ON SELECTING THE MOST RELIABLE COMPONENTS

Consider a series system consisting of n components of k types. Whenever a unit fails, it is replaced immediately by a new one to keep the system working. Under the assumption that all the life lengths of the components are independent and exponentially distributed and that the replacement policies depend only on the present state of the system at each failure, the system may be represented by a birth and death process. The existence of the optimum replacement policies are discussed and the e-optimal policies axe derived. If the past experience of the system can also be utilized, the process is not a Markov process. The optimum Bayesian policies are derived and the properties of the resulting process axe studied. Also, the stochastic processes are simulated and the probability of absorption, the mean time to absorption and the average proportion of the retrograde motion are approximated.

. Introduction Suppose we have k types of components.Assume that all the life lengths of the components are independent and exponentially distributed with unknown constant failure rates , ),..-, ).Since the rank order of these k failure rates is assumed to be completely unknown, the natural problem arises of how to select the most reliable components associated with the smallest failure rate.
The decision problem could be classiiied in the areas of multiple comparisons, ranking and selection and reliability theory, but especially in the area of ranking and selection.The simplest model hereby is to operate separately k components, one from each type.When a component fails, we replace it immediately by a new one of the same type.The failure counting processes in time are k independent Poisson processes with failure rates j, j 1,2,-.-,k.Dixon and Bland (1971) have observed the k Poisson processes over time periods of a common fixed length and derived a Bayes solution to the problem of complete ranking through combined paired comparisons.For the same model, Goel (1972) has shown that the natural rule for selecting the population with the largest ,-value cannot be implemented in the indifference zone approach at predetermined P*-values.Subset selection procedures, however, can be established under the P*-criterion, as it has been done by Gupta and Huang (1975) and Gupta and Wong (1977).The alternative problem, i.e. the one concerning subset selection with emphasis on the smallest ,value, has been treated in Gupta, Leong, and  Wong  (1979).Results on selecting the population with the largest ,-value in terms of the highest posterior probability of .smbeing the best population have been derived by Wilson  (1992).Lists of the relevant literature in this area can be found in Gupta and Panchapakesan (1979) and Miescke and Shi (1995).
Recently, Miescke (1990) and Miescke and Shi (1995) have treated the problem in a new dynamic approach.Instead of observing the k parallel running, independent experiments, the k types of components are examined in a series operating system where it is emphasized that at each failure, an immediate replacement is made in some optimum way to keep the system running.Under this model Miescke (1990) has derived some natural 'look-ahead-one-failure' Bayes replacement policies to maximize, at each failure, the expected waiting time for the next failure in the system, or the posterior probability that the waiting time for the next failure exceeds a given value.These Bayes rules have been obtained for both cases, when the past experiences of the system could be utilized or not.In the latter case, the system can be represented by a Markov process.Miescke and Shi (1995) have studied the process for k = 2 and, under certain conditions, derived Markov replacement policies in some optimum way to ensure that, with maximum absorption probability, the system will finally be composed of only the components of the better type, possessing the smaller failure rate.
The first objective in the present paper is to study the existence of the optimum replacement policies and define the e-optimal replacement policies for the Markov process.If the past information of the system could also be utilized, the process is not a Markov process.Our second objective is to derive the optimum Bayes replacement policies through the generalized maximum likelihood method for the non-Markov processes.Finally, the absorbing property of the non-Markov processes will be studied and the results of the computer simulation of the processes will be given at the end.
2. Policies Based only on the Present Information 2.1.Model Assumptions Consider a series system consisting of n components which axe all kept permanently in working condition.Each time a failure occurs, the failed component is replaced immediately by a fresh one according to some replacement policy.Available for use axe two types of components.Assume that all life lengths axe independent and each unit of type k has a cumulative life distribution function G($) 1 e-, >_ 0, k 0, 1, where A0, A1 > 0 are unknown parameters.
The replacement policies considered in this section depend only on the present state of the system, i.e. which type of component caused the failure and how many components of each type were in use in the system.To be more specific, let F, 0 < i < n (S, 0 _< < n) denote the probability of making a replacement with a component of type 0(1), after a component of type 1(0) fails when the system contains i components of type 1, and let $'i 1 F and i 1 S. Assume that F > 0, S > 0, 0 < i < n and Fn So 0, i.e. the state of the system will not be changed if all the components in use are of the same type.Particular policies will be denoted by a (F1,..., F,; So,.-., S,,-1) while the set of all possible replacement policies under consideration will be denoted ,4.Note that if type 1 is better than type 0, i.e. if A1 < A0, then S and F are suggesting a desirable replacement and S and F a wrong replacement.
Let X(t) denote the number of units of type 1 in the system, operating at time t.
Then, the motion of X(t) can be described as follows: if, at time t, the process is in state i, 0 < i < n, then, after staying at state for an exponentially distributed length of time, it may move to state i + 1 (a unit of type 0 fails and is replaced by a new one of type 1) or state i 1 (a type 1 fails and is replaced by a type 0), or stay at state i (any one type fails and is replaced by the same type).If X (t) 0 or n, then the state of the process will never be changed.Due to the lack of memory of the exponential law, the process {X(t), t > 0} is a birth and death process with fte state space {0,1, 2,---, n} and absorbing states 0 and n. (Miescke and Shi, 1995).The first natural step is to find a replacement policy which maximizes the probability that the process ends up in the better of the two absorbing states, i.e. all the components in use are of the same type which has the smaller -value.To this end, we introduce an embedded Markov chain as follows.Let T0 0 < T1 < T2 < < Tr < be the successive transition times of the birth and death process, i.e. the successive moments when a component fails and is replaced immediately, according to the replacement policy, by a unit of different type.Let X0 X(0), Xr X(Tr), r >_ 1.Then, {X, r >_ 0} is an imbedded Markov chain associated with the given Markov process {X(t), t >_ 0}.This Markov chain records only the jumps of the birth and death process at the transition times, but the probabilities of absorption into states 0 or n for this Markov chain are the same as for the birth and death process since both of them execute the same transitions.
Let aij denote the probability that, starting at a transient state i E {1, 2, ..., n- 1}, the Markov chain ends up in an absorbing state j E {0, n}.Then, for any replacement policy a 6,4, the absorption probabilities aij can be written in the following concise form: a, a, where 6 Az c = 0 < i < n, So = I. (Miescke and Shi, 1995) S1S2...Si The symmetric replacement policies Our goal is to find a replacement policy which maximizes the probability that the process ends up in the better of the two absorbing states.Since we do not know which type of components is better we should find a policy that maximizes ain, 0 < i < n when type 1 is better ($ < 1), and simultaneously, minimizes ain (maximizes aio) when type 0 is better ( > 1).Unfortunately, the problem has no solution if the replacement policies are not symmetric or the system is not completely sl/mmetric, i.e. if the fixed number n of components in the system is odd, n 2m + 1, say, or n is even, n 2m, say, but the starting state is not m, i.e. the system starts with unequal numbers of components of each type.
The first statement is obvious since 0 < an < 1 for 11 > 0 and all a E and the limit of ain is 0 (1) as Sn-1 (F1) approaches 0, that is, infA ai, 0 and supA ain 1 for all 6 > 0. This indicates that there does not exist any optimum replacement policies in .A, nor even in the extended policy set 4 {a Fi > 0, 0 < i < n,F,= 0; Si > 0, 0 < < n, So 0}.To overcome the obstacles, it is reasonable to consider only those replacement policies which treat the two types components without any bias towards their given labels, i.e. which act analogously if the labels 'type 0' and 'type 1' are exchanged.Thus, throughout the sequel, it is assumed that the replacement policies are symmetric in the following way: Fi Sn-i, i 1, 2,---, n.The set of all symmetric replacement policies and its extension will be denoted by fl.i and Jtl {a]F Sn->_ O, 0 < i < n, Fn So 0}, respectively.
To prove the second statement and compare the replacement policies, let us con- sider the extreme case where F S=_i approaches 0. It follows that ci approaches 0, i 1, 2, n 2, and consequently, a approaches a* l_b., .1If the system is not completely symmetric, then sup4 an for < 1 and inf4 a for > 1 are not the same.To see this, let us investigate the case where n 2m + 1 and _< m, for instance.
After some standard steps of calculation, the difference ai a can be written ts -E =o -J Its denominator is positive, while its numerator can be written as -1 aj6n-l-J, which is negative if $ < 1.That is, a. < a* for all $ < 1.Note that the limit of ai. is an as Ft approaches 0. It follows that, in the present case, sup.41 a. a. for all 5 < 1.
On the other hand, if we multiply both the numerator and the denominator of ain in Eq.( 1) by St $2"" S., we can see that each term of the numerator and the denominator contains a common factor S,., except for the middle term of the denominator, which is ( ,. )-tFtF2... F.$".It follows that an approaches 0 as Sm approaches 0 and consequently, in the present case, infAl ai, 0 for 5 > 1.
This indicates that there does not exist any optimum replacement policies in At, nor in its extension Jh.
..3.The completely symmetric system For the case n 2m > 2 and i m, due to the complete symmetry, it can be seen that if J < 1, then a.. < a., < a,., for all a .Aa; (2) and if > 1, then w for all a E At, where a.,w _Atl+.Since the limit of am. is am.as F1 approaches 0 we see that a. sup a,. for all < 1 and a. infh a.. for all > 1. Analogously, as an interesting by-product, it is readily seen that a. inft a,.., for all < 1 and a,, supA a..., for all > 1 since the limit of a.. is a,.. as S.-t approaches 0 for all > 0.
It is now obvious that there is no optimal policy even for the completely symmetric system, since the ranges of the absorption probability am. are open intervals.But it can also be seen that, in this case, e-optimal policies do exist.

The e-optimal policies
A policy a is called an e-optimal replacement policy in .At if, simultaneously, for all 6>1, a,,, > a,.-e for all 6 < 1.
To see the existence of the e-optimal policies, let us have a further investigation of * which can be written as the difference d am.am.
The denominator of d is a polynomial of $ with degree 2n-2 and is always greater than 1, while the numerator is a polynomial of with degree n / m 2 and each term of the numerator contains a common factor F1. Hence, for all a E 41, d/F1 is a continuous function of in (0, c).Since lim_0 d 0 and lim.., d 0, we see that d/F1 is bounded from above on (0, o).That is, Idl <_ F1 M, 0 < < c, for some positive number M. This implies that a,nn approaches an uniformly in J as F1 approaches zero.Consequently, the following theorem has been shown to be true.
THEOREM 1 The best replacement policy, which maximizes the probability that the process ends up in the better of the two absorbing states does not exist, but e-optimal policies do exist if the system is completely symmetric.
The s-optimal policie described above suggest that if the system contains only one component of type 1 (or type 0) and it fails, then it should be replaced by a unit of the same type with a probability as close to one as possible.The process then stops very slowly to ensure that it will end up in the better absorbing state with the absorption probability a,, as close to a as desired.
The remaining parts of the above policies, Fi Sn-i, i 2, 3,-.-, n 1, are completely free.We could apply, for example, the Play the Winner policy, denoted by P: Fi Si 1, i 1, 2,.-., n-1, i.e. if the failure is caused by a component of type 1 (0), then it should be replaced by a component of type 0 (1).The probability of absorption for 7 turns out to be i=0 i=0 (Miescke and Shi, 1995).Now we could recommend an e-optimal policy, :P as follows: F S,_ a and Fi Sn-i 1, 2, 3,---, n-1.
The probability of absorption into the better state for = 1/2 and n 4, 8, 12 are listed in Table 1, where B (/Y) stands for the best (worst) limiting value (a,) and :P0.for an e-optimal policy :P, with a 0.1.Table I.Absorption Probabilities 4 0.8889 0.8841 0.8485 0.66'67 8 0.9922 0.9918 0.9879 0.6667 0.9995 0.9995 0.9993 0.6667 3. Policies based on both the present and the past information Suppose that the past information of the system is available and, from now on, that the replacement policies depend not only on the present information, but also on the past experience of the system.Then, the process {X(t), t _> 0} is not a Markov process any more.To find an optimum replacement policy, we employ the Bayesian approach.
3.1.The likelihood Let ntt (not), > 1 be the number of components of type 1 (0), operating in the system between the (1 1)th and/th failure, ntl not m n/2.Let It denote the type of component which causes the/th failure, Rt the type of component which replaced the lth failure, It (I0) the number of fures caused by components of type 1 (0) during the first failures and Rlt (Rot) the number of components of type 1 (0), which replaced the failed units during the first failures, > 1.Then, it is obvious that It 'j=t Ij, Iol l-Ill, Rt j= Rj, R0 l-Rt and nt=m-It_t+Rtl_t, n0t=m+It-t-Rt-t, l>l.
The probability density of I, given ntt, is ft(kl), m) RIIA1 + n0tA0' k 1, 0, which depends on the unknown parameter A (At, Ao), the type of components which caused the first l-1 failures, I1, I2,.-., It-l, and the type of components which made the first l-1 replacements, R,R2,... Note that a replacement policy is just a series of probability densities of Rt, > 1.It is easily seen that at the first stage, the policy depends only on It.At the second stage, the policy depends only on It, Rt, and I2.Since Rt is determined by 11, we see that, at the second stage, the policy depends only on It and I2.Thus, we conclude that, at the/th stage, the replacement policy under consideration depends only on It, I2,.-., It and :.As a consequence, we may obtain, for instance, that P{I i[ I =,i} P{I i=l z where nt2 m-it + r, noz m + it -rl, and r depends on i.
It is now readily seen that the likelihood of I (I,I2,...,It)

Bayes replacement policy
As usual, if there is no a priori known preference between the two types of com- ponents, symmetric priors are the natural choice.Let us assume, throughout the sequel, that (0, 1) is a realization of a random vector A (Ao, A1) with a prior density 7r() which is permutation symmetric.Then, the joint density of I (I1,I2,'",I) and A is (jI-[ % "= Hj=I (nlj'l + nOj.O) the marginal density of I is j=l and hence, the posterior density of A, given ( )

1-I= (n + Ano)m
Under 0 1 loss, the posterior expected losses of actions R 1 and 0 are and where the lt equity is obtained by exchng d 0. Hence, a sufficient condition for E=(OL(, 1) < E=(XO L(A, O) is, for aft 0 < A < 1,

H-'I (lj "-A0j)
To summarize our results, let us define d d;(z) x -.= z n n x ++znno _> 1. (6) The following has been shown to be true.THEOREM 2 Assume that A (A0, A1) has a permutation symmetric probability density 7r.Then, under 0-1 loss, the Bayes replacement policy which minimizes the posterior expected loss is as follows: after the th failure, replace the failed unit by a component of type 1 (0) if dl > (<) 1 for all 0 < x < 1. COROLLARY 1 Suppose that A (A0, A1) has two possible values (u,v) and (v, u), u < v, say, with both values being equally likely, i.e. r(u, v) 7r(v, u) 1/2.Then, under 0 1 loss, the Bayes rule for the system is R; 1 (0) if d; > (<) 1.

An Example
Example 1: Suppose that A (A0, A1) has two possible values (1,2) and (2,1) with 7r(1, 2) 7r(2, 1) 1/2.Then, the Bayes rule is that R 1 (0) if 2nj 4- (7) .=nj / 2no/ Assume that n 8 and I (0,1,1,0,1,0,0,0,0,0),i.e.I 0, I2 1, etc.Then, following the Bayes rule we obtain that R (1, 1, 0, 1, 0, 1,1, 1, 1, 1).The computations are very simple and hence omitted.Remark.From the example, one sees that after the 9th replacement, all the components in use are of type 1.Then, definitely, the next failure will be caused by a component of type 1.According to the replacement rule, we see that the failed unit must be replaced by a component of the same type, i.e. type 1. Hence after the next replacement, all the components in use are still of type 1.This is always true for the successive replacements after the 9th.We conclude that the process has two absorbing states.The proof of the absorption property of the process will be given in the next section.
4. Simulation of the process 4.1.The absorbing property of the process THEOREM 3 Assume that the Bayes replacement policy stated in Theorem 2 is adopted.Then, the process {X(t), t >_ O) has two absorbing states 0 and n, i.e. the system will not change its state as soon as all the components in use are of the same type.
Proof: Let to 0 < tl < < tz < ..., be the successive failure times of the process and Xz nu+.Suppose, without loss of generality, that stage is the first time when all the components in use are of the same type, type 0, say, for explicitness, i.e.X=O; O<X<n, O<j</.
The proof can be completed by standard induction.

Simulation of the process
Consider the system discussed in Section 3 with the assumption that A (A0, A1) has two possible values (u, v) and (v, u) with u < v and vr(u, v) 7r(v, u) 1/2.Let X(t) denote the number of components of type 1, operating in the system at time t.Let to 0 < t < < tt < be the successive failure times and Xz nt+, _> 0, the number of components of type 1, operating in the system between the/th and (l + 1)th failure.Then, the discrete time stochastic process {X, >_ 0} may be simulated by using computer experiments.The initialization step and the main steps are as follows.
Step 0. Let Xo rn = n/2 and fix the ratio of A A0/A > 1, say.Let 0, do 1 and go to Step 1.
Step 1. Generate It+l, which may be 0 or 1, by using random numbers such that

P{h+
x } xz + (n-zz)A and go to Step 2.
Step 3. Determine Rt+ by Rz+ 1 (0), if d+x > (<) Step 5. Decide whether to continue or to stop: if 0 < z+t < n, then replace by + 1 and go back to Step 1; otherwise, if Z+l 0 or n, then stop.
By using a supercomputer to simulate the process {X, >_ 0}, we can estimate the relevant quantities of the process we are interested in.For example, we may estimate the probability of absorption into the better of the two absorbing states, the mean time to absorption, and the average proportion of the retrograde or forward motion.

The retrograde motions
To further compare the Bayes replacement policy derived above with the -optimal policies studied in Section 2, let us investigate the mean time to absorption and 144 D. Sttl  the average proportion of correct and wrong replacements for each policy.Assume, from now on, that the components of type 1 are better.Then, at each failure, a new component of type 1 should be the right choice for the replacement.This lets the Markov process move forward to the better absorbing state n.On the other hand, a component of type 0 is a wrong choice for the replacement, which leads to a retrograde motion towards the inferior absorbing state 0.
For a given replacement policy and the resulting stochastic process, let the ran- dom variable T be the total number of replacements until absorption.Then, R T=I R and W = T-R are the number of right and wrong replacements, which are made by components of type 1 and 0, respectively.Let the ratio of B WIT (R/T) be called the retrograde (]orward) proportion o the policy and its expectation E(B) (E(R/T)) the average retrograde (forward) proportion o.f the policy.
Since no closed-form representations are available, computer simulations are em- ployed to approximate E(T) and E(B).The results are given in Table 6 and 7.
From Table 5 one can see that the e-optimal policies for the absorbing Markov processes possess greater probabilities of absorption into the better terminal state.

SELECTING RELIABLE COMPONENTS 145
On the other hand, from Table 6 and Table 7, the Bayes replacement policy de- rived above requires shorter duration of time for the process to get absorbed, has much greater average proportion of forward motions, and hence, much less inferior components to be used in the course of time.

Acknowledgements
The author thanks the Managing Editor of JAMDS and two anonymous referees for their helpful comments and suggestions, which improved the presentation of the paper.Special thanks go to Professor K. J. Miescke for his helpful comments and encouragements.Research support from Indiana University Southeast is gratefully acknowledged.

Table 3 .
Mean Time to Absorption (Af)

Table 4 .
Average Forward Proportion

Table 7 .
Average Forward Proportion