Complexity of Scheduling Problem in Single-Machine Flexible Manufacturing System with Cyclic Transportation and Unlimited Buffers

The paper concerns complexity studies on the scheduling problem arising in a simple flexible manufacturing system. The system consists of a single machine, one depot (both with unlimited buffers), and one vehicle (automated guided vehicle). The vehicle operates according to the regular metro strategy. This means that it travels in cycles of the constant length, without stops, transporting at most one job at a time between the depot and the machine. The machine executes available jobs in the non-preemptive way. The goal is to minimize the schedule length, i.e., to minimize the number of vehicle cycles necessary to transport and execute all jobs in the system. We prove the strong NP-hardness of this problem and show that any list algorithm has the worst-case performance ratio equal to 2. Moreover, we mention that special cases of the considered problem, with zero transportation times from the depot to the machine and from the machine to the depot, are polynomially solvable.


Introduction
Flexible manufacturing systems (FMS) play an extremely important role in the contemporary industry. Most factories are equipped with machine stations and storage areas connected with a transportation system of various types such as robots, automated guided vehicles (AGV), and conveyors [1]. The production process in such systems is described by many parameters as well as constraints, and its quality is estimated with regard to many objective functions, from the classical ones, like the schedule length, to the more specialized ones, like the late work [2]. On the one hand, practical applications inspired intensive studies on flexible manufacturing systems within operational research, which resulted in numerous papers on specific problems [3][4][5][6][7][8][9], including robotic cells problems [10,11] and surveys of results [12][13][14][15][16]. On the other hand, closely related problems, concerning automated guided vehicles traffic control systems, could be found in logistics, e.g., in automated container terminals [17][18][19][20][21].
Due to the strong connection to practice, studies on flexible manufacturing system focus mainly on complex models, since their main goal is to propose efficient methods for solving real optimization problems. For this reason, no systematic research on particular FMS models has been done, as one can observe for "classical" scheduling models. In this latter case, researchers have intensively explored the borderline between easy and intractable problems. From the theoretical point of view, it is interesting to determine the complexity status of the simplest FMS models, which are in general NP-hard.
In this paper, we considered a basic flexible manufacturing system which consists of one machine, one depot (both with unlimited buffers), and one vehicle (AGV) moving in cycles of the constant length. Such a model is the simplest system with the AGV-based transportation, and its complexity status has not been studied yet. Most literature has been devoted to FMS problems with many vehicles [22][23][24] and with the unidirectional single-loop transportation system [25]. A single vehicle has been analyzed in the context of more complex systems: with buffers [26], with many machines (station) [27], or with dedicated machines [23]. Particularly, Suri and Desiraju [28] considered the production environment with only one vehicle, but in a more complex FMS, namely in the FMS with a single discrete material handling device (MHD). Lacomme et al. [23] deal with the problem of simultaneously scheduling jobs on machines and identical automated guided vehicles, analyzing many machines and many AGVs.
In this work, we show that the basic problem of minimizing the schedule length on a single machine, one depot, unlimited buffers, and one AGV with the constant cycle length is intractable. In Sect. 2, we define the problem formally, and we provide practical motivations for it. In Sect. 3, we present the strong NP-hardness proof for the considered model. In Sect. 4, we show that the worst-case performance ratio for any list heuristic is equal to 2, and we comment on some polynomially solvable special

Problem Definition
We consider the problem of simultaneously scheduling and transporting jobs in the flexible manufacturing system (FMS), consisting of a single machine, a depot (storage), and one automated guided vehicle (AGV) (cf. Fig 1).
All jobs are available at time zero at depot D. They have to be delivered by the AGV from depot D to machine M for non-preemptive processing. A job, after being completed, has to be transported by the AGV from the machine back to the depot. The AGV travels between the machine and the depot, in cycles of the constant length, picking up and dropping off jobs, if they are available. The AGV keeps moving even if there is no job to be transported, unless the last job is delivered to the depot. The goal is to minimize the length of the whole schedule, i.e., to minimize the delivery time of the last job to depot D.
More formally speaking, we study the problem of executing a set of n jobs on machine M, J = {J 1 , J 2 , . . . , J j , . . . , J n }. Each job J j is described by processing time p j necessary for its completion by the machine. The transportation times from/to the depot are job independent. The AGV moves without any break from the depot to the machine in time t 1 , and from the machine to the depot in time t 2 , so its constant cycle time is equal to T = t 1 + t 2 . The loading and unloading times are included in the transportation times t 1 and t 2 . Moreover, we assume that both the depot and the machine are equipped with buffers of unlimited capacity, the AGV can take only one job at a time, and it cannot stop either at the machine or at the depot to wait for a job. To construct a schedule for the machine and the AGV simultaneously, one has to determine for each job J j three time moments: -the depart time from the depot to the machinet D j , -the completion time on the machinect M j , -the delivery time from the machine to the depot (i.e., the completion time at the depot)ct D j . Obviously, the processing of job J j on machine M can start only after its delivery from depot D, i.e., ct M j − p j ≥ t D j +t 1 . Similarly, the completion time of the job on the can take only integer values, being multiples of the cycle length T . This means that for each job J j one has to determine two cycle numbers (slots), in which this job is taken by the AGV to the machine k M j , and delivered back to the depot k D j . Hence, depart and delivery times are equal to t D j = k M j T and ct D j = k D j T , respectively, where k M j is a nonnegative integer and k D j is a positive integer. Problem parameters for one job J j are depicted in Fig 2. The quality of a solution is estimated with regard to the schedule length (makespan). Actually, it is possible to determine two makespan values: at the machine and at the depot. The machine makespan is defined as the maximum job completion time at the machine, i.e., C M max = max j=1...n {ct M j }. This makespan is important, since it allows evaluating the machine utilization, but it does not take into account the necessity of delivering all jobs to the depot after their processing. The length of the whole schedule in the considered system is determined by the depot makespan, defined as the delivery time of the last job, C D max = max j=1...n {ct D j }, or as the total duration of vehicle cycles necessary to deliver and process all jobs, C D max = max j=1...n {k D j }T . The problem of minimizing makespan on a single machine without any constraint is obviously easy. However, the existence of the transportation system makes the case intractable, as shown in the next section. The transportation system considered in the paper is a basic one: It works according to the metro transportation strategy, where a vehicle travels the system cyclically and continuously within the constant cycle time. A more flexible and popular strategy-transport on demand-allows AGVs to stop and wait at the machine and/or at the depot. Despite its inflexibility, the metro strategy is also an interesting subject of research. It can be found in production systems, namely subsystems, equipped with robot arms making cyclic movements or specialized types of conveyors. Furthermore, a very natural application field for this model comes from micro logistic systems. The AGV can represent a shuttle delivering and picking up groups of employees to/from a certain location (machine). The shuttle station repre-sents a depot. Groups have to perform jobs of given lengths. The machine makespan models the time at which the whole project (the set of jobs) is finished, while the storage makespan represents the time in which all groups of employees return to the starting point. Similar problems can be observed in mines, where workers and machines are transported with lifts. We can meet them also in factories or warehouses in the process of loading/unloading goods/parts/material, where tracks cyclically deliver empty containers for loading, and pick up filled ones.

Problem Intractability
The problem of minimizing the schedule length in the simple FMS system, consisting of one machine, one depot, and one AGV, operating according to the metro strategy, is strongly NP-hard. Within this section, we will show the transformation from the strongly NP-complete three partition problem [29] to the decision counterpart of the considered optimization problem. We will show, in Theorem 3.1, that even a special case, with equal transportation times from/to the depot (i.e., t 1 = t 2 = t), is intractable, resulting in the intractability of the general problem with distinct times (i.e., t 1 = t 2 ).

Depot Makespan Minimization Problem (DMMP)
Instance: A set of n jobs J = {J 1 , J 2 , . . . , J j , . . . , J n } characterized by processing time p j , the AGV cycle length T = t + t, constant C max . Question: Is it possible to find a schedule on the machine and the AGV simultaneously, with the depot makespan C D max not exceeding C max ? Proof Obviously, DMMP belongs to the class NP, since for a given schedule we can verify in polynomial time whether C D max ≤ C max . Furthermore, we can transform each instance of the strongly NP-complete 3PP to an instance of DMMP in the following way. We construct a set of n = 4m + 1 jobs J , which includes 3m jobs {J 1 , . . . , J 3m } with the processing times determined by the size of elements from 3PP ( p j = a j ), and additional m + 1 jobs {J 3m+1 , . . . , J 4m+1 }, each with processing time equal to 3B. The AGV cycle length T is set to B, and threshold C max takes value 4B(m + 1). We ask whether it is possible to construct a schedule for DMMP with the depot makespan C D max ≤ 4B(m + 1). Summing up, the instance transformation is as follows: n = 4m + 1, p j = a j for j = 1, . . . , 3m, p j = 3B for j = 3m + 1, . . . , 4m + 1, First, we will show that if the answer for DMMP is YES, i.e., there exists a schedule with C D max ≤ 4B(m + 1), then 3PP has a solution too. Obviously, the depot makespan C D max cannot be shorter than the time necessary to deliver the first job to the machine (t), increased by the total job processing time n j=1 p j , and the time necessary to deliver the last job to the depot (t): Thus, C D max = 4B(m + 1), and there is no idle time between two consecutive jobs on the machine. To construct a schedule on the machine without idle time, we cannot start it with any job J j ∈ {J 1 , . . . , J 3m }. Such a job would finish at the machine M at time t + p j < t + B 2 , while the second job can be delivered by the AGV to the machine at earliest at time t + T = t + B. The machine would be idle at least between t + B 2 and t + B. The first job J [1] in the sequence has to be taken from the subset {J 3m+1 , . . . , J 4m+1 }, since it finishes at the machine at ct M [1] = t + p [1] = t + 3B, giving the AGV enough time to deliver the next jobs for continuous processing by the machine.
The first job J [1] , after processing by the machine, is delivered to the depot at time ct D [1] = t + 3B + t = 4B. There are still 4m jobs to be delivered to the depot in the following time slots (i.e., the following AGV cycles). Since the AGV takes only one job at a time, it needs at least 4mT = 4m B time units to transport the remaining jobs from the machine to the depot. The last delivery to the depot finishes at earliest at time ct D [1] + 4mT = 4B + 4m B = 4B(m + 1). Taking into account that C D max ≤ 4B(m + 1), the AGV has to deliver jobs from the machine to the depot at each time slot between 4B and 4B(m + 1).
To achieve this goal, the second job cannot be taken from {J 3m+1 , . . . , J 4m+1 }. Such a job would be completed on the machine at ct M [1] + 3B = t + 3B + 3B = t + 6B, delivered to the depot at t + 6B + t = 7B, and there would be no job to be delivered to the depot from the machine at time moments 5B and 6B.
The decision counterpart of the depot makespan minimization problem is strongly NP-complete, so the considered optimization problem is strongly NP-hard. The result holds even for a special case of the problem with equal AGV transportation times from the depot (t 1 ) and to the depot (t 2 = t 1 = t). The case with t 1 = t 2 is obviously intractable too.

Worst-Case Performance Ratio
The analyzed problem of scheduling a set of jobs on a single machine with a cyclic metro transportation system is strongly NP-hard, so there exists no exact algorithm solving it, in either polynomial or pseudo-polynomial time (unless P = NP). Obviously, we can apply heuristic methods to construct a feasible schedule. List heuristics are basic and commonly used approaches to tackle with hard scheduling problems (e.g., [30]). The analysis of their behavior often delivers useful hints for constructing efficient exact and heuristic methods (e.g., [31,32]). For the considered problem, any list method is a 2-approximation algorithm. In the following theorems, we will show that the worstcase performance ratio for any list algorithm is equal to 2 and that this bound is asymptotically tight (obviously, excluding exponential time algorithms browsing all possible solutions of the problem). Proof First, we will specify the lower bound for the depot makespan in the considered system. In any feasible schedule, all n jobs have to be delivered from the depot to the machine, and-after processing-delivered from the machine to the depot. Since the vehicle transports only one job at a time, it needs at least n cycles of length T , i.e., it needs at least nT time units, to transport all jobs. Moreover, since the AGV cannot wait at the machine, it needs one more cycle to deliver the last job to the depot. Thus, from the AGV's point of view, the optimal depot makespan C D * max cannot be smaller than nT + T . On the other hand, executing all jobs requires delivering the first job to the machine, processing all jobs at least in time n j=1 p j , and delivering the last job from the machine to the depot, where both mentioned deliveries consume T time units. Thus, from the machine's point of view, the optimal depot makespan C D * max cannot be smaller than n j=1 p j + T . Summing up, the following bound holds: Now, we will determine the upper bound of the depot makespan in any list schedule.
As we have mentioned, in any feasible schedule, all jobs have to be delivered from the depot to the machine, processed and delivered from the machine back to the depot. During executing job J j by the machine, the AGV makes at most Thus, we can state that Therefore, if nT ≥ n j=1 p j , then Otherwise, if nT ≤ n j=1 p j , then Summing up, the bound

Theorem 4.2
The worst-case performance ratio for any list algorithm, Proof We will consider the instance of the depot makespan minimization problem with n = k 2 + k + 1 jobs. The set of jobs contains (k + 1) long jobs with p j = kT for j = 1 . . . (k + 1) and k 2 short jobs with p j = T k for j = k + 2, . . . (k 2 + k + 1). The optimal schedule for this instance is shown in Fig. 5. Since there is no idle time, the optimal depot makespan is equal to Figure 6 shows the list schedule generated according to SPT (shortest processing time) rule. This schedule contains idle times, since the machine has to wait for delivering short jobs with processing times T k . The depot makespan of SPT solution is equal to C D max = t 1 + k 2 T + (k + 1)kT + t 2 = 2k 2 T + kT + T.
Comparing both schedules, we have: Obviously both Theorems 4.1 and 4.2 hold also for a simpler transportation system with t 1 = t 2 = t. The general depot makespan minimization problem, with t 1 = 0 and t 2 = 0, and its special case with t 1 = t 2 are strongly NP-hard, and a list algorithm is 2-approximation heuristic. On the other hand, the special cases, with t 1 = 0 or t 2 = 0, are polynomially solvable. If t 1 = 0, i.e., all jobs are available at the machine at time zero, then the optimal schedule can be constructed by the shortest processing time list algorithm. In the SPT schedule, jobs are completed as soon as possible (the sum of job completion times on the machine is minimum), allowing the AGV for delivering jobs from the machine to the depot. If t 2 = 0, i.e., all jobs remain after their processing at the machine, then the optimal schedule can be constructed by the longest processing time list algorithm (LPT). In the LPT schedule, the machine processes longer jobs k-th group of k short jobs 2-nd group of k short jobs 1-st group of k short jobs Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.