A multi-component and multi-failure mode inspection model based on

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A multi-component and multi-failure mode inspection model based on the delay time concept Wenbin Wang a,b,, Dragan Banjevic c, Michael Pecht b,d a

Salford Business School, University of Salford, Salford, M5 4WT, UK PHM Centre of City University of Hong Kong, Hong Kong c Department of Mechanical and Industrial Engineering, University of Toronto, Canada d CALCE Electronic Products and Systems, University of Maryland, College Park, MD 20742, USA b

a r t i c l e in fo

abstract

Article history: Received 26 June 2009 Received in revised form 26 March 2010 Accepted 4 April 2010

The delay time concept and the techniques developed for modelling and optimising plant inspection practices have been reported in many papers and case studies. For a system comprised of many components and subject to many different failure modes, one of the most convenient ways to model the inspection and failure processes is to use a stochastic point process for defect arrivals and a common delay time distribution for the duration between defect the arrival and failure of all defects. This is an approximation, but has been proven to be valid when the number of components is large. However, for a system with just a few key components and subject to few major failure modes, the approximation may be poor. In this paper, a model is developed to address this situation, where each component and failure mode is modelled individually and then pooled together to form the system inspection model. Since inspections are usually scheduled for the whole system rather than individual components, we then formulate the inspection model when the time to the next inspection from the point of a component failure renewal is random. This imposes some complication to the model, and an asymptotic solution was found. Simulation algorithms have also been proposed as a comparison to the analytical results. A numerical example is presented to demonstrate the model. & 2010 Elsevier Ltd. All rights reserved.

Keywords: Delay time Inspection Optimal inspection interval Maintenance

1. Introduction Inspections are important activities in any preventive maintenance program. When plant items are being inspected, potential defects may be identified and removed to prevent future failures from occurring. For any planned maintenance shutdown, it is estimated that up to 80% of activities result from defects identified or previously reported, [1]. Thus, the determination of inspection intervals is one of the key decisions of a maintenance manager. Traditionally, for time-based maintenance, the interval between planned maintenance interventions is determined by the managers’ experience or by the original equipment manufacturer’s recommendation [2]. Evidence has been provided that such practice is sub-optimal and conservative [3]. On the other hand, many researchers have developed models to optimise inspection intervals under various modelling scenarios (see the reviews of [4,5]), but few have actually been used. The delay time concept and associated models originally proposed by Christer [6] are exceptional, in that case studies have been conducted and

 Corresponding author.

E-mail address: [email protected] (W. Wang).

reported, e.g. [7–13]. For a recent review of the delay time inspection models, see [14]. The delay time concept considers the failure process as a two stage process from new to an initial point that a defect can be identified by an inspection, and then from that point to failure if the defect is unattended. The time from the initial point of an identifiable defect to failure is called the delay time of the defect. If an inspection is carried out during the delay time of the defect, the defect should/could be removed, depending on the quality of the inspection and defect removal. The delay time models can be divided into two categories: a complex system model and a component tracking model, where the former refers to a system with many components and failure modes and the latter refers to a single component subject to a single failure mode. A majority of the delay time models previously reported were complex system-based (see [3,7,8,10,11] and [15–17] for some examples), but few have dealt with single component models [18–20]. For complex systems, an approximation was made so that defect arrivals from all components are grouped and modelled by a stochastic point process, such as an HPP or an NHPP. It was also assumed that the delay times of all defects follow one identical distribution. If the numbers of components and failure modes are large, this

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approximation is good and can be justified mathematically [21]. If, however, a system has only a few major components with a few major failure modes, then the above approximation will be invalid. We propose in this paper a multi-component-based delay time model, which is different from the previous complex system and single component models, in that (1) several components are considered individually but at the same time to form the subsystem or system model, (2) there can be more than one failure mode of a component, (3) inspections are scheduled for the subsystem or system rather than for each individual component, and (4) there could be possible opportunity inspections at the component failure times to check other components within the same subsystem. None of these situations have been considered before, and thus this model offers maintenance managers a useful tool for determining the optimal plant inspection intervals for cases similar to those considered here. In the following, we start with some necessary modelling assumptions and notation. Then, we present an analytical model of the intended system assuming all components and failure modes are independent. If there are opportunistic inspections at the time of a component’s failure to check other components, then the model becomes very complicated, and therefore, we resort to a simulation-based algorithm. Finally, we illustrate the model in a comprehensive numerical example.

2. Modelling assumptions and notation 2.1. Assumptions 1. Consider a multi-component system with several subsystems. Each subsystem may have several components. 2. A component is the lowest level of the product tree and may have several failure modes. 3. Each failure mode has its own initial and delay time probability density functions (pdfs). 4. If a failure occurs according to a failure mode, then the complete component is replaced or repaired to an ‘as new’ condition. This corrective maintenance implies that the component is renewed. This is termed a failure renewal. 5. All failure modes are independent. 6. The failure cost is the same for a component, regardless of the particular failure mode. This is a consequence of the renewal assumption, though this can be easily relaxed.

7. Inspections are carried out periodically on the subsystem, and then on the system as a whole but they may have different intervals. 8. Inspections are not scheduled according to the lapsed time since the last renewal of a component, but according to a fixed plan across the subsystem. 9. At an inspection, if a component is found to be in a defective state due to one or more of the failure modes, the component is renewed. This renewal is referred to as an inspection renewal. 10. The inspection process is perfect, in that any defects present will be identified. 11. If one component has failed, then there could be an opportunity to check the other components within the subsystem, while conducting the failure repair or replacement.

Most of the above assumptions are consistent with plant inspection practices and some can be relaxed for more realistic considerations, particularly assumptions 9 and 10 which can be relaxed at the expense of more complicated formulations. However, assumption 8 needs more explanation since it increases the modelling and computing complexity. Assumption 8 actually follows a common practice observed in industry, [3,7,8]. The main advantages of this practice are managerial convenience and a possible cost reduction by sharing a common set-up time for all components concerned. It is usually impractical for a maintenance planner to plan an age-based individual preventive maintenance (PM) schedule for each component, unless the component is critically important. Age-based PM for components has been well-documented in the literature [4] and [5]. For now, we assume that the system is the one shown in Fig. 1 as an example, though there could be more subsystems, components and failure modes. Fig. 2 gives a definition of the initial and delay

Initial time, u Delay time, h

New

Failed Time

Fig. 2. The initial and delay times of a component failure process (J defect arrival,  failure).

System

Subsystem 1

Comp11

M111

M112

Comp12

M121

Subsystem 2

Comp13

M131

Comp21

M132

M211

Comp22

M221

M222

Fig. 1. A hypothetical system with two subsystems: 5 components and 8 failure modes.

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¼

Comp21 M211

¼

YK k¼1

YK k¼1

Z

1

Z

tn1 1

½ ½ tn1

3

Z gk ðuÞdu

tn

gk ðuÞFk ðtuÞdu

tn1 t

Z gk ðuÞdu

gk ðuÞFk ðtuÞdu:

tn1

ð1Þ

M221

The pdf of L at intervals of continuity is then given by

Comp22 M222

pL ðtÞ ¼ 

@ PðUmin 4 tn1 ,Tmin 4 tÞ, @t

ð2Þ

for tn  1 ototn, n ¼1, 2, y. It follows from Eq. (1) that

t21

t22

t23

Fig. 3. A schematic representation of the failure, inspection and defect identification process for subsystem 2 (J defect arrival,  failure, } defect removal, t21, etc. inspection time).

times and Fig. 3 shows a scenario of defect arrivals, failures and inspections of subsystem 2 using the delay time concept. 2.2. Notation

RL ðtn1 Þ ¼ PðL 4tn1 Þ ¼ PðUmin 4tn1 Þ ¼

Z

YK k¼1

1 tn1

gk ðuÞdu,

ð3Þ

and RL ðtn Þ ¼ limPðUmin 4tn1 ,Tmin 4 tÞ ¼ PðUmin 4tn1 ,Tmin 4 tn Þ tstn Z 1 Z tn YK ½ g ðuÞdu gk ðuÞFk ðtn uÞdu: ¼ k k¼1 tn1

ð4Þ

tn1

Then for a renewal after an inspection, from Eqs. (3) and (4)

For now, we start first with a simple case of one component with several failure modes, which is the building block for the cost model at the subsystem and system level. K number of the failure modes. gk(u), Gk(u) initial time pdf and cumulative probability function (cdf) of the kth failure mode of a component. fk(h), Fk(h) delay time pdf and cdf of the kth failure mode of a component. the time of the nth inspection epoch. tn random time to defect initiation of the kth failure mode Uk of a component. random time to the first defect initiation; min{U1, Umin U2, y, Uk}. random delay time of the kth failure mode of a Hk component. random time to failure of the kth failure mode of a Tk component; Tk ¼Uk + Hk. random time to the first failure; min{T1, T2, y, Tk}. Tmin L random time to the first renewal, either at an inspection renewal or a failure renewal. an event that the component failed in (tn  1, tn) before Bn any defect is identified. an event that a defect is identified at tn, before failure Dn (failure at tn can be ignored, being of probability zero).

PðDn Þ ¼ PðL ¼ tn Þ ¼ RL ðtn ÞRL ðtn Þ ¼ PðUmin 4 tn1 ,Tmin 4tn ÞPðUmin 4 tn Þ Z 1 Z tn YK ¼ ½ gk ðuÞdu gk ðuÞFk ðtn uÞdu k¼1 tn1 tn1 Z 1 YK  k¼1 gk ðuÞdu,

ð5Þ

tn

and for a renewal after a failure PðBn Þ ¼ Pðtn1 oL otn Þ ¼ Pðtn1 o Lr tn ÞPðL ¼ tn Þ ¼ RL ðtn1 ÞRL ðtn ÞPðL ¼ tn Þ ¼ PðUmin 4 tn1 ÞPðUmin 4 tn1 ,Tmin 4 tn Þ Z 1 Z 1 YK YK ¼ gk ðuÞdu k ¼ 1 ½ gk ðuÞdu k¼1 Z 

tn1

tn1

tn tn1

gk ðuÞFk ðtn uÞdu:

ð6Þ

If n¼1, P(Uk 40) ¼1 and Eqs. (5) and (6) become  YK   Z t1 Z 1 YK g ðuÞF ðt uÞdu  g ðuÞdu : 1 PðD1 Þ ¼ 1 k k k k¼1 k¼1 t1

0

PðB1 Þ ¼ Pð0 oTmin r t1 Þ ¼ 1

YK k¼1

PðTk 4 t1 Þ

For a component with one failure mode [18] Z tn g1 ðuÞF1 ðtn uÞdu: PðBn Þ ¼ Pðtn1 oU1 rtn ,tn1 oT1 rtn Þ ¼

ð7Þ

tn1

3. Analytical model PðDn Þ ¼ Pðtn1 oU1 rtn ,T1 4 tn Þ ¼

Z

tn

g1 ðuÞ½1F1 ðtn uÞdu:

ð8Þ

tn1

3.1. Probabilities of failures and defect identifications We then have, in total In general, either a component fails between inspections before any defect is identified, or a defect is identified at an inspection before failure by any failure mode. By our assumptions, the component is completely renewed in both cases. For a component with K failure modes, and following the assumption that failure modes are independent, for the time to the first renewal L, we have, for tn  1pt otn YK RL ðtÞ ¼ PðL 4 tÞ ¼ PðUmin 4 tn1 ,Tmin 4tÞ ¼ PðUk 4 tn1 ,Tk 4tÞ k¼1 ¼ ¼

YK

PðUk 4 tn1 ,Hk 4tUk Þ k¼1

YK

k¼1

½PðUk 4 tn1 ÞPðUk 4tn1 ,Hk rtUk Þ

PðComponent failedÞ ¼ PðBÞ ¼

1 X

PðBn Þ,

n¼1

PðDefect identified before failureÞ ¼ PðDÞ ¼

1 X

PðDn Þ:

ð9Þ

n¼1

and P(B)+ P(D)¼1. 3.2. The cost model Now we consider the cost model. We will use the renewal reward theorem, [22], and for now we will make the assumption/

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approximation that a new cycle always starts at an inspection, or, equivalently, that the new cycle of inspections starts at a renewal. Eqs. (1)–(9) are derived using this assumption. The assumption is true only for the first renewal cycle and the cycles started at inspections with defect removals. For the cycles started at failure renewals, if we follow assumption 8, the time until the first inspection of a component since the failure renewal would not be the same as the subsequent inspection intervals. This disparity is due to the fact that a failure can occur anytime within the inspection interval. The exact situation will be modelled analytically, either by using the limiting result or by a simulation which will be described later. A comparison between these cases will also be shown in the numerical example section. When we formulate the expected renewal cycle cost and length, an implied assumption is that the downtime caused by inspections, inspection and failure renewals is so small compared with the inspection intervals and the component life times that they can be ignored. This is reasonable since the maintenance downtimes are usually in hours, but the inspection interval and component lives can be measured in months. We also need to introduce subscripts i and j to denote the ith subsystem and the jth component to all previous notation, in addition to the kth failure mode, in Section 2.2. The decision variable is the inspection interval for the ith subsystem. We first consider the expected cost for the jth component within a renewal cycle. There are two renewal possibilities that is a failure renewal or an inspection renewal with defect identification and removal. The expected cost caused by a failure renewal within (ti,n  1, tin) of the jth component within the ith subsystem is ½ðn1ÞCijs þ Cijf PðBijn Þ

ð10Þ

where Cijs denotes the inspection cost of checking the jth component within the ith subsystem, and Cijf denotes the failure renewal cost of the jth component within the ith subsystem. It follows that PðBijn Þ ¼ PðUijmin 4ti,n1 ,ti,n1 o Tijmin rtin Þ YKij YKij PðUijk 4 ti,n1 Þ k ¼ 1 PðUijk 4ti,n1 ,Tijk 4 tin Þ ¼ k¼1

ð11Þ

from Eq. (6). We use notation Bijn for a failure renewal within (ti,n  1, tin) and Dijn for an inspection renewal at tin. Summing over all inspection intervals we have 1 X

1 X

½ðn1ÞCijs þCijf PðBijn Þ ¼ Cijs

n¼1

ðn1ÞPðBijn Þ þ Cijf PðBij Þ:

ð12Þ

n¼1

Similarly, the expected cost due to an inspection renewal because of identified defects is 1 X

ðnCijs þ Cijd ÞPðDijn Þ ¼ Cijs

n¼1

1 X

nPðDijn Þ þ Cijd PðDij Þ,

ð13Þ

n¼1

where

where Cij is the renewal cycle cost of the jth component within the ith subsystem and Cijd is the renewal cost of removing the defect identified on the jth component within the ith subsystem. The expected cycle length is obtained similarly and is given by 1 Z tn X ½ tpL ðtÞdt þ tin PðDijn Þ EðLij ; ti Þ ¼ ¼

tn1 n¼1 1 Z tn X

EðCij ; ti Þ ¼

¼

1 X

½ðn1ÞCijs þ Cijf PðBijn Þ þ

RL ðtÞdt,

ð16Þ

0

ð17Þ

If the optimal inspection interval for the subsystem i is t i , then 0 1   J i X EðCij ; ti Þ A: ti ¼ argminti @ ð18Þ EðLij ; ti Þ j¼1 If the system can be inspected as a whole, then the optimal inspection interval, t*, is 0 1  Ji  I X X EðC ; t Þ ij  A: t ¼ argmint @ ð19Þ EðLij ; tÞ i¼1j¼1 If the number of the subsystems is not large, a separate inspection schedule for each subsystem is feasible from a planning point of view. Otherwise, grouping algorithms must be employed to reduce the number of different inspection schedules and to take the advantage of the shared set-up time, [5]. This is also why we observe in practices that a single inspection schedule is applied to the whole system, [8].

4. Exact formulation of failure distribution with start of the renewals at failures In Eqs. (15) and (16), we have used an assumption that all renewal cycles start at an inspection; i.e., as if the start time is zero and inspections are performed at tin ¼nti, n¼1, 2, 3 y. The assumption actually only applies to a case, where the previous renewal is an inspection renewal due to a defect removal. But for a failure renewal, the inspections start again at ti1 ¼ ti  x, where x (opx o ti) is the time of the previous failure measured from the inspection just before the failure (let us call it the ‘‘forward time’’), and continue at tin ¼nti  x, n¼2, 3, y. Thus, the expected cycle cost and cycle length all dependent on x is 1 X

½ðn1ÞCijs þ Cijf PðBijn jxÞ þ

1 X

½nCijs þ Cijd PðDijn jxÞ,

n¼1

ð20Þ ð14Þ and

1 X

ðnCijs þCijd ÞPðDijn Þ n¼1 n¼1 1 X Cijs n½PðBijn Þ þ PðDijn Þ þðCijf Cijs ÞPðBij Þ þ Cijd PðDij Þ, n¼1

1

EðCij ; ti Þ EðLij ; ti Þ

n¼1

P P1 from Eq. (5) and PðBij Þ ¼ 1 n ¼ 1 PðBijn Þ and PðDij Þ ¼ n ¼ 1 PðDijn Þ from Eq. (9). Let the inspection interval for the ith subsystem be ti, i.e., tin ¼nti. Then the expected cost per renewal cycle for the jth component is EðCij ; ti Þ ¼

Z

where Lij is the renewal cycle length of the jth component within the ith subsystem. Finally, the expected cost per unit time for the jth component, given ti is given by

EðCij ; ti jxÞ ¼

PðDijn Þ ¼ Pðti,n1 o Uijmin rtin ,Tijmin 4 tin Þ YKij YKij ¼ PðUijk 4 ti,n1 ,Tijk 4 tin Þ k ¼ 1 PðUijk 4 tin Þ, k¼1

tn1

n¼1

PðUmin 4tn1 ,Tmin 4 tÞdt ¼

ð15Þ

EðLij ; ti jxÞ ¼

Z

1 0

RLij ðtjxÞdt ¼

1 Z X

ti,n

n¼1

ti,n1

PðUijmin 4 ti,n1 ,Tijmin 4 tjxÞdt, ð21Þ

where P(Bijn|x) and P(Dijn|x) are calculated from Eqs. (11) and (14) using tin ¼nti x as well as PðUijmin 4 ti,n1 ,Tijmin 4 tjxÞ in Eq. (16), using Eq. (1). The forward time x is a realization of a random variable which depends on previous removals, and therefore its distribution cannot be easily found. If we prove that there exists a limiting distribution for forward time, pij(x), then we can calculate the unconditional expected cycle cost and length. Notice that

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pij(x) has a positive mass at x ¼0, which will be discussed later. The unconditional expected cycle cost and the unconditional cycle length are then Z ti pij ðxÞEðCij ; ti jxÞdxþ pij ð0ÞEðCij ; ti j0Þ, ð22Þ EðCij ; ti Þ ¼ 0

Z ti 0

pij ðxÞEðLij ; ti jxÞdx þ pij ð0ÞEðLij ; ti j0Þ:

ð23Þ

pðxjyÞdx ¼ PðXl þ 1 A ½x,x þ dxjXl ¼ yÞ, pð0jyÞ þ

where y is effectively the x at a previous renewal, yA[0,t]. Then,

¼

n¼1 1 X n¼1

Pððn1Þty o U rnty,U þ H 4 ntyÞ Z nty

gðuÞð1FðntyuÞÞdu,

Eq. (28) is an ordinary type of integral equation and can be solved numerically by solving a system of linear equations, or recursively as follows. Defining pl(x) as the lth approximation of p(x), then letting p0(x) ¼0 or any arbitrary value bounded within (0,1), and starting from p1(x)¼p1(x|0), the recursive approximation is performed via the following sequence p0 ðxÞ ¼ 0, p1 ðxÞ ¼ p1 ðxj0Þ, pl þ 1 ðxÞ ¼ p1 ðxj0Þ þ

ð24Þ

ðn1Þty

pð0jyÞ ¼

and pðxjyÞ ¼

¼ 1 X

Pððn1Þty oU r nty,U þH A ½ðn1Þty þ x,

n¼1

ðn1Þtyþ x þ dxÞ=dx ¼

1 Z nty X n¼1

ð28Þ

0

Z t 0

pl ðvÞ½p1 ðxjvÞp1 ðxj0Þdv,

ð29Þ

until the required precision is obtained, such as - when jpl þ 1 ðxÞpl ðxÞj o e for all x,0px o t, given precision e. The unconditional expected cycle cost and length will be taken with respect to the forward time x, once its distribution is calculated for the multiple failure modes, as given above in Eqs. (22) and (23), where p(x)¼ pij(x) is calculated following Eqs. (24)– (29), except that for the one-step transition probability distribution formulated in Eqs. (24) and (25), we use

pðxjyÞdx ¼ 1,

1 X

from Eq. (27), we have that Z t pðvÞ½p1 ðxjvÞp1 ðxj0Þdv: pðxÞ ¼ p1 ðxj0Þ þ

0 ox o t,

0

pð0jyÞ ¼

which does not require calculating pl(0|y) separately. It can be proven from the ergodic property of the regular Markov chain that lim pl ðxjyÞ ¼ pðxÞ exists and is independent of y (see [23]). Then, l-1

What we need now is the expression for pij(x), which can be formulated using the limiting result. For simplicity, we remove all subscripts i, j and k. We will first formulate p(x) for the case of one failure mode. The case for multiple failure modes can be readily developed following the same principle. We will use t as the length of the inspection interval and let U¼Uk, k¼1 and H¼Hk, k¼1. Let Xl denote the forward time for the lth renewal either at a failure or at an inspection, l ¼1, 2, y. Define X0 ¼x0, 0px0 o t, as the actual initial starting time of the component, if the component starts at an inspection time, x0 ¼0. Then, X0, X1, X2, y form a homogeneous Markov chain with state space [0,t]. The defect can be found at an inspection, so that Xl, l ¼1, 2, y has a positive probability at x ¼0, and a pdf for 0 oxo t. One-step transition probability is defined by p(0|y)¼p(Xl + 1 ¼0|Xl ¼y) and by Z t

where vA[0,t] represents the time of a renewal since the PM prior to it. Rt Since pl ð0jyÞ ¼ 1 0 pl ðvjyÞdv, we have Z t pl þ 1 ðxjyÞ ¼ p1 ðxj0Þ þ pl ðvjyÞ½p1 ðxjvÞp1 ðxj0Þdv, ð27Þ 0

and EðLij ; ti Þ ¼

5

¼

1 X

Pððn1Þti yo Umin r nti y,Tmin 4nti yÞ

n¼1 1 X

PðDn jyÞ

n¼1 1 X n¼1

Z YK ½ k¼1

Z nti y 

gðuÞf ððn1Þty þ xuÞÞdu:

ðn1Þty

ðn1Þti y

1 ðn1Þti y

gk ðuÞdu

Z  YK gk ðuÞFk ðnti yuÞdu  k ¼ 1

1 nti y

ð25Þ (see Fig. 4). Let the l—step transition probability distribution be defined by pl(0|y) ¼P(Xl ¼0|X0 ¼y) and pl(x|y)dx¼P(XlA[x,x+ dx]|X0 ¼y)dx. Then Z t pl ðvjyÞp1 ð0jvÞdv, pl þ 1 ð0jyÞ ¼ pl ð0jyÞp1 ð0j0Þ þ Z t0 pl þ 1 ðxjyÞ ¼ pl ð0jyÞp1 ðxj0Þ þ pl ðvjyÞp1 ðxjvÞdv: ð26Þ

gk ðuÞdu: ð30Þ

and pðxjyÞ ¼

1 X

Pððn1Þti y o Umin rnti y,Tmin A ððn1Þti

n¼1

y þ x,ðn1Þti y þx þ dxÞÞ=dx 1 X pL ððn1Þti y þ xjyÞ: ¼

ð31Þ

n¼1

0

u+h

y

t1 = τ

… t 2 = 2τ

t n−1 = (n − 1)τ x

t n = nτ

Fig. 4. Two consecutive failures at y and x.

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where Z 1 @ YK pL ððn1Þti y þxjyÞ ¼  ½ g ðuÞdu @x k ¼ 1 ðn1Þti y k Z nti y gk ðuÞFk ððn1Þti y þ xuÞdu:  ðn1Þti y

ð32Þ Eqs. (30)–(32) are derived analogously to the derivation of Eqs. (5), (1) and (2). If we use the recursive approach of Eq. (29) to calculate p(x), we have to numerically evaluate pl(x). This can cause computation difficulties if the partition number over x is large. Practically, we can try various approximate options. Option 1: ignore the p(x), and assume each renewal starts at an inspection, as we did in Eqs. (15) and (16). Option 2: let pðxÞ  p1 ðxÞ ¼ p1 ðxj0Þ, i.e., we assume the previous renewal was always at the start of an inspection, but the current renewal was at x, the first order approximation. Rt Option 3: let pðxÞ  p2 ðxÞ ¼ p1 ðxj0Þ þ 0 p1 ðyj0Þ½p1 ðxjyÞ 1 p ðxj0Þdy, the second order approximation. In the numerical example section, we used each of the options.

Below, we describe a simulation procedure for a subsystem with opportunistic inspections at failures. In the following, we will suppress subscript i. Let C be the total cost, t—the inspection interval, tmax—the maximum simulation length, Tcjk—cumulative failure time of the kth failure mode of the jth component, j¼1, y, J and k¼1, y, Kj (Kj is the number of failure modes of the jth component), Ujk—random time of the initiation of the kth failure mode of the jth component, Hjk—random delay time of the kth failure mode of the jth component, Tcmin—cumulative minimum failure time, Cjf —failure cost of the jth component, Cs—inspection cost of the subsystem (Cs is simply the sum of the inspection costs of each

5. Simulation analysis In this section, we use simulations to validate our previous modelling developments. Because of the flexibility and the powerful nature of simulation, we further assume that there are opportunistic checks at the time of a component failure on the other components within the subsystem. Then the random time of failure becomes an opportunistic inspection and an analytical solution is even more difficult, though possible.

Fig. 6. The opportunistic checking routine.

Fig. 5. The main simulation routine for t ¼10.

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Expected cost per unit time

W. Wang et al. / Reliability Engineering and System Safety ] (]]]]) ]]]–]]]

7

0.92 0.91

Analytical

0.9

Simulated

0.89 0.88 0.87 0.86 0.85 0.84 7.5

10

15 20 25 Inspection interval

30

35

Fig. 8. Expected cost per unit time of subsystem 2 from analytical and simulated computing ignoring failure mode M222.

Table 1 The distribution parameters.

a211

b211

a221

b221

a222

b222

0.01

0.1

0.05

0.2

0.01

0.01

Table 2 The cost parameters. C1f

Expected cost per unit time

Fig. 7. The inspection routine.

Beta222 = 0.01

0.905

Beta222 = 0.1

0.895 0.885 0.875 0.865

C1s

C2f

C2s

C1d

10

C2d

15

20

25

30

35

Inspection interval 10

20

1

1.5

2

4 Fig. 9. Expected cost per unit time when failure mode M222 was added with two different scale parameters (using simulation).

component within the subsystem), Cjd —defect removal cost of the jth component. Figs. 5–7 show the flow charts of the simulation algorithms.

350

The example below represents subsystem 2 shown in Fig. 1. Initially, we assume each component has only one failure mode for the purpose of a comparison between the approximated analytical and exact simulation results. We then add one more failure mode to component 22 to observe the effect of competing failure modes. To enable a quick comparison, we assume that both the initial and the delay time distributions for all failure modes of components 21 and 22 are exponential i.e., g211 ðuÞ ¼ a211 ea211 u , a221 u

g221 ðuÞ ¼ a221 e

,

f211 ðhÞ ¼ b211 eb211 h , f221 ðhÞ ¼ b221 eb221 h ,

Number of failures

300

6. Numerical example

Expected value calculated Simulated

250 200 150 100 50 0 1

2

3

4

5 6 Time to failure

7

8

9

10

Fig. 10. Comparison of the simulated and numerically calculated values of the numbers of failures within an inspection interval of 10 based on component M21.

g222 ðuÞ ¼ a222 ea222 u and f222 ðhÞ ¼ b222 eb222 h However, the distribution can either be Weibull or any other distribution since all integrals and differentiations are manipulated numerically. The distribution parameters are shown in Table 1 and the cost parameters are shown in Table 2. Fig. 8 shows the results in terms of the expected cost from analytical and simulated computing. The difference between them in the case of smaller inspection intervals is due to the fact that we used option 1 as an approximation. This is explained in Section 4. However, it can be seen from this particular example that the difference is not large. Fig. 9 shows the results when failure mode M222 was added. The experiment was conducted using simulation. Initially, we set the scale parameter (Beta222) of the delay time to 0.01, which

produces curve in the solid line and the minimum can be found at an inspection interval ¼20. Compared with the result from Fig. 8, we can see that the optimal inspection interval is unchanged. By changing the scale parameter to 0.1, which gives a relatively shorter delay time, the optimal inspection is larger than 25. This is interesting since one may always try to shorten the inspection interval once more failure modes are present. However, the outcome depends on the combination of the cost parameters and the initial and delay time distributions. Before we took p(x) into account, we first tested our recursive formulation to approximate p(x). We used the Simpson rule to calculate the numerical integration. Fig. 10 shows that the two results are very close, which validates our formulation.

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We also used options 2 and 3 to approximate p(x) in the expected cost and length calculations (see Fig. 11). All expected cost and length are reformulated as required to take p(x) into account (an example is shown in Eq. (26)). Fig. 11 shows that option 3 is the best, as expected. It also shows that the impact of p(x) must be taken into account since it can lead to different recommendations, in terms of the optimal inspection interval. In this particular case, options 2, 3 and the simulation recommended the same optimal inspection interval, i.e., 20, but option 1 recommended 15, which is inaccurate based on our simulation benchmark. To complete the example, we now add the opportunistic check at the failure time of a component to inspect other components within the subsystem and renew them if necessary. We make a further assumption here that the cost for inspecting the other components and the cost for renewing the faulty components within the same subsystem of the failed component are reduced because of the shared downtime. This assumption is reasonable since the system is already down for a failure renewal. Of course, the reduction depends on the length of the failure renewal and the lengths required to perform opportunity inspections and renewals. This policy may also lengthen the inspection interval since failure times are all considered inspection epochs, so the need for normal inspections is reduced (see Fig. 12). Curves obtained in Fig. 12 are based on three cost estimates for opportunity checks and faulty component renewals, i.e., a 20%, 50% and 70% reduction of the original costs if the component is opportunistically checked and renewed (if defects are found) at the time of the other component failure. The reduction applies to both costs of inspection and renewal, if defects are found. This is just an illustration, since other ways to consider the cost reduction exist. The solid lined curve in Fig. 9 is also presented

Expected cost per unit time

0.91 0.9

Simulation, Beta222 = 0.01

0.89

Analytical_option 1

0.88

Analytical_option 2

0.87

Analytical_option 3

0.86 0.85 10

15

20 25 Inspection interval

30

35

Fig. 11. Expected cost per unit time for subsystem 2.

Expected cost per unit time

0.92

r = 0.2 r = 0.5 r = 0.7 No opportunity check

0.91 0.9

in Fig. 12 to compare the result, which depicts the case of no opportunity check. In Fig. 12, r is the cost reduction ratio. Fig. 12 shows that as r increases, both the expected cost per unit time and the optimal inspection interval (all longer than the original 20) decreases. It shows that only the case of r ¼0.7 makes the opportunity check beneficial.

7. Conclusions In this paper, we develop a delay-time-based model to determine the optimal inspection interval for a system with a number of components and failure modes. The model fills the gap in theory that is not covered by previous delay-time-based studies. This model, if programmed into a software package, can be readily used to advise plant maintenance managers of the best inspection service interval. Theoretically speaking, there is no limitation on the number of components and failure modes given that we know the individual initial and delay distributions. One of the difficulties in the derivation of the expected cost per unit time is the assumption that the inspection is fixed, not resumed after the random failure of a component. This is often seen in practice since an inspection cannot be re-scheduled specifically for an individual component, unless it is critically important. Inspection is usually conducted on a scheduled basis for whole systems, subsystems or critical components. This practice causes a problem for us, in that the time from a failure renewal to the next PM is a random point, which must be modelled. To solve this issue, we developed a recursive formula and demonstrated its correctness through simulation. The closed form of an expected cost per unit time model is developed assuming all components and failure modes are independent. A more realistic scenario of opportunistic inspections is modelled using a simulation algorithm. The numerical examples demonstrate the results of the proposed model. Both analytical and simulated outcomes match reasonably well. Three approximated options are considered for p(x) for the calculation of the expected cost per unit time and compared with simulations. The result shows that different approximations do produce different optimal results and that at least option 3 should be used. There is a need to prove theoretically that the process of x is a Markov chain and ergodic, but this will be left to a separate paper. It must be pointed out that although simulations are relatively easier to run than the analytical counterparts, the analytical form has the advantage of insight analysis and for parameters estimation. The model can also be developed further to take the dependency between components and failure modes into account.

Acknowledgements The research reported here is partially supported by EPSRC under Grant nos. EP/C54658X/1 and EP/G023042/1 and by the Ontario Centre of Excellence and Natural Sciences and Engineering Research Council of Canada. The authors also want to thank Dr. Matthew Carr for his valuable comments on an earlier version of this paper.

0.89 0.88 0.87

References

0.86 10

15

20 25 Inspection interval

30

35

Fig. 12. Expected cost per unit time with opportunistic checking at failures using simulation.

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Please cite this article as: Wang W, et al. A multi-component and multi-failure mode inspection model based on the delay time concept. Reliab Eng Syst Safety (2010), doi:10.1016/j.ress.2010.04.004