problem. Woo et al. [38] continued this research on the unrelated parallel machine and presented a MILP model and genetic algorithm to obtain an optimal and near-optimal solution, respectively.

Noteworthily, LDJSP has received sufficient attention in recent years, especially in the flowshop environment. But the existing researches addressed the problem with linear deterioration function. Notably, there are two or more series machines similar to the flowshop environment in many real production situations such as steel rolling and lathing processes. On the other hand, the actual processing time of each job in real-world applications is a piecewise linear function. Deterioration happens in a period after the start of the process, leading to an increase in the actual processing time. This increment there is not at first and will not continue to the end. The deterioration will be started at a specific time and after another specific time will be constant until the end of the process. The piecewise linear deterioration function will cover all the possible forms of linear deterioration functions. Therefore, this function is addressed in the paper. No research can be found on the PLDJSP in the flowshop environment in the literature review; thus, the problem is considered in the study. The main contribution of the paper is the use of the piecewise linear deterioration function in the flowshop scheduling problem. Dominance rules and lower bounds in the B&B algorithm are other contributions of the research.

This paper is organized as follows. In Sect. 2, the problem is described and the mathematical model is presented. In Sect. 3, the complexity of the problem is discussed. The B&B algorithm and several simple heuristic procedures are established in Sects. 4 and 5, respectively. In Sect. 6, the computational experiments are provided to assess the efficiency of the proposed algorithms. Finally, the conclusion and directions for future researches are mentioned in Sect. 7.

Problem Description and Mathematical Model

In this section, we consider the PLDJSP in the flowshop environment. At first, a mathematical model is presented for the problem, and then, this problem is solved by B&B and heuristic algorithms. The indexes, parameters, and variables needed to define the problem are given as follows.

Indexes:
i: Index of jobs.......... i = 1, 2,..., n
j: Index of machines.. j = 1, 2
k: Index of the area of deterioration function.. k = 1, 2, 3
l: Index of the position of the job.. l = 1, 2,.., n
Parameters:
n: Number of jobs
Ji: ith job
Mj: jth machine
\(N=\{J_1,J_2,...,J_n\}\): The set of jobs to be scheduled
\(a_{ij}\): Normal processing time of Ji on Mj
\(b_{i}\): Deterioration rate of Ji
\(y_{1},y_{2}\): The parameters that determine the area of deterioration function
M: The large number
Variables:
\(P_{ij}\): Actual processing time of Ji on Mj
\(S_{ij}\): Actual starting time of Ji on Mj
\(C_{ij}\): Completion time of Ji on Mj
\(C_{\lbrack l\rbrack j}\): Completion time of lth position on Mj
\(X_{i\lbrack l\rbrack}\): The binary variable is equal to 1 if Ji is scheduled in position l; otherwise, it equals 0
\(a_{ik}\): The binary variable is equal to 1 if Ji is scheduled according to the kth area of deterioration function; otherwise, it equals 0

It is assumed that all the jobs are available for processing at time T0 ≥ 0 without interruption or preemption. Machines are supposed to be available all the times, and each one can handle one job at the moment. Each job can be processed on only one machine at a time. Pij is calculated based on a piecewise linear function of start time Sij, as Eq. (2) and Fig. 1, in which bi is deterioration rate and y1 and y2 are considered the input parameters for computation of Pij. Parameters y1 and y2 get their values independent of machines. Notably, if they are dependent upon the machines, some dominance rules need only to be added and the proposed algorithms in the following sections will still be valid.

Fig. 1
figure1

Actual processing time of Ji

Let Cij(S) and C[l]j(S) denote the completion time of Ji on Mj and the completion time of the job scheduled in the lth position on Mj under schedule S, respectively. The objective is to earn an optimal schedule S* so that for any schedule S, we have:

\(Max\left\{{C}_{12}({S}^{*}),{C}_{22}({S}^{*}),...,{C}_{n2}({S}^{*})\right\}\le Max\left\{{C}_{12}(S),{C}_{22}(S),...,{C}_{n2}(S)\right\}\) or \({C}_{[n]2}({S}^{*})\le {C}_{[n]2}(S)\)

The problem can be demonstrated \(F2\left|{P}_{ij}={a}_{ij}+{b}_{i}\left({S}_{ij}-{y}_{1}\right), {y}_{1}>0,{y}_{2}>{y}_{1}\left|{C}_{max}\right.\right.\) using the three-field notation.

$$P_i=\left\{\begin{array}{cc}a_{ij}&if\;S\;\leq\;y_1\\a_{ij}+b_i(S_{ij}-y_1)&if\;y_1\;<\;S_{ij}<y_2\\a_{ij}+b_1(y_2-y_1)&if\;S_{ij}\geq\;y_2\end{array}\right.$$
(2)

For the problem \(F2\left|{P}_{ij}={a}_{ij}+{b}_{i}\left({S}_{ij}-{y}_{1}\right), {y}_{1}>0,{y}_{2}>{y}_{1}\left|{C}_{max}\right.\right.\), the Mixed Integer Non-Linear Programming (MINLP) is provided as follows.

$$Min\qquad{C}_{n2}$$
(3)
$$S.t:$$
$$\sum_{i=1}^nX_{i\lbrack l\rbrack}=1\qquad\qquad l=1,2,...,n$$
(4)
$$\sum_{l=1}^{n}{X}_{i\left[l\right]}=1\qquad\qquad i=\mathrm{1,2},\dots ,n$$
(5)
$$P_{ij}=\alpha_{il}\alpha_{ij}+\alpha_{i2}(\alpha_{ij}+b_{i}(S_{ij}-y_{1}))+\alpha_{i3}(\alpha_{ij}+b_{i}(y_{2}-y_{1}))\qquad\qquad i=1,2,\dots,n\quad j=1,2$$
(6)
$$S_{ij}\leq y_1+M(1-\alpha_{i1})\qquad\qquad i=1,2,\dots,n\qquad j=1,2$$
(7)
$$S_{ij}>y_1-M(1-\alpha_{i2})\qquad\qquad i=1,2,\dots,n\qquad j=1,2$$
(8)
$$S_{ij}\leq y_2+M(1-\alpha_{i2})\qquad\qquad i=1,2,\dots,n\qquad j=1,2$$
(9)
$$S_{ij}>y_2-M(1-\alpha_{i3})\qquad\qquad i=1,2,\dots,n\qquad j=1,2$$
(10)
$$\alpha_{i1}+ \alpha_{i2}+\alpha_{i3}=1\qquad i=1,2,\dots,n$$
(11)
$$S_{ij}\geq C_{[l]j-1}X_{i[l]} \qquad\qquad l=1,2,\dots,n\qquad i=1,2,\dots,n\qquad j=2$$
(12)
$$S_{ij}\geq C_{[l-1]j}X_{i[l]} \qquad\qquad l=2,3,\dots,n\qquad i=1,2,\dots,n\qquad j=1,2$$
(13)
$$C_{\lbrack l\rbrack j}\geq S_{ij}X_{i\lbrack l\rbrack}+P_{ij}X_{i\lbrack l\rbrack}\qquad\qquad l=1,2,\dots,n\qquad i=1,2,\dots,n\qquad j=1,2$$
(14)
$$x_{i\lbrack l\rbrack},\alpha_{i1},\alpha_{i2},\alpha_{i3}\in\{0,1\}\qquad\qquad i=1,2,\dots,n\qquad l=1,2\dots,n$$
(15)
$$C_{\lbrack l\rbrack j}\geq0\qquad\qquad i=1,2,\dots,n\qquad l=1,2$$
(16)
$$S_{ij}\geq0\qquad\qquad i=1,2,\dots,n\qquad l=1,2$$
(17)
$$P_{ij}\geq0\qquad\qquad i=1,2,\dots,n\qquad l=1,2$$
(18)

The objective function in the model is to minimize the makespan, which is equal to the completion time of nth position on M2 as. \({C}_{[n]2}\) Constraint (4) shows that each position like l must include one job. Equation (5) ensures that each job like i must be scheduled in sequence exactly one time. The actual processing time of each job on each machine is calculated by Constraint (6), according to Eq. (2). Constraints (7), (8), (9), and (10) determine the area of deterioration function by comparison Sij with y1 and y2 so that at any comparison, exactly one area is selected and controlled by Eq. (11). In the flowshop problems, the actual starting time of each job on Mj in position l depends on two factors. The first one is the completion time of the scheduled job in position l on Mj-1 is calculated by Constraint (12). Another one is the completion time of the scheduled job in the previous position on Mj is calculated by Constraint (13). Equation (14) determines the completion time of each scheduled job in position l on Mj. The conditions on the variables are controlled by constraints (15), (16), (17), and (18).

Linearization of the Model

Constraints (12), (13), and (14) are non-linear due to the multiplication of a binary variable by a non-negative variable which leads to the complexity of the model. Using the auxiliary variables and applying some additional constraints, the model can be converted to a linear model. A non-negative variable \(C{X}_{[l](j-1)i[l]}\) is used instead of the term \({C}_{[l]j-1}{X}_{i[l]}\) in constraint (12) and below constraints are defined for the linearization of constraint (12).

$$C{X}_{[l](j-1)i[l]}\le M{X}_{i[l]}$$
(19)
$$C{X}_{[l](j-1)i[l]}\le {C}_{[l]j-1}$$
(20)
$$C{X}_{[l](j-1)i[l]}\ge {C}_{[l]j-1}-M(1-{X}_{i[l]})$$
(21)

In constraint (13) using a non-negative variable \(C{X}_{[l-1]ji[l]}\) instead of \({C}_{[l-1]j}{X}_{i[l]}\) and non-negative variables \(S{X}_{iji[l]}\) and \(P{X}_{iji[l]}\) instead of \({S}_{ij}{X}_{i[l]}\) and \({P}_{ij}{X}_{i[l]}\) in constraint (14), the above procedure is repeated for the linearization of constraints (13) and (14).

$$C{X}_{[l-1]ji[l]}\le M{X}_{i[l]}$$
(22)
$$C{X}_{[l-1]ji[l]}\le {C}_{[l-1]j}$$
(23)
$$C{X}_{\left[l-1\right]ji\left[l\right]}\ge {C}_{\left[l-1\right]j}-M\left(1-{X}_{i\left[l\right]}\right)$$
(24)
$$S{X}_{iji[l]}\le M{X}_{i[l]}$$
(25)
$$S{X}_{iji[l]}\le {S}_{ij}$$
(26)
$$S{X}_{iji[l]}\ge {S}_{ij}-M(1-{X}_{i[l]})$$
(27)
$$P{X}_{iji[l]}\le M{X}_{i[l]}$$
(28)
$$P{X}_{iji[l]}\le {P}_{ij}$$
(29)
$$P{X}_{iji[l]}\ge {P}_{ij}-M(1-{X}_{i[l]})$$
(30)

The Complexity of the Problem

This section establishes the PLDJSP in a two-machine flowshop environment (i.e., machines M1 and M2) as \(F2\left|{P}_{ij}={a}_{ij}+{b}_{i}\left({S}_{ij}-{y}_{1}\right), {y}_{1}>0,{y}_{2}>{y}_{1}\left|{C}_{max}\right.\right.\). At first, we analyze the complexity of the problem based on relevant researches in literature. If any problem P is reduced to problem Q and problem P is NP-hard, then problem Q will also be NP-hard [39]. Kubiak and Velde [18] proved that the problem \(1\left|{P}_{i}={a}_{i}+{b}_{i}\left({S}_{i}-{y}_{1}\right), {y}_{1}>0,{y}_{2}=\infty \left|{C}_{max}\right.\right.\) is NP-hard. It is shown that any scheduling problem on a single machine can be reduced to the same problem on two-machine flowshop [39]. As a result the problem \(1\left|{P}_{i}={a}_{i}+{b}_{i}\left({S}_{i}-{y}_{1}\right), {y}_{1}>0,{y}_{2}=\infty \left|{C}_{max}\right.\right.\) is reducible to \(F2\left|{P}_{ij}={a}_{ij}+{b}_{i}\left({S}_{ij}-{y}_{1}\right), {y}_{1}>0,{y}_{2}>{y}_{1}\left|{C}_{max}\right.\right.\); therefore, the problem \(F2\left|{P}_{ij}={a}_{ij}+{b}_{i}\left({S}_{ij}-{y}_{1}\right), {y}_{1}>0,{y}_{2}>{y}_{1}\left|{C}_{max}\right.\right.\) is also known as NP-hard. A B&B algorithm will be proposed to obtain the optimal solution to the problem \(F2\left|{P}_{ij}={a}_{ij}+{b}_{i}\left({S}_{ij}-{y}_{1}\right), {y}_{1}>0,{y}_{2}>{y}_{1}\left|{C}_{max}\right.\right.\) according to the complexity of the problem. In addition, several heuristics are provided and the best one can be used as an upper bound for the B&B algorithm. They are also utilized to find the near-optimal solution for large-sized instances.

B&B Algorithm

A B&B algorithm using the depth first strategy is presented in this section, which requires very little memory and computational effort. This algorithm assigns the jobs to the positions in sequence in a forward manner. Each time, a branch is chosen and is systematically worked down the searching tree until dominance rules, and lower bounds fathom it or the final node is reached. The latter case is used as a substitute for the upper bound if the makespan is decreased. Therefore, an upper bound for the makespan and several dominance properties are established in Sects. 4.1 and 4.2, respectively. Besides, nine lower bounds are proposed in Sect. 4.3 to fathom the nodes of the searching tree.

Upper Bound

The best heuristic algorithm described in the next section is considered an upper bound for the makespan and its final sequence will be supposed to make the searching tree in the B&B algorithm.

Dominance Rules

Dominance rules are essential tools for node elimination. Let \({\delta }_{1}= (\pi ,{J}_{i},{J}_{j},\pi )\) where \(\pi\) and \({\pi }^{^{\prime}}\) are partial sequences. A new sequence \({\delta }_{2}= (\pi ,{J}_{j},{J}_{i},\pi )\) is created if a pairwise interchange of Ji and Jj in δ1 is performed. It is assumed that the completion time of \(\pi\) on M1 and M2 in δ1 is S1 and S2, respectively. It is obvious S1, and S2 are similar to δ1 in δ2. Hence, to prove that δ1 dominates δ2, it is sufficient to show \(Max\left\{{C}_{k1}({\delta }_{1}),{C}_{j2}\left({\delta }_{1}\right)\right\}\le Max\left\{{C}_{k1}({\delta }_{2}),{C}_{i2}\left({\delta }_{2}\right)\right\}\) where JK is the first job in \({\pi }^{^{\prime}}\).

Dominance rules are defined according to the completion time of jobs. Also, the completion time of jobs is calculated based on their actual processing times. As previously discussed, the actual processing times of jobs are computed based on Eq. (2) by comparing the start times with y1 and y2. Therefore, there will be several dominance rules according to y1 and y2 presented in the forms of 36 situations in the Appendix.

Each situation includes a set of conditions; at first, there are some equations called basic equations. After that, a set of equations are classified into three categories A, B, and C. Notably, the categories A, B, and C may not simultaneously exist in some situations, e.g., situation 14. The basic equations, along with exactly one equation from each existing category, will form a dominance rule. In situation 16, for example, four dominance rules according to the basic equations and exactly one equation from categories A and C will be made as follows.

Basic equations:

$$Min\left\{{S}_{1}+{a}_{i1},{S}_{1}+{a}_{j1}\right\}\ge {y}_{2}\;Max\left\{{S}_{1},\right.\left.{S}_{2}\right\}\le {y}_{1}\;{b}_{j}\le {b}_{i}$$

Exactly one equation from case A:

$${b}_{j}({y}_{2}-{y}_{1})+{a}_{i1}\le {a}_{i2}+{b}_{i}({y}_{2}-{y}_{1})$$

Exactly one equation from case C:

$${a}_{i1}\le {a}_{j1}$$

Since there is no case B in situation 16, only cases A and C are considered. Accordingly, the following four properties are derived.

Property 1

If, \(Min\left\{{S}_{1}+{a}_{i1},{S}_{1}+{a}_{j1}\right\}\ge {y}_{2}\), \(Max\left\{{S}_{1},\right.\left.{S}_{2}\right\}\le {y}_{1}\), \({b}_{j}\le {b}_{i}\) \({b}_{j}({y}_{2}-{y}_{1})+{a}_{i1}\le {a}_{i2}+{b}_{i}({y}_{2}-{y}_{1})\) and, \({a}_{i1}\le {a}_{j1}\) then δ1 dominates δ2.

Proof

From \(Min\left\{{S}_{1}+{a}_{i1},{S}_{1}+{a}_{j1}\right\}\ge {y}_{2} \mathrm{and}\;Max\left\{{S}_{1},\right.\left.{S}_{2}\right\}\le {y}_{1}\) we have:

$$Min\left\{{S}_{1}+{a}_{i1},{S}_{1}+{a}_{j1}\right\}\ge Max\left\{{S}_{1},\right.\left.{S}_{2}\right\}$$
$$Min\left\{a_{j1}+b_j(y_2-y_1)+S_1+a_{i1},\right.\left.a_{i2}+b_i(y_2-y_1)+S_1+a_{i1}\right\}\geq y_2$$
$$Min\left\{{a}_{j2}+{b}_{j}({y}_{2}-{y}_{1})+{S}_{1}+{a}_{j1},\right.\left.{a}_{i1}+{b}_{i}({y}_{2}-{y}_{1})+{S}_{1}+{a}_{j1}\right\}\ge {y}_{2}$$

Thus, the completion times of Ji and Ji on M2 in \({\delta }_{1}\) and \({\delta }_{2}\) are, respectively, calculated as follows:

$$\begin{aligned}C_{j2}(\delta_1)& =a_{j2}+b_j(y_2-y_1)+Max\left\{a_{j1}+b_j(y_2-y_1)+S_1+a_{i1},\right.\left.a_{i2}+b_i(y_2-y_1)+S_1+a_{i1}\right\}\\ &=\;Max\left\{a_{j2}+b_j(y_2-y_1)+a_{j1}+b_j(y_2-y_1)+S_1+a_{i1},\right.\left.a_{j2}+b_j(y_2-y_1)+a_{i2}+b_i(y_2-y_1)+S_1+a_{i1}\right\}\end{aligned}$$
(31)

And

$$\begin{aligned}C_{i2}(\delta_2)& =a_{i2}+b_i(y_2-y_1)+Max\left\{a_{j2}+b_j(y_2-y_1)+S_1+a_{j1},\right.\left.a_{i1}+b_i(y_2-y_1)+S_1+a_{j1}\right\}\\ &=\;Max\left\{a_{i2}+b_i(y_2-y_1)+a_{j2}+b_j(y_2-y_1)+S_1+a_{j1},\right.\left.a_{i2}+b_i(y_2-y_1)+a_{i1}+b_i(y_2-y_1)+S_1+a_{i1}\right\}\end{aligned}$$
(32)

Owing to the relations \({b}_{j}({y}_{2}-{y}_{1})+{a}_{i1}\le {a}_{i2}+{b}_{i}({y}_{2}-{y}_{1})\) and, \({a}_{i1}\le {a}_{j1}\) we have:

the first term in Eq. (31) ≤ the first term in Eq. (32).

the second term in Eq. (31) ≤ the first term in Eq. (32).

Hence, \({C}_{j2}({\delta }_{1})\le {C}_{i2}({\delta }_{2})\).

The completion time of Jk on M1 in \({\delta }_{2}\) and \({\delta }_{2}\) is determined as follows, respectively:

$${C}_{k1}({\delta }_{1})={a}_{k1}+{b}_{k}({y}_{2}-{y}_{1})+{a}_{j1}+{b}_{j}({y}_{2}-{y}_{1})+{S}_{1}+{a}_{i1}$$
(33)

And

$${C}_{k1}({\delta }_{2})={a}_{k1}+{b}_{k}({y}_{2}-{y}_{1})+{a}_{il}+{b}_{i}({y}_{2}-{y}_{1})+{S}_{1}+{a}_{jl}$$
(34)

From \({b}_{j}\le {b}_{i}\) it implies that \({C}_{k1}({\delta }_{1})\le {C}_{k1}({\delta }_{2})\).

Because of establishing the relations \({C}_{k1}({\delta }_{1})\le {C}_{k1}({\delta }_{2})\) and \({C}_{j2}({\delta }_{1})\le {C}_{i2}({\delta }_{2})\), the relation \(Max\left\{{C}_{k1}({\delta }_{1}),{C}_{j2}\left({\delta }_{1}\right)\right\}\le Max\left\{{C}_{k1}({\delta }_{2}),{C}_{i2}\left({\delta }_{2}\right)\right\}\) is satisfied; so, δ1 dominates δ2.

Property 2

If, \(Min\left\{{S}_{1}+{a}_{i1},{S}_{1}+{a}_{j1}\right\}\ge {y}_{2}\), \(Max\left\{{S}_{1},\right.\left.{S}_{2}\right\}\le {y}_{1}\), \({b}_{j}\le {b}_{i}\) \({b}_{j}({y}_{2}-{y}_{1})+{a}_{i1}\le {a}_{i2}+{b}_{i}({y}_{2}-{y}_{1})\) and, \({a}_{j2}+{b}_{j}({y}_{2}-{y}_{1})\le {a}_{j1}+{b}_{i}({y}_{2}-{y}_{1})\) then δ1 dominates δ2.

Property 3

If, \(Min\left\{{S}_{1}+{a}_{i1},{S}_{1}+{a}_{j1}\right\}\ge {y}_{2}\), \(Max\left\{{S}_{1},\right.\left.{S}_{2}\right\}\le {y}_{1}\), \({b}_{j}\le {b}_{i}\) \({a}_{j2}+2{b}_{j}({y}_{2}-{y}_{1})\le {a}_{i2}+2{b}_{i}({y}_{2}-{y}_{1})\) and, \({a}_{i1}\le {a}_{j1}\) then δ1 dominates δ2.

Property 4

If, \(Min\left\{{S}_{1}+{a}_{i1},{S}_{1}+{a}_{j1}\right\}\ge {y}_{2}\), \(Max\left\{{S}_{1},\right.\left.{S}_{2}\right\}\le {y}_{1}\), \({b}_{j}\le {b}_{i}\) \({a}_{j2}+2{b}_{j}({y}_{2}-{y}_{1})\le {a}_{i2}+2{b}_{i}({y}_{2}-{y}_{1})\) and, \({a}_{j2}+{b}_{j}({y}_{2}-{y}_{1})\le {a}_{j1}+{b}_{i}({y}_{2}-{y}_{1})\) then δ1 dominates δ2.

Notably, the proofs of the other properties are omitted since they are similar to that of property 1. They will be available upon request of the interested readers.

Lemma 1

In the problem \(F2\left|{P}_{ij}={a}_{ij}+{b}_{i}\left({S}_{ij}-{y}_{1}\right), {y}_{1}>0,{y}_{2}>{y}_{1}\left|{C}_{max}\right.\right.\), if S1 ≥ y2, then the optimal sequence after y2 will be obtained by executing the well-known Johnson’s rule (JR) for set δ′.

Proof

After y2, the actual processing time of Ji on Mj in the set δ’ will be constant. \({P}_{ij}={a}_{ij}+{b}_{i}({y}_{2}-{y}_{1})\) Thus, the problem after y2 is equivalent to the basic form \(F2||{C}_{max}\) where the JR algorithm gives the optimal solution.

Lower Bounds

The lower bounds help further speeding up the searching process in the B&B algorithm. In this subsection, nine lower bounds are designed on M1 and M2. Suppose that δ is a partial sequence where the positions of the first k jobs have been determined and δ’ denotes the set of r remaining unscheduled jobs (note that \(k+r=n\)). Furthermore, let S1 and S2 denote the completion time of the last job in δ on M1 and M2, respectively. In each node of the searching tree, the objective function of the partial sequence δ is considered as and its lower bound is shown by LB*. The following theorems are used to obtain LB*.

Theorem 1

The following relation calculates the lower bound of the partial sequence \(\delta\) according to M1:

$$LB=P+Min\left\{a_{i2}+b_i(Max\{y_1,Min\{y_2,P\right\}\}-y_1\}$$
(35)
$$where\;P=S_1+\sum_{i\in\delta'}(a_{il}+b_iMax\left\{y_{1,}Min\left\{S_1,y_2\right\}\right\}-y_1))$$

Proof

According to Eq. (2) and Fig. 1, the actual processing time of Ji on Mj is:

$$P_{ij}=a_{ij}+b_i(Max\left\{y_1,Min\{S_{ij},y_2\}\right\}-y_1$$

Similarly, for each Ji in, δ’ the following relations are valid:

$$P_{il}\geq a_{il}+b_i(Max\left\{y_1,Min\{S_{1},y_2\}\right\}-y_1$$
$$\begin{aligned}{P}_{\left[n\right]2}=&{a}_{\left[n\right]2}+{b}_{\left[n\right]}\left(Max\left\{{y}_{1},Min\left\{{y}_{2},Max\left\{{C}_{\left[n-1\right]2},{C}_{\left[n\right]1}\right\}\right\}\right\}-{y}_{1}\right)\ge {a}_{\left[n\right]2}\\ &+{b}_{\left[n\right]}\left(Max\left\{{y}_{1},Min\left\{{y}_{2},{C}_{\left[n\right]1}\right\}\right\}-{y}_{1}\right)\end{aligned}$$

Based on the definition of makespan, we have:

$$C_{max}=C_{\lbrack n\rbrack2}=P_{\lbrack n\rbrack2}+Max\left\{C_{\lbrack n\rbrack2},C_{\lbrack n-1\rbrack2}\right\}\geq P_{[n]2}+C_{[n]1}\geq Min\left\{P_{[n]2}\right\}+C_{[n]1}$$
$$C_{\lbrack n\rbrack1}=S_1\sum_{ie\delta}P_{il}\geq S_1+\sum_{ie\delta'}(a_{il}+b_i(Max\left\{y_1,Min\left\{S_1,y_2\right\}\right\}-y_1))$$
$$C_{max}\geq Min\left\{P_{\lbrack n\rbrack2}\right\}+C_{\lbrack n\rbrack1}\geq Min\left\{P_{\lbrack n\rbrack2}\right\}+S_1+\sum_{ie\delta'}(a_{il}+b_i(Max\left\{y_1,Min\{S_1,y_2\}\right\}-y_1))=LB\qquad where$$
$$Min\left\{P_{\lbrack n\rbrack2}\right\}=Min\left\{a_{i2}+b_i(Max\left\{y_1,\;Min\left\{y_2,S_l+\sum_{ie\delta'}(a_{il}+b_i(Max\left\{y_1,\right\}Min\left\{y_2,S_l\right\}\right\}\right\}-y_1\right\}.$$

According to theorem 1 and, S1 three lower bounds can be derived as follows:

$$LB_1=Min\left\{a_{i2}+b_i(Max\left\{y_1,\;Min\left\{S_1\sum_{ie\delta'}a_{il},y_2\right\}\right\}-y_1)\right\}+S_1+\sum_{ie\delta'}a_{il}\qquad(If\;S_{1}\leq\;y_{1})$$
$$LB_2=Min\left\{a_{i2}+b_i(Min\left\{y_2,S_1+\sum_{ie\delta'}(a_{il}+b_1(S_1-y_1))\right\}-y_1)\right\}+S_1+\underset{ie\delta'}{\sum(}a_{il}+b_i(S_1-y_1))\qquad(If\;y_1<S_1<y_2)$$
$$LB_3=Min\left\{a_{i2}+b_i(y_2-y_1)\right\}+S_1+\underset{ie\delta'}{\sum(}a_{il}+b_i(y_2-y_1))\qquad(If\;S_1\geq y_2)$$

Theorem 2

The following relation calculates the lower bound of the partial sequence δ according to M2:

$$LB=P+\sum_{i\in {\delta }^{^{\prime}}}({a}_{i2}+{b}_{i}(Max\left\{{y}_{1},Min\left\{P,{y}_{2}\right\}\right\}-{y}_{1}))$$
(36)
$$where\;P=Max\left\{S_1+Min\left\{a_{il}+b_i(Max\left\{y_1,\;Min\;\left\{S_1,y_2\right\}\right\}-y_1)\right\},S_2\right\}$$

Proof

According to Eq. (2) and Fig. 1, the actual processing time of Ji on Mj is:

\({P}_{ij}={a}_{ij}+{b}_{i}(Max\left\{{y}_{1},Min\left\{{S}_{ij},{y}_{2}\right\}\right\}-{y}_{1})\)

Similarly, the actual processing time of each Ji in δ’ is:

$${P}_{i2}\ge {a}_{i2}+{b}_{i}\left(Max\left\{{y}_{1},Min\left\{{S}_{\left[k+1\right]2},{y}_{2}\right\}\right\}-{y}_{1}\right)$$

Based on the definition of makespan, we have:

$${C}_{\mathrm{max}}={C}_{\left[n\right]2}={P}_{\left[n\right]2}+Max\left\{{C}_{\left[n\right]1},{C}_{\left[n-1\right]2}\right\}\ge {P}_{\left[n\right]2}+{C}_{\left[n-1\right]2}$$

Also, the following relation is satisfied.

$${C}_{\left[n-1\right]2}\ge {S}_{\left[k+1\right]2}+\sum_{i\in {\delta }^{3}}{P}_{i2}-{P}_{\left[n\right]2}$$

Thus, we have the below relation

$$\begin{aligned}{C}_{\mathrm{max}}=&{C}_{\left[n\right]2}\ge {P}_{\left[n\right]2}+{C}_{\left[n-1\right]2}\ge {S}_{\left[k+1\right]2}+\sum_{i\in {\delta }^{^{\prime}}}{P}_{i2}\ge {S}_{\left[k+1\right]2}\\ &+\sum_{i\in {\delta }^{^{\prime}}}({a}_{i2}+{b}_{i}(Max\left\{{y}_{1},Min\left\{{S}_{\left[k+1\right]2},{y}_{2}\right\}\right\}-{y}_{1}))=LB\end{aligned}$$

Where \({S}_{[k+1]2}=Max\left\{{S}_{1}+Min\left\{{a}_{i1}+{b}_{i}(Max\left\{{y}_{1},Min\left\{{S}_{1},{y}_{2}\right\}\right\}-{y}_{1})\right\},{S}_{2}\right\}\)  

According to theorem 2 and, S2 three lower bounds can be derived as follows:

$$LB_4=Max\left\{S_1+Min\left\{a_{il}\right\},S_2\right\}+\sum_{i\in\delta'}a_{i2}\qquad(If\;S_2\leq y_1)$$
$$LB_5=Max\left\{S_1+Min\left\{a_{il}+b_i\right\}(Max\left\{S_1,y_1\right\},-\;y_1)\right\},S_2\}+\sum_{i\in\delta'}(a_{i2}+b_i(Min\left\{P,y_2\right\}-y_1))\qquad(If\;y_1\leq S_2<y_2)$$
$$LB_6=Max\left\{S_1+Min\left\{a_{il}+b_1\right\}(Max\left\{y_1,\;Min\left\{S_1,y_2\right\}\right\}-y_1),S_2\right\}+\sum_{i\in\delta'}(a_{i2}+b_i(y_2-y_1))\qquad(If\;S_2\geq y_2)$$

Theorem 3

The lower bound of the partial sequence \(\updelta\) according to M1 is calculated by the following relation where C1 is the least possible makespan on M1

$$L{B}_{7}=Min\left\{{a}_{i2}+{b}_{i}(Max\left\{{y}_{1},Min\left\{{C}_{1},{y}_{2}\right\}\right\}-{y}_{1})\right\}+{C}_{1}$$

Proof

From, \({P}_{i2}={a}_{i2}+{b}_{i}(Max\left\{{y}_{1},Min\left\{{S}_{i2},{y}_{2}\right\}\right\}-{y}_{1})\) it is derived that.

$${P}_{\left[n\right]2}={a}_{\left[n\right]2}+{b}_{\left[n\right]}\left(Max\left\{{y}_{1},Min\left\{{S}_{\left[n\right]2},{y}_{2}\right\}\right\}-{y}_{1}\right)$$

Since \({S}_{[n]2}=Max\left\{{C}_{[n]1},{C}_{[n-1]2}\right\}\) and \({C}_{[n]1}\ge {C}_{1}\) the relation \({S}_{[n]2}\ge {C}_{1}\) is valid; therefore, we have:

$$Min\left\{P_{\lbrack n\rbrack2}\right\}\geq Min\left\{a_{i2}+b_i(Max\left\{y_1,Min\left\{C_1,y_2\right\}\right\}-y_1)\right\}$$

On the other hand, we have:

$${C}_{\mathrm{max}}\ge {P}_{\left[n\right]2}+{C}_{\left[n\right]1}\ge Min\left\{{P}_{\left[n\right]2}\right\}+{C}_{\left[n\right]1}\ge Min\left\{{a}_{i2}+{b}_{i}\left(Max\left\{{y}_{1},Min\left\{{C}_{1},{y}_{2}\right\}\right\}-{y}_{1}\right)\right\}+{C}_{1}$$

If we can get the amount of C1, then a lower bound for the problem will be derived.

According to the optimality of the created sequence based on the non-decreasing order of \(\frac{{a}_{i}}{{b}_{i}}\) in the problem \(1\left|{P}_{i}={a}_{i}+{b}_{i}S\left|{C}_{\mathrm{max}}\right.\right.\) [40], if this rule is utilized for any position L after the partial sequence δ on M1 in the problem \(F2\left|{P}_{ij}={a}_{ij}+{b}_{i}\left({S}_{ij}-{y}_{1}\right), {y}_{1}>0,{y}_{2}>{y}_{1}\left|{C}_{max}\right.\right.\), then the least completion time for this position will be achieved. Since there is no deterioration for the positions with the start time before y1, the ratio \(\frac{{a}_{i1}}{{b}_{i}}\) reduces to \({a}_{i1}\). Hence, the non-decreasing order of \({a}_{i1}\) in the set \({\delta }^{^{\prime}}\) makes this rule until the completion time of a position reaches y1 or the completion time of the nth position is calculated. If the start time of the position is larger than y1, then its completion time is calculated based on the following procedure:

For any position L after δ on M1, a remaining job with the least value of ai1 in the set \({\delta }^{^{\prime}}\) is placed in the position. Therefore, the following relation holds:

$${a}_{\left[L\right]1}\ge {a}_{\left[H\right]1}\;H=\mathrm{1,2},\dots ,L-1$$

We aim at satisfying \(\frac{{a}_{[1]1}}{{b}_{[1]}}\le \frac{{a}_{[2]1}}{{b}_{[2]}}\le ...\le \frac{{a}_{[L]1}}{{b}_{[L]}}\) where \({a}_{[i]1}\) and \({b}_{[i]}\) are not necessarily related to the same job. Thus, the following relation should be valid:

$${b}_{\left[L\right]}\le {b}_{\left[H\right]}\;H=\mathrm{1,2},\dots ,L-1$$

Consequently, from the deterioration rates in the set, δ’L of the least deterioration rates should be assigned to positions 1 to L in non-increasing order. After that, the completion time of position L is considered as the start time of position L + 1. This process continues until the completion time of the nth position on M1 is calculated (called C1). Notably, as previously mentioned, it is the least possible makespan on M1.

Theorem 4

The lower bound of the partial sequence δ according to M2 is C2 (LB8 = C2) where C2 is the least possible makespan on M2 if the least start time of position k + 1 on M2 is considered based on the following relation:

$${S}_{\left[k+1\right]2}=Max\left\{{S}_{1}+Min\left\{{a}_{i1}+{b}_{i}\left(Max\left\{{y}_{1},Min\left\{{S}_{1},{y}_{2}\right\}\right\}-{y}_{1}\right)\right\},{S}_{2}\right\}$$

Proof

As previously discussed, the start time of position k + 1 on M2 is as follows:

$$\begin{aligned} S_{\{k+1\}2}&=Max\left\{C_{\lbrack k+1\rbrack,}C_{\lbrack k+2\rbrack}\right\}=Max\left\{C_{\lbrack k+1\rbrack,}S_2\right\}\\ &=Max\left\{S_1+a_{\lbrack k+1\rbrack l}+b_{\lbrack k+1\rbrack}(Max\left\{y_1,\;Min\left\{S_1,y_2\right\}\right\}-y_1),S_2\right\}\\ &\geq Max\left\{S_1+Min\left\{a_{\lbrack k+1\rbrack l}+b_{\lbrack k+1\rbrack}(Max\left\{y_1,\;Min\left\{S_1,y_2\right\}\right\}-y_1)\right\},S_2\right\}\end{aligned}$$

Theorem 5

The lower bound (LB9) of the partial sequence δ’ is obtained by using JR algorithm if the actual processing time of Ji on Mj in the set δ’ is calculated as follows:

$$P_{il}^{\prime}=a_{il}+b_i(Max\left\{y_1,\;Min\left\{S_{1},y_{2}\right\}\right\}-y_{1})$$
$$P_{i2}^{\prime}=a_{i2}+b_i(Max\left\{y_{1,\;}Min\left\{y_2,\;Max\left\{S_2,S_1+Min\left\{P_{il}^{\prime}\right\}\right\}\right\}\right\}-y_1)$$

Proof

For each Ji in,values of Pi1 and Pi2 are constant, and the following relations are valid:

$$P_{il}\geq\;P^{\prime}_{il}\;and\;P_{i2}\geq\;P^{\prime}_{i2}$$

Furthermore, since the makespan for \(F2||{C}_{max}\) is optimized by JR algorithm, considering Pi1 and Pi2 for all the jobs in the set, δ’ the makespan of any complete sequence will not be less than LB9.

Finally, in order to make the lower bound tighter, their maximum value is chosen by

$$L{B}^{*}=Max\left\{L{B}_{1},L{B}_{2},L{B}_{3},L{B}_{4},L{B}_{5},L{B}_{6},L{B}_{7},L{B}_{8},L{B}_{9}\right\}$$

Heuristic Algorithm

Since the computation time and memory required in the exact methods are vast, an alternative approach to the NP-hard problems is to provide the heuristic algorithms. However, they do not necessarily generate an optimal solution. Thus, several simple heuristics are proposed for getting an excellent answer to the described problem.

JR gives the optimal answer for the classical two-machine scheduling problem where the deterioration is not considered. Though JR can not present the optimal solution for our problem, we use it as the first heuristic algorithm. The main idea behind the second one is to lessen the idle times on M2; therefore, the jobs are arranged according to the shortest normal processing time on M1 (SNPT1). The third heuristic focuses on reducing the queuing time of jobs on M2 (SNPT2). By applying SNPT to the sum of the normal processing times on M1 and M2, the fourth heuristic (SNPT1+2) is formed.

The deterioration rates of jobs are unequal; therefore, the lowest deterioration rate rule (LDR) and the highest deterioration rate rule (HDR) can be considered the fifth and sixth heuristics, respectively. Also, the sequences formed according to the non-decreasing rates of \(\frac{{a}_{i1}}{{b}_{i}}\), \(\frac{{a}_{i2}}{{b}_{i}}\) and \(\frac{{a}_{i1}+{a}_{i2}}{{b}_{i}}\) are proposed as the seventh, eighth, and ninth heuristics denoted as Ratio1, Ratio2, and Ratio1+2, respectively. Moreover, the efficiency of the mentioned heuristic algorithms may be increased using a pairwise interchange algorithm (PI). The procedures of the heuristic algorithms are demonstrated as follows:

figurea
figureb

Computational Experiments

In this section, computational experiments are conducted to evaluate the performance of the B&B algorithm and the accuracy of heuristics. All the algorithms were coded in C + + and run on a PC with CPU Intel Core i5-9400 and RAM ADATA Premier 8 GB. Notably, in the B&B algorithm, a time limit of 4000 s for each problem was considered so that if a problem could not get the optimal solution in this time limitation, then the B&B procedure was stopped. Parameters were randomly generated according to the related researches in the literature. The normal processing time of each job on Mj was generated [12] from a uniform distribution over the integers between 1 and 10. Variables y1 and y2 were obtained [6] from the continuous uniform distributions over intervals [0,A/3] and [0,2A/3] for y1 and [A/3,2A/3] and [2A/3,A] for y2 where A is assumed to be as \(A=\sum_{i=1}^{n}{\sum }_{j=1}^{2}{a}_{ij}\). Since y2 > y1, there are three distinctive conditions for the different combinations of y1 and y2 called Set1, Set2, and Set3.

The computational analyses include three experiments. The first experiment is designed to study the effect of deterioration rate and variables y1 and y2 on the performance of B&B and heuristic algorithms; the results are summarized in Table 1. The job size was fixed on n = 9 and 10, and the deterioration rate b was segmented into three groups of small, medium, and large called bS, bM, and bL. They were generated from continuous uniform distributions on three intervals (0, 0.3], (0.3, 0.6] and (0.6, 1), respectively. Accordingly, 18 conditions were examined in the first experiment. 20 replications were randomly generated for each condition so that 360 (18 × 20) problems were generated and solved totally.

Table 1 Results for B&B and heuristic algorithms for n = 9 and 10

From Table 1, the number of nodes in the B&B algorithm increases when the deterioration rate increases, especially from the small to medium group. The same trend is observed in the average and maximum CPU time. The increment trend in the number of nodes and CPU time becomes less obvious when the deterioration rate is higher. In other words, the problems with the larger deterioration rate are more complicated, and the problems with the smaller deterioration rate are easier. The main reason is that the problems with smaller deterioration rates are close to the traditional two-machine flowshop scheduling problem where JS procedure provides the optimal solution. By incrementing the deterioration rate, the problem becomes further away from the traditional problem and tends to be complicated. The second reason is because of the completion time of jobs and the objective function. The completion time of jobs and objective function is small for the problems with the smaller deterioration rate, which leads to the easier implementation of the lower bounds, but the lower bounds become tight and very tight when the deterioration rate is medium or high, respectively. The results in Table 1 show that if values of y2 are increased, the problems tend to be hard because it leads to a decrease and an increase in the number of utilizations of Lemma 1 and lower bounds, respectively. Large values of y1 may also make the problems hard because the jobs have no deterioration before y1, and growing the completion times of jobs are decreased.

Performances of the heuristic algorithms that algorithm PI is utilized in all of them are presented in Table 1 by the number of the best solutions and average error percentage. The best solution is defined for the heuristic algorithms, and it means which algorithm finds the best answer compared to the optimal solution. The number of the best solutions indicates that each heuristic algorithm has obtained the best solution for how many problems. The error percentage of the solution obtained by a heuristic algorithm is calculated as follows where Z is the makespan obtained from the heuristic method, and Z* is the makespan of the optimal schedule.

$$\%Error=\frac{(Z-{Z}^{*})}{{Z}^{*}}\times 100\%$$

Noteworthily, some heuristic algorithms, e.g., LDR, are not presented in Table 1 due to a better solution than the others or a high error. Also, the CPU time of heuristic algorithms is zero; therefore, they are not reported in Table 1. The results demonstrate the performances of Ratio1+2 + PI and Ratio1 + PI are the best in terms of the number of the best solutions and the average error percentage. For example, the number of the best solutions for Ratio1 + PI, Ratio1+2 + PI, and HDR + PI is 11, 8, and 1 in the first row of Table 1, respectively. It means Ratio1 + PI algorithm has obtained the best solution for 11 problems from 20 problems (as previously mentioned, for each condition, 20 problems were randomly generated). For Ratio1+2 + PI and HDR + PI, it is equal to 8 and 1 and for the other algorithms is 0. Although JR creates the optimal solution for the traditional two-machine flowshop, it has the worst performance among the algorithms in Table 1. Also shown, the average error percentage increases when the deterioration rate increases.

In the second experiment, the algorithms are tested with different job sizes \(n=\mathrm{6,7},\mathrm{8,9},\mathrm{10,11,12,13,14,15,16,17,18}\;and\;19\) and the deterioration rate is generated [14, 15] from a continuous uniform distribution over (0,1). Evaluating the efficiency of the best heuristic algorithm and the B&B algorithm is the goal of this experiment, where the results are given in Table 2. Consequently, 42 conditions were examined in the second experiment, and 840 (42 × 20) problems were generated and solved totally.

Table 2 Results for B&B and heuristic algorithms for various n values

For the B&B algorithm, the average execution time (in seconds), the efficiency, and the number of optimal solutions are reported in Table 2. The efficiency of the B&B approach is calculated by comparing the number of nodes explored to the total number of nodes. As previously emphasized, if a problem could not find the optimal solution in 4000 s, then the B&B procedure will be stopped, and its result will be removed. Table 2 and Fig. 2 show that the B&B algorithm can solve most problems with job sizes of less than or equal to 19 in a reasonable time. Ng et al. [10] and Yang and Wang [13] proposed a B&B algorithm to solve the problems \(F2\left|{P}_{i}={{a}_{i}+b}_{i}S\left|\sum {C}_{i}\right.\right.\) and \(F2\left|{P}_{i}={b}_{i}S\left|\sum {w}_{i}{C}_{i}\right.\right.\) which could handle the problems up to 15 and 14 jobs, respectively. Since piecewise linear deterioration is much harder than linear deterioration, we can claim that the proposed algorithm in the paper has solved problems with appropriate job sizes in a reasonable time optimally.

Fig. 2
figure2

The average execution time of the B&B algorithm in 3 sets for various n values

The average execution times increase dramatically when the job sizes increase since it is an NP-hard problem. According to Fig. 2, this increment is not reported to Set1, Set2, and Set3 for job sizes 17, 18, and 19 because the B&B algorithm cannot solve all the problems in these groups in 4000 s. The number of problems solved by B&B in 4000 s is shown as the number of optimum samples in Table 2; e.g., the B&B algorithm could solve 12 problems (from 20 problems) with job size of 17 in Set1. As given in Table 2, the efficiency of the B&B algorithm in all the groups increases as the job size increases. Since the number of explored nodes is compared with the total number of nodes, the smaller value of the index confirms the better performance. Also, the performance of the B&B algorithm increases when increasing the job size. The same trend is also observed in the average percentage of entire fathomed nodes by the dominance rules, lower bounds, lemmas, etc. These performance measures prove the fantastic performance of the proposed B&B algorithm.

In addition, for the heuristic algorithms, the number of the optimum solutions, the average error percentage of the best heuristic algorithm, and the number of the best solutions are noted in Table 2. The running time of the heuristic algorithms is zero, and it is omitted from Table 2. The number of the optimum solutions indicates that the heuristic algorithms can optimally solve only some problems showing the complexity and difficulty of the studied problem in this paper. The number of the best solutions shows that the heuristic algorithms Ratio1+2 + PI and Ratio1 + PI are the best. Also, the average error percentage of the best heuristic algorithms (i.e., Ratio1+2 + PI and Ratio1 + PI) is less than 1.6%, as illustrated in Fig. 3. As previously discussed and presented in Table 2, JR has the worst performance among the considered algorithms.

Fig. 3
figure3

The average error percentage of the best heuristic algorithm in 3 sets for various n values

In the last experiment, the performance of the heuristic algorithms is compared for different large-sized instances, namely \(n=\mathrm{20,40,60},...,\text{\hspace{0.17em}and\hspace{0.17em}}200\) (similar to Lee et al. [7]), where the results are given in Table 3. Accordingly, 30 conditions were examined in the third experiment, and 600 (30 × 20) problems were generated and solved totally.

Table 3 Results of heuristic algorithms for large n values

As seen in Table 3, Ratio1+2 + PI and Ratio1 + PI are the best since they have the greatest number of the best solutions. To further analyze the efficiency of heuristic algorithms for large-sized instances, the average values of the best ratios (i.e., \({V}_{1}/{V}_{0},{V}_{2}/{V}_{0},{V}_{3}/{V}_{0},{V}_{4}/{V}_{0}\), and \({V}_{5}/{V}_{0}\)) are reported where, \({V}_{0},{V}_{1},{V}_{2},{V}_{3},{V}_{4}, and {V}_{5}\) denote the solutions of Ratio1+2 + PI, SNPT1 + PI, HDR + PI, Ratio1 + PI, Ratio2 + PI, and SNPT1+2 + PI, respectively. Notably, the ratio for JR + PI is higher than the mentioned algorithms. The results indicate that Ratio1+2 + PI with value V0 yields the best solution and it is the recommended algorithm to obtain a near-optimal solution for our problem.

Concluding Remarks and Future Research

Although scheduling the deteriorating jobs has been recently studied, those studies are rare in the flowshop environment. In this paper, two-machine flowshop scheduling problem was considered where the actual processing time of each job is deteriorated based on a piecewise linear function of its start time. The objective was to minimize the makespan. At first, a mathematical model was presented for the problem and linearization of it is performed. After that, a B&B algorithm was established to solve the problem optimally. Several simple heuristics were also proposed to obtain the near-optimal solution. The experimental results showed the high performance of the proposed B&B algorithm as it could solve most of the problems with the job sizes less than or equal to 19 in a reasonable time. Moreover, it was shown that the average error percentage of heuristics is less than 1.6%. Finally, the recommended heuristic algorithm to obtain a near-optimal solution for large-scale problems was discussed. Future researches may be focused on heuristic and metaheuristic algorithms to solve the problem. Furthermore, some realistic assumptions, like machine availability constraint or release times, might be considered. It is also suggested that the other types of deterioration function, such as the exponential by assuming the learning or forgetting effects, is investigated.

Data Availability

Data will be made available on reasonable request.

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Jafari-Nodoushan, A., Zare, H.K., Lotfi, M. et al. Scheduling Piecewise Linear Deteriorating Jobs to Minimize Makespan in a Two-Machine Flowshop. Oper. Res. Forum 2, 49 (2021). https://doi.org/10.1007/s43069-021-00096-7

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Keywords

  • Scheduling
  • Flowshop
  • Deteriorating jobs
  • Branch and bound
  • Heuristic