Proceedings of the 2012 International Conference on Industrial Engineering and Operations Management Istanbul, Turkey, July 3 – 6, 2012
336
Integrating Production Scheduling and Maintenance:
Practical Implications
Laith A. Hadidi and
Umar M. Al-Turki
Department of Systems Engineering
King Fahd University of Petroleum & Minerals
Dhahran 31261, Saudi Arabia
M. Abdur Rahim
Faculty of Business Administration
University of New Brunswick
Fredericton, NB E3B 5A3, Canada
Abstract
This work discusses the integration of the scheduling decisions regarding production and maintenance operations for minimizing total cost of production including product holding cost and maintenance costs. Delays in production occur whenever the machine goes through preventive maintenance or corrective maintenance activities which in effect increases the total production cost. Preventive maintenance may be done between job processing with the purpose of reducing the chance of unplanned machine failures while processing a job. In this paper we discuss the case involving a single machine processing a number of jobs differing in processing time requirements. In practice, decisions regarding job scheduling are usually dealt with disjointly from maintenance related decisions and models that optimize such decisions are scarce in the literature. In this paper we present a modified version of the integrated model developed by Hadidi, et al. (2011) and demonstrate its utilization under various conditions to emphasize the practical implications of the model
.
Key words
Production scheduling; maintenance scheduling; preventive maintenance; deteriorating process; integrated models; unreliable machine.
1. Introduction
In this paper we consider a production system consisting of a single machine or a single production line that produces a certain type of product in batches (jobs) of different sizes and processing requirements. A fixed number of different jobs ready for processing at the beginning of the planning horizon, are to be scheduled in a way that minimizes total delays and hence total in process inventory (holding) costs. The machine may be interrupted by breakdowns that causes unplanned delays in production and hence, missing deadlines and losing customer goodwill. To reduce such unplanned interruptions, planned preventive maintenance (PM) operations is chosen to be the strategy for reducing delays that are not accounted for. Such maintenance operations are assumed to return the machine back to its original condition (as good as new) and has to be conducted between job processing, i.e., without interrupting processing a job. However, PM does not eliminate breakdowns completely and random machine breakdown may still occur. In such case, minimal repair is applied to return the machine back to its condition just before the start of processing the current job. All stoppages, planned or unplanned, adds to the total cost of production. The objective is to construct, at the beginning of the planning horizon, a schedule for processing jobs and preventive maintenance operations simultaneously such that the total expected cost is minimized. Once the schedule is set and production starts, no changes are made on that schedule.
The case considered above is quite common and exists in almost all types of production systems. The literature includes several practical cases establishing the link between production scheduling and maintenance planning such as Aksoy and Ozturk (2010) and Karamatsoukis and Kyriakidis (2010). However, the production scheduling and PM are, in practice, handled disjointly and in most cases by different entities within the organization. The
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production department, for example, constructs a production schedule, without taking machine interruptions into consideration. The schedule is forwarded to the maintenance department for scheduling PMs without altering the production schedule. Alternatively, the maintenance department sets a schedule for PM operations and leaves it to the production department to schedule their production within the time slots in which the machine is available. Such disjoint scheduling results a suboptimal schedule in terms of its total cost and also conflicts between two functional departments within the organization. Integrating the two scheduling decisions into a single global schedule resolves conflicts and minimizes the total cost.
The resulting plans of a specific function may disrupt the other function plans. For example, the maintenance function assigns scheduled shut-down intervals. These intervals will be communicated to the production unit. The suggested maintenance intervals may maximize the machine availability, but they will affect production plans. Similarly, production schedulers may have the tendency to utilize machines to their full capacity to meet demand. Under this condition, productivity may increase, but machine availability will decrease, due to having more breakdowns.
In the literature, Models developed for optimizing decisions related to production scheduling and maintenance planning simultaneously is scarce. Cassady and Kutanoglu (2003, 2005) were among the first to consider such models. Kuo and Chang (2007) considered the problem under cumulative damage failure process. Yulan et al. (2008) considered multiple performance criteria in their objective function in the integrated problem. In their review, Hadidi, et al. (2012) developed a classification scheme for problems related to production scheduling and maintenance planning and reviewed the literature related to both.
Hadidi, et al. (2011) formulated a cost model that simultaneously consider, maintenance, minimal repair and holding costs for several production jobs with the objective to minimize the expected total costs. The maintenance is assumed to be perfect and restores the machine to an ‘as good as new’ condition. Hence, after each maintenance action a new cycle will commence without being affected by the machine condition before maintenance. The expected cycle length includes: average holding time, average maintenance time, and average minimal repair time. Similarly, total expected cycle cost includes: expected maintenance cost, expected minimal repair cost and expected holding cost.
The focus of this work is to show the practical implication of the integrated model. This is done by considering two special cases. The organization of the rest of this paper is as follows. Section 2 presents model notations. Section 3 defines, in detail, the integrated problem and it’s assumptions. Section 4 presents and solves an example for the integrated problem. Section 5 shows practical implications for the model and finally Section 6 discusses the conclusions.
2. Notations
n Number of jobs to be scheduled
Pj Processing time of job j
tj Time needed to perform PM on the machine
a[0] Age of the machine at the beginning of the planning horizon
a[j] Age of the machine after the ith job in the sequence
tr Time required to repair the machine
c[j] Completion time of the ith job in the sequence
P[j] Processing time of the i
th
job in the sequence
N(τ) Number of failures in τ time units of continuous operation
m(τ) Expected number of failures in τ time of continuous operation
cp Cost of conducting PM
Cm Minimal repair cost
Qi Job order size
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3. Problem Formulation
Consider n job orders to be processed on a single machine that is subject to a Preventive Maintenance (PM)
requirement. Each job order will represent processing one batch of size Qi. Raw materials for all job orders are
assumed to be released to the shop floor at time 0. These job orders are available at time zero with no precedence constraints. Moreover, the machine is available continuously along the time horizon unless a machine breakdown occurs during job processing. If so, a minimum repair that restores the machine to its condition before breakdown is conducted. Also, the job that was interrupted by the breakdown should resume the remaining portion of the job, after
machine repair. Each breakdown will delay completion time of successive jobs by the needed time to repair tr
(assumed to be constant). To reduce the chance for machine breakdown, a PM activity can be performed before starting any job. If so, it will restore machine condition to ‘as good as new’. This PM activity will delay successive
jobs by the time of PM tp (assumed to be constant). The machine may or may not fail, causing the completion time
for each job to be stochastic. The PM decisions affect the stochastic process governing machine failure. Hence, change the expected value of job completion time E(ci).
Each maintenance action costs a fixed PM cost cp. Similarly, each break down will cost a fixed minimal repair cost
cm. It is assumed that all jobs are released to the shop floor ready to be processed at the start of the schedule. Hence,
given that Qi is the average work-in-process inventory, each job order i waiting on the shop floor will incur a
holding cost of h per unit time.
The problem would be to identify a set of PM decisions, as well as a set of job sequencing decisions in a way that reduces the total weighted expected completion times. Prior to the job starting, a decision has to be made whether to perform PM or not. If so, PM will take a constant time that will delay consequent jobs by tp. If not, the job will start at the completion time of the previous job. However, the single machine can fail during job processing. The number of failures, during job processing, is strongly affected by the machine age i.e., when the machine ages it has a higher probability to fail.
The following assumptions are also considered to complete the description of the problem. • Jobs cannot be preempted
• Jobs interrupted by failure are resumed after repair
• Machine has increasing hazard rate
• Upon failure, minimum repair is conducted without change in its age status • PM restores the machine to a `as good as new' condition
• PM and Repair times are deterministic and known in advance
• Machine breakdowns (failures) may occur randomly and independent from each other
4. Problem formulation
Let N(τ) be the number of machine failures in τtime units of continuous machine operation. Given that τpiis the time
units of machine operation over job i, then E[N(τpi)] is the expected number of machine failures during its
processing of job i. The decision of whether to perform a PM or not before the start of processing a job i, can be modeled by a binary variable yi. Let
1 if PM is performed before job
1 2, , , 0 i i y i n Otherwise = = (1)
The Gantt chart in Figure 1 resembles the time span for machine activities while processing jobs in a given sequence j1, j2, …, jn-1, jn, with hatched area representing PM decisions and unplanned breakdowns may or may not occur
during processing jobs. For a given sequence of jobs the problem reduces to finding the optimum values of yi that
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. . .
. . .
1 J J2 Ji Jn 1 c 1 c 2 c ci cn 1 J 2 c 2 J i c i J. . .
n c n J. . .
Figure 1 Gantt charts for job processing with and without maintenance for a given sequence
Let a[i] be the machine age and under the assumption that PM restores the machine to an `as good as new' condition,
i.e., machine age after PM will reduce to 0, then
a[i] = (1-yi ) a[i-1] + P[i] i=1,2, …,n (2)
Suppose the machine used to process the jobs is subject to failure, and the time to failure for the machine, is governed by a Weibull probability distribution, having shaping parameter β greater than 1. When the machine fails, it is assumed that the machine is minimally repaired i.e., the machine is restored to an operating condition, but machine age is not altered. This implies that, upon machine failure, the machine operator does just enough maintenance to resume machine function. The assumption that PM restores the machine to `as good as new' implies that PM is a more comprehensive action than repair that may include the replacement of many key parts in the machine. Also, the operation and maintenance of the machine (between two successive PM's) can be modeled as a renewal process. Due to the minimal repair, the occurrence of failures during each `cycle' of the renewal process can be modeled using a non homogeneous Poisson process. Given that the job I starts at age (1-yi ) a[i-1] and ends at age
a[i] then [ ] [ ] [ ] [ 1] [ 1] 1 (1 ) (1 ) [ ] [ 1] [ ] [ 1] [ ( )] ( ) ( ) ((1 ) ) (1 ) i i i i i i i a a P y a y a i i i i i i E N z t dt t dt m a m y a a y a β β β β β τ η η η − − − − − − − = = = − − − = −
∫
∫
(3)
Where z(t) corresponds to the hazard function. Figures 2, presents the effect of PM on the age of the machine and on the completion time of job i.
. . .
. . .
Machine Age 0 [i 1] a − [2] a a[ ]i a[ ]n. . .
1 c 1 J 2 c 2 J i c i J. . .
n c n J [0] a340
A general formula for E[c[i]] can be found as follows
{
}
{
}
[ ] [ ] [ ] 1 [ ] [ 1] 1 [ ] [ ( )] 1, 2, , ( ) ((1 ) ) 1, 2, , k i i p k k r P k i p k k r k k k k E c t y P t E N i n t y P t m a m y a i n τ = − = = + + = = + + − − =∑
∑
(4)The objective is to provide jobs sequence, as well as PM schedules to minimize the expected total cost that includes: expected holding cost (HC), expected minimal repair cost (MRC), and PM cost (PMC). Figure 3 shows the inventory level for job order i. The average holding time for a job i depend on it’s starting and completion times.
The average holding amount per unit time is equal to the sum of areas I and II (in Figure 3) over E[c[i]], i.e.
[ ] [ ] ( ) i i I II Q E c + = Batch size 0 I II [ ]i Q
. . .
1 c 1 J 2 c 2 J i c i J. . .
n c n JFigure 3 holding amount of job i
[ ] [ ] [ 1] [ ] [ ] [ ] [ ] [ ] [ ] [ 1] [ ] [ ] [ 1] [ ] [ ] [ ] [] [] [ 1] [ [ ( ) ((1 ) )]] ( ) 2 ( ) [ [ ( ) ((1 ) )]] [ [ ( ) ((1 ) )]] 1 1 2 ( ) 2 ( ( ( ) ((1 ) )) i r i k i i i i i i r i i i i r i i i i i i p k k r k k k P t m a m y a Q E c Q E c P t m a m y a P t m a m y a Q Q E c t y P t m a m y a − − − − + − − − = + − − + − − = − = − + + − −
1 i k=
∑
(5)
The holding cost is usually estimated by holding cost unit h, which is expressed in monetary unit per work piece unit per time unit. Let E(c[n]) be the expected time to complete all n jobs. Then, the expected holding cost will be,
[ ] [ ] 1
(
)
n n i iHC
h E c
Q
== ×
×
∑
(6)
Given that cm is the expected minimal repair cost per failure. Then expected minimal repair cost (MRC) is,
[ ] [ 1] 1 [ ( ) ((1 ) )] n m i i i i MRC c m a m y a − = =
∑
− −(7) Given that
c
pis the cost for each PM with a total cost of, PMC = cp ∑yi341
[ ] [ ] [ ] [ 1] 1 1 1 ( ) [ ( ) ((1 ) )] n n n n i m i i i p i i i i Min hE c Q c m a m y a − c y = = = + − − + ∑
∑
∑
Subject to [ ]i (1 i) [i 1] [ ]i 1, 2, , a = −y a − +P i = n(8)
{
}
[ ] [ ] [ ] [ 1] 1 ( ) ( ) ((1 ) ) n n p k k r k k k k E c t y P t m a m y a − = =∑
+ + − − (9) [ ] [ ] [ 1] [ ] [ ] [ ] [ ] [ 1] 1 [ [ ( ) ((1 ) )]] 1 1, 2, , 2 ( ( ( ) ((1 ) )) i r i i i i i i p k k r k k k k P t m a m y a Q Q i n t y P t m a m y a − − = + − − = − = + + − −
∑
(10) [ ] [ ] 1 1 ( ), ( ) 1, 2, , n n i j ij i j ij j j P P x Q Q x i n = = =
∑
=∑
= (11)
(12) [0] a a = 1, 2, , 0 1; , ; 1, 2, , ij i x y = or i = n j = n
The production scheduling is modeled by the decision variable xij where, th
1 if the i job performed is job j
0 Otherwise
ij x =
Constraint (8) keeps track of the age of the machine as it processes jobs and as it goes through maintenance operations. Constraints (9) and (10) define the objective function. Constraint (11) defines the ordered set of jobs with respect to their processing times and completion times. The model has two logical sets in constraint (12), states that job i cannot seize two positions at the same time and the other states that one position cannot hold more than one job. The developed model is illustrated by an example in the next section.
5. Solving the Integrated Problem
The problem formulation can be solved through one of the mathematical programming languages. In the following example GAMS language was used to input the model and the BARON solver was used to reach the optimal solution. The BARON solver is a computational system designed for solving non-convex NLP optimization problems to global optimality. When β>1, it may be practical to perform preventive maintenance on the machine in order to reduce the increasing risk of machine failure. We will illustrate the formulation and its solution in the same example presented in Hadidi et al. (2011).
Consider processing 3 job orders consisting of Q1 = Q2 = Q3 = 500 work piece. Job order 1 needs 6 minutes for each
work piece. Job order 2 needs 3 minutes for each work piece. Job order 3 needs 2 minutes for each work piece. The
machine age is a[0] =88 hours. The preventive maintenance time is tp = 5 hours. The machine failure rate follows a
Weibull distribution with the following parameters β=2, η=100. Upon failure, a minimal repair is conducted with a repair time tr=15 hours. For the Weibull distribution:
( ) 1 1 1 0 ( ) 1.50$ / / , $1500 $500 t m p t m t t dt h workpiece hour c and c β β β β η η − = = = = =
∫
1 1 1, 1 1, 2, , ; 1, 2, , n n ij ij j i x x i n j n = = = = = =∑
∑
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Job Order order size
(work piece)
Processing time per work piece (minute)
Processing time per job order (hour)
1 500 6 50
2 500 3 25
3 500 2 16.67
The Formulation will be:
3 3 3 [3] [ ] [ ] [ 1] 1 1 1 ( ) i m [ ( i ) ((1 i) i )] p i i i i Min hE c Q c m a m y a − c y = = = + − − +
∑
∑
∑
Subject to [1] 11 12 13 [2] 21 22 23 [3] 31 32 33 [1] 11 12 13 [2] 21 22 23 [3] 31 32 33 [0] [1] [1] 1 [2] [2] 2 [1] 53 26 18.67 53 26 18.67 53 26 18.67 500 500 500 500 500 500 500 500 500 88 (1 )88 (1 ) (1 P x x x P x x x P x x x Q x x x Q x x x Q x x x a a P y a P y P y = + + = + + = + + = + + = + + = + + = = + − = + − + −{
}
1 [3] [3] 3 [2] 2 [1] 1 3 [3] [ ] [ ] [ 1] 1 )88 (1 ) (1 ) (1 )88 ( ) p k k r ( k ) ((1 i) k ) k a P y P y P y E c t y P t m a m y a − = = + − + − + − =∑
+ + − − [1] [1] 1 1 [1] [1] 1 [1] [1] 1 1 [ 2 ] [ 2 ] 2 [1] [ 2 ] [ 2 ] 2 [ ] [ ] [ 1] 1 [ [ [ ( (1 )88) ((1 )88)]] 1 2( ( ( (1 )88) ((1 )88))) [ [ ( ) ((1 ) )]] 1 2 ( ( ( ) ((1 ) )) r p r r p k k r k k k k P t m P y m y Q Q t y P t m P y m y P t m a m y a Q Q t y P t m a m y a Q − = + + − − − = − + + + − − − + − − = − + + − −
∑
[ 3 ] [ 3 ] 3 [ 2 ] 3 ] [ 3 ] 3 [ ] [ ] [ 1] 1 [ [ ( ) ((1 ) )]] 1 2 ( ( ( ) ((1 ) )) r p k k r k k k k P t m a m y a Q t y P t m a m y a − = + − − = − + + − −
∑
11 12 13 21 22 23 31 32 33 11 21 31 12 22 32 13 23 33 11 12 13 21 22 23 31 32 33 1 2 3 1 1 1 1 1 1 , , , , , , , , , , , x x x x x x x x x x x x x x x x x x binary variables x x x x x x x x x y y y + + = + + = + + = + + = + + = + + =Solution: x13=x22=x31=1 and x11=x12=x21=x23=x32=x33=0 and y1=y3=1 and y2=0 with a total expected cost equal to
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6. Practical Implications of the Model
In this section we will discuss two extreme cases of the model.
Case I: Consider the case where the holding cost is so small to be negligible. The problem can be approximated with minimizing the total expected weighted completion times which was presented in Cassady and Kutanoglu (2005). The following example was shown in their work.
Consider processing 3 jobs of processing times 8, 48 and 41. Each job has a different weight w1=2, w2=10, w3=10.
Processing starts when the machine has an age of a[0]=88 time unit and preventive maintenance time tp=5 and
corrective maintenance (minimal return) time tr=15 units of time. The machine failure rate follows a Weibull
distribution with following parameters β=2, η=100. Upon failure a minimal repair is conducted. They used full
enumeration to find the optimal solution. The model results the schedule, x13=x21=x32=1 is the optimal sequence and
y1=y3=1 and total cost of 1741.
Case II: We consider as the case where maintenance cost parameters cm is much higher than other cost parameters
cp and h. It worth to mention that βis greater than 1, otherwise PM will have no effect. In this case y1=y2=…=yn=1,
to minimize repair cost to the minimum. The problem will reduce to a typical scheduling problem of finding the minimum average completion time with WSPT (Waited Shortest Processing Time) as an optimal schedule.
7. Conclusions and Future Research
Integrated models are expected to provide better savings over disjoint models. Research in integrated modeling still has great potential to further contribute to more efficient utilization of resources justified by the expected savings provided by integrated modeling. This work presented a revised version of the previous work of Hadidi et al. (2011) and demonstrated its utilizations through two practical special cases. A mathematical model was developed that gives an optimal solution for both production scheduling and preventive maintenance policy with respect to minimizing product holding cost and maintenance costs simultaneously. Both preventive maintenance and corrective maintenance are assumed as an extension to this work, this assumption can be generalized where repair and preventive maintenance times are random variables following a general distribution. The model can be studied under the effect of different repair and preventive maintenance policies. This effect can be investigated in future work.
Acknowledgment
The authors acknowledge the support of King Fahd University of Petroleum and Minerals and University of New Brunswick, Fredericton, Canada.
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