A certain model of reliability diagnostic of renewal objects

Microelectron. Reliab., Vol. 25, No. 4, pp. 695~fi98, 1985. Printed in Great Britain.

0026-2714/8553.00 + .00 © 1985 Pergamon Press Ltd.

A CERTAIN MODEL OF RELIABILITY DIAGNOSTIC OF RENEWAL OBJECTS JACEK M. CZAPLICKI Mining Mechanization Institute, Silesian Technical University, Pstrowskiego 2,44-101 Gliwice, Poland

(Received for publication 18 December 1984) Abstract--In the paper the model of reliabilitydiagnostic of renewal objects which exploitation process can

be described as the interrupting renewal process with the finite time of repair is discussed. The essential relationships between reliabilityof object and diagnostic parameters are presented.

INTRODUCTION Technical diagnostic is the discipline of science which consists of an estimation of technical object state. This estimation is possible to get as a result of investigation of the properties of an object and investigation of accompanying processes of its exploitation. Obtained estimation is a base for decision-making to further treatment of object. Hence, we can decide its operation, maintenance, change of construction or change of technology of its production. The purpose of decision of maintenance contracts of disorganisation process is the main injurious process of each object during its exploitation. The purpose of changes of construction or technology of the object is to reduce the sensitiveness of the influence of outside and inside force agents. The above decisions can change as a consequence of the reliability of an object. If we make preventive or therapeutic actions, the reliability of the exploited object will change. If we make decisions about changes of construction or technology, the object can have quite a different reliability--different values of indices and characteristics. The subject of consideration of this paper is a problem of changes in reliability of an object which exploitation process is unequivocally determined. These changes of reliability are consequences of regular diagnosis of object, prophylactic and therapeutic actions and renewal of wearing elements of the object inclusively. This case is interesting obviously when these changes are positive from an economic point of view and exploitation of the object is more profitable than without diagnostic. FORMULATION OF PROBLEM

There is given an object in which the exploitation process is as follows: there is given time Td per day in which the object operates (is utilized); in operating time the object can be in two states: work or repair;

in the remaining time (24h--operating time) the object is in standstill state. The exploitation process of the object operating in that way can be called the interrupting renewal process with the finite time of repair. Let us assume: distribution time of repair G y (t); distribution time of work between two neighbouring repairs F x (t); mean value of losses caused by one repair S = cE(Y)+K,

(1)

where: c--proportional coefficient, E(Y)--expected value of repair time, Kn--mean value of repair cost, are known. In order to reduce losses we decide to organize a technical survey of object irt every z unit of work time, counting from the moment of end of repair. In the survey time renewal of wearing elements both with prophylactic action or the prophylactic action only can be done. The mean cost of renewal during single survey is Kw, the mean cost of prophylactic action is K I. If renewal is done we assume that the workrepair process is regenerated. It is like the repair time is finished and the work time begins. Prophylactic action prolongs, on average, work time for V time units. Some of actions can be done partly incorrect so the work time will be shortened. This case is not taken into account because it is assumed that it is enough if we do not take into consideration prolongation of work time in the renewal moment in recompense of incorrect action. Let Pw(kz); k = 1, 2.... be a probability of renewal of element of object in survey time in moment kr. Let us assume: E ( X ) >> Td, (this expression means the object is of high reliability); the survey time Tp is not longer than standstill time in24h;Tp~< T k.

These two assumptions allow us to propose that surveys can be done always in the standstill time of 695

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JACEKM. CZAPLICKI

the object and prolongation or shortening of time (so that every survey is possible to realize in the standstill time) will be small. Hence, the survey for the work-repair process will be stochastic impulse with zero length. Furthermore, the sum of prolongation and shortening of time r will be zero in a long period of exploitation. Our problem is to find such time r that the total sum of losses in costs, costs of renewal and prophylactic actions are minimal. Moreover, we must test whether overall costs of exploitation in long time are smaller than the overall costs of exploitation of objects in this time, not taking into consideration surveys. SOLUTION

Consider the function Pw(kr). In the survey moment we can recognize elements (not all, of course) of an object which are worn in a certain way and soon they are going to fail. Let us assume the range of our information about wearing elements is on average ~ units of time. Therefore we can propose that P~(kr) is as follows P.,(kr) = P { ( k + l ) z - k V

> X > kz-(k-1)V}p¢ = (Fx[(k + 1)z-kV] - F x [ k r - ( k - 1) V])p¢ (2)

where p¢ = probability that the worn element will be found out. Let us assume moreover >> V.

(3)

Let us notice that for the process with surveys

Fx (r) is the probability of an event that the work time between two neighbouring repairs is shorter than l".

P~(r)p~ is the probability of an event that the work times between two neighbouring repairs will be finished by renewal and equals v. P~(r)(1 -PC) is the probability of an event that the work time between two neighbouring repairs is within time interval (r, 2T). P2(~)p~ is the probability of an event that the work time between two neighbouring repairs is finished by renewal and equals 2z, and so on. In the light of the above F x ( z ) + ~, Pk('r)- 1,

Remember, in the exploitation process with surveys work and repair states are not alternate (without counting the standstill) because after work state can be work state again, if in the survey renewal is realized. Let us count the number of work states / which fall to one repair state. We have

I ~ [ F x ( z ) + P l ( r ) ( l - p ~ ) + P2(r)p~ + "" "] 1 =

[

Fx(rj+(1-pe)

11

Pk (I") k=l

(6)

Taking (4) we obtain

OF PROBLEM

= PkU)P¢

where )( = random variable, work time in the process with surveys.

/~ {l-p,~[1-Fx(z)][ ~

(7)

Let us notice that for for

p~=l r~vo

/=Fx~(r); 141;

for for

p~=0 1=1: r~O l~{1-p¢)

~.

The diagram of the function 1-- l(p¢) for a few values of parameter a = 1 - Fx (r) is shown in Fig. 1. Looking at the plots shown in Fig. 1 and taking into consideration symmetric construction of pattern (7) with regard of probability p~ and parameter a = 1 - Fx (r) we can state that: if probability p: is small or time r is long then, even for a high value of second magnitude (a in the first case, e.g. r short, high value of pC in the second case), gain connected with the survey expressed by extension of work time of the object, is small and, taking into consideration survey cost, diagnostic will be usually not profitable. Let us define the mean total costs /(ZT of exploitation of object with surveys in time

--

T = E()()+I

~E(Y).

~

9

0

(8)

07

(4)

k=l

with the accuracy given by result of the condition (3). Let us define the expected value of work time. We have:

05 OI P

E()() = f l xJ(x)dx + p¢r k=x ~ kPk(Z) co

+(1-Pc) E k=t

0

f

P~

(~(k + l)z

/ dkr+V

x )ix (x = y - k V ) d y

(5)

Fig. I. Relation between number of work states I which fall to one repair state and probability p~ at different values of parameter a = I -F~ (z).

Reliability diagnostics This time corresponds to the mean time of stochastic cycle work-repair in the exploitation process with surveys. We have I(ET =

697

AKzT T

(9)

Solving the equation - -

&

=

0

(10)

we can find the minimum. It means that

Sfx(z)+ P'l(z)p¢(Kf + Kw)+ P'l(z)(1-p¢)(K f + S) +P'2(z)p¢(2Ki+Kw)+ . . . . 0 (11) and

Sfx(z)+ ~ P'k(z)[p¢(Kw- S)+ S +kKi] = O. (12) k=a

K X T -- I ( Z T

(17)

T

where

Fx (z)S + Pl(z)p~(Kf + Kw) +Pl(Z)(1 -p~)(Kf + S)+ Pz(z)P~(2Kf + Kw) + P2(z) (1 - p~) (2Kf + S) + "" ".

-

Kzr

TS E(X)+E(Y)

(18)

means the average cost in the exploitation process without surveys in time 7". If r / > 0 then surveys are profitable. The magnitude r/shows how many units of money in the time unit will be saved when exploitation of the object with surveys is realized. At the end, let us recall that to obtain the above approximate solution of this problem fulfilment of four conditions is necessary: exploitation process of object is the interrupting renewal process with the finite time of repair, E(X) >> Td,

T. <<.Tk, z>> V.

Let us notice that lim P'k(Z)= 0,

(13)

Magnitudes whose values we have to know in order to obtain the solution are: costs: K., K y, K w, c, time: E(X), E(Y ), V, probabilities: pc, Fx (t), Gv (t).

SO we can perform the equation (12) as f• (z) [S(p~ - p~ + 1) - p~ (pc Kw + Kj-)]

-peKy ~ kfx[(k+l)z-kV] = O. (14)

EXAMPLE

k=l

Let us call this equation the essential one. Analysing the essential equation we can get the condition for profitability of surveys. Let us note, the roots of equation (14) will be within the interesting interval (0, oo) if the first component is more than zero. The second component is less than zero so we obtain that the mean survey cost without renewal Ky ~s limited from the top by the relationship

Kf < S ( p 4 - l + l ) - p ¢ K w .

There is given an object of which the exploitation process is according to the above mentioned assumptions. The density functionfx (t) is exponential

fx(t) = 2 e x p ( - 2 t ) , 2 -1 = E(X). Let us introduce a new variable u = exp ( - 2 z ) .

(20)

The essential equation (14) can be shown as follows

(15)

Zo- ~, kexp(2kV) uk = 0 Usually S > K w. Let us analyse the above inequality. The limit of the mean survey cost without renewal K I depends on the loss function S, renewal cost K w and the probability p~. It is easy to get that for S = Kw even the perfect diagnostic (it means pc = 1) is not profitable. Let us come back to analyses of the essential equation (14). We can find r = %vt which fulfils the equation (14) if we know the density functionfx (t). Verification of the condition -

-

=

0

(19)

(21)

k=l

where

S

1

Kw

Z0 = ~ff ( p ~ - - l + p ~ - ) - p ~ f f

--1.

Let us assume the a priori number N of components of the sum in equation (21) which will be taken into consideration to obtain the estimate Zov,. Hence the solution of the equation (21) is equivalent to find the roots of the multinomial of order N under condition 0 < u < 1.

(22)

(16)

SOME CONCLUSIONS

for z = zot,t is necessary. In the case when few roots of equation (14) exist verification of which root is really %v, is indispensable. For the investigation whether object exploitation with surveys will be more profitable than without we can consider the ratio

The model of reliability diagnostic of renewal objects of which the exploitation process is the interrupting renewal process with the finite time of repair is a trial of formal description of reliability change due to survey reason--renewal and prophylactic actions. It is known that this kind of problem can be solved

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by simulation methods. However, the Monte Carlo methods have their own disadvantages. The essential advantage of this model is that it allows investigation of the mechanism of object reliability due to change of diagnostic parameters and costs. Let us emphasise, at the end, three conclusions: (1) Availability of the object of which the exploitation process comprises surveys is described by the pattern

E(X) I(° - E(X)+E(Y){I-p¢[1-Fx(z)]} "

(23)

(2) If

E(2) 1 -p~[l

- Fx(z)]

>E(X)

then

/(~ > Kg (24)

where K o = availability of object of which exploitation process is without surveys. (3) If /(g > Kg, we cannot say the model with surveys is better than without. The only criterion is an economic one. Sometimes if/(9 > Kg we can convince ourselves that the exploitation with surveys is more expensive than without. (4) Pay attention for high precision of counting! It gives you adequate estimates of interesting values.