Sequential inspection strategy for multiple systems ... - Semantic Scholar
European Journal of Operational Research 155 (2004) 170–177 www.elsevier.com/locate/dsw
Stochastics and Statistics
Sequential inspection strategy for multiple systems under availability requirement L.R. Cui, H.T. Loh, M. Xie
*
National University of Singapore, 10 Kent Ridge Crescent, Singapore 119260, Singapore Received 5 June 2001; accepted 25 September 2002
Abstract System failures are usually observed during regular maintenance or inspection and this is especially the case for systems in standby or storage, which is common for safety critical systems. A periodic inspection policy is usually adopted. However, during the inspection, a lot of information is gained about the status of the system. Such information should be used in deciding upon the time for the next inspection. Hence sequential inspection is more appropriate, especially when the aging property of the system is unknown, and has to be estimated with the information from inspection. In this paper, a model is developed and sequential inspection strategies are studied in this situation. The focus is on the case when there are multiple systems inspected at the same, but discrete times. We also do not assume a known distribution of the system life time, and the estimation of that is incorporated into the analysis and decision making. Different availability criteria are considered and numerical examples are provided to illustrate the procedure. Ó 2003 Elsevier B.V. All rights reserved. Keywords: Sequential inspection; Availability; Reliability; Repair; Replacement
1. Introduction A system can either be in an operational state or a down state. However, the state of the system may actually be usually unknown unless it is inspected (see, e.g., Cerone, 1993; Wortman et al., 1994; Vaurio, 1999). This is typical for systems in
*
Corresponding author. Address: Department of Industrial and Systems Engineering, National University of Singapore, 10 Kent Ridge Crescent, Singapore 119260, Singapore. Fax: +656777-1434. E-mail address: [email protected] (M. Xie).
storage and for systems that can still perform a limited function after failure of some of its components. Although continuous monitoring and inspection is possible, periodic or discrete time inspection is usually employed, due to the cost and other practical constrictions. Here the term of periodic inspection used does not necessarily mean that the system is inspected after every fixed period of time regularly, but rather that the system is inspected at some discrete points in time. For systems for which failures are only detected at the time of inspection, it is important to be able to determine the optimal time of inspection. Fewer inspections will lead to lower availability upon
0377-2217/$ - see front matter Ó 2003 Elsevier B.V. All rights reserved. doi:10.1016/S0377-2217(02)00822-6
L.R. Cui et al. / European Journal of Operational Research 155 (2004) 170–177
171
demand, and frequent inspection will lead to higher cost. When there is an availability requirement, the problem is usually to develop an inspection policy that meets the availability requirements. The problem is formulated for periodic inspection to minimize the cost with respect to the time interval for inspection (Cerone, 1993; Ito and Nakagawa, 2000; Hariga, 1996; Vaurio, 1999). Most of the optimal policies are derived based on average cost or reliability that are valid only asymptotically. In general, the system reaching the limiting or steadystate, for example, requires a long period of usage and usually it is not clear when its asymptotic results are accurate enough. Yang and Klutke (2000) studied some inspection schemes, in which they focused on steadystate availabilities for several models under some inspection schemes. The inspection policy defined is based on the availability of the system when it is required. It is given as follows. Fix 0 < a < 1 and let
strategies based on a finite-state continuous-time Markov model, see also Yeh (1997). Brint (2000) discussed the problem of sequential inspection sampling for maintained system and presented a Bayesian formulation. Chelbi and Ait-Kadi (2000) also considered some general inspection strategies. These papers all deal with the case of a single system. In this paper, we develop a sequential inspection policy for the case when there are a number of systems on the site or in storage. The condition of the systems is inspected and failed systems are replaced or repaired in such a way that they become as good as new. The time to the next inspection is determined based on the results of previous inspections. Furthermore, the system lifetime distribution is not assumed to be known, but incorporated into the analysis and decision making. Different availability criteria are discussed and numerical examples are provided to illustrate the procedure.
s1 ¼ supft > 0 : P fL > tg P ag; sN ¼ supft > sN 1 : P fL > tjL > sN 1 g P ag;
2. The model and assumptions
N P 2; where fsi gi¼1;2;... are inspection time, L, a random variable representing the lifetime of the system. This is named as ‘‘quantile-based inspection’’. It is clear that if F , the system lifetime distribution, is continuous and strictly increasing, the definition is equivalent to rN ¼ R1 ðaN Þ; N ¼ 1; 2; . . .. RðxÞ ¼ 1 F ðxÞ. This inspection policy makes use of the information about the remaining life that is inherent in the sequence of previous inspection times. The value of a can be seen as ‘‘minimum availability required’’ during the next period when the system was still operational at last inspection time. However, they assumed that the distribution of system lifetime is known and the policy is used for a single system, that is, only one item is inspected. In many cases, there can be several similar systems operating simultaneously on one site or under the same environment. The distribution of system lifetime, in most cases, is also unknown. Lam and Yeh (1994) discussed a sequential policy and compared it with some continuous
We assume that there are n items (or systems) in the field and they are inspected at the same, but discrete times. The following assumptions are used. Assumptions 1. The inspections are carried out at times s1 ; s2 ; . . .; and all n systems are inspected each time. 2. Failures are observed only by inspection, and replacement or perfect repair is carried out for failed systems. 3. The inspection action does not intervene with the system if failure is not found. 4. The inspection time and repair/replacement time are negligible. 5. The lifetimes of all systems have the same distribution with cdf F ðxÞ, but the parameters are not known. Note that a is the required availability or the probability that the system is still functioning. One
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L.R. Cui et al. / European Journal of Operational Research 155 (2004) 170–177
System n System n-1
System 2 System 1 τ2
τ1
τ3
Time τ5
τ4
Failure periods
Working periods
Fig. 1. One sample of the inspected system state.
sample path of the inspected system state is shown in Fig. 1. Let 8 < 1; if the failure for system i ðlÞ Xi ¼ is found at the lth inspection; : 0; otherwise: When the reliability function is known and there is only one system i, then the next inspection time is ðiÞ mj which is the solution of ðiÞ
Rj1 ðsj1 sjk þ mj Þ ¼ aRj1 ðsj1 sjk Þ ðj1Þ Xi
ðjkþ1Þ Xi
ðjkÞ Xi
¼ ¼ ¼ 0; ¼ 1, assuming that is, the last failure for system i found at the ðj kÞth inspection and is still operational at this inspection. When adopting a sequential strategy in practice, the reliability function is usually unknown. Denote b j1 ðsj1 sjk Þ the estimated reliability funcby R tion. The inspection time for system i is estimated as ðiÞ b j1 ðsj1 sjk þ mðiÞ mj which is the solution of R j Þ ¼ ðj1Þ ðjkÞ ðjkÞ b a R j1 ðsj1 sjk Þ if Xi ¼ ¼ Xi ¼ 0; Xi ¼ ðj1Þ b j1 ðvðiÞ 1; k P 2 or R Þ ¼ a if X ¼ 1. Also, denote j i ðiÞ
ðiÞ
by sj ¼ sj1 þ mj , where sj1 is the (j 1)th inðiÞ spection time; mj is a required inspection period for ðiÞ system i in the jth inspection and sj is the jth inspection time for system i to meet its required availability. Different criteria can be used to determine the time to the next inspection: ðiÞ
sj ¼ min fsj g; 16i6n
ð1aÞ
sj ¼
n 1X ðiÞ s ; n i¼1 j
ð1bÞ ðiÞ
sj ¼ v%-percentile of descending ordered sj ; ð1cÞ or sj ¼ sj1 þ pj1 ;
ð2Þ
where pj satisfies the following equation: min
n X
b j1 ½pj I ðj1Þ þ ðsj1 sjk þ pj ÞIfEg fR fX ¼1g i
i¼1
b j1 ðsj1 sjk ÞIfEg g2 aIfX ðj1Þ ¼1g a R i
and the event ðj1Þ
E ¼fXi
ðjkþ1Þ
¼ ¼ Xi
ðjkÞ
¼ 0; Xi
¼ 1;
k > 2g; where pj is a required inspection period for systems in the jth inspection. The explanations of the above criteria are as follows. Criterion (1a) is a conservative inspection strategy, since every system in the field satisfies the minimum availability not less than a when the system is operational on the last inspection time. Criterion (1b) is a reasonable inspection strategy which takes a balance among all the systems as a whole. The criterion (1c), inspection strategy ensures that the availabilities of v% of the systems is not less than a. For the criterion (2), inspection strategy makes the sum of squares of the difference
L.R. Cui et al. / European Journal of Operational Research 155 (2004) 170–177
between system availability and requirement availability a, minimal. All criteria result from the basic inequality P fL > si g P aP fL > si1 g.
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Step 4: Determine the next inspection time based on an appropriate criteria; Step 5: Summarize the information as Fi ¼ finformation obtained by time si g
3. Sequential inspection procedure The main tasks for the sequential inspection policy are: (1) to provide an estimate of the parameters of the lifetime distribution; (2) to derive optimal inspection times based on information on the remaining life that is inherent in the sequence of previous inspection times. The sequential inspection procedure is first stated as follows. The explanations are presented later. We also discuss the issues related to the estimation of the parameters. 3.1. The procedure The sequential inspection procedure can be summarized as follows: Step 1: The first inspection time s1 is based on some information or standard guidelines such as contractual requirement; Step 2: After the first inspection at time s1 , the state of each system is observed and the state ð1Þ vector is denoted by fX1 ; . . . ; Xnð1Þ g; Step 3: Estimate the parameters based on the ð1Þ sample fX1 ; . . . ; Xnð1Þ g. If there is one parameter in the distribution, we b 1 ðs1 Þ ¼ s1 =n, where s1 is the number of can get R functional systems, from which, parameter may be easily estimated and s1 is the initial inspection time that can be determined based on the experience or contractual requirement. If there are two parameters in the distribution, we need a second inspection time s2 , then we get b 2 ðs1 Þ; R b 2 ðs2 ÞÞ. Two parameters are evaluated ðR based on this. Furthermore, for the three parameters case, if the parameters can be derived from this, then go to step 4. Otherwise, we need a third inspection time, and so on.
[fthere are ri ðsi Þ failed ðsuccessÞ systems during interval ½0; bi g where bi 2 B ¼ fDs1 ; . . . ; Dsm ; Ds1 þ Ds2 ; . . . ; Dsm1 þ Dsm ; . . . ; Ds1 þ Dsm g and repeat step 4 as long as it is necessary. Here Fi denotes the information on the whole system just after the ith inspection, ri ðsi Þ is the number of failed (successed) systems during the interval ½0; bi , bi is possible inspection times, and Dsj ¼ sj sj1 , where s0 ¼ 0. 3.2. Parameter estimation Standard statistical procedures can be used to estimate the parameters and it can be done after each inspection to incorporate the latest information. However, as the information is censored and we do not assume that we have detailed information for each system, it is tedious to adopt the traditional method such as maximum likelihood estimation. An adapted Weibull plotting approach is suggested. Weibull distribution is widely used in reliability, since it is suitable to describe the deteriorating system in many cases. We use Fi to obtain an estimation of the parameters. In general, the following least-square method can be used. First, we estimate Rðbi Þ by b ðbi Þ ¼ si =ðsi þ ri Þ. If some si ¼ 0 (or ri ¼ 0), then R b ðbi Þ ¼ 1=ð1 þ ri Þ (or R b ðbi Þ ¼ we can use R si =ðsi þ 1Þ) or any other empirical estimates. Although this is a rough estimate, it is reasonable and simple and hence adopted here. Second, we linearize RðtÞ to transform the parameters to regression parameters of a linear equation. For example, for the two-parameter b Weibull distribution, RðtÞ ¼ expððt=dÞ Þ, and after the linearization, we have y ¼ a1 x þ a0 , where y ¼ lnð lnðRðtÞÞÞ, a0 ¼ b, x ¼ lnðtÞ, a1 ¼ b lnðdÞ. Thus we can use the least squares method to estimate the parameters in distribution. For Weibull distribution parameter ðb; dÞ we use the following to estimate the parameters. Namely,
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min
L.R. Cui et al. / European Journal of Operational Research 155 (2004) 170–177 jBj X
ðyi a1 xi a0 Þ
Proof. For every i since bi < biþ1 , there are ri failed (siþ1 success) systems in the interval ½0; bi ð½0; biþ1 Þ. These ri failed (siþ1 success) systems must be counted into the interval ½0; biþ1 ð½0; bi Þ.
2
i¼1
where yi ¼ lnð lnðsi =ðsi þ ri ÞÞÞ, xi ¼ lnðbi Þ. From the standard regression analysis, we have that the estimates can be obtained as Pn Pn Pn 8 n xi yi xi yi > i¼1 i¼1 i¼1 > P P ^ a ¼ ; > 2 n n < 0 n x2i ð xi Þ i¼1 i¼1 Pn Pn 2 Pn Pn > yi xi xi xi yi >^ i¼1 > : : a1 ¼ i¼1 Pni¼1 2 Pi¼1n 2 n x ð xi Þ i¼1 i i¼1 That is, b^ ¼ a^0 , d^ ¼ expð^ a1 =b^Þ, where n ¼ jBj is the number of distinct elements in B. OfPcourse, one can adopt any other methods b ðÞÞ2 to get an estimate of model min ðRðÞ R parameters. The following results can be useful. Proposition 1. jBj 6 space Fm .
mðmþ1Þ 2
4. Some illustrative examples Here some numerical examples are shown in order to illustrate how the procedure works. The data is simulated from the Weibull distribution b F ðxÞ ¼ 1 expððx=dÞ Þ, b ¼ 2:0, d ¼ 20:0. The number of systems in the field is ten, i.e., n ¼ 10, and the required ‘‘availability’’ is a ¼ 0:6. We shall discuss the results for all four criteria. The cases below refer to the corresponding criteria. ðjÞ
Case 1a. For this case, si ¼ min1 6 j 6 n fsi g. Let Dsi ¼ si si1 and s0 ¼ 0. The steps can be described as follows. Step 1: Assume that based on other information we first determine Ds1 ¼ 15. After the inspection the simulated data shown is in the second column of Table 1. From that we have F1 ¼ fn ¼ 10; ðr1 ; s1 Þ ¼ ð5; 5Þg, then an estimate of b 1 ð15Þ ¼ 0:5. R Step 2: Since the model has two parameters, we need information from the second inspection, and take this as Ds2 ¼ 12. After this inspection, the data is as that shown in the third column of Table 1. Thus,
for the information
Proof. There are m þ ðm 1Þ þ þ 1 ¼ mðm þ 1Þ=2 combinatorial cases in Fm . However, some cases may lead to the same element. Hence the number of distinct elements is less than mðm þ 1Þ=2. Proposition 2. If we arrange bi in B in ascending order, then ri 6 riþ1 ; si P siþ1 for every i. Table 1 Data from inspections and the sequential inspection time (Criterion 1a) No.
1
2
3
4
5
6
7
8
9
1 2 3 4 5 6 7 8 9 10 Dsi b^i d^i
–– –– –– –– –– 15
–– –– –– –– –– 12 2.151 20.265
–– –– –– –– –– –– –– 3.235 1.629 17.775
–– –– –– –– –– –– –– –– –– –– 3.840 1.841 18.572
–– –– –– –– –– –– –– –– –– 2.984 1.817 20.195
–– –– –– –– –– –– –– –– 4.852 1.817 20.195
–– –– –– –– –– –– –– –– –– 4.222 1.817 20.195
–– –– –– –– –– –– 5.336 1.817 20.195
–– –– –– –– –– –– –– 4.516 1.817 20.195
Note that: throughout this section: denotes the system is down before the inspection; –– denotes the system is still up after the inspection.
L.R. Cui et al. / European Journal of Operational Research 155 (2004) 170–177
F2 ¼fn ¼ 15; ½0; b1 ¼ ½0; 12 ) ðr1 ; s1 Þ ¼ ð2; 8Þ; ½0; b2 ¼ ½0; 15 ) ðr2 ; s2 Þ ¼ ð7; 5Þ; ½0; b3 ¼ ½0; 27 ) ðr3 ; s3 Þ ¼ ð10; 2Þg: Thus, using the empirical estimate, we have b 2 ð12Þ ¼ 8=10; R b 2 ð15Þ ¼ 5=12; R b 2 ð27Þ ¼ 2=12. R Based on the formula in the details of estimation of Section 3.2, we get ðb^2 ; d^2 Þ ¼ ð2:15; 20:265Þ and ð1Þ b 2 ðm3ð1Þ Þ ¼ 0:6 s3 ) R
ð2Þ
s3
ð3Þ
s3
ð1Þ
ð4Þ
ð6Þ
ð2Þ
ð7Þ
ð8Þ
ð3Þ
ð5Þ
ð9Þ
ð10Þ
) m3 ¼ m3 ¼ m3 ¼ m3 ¼ m3 ¼ 13:846; b 2 ðDs2 þ m3ð2Þ Þ ¼ 0:6 R b 2 ðDs2 Þ )R ) m3 ¼ m3 ¼ m3 ¼ 6:633; b 2 ðs2 þ m3ð3Þ Þ ¼ 0:6 R b 2 ðs2 Þ )R ) m3 ¼ m3 ¼ 3:235;
s3 ¼ s2 þ min
1 6 i 6 10
ðiÞ fm3 g
¼ 27 þ 3:235 ¼ 30:235:
Step 3: An inspection is carried out after Ds3 ¼ 3:235, and the data is shown in fourth column of Table 1. We then obtain F3 ¼fn ¼ 20; ½0; b1 ¼ ½0; 3:235 ) ðr1 ; s1 Þ ¼ ð0; 13Þ; ½0; b2 ¼ ½0; 12 ) ðr2 ; s2 Þ ¼ ð2; 8Þ; ½0; b3 ¼ ½0; 15 ) ðr3 ; s3 Þ ¼ ð7; 6Þ; ½0; b4 ¼ ½0; 15:235 ) ðr4 ; s4 Þ ¼ ð9; 3Þ; ½0; b5 ¼ ½0; 27 ) ðr5 ; s5 Þ ¼ ð12; 2Þ; ½0; b6 ¼ ½0; 30:235 ) ðr6 ; s6 Þ ¼ ð13; 1Þg:
In a similar way, we get that ðb^3 ; d^3 Þ ¼ ð1:629; 17:775Þ and Ds4 ¼ 3:840. Step 4: Another inspection is carried out after Ds4 ¼ 3:840. The data is shown in the fifth column of Table 1. We also obtain that F4 ¼fn ¼ 23; ½0; b1 ¼ ½0; 3:235 ) ðr1 ; s1 Þ ¼ ð0; 16Þ; ½0; b2 ¼ ½0; 3:840 ) ðr2 ; s2 Þ ¼ ð0; 16Þ; ½0; b3 ¼ ½0; 7:075 ) ðr3 ; s3 Þ ¼ ð0; 13Þ; ½0; b4 ¼ ½0; 12 ) ðr4 ; s4 Þ ¼ ð2; 8Þ; ½0; b5 ¼ ½0; 15 ) ðr5 ; s5 Þ ¼ ð7; 6Þ; ½0; b6 ¼ ½0; 15:235 ) ðr6 ; s6 Þ ¼ ð9; 3Þ; ½0; b7 ¼ ½0; 19:075 ) ðr7 ; s7 Þ ¼ ð9; 3Þ; ½0; b8 ¼ ½0; 27 ) ðr8 ; s8 Þ ¼ ð12; 2Þ; ½0; b9 ¼ ½0; 30:235 ) ðr9 ; s9 Þ ¼ ð13; 1Þ; ½0; b10 ¼ ½0; 34:075 ) ðr10 ; s10 Þ ¼ ð13; 1Þg:
175
We get ðb^4 ; d^4 Þ ¼ ð1:841; 18:572Þ and Ds5 ¼ 2:984. Step 5: We then get F5 , ðb^5 ; d^5 Þ ¼ ð1:817; 20:195Þ and Ds6 ¼ 4:852, and we can continue with this procedure. It is tedious to deal with Fi . In fact, based on the calculated results, as the information increases, the estimate of ðb; dÞ—ðb^; d^Þ changes very little, that is, F is saturated in inferring the parameters of Weibull distribution after only a few steps. We can get the estimate of ðb^; d^Þ ¼ ð1:817; 20:195Þ. Based on ðb; dÞ ¼ ð1:817; 20:195Þ, we get Dsi . For example, Ds6 ¼ 4:852, Ds7 ¼ 4:222, Ds8 ¼ 5:336, Ds9 ¼ 4:516. Pn ðjÞ Case 1b. For the case of si ¼ ð1=10Þ j¼1 si , we get Table 2 by using a similar procedure. Case 1c. For the case of si in the top 80% of ðjÞ fsi ; j ¼ 1; 2; . . . ; 10g, that is v ¼ 80%, we can similarly obtain Table 3. Case 2. For this case, si is obtained by pj1 , in the second step. We minimize the following function in order to get p2 : b 2 ð12 þ p2 Þ 0:6 R b 2 ð12ÞÞ2 b 2 ðp2 Þ 0:6Þ2 þ 3ð R 5ð R b 2 ð27 þ p2 Þ 0:6 R b 2 ð27ÞÞ2 : þ 2ð R In this case p2 ¼ 12:780. Similar arguments as before are used in order to obtain other values of pi . After these steps, we have the results in Table 4. It should be pointed out that in the above tables, some b^i , the estimation of b, are biased. This is partly due to the simple method used, but also due to the fact that the actual failure time data is not available, and it is neither grouped nor simply censored data. Because of this, methods such as maximum likelihood are difficult to be employed. On the other hand, it is a fact that the distribution of the lifetime in actual use is usually unknown and most models need to be estimated with field data.
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L.R. Cui et al. / European Journal of Operational Research 155 (2004) 170–177
Table 2 Data from inspections and the sequential inspection time (Criterion 1b) No.
1
2
3
4
5
6
7
8
9
1 2 3 4 5 6 7 8 9 10 Dsi b^i d^i
–– –– –– –– –– 15
–– –– –– –– –– 12 2.151 20.265
–– –– –– –– –– –– 10.051 2.612 21.506
–– –– –– –– –– –– 11.089 2.695 23.472
–– –– –– –– 11.748 2.369 24.107
–– –– –– –– 13.647 2.457 22.338
–– –– –– 12.617 2.457 22.338
–– –– –– –– –– –– 14.093 2.457 22.338
–– –– –– –– –– 10.113 2.457 22.338
Table 3 Data from inspections and the sequential inspection time (Criterion 1c) No.
1
2
3
4
5
6
7
8
9
1 2 3 4 5 6 7 8 9 10 Dsi b^i d^i
–– –– –– –– –– 15
–– –– –– –– –– 12 2.151 20.265
–– –– –– –– –– –– –– 6.633 2.306 19.624
–– –– –– –– –– –– 9.010 2.407 19.983
–– –– –– –– –– –– –– 4.873 2.224 20.847
–– –– –– –– –– –– –– 6.115 2.272 20.180
–– –– –– –– –– –– 6.899 2.272 20.180
–– –– –– –– –– –– 6.046 2.272 20.180
–– –– –– –– –– –– –– 6.088 2.272 20.180
Table 4 Data from inspections and the sequential inspection time (Criterion 2) No.
1
2
3
4
5
6
7
8
9
1 2 3 4 5 6 7 8 9 10 Dsi b^i d^i
–– –– –– –– –– 15
–– –– –– –– –– 12 2.151 20.265
–– –– –– –– 12.780 2.091 20.945
–– –– –– –– –– 13.540 2.010 22.151
–– –– –– –– –– 12.745 2.145 21.436
–– –– –– 14.120 2.099 21.199
–– –– –– 14.140 2.099 21.199
–– –– –– –– –– –– –– 14.142 2.099 21.199
–– –– –– –– –– 10.539 2.099 21.199
L.R. Cui et al. / European Journal of Operational Research 155 (2004) 170–177
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5. Conclusions and discussions
Acknowledgements
In this paper the model in Yang and Klutke (2000) is extended to the case of multiple systems. Under different criteria, sequential plans are obtained to ensure that the availability is at the required level. Sequential inspection is important, especially during the set-up and installation stage. Frequent inspection leads to a high cost and infrequent inspection will lead to low availability of the system upon demand. Although the cost might be an issue in this type of analysis, the focus here is to meet the availability requirement with an appropriate time to next inspection. The sequential inspection procedure and decision making procedure studied in this paper allows an appropriate level of availability to be reached with minimum cost as well. Furthermore, we do not assume the distribution of system lifetime to be completely known which is usually the case. The information from the inspection can be used to determine the parameters of system lifetime distribution. Hence, such a combined estimation and decision-making analysis is important and useful in practice. Although the procedure proposed in this paper can be implemented easily, there are several interesting questions that could be raised. Since the estimation for parameters of the lifetime distribution is required at the beginning, one could investigate different estimation methods and also investigate the effect of the estimation error. A common assumption is the independence of the failures and the case of dependence is of interesting. Modelling of the dependence is generally difficult as specific models describing the degree of dependence will be needed.
This research is supported by a research grant from the National University of Singapore for the project ‘‘Reliability analysis of engineering systems’’ (RP3992679). The authors would like to thank two anonymous referees for their helpful comments and suggestions, which improved the paper. References Brint, A.T., 2000. Sequential inspection sampling to avoid failure critical items being in an at risk condition. Journal of the Operational Research Society 51, 1051–1059. Cerone, P., 1993. Inspection interval for maximum future reliability using the delay-time model. European Journal of Operations Research 68, 236–250. Chelbi, A., Ait-Kadi, D., 2000. Generalized inspection strategy for randomly failing systems subjected to random shocks. International Journal of Production Economics 64, 379– 384. Hariga, M.A., 1996. A maintenance inspection model for a single machine with general failure distribution. Microelectronics and Reliability 36, 353–358. Ito, K., Nakagawa, T., 2000. Optimal inspection policies for a storage system with degradation at periodic tests. Mathematical and Computer Modelling 31, 191–195. Lam, C.T., Yeh, R.H., 1994. Comparison of sequential and continuous inspection strategies for deteriorating systems. Advances in Applied Probability 26, 423–435. Vaurio, J.K., 1999. Availability and cost functions for periodically inspected preventively maintained units. Reliability Engineering and System Safety 63, 133–140. Wortman, M.A., Klutke, G.A., Ayhan, H., 1994. A maintenance strategy for systems subjected to deterioration governed by random shocks. IEEE Transactions on Reliability 43, 439–445. Yang, Y., Klutke, G.A., 2000. Improved inspection schemes for deteriorating equipment. Probability in the Engineering and Informational Sciences 14, 445–460. Yeh, R.H., 1997. Optimal inspection and replacement policies for multi-state deteriorating systems. European Journal of Operations Research 96, 248–259.
Stochastics and Statistics
Sequential inspection strategy for multiple systems under availability requirement L.R. Cui, H.T. Loh, M. Xie
*
National University of Singapore, 10 Kent Ridge Crescent, Singapore 119260, Singapore Received 5 June 2001; accepted 25 September 2002
Abstract System failures are usually observed during regular maintenance or inspection and this is especially the case for systems in standby or storage, which is common for safety critical systems. A periodic inspection policy is usually adopted. However, during the inspection, a lot of information is gained about the status of the system. Such information should be used in deciding upon the time for the next inspection. Hence sequential inspection is more appropriate, especially when the aging property of the system is unknown, and has to be estimated with the information from inspection. In this paper, a model is developed and sequential inspection strategies are studied in this situation. The focus is on the case when there are multiple systems inspected at the same, but discrete times. We also do not assume a known distribution of the system life time, and the estimation of that is incorporated into the analysis and decision making. Different availability criteria are considered and numerical examples are provided to illustrate the procedure. Ó 2003 Elsevier B.V. All rights reserved. Keywords: Sequential inspection; Availability; Reliability; Repair; Replacement
1. Introduction A system can either be in an operational state or a down state. However, the state of the system may actually be usually unknown unless it is inspected (see, e.g., Cerone, 1993; Wortman et al., 1994; Vaurio, 1999). This is typical for systems in
*
Corresponding author. Address: Department of Industrial and Systems Engineering, National University of Singapore, 10 Kent Ridge Crescent, Singapore 119260, Singapore. Fax: +656777-1434. E-mail address: [email protected] (M. Xie).
storage and for systems that can still perform a limited function after failure of some of its components. Although continuous monitoring and inspection is possible, periodic or discrete time inspection is usually employed, due to the cost and other practical constrictions. Here the term of periodic inspection used does not necessarily mean that the system is inspected after every fixed period of time regularly, but rather that the system is inspected at some discrete points in time. For systems for which failures are only detected at the time of inspection, it is important to be able to determine the optimal time of inspection. Fewer inspections will lead to lower availability upon
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L.R. Cui et al. / European Journal of Operational Research 155 (2004) 170–177
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demand, and frequent inspection will lead to higher cost. When there is an availability requirement, the problem is usually to develop an inspection policy that meets the availability requirements. The problem is formulated for periodic inspection to minimize the cost with respect to the time interval for inspection (Cerone, 1993; Ito and Nakagawa, 2000; Hariga, 1996; Vaurio, 1999). Most of the optimal policies are derived based on average cost or reliability that are valid only asymptotically. In general, the system reaching the limiting or steadystate, for example, requires a long period of usage and usually it is not clear when its asymptotic results are accurate enough. Yang and Klutke (2000) studied some inspection schemes, in which they focused on steadystate availabilities for several models under some inspection schemes. The inspection policy defined is based on the availability of the system when it is required. It is given as follows. Fix 0 < a < 1 and let
strategies based on a finite-state continuous-time Markov model, see also Yeh (1997). Brint (2000) discussed the problem of sequential inspection sampling for maintained system and presented a Bayesian formulation. Chelbi and Ait-Kadi (2000) also considered some general inspection strategies. These papers all deal with the case of a single system. In this paper, we develop a sequential inspection policy for the case when there are a number of systems on the site or in storage. The condition of the systems is inspected and failed systems are replaced or repaired in such a way that they become as good as new. The time to the next inspection is determined based on the results of previous inspections. Furthermore, the system lifetime distribution is not assumed to be known, but incorporated into the analysis and decision making. Different availability criteria are discussed and numerical examples are provided to illustrate the procedure.
s1 ¼ supft > 0 : P fL > tg P ag; sN ¼ supft > sN 1 : P fL > tjL > sN 1 g P ag;
2. The model and assumptions
N P 2; where fsi gi¼1;2;... are inspection time, L, a random variable representing the lifetime of the system. This is named as ‘‘quantile-based inspection’’. It is clear that if F , the system lifetime distribution, is continuous and strictly increasing, the definition is equivalent to rN ¼ R1 ðaN Þ; N ¼ 1; 2; . . .. RðxÞ ¼ 1 F ðxÞ. This inspection policy makes use of the information about the remaining life that is inherent in the sequence of previous inspection times. The value of a can be seen as ‘‘minimum availability required’’ during the next period when the system was still operational at last inspection time. However, they assumed that the distribution of system lifetime is known and the policy is used for a single system, that is, only one item is inspected. In many cases, there can be several similar systems operating simultaneously on one site or under the same environment. The distribution of system lifetime, in most cases, is also unknown. Lam and Yeh (1994) discussed a sequential policy and compared it with some continuous
We assume that there are n items (or systems) in the field and they are inspected at the same, but discrete times. The following assumptions are used. Assumptions 1. The inspections are carried out at times s1 ; s2 ; . . .; and all n systems are inspected each time. 2. Failures are observed only by inspection, and replacement or perfect repair is carried out for failed systems. 3. The inspection action does not intervene with the system if failure is not found. 4. The inspection time and repair/replacement time are negligible. 5. The lifetimes of all systems have the same distribution with cdf F ðxÞ, but the parameters are not known. Note that a is the required availability or the probability that the system is still functioning. One
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System n System n-1
System 2 System 1 τ2
τ1
τ3
Time τ5
τ4
Failure periods
Working periods
Fig. 1. One sample of the inspected system state.
sample path of the inspected system state is shown in Fig. 1. Let 8 < 1; if the failure for system i ðlÞ Xi ¼ is found at the lth inspection; : 0; otherwise: When the reliability function is known and there is only one system i, then the next inspection time is ðiÞ mj which is the solution of ðiÞ
Rj1 ðsj1 sjk þ mj Þ ¼ aRj1 ðsj1 sjk Þ ðj1Þ Xi
ðjkþ1Þ Xi
ðjkÞ Xi
¼ ¼ ¼ 0; ¼ 1, assuming that is, the last failure for system i found at the ðj kÞth inspection and is still operational at this inspection. When adopting a sequential strategy in practice, the reliability function is usually unknown. Denote b j1 ðsj1 sjk Þ the estimated reliability funcby R tion. The inspection time for system i is estimated as ðiÞ b j1 ðsj1 sjk þ mðiÞ mj which is the solution of R j Þ ¼ ðj1Þ ðjkÞ ðjkÞ b a R j1 ðsj1 sjk Þ if Xi ¼ ¼ Xi ¼ 0; Xi ¼ ðj1Þ b j1 ðvðiÞ 1; k P 2 or R Þ ¼ a if X ¼ 1. Also, denote j i ðiÞ
ðiÞ
by sj ¼ sj1 þ mj , where sj1 is the (j 1)th inðiÞ spection time; mj is a required inspection period for ðiÞ system i in the jth inspection and sj is the jth inspection time for system i to meet its required availability. Different criteria can be used to determine the time to the next inspection: ðiÞ
sj ¼ min fsj g; 16i6n
ð1aÞ
sj ¼
n 1X ðiÞ s ; n i¼1 j
ð1bÞ ðiÞ
sj ¼ v%-percentile of descending ordered sj ; ð1cÞ or sj ¼ sj1 þ pj1 ;
ð2Þ
where pj satisfies the following equation: min
n X
b j1 ½pj I ðj1Þ þ ðsj1 sjk þ pj ÞIfEg fR fX ¼1g i
i¼1
b j1 ðsj1 sjk ÞIfEg g2 aIfX ðj1Þ ¼1g a R i
and the event ðj1Þ
E ¼fXi
ðjkþ1Þ
¼ ¼ Xi
ðjkÞ
¼ 0; Xi
¼ 1;
k > 2g; where pj is a required inspection period for systems in the jth inspection. The explanations of the above criteria are as follows. Criterion (1a) is a conservative inspection strategy, since every system in the field satisfies the minimum availability not less than a when the system is operational on the last inspection time. Criterion (1b) is a reasonable inspection strategy which takes a balance among all the systems as a whole. The criterion (1c), inspection strategy ensures that the availabilities of v% of the systems is not less than a. For the criterion (2), inspection strategy makes the sum of squares of the difference
L.R. Cui et al. / European Journal of Operational Research 155 (2004) 170–177
between system availability and requirement availability a, minimal. All criteria result from the basic inequality P fL > si g P aP fL > si1 g.
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Step 4: Determine the next inspection time based on an appropriate criteria; Step 5: Summarize the information as Fi ¼ finformation obtained by time si g
3. Sequential inspection procedure The main tasks for the sequential inspection policy are: (1) to provide an estimate of the parameters of the lifetime distribution; (2) to derive optimal inspection times based on information on the remaining life that is inherent in the sequence of previous inspection times. The sequential inspection procedure is first stated as follows. The explanations are presented later. We also discuss the issues related to the estimation of the parameters. 3.1. The procedure The sequential inspection procedure can be summarized as follows: Step 1: The first inspection time s1 is based on some information or standard guidelines such as contractual requirement; Step 2: After the first inspection at time s1 , the state of each system is observed and the state ð1Þ vector is denoted by fX1 ; . . . ; Xnð1Þ g; Step 3: Estimate the parameters based on the ð1Þ sample fX1 ; . . . ; Xnð1Þ g. If there is one parameter in the distribution, we b 1 ðs1 Þ ¼ s1 =n, where s1 is the number of can get R functional systems, from which, parameter may be easily estimated and s1 is the initial inspection time that can be determined based on the experience or contractual requirement. If there are two parameters in the distribution, we need a second inspection time s2 , then we get b 2 ðs1 Þ; R b 2 ðs2 ÞÞ. Two parameters are evaluated ðR based on this. Furthermore, for the three parameters case, if the parameters can be derived from this, then go to step 4. Otherwise, we need a third inspection time, and so on.
[fthere are ri ðsi Þ failed ðsuccessÞ systems during interval ½0; bi g where bi 2 B ¼ fDs1 ; . . . ; Dsm ; Ds1 þ Ds2 ; . . . ; Dsm1 þ Dsm ; . . . ; Ds1 þ Dsm g and repeat step 4 as long as it is necessary. Here Fi denotes the information on the whole system just after the ith inspection, ri ðsi Þ is the number of failed (successed) systems during the interval ½0; bi , bi is possible inspection times, and Dsj ¼ sj sj1 , where s0 ¼ 0. 3.2. Parameter estimation Standard statistical procedures can be used to estimate the parameters and it can be done after each inspection to incorporate the latest information. However, as the information is censored and we do not assume that we have detailed information for each system, it is tedious to adopt the traditional method such as maximum likelihood estimation. An adapted Weibull plotting approach is suggested. Weibull distribution is widely used in reliability, since it is suitable to describe the deteriorating system in many cases. We use Fi to obtain an estimation of the parameters. In general, the following least-square method can be used. First, we estimate Rðbi Þ by b ðbi Þ ¼ si =ðsi þ ri Þ. If some si ¼ 0 (or ri ¼ 0), then R b ðbi Þ ¼ 1=ð1 þ ri Þ (or R b ðbi Þ ¼ we can use R si =ðsi þ 1Þ) or any other empirical estimates. Although this is a rough estimate, it is reasonable and simple and hence adopted here. Second, we linearize RðtÞ to transform the parameters to regression parameters of a linear equation. For example, for the two-parameter b Weibull distribution, RðtÞ ¼ expððt=dÞ Þ, and after the linearization, we have y ¼ a1 x þ a0 , where y ¼ lnð lnðRðtÞÞÞ, a0 ¼ b, x ¼ lnðtÞ, a1 ¼ b lnðdÞ. Thus we can use the least squares method to estimate the parameters in distribution. For Weibull distribution parameter ðb; dÞ we use the following to estimate the parameters. Namely,
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min
L.R. Cui et al. / European Journal of Operational Research 155 (2004) 170–177 jBj X
ðyi a1 xi a0 Þ
Proof. For every i since bi < biþ1 , there are ri failed (siþ1 success) systems in the interval ½0; bi ð½0; biþ1 Þ. These ri failed (siþ1 success) systems must be counted into the interval ½0; biþ1 ð½0; bi Þ.
2
i¼1
where yi ¼ lnð lnðsi =ðsi þ ri ÞÞÞ, xi ¼ lnðbi Þ. From the standard regression analysis, we have that the estimates can be obtained as Pn Pn Pn 8 n xi yi xi yi > i¼1 i¼1 i¼1 > P P ^ a ¼ ; > 2 n n < 0 n x2i ð xi Þ i¼1 i¼1 Pn Pn 2 Pn Pn > yi xi xi xi yi >^ i¼1 > : : a1 ¼ i¼1 Pni¼1 2 Pi¼1n 2 n x ð xi Þ i¼1 i i¼1 That is, b^ ¼ a^0 , d^ ¼ expð^ a1 =b^Þ, where n ¼ jBj is the number of distinct elements in B. OfPcourse, one can adopt any other methods b ðÞÞ2 to get an estimate of model min ðRðÞ R parameters. The following results can be useful. Proposition 1. jBj 6 space Fm .
mðmþ1Þ 2
4. Some illustrative examples Here some numerical examples are shown in order to illustrate how the procedure works. The data is simulated from the Weibull distribution b F ðxÞ ¼ 1 expððx=dÞ Þ, b ¼ 2:0, d ¼ 20:0. The number of systems in the field is ten, i.e., n ¼ 10, and the required ‘‘availability’’ is a ¼ 0:6. We shall discuss the results for all four criteria. The cases below refer to the corresponding criteria. ðjÞ
Case 1a. For this case, si ¼ min1 6 j 6 n fsi g. Let Dsi ¼ si si1 and s0 ¼ 0. The steps can be described as follows. Step 1: Assume that based on other information we first determine Ds1 ¼ 15. After the inspection the simulated data shown is in the second column of Table 1. From that we have F1 ¼ fn ¼ 10; ðr1 ; s1 Þ ¼ ð5; 5Þg, then an estimate of b 1 ð15Þ ¼ 0:5. R Step 2: Since the model has two parameters, we need information from the second inspection, and take this as Ds2 ¼ 12. After this inspection, the data is as that shown in the third column of Table 1. Thus,
for the information
Proof. There are m þ ðm 1Þ þ þ 1 ¼ mðm þ 1Þ=2 combinatorial cases in Fm . However, some cases may lead to the same element. Hence the number of distinct elements is less than mðm þ 1Þ=2. Proposition 2. If we arrange bi in B in ascending order, then ri 6 riþ1 ; si P siþ1 for every i. Table 1 Data from inspections and the sequential inspection time (Criterion 1a) No.
1
2
3
4
5
6
7
8
9
1 2 3 4 5 6 7 8 9 10 Dsi b^i d^i
–– –– –– –– –– 15
–– –– –– –– –– 12 2.151 20.265
–– –– –– –– –– –– –– 3.235 1.629 17.775
–– –– –– –– –– –– –– –– –– –– 3.840 1.841 18.572
–– –– –– –– –– –– –– –– –– 2.984 1.817 20.195
–– –– –– –– –– –– –– –– 4.852 1.817 20.195
–– –– –– –– –– –– –– –– –– 4.222 1.817 20.195
–– –– –– –– –– –– 5.336 1.817 20.195
–– –– –– –– –– –– –– 4.516 1.817 20.195
Note that: throughout this section: denotes the system is down before the inspection; –– denotes the system is still up after the inspection.
L.R. Cui et al. / European Journal of Operational Research 155 (2004) 170–177
F2 ¼fn ¼ 15; ½0; b1 ¼ ½0; 12 ) ðr1 ; s1 Þ ¼ ð2; 8Þ; ½0; b2 ¼ ½0; 15 ) ðr2 ; s2 Þ ¼ ð7; 5Þ; ½0; b3 ¼ ½0; 27 ) ðr3 ; s3 Þ ¼ ð10; 2Þg: Thus, using the empirical estimate, we have b 2 ð12Þ ¼ 8=10; R b 2 ð15Þ ¼ 5=12; R b 2 ð27Þ ¼ 2=12. R Based on the formula in the details of estimation of Section 3.2, we get ðb^2 ; d^2 Þ ¼ ð2:15; 20:265Þ and ð1Þ b 2 ðm3ð1Þ Þ ¼ 0:6 s3 ) R
ð2Þ
s3
ð3Þ
s3
ð1Þ
ð4Þ
ð6Þ
ð2Þ
ð7Þ
ð8Þ
ð3Þ
ð5Þ
ð9Þ
ð10Þ
) m3 ¼ m3 ¼ m3 ¼ m3 ¼ m3 ¼ 13:846; b 2 ðDs2 þ m3ð2Þ Þ ¼ 0:6 R b 2 ðDs2 Þ )R ) m3 ¼ m3 ¼ m3 ¼ 6:633; b 2 ðs2 þ m3ð3Þ Þ ¼ 0:6 R b 2 ðs2 Þ )R ) m3 ¼ m3 ¼ 3:235;
s3 ¼ s2 þ min
1 6 i 6 10
ðiÞ fm3 g
¼ 27 þ 3:235 ¼ 30:235:
Step 3: An inspection is carried out after Ds3 ¼ 3:235, and the data is shown in fourth column of Table 1. We then obtain F3 ¼fn ¼ 20; ½0; b1 ¼ ½0; 3:235 ) ðr1 ; s1 Þ ¼ ð0; 13Þ; ½0; b2 ¼ ½0; 12 ) ðr2 ; s2 Þ ¼ ð2; 8Þ; ½0; b3 ¼ ½0; 15 ) ðr3 ; s3 Þ ¼ ð7; 6Þ; ½0; b4 ¼ ½0; 15:235 ) ðr4 ; s4 Þ ¼ ð9; 3Þ; ½0; b5 ¼ ½0; 27 ) ðr5 ; s5 Þ ¼ ð12; 2Þ; ½0; b6 ¼ ½0; 30:235 ) ðr6 ; s6 Þ ¼ ð13; 1Þg:
In a similar way, we get that ðb^3 ; d^3 Þ ¼ ð1:629; 17:775Þ and Ds4 ¼ 3:840. Step 4: Another inspection is carried out after Ds4 ¼ 3:840. The data is shown in the fifth column of Table 1. We also obtain that F4 ¼fn ¼ 23; ½0; b1 ¼ ½0; 3:235 ) ðr1 ; s1 Þ ¼ ð0; 16Þ; ½0; b2 ¼ ½0; 3:840 ) ðr2 ; s2 Þ ¼ ð0; 16Þ; ½0; b3 ¼ ½0; 7:075 ) ðr3 ; s3 Þ ¼ ð0; 13Þ; ½0; b4 ¼ ½0; 12 ) ðr4 ; s4 Þ ¼ ð2; 8Þ; ½0; b5 ¼ ½0; 15 ) ðr5 ; s5 Þ ¼ ð7; 6Þ; ½0; b6 ¼ ½0; 15:235 ) ðr6 ; s6 Þ ¼ ð9; 3Þ; ½0; b7 ¼ ½0; 19:075 ) ðr7 ; s7 Þ ¼ ð9; 3Þ; ½0; b8 ¼ ½0; 27 ) ðr8 ; s8 Þ ¼ ð12; 2Þ; ½0; b9 ¼ ½0; 30:235 ) ðr9 ; s9 Þ ¼ ð13; 1Þ; ½0; b10 ¼ ½0; 34:075 ) ðr10 ; s10 Þ ¼ ð13; 1Þg:
175
We get ðb^4 ; d^4 Þ ¼ ð1:841; 18:572Þ and Ds5 ¼ 2:984. Step 5: We then get F5 , ðb^5 ; d^5 Þ ¼ ð1:817; 20:195Þ and Ds6 ¼ 4:852, and we can continue with this procedure. It is tedious to deal with Fi . In fact, based on the calculated results, as the information increases, the estimate of ðb; dÞ—ðb^; d^Þ changes very little, that is, F is saturated in inferring the parameters of Weibull distribution after only a few steps. We can get the estimate of ðb^; d^Þ ¼ ð1:817; 20:195Þ. Based on ðb; dÞ ¼ ð1:817; 20:195Þ, we get Dsi . For example, Ds6 ¼ 4:852, Ds7 ¼ 4:222, Ds8 ¼ 5:336, Ds9 ¼ 4:516. Pn ðjÞ Case 1b. For the case of si ¼ ð1=10Þ j¼1 si , we get Table 2 by using a similar procedure. Case 1c. For the case of si in the top 80% of ðjÞ fsi ; j ¼ 1; 2; . . . ; 10g, that is v ¼ 80%, we can similarly obtain Table 3. Case 2. For this case, si is obtained by pj1 , in the second step. We minimize the following function in order to get p2 : b 2 ð12 þ p2 Þ 0:6 R b 2 ð12ÞÞ2 b 2 ðp2 Þ 0:6Þ2 þ 3ð R 5ð R b 2 ð27 þ p2 Þ 0:6 R b 2 ð27ÞÞ2 : þ 2ð R In this case p2 ¼ 12:780. Similar arguments as before are used in order to obtain other values of pi . After these steps, we have the results in Table 4. It should be pointed out that in the above tables, some b^i , the estimation of b, are biased. This is partly due to the simple method used, but also due to the fact that the actual failure time data is not available, and it is neither grouped nor simply censored data. Because of this, methods such as maximum likelihood are difficult to be employed. On the other hand, it is a fact that the distribution of the lifetime in actual use is usually unknown and most models need to be estimated with field data.
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Table 2 Data from inspections and the sequential inspection time (Criterion 1b) No.
1
2
3
4
5
6
7
8
9
1 2 3 4 5 6 7 8 9 10 Dsi b^i d^i
–– –– –– –– –– 15
–– –– –– –– –– 12 2.151 20.265
–– –– –– –– –– –– 10.051 2.612 21.506
–– –– –– –– –– –– 11.089 2.695 23.472
–– –– –– –– 11.748 2.369 24.107
–– –– –– –– 13.647 2.457 22.338
–– –– –– 12.617 2.457 22.338
–– –– –– –– –– –– 14.093 2.457 22.338
–– –– –– –– –– 10.113 2.457 22.338
Table 3 Data from inspections and the sequential inspection time (Criterion 1c) No.
1
2
3
4
5
6
7
8
9
1 2 3 4 5 6 7 8 9 10 Dsi b^i d^i
–– –– –– –– –– 15
–– –– –– –– –– 12 2.151 20.265
–– –– –– –– –– –– –– 6.633 2.306 19.624
–– –– –– –– –– –– 9.010 2.407 19.983
–– –– –– –– –– –– –– 4.873 2.224 20.847
–– –– –– –– –– –– –– 6.115 2.272 20.180
–– –– –– –– –– –– 6.899 2.272 20.180
–– –– –– –– –– –– 6.046 2.272 20.180
–– –– –– –– –– –– –– 6.088 2.272 20.180
Table 4 Data from inspections and the sequential inspection time (Criterion 2) No.
1
2
3
4
5
6
7
8
9
1 2 3 4 5 6 7 8 9 10 Dsi b^i d^i
–– –– –– –– –– 15
–– –– –– –– –– 12 2.151 20.265
–– –– –– –– 12.780 2.091 20.945
–– –– –– –– –– 13.540 2.010 22.151
–– –– –– –– –– 12.745 2.145 21.436
–– –– –– 14.120 2.099 21.199
–– –– –– 14.140 2.099 21.199
–– –– –– –– –– –– –– 14.142 2.099 21.199
–– –– –– –– –– 10.539 2.099 21.199
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5. Conclusions and discussions
Acknowledgements
In this paper the model in Yang and Klutke (2000) is extended to the case of multiple systems. Under different criteria, sequential plans are obtained to ensure that the availability is at the required level. Sequential inspection is important, especially during the set-up and installation stage. Frequent inspection leads to a high cost and infrequent inspection will lead to low availability of the system upon demand. Although the cost might be an issue in this type of analysis, the focus here is to meet the availability requirement with an appropriate time to next inspection. The sequential inspection procedure and decision making procedure studied in this paper allows an appropriate level of availability to be reached with minimum cost as well. Furthermore, we do not assume the distribution of system lifetime to be completely known which is usually the case. The information from the inspection can be used to determine the parameters of system lifetime distribution. Hence, such a combined estimation and decision-making analysis is important and useful in practice. Although the procedure proposed in this paper can be implemented easily, there are several interesting questions that could be raised. Since the estimation for parameters of the lifetime distribution is required at the beginning, one could investigate different estimation methods and also investigate the effect of the estimation error. A common assumption is the independence of the failures and the case of dependence is of interesting. Modelling of the dependence is generally difficult as specific models describing the degree of dependence will be needed.
This research is supported by a research grant from the National University of Singapore for the project ‘‘Reliability analysis of engineering systems’’ (RP3992679). The authors would like to thank two anonymous referees for their helpful comments and suggestions, which improved the paper. References Brint, A.T., 2000. Sequential inspection sampling to avoid failure critical items being in an at risk condition. Journal of the Operational Research Society 51, 1051–1059. Cerone, P., 1993. Inspection interval for maximum future reliability using the delay-time model. European Journal of Operations Research 68, 236–250. Chelbi, A., Ait-Kadi, D., 2000. Generalized inspection strategy for randomly failing systems subjected to random shocks. International Journal of Production Economics 64, 379– 384. Hariga, M.A., 1996. A maintenance inspection model for a single machine with general failure distribution. Microelectronics and Reliability 36, 353–358. Ito, K., Nakagawa, T., 2000. Optimal inspection policies for a storage system with degradation at periodic tests. Mathematical and Computer Modelling 31, 191–195. Lam, C.T., Yeh, R.H., 1994. Comparison of sequential and continuous inspection strategies for deteriorating systems. Advances in Applied Probability 26, 423–435. Vaurio, J.K., 1999. Availability and cost functions for periodically inspected preventively maintained units. Reliability Engineering and System Safety 63, 133–140. Wortman, M.A., Klutke, G.A., Ayhan, H., 1994. A maintenance strategy for systems subjected to deterioration governed by random shocks. IEEE Transactions on Reliability 43, 439–445. Yang, Y., Klutke, G.A., 2000. Improved inspection schemes for deteriorating equipment. Probability in the Engineering and Informational Sciences 14, 445–460. Yeh, R.H., 1997. Optimal inspection and replacement policies for multi-state deteriorating systems. European Journal of Operations Research 96, 248–259.
