The stability radius of an optimal line balance with maximum efficiency for a simple assembly line

Accepted Manuscript

The stability radius of an optimal line balance with maximum efficiency for a simple assembly line Tsung-Chyan Lai, Yuri N. Sotskov, Alexandre Dolgui PII: DOI: Reference:

S0377-2217(18)30864-6 https://doi.org/10.1016/j.ejor.2018.10.013 EOR 15405

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European Journal of Operational Research

Received date: Revised date: Accepted date:

29 May 2017 5 October 2018 8 October 2018

Please cite this article as: Tsung-Chyan Lai, Yuri N. Sotskov, Alexandre Dolgui, The stability radius of an optimal line balance with maximum efficiency for a simple assembly line, European Journal of Operational Research (2018), doi: https://doi.org/10.1016/j.ejor.2018.10.013

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Highlights • A criterion for a zero stability radius of the line balance is proven • A criterion for an infinite stability radius of the line balance is proven • Lower and upper bounds on the stability radius are derived

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• An algorithm for calculating the stability radius is developed

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• Formulae for calculating the stability radius are derived

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The stability radius of an optimal line balance with maximum efficiency for a simple assembly line Tsung-Chyan Laia,1 , Yuri N. Sotskovb,1 , Alexandre Dolguic a School

of Economics and Management, Harbin Engineering University, 145 Nantong Street, Harbin, 150001 China Institute of Informatics Problems, National Academy of Sciences of Belarus, Surganova Str. 6, Minsk, 220012 Belarus c Ecole des Mines de Nantes, IRCCyN, UMR CNRS 6597, La Chatrerie, 4, rue Alfred Kastler - B.P. 20722, F-44307 Nates Cedex 3, France

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Abstract

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We consider a simple assembly line balancing problem in which each element of the partially ordered set of assembly operations must be assigned to one element of the set of workstations used for processing the operations. An objective is minimizing the product of the number of workstations used in the line balance and the cycle time of the line balance among all admissible line balances. An admissible line balance is a partition of all assembly operations into at least two workstations without violating the precedence relations among the assembly operations. We assume that during the lifespan of the assembly line, the duration of each manual operation may deviate from an initially estimated value, while the duration of each automated operation is deterministic. We conduct the stability analysis of an optimal line balance. First, we derive a sufficient and necessary condition for an optimal line balance to be stable. Second, we show that the stability radius of an optimal line balance could be infinitely large. We also establish some lower and upper bounds for a finite stability radius. Third, we derive formulae that are needed and develop an algorithm for obtaining the stability radius of an optimal line balance.

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Keywords: Scheduling, Assembly line balance, Variable durations, Stability analysis

1. Introduction

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The assembly line that we consider consists of a linearly ordered set S = {S 1 , S 2 , . . . , S m } of the workstations, where all or a part of the m workstations will be used for processing the set V = {1, 2, . . . , n}, n ≥ 2, of n indivisible assembly operations to produce a final product. The workstations are linked by a conveyor belt (or another moving equipment), which moves an inprocess product from one workstation to the next at a constant pace. This type of the assembly line is called the single-model paced assembly line which is designed for continuously manufacturing a homogeneous product in a large quantity. By letting V denote the set of all assembly operations and A denote the set of arcs, the precedence digraph G = (V, A) defines the partial order relations on the set V for processing the operations and for assigning them to the used workstations from the set S . For such an assembly line, we investigate a Simple Assembly Line Balancing Problem that is denoted by SALBP-E in (Baybars, 1986; Scholl, 1999). Given durations of all assembly operations, the deterministic counterpart of the problem SALBPE is to find a line balance br among all line balances, which minimizes the product of the number

1 Corresponding author. Tel: 217-375-2842125; fax: +375-217-2318403. E-mail addresses: [email protected], [email protected] (T.-C. Lai), [email protected] (Yu.N. Sotskov), [email protected] (A. Dolgui)

Preprint submitted to European Journal of Operational Research

October 15, 2018

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m(br ) of the workstations used and the cycle time c(br , t) of the line balance br . The cycle time of the line balance is a total duration of the operations assigned to the bottleneck workstation. We note that minimizing the product of the number of the workstations used in a line balance and the cycle time of a line balance is equivalent to maximizing the efficiency of this line balance. The efficiency of the line balance br is defined as follows: E(br , t) :=

tP , m(br ) · c(br , t)

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P where tP denotes a total duration of all assembly operations, tP := ni=1 ti . Maximizing the efficiency (1) is equivalent to minimizing the total idle time m(br )·c(br , t)−tP of the workstations used. e⊆V Assume that the set V consists of two types of the assembly operations, where the subset V e consists of all manual operations performed by human operators and the subset V \ V consists of all automated operations performed by automatic machines (robots) installed on the workstations. The initial vector t = (t1 , t2 , . . . , ti , . . . , tn ) of the operation durations is given before solving the e is deterministic during problem SALBP-E. The duration ti of each automated operation i ∈ V \ V e may vary due to the assembly line lifespan, while the duration t j of each manual operation j ∈ V various factors such as the operator’s skill, motivation, fatigue, learning effect, etc. e = {1, 2, . . . , n˜ } and V \ V e = {˜n +1, n˜ +2, . . . , n}, Let the operations V be enumerated as follows: V where 0 ≤ n˜ ≤ n. The given vectors of the operation durations are denoted as follows: e t = e (t1 , t2 , . . . , tn˜ ), t = (tn˜ +1 , tn˜ +2 , . . . , tn ), t = (t, t) = (t1 , t2 , . . . , tn ). We investigate the following question. How much can all or some components of the vector e t be simultaneously and independently modified such that the line balance br , which is optimal for the initial durations t = (e t, t), remains optimal for the modified durations t0 = (e t0 , t)? We define a stability radius of an optimal line balance in a similar manner to that of an optimal schedule used in (Br¨asel et al., 1996; Kravchenko et al., 1995; Lai et al., 2004; Sotskov, 1991). If the stability radius of the optimal line balance br is strictly positive, then the line balance br remains optimal for any variation of each duration e within the ball with this radius and the center e t j , j ∈ V, t. If the stability radius of the optimal line balance br is equal to zero, then the line balance br may no longer be optimal even for some infinitely small variations of the durations of the manual operations. The rest of this paper is organized as follows. Preliminaries including descriptions of the problems, a formal definition on the stability radius, and main notations are given in Section 2. A survey of the related literature on the stability analysis of the optimal line balance is presented in Section 3. A sufficient and necessary condition for a zero stability radius of the line balance is established in Section 4. In Section 5, we first show that the stability radius may be infinitely large and then establish the upper and lower bounds on the stability radius. Some formulae needed for obtaining the stability radii of the optimal line balances are derived in Section 6. Three numerical examples are used for illustrating on how to calculate for the stability radii in Subsections 4.2, 6.1, and 6.2. Our algorithm for calculating the stability radius is described in Section 7. The necessity proofs for Theorems 2 and 4 are given in Appendixes A and B. We conclude with Section 8. 2. Preliminaries

In what follows, we assume that there are given at least two assembly operations, n ≥ 2, and the given digraph G = (V, A) has no circuits. Let a non-empty subset Vkbr , ∅ of the assembly 3

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operations, Vkbr ⊂ V, be assigned to the workstation S k ∈ S , where k ∈ {1, 2, . . . , m(br )}. The following partition br : [ [ [ br V = V1br V2br ... Vm(b (2) r)

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of the operation set V into m(br ) non-empty subsets Vkbr , ∅, where k ∈ {1, 2, . . . , m(br )}, T Vkbr Vlbr = ∅, 1 ≤ k < l ≤ m(br ), is called a line balance if the following condition holds. Condition 1. The partition br does not violate the partial order imposed on the operation set V by the precedence digraph G = (V, A), i.e. each arc (i, j) ∈ A implies that operation i ∈ V is assigned to the workstation S k and operation j ∈ V is assigned to the workstation S g in a way such that the following inequalities hold: 1 ≤ k ≤ g ≤ m(br ). Let B(G) = {b1 , b2 , . . . , br , . . . , bh } denote a set of all line balances for the problem SALBP-E with the precedence digraph G = (V, A) and each number m(br ) ≥ 2 of the workstations used. Since the line balance br ∈ B(G) is a partition (2) of the set V, i.e. each subset Vkbr of the set V is not empty, we obtain the upper bound m(br ) ≤ n on each admissible number m(br ) of the workstations used. The cycle-time c(br , t) of the line balance br ∈ B(G) with the vector t = (e t, t) m(br ) P of the operation durations is determined as follows: c(br , t) := maxk=1 i∈V br ti . The efficiency k E(br , t) of the line balance br ∈ B(G) is determined in (1). The maximum efficiency E(t) := max{E(bu , t) : bu ∈ B(G)}

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of the assembly line with the vector t of operation durations is achieved by the line balance br ∈ tP B(G), E(t) = E(br , t), if the product m(br ) · c(br , t) is minimal, i.e. m(br ) · c(br , t) = E(t) . Thus, the line balance br is optimal for the problem SALBP-E with the precedence digraph G = (V, A) and the vector t = (e t, t) of the operation durations if Condition 2 folds. Condition 2. The line balance br ∈ B(G) is optimal, if thePfollowing equalities hold: t m(br ) · c(br , t) = min {m(bu ) · c(bu , t) : bu ∈ B(G)} = E(t) . Let B(G, t) denote a set of all optimal line balances for the assembly line with the precedence digraph G = (V, A) and the vector t = (e t, t) = (t1 , t2 , . . . , tn ) of the operation durations. The inequality m(br ) ≥ 2 must hold for each admissible line balance br ∈ B(G) = {b1 , b2 , . . . , bh } in order to exclude a trivial line balance with only one workstation, i.e. all operations V are assigned to the same workstation S 1 ∈ S , m(bk ) = 1, and there is no idle time for this workstation. It is clear that such a line balance makes no sense in the assembly industry. Condition 3. For each line balance br ∈ B(G) = {b1 , b2 , . . . , bh } under consideration, the inequality m(br ) ≥ 2 must hold. e the operation duration ti is deterministic, we can Since for each operation i in the set V \ V, e We also assume that the initial duration assume that ti > 0 for each automated operation i ∈ V \ V. ti is a strictly positive real number ti > 0 for each operation i ∈ V. The value of the duration t j > 0 e ⊆ V can vary during the assembly line lifespan. Let V ebr denote a of each manual operation j ∈ V k subset of the set Vkbr of the manual operations as follows: \ e ebr := V br V, (4) V k k P where k ∈ {1, 2, . . . , m(br )}. The workstation duration t(Vkbr ) := i∈V br ti consists of the determink P P ebr ). A modified duration t0 may be even istic part i∈V br \Vebr ti and the variable part i∈Vebr ti =: t(V j k k

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ebr is processed by an equal to zero: t0j ≥ 0. A zero duration t0j = 0 means that the operation j ∈ V k additional operator in parallel with the other operations processed on the workstation S k ∈ S . Due to the additional operator, processing the operation j does not increase the workstation duration X X t0 (Vkbr ) := ti0 = ti0 . (5) i∈Vkbr

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Hereafter, t0 indicates the vector t0 = (e t0 , t) = (t10 , t20 , . . . , tn0˜ , tn˜ +1 , tn˜ +2 , . . . , tn ) = (t10 , t20 , . . . , tn0˜ , tn0˜ +1 , tn0˜ +2 , P . . . , tn0 ), for which the workstation duration t0 (Vkbr ) = i∈V br ti0 is calculated. The second equality in k (5) is valid due to the equality t0j = 0. Thus, the following condition is assumed. Condition 4. The initial duration ti is a strictly positive real number ti > 0 for each operation e , ∅ can vary during the i ∈ V. The value t j > 0 of the duration of each manual operation j ∈ V 0 assembly line lifespan. The modified duration t j may be equal to zero: t0j ≥ 0. Let Rn˜ denote space of all real n˜ -vectors e t = (t1 , t2 , . . . , tn˜ ) with the following metric: The distance d(e t, e t0 ) between the vector e t = (t1 , t2 , . . . , tn˜ ) ∈ Rn˜ and the vector e t0 = (t10 , t20 , . . . , tn0˜ ) ∈ Rn˜ is e where |ti − t0 | denotes an absolute value of the defined as follows: d(e t, e t0 ) := max{|ti − ti0 | : i ∈ V}, i 0 n˜ n˜ difference ti − ti . Let R+ ⊂ R denote a set of all non-negative real n˜ -vectors.

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Definition 1. A closed ball Oρ (e t) in space Rn˜ with the radius ρ ∈ R1+ and the center e t ∈ Rn+˜ is called a stability ball of the balance br ∈ B(G, t), if for any vector t0 = (e t0 , t) of the operation T line 0 n ˜ durations with e t ∈ Oρ (e t) R+ , the line balance br remains optimal, i.e. br ∈ B(G, t0 ). The largest radius ρbr (t) of the radii of all stability balls is called a stability radius of the line balance br .

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In Definition 1, the initial vector t = (e t, t) of the operation durations T is given and the vector e t0 of the durations of manual operations may vary within the set Oρ (e t) Rn+˜ . The above problem SALBP-E is a generalization of the assembly line balancing problems SALBP-1 and SALBP2 (Baybars, 1986; Morrison et al., 2014; Otto et al., 2013; Scholl, 1999). If the cycle time is predetermined for all line balances, i.e. the equality c(br , t) = c∗ holds for each admissible line balance br ∈ B(G), then the problem SALBP-E reduces to the problem SALBP-1. If the number of the workstations is predetermined, i.e. the equality m(br ) = m∗ holds for each admissible line balance br ∈ B(G), then the problem SALBP-E reduces to the problem SALBP-2.

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3. Related literature on the stability analysis of optimal line balances

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Since the production conditions may change over time in response to the changing demands of the customers, the need of re-balancing an assembly line may arise. Assembly re-balancing entails tedious procedures requiring significant costs and amounts of manpower (Chen et al., 2004; Chica et al., 2013; Gamberini et al., 2006). It is the stability analysis that can help us to identify the right time for re-balancing. In spite of its practical importance, the OR literature on the stability analysis of the line balancing is scanty (Chica et al., 2013; Gurevsky et al., 2012, 2013; Lai et al., 2016; Sotskov and Dolgui, 2001; Sotskov et al., 2005, 2006, 2015). We next discuss the stability and robustness results in the OR literature established especially for the assembly lines. In contrast to a stochastic assembly line (Dong et al., 2014; Erel and Sarin, 1998; Gamberini et al., 2006; Kahan et al., 2009), in the stability analysis, a probability distribution is not assumed to be known for e Note also that each operation duration ti , i ∈ V, is each duration t j of the manual operation j ∈ V. 5

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assumed to be a real number instead of an integer number as assumed by Scholl (1999) and many other authors. The deterministic counterpart of the problem SALBP-E is NP-hard (Gutjahr and Nemhauser, 1964; Wee and Magazine, 1982) since the NP-hard bin-packing problem is a special case of the problem SALBP-E, where the precedence digraph G = (V, A) has no arcs. As it is proven by Ramaswamy and Chakravarti (1995), if the deterministic problem is NPhard, then to find the stability radius of the given solution to this problem is NP-hard as well even if n˜ = 1 (this is discussed by Chakravarti and Wagelmans (1998)). The problem SALBP-E with variable operation durations was considered by Gurevsky et al. (2012), where a formula was established for the maximum variation of each operation duration preserving the feasibility of the line balance br provided that the upper bounds m∗ and c∗ on the used workstations and feasible cycle times were predetermined and given as a part of the input data, i.e. both inequalities m(br ) ≤ m∗ and c(br , t) ≤ c∗ must hold for all admissible line balances. The authors established a sufficient and necessary condition for a zero stability radius provided that c(br , t) ≤ c∗ . The stability analysis for a more general assembly line is developed in (Gurevsky et al., 2013), where there are several workplaces associated with a workstation, the set of operations assigned to the same workstation is partitioned into blocks, and all operations of the same block are processed simultaneously. A total duration of the operations from the block is assumed to be equal to the largest duration of the operation in this block. The authors derived a formula for the maximum variation of each operation duration preserving the feasibility of a line balance and established a lower bound on the stability radii of the optimal line balances. Hamta et al. (2013) considered a multi-objective assembly line balancing problem in which each operation duration can assume any value between the known lower and upper bounds. The three objectives that the authors considered were the minimization of the cycle-time, the total equipment cost, and the smoothness index. A heuristic algorithm was proposed based on the combination of a particle swarm optimization with a neighborhood search. Chica et al. (2013) studied the time and space assembly line balancing problem with a joint minimization of the cycle time, the number of workstations, and the workstation area. It was assumed that the demands of the final products were uncertain. The authors investigated how robust the assembly line configuration is when the production plan is changed as demands change. The robustness functions were based on the workstation overload under a future demand pattern and used as a posteriori information on the set of non-dominant solutions. In (Corominas et al., 2008), the authors studied how to re-balance the assembly line at a motorcycle plant for increasing the production in spring and summer. The motorcycle company used temporary operators to increase the production capacity. The goal was to minimize the number of temporary operators required without increasing the number of permanent operators. The problem was modeled as a linear program and solved by means of the ILOG CPLEX 9.0 optimizer. For the problem SALBP-1, a sufficient and necessary condition for a zero stability radius and formulae for obtaining the stability radius have been established in (Sotskov et al., 2006). The enumerative algorithms and the programs for constructing admissible, optimal, and stable line balances have been developed in (Sotskov et al., 2015). The stability analysis for the problem SALBP-2 with variable operation durations was developed in (Sotskov et al., 2005; Lai et al., 2016), where formulae, algorithms, and the software for obtaining the stability radius were developed and tested on the moderate sized benchmark instances. The following criterion for a zero stability radius has been established in (Sotskov et al., 2005). 6

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Theorem 1. For the problem SALBP-2, the optimal line balance b1 has a zero stability radius, if and only if there exists another optimal line balance br , b1 such that the inclusion (6)

W(b1 , t) ⊆ W(br , t)

ebr of the sets Vk br , defined in (4), for does not hold, where W(br , t) denotes a set of all subsets V k which the equality t(Vkbr ) = c(br , t) holds, k ∈ {1, 2, . . . , m}.

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In Section 4, we derive a criterion for a zero stability radius of the line balance br ∈ B(G, t) for the problem SALBP-E using the corresponding analog of the above set W(br , t). 4. A zero stability radius of the optimal line balance for the problem SALBP-E

To present a sufficient and necessary condition for a zero stability radius of the optimal line balance for the problem SALBP-E, we need the following definition.

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ebr of the sets V br , k ∈ {1, 2, . . . , m(br )}, Definition 2. Let W(br , t) denote a set of all subsets V k k satisfying the following inequalities: m(br ) · t(Vkbr ) = m(br ) · c(br , t) = min {m(bu ) · c(bu , t) : bu ∈ B(G)} =

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It is clear that the set W(br , t) defined for the line balance br ∈ B(G) for the problem SALBP-E using the equalities (7) reduces to the set W(br , t) used in Theorem 1 for the problem SALBP-2. If the line balance br is not optimal for the vector t of the operation durations, br ∈ B(G)\B(G, t), then the set W(br , t) is empty. On the other hand, since the inclusion br ∈ B(G, t) implies the equivalence of the condition (7) to equality t(Vkbr ) = c(br , t), the set W(br , t) cannot be empty for the optimal line balance br ∈ B(G, t), i.e. W(br , t) , ∅. We thus have the following claim.

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Lemma 1. The set W(b s , t) is empty, if and only if b s ∈ B(G) \ B(G, t).

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The set W(br , t) associated with the optimal line balance br ∈ B(G, t) may contain an empty ebr = ∅ as its element, ∅ ∈ W(br , t), if there is no manual operation assigned to the bottleneck set V k workstation S k for which the equality t(Vkbr ) = c(br , t) holds. Thus, the equality W(br , t) = {∅} may hold, i.e. the set W(br , t) may be a singleton containing ebr = ∅. Note that a singleton W(br , t) = {∅} containing the set V ebr = ∅ is a proper only the set V k k subset of any non-empty set W(b s , t) , {∅}, i.e. {∅} = W(br , t) ⊂ W(b s , t) , {∅}. 4.1. A sufficient and necessary condition for a zero stability radius

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We next establish a criterion for the optimal line balance b1 ∈ B(G, t) to be stable, i.e. ρb1 (t) > 0.

Theorem 2. For the problem SALBP-E, the optimal line balance b1 ∈ B(G, t) is stable, ρb1 (t) > 0, if and only if for each optimal line balance br ∈ B(G, t) \ {b1 }, the following inclusion holds: W(b1 , t) ⊆ W(br , t) along with one of the following three conditions: 1) m(b1 ) = m(br ); 2) m(b1 ) , m(br ), ∅ ∈ W(br , t), W(b1 , t) = {∅}; 3) m(b1 ) < m(br ), ∅ ∈ W(br , t), W(b1 , t) , {∅}.

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  S ALBP − E H   H  H  HH      j H ∃ br ∈ B(G, t) \ {b1 } ∀ br ∈ B(G, t) \ {b1 }

W(b , t) ⊆ W(br , t) 1 H   H   H   j H    m(b1 ) = m(br ) m(b1 ) , m(br )      Case 1  ρb1 (t) > 0         ? ∅ < W(br , t) ∅ ∈ W(br , t)      Case 7   ρb1 (t) = 0       ?    W(b1 , t) = {∅} W(b1 , t) , {∅}      Case 2   ρb1 (t) > 0     ?    m(b1 ) > m(br ) m(b1 ) < m(br )     Case 8 Case 3 ρb1 (t) = 0 ρb1 (t) > 0

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W(b , t) * W(b , t) 1 H r   H     j H   H m(b1 ) = m(br ) m(b1 ) , m(br )       Case 4  ρb1 (t) = 0       ?   W(b1 , t) = {∅} W(b1 , t) , {∅}     Case 5 Case 6 ρb1 (t) = 0 ρb1 (t) = 0

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Figure 1: The cases in the problem SALBP-E needed to be considered in the proof of Theorem 2.

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Proof. Sufficiency. There are nine cases needed to be considered in the problem SALBP-E, where in case 0 there is only one optimal line balance, while there are two or more optimal line balances in each of the other eight cases, cases 1 to 8. We start with case 0. Case 0: B(G, t) = {b1 }. We need to compare the optimal line balance b1 ∈ B(G, t) with an arbitrary line balance b s ∈ B(G) \ B(G, t) = B(G) \ {b1 }. Since the line balance b s is not optimal for the vector t of the operation durations, the strict inequality m(b s ) · c(b s , t) > m(b1 ) · c(b1 , t) holds. Therefore, for any vector e tδ ∈ Ren+ with the distance d(e t, e tδ ) = δ > 0, the opposite inequality

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m(b s ) · c(b s , tδ ) ≤ m(b1 ) · c(b1 , tδ )

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may hold for the vector tδ = (e tδ , t) only if the following condition holds: δ≥

m(b s ) · c(b s , t) − m(b1 ) · c(b1 , t) =: β(b1 , b s ), e n · max{m(b1 ), m(b s )}

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where the notation β(b1 , b s ) is useful for a later analysis. It is clear that one can compensate the difference m(b s ) · c(b s , t) − m(b1 ) · c(b1 , t) > 0 via an appropriate modification (increasing or e 1, bs) ⊆ V e of the manual operations, where decreasing) of the durations of a suitable subset V(b 0 e |V(b1 , b s )| ≤ e n ≥ 1. The value |ti − ti | of such a modification of the duration ti of the manual e operation i ∈ V(b1 , b s ) may reduce the difference m(b s ) · c(b s , t) − m(b1 ) · c(b1 , t) 8

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at most by the following value: |ti0 − ti | · max{m(b1 ), m(b s )}. Therefore, the minimum value of δ e 1 , b s ) attaining the inequality (9), must of the modification of durations ti of the operations i ∈ V(b satisfy the inequality given in (10). Due to the lower bound (10) imposed on the δ, the desired vector e tδ ∈ Ren+ for which the inequality (9) holds, cannot be arbitrarily close to the vector e t. If the inequality (9) holds, the line balance b s is called a competitor for the optimal line balance b1 ∈ B(G, t). Since the bound (10) must hold for any non-optimal line balance b s ∈ B(G) \ B(G, t), which may be a competitor for the optimal line balance b1 ∈ B(G, t), one can conclude that the following condition holds for the desired vector e tδ ∈ Ren+ in case 0: d(e t, e tδ ) > β(b1 ) := min{β(b1 , b s ) : b s ∈ B(G) \ B(G, t)} > 0.

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Thus, it is proven that the following inequalities hold: ρb1 (t) > β(b1 ) > 0, where β(b1 ) is determined in (12) and β(b1 , b s ) in (10). In the remaining sufficiency proof, we consider cases 1, 2, and 3 (see Fig. 1) for which the set B(G, t) is not a singleton, i.e. |B(G, t)| ≥ 2, and the condition (8) holds for each optimal line balance br ∈ B(G, t) \ {b1 }. Case 1: There is an optimal line balance br ∈ B(G, t) \ {b1 } , ∅ such that the condition (8) holds along with the equality m(b1 ) = m(br ). If the number of used workstations is predetermined, i.e. m(bu ) = m∗ for any admissible line balance bu ∈ B(G), then the problem SALBP-E reduces to the problem SALBP-2. The definition of the set W(br , t) used in Theorem 1 for the problem SALBP-2 coincides with the definition of the set W(br , t) used in Theorem 2 for the problem SALBP-E provided that the number of used workstations is predetermined. Taking into account that for case 1 in the problem SALBP-E, the equality m(b1 ) = m(br ) and the inclusion W(b1 , t) ⊆ W(br , t) hold for any optimal line balance br , r , 1, we conclude that the necessity of Theorem 1 established for the problem SALBP-2 in (Sotskov et al., 2005) implies the inequality ρb1 (t) > 0 for case 1 in the problem SALBP-E. Before considering cases 2 and 3 (see Fig. 1), we make the following observation. If the optimal line balance b1 is compared with the non-optimal line balance b s ∈ B(G) \ B(G, t), m(b s ) · c(b s , t) > m(b1 ) · c(b1 , t), then the inequality (9) may be attained for some vector e tδ = (t1δ , t2δ , . . . , tenδ ), where |tiδ − ti | ≤ δ =: δbb1s , only if the condition (10) holds. Hence, for cases 2 and 3, we obtain the lower bound (12) established for the distance d(e t, e tδ ) = δ for any vector e tδ ∈ Ren+ , which destroies the optimality of the line balance b1 using the non-optimal line balance b s ∈ B(G) \ B(G, t) as a competitor for the line balance b1 . To prove the inequality ρb1 (t) > 0 in cases 2 and 3, we need to compare the line balance b1 with other line balances br ∈ B(G) \ B(G, t). Case 2: There is an optimal line balance br ∈ B(G, t) \ {b1 } , ∅ such that the following conditions hold: W(b1 , t) ⊆ W(br , t), m(b1 ) , m(br ), ∅ ∈ W(br , t), W(b1 , t) = {∅}. We introduce the following notation: 1 ebr < W(br , t)}], if k ∈ {1, 2, . . . , m(br )} with  · [c(br , t) − max{t(Vkbr ) : V  k e n      t(Vkbr ) < c(br , t);  (13) γ(br ) :=   e  min{ti : i ∈ V}, if t(Vkbr ) = c(br , t) for each      k ∈ {1, 2, . . . , m(br )}. Let br be any optimal line balance from the set B(G, t) \ {b1 }, i.e. the following equality holds: m(b1 ) · c(b1 , t) = m(br ) · c(br , t). 9

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Due to conditions W(b1 , t) ⊆ W(br , t), ∅ ∈ W(br , t), and W(b1 , t) = {∅}, any decrease tiδ = ti − δ e 1 , br ) of the set V e of the manual operations of the duration ti by the value of δ for any subset V(b will not affect the equality (14) via substituting the vector t by the vector tδ = (e tδ , t), where e tδ = δ δ δ δ δ e 1 , br ) and t = t j for each operation (t1 , t2 , . . . , ten ) with ti = ti − δ for each operation i ∈ V(b j e e j ∈ V \ V(b1 , br ). Thus, the following equality holds: m(b1 ) · c(b1 , tδ ) = m(br ) · c(br , tδ )

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e t, e tδ ) = δ ≤ min{ti : i ∈ V}. for any variation tδ = (e tδ , t) of the vector t provided that d(e Due to the equality W(b1 , t) = {∅} and the inequality e n ≥ 1, the value of γ(b1 ) determined in (13) is positive: γ(b1 ) > 0. It is clear that any increase tiα = ti + α of the duration ti by the value of e 1 , br ) of the manual operations V e implies the inequality α ≤ γ(b1 ) for any subset V(b m(b1 ) · c(b1 , tα ) ≤ m(br ) · c(br , tα )

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for any variation tα = (e tα , t) of the vector t such that d(e t, e tα ) ≤ γ(b1 ), where e tα = (t1α , t2α , . . . , tenα ) with e \ {i}. From this and the inequality in (12), it follows tiα = ti + α and tαj = t j for each operation j ∈ V that the following inequalities hold for case 2: ρb1 (t) > min{β(b1 ), γ(b1 )} > 0,

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where β(b1 ) is determined in (12) and γ(b1 ) in (13). Case 3: There is an optimal line balance br ∈ B(G, t) \ {b1 } , ∅ such that the following conditions hold: W(b1 , t) ⊆ W(br , t), ∅ ∈ W(br , t), W(b1 , t) , {∅}, m(b1 ) < m(br ). Let br be any optimal line balance from the set B(G, t) \ {b1 }. Hence, the equality (14) holds for the line balances b1 and br . Due to Lemma 1, the set W(b1 , t) cannot be empty: W(b1 , t) , ∅. Due to conditions W(b1 , t) ⊆ W(br , t), W(b1 , t) , {∅}, and W(b1 , t) , ∅, any increase tiψ = ti + ψ of the eb1 ∈ W(b1 , t) implies the following inequality: duration ti of any operation i ∈ V k (18)

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m(b1 ) · c(b1 , tψ ) < m(br ) · c(br , tψ ),

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since the inequality m(b1 ) < m(br ) and the equality (14) hold. In the inequality (18), the variation tψ = (e tψ , t) of the vector t is determined by the vector e tψ = (t1ψ , t2ψ , . . . , tenψ ), where tiψ = ti + ψ and e \ {i}. tψj = t j for each operation j ∈ V Due to the definition of the set W(b1 , t) , {∅}, any increase tψj of the duration t j , where j ∈ o n b e \ Sm(br ) V e1 :V eb1 ∈ W(b1 , t) , by the value of ψ ≤ γ(b1 ) will not increase the cycle time of V k

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the line balance b1 . Thus, the equality c(b1 , t) = c(b1 , tψ ) holds provided that tψj = t j + ψ, where ψ ≤ γ(b1 ). Therefore, the following inequality holds for any increase e tψ of the vector e t: m(b1 ) · c(b1 , tψ ) ≤ m(br ) · c(br , tψ ).

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It is shown that the inequality (19) holds for any increase e tψ of the durations of manual operae if the value of ψ is not greater than the value of γ(b1 ) determined in (13). tions V, e can reduce the cycle time Any decrease ti = ti − of the duration ti of any manual operation i ∈ V of the line balance b1 , i.e. the inequality c(b1 , t) ≥ c(b1 , t ) may hold, where e ti = (t1 , t2 , . . . , ten ) and e However, any decrease t of the duration ti of any manual operation i ∈ V e 0 <  ≤ min{ti : i ∈ V}. i 10

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@

t3 = 8 - l 3 @ R @ t4

? 4l =8

Figure 2: The precedence digraph G = (V, A) determining the partial order on the set V for Example 1.

cannot reduce the cycle time of the line balance br due to the inclusion ∅ ∈ W(br , t). Therefore, the equality c(br , t) = c(br , t ) holds and we obtain the following inequality:

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m(b1 ) · c(b1 , t ) ≤ m(br ) · c(br , t )

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for any variation t = (e t , t) of the vector t such that the inequality d(e t, e t ) ≤ γ(b1 ) holds. Thus, it is proven that for any value of δ > 0, which is less or equal to the γ(b1 ), the inequality (20) holds for any line balance br ∈ B(G, t) \ {b1 } and for any vector t = (e t , t) with the distance  d(e t, e t ) ≤  > 0. From this and the inequality (12), it follows that the inequalities (17) hold for case 3 in the problem SALBP-E. This completes the sufficiency proof. The necessity proof of Theorem 2 is given in Appendix A.  The above sufficiency proof of Theorem 2 (see case 0) implies the following claim. Corollary 1. If B(G, t) = {b1 }, then ρb1 (t) > 0.

4.2. Example 1 with two unstable optimal line balances

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We consider the following example, Example 1. The set V = {1, 2, 3, 4} of the assembly operations must be processed on a simple assembly line, where there are two manual operations, e n = 2, and two automated operation. The operation durations are determined by the following vector t = (e t, t) = (t1 , t2 , t3 , t4 ) = (3, 7, 8, 8). The precedence digraph G = (V, A) S is presented in Fig. 2. SFor Example 1, thereSare 11 line balances as follows: b1 : V = {1, 3} S S {2,S4}, b2 : V = {1, 2} {3, 4}, b : V = {1} {2, 3, 4}, b : V = {1, 2, 3} {4}; b : V = {1} 3 4 5 S S S S S S {2} {3, 4}; b6 : VS= {1} : V S = {1}S {2, 3} {4}, b8 : SV =S{1, 2} S {3} {2, 4}, b7S S {3} {4}, b9 : V = {1, 3} {2} {4}; b10 : V = {1} {2} {3} {4}, b11 : V = {1} {3} {2} {4}. The products m(br ) · c(br , t) for all line balances br ∈ B(G) are determined as follows: m(b1 ) · c(b1 , t) = 2 · 15 = 30, m(b2 ) · c(b2 , t) = 2 · 16 = 32, m(b3 ) · c(b3 , t) = 2 · 23 = 46, m(b4 ) · c(b4 , t) = 2 · 18 = 36; m(b5 ) · c(b5 , t) = 3 · 16 = 48, m(b6 ) · c(b6 , t) = 3 · 15 = 45, m(b7 ) · c(b7 , t) = 3 · 15 = 45, m(b8 ) · c(b8 , t) = 3 · 10 = 30, m(b9 ) · c(b9 , t) = 3 · 11 = 33; m(b10 ) · c(b10 , t) = 4 · 8 = 32, m(b11 ) · c(b11 , t) = 4 · 8 = 32. There are two optimal line balances B(G, t) = {b1 , b8 } since m(b1 ) · c(b1 , t) = m(b8 ) · c(b8 , t) = min {m(br ) · c(br , t) : br ∈ B(G)} = 30 < 32 = min {m(br ) · c(br , t) : br ∈ B(G) \ {b1 , b8 }} . We next show that equalities ρb1 (t) = 0 = ρb8 (t) hold due to Theorem 2 (case 6 in Fig. 1). First, we determine W(b1 , t) = {{2}} and W(b8 , t) = {{1, 2}}. For the optimal line balance b1 , there is another optimal line balance b8 ∈ B(G, t) \ {b1 } such that the inclusion (8) does not hold, i.e. W(b1 , t) = {{2}} * {{1, 2}} = W(b8 , t), and the following conditions hold: m(b1 ) = 2 , 3 = m(b8 ), W(b1 , t) = {{2}} , {∅}. Thus, we obtain case 6 for the optimal line balance b1 . Due to Theorem 2, the optimal line balance b1 is unstable, ρb1 (t) = 0. Similarly, we see that for the optimal line balance b8 , there is another optimal line balance b1 ∈ B(G, t) \ {b8 } such that the inclusion (8) does not hold, W(b8 , t) = {{1, 2}} * {{2}} = W(b1 , t), and both conditions m(b8 ) = 3 , 2 = m(b1 ) and W(b8 , t) = {{1, 2}} , {∅} hold. We obtain case 6 for the optimal line balance b8 . Due to Theorem 2, the optimal line balance b8 is unstable, ρb8 (t) = 0. 11

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We can also show the instability of the optimal line balance b1 using Definition 1 directly. To this end, we set t20 = t2 +  = 7 − , where  ∈ R1+ is any arbitrary small real number. We set t10 = t1 = 3. For the modified vector t0 = (e t, t) = (t10 , t20 , t3 , t4 ) = (3, 7 − , 8, 8) of the operation durations, we obtain m(b1 ) · c(b1 , t0 ) = 30 − 2 > 30 − 3 = m(b8 ) · c(b8 , t0 ). Thus, we have b1 < B(G, t0 ) = {b8 } and ρb1 (t) = 0. 5. Lower and upper bounds on the stability radius

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In (Sotskov et al., 2005), the following lower bound was established on the non-zero stability radius ρb1 (t) of the optimal line balance b1 for the problem SALBP-2: ρb1 (t) ≥ min{β(b1 ), γ(b1 )} > 0. Using this bound for the problem SALBP-2 that remains correct for case 1 in the problem SALBP-E, and due to the above proof of the sufficiency of Theorem 2 (cases 2 and 3), we obtain the following claim for the problem SALBP-E.

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Corollary 2. If ρb1 (t) > 0, then ρb1 (t) ≥ min{β(b1 ), γ(b1 )} > 0, where γ(b1 ) is defined in (13) and β(b1 ) is defined in (10) and (12). Corollary 2 gives the lower bound on the stability radius of the line balance b1 ∈ B(G, t) for the problem SALBP-E. The following sufficient and necessary condition for the existence of an infinite stability radius for the problem SALBP-2 has been established in (Lai et al., 2016).

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e there exists a line balance bu such that Theorem 3. Assume that for each manual operation i ∈ V, bu the equality Vk = {i} holds. Then, the line balance b1 , which is optimal for the problem SALBP-2 with the vector t ∈ Rn+ of the operation durations, has an infinite stability radius if and only if the e and the line balance b1 remains optimal for the equality Vkb1 = {i} holds for each operation i ∈ V ∗ ∗ e modified vector t = (t , t) = (0, 0, . . . , 0, ten+1 , ten+2 , . . . , tn ) ∈ Rn+ of the operation durations.

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We next present a criterion for the existence of an infinite stability radius ρb1 (t) = ∞ of the line balance b1 ∈ B(G, t) for the problem SALBP-E.

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Theorem 4. For the problem SALBP-E, there exists an optimal line balance b1 ∈ B(G, t) with an infinite stability radius ρb1 (t) = ∞ if and only if either the equality e n = 0 holds or both inequality e n ≥ 1 and equality n = 2 hold. S Proof. Sufficiency.SLet the equality n = 2 hold. Then there exist two line balances b1 : V = {1} {2} and b2 : V = {2} {1} for the problem SALBP-E. For each line balance br ∈ B(G), we obtain m(br ) ≤ n = 2 due to condition Vkbr , ∅. Since the inequality m(br ) ≥ 2 holds (Condition 3), we obtain the equality m(br ) = 2 for each line balance br ∈ B(G). Hence, B(G) = {b1 , b2 } and both line balances b1 and b2 are optimal for the vector t = (e t, t) of the operation durations. Furthermore, both line balances b1 and b2 remain optimal for any modified vector t0 = (e t0 , t) with the durations e t0 ∈ Ren of the manual operations. Hence, both line balances b1 and b2 have the infinite stability radii, ρb1 (t) = ∞ = ρb2 (t). Since the precedence digraph G = (V, A) has no circuits and the inequality n ≥ 2 holds, the set B(G) is not empty and so there is at least one optimal line balance br ∈ B(G, t). Let the equality e n = 0 hold. Then the initial vector t = (e t, t) = (t1 , t2 , . . . , tn ) of the operation durations is fixed e = ∅ and so any optimal line balance during the lifespan of the assembly line due to the equality V br ∈ B(G, t) cannot lose its optimality, i.e. ρbr (t) = ∞. The sufficiency has been established. 12

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The necessity proof of Theorem 4 is given in Appendix B.  It takes O(1) time to test the condition of Theorem 4. In what follows, we use the following notation k(i, br ) for the index u of the workstation S u . If operation i is assigned to the workstation S u ∈ S in the line balance br ∈ B(G), then we denote the index u of the workstation S u as follows: u = k(i, br ). Let b1 ∈ B(G, t) and there exist a workstation S k ∈ S such that the set Vkb1 contains only manual operations, i.e. \ e=V eb1 = V b1 , ∅. Vkb1 V (21) k k

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Let S (b1 , t) denote a set of all workstations S k ∈ S such that the condition (21) holds. The e holds for inclusion S k ∈ S (b1 , t) means that Vkb1 = {ik1 , ik2 , . . . , ikn(k) }, where the inclusion ik j ∈ V ∗ each index j ∈ {1, 2, . . . , n(k)} and n(k) ≥ 1. If S k ∈ S (b1 , t), we set tk j = 0 for each index e \ V b1 . Due to the condition (21), we j ∈ {1, 2, . . . , n(k)}. We assume t∗ = tl for each operation l ∈ V l

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can reassign all operations Vkb1 either to the workstation S k−1 (if k ≥ 2) or to the workstation S k+1 (if k ≤ m(b1 ) − 1) without violating Condition 1 for any given precedence digraph G = (V, A). Due to this reassignment of the operations from set Vkb1 , the line balance b1 reduces to the line balance br , which does not use the workstation S k . We obtain the equality m(bk ) = m(b1 ) − 1. e \ {i}. These settings imply the equality We set ti∗ = 0 and t∗j = t j for each manual operation j ∈ V ∗ ∗ c(b1 , t ) = c(br , t ) along with the following equalities: m(br ) · c(br , t∗ ) = m(br ) · c(b1 , t∗ ) = (m(b1 ) − 1) · c(b1 , t∗ ) = m(b1 ) · c(b1 , t∗ ) − c(b1 , t∗ ),

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which imply the inequality m(br )·c(br , t∗ ) < m(b1 )·c(b1 , t∗ ), since the inequality c(b1 , t∗ ) > 0 holds due to the assumption t j > 0 for each operation j ∈ V (Condition 4). We obtain b1 (t∗ ) < B(G, t∗ ) ∗ and ρ n b1 (t) ≤ ti − ti = tib. It ois shown that the stability radius ρb1 (t) is bounded as follows: ρb1 (t) ≤ e {i} = V i min ti : i ∈ V, k(i,b1 ) . The following upper bound on the stability radius is established. n o b1 Lemma 2. If n ≥ 3 and S (b1 , t) , ∅, then ρb1 (t) ≤ minS k ∈S (b1 ,t) maxn(k) j=1 ti j : i j ∈ Vk .

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Obviously, it takes O(e n) time to obtain the upper bound on the finite stability radius given in Lemma 2. We next show how to calculate the stability radius if ρb1 (t) < ∞.

6. An exact value of the finite stability radius

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Due to Definition 1, in order to obtain an exact value of the finite stability radius ρb1 (t) < ∞ of the optimal line balance b1 ∈ B(G, t), one needs to find a competitor br ∈ B(G) \ {b1 } for the line balance b1 and the vector e t = (t1 , t2 , . . . , tn˜ ) ∈ Rn+˜ such that the following inequality holds: m(b1 ) · c(b1 , t ) > m(br ) · c(br , t ),

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t , e t) is minimal. Since the value of c(b1 , t) depends linearly where t = (e t , t) and the distance d(e on the components of the modified vector e t = (t1 , t2 , . . . , tn˜ ), before attaining the inequality (22) by continuously changing the components of the vector e t, the following equality: m(b1 ) · c(b1 , tδ ) = m(br ) · c(br , tδ )

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will be attained for some modified vector t0 = (t10 , t20 , . . . , tn0 ) ∈ Rn+ of the operation durations. In order to attain the equality (23), we will use the following claim. 13

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t1 = 5 1l H

t4 = 8 t3 = 10 HH j l - l 3 4 *    2l

t2 = 3

Figure 3: The precedence digraph G = (V, A) (without transitive arcs) determining the partial order on the set V for Example 2.

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Lemma 3. Let inclusions b1 ∈ B(G, t) and br ∈ B(G)\{b1 } hold. For the vector t(uk) = (e t(uk) , t) ∈ Rn+ of the operation durations, the inclusion {b1 , br } ⊆ B(G, t(uk) ) holds only if there exist workstations S u and S k with 1 ≤ u ≤ m(b1 ) and 1 ≤ k ≤ m(br ) such that the following three equalities hold: (a) t(uk) (Vub1 ) = c(b1 , t(uk) ); (b) t(uk) (Vkbr ) = c(br , t(uk) ); tP (c) m(b1 ) · t(uk) (Vub1 ) = m(br ) · t(uk) (Vkbr ) = E(t(uk) ) , where E(t(uk) ) is the maximum efficiency (3) of the assembly line with the vector t(uk) of the operation durations. The correctness of Lemma 3 follows directly from the definition (2) of the line balance br ∈ B(G) and Condition 2 of the line balance optimality. We next demonstrate on how to calculate for the stability radius ρb1 (t) using Lemma 3 and Theorem 5 described in Subsection 6.1. 6.1. How to calculate the stability radius for a small example

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We consider the following Example 2. The set V = {1, 2, 3, 4} of four assembly operations must be processed on a simple assembly line, where there are 3 manual operations, e n = 3, and one automated operation. The operation durations are determined by the vector t = (e t, t) = (t1 , t2 , t3 , t4 ) = (5, 3, 10, 8). The precedence digraph G = (V, A) is presented in Fig. 3, where S S the transitive arcsS are omitted. For Example 2, S thereS are 11 line balances: b1S : V = {1, S S2} {4} {3}, b2 : SV = {1} {2} {3, 4}, b : V = {2} {1} {3, 4}, b : V = {1} {2, 4} {3}, 3S S S 4 S b5 : V = {2} {1, 4} {3}; : V = {2} {1, 3, 4},S b8 :SV =S{1, 2} {3, 4}, b9 : S b6 : V = {1} {2, S 3, 4}, S b7S V = {1, 2, 4} {3}; b10 : V = {1} {2} {4} {3}, b11 : V = {2} {1} {4} {3}. The products m(br ) · c(br , t) for all line balances br ∈ B(G) are determined as follows: m(b1 ) · c(b1 , t) = 3 · 10 = 30, m(b2 ) · c(b2 , t) = 3 · 18 = 54, m(b3 ) · c(b3 , t) = 3 · 18 = 54, m(b4 ) · c(b4 , t) = 3 · 11 = 33, m(b5 ) · c(b5 , t) = 3 · 13 = 39; m(b6 ) · c(b6 , t) = 2 · 21 = 42, m(b7 ) · c(b7 , t) = 2 · 23 = 46, m(b8 ) · c(b8 , t) = 2 · 18 = 36, m(b9 ) · c(b9 , t) = 2 · 16 = 32; m(b10 ) · c(b10 , t) = 4 · 10 = 40, m(b11 ) · c(b11 , t) = 4 · 10 = 40. The line balance bP1 is uniquely optimal since m(b1 ) · c(b1 , t) = min {m(br ) · c(br , t) : br ∈ B(G)} t = 30 = 26 26 = E(t) < 32 = min {m(br ) · c(br , t) : br ∈ B(G) \ {b1 }} . Due to Corollary 1, the 30 equality B(G, t) = {b1 } implies the inequality ρb1 (t) > 0. In order to obtain the stability radius ρb1 (t), we will look for a competitor br ∈ B(G) \ {b1 } for the line balance b1 and workstations S u and S k such that the equalities (a), (b), and (c) hold and the distance d(e t, e t(uk) ) is minimal. We compare the line balance b1 with each line balance br ∈ B(G) \ {b1 } in the non-decreasing order of the products m(br ) · c(br , t). We first compare the line balance b1 with the line balance b9 , whose product m(b9 ) · c(b9 , t) is equal to 32. The equality (a) given in Lemma 3 holds for the workstation S u = S 3 and the equality (b) holds for the workstation S k = S 1 . Since m(b1 ) = 3 > 2 = m(b9 ), in order to attain the equality (23), we 14

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rk calculate for the minimum distance d(e t, e t(uk) ) =: δ>b b1 u using the following formula: rk δ>b b1 u =

m(br ) · t(Vkbr ) − m(b1 ) · t(Vub1 ) . ebr \ V eub1 | − |V ebr ∩ V eub1 |] + m(b1 ) · |V eub1 | m(br ) · [|V k

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k

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rk e e(uk) ) = It is easy to convince that the above value of δ>b b1 u is equal to the minimum distance d(t, t br b1 rk δ>b b1 u since the difference m(br ) · t(Vk ) − m(b1 ) · t(Vu ) can be compensated by increasing each br b1 rk duration ti , where i ∈ Vub1 , by the value δ>b b1 u and decreasing each duration ti , where j ∈ Vk \ Vu , rk e(uk) of the vector e by the same value δ>b t, we obtain the equalities (c). br k . Using the modification t >b7 1 Based on the formula (24), we can obtain the following value of δb1 3 =

m(b7 ) · t(V1b7 ) − m(b1 ) · t(V3b1 ) 2 2 2 · 16 − 3 · 10 = = . = b7 e b1 b7 b1 b1 e e e e 2 · [|{1, 2} \ {3}| − |{1, 2} ∩ {3}|] + 3 · |{3}| 2 · [2 − 0] + 3 · 1 7 m(b7 ) · [|V1 \ V3 | − |V1 ∩ V3 |] + m(b1 ) · |V1 |

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In order to attain the equality (23), we modify the vector e t of the durations of manual operations as follows: t1(3,1) = t1 − 72 = 5 57 , t2(3,1) = t2 − 27 = 1 75 , t3(3,1) = t3 + 27 = 10 72 . We obtained the modified vector t(3,1) = (e t(3,1) , t) ∈ Rn+ of the operation durations for which the equalities (a), (b), and (c) hold. For the modified vector t(3,1) = (e t(3,1) , t) of the operation durations, there are two optimal line balances: B(G, t(3,1) ) = {b1 , b9 }. Due to Theorem 2 (see case 6 in Fig. 1), the equality ρb1 (t(3,1) ) = 0 holds, since there exists the line balance b9 ∈ B(G, t(3,1) ) \ {b1 } such that the inclusion (8) does not hold, i.e. W(b1 , t(3,1) ) = {{3}} * {{1, 2}} = W(b9 , t(3,1) ), along with conditions m(b1 ) = 3 , 2 = m(b9 ) and W(b1 , t) , {∅}. Thus, using the line balance b9 as a competitor for the line balance b1 , we obtain the following upper bound on the stability radius:

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2 91 ρb1 (t) ≤ δ>b b1 3 = . 7

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CE

PT

We next compare the line balance b1 with the line balance b4 for which the equality (b) holds for the workstation S k = S 2 and m(b4 ) · c(b4 , t) = 33. eub1 and V ebr as follows: V eub1 ⊕ V e br the sign ⊕ denote oa direct summation of the two sets V k k nLet o n S T eub1 V ebr \ V eub1 V ebr . Since the equality m(b1 ) = m(b2 ) holds, in order to attain the equality = V k k rk (23), we calculate for the minimum distance d(e t, e t(uk) ) =: δ=b b1 u using the following formula: rk δ=b b1 u

t(Vkbr ) − t(Vub1 ) . = eub1 | ebr ⊕ V |V

(26)

AC

k

rk rk e e(uk) ) = δ=b The above value of δ=b b1 u is equal to the minimum distance d(t, t b1 u since the difference m(br )·t(Vkbr )−m(b1 )·t(Vub1 ) = m(b1 )·[t(Vkbr )−t(Vub1 )] can be compensated by increasing each duration br b1 rk ti , where i ∈ Vub1 \ Vkbr , by the value δ=b b1 u and decreasing each duration ti , where j ∈ Vk \ Vu , rk e(uk) of the vector e by the same value δ=b t, we obtain the equalities (c). br k . Using the modification t Based on the formula (26), we have that 42 δ=b b1 3 =

t(V2b4 ) − t(V3b1 ) 11 − 10 1 1 = = . = b b 4 1 e ⊕V e | |{2} ⊕ {3}| |{2, 3}| 2 |V 2

3

15

ACCEPTED MANUSCRIPT t5 = 9 5l

t1 = 1 - 1l

t2 = 2 - 2l

t3 = 4 - 3l

t4 = 6 - 4l

t6 = 8 - 6l

Figure 4: The precedence digraph G = (V, A) (without transitive arcs) determining the partial order on the set V for Example 3.

CR IP T

In order to attain the equality (23), we modify the vector e t of the durations of manual operations as follows: t2(3,2) = t2 − 12 = 2 12 , t3(3,2) = t3 + 21 = 10 12 . The equalities (a), (b), and (c) hold for the constructed vector t(3,2) = (e t(3,2) , t) ∈ R4+ of the operation durations. We next compare the line balance b1 with the line balance b8 for which the equality (b) holds 82 for the workstation S k = S 2 and m(b8 ) · c(b8 , t) = 36. Using the formula (24), we have that δ>b b1 3 = m(b8 ) · t(V2b8 ) − m(b1 ) · t(V3b1 ) 6 2 · 18 − 3 · 10 = = 6. = b8 e b1 b8 b1 b1 e e e e 2 · [|{3} \ {3}| − |{3} ∩ {3}|] + 3 · |{3}| 2 · [0 − 1] +3·1 m(b8 ) · [|V \ V | − |V ∩ V |] + m(b1 ) · |V | 2

3

2

3

1

M

AN US

We note that it is unnecessary to compare the line balance b1 with all remaining line balances br ∈ {b2 , b3 , b5 , b6 , b7 , b10 , b11 } from the set B(G) \ {b1 , b4 , b8 , b9 }, since due to the upper bound (10), the comparison of the line balance b1 with any line balance br ∈ {b2 , b3 , b5 , b6 , b7 , b10 , b11 } >b9 1 rk gives at least the value of β(b1 , br ) = 10 , which is the lower bound on the value of δ>b b1 u = δb1 3 ≥ 12 β(b1 , b9 ) = β(b1 , br ) and which is greater than the already obtained value of 72 ≥ ρb1 (t) given in 2 2 1 91 (25). Therefore, the minimum value δ>b b1 3 = 7 = min{ 7 , 2 , 6} of the calculated upper bounds on the stability radius ρb1 (t) is attained for the competitor b9 for the line balance b1 on the workstation S 3 in the line balance b1 and the workstation S 1 in the line balance b9 since for the modified vector t(3,1) = (e t(3,1) , t) ∈ R4+ of the operation durations, the equalities (a), (b), and (c) hold. 2 91 To show that equalities ρb1 (t) = d(t, t(3,1) ) = δ>b b1 3 = 7 hold, we can test the inequality (d) of Theorem 5 that follows directly from Definition 1, Lemma 3, and the inequality (22).

PT

ED

Theorem 5. Let the inclusions b1 ∈ B(G, t) and br ∈ B(G) \ {b1 } hold. Then, the equality ρb1 (t) = d(t, t(uk) ) holds if and only if there exist workstations S u and S k , 1 ≤ u ≤ m(b1 ), 1 ≤ k ≤ m(br ), such that the equalities (a), (b), and (c) hold, and for any small real number  ∈ R1+ , there exists a vector e t ∈ Ren+ with d(t , t(uk) ) =  and t = (e t , t) for which the following inequality holds: b1 br   (d) m(b1 ) · t (Vu ) > m(br ) · t (Vk ).

CE

Let us test the condition (d) of Theorem 5 for Example 1. By assuming that t1 = t1(3,1) , t2 = t2(3,1) , and t3 = t3(3,1) +  for any given small real number  ∈ R1+ , the condition (d) holds as follows: m(b1 ) · t (V3b1 ) = 3 · (10 72 + ) = 30 67 + 3 ·  > 30 67 = 2 · (8 + 1 57 + 5 57 ) = 2 · t (V1b7 ) = m(b7 ) · t (V1b7 ). Hence, b1 < B(G, t ). Due to Theorem 5, the equalities ρb1 (t) = d(t, t(3,1) ) = 72 hold.

AC

6.2. Other formulae needed for calculating finite stability radii

To derive other formulae, which may be used for obtaining the stability radius ρb1 (t), we consider the following Example 3. The set V = {1, 2, 3, 4, 5, 6} of the assembly operations must be processed on the simple assembly line, where there are four manual operations, e n = 4, and two automated operations. The operation durations are determined by the vector t = (e t, t) = (t1 , t2 , t3 , t4 , t5 , t6 ) = (1, 2, 4, 6, 9, 8). The digraph G = (V, A) is presented in Fig. 4, where the transitiveSarcs are omitted for simplicity. S In total, there are 30 line S balances as follows: b1 :SV = {1, 2, 3, 5} {4, 6}, b2 : V = {1, S 2, 5} {3, 4, 6}, b3 : V S = {1, 5} S {2, 3, 4, 6}, b4 : V = {5} S{1, 2, 3,S4, 6}, b5 : V = {1, 2, 3, 4, 5} {6}; b : V = {1, 5} {2, 3, 4} 6 S S S S {6}, b7 : V = {1, 2, 5} S {3, 4} S {6}, b8 : V = {5} {1, 2, 3, 4} {6}, b9 : V = {5} {1, 2, 3} {4, 6}, b10 : V = {1, 5} {2, 3} {4, 6}, b11 : V = 16

ACCEPTED MANUSCRIPT

M

AN US

CR IP T

S S S S S S {1, 2, 5} {3} {4, 6}, b : V = {1, 2, 3, 5} {4} {6}, b : V = {5} {1, 2} {3, 12 13 S S S S S S S 4, 6},Sb14 : V = {5} {1} S {2, 3, 4,S 6}; bS15 : V = {5} {1, 2, 3}S {4}S {6}, Sb16 : V = {5} {1, 2} S {3,S4} {6}, Sb17 : V = {1, 5} {2, 3}S {4}S {6},Sb18 : V = {1, 5} {2} {3, 4} {6}, b : V = {5} {1}S {2, 3, S 4} {6}, 19 S S S S b20 : V = {1, 2, 5} {3}S {4}S {6}, Sb21 : V = {5} {1} S{2, 3}S {4,S6}, b22 : V = {5} {1, 2} S {3} S {4, 6}, : V = {1, 5} {2} {3} V = {5}S {1, 2} S b23S S {4, S 6}, b24S: V =S{5} {1} {2} {3,S4, 6};Sb25 : S {3} {4} S {6}, bS 26 : VS= {5} S {1} {2, 3} {4}S {6},Sb27 : SV =S{5} {1} {2} {3, 4} S {6},Sb28 : V = {1, 5} {2} {3} {4} {6}, b29 : V = {5} {1} {2} {3} {4, 6}; b30 : V = {5} {1} {2} S S S {3} {4} {6}. The products m(br ) · c(br , t) for all line balances br ∈ B(G) are obtained as follows: m(b1 ) · c(b1 , t) = 2 · 16 = 32, m(b2 ) · c(b2 , t) = 2 · 18 = 36, m(b3 ) · c(b3 , t) = 2 · 20 = 40, m(b4 ) · c(b4 , t) = 2 · 21 = 42, m(b5 ) · c(b5 , t) = 2 · 22 = 44; m(b6 ) · c(b6 , t) = 3 · 12 = 36, m(b7 ) · c(b7 , t) = 3 · 12 = 36, m(b8 ) · c(b8 , t) = 3 · 13 = 39, m(b9 ) · c(b9 , t) = 3 · 14 = 42, m(b10 ) · c(b10 , t) = 3 · 14 = 42, m(b11 ) · c(b11 , t) = 3 · 14 = 42, m(b12 ) · c(b12 , t) = 3 · 16 = 48, m(b13 ) · c(b13 , t) = 3 · 18 = 54, m(b14 ) · c(b14 , t) = 3 · 20 = 60; m(b15 ) · c(b15 , t) = 4 · 9 = 36, m(b16 ) · c(b16 , t) = 4 · 10 = 40, m(b17 ) · c(b17 , t) = 4 · 10 = 40, m(b18 ) · c(b18 , t) = 4 · 10 = 40, m(b19 ) · c(b19 , t) = 4 · 12 = 48, m(b20 ) · c(b20 , t) = 4 · 12 = 48, m(b21 ) · c(b21 , t) = 4 · 14 = 56, m(b22 ) · c(b22 , t) = 4 · 14 = 56, m(b23 ) · c(b23 , t) = 4 · 14 = 56, m(b24 ) · c(b24 , t) = 4 · 18 = 72; m(b25 ) · c(b25 , t) = 5 · 9 = 45, m(b26 ) · c(b26 , t) = 5 · 9 = 45, m(b27 ) · c(b27 , t) = 5 · 10 = 50, m(b28 ) · c(b28 , t) = 5 · 10 = 50, m(b29 ) · c(b29 , t) = 5 · 14 = 70; m(b30 ) · c(b30 , t) = 6 · 9 = 54. The line balance bP1 is uniquely optimal, since m(b1 ) · c(b1 , t) = min {m(br ) · c(br , t) : br ∈ B(G)} t 26 = 32 = 26 32 = E(t) < 36 = min {m(br ) · c(br , t) : br ∈ B(G) \ {b1 }} . Due to Corollary 1, the equality B(G, t) = {b1 } implies the strict inequality ρb1 (t) > 0 for Example 3. We next compare the line balance b1 with line balances br ∈ B(G) \ {b1 } in the non-decreasing order of the products m(br ) · c(br , t). We first compare the line balance b1 with the line balance b2 in which m(b2 ) · c(b2 , t) = 36. For the line balance b2 , the equality (b) holds for the workstation S k = S 2 . Based on the formula (26), we have that t(V2b2 ) − t(V1b1 ) 18 − 16 2 2 = = = = . b2 b1 e e |{3, 4} ⊕ {1, 2, 3} |{1, 2, 4}| 3 |V ⊕ V |

ED

22 δ=b b1 1

2

1

AC

CE

PT

In order to attain the equality (23), we modify the vector e t of the durations of manual operations as follows: t1(1,2) = t1 + 32 = 1 23 , t2(1,2) = t2 + 32 = 2 32 , t4(1,2) = t4 − 23 = 5 13 . The equalities (a), (b), and (c) hold for the constructed vector t(1,2) = (e t(1,2) , t) ∈ R6+ of the operation durations. We next compare the line balance b1 with the line balance b6 for which the equality (b) holds for the workstation S k = S 2 and m(b6 ) · c(b6 , t) = 36. Since m(b1 ) < m(b6 ), to attain the equality rk (23), we calculate for the minimum distance d(e t, e t(uk) ) =: δ
m(br ) · t(Vkbr ) − m(b1 ) · t(Vub1 ) . = ebr | + m(b1 ) · [|V eub1 \ V e br | − | V eub1 ∩ V ebr |] m(br ) · |V k k k

(27)

rk rk e e(uk) ) = δ
17

ACCEPTED MANUSCRIPT

62 Using the formula (27), we have that δ
b

b

m(b6 )·t(V2 6 )−m(b1 )·t(V1 1 )

eb6 |+m(b1 )·[|V e b1 \V eb6 |−|V e b1 ∩ V eb6 |] m(b6 )·|V 2 1 2 1 2

=

4 4 4 3 · 12 − 2 · 16 = = = . 3 · |{2, 3, 4}| + 2 · [|{1, 2, 3} \ {2, 3, 4}| − |{1, 2, 3} ∩ {2, 3, 4}|] 3 · 3 + 2 · [|{1} − |{2, 3}] 9 + 2 · [1 − 2] 7

CR IP T

In order to attain the equality (23), we construct the vector e t of the durations of the manual (1,2) (1,2) (1,2) 4 4 4 3 operations as follows: t1 = t1 + 7 = 1 7 , t2 = t2 − 7 = 1 7 , t3 = t3 − 74 = 3 37 . t4(1,2) = t4 − 74 = 5 37 . We obtain the modified vector t(1,2) = (e t(1,2) , t) ∈ R6+ of the operation durations for which the equalities (a), (b), and (c) hold. For the modified vector t(1,2) = (e t(1,2) , t) of the operation durations, there are two optimal line balances: B(G, t(1,2) ) = {b1 , b6 }. Due to Theorem 2 (see case 6 in Fig. 1), the equality ρb1 (t(1,2) ) = 0 holds, since there exists the line balance b6 ∈ B(G, t(1,2) ) \ {b1 } such that the inclusion (8) does not hold, i.e. W(b1 , t(1,2) ) = {{1, 2, 3}} * {{2, 3, 4}} = W(b6 , t(1,2) ), along with conditions m(b1 ) = 2 , 3 = m(b6 ) and W(b1 , t) , {∅}. Thus, using the line balance b6 as a competitor for the line balance b1 , we obtain the following upper bound on the stability radius:

m(b7 ) · t(V1b7 ) − m(b1 ) · t(V1b1 ) 3 · 12 − 2 · 16 = b b b b b 7 7 7 1 1 e | + m(b1 ) · [|V e \V e | − |V e ∩V e |] 3 · |{1, 2}| + 2 · [|{1, 2, 3} \ {1, 2}| − |{1, 2, 3} ∩ {1, 2}|] m(b7 ) · |V 1 1 1 1 1 =

4 4 4 = = = 1. 3 · 2 + 2 · [|{3}| − |{1, 2}|] 6 + 2 · [1 − 2] 4

M

71 δ
AN US

4 62 (28) ρb1 (t) ≤ δ
15 1 δ
ED

We compare the line balance b1 with the line balance b15 for which the equality (b) holds for the workstation S k = S 1 and m(b15 ) · c(b15 , t) = 4 · 9 = 36. Using the formula (27), we have that m(b15 ) · t(V1b15 ) − m(b1 ) · t(V1b1 ) 4 · 9 − 2 · 16 4 2 = = = . b b b b b 15 15 15 1 1 e | + m(b1 ) · [|V e \V e | − |V e ∩V e |] 4 · |∅| + 2 · [|{1, 2, 3} \ ∅| − |{1, 2, 3} ∩ ∅|] 0 + 6 3 m(b15 ) · |V 1 1 1 1 1

AC

CE

PT

We note that it is unnecessary to compare the line balance b1 with any line balance br ∈ {b8 , b9 , b10 , b11 , b12 , b13 , b14 }, where m(br ) = 3, since due to the upper bound (10), the comparison of the line balance b1 with any line balance br ∈ {b8 , b9 , b10 , b11 , b12 , b13 , b14 } gives at least 39−32 7 the value of β(b1 , br ) = 4·max{m(b = 4·3 = 127 which is the lower bound on the value of 1 ),m(br )} 4 rk δ
b

b

m(b16 )·t(V3 16 )−m(b1 )·t(V1 1 )

eb16 |+m(b1 )·[|V e b1 \V eb16 |−|V eb1 ∩V eb16 |] m(b16 )·|V 3 1 3 1 3

=

4 · 10 − 2 · 16 40 − 32 8 4 = = = . 4 · |{3, 4}| + 2 · [|{1, 2, 3} \ {3, 4}| − |{1, 2, 3} ∩ {3, 4}|] 4 · 2 + 2 · [|{1, 2}| − |{3}|] 8 + 2 · [2 − 1] 5 18

ACCEPTED MANUSCRIPT

CR IP T

To attain the equality (23), we construct the vector e t of the durations of the manual operations 4 4 4 4 0 0 0 as follows: t1 = t1 + 5 = 1 5 , t2 = t2 + 5 = 2 5 , t3 = t3 − 45 = 3 15 , t40 = t4 − 45 = 5 15 . For the obtained vector t0 , the equality (b) is violated as follows: c(b16 , t0 ) = t0 (V1b16 ) = 9 > 8 52 = t30 + t40 = t0 (V3b16 ) = t0 (Vkbr ). We denote t0 (V bf r ) := t0 (V1b16 ) = c(br , t0 ). To preserve the equality (b) for the modified vector t(1,3) = t(uk) , the variation x of the duration t3(1,3) must be subject to the b16 b16 16 3 following equation: x + δ
17 1 δ
AN US

16 3 The solution of the system (29) with br = b16 , k = 3, and u = 1 is as follows: δ
m(b17 ) · t(V1b17 ) − m(b1 ) · t(V1b1 ) 4 · 10 − 2 · 16 = = b17 b17 b1 eb17 b1 e e e e 4 · |{1}| + 2 · [|{1, 2, 3} \ {1}| − |{1, 2, 3} ∩ {1}|] m(b17 ) · |V1 | + m(b1 ) · [|V1 \ V1 | − |V1 ∩ V1 |] =

40 − 32 8 1 = =1 . 4 · 1 + 2 · [|{2, 3}| − |{1}|] 4 + 2 · [2 − 1] 3



ebr : ti < δ

CE


where V k

=


m(br ) · t(Vkbr ) − m(b1 ) · t(Vub1 ) − m(br ) · t(V k )

PT

rk δ
ED

M

To attain the equality (23), we modify the vector e t of the durations of the manual operations as follows: t10 = t1 − 1 31 = − 13 , t20 = t2 + 1 13 = 3 31 , t30 = t3 + 1 31 = 5 13 . However, we obtain a e Therefore, instead of using the formula (27), we use negative duration − 31 of the operation 1 ∈ V.

AC

17 1 δ

m(b17 ) · t(V1b17 ) − m(b1 ) · t(V1b1 ) − m(b17 ) · t(V 1

eb17 \ m(b17 ) · |V 1


)

17 e b1 \ V eb17 | − |V e b1 ∩ { V eb17 \ V
,

(30)

=

4 · 10 − 2 · 16 − (4 − 2) · 1 40 − 32 − 2 6 1 = = =1 . 4 · |{1} \ {1}| + 2 · [|{1, 2, 3} \ {1}| − |{1, 2, 3} ∩ {{1} \ {1}}|] 4 · |∅| + 2 · [|{2, 3}| − |{1, 2, 3} ∩ ∅|] 4 · 0 + 2 · 2 2

We construct the following vector: t1(1,1) = t1 − 1 = 0, t2(1,1) = t2 + 1 21 = 3 12 , t3(1,1) = t3 + 1 12 = 5 12 for which the equalities (a), (b), and (c) hold. We further compare the line balance b1 with the line balance b18 in which the equality (b) holds for the workstation S k = S 1 and m(b18 ) · c(b18 , t) = 40. Using the formula (27), we have that 18 1 δ
m(b18 ) · t(V1b18 ) − m(b1 ) · t(V1b1 ) 4 · 10 − 2 · 18 = b b b b b 18 18 18 1 1 e | + m(b1 ) · [|V e \V e | − |V e ∩V e |] 4 · |{1}| + 2 · [|{1, 2, 3} \ {1}| − |{1, 2, 3} ∩ {1}|] m(b18 ) · |V 1 1 1 1 1 19

ACCEPTED MANUSCRIPT

=

40 − 32 8 1 = =1 . 4 · 1 + 2 · [|{2, 3}| − |{1}|] 4 + 2 · [2 − 1] 3

To attain the equality (23), we modify the vector e t of the durations of the manual operations as 1 1 1 1 0 0 0 follows: t1 = t1 − 1 3 = − 3 , t2 = t2 + 1 3 = 3 3 , t3 = t3 + 1 31 = 5 13 . However, we not only obtain a negative duration − 13 of the operation 1 but also violate the equality (b). So instead of using the

=

m(b18 ) · t(V1b18 ) − m(b1 ) · t(V1b1 ) − [m(b18 ) − m(b1 )] · t(V 1 )

CR IP T

18 1 δ
b18

b18

eb18 \ V 1 | + m(b1 ) · [|V e b1 \ V eb18 | − |V e b1 ∩ { V eb18 \ V 1 }|] m(b18 ) · |V 1 1 1 1 1

=

40 − 32 − 2 6 1 4 · 10 − 2 · 16 − (4 − 2) · 1 = = =1 . 4 · |{1} \ {1}| + 2 · [|{1, 2, 3} \ {1}| − |{1, 2, 3} ∩ {{1} \ {1}}|] 4 · |∅| + 2 · [|{2, 3}| − |{1, 2, 3} ∩ ∅|] 4 · 0 + 2 · 2 2

m(b18 ) · t(V3b18 ) − m(b1 ) · t(V1b1 ) 4 · 10 − 2 · 16 = b b b b b 18 18 18 1 1 e | + m(b1 ) · [|V e \V e | − |V e ∩V e |] 4 · |{3, 4}| + 2 · [|{1, 2, 3} \ {3, 4}| − |{1, 2, 3} ∩ {3, 4}|] m(b18 ) · |V 3 1 3 1 3

M

18 3 δ
AN US

We set the following durations of the operations 1, 2, and 3: t1(1,1) = t1 − 1 = 0, t2(1,1) = t2 + 1 21 = 3 12 , t3(1,1) = t3 + 1 12 = 5 12 . To attain the equality (b), we need to decrease the duration t4 of the eb18 \ V eb1 by the value of t(1,1) (V b18 ) − t(1,1) (V b18 ) = 11 1 − 9 = 2 1 . Hence, we have operation 4 ∈ V 3 3 1 1 2 2 that t4(1,1) = t4 − 2 21 = 3 21 . The equalities (a), (b), and (c) hold for the duration vector t(1,1) . The calculated value of δb
40 − 32 8 4 = = . 4 · 2 + 2 · [|{1, 2}| − |{3}|] 8 + 2 · [2 − 1] 5

ED

=

AC

CE

PT

To attain the equality (23), we modify the vector e t of the durations of the manual operations as follows: t10 = t1 + 54 = 1 45 , t20 = t2 + 45 = 2 45 , t30 = t3 − 45 = 3 51 . t40 = t4 − 45 = 5 15 . However, we violate the equality (b) as follows: t0 (V1b18 ) = 9 + 1 54 = 10 45 > 8 25 = 3 51 + 5 15 = t0 (V3b18 ). Therefore, an admissible increase x of the duration t1 must be subject to the following equalities:
m(br ) · t(Vkbr ) − m(b1 ) · t(Vub1 ) , = ebr ∩ V eub1 |] eub1 \ V| + m(br ) · [|V ebr \ V eub1 | − |V m(b1 ) · |V k k

(31)

e : t(uk) = ti }. Using the formula (31), we have that where V = {i ∈ V i 18 3 δ
=

m(b18 ) · t(V3b18 ) − m(b1 ) · t(V1b1 ) eb1 \ V| + m(b18 ) · [|V eb18 \ V e b1 | − | V eb18 ∩ V eb1 |] m(b1 ) · |V 1 3 1 3 1

40 − 32 8 4 · 10 − 2 · 16 = = = 2. 2 · |{1, 2, 3} \ {1}| + 4 · [|{3, 4} \ {1, 2, 3}| − |{3, 4} ∩ {{1, 2, 3}|] 2 · |{2, 3}| + 4 · [|{4}| − |{3}|] 2 · 2 + 4 · 0 20

ACCEPTED MANUSCRIPT

PT

ED

M

AN US

CR IP T

Hence, we construct the following duration vector: t1(1,3) = t1 = 1, t2(1,3) = t2 + 2 = 4, t3(1,3) = 18 t3 + 2 = 6, t4(1,3) = t4 − 2 = 4 for which the equalities (a), (b), and (c) hold. The value of δb b1 1 = 7 = min{ 3 , 7 , 1, 3 , 1, 1 2 , 2} of the obtained upper bounds on the stability radius ρb1 (t) is attained for the competitor b6 for the line balance b1 on the workstation S 1 in the line balance b1 and the workstation S 2 in the line balance b6 . We showed that for the modified vector t(1,2) = (e t(1,2) , t) ∈ R6+ of the operation durations, the 4 61 equalities (a), (b), and (c) hold. In order to show that equalities ρb1 (t) = d(t, t(1,2) ) = δ 30 67 = 3 · (1 73 + 3 37 + 5 37 ) = 3 · 10 27 = 3 · t (V2b6 ) = m(b6 ) · t (V2b6 ). Hence b1 < B(G, t ) = {b6 } and due to Theorem 5, the equalities ρb1 (t) = d(t, t(1,2) ) = 74 hold.

CE

7. An algorithm and formulae for calculating the stability radius

AC

As demonstrated in Subsection 6.1 on Example 2 and in Subsection 6.2 on Example 3, to find the stability radius ρb1 (t) of the optimal line balance b1 ∈ B(G, t), one can perform a sequence of comparisons of the workstation time t(Vub1 ) in the line balance b1 ∈ B(G, t) with the workstation time t(Vkbr ) in line balances br ∈ B(G) \ {b1 }. Based on such comparisons, one can calculate for the =br k >br k (uk) rk minimum distances δ
21

ACCEPTED MANUSCRIPT

7.1. Calculating the finite stability radius ρb1 (t)

In order to derive a theorem and an algorithm for obtaining the stability radius ρb1 (t) < ∞, we =br r introduce notations δ
(32)

k=1

CR IP T

rk where values of δ
o m(b1 ) m(br ) n =br k r eub1 ∪ V ebr , ∅, m(br ) = m(b1 ) , δ=b := min max δ : V b1 b1 u k u=1

k=1

=br

where values of δb=b1 ur k are calculated by using the formula (26). If the set V k

(33)

ebr \ V eub1 : = {i ∈ V k =br

AN US

(uk) rk ti < δ=b are negative for every operation i ∈ V k , b1 u } is not empty and so the obtained values ti =br then the value of δb1 must be recalculated by using the following analog of the formula (30): br

rk δ=b b1 u

=

eub1 ) t(Vkbr ) − t(Vub1 ) − t(V k \ V br

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ebr ⊕ V eub1 | eub1 | − |V k \ V |V k

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o m(b1 ) m(br ) n >br k r eub1 , V ebr , m(br ) > m(b1 ) , max := min δ : V δ>b b1 u k b1 u=1

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.

(35) >br

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rk where values of δ>b b1 u are calculated by using the formula (24). If the set V k

ebr \ V eub1 : = {i ∈ V k >br

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(uk) rk ti < δ>b are negative for every operation i ∈ V k , b1 u } is not empty and so the obtained values ti >br then the value of δb1 must be recalculated by using the following analog of the formula (30):

=

>br

eub1 ) m(br ) · t(Vkbr ) − m(b1 ) · t(Vub1 ) − m(br ) · t(V k \ V

>br r ebr \ {V eub1 ∪ V >b eb1 eb1 m(br ) · [|V k }| − |{Vu ∩ V k }|] + m(b1 ) · |Vu | k

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rk δ>b b1 u

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>br k rk If the equality (b) is violated for the calculated values δ=b b1 u or δb1 u , then these values must be recalculated by using the corresponding analogs of the formula (31) or system (29) given in Subsection 6.2. There are different analogs of the formula (31) and system (29) since there are several types for the durations of the workstations to cause the violations of the equality (b). =br k >br k rk Nevertheless, we assert that calculating or recalculating for any value δ (G), where B< (G) = {br ∈ B(G) \ {b1 } : m(br ) < m(b1 )}, B= (G) = {br ∈ B(G) \ {b1 } : m(br ) = m(b1 )}, and B> (G) = {br ∈ B(G) \ {b1 } : m(br ) > m(b1 )}. Following the above discussion and using the notations (32), (33), and (35), we present the following formula (37) for obtaining the stability radius.

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Theorem 6. If b1 ∈ B(G, t) and ρb1 (t) < ∞, then ( ) =b s >be
br ∈B (G)

b s ∈B (G)

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Once the set B(G) of all admissible line balances has been constructed, one can find an optimal line balance by choosing the line balance b1 from the set B(G) with the maximum efficiency E(b1 , t). By using Theorem 4, one can test whether the optimal line balance b1 ∈ B(G, t) has an infinite stability radius. By using Theorem 2, one can test whether the line balance b1 has a zero stability radius. We next present our algorithm for obtaining the stability radius ρb1 (t). The first step of this algorithm may adopt the algorithm described in (Sotskov et al., 2015) for constructing the set of all admissible line balances for the problem SALBP-2.

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Algorithm STABRAD-E e of the manual operations, set V \ V e of the automated operations, precedence diInput: Set V graph G = (V, A), and vectors t = (e t, t) = (t1 , t2 , . . . , tn˜ , tn˜ +1 , tn˜ +2 , . . . , tn ) of the operation durations. Output: Stability radius ρb1 (t) of the optimal line balance b1 ∈ B(G, t).

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Step 1: Construct the set B(G) of all admissible line balances. Step 2: Calculate for the product m(br ) · c(br , t) for each line balance br ∈ B(G). Step 3: Index all line balances in the set B(G) in the non-decreasing order of the products m(br ) · c(br , t), i.e. B(G) = {b1 , b2 , . . . , bh }, where the inequality m(br ) · c(br , t) ≤ m(br+1 ) · c(br+1 , t) holds for each index r ∈ {1, 2, . . . , h − 1}. Step 4: Verify the conditions of Theorem 4 for the optimal line balance b1 ∈ B(G, t) in order to test whether the stability radius ρb1 (t) is infinite IF ρb1 (t) = ∞ THEN STOP Step 5: ELSE ρb1 (t) < ∞. Verify the conditions of Theorem 2 for the line balance b1 in order to test whether it is unstable IF ρb1 (t) = 0 THEN STOP Step 6: ELSE ρb1 (t) > 0. Set B := B(G) \ {b1 } = {b2 , b3 , . . . , bh } and set δb1 := min{β(b1 ), γ(b1 )}, where γ(b1 )} is defined in (13) and β(b1 ) in (10) and (12). Step 7: IF B = ∅ THEN GOTO step 21 ELSE Step 8: Choose a line balance br from the set B with the minimum index r. Step 9: Calculate for the value of β(b1 , br ) defined in (10) IF β(b1 , br ) ≥ δb1 THEN set B := B \ {b s ∈ B(G) : m(b s ) = m(br ), s ≥ r} GOTO step 10 ELSE Step 10: IF m(b1 ) < m(br ) THEN GOTO step 16 ELSE Step 11: IF m(b1 ) = m(br ) THEN GOTO step 14 ELSE m(b1 ) > m(br ). >br >br k r Step 12: Calculate for δ>b b1 determined in (35), where δb1 = δb1 u . br k >br k r Step 13: Set B := B \ {br }, δbbr1 := δ>b b1 , and δb1 u := δb1 u GOTO step 18 =br =br k r Step 14: Calculate for δ=b b1 determined in (33), where δb1 = δb1 u . br k =br k r Step 15: Set B := B \ {br }, δbbr1 := δ=b b1 , and δb1 u := δb1 u GOTO step 18
δbbr1 THEN verify the conditions of Theorem 2 for the optimal line balance b1 ∈ B(G, t(uv) ) 3 br in order to test whether the line balance b1 is stable for the modified vector t(uk) ∈ Rn+ of the operation durations. 23

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Step 20: IF ρb1 (t(uk) ) > 0 THEN GOTO step 7 ELSE set δb1 := δbbr1 GOTO step 7 Step 21: Set ρb1 (t) = δb1 STOP We next estimate the running time of the above algorithm. 7.2. A bound on the running time of Algorithm STABRAD-E

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Let set B(G, m) = {bl1 , bl2 , . . . , blh } be a subset of the set B(G) of all admissible line balances, where m workstations are used. If the set of operations V is unordered, i.e. the precedence digraph is empty, G = (V, ∅), then the cardinality of the set B(G, m) is maximal for the fixed numbers n = |V| and m. The cardinality of the set B(G, m) is given in Lemma 4 proven in (Sotskov et al., 2015).

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Lemma 4. If the precedence digraph G is empty, G = (V, ∅), and the inequality m ≤ n holds, then ! n |B(G, m)| = · m! · mn−m . (38) m

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Due to Conditions 1 and 3, the inequalities 2 ≤ m ≤ n hold for the problem SALBP-E under ! Pn n · m! · mn−m must hold consideration. Therefore due to Lemma 4, the equality |B(G)| = m=2 m if the precedence digraph G is empty and m ≤ n. Since the line balance b1 ∈ B(G, t) has to be compared with all possible competitors br ∈ B(G) \ {b1 }, Algorithm STABRAD-E makes at most ! n comparisons of the line balance b1 with line balances from the set B(G) \ {b1 }. Each such 2 a comparison needs O(n) time (see Subsection 7.1, where n is defined). Thus, the complexity of " ! #! P n · m! · mn−m provided Algorithm STABRAD-E is estimated by the value of O n · n2 · nm=2 m that the given digraph G is empty, G = (V, ∅), and m ≤ n. The obtained asymptotic bound remains correct for the complexity of Algorithm STABRADE for any given precedence digraph G = (V, A) since |B(G, m)| ≤ |B((V, ∅), m)|. 7.3. How large may the problem SALBP-E be for the stability analysis

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The complete stability analysis of the optimal line balance based on Algorithm STABRADE is time-consuming for the problem SALBP-E with a large size since Algorithm STABRAD-E includes the constructing of line balances B(G). The cardinality |B(G)| of the set B(G) depends on the following Order Strength (OS ) of the digraph G = (V, A): OS = |A|/|AC |, where |AC | = [n · (n − 1)]/2 is the maximal number of arcs in the circuit-free digraph G = (V, A) of the order n. The computational results for the benchmark problems SALBP-1 and SALBP-2 presented in (Lai et al., 2016; Sotskov et al., 2015) showed that the CPU-time of the laptop needed for constructing the set B(G) and the complete stability analysis of the problems SALBP-1 and SALBP-2 is not large for small assembly lines with up to 50 operations and for the moderate assembly lines with up to 150 operations and rather large order strengths OS of the digraphs G. Note that the complete stability analysis described in (Lai et al., 2016; Sotskov et al., 2015) included constructing the set B(G) and calculating the stability radii for all optimal line balances B(G, t). For constructing the set B(G) and the complete stability analysis for a larger assembly line, a faster computer is required. If it is impossible to construct the set B(G), one can construct its subset 24

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and then save it in the bank of the line balances. In order to increase a size of the problem SALBPE, for which the stability analysis may be applied, one can decompose the problem SALBP-E into several subproblems SALBP-E due to the given precedence constraints. If the running time of Algorithm STABRAD-E exceeds one’s budget of time, one can still obtain the lower and upper bounds on the value of ρb1 (t) by terminating the running of Algorithm STABRAD-E. It is worthwhile to mention that to reduce computational burden, one does not need to distinguish the line balances having different orders of the subsets Vhbr in the partition (2): V = S S S br br V1br V2br . . . Vm(b provided that the sets of subsets {V1br , V2br , . . . , Vm(b } are the same for these r) r) line balances. Note also, that in practice not all line balances are suitable for a realization in the assembly line since not only the precedence constraints determined by the digraph G = (V, A) have to be taken into account. Due to this, the cardinality of the set of admissible line balances may be considerably smaller than the cardinality |B(G)| of the set B(G). Since the speed of the computer will be more and more fast, it will be possible to construct several optimal line balances or even the whole set B(G, t) for a practical problem SALBP-E with moderate and even large sizes. Using a powerful computer or supercomputer, one can construct a set of the potentially optimal line balances, i.e. optimal line balances or k-best ones for the initial vector t of the operation durations. If the durations of the operations will change considerably, one can select an appropriate line balance from the bank of the potentially optimal line balances. The problem SALBP-E arises when an assembly line is installed or when a change in the cycle time of an assembly line is justified due to a change in the demand of the assembled product. For the whole lifespan of the assembly line, our stability analysis needs only to be conducted once at the planning phase of designing a new line balance when one has much more time for computing. The results of our stability analysis conducted at the planning stage can then be used at the on-line scheduling phase of operating the assembly line when one has less time for computing.

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8. Conclusion

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The stability analysis of the optimal line balance developed in this paper is useful for reducing the number of re-balancing times. As long as the deviations of the manual operation durations are within the stability radius, it is unnecessary to re-balance the assembly line. Algorithm STABRAD-E for obtaining the stability radius is given in Subsection 7 based on Theorems 2, 4, 5, and 6. Due to Theorem 2, one can test whether there exists a stable optimal line balance or not. Due to Theorem 4, one can test whether there exists an optimal line balance with an infinite stability radius. The lower and upper bounds on the stability radius established in Sections 4 and 5 may be calculated for middle-size problems SALBP-E and for a large-size problem SALBP-E provided that the order strength of the given digraph G = (V, A) is large. For a large-size problem SALBP-E, Algorithm STABRAD-E seems computationally impractical since in computing for the stability radius ρbr (t), br ∈ B(G, t), in the worst case one needs to compare the line balance br with the most line balances in the set B(G) \ {br }. Hence, one further research avenue may focus on algorithmic improvements. Another future research avenue is to adopt the stability analysis developed in this paper to deal with more general and practical assembly line balancing problems.

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Acknowledgment

The authors are grateful to the anonymous reviewers for their useful remarks and suggestions on the early draft of this paper. References

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Baybars, I. (1986). A survey of exact algorithms for the simple assembly line balancing problem. Management Science, 32(8):909–932. Br¨asel, H., Sotskov, Y., and Werner, F. (1996). Stability of a schedule minimizing mean flow time. Mathematical and Computer Modelling, 24(10):39–53. Chakravarti, N. and Wagelmans, A. (1998). Calculation of stability radii for combinatorial optimization problems. Operations Research Letters, 23(1):1–7. Chen, R.-S., Lu, K.-Y., and Tai, P.-H. (2004). Optimizing assembly planning through a three-stage integrated approach. International Journal of Production Economics, 88:243–256. Chica, M., Gordon, O., Damas, S., and Bautista, J. (2013). A robustness information and visualization model for time and space assembly line balancing under uncertain demand. International Journal of Production Economics, 145:761–772. Corominas, A., Pastor, R., and Plans, J. (2008). Balancing assembly line with skilled and unskilled workers. Omega, 36:1126 1132. Dong, J., Zhang, L., Xiao, T., and Mao, H. (2014). Balancing and sequencing of stochastic mixed-model assembly u-lines to minimise the expectation of work overload time. International Journal of Production Research, 52(24):7529–7548. Erel, E. and Sarin, S. (1998). A survey of the assembly line balancing procedures. Production Planning & Control, 9(5):414–434. Gamberini, R., Grassi, A., and Rimini, B. (2006). A new multi-objective heuristic algorithm for solving the stochastic assembly line re-balancing problem. International Journal of Production Economics, 102:226–243. Gurevsky, E., Battaia, O., and Dolgui, A. (2012). Balancing of simple assembly lines under variations of task processing times. Annals of Operations Research, 201:265–286. Gurevsky, E., Battaia, O., and Dolgui, A. (2013). Stability measure for a generalized assembly line balancing problem. Discrete Applied Mathematics, 161:377–394. Gutjahr, A. and Nemhauser, G. (1964). An algorithm for the line balancing problem. Management Science, 11(2):308–315. Hamta, N., Fatemi Ghomi, S., Jolai, F., and Akbarpour Shirazi, M. (2013). A hybrid pso algorithm for a multi-objective assembly line balancing problem with flexible operation times, sequence-dependent setup times and learning effect. International Journal of Production Economics, 141:99–111. Kahan, T., Bukchin, Y., Menassa, R., and Ben-Gal, I. (2009). Beckup strategy for robots’ failures in an automotive assembly system. International Journal of Production Economics, 120:315–326. Kravchenko, S., Sotskov, Y., and Werner, F. (1995). Optimal schedules with infinitely large stability radius. Optimization, 33:271–280. Lai, T.-C., Sotskov, Y., Dolgui, A., and Zatsiupa, A. (2016). Stability radii of optimal assembly line balances with a fixed workstation set. International Journal of Production Economics, 182:356–371. Lai, T.-C., Sotskov, Y., Sotskova, N., and Werner, F. (2004). Mean flow time minimization with given bounds of processing times. European Journal of Operational Research, 159:558–573. Morrison, D., Sewell, E., and Jacobson, S. (2014). An application of the branch, bound, and remember algorithm to a new simple assembly line balancing dataset. European Journal of Operational Research, 236(2):403–409. Otto, A., Otto, C., and Scholl, A. (2013). Systematic data generation and test design for solution algorithms on the example of salbpgen for assembly line balancing. European Journal of Operational Research, 228(1):33–45. Ramaswamy, R. and Chakravarti, N. (1995). Complexity of determining exact tolerances for min-sum and min-max combinatorial optimization problems. Technical Report WPS-247/95, Indian Institute of Management, Calcutta. Scholl, A. (1999). Balancing and Sequencing of Assembly Lines. Physica-Verlag, A Springer-Verlag Company, Heidelberg. Sotskov, Y. (1991). Stability of an optimal schedule. European Journal of Operational Research, 55:91–102. Sotskov, Y. and Dolgui, A. (2001). Stability radius of the optimal assembly line balance with fixed cycle time. In Proceedings of the IEEE Conference ETFA2001, pages 623–628, Troyes, France. Sotskov, Y., Dolgui, A., Lai, T.-C., and Zatsiupa, A. (2015). Enumerations and stability analysis of feasible and optimal line balances for simple assembly lines. Computers & Industrial Engineering, 90:241–258. Sotskov, Y., Dolgui, A., and Portmann, M.-C. (2006). Stability analysis of optimal balance for assembly line with fixed cycle time. European Journal of Operational Research, 168(3):783–797. Sotskov, Y., Dolgui, A., Sotskova, N., and Werner, F. (2005). Stability of optimal line balance with given station set. In Fiat, A. and Woeginger, G., editors, Supply Chain Optimization.Applied Optimization, volume 94, pages 135–149. Springer, USA, New York. Wee, T. and Magazine, M. (1982). Assembly line balancing as generalized bin packing. Operations Research Letters, 1:56–58.

Appendix A The necessity proof of Theorem 2 To prove the necessity of Theorem 2, we need to show that the equality ρb1 (t) = 0 holds in cases 4 – 8 (see Fig. 1). Case 4: There exists an optimal line balance br ∈ B(G, t) \ {b1 } , ∅ such that the condition (8) does not hold, while the equality m(b1 ) = m(br ) holds. 26

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If the number of the used workstations is predetermined, i.e. m(bu ) = m∗ for any line balance bu ∈ B(G), then the problem SALBP-E reduces to the problem SALBP-2. By taking into account that the equality m(b1 ) = m(br ) holds for case 4 in the problem SALBP-E, while the inclusion W(b1 , t) ⊆ W(br , t) does not hold for each optimal line balance br , r , 1, we conclude that the sufficiency of Theorem 1 established for the problem SALBP-2 in (Sotskov et al., 2005) implies the equality ρb1 (t) = 0 for case 4 in the problem SALBP-E. Case 5: There exists an optimal line balance br ∈ B(G, t) \ {b1 } , ∅ such that m(b1 ) , m(br ), W(b1 , t) = {∅}, and the condition (8) does not hold for the line balance br . Since W(b1 , t) = {∅} and the condition (8) does not hold for the line balance br , we obtain ∅ < W(br , t). Due to Lemma 1, the inclusion br ∈ B(G, t) implies W(br , t) , ∅. Therefore, at least one manual operation is assigned to each workstation S k in the line balance br such that the equality t(Vkbr ) = c(br , t) holds. For any arbitrarily small real e Since at least one manual operation is assigned to number  > 0, we set ti0 = ti −  for each manual operation i ∈ V. br each workstation S k with the equality t(Vk ) = c(br , t), we obtain the inequality c(br , t0 ) < c(br , t), where t0 = (e t0 , t) 0 0 0 0 0 and e t = (t1 , t2 , . . . , ten ). The equality W(b1 , t) = {∅} implies the equality c(b1 , t ) = c(b1 , t). The inclusions bi ∈ B(G, t) and br ∈ B(G, t) imply the equality (14). Using conditions c(br , t0 ) < c(br , t), c(b1 , t0 ) = c(b1 , t), and the equality (14), we obtain the following sequence of comparisons: m(b1 ) · c(b1 , t0 ) = m(b1 ) · c(b1 , t) = m(br ) · c(br , t) > m(br ) · c(br , t0 ). t, e t0 ) =  > 0 as small as desired such that the inequality Thus, there exists a vector t0 ∈ Ren+ with the distance d(e 0 0 0 m(b1 ) · c(b1 , t ) > m(br ) · c(br , t ) holds. Hence, b1 < B(G, t ) and the equality ρb1 (t) = 0 holds for case 5. Case 6: There exists an optimal line balance br ∈ B(G, t) \ {b1 } , ∅ such that m(b1 ) , m(br ), W(b1 , t) , {∅}, and the condition (8) does not hold for the line balance br . eb1 ∈ W(b1 , t), which does not belong to the set W(br , t), Since the condition (8) does not hold, there exists a set V k b 1 e < W(br , t). Since {b1 , br } ⊆ B(G, t), due to Definition 2, we obtain the following equalities: V k m(b1 ) · t(Vkb1 ) = m(b1 ) · c(b1 , t) =

tP = m(br ) · c(br , t) = m(br ) · t(Vebr ). E(t)

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We need to consider the following three sub-cases. eebr ∈ W(br , t) which is a proper subset of the set V e b1 , V eebr ⊂ V eb1 , and the inequality Sub-case 6-i: There exists a set V k k m(b1 ) > m(br ) holds. eebr ⊂ V eb1 and the condition W(b1 , t) , {∅} imply V e b1 \ V eebr , ∅, i.e. there exists a manual operation The inclusion V k k b b ε ε ε ε ε 1 r e \V ee . We construct the vector e e \ {i}. i∈V t = (t1 , t2 , . . . , ten ), where ti = ti + ε and tεj = t j for each operation j ∈ V k ε ε e Let ε denote any arbitrarily small real number, ε > 0. For the vector t = (t , t), we obtain m(b1 ) · c(b1 , tε ) = m(b1 ) · c(b1 , t) + m(b1 ) · ε > m(br ) · c(br , t) + m(br ) · ε = m(br ) · c(br , tε ), where the inequality holds due to the equality (14) and the inequality m(b1 ) > m(br ). We conclude that there exists a vector e tε ∈ Rn+˜ such that d(e t, e tε ) = ε and the line balance b1 is not optimal for the ε ε vector t of the operation durations, b1 < B(G, t ). Since the vector e tε may be chosen as close to the vector e t as desired, we obtain the equality ρb1 (t) = 0 for sub-case 6-i. eebr ∈ W(br , t) which is a proper subset of the set V e b1 , V eebr ⊂ V eb1 , and the Sub-case 6-ii: There exists a set V k k inequality m(b1 ) < m(br ) holds. eebr ⊂ V eb1 and the condition W(b1 , t) , {∅} imply that both sets V e b1 ∩ V eebr and V e b1 \ V eebr are non-empty. The inclusion V k k k e b1 ∩ V eebr and j ∈ V e b1 \ V eebr . For any arbitrarily small real number Therefore, there exist two manual operations i ∈ V k k (ε) (ε) (ε) (ε) (ε) ε ε > 0, we construct the vector e t(ε) = (t1 , t2 , . . . , ten ), where ti = ti + m(bε r ) , t(ε) j = t j − m(br ) , and tl = tl for each e \ {i, j}. For the vector t(ε) = (e operation l ∈ V t(ε) , t), we obtain m(b1 ) · c(b1 , t(ε) ) = m(b1 ) · c(b1 , t) + m(b1 ) · m(bε ) > r

m(br ) · c(br , t) + m(br ) · m(bε r ) − m(br ) · m(bε r ) = m(br ) · c(br , t(ε) ), where the inequality holds due to the equality (14) and the inequality m(b1 ) · m(bε r ) > 0. Hence, there exists a vector e t(ε) ∈ Rn+˜ such that the equality d(e t, e tε ) = ε holds and the line balance b1 is not optimal for the vector t(ε) of the operation durations, b1 < B(G, t(ε) ). Since the vector e t(ε) may be e chosen as close to the vector t as desired, we obtain the equality ρb1 (t) = 0 for sub-case 6-ii. eebr ∈ W(br , t) which is a proper subset of the set V e b1 . Sub-case 6-iii: There is no set V k b b b 1 1 e =∅⊆V ee r ∈ W(br , t)). For any arbitrarily e , ∅ (otherwise, V Due to the condition W(b1 , t) , ∅, we obtain V k k 27

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small positive number  ≤ δ(br ), we construct the vector e t[ε] = (t1[ε] , t2[ε] , . . . , tn[ε] ), where   eb1 ,  ti + ε, if i ∈ V [ε] k ti =   e\V e b1 . ti , if i ∈ V k

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For the vector t[ε] = (e t[ε] , t), we obtain the following sequence of comparisons: m(b1 )·c(b1 , t[ε] ) = m(b1 )·ti[ε] (Vkb1 ) = b 1 e |] > m(br ) · c(br , t[ε] ), where the inequality follows from the fact that the set V eb1 is not a subset m(b1 ) · [c(b1 , t) + ε|V k k [ε] b b ee r ∈ W(br , t) and the equality t = ti holds for each manual operation i ∈ V e\V e 1. of any set V i k Using similar arguments for the vector t[ε] as those used for the vector tε in sub-case 6-i, we conclude that b1 < B(G, t[ε] ) and ρb1 (t) = 0 in sub-case 6-iii. Thus, it is proven that ρb1 (t) = 0 for case 6. Case 7: For any optimal line balance br ∈ B(G, t) \ {b1 } , ∅, the following conditions hold: W(b1 , t) ⊆ W(br , t), m(b1 ) , m(br ), ∅ < W(br , t). Let ε be any arbitrarily small real number such that 0 < ε ≤ min{γ(b1 ), γ(br )}. Due to the condition m(b1 ) , m(br ), we need to consider the following sub-cases: Either m(b1 ) < m(br ) or m(b1 ) > m(br ). Sub-case 7-i: m(b1 ) < m(br ). We construct the vector e tε = (t1ε , t2ε , . . . , tenε ) by setting tiε = ti − ε for each operation ε e 1 , br ) and t = t j for each operation j ∈ V e \ V(b e 1 , br ), where the set V(b e 1 , br ) includes at least one operation i ∈ V(b j br br e ebr cannot from each set Vk such that the equality t(Vk ) = c(br , t) holds. Due to the condition ∅ < W(br , t), each set V k be empty. Therefore, the following equality holds: c(br , tε ) = c(br , t) − ε,

(40)

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Multiplying the equality (41) by m(b1 ), we obtain the first equality in the following sequence of comparisons: m(b1 ) · c(b1 , tε ) = m(b1 ) · c(b1 , t) − ε · m(b1 ) = m(br ) · c(br , t) − ε · m(b1 ) > m(br ) · c(br , t) − ε · m(br ) = m(br ) · c(br , tε ), where the second equality holds due to the equality (14), the inequality holds due to the inequality m(b1 ) < m(br ), and the last equality holds due to the equality (40). Thus, there exists a vector tε ∈ Ren+ with the distance d(e t, e tε ) =  > 0 as ε ε ε small as desired such that m(b1 ) · c(b1 , t ) > m(br ) · c(br , t ), i.e. b1 < B(G, t ). We obtain the equality ρb1 (t) = 0. Sub-case 7-ii: m(b1 ) > m(br ). We construct the vector e t[ε] = (t1[ε] , t2[ε] , . . . , ten[ε] ) by setting ti[ε] = ti + ε for operation b b 1 1 e such that the equality t(V ) = c(br , t) holds, and setting t[ε] = t j for any other manual operation j ∈ V e \ {i}. i∈V j k k e b1 The conditions W(b1 , t) ⊆ W(br , t) and ∅ < W(br , t) imply the condition ∅ < W(b1 , t). Therefore, the chosen set V k cannot be empty and the following equality holds: c(b1 , t[ε] ) = c(b1 , t) + ε,

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(43)

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Multiplying the equality (42) by m(b1 ), we obtain the following sequence of comparisons: m(b1 ) · c(b1 , t[ε] ) = m(b1 ) · c(b1 , t) + ε · m(b1 ) = m(br ) · c(br , t) + ε · m(b1 ) > m(br ) · c(br , t) + ε · m(br ) = m(br ) · c(br , t[ε] ), where the second equality holds due to the equality (14), the inequality holds due to the inequality m(b1 ) > m(br ), and the last equality holds due to the equality (43). Thus, there exists a vector t[ε] ∈ Ren+ with the distance d(e t, e t[ε] ) =  > 0 as small as [ε] [ε] [ε] desired such that m(b1 ) · c(b1 , t ) > m(br ) · c(br , t ). Hence, b1 < B(G, t ) and ρb1 (t) = 0 for case 7. Case 8: For any optimal line balance br ∈ B(G, t) \ {b1 } , ∅, the following conditions hold: W(b1 , t) ⊆ W(br , t), ∅ ∈ W(br , t), W(b1 , t) , {∅}, m(b1 ) > m(br ). Let ε denote any arbitrarily small positive real number. We construct the vector e t[ε] = (t1[ε] , t2[ε] , . . . , ten[ε] ) by setting [ε] b b e 1 such that the equality t(V 1 ) = c(b1 , t) holds. We set t[ε] = t j for any other manual ti = ti + ε for an operation i ∈ V j k k e b1 e \ {i}. Due to the condition W(b1 , t) , {∅}, the chosen set V eb1 is not empty. Since the inclusion i ∈ V operation j ∈ V k

and the equality t(Vkb1 ) = c(b1 , t) hold, we obtain

c(b1 , t[ε] ) = c(b1 , t) + ε, 28

k

(44)

ACCEPTED MANUSCRIPT

eb1 ∈ W(b1 , t) ⊆ W(br , t) imply where t[ε] = (e t[ε] , t). The inclusions i ∈ V k c(br , t[ε] ) = c(br , t) + ε.

(45)

CR IP T

Multiplying the equality (44) by m(b1 ), we obtain the first equality in the following sequence of comparisons: m(b1 ) · c(b1 , t[ε] ) = m(b1 ) · c(b1 , t) + ε · m(b1 ) = m(br ) · c(br , t) + ε · m(b1 ) > m(br ) · c(br , t) + ε · m(br ) = m(br ) · c(br , t[ε] ), where the second equality holds due to the equality (14), the inequality holds due to the inequality m(b1 ) > m(br ), and t, e t[ε] ) =  > 0 the last equality holds due to the equality (45). Thus, there exists a vector t[ε] ∈ Ren+ with the distance d(e [ε] [ε] as small as desired such that the inequality m(b1 ) · c(b1 , t ) > m(br ) · c(br , t ) holds. Hence, b1 < B(G, t[ε] ) and the equality ρb1 (t) = 0 holds for case 8. We conclude that for cases 1 – 3, the inequality ρb1 (t) > 0 holds, while for cases 4 – 8, the equality ρb1 (t) = 0 holds. Since we have investigated each of cases 1 – 8 with |B(G, t)| ≥ 2 (see Fig. 1) and have shown that ρb1 (t) > 0 for case 0 with |B(G, t)| = 1, this completes the proof of Theorem 2. Appendix B

PT

ED

M

AN US

The necessity proof of Theorem 4 Assume that neither equality e n = 0 holds nor both inequality e n ≥ 1 and equality n = 2 hold. Since n ≥ 2 for the problem SALBP-E, the above assumption is equivalent to the following inequalities: e n ≥ 1 and n ≥ 3. We need to show that there is no optimal line balance b1 ∈ B(G, t) with an infinite stability radius, i.e. ρb1 (t) < ∞. e We next consider two possible cases. Due to the inequality e n ≥ 1, there exists a manual operation i ∈ V. e b1 ∈ B(G, t), and u = k(i, b1 ). Case A: Let the equality {i} = Vub1 hold, where i ∈ V, Since n ≥ 3 and only operation i is assigned to the workstation S k(i,b1 ) , we can reassign operation i either to the workstation S u−1 (provided that u ≥ 2) or to the workstation S u+1 (provided that u ≤ m(b1 ) − 1) without violating Condition 1 for any given precedence digraph G = (V, A). After such a reassignment of operation i, the line balance b1 reduces to the line balance br , which does not use the workstation S u . Hence, the equality m(br ) = m(b1 ) − 1 holds. e \{i}. These settings imply the equality c(b1 , t∗ ) = c(br , t∗ ) We set ti∗ = 0 and t∗j = t j for each manual operation j ∈ V ∗ ∗ along with the equalities m(br ) · c(br , t ) = m(br ) · c(b1 , t ) = (m(b1 ) − 1) · c(b1 , t∗ ) = m(b1 ) · c(b1 , t∗ ) − c(b1 , t∗ ), which imply the inequality m(br ) · c(br , t∗ ) < m(b1 ) · c(b1 , t∗ ) since the inequality c(b1 , t∗ ) > 0 holds due to the assumption t j > 0 for each operation j ∈ V (Condition 4). Hence, we obtain b1 (t∗ ) < B(G, t∗ ) and ρb1 (t) ≤ ti − ti∗ = ti . It is shown that in case A the stability radius ρb1 (t) cannot be infinitely large since the value of ρb1 (t) is bounded as follows: n o e {i} = V bi ρb1 (t) ≤ min ti : i ∈ V, (46) k(i,b1 ) .

CE

b1 e with b1 ∈ B(G, t). Case B: Let the condition {i} , Vk(i,b hold for each manual operation i ∈ V 1) − We next consider three possible sub-cases using the notation V (i) (the notation V + (i), respectively) for the set of all predecessors to the vertex i ∈ V (for the set of all successors of the vertex i ∈ V) in the digraph G = (V, A). e and the line balance b1 ∈ B(G, t): Sub-case B-i: Let the following conditions hold for the manual operation i ∈ V n b o\n o 1 e , ∅; Vk(i,b \ {i} V \ V (47) ) 1

AC

n b o 1 Vk(i,b \ {i} ⊆ V + (i); 1) k(i, b1 ) = m(b1 ).

Let the following inequality hold:

b1

b1

t(V k(i,b1 )−1 ) < t(V k(i,b1 ) ).

(48)

(49) (50)

We construct the line balance br ∈ B(G) with m(br ) = m(b1 ) from the line balance b1 except that: The operation i is S br b1 br b1 now assigned to the workstation S m(br )−1 = S m(b1 )−1 , i.e. both equalities Vm(b = Vm(b {i} and Vm(b = Vm(b \ {i} 1 )−1 r) 1) r )−1 b1 0 0 e \ {i}. Due to the inequality (50), we ) and t = t j for each operation j ∈ V hold. We set t = ti + c(b1 , t) − t(V i

m(b1 )

j

obtain the inequality c(b1 , t0 ) > c(br , t0 ), which implies m(b1 ) · c(b, t0 ) > m(br ) · c(br , t0 ) since m(br ) = m(b1 ). Hence, b1 b1 (t0 ) < B(G, t0 ) and the stability radius ρb1 (t) cannot be infinitely large: ρb1 (t) ≤ ti − ti0 = c(b1 , t) − t(Vm(b ). 1) 29

ACCEPTED MANUSCRIPT

b1

b1

Let the following inequality hold: t(V k(i,b1 )−1 ) ≥ t(V k(i,b1 ) ), which is opposite to the inequality (50). If the inequality b1 b1 c(b1 , t) > t(Vm(b ) holds, then we use the above modification ti0 = ti + c(b1 , t) − t(Vm(b ) and t0j = t j for each operation 1) 1) e \ {i}. As a result, we obtain the equality c(b1 , t0 ) = t0 (V b1 ). j∈V m(b1 ) We construct the competitor bu ∈ B(G) for the line balance b1 from the line balance b1 except that: All operations b1 b1 \ {i} are now assigned to the workstation S m(bu ) , i.e. the following equalities hold: from the set V m(b1 ) = Vm(b 1) b1

bu bu Vm(b = {i}, Vm(b = V m(b1 ) , and m(bu ) = m(b1 ) + 1. We next calculate for a value of ∆ such that the equality u )−1 u)

(51)

CR IP T

m(b1 ) · c(b1 , t∗ ) = m(bu ) · c(bu , t∗ )

e \ {i}. t∗ = (t1∗ , t2∗ , . . . , ten∗ ), ti∗ = ti0 + ∆, and t∗j = t0j for each operation j ∈ V holds for the modified vector t∗ = (e t∗ , t), where e The equality (51) is equivalent to the following one:  m(b1 ) · [c(b1 , t∗ ) + ∆] = [m(b1 ) + 1] · max c(bu , t∗ ), ti∗ + ∆ . (52)

AN US

Let the inequality c(bu , t∗ ) ≤ ti∗ + ∆ hold. We obtain the following maximum value of ∆ = ∆max from the equality (52): ∆max := m(b1 ) · [c(b1 , t∗ ) − ti∗ ] − ti∗ . Let  > 0 be a sufficiently small positive real number. We set ti(∗) = ti∗ −  and ∗ (∗) (∗) e t(∗) j = t j for each operation j ∈ V \ {i}. We obtain the following inequality: m(b1 ) · c(b, t ) > m(br ) · c(br , t ). Hence, (∗) (∗) b1 (t ) < B(G, t ) and the stability radius ρb1 (t) cannot be infinitely large: b1 ρb1 (t) ≤ ti − ti(∗) = c(b1 , t) − t(Vm(b ) + ∆. 1)

(53)

Let the inequality c(bu , t∗ ) > ti∗ + ∆ hold. We obtain the following minimum value of ∆ = ∆min from the equality (52): ∆min := c(br , t∗ ) +

c(br , t∗ ) − c(b0 , t∗ ). m(b1 )

(54)

M

e \ {i}. The obtained inequality m(b1 ) · c(b, t[∗] ) > We set ti[∗] = ti∗ +  and t[∗] = t∗j for each operation j ∈ V j m(br ) · c(br , t[∗] ) implies b1 (t[∗] ) < B(G, t[∗] ). Thus, the stability radius ρb1 (t) is finite:

ED

b1 ρb1 (t) ≤ ti − ti(∗) = c(b1 , t) − t(Vm(b ) + ∆min . 1)

PT

Sub-case B-ii: Let the condition (47) hold along with the following ones: n b o 1 Vk(i,b \ {i} ⊆ V − (i); 1) k(i, b1 ) = 1.

(55)

(56) (57)

AC

CE

Due to the symmetry of workstations S m(b1 )−1 and S m(b1 ) with workstations S 1 and S 2 and due to the symmetry of the conditions (48) and (49) with the conditions (56) and (57), we can argue similarly as in sub-case B-i to show that the stability radius ρb1 (t) cannot be infinitely large in sub-case b-ii, since either the upper bound (53) holds or the upper bound (55) holds. Sub-case B-iii: Let the following equality hold: n b o\n o 1 e =∅ Vk(i,b \ {i} V \V (58) 1)

and neither both conditions (48) and (49) nor both conditions (56) and (57) hold. b1 We show that in sub-case B-iii, there exists at least one operation in the set Vk(i,b , which may be reassigned 1) either to the workstation S k(i,b1 )−1 or to the workstation S k(i,b1 )+1 without violating Condition o n 1 for oany given digraph n b b1 1 e . If the equality G = (V, A). The equality (58) holds if either Vk(i,b1 ) \ {i} = ∅ or ∅ , Vk(i,b1 ) \ {i} * V \ V b1 Vk(i,b1 ) \ {i} = ∅ holds, then we end up with case A, where operation i may be reassigned either to the workstation S k(i,b1 )−1 or to the workstation S k(i,b1 )+1 without violating Condition 1 and n b o wen have oshown that the upper bound (46) 1 e holds, then there exists a manual holds for the stability radius ρb1 (t) in case A. If condition ∅ , Vk(i,b1 ) \ {i} * V \ V operation j , i, which may be reassigned either to the workstation S k(i,b1 )−1 or to the workstation S k(i,b1 )+1 without 30

ACCEPTED MANUSCRIPT

violating Condition 1 for any given digraph G = (V, A). Let the equality (58) hold and either the condition (48) or (49) b1 do not hold. If the condition (48) does not hold, then there exists an operation j ∈ Vk(i,b such that there is no arc ( j, i) 1) e in the digraph G = (V, A), i.e. ( j, i) < A. Therefore, operation j ∈ V \ V may be reassigned to the workstation S k(i,b1 )−1 without violating Condition 1 for any given precedence digraph G = (V, A). Let both conditions (48) and (58) hold and let the condition (49) do not hold. In this case, the inequality k(i, b1 ) < b1 m(b1 ) holds and there exists an operation j ∈ Vk(i,b \ {i} such that there is no arc ( j, s) < A for each operation 1) b1 e s∈V \ {i}. Therefore, the operation j ∈ V \ V may be reassigned to the workstation S k(i,b1 )−1 without violating k(i,b1 )

CR IP T

b1 Condition 1 for any given digraph G = (V, A). Due to this reassignment of the operation j ∈ Vk(i,b , the line balance 1) b1 is reduced to the line balance br , which has the same number of the workstations and in which the inequality b1 t∗∗ = (t1∗∗ , t2∗∗ , . . . , ten∗∗ ), ti∗∗ = ti +∆min , Vk(i,b > Vkbr holds, where k = k(i, b1 ). We construct the vector t∗∗ = (e t∗∗ , t), where e 1) ∗∗ e \ {i} and the value of ∆min is determined in (54). The inequality V b1 > V br implies t = t j for each operation j ∈ V j

k(i,b1 )

k

AC

CE

PT

ED

M

AN US

the inequality c(b1 , t∗∗ ) > c(br , t∗∗ ). Hence, b1 < B(G, t[∗∗] ) and the stability radius ρb1 (t) cannot be infinitely large: b1 ρb1 (t) ≤ ti − ti(∗) = c(b1 , t) − t(Vm(b ) + ∆min . This completes the proof of Theorem 4. 1)

31