Opl Res. Q., Pergamon Press 1976. Vol. 27, I, i, pp. 101 to 110. Printed in Great Britain
Reliability Analysis of a One-unit System with Unrepairable Spare Units and its Optimization Applications TOSHIO NAKAGAWA Department of Mathematics, Meijo University, Nagoya 468, Japan
and SHUNJI OSAKI Department of Industrial Engineering, Hiroshima University, Hiroshima 730, Japan It is assumed that a unit is either in operation or is in repair. When the main unit is under repair, spare units which cannot be repaired are used. In this system the following quantities are of interest: (i) The time distribution and the mean time to first-system failure, given that the n spare units are provided at time 0. (ii) The probability that the number of the failed spare units are equal to exactly n during the interval (0, t], and its expected number during the interval (0, t]. These quantities are derived by solving the renewal-type equations. Two optimization problems are discussed using the results obtained, viz.: (i) The expected cost of two systems, one with both a main unit and spare units and the other with only spare units is considered. (ii) A preventive maintenance policy of the main unit is considered in order to minimize the expected cost rate. Some policies of the two problems are discussed under suitable conditions. Numerical examples are also presented.
INTRODUCTION RELIABILITY theory is often concerned with the stochastic behaviour of redundant repairable systems. Barlow and Proschan2 have summarized these problems as repairman problems and have also derived some new interesting results. In particular, SrinivasanP Natarajan,lS Bhat, 3 and Nakagawa10 have discussed standby redundant systems with spare units but only for the case where any failed units are repaired and become as good as new upon repair completion. In the real world, it may be worth while to scrap some failed units without repairing, depending on the nature of the failed unit. For instance, we should scrap a failed unit in the case where the cost of a necessary repair exceeds the resulting gain in its value. From this point of view, the repair limit problems have been considered by Hastings et a/., 5- 7 who took army trucks as an example, and by several other authors. 4• 8• 11 In this paper, we treat a modified model of a standby redundant system, with repair limit policies. We consider the model with a main unit and n spare 101
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Operational Research Quarterly Vol. 27 No. 1, i sub-units which are statistically not identical to each other but the spare units have the same functions as the main unit if they take over operation. The system functions in such a way that if the main unit fails then it undergoes repair immediately and one of the spare units replaces it. As soon as repair of the main unit is completed it begins to operate and the operating spare unit is available for further use. A failed spare unit is scrapped. The system functions until the last (the nth) spare unit fails, i.e. a system failure occurs when the last spare unit fails while the main unit is under repair. This often occurs when something is broken or lost, and we temporarily use a substitute until it is repaired or replaced. We believe that this model can be applicable to other fields. The operating characteristics of this model require knowledge of: (i) The time distribution and the mean time to first-system failure, given that then spare units are provided at time 0. (ii) The probability that the number of the failed spare units are exactly equal to n during the interval {0, t], and its expected number during the interval {0, t]. These quantities are derived from the renewal equations and used for two optimization problems and to determine the initial number of spares to stock. For two optimization problems we adopt the "expected cost rates" as the appropriate objective functions. In the first, we compare two systems, viz. the use of both main and spare units or the use of only unrepairable spare units. The problem is to select which kinds of units should be scrapped or repaired on failure. In the second we consider preventive maintenance of the main unit. When the main unit works for a specified time t0 without failure, the operating of the working unit is stopped and one of the spare units takes over operation. The main unit is serviced on failure or its age t0 , whichever occurs first. The costs suffered for each failed unit and for each preventively maintained unit are considered. Then, the optimum preventive maintenance policy minimizing the expected cost rate is derived explicitly under suitable conditions. Similar optimization problems have been discussed by Morse, 9 Rozhdestvenskiy and Fanarzhi,l6 Osaki and Asakura14 and Nakagawa and Osaki.12 Finally, we state some other problems and show a numerical example. ANALYSIS Assume that the failure time of the main unit has an arbitrary distribution F(t) with mean 1/'A and its repair time has an arbitrary distribution G(t) with mean 1/p.. Also assume that the failure time of each operating spare unit has an arbitrary distribution K(t) with mean 1/0, even if it has been used before, i.e. the life of a spare unit, which has already been in service, is not affected by the past. All random variables considered here are independent. Unless otherwise stated, we assume that all units are good at time 0. It is further assumed that
102
T. Nakagawa and S. Osaki- Reliability Analysis failures are instantly detected and repaired (or scrapped), and each switchover is perfect and each switchover time is instantaneous. First suppose that n spare units are provided at time 0. Let H.,(t) denote the first-passage time to system failure. Then, we have the following renewal equation: H.,(t)
= F(t)*
{1
1G(t)dKCn>(t)+ n-1 ~ H.,_:i(t)*
0
j=O
i' 0
}
[KC:i>(t)-KCi+1 >(t)]dG(t)
(n = 1,2, ...),
(1)
where G(t) = 1- G(t) and the asterisk represents the convolution of any two functions a(t) and b(t), i.e. a(t)*b(t) =
J~b(t- u) da(u),
and K(i)(t) U = I, 2, ...) represents the j-fold Stieltjes convolution of K(t) with itself and KW>(t) =I for t;;::::O. For, the first term of the bracket on the righthand side is the time distribution that all of n spares fail before the first repair completion of the failed main unit, and the second term is the time distribution that exactly j spares U = 0, I, ..., n-I) fail until the first repair completion of the main unit, and then the system with n-j spares operates again. The firstpassage time distribution to system failure, H.,(t), can be calculated recursively and uniquely determined from (I). Introducing the notation of the generating function of the Laplace-Stieltjes (LS) transformation,
"" i""
H(z,s) = ~ zn e-sldH.,(t). n=l 0
Then, taking the LS transformation on both sides of (I) and using the generating function H(z,s), we have
H(z,s) =
co '=ilco o ' e-si[KCi>(t)- KCi+l>(t)] dG(t) i=O 0
I -f(s) ~ z:l
(2)
where f(s)= f0 e- 81 dF(t). {In general, the LS transformation of the function is denoted by the corresponding lower letter throughout this paper.) Moreover, let!., denote the mean first-passage time to system failure. Then, from {1), !., = If'A+ ( 00 [I-KCn>(i)]G(t)dt+ ""'£.11.,_1
Jo
:1=0
(
00
Jo
[KU>-KCi+l>(t)]dG(t)
(n = I, 2, ...),
103
(3)
Operational Research Quarterly Vol. 27 No. 1, i and hence, the generating function is given by
L(z) =
00
"£. zn ln
n=l
= (1/.\)[z/(1-z)]+j~/ 1- j~/
L 00
L 00
[1-K
(t)]G(t)dt (4)
[K(t)-K(t)]dG(t)
In a similar way, we derive the expected number of the failed spare units during (0, t]. Let Mn(t) be the probability that the total numbers of the failed spares during the interval (0, t] are exactly n and let M(t) = "£.~= 1 nMn(t). Then, we obtain M0 (t) = F(t) + F(t)* [.K(t) G(t) + M0 (t)* f~.K(t) dG(t)],
(5)
Mn(t) = F(t)*{ [K(t)- K(t)] G(t)
+1~0Mn_j(t)* J)K(t)-K(t)]dG(t)}
(n = 1,2, ...),
(6)
where note that K(t)- K(t) represents the probability that the total numbers of the failed spares are n during (0, t]. Introducing the notation
M(z,s) =
n~Ozn fooo e-sldMn(t),
we have, from (5) and (6),
1-f(s) [ 1-j~ zi M(z,s) =
oo
J
00
0
foo
e-81 d{[K(t)-K
1-f(s)j~o zi J0 e-81 [K(j)(t)- K
(7)
We can prove that limz...,1 M(z,s) = 1. The LS transformation of the expected number of the failed spares during (0, t] is
m(s) = oM(z,s) OZ
Z=1
= f(s) J 0 e-81 G(t)drx(t)/ 00
1-f(s)g(s),
(8)
where rx(t) = "£.~ 1 K(j)(t) which is known as the renewal function (see Barlow and Proschan2). Further, the limit of the expected number of the failed spares
104
T. Nakagawa and S. Osaki- Reliability Analysis per unit of time is M= IimM(t) t~co
t
(9)
Result (9) can be intuitively derived because the numerator represents the total expected number of failed spares during the repair time of the main unit and the denominator the mean time from the starting point of the operating main unit to the repair completion. Example 1. G(t) = 1 - e-td In this case, from (4), the mean time to system failure, given that n spare units are provided at time 0, is
Note that, by adding one spare unit to the system, the mean time to system failure increases by a constant, a, independent of the number of the former spare units, where a= (1/A.+ 1/p.) [1-K(p.)]/[K(p.)]. Further, the LS transform of the expected number of failed spares during (0, t] is f(s)k(s+ p.) m(s) = [1-k(s+ p.)]{1- [p./(s+ p.)]f(s)}'
and its limit per unit of time is M
= 1/{(1/A.+ 1/p.)[1-K(p.)]/[K(p.)]},
which is also equal to 1/a. Other quantities are H(z s) = zf(s)k(s+p.) ' 1-zk(s+p.)-f(s)[p.f(s+p.)] [1-k(s+ p.)]' M(
) _ [1-f(s)]£1-zk(s+p.)] +f(s)[sf(s+p.)][1-k(s+ p.)] z,s 1-zk(s+p.)-f(s)[p.f(s+p.)]£1-k(s+p.)] •
EXPECTED COST PER UNIT OF TIME AND ITS OPTIMIZATION PROBLEM First we derive the expected cost per unit of time for an infinite time span, by introducing costs incurred for each failed main unit and each failed spare. This expected cost is easily deduced from the expected numbers of failed units. Then, we compare the expected cost for the system of both main unit and spares with that for the system of spares only, and determine which of the systems are more economical.
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Operational Research Quarterly Vol. 27 No. 1, i Each failed main unit which is repaired costs c,. which includes all costs resulting from its failure and repair. Each failed spare which is replaced and scrapped costs c8 including the cost of itself. Let C(t) be the total expected cost incurred during the interval (0, t]. Then, for an infinite time span, the expected cost per unit of time is given by C =lim C(t)/t = c,.M,.+c8 M, Hex>
(10)
(see Barlow and Proschan2), where M,. is the expected number of the failed main unit per unit of time and hence M,. = 1/(1/"A+ 1/p.). Thus, from (9), C = [c,.+c8 fo
(11)
which is also equal to the total expected cost of one cycle from the beginning of the operating main unit to its repair completion. If only spare units are allowed, then the expected cost C8 per unit of time is C8 = 8c8 •
Therefore, comparing (11) with (12), we have
(12) C~ C8
if and only if
c,.~c8 [8(1/"A+1/p.)- fo
(13)
It is hard to compute the above costs in general but for particular cases, results which are useful in practical situations can be obtained, e.g. cx(t) ~ 8t-1 for t~O (Barlow and Proschan2) and hence, if c,.~c8(8f"A+1) then C~C8 • On the other hand, if K(t) is IFR, i.e. K(t) has a non-decreasing failure rate, and c,.:::;; esC 8/ >..) then C:::;; C8 • In this case, where K(t) = 1- e-ot then the reverse inequality is also satisfied, i.e. if c,.~cs(8f>..) then C~C8 • Finally, in the model of Example 1, we have the relation C ~ C8 if and only if
c,. ~ C8{ 8(1/"A+ 1/p.)-K{p.)/[1-K(p.)]}. PREVENTIVE MAINTENANCE POLICY MINIMIZING THE EXPECTED COST Consider_ the preventive maintenance policy where the operating main unit is maintained at t0 hours (a constant) after its installation or is repaired at failure, whichever occurs first. The spare units work temporarily during the interval of the repair time or the maintenance time of the main unit. We assume that the time to maintenance completion has the same distribution G(t) as the repair time. The main unit is as good as new upon repair or maintenance and begins to operate immediately. The costs suffered for each failed main unit and each failed spare are the same c,. and c8 , respectively, as the preceding ones. The preventive maintenance cost cP where cP < c,. occurs for each non-failed main unit which is maintained. 106
T. Nakagawa and S. Osaki- Reliability Analysis The total expected cost of one cycle from the beginning of the operation of the main unit to repair (or preventive maintenance completion) is F(t0) [cp+cs
fo"' cx(t) dG(t)J +F(t0) [cr+C fo"' cx(t)dG(t)J, 8
(14)
and the mean time of one cycle is F(t0 ) [1/p.+t0]+F(t0)(1/p.)+
Jo['· tdF(t) =
1/p.+
Jo['·F(t)dt.
(15)
In this case, it can be easily shown (Barlow and Hunter1 or Ross15) that the expected cost per unit of time for an infinite time span is equal to the total expected cost of one cycle over its mean time and is given by (16)
which is a function of the time to make preventive maintenance, where c~=cp+c8 f~cx(t)dG(t) and c;=cr+c8 f~cx(t)dG(t). We have c~
t:
r(t0)[J;'F(t)dt+1fp.] -F(t0 ) =
c~f(c;-c~).
(17)
Thus we have the following optimum preventive maintenance policy (see the Appendix): (1) If r(t) is continuous and monotonically increasing then (i)
t:
is a finite and positive solution satisfying (17) uniquely for r(O)D,
t; = 0 for r(O);;::: d, (iii) t; = oo for r(oo) ~ D, (ii)
where d= p.c~f(c;-c~), D = [1/(1/A+ 1/p.)] [c;;cc;-c~)].
t;
(2) If r(t) is continuous and non-increasing then = 0 or In case (i) of (1), the expected cost per unit of time is C(tt) = r(tt)Ccr-Cp).
107
t; = oo. (18)
Operational Research Quarterly Vol. 27 No. 1, i and
r(to) [J:'.F(t)dt+ 1/p.] -F(t0 ) >r(t0)(1/>t+ 1/p.)-1
(19)
for 0::;;,;; t0 < oo. Let /0 satisfy the equation
r(i0) =D.
(20)
r;
Then there exists a unique f0 such that < f0 , which might be a useful upper limit of when we compute numerically. Until now we have assumed that the time to preventive maintenance completion has a distribution G(t). In reality, it would be different from the repair time in many cases and would probably also be smaller. So, we suppose that the time to preventive maintenance completion has an arbitrary distribution Gp(t) with mean 1/P.p· Then, in a similar way, the expected cost per unit of time for an infinite time span is
tt
r;
C(t ) _ F(to) [Cp + C8 LX> a:(t) dGp(t)] + F(t0) [ Cr + C8 0
-
f
00
0
a{t) dG(t)]
(21)
F(t0 ) / p.p+F(t0 ) / p.+ L''F(t)dt
which is equal to (16) for Gp(t) = G(t). We can also obtain the optimum preventive maintenance policy minimizing C(t0) in (21) by a similar method. CONCLUDING REMARKS For other optimization problems and numerical examples, consider the problem of ensuring that sufficient numbers of spares are provided initially to protect against shortage, i.e. if the probability, a:, of occurrences of no shortage during (0, T] is given a priori, we find the minimum number of spares to maintain this level of confidence. The answer can be shown by computing the minimum n such that 1- Hn(T);;;: a: or Hn(T)::;;,;; 1-a:. Consider the minimum number of the initial stocks during (0, T] "on the average" without probabilistic guarantee then we can easily obtain the minimum n such that ln;;;: T. Other kinds of optimization problems could be considered by using the results in this paper. We conclude with the following numerical example: K(t) = 1-e-Ot, G(t) = 1- e-pt and dF(t)fdt = {32 t e-Pt. In this case, the expected cost per unit of time is C = Cr+c8(8fp.)
2/fJ+ 1/p. and C8 = 8c8 , and hence, we have C ~ C8 if and only if Cr ~ cs(28/{!). Further, we also have that the failure rate r(t) is {32 t/(1 +{Jt) which is continuous and monotonically increasing. Thus, from the results (1) in Section 4,
108
T. Nakagawa and S. Osaki- Reliability Analysis if fJ::::;; D, i.e. fJ(c;- c~) ::::;; p.(2c~- c;), we should make no preventive maintenance of the main unit, where c~ = cp+c8(8fp.) and c; = cr+c8(8fp.). If fJ(c;-c~)> p.(2c~- c;), we should adopt a finite and unique t;, as an optimum preventive maintenance policy, which satisfies
(f12t
] = --,-1!-,. c'
-1- _o+fJto-(1-e-Pio) 1+fJt0 p.
cr-cp
In this case, the expected cost per unit of time is C(tt)
Further, from the inequality
= [fJ2 t; /(1 + fJtt)](cr- cp).
t; < i
0
in (20), we have
fJtt < p.c;f[fJ(c;- c~)- p.(2c~- c;)],
t;
which is very useful for computing numerically. Table 1 shows the numerical examples where we have applied the successive approximations for solving the nonlinear equation. TABLE 1. DEPENDENCE OF THE MEAN FAILURE TIME 2/{3 IN THE OPTIMUM PREVENTIVE MAIN· TENANCE TIMEt~ AND ITS ASSOCIATED EXPECTED COST PER UNIT OF TIME (1f{J 1, 1/p. 5,
=
c., = 1, c. = 2, c, = 10)
Optimum preventive maintenance time 2/13
to*
=
Expected cost per unit of time
C(t~)
1.00
0.06
2.18
2.00
0.31
2.13
3.00
0.78
2.06 1.94
4.00
1.54
s.oo
2.63
1.84
6.00
4.08
1.72
7.00
5.91
1.61
8.00
8.14
1.50
9.00
10.78
1.41
10.00
13.88
1.32
REFERENCES R. E. BARLOW and L. C. HUNTER (1960) Optimum preventive maintenance policies. Opns Res. 8, 90. 2 R. E. BARWW and F. PROSCHAN (1965) Mathematical Theory of Reliability. Wiley, New York. 3 U. N. BHAT (1973) Reliability of an independent components, s-spare system with exponential life times and general repair times. Technometrics 15, 529. 1
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Operational Research Quarterly Vol. 27 No. 1, i J. C. DivEROL (1974) Optimal continuous policies for repair and replacement. Opl. Res. Q. 25, 89. 5 R. W. DRINKWATER and N. A. J. HASTINGS (1967) An economic replacement model. Op/ Res. Q. 18, 121. 6 N. A. J. HASTINGS (1968) Some notes on dynamic programming and replacement. Opl Res. Q. 19, 453. 7 N. A. J. HASTINGS (1969) The repair limit replacement method. Opl Res. Q. 20, 337. 8 T. A. LAMBE (1974) The decision to repair or scrap a machine. Opl Res. Q. 25, 99. 9 P.M. MoRSE (1958) Queues, Inventories, and Maintenance. Wiley, New York. 10 T. NAKAGAWA (1974) The expected number of visits to state k before a total system failure of a complex system with repair maintenance. Opns Res. 22, 108. 11 T. NAKAGAWA and S. OsAKI (1974) The optimum repair limit replacement policies. Op/ Res. Q. 25, 311. 111 T. NAKAGAWA and S. OsAKI (1974) Optimum preventive maintenance policies for a 2-unit redundant system. IEEE Trans. Reliability R/23, 86. 13 R. NATARAJAN (1968) A reliability problem with spares and multiple repair facilities. Opns Res. 16, 1041. 14 S. OsAKI and T. AsAKURA (1970) A two-unit standby redundant system with repair and preventive maintenance. J. Appl. Prob. 7, 641. 15 S. M. Ross (1970) Applied Probability Models with Optimization Applications. HoldenDay, San Francisco. 16 D. V. ROZHDESTVENSKIY and G. N. FANARZID (1970) Reliability of a duplicated system with renewal and preventive maintenance. Engng Cybernetics 8, 475. 17 V. S. SRINIVASAN (1968) First emptiness in the spare parts problem for repairable components. Opns Res. 16, 407. 4
APPENDIX We shall prove the result of Section 4 under the assumption that cP
Then, it is easily shown that q(t0 ) is monotonically increasing (decreasing) if r(t0) is monotonically increasing (decreasing). Further, q(O) = (1/IL)r(O), q(oo) = (1/.\+ 1/IL)r(oo)-1. First suppose that r(t) is monotonically increasing and define A= c~/(c;-c~). If r(O)D, then q(O)A. Thus, from the monotonicity of q(t0 ), there exists a unique and finite tt satisfying (17), which minimizes C(to). If r(O)~d, then q(O)~A. Thus, q(to)>A and C'(to)>O for any positive t0 giving tt = 0 because C{t0) is monotonically increasing. If r(oo)~ D, then q(oo)~A, i.e. q(to) C(oo) then we adopt tt = oo, and vice versa.
110
