cost analysis of a three-unit standby system subject to random shocks and linearly increasing...

Reliability Engineering and System Safety 33 ( 1991) 249-263 ,- ,~ ,Z ̧

Cost Analysis of a Three-Unit Standby System Subject to Random Shocks and Linearly Increasing Failure Rates

Rakesh Gupta & Sachendra Bansal

Department of Statistics, Institute of Advanced Studies, Meerut University, Meerut--250005, India

(Received 19 April 1990; accepted 17 June 1990)

ABSTRACT

This paper deals with the cost analysis of a single-server three identical unit cold standby system. The operating unit is subject to random shocks occurring from time to time. Due to each shock it may happen with a fixed known probability that (i) the operating unit is not at all affected, (if) the failure rate of the unit increases and the unit is said to work in quasi-normal mode, and ( iii ) the operating unit fails totally. Failure time distributions of the operative unit are taken to be Rayleigh with different parameters while the repair and shock time distributions are negative exponentials. The system is observed at suitable regenerative epochs in order to obtain various interesting measures of system effectiveness.

INTRODUCTION

Various authors 1-4 have studied the reliability systems subject to random shocks under the assumption that a unit can fail only due to shocks and have obtained life time distributions of the system. Murari and AI-AIi 5.6 analysed a standby system model subject to random shocks assuming that the operating unit can fail either due to operation or due to shocks. Recently AI-Ali 7 extended the above work by taking the three-unit system and obtained only the mean time to system failure taking constant failure and repair rates and general shock time distribution. In real situations there are many systems where the failure rate of unit increases with time. In the literature of reliability such types of work have not been seen so far.

249 Reliability Engineering and System Safety 0951-8320/91/$03.50 (~ 1991 Elsevier Science Publishers Ltd, England. Printed in Great Britain

250 Rakesh Gupta, Sachendra Bansal

In the present paper we investigate a three-unit system subject to r andom shocks and linearly increasing failure rates of the units. Initially one unit is operative and the other two are kept as cold standby. It is assumed that due to shock (i) the normal operat ing unit may not be affected, (ii) the failure rate of the unit may increase and as such the unit is said to work in a quasi- normal mode, and (iii) the unit may fail completely. Further, due to shock, an operative unit in quasi -normal mode either remains unaffected or fails totally. In the case where the failure rate of a normal operat ing unit remains unaffected due to shock, it is assumed that its further operat ion is fresh, i.e. the operative time already spent before the shock goes to waste. A single repair man is available in the system, who immediately starts the repair o f a unit that has failed, either due to shocks or due to operation. After

90

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55

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I I ! I f I ! l 2~ I 1.2 1.4 1.6 1.8 2.0 2.2 2A 2.6 3.0

Bchaviour o f M T S F w.r.t, p for different values o f ft.

Cost analysis of a three-unit standby system 251

280

240

200

~ 160 I-,-

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120 I,,,- 113 0 U

80

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Fig. 2. Behaviour of M T S F w.r.t. P for different values of /L

completion of the repair, the unit becomes as good as new. Failure time distributions of the operative unit in normal and quasi-normal modes are taken to be Rayleigh while the shock time (time between two consecutive shocks) and repair time distributions are negative exponentials. We analyse the system by using regenerative points and obtain the following economic related measures of system effectiveness:

(i) Reliability analysis and mean time to system failure (MTSF). (ii) Mean up-time of the system during (0, t].

(iii) Expected busy period of the repair man during (0, t]. (iv) Expected number of repairs during (0, t]. (v) Expected number of visits by the repair man in (0, t].

The MTSF and cost functions are also studied graphically in a particular case in respect of different parameters in Figs 1 and 2.


STATES OF THE SYSTEM AND NOTATION

Symbols for the modes

F Failure mode N Normal mode N' Quasi-normal mode

Subscripted symbols

o Operative s Standby r Under repair w Waiting for repair

Thus, considering the above symbols, the possible states of the system (comprising three similar units) are:

So: (N o, N~, N~) $3: (F~, No, N,) Ss: (F r, F w, N,~)

S,: (No, N~,N,) S,,: (F. Fw, No) $6: (F~,rw, rw) S 2: (F r, N o, N,)

Of these, all the states are regenerative. The state S 6 is the failed state while the rest are up states. The possible transitions between the states are shown in Fig. 3.

ql q, q

( ~ : UP STATE I ] :FAILED STATE

Fig. 3. Transition diagram.


Other symbols

Ai(t) B~(t) E Eo Ni(t) PO Po

qij(" )Qij(" )

qo,qt

ro, rl

Ri(t)

Vi(t)

~,# iff t, tff' t

Erf x

©,®

Pl-system is up at time tiE o = Si] P[system is busy at epoch t starting from state Sic E] Set of regenerative states = So to $6 State of the system at t = 0 Expected number of repairs during (0, t] lEo = S i e E Transition probability from regenerative state S, to S j - Qo(oo) Probability that the failure rate of the unit increases due to shock P.d.f. and c.d.f, of transition time from regenerative state S~ to Sj Probability that the unit is not at all affected by the shock in normal and quasi-normal modes Probability that the unit fails due to shocks in normal and quasi-normal modes; note that Po + qo + ro = 1, qt + rt = 1 Reliability of the system when Eo = S~eE Mean sojourn time in state S~ Expected number of visits by the repair man in (0, t] [Eo = S ~ E

Constant shock rate and repair rate for a unit Linearly increasing failure rate of the operating unit in normal and quasi-normal modes (13' > 8)

Error function, defined as .I2 e-U2 du

Symbols for Laplace and Laplace-Stieltjes transforms Symbols for ordinary and Stieltjes convolutions, i.e.

Io A(t) © B(t) = a(u)B(t - u) du

A(t) ® B(t) = I' da(u)B(t - u) Jo

TRANSITION PROBABILITIES AND SOJOURN TIMES

The linear hazard model is the simplest time-dependent model and has the form

Z(t) = bt t > 0 b is a constant

The reliability and density functions for this model are given by

R( t )=exp ( -b t2 /2 ) and f ( t ) = b t e x p ( - b t 2 / 2 )


It may be noted t ha t f ( t ) is a Rayleigh density function. The mean for this density function is given by

f 1/2 E(T) = e -b''/2 dt = 2~/b/----~ = ~/(n/2b)

Let To(-. = 0), Tt . . . . denote the epochs at which the system enters any state SieE. Let X. denote the state visited at epoch T.+, i.e. just after the transition at T.; then

Qq(t) = P[X.+x =j, T.+ I - T. < t lX . = i]

The transit ion probabil i ty matrix (t.p.m.) is given by

P = (pq) = (Qo(~)) = Q ( ~ )

with non-zero elements: s

Poo = qo j" ~ e-('+#'/2)' dt = ~qo e~ [~ /n - 2 E r fx l ] / x /2 fl

Pol = Po j" ~ e - (" + #'/2), d t = ~Po eX~[ x / n - 2 Erf X x ]/~/2fl

P02 = r o ~ ~ e - (~ + i~t/2)t dt + ~ fit e - (~ + l~t/2)t dt

: + qo /

Pxx = qt ~ ~t e - (~ + #,t/2)t d t = ~tql e ~] [ x / n - 2 Erfx3]/x/2 f f

P12 = rl ~ ct e-(~+a't/2)t dt + ~ fl' t e-(~+a't/2)t dt = (ql - P i t + rlPxx)qt

P22 = qo ~ ~t e -('+~'+p'/2)t dt = ~tqo eX22 [.,/n - 2 Erfx2]/v/2 fl

P20 = ~ ~ e - (~ + u +#t/2)t dt = I, tP22/~qo

P23 = Po ~ ~ e - t ~ + u + #t/2)t dt = P o P 2 2 / q o

p2, ,=ro S~e-~'+~'*a'/2)tdt + ~ [3te-('+u+#'/2'tdt

_ roP2_______22 F [ I (" + ")P22.]

qo ~qo J

P33 = ql .~ ~X e -{~+#+#'t/2)t dt = otql e~2[~/7c -- 2 Erfx,]/~/2fl '

Pal ---- J"/,t e -(~+#+#'t/2)t dt = l t P a a / # q l

P3, ---- rl .~ u e-('+# + #',/2), dt + ~ fit e -('+" + #',/2), dt

_ r , p 3 , + [ 1 (~ + U)__P3,.] ql ~tqx d

P*,~ =P22 P42 =P20 P4s =P23 P46 =P24

Pss =P33 Psa = P a t P s 6 = P a , p64 = S p e - # t d t = 1


where

x , =

It can easily be verified that

Poo +Pot -I-Po2 = 1 Pit q - P l 2 = 1

P2o -I- P22 q-P23 -I- P24 = 1 P31 -I- P33 q-P34 -- 1

=

x , = +

P.t2 -t- P44 + P45 + P46 = 1

P53 +P55 q-P56 = 1 P64 = 1

To calculate the mean sojourn time To in state So, we observe that so long as the system is in So there is no transition to St and $2. Hence, if Uo denotes the sojourn time in So, then

Similarly,

To = S P [ V o > t] dt = S e-~+a'/2't dt =Poo/~qo

1"1 = P l l / ~ q l 1"4 = 7"2

T2 = P22/Otqo T5 = T 3

T3 = P33/otql Z 6 = 1/11 = m (say)

ANALYSIS OF RELIABILITY A N D M E A N TIME TO SYSTEM F AILURE (MTSF)

Let the random variable W~ denote the time to system failure when the system starts from state Si E E. Then the reliability of the system is given by

Rift) = P[ W~ > t]

To determine the reliability of the system we regard the failed state of the system ($6) as absorbing. By probabilistic arguments we have

Ro(t) = Zo(t) + qoo(t) © Ro(t) + qol(t) © R d t ) + qo2(t) © R2(t) (1)

Rz(t) = Zt(t) + ql 1(0 © Rl( t ) + ql2(l) © R2(t) (2)

R2(t ) ----- Z2(t ) + qao(t) © Ro(t) + qzz(t) © Rz(t) + q23(t) (~) R3(t) + q2a(t) © R4(t) (3)

R3(t) = Z 3 ( I ) + q31(l) © Rl(t) + q33(/) © R3(I) 4" qaa(t) © R4(l) (4)

R4(t) = Z4(t) + qa2(t) © R2(t ) + q44(t) © Ra(t ) + q45(t) © Rs(t ) (5)

R5(I) = Zs(t) + qsa(t) © R3(t ) + q55(t) © Rs( t ) (6)


where

Zo(t ) = e-{a +#t/2J t Z3(/) = e-I~+u+o't/z}t

Zl( t) = e -{a+o''/2}t Za(t) = Z2(/) Z 2 ( l ) = e-t~+u+#t/2)t Z s ( / ) = Z 3 ( / )

T a k i n g the Lap l ace t r a n s f o r m o f re la t ions (1) to (6) a n d s impl i fy ing for R'd(s), a n d o m i t t i n g the a r g u m e n t ' s ' fo r brevi ty , we get

R~(s) = Nl(s)/Ol(s ) (7)

where

N~(s) = [Z~(1 - q~'~) + Z t qo~]

x [(1 - q~2){(1 - q$3Xl - q,~4)(1 - q~5) - q34q45q53} - q$2( 1 - q55){q23q3, + q~,( l - q~3)}]

- q3 t (Zo q t 2 - Z t qo 2)[q2,,q45qs a + q~3(1 - q*,d( 1 - q~s)]

+ [qo lq l2 + q~2( 1 - q~'t)] x [ { Z * ( l - q~5) + Z~q*5}{q'~aq'~, + q~',(1 - q~3)}

+ Z~'{(1 - q~3)(l - q~,,d(1 - q~5) - qa,~q,,sqsa}

+ ZJ'{q~3(1 - q*4)(1 - q~5) + q2,q,sq53}]

a n d

Dt(s) = [(1 - q~o)(1 - q~'t)(1 - q~2) - - ~,/0"* t~/1"*2~/20"* - qo2q20]* *

x [(1 -- q~3)(1 -- q*4)(1 -- q~5) - q34q45qs3] - (1 -- q~o)

x (1 -- qt l)q42[q23q34( 1 -- q~5) + q~4(1 -- q~3)(1 -- q~'5)]

- - (1--qoo)q12q3t[q~3(1--q*4)(1--q~s)+ q24q45qs3]* * *

T a k i n g the inverse L a p l a c e t r a n s f o r m o f express ion (7), we can get the re l iabi l i ty o f the sys t em s t a r t i ng f r o m So. To get the M T S F , we h a v e a f o r m u l a

E(Wo) = S Ro(t) dt = lira R'~(s) = NI /DI (8) $~0

where

N t = [To(1 - P t l ) + T t p o t ]

x [ ( 1 - P 2 2 ) { ( 1 - P 3 3 ) ( 1 - P 4 4 ) ( 1 - P s s ) - P 3 4 P 4 s P s 3 }

--,042( 1 --Pss){P23P3,t +P24( 1 --.1033)}] --/73 l(Toff! 2 - - Tt PO2)Uff24P45P 53 "4- P 2 3(1 -- P4,*)(1 -- P55)]

+ [ P o t P l 2 +P02( 1 - - P l t ) ]

x [ / ' 2 { ( 1 - - [ 3 3 ) ( 1 - p44)(1--Pss)--P34.P45P53} + {T2(1 - p , , ) + + p,,(1 + T3{P2¢P45P53 "t" P23 (1 - - p4a.) ( l - - P s s ) } ]

Cost analysis o f a three-unit standby system 257

and

D t = [(1 - Poo)(1 - Pl 1)(1 - P22) - Po 1Pl 2P20 - Po2P20] x [(1 --P33)(1 -- p44)(1 - -Pss) - -P34P 45P s3] -- (1 --Poo) x (1 -- Pt I)P42[P23P3,*( 1 -- P55) + P24( 1 -- P33)( 1 -- P55)] - (1 - Poo)Px 2P31 UP23(1 -- P,*4)(1 -- P55) + P24P,tsP53]

C O S T F U N C T I O N A N A L Y S I S

In o rde r to find the cost funct ions C~(t) and C2(t), we first obta in the following.

(i) Availability analysis

We have defined Ai(t) as the probabi l i ty that the system is up at t ime t iE 0 = Sie E. Elemen ta ry probabi l i ty a rgumen t s yield the fol lowing relations:

Ao(t ) = Zo(t ) + qoo(t) © Ao(t) + qox(t) ©

Ax(t) = Zl(t) + qt l(t) © Al(t) + ql2(t) ©

A2(t ) = Z2(t ) + q20(t) © Ao(t) + q22(t) © + q2,,(t) © A4(t)

A3(t ) = Z3(t ) + q31(t) © Al(t) + q33(t) ©

A4(t) = Z4(t) + q42(t) © A2(t) + q44(t) © + q46(t) © A6(t)

As(t) = Zs(t ) + q53(t) © A3(t) + q55(t) ©

As(t) = q64(t) © A4(t)

A l(t) + qo2(t)© A2(t ) (9)

A2(t ) (10)

A2(t) + q23(t) © A3(t)

A3(t) + q34(t) © A4(t)

A4(t) + q,,5(t) © As(t)

As(t) + qs6(t) © A6(t)

(11) ( 1 2 )

(13) (14) (15)

Taking the Laplace t r a n s f o r m o f the relat ions (9) to (15) and simplifying for A](s), we get

A'd(s) = N2(s)/O2(s) (16) where

N2(s ) = [Z~(1 - q~'l) + Z t qol] x [(1 - q~2)(1 - q~3)(1 - q*,)(1 - q~5) - q~4(1 - q~2)(1 - q~3)

* * m * * * x {q,,sqs6 + q~.6( 1 q~s)} - q~.2( 1 - qss){q23qa,, + q~',,(1 -- q~3)} * * * * * * *

-- (1 q22)q34q,*sqs3] + q31(Zo ql2 -- Z*q~2)

× [q23q64{q4sqs6 + q,~6( 1 - q~'s)}

q2,*q4sqs3] - q ~ 3 ( 1 - q , ~ . X l - q ~ s ) - * * * + [q~2( 1 - q~'t) + qolq12-1

* * * * * * { Z t ( l * * Z2 q34)q4sqs3 -- q~3) + Z3 q23} × [(Z3 q2,* - + × { ( 1 - - - * * ~,Ls~s6'164 - q46q64(1 - q~s)} + {q~3q~, + q~4( 1 - q~3)}{Z*( 1 - q~s) + Z*q*s}]


a n d

D2(s ) = (1 - q~o)(1 - q]'l)

x [(1 -- q~2)(1 -- q~3)(1 -- q*4)(1 -- q~5) -- q~4(1 -- q~2)(1 -- q~3)

x {q4sqs6 + q~6(1 - q~5)} - q~2( 1 - q~'5) * * * * * *

x {q23q34 + q~4( 1 -- q~'3)} --(1 -- q22)qaaq4sq53] * * __ * * * * *

+ qt2qat(1 qoo)[q23q64{q45qs6 + q:6(1 - q~'5)}

q~3(1 - q*4)( 1 - q~5) - q2,qasq53] + [q~2( 1 - q~'l) + qo tq t2 ]

x [q20q3,tq,sq53 - q~o( 1 - q~3) x {(1 - q%)(1 - q ~ 5 ) - "* "* "* * * ,145v56~t64 - q46q64(l - q~'5)}]

T a k i n g the inverse Lap l ace t r a n s f o r m o f (16), we can get the po in twi se avai labi l i ty . T h e s t eady- s t a t e ava i l ab i l i ty is g iven by

A o = l im sA~(s) = N 2 / D 2 (17) s ' -*0

where in t e rms o f

At = P o t P l 2 +P02( 1 - - P i t ) K t = (1 - p44)(1 - P 5 5 )

A2 =P23P34 +P24( 1 --P33) K2 =P45P56 +]746( 1 - P 5 5 )

we have

N2 = [ToO - P I t ) + TIPol] x [(1 - P22)(1 - p 3 3 ) ( K t - K 2 )

- - A 2 P 4 2 ( 1 - - P 5 5 ) --P34P45Ps3( 1 - - P 2 2 ) ]

- - P 3 I ( T o P l 2 - - TIPo2)[P23(Kl - K 2 ) d- P24P45P53]

+ AI[(Tap2,~ - T2p34)p45p53 + (Kt - K2)

x {T2(1 - P 3 3 ) + T3p23} + A2{T2(1 - P s 5 ) + T3P45}] (18)

a n d

D 2 = [ (K 1 - K 2 ) (1 - P33) - Pa4P45P53] X [P2o(ToPt2 + T t p o 0 + T2At] + [ P 2 3 ( K l - / ( 2 ) + P 2 4 P 4 5 P 5 3 ]

x (Tip31 + T3Px2)(1 - P o o )

+ TAP45 [Pa4(I - Poo)(1 - Pl 1)(1 - - P 2 2 ) + P3 tP l 2P2,~( "1 - - P 0 0 ) - - , 4 1 P 2 0 P 3 4 ]

+ T2A2(I -- Poo)(1 -P11)(1 - -Ps s )

+ m K 2 [ A 1 - Pl 2P23P31( 1 -- Poo) + (1 -- Poo)(1 - P l 1)(1 - - P22 ) (1 - - P 3 3 ) ]

N o w m e a n u p - t i m e o f the sys t em d u r i n g (0, t] is

la.p(t) = I ' Ao(u) du do

so t h a t

I~up(s) - A'~(s)/s (19)


(ii) Busy period analys i s

Bi(t) is defined as the probability that the system is busy at epoch t starting from state Si~ E. We have the following recursive relations:

Bo(t) = qoo(t) © Bo(l) + qot(t) © Bl(t ) + qo2(t) © B2(t ) (20)

Bl(t) = ql t(t) © Bl(t) + qx2(t) © B2(t) (21)

B2(t) = Z2(t) + q2o(/) © Bo(t) + q22(t) © B2(I) + q23(/) © B3(t) + q2,(l) © B,(t) (22)

B3(/) = Z3(l) + q3 t(l) © Bt(l) + q33(/) © B3(/) + q3,,(t) © Ba(t) (23)

B4(I) = Z,(I) + q,2(/) © B2(t) + q44(l) © B,(t) + q45(t) © Bs(/) -Jr q46(t) © B6(I ) (24)

Bs(t) = Zs(/) + qsa(t) © B3(/) + q55(t) © Bs(t ) + q56(t) © B6(t ) (25)

B6(t) = Z6(t ) + q6,,(t) © B4(t ) (26)

Taking the Laplace transform of relations (20) to (26) and simplifying for B'd(s), we get

B'~(s) = N3(s)/O2(s ) (27) where

N3(s) = [qolql2 + q*2( 1 - q~'l)] x [(Zaq2, * * * * {Z~' (1 - * * -- Z2 q34)q45q53 + q~3) "+" Z3 q23}

x {(1 - q*,)(1 q~'5) - "* "* "* * * - - "/45'/56"/64 -- q46q64(1 -- q~5)}

+ {q23q34 + q~4(1 -- q~3)} x{Z~(1 q~5)+ * * * * * +Z~'q~6(1 q~5)}]

- Zs q45 + Z6 q45q56 -

In the steady state, the probability that the repair man will be busy is given by

B o = lim Bo(t ) = lim sB'~(s) = N3/D 2 (28) t~X~ $ ~ 0

where

N3 = AI[P45Ps3(T3P24 - T2P3,) + (KI - / ( 2 ) × {r2(1 - p 3 3 ) + r3p23} + a2{Te(X - p . ) + 7~p,s + mK2}]

a nd D 2 is the s a m e as g iven in eqn (18). The expected duration of the busy time of the repair man in (0, t] is

Pb(t) = I t Bo(U) du do

so that

p*(s) = B~(s)/s (29)


(iii) Expected number of repairs during (0, t]

Let us consider the definition

Ni(t ) -expected number of repairs during (0, t] IE O = SieE

By probabilistic reasoning, we have the following recursive relations:

No(t) = Qoo(t) ~) No(t ) + Qol(t) ® Nt(t ) + Qo2(t) ~ N2(t ) (30)

Nt(t) = QI l(t) ~ Nl(t) + Qtz(t) ® N2(t) (31)

N2(/) = Q2o(t) ® [1 + No(t) ] + Qz2(t) ® N2(t) + Q23(t) ® N3(t) + Q2,(t) ® N i t ) (32)

N3(t) = Q3t(t) ® [1 + Nt(t)] + Q33(t)®N3(t)+ Q34(t)®N4(t) (33)

N4(t) = Q`*2(t) ® [1 + N2(t) ] + Q,,(t) ® N4(t) + Q45(t) ® N5(t) + Q46(t) ® N6(t ) (34)

Ns(t ) = Q53(t) ® [1 + N3(t) ] + Qs5(t) ® Ns(t) + Q56(t) ® N6(t) (35)

N6(/) = Q6`*(t)® [1 + N`*(t)] (36)

Taking the Laplace-Stieltjes transform and simplifying for .~o(S), we get

No(s) = N`*(s)/O3(s) (37)

where

N`*(s) = [(~o~(~2 + 0o2 (1 - (~ ~)] x e(o3 t ~z`*- 02003`*)O`*sO53 -I- {020(1- {~33) + Q31023}

X {(l -- ~`*`*)(1 - - ~ 5 5 ) - - Q , I .5~5606a - Oa606~(1 - Q55) }

+ {(~z3(2a`* + (~2`*(1-033)} x { (~42(1 -(-955)+ (~53(~45 + Q45Q56Q6`* + (~`.6(~6`*( 1 - (~55)}]

and

D3(s ) = (1 - Qoo)(1 - Qt l) × [(1 - (~22)(1 - L933)(1 - (2,,`*)(1 - (255)- L96,,(1 - L922)(1 - (.933)

x {(~,,5(~56 + (-9,,6(1- (~55)}- Q`*z(1-055) X {~23Q3̀*-1" ~2 *̀(1--~33)}--(I--Q22)~34.Q *̀sQ53]

"[- Q 1 2 ~ 3 1 ( 1 - 0 0 0 ) F ~ 2 3 Q 6 ` * { O 4 5 ~ 5 6 -at- 0,,6( 1 - - 0 5 5 ) } - (~23(i - 0,`*)(1 - (255)- Qz`*Q`*sQ53]

+ [~o2(1- (2,,)+ (~o, ~,z] x [020~3`*L7`*,053- ~20(1 - 033)

x {(I - L:)̀ *`*)(I - L'.')55) - L.~,, 5 L')s6Q~ *̀ - L7`.606,,(I - (~55)}]

In the steady state, the per-unit time expected number of repairs is given by

No = lim [No(O/t ] = N J D 2 (38) t ~ O 0


where

N4 = Al[p45Psa(PalP2,~ - P20P3,*) + (Kt - K2) x {P20(l -P33) +P23P3,} + A2{P,~2(I --P55) + P,~sP53 + K2}]

and D 2 is the same as given in eqn (18).

(iv) Expected number of visits by the repair man in (0, t]

According to the definition of V~(t), by elementary probability arguments we have the following relations:

Vo(t)= Qoo(t)® Vo(t)+ Qo2(t)®

vl(t)= Q~( t )® v~(t)+ Qt2(t)@

V2(t)= Q20(t)® Vo(t)+ Q22(/)®

v3(t)= Q3t(t)~ vt(t)+ Q33(t)®

va(t)= Q42(t)® v2(t)+ Q,,( t ) (~

Vs(t)= Q53(t)@ v3(t)+ Q55(t)®

v6(t)= Q6,(t)® v4(t)

[1 + Vz(t)] + Qo,( t )® Vt(t ) (39)

[1 + V2(t)] (40)

V2(t) + Q23(t)® v3(t)+ Q2~(t)(fi) v4(t) (41)

v3(t) + Q34(t) ® v4(t) (42)

v4(t) + Q45(t) ® vs(t) + Q46(t) ® v6(t) (43)

Vs(t ) + Q56(t)® v6(t ) (44)

(45)

Taking the Laplace-Stieltjes transform and simplifying for Vo(s), we get

k'o(S) = Ns(s)/O3(s) (46) where

Us(s) = EOo, 0,2 + 0o2( 1 -- 011)3 x [ (1 - 022)(1- Q33)(1- Q44)(1- 055 ) -Q64(1 - 622)(1- 633)

× {6,56s6 + 6,6(I - 6ss)} - (I - 622)03 ,0 , ,6s3 -- Q 2 3 6 3 4 6 4 2 ( 1 -- 0 5 5 ) - 02,{242( 1 - 633)(1 - 6s5)]

In the steady state, the number of visits per unit time is given by

V 0 = lira [ Vo(t)/t ] = Ns/D2 (47) ¢'-* oO

where

N 5 = A 1 [-(g I - / ( ' 2 ) (1 - t722)( 1 - P a 3 ) -Pa4P45P53( 1 - P 2 2 ) - A 2Pa.2( 1 - P 5 5 ) ]

We are now in a position to obtain the two cost functions of the system considering the characteristics obtained above. The net expected total costs incurred in (0, t] are

Ct(t) = expected total revenue in (0, t] - expected total expenditure during (0, t]

= Pol~up(t) -- P, Pb(t) -- P3 Vo(t) (48)

262 Rakesh Gupta. Sachendra Bansal

and

C2(t) = Polluo(t) - - PzNol t) - - P3 ~o(t) (49)

The expected total costs per unit time in the steady state is given by

C, = lira C, ( t j / t = lim sZC~(s) = PoAo - P~ B o - P3 Vo (50) t ~ ~ : s ~ O

and

C , = lim C2{t)/t = l i m s Z C * ( s ) = PoA o - PzNo - P3 Vo (5I) t ~ ~ s ~ O

where Po is the revenue per unit up-time, ,°1 is the amount paid per unit of time when the repair man is busy, P2 is the per-unit repair cost, and P3 is the cost per visit by the repair man.

P A R T I C U L A R CASE

When there is no shock at all, i.e. ~ = 0, then in terms of T O = ,/'rc/'2fl, 7_, = e*"[,.'rc - 2 Erfx]/,~ 2fl, P42 = I~T2 and/046 = 1 - P 4 2 , we have

E(~o) = [To(1 -P42P46)+ T2(I +P46)]/(P46) 2

Ao = [To(P,,2) 2 + T2]/D

B 0 = [m(Pa6) 2 + T2]/'D

N O = (1 - p42P46)/D

I 0 = (p~2)Z/o

c , = eoAo - e18o - P 3 Vo

C., = t ' oAo -- P2No - e3 Vo

where

D = 2 m p 4 6 + T 2 and x = p / v / 2 f l

Study of system behaviour through graphs

For a more concrete study of M T S F and cost functions (C t and C2), we plot these characteristics w.r.t./a (repair rate) alternatively by changing the parameter ft. The curves so obtained are shown in Figs 1 and 2, respectively, for the M T S F and cost functions.

In Fig. 1 each curve represents the graph between/~ and the M T S F for/~ = 1.44, 1.96 and 2"56. These curves clearly indicate that the M T S F increases uniformly as the repair rate increases. Also, with the increase in parameter fl, the M T S F decreases.


Figure 2 represents the behaviour of the cost functions C1 and C2 w.r.t, p for different values of /3 (= 1-44, 1-96 and 2-56) when P0 = 150, Pt = 100, P2 = 50 and P3 = 25. Each smooth curve of this figure shows the behaviour of cost C1 and the dotted curves indicate the behaviour of cost C2. From the figure it is clear that C~ and C 2 increase as p increases. Also, with the increase in fl, both of the cost functions decrease. Further, it is important to note that for F~ < 2.0 the cost function C2 is better than C~ for all fl, while for/a > 2.0 the cost function Ct is better than C2.

R E F E R E N C E S

1. Esary, J. D. & Marshall, A. W., Shock models and wear processes. Ann. Probab., 4 (1973) 627-49.

2. Feldman, R. M., Optimal replacement for systems governed by Markov additive shock processes. Ann. Probab., 3 (1977) 413-29.

3. Zuckerman, D., Optimal stopping in a semi-Markov shock model. Appl. Probah., 15 (1978) 629-34.

4. Feldman, R. M., Optimal replacement with semi-Markov shock models using discounted cost. Math. Ops Res., 2 (1977) 78-90.

5. Murari, K. & AI-Ali, A. A., A standby reliability model subject to random shocks. In Proceedings of Seventy-fourth Session of the Indian Science Congress, Bangaiore, India, 1987.

6. Murari, K. & A1-Ali, A. A., One unit reliability system subject to random shocks and preventive maintenance. Microelectron. Reliab., 28(3)11988) 373-7.

7. AI-Aii, A. A., Three-unit reliability system subject to random shocks. Int. J. of Management and Systems, 5(2) (May-August 1989).

8. Bateman, H., Higher Transcendental Functions, Vol. II. McGraw-Hill Book Company, Inc., 1953.

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