A Renewal Problem with Bulk Ordering of Components Author(s): D. R. Cox Source: Journal of the Royal Statistical Society. Series B (Methodological), Vol. 21, No. 1 (1959), pp. 180-189 Published by: Wiley for the Royal Statistical Society Stable URL: http://www.jstor.org/stable/2983940 Accessed: 14-02-2017 10:15 UTC JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact support@jstor.org. Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at http://about.jstor.org/terms Royal Statistical Society, Wiley are collaborating with JSTOR to digitize, preserve and extend access to Journal of the Royal Statistical Society. Series B (Methodological)
180 [No. 1, A RENEWAL PROBLEM WITH BULK ORDERING OF COMPONENTS By D. R. Cox Birkbeck College, University of London [Received July, 1958] SUMMARY SUPPOSE that k components are in use simultaneously, that batches of n items are ordered, and that all items in use at one time must be from the same batch. After n - k + 1 failures have occurred the remaining k - 1 items are wasted. Properties of the system are investigated and different strategies of use compared. 1. INTRODUCTION Consider a system in which k similar components are in use together. Suppose further that a batch of n components, n > k, is obtained and that when n - k + 1 failures have occurred, so that only k - 1 unbroken components remain, a new batch must be secured. The remaining k - 1 components of the first batch are rejected. That is, the k components in use at one time must be alike, taken from the same batch. This set-up was suggested to me by Mr. A. Glaskin, then of the Hosiery and Allied Trades Research Association, in connection with women's stockings. Here k = 2, and it is assumed that any number, n, of identical stockings can be bought together and any two of them worn as a pair, but that when only one of the batch remains fit for use, it cannot be worn with stockings from a new batch and is therefore scrapped. Another application is to situations where, when the system is first manufactured, spare components must be ordered for the whole life of the system, and where, if not enough spares are ordered, a completely new batch of components must be obtained. The two aspects of this situation studied in this paper are first the appropriate choice of n, and second the advantages of trying to equalize wear on all unbroken components. More precisely we consider the following two strategies: A. Each component in turn is used for a very short period, the choice of components for use being such that at any time each component that has not failed has been in service the same total time. B. From the batch of n components, k are chosen randomly for use. When one fails, it is replaced by an unused component, and so on until no more remain. Of course, in practice it might very well be impossible to adhere at all strictly to A, but it is reasonable to consider it for theoretical work. It will be assumed that the life-times of different components are mutually independent random variables Xi,..., X., all with probability density function f(x) and mean [, and that the life-times are unaffected by the strategy of use; this last assumption may well be seriously wrong in some applications. Let T denote the life of the batch, i.e. the time up to the (n - k + 1)th failure.
1959] Bulk Ordering of Components 181 2. EXPONENTIAL DISTRIBUTION OF LIFE-TIME When the distribution of life-time is exponential, f(x) = -le-x/, a very simple solution can be given. At any time, all components that have not failed are equivalent; their future life is independent of the length of time for which they have been used in the past. Therefore strategy of use is irrelevant. In any time interval (t, t + h), there is a constant chance kh/,u + o(h) of a failure on one of the k components in use. Thus T is the time up to the (n - k + 1)t event in a Poisson process with parameter k/,u. Therefore 2kT/u is distributed as X2 and in particular E(T)=-{ [(n -k + 1)/k. (1) In some applications, such as the second mentioned in section 1, we may wish to choose n so that T exceeds a given value to with high probability. If [t is known, a suitable n is easily found from tables of the x2 distribution. In other applications, such as the one to stockings, long-run behaviour is of most interest and (1) implies that components are used at a rate n/e(t)=-nk/[(n- k + 1)?] compared with a rate k/,u if the restriction to the use only of components from the batch were removed. Quite generally, for any3distribution of life-time with mean may define an index of efficiency I ke(t) (2) n,u measuring the ratio of the rate of use if the batch restriction were removed to that with batch size n. Equivalently I measures the fraction of the total life of all components that is utilized. For the exponential distribution, the index becomes Ie n-k 1 (3) n Equation (3) can be obtained directly from the consideration that at the end of a batch, the k - 1 components rejected are, with an exponential distribution of life-time, "as good as new". The loss measured by Ie has to be set against the advantage of spreading expenditure, the mean interval between purchases being (1), and against the cost of storing spares, the number of which varies between 0 and n - k, all values being equally frequent in the long run. In some applications the cost will depend on the number of batches bought as well as on the total number of components. If total costs can be assessed quantitatively, the optimum n can be found. 3. THE STRATEGY OF EQUALIZED WEAR Suppose now that the p.d.f. of life-time, f (x), is not necessarily exponential and that strategy A of section 1 is adopted. Denote by X(1),..., X(n) the order statistics of Xi,..., X., i.e. the n life-times rearranged in decreasing order. Now the first failure will occur after component life-time X(.), i.e. after clock-time nx(fl)/k. The second
182 Cox-A Renewal Problem with [No. 1, failure will occur after a further componen a further clock-time of (n - 1)(X(n-1) - X(n))/k. Finally the interval in life-time and the corresponding clock-time between penultimate and final failure are both X(k) - X(k+l). Therefore, if X(n+l) is defined to be 0 n-k n kt =, (n -j)(x(n-j) - X(n+l) X() + kx(k). (4) 1=0 1=k+1 The distribution o is a normal density function, the mean and variance of T can be found from the means, variances and covariances of the X(W), which are tabulated up to n = 20 (Teichroew, 1956; Greenberg and Sarhan, 1956); values of the means to two decimal places are given up to n 50 by Fisher and Yates (1957). If in a random sample of size n from a normal population of mean,t and standard deviation a, the expected value of X(W) is denoted by Z + mna, (5) then the index of efficienc where I ke(t) 1 n-g./., (6) n,u k-1 ng mn - (k -1) mkn. (7) ngk To obtain (7) the result n has been used. The entries (a) in Table 1 give gk for n < 8. The table is discussed later in the section. In order to examine how seriously the factors depend on the assumption of normality, it is worthwhile examining first a E(X(>) distribution of life-time +n-r+ that is rectangular 1 (8) over (- A, U + A). It is well known that in random samples of n from this distribution n and some simple calculation from (4), together with the fact that for the rectangular distribution a = V3 A, leads to the index of efficiency Ir, where Ir 1 -/V3 k(k - 1) a (9) n(n + 1) I- as the analogue of (6). A selection of values of the factor multiplying ai> in (9) is g in Table 1 (b).
1959] Bulk Ordering of Components 183 TABLE 1 Factor Determining the Index of Efficiency I for Various Distributions of Life-Time n 2 3 4 5 6 7 8 k 2 (a) 0 564 0 282 0-183 0.134 04104 0.085 0-071 (b) 0 577 0.289 0.173 0-116 0-082 0.062 0-048 (c) 0 530 0-327 0.231 0-184 (d) 0 500 01333 0 250 0.200 0-167 0143 0.125 3 (a) * 0-846 0-480 0-332 0.251 0.201 0.166 (b) * - (c) 0733 0510 0-391 (d) 0 667 0 500 0 400 0 333 0 286 0 250 4 (a) - 1*029 0 629 0 452 0 352 0 286 (b). 1*040 0 693 0 495 0*371 0 289 (c).. 0845 0 627 (d).. 0750 0 600 0 500 0 429 0*375 5 (a)...... 1163 0 746 0.553 0 439 (b) (c).... 0918 (d).. 0800 0 667 0.571 0500 6 (a).... 1267 0 842 0 639 (b) * 1*237 0 928 0722 (c)..... (d).... 0*833 0.714 0*625 7 (a).. 1352 0 924 (b) * *. (c)....... (d) *....... 0 857 0*750 8 (a) *. ** 1.424 (b).4..........7 (d) *- *- * 0 875 To calculate the index I, (a) Normal distribution; factor is glk (b) Rectangular distribution; factor is V3 k(k - 1)/[n(n + 1)]. (c) Distribution a2xe-az; factor is h'. (d) Exponential distribution; factor is (k - 1)/n. Over the range tabulated, the factors for the normal and rectangular distributions are in quite close agreement. Bearing in mind the sort of accuracy that is likely to be required in applications, it seems reasonable to conclude that for these n and k, the exact shap of the frequency distribution of life-time, if roughly symmetrical, is for the present purpo not very important. A family of skew distributions, for which the expected values of order statistics are tabulated or easy to calculate does not seem to be available. If life-time X is log-normal, so that Y = log X is normal, it should be adequate, unless r or n - r is small and V(Y) large, to use E(X(r)) E(ey(r)) exp {E(Y(r)) + 1 V( Y(r))}, (10) where E(Y(r)) and V( Y(r)) refer to a normal sample and are
184 Cox-A Renewal Problem with [No. 1, (10) would be exact if Y(r) was normally distributed and is a good approximation in any case if V( Y(r)) < 1. This method is probably the best for numerical work with a particular application, because it is likely that many empirical distributions of life-time can be reasonably well approximated by a log-normal curve. It will, however, not be followed up further here. Instead, to obtain results directly comparable with those obtained in section 4 for strategy B, we consider first the p.d.f. a2xe-a, proportional to that of X2 and with me 2/a. In samples of n from this distribution, the ratio of the expected value of an statistic to the population mean is 00 E(X(r))/(2a-1) n 2(r-1)!(n-r)! 1. [1 - (x + 1) e-0]n-r (x + 1)r-le-rxx2dx 0 n! n-rs+r-1 2 * - E E (_ )s X 2( ) =0 t=o (r + s - 1)! (t + 1)(t + 2) (l ) s!(n- r -s)!(r+s- 1 -t)!(r+s)t+3 ( The series can be expressed in terms of confluent hypergeometric functions, but this does not seem helpful for numerical work, which- is straightforward from (11). Since the expected values may possibly be of use in some other application they are given for n < 5 in Table 2. The values are checked by noting that each column adds up to n. TABLE 2 Ratio to Population Mean of Expected Value of rth Order Statistic in Samples of n from a2xe-x n 2 3 4 5 r 1 1.375 1.606 1*774 1 904 2 0*625 0 912 1 105 1*251 3.. 0.482 0*719 0 886 4.... 0 402 0*608 5..... 0'351 The corresponding values of E(T) and hence the indices of efficiency, Ig, are easily calculated from (3). The indices themselves are given in Table 6 in section 5, but for comparison with the results for the normal and rectangular distributions, the facto such that Ig = 1 - hkj, (12) where a = 1 /V2, is given in Table responding factor, (k - 1)/n, for the exponential distribution. Again, there is moderately good agreement over this range with the other factors in the table, bearing in mind the wide range of distributional form. One further numerical result has been obtained for the special distribution (2e-x + -Lje-xI5), with n = 3, k = 2, I = 0 548. This will be used in section 5.
1959] Bulk Ordering of Components 185 A final remark is that if n is large, the asymptotic theory of order statistics (Cramer, 1946, p. 367) can be used to approximate to L 4. THE STRATEGY OF REPLACEMENT ON FAILURE Suppose now that strategy B of section 1 is followed, components being replaced only on failure. We then have in effect k independent simple renewal processes in operation simultaneously, one arising from each of the k components in use originally. A result for such systems given by Cox and Smith (1954) can be used to obtain a first approximation to the index of efficiency* I' valid for large values of n/k. We consider the system at the (n - k + 1) th failure. For each of the k - 1 processes on which this last failure does not occur, there is a residual life-time, defined as the time remaining before the next failure on that process. Provided that the number of failures that have occurred on each process is not too small, these residual life-times will be those of an equilibrium renewal process, i.e. an ordinary renewal process considered at a random point a long time from the start of the process. The expectation of such a residual lifetime is ([2 + g2) /(2M). Thus the expected wasted time is (k - 1)([2 + cr2) /(2p) therefore I 1 (k- 1)(IL2 + g2) (13) 2nK2 (3 This formula is exact for exponentially distributed life-times. It seems likely that (13) is a reasonable approximation even when k/n is no a/i is fairly small. If so, and if n - k is large, there will be an appreciable gain from the use of strategy A. For then I is near unity, whereas I' is appreciably less than unity. To see what is involved in the exact evaluation of I', we use methods standard in renewal theory. Let F(x) be the distribution function and f(x) the density function of life-time, and let Fr(x) be the r-fold convolution of F(x) with itself. Let Nt be the number of failures in (0, t) in a single renewal process and let Zr denote the sum of a fixed number r of independent life-times. Then pr (Nt < r) = pr (Zr > t) = 1 Fr(t). (14) Thus the probability generating function of Nt, G(4, t), say is given by G(z; t) E- z rfr(t) + 1(15) r=o If an asterisk is used to denote a Laplace transform with respect to t, so that 00 G*(4, s) G(4, t) e-st dt, (16) we have the well-known equation that G*(, s) 1 - f*(s) (17) * A prime will be used for indices of eff
186 Cox-A Renewal Problem with [No. 1, on using the result that 0 e-stfr(t) dt = [f()]. (18) S Now let A(7) be the number of failures in (0, t) in the combined set of k renewal processes. Then G(k)(4, t), its probability generating function, is given by G(k)(4, t) = [G(4, t)]k, (19) since N(k) is the sum of k independent rando Denote by Tm the time in the combined sequence measured up to the mth failure; we require m - n - k + 1. Then by a reversal of the argument of (14) pr (Tm > t) = pr (N(7) < m), (20) from which it follows easily that 00 E Cm pr (Tm > t) = G(k)(4, t). (21) m=1 1 Now we are particularly interested in E(Tm) whxich can be written Go E(Tm) = f pr (Tm > t) dt. (22) Therefore 0 00 0 0 E mnle m=1 0 an equation for the moment generating function of Tm can be obtained in a similar way. Thus, to determine E(Tn-k+l), and hence I', we need to invert (17) to obtain G(4, t), to raise this to the kth power, to integrate with respect to t and finally to pick out the appropriate power of C in (23). These formulae can be evaluated explicitly whenf*(s) is a simple rational function of s. Thus whenf(x) = xe-,f*(s) = 1/(1 + S)2, and by (17) from which and finally G*(4, s) -s+12_4 (24) G(4, t) =2Vg {(A/4 + 1) evl e-kt k G(k)(t (4)k/2 t = (r(v + )r (ly - I)k-r e(2rvg-kv\ E VnmE(T'm)= (k w +1) V -1)- (25) m=1 (4)Ok/2 (1 - ) r=o r k(l + C)2r,\,/ In particular, when k = 2, the right-hand side of (25) reduces to 4(5-4) /[4(1 -
1959] Bulk Ordering of Components 187 from which E(Tm) l m + + (26) It follows that the index I', which is given in general by f- ke(tn-k+l) (27) nflp where M is the mean life-time (and is equal to 2 for the above distribution), is ' 1-3. (28) 4n The approximate formula (13) with k = 2, g2/2 = -, is exact in this case, a rather surprising result. When k = 3, the corresponding generating function (25) becomes 2~(13-5) (29) 3(1-0)2 (9 - The approximate formula (13) reduces in this case to I' 1-3. (30) 2n The exact values of I' for n = 3, 4, 5 are 0 4815, 0 6235 and 0 6999, whereas the corresponding approximate values from (30) are 0 5, 0 625 and 0 7 indicating excellent agreement even for the smallest value of n. The special case k = 2, n = 3, f(x) - Ie-x + -Lj-e-0/5 has also been investigated, the value of I' being 5/9. The general approximate formula (13) can be derived as a limit as n -o 00 with k fixed by arguments similar to those leading to (23), but the details are complicated and very heavy algebra seems necessary to get a closer approximation. If results are required for a particular distribution of life-time, and it is thought that the approximation (13) may not be adequate, the best procedure is either to fit a distribution with a rational Laplace transform and to work through the exact theory, or to use an empirical sampling method. The theoretical method is difficult if the fitted Laplace transform is of high degree, and it is therefore worth discussing very briefly suitable sampling methods. The best methods to use depend on the particular values of n and k and on the properties of the distribution being sampled. Consider, as an example, a normal distribution of life-times and suppose that n - k is sm4ll. Let X1,..., Xn be the n life-times i order of use, and let Tr = Tr(Xi,..., Xn) be the time up to the rth failure in the combined process. Now it will happen frequently that Tr is equal to the rth largest observation among X1,..., Xk, a random variable, Ur, of known expectation. We therefore determine by sampling the expectation of Zr = Ur - Tr. To do this, first take random sampling numbers for X1,..., Xk and find Ur. In simple cases, we can determine exactly pr (Zr = 0) and then can sample for Xk+?,..., Xn conditionally on Zr =# 0 Additional precision can be obtained, (a) by "splitting", i.e. by taking several sets of
188 Cox-A Renewal Problem with [No. 1, Xk+l,..., X. for those configurations for which a large contribution to E(Zr) is likely, and (b) by stratified sampling for X1,..., Xk, or alternatively by weighted sampling based on the range of X1,..., Xk. These remarks are intended simply as a general indication of the methods to be followed. As an example, E(T2) has been determined for a normal distribution of mean 10 and standard deviation 3, taking k = 2, n = 3. The mean of 20 determinations was 11 62? 0 032, giving I'-0 775? 0 021. The value from the approximation (13) is 0 82. The agreement is good considering that (13) is based on the assumption that n/k is large. A rough estimate is that the devices described above lead to a several hundredfold increase in precision, as compared with simple random sampling. 5. THE COMPARISON OF STRATEGIES The following results have been reached relevant to the comparison of strategies. (i) For the exponential distribution strategy does not matter. (ii) For the distribution a2xe-ax, the results collected in Table 3 show that for small n and k, strategy A gives a higher index of efficiency, but that the difference between strategies is rather small. (iii) The general approximate formulae of sections 3 and 4 indicate that strategy A should, in general, give the higher index of efficiency when g/u < 1. The index of efficiency is however near 1 when n> k, so that the main situation in which an appreciable difference is likely is when k/n is not small. (iv) The calculation for the distribution le-x + -L-je-xl5 shows that it is possible for strategy B to give the higher index. TABLE 3 Indices of Efficiency when the Distribution of Life-Time is a2xe-x n 2 3 4 5 k 2 (a) 0 625 (a) 0 768 (a) 0 833 (a) 0 870 (b) 0 625 (b) 0 750 (b) 0-812 (b) 0 850 (c) 0625 (c) 0 750 (c) 0812 (c) 0 850 3.. (a) 0 482 (a) 0640 (a) 0 723 (b) 0482 (b) 0 624 (b) 0 700 (c) 0 5 (c) 0 625 (c) 0700 4.... (a) 0402 (a) 0*557 (b) 0 402 (b) (c) 0 438 (c) 0*550 5...... (a) 0 351 (b) 0-351 (c) 0400 (a) Strategy A: exact. (b) Strategy B: exact. (c) Strategy B: approximate. Now the optimum strategy need be neither A nor B. When the life of the system ends at the (n - k + 1)th failure, there are left the (k - 1) residual life-times for the items that have not failed. The best strategy will be the one that minimizes the expected value of
1959] Bulk Ordering of Components 189 the sum of these residual life-times. Now if the distribution function of life-time X is F(x) and the survivor curve, giving the probability of surviving to at least age x is FC(x) = - F(x), we can define the expectancy of life at age x to be 00 ef (x) E[X - x X > x] J' ydf(y) /Fc(x) -x x 00 = Fc(y) dy/fc(x), (31) x [ ~L {log je FJ(y) dy (32) An intuitively very reasonable strategy is to use at any instant those comp greatest expectancy of life. Now ef(x) is a decreasing function of x if and only if d2 00 j2 log F0(y) d < 0, dxx i.e. if and only if the integral of F0(y) is logarithmically concave. This condition is satisfied by many common distributions such as the normal and the x2 distributions with more than two degrees of freedom, so that strategy A would be expected to be optimum in these cases. The mixed exponential distribution that has been investigated has ef(x) an increasing function of x, so that strategy B would be expected to be the better in this case, as was indeed found. Finally, it must be repeated that all these comparisons depend on the assumption that the life of individual components is unaffected by strategy of use, so that, for example, there are no fatigue effects. REFERENCES Cox, D. R. & SMITH, W. L. (1954), "On the superposition of renewal processes", Biometrika, 41, 91-99. CRMEmR, H. (1946), Mathematical methods of statistics. Princeton: University Press. FISHER, R. A. & YATES, F. (1957), Statistical tables for biological agricultural and medical research. Edinburgh: Oliver & Boyd. 5th ed. GREENBERG, B. G. & SARHAN, A. E. (1956), "Estimation of location and scale parameters by ord from singly and doubly censored samples, I", Ann. Math. Statist., 27, 427-451. TEICHROEW, D. (1956), "Tables of expected values of order statistics and products of order st samples of twenty and less from the normal distribution", Ann. Math. Statist., 27, 410-426. VOL. XXI. NO. 1. G