Integrated Production Scheduling and Preventive Maintenance Planning for a Single Machine Under a Cumulative Damage Failure Process

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1 Integrated Production Scheduling and Preventive Maintenance Planning for a Single Machine Under a Cumulative Damage Failure Process Yarlin Kuo, Zi-Ann Chang Department of Industrial Engineering and Management, Yunlin University of Science and Technology, Yunlin, Taiwan 640 Received 29 August 2005; revised 28 November 2006; accepted 3 March 2007 DOI /nav Published online 4 June 2007 in Wiley InterScience Abstract: This paper finds the optimal integrated production schedule and preventive maintenance plan for a single machine exposed under a cumulative damage process, and investigates how the optimal preventive maintenance plan interacts with the optimal production schedule. The goal is to minimize the total tardiness. The optimal policy possesses the following properties: Under arbitrary maintenance plan when jobs have common processing time, and different due dates, the optimal production schedule is to order the jobs by earliest due date first rule; and when jobs have common due date and different processing times, the optimal production schedule is shortest processing time first. The optimal maintenance plan is of control limit type under any arbitrary production schedule when machine is exposed under a cumulative damage failure process. Numerical studies on the optimal maintenance control limit of the maintenance plan indicate that as the number of jobs to be scheduled increases, the effect of jobs due dates on the optimal maintenance control limit diminishes Wiley Periodicals, Inc. Naval Research Logistics 54: , 2007 Keywords: production scheduling; maintenance planning; dynamic programming; cumulative damage processes 1. INTRODUCTION Production scheduling focuses on how to allocate machine capacity to jobs, and maintenance planning focuses on how to maintain machine capacity. Clearly, these two problems are closely related and should be solved jointly in order to achieve true optimality. However, most researches in production management study these two problems separately. Unrealistic assumption that the machine never fails is often made in production scheduling researches and, on the other hand, maintenance planning studies seldom consider the impact of maintenance on system s due date performance. Starting in 1980s, researches on production scheduling with fallible machines or varied machine capacity emerged. A major part of these researches, such as Glazebrook [7, 8], Birge and Glazebrook [2], Pinedo and Rammouz [15], Adiri et al. [1], Hirayama and Kijima [10], Federgruen and Mosheiov [5], Leung and Pinedo [13], and Cai et al. [3], take a passive approach toward machine unavailability and focus on how to adjust the production schedule to circumvent the time when machines are unavailable. It is well known that conducting preventive maintenance on a machine with increasing failure rate can effectively reduce the occurrence of machine failures and increases the machine availability. Thus, Graves and Lee [9], and Lee and Chen [12] schedule the jobs Correspondence to: Y. Kuo kuoyl@yuntech.edu.tw) and one preventive maintenance simultaneous. Both papers assume that each machine must be maintained once during the planning horizon. Implicitly, their approaches on production and preventive maintenance scheduling consist of two stages: 1) find the interval during which a machine has to be maintained once to increase machine availability this stage of task is assumed to have completed in their papers) and 2) within the interval found in 1) schedule jobs and one preventive maintenance simultaneously. Cassady and Kutanoglu [4] compares the optimal total weighted tardiness under integrated production scheduling and preventive maintenance planning and that under separate production scheduling and preventive maintenance planning numerically. They assume that the up time of a machine follows Weibull distribution; the machine is minimally repaired when it fails; and the preventive maintenance restores the machine to good as new state. Their results indicate that there is an average of 30% reduction in the expected total weighted tardiness when the production scheduling and preventive maintenance planning are integrated. Thus the need to integrate production scheduling with preventive maintenance planning is clearly demonstrated in their paper. Similar to Cassady and Kutanoglu [4], this paper finds the optimal integrated production schedule and preventive maintenance plan to minimize the total tardiness of a fixed number of jobs processed on a single machine, and investigates how the optimal preventive maintenance plan interacts with the optimal production 2007 Wiley Periodicals, Inc.

2 Kuo and Chang: Production Scheduling and Maintenance Planning 603 schedule. Unlike Cassady and Kutanoglu [4], the optimal policy of the integrated production schedule and preventive maintenance plan problem is found analytically. This study focuses on production scheduling of a fixed number of jobs processed on a single machine and the machine maintenance planning during the processing of these jobs. There is no future job arrival. Thus, the scheduling problem we study here is a static one. We assume that release times of all jobs are zero and a machine will process jobs when it is operative and when there are jobs waiting to be processed. Thus production schedule will not affect the failure process of a machine. The rest of this paper is organized as follows. Section 2 studies the optimal production schedule of jobs with common processing time and different due dates. First we find the optimal production schedules for systems with no preventive maintenance and those under an arbitrary preventive maintenance plan, respectively. Then, the optimal maintenance plan of a machine exposed under a cumulative damage process and under an arbitrary production schedule is studied. Section 3 follows the same steps as Section 2 but studies systems of jobs with different processing times and different due dates. On the optimal maintenance plan, we consider two cases. In case 1, we assume that preventive maintenance can be conducted at the start of processing of each elementary job to be defined later). In case 2, we assume that we can conduct preventive maintenance on a machine only at the start of processing of each job. Section 4 studies the optimal preventive maintenance control limit when the preventive maintenance can be implemented at the start of processing of each elementary job. Section 5 conducts numerical studies on the properties of the optimal maintenance control limit. Section 6 concludes the paper. 2. JOBS WITH COMMON PROCESSING TIME AND DIFFERENT DUE DATES There are N unprocessed jobs, J 1, J 2,..., J N }, in the system at time zero. The processing time and due date of job i are p i and d i, respectively. p i = p for all i = 1, 2..., N, p R +. The machine is susceptible to failures. When the machine fails, the job under service is preempted and repair on the machine is initiated immediately. The repair, which takes a fixed t r units of time, restores the state of the machine to good as new state. At the repair completion epoch, the preempted job resumes its processing from where it was left off Optimal Production Schedule Without Preventive Maintenance When there are n jobs remaining to be processed, consider an arbitrary production schedule n =J [n], J [n 1],..., J [1] }, where J [i] is the n i + 1)th job to be processed under n. Let d [i] be due date of job J [i]. Define V n T, n ) as the total tardiness under schedule n given that the total elapsed time is T and there are n jobs remaining to be processed. To derive the properties of the optimal production schedule, we need the following Lemmas. LEMMA 1: If T 1 T 2 and d a d b then T 2 d a + T 1 d b T 2 d b + T 1 d a where x = maxx,0}. PROOF: By the convexity of x. LEMMA 2: Consider an arbitrary sample path of machine failure process and two schedules: = J [N],..., J [n+1], J [n], J [n 1],..., J [m+1], J [m], J [m 1],..., J [1] }, and = J [N],..., J [n+1], J [m], J [n 1],..., J [m+1], J [n], J [m 1],..., J [1] }. a. The sufficient condition for V N T, ) V N T, ) is T + n 0 t r + N n + 1)p d [n] + T + n 0 t r + N m + 1)p d [m] T + n 0 t r + N n + 1)p d [m] + T + n 0 t r + N m + 1)p d [n] where n 0 is number of failures occur in N n + 1)p and n 0 is number of failures occur in N m + 1)p. n 0 Z + 0} and n 0 = n 0 + u, u Z + 0}. b. When d [n] d [m], V N T, ) V N T, ). PROOF: By pairwise interchange argument. Lemma 2 holds for any arbitrary sample path of the machine failure process, by Lemma 2 and interchange arguments we have THEOREM 1: The optimal production schedule to minimize the total tardiness of a system with an arbitrary machine failures process and jobs of common processing time, common release time and different due dates is to schedule the jobs according to EDD Earliest Due Date) rule Optimal Production Schedule Under an Arbitrary Preventive Maintenance Plan Assume that at the start of processing of each job, we have the option of conducting preventive maintenance, which lasts t p, a fixed value t p t r ), to restore the machine state to good as new state. Let V n T, n) be the total tardiness under an arbitrary schedule n and an arbitrary preventive maintenance plan given that the elapsed system time is T at the scheduling epoch.

3 604 Naval Research Logistics, Vol ) LEMMA 3: Consider an arbitrary sample path of the machine failure process, an arbitrary preventive maintenance plan, and two jobs schedules: =J [N],..., J [n+1], J [n], J [n 1],..., J [m+1], J [m], J [m 1],..., J [1] } and =J [N],..., J [n+1], J [m], J [n 1],..., J [m+1], J [n], J [m 1],..., J [1] }. a. The sufficient condition for V N T, ) V N T, ) is T + n 1 t p + n 2 t r + N n + 1)p d [n] + T + n 1 t p + n 2 t r + N m + 1)p d [m] T + n 1 t p + n 2 t r + N n + 1)p d [m] + T + n 1 t p + n 2 t r + N m + 1)p d [n] ) where n 1 n 1 ) and n 2 n 2 ) are numbers of preventive maintenance and repairs in N n+1)pn m+1)p), respectively. n 1, n 2 Z + 0} and n 1, n 2 ) = n 1 + u 1, n 2 + u 2 ), u 1, u 2 Z + 0}. b. When d [n] d [m], V N T, ) V N T, ). PROOF: By pairwise interchange argument. By Lemma 3 and simple interchange argument, we have THEOREM 2: Consider a single machine system with an arbitrary machine failure process and under an arbitrary preventive maintenance plan. The optimal production schedule to minimize the total tardiness of jobs with common processing time, common release time and different due dates, is to process the jobs according to EDD rule Optimal Preventive Maintenance Plan for a Cumulative Damage Process Assume that machine failures are caused by a cumulative damage process: The machine is exposed under a stream of shocks and each shock causes some damage on the machine. The damages are cumulative and when the accumulated damage on the machine exceeds a random threshold value, the machine fails. This kind of failure process is commonly used in modeling fatigue failures of mechanical part [11, 14, 16, 17]. The damage increments are assumed to be independently and identically distributed iid) random variables and the stream of shocks is assumed to be a stationary point process. Thus, the incremental damages during identical processing times are iid. Let the incremental damage during the unit processing time, p, bea. When the accumulated damage exceeds a random threshold value D, a failure occurs and preempts the job in processing. Repair is initiated immediately after failure and restores the machine state to new. Define the state of the machine as k when the accumulated processing time of the machine since new is kp. Let ξ k be the conditional probability of having a failure in the next p processing time given that the machine has operated kp units of time since new without failure. Thus, ξ k = P k+1 j=1 a j D k j=1 a j <D). To exclude the possibility of having multiple failures during a unit processing time, p, we assume that ξ 0 = 0. This assumption is not too restrictive when both p and the initial failure rate of a machine are small. Let n = J [n], J [n 1],..., J [1] ), an arbitrary production schedule for n remaining jobs. To find the optimal preventive maintenance plan to minimize total tardiness under n, we formulate the problem as a dynamic programming with state k, T, ), where k is machine state, T is elapsed system time, and is the production schedules of unprocessed jobs at the decision epoch. Let Ṽ n k, T, n ) be the optimal total tardiness-to-go given that system state is k, T, n ) and there are n jobs remaining to be processed. Define U n k, T, n ) as the total tardiness-to-go after the preventive maintenance has been performed, if any, given that the state is k, T, n ) after the preventive maintenance and there are n jobs remaining to be processed. Recall that we assume ξ 0 = 0 thus, U n k, T, n ) = 1 ξ k ) [ T + p d [n] + Ṽ n 1 k + 1, T + p, n 1 ) ] + ξ k [ T + tr + p d [n] + Ṽ n 1 1, T + t r + p, n 1 ) ] The DP recursion becomes Ṽ n k, T, n ) = minu n k, T, n ), U n 0, T + t p, n )} n = 1, 2,..., N Ṽ 0 k, T, n ) = 0 where U n 0, T +t p, n )U n k, T, n )) is the total tardinessto-go when we choose to perform preventive maintenance not to perform preventive maintenance) on the machine at the decision epoch n. Wehave LEMMA 4: If T 1 T 2 then U n k, T 1, n ) U n k, T 2, n ) n, k. PROOF: By induction on n. LEMMA 5: Under an arbitrary production schedule, U n k, T, n ) is nondecreasing in k for any fixed n, T. PROOF: See Appendix. THEOREM 3: There exists an optimal integrated machine preventive maintenance and production schedule for a machine with failures caused by a cumulative damage process and jobs of common processing time and different due dates such that 1) jobs are scheduled according to EDD rule and 2) the optimal machine maintenance policy is of control limit type.

4 Kuo and Chang: Production Scheduling and Maintenance Planning 605 PROOF: 1) follows from Theorem 2. Let schedule E n be the EDD schedule. By Lemma 5, and Ṽ n k, T, E n ) = minu n 0, T + t p, E n ), U n k, T, E n )}, there exists a k E n, T) = infk U n k, T, E n ) U n 0, T + t p, E n )}. Thus, when the state of the system is k, T, E n ) at decision epoch n, and k k E n, T), it is optimal to maintain the machine preventively. 3. JOBS WITH DIFFERENT PROCESSING TIMES AND DIFFERENT DUE DATES Consider a set of jobs Ɣ 1, Ɣ 2,..., Ɣ N }. Processing of job Ɣ i consists of k i elementary jobs and each elementary job lasts p units of time. Due date of job Ɣ i is d i. To extend the results of Section 2, let us divide job Ɣ i into k i elementary jobs and process them in the order of Ɣ i,ki, Ɣ i,ki 1,..., Ɣ i,1 }. Let C i be the completion time of job Ɣ i and C ij be the completion time of elementary job Ɣ i,j. The completion time of Job Ɣ i is the completion time of job Ɣ i,1. To make the sum of tardiness of elementary jobs Ɣ i,ki, Ɣ i,ki 1,..., Ɣ i,1 } equal to the tardiness of job Ɣ i, let us set the due date of elementary job Ɣ i,1, d i,1 to be d i and due dates of other elementary jobs Ɣ i,j, d i,j to be M, a very large number, j = k i, k i 1,..., 2. Then the tardiness of elementary jobs Ɣ i,j, j = k i, k i 1,...,2 are all zero under any arbitrary schedule and tardiness of job Ɣ i, C i d i, equals to k i j=2 C ij M + C i1 d i, total tardiness of all elementary jobs of job Ɣ i. Essentially, we transform the original scheduling problem of N jobs with different processing times and different due dates to a revised scheduling problem of N i=1 k i jobs with common processing time and different due dates. However these two problems are not equivalent, since in the original problem, the elementary jobs of a job, say Ɣ i, are processed consecutively and in the order of Ɣ i,ki, Ɣ i,ki 1,..., Ɣ i,1 }. In the revised problem, all the elementary jobs can be freely sequenced. To make these two problems equivalent, let us proceed as follows. An elementary job, say job Ɣ i,j is said to be in sequence in schedule if the position of job Ɣ i,j in is such that schedule processes Ɣ i,ki, Ɣ i,ki 1,..., Ɣ i,1 } consecutively. Now, let the objective function of the revised problem be N ki i=1 j=1 1S ij )M + C i1 d ij, where 1S ij ) = 1if elementary job Ɣ i,j is not in sequence, and 1S ij ) = 0, otherwise. The N ki i=1 j=1 1S ij )M terms in the objective function force us to move elementary jobs Ɣ i,ki, Ɣ i,ki 1,..., Ɣ i,1 } together when searching for the optimal schedule for the revised problem. Thus, under this objective function, the optimal schedule of the revised problem has the property that all the elementary jobs are in sequence and the optimal value and sequence of the elementary jobs in the original problem and the revised problem are the same. In the following, in stead of working on the original problem, we will solve the revised problem with objective function N ki i=1 j=1 1S ij )M + C i1 d ij Optimal Production Schedule Without Preventive Maintenance Consider an arbitrary schedule of N jobs: = Ɣ [N], Ɣ [N 1],..., Ɣ [1] }. The system is said to be in stage n, m) of schedule if the machine is processing elementary job Ɣ [n],m, n = 1, 2,..., N, m = 0, 1, 2,..., k [n] i.e., there are n jobs not yet completed their processing and job Ɣ [n] has m un-processed elementary jobs), Under such definition, we have Ɣ [i],0 = Ɣ [i 1],k[i 1]. Define ˆV n,m) T, ) as the total-tardiness-to-go given that the system is in stage n, m); the elapsed time is T ; and schedule is followed. Thus, ˆV n,0) T, ) = ˆV n 1,k[n 1] )T, ). Define ˆV n T, ) = ˆV n,k[n] )T, ) and ˆV 1,0) T, ) = 0. To find the optimal production schedule we need the following two Lemmas: LEMMA 6: Consider an arbitrary sample path of machine failure process and two jobs schedules: = Ɣ [N],..., Ɣ [n+1], Ɣ [n], Ɣ [n 1], J [n 2],..., J [1] } and = Ɣ [N],..., Ɣ [n+1], Ɣ [n 1],,Ɣ [n],..., Ɣ [1] }. Let Ɣ a = Ɣ [n], Ɣ b = Ɣ [n 1] with processing times p a = k a p and p b = k b p, due dates d a and d b, respectively. Assume p a p b and the elapsed system at the start of production scheduling is T 0. I. The sufficient condition for ˆV N T 0, ) ˆV N T 0, ) is T + n 0 t r + p b M + T + n 0 t r + p a d a + T + n T + n 0 t r + p b d b + T + n 0 t r + p a M + T + n 0 t n0 k b, n 0 Z + 0} n 0 = n 0 + u, u k a k b, u Z + 0} n 0 = n 0 + v, v k b, v Z + 0} where T = T 0 + mt r + Kp, m K, m Z + 0}, K = N i=n+1 k [i], m is the number of repairs in Kp. See Fig. 1 for an illustration.) II. a. i. If d b d a or d a d b T + p a then ˆV N T 0, ) ˆV N T 0, ) ii. If d a T + k a + k b )t r + p a + p b then ˆV N T 0, ) ˆV N T 0, ) d b

5 606 Naval Research Logistics, Vol ) Figure 1. ˆV N T 0, ) vs. ˆV N T 0, ). [Color figure can be viewed in the online issue, which is available at b. i. If d b d a T + k a t r + p a then ˆV N T 0, ) ˆV N T 0, ) ii. If d b T + k a + k b )t r + p a + p b then ˆV N T 0, ) ˆV N T 0, ) d a PROOF: See Appendix. THEOREM 4: Consider N jobs Ɣ 1, Ɣ 2,..., Ɣ N } where job Ɣ i has due date d i and processing time p i = k i p and is consisted of k i elementary jobs Ɣ i,ki, Ɣ i,ki 1,..., Ɣ i,1 }. Assume d N d N 1... d 1 and p N p N 1... p 1. If the machine is maintained only when it fails, then the optimal production schedule to minimize the total tardiness is to order the jobs according to the schedule S =Ɣ N, Ɣ N 1,..., Ɣ 1 }. Thus, if d N = d N 1 =... = d 1 the shortest processing time first SPT) rule is optimal; if p N = p N 1 =... = p 1 the earliest due date EDD) rule is optimal. PROOF: Consider two jobs Ɣ a and Ɣ b in the N jobs. By Lemma 6 a)i), processing job Ɣ b then immediately followed by job Ɣ a results in smaller total tardiness than processing job Ɣ a then immediately followed by job Ɣ b if p a p b and d a d b. Since p N = min 1 i N p i } and d N = min 1 i N d i }, thus schedule job Ɣ N as the first job is optimal. At the completion time of job Ɣ N, p N 1 = min 1 i N 1 p i } and d N 1 = min 1 i N 1 d i }, thus the schedule job Ɣ N 1 as the second job is optimal. Continue in this manner, we have S =Ɣ N, Ɣ N 1,..., Ɣ 1 } is optimal Optimal Production Schedule Under an Arbitrary Preventive Maintenance Plan with Preventive Maintenance at the Start of Processing of Each Elementary Job Assume that we are allowed to conduct preventive maintenance on the machine at the start of each elementary job. Then under any arbitrary preventive maintenance plan and any arbitrary sample path of the machine failure process, we have LEMMA 7: Consider an arbitrary sample path of the machine failure process and two jobs schedules: =Ɣ [N],..., Ɣ [n+1], Ɣ [n], Ɣ [n 1], J [n 2],..., J [1] } and =Ɣ [N],..., Ɣ [n+1], Ɣ [n 1], Ɣ [n],..., Ɣ [1] }. Let Ɣ a = Ɣ [n], Ɣ b = Ɣ [n 1] with processing times p a = k a p and p b = k b p, due dates d a and d b, respectively. Assume p a p b and the system starts with elapsed time T 0. I. The sufficient condition for ˆV N T 0, ) ˆV N T 0, ) is T + n 1 t p + n 2 t r + p b M + T + n 1 t p + n 2 t r + p a d a + T + n 1 t p + n 2 t r + p a + p b d b T + n 1 t p + n 2 t r + p b d b + T + n 1 t p + n 2 t r + p a M + T + n 1 t p + n 2 t n 1 + n 2 k b, n 1, n 2 Z + 0} n 1, n ) 2 = n1 + u 1, n 2 + u 2 ), u 1 + u 2 k a k b, u 1, u 2 Z + 0} n 1, 2) n = n 1 + v 1, n 2 + v 2), v1 + v 2 k a k b, v 1, v 2 Z + 0}

6 Kuo and Chang: Production Scheduling and Maintenance Planning 607 where T = T 0 + mt p + ut r + Kp, m + u K, m, u Z + 0}, K = N n+1 k [i], m u) is the number of preventive maintenance repairs) in Kp. II. a. i. If d b d a or d a d b T + p a then ˆV N T 0, ) ˆV N T 0, ) ii. If d a T + k a + k b )t r + p a + p b then ˆV N T 0, ) ˆV N T 0, ) d b b. i. If d b d a T + k a t r + p a then ˆV N T 0, ) ˆV N T 0, ) ii. If d b T + k a + k b )t r + p a + p b then ˆV N T 0, ) ˆV N T 0, ) d a. PROOF: The proof is similar to that of Lemma 6 and is skipped. THEOREM 5: Consider N jobs Ɣ 1, Ɣ 2,..., Ɣ N } where job Ɣ i has due date d i and processing time p i = k i p and is consisted of k i elementary jobs Ɣ i,ki, Ɣ i,ki 1,..., Ɣ i,1 }. Assume d N d N 1 d 1 and p N p N 1 p 1. If the machine can be maintained preventively at the start of processing of each elementary job. Then under any arbitrary preventive maintenance plan, the optimal production schedule to minimize the total tardiness is to order the jobs according to the schedule S =Ɣ N, Ɣ N 1,..., Ɣ 1 }. Thus, if d N = d N 1 = = d 1 the shortest processing time first SPT) rule is optimal and if p N = p N 1 = = p 1 the EDD rule is optimal. PROOF: The proof is similar to that of Theorem 4 and is skipped. Note that when we are allowed to conduct preventive maintenance at the start of the processing of each elementary job, since all elementary jobs are of common processing times, results from Section 2.3 apply here. Thus we have THEOREM 6: Consider systems of jobs with different processing times and different due dates, and machine failures are caused by a cumulative damage process. When preventive maintenances are allowed at the start of processing of each elementary job, the optimal machine maintenance policy is of control limit type Production Scheduling with Preventive Maintenance at the Start of Processing of Each Job Sometimes, it is not possible or economical feasible to conduct preventive maintenance at the start of processing of each elementary job and for the easiness of management, preventive maintenances are allowed only at the start of processing of each job. If all jobs have common processing time and different due dates, the result of Section 2 applies, thus the optimal production schedule is EDD and the optimal preventive maintenance plan is of control limit type. If all jobs have different processing time and common due dates, the optimal jobs schedule is not SPT. We illustrate this result by an example. Consider a system of 5 jobs: Ɣ 5, Ɣ 4, Ɣ 3, Ɣ 2, Ɣ 1 } with processing times p 5, p 4, p 3, p 2, p 1 }=p,2p,3p,4p,5p}, p = 6 and common due date d = 90. The initial system state k, T)= 3, 0). Values of other parameters are D = 10.25, Ea) = 2.12, t r = 20, t p = 6. Then under SPT rule Ɣ 5, Ɣ 4, Ɣ 3, Ɣ 2, Ɣ 1 }) the total tardiness is The optimal schedule of this problem is Ɣ 5, Ɣ 2, Ɣ 4, Ɣ 3, Ɣ 1 } and the total tardiness is Thus, SPT is not optimal. 4. OPTIMAL CONTROL LIMIT Assume that machine failures are caused by a cumulative damage process. If we are allowed to conduct preventive maintenance at the start of each elementary job, then under arbitrary production schedule the optimal preventive maintenance are proved to be of control limit type with the control limit dependent on jobs schedule. In this section, we investigate how the optimal control limit interact with jobs schedule. Consider systems with jobs of common processing time and different due dates, assume that d n d n 1..., d 1 and p i = p, i = 1, 2,..., N. WeuseE n to denote both the EDD schedule of n jobs and its corresponding due dates. Define k E n, T) = infk U n k, T, E n ) U n 0, T + t p, E n )}, the control limit of the optimal preventive maintenance plan under EDD schedule. In the following we give a relationship between k E n, T)and d n. THEOREM 7: If k E n 1, T + p) k E n, T) then for fixed d n 1, d n 2,..., d 1, the dependence of k E n, T) = k d n, d n 1,..., d 1 ), T)on d n is as follows: a. When d n T + p, k E n, T) is a fixed value, say k 1. b. When T + p d n T + t p + p, k E n, T) is nonincreasing in d n. c. When T + t p + p d n T + t r + p, k E n, T) is nondecreasing in d n. d. When T + t r + p d n, k E n, T)is a fixed value, say k 2. e. When d n = t p + p, the value of k E n, T) is at its minimum, say k 3.

7 608 Naval Research Logistics, Vol ) Figure 2. k E n,0) vs. d n. [Color figure can be viewed in the online issue, which is available at t p t r f. If, in addition, Ṽn 11, T + t p + p, E n 1 ) U n 1 0, T + t p + p, E n 1 ) Ṽ n 1 1, T + t r + p, E n 1 ) U n 1 0, T + t p + p, E n 1 ) holds, then k 1 is the maximum value of k E n, T). Thus, k 3 k E n, T) k 1 d n. See Fig. 2 for an illustration.) PROOF: See Appendix. 5. NUMERICAL STUDIES ON THE OPTIMAL CONTROL LIMIT In this section, we study how the optimal control limit k E n,0) is affected by number of jobs, n, job due dates E n and common processing time p. In the numerical experiments, we adopt the approach designed by Fisher [6] in setting up the experiments. The due dates of jobs are independently sampled from uniforms1 τ R/2), S1 τ + R/2)), where τ is the tardiness factor, R is the due date range and S = Np is the total processing times. S1 τ) is the mean of the due dates. When S is fixed, larger τ indicates the due dates of all jobs are closer to the scheduling time and it is more likely to have tardy jobs. RS is the range of due dates. For fixed S, larger R indicates that the dispersion of due dates is larger. In the experiments, we set the value of τ at three levels: 0.25, 0.5, 0.75, and the value of R at four levels: 0.25, 0.5, 0.75, 1 and use different random number generator seeds to generate several sets of due dates for each τ, R) combination. We conduct the experiments on several parameter data sets. Because of limited space here, we present the results of several experiments only. Other experiments exhibit similar behavior as those we present below On n, E n and Optimal Control Limit For each τ, R) combination, we generate 17 due dates and order them from large to small. Depending on the value of n, we pick up the first n due dates from left to right forward picking) and right to left backward picking) and find the corresponding k E n, T)value. For example, if E = 85, 84, 81, 78, 77, 75, 75, 73, 71, 70, 68, 68, 67, 67, 66, 64) and n = 3, forward picking results in 85, 84, 81) and backward picking obtains 64, 66, 67). Table 1 gives result of one of the numerical experiments. Figure 3 summarizes how k E n, T) is affected by n and E n. The result indicates that as n increases, the effect of E n on k E n, T)diminishes. For n large enough, k E n, T)does not depends on E n at all. Thus when there are many jobs to be scheduled, the optimal maintenance control limits is independent of the due dates of all jobs Table 1) On p and Optimal Control Limit Now we study how k E n, T)is affected by p. We consider two cases here. Case 1, the damage model, the incremental damage a is independent of p. Thus, damage to a machine is mainly due to number of jobs it has processed. This case is applicable when damages occur mainly during the machine startup and/or shutdown. Case 2, the wear model, the incremental damage is proportional to p. This model is more appropriate when the damage of a machine is related accumulated machine operation time. Table 2 presents an experiment result of case 1. As p increases, eventually d n T p, thus k E n, T)converges to k1. Table 3 lists experiment results of case 2. We observe that as p increases, some k E n, T)first increase then decrease. 6. CONCLUSION In this paper, we study the integrated production scheduling and preventive maintenance planning of a single machine. When each job is consisted of some elementary jobs of identical processing time, p, we prove that under an arbitrary Figure 3. n vs. k E n, T). [Color figure can be viewed in the online issue, which is available at

8 Kuo and Chang: Production Scheduling and Maintenance Planning 609 Table 1. E n vs. k = k E n, T). D, Ea i )) = 10.25, 2.12), tr, tp, p) = 12, 8, 6), T = 0 Ñ, p) = 17, 6) k d 1, d 2,..., d n }, T = 0) τ, R) Set of due-dates 5 a , 0.25) 89, 85, 84, 81, 78, 77, 75, 75, 73, 71, 70, 68, 68, 67, 67, 66, 64) 0.25, 0.50) 97, 95, 95, 92, 88, 86, 84, 80, 78, 78, 75, 71, 71, 63, 59, 56, 53) 0.25, 0.75) 113, 113, 110, 101, 84, 80, 72, 61, 61, 58, 56, 48, 45, 42, 40, 38, 38) 0.25, 1.00) 126, 124, 119, 101, 100, 98, 88, 72, 68, 67,54, 53, 40, 39, 37, 31, 25) 0.50, 0.25) 62, 62, 57, 57, 56, 54, 54, 53, 52, 51, 50, 49, 49, 47, 45, 40, 40) 0.50, 0.50) 73, 70, 69, 64, 63, 59, 58, 58, 55, 55, 54, 53, 48, 44, 35, 28, 28) 0.50, 0.75) 86, 79, 79, 78, 77, 76, 66, 52, 51, 50, 40, 37, 35, 17, 17, 17, 16) 0.50, 1.00) 100, 98, 95, 89, 77, 75, 72, 70, 65, 65, 57, 45, 41, 39, 31, 7, 0) 0.75, 0.25) 38, 36, 33, 32, 32, 31, 29, 29, 27, 25, 23, 22, 22, 20, 14, 14, 14) 0.75, 0.50) 48, 48, 47, 43, 42, 33, 32, 26,26,25,24,23, 21, 19, 15, 8, 4) 0.75, 0.75) 62, 59, 53, 48, 41, 38, 37, 35, 35, 25, 21,18,7, 4, 3, 1, 1) 0.75, 1.00) 77, 73, 59, 51, 43, 43, 43, 38,37,36,32,21, 20, 13, 2, 1, 0) a The numbers 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, and 17 are the n values. F B F B F B F B F B F B F B F B F B F B F B F B machine failure process and arbitrary preventive maintenance plan, when all jobs have common processing time, the optimal production schedule is EDD. When preventive maintenance is allowed at the start of processing of each elementary job and when all jobs have common due date, the optimal production schedule is SPT. When machine failures are caused by a cumulative damage failure process and preventive maintenance on the machine at the start of processing of each elementary job is allowed, the optimal preventive maintenance plan is of control limit type with the control limit dependent on production schedule. Thus, if the machine can be maintained at any arbitrary epoch during processing, then as p approaches zero, the results above holds. Note that as p approaches zero, the restrictive assumption ξ 0 = 0 holds for all machines that are not badly designed and/or manufactured) and we can relax the assumption that the processing times of all jobs have to be multiples of p. However, when preventive maintenance is allowed only at the start of Table 2. p and k = k E n, T)damage model). D, Ea i )) = 6.25, 1.12), tr, tp, p) = 12, 8), T = 0 n, p) = 10, 6) k d 1, d 2,..., d 10 }, T = 0) Set of due-dates 1 a τ, R) = 0.25, 0.50) k , 54, 51, 50, 47, 46, 46, 44, 42, 34) A mind j }=34 B τ, R) = 0.50, 0.50) k , 37, 29, 28, 27, 23, 22, 19, 18, 15) A mind j }=15 B τ, R) = 0.75, 0.50) k , 23, 22, 19, 19, 18, 17, 16, 13, 1) A mind j }=1 B a The numbers 1, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, and 39 and the p values.

9 610 Naval Research Logistics, Vol ) Table 3. p and k = k E n, T)wear model). D = 12.25, tr, t p ) = 12, 8), T = 0 n, p) = 10, 12) k d 1, d 2,..., d 10 }, T = 0) 5 a τ, R) Set of due-dates 1.12 b , 0.25) 100, 88, 86, 84, 82, 81, 80, 79, 78, 75) , 0.50) 118, 108, 101, 99, 95, 92, 92, 87, 84, 69) , 0.75) 122, 117, 109, 107, 96, 87, 81, 60, 54, 48) , 1.00) 142, 137, 132, 102, 92, 66, 65, 60, 59, 41) , 0.25) 75, 75, 70, 61, 58, 48, 46, 45, 45, 45) , 0.50) 88, 75, 58, 55, 54, 46, 44, 38, 37, 30) , 0.75) 103, 98, 81, 79, 77, 70, 54, 40, 28, 20) , 1.00) 87, 85, 73, 72, 68, 59, 53, 32, 12, 10) , 0.25) 43, 43, 43, 34, 32, 31, 29, 26, 25, 20) , 0.50) 53, 45, 44, 39, 39, 35, 33, 33, 26, 3) , 0.75) 63, 63, 62, 61, 31, 30, 18, 11, 9, 10) , 1.00) 85, 72, 70, 53, 28, 9, 21, 22, 23, 24) a The numbers 5, 10 15, and 20 are the p values. b The numbers 1.12, 2.24, 3.36, and 4.48 are the E p a j ) values. processing of each job, we give a counterexample to show that SPT rule is no longer optimal when jobs have common processing time and different due dates. We also give the relationship between the optimal preventive maintenance control limit and the minimum due date of unscheduled jobs. Furthermore, we conduct numerical studies on the relationships between optimal preventive maintenance control limit and the number of jobs to be scheduled, jobs due dates and p. The numerical results indicate that as the number of jobs to be scheduled increases, the optimal maintenance control limit is independent of due dates of these jobs. Thus the production scheduling problem and preventive maintenance planning problem can be solved separately where the number of jobs to be scheduled is large. ACKNOWLEDGMENTS We would like to thank two anonymous referees for comments and suggestions that led to improved contents and simplified proofs. PROOF OF LEMMA 5: APPENDIX [ 1 ξk+1 )[T + p d U n k + 1, T, n ) U n k, T, n ) = [n] ] [ + Ṽ n 1 k + 2, T + p, n 1 ) 1 ξk )[T + p d +ξ k+1 [T + t r + p d [n] [n] ] + Ṽ n 1 k + 1, T + p, n 1 ) + Ṽ n 1 1, T + t r + p, n 1 ξ k [T + t r + p d [n] + Ṽ n 1 1, T + t r + p, n 1 ) = 1 ξ k+1 )Ṽ n 1 k + 2, T + p, n 1 ) 1 ξ k )Ṽ n 1 k + 1, T + p, n 1 ξ k+1 ξ k )[T + t r + p d [n] T + p d [n] + Ṽ n 1 1, T + t r + p, n 1 )] We will prove U n k + 1, T, n ) U n k, T, n ) 0 k, T by induction. When n = 1, U 1 k + 1, T, 1 ) U 1 k, T, 1 ) = ξ 1 ξ 0 )[T + t r + p d [1] T + p d [1] ] 0 k, T. Assume when n = m, U m k + 1, T, m ) U m k, T, m ) 0 k, T. When n = m + 1 U m+1 k + 1, T, m+1 ) U m+1 k, T, m+1 ) = 1 ξ k+1 )[Ṽ m k + 2, T + p, m ) 1 ξ k )Ṽ m k + 1, T + p, m )] + ξ k+1 ξ k )[T + t r + p d [m+1] T + p d [m+1] where Ṽ m k + 2, T + p, m ) = minu m 0, T + p + t p, m ), + Ṽ m 1, T + t r + p, m )] U m k + 2, T + p, m )} Ṽ m k + 1, T + p, m ) = minu m 0, T + p + t p, m ), U m k + 1, T + p, m )} Ṽ m 1, T + t r + p, m ) = minu m 0, T + t r + p + t p, m ), U m 1, T + t r + p, m )} minu m 0, T + t r + p + t p, m ), U m 0, T + t r + p, m )} U m 0, T + t r + p, m )by Lemma 4) Since U m k + 2, T + p, m ) U m k + 1, T + p, m ) k, T, consider the following scenarios: a. U m k + 2, T + p, m ) U m k + 1, T + p, m ) U m 0, T + t p + p, m ) b. U m k + 2, T + p, m ) U m 0, T + t p + p, m ) U m k + 1, T + p, m ) c. U m 0, T + t p + p, m ) U m k + 2, T + p, m ) U m k + 1, T + p, m )

10 Kuo and Chang: Production Scheduling and Maintenance Planning 611 Under a) we have Ṽ m k + 2, T + p, m ) = U m 0, T + p + t p, m ) Ṽ m k + 1, T + p, m ) = U m 0, T + p + t p, m ) Thus U m+1 k + 1, T, m+1 ) U m+1 k, T, m+1 ) = 1 ξ k+1 )[Ṽ m k + 2, T + p, m ) 1 ξ k )Ṽ m k + 1, T + p, m )]+ξ k+1 ξ k )[T + t r + p d [m+1] T + p d [m+1] + Ṽ m 1, T + t r + p, m )] 1 ξ k+1 )U m 0, T + t p + p, m ) 1 ξ k )U m 0, T + t p + p, m ξ k+1 ξ k )[T + t r + p d [m+1] T + p d [m+1] + U m 0, T + t r + p, m )] ξ k ξ k+1 )U m 0, T + t p + p, m ξ k+1 ξ k )[T + t r + p d [m+1] T + p d [m+1] + U m 0, T + t r + p, m )] ξ k+1 ξ k )[T + t r + p d [m+1] T + p d [m+1] + U m 0, T + t r + p, m ) U m 0, T + t p + p, m )] 0 b) and c) can be similarly proved, thus we have U m+1 k + 1, T, m+1 ) U m+1 k, T, m+1 ) 0 k, T PROOF OF LEMMA 6: I. Let B be the total tardiness of first N n jobs, Ɣ [N],..., Ɣ [n+1], T be the system at the completion of first N n jobs, and A be the total tardiness of the last n 2 jobs, J [n 2],...,,..., J [1]. Then ˆV N T 0, ) = B + k a 1 j=1 T + jp + n j r t r M + k T + p a + n ka r t ) a+k + b 1 r d a + T + jp + n j r t r M = B + k a+k b 1 j=1 j=k a+1 + T + p a + p b + n ka+k ) b + r t r d b + A T + jp + n j r t r M T + pa + n ka r t r M + T + p a + n ka r t r d a + T + pa + p b + n ka+k ) b + r t r d b + A where n j r is the number of repairs during the processing of j elementary jobs. Similarly, ˆV N T 0, ) = B + k a+k b 1 j=1 T + jp + n j r t r M T + p b + n k b r t r M + T + pa + n ka r t r d a + T + p a + p b + n ka+k ) b + r t r d b + A Let n 0 = n k b r, n 0 = nka r = n 0 + u and n 0 = nka+k b r = n 0 comparing ˆV N T 0, ) with ˆV N T 0, ), we have the result. + v. By II. a. By I), the sufficient condition for ˆV N T 0, ) ˆV N T 0, ) is T + n0 t r + p b M ) + T + n + 0 t r + p a d a + T + n T + n 0 t r + p b d b + T + n 0 t r + p a M + T + n 0 t n 0 k b, n 0 Z + 0} n 0 = n 0 + u, u k a k b, u Z + 0} n 0 = n 0 + v, v k b, v Z + 0} i. Since n 0 = n 0 + v, n 0 = n 0 + u and n 0, u, v 0, then n 0 n 0 n 0. Thus, T + n 0 t r + p a + p b T + n 0 t r + p a T + n o t r + p b. When M d a d b,wehave T + n 0 t r + p b M + T + n 0 t r + p a d a + T + n T + n0 t r + p b d a + T + n 0 t r + p a M + T + n since M d a ) T + n 0 t r + p b d b + T + n 0 t r + p a M + T + n 0 t since d a d b ) Thus ˆV N T 0, ) ˆV N T 0, ). When d a d b T + p a, define AB = [ T + n 0 t r + p b M + T + n 0 t r + p a d a + T + n ] BA = [ T + n 0 t r + p b d a + T + n 0 t r + p a M + T + n 0 t ]

11 612 Naval Research Logistics, Vol ) then AB BA = [ 0 + T + n 0 t r + p a d a + T + n 0 t ] [ r + p a + p b d b T + n0 t r + p b d b T + n 0 t ] = [ 0 + T + n 0 t ) r + p a d a + T + n 0 t )] [ r + p a + p b d b T + n0 t r + p b d b T + n 0 t )] = T + n 0 t ) r + p a d b + T + n0 t r + p b d b = T + n 0 t r + p a d b + T + n0 t r + p b d b since T + n 0 t r + p a d b 0 ) ) 0 since T + n 0 t r + p a T + n 0 t r + p b since d a d b T + p a ) Thus ˆV N T 0, ) ˆV N T 0, ) ii. If d a T + k a + k b )t r + p a + p b, then AB BA = [ 0 + T + n 0 t r + p a d a + T + n 0 t ] [ r + p a + p b d b T + n0 t r + p b d b T + n 0 t ] = [ T + n 0 t ] r + p a + p b d b [T + n0 t r + p b d b ] since T + n 0 t r + p a T + n 0 since T + n0 t r + p b T + n 0 t r + p a + p b ) Thus ˆV N T 0, ) ˆV N T 0, ) b. The sufficient condition for ˆV N T 0, ) ˆV N T 0, ) is 0 t r + p a + p b T + k a + k b )t r + p a + p b d a ) T + n 0 t r + p b M + T + n 0 t r + p a d a + T + n i. If d b d a T + k a t r + p a then T + n 0 t r + p b d b + T + n 0 t r + p a M + T + n 0 t n 0 k b, n 0 Z + 0} n 0 = n 0 + u, u k a k b, u Z + 0} n 0 = n 0 + v, v k b, v Z + 0} AB BA = [ 0 + T + n 0 t r + p a d a + T + n ] [ T + n0 t r + p b d b T + n 0 t ] = [ T + n 0 t ] [ r + p a + p b d b T + n 0 t ] since T + n0 t r + p b T + n 0 t ) r + p a T + k a t r + p a 0 since d b d a ) Thus ˆV N T 0, ) ˆV N T 0, ) ii. If d b T + k a + k b )t r + p a + p b then AB BA = [ 0 + T + n 0 t r + p a d a + T + n 0 t ] r + p a + p b d b [T + n0 t r + p b d b T + n 0 t ] = [ 0 + T + n 0 t ] [ r + p a d a T + n 0 t ] since T + n0 t r + p b T + n 0 since T + n 0 t r + p a T + n 0 t r + p a + p b ) 0 t r + p a + p b T + k a + k b )t r + p a + p b d b ) Thus ˆV N T 0, ) ˆV N T 0, ) PROOF OF THEOREM 7: Let us prove the result for T = 0 first. Ṽ n k,0,e n ) = minu n 0, t p, E n ), U n k,0,e n )} where Thus, U n 0, t p, E n ) = t p + p d n + Ṽ n 1 1, t p + p, E n 1 ) U n k,0,e n ) = 1 ξ k )[p d n + Ṽ n 1 k + 1, p, E n 1 )]+ξ k [t r + p d n + Ṽ n 1 1, t r + p, E n 1 )] =[p d n + Ṽ n 1 k + 1, p, E n 1 )]+ξ k [t r + p d n p d n + Ṽ n 1 1, t r + p, E n 1 ) Ṽ n 1 k, t p + p, E n 1 )] k E n,0) = infk U n k,0,e n ) U n 0, t p, E n )}=infk ξ k [t r + p d n p d n + Ṽ n 1 1, t r + p, E n 1 ) Ṽ n 1 k + 1, p, E n 1 )] t p + p d n p d n + Ṽ n 1 1, t p + p, E n 1 ) Ṽ n 1 k + 1, p, E n 1 )} = inf k ξ k [t } p + p d n p d n ]+[Ṽ n 1 1, t p + p, E n 1 ) Ṽ n 1 k + 1, p, E n 1 )] [t r + p d n p d n ]+[Ṽ n 1 1, t r + p, E n 1 ) Ṽ n 1 k + 1, p, E n 1 )]

12 Kuo and Chang: Production Scheduling and Maintenance Planning 613 Since the right-hand-side of inequality within the }above is a function of k, it is difficult to find the value of k E n,0) and its properties. To circumvent this difficulty, let us proceed as follows. Define By the definition of k E n,0) we have Q = [t p + p d n p d n ]+[Ṽ n 1 1, t p + p, E n 1 ) U n 1 0, t p + p, E n 1 )] [t r + p d n p d n ]+[Ṽ n 1 1, t r + p, E n 1 ) U n 1 0, t p + p, E n 1 )] ξ k E n,0) [t p + p d n p d n ]+[Ṽ n 1 1, t p + p, E n ) Ṽ n 1 k E n,0,n 1, p, E n 1 )] [t r + p d n p d n ]+[Ṽ n 1 1, t r + p, E n 1 ) Ṽ n 1 k E n,0,n 1, p, E n 1 )] = [t p + p d n p d n ]+[Ṽ n 1 1, t p + p, E n 1 ) U n 1 0, t p + p, E n 1 )] [t r + p d n p d n ]+[Ṽ n 1 1, t r + p, E n 1 ) U n 1 0, t p + p, E n 1 )] = Q Since k E n 1, p) k E n,0), thus Ṽ n 1 k E n,0 1, p, E n ) = U n 1 0, t p + p, E n 1 )) and ξ k E n,0) 1 < [t p + p d n p d n ]+[Ṽ n 1 1, t p + p, E n 1 ) Ṽ n 1 k E n,0), p, E n 1 )] [t r + p d n p d n ]+[Ṽ n 1 1, t r + p, E n 1 ) Ṽ n 1 k E n,0,n), p, E n 1 )] = [t p + p d n p d n ]+[Ṽ n 1 1, t p + p, E n 1 ) U n 1 0, t p + p, E n 1 )] [t r + p d n p d n ]+[Ṽ n 1 1, t r + p, E n 1 ) U n 1 0, t p + p, E n 1 )] = Q Since ξ k E n,0) Q, ξ k E n,0) 1 <Q, and ξ k is nondecreasing in k, we can re-express k E n,0) as k E n,0) = infk ξ k Q}. Since Q is independent of k, k E n,0) can be easily determined. By Lemmas 4 and 5, U n k, T, E n ) is nondecreasing in T for fixed n, k and nondecreasing in k for fixed n, T, Thus Hence, Ṽ n 1 1, t r + p, E n 1 ) = minu n 1 0, t r + t p + p, E n 1 ), U n 1 1, t r + p, E n 1 )} minu n 1 0, t r + t p + p, E n 1 ), U n 1 0, t r + p, E n 1 )} U n 1 0, t p + p, E n 1 ) Ṽ n 1 1, t p + p, E n 1 ) = minu n 1 0, 2t p + p, E n 1 ), U n 1 1, t p + p, E n 1 )} minu n 1 0, 2t p + p, E n 1 ), U n 1 0, t p + p, E n 1 )} = U n 1 0, t p + p, E n 1 ) Ṽ n 1 1, t r + p, E n 1 ) U n 1 0, t p + p, E n 1 ) Ṽ n 1 1, t p + p, E n 1 ) U n 1 0, t p + p, E n ) 0 Define then A = Ṽ n 1 1, t r + p, E n 1 ) U n 1 0, t p + p, E n 1 ) B = Ṽ n 1 1, t p + p, E n 1 ) U n 1 0, t p + p, E n 1 ), A B 0) Q = [t p + p d n p d n ]+B [t r + p d n p d n ]+A Consider the following conditions: a. If d n p, then k E n,0) = inf k ξ k [t p + p d n ) p d n )]+B [t r + p d n ) p d n )]+A = inf k ξ k t p + B t r + A } = k 1 } b. If p d n t p + p then k E n,0) = inf k ξ k t } p + p d n + B t r + p d n + A = inf k ξ k t } p + B d n p) t r + A d n p) Since ξ k increases in k, t r + A t p + B and d n p 0, as d n p, t p +p) increases t p +B d n p))/t r +A d n p)) decreases. Thus k E n,0) decreases in k. c. If t p + p d n t r + p, then k E n,0) = inf k ξ k } B t r + p d n A = inf k ξ k B A + t r + p d n ) B t r + p d n 0, thus as d n t r + p, t p + p) increases, A+t r +p d n) and k d 1, d 2,..., d n },0) decrease. d. If t r + p d n then k E n,0) = infk ξ k B A }=k 2 a constant). e. by b. and c. f. This condition is equivalent to tp t r B A. Since k 1 = infk ξ k t p+b t r +A }, k 2 = infk ξ k B tp+b A } and t r +A B A = Atp Btr t r +A)A thus when At p Bt r 0 tp t r B tp+b A ), t r +A B A, then k 1 k 2. Thus, k3 k E n,0) k1 for all d n. To prove the result for T>0, define A T = Ṽ n 1 1, T + t r + p, E n 1 ) U n 1 0, T + t p + p, E n 1 ) B T = Ṽ n 1 1, T + t p + p, E n 1 ) U n 1 0, T + t p + p, E n 1 ), A T B T 0) then k E n, T)= inf k ξ k [T + t } p + p d n ) T + p d n )]+B T [T + t r + p d n ) T + p d n )]+A T = inf k ξ k [t p + p d n T ) p d n T } )]+B T [t r + p d n T ) p d n T )]+A T Thus k d 1, d 2,..., d n }, T)= k d 1 T, d 2 T,..., d n T },0) and the result of T = 0 can be suitably extended to T>0. }

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