A Heuristic Based on LPT rule and a restricted B&B Method to Minimizing the Expected Makespan on a Single Machine

Considering the production and maintenance schedule simultaneously has a great attention for more than two decades. At the beginning, most of these researches treated maintenance activities as a fixed hole in the study period. Later, in the majority of them, the maintenance periods starting time were fixed but their intervals may variate. In the non-preemptive case, only few of articles that deals with the maintenance activities starting time as a decision variable. In this article, the time to starting the planed maintenance activities (PM) is a decision variable and the expected repair time is estimated according to nonhomogeneous Poisson process (NHPP). The failure function is Weibully distributed and the repair due to failure is minimal while the PM repair is perfect. Two cases were considered for the model which is aiming to minimizing the expected makespan (EMS), first, the preemptive case which was solved optimally. Secondly, a heuristic based on longest processing time rule (LPT) and a restricted branch and bound method (B&B) are provided to solve the problem for the non-preemptive case. Each of the proposed procedures uses the optimal solution properties of the preemptive case. The efficiency of proposed method in case of preemptive model is high where it gave the optimal solution. The optimal solution for the non-preemptive model can be reached using LPT in case of 𝑁 ≤ 2 ( 𝑘 + 1 ) . Finally, the results showed the robustness for B&B which their lower bound obtained by the optimal solution of the preemptive case and the upper bound is the first solution obtained by LPT.


INTRODUCTION
The traditional assumption that the machine is available during the entire time horizon may not satisfy the conditions of the real production environment.The machine unavailability has many causes, such as machine breakdowns or PM, etc.Recently, many researches have considered the machine unavailability periods in the production scheduling.However, the deterministic case, where the unavailability period's starting times and lengths are known in advance, is the main assumption in most of these studies.Actually, the random effects such as breakdowns are other causes for the machine unavailability.Taking them into account makes the solutions for production scheduling problems more practical and reasonable.On the other hand, considering both PM (deterministic) and breakdowns (probabilistic) in the production environments makes the problem more complex, especially if the PM time segment in the study period is a decision variable too.Schmidt was the first who studied the unavailability periods on parallel machines [1], and Lee introduced a wide range study for a single machine problem with several performance measures [2].The scheduling of production with unavailability constraints still has a great interest by researchers in this field so far.Graves and Lee supposed only one maintenance action during a certain period (T) which is the unavailability period for the machine.The model is semiresumable while the objectives are minimizing the total weighted completion times and maximum lateness.The problem is NP hard when the planning horizon is long in relation to T, and dynamic programming algorithms for both measures were presented.However, it is NPcomplete when the planning horizon is short in relation to T [3].A mathematical model for minimizing the total weighted tardiness on a single machine is introduced by [4].The model is resumable and the fails are predicted depending on the Weibull function.The repair restores the machine to a functional status only (minimal repair), besides, the PM restores the machine to as good as new.The PM cannot interrupt the job processing and the enumerative method has been used to solve the problem.Later, a genetic algorithm (GA) heuristic is proposed by Sortrakul, et al. to solve the model in [4] for small, medium and large problems [5].The jobs are deteriorating linearly, and the unavailability periods are fixed in [6].The problem is formulated as binary integer programming model for minimizing the makespan on a single machine and it is solved optimally by Lingo.8.Additionally, five heuristics that depend on bin packing concepts are proposed.Two mixed integer binary models, which consider the job scheduling on a single machine with constant periods of time for maintenance were discussed in [7].The objective is to minimize the makespan and a heuristic method was provided to solve the large size problems.Wang and Liu introduced a B&B method to determine the solution for a model aiming to integrating the production and PM schedule for minimizing the total weighted completion time.The failures are expected according to Weibull distribution [8].Some measures which are makespan, total completion time, maximum tardiness and number of tardy jobs are analyzed by [9].The maintenance should be performed before a certain deadline and its duration is an increasing function in the starting time.Finally, a polynomial-time algorithm for minimizing the above mentioned measures was provided.A model for minimizing the total completion time on a single machine with non-preemptive case is presented in [10].The starting time of the maintenance is known and its duration is an increasing function of the machine's load ( f ).In case of the  ′ ≥ 1, a polynomial time optimal algorithm is proposed and if  ′ < 1, an approximation scheme is suggested.An exact algorithm to solve a model, for minimizing the total expected completion time on a single machine by integrating the production and PM schedule, is addressed in [11].Failures occur according to NHPP with Weibull failure rate ( > 1).Repairs for the failures and PM are minimal and perfect respectively.The case is non-preemptive, and PM performing time is a decision variable.An integrated decision model for the production and predictive maintenance based on prognostics information so that the total expected cost is minimized.The health status and the dummy age which is subjected to the machine degradation were considered by the proposed model.The applied case study shows the efficiency of the integrated model in front of scheduling both activities separately [12].Assia, et al. addressed the integrating of production schedule, preventive maintenance planning and corrective maintenance on a single machine to minimize total energy consumption.They propose a binary mixed model that is solved by the exact B&B method and GA.The B&B method was more accurate compared with GA [13].To minimize the total completion time, Abdul Halim and Shariff combined the genetic algorithm with Monte Carlo simulation for single machine scheduling, considering preventive maintenance planning and random failure.The problem is divided into two stages: predicting the possible failures by the simulation is the first, and scheduling the jobs on the machine using a metaheuristic based on a genetic algorithm is the second.The savings achieved in the total completion time were significant [14].For a single machine with periodical unavailability periods and learning effects, Wu and Zheng proposed a two-stage binary integer programming (BIP) model to minimize the makespan in small-size problems.Further, a B&B method as a solution procedure for the mentioned model was suggested as well.A GA with the tabu search technique was employed for medium-and large-scale problems.The results show that the robustness of the two-stage BIP model is high, and the combined GA with tabu search method has strong global search ability and powerful local search ability [15].Jiag, et al. suggested a model based on game theory for integrating the production schedule and preventive maintenance plan.The failure function is according to Weibull distribution, and the corrective maintenance is imperfect.The state of the machine after repair is degraded with rate α (0 ≤ α ≤ 1).The model optimizes two performance measures simultaneously: the earliness or tardiness cost for production and the total maintenance cost.In order to solve the model, a dynamic game with three stages was applied [16].Scheduling with availability constraints has also been discussed in other shop scheduling models such as the two machines flowshop and multimachine setting.More details can be obtained from, but not limited to [17, 18, and 19].Most results show that considering the unavailability periods for production systems having more than one processor is difficult.However, [19] can solve large scale problems for the special case of identical parallel machines.In this paper, the discussed case in [11] will be studied for minimizing the expected makespan (EMS) in both preemptive and non-preemptive cases.A heuristic based on LPT rule and a restricted branch and bound method are provided to solve the problem for the non-preemptive case.Each of the proposed procedures uses the optimal solution properties of the preemptive case.

PREEMPTIVE CASE
In this case, the PM can interrupt the job processing and the optimal time to carryout it is  ̇ when the failure function of the machine is Weibully distributed as in [4] is: Where β (β > 1), η,   and   are shape parameter, scale parameter, PM time and repair time respectively.Then, the total age of the machine (total processing time plus the machine age at the beginning of the study period) can be divided into k segments with length equal to ̇ for each.However, k is most probably not integer and the last segment is a part of .Therefore, a decision about it should be made.

Lemma 1
Suppose that, the machine age is zero and T0, T1 and T2 are successive times.Let  1 be the overall machining time of jobs between T0 and T1, and  2 be the overall machining time of jobs between T1 and T2.Furthermore, suppose that the jobs are rearranged into two other different portions, such that the overall machining times in each portion are  ̅ 1 and  ̅ 2 respectively (See Figure 1).Let, Then the produced schedule by replaced the jobs in the [T0, T1] period by the jobs in the portion having total machining time  ̅ 1 , and the jobs of the [T1, T2] period by the jobs of the portion having total machining time  ̅ 2 , and nothing else is changed, has shorter EMS.
So, it equivalently to: It can be reduced farther to: Let, () =   + ( − )  Where,  is the total machining time of the jobs that sequenced in the interval  and ( -) represents the total machining time of the rest of jobs.
So, the EMS is shorter as long as the difference between the two jobs families is minimal.∎Depending on the above conclusion, the optimal EMS value can be determined for the preemptive case as follows: Let, at the beginning of the study period the age of the machine is  0 and; − 1.However, if it is not, then the number of PM times is k 1 or k 2 where: So, the time span between two consecutives preventive maintenances is The EMS which is a convex function in , will have the minimum estimated value at  1 or  2 .

𝐸𝑀𝑆 = min
Where,   and   are PM time and expected repair time, respectively.

NON-PREEMPTIVE CASE
Even though the jobs cannot be interrupted in this case, the property of the best solution when the case is preemptive is still true.The ideal time between two consecutive preventive maintenances (T) does not change.However, in the non-preemptive case, it is difficult to fulfill this period exactly.The time periods for the jobs segments (sum of jobs processing times between two preventive maintenances) will be varied from one to another, and minimizing this variation will lead to minimizing the EMS.

LPT Rule
The LPT rule introduced by Graham in [20] can play a worthy role, but it cannot guarantee the optimal solution.To illustrate the using of LPT rule in this work let: Then, the jobs including  0 will be assigned to the (k+1) segments as follows: Whenever the segment has minimum total processing times, assign that job which is not yet assigned and has the largest processing time to that segment.Two main cases can be discussed here:

𝑵 > 𝟐(𝒌 + 𝟏)
The following condition is sufficient but not necessary for the optimality of a solution.Let u be a nonnegative integer such that: If there is a pairwise disjoint partition of the jobs and the machine age at the beginning with index sets J1, J2,…,Jk+1 such that: and; where,  0 =  0 , then a solution such that the PM are between the parts, is optimal.
The LPT rule can be also used to find the upper bound for the restricted branch and bound method which will be introduced in the following section.Also, it can help to eliminate some unpromising branches.

Restricted Branch and Bound Method
The restricted branch and bound is a search method confined to portions of possible and feasible solutions which most probably contains good solutions.The target area of the solution space is the set of consecutive segments that can be generated by two different references ,  according to the following statements: • Branching Procedure: the sub-branches of a branch are defined as follows.The problems H1, H2, G1 and G2 are solved.The set of jobs up to the PM is one of the optimal solutions.T hus, at least one and at most four sub-branches are generated.One sub-branch is possible if for all j   > ,  or ̇=  and there is a subset J of jobs such that ∑   =  ∈ .
• Lower bound: node zero has the lower bound of the EMS which is the optimal solution of the preemptive case.
• Upper bound: the initial upper bound is the value of the LPT solution which is generated p rior to the beginning of the restricted branch and bound method.It is up-dated when a bette r feasible solution is obtained.
• Fathoming procedure: the node is fathomed if: a) The lower bound of the EMS is greater than the upper bound.
b) The LPT rule gives the optimal solution for the remaining jobs and then, the EMS (lowe r bound) is greater than the upper bound.
• Expected makespan at each node: the EMS at each node can be calculated as follows: where,  is the node number and; : The EMS for non-preemptive jobs in consecutive segments.
: The EMS for non-preemptive jobs in the previous node.: Time segment or the sum of processing times of the jobs assigned in one of G1, G2, H1 or H2 at node i.
: The optimal value of the EMS for the remaining loads () at node i (jobs not included in   ) as a preemptive job(s).and k is the best of  1 and  2 for the remaining load ().
Note: if the initial age of the machine is larger than zero ( 0 > 0), it is assumed as a job during construction of the 4 statements above (G1, G2, H1 and H2).Then, we deal with it as an age in EMS calculation.So, if it is assigned to a segment at node i, then By applying the rule in section 3.1, the optimal solution can be obtained by  = 3 .Thus, the number of segments are  + 1 = 4, {35+10, 30+15, 29+18,24+22} and the EMS = 200.156.

Restricted branch and bound
Lower bound obtained by the optimal solution of the preemptive case and the upper bound is the first solution obtained by LPT (EMS = 200.2338).Node (0): EMS = 200.02and infeasible.The statements in G1, H1, G2 and H2 will produce just two branches from the jobs list which are ordered in nonincreasing order.First, j1 =35; which satisfy G1 and G2.Second, j1 + j7 = 45 which satisfy H1 and H2.Therefore, the previous two segments will constitute two branches at node (1) with 35 in its segment and node (2) with 45.

CONCLUSION
In this article, the target is to minimize the EMS in preemptive and non-preemptive cases.The optimal solution in the first case was reached mathematically in a simple way and then the property of this solution has been used to get the optimal or a near optimal solution for the second case.A heuristic based on LPT rule, which can determine the optimal solution (for nonpreemptive case) in case of  ≤ 2( + 1), is provided.Additionally, a restricted branch and bound that can provide the global optimal solution is suggested.The results show the robustness of B&B method, where the lower and upper bounds were obtained by the optimal solution of the preemptive case and the first solution for the LPT rule, respectively.Finally, in the middle and large-size problems in the non-preemptive case, the heuristic based on LPT is recommended.

Figure. 1 :
Figure.1: Descriptive of case in lemma 1 =1 : be the total processing times.Assume that τ̇<< ∑    =1 .If ∑    =1is an integer multiplier of τ̇ then the PM will be carried out  times where k = (