BAYESIAN NONPARAMETRIC ANALYSIS OF SINGLE ITEM PREVENTIVE MAINTENANCE STRATEGIES

Transcription

1 Proceedings of 17th International Conference on Nuclear Engineering ICONE17 July 1-16, 9, Brussels, Belgium ICONE BAYESIAN NONPARAMETRIC ANALYSIS OF SINGLE ITEM PREVENTIVE MAINTENANCE STRATEGIES Dmitriy Belyi Operations Research and Industrial Engineering Cockrell School of Engineering Paul Damien Department of Information, Risk, and Operations Management Red McCombs School of Business Ernie Kee South Texas Project Nuclear Operating Company Risk Management Wadsworth, TX, USA David Morton Operations Research and Industrial Engineering Cockrell School of Engineering Elmira Popova Operations Research and Industrial Engineering Cockrell School of Engineering Drew Richards South Texas Project Nuclear Operating Company Wadsworth, TX, USA ABSTRACT This work addresses the problem of finding the minimal-cost preventive maintenance schedule for a single item. We develop an optimization algorithm that reduces the computational effort to find the optimal schedule. This approach relies on the item having an increasing failure rate, which is typical, and employs a Gibbs sampling algorithm to simulate from the failure rate distribution using real data. We also analyze the case when the item has a bathtub failure rate; we develop techniques that lead to an algorithm that finds an optimal schedule for this case as well. We then analyze the effectiveness of our approach on both artificial and real data sets from the South Texas Project nuclear power plant. NOMENCLATURE PM Preventive maintenance (restore functioning item to new condition CM P trip C cm C trip Corrective maintenance (restore failed item to its condition just before failure Probability of item failure causing a loss of production Cost of preventive maintenance Cost of corrective maintenance Downtime cost INTRODUCTION Proper and safe operation is a major concern for any nuclear power plant. An inefficiently functioning plant costs more to operate and produces less energy than it potentially could. One of the main factors that greatly influences a plant s efficiency is the maintenance of its subsystems and components. Should some components fail too frequently, the plant will incur losses resulting from a drop in production. On the other hand, replacing or repairing components too often may result in a substantial increase to the cost of operating the plant. A maintenance policy 1 Copyright c 9 by ASME

2 is a decision-making rule that defines when and how, an item, or system, will be maintained. We seek an optimal maintenance policy from a class of time-based maintenance policies. In other words, an optimal maintenance policy for a plant balances safe and consistent operations with cost-effective upkeep by using a pre-specified utility function that may consider operating cost, production levels, safety standards, as well as many other factors. Finding an optimal maintenance policy may be difficult. Decision makers have to deal with the fact that the behavior of many components may be stochastic. Also, the system requiring maintenance may be very complex and involve many interdependencies, which may be hard to model and even harder to optimize its maintenance. This paper addresses modeling the stochasticity of a single component without delving too much into modeling component interdependencies. Specifically, we will discuss modeling a component s failure distribution and making optimal maintenance decisions. item has survived until some time t; the probability that the item fails before t + t is approximately z(t t. In other words, z(t t Prob(t < Failure Time < t + t Failure Time > t, (3 with a more precise approximation as t decreases. Now, let N(t 1,t represent the number of item failures between times t 1 and t, and let E[N(t 1,t ] be the expected number of such failures. An interesting property of the failure rate function (see [1] is that t t 1 z(udu = E[N(t 1,t ]. (4 For more in-depth background information, see a textbook on reliability theory, e.g. []. BACKGROUND Consider the problem of developing the optimal maintenance policy for a single item. We say that the item has failure distribution f if the time to failure from when the item is new has probability density f, i.e. the probability that the item fails before time t is t Prob(Failure Time t = f (udu = F(t. (1 In this paper we assume that the item s failure distribution is time-stationary, meaning the failure distribution does not change with time. Then, since PM restores the item to a new condition and CM restores the item to a good-as-old condition, the item s failure distribution f depends only on the last time the item was new. The assumption of a stationary failure rate implies that the failure distribution f does not change throughout the time horizon L, which may not hold true if L is very long. The focus of this study, however, is to provide the optimal maintenance interval for a number of months in advance. As the plant ages and more failures are recorded, the maintenance intervals will be re-estimated using newly available data to accommodate the changing failure distribution. Our plans for future research include the possibility that this failure time is not stationary so that we can incorporate plant aging in our model. The failure rate function, z(t, is defined as f (t z(t = t f (udu. ( A more intuitive definition of the failure rate is a conditional probability of an item s failure in a time interval. Say that an 1 INCREASING FAILURE RATE Consider the problem of scheduling exactly one PM for an item with a strictly increasing, stationary, and continuous failure rate between times and L, where L is the time horizon (i.e., we don t care what happens to the item after time L. We assume that the item is in a new state at time, and is again restored to a new state by PM. Should an item fail, it is restored to its condition just before the failure (i.e., good-as-old state by corrective maintenance (CM. During a failure, an item may trigger a larger system failure or trip with probability P trip. Should this happen, instead of a simple CM we must address the entire system trip and incur the cost of downtime, which is more expensive than CM. We also assume that PM, CM, and restoration of a downed system are performed instantaneously. Suppose that we wish to optimize our maintenance schedule by minimizing the expected cost, and we can only perform a single maintenance. Letting C = [P trip C trip + (1 P trip C m ], we can express this problem as: [ min X(T = Cpm +C (E[N(,T ] + E[N(,L T ] ], (5 T A where, nominally, A = [,L]. However, the set A can also impose constraints on the optimal maintenance time, for instance, it can contain only time periods corresponding to planned outages. In our current presentation we do not assume any restrictions. Model (5 requires that we perform a single maintenance. If we find the optimal T, which solves model (5 then we can compare X(T with E[N(,L], the expected cost of not performing a PM and choose the alternative with smaller expected cost. The objective function in model (5 is convex in T. We can Copyright c 9 by ASME

3 failure rate cannot have varying intervals between PM, and must therefore have equal intervals between PMs. If we write the problem for a general number of PM intervals, n 1, we can formulate: Figure 1. PM OVER INTERVAL [,L]. show this by rewriting it in terms of the failure rate, z(t. The objective function in (5 becomes: ( T L T X(T = +C z(tdt + z(tdt. (6 Taking the second derivative with respect to T yields: d X(T dt = C[z (T + z (L T ], (7 which must be greater than as C >, and z (t > for any t since the item has increasing failure rate. Since the objective function is convex, we can solve for the optimal T by taking the derivative of X(T and setting it to : dx(t = dt (8 C(z(T z(l T = (9 z(t = z(l T. (1 Since z(t is increasing, z(t = z(l T is satisfied if and only if T = T = L/. Therefore, if it is profitable to perform PM on the time interval [,L], the optimal time is directly in the middle of the interval. Now suppose that it is profitable to schedule multiple PM during the time horizon L. We can show that the optimal schedule will have equally spaced PMs. Consider a schedule with at least two adjacent but different intervals between PM. In other words, suppose that PM is performed at times T 1 and T after a renewal at some time T (see Figure 1. Since we assume that the failure rate is stationary, if we only change the time of the PM performed at T 1 and keep the renewals at times T and T, we will only affect the interval from T to T. Using the result that the optimal time for a single PM in a given interval is directly in the middle, we can shift the PM at T 1 to be directly in the middle of [T,T ] and arrive at a lower cost schedule over that subinterval. But since this does not affect the cost outside this subinterval, then shifting PM at T 1 decreases the total cost of the maintenance schedule. Thus, the optimal schedule with multiple PMs (assuming an increasing, stationary min (n 1 + n 1 n min T 1,...,T n, i T i =L i=1 E(,T i. Knowing the PMs are equally spaced at optimality reduces this problem to: [ min (n 1Cpm + n E(,L/n ]. n 1 Ignoring the fact that we must have T = L/n, we can write a relaxation of this as: [ min (LT LT 1 E(,T ]. T Differentiating with respect to T, and setting the result to zero yields the condition: = T z(t EN(,T, (11 which has a single root as the right-hand side of (11 is increasing given that z (T >. Given the failure rate function, z(t, we may be able to solve (11 analytically for T. If not, we can numerically compute T using the bisection algorithm that follows: 1. Set T = L.. Evaluate the right-hand side of Eqn.(11. (a If < T z(t EN(,T (1 then PM is profitable in an interval of length T. If T T / < ε, return T as the optimal interval. Otherwise, set T = T, and repeat step. (b If > T z(t EN(,T (13 then PM is not profitable in an interval of length T. If T = L, then it is not profitable to perform PM in the given time horizon at all. If T T +T < ε, return T as the optimal interval. Otherwise, set T = T +T, and repeat step. 3 Copyright c 9 by ASME

4 (c If Table 1. PARAMETERS OF EXAMPLE COMPONENT = T z(t EN(,T (14 then PM is profitable in an interval of length T, and T happens to be the optimal interval; return T. There has been work employing similar models (see, for example, [3] or [4], where the authors use various techniques to decide when to schedule PM. However, if an item has an increasing failure rate (IFR, this result (that optimally scheduled PM will be equally spaced greatly reduces the difficulty in solving this problem. For more details on the proofs for Increasing Failure Rate case, see [5]. Parameter Value $8, C cm $97,997.7 C trip $434,978.5 P trip.73 Modeling Increasing Failure Rate For a Single Item To model an increasing failure rate as an extended gamma process, we divide our time horizon into n subintervals s i, 1 i n, and consider δ i, the increment of the hazard rate over the interval s i. In other words, z(t k i=1 δ i, where k is the largest integer such that k i=1 s i t. [6] show that if we denote as the vector of all the δ i, we can use the following Gibbs sampling technique to sample from : [ data,e,m] gamma ID distribution [e data,,m] truncated exponentials [m data,e, ] multinomials where e and m are auxiliary variables. A more detailed description of this technique is available in Damien et al. (1996. This Gibbs sampling technique falls under the umbrella of Bayesian non-parametric methods. It gives a probabilistic estimate of the failure rate distribution; the mean and median estimates are very close to the maximum likelihood estimation of the failure rate, but this technique also lets us consider, for example, the worst 5% scenario of the failure rate. We turn to an example to illustrate these techniques. Figure shows the simulated failure rate for a component from STP. Here, we use the median estimate of the failure rate, and we see that it is very close to the maximum likelihood estimate of the failure rate. Using,55 days as L, and using parameters specified in Table 1, we estimate our expected cost to be approximately $84,43 over the course of L. The optimal schedule is to perform one PM after 1,75 days. BATHTUB FAILURE RATE We now consider the case when a component has a bathtub failure rate. For our analysis, we say that an item has a bathtub Figure. ACTUAL AND SIMULATED FAILURE RATES failure rate if the item has a strictly decreasing failure rate on the subinterval [, I], I < L, and has a strictly increasing failure rate after time I. I is a time such that the failure rate is decreasing before I and increasing afterward. We can once again express the problem of scheduling a single PM as: min T X(T = +C (E[N(,T ] + E[N(,L T ]. (15 Taking the second derivative again yields d X(T dt = C[z (T + z (L T ] (16 and since z (t is not always greater than, we cannot use the convexity result as in the increasing failure rate case. Proposition 1. Consider ˆT [, L/]. Then, scheduling a PM at L ˆT yields an identical cost; i.e. X( ˆT = X(L ˆT Proof: Writing out the cost function ( in terms of the hazard rate z(t, we get X( ˆT = +C ˆT z(tdt + L ˆT z(tdt, and ( X(L ˆT = +C L ˆT z(tdt + ˆT z(tdt = X( ˆT. Proposition. At the optimal solution T, z(t = z(l T. 4 Copyright c 9 by ASME

5 Proof: Suppose that z(t z(l T. Without loss of generality (see Proposition 1, we will assume that z(t < z(l T. Then, since z(t is continuous, there exists t > such that z(t + t < z(l T t, and therefore: and T j 1 and decreasing the interval between T i and T i 1 by the same amount will not affect the expected maintenance cost of the rest of the intervals. Since z(t is continuous, there exists t > such that z(t j T j 1 + t < z(t i T i 1 t, and therefore: T + t T z(tdt < L T L T t z(tdt. (17 Tj T j 1 + t T j T j 1 z(tdt < Ti T i 1 T i T i 1 t z(tdt. (1 Rewriting the costs in terms of z(t yields: Rewriting the costs of the two subintervals in terms of z(t yields: ( T L T Cost T = +C z(tdt + z(tdt (18 ( Ti T i 1 Tj T j 1 Cost Ti,T j = +C z(tdt + z(tdt ( ( T T + t > +C z(tdt + z(tdt T ( Ti T i 1 Ti T i 1 > +C z(tdt z(tdt T i T i 1 t ( L T L T +C z(tdt z(tdt L T t (19 ( Tj T j 1 Tj T j 1 + t +C z(tdt + z(tdt T j T j 1 (3 = Cost T + t, ( which contradicts the assumption of T being the optimal solution. Since z(t = z(l T at the optimal T and assuming the strict bathtub failure rate, the optimal T can occur in two possible ways: 1. T = L/ = L T, where L/ I, or. T1 < I, and L T 1 > L I (this is identical to T > L I, and L T < I by Proposition 1. Note that we don t consider the case when T = L/,L/ < I. This is because this case will provide a suboptimal solution. We can use a proof similar to the proof for Proposition ( to show that extending one of the intervals at the expense of another will provide a less expensive maintenance schedule. We now consider the case when multiple PM are to be performed in the interval [,L]. Let us denote the scheduled PM times as T 1, T,...,T n (see Figure 3. Proposition 3. At the optimal schedule, z(t 1 = z(t T 1 = z(t 3 T =... = z(t n T n 1. Proof: Suppose we have an optimal schedule where for some PM times T i and T j, z(t i T i 1 > z(t j T j 1. By our assumption of a stationary failure rate, increasing the interval between T j = Cost Ti t,t j + t, (4 which contradicts the notion of the supposed schedule being optimal. When an item has a bathtub failure rate, it may have two times, t 1 and t, when z(t 1 = z(t. Therefore, it may be possible that we have two types of intervals in the optimal solution, one of length t 1 and another of length t. Without loss of generality, let s assume t 1 < t. Proposition 4. Two types of optimal schedules are possible: 1. All of the intervals are of length T, where T I.. All but at most one interval are of length T ; the other interval is of length T, where T < I < T. Proof: Suppose the optimal schedule contains more than one interval of length T. Let us consider two of these subintervals; one happens to occur between renewal times T i and T i 1, the other between T j and T j 1. We know that if we extend one by t and shorten the other by t, it will not affect the cost of the rest of the schedule. Since T < I, we know that at the endpoints of both subintervals, both failure rates are decreasing. Therefore, Ti T i 1 T i T i 1 t Tj T j 1 + t z(tdt > z(tdt (5 T j T j 1 5 Copyright c 9 by ASME

6 Figure 3. MULTIPLE PM OVER INTERVAL [,L]. and rewriting the cost of both subintervals in terms of z(t yields: ( Ti T i 1 Tj T j 1 Cost Ti,T j = +C z(tdt + z(tdt ( Ti T i 1 t Tj T j 1 + t > +C z(tdt + z(tdt (6 (7 = Cost Ti t,t j + t. (8 This violates our assumption of our solution being optimal, and therefore we know that at most one interval can be of length T < I. Similar to the increasing failure rate case, sometimes we may be able to explicitly solve for the optimal interval T. We know that the optimal T will satisfy the following expression: = T z(t EN(,T (9 as well as the condition that T I. If we have can analytically express z(t, we may be able to explicitly solve Eqn. (9 for T. If we cannot analytically express z(t but can somehow model it, we can use a bisection search similar to the IFR case to numerically estimate T. Taking advantage of the fact that T > I, we can estimate T to an arbitrary accuracy ε > as follows: 1. Set T = max( L,I.. Evaluate the right-hand side of Eqn. (9. (a If < T z(t EN(,T (3 then PM is profitable in an interval of length T. If T T / < ε, return T as the optimal interval. Otherwise, set T = max( T,I, and repeat step. (b If > T z(t EN(,T (31 then PM is not profitable in an interval of length T. If T L, then it is not profitable to perform PM in the given time horizon at all. If T T +T < ε, return T as the optimal interval. Otherwise, set T = T +T, and repeat step. (c If = T z(t EN(,T (3 then PM is profitable in an interval of length T, and T happens to be the optimal interval; return T. 3 CONCLUSION In this paper, we address the problem of optimizing a maintenance schedule for a single item where maintenance actions consist of corrective and preventive maintenance. We discuss an algorithm to approximate the optimal interval between preventive maintenances when the item has increasing failure rate; this algorithm uses Bayesian models to approximate the failure rate and takes advantage of constant time intervals between preventive maintenances in the optimal schedule. We then show that if the item has a bathtub failure rate, a similar result holds where all or almost all intervals between preventive maintenances will be the same. This result can lead to algorithms that can compute or approximate optimal maintenance schedules. In future work, we will implement a Bayesian model to approximate a bathtub failure rate (see [7], and use it to implement such algorithm. REFERENCES [1] Barlow, R., and Hunter, L., 196. Optimum preventive maintenance policies. Operations Research, 8, pp [] Rausand, M., and Høyland, A., 4. System Reliability Theory: Models, Statistical Methods, and Applications. John Wiley & Sons, Inc. [3] Damien, P., Galenko, A., Popova, E., and Hanson, T., 7. Bayesian semiparametric analysis for a single item maintenance optimization. European Journal of Operational Research, 18, pp [4] Grall, A., Dieulle, L., Brenguer, C., and Roussignol, M.,. Continuous-time predictive-maintenance scheduling for a deteriorating system. IEEE Transactions on Reliability, 51, pp [5] Belyi, D., 1. Optimization models and methods under nonstationary uncertainty. PhD Thesis, University of Texas at Austin, Austin, TX. forthcoming. [6] Damien, P., Smith, A., and Laud, P., Monte Carlo methods for approximating a posterior hazard rate process. Statistics and Computing, 6, pp Copyright c 9 by ASME

7 [7] Laud, P., Damien, P., and Walker, S., 6. Computations via auxiliary random functions for survival models. Scandinavian Journal of Statistics, 33, pp Copyright c 9 by ASME