Optimal allocation of cost to detectors in a two-unit series system

Microelectron. Relia6., Vol. 24, No. 4, pp. 633-636, 1984. Printed in Great Britain.

0026-2714/'8453.00+ .00 © 1984 Pergamon Press Ltd.

OPTIMAL ALLOCATION OF COST TO DETECTORS IN A TWO-UNIT SERIES SYSTEM ASHOK KUMAR Training Division, Defence Science Centre, Metcalf House, Delhi-110054, India and M. C. GUPTA 10/2, Shakti Nagar, Delhi-110007, India

(Received for publication 1 February 1984) Abstract--A two-unit series system, in which each unit is equipped with a separate detector to detect failure, is considered. The probability that a detector operates successfully at the time of need (i.e. when system fails) is a function of cost spent on the detector. The problem is that of allocation of total resources (cost) to the two detectors such that the overall expected profit is maximized. The optimization problem has been formulated. Two examples are included to show the uses of results. INTRODUCTION

Most of the papers that have appeared in the field of reliability theory concerning redundant/series system models assume that failure is detected as and when it occurs. However, in practice there are systems which make use of fault detection devices, e.g. power systems and electronic systems. Gross [1] has introduced the concept of a fault detector in a two-unit series system. He has assumed that the amount spent on the detector has direct bearing on the probability of correct detection of a failure by a detector and obtained the sufficient condition under which monitoring of detection mechanism would be economically feasible. K u m a r [2] generalized the results of Gross [1] for an n-unit series system. Takami et al. [3] considered the problem of allocation of fault detectors in an n-unit series system. The failure and repair times of units follow exponential distributions. They used 0-1 programming for obtaining the results. K u m a r and K a p o o r [4] analyzed a two-unit maintained series system with two fault detectors and obtained the s-expected profit of the system. The fault detector is subject to two types of failure modes: not detecting a failed unit and giving false alarms. Since the fault detection device always involves cost whenever they are monitored in the system and as the probability of correct detection of a failed unit depends upon the cost spent on the detector, the problem of cost allocation in an optimal manner maximizing some appropriate objective function like expected profit assumes significance. The present paper is an attempt in the direction of allocation of cost to two detectors in a two-unit series system maximizing expected profit under cost constraint for detectors.

(2) Each unit is equipped with a separate detector to detect failure. As soon as the system failure occurs, i.e. one of two-units fails, detectors will detect the failures with some prespecified probabilities. (3) There is no interconnection between the detectors. In fact, probability of correct detection is directly related to the cost spent on detectors, i.e. the more the cost spent on the detector, the better probability of correct detection (smaller probability of wrong detection) will be ensured. Some of the properties of these probabilities are discussed in K u m a r [2]. (4) When a unit is wrongly detected as failed, it enters for repair. It is assumed here that the repair is not carried out completely but after a short (random) duration it is identified that there was wrong detection. As such the failed unit goes to repair without any ambiguity now. (5) When the system is down, the operative unit cannot fail. (6) It is assumed that detectors do not fail at the time of need. (7) Failure time distributions of units are assumed to be exponential whereas repair time distributions are general. (8) The system earns (loses) at a fixed rate in each state which can be different for each state. There is a fixed transition reward (cost) whenever the system changes its state. Define the following system states at any instant.

System States and Transitions S o both the units 1 and 2 are in operative state. S 1 unit 2 fails but unit 1 is detected as failed and goes for repair S 2 unit 1 fails but unit 2 is detected as failed and goes for repair. S 3 unit 2 fails and is detected as failed and goes for repair. $4 unit 1 fails and is detected as failed and goes for repair.

D E V E L O P M E N T O F SYSTEM M O D E L

(1) There is a two-unit series system. Let the units be identified as 1 and 2. 633

634

ASHOK KUMARand M. C. GUPTA

T h e s y s t e m is u p in S o a n d is d o w n in Si(i = 1,2,3,4). T r a n s i t i o n s b e t w e e n v a r i o u s states are given in Fig. 1. NOTATION

21 22 mt m2 rn~2

m~

Ptz(xl)

P21(x2)

P11(x1)

P22(x2)

i,j Pij

P=

(Pij)

1

constant failure rate of unit l constant failure rate of unit 2 mean repair time of unit 1 mean repair time of unit 2 mean time for which the repair is carried out on unit 1 in S 1 wrongly whereas unit 2 was actually in failed state or mean time taken for detecting unit 2 as failed mean time for which the repair is carried out on unit 2 in S 2 wrongly whereas unit 1 was actually in failed state or mean time taken for detecting unit 1 as failed probability of detecting unit 2 as failed whereas unit 1 has failed when x 1 units of money are spent on detector attached to unit 1 probability of detecting unit 1 as failed whereas unit 2 has failed when x 2 units of money are spent on detector attached to unit 2 probability of detecting unit 1 as failed when unit 1 fails and xl units of money are spent on detector attached to unit 1 probability of detecting unit 2 as failed when unit 2 fails and x z units of money are spent on detector attached to unit 2 subscripts that imply system states i,j = 0,1,2,3,4 one step transition probability for making a transition from S~ to Sj transition probability matrix identity matrix of order 5

D 1-P Di

Fii r0

g

Fig. 1. Transitions between various states. Obviously P13 = P24 = P3o = P*o = 1.

F u r t h e r , it is easy to see t h a t

To =

21e-;"' e -'ht dt - 21 + 2 2 TI = m l l , _

T3

T2 = m 1 2 ~ _ ( •

m2, T4

mt )

(2)

Also P12(xl)+Pll(xl) = i} P21(x2) + P22(x2 ) .

(3)

F o l l o w i n g B a r l o w a n d P r o s c h a n [6], we k n o w

ith subdeterminant of D, deleting ith row, ith column / steady-state probability in S~ = d~/Y~ dj --j row vectors, (no, n t , . . . ,rt,) mean unconditional waiting time in S~ earning rate in S~ transition reward (cost) for making a transition from S~ to Ss total expected earning per unit time in steady-state or steady-state expected profit.

n =nP

(4)

n(l - P ) = 0

(5)

so, we o b t a i n d~ as follows

do = 1,dx = Pol,d2 = Po2, d3 = Pol + P o 3 , d , = Po2 + P o 4 .

(5)

Also di

ni = • d~"

(7)

J F o l l o w i n g H o w a r d [5], s t e a d y - s t a t e e x p e c t e d profit, 9 c a n be given by

EXPECTED PROFIT OF THE SYSTEM

O b s e r v i n g state t r a n s i t i o n s f r o m Fig. 1, various state t r a n s i t i o n probabilities, P o c a n be w r i t t e n as Pol = P n ( x D

f

~

22e-~'e-~dt

bc°

Po2 = P12(xl)

o -

22

-- P21(x2) ~ 2 2

"

i

Z,~,v,

Where

~1

;tie .... te-X~tdt = P l a ( x l ) 2 ~ 2 2

1

qi = rij + ~ ~

0 ,).2e-x~te-~'~' d t = P 2 2 ( x 2 ) j . l ~ ) ~ 2 b°°

Pog=PII(x1)

21e-a'te-'htdt=Pll(Xl)21 o

(8)

i

o

Po3 = P 2 2 ( x 2 )

E nffiqi

P i j " rij.

S u b s t i t u t i n g (7) in (8), we get

/]-1

~ diTiq,

+22 " (1)

g=

i

Z 4r, i

(9)

Two-unit series system O n making appropriate substitutions in (9) and taking r i i = 0 for i C j , also using (3) and after simplification g is obtained as roo +/~2r33m2 +

21r44ml+ ) ` 2 P 2 1 (x2)rl lmi2

Observing the equality, xl + x 2 = x, (15) reduces to Ax2 + Dxl + E g A=

g = 1 +,~.2(P21(x2)m12+m2)+)`l(P12(xl)mll+rnl)"

If no cost is spent on detectors, it is reasonable to think that on the failure of the system, any unit (1 or 2) can go for repair with equal probability, i.e. P21(0) = P11(0) = P22(0) = ½.

1 - E 1,

F = 2 B l x + F 1 - G 1, G = (1 + x ) ( 2 B 1 + G 1 ) + F 1 and A 1 = r0o +,~2r33m2 -+-,~lr44ml, D1 = )`2rl 1mr2,

(11)

E 1 = )`lr22m11, B 1 = 1 +22m 2 +21ml,

Then

F i

),2 21 roo W J.2r33m2-+-~,lr44ml + ~ r l l m 1 2 - } - ~ r 2 2 m l l

-2A t,D=2Alx+D

E -- (1 + x ) ( 2 A 1 + E 1 ) + D 1 , B = - 2 B 1 ,

(10)

=

(16)

Bx2+Fxl+G

Where

+ ),1P12(xl)r22mll

P12(0)

635

= J,2m/2, G1 = 21rntl.

To maximize g in (16) we differentiate g with respect to xl and put d ~ = 0, which gives

g=

1+),2(~+mz)

+)`1 ( ~

+ rex) xl

(12)

-

L

+

~

(17)

K

Where

OPTIMAL COST ALLOCATIONTO DETECTORS NOW, if one is posed with the problem of optimal cost allocation to detectors attached to units 1 and 2 with a view to maximize expected profit, g, of the system under given cost constraint, then the problem reduces to:

K=AF-DB,

L = AG-EB,

M = GD-EF.

As x 1 is cost we consider only positive values of x 1, and denote it by x*. To examine if x* is maximum point, it is ethical to verify

maximize g in (10)

d2g2 x~ = xr < 0. dx

subject to x 1 +x2 = x x x , x 2 , x > O.

(13)

Where xl, x : are the costs to be spent on detectors attached with units 1 and 2 respectively, x is the total amount available. We now discuss the above problem in the light of two cost structures used for failure detection mechanism. Failure detection cost structure Let

1} .

Pz1(X2)

To illustrate the results numerically, let us specify the various parameters for the problem as under 21 = 0.2, ),2 = 0.5, m 1 = 0.8, m 2 = 0.4, rn/i = 0.1,

r33

=

--25.0,

r,4 = -27.0, x = 50.0.

When the above values are substituted in (17), we get x~' = 15.242, x* = 34.758 (14)

1 ~2(1+X2

It may be noted that (14) satisfies properties given in K u m a r [2]. O n substituting (14) in (13), the problem is to maximize g subject to x 1 + x 2 = x where g is given by

g=

(18)

x* = x - x L

m/2 = 0.2, roo = 20.0, rla = - 2 2 . 0 , r22 = -23.0,

P12(X1) -- 2(1 + x l ~ and

Once x* is determined, x* (the optimal cost spent on the second detector) is obtained as

J,2rllm12 ,~qr22mll roo +)`2r33m2 + )`1r4.4m1 -I - -I 2(1 + x 2 ) 2(1 + x l )

and

g = 7.8083.

On the other hand, if no cost is spent on detectors, (12) gives g = 6.5845. Thus inclusion of detectors ensures an increase of 18.58 % in expected profit. Failure detection cost structure 2 Let P12(xl) = 2e . . . . .

),2m/2 ,~Im/l l+22m2+2xmlq--F 2(1+x2) 2 ( 1 + x l )

and (15)

. e12(xz) = ½e . . . . .

(19)

636

ASHOK KUMARand M. C. GUPTA

Where a 1 a n d a 2 are c o n s t a n t s depending u p o n the requirement of correct detection assurance. O n substituting (19) in (13), the problem is to maximize 9 subject to x~ + x 2 = x where 9 is given by

To decide t h a t x* actually maximizes g in (21) it is ethical to verify d2g

< 0.

dx~ x, = x* /[2

a x

roo + A2r33 + 21r44ml + ~ e

- 2 2rllmt2

fl~l

fllXl r22mll

q---e2

9-

2z

1+

+~

~2m2 + ,~lml

Once x* is determined, x* (the optimal cost to be spent on second detector) is o b t a i n e d as x* = x-x*.

1 e -a2X2m12 + ~ e -a~x~mll

To illustrate the results numerically, let us specify the various p a r a m e t e r s as under (20)

~-i = 0.02, m 1 = 0.8, tEe = - 2 3 0 . 0

Observing the equality x~ + x 2 = x, (20) reduces to A + B e - a ~ x ' +Hea2x, (J = E + F e _ a , x , + Kea2Xl •

21 = 0.05, m 2 = 0.4, r33 mn=

(21)

=

-240.0

0.1, roo = 200.0, r44 = - 2 5 0 . 0

rote = 0.2, rl1 = - 2 2 0 . 0 , x = 1000.0. Where A = r o o + 2 2 r 3 3 m 2 + 2 t r a 4 m t , E = 1 +22rn2 +,~tml, B

H-

/'lr22mtl F -~'lmll 2 ' 2 ~-2rl lml2e -tax

2

,K-

in the

x~ = 305.9075, '

x~ = 694.0925

).2mi2 e -a2x

and

2

The value of a2 x has been b o u n d e d for the traceability of results, i.e. la2x I ~< 4. To maximize g in (21) we differentiate 9 with respect to x~ a n d put ~

W h e n the above substitutions are made solution, we get

O n the other h a n d if no cost is spent on detectors, we find from (12) 9 = 182.2168.

= 0, which gives

sl e .... +s2ea2X~+s3e (a~-aOxl = 0.

9 = 184.3361.

(22)

Thus we observe that employing detectors increases the expected profit of the system.

Where S1 = a l ( A F - E B ) ,

S2 = a 2 ( E H - A K ) ,

Acknowledgement--The authors are thankful to Prof. S. M.

Sinha, Department of Operational Research, University of Delhi, Delhi, for his keen interest, constant encouragement and help in the preparation of this paper.

s3 = (al + a 2 ) ( F H - K B ) . P u t t = e .... in (22). Then s2t(q, +qz/q2) ~_ S3 t + Sl = 0.

(23) REFERENCES

Further, i f a 1 = a2, (23) reduces to S2 t2 + s 3 t + s 1 = 0 .

(24)

Solving (24) for t, we get t=

- s 3 + x/s 2 -4sls2 2s2

Since x 1 is cost, we consider only that value of t which gives xl > 0 f r o m log t X1 ~

• a2

W h e r e log is t a k e n to the base e. Let us denote this value of xl by x*.

1. A. J. Gross, Minimization of mis-classification of component failures in a two-component system, IEEE Trans. Reliab. R-19, 120-122 (1970). 2. A. Kumar, Optimum detection of failures in an n-component system. Z. anyew. Math. Mech. 55, 679 680 (1975). 3. I. Takami, T. Inagaki, E. Sakino and K. Inoue, Optimal allocation of fault detectors, IEEE Trans. Reliab. R-27, 360-362 (1978). 4. A. Kumar and V. B. Kapoor, Optimal Classification of failures in a two-unit series system. Int. J. Systems Sci. 12, 127-132 (1981). 5. R. A. Howard, Dynamic Probabilistic Systems, Vol. II. John Wiley, New York (1971). 6. R. E. Barlow and F. Proschan, Mathematical Theory of Reliability. John Wiley, New York (1965).