BAYESIAN MAINTENANCE POLICIES DURING A WARRANTY PERIOD

Transcription

1 Communications in Statistics-Stochastic Models, 16(1), (2000) 1 BAYESIAN MAINTENANCE POLICIES DURING A WARRANTY PERIOD Ta-Mou Chen i2 Technologies Irving, TX 75039, USA Elmira Popova 1 2 Graduate Program in Operations Research and Industrial Engineering Department of Mechanical Engineering The University of Texas at Austin Austin, TX 78712, USA ABSTRACT Most of the brand new items are released on the market with a certain type of warranty. A fixed length warranty period is assumed in this paper. A set of maintenance policies which consist of minimal repair and preventive maintenance is analyzed for the case of known and unknown failure parameters of the item s lifetime distribution. For the second case, two types of Bayesian policies are considered. An extensive simulation study comparing the performance of these maintenance policies is performed. Keywords: Warranty, Maintenance, Stochastic failures, Bayesian analysis. 1 This material is based upon work supported by the Texas Advanced Technology Program under Grant No The authors would like to thank the anonymous referee whose recommendations greatly improved the paper

2 1 Introduction Most of the new products are released to the market with a warranty. One of the widely spread warranty policies is the one with a fixed length. During that time, the manufacturer either replaces the item upon failure or repairs it for free. Designing a proper warranty program is an important marketing tool. This is especially true when a brand new item is introduced. The consumer, in general, will be uncertain about the quality of the new item. A good warranty will increase consumer s confidence. Upon purchase of the new product, a warranty is given to the consumer together with a recommended maintenance schedule (consider as an example leasing a car). From manufacturer s point of view, designing a good maintenance program for the duration of the warranty should be an important task. For instance, if one leases a car, she receives a fixed length warranty (which equals to the duration of the lease). The manufacturer is interested in providing a maintenance schedule which minimizes the total cost incurred for the length of the lease period. There are several authors who consider the maintenance, replacement or repair during a warranty period. Nguyen and Murthy [10] consider the following strategy: an item is replaced by a new one if it fails during the interval (0,W α) andis repaired if it fails during the interval (W α, W ), where W is the fixed warranty period, and α is such that minimizes the total expected cost. Jack and Dagpunar [8] consider an imperfect maintenance policy during a warranty period. They assume that each preventive maintenance can reduce the age of the machine by x units. Under a constant maintenance cost, an optimal policy is derived. In [6], the constant cost assumption is relaxed. Several authors consider the maintenance/repair policy following the expiration of the warranty from the consumer s point of view (see [4], [5] and [11]). For a system of one unit which fails randomly in time, Boland and Prochan [3] provided optimality results for a policy which preventively replaces the item every y units of time and minimally repairs it if failure occurs between scheduled 2

3 maintenance interventions. The objective function is the total expected cost over a finite horizon. We will use their result throughout this paper. There are numerous authors who analyzed optimal maintenance policies for a single-unit system (see Valdez-Flores and R. Feldman [13] for an extensive review). All of the above literature assume that the failure parameters are known with certainty. This assumption will not necessarily be correct when the product is new to the market. Only few failure data will be available. In order to plan maintenance activities, reliability engineers need to know the failure behavior of the system under study. While designing, testing and operating the equipment, they have developed an intuition about its failure behavior. This information combined with the actual observations can provide better assessment of the failure rate. Bayesian analysis is one way to incorporate this information into the decision making process. Another reason we propose this approach is that, under some reasonable assumptions, using a Bayesian scheme provides for optimal information processing (see Zellner [15]). There are few papers which investigate Bayesian maintenance policies. Most of them consider infinite time horizon problem and propose different adaptive maintenance strategies, see Wilson and Popova [14] for a detailed survey. No optimality results are obtained so far. The purpose of this paper is to present the interaction between the warranty and maintenance policies. Two new classes of Bayesian maintenance policies are proposed. The first one calls for replacement every y units of time and perform minimal repair if the item fails; every y units of time the Bayesian estimates of the failure parameters are obtained and new maintenance plan is computed based on the updated estimates. The second one also calls for replacement every y units of time and minimal repair if failure occurs but updates the Bayesian estimates and maintenance plan at each failure and renewal time. The objective function is the total expected cost for the duration of the warranty period. The manufacturer can then test different warranty length policies with their associated maintenance schedules, 3

4 compare total expected cost and decide which warranty/maintenance combination to choose. The same methodology can be applied to obtain a maintenance schedule (which is a combination of minimal repair and replacement) for any finite interval of time. Results for a Weibull failure time with known parameters are obtained in 2. Bayesian maintenance policies are presented in 3. Numerical results and the simulation study are given in 4. 2 Maintenance policy for Weibull failure time with known parameters Assume a fixed warranty period, W, defined by the manufacturer. During that time all servicing activities are free for the customer. For an item whose lifetime has an increasing failure rate (i.e. with age the probability of failure increases), it is intuitively appealing to design a maintenance program during that period which minimizes the expected total cost. We will consider the set of maintenance policies that are a combination of preventive maintenance and minimal repair. Preventive maintenance is when the item is renewed (or brought to as good as new state), whereas minimal repair is when it is brought to as good as old state. Any of the proposed policies is an element of the set of policies given by n +1 random points S 1,...,S n,s n+1,wherep [S 1 = s 1 ]=1,P [S n+1 = W ] = 1, and the following rule: at time points S i,i =1,...,n the item is renewed (for a cost of C m per item) and any failure is removed by a minimal repair (for a cost of C r ). Thus, any policy is completely determined by the sequence S 1,...,S n+1,orequivalently, by the sequence Y 1,...,Y n+1, Y i = S i S i 1, S 0 0. Let X is a random variable for the time to failure of an item. We will assume that X follows a Weibull distribution with parameters α and λ and f(x) andf (x) are its density and distribution functions, correspondingly. The expected number of minimal repairs in an interval of length y equals to R(y) = y 0 r(u)du where r(u) = f(u)/[1 F (u)] is the hazard rate (see Feldman [7],p.208). The objective 4

5 is to obtain the number of preventive maintenance points, n, and consequently, the interval lengths, Y i, between each maintenance (renewal times) such that to minimize the total expected cost over the finite horizon W. The objective function equals to z(y 1,Y 2,...,Y n,y n+1 )={C r [R(Y 1 )+R(Y 2 )+ + R(Y n )+R (Y n+1 )] + nc m } (1) A maintenance policy Y 1,...,Y n+1 is called deterministic, if P [Y i = y i ] = 1 for i =1,...,n+ 1. Boland and Proschan [3] proved that for the deterministic case the policy given by y1 = y 2 = = y n = y n+1 = W/(n + 1) is optimal. In this case we can rewrite the objective function in the following way: ( ) W z(n) =(n +1)C r R + nc m (2) n +1 For known failure parameters of the Weibull distribution, R(t) =λt α. Therefore, ( ) ( R(y) =R W n+1 = λ W α. n+1) The objective function becomes: ( ) W α z(n) =(n +1)λC r + nc m (3) n +1 It is a nonlinear objective function whose minimum is sought over the set of all integers. The optimal number of maintenance interventions equals to { n = min n N n n 1 α (n +1) 1 α C m λc r W α }. (4) The length of the maintenance interval will then equal to y = W n +1. We will refer to this policy as Policy 1. Figure 1 illustrates the above procedure when W = 8, C r =1,C m = 3, and Weibull(13/6,2) failure time. The straight line corresponds to the right sand side of the inequality given in (4) whereas the decreasing curve is its left hand side. It is obvious that the optimal n =6. Insert Figure 1 here 3 Bayesian maintenance policies The previous section assumes that the failure parameters are known with certainty and that once decision (i.e. maintenance plan) is made no corrections are done to it 5

6 over time. Such policy makes sense, if the lifetime distribution is known exactly. If this is not the case, than it should be possible to exploit the experience made during the warranty period in order to improve the policy. It would have been perfect to design a maintenance policy which hedges against all future uncertainties. One step in this direction is to approximate this optimal policy with a policy which changes/adapts as soon as the uncertainty unfolds. Such a policy allows us to change the plan one step at a time. From a statistical point of view it will be better to collect as much information as often as possible. From optimization point of view we would like to change the maintenance plan at optimal times. We will assume that the customers are cooperative and therefore no additional cost occurs at the time of maintenance plan change. We now describe the first Bayesian maintenance policy. Suppose that a customer buys a new item and a maintenance plan is given to her at that time. This initial maintenance schedule is derived using the prior opinion of design and reliability engineers. The methodology for this analysis is presented in 3.1. At the time of next scheduled preventive maintenance, the item is brought to as good as new state for a cost of C m. If the item has failed during that time, it was minimally repaired for a cost of C r. At renewal point, the posterior distribution of the failure parameters is updated given the observed failure times and a new maintenance plan is calculated. This rolling horizon approach is repeated until the end of the warranty period. The proposed methodology will produce a lower bound of the objective function for the policy which is optimal over all sample paths. More formally, we will call (Y 1,Y 2,...,) a Bayes Policy for Renewal Points, if Y 1 is determined by minimizing (1) for the given W and based on the assumed prior distribution for the parameters of the lifetime distribution. Y 2 is determined by minimizing (1) for W Y 1 based on the posterior distribution of the lifetime parameters given the observed failure times in the interval of length Y 1. 6

7 ... Y i is determined by minimizing (1) for W i 1 k=1 Y k based on the posterior distribution function of the lifetime parameters given the observed failure times in the interval of length Y i 1. It is obvious that the number of renewals following this policy is a random variable. We will refer to this policy as Policy 2. There are several issues related to any Bayesian analysis. The first one is the choice of a prior distribution. It will be assumed that through existing methods (see Bernardo and Smith [2], Chapter 3) the decision maker can come up with a good prior. We decided to choose a conjugate prior in this model for mathematical convenience. However, where possible, informative prior distribution could be elicited and incorporated into the analysis (see Kadane, Dickey, Winkler, Smith and Peters [9].) The second issue is when to update the parameters of the prior distributions, and finally, the third one is when to change the maintenance plan. In 4 wecomparethe performance of the proposed policies for several prior distributions. To investigate the second and third issues we include the following policy in our analysis. We will call (Y 1,Y 2,...,) a Bayes Policy for Failure and Renewal Points if: Y 1 is determined by minimizing (1) for the given W and based on the assumed prior distribution for the parameters of the lifetime distribution (same as Policy 2). The next decision point is min(y 1,X y 1 1 ), where Xy 1 1 is the time of first failure in (0,Y 1 ). Y 2 is determined by minimizing (1) for W min(y 1,X y 1 1 ). The next decision point is min(y 2,X y 2 1 ), where Xy 2 1 is the time of the first failure in the interval [min(y 1,X y 1 1 ),Y 2].... Y i is determined by minimizing (1) for W min(y i 1,X y i 1 1 ). The next decision point is min(y i,x y i 1 ), where Xy i 1 is the time of the first failure in the 7

8 interval [ min(y i 1,X y i 1 1 ),Y i ]. We will refer to this policy as Policy 3. The comparison between the two different strategies for update of the failure parameters and maintenance plan is presented in Determining an initial maintenance plan Assume that the lifetime of an item follows Weibull distribution with unknown parameters α and λ. We will follow the recommendation by Soland [12] for a joint conjugate prior distribution of the Weibull parameters. The prior distribution of α is assumed discrete p 0 i = Pr( α = α i), i =1,...,m where m i=1 p 0 i = 1. The prior distribution of λ given α = α i is Gamma with known parameters r 0 i and b 0 i. We will assume that the values of the hyperparameters r0 i and b 0 i are known to the decision maker. For detailed discussion of methods for estimating or obtaining them, see Bernardo and Smith [2], p.372. The probability density function of λ given α i is ( ) ( ) b g(x α i )=g x ri 0 0 r 0 i,b0 i = i x r0 i 1 e xb0 i Γ(ri 0),i=1,...,m, where 0 x<, 0<r 0 i,b0 i <, andγ( ) is the Gamma function. Now follow the procedure from 2, to obtain the initial replacement plan with a length of t 0. It should be noted that each scheduled maintenance time is a renewal point whereas each failure within a maintenance interval is not a renewal point due to the minimal repair assumption. The new objective function z 1 (n) is: (n +1)E [ ( )] W R C r + nc m, (5) n +1 where the expectation is taken with respect to the prior distributions of the parameters. 8

9 By definition: [ ( )] W E R n +1 = = m ( ) p 0 b 0 r 0 i i λ r0 i 1 i e λ ib 0 i i Γ(r 0 i=1 i ) λ i m p 0 r 0 ( ) i W αi i b 0 i=1 i n +1 ( ) W αi dλi (6) n +1 Then the objective function becomes: (n +1)C r m p 0 ri 0 i b 0 i=1 i ( ) W αi + ncm (7) n +1 We can now easily obtain the optimal number of maintenance interventions by minimizing the above objective function. This will be our initial maintenance plan. 3.2 Bayesian policy at each renewal point Here we describe in more detail the steps involved in determining the Bayes Policy for Renewal Points. The maintenance policy of interest in this section is the one which (i) performs Bayesian update of the failure parameters at each renewal point. (ii) changes (updates given the past information) the maintenance plan at each renewal point (see Figure 2). Insert Figure 2 here Bayesian update of the failure parameters The Bayesian statistical procedure to obtain estimates at each renewal point is as follows. Let x i bethetimeofthei th failure in the interval (0,y)andletu be the number of failures in this interval. At time y the likelihood equals to (see Bassin [1]): u l(x 1,x 2,...,x u λ, α) = f(x j λ, α) [(1 F (y λ, α)] j=1 = α u λ u (x 1 x 2...x u ) α 1 e λyα (8) 9

10 The posterior distribution of λ and α is proportional to the likelihood times the prior (by following the standard Bayesian scheme). The posterior distribution of the parameters is the same as the prior but with different hyperparameters (see [12] for details). Follow [12] and [1] to obtain: Let ν u j=1 x j and γ i y α i, i =1,...,m. The posterior distribution has the same form as the prior distribution with P ( α = α i x 1,x 2,...,x u ) p 1 i = p 0 ( i αu i να i b 0 ) r 0 [ i (b i Γ(ri 1)/ ) 1 r 0 i i Γ ( ri 0 ) ] mi=1 p 0 ( ) i αu i να i b 0 r 0 i i Γ ( ri 1 ) [ (b ) / 1 r 0 i i Γ ( ri 0 ) ] (9) for i =1,...,m and f(λ α i,x 1,x 2,...,x u )=f(λ r 1 i,b1 i ), where r1 i = r 0 i + u and b 1 i = b0 i + γ i. The above procedure will be applied at each renewal point. Change of the maintenance plan After getting the updated failure distribution, we will adjust the maintenance activities accordingly. Thus, the interval between future preventive maintenance activities will not be the same. The objective is to find the number of maintenance (renewals) such that the function (n +1)C r E [R(t)] + nc m is minimized. It is a rolling horizon approach and therefore we will pretend as if we would follow this new schedule for the end of the warranty period. Let y i be the interval between the (i 1) th and i th maintenance and n 0 be the number of renewals up to the updating time. At the (n 0 +1) th renewal point, the new objective function is: z(n) =C r E [R( W n 0 i=1 y i n +1 ] ) + nc m The above objective is exactly the same objective function as the original one except now it is the remaining warranty length instead of the total warranty length. Let n be the optimal solution for the objective function and the maintenance interval y new be (W n 0 i=1 y i) / (n + 1). Perform maintenance after y new more time units. 10

11 3.3 Bayesian policy at each failure and renewal time In this section we describe the steps involved in the Bayes Policy for Failure and Renewal Points. It includes the following steps: perform Bayesian update of the failure parameters and change the maintenance plan at the first failure or renewal point, whichever comes first. Assume that we obtain an initial maintenance plan following the procedure in 3.1 and let y 1 be the resulting length of the renewal cycle. If the item fails at time x 1 <y 1 then we update the failure distribution as follows. If the prior distribution is ( ) P [α = α 1 ]=p 0 1, λ 1 Γ r1 0,b0 1 ( ) P [α = α 2 ]=p 0 2, λ 2 Γ r2 0,b0 2 then the posterior parameters are r1 1 = r1 0 +1,b1 1 = b0 1 + xα 1 1 ( ) b 0 r 0 [ 1 (b 1 Γ(r1 1)/ ) 1 r Γ ( r1) ] 0 p 1 1 = p 0 1 α1 1 xα 1 1 2i=1 p 0 i α1 i xα i 1 ( ) b 0 r 0 i i Γ ( ri 1 ) [ (b ) / 1 r 1 i i Γ ( ri 0 ) ] r2 1 = r2 0 +1,b 1 2 = b x α 2 1 ( ) b 0 r 0 [ 2 (b 2 Γ(r2 1)/ ) 1 r Γ ( r2) ] 0 p 1 2 = p 0 2 α1 2 xα 2 1 2i=1 p 0 i αu i xα i 1 ( ) b 0 r 0 i i Γ ( ri 1 ) [ (b ) / 1 r 1 i i Γ ( ri 0 ) ] At this point in time, we also change the maintenance plan by minimizing [ ] W x1 y 2 E [R (x 1 + y 2 ) R (x 1 )] C r + ne C r + nc m. n with respect to t 1. Let y 2 be the time interval from the failure point to the next scheduled maintenance and x 2 be the time between the first failure x 1 and the next failure. If x 2 <y 2 then we repeat the above procedure. We should add that at time x 2 we pretend as we did not update the failure distribution at time x 1 due to the minimal repair assumption.; if x 2 >y 2 then we follow the procedure from 3.2 and change the maintenance plan at time x 1 + y 2 (which is a renewal point). Figure 3 shows how this policy operates. 11

12 4 Simulation study Insert Figure 3 here In 2 we illustrated Policy 1 for Weibull(13/6,2) lifetime, C r =1,C m =3,and W = 8. The optimal number of maintenance intervals was 6 with a corresponding expected cost of We now perform a simulation where we simulated failure times from Weibull(13/6,2) distribution and computed an estimate of the expected cost for Policy 1. For 300 simulation runs that estimate equals to with a standard deviation of 4.4. Next we compare the performance of the previously proposed policies: Policy 1: Use the average value of the parameters from the associated prior distribution to determine the maintenance plan and never change it. Policy 2: Use the prior distribution to determine an initial maintenance plan and update the plan at each renewal time. Policy 3: Use the prior distribution to determine an initial maintenance plan and update the plan at each failure and renewal time. We have three goals: the first one is to present in detail the simulation optimization steps we follow in the study, the second one is the compare the policies in the case when the real lifetime distribution is with larger (smaller) failure rate than our prior assessment, and the third one is to see whether more(less) knowledge about the failure time parameters has an impact on the policies performance. 4.1 Simulation-optimization steps We first illustrate the simulation-optimization steps which we perform in our simulation study. The example bellow follows the Bayes Policy for Renewal Points (Policy 2). Assume that the length of the warranty period W = 8, the minimal repair cost is C r = 1 and the preventive maintenance cost is C m = 3. Also assume that the 12

13 prior distribution is: Pr(α 1 =4/3) = 0.5 =p 1,λ 1 Γ(4, 2), Pr(α 2 =3)=0.5 =p 2, λ 2 Γ(4, 2). The corresponding objective function will equal to z(n) = p 1 C r (n +1)R(y α 1,r 1,b 1 )+p 2 C r (n +1)R(y α 2,r 2,b 2 )+nc m 8 8 = (0.5)(1)(n +1)( n +1 )4/3 (2) + (0.5)(1)(n +1)( n +1 )3 (2) + 3n. The initial maintenance plan is to renew every units of time and the number of renewals equals to 6. We simulate failure times from Weibull(α = 4/3, λ=2). We would like to observe and record the change of the maintenance interval. From the simulation, failures occur at times x 1 =0.297 and x 2 =0.494 (and minimal repairs are performed). The scheduled maintenance is at time (which is the renewal point). Follow the procedure from 3.2 to obtain the following posterior distribution: Pr(α 1 = 4/3) = is f(λ 1 α 1,x 1,x 2 )=Γ(6, 3.195) Pr(α 2 =3)=0.108 f(λ 2 α 2,x 1,x 2 )=Γ(6, 3.493). At the time of the first scheduled renewal (t = 1.143) the new objective function z(n) =(0.892)(1)(n +1)( n+1 )4/ (0.108)(1)(n +1)( n+1 ) n. The minimum is obtained for n=3 and the new maintenance interval is ( )/(3+1) = The interval is lengthened from to For the second maintenance interval, the simulated failure times since the last renewal point are x 3 =0.085, x 4 =0.328, x 5 =0.682, x 6 =1.128, x 7 =1.324, x 8 =1.538 and x 9 = At the end of the second interval, the new posterior distribution is Pr(α 1 =4/3) = f(λ 1 α 1,x 3,...,x 9 )=Γ(13, 5.246) Pr(α 2 =3)=

14 f(λ 2 α 2,x 3,...,x 9 )=Γ(13, 8.531) The new minimum is when n equals 1 and the next maintenance interval length is Follow the same procedure till the end of the warranty period. From the simulated failure times, there are 3 maintenance interventions and 25 minimal repairs within the warranty length. The total warranty cost for this sample path is = Policies comparison when the real failure rate is different from the prior rate We will first assume that we know the prior distribution (for instance, it is given to us by the reliability engineers) but the true (i.e. the simulated lifetime distribution) failure rate increases slower than the expected prior failure rate. Case 1: The simulated failure rate increases slower than the expected failure rate associated with the prior distribution Assume that the maintenance cost C m =3, minimal repair cost C r =1 and warranty length W = 8. The failure distribution actually simulated is Weibull (α = 4/3,λ = 2) and the number of simulation runs is N = 300. The corresponding hazard function equals to 8/3t 1/3. Assume that the prior distribution obtained from the reliability engineer is Pr(α 1 =4/3) = 0.5, λ 1 Γ(4, 2) Pr(α 2 =3)=0.5, λ 2 Γ(4, 2) Due to the prior distribution assumption we cannot compute the exact hazard function, but we can provide the expected hazard function. It equals to 4/3t 1/3 +3t 2. For t>0.2 the failure rate of the simulated distribution increases slower than the expected prior failure rate. The average cost for each policy is listed in Table 1. We performed a t-test to investigate the significance of the difference between the average cost of Policy 1 and the average costs of Policies 2 and 3. The t-statistics 14

15 Policy 1 Policy 2 Policy 3 Avg. Cost Std. Deviation Avg. of Maintenance Std. Deviation Table 1: Summary of the simulation results for Weibull(4/3,2) lifetime. Policy 1-2 Policy 1-3 t-statistic % Significant Yes Yes Table 2: Test for the significance of difference between the maintenance costs of Policy 1 and Policies 2 and 3 and Weibull(4/3,2) lifetime. are given in Table 2. We can see that the cost of Policy 1 is significantly higher than the cost of Policy 2 and Policy 3. The number of maintenance interventions performed for Policy 2 and Policy 3 is also less than Policy 1. For Policy 1, once the initial maintenance plan is determined, it is never changed. Policies 2 and 3 update the posterior distribution and adjust the maintenance plan accordingly. The actual failure rate increases slower than the failure rate of the prior distribution. The failure rate of the prior distribution is overestimated. The number of the needed maintenance interventions is also overestimated. If the actual simulated failure distribution is used to determine the maintenance plan, the optimal number of maintenance is 2 with expected service cost of Policy 2 and Policy 3 can learn from the observations and adjust the plan. That is the reason that their number of maintenance interventions is less than the number of maintenance of Policy 1. In this case Policy 2 seems to outperform Policy 3. It is intuitively correct as the simulated failure rate is relatively flat and therefore the item does not age rapidly. 15

16 Policy 1 Policy 2 Policy 3 Avg. Cost Std. Deviation Avg. of Maintenance Std. Deviation Table 3: Summary of the simulation results for Weibull(3,2) lifetime. Policy 1-2 Policy 1-3 t-statistic % Significant Yes Yes Table 4: Test for the significance of the difference between the maintenance costs of Policy 1 and Policies 2 and 3 and Weibull(3,2) lifetime. Case 2: The simulated failure rate increases faster than the expected failure rate of the prior distribution Like the previous case, assume that the maintenance cost is C m =3, minimal repair cost is C r =1 and warranty length is W = 8. However, the failure distribution that is actually simulated in this case is Weibull(α =3,λ = 2) and the number of simulation runs is N = 300. In this case the actual failure rate equals to 6t 2. The prior distribution is Pr(α 1 =4/3) = 0.5, λ 1 Γ(4, 2) Pr(α 2 =3)=0.5, λ 2 Γ(4, 2), with expected failure rate of 4/3t 1/3 +3t 2. The estimates of the expected costs for each policy are listed in Table 3 and the t-test results are given in Table 4. As in Case 1, the cost of Policy 1 is significantly higher than the cost of Policy 2 and 3. The actual failure rate increases faster than the expected failure rate associated with the prior distribution. As a result, the expected failure rate of the prior distribution is underestimated. The number of the needed maintenance is also underestimated. If we used the simulated failure distribution to determine the maintenance plan, the optimal number of maintenance interventions is 8 with 16

17 Policy 1 Policy 2 Policy 3 Avg. Cost Std. Deviation Avg. of Maintenance Std. Deviation Table 5: Summary of the simulation results for set I and Weibull(13/6,2) lifetime. expected warranty cost of In this case Policy 3 seems to be the best one to choose. The item ages rapidly and therefore it will be better to change the maintenance plan faster in order to capture these quick changes. 4.3 The effect of the degree of knowledge about the prior distribution In this case we are investigating the impact of more or less knowledge about the prior distributions on the performance of Policy 2 and Policy 3 when C m =3, C r =1, and W = 8. The comparison is made using two different sets of prior distributions. Set I Set II Pr(α 1 =4/3) = 1/3,λ 1 Γ(4, 2) Pr(α 1 =4/3) = 1/8,λ 1 Γ(20, 10) Pr(α 2 =13/6) = 1/3,λ 2 Γ(4, 2) Pr(α 2 =13/6) = 3/4,λ 2 Γ(20, 10) Pr(α 3 =3)=1/3,λ 3 Γ(4, 2) Pr(α 3 =3)=1/8,λ 3 Γ(20, 10) The choice of these prior distributions is not arbitrary. Both of them have the same expected value. That way, we can think that on average both prior opinions are the same. The first prior distribution is relatively flat which might be an indication of less confidence in the prior knowledge and the second one is much steeper showing more confident prior knowledge about the system. (i) Failures are simulated from Weibull(α = 13/6,λ = 2) In this case, the failure distribution is simulated from a Weibull distribution with parameters α =13/6 andλ = 2 and the number of simulation runs is N = 300. Table 5 lists the average costs for each policy when we use the prior set I and Table 6 the corresponding results for set II. The costs of Policies 2 and 3 for set I are larger than the corresponding costs 17

18 Policy 1 Policy 2 Policy 3 Avg. Cost Std. Deviation Avg. of Maintenance Std. Deviation Table 6: Summary of the simulation results for set II and Weibull(13/6,2) lifetime. Policy 1 Policy 2 Policy 3 Avg. Cost Std. Deviation Avg. of Maintenance Std. Deviation Table 7: Summary of the simulation results for set I and Weibull(4/3,2) lifetime. for set II. This shows that if we are more confident about parameters of the failure distribution and the true distribution is closer to what we think, the cost would be less than the cost of the flat prior distribution. This makes perfect sense since the fluctuations of samples would have a larger impact on the prior distribution with less confidence, the new adjusted maintenance plan would be more volatile, and move it farther away from what it should be than the one with higher confidence. (ii) Failures are simulated from Weibull(α = 4/3,λ = 2) Table 7 shows the average cost for each policy with set I prior distributions. For set II, the average cost for each policy is given in Table 8. In this case we get the opposite relationship to the one in (i). This shows that if we are confident about parameters of the failure distribution but the true Policy 1 Policy 2 Policy 3 Avg. Cost Std. Deviation Avg. of Maintenance Std. Deviation Table 8: Summary of the simulation results for set II and Weibull(4/3,2) lifetime. 18

19 distribution is not closer to what we think, the cost would be more than the cost of the flat prior distribution. From the simulation study, it is clear that neither Policy 2 nor Policy 3 is dominant. Both policies use Bayesian method to update the failure distribution and adjust the maintenance plan. For Policy 2, the maintenance plan is updated at renewal time only, while the maintenance plan is updated at each failure and renewal time for Policy 3. To implement the maintenance plan under warranty period, customers cooperation is very important. From the consumer s point of view, it is preferable to know the time for the next maintenance intervention well in advance so that they can prepare for it. Policy 2 should be a better choice in this regard. 5 Summary We introduce an intuitive set of policies which learn from past history and adapt accordingly. To compare their performance we made an extensive simulation study. A method to generate failures under minimal repair assumption is presented. Inverse transform algorithm and restricted sampling techniques are also used. Several examples with different failure rates and prior distributions are given. When the true failure distribution is different from the prior distribution, Bayesian maintenance policies would update the posterior distribution and change the maintenance plan adaptively. The effect of the degree of knowledge about the prior distribution is illustrated in Case 3. If the reliability engineers have more confidence about the prior distribution and the failure distribution is close to what reliability engineers think, then the cost is less than the cost with a flat prior. However, if the original prior distribution is not close to the true failure distribution, the cost for a flat prior distribution is less than the centered prior distribution. The empirical study suggests that there exists a trade-off between knowledge accumulation and optimal decision making. The natural conjecture that changing the maintenance plan continuously while learning system s failure behavior should 19

20 be a superior policy appears not to be true. From an implementor s point of view, Policy 2 should be a better choice since it requires less updates and it would be more easily accepted by the customers. References [1] William M. Bassin. A bayesian optimal overhaul interval model for the Weibull restoration process case. Journal of the American Statistical Association, pages , [2] J. M. Bernardo and A. F. M. Smith. Bayesian Theory. John Wiley and Sons, [3] P. J. Boland and F. Proschan. Periodic replacement with increasing minimal repair costs at failure. Operations Research, pages , [4] K-J. Chung. Optimal repair cost limit for a customer following expiration of a warranty. Microelectronics and Reliability, pages , [5] J. S. Dagpunar and N. Jack. Optimal repair cost limit for a consumer following the expiration of a warranty. IMA Journal of Mathematics Applied in Business and Industry, pages , [6] J. S. Dagpunar and N. Jack. Preventive maintenance strategy for equipment under warranty. Microelectronics and Reliability, pages , [7] R. Feldman and C. Valdez-Florez. Applied probability and stochastic processes. PWS Publishing Company, [8] N. Jack and J. S. Dagpunar. An optimal imperfect maintenance policy over a warranty period, [9] J. B. Kadane, J. M. Dickley, R. L. Winkler, W. S. Smith, and S. C. Peters. Interactive elicitation of opinion for a normal linear model. Journal of the American Statistical Association, pages ,

21 [10] D. G. Nguen and D. N. P. Murthy. An optimal policy for servicing warranty. Journal of the Operational Research Society, pages , [11] I. Sahin and H. Polatoglu. Maintenance strategies following the expiration of warranty. IEEE Transactions on Reliability, pages , [12] R. M. Soland. Bayesian analysis of the Weibull process with unknown scale and shape parameters. IEEE Transactions on Reliability, pages , [13] C. Valdez-Florez and R. Feldman. A survey of preventive maintenance models for stochastically deteriorating single-unit systems. Naval Research Logistics, pages , [14] J. G. Wilson and E. Popova. Bayesian approaches to maintenance intervention. In Proceedings of the Section on Bayesian Science of the American Statistical Association, pages , [15] A. Zellner. Optimal information processing and Bayes theorem. American Statistician, pages ,

22 n (1-a) - (n+1) (1-a) C m /(λc r W α ) n n(a) Boundary

23 Initial plan y y y... 0 W observe Change the plan at time y y new y new Figure 2. Bayesian maintenance policy at each renewal point.

24 Initial plan y 1 y 1 y W Failure occurs x 1 y 1.. Update the plan y 2.. Figure 3. Bayesian maintenance policy at each failure and renewal time, whichever occurs first.