RV Model for Condition Based Maintenance

The three possibilities that arise at any time of inspection are: do nothing, PM or renewal, as shown in the maintenance decision tree in Figure 7.

Figure 7: CBM decision tree for the RV model

At the time of first inspection tb if we observe X(q) < cp we do nothing and the time of the future preventive maintenance can be predicted as tPM = cpt: / X (tI) , since one inspection suffices to determine the sample path in the linear RV deterioration. This is referred to as delayed PM and note that no additional inspection is required before the time of replacement. The other two situations are straightforward. The PM is immediately conducted when cp < X(tI) < p, and it is replaced when X(tI) > p. A renewal is thus a delayed PM, immediate PM, or a CM.

The most important point is that the under assumption of RV deterioration model, only one inspection is required for the implementation of CBM strategy.

Using the maintenance decision tree and associated costs shown in Figure 7, it is easy to set up expressions for the mean renewal cycle cost, C, and length, L in terms of the time of first inspection

interval tj and PM ratio c. The renewal cycle cost is simply evaluated as

E[C(tj, c)] = (Cj + CP)P{X (tj) < p} + CF P{X(tj) > p} = (Cj + Cp — CF)Fa (p /1j) + CP (15) Note that the mean cycle cost is independent of the PM ratio, c. The renewal cycle length is evaluated as

E[L{tj, c)] = £tj P{X(t) < p}dt + J>{X(t) < cp}dt = £tj PA (p/1)dt + J~PA (cp/1)dt (16)

The derivation of Eq. (16) uses the mathematical identity for the expectation of a non-negative random variable, i. e., E[T] = J [1 — PT (t)]dt.

Eq. (15) and (16) can be easily evaluated given the parameters of the gamma distribution of the random rate A. According to renewal theory, the mean cost rate is given as

к (,„c)=TiaM

E[L(tj, c)]

When c = 1, the renewal cycle length equals the mean lifetime for any inspection interval. As the inspection interval tj ^ ^, the CBM policy becomes equivalent to the age-based replacement policy.

E[C(tj, c)] = (C + CP)P{PM at ntj} + [(n — 1)Cj + CF ]P{Failed in ((n — 1)tj, ntj ]}) (18)

The mean cycle length can be obtained as

E[L(tj, c)] = (ntjP{PM at ntj} + E[Tf | (n — 1)tj < TF < ntj ]) (19)

Expressing each probability above in terms of the probabilities of the deterioration at corresponding times, we calculate numerically the mean cycle cost and mean cycle length using the following expressions (Park 1988):

E[C(t,,c)] = CP + (CP – Cp)[1 + 5(cp)]

cp (20)

-(CP – C, – Cp)[GA(p;at,, в) + | s(x)GA(p-x;at,,p)dx]

E[L(t,, c)] = GA(p; a t, fi)dt + s(x)GA(p – x;at, fi)dtdx (21)

where S(x) = ^ GA(x;nat,,p) and s(x) = ga(x;nat,,p) .

There are two limiting cases of the CBM policy. As the inspection interval p becomes large, the mean cost rate converges to CP / /LlT. Secondly, as the PM ratio c approaches 1, t, converges to the optimal replacement age similar to that in the age-based maintenance policy.