Two-machine flow shop scheduling with convex resource consumption functions

We consider a two-machine flow shop scheduling problem in which the processing time of each operation is inversely proportional to the power of the amount of resources consumed by it. The objective is to minimize the sum of the makespan and the total resource consumption cost. We show that the problem is NP-hard, and its constrained version remains so. Then, we develop 1.25- and 2-approximation algorithms for the problem and its constrained version, respectively.


Introduction
Scheduling problems with controllable processing times have been extensively studied since Vickson [13]. Refer to [6,10] for the comprehensive surveys. In most scheduling problems with controllable processing times, it is assumed that the processing time of job j is determined by a linear resource consumption function, which is described as where u i,j is the resource consumption amount of job j on machine i, and p i,j and ū i,j are the initial processing time and upper bound on the resource consumption amount (1) 1 3 of job j on machine i, respectively. However, linear resource consumption functions in (1) cannot reflect the law of diminishing marginal returns, which can be found in many resource allocation problems in physical or economic systems. The law means that productivity increases at a decreasing rate with the resource consumption amount. In this paper, to reflect this, assume that each job processing time decreases at an increasing rate with the amount of resource consumption, which is described as where w i,j > 0 is the workload of operation O i,j and k is a positive constant. Thus, we mainly focus on introducing the previous research with convex resource consumption functions in (2). Table 1 shows a summary of complexity results for scheduling problems with convex resource consumption functions. Please refer to [10] for any missing definitions.
For a single-machine case, Kayan and Akturk [7] and Cheng and Janiak [2] introduced the application of convex resource consumption functions. In a CNC machine scheduling problem [7], the job processing time is determined by a convex function in (2) of the feed rate and spindle speed used for each operation. In a steel mill industry [2], the time to preheat each batch of ingots in each soaking pit to a certain temperature is determined by a convex decreasing function of the gas flow intensity. Shabtay and Kaspi [8] where v j is the weight of job j. Although the computational complexity remains open, they revealed a closed-form of the optimal resource allocation for a given job sequence, and presented polynomially Table 1 Complexity results for scheduling with convex resource consumption functions 1�conv, up, ∑ j c j u j ≤ K�C max NP-hard, 1.5-approx, FPTAS [12] 1�conv, up�C max + ∑ j c j u j NP-hard, FPTAS [4] Pm�conv, pmtn, Pm�conv, ∑ j u j ≤ K�C max NP-hard [9] Pm�conv, Fm�conv, prmu, F2�conv, nw,

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Two-machine flow shop scheduling with convex resource… solvable cases. Xu et al. [14] showed that 1�conv, d j = d, . Recently, Shabtay and Zofi [12] considered 1�conv, ∑ j u j ≤ K�C max with one unavailability period, and showed that it is NP-hard and has a 1.5-approximation algorithm and an FPTAS. Choi and Park [4] extended the results into the case with multiple unavailability periods. For a parallel-machine case, Shabtay and Kaspi [9] showed that Pm�conv, ∑ j u j ≤ K�C max is NP-hard while its preemption vesion and Pm�conv, ∑ j u j ≤ K� ∑ j C j are polynomially solvable. For a flow shop case, Cheng and Janiak [2] considered Fm�conv, ∑ i,j u i,j ≤ K�C max such that permutation schedules are only considered and the bound constraint for each operation, that is, i,j ≤ u i,j ≤ i,j , exists. They proved the NP-hardness of the case with m = 2 , and developed a branch-and-bound algorithm and three m-approximation algorithms whose performances were effective through numerical experiments. Shabtay et al. [11] considered F2�conv, ∑ i,j u i,j ≤ K�C max such that no-wait constraints exist, that is, no idle time exists between the first and the second operations of each job. They proved its strong NP-hardness, developed a 2 1∕k+1 -approximation algorithm, and introduced three polynomially solvable cases. Furthermore, they developed two heuristics whose performance were effective through numerical experiments.
To the best of our knowledge, there has been no study on the complexity of F2�conv, ∑ i,j u i,j ≤ K�C max without the bound and no-wait constraints. Note that the optimality property obtained from the bound constraints is to fully compress or not to compress each job under an optimal schedule, which had been used for the NP-hardness proof of [5]. Cheng and Janiak [2] also proved the NP-hardness of F2�conv, ∑ i,j u i,j ≤ K�C max with i,j ≤ u i,j ≤ i,j based on this optimality property. Since the convex resource consumption functions in (2) are not locally bounded, however, this optimality property cannot be used for the NP-hardness of F2�conv, ∑ i,j u i,j ≤ K�C max . Furthermore, two cases with and without no-wait constraint are completely different problems. Thus, it is not straightforwardly that the NP-hardness result of [2,5,11] holds in F2�conv, ∑ i,j c i,j u i,j ≤ K�C max . The contributions of this paper are twofold. First, we prove the NP-hardness of F2�conv�C max + ∑ i,j c i,j u i,j and F2�conv, whose complexity remains open in [10]. Second, we develop 1.25-and 2-approximation algorithms for Since an optimal schedule exists among the set of the permutation schedules in the twomachine flow shop scheduling problem with makespan criterion, an m-approximation algorithm of Cheng and Janiak [2] becomes a 2-approximation algorithm for F2�conv, The remainder of this paper is organized as follows: Sections 2 and 3 introduce the problem definition and some optimality properties, respectively. In Sects. 4 and 5, we prove the NP-hardness and develop approximation algorithms, respectively.

Problem definition
Our problem can be formally stated as follows. Let J = {1, 2, ..., n} and M = {1, 2} be the sets of jobs and machines, respectively. (2), ..., (n) is the job sequence on both machines, where (j) is the jth job to be processed on both machines in permutation ; Note that ⋅ It is known from [1] that it suffices to consider only the permutation schedule with respect to any regular performance measures (e.g., makespan) in the twomachine flow shop model; For simplicity, we will use p i,j instead of p i,j (u i,j ) for O i,j ∈ O when no confusion exists. For O i,j ∈ O , let C i,j ( ) and S i,j ( ) be the completion and start times of O i,j in , and C max ( ) = C 2, (n) ( ) be referred to as the makespan. The objective is to find a schedule with the minimum sum of the makespan and the total resource consumption cost, that is, Let the problem above be referred to as Problem P. Furthermore, let the constrained version of Problem P be stated as follows: where K is a given threshold.
Two-machine flow shop scheduling with convex resource…

Optimal properties of an optimal schedule
In this section, we introduce some optimality properties of Problem P. First, we introduce a new terminology. Let a job j be referred to as a pivot in , if C 1,j ( ) = S 2,j ( ) . Then, we have the following optimality properties.

Proposition 1
The first and last jobs are pivots in any optimal schedule.
Proof Let f and l be the first and last jobs in the optimal schedule, respectively. Without increasing the makespan, we can decrease u 2,f and u 1,l until jobs f and l become pivots. Thus, Proposition 1 holds.

Proposition 2 No idle time exists between consecutive jobs in any optimal schedule.
By Propositions 1 and 2, henceforth, we consider only a schedule with which implies that Let * = * ;u * be an optimal schedule, and For simplicity, let Since we have By Eq. (4) and the strict convexity of and, furthermore, we have b ( a,b ) and Then, by inequalities (5) and (7), This is a contradiction. For simplicity, we introduce the following notations: When jobs (h) and (m) are the consecutive pivots in , let where the first term is the total processing times on machines 1 and the second and third terms are the total resource consumption costs of these operations, respectively, and the following equation holds:

Lemma 3 z h,m ( ) is minimized when two constants ̂ and ̃ exist such that
Proof Suppose that ∑ j∈A h,m p 1,j (u 1,j ) = , where > 0 is some value. Then, z h,m ( ) is minimized when is optimal for the following problem: Since Lagrangian L(u; ) for the above problem is expressed as follows: By Karush-Kuhn-Tucker (KKT) necessary and sufficient conditions, there exists a constant such that Note that by Eq. (4), c 1,j = −p � 1,j ( 1,j ) . Then, since p � 1,j (u 1,j ) ≠ 0 , Eq. (10) can be rewritten as follows: which implies that for j ∈ A h,m , By Eq. (11) and setting ̂= 1 k+1 , By Eqs. (8) and (9), z h,m ( ) is also minimized when is optimal for the following problem:  where P h,m ( ) is the total processing time of operations in A h,m .

Corollary 1 In * ,
Proof Let Q * be the set of the pairs of the consecutive pivots in * . Then, by Lemma 2 and relations (6) and (17), we have ◻

Computational complexity
In this section, we show that Problem P and its constrained version are NP-hard by using the optimality properties in Sect. 3.

Theorem 1
Problem P is NP-hard, even when k = 1 and w 1,j = w 2,j for j ∈ J .
Proof We prove it by reduction from the partition problem, which is known to be NP-complete: Given g integers in {a 1 , a 2 , ..., a g } with ∑ g j=1 a j = 2A , is there a subset A ⊂ {1, 2, ..., g} such that Given an instance of the partition problem, we can construct an instance of Problem P as follows: It is observed that no idle time exists between the consecutive jobs in ̄ on both machines. Thus, since and we have (⇐ ) Suppose that there exists an optimal schedule * with z( * ) ≤ Z . Then, we can obtain ̃= * ;û by setting where ̂ is a sequence by Johnson's rule. Note that C max (̂) ≤ C max (̃) . Thus, (p 1,j ,p 2,j ) = Now, we will see the structure of ̂ in Claims 1 and 2. Claim 1 Jobs (g + 2) and (g + 3) are the first and last jobs in ̂ , respectively.
Proof Suppose that, in ̂ , job (g + 1) is not a pivot, it is the lth job, and it is sequenced between the consecutive pivots ̂(h) and ̂(m) , where h < l < m . By (18) and Lemma 3, This implies that job ̂(l) becomes a pivot, which is a contradiction.
This implies that jobs ̂(m) is not a pivot, which is a contradiction. By cases i) and ii), Claim 2 holds. ◻ Now, we will derive a lower bound of z(̂) . Let L and R be the sets of jobs in {1, 2, ..., g} before and after job (g + 1) in ̂ , respectively.

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Two-machine flow shop scheduling with convex resource… 1 + k (k+1) 2 has the maximum of 1.25 at k = 1 , and it converges to 1 when k goes to 0 or ∞ . Henceforth, we will prove a 2-approximablity for the constrained version. Let ̂= ; be a schedule with an arbitrary sequence , where Note that ̂ can be obtained in O(n) time. For simplicity, let û i,j = i,j and p i,j = p i,j (û i,j ) for O i,j ∈ O.

Theorem 4
The constrained version of Problem P has 2-approximability.
Proof Consider the following problem.
Lagrangian L(u; ) for problem (25) is described as and KKT necessary and sufficient conditions are as follows: and Since = 1 k+1 and u i,j =û i,j for O i,j ∈ O satisfy Eqs. (26) and (27), (û i,j ) O i,j ∈O becomes an optimal solution to problem (25). Since (u * i,j ) O i,j ∈O is a feasible solution of problem (25), we have