Multi-machine scheduling with interval constrained position-dependent processing times

This paper investigates multi-machine scheduling problems with interval constrained actual processing times. The actual processing time of each job is assumed to be restricted in a given interval otherwise the extra earliness or tardiness time should be used to patch up the flaw of job. The objectives are to find the optimal job sequence to minimize the total load of machines, the number of exceeding-interval jobs and the makespan of job schedule, respectively. This paper shows that both of the total load minimization problem and the exceeding job number minimization problem are polynomially solvable. For the makespan minimization problem, this paper proves that it is NP-hard, and proposes a fully polynomial time approximation scheme (FPTAS) for the case with two parallel machines.


1.
Introduction. In the steel rolling mill, the temperature of an ingot while waiting to enter the rolling machine drops below a certain level, requiring the ingot to be reheated before rolling in Gupta and Gupta [10]. Motivating from the variable processing times in various effects such as the reheat requirement of ingots in steel rolling mills, many scheduling literatures paid attention to the scheduling problems in which the processing time of a job is not constant. References Gupta and Gupta [10] and Browne and Yechiali [3] introduced the variable job processing times due to deteriorating effects into scheduling studies, and proposed the optimal sequencing principle of minimizing the makespan. The total flow time minimization problem with linear actual processing times of jobs was considered by Mosheiov [16], and its optimal schedule was proved to be V-shaped of basic processing times of jobs.
Based on general job-dependent learning curves, Mosheiov and Sidney [18] established several models of job processing times and showed that two single machine problems and two multi-machine problems remain polynomially solvable. Liu et al. [15] studied parallel-machine scheduling problem with linear deteriorating effect of minimizing the makespan of jobs, and construct an FPTAS to solve the problem, which shows that the considered problem is NP-hard in the ordinary sense. Many other scheduling studies with variable processing times of jobs can be found in Biskup [1], Kuo and Yang [14], Mosheiov [17], Ji et al. [13], Yin et al. [28], Zhang et al. [31], etc.
Another interesting topic is scheduling with interval constraints of processing times. Chen et al. [4] explored to propose the algorithm for generating optimal cyclic schedules of hoist where the time of a printed circuit board spend in a tank is upper and lower bounded. Yan et al. [27] proposed a branch and bound algorithm for optimal cyclic scheduling in a robotic cell where the exact processing time of each part is limited in a time window. Several recent papers (see, e.g., Zhou et al. [32], Yan et al. [26] and Yan et al. [25] also paid their attention to explore the cyclic robotic cell scheduling problems with time constraints. To cope with the quality requirement of job processing, the actual processing times of jobs are required to be fallen in given intervals in several recent studies, else if the processing time of a job is not within its time interval may lead to quality flaws, the extra cost or repair time should be paid. Yu and Zhang [29] studied single-machine scheduling problems with general position-dependent aging effect and machine maintenance, and assumed that the actual processing times of jobs should not exceed the upper-bounds, and proved that the considered problems remain polynomial solvable. Xue et al. [24] considered the single machine scheduling problem with machine maintenance and interval constrained processing times, where each job has a processing interval, the optimal polynomial time algorithms are designed to the considered problem. Finke et al. [8] showed the single machine problem with machine maintenance and interval constraints of processing times remains polynomial solvable if the objective is extended from the makespan of schedule to another four criterions. To the best of our knowledge, it is rare to see the research of multi-machine scheduling with both general position-dependent deteriorating effect and interval constraints of job processing times, which it is the common scenario in the actual production process. For an example, in the porcelain firing workshop with multiple furnaces, the actual processing time of porcelain on each furnace is in deteriorating effect due to the wear of furnace, the firing time of each porcelain batch should be in a given time interval to guarantee the quality of porcelain. Three objectives are considered in this paper. Firstly, in order to minimize the cost of machine running, minimizing the total processing times on machines (i.e., the total load of machines) is the first considered objective. Second, the number of jobs exceeding the time interval should be minimized for the reason that the processing time of job exceeding the time interval lead to the bad quality product. The third considered objective is to minimize the makespan of schedule to obtain the most efficient schedule. This paper is organized as follows. In section 2, the problem formulation is introduced. In section 3, the total load minimization problem and the exceeding job number minimization problem are considered. In section 4, we propose the NP-hard analysis for the makespan minimization problem, and design an FPTAS to solve a special case. Finally, we conclude the paper.

2.
Notation and problem formulation. The notation and terminology throughout this paper is proposed as follows. There are n independent jobs ready to be processed on m machines at time zero. Preemption is not allowed and a machine is only able to process one job at a time. The number of jobs scheduled on machine i is denoted as n i , which can be inferred to satisfy m i=1 n i = n, (i = 1, 2, · · · , m) . (1) Job j has a normal processing time p j (j = 1, 2, . . . , n). If job j is assigned to the rth position on machine i, its the actual processing time p r ij is equal to where f ij (i, j, r) is not restricted to be specific and monotone, and can vary with machine i and job j. When job j is assigned to the first position on a machine, its actual processing time is equal to its normal processing time, i.e., f ij (i, j, 1) = p j . The actual processing time of job j is required in the time interval [a j , b j ]. If the actual processing time of a job exceed its corresponding interval, the repair time should be added to restore the quality of the job. Concretely, the earliness repair time of job j according to the time interval [a j , b j ] is denoted as and the tardiness repair time of job j according to the time interval [a j , b j ] is denoted as where α 1 and α 2 are the parameters of the tardiness repair time and the tardiness repair time, respectively. The normal processing time p j is assumed to fall in the interval [a j , b j ], i.e., a j ≤ p j ≤ b j .
The completion time of job j on machine i is denoted by C ij . The total load on machines can be denoted as Based on (2), (3) and (4), the total load T L can be transferred as when job j is scheduled in the rth position on machine i, x ijr = 1, otherwise x ijr = 0. The number of job j exceeding the time interval [a j , b j ] is Based on (7), the number of all the exceeding intervals jobs can be denoted as when job j is scheduled in the rth position on machine i, x ijr = 1, otherwise x ijr = 0. The completion time of job j on machine i is denoted by C ij , and the last completion time of jobs on machine i is denoted by C i , i.e. C i =max{C ij , j = 1, 2, · · · , n i }. The makespan of the schedule is C max = max{C i , i = 1, 2, · · · , m}. Based on (2), (3) and (4), the makespan can be transferred as Let IC denote the character of interval constraint. By using the three-field notation scheme for scheduling problems introduced by Graham [9], the total load minimization problem is denoted as R m |p r ij = f ij (i, j, r), IC|T L, the problem of minimizing the the number of all the exceeding intervals jobs can be denoted as R m |p r ij = f ij (i, j, r), IC| U j , and the problem of minimizing the the number of all the exceeding intervals jobs can be denoted as R m |p r ij = f ji (j, i, r), IC|C max .
3. Multi-machine scheduling problem of minimizing the total load. In this section, the problem R m |p r ij = f ij (i, j, r), IC|T L is firstly analyzed.
Proof. The two-step procedure in Mosheiov [17] can be used to optimally solve the problem R m |p r ij = f ji (j, i, r), IC|T L . The first step is giving an allocation of the numbers of jobs on machines (n 1 , n 2 , . . . , n m ) . Based on m i=1 n i = n and the results in Ji and Cheng [12], it can be obtained that the above upper bound for the number of different allocations for n jobs m machines is O(n m−1 ).
In the second step, for a given allocation on machines (n 1 , n 2 , . . . , n m ), we can transfer the problem R m |p r ij = f ij (i, j, r), IC|T L as a standard (n×n) assignment problem as follows.
Combining two steps, it can be seen that the total computational complexity of solving the problem R m |p r ij = f ij (i, j, r), IC|T L is O(n m+2 ). In what follows, three special cases for the problem R m |p r ij = f ij (i, j, r), IC|T L are considered. In the first special case, the actual processing time f ij (i, j, r) is non-decreasing in r, and the considered problem can be denoted as Then we only need to analyze the tardiness repair time Then the total load in this case can be obtained when job j is scheduled in the rth position on machine i, x ijr = 1, otherwise x ijr = 0.
For the second special case, the machine setting is parallel-machine. Then the actual processing time of job j can be denoted as p r j = f j (j, r), which is also assumed to be non-decreasing in r. This special case can be denoted as P m |p r j = f j (j, r), IC, N D|T L, where the total load T L can be calculated by Proof. Since f ij (i, j, r) is non-decreasing in r, it is easy to obtained that f j (i, j, r)+ α 2 max(0, f j (j, r) − b j ) is also non-decreasing in r. Due to the monotonicity of f j (i, j, r) + α 2 max(0, f j (j, r) − b j ) and the parallel-machine setting, we can use the method of changing job position in Yu et al. [30] to prove that the number of jobs on machine i satisfies a balance principle, i.e., [n/m] ≤ n i ≤ [n/m] + 1, i = 1, 2, · · · , m. It is easy to obtain the possible positions for job allocation in the optimal schedule based on the balance principle. Then the problem P m |p r j = f j (j, r), IC, N D|T L can be transferred as an standard n × n assignment problem, which can be optimally solved in O(n 3 ) time.
In the third special case, if job j is assigned to the rth position on a machine, its actual processing time is where the functions h(p j ) and l(r) are non-decreasing in p j and non-decreasing in r, respectively. The actual processing time of job j should be constrained in the time interval [p j , h(p j )b 0 ] otherwise the extra repair time α 2 max(0, h(p j )(l(r) − b 0 )) should be added. This special case can be denoted by P m |p r j = h(p j )l(r), IC, N D|T L. Similar with the two previous special cases, the total load of this case can be calculated as follows: when job j is scheduled in the rth position on machine i, x ijr = 1, otherwise x ijr = 0. Before analyzing the solving algorithm of the problem P m |p r j = h(p j )l(r), IC, N D |T L, we firstly introduce a useful lemma.  Proof. Since p r j = h(p j )l(r) is a specific form of p r j = f j (j, r), the problem P m |p r j = h(p j )l(r), IC, N D|T L is a special case of the problem P m |p r j = f j (j, r), IC, N D |T L. Based on the analysis of the problem P m |p r j = f j (j, r), IC, N D|T L, it can be obtained that the number of jobs on machine i satisfies a balance principle, i.e., [n/m] ≤ n i ≤ [n/m] + 1, i = 1, 2, · · · , m for the problem P m |p r j = h(p j )l(r), IC, N D|T L. Then, the possible positions for job allocation in an optimal schedule can be proposed by the balance principle.
Since l(r) is non-decreasing in r, it is easy to get that l(r)+α 2 max(0, (l(r)−b 0 )) is also non-decreasing in r. Because of (12), the total load for this considered case can be calculated as follows: when job j is scheduled in the rth position on machine i, x ijr = 1, otherwise x ijr = 0. Due to Lemma 1 and the monotonicity of h(p j ) and l(r), an optimal schedule of the problem P m |p r j = h(p j )l(r), IC, N D|T L can be constructed by sequencing p j and l(r) + α 2 max(0, (l(r) − b 0 )) in the opposite monotonicity in O(n log n) time.
4. Multi-machine scheduling problem of minimizing the number of exceeding jobs. In the section, the exceeding-job-number minimization problem R m |p r ij = f ij (i, j, r), IC| U j will be analyzed. Moreover, we will consider its three special cases based on the similar assumptions for R m |p r ij = f ij (i, j, r), IC, N D|T L, P m |p r j = f j (j, r), IC, N D|T L and P m |p r j = h(p j )l(r), IC, N D|T L, respectively. In turn, the considered problems in this section can be denoted as R m |p r ij = f ij (i, j, r), IC, N D| U j , P m |p r j = f j (j, r), IC, N D| U j and P m |p r j = h(p j )l(r), IC, N D| U j , respectively. Theorem 4.1. The problem R m |p r ij = f ij (i, j, r), IC| U j can be optimally solved in O(n m+2 ) time.
Proof. After changing the weight w ijr from f ij (i, j, r) + α 1 max(0, a j − f ij (i, j, r)) + α 2 max(0, f ij (i, j, r)−b j ) to g ij (i, j, r), the two-step method in the proof of Theorem 3.1 can be used to solve the problem R m |p r ij = f ij (i, j, r), IC| U j in O(n m+2 ) time.
Similarly, by changing the weight w ijr from f ij (i, j, r)+α 2 max(0, f ij (i, j, r)−b j ) to g ij (i, j, r), we can use the proof method in Theorem 3.2 to obtain the following theorem. When the machine setting is parallel machine, the number of job j exceeding the time interval [a j , b j ] is Then the number of exceeding-interval jobs can be denoted as U j = n j=1 m i=1 ni r=1 g j (j, r)x ijr when job j is scheduled in the rth position on machine i, x ijr = 1, otherwise x ijr = 0. Similarly, after changing the weight w ijr from f j (j, r) + α 2 max(0, f j (j, r) − b j ) to g j (j, r), we can use the proof method in Theorem 3 to obtain the following theorem. Comparing with the problem P m |p r j = h(p j )l(r), IC, N D|T L, the following theorem shows that the computational complexity of solving P m |p r j = h(p j )l(r), IC, N D | U j can be reduced from O(n log n) to O(n). Proof. For the problem P m |p r j = h(p j )l(r), IC, N D| U j , the time interval of the actual processing time of job j is [p j , h(p j )b 0 ], the number of job j exceeding the time interval [p j , h(p j )b 0 ] can be obtained by comparing l(r) and b 0 , and then the number of all the exceeding-interval jobs is the number of jobs with l(r) > b 0 . Since l(r) is non-decreasing in r, we have U j = n − r 0 , where r 0 = min{r|l(r) > b 0 }. Obviously, the computational complexity is at most n times comparing calculation is O(n).
Based on Theorem 8, we can obtain the following corollary for the case that l(r) is strictly increasing in r,. Corollary 1. If l(r) is strictly increasing in r, the problem P m |p r j = h(p j )l(r), IC, N D| U j can be optimally solved in O(1) time, and the optimal objective value U j is equal to n − ( l −1 (b 0 ) + 1).
Proof. Based on the proof in Theorem 8, the number of all the exceeding-interval jobs is the number of jobs with l(r) > b 0 . Since l(r) is strictly increasing in r, l(r) > b 0 implies that r > l −1 (b 0 ). Because that r is a integer, r > l −1 (b 0 ) is equivalent with r ≥ l −1 (b 0 ) + 1, then the number of all the exceeding-interval jobs U j is n − ( l −1 (b 0 ) + 1).

5.
Multi-machine scheduling problem of minimizing the makespan. In this section, we analyze the problem R m |p r ij = f ij (i, j, r), IC|C max and its special case P 2 |p r 1j = h(p j )l(r), p r 2j = p j , IC, N D|C max , where the time constrained interval is [p j , h(p j )b 0 ] based on the analysis in Theorem 3.5, and the actual processing times p r 1j and p r 2j denote the actual processing time of job j on machine 1 and machine 2, respectively.
If the number of machines m is not restricted, we will show that the problem R m |p r ij = f ij (i, j, r), IC|C max is NP-hard in strong sense in the following. Theorem 5.1. The problem R m |p r ij = f ij (i, j, r), IC|C max is NP-hard in a strong sense.
Proof. We propose an instance of the problem R m |p r ij = f ij (i, j, r), IC|C max as follows: m = K, n = 3K, the actual processing time of job j is A j , i.e., p A j = A j such that B/4 < A j < B/2 and 3K j=1 A j = KB for j = 1, 2, · · · , n. To know wether the problem R m |p r ij = f ij (i, j, r), IC|C max have the objective value B (i.e., C max = B) or not, it is only necessary that there exist K antipathetic triplets S 1 , S 2 , · · · , S K such that Si A j = B, i = 1, 2, · · · , K. In this instance, the problem R m |p r ij = f ij (i, j, r), IC|C max is reduced to the 3-PARTITION Problem, which is known as a NP-hard problem in strong sense. This implies that the problem R m |p r ij = f ij (i, j, r), IC|C max is also NP-hard in strong sense. Due to the strong NP-hardness of the problem R m |p r ij = f ij (i, j, r), IC|C max , we let the analysis of its solving algorithm as an open problem in this paper. In what follows, we study its special case P 2 |p r 1j = h(p j )l(r), p r 2j = p j , IC|C max and analyze its solving algorithm. If p r 1j = p r 2j = p j and IC = [p j , h(p j )b 0 ], the problem P 2 |p r 1j = h(p j )l(r), p r 2j = p j , IC, N D|C max is reduced to the problem P 2 ||C max , which is shown to be a NP-hard problem in Pinedo [20]. Then, we will obtain the following lemma.
Lemma 5.2. The problem P 2 |p r 1j = h(p j )l(r), p r 2j = p j , IC, N D|C max is NP-hard. By using the standard interchange technology of jobs, the following lemma is easy to prove. Lemma 5.3. The optimal sequencing with respect to the makespan of jobs scheduled on machine 1 is the LPT (longest processing times first) order such that jobs with bigger p j are processed earlier.
This lemma indicates a dynamic programming algorithm, which starts from an initial job sequence with n jobs in the LPT order and can eventually find an optimal schedule by making either processing on machine 1 or processing on machine 2 decision on jobs one-by-one. For convenience, we let the jobs be indexed in the LPT order such as p 1 ≥ p 2 ≥ · · · ≥ p n . Let (j, C 1 , r, C 2 , C j max ) be the state representative denoting a partial schedule on the first j jobs, where j ≤ n, C i is the completion time of the last processed job on machine i (i = 1, 2), and r is the number of jobs processing on machine 1, and C j max is the makespan. If j = n, the state (j, C 1 , r, C 2 , C j max ) denotes a full schedule and also a feasible solution to the problem. For j < n, the next unscheduled job J j+1 can be scheduled in the following ways.
• Case 1. Job J j+1 is scheduled on machine 1. State (j + 1, . Before developing an efficient dynamic programming algorithm, the following easy-to-prove lemma is presented. Lemma 5.4. Consider two states (j, C 1 , r, C 2 , C j max ) and (j , C 1 , r , C 2 , C j max ) with 0 < j = j < n, r = r , C j max = C j max , C 1 = C 1 , and C 2 < C 2 (or with 0 < j = j < n, r = r , C j max = C j max , C 2 = C 2 , and C 1 < C 1 ), as any later schedules generated from (j , C 1 , r , C 2 , C j ) cannot be advantage to the corresponding schedules generated from (j, C 1 , r, C 2 , C j max ), eliminating state (j , C 1 , r , C 2 , j ) won't lead to any non-optimal solutions for the problem P 2 |p r 1j = h(p j )l(r), p r 2j = p j , IC, N D|C max .

[Elimination]
If j < n, for any two states (j, C 1 , r, C 2 , C j max ) and (j , C 1 , r , C 2 , C j max ) with 0 < j = j < n, r = r , C j max = C j max , C 1 = C 1 , and C 2 < C 2 (or with 0 < j = j < n, r = r , C j max = C j max , C 2 = C 2 , and C 1 < C 1 ), eliminate state (j , C 1 , r , C 2 , C j max ) from T , and then set S (j+1) = T after all the possible eliminations are conducted. Otherwise, set S (n) = T . [Optimization] If S (n) = ∅, find the state (n, C 1 , r, C 2 , C n max ) with the smallest makespan over all states in S (n) . Obtain the optimal solution value and trace back to get the corresponding schedule as the optimal schedule. n j=1 p j and C j max + max(0, C 2 + p j+1 − C j max ) are no more than In the following, a fully polynomial time approximation scheme (FPTAS) for the considered problem P 2 |p r 1j = h(p j )l(r), p r 2j = p j , IC, N D|C max is developed by trimming the state space of the Algorithm A1 by fusing states that are "close' to each other. Woeginger [23] developed a classic general technique, where the fusion of states is obtained by constraining the jobs in any schedule to finish at times of the form τ i = (1 + ε ) j for j ≥ 0 and ε > 0. Engels et al. [7] and Cheng and Sun [5] used the technique in Woeginger [23] to develop pseudo-polynomial algorithms and FPTASs for the considered single machine scheduling problems with rejection cost. Steiner and Zhang [22] used the static partitioning method (SPM) introduced in Sahni [21] and the bound improvement procedure (BIP) developed in Chubanov et al. [6] to convert their dynamic programming algorithm into an FPTAS and this FPTAS runs strongly in polynomial time. This section will the similar analysis method in Steiner and Zhang [22] and Sahni [21] to convert the FPTAS.
Similar with Steiner and Zhang [22], we propose the bound analysis of the makespan before conducting the FPTAS. In the above analysis of Algorithm A1, it is shown that C * max ≤ n j=1 p j , where C * max denotes the optimal solution value of the problem P 2 |p r 1j = h(p j )l(r), p r 2j = p j IC, N D|C max . Because that C max = max{C 1 , C 2 } ≥ (C 1 + C 2 )/2 ≥ n j=1 p j /2, it can be obtained that C * max ≥ ( n j=1 p j )/2.
Then the bound of the optimal makespan is obtained in the following lemma.
[StateGeneration] Do the same as in Algorithm 1.
[StateElimination] If j < n, for the states (j, C 1 , r, C 2 , C j max ) with the same r value and C j max ≤ U values falling in the same subinterval, keep only the one with the smallest C 1 or C 2 value, and set S (j+1) = T after all of these operations. Otherwise, set S (n) = T . [StateSelection] In S (n) , find the state (n, C 1 , r, C 2 , C n max ) with the smallest C n max value. Set C n max as the solution value and trace back to obtain the corresponding schedule).
Lemma 5.6. The optimal makespan of the problem P 2 |p r 1j = h(p j )l(r), p r 2j = p j , IC, N D|C max is in the interval [L, 2L], where L = n j=1 p j /2.
Implementing SPM in Sahni [21] on Algorithm A1, the above bounds [L, U ] will lead to an approximation algorithm, which will find an (1 + ε)-approximation solution to the P 2 |p r 1j = h(p j )l(r), p r 2j = p j IC, N D|C max problem in O( 2n 2 ε ) time, where U = 2L and ε > 0. In the following, Algorithm 2 is proposed.
Remark 1. In [Partitioning], the interval [0, 2L] is divided into 2n ε subintervals as the length of each subinterval has to be no greater than εL n , which is the maximum error allowed to be introduced in each iteration. This gives that the total cumulative error after n iteration won't be greater than εL, which achieves the designated approximation ratio ε.
Theorem 5.7. For any given ε > 0, the FPTAS A2 finds an (1 + ε)-approximation solution value such as v ≤ (1 + ε)v * for the problem P 2 |p r 1j = h(p j )l(r), p r 2j = p j IC, N D|C max . The run time is O( n 2 ε ) time. Proof. The correctness of Algorithm A2 directly follows the above discussions and the proof in Sahni [21]. For run time, Sahni [21] showed that A(L, n, ε) runs in O( n 3 ε ) time for the bound interval of objective [L, nL]. In Algorithm A2, the bound interval of objective is [L, 2L], its overall run time is reduced to O( 2n 2 ε ), which is indeed O( n 2 ε ).
6. Conclusions. In this paper, three multi-machine scheduling problems with interval constrained actual processing times are analyzed. For the first problem of minimizing the total load on machines, four cases are proved to be polynomially solved in the decreasing computational complexities when constraints are added; For the second problem, the objective is to minimize the number of jobs exceeding intervals, this paper shows that five cases of this problem can be also solved in polynomial time. For the makespan minimization problem, this paper proved that the problem is NP-hard in strong sense if the machine number is not restricted, and proposed a dynamic programming algorithm and an FPTAS to solve the special case with two parallel-machine setting, which is showed as NP-hard in ordinary sense.