A maintenance inspection model for a single machine with general failure distribution

Microelectron. Reliab., Vol. 36, No. 3, pp. 353--358, 1996

Pergamon

Copyright © 1995 ElsevierScience Ltd Printed in Great Britain. All rights reserved 0026-2714/96 $15.00+ .00 0026-2714(95)00094-1

A MAINTENANCE INSPECTION MODEL FOR A SINGLE MACHINE WITH GENERAL FAILURE DISTRIBUTION M. A. H A R I G A Industrial Engineering Program, College of Engineering, King Saud University, P.O. Box 800, Riyadh 11421, Saudi Arabia

(Received for publication 12 April 1995) Abstract--ln this paper we develop a mathematical model for determining a periodic inspection schedule in a preventive maintenance program for a single machine subject to random failure. We formulate the problem as a profit maximization model with general failure time distribution. We show that under certain conditions on the probability density function of failure, a unique optimal inspection interval can be obtained. When the failure times are exponentially distributed, we propose alternative optimal and heuristic procedures to find exact and approximate inspection intervals. Our heuristic solution method is shown numerically to be more efficient than an earlier published heuristic procedure. We also investigated the sensitivity of the optimal inspection interval and expected profit per unit of time with respect to the changes in the two parameters of the Weibull time to failure distribution.

1. INTRODUCTION Suppose a machine which is used as part of a manufacturing process is subject to random failures that can only be detected through inspection. In a preventive maintenance program, the scheduling of periodic inspections for this machine will certainly reduce the number of breakdowns and the associated costs of machine downtime and emergency repairs. These periodic inspections will also increase the expected production revenue from operating the machine. On the other hand, materials and labor costs and cost of production loss due to scheduled downtime will be incurred as a result of these inspections. Therefore, the problem is to determine an optimal periodic inspection policy that maximizes the expected profit per unit of time. The machine inspection problem has been given considerable attention in the reliability and maintenance literature [-1-5]. Recently, Baker I-6] discussed the problem of how often a machine with exponential failure time should be inspected to generate maximum profit. Baker used the N e w t o n - R a p h s o n method to develop a table that can be employed to find the optimal inspection interval. After showing the limited use of Baker's table and with the help of some mathematical approximations, Chung 1-7] developed an explicit equation for a near-optimal inspection interval. Using only the two example problems of Baker, he concluded that his heuristic procedure is an efficient one. In this paper, we again address the same machine inspection problem and develop a general profit maximization model. For a general class of time to failure distributions, we show that there exists at least an optimal inspection interval, maximizing the expected profit per unit of time. Moreover, under certain conditions on the failure time density function, this 353

optimal inspection interval is unique. Using an exponential time to machine failure, we propose alternative optimal and heuristic procedures to get exact and approximate inspection intervals. Our heuristic procedureis shown numerically to be more efficient than the one proposed by Chung. We also studied the sensitivity of the optimal inspection interval and expected profit per unit of time, with respect to the changes in the two parameters of the Weibull time to failure distribution.

2. NOTATION AND ASSUMPTIONS The following notation will he used throughout the paper. Additional notation will be introduced when needed. p Cr

Profit per unit of time when the machine is operating Cost of repairing or replacing the machine, if it is found to have failed C~ Cost of a scheduled inspection f(t) Probability density function of the time to failure during the inspection interval

F(t)

t l f f u ) du cumulative distribution function of failure

R(t) r(t) T

P(T) Z(T)

1 - F(t), reliability function f(t)/R(t), failure rate Length of the inspection interval Expected profit per inspection cycle Expected profit per unit of time Mean time to failure.

Moreover, the mathematical model of the single machine inspection problem is developed under the following assumptions: (1) the times for inspection and repair are negligible; (2) all failures are equally expensive to repair; (3) the inspection operation is error-free; (4) failure completely halts production; (5) the machine is as good as new after repair, so that the probability density function of the failure time is again

f(t).

354

M.A. Hariga 3. MATHEMATICAL MODEL

In this inspection problem, there are two possible cycles of operations. In the first cycle, the machine is found to be in a good state upon inspection at time T. However, in the second cycle, a failure is detected after inspecting the machine at the end of the inspection interval. These two possible cycles are illustrated in Fig. 1. In this case, the expected profit per cycle can be written as P = profit of a good cycle x probability of a good cycle + profit of a failure cycle

~T

Proposition 1 There exists at least a break-even inspection interval for any probability density function of failure time.

The proof is straightforward since g(0) = - Ci and

It is clear that the profit of a good cycle is equal to ( p T - Ci) since the machine is found to be in a good state after the inspection at time T. However, in a failure cycle, the machine can break down at any time t < T. Thus, the profit of a failure cycle would be C r --

g(T) = p (T R(t) d t + C,R(T) - Ci - - Cr. (4) dO

Proof

x probability of a failure cycle.

E[pt/t < T] - Ci - -

exists at least a break-even inspection interval which makes the profit equal to zero. This is shown in the next proposition. However, before stating and proving this proposition, let us define the numerator in the right-hand side of eqn (3) as g(T), i.e.

ptf(t) dt

Ci - Cr, (1)

g ( - [ - 0 0 ) = p~A - - C i - - C r > O.

Proposition 1 shows that there is at least (not necessarily unique) a break-even inspection interval, Tb. However, under certain conditions on the failure rate, r(t), it can be shown that Tb is unique. This result is shown in Lemma 1 below.

F(T)

where E[ ] denotes the expectation. The expected profit per cycle can then be written as

P(T) = ( p T - Ci)R(T) + f : ptf(t) dt

Substituting F ( T ) = 1 - R(t) and manipulating yield

P j~ R(t) d t + CrR(T) - C~- Cr. (2)

The expected profit per unit of time is simply P/T. Thus,

Z ( T ) = ( p f f R(t)dt + C r R ( T ) - C i - C r ) / T .

The break-even inspection interval, Tb, is unique in each of the following types of failure distributions: (i) constant failure rate; (ii) decreasing failure rate, DFR; (iii) increasing failure rate, IFR.

-- (C~ + Cr)F(T).

P(T)

Lemma 1

(3)

Note that p/~ is the expected profit till failure. Therefore p# - Ci - Cr should be positive in order to be profitable to use the machine at all. Moreover, it is evident that Z(0 +) = - C i / ( + o o ) = - ~ and Z ( - . ~ - o o ) = ( p # - - C i - - C r ) / ( - ' ~ - o o ) = 0 +. Thus, there

Proof By differentiating 9(T) with respect to T, one obtains dg(T)/dT

T

Good

- C,f(T)

= C,(p/C, - r(T))/R(T).

Therefore, the solution, To, to g'(T) = 0 satisfies

r(To) = p/Cr. Using the results of proposition 1, namely g(0) < 0 and g ( + ~ ) > 0, we will show for each of the three cases that there exists a unique break-even inspection interval.

Failure

Inspect

0

= g'(T) = p R ( T )

0

t Failure cycle

cycle Fig. 1. Possible cycles of operations.

Inspect

T

A maintenance inspection model (i) Constant failure rate (exponential failure time). In this case, it is clear that g'(T) > 0 for any value of T since r(T) = 1/# and p/~ - Cr > 0. (ii) Decreasing failure rate. Two cases have to be distinguished for monotonically decreasing failure rates. First, when p/C~ > f(0) for all values of T, it is clear that g'(T) > 0. In the second case, when p/C~ intersects r(T) at TO, g ' ( T ) < 0 for all T < TO and 9'(T) > 0 elsewhere. Therefore, we need to show that g(To) < g(0). In fact, since the failure rate is monotonically decreasing, we have

p/C~ = r(To) < r(T)

for all T < To.

time. The resulting optimal Z* is given by

Z* = pg(T*) - Crf(T*).

h(Tb) = (CrTb/R(Tb))(p/Cr - r(Tb)) > 0

or

f? p

for both increasing and decreasing failure rates. Expression (8) will be useful in the derivation of a search interval for optimal T*. The first derivative of h(T) with respect to T is h'(T) = - TCJ(T)(p/C~ - k(T)),

(9)

k(T) = - f ' ( T ) / f ( T ) .

(10)

Next, under each of the three conditions stated in the main Theorem below, one can determine a unique optimal inspection interval. Note that a function, f ( T ) , is said to be log-convex (log-concave) if the first derivative of l o g ( f (T)) is a strictly increasing (decreasing) function of time. Log-convexity and log-concavity will ensure the uniqueness of the optimal inspection interval and will guarantee that the solution to h ( T ) = 0 corresponds indeed to a maximum expected profit per unit of time.

for all T < To .

R(t) dt < CrF(To) = C~(1 - R(To)),

~0T°R(t) dt + CrR(To) -

(8)

where

Integrating both sides yields,

p

(7)

Furthermore, it can be easily shown that

Therefore,

pR(T) < Crf(T)

355

C~ < O.

Substracting C~ to both sides, gives

Theorem

g(To) < g(O).

Under each of the following three conditions:

The above shows that g(T) is strictly increasing in the interval [ T o, + ~ ) from a negative value of g(To) to a positive value of g ( + ~ ) . Hence, there exists a unique break-even inspection interval in the interval i-T0, + ~ ) . (iii) Montonically increasing failure rate. In this case, it is clear that the function g'(T) is positive for all T < To and it is negative elsewhere. Proceeding in a similar manner as in the previous case of monotonically decreasing failure rate, it can be shown that g ( T o ) > ( g + ~ ) . Therefore, there exists a unique break-even inspection interval in the interval [0, To]. The purpose of the periodic single machine inspection problem is to determine the optimal inspection interval which maximizes the expected profit per unit of time. Differentiating Z ( T ) with respect to T and manipulating the resulting expression, we get

Z'(T) = h(T)/T 2, where

h(T)

Ci + Cr - p f ~ tf(t) dt - Cr(R(T) + Tf(T)),

(5) which can be rewritten as

h(T) = C~ - f ro t(pf(t) + Crf'(t)) dt.

(i) exponential time to failure distribution; (ii) log-convex probability density function of the time to failure; (iii) log-concave probability density function of the time to failure; there exists a unique optimal T* maximizing the expected profit per unit of time.

Proof (i) Exponential time to failure distribution. Substitutingf(T) = (1/#) e-r/u into eqn (7), yields k(T) = 1/l~. Thus h'(T) = - ( T f ( T ) / l O ( p p - Cr) < 0 since p/~ Cr - Ci > 0. This implies that h(T) is strictly decreasing in the interval [0, + ~ ) from C i > 0 to - p # + Cr + C i < 0. Hence, there exists a unique optimal inspection interval. (ii) Log-convex probability density function of the time to failure. Using the definition of log-convexity, it is clear that k(T) is strictly decreasing. In this case, if p/Cr > k(0) then h(T) is also decreasing and T* is unique. However, ifp/C r < k(0), then h'(T) > 0 for all T < Tf and h ' ( T ) < O for T > Tf, where Tf is the solution to p/Cr = k(T). It remains to show that h(Tf) > h ( 0 ) = Ci. In fact, for all T < Tf, we have h'(T) > 0, or equivalently

(6)

ro't(pf(t) + Crf'(t)) dt > 0

From eqn (6), it is evident that h(0) = C~ > 0 and h(+ oo) = - p ~ + C~ + Cr < 0. Therefore, there exists at least (not necessarily unique) an optimal inspection interval, T*, maximizing the expected profit per unit of

which after adding Ci to both sides gives h(Tf) > C i = h(0) > 0. Therefore, there exists a unique optimal inspection

356

M.A. Hariga

interval in [Tf, + ~ ) since h ( T ) is strictly decreasing from h(T0 > 0 to h ( + , ~ ) < O. (iii) Lo~]-concave probability density .]'unction o f the time to failure. Here, log-concavity implies that k ( T ) is strictly increasing. Then if p/C~ > k(+~z) then h ' ( T ) < 0 and a unique T* exists. On the other hand, if p / C r < k ( + ~ ) , then there exists Tf such that P/Cr = k(T0. Following a similar analysis to the one used in the previous case, it can be shown that there exists a unique T* in the interval [0, Tf]. 4. SPECIAL CASES

Exponential time to failure distribution

In this case the p.d.f, of the time to failure is given by t > O,

T 1 = 2#(q/2)~1/2)/(1 - (q/2)ll/2~).

~ > 0

and Z(T)=[(pp-C~-C,)-(plt-C)e

r;"]/T.

(11)

The break-even inspection interval, Tb, is Tb = - ~ log(d),

(12)

where d = (pl~ - Cr - C~)/(pFt - C~) < 1.

The optimal inspection interval is the solution to e-T"u(T/# + 1) = d.

(13)

Baker [6] proposed to solve eqn (13) using the N e w t o n - R a p h s o n line search method. Chung [7] approached the problem heuristically and suggested an explicit equation for a near-optimal inspection interval. After using the following mathematical approximation e -r/~ = (2 - T//1)/(2 + T/~)

(14)

for every small value of T/~, he developed an approximate inspection interval given by Ta = 0.5(q + (q2 + 8q)(l/Z))j/

(16)

Another approximate inspection interval can be obtained by using the first three terms of the Maclaurin series of e - f l u in eqn (11). It is simply given by T2 =/t(2q) ~1j2~.

In this section, the cases of exponential and Weibull distributions of the time to failure will be used to illustrate the mathematical model developed above.

f(t) = l/~e-'""

inspection problem when the time to failure is exponentially distributed. A p p r o x i m a t e procedure. After substituting eqn (14) into eqn (11), one can easily show that the inspection interval maximizing the new expression of Z ( T ) is given by

(17)

However, in some cases T2 may be smaller than Th. Therefore, we let T2 = max[/~(2q) (~'2), Tb]. Finally, we suggest to use the average of 7"1 and T2 as an approximate value for the inspection interval, i.e. T n = (T 1 + 7"2)/2.

(18)

Using the above example, we found 7'1 = 441.5339, T2 = max[137.649, 294.44], T, = 367.986, and Z ( T , ) 7.075 which is only 19.6~o lower than the optimal expected profit per unit of time. We also tested our approximate inspection interval using the other two numerical examples solved in Baker [6] and obtained the same optimal solution in each of them, Optimal procedure. The optimal procedure we propose to solve eqn (13) works as follows: (0) let T = 7",; (1) compute T new = 1t Log((T/'kl + 1)/d);

(19)

(2) if IT . . . . T I < 10 -6 , let T* = T and stop; (3) if IT new - TI > 10 -6, let T = T n~w and go to step 1. Note that eqn (19) is derived from eqn (13). Our experience with this simple procedure revealed that only a few iterations are required before the convergence to the optimal inspection interval. Weibull time to failure distribution

In this case the p.d.f, of the time to failure is given by f ( t ) -- cb(bt) c i e ~')°,

(15)

where q = 1 - d = Ci/(pl~ - C~). However, he missed the fact that his approximate inspection interval may yield a negative profit as is shown in the following example. Let/t = 100, p = 1000, C~ = 5000, Ci = 90,000. This example was solved in Baker's paper but not in the one by Chung. In this case q = 0.9474, d = 0.0526, T~ = 192.94 < Tb = 294.44. The optimal inspection interval and expected profit per unit of time are T* = 468.168 and Z* = 8.8. The expected profit per unit of time for Chung's solution is Z(T~) = -45.596 which is 618}g below the optimal one. We will next provide alternative simple approximate and optimal procedures for the single machine

where (l/b) and c are the scale and shape parameters, respectively. The mean time to failure is /~ = (1/b)F[1 + 1/c]. The reliability function and the failure rate are given by R(t) = e -(bOo, r(t) = cb(bt) c- 1

The expected profit per unit of time is Z ( T ) = g ( T ) / T , where g ( T ) = p Jo' e-(bt)c d t + Cr e -~brv - Ci - Cr.

The function k ( T ) is given by k ( T ) = b[(1 - c ) ( b T ) - ~ + c ( b T ) c- ~]

(20)

A maintenance inspection model

-6 O

0

Q

M

357

358

M.A. Hariga

and TO is To = (1/b)(p/(cbCr)) ~1/~c- 1,.

The optimal inspection interval is the solution to h(T) = Ci + Cr - Cr e-lbrlC(l + c(bT) c) - pc

f:

(bt) c e -(bt~cdt = 0.

(21 )

The optimal expected profit per unit of time, Z*, is Z* = (p - Crcb(bT*) c- 1) e (br*~o.

It can be easily shown that for c < 1 the failure rate, r(T), is decreasing and f ( T ) is log-convex. Moreover, for c > 1 the failure rate is increasing and the density function is log-concave. Therefore, using eqn (8), the search interval for the optimal T* is [max(Tf, Tb), + :~) for decreasing failure rates and [ Tb, Tf] for increasing failure rates. A BASIC program was written and implemented on an IBM personal computer to find Tb and the optimal T* and Z* for Weibull time to failure distributions. The program is composed of three segments each corresponding to the search for Tb, Tf, and T*, respectively, In the first segment, the program determines the interval [j, j + 1] such that g ( j ) g ( j + 1) < 0 and then uses the bi-section method in this interval to find Tb. A similar procedure is employed in the other two segments to determine Tf and T*. The program also has access to two different subroutines for the evaluation of the integrals in eqns (20) and (21) for any value of 7-. Numerical example .]or log-convex Weibull distribution. 1 / b = 5 0 , c = 0 . 5 , p = 1000, C r = 5 0 0 0 and

c,

=

100.

Then p = 100 and To = 0.125. Note that g(To) = - 6 7 . 6 and g(1) = 151.154. Therefore, we conducted a line search, using the bi-section method, in the interval [0.125, 1] to find that Tu = 0.721. Moreover Tf = 3.125 and T* is in the interval [11, 12] since h ( l l ) = 23.499 and h(12) -= -77.169. Using again the bi-section method for h(T) in the interval [11, 12], we obtained T * = 11.238. The corresponding optimal expected profit per unit of time is Z* = 556.79. Numerical example Jbr log-concave Weibull distribution. 1/b = 112.74, c = 2, p = 1000, C, = 5000 and c~ = 100.

Then i~ = 100 and To = 1273.23. Tu = 0.100004, Tf = 1278.5, T* -=- 10.371 and Z* --= 983.48. We also solved 30 more problems to study the effect of the mean time to failure and the scale and shape parameters of the Weibull distribution on the optimal inspection policy. We considered four possible values of the mean time to failure (100, 50, 25, 10) with eight combinations of the scale and shape parameters for each of these values. The values of the remaining problem parameters (p, C~, Ci) were kept the same as in the two examples above. The results of this investigation are reported

in Table 1. It can be observed from this table that for a given shape parameter smaller than one (decreasing failure rate, DFR), the optimal inspection interval is increasing as the mean time to failure decreases. However, a reversal effect of ~ on T* can be noticed for a fixed shape parameter larger than one (increasing failure rate, IFR). It can also be observed that for a given shape parameter the optimal expected profit per unit of time is an increasing function of the mean time to failure. Moreover, for a given value of the mean time to failure, it can be seen in Table 1 that the optimal inspection interval increasis for D F R distributions and decreases for IFR distributions as the shape parameter increases. The same parameter has an increasing effect on the optimal expected profit per unit of time for fixed mean time to failure. These results indicate that the shape parameter and the mean time to failure (scale parameter) have opposite impacts on the optimal inspection interval and expected profit per unit of time for both D F R and IFR Weibull distributions. Finally, it can be observed from Table 1 that the two Weibull parameters do not have significant effects on the break-even inspection interval for IFR distributions. 5. CONCLUSION In this paper, we formulated a generalized model for the single machine inspection problem to determine an optimal periodic inspection policy. The optimal inspection interval is the one that maximizes the expected profit per unit of time. For the case of exponential time to failure distribution, we proposed alternative optimal and heuristic solution procedures to find exact and approximate inspection intervals. The results of a sensitivity analyses study carried out for Weibull time to failure distributions show that greater expected profit per unit of time is obtained for larger shape parameter and mean time to failure. Moreover, the shape and scale parameters have opposite effects on the optimal inspection interval and expected profit per unit of time for both D F R and IFR distributions. REFERENCES

1. A. S. Goldman and T. B. Slattery, A Major Elements of System Effectiveness. Kreiger, Huntington, New York (1977). 2. M. Kamins, Determining checkout intervals for systems subject to random failures, RAND Corporation, RM2578, Santa Monica, CA (1960). 3. D. M. Brender, A surveillance model for recurrent events, IBM research report RC-837, Yorktown Heights, New York (1962). 4. A. K. S. Jardine, Maintenance, Replacement and Reliability. Pitman, London (1973). 5. A. H. S. Ang and W. H. Tang, Probability Concepts in Engineering Planning and Design, Vol. 2. Wiley, New York (1984). 6. M.J.C. Baker, How often should a machine be inspected?, Int. J. Quality Reliab. Mgmt. 7, 14-18 (1990). 7. K. J. Chung, A note on the inspection interval of a machine, Int. J. Quality Reliab. Mgmt. 10, 71-73 (1993).