Flowshop Scheduling Problems with a Position-Dependent Exponential Learning Effect

Abstract

We consider a permutation flowshop scheduling problem with a position-dependent exponential learning effect. The objective is to minimize the performance criteria of makespan and the total flow time. For the two-machine flow shop scheduling case, we show that Johnson’s rule is not an optimal algorithm for minimizing the makespan given the exponential learning effect. Furthermore, by using the shortest total processing times first (STPT) rule, we construct the worst-case performance ratios for both criteria. Finally, a polynomial-time algorithm is proposed for special cases of the studied problem.

1. Introduction

Many researchers have studied flowshop scheduling problems under various assumptions and with different objective functions. Minimizing the makespan of the classical flowshop is known to be an NP-hard problem except for Johnson’s [1] two-machine case. Dannenbring [2], Gonzalez and Sahni [3], Smutnicki [4], and Cepek et al. [5] have developed approximation algorithms for some special cases of 𝑚-machine flowshop problems.

In traditional machine scheduling theory, the processing time of a job is independent of its processed position. However, due to workers’ learning ability, working attitude, and their continuously improved skills with the passage of time, the processing time of a job is shorter if it is scheduled later in the production sequence. This phenomenon is known as a learning effect, which has been employed in management science since its discovery by Wright [6]. Although the learning theory was first applied to industry more than 70 years ago, it remains an interesting important topic in scheduling research. Biskup [7] and Cheng and Wang [8] investigated the effect of learning in the framework of scheduling. Since then, scheduling with learning effect has received growing attention. Mosheiov [9] investigated Biskup’s learning effect model in various scheduling problems and through several examples showed that the optimal schedule is very different from that of the classical versions. Mosheiov and Sidney [10] considered a single-machine scheduling problem with job-dependent learning effects with objectives such as makespan and total flow time, which are proved to be polynomial-time solvable. Bachman and Janiak [11, 12] investigated several single machine scheduling problems with position-dependent processing time.

Many researchers have extended several kinds of learning effect model in machine scheduling problems. Wang [13] studied flowshop scheduling problems with job processing times dependent on their positions and suggested Johnson’s rule as a heuristic algorithm to analyze the worst-case ratio of the makespan and special cases of 𝑚-machine flowshop. Wang and Xia [14] and Xu et al. [15] extended Biskup’s learning effect to minimize makespan and the total completion time in a flowshop setting. Biskup [16] provided an extensive review of scheduling with learning effects. Janiak and Rudek [17] considered a learning effect model in which the learning curve is S-shaped and provided NP-hard proofs for two cases of the problem to minimize the makespan. S.-J. Yang and D.-L. Yang [18] investigated a single-machine scheduling with a position-dependent aging effect described by a power function and variable maintenance duration. Li and Hsu [19] considered the case of two agents competing for a common single machine with learning effect. Lee et al. [20] studied a uniform parallel machine problem to jointly find an optimal assignment of operators to machines and an optimal schedule to minimize the makespan. Jiang et al. [21] introduced an actual time-dependent and job-dependent learning effect into single-machine scheduling problems. Wu et al. [22] proposed two truncated learning models in single-machine scheduling problems. The related literature also includes Koulamas and Kyparisis [23], Mosheiov [24], Janiak et al. [25], Lee and Wu [26], Huang et al. [27], Zhang and Yan [28], Koulamas [29], J.-B. Wang and J.-J. Wang [30], and Kuo [31] and further references.

We know that the model 𝑝_𝑗𝑟=𝑝_𝑗𝑟^𝛼 is proposed by Biskup [16], the effect for the given job processed in different positions is not stable; that is, if job 𝐽_𝑗 is processed at 𝑟th and (𝑟+1)th position, respectively, then 𝑝_{𝑗𝑟+𝑘}/𝑝_𝑗𝑟=((𝑟+1)/𝑟)^𝛼, and the processing time decreases quickly if 𝛼<0 and far from zero. However, in many realistic settings, the learning process of workers should be stable, which is of utmost importance to guarantee the quality of the product. Therefore, we propose a position-dependent exponential learning effect model 𝑝_𝑗𝑟=𝑝_𝑗𝛼^𝑟−1, where 𝛼∈(0,1] is a learning index. In such model, the processed position effect for a given job in some schedule is a constant because of 𝑝_{𝑗𝑟+𝑘}/𝑝_𝑗𝑟=𝛼, and the processing time decreases slowly if 𝛼 is near 1, and the proposed learning effect model can reflect reasonably the stability of the manufacturing process.

In this paper, we study permutation flowshop scheduling problems with a position-dependent exponential learning effect to minimize one of the following two regular performance criteria: makespan and the total flow time. An example is constructed to show that the classical Johnson rule is not optimal for the two-machine case with minimizing makespan under such a position-dependent exponential learning effect. We use the shortest total processing time first (STPT) rule (Dannenbring [2]) to solve our problem and obtain the same worst-case performance ratio for both criteria and prove this performance ratio to be tight. Furthermore, a polynomial-time algorithm is proposed for two special cases: identical processing times on all the machines for any given jobs and the flowshop scheduling problem with dominant machines.

The rest of the paper is organized as follows. In Section 2, we formalize the problem. We develop the worst-case performance ratios of the STPT algorithm and use it to solve our problem in Section 3. In Section 4, we analyze several special cases and prove that all the problems are solvable in polynomial time. Finally, concluding remarks are given in Section 5.

2. Formulation of the Problem

𝑁(𝐽₁,𝐽₂,…,𝐽_𝑛) jobs are to be processed on an 𝑚-machine flowshop. We assume that the normal processing time of job 𝐽_𝑖 on machine 𝑀_𝑗 is denoted as 𝑝_𝑖,𝑗 and the actual processing time of job 𝐽_𝑖 on machine 𝑀_𝑗 at the 𝑟th position is denoted as 𝑝_{𝑖,𝑗,𝑟}. Namely, the actual processing time of a job is characterized by a position-dependent exponential function: 𝑝_{𝑖,𝑗,𝑟}=𝑝_𝑖,𝑗𝛼^𝑟−1, 𝑖=1,2,…,𝑚; 𝑟,𝑗=1,2,…,𝑛, where 𝛼∈(0,1] denotes the common learning index for all jobs. The aim in this paper is to find a schedule so that 𝑛 jobs should be processed on the 𝑚 machines to minimize a given objective function.

For a given schedule 𝜎, 𝐶_𝑖(𝜎) represents the completion time of job 𝐽_𝑖, 𝐶_max(𝜎)=max{𝐶_𝑖(𝜎)∣𝑖=1,2,…,𝑛} denotes the makespan, and 𝜎=([1],[2],…,[𝑛]) denotes a schedule, where [𝑟] denotes a job that occupies the 𝑟th position in 𝜎. In the remaining part of this paper, all problems considered are denoted by the three-field notation scheme 𝛼|𝛽|𝛾 introduced by Graham et al. [32]; that is, 𝐹𝑚∣𝑝_{𝑖,𝑗,𝑟}=𝑝_𝑖,𝑗𝛼^𝑟−1∣𝑓(𝜎), where 𝑓(𝜎)∈{𝐶_max,∑𝐶_𝑖}.

3. Worst-Case Ratio for the General Case

It is well known that problems 𝐹𝑚∣𝑝_{𝑖,𝑗,𝑟}=𝑝_𝑖,𝑗𝛼^𝑟−1∣𝐶_max(𝑚≥3) and 𝐹𝑚∣𝑝_{𝑖,𝑗,𝑟}=𝑝_𝑖,𝑗𝛼^𝑟−1∣∑𝐶_𝑗(𝑚≥2) are all strong NP-hard (Garey and Johnson [33]) even for 𝛼=1. Johnson [1] proved that 𝐹2‖𝐶_max can be optimally solved by Johnson’s rule. However, in the following example, we show that this policy is not optimal for problem 𝐹2∣𝑝_{𝑖,𝑗,𝑟}=𝑝_𝑖,𝑗𝛼^𝑟−1∣𝐶_max, where 0<𝛼≤1 is a constant learning index.

Example 1. Let 𝑛=3, 𝑚=2, 𝑝_1,1=2, 𝑝_1,2=3, 𝑝_2,1=1, 𝑝_2,2=20, 𝑝_3,1=7, 𝑝_3,2=5, and learning index 𝛼=0.85. The schedule given by Johnson’s rule is (𝐽₂,𝐽₁,𝐽₃) with the value 𝐶_max=27.16. However, the optimal permutation schedule is (𝐽₁,𝐽₂,𝐽₃), and the optimal value 𝐶^∗_max=25.61.

Before analyzing the worst-case ratio of the problems, we provide two lemmas. By simple interchange technique, the following results can be easily obtained.

Lemma 2. For problem 1∣𝑝_𝑖,𝑟=𝑝_𝑖𝛼^𝑟−1∣𝐶_max, an optimal schedule can be obtained by shortest processing time first (SPT) rule.

Lemma 3. For problem 1∣𝑝_𝑖,𝑟=𝑝_𝑖𝛼^𝑟−1∣∑𝐶_𝑗, an optimal schedule can be obtained by SPT rule.

In the following, we turn our attention to obtain quasi-optimal schedule. For the 𝐹𝑚‖∑𝐶_𝑗 problem, Dannenbring [2] gave the approximation algorithm shortest total processing time first (STPT) (a nondecreasing order of {𝐿_𝑗}(𝐿_𝑗=∑^𝑚_𝑖=1𝑝_𝑖,𝑗)) and showed that it has worst-case performance ratio 𝑚.

Similarly, we use the STPT rule to construct an approximate algorithm for problem 𝐹𝑚∣𝑝_{𝑖,𝑗,𝑟}=𝑝_𝑖,𝑗𝛼^𝑟−1∣𝐶_max.

Theorem 4. If 𝜎^∗ is an optimal schedule and 𝜎 is an STPT schedule for 𝐹𝑚∣𝑝_{𝑖,𝑗,𝑟}=𝑝_𝑖,𝑗𝛼^𝑟−1∣𝐶_max, then 𝐶_max(𝜎)/𝐶_max(𝜎^∗)≤𝑚, and the bound is tight.

Proof. Without loss of generality, we assume that 𝐿₁≤𝐿₂≤⋯≤𝐿_𝑛. Let 𝐶_𝑖(𝜎) be the completion time of job 𝐽_𝑖 in schedule 𝜎; then we have 𝐶_𝑖(𝜎)≤∑^𝑖_𝑗=1𝐿_𝑗𝛼^𝑗−1 and subsequently 𝐶_max(𝜎)≤∑^𝑛_𝑗=1𝐿_𝑗𝛼^𝑗−1. Let 𝜎^∗=(𝐽_[1],𝐽_[2],…,𝐽_[𝑛]) be the optimal schedule. We will have 𝐶_max(𝜎^∗)≥(∑^𝑛_𝑗=1𝐿_[𝑗]𝛼^𝑗−1)/𝑚; hence, 𝐶_max(𝜎^∗)≥(∑^𝑛_𝑗=1𝐿_[𝑗]𝛼^𝑗−1)/𝑚≥(∑^𝑛_𝑗=1𝐿_𝑗𝛼^𝑗−1)/𝑚, and the term (∑^𝑛_𝑗=1𝐿_[𝑗]𝛼^𝑗−1)/𝑚 is minimized by the nondecreasing order of 𝐿_𝑗 (Lemma 2 corresponding to the last machine). Therefore, we have 𝐶_max(𝜎)/𝐶_max(𝜎^∗)≤𝑚.
We show that the bound is tight for the following example. Assume that there are 𝑚 jobs 𝐽₁,𝐽₂,…,𝐽_𝑚 with normal processing times 𝑝_𝑖,𝑖=𝑡,𝑝_𝑖,𝑗=𝜀(𝜀≪𝑡),(𝑖≠𝑗=1,2,…,𝑚), and the learning index 0<𝛼≤1. Since 𝐿_𝑗=∑^𝑚_𝑖=1𝑝_𝑖,𝑗=𝑡+(𝑚−1)𝜀 for all jobs, the proposed heuristic may schedule the jobs in any order. Suppose that we process the jobs in the order of (𝐽₁,𝐽₂,…,𝐽_𝑚) on all machines. The makespan of 𝜎 is 𝐶_max(𝜎)=𝑡((1−𝛼^𝑚)/(1−𝛼))+𝜀((1−𝛼^𝑚−1)/(1−𝛼)). The optimal schedule is 𝜎^∗=(𝐽_𝑚,𝐽_𝑚−1,…,𝐽₁), and the makespan of 𝜎^∗ is 𝐶_max(𝜎^∗)=𝑡+(𝑚−2)𝜀+𝜀((1−𝛼^𝑚−1)/(1−𝛼)). Hence, 𝐶_max(𝜎)/𝐶_max(𝜎^∗)→𝑚(𝛼→1), and this bound is tight.

Similarly, we can use the STPT rule to put together an approximation algorithm for problem 𝐹𝑚∣𝑝_{𝑖,𝑗,𝑟}=𝑝_𝑖,𝑗𝛼^𝑟−1∣∑𝐶_𝑖.

Theorem 5. Let 𝜎^∗ be an optimal schedule, and let 𝜎 be the STPT schedule for 𝐹𝑚∣𝑝_{𝑖,𝑗,𝑟}=𝑝_𝑖,𝑗𝛼^𝑟−1∣∑𝐶_𝑖; then ∑𝐶_𝑖(𝜎)/∑𝐶_𝑖(𝜎^∗)≤𝑚, and the bound is tight.

Proof. Without loss of generality, we assume that 𝐿₁≤𝐿₂≤⋯≤𝐿_𝑛. Let 𝐶_𝑖(𝜎) be the completion time of job 𝐽_𝑖 in the STPT schedule 𝜎; then 𝐶_𝑖(𝜎)≤∑^𝑖_𝑗=1𝐿_𝑗𝛼^𝑗−1 and ∑𝐶_𝑖(𝜎)≤∑^𝑛_𝑖=1∑^𝑖_𝑗=1𝐿_𝑗𝛼^𝑗−1. Let 𝜎^∗=(𝐽_[1],𝐽_[2],…,𝐽_[𝑛]) be the optimal schedule. For 𝜎^∗, we have 𝐶_[𝑖](𝜎^∗)≥(∑^𝑖_𝑗=1𝐿_[𝑗]𝛼^𝑗−1)/𝑚; hence 𝐶_𝑖𝜎^∗≥∑^𝑛_𝑖=1∑^𝑖_𝑗=1𝐿_[𝑗]𝛼^𝑗−1𝑚≥∑^𝑛_𝑖=1∑^𝑖_𝑗=1𝐿_𝑗𝛼^𝑗−1𝑚, (1) while the term (∑^𝑛_𝑖=1∑^𝑖_𝑗=1𝐿_𝑗𝛼^𝑗−1)/𝑚 is minimized by the nondecreasing order of {𝐿_𝑗} (Lemma 3 corresponding to the last machine). Consequently, we have that ∑C_𝑖(𝜎)/∑𝐶_𝑖(𝜎^∗)≤𝑚.
We now show that the bound is tight. Assume that there are 𝑚 machines and 𝑚𝑛 jobs 𝐽₁,𝐽₂,…,𝐽_𝑚𝑛 with processing times 𝑝_1,𝑗=1, 𝑝_𝑖,𝑗=𝜀, 𝜀≪1, 𝑖=2,…,𝑚, 𝑗=1,2,…,𝑛; 𝑝_1,𝑗=𝑝_𝑖,𝑗=𝜀, 𝑝_2,𝑗=1, 𝑖=3,…,𝑚, 𝑗=𝑛+1,𝑛+2,…,2𝑛; 𝑝_𝑖,𝑗=𝜀, 𝑝_𝑖,𝑗=1, 𝑖=1,…,𝑚−1, 𝑗=(𝑚−1)𝑛+1,(𝑚−1)𝑛+2,…,𝑚𝑛, and the learning index 0<𝛼≤1. Since 𝐿_𝑗=∑^𝑚_𝑖=1𝑝_𝑖,𝑗=1+(𝑚−1)𝜀 for all jobs, our heuristic may schedule the jobs in any order. Suppose that we arrange the jobs in the order of 𝜎=(𝐽₁,𝐽₂,…,𝐽_𝑚𝑛) on 𝑚 machines. The cost of the schedule is ∑𝐶_𝑖(𝜎)=𝑚𝑛/(1−𝛼)−𝛼(1−𝛼^𝑚𝑛)/(1−𝛼)²+𝑜(𝜀). However, the optimal schedule 𝜎^∗ is such a one in which the jobs are executed in the order of (𝐽_{(𝑚−1)𝑛+1},𝐽_{(𝑚−2)𝑛+1},…,𝐽_𝑛+1,𝐽₁,𝐽_{(𝑚−1)𝑛+2},…,𝐽_𝑛+2,𝐽₂,…,𝐽_𝑚𝑛,…,𝐽_𝑛), and the cost is 𝐶_𝑖(𝜎^∗)=𝑚𝑛/(1−𝛼)−𝑚𝛼(1−𝛼^𝑛)/(1−𝛼)²+𝑜(𝜀). Hence, ∑𝐶_𝑖(𝜎)/∑𝐶_𝑗(𝜎^∗)=(𝑚𝑛/(1−𝛼)−𝛼(1−𝛼^𝑚𝑛)/(1−𝛼)²+𝑜(𝜀))/(𝑚𝑛/(1−𝛼)−𝑚𝛼(1−𝛼^𝑛)/(1−𝛼)²+𝑜(𝜀)). As 𝛼 approximates 1, we have lim_𝛼→1(∑𝐶_𝑖(𝜎)/∑𝐶_𝑖(𝜎^∗))=(𝑚𝑛+1)/(𝑛+1)≈𝑚.

4. Special Cases Solvable in Polynomial Time

We now consider the special case where a job has the same processing times on all machines for any given job; that is, 𝑝_𝑖,𝑗=𝑝_𝑖. We know that for the 𝐹𝑚∣𝑝_𝑖,𝑗=𝑝_𝑖∣𝐶_max problem, the completion time of 𝐽_[𝑖] is 𝐶_[𝑖](𝜎)=∑^𝑖_𝑘=1𝑝_[𝑘]+(𝑚−1)max{𝑝_[1],𝑝_[2],…,𝑝_[𝑖]} (Pinedo [34]), where 𝜎 denotes a schedule and [𝑖] denotes the job that occupies the 𝑖th position in 𝜎. Hence, for the 𝐹𝑚∣𝑝_{𝑖,𝑗,𝑟}=𝑝_𝑖𝛼^𝑟−1∣𝐶_max problem, the completion time 𝐶_[𝑖] of 𝐽_[𝑖] is 𝐶_[𝑖](𝜎)=^𝑖_𝑘=1𝑝_[𝑘]𝛼^𝑘−1+(𝑚−1)max𝑝_[1]𝛼⁰,𝑝_[1]𝛼¹,…,𝑝_[𝑖]𝛼^𝑖−1. (2)

Theorem 6. For problem 𝐹𝑚∣𝑝_{𝑖,𝑗,𝑟}=𝑝_𝑖𝛼^𝑟−1∣𝐶_max, an optimal schedule can be obtained by the SPT rule.

Proof. Consider a schedule 𝜎. Suppose that job 𝐽_[𝑖] occupy the 𝑖th position in 𝜎 and that 𝑝_[𝑖] is defined accordingly. The schedule 𝜎^′ is obtained from the schedule 𝜎 by swapping jobs 𝐽_[𝑖] and 𝐽_[𝑖+1]. Assume that the schedule 𝜎, in which 𝑝_[𝑖]>𝑝_[𝑖+1], is optimal. The difference in makespan for the two schedules is given as follows: 𝐶_max(𝜎)−𝐶_max𝜎^′=𝑝_[𝑖]𝛼^𝑖−1+𝑝_[𝑖+1]𝛼^𝑖+()𝑝𝑚−1×max_[1],𝑝_[2]𝛼,…,𝑝_[𝑖]𝛼^𝑖−1,𝑝_[𝑖+1]𝛼^𝑖,…,𝑝_[𝑛]𝛼^𝑛−1−𝑝_[𝑖+1]𝛼^𝑖−1−𝑝_[𝑖]𝛼^𝑖−()𝑝𝑚−1×max_[1],𝑝_[2]𝛼,…,𝑝_[𝑖+1]𝛼^𝑖−1,𝑝_[𝑖]𝛼^𝑖,…,𝑝_[𝑛]𝛼^𝑛−1=𝑝_[𝑖]−𝑝_[𝑖+1]𝛼^𝑖−1−𝛼^𝑖+()×𝑝𝑚−1max_[1],𝑝_[2]𝛼,…,𝑝_[𝑖]𝛼^𝑖−1,𝑝_[𝑖+1]𝛼^𝑖,…,𝑝_[𝑛]𝛼^𝑛−1−max𝑝_[1],𝑝_[2]𝛼,…,𝑝_[𝑖+1]𝛼^𝑖−1,𝑝_[𝑖]𝛼^𝑖,…,𝑝_[𝑛]𝛼^𝑛−1. (3) Since 𝑝_[𝑖]>𝑝_[𝑖+1]𝑝,0<𝛼≤1,_[𝑖]𝛼^𝑖−1>𝑝_[𝑖+1]𝛼^𝑖−1,𝑝_[𝑖]𝛼^𝑖−1>𝑝_[𝑖]𝛼^𝑖, (4) we have max𝑝_[1],𝑝_[2]𝛼,…,𝑝_[𝑖]𝛼^𝑖−1,𝑝_[𝑖+1]𝛼^𝑖,…,𝑝_[𝑛]𝛼^𝑛−1−max𝑝_[1],𝑝_[2]𝛼,…,𝑝_[𝑖+1]𝛼^𝑖−1,𝑝_[𝑖]𝛼^𝑖,…,𝑝_[𝑛]𝛼^𝑛−1≥0. (5) That is, 𝐶_max(𝜎)>𝐶_max(𝜎^′), which contradicts our assumption that 𝜎 is an optimal schedule. Therefore, the schedule obtained by using the SPT rule is optimal.

Theorem 7. For problem 𝐹𝑚∣𝑝_{𝑖,𝑗,𝑟}=𝑝_𝑗𝛼^𝑟−1∣∑𝐶_𝑗, an optimal schedule can be obtained by the SPT rule.

Proof. The proof is similar to that of Theorem 6 and is omitted.

We now consider another special case. Assume that the flowshop has dominant machines. Following Ho and Gupta [35], machine 𝑀_𝑟 is dominated by 𝑀_𝑘, or 𝑀_𝑘 dominates 𝑀_𝑟 iff min{𝑝_𝑖,𝑘∣𝑖=1,2,…,𝑛}≥max{𝑝_𝑖,𝑟∣𝑖=1,2,…,𝑛} (denoted by 𝑀_𝑘>𝑀_𝑟). Refer to van den Nouweland et al. [36] and Cepek et al. [5] for such special cases.

For the dominant machine case, we form the machine in an increasing sequence of dominating machines (𝑖𝑑𝑚); that is, 𝑀₁<𝑀₂<⋯<𝑀_𝑚. Recall that for the 𝐹𝑚|𝑖𝑑𝑚|𝑓(𝜎) problem with 𝑓(𝜎) measure, the schedule 𝜎=(𝐽_[1],𝐽_[2],…,𝐽_[𝑛]) has the completion time 𝐶_[𝑖] of job 𝐽_[𝑖] equal to 𝐶_[𝑖]=∑^𝑚−1_𝑗=1𝑝_[1],𝑗+∑^𝑖_𝑘=1𝑝_[𝑘],𝑚 (Pinedo [34]). Therefore, for the 𝐹𝑚∣𝑝_{𝑖,𝑗,𝑟}=𝑝_𝑖,𝑗𝛼^𝑟−1,𝑖𝑑𝑚∣𝑓(𝜎) problem, the completion time 𝐶_[𝑖] of job 𝐽_[𝑖] is 𝐶_[𝑖]=^𝑚−1_𝑗=1𝑝_[1],𝑗+^𝑖_𝑘=1𝑝_[𝑘],𝑚𝛼^𝑘−1. (6) Therefore, an optimal algorithm can be developed as follows.

Algorithm  A

Step  1. Construct the SPT sequence (𝐽_[1],𝐽_[2],…,𝐽_[𝑛]) of processing time 𝑝_𝑖,𝑚,𝑖=1,2,…,𝑛, on machine 𝑀_𝑚. That is, 𝑝_[1],𝑚≤𝑝_[2],𝑚≤⋯≤𝑝_[𝑛],𝑚.

Step  2. Let 𝜎_𝑖=(𝐽_[𝑖],𝐽_[1],𝐽_[2],…,𝐽_[𝑖−1],𝐽_[𝑖+1],…,𝐽_[𝑛]). Compute 𝑓(𝜎_𝑖).

Step  3. Let 𝑓(𝜎_𝑖^∗)=min{𝑓(𝜎_𝑖)∣𝑖=1,2,…,𝑛}; then an optimal schedule is 𝜎_𝑖^∗=(𝐽_[𝑖^∗_],𝐽_[1],𝐽_[2],…,𝐽_[𝑖^∗_−1],𝐽_[𝑖^∗_+1],…,𝐽_[𝑛]).

Theorem 8. For problem 𝐹𝑚∣𝑝_{𝑖,𝑗,𝑟}=𝑝_𝑖,𝑗𝛼^𝑟−1,𝑖𝑑𝑚∣𝐶_max, the schedule generated by Algorithm A is optimal.

Proof. For any feasible solution 𝜎^′=(𝐽_[1]′,𝐽_[2]′,…,𝐽_{[𝑖−1]′},𝐽_[𝑖]′,𝐽_[𝑖+1]′,…,𝐽_[𝑛]′), let 𝐽_[1]′=𝐽_[𝑘], where 𝐽_[𝑘] is the 𝑘th element of 𝜎=(𝐽_[1],𝐽_[2],…,𝐽_[𝑖−1],𝐽_[𝑖],𝐽_[𝑖+1],…,𝐽_[𝑛]) obtained by Step  1 of Algorithm A. We have 𝐶_max𝜎^′=^𝑚−1_𝑖=1𝑝_[1]′,𝑖+^𝑛_𝑖=1𝑝_{[𝑖]′,𝑚}𝛼^𝑖−1=^𝑚−1_𝑖=1𝑝_[𝑘],𝑖+^𝑛_𝑖=1𝑝_{[𝑖]′,𝑚}𝛼^𝑖−1≥^𝑚−1_𝑖=1𝑝_[𝑘],𝑖+^𝑘−1_𝑖=1𝑝_[𝑖],𝑚𝛼^𝑖+^𝑛_𝑖=𝑘+1𝑝_[𝑖],𝑚𝛼^𝑖−1=𝐶_max𝜎_𝑘≥𝐶_max𝜎_𝑗^∗. (7) This completes the proof.

Algorithm A requires 𝑂(𝑚𝑛), and the computation of an SPT schedule needs 𝑂(𝑛log𝑛), so the computation of 𝑓(𝜎_𝑗^∗) needs 𝑂(𝑚𝑛).

From (6), we have ∑^𝑛_𝑖=1𝐶_[𝑖]=∑^𝑛_𝑖=1∑^𝑚−1_𝑘=1𝑝_[1],𝑘+∑^𝑛_𝑖=1∑^𝑖_𝑘=1𝑝_[𝑘],𝑚𝛼^𝑘−1, where the first term is only dependent on job 𝐽_[1], while the second term is the total flow-time of all the jobs on machine 𝑀_𝑚 and is minimized by the SPT rule (Lemma 3). Therefore, an optimal schedule can be constructed for the problem 𝐹𝑚∣𝑝_{𝑖,𝑗,𝑟}=𝑝_𝑖,𝑗𝛼^𝑟−1∣∑𝐶_𝑗 by Algorithm A.

Theorem 9. For problem 𝐹𝑚∣𝑝_{𝑖,𝑗,𝑟}=𝑝_𝑖,𝑗𝛼^𝑟−1,𝑖𝑑𝑚∣∑𝐶_𝑗, the schedule generated by Algorithm A is an optimal one.

Proof. The proof is similar to that of Theorem 8 and is omitted.

5. Conclusion

In order to overcome the lack of stability in Biskups model in a manufacturing setting, we focus on flowshop scheduling problems with position-dependent exponential learning effect to minimize the makespan and the total flow time. For the two-machine case, we show that Johnson’s algorithm does not obtain optimal result. We thus develop the worst-case ratios of the STPT algorithm to minimize the makespan and the total flow time, respectively, and empirically prove that both bounds are tight by several examples. Finally, for special cases, we show that the problems remain polynomially solvable.

Notice that the complexity of the studied problem remains unreciprocated and the worst-case ratios we developed heavily depend on the number of machines. The dependency of machine number in the flowshop is a limitation of our algorithm as it may create notable discrepancy between the optimal solution and that derived by our method. Therefore, our future research includes proving the complexity of the flowshop scheduling problem with position-dependent exponential learning effect and developing generalizable and effective algorithms for diverse problems.

Acknowledgments

This work was supported partly by Humanities and Social Science Fund of Ministry of Education of China (no. 11YJCZH019), Humanities and Social Science Key Research Base of Major Projects of Guangdong Province Universities (no. 11jdxm63001), Breeding Project of Department of Education of Guangdong Province (no. wym09116), and Higher Education Research Fund Projects of GDUT (no. 2013WT05).