An heuristic approach for finite time maintenance policy

An heuristic approach for finite time maintenance policy

251 International Journal ofProduction Economics, 27 ( 1992) 25 1-256 Elsevier An heuristic approach for finite time maintenance policy V. Jayabalan...

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251

International Journal ofProduction Economics, 27 ( 1992) 25 1-256 Elsevier

An heuristic approach for finite time maintenance policy V. Jayabalan” and Dipak Chaudhurib “Industrial Engineering Division, College of Engineering, Anna University, Madras 600 025, India blndustrial Engineering and Management Division, Indian Institute of Technology, Madras 600 036, India

Abstract While treating finite time models, the computational time required to obtain the optimal solution increases as the planning horizon increases. A similar effect is also seen for variations in some of the parameters. If preventive maintenance (PM) is imperfect, the expected time of a next failure of a repaired/maintained system is less than that of a new system. The optimum PM intervals in such situations are not constant and the mathematical model to determine the optimum PM/replacement points is more complex than that associated with a perfect PM. In such situations, it is advisable to use a heuristic algorithm to obtain solutions of good quality with less computational efforts. This paper considers a simple and easy to understand heuristic algorithm which gives optimal and near optimal solutions most of the times with less CPU time. The solutions are compared for a wide variety of parameters with the solutions of branching algorithms through numerical examples.

1. Introduction Conventional preventive maintenance (PM ) policies assume that the system after each PM intervention is restored like new. This assumption does not hold true in many real situations since any PM performed on the system will improve the system condition, but inferior to the original state. In general, if PM does not return the system to its original state, it is known as imperfect PM. Several assumptions have been made to analyze imperfect PM: (i) The state of the system after PM is “bad as old” with probability p and “good as new” with probability 1 -p [ 1,2 1. PM reduces the failure rate, but does not re(ii) store the age to zero [ 3-5 1. (iii ) The age of the system is reduced at each PM intervention introducing the concept of an improvement factor [ 3,6- 111.

Correspondence to: V. Jayabalan, Industrial Engineering Division, College of Engineering, Anna University, Madras 600 025, India.

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The system has different time to failure distribution between PMs [ 3,4]. The age of the system is reduced and the failure rate of the system is adjusted incorporating an improvement factor [3,6-lo]. The system degrades with time. PM restores the failure function to the same shape while the level remains unchanged and the failure rate function is monotone, etc. [ 12 1. In many areas of modern society, it is extremely desirable to have highly reliable systems [ 13 ] and it is common practice to perform PM whenever the system reaches the maximum acceptable level of failure rate, or minimum acceptable level of reliability [3,6-lo]. In reality, as the system ages, the post maintenance state of the system lies between “bad as old” and “good as new”. Also operation of such systems causes stress which results in system degradation and, hence, an increase in the level of failure rate with time. Though PM improves the condition of the system, there is a gradual deterioration over time requiring replacement of the system after some time [ 141. The expected time of next failure of a repaired/maintained system is less that of a new

0 1992 Elsevier Science Publishers B.V. All rights reserved.

252 system. The failure rate at time t during interval i is greater than the failure rate at time t during the interval i- 1. A branching algorithm to determine the schedule of PM and replacement times of the deteriorating system for a given planning period has been developed by Jayabalan and Chaudhuri [ 6 1. The downtimes are assumed to be negligible for PM and replacement, and optimal schedules have been derived to minimize the total cost incurred during the given planning period. If the downtime for PM and replacement is not negligible compared with the interval between PMs [ 141) Jayabalan and Chaudhuri [ 71 extended the model [ 61 by considering downtimes. This paper gives a heuristic algorithm for the problem [ 7 ] with similar assumptions, except that the downtime for maintenance and replacement are negligible. The solutions are compared for a wide variety of parameters for their quality and computational capability. The assumptions made are given below: ( 1) Replacement/PM is performed only when the system reaches /2,,. (2) intervening failures are removed by minimal repairs. (3 ) Acquisition cost of the system increases at a constant rate. (4) Downtime for minimal repair, PM, and replacement are negligible. (5) The scrap value of the system is negligible. 2. Notation = maximum acceptable failure rate. = age of the system before ith PM. = age of the system after ith PM. =time of ith PM intervention for jth system. = acquisition cost ofjthsystem. = rate of increase in acquisition cost of the system per year. =rate of increase in maintenance cost factor. ~maintenance cost factor for ith PM,

TC, m

SC ST VI;

= total cost incurred for j systems. =number of systems required during planning period. = total cost incurred for m systems. = cumulative usage time for m systems. = reduction in cost rate for systemj.

3. Scheduled PM times and cost equations If zi is the age of the system before ith PM and y, is the post-maintenance age of the system, it can be shown for a constant improvement factor /I, that (7j_l--Zi_z),

71=7j-_l+(l-l/p)

Y,=7,-(l-l/lp)

(7,-7,-l>.

(1) (2)

If the system reaches A,,, for the first time at 71, then the scheduled PM times 7, (i> 1) are Zj=Z,_]

+

= c

(l-l,/,B)‘-‘71

(3)

(1--1//?)“-‘7i.

k=l

The acquisition cost C, of the jth system at Tj will be Ct ( 1 +rTj), where the replacement time T, = Xi:‘, tnk. If system j works for t, time units before replacement, the cumulative total cost incurred for j systems is given by

k=l

\

ifl

/

(4)

where ri, the maintenance cost factor for ith PM, is b’- ‘rl, and b> 1. The algorithm determines the number of systems used (m) to minimize the expected total cost incurred during the planning period. 4. Heuristic algorithm For problems where the CPU time increases at higher rate with respect to change in some pa. . rameters like T,,,, /I, r, ‘I,, etc., it 1s reasonable to seek for some heuristic methods which give optimal or near optimal solutions with less CPU times. 4.1. Deseri~ti~~ ofthe al~~~ithrn

= = = =

improvement factor. planning period. period of operation ofjth system. time ofjth replacement.

The initial number of systems required (m), the system operational time (t,; j= 1,2,...,m), and the total cost (SC ) are determined. For sys-

253 tern j, the maximum number of PM interventions to be performed is n,- 1. By reducing the number of PM interventions by one for system j (i.e. system usage time t, is reduced to t, = t,,_, ,,), without violating the time constraint, and determining the total cost Qj owing to this reduction, the cost rate @jis found from the following relationship:

v

1+rT, I

CjxCt(

1

P, .

0

4 p,=l

p+ll, I

BP,

COMPUTE

-4 q,:(p+l), 1 COMPUTE

Eiq,

COMPUTE

TC,

YES

SET tnj: tq,

I

Fig. 1. Flow chart and number of PM interventions.

@,=

(SC-Q,Mtn,--&,I.

(5)

If r$,+is the maximum cost rate among the @j’iS, then the usage time of the system k is updated such that t,, = t,,. This process continues until Gk equals zero, i.e. all cost rates are zero and further reduction in the number of PM interventions for any system violates the time constraint. The total cost incurred for each system and for the m system are determined. The required number of systems is increased from m to m+ 1, and by using the same procedure the total cost is found. The number of systems which gives minimum total cost is determined. The steps for the algorithm are given below: (1) Initially set TCo = 0, t,, = 0, cost = 03, j = 1, and T,=O. (2) C,=C,( l+rT,). (3) Set p, = 0. (4) p/z (p+ 1 ),. Compute BP, = Cl [ 1 + r( q -t &,) ISet q,>p, and compute B, such that B, = C C,%;,“v;>B . I/t, < tP, + z, rekrn to step (4). Otherwise (5) compute TC, = TCj_ 1+A,, where Aj is the total cost incurred for the system j= C,+ C, C ‘27 ’ )I r,, and t, = t,. Compute T, + t,. If and T,=Cj=it,,. T,‘+ t, < T,,, > j=j+l, Return to step 2. If T,+ t, > T,,, , the initial number of systems required m =j, total cost SC = TC,, total time ST =T, + t,, . Proceed to step ( 6 ) . (6) Set j=O. (9),otherwisej=j+l. (7) Ifj=mgoto Compute Qj= C ,“=1A,, and (8) Let t,=tc,_,,,. Rj=CEl,;+z/t,, +t,,. IfR, cost, then go to step ( 11). Otherwise cost = SC and time = ST. Increase the number of system m = m + 1 and let period of operation of the system m, t,, = t,. Compute Qm = 1; ,A, and R, = 12 11,. Set SC = Qm and ST = R,; return to step (6 ) . (11) The total cost incurred during the planning

254

a 1

r

I

COST = SC TIME

: ST

TOTAL

COST

: CC&l

m:m+l

I%‘,“,

COMPUTE Qm L

Rm

SC : Qm ST : Rm

:

Fig. 2. Flow chart for the algorithm to find the total cost and PM intervention

period is equal to cost and the total time equals time. Using steps ( 1) to (5 ), the initial number of systems and the maximum number of PM interventions for each system are found. The updat-

times.

ing of the number of PM interventions for each system and the total cost owing to this updating are done by steps ( 7 ) to ( 9 ) . The number of systerns which give minimum total cost is obtained by step ( 10). The flow chart to estimate the ini-

255 tial number of systems and the number of PM interventions for each system (steps ( 1) - ( 5 ) ) is given in Fig. 1. The flow chart for the algorithm to find the total cost and the PM intervention times is given in Fig. 2. 5. Numerical example The model has been tested for the parameters r,=4.0 yr, /3=3, C,= 1000, r=O.l, r,=0.3, b= 1.1, and t,,,= 20 yr. The values of r, (i= 2 to IO) from the start of the system are 6.67, 8.44, 9.63, 10.42, 10.95, 11.30, 11.53, 11.69, and 11.79 yr, respectively. By substituting the parameters through steps ( 1 )-( 5), the minimum number of systems required (m) is found to be two and t,, = 11.30 yr and t nz = 10.95 yr. The cost rates are computed by step (8 ) and Table 1 gives the system k which has the maximum cost ratio Qlkand the updated time t,,. This updating is continued such that the cumulative operational time of the systems is not less than T,,,. For 172~2, the total cost X=6491.96 and the total time ST = 20.05 yr, and further reduction in TABLE 1 Maximum cost rate gk and the updated usage times for m = 2 System No. k

I,~ (Yr)

2 1

10.95 11.30 IO.95 10.42

1 2

@k

1775.91 1615.08 1073.08 1031.94

Updated t,, = t, (Yr) 10.420 10.947 10.420 9.630

TABLE 2 Maximum cost rate & and the updated usage times for m= 3 System No. k

tn4

3 3 1 2 1

10.42 9.63 10.42 9.63 9.63 8.44 8.44 8.44

1 2 3

@k

WI 1518.59 920.36 867.66 764.22 632.28 511.63 472.38 433.13

Updated t,* = t, (yr) 9.630 8.444 9.630 8.444 8.444 6.667 6.667 6.667

the number of PM interventions violates the time constraint. Since SC
solution by heuristic-optimal optimal solution

solution (7)

x 100. For the solutions of the set of problems (/I= 3.0, r=O.lO,z,=4.0yr,r,=0.3,and6=l.l,withT,,, varying from 5 to 50 in steps of 5 yr ), the CPU time used for the set was 0.1740 s, whereas the branching algorithm takes 1.0649 s. For the above set of problems, the heuristic algorithm gives solutions with maximum and mean % error of 1.595 and 0.174, respectively. The algorithms (branching and heuristic) have been programmed in FORTRAN and run on Siemens 7580B system for the following range of parameters: = 3.00 to 5.00 in steps of 1.OO. B r =O. 10 to 0.20 in steps of 0.05. = 3.00 to 5.00 in steps of 1.OOyr. r1

256 Tmax = 5.00to 50.00 in steps of 5.00 yr.

7. Conclusions

One set contains ten problems of particular values of 8, r, and 7l for different 7’,,, values from 5 to 50 yr in steps of 5 yr. Thus there are totally 27 sets of problems. The mean % error for each set of problems for the entire range of parameters and the number of optimal solutions obtained N for each set are tabulated in Table 3. The heuristic algorithm performs well in respect of the CPU time required and more than 70% ( 197 problems) of the time the optimal solution is obtained. The mean and maximum % error are 0.64 and 6.92, respectively. The total CPU time used by this algorithm for solving 270 problems is 5.1836 s, whereas the branching aigorithm requires 30.3427 s.

References 1 2

3

TABLE 3 Mean % error of the heuristic algorithms P

Using a simple and easy to understand heuristic algorithm, more than 70% of the time the optimal solutions are obtained with very low mean % error (0.64%). Attempts are made to improve the algorithm for better solutions with less CPU times. The algorithm could easily be extended, incorporating the downtime for maintenance and replacement, and the downtime cost rate.

r

71

4

Heuristic algorithm Mean % error

N

5

6

3.0 3.0 3.0 3.0 3.0 3.0 3.0 3.0 3.0

0.10 0.10 0.10 0.15 0.15 0.15 0.20 0.20 0.20

3.0 4.0 5.0 3.0 4.0 5.0 3.0 4.0 5.0

0.12 0.17 1.19 0.22 0.32 0.67 0.38 0.33 0.87

9 8 7 8 8 8 8 8 7

4.0 4.0 4.0 4.0 4.0 4.0 4.0 4.0 4.0

0.10 0.10 0.10 0.15 0.15 0. I 5 0.20 0.20 0.20

3.0 4.0 5.0 3.0 4.0 5.0 3.0 4.0 5.0

0.97 0.43 0.34 1.04 0.55 0.42 1.21 0.34 0.51

5 8 9 6 8 9 5 8 8

5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0

0.10 0.10 0.10 0.15 0.15 0.15 0.20 0.20 0.20

3.0 4.0 5.0 3.0 4.0 5.0 3.0 4.0 5.0

0.37 0.82 1.28 0.52 0.37 1.31 0.91 0.49 1.25

7 7 I 7 9 7 7 7 6

7

8

9

10 11 12

13 14 15

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