The Theory of Delayed Repair for all Systems with 2 or 3 Components

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Mathematik und

Informatik

Informatik-Berichte 64 – 12/1986

Winfrid G. Schneeweiss, Janusz Karpinski

The Theory of Delayed Repair for all Systems

with 2 or 3 Components

THE THEORY OF DELAYED REPAIR

FOR ALL SYSTEMS WITH 2 OR 3 (OMPONENTS

Winfrid G. Schneeweiss, Janusz Karpinski ¹¹

Abstract - This is a report on the derivation of the probability distribution of residual life time after certain (hardware) compo- nent failures. For redundant systems, typically fault tolerant com- puters,it may be necessary or economical to start repairs only after certain tolerable faults, knowing that in most cases repairs will be finished prior to system failure. In such cases the distribution function of residual life is the probability of not meeting this goal.

All systems with 2 or 3 components are discussed at length to give 1) easy to understand results for such simple systems, 2) many examples to test more general methods.

Complete results are given for the case of s-~ndependent components, especially for the exponential case, and for cold standby, where appropriate. For the general case of s-dependent components, a general approach is outlined and shown to work with two selected examples.

Key Words - Delayed repair, Fault tolerance, Residual life

1) DAAD-Scholar from the Systems Research Institute of the Polish Academy of Sciences

2

(ONTENTS

1 Introduction

2 The distribution of residual life of all non-trivial 2- and 3- cornponent systerns of s-independent cornponents

2. 1 The 1-out-of-2: G systern ¹

»

2.2 The sirnplest 2-out-of-2:G systern of 2 subsysterns, one of which is a 1-out-o~2:G systern

2.3 The sirnplest 1-out-of-2:G systern of 2 subsysterns, one of which is a 2-out-of-2:G systern

2.4 The 2-out-of-3:G/F systern 2.5 The 1-out-of-3:G systern

3 The case of s-dependent cornponents 3.1 The general 1-out-of-2:G systern 3.2 The general 2-out-of-3:G/F systern Appendix

A.1 Checks for diverse F 's tobe Cdf's r A.2 Sorne auxiliary integrals

Cornrnent: The authors found no non-trivial literature to cite.

1) A k-out-of-n:G/F system is 51.ood/!_ails if at least k of its n components are good/faulty.

3

1 lNTRODUCTION

In many engineering systems the cost of maintenance depends strongly on the maximum allowed system down-time. Maintenance costs decrease with increased down-time. (This is no contradiction to the well-

known fact that long repair times mean high maintenance costs . . Rather, long down-times, first of all, mean waiting for a proper moment to start repair.) Hence, i t is also of practical interest to determine the probability with which a system fails within a certain time interval after a given event has occurred (inside the system). The concept of residual life is readily visualized via the state tran- sition graph well-known from Markov analysis. Figure 1 shows the set of the 8 elementary states of a 3-components system for the case of

no repairs. Obviously, i t is the 1-out-of-3:G system. Residual life can be the time the system spends between the "k th component

- all components good _ .. _1_⁵~-. component faifure .

+- onfy 1 component bad

...

?~~

.

~'?':!P.· __ fa_i fure

+-only 1 component good

-all components bad

Fig. 1 Transition graph of a 3-component system.

failure" line and the "system failure" line. Of course, i t can also

4

be defined, eg., tobe the time from leaving state 100 until system failure.

In this report, cases where the residual life time Tr is always zero are not discussed. But cases, where sometimes T =0 are included. All r component lives are assumed tobe stochastically (s) independent of each other. Except for trivial cases, the distribution function of T is always also given for exponentially distributed component

r lives.

The reader is advised to read the notation list prior to the rest of the report.

Note that since all Cdf's discussed are those of non-negative ran- dom variables, the expressions given for them are only correct for non-negative arguments.

Because of the ease of handling F. in the exponential case, F. is

i i

preferred to Fi whereever possible.

5

NOTATION

M

A _ A(t' ,t" ,, ,S,L)

- [ V fl

(T, ;;, t ^{1 ) ]} n (Ts > t ¹¹ + 1) ] , t ^{1 ;;,} t"

• 0 J

i=1 J:C,E~.

J l.

F. , f.

l. l.

-

component i

Cdf, pdf of Ti ¹⁾

F 1-F (survivor function) i, j indices

l. residual life set j, i.e. a set j of certain c. 's such

J 1

that, when the last o[ them fails, Tr begins L

=

{l

1, ... ,lM} set of all residual life sets of S under consideration Ai failure rate of ci (in the exponential case)

Pr{a} probability of (random) event a s system (under investigation) Ti life (time) of ci

T residual life of S r

t,T,t' ,t" time variables

1 ( 1 ) unit step function 1 for ,~O, and O elsewhere

{j:cjEli} is the set of all j whose corresponding cj are elements o~

,f_i

1) Cdf - Cumulative distribution function; pdf _ probability density function.

6

2 THE DISTRIBUTION OF RESIDUAL LIFE OF ALL NON-TRIVIAL 2-AND-3-COMPONENT SYSTEMS Ot s-INDEPENDENT COMPONENTS

Whichever residual life set L is defined for a series (n-out-of-n:G) system, T will always be O. Hence, the 2-out-of-2:G system and the

r

3-out-of-3:G systemare not considered in this section; they are

"trivial" in this context.

In all cases only the Cdf of T will be given. Other quantities are r

easily derived from Fr' e.g. the mean as

(X)

F

^(T)dT

r ^{( 2-1 )}

In this section hot standby and cold standby are always discussed.

The component lives T

1,T2,T

3 are usually assumed tobe stochastically (s-) independent (of each other). To illustrate the results for

general distributions of component lives also the exponential case (with exponential pdf's of Ti's) is always presented. Furthermore, the differences of rnodelling resulting from assuming more or less know-

ledge of component failures are amply discussed.

7

2.1 The 1-out-of-2:G system

In order to determine Fr for the case where Tr starts when c 1 fails, we have as ''residual life set" L:

L = { -t 1 } ; l.₁ = { c

1 } • ( 2-2)

Let c

2 be standby for c

1• For hot standby

Otherwise Tr can be regardeu as being undefined. In that case Fr is the conditional probability

(2-3) (The alternative case, that Tr=O if c

2 fails first, is discussed below.) By the definition of conditional probability, (2-3) yields

= Pr {O < T

2 - T

1 ^~ , } /Pr {T

1 < T 2 } . Now, (for hot standby and) s-independent T

1 and T 2

00

Pr {o<T

2^-T1 ^~ 1} = f f

1 (t) [F

2 (t) - F

2 (t+•) ]dt ,

0

where

F2(t) - F2 (t+1) = Pr{c 2 fails between t and t+1}

and

00 00

Pr {T

1 < T } = f f

1 (t)F

2(t)dt = ^f f

2(t)F

1 (t)dt

2 0 0

Finally, (2-4) yields

00 00

F ( 1) = r ^J f1 (t) [F2(t)-F2(t+1) ]dt/ f

t,

(t)F2(t)dt

0 0

00 00

= 1 - J f1 (t)F2 (t+1)dt/ J f1 {t)F~ (t)dt.

. L

0 0

(2-4)

( 2-5)

;

(2-6)

.

(2-7)

8

(Check : Obviously Fr(O)=O and Fr(⁰⁰)=1 since F2 (⁰⁰)=0.) In the exponential case, ie. if

f.(t) =;,..,. exp(-;,..,.t), F.(t) = 1-exp(-t-..

1. t ) ,

1 1 1 1

(2-7) yields

0:,

t-..1 ^! exp(-t-..

1t)exp[-t-..

2(t+T)]dt

0

F ( T) = 1 _r

-

00

t-.. 1 ^f

0

exp(-;,..,

1t)exp(-t-..

2t)dt

= 1

-

^{exp (-t-..}₂^T).

( 2-8)

(2-9)

(This shows the so-called memorylessness of the exponential distri- bution.) In the above investigation i t was assumed that the case T2<T

1 makes no sense. However, if the failure of c 2 is not reported, it makes sense to assume Tr=O, if c 2 fails first. Consequently, then

Specifically, for s-independent T 1,T

2, by (2-6), (2-5)

00 00

0 0

00

= 1- f f1 (t)F2 (t+T)dt,

0

as shown in the appendix.

Obviously, here by (A2-3) ( see the appendix), in general

(2-10)

(2-11)

0 < F ( +O ) _r < 1 . ( 2 - 1 ^{.!. )}

Moreover, for equal components, as shown in the appendix [see (A2-6) J F _r (+0)

=

1/2 .

(Check : Obviously Fr(⁰⁰)=1 .)

(2-12a)

9

In the exponential case ( 2-8) , by ( 2-11)

F (·r) /\

= 1- exp(-'A.

2,), r

"-1 + "-2

(2-13)

v,hence

F ( +O) "-2 r =

+ "-2

"-1

(2-13.::)

ln case of cold standby (of c

2 for c

1), trivially

(2-14) This result can be derived formally from (2-11) by replacing T

2 by T 2-T 1 :

00

Fr(,) = 1-

F

₂(,) f f

1 (t)dt ,

0

X2-14a) :Erom which, by (A2-1), (2-14) results. The case L={.t

1},.t 1={c

2} is clear.

Now, let us investigate Fr for Tr beginning at the first component failure. Here, obviously

(2-15) In case of hot standby, since first failures of c

1 and c₂ exclude each other, for s-independent T

1 and T 2

00

Fr ( 1)

=

f {f

1 ( t) [

F

2 ( t) -

F

1 ( t +-r ) J

0

+ f2 (t) [F1 (t) - F1 (t+,) ]}dt . (2-16)

r

10

Note that in any case of systern operation,Tr is well defined, so that here F (,) is not a conditional probability.(A check for F (oo)=1

r r

is given in the appendix.)

In the exponential case, using (A2-7)

+ [ 1 - exp ( - ;,..

1 , ) ] A.2 + A.1

= 1 - 1

[ ;,..

1 exp ( - ;,..

2 ^{-r )} + ;,..

2 exp ( - ).._ 1 , ) ] .

"1 + "2

For equal cornponents with ;,..

1=,..

2=).._;.as should be expected

(2-17)

(2-17a)

, 1

2.2 The sirnplest 2-out-of-2:G systern of 2 subsysterns, one of which is a 1-out-of-2:G systern

Figure 2 shows the reliability block diagrarn of this systern.

Fig. 2 Reliability block diagrarn of the systern of section 2.2.

First, let

L = { -t₁ } , .t₁ = { c1 } . (2-18)

With T1 ending first, Tr=rnin(T2,T3) the condition of c

1 failing first

Hence, sirnilar to (2-4), under

Now, for hot standby (of c

2 for c1) and s-indpendent T 1,T

2,T 3

0:,

= f f, (t) [F2(t)F3(t) - F2(t+T)i\(t+T) ]dt

0

where

(2-19)

(2-20)

12

= Pr{c

2 and c

3 fail after t} - Pr{c

2 and c

3 fail after t+1}

=

Pr{c

2 or c

3 (or both of thern) fail between t and t+,}.

Furtherrnore, since a<rnin(b,c)=(a<b)t-.(a<c) :

CD

Pr { T

1 < rni n ( T 2 , T

3 ) }

=

J f 1 ( t)

i\ (

^{t )}

i\ (

^{t) d t .}

0

Finally, (2-19) is transforrned to

00 ""

(2-21)

(2-22)

Fr(,) = J f 1 (t) [F2 (t)F

3 {t)-F2(t+,)F

3 (t+,) ]dt/ff

1 (t)F

2 (t)F

3 (t)dt

0 0

00 00

= 1 - J f1 (t)F2(t+1)F3(t+1)dt/J f1 (t)F2(t)F3(t)dt.

0 0

(Check Obviously, Fr(O)=O and Fr(⁰⁰)=1 .) In the exponential case (2-8) :

00

Fr{T)

=

1-A, J exp(-A1t)exp[-(A2 + A3)t]exp[-(A2 + A3)T]dt

0

00

/{A1 J exp[-(A

1 + A2 + A

3)t]dt}

0

(2-23)

(2-24)

~his is plausible, since, after c1 failing first, the systern failure rate is A

2 + A 3.

If the condition of c1 failing first is rernoved, then also c2 or

c

₃

"can" fail before c1, in which case Tr=O. Hence, sirnilar to (2-10) for hot standby

13

(2-25) For , ~ O we have, by (2-22) and (2-20), for s-independent T

1,T 2,T

3

CX) CX)

0 0

CX)

+ J f

1 (t) [F2 (t)F3 (t) -

.r

₂ (t+,)i\ (t+,) ]dt,

0

!The check result F (cx,)=1 is derived in the appendix.) r In the exponential case

F (,)

= - - - -

1 r

=

1 -

where

F (+O) "2 + "3 r =

A. 1 + ^A. 2 + A. 3

(2-26)

(2-27)

(2-27a)

In case of cold standby (of c

2 for c1) and s-independent T 1,T

2,T 3,

•r

₂ must simply be replaced by T

2-T1. Hence, from (2-23)r i.e. under the condition of c1 failing first :

CX)

Fr(,)= 1-J f 1 (t)F

2(,)F

3 (t+,)dt/J f

1 (t)F

3(t)dt. ^(2-2°)

0 0

(Check Obviously Fr(O)=O and Fr(cx,)=1.)

1 4

In the exponential case, here

CX)

Fr(,) = 1- A1 f exp(-A1t)exp(-A

2,)exp[-A

3(t+,) ]dt

0

CX)

/A 3 J exp(-A₁t)exp(- A

3t)dt

0

(2-29)

there is no difference between hot and cold standby in this case.

Alternatively, if the case T

3 < T

1 is not excluded

CX)

(2-30)

which checks with (2-26) with the first integral there deleted and T2 replaced by T2-T1 . Again, in the exponential case

Fr(.) A3

+ A1

{1- exp[-(A

2 + A3^{h ] }}

=

A1 + A3 A1 + A3

1- A1

exp[-(A

2 + A3 h],

=

A1 + A3 (2-31)

where

Fr ( +O) A3

=

A1 + A3 (2-31a)

The case L={l 1}, l 1={c 2 } is trivially solved by interchanging indices 1 and 2 in the case of (2-18).

1 5

Cornment: In the case L={.t 1}, .t 1={c

3}, trivially Fr(t)=1 (1).

'1'he case of Tr starting at the moment of the first component failure is implied by

(2-32) If c

3 fails first, Tr=O. However, since, generally, also c

1 and c 2 can fail first, i t makes sense to ask for the Cdf of Tr. Similar to

(2-16), for hot standby and s-independent T 1,T

2,T

3, using (2-21)

CO

Fr(t) = f f

1(t)

[F

₂(t)F\(t) - F

2(t+t)F

3(t+1)]dt

0

CO

+ J f2 (t) [p1 (t)F3 (t) - F1 (t+t)i\ (t+t) ]dt

0

CO

+ f f3(t)F1 (t)F2(t)dt.

0

(2-33)

It is correct that the last integral of (2-33) does not depend on 1 , since there is no standby for c

3. Obviously, Fr(+O) = Pr{c

3 fails first}, (2-34)

and, as shown in the appendix,Fr(⁰⁰)=1. It is also shown in the appen- aix [see (A1-6)] that, plausibly, for equal components

(2-34a) In the exponential case (2-33) becomes

F (1) =

r

=

Obviously, F (+O) = r

1- /\ 1

here

/\ 1 +

1 6

exp[-(;\

2 + ;\3)T ]+ /\2 exp(-(;\ 1 + /\3 )T]

.

(2-35) /\ 1 + /\2 + /\ 3

/\ 1

/\ 2 + /\3 (2-35a)

Cornrnent: Here the investigation of l. 's with 2 components makes no

1

sense, since in all these cases Tr=O.

17

2.3 The simplest 1-out-of-2:G system of 2 subsystems, one of which is a 2-out-of-2:G systern

Figure 3 shows the reliability block diagrarn of this systern.

1 2

3 1 - - - '

Fig. 3 Reliability block diagrarn of the systern of section 2.3.

First, let

(2-36) If c 1 fails first, then fcr hot standby {cf c

3 for the series system of c 1 and c2) Tr=T3-T1. Hence, under the condition of c

1 failing first

As in similar cases above, for s-independent T 1 ,T

2,T 3 Pr { ( 0 < T 3 - T 1 ^{~ 1)} n ( T

1 < T 2) }

00

= J t ₁(t)F

2 ^(t)[F3 ^{(t) - F}3 (t+,)Jdt,

0

and by (2-22)

(2-37)

(2-38)

1 8

00

- -

=

J f

1(t)F

2(t)F

3(t)dt. (2-39)

0

Insertion in (2-37) yields here

00 00

= 1- f

t,

(t)F2(t)F3(t+1)dt/ J

t,

(t)F2(t)F3(t)dt. ^(2-40)

0 0

(The checks for Fr(0)=0 and Fr(⁰⁰)=1 are trivial.) In the exponential case (2-8), as should be expected

(2-41)

For cold standby (of c 3) T

3 must be replaced by T 3-T

1. Hence, from (2-40)

00 00

0 0

(2-42)

a result that should be expected (under the condition of c

1 failing first). This finishes the conditional case.

If, in case of hot standby, T 3<T

1, then (on the failure of c

1) Tr=0, irrespective of T

2. Hence, as in the 1-out-of-2:G case [see (2-10) with index 2 replaced by 3] :

For s-independent T 1 ,T

3 , by (2-11)

00

=

1 - J f1 (t)F3(t+T)dt.

0

For exponential component lives, here

(2-43)

F ( t) = 1 - r

1 9

(2-44)

In case of cold standby (of c

3) and T

2<T1 ,T3 will start when c2 fails and not only when c

1 fails. Hence, here

CO

Fr(t) = J f 1(t)F

2(t)F3(t)dt

0

CO CO

+ f f

2(t) J f

1(t+t')F

3(t')dt'dt

0 0

CO CO

+ f f

2 ( t) f f

1 ( t +t ^{1 ) [} F

3 ( t ¹ + t ) - F

3 ( t ' ) ] d t ^I d t ,

0 0

(2-45)

where the second term covers the case where c2 and c3 fail before c₁. Simplifying (2-45) yields

CO CO 00

(2-45a) (The check result Fr(⁰⁰)=1 is derived in the appendix.)

In the exponential case,by ^(A1-9) (see the appendix)

II. 1 11.1 + 11.2 + _11.3

Fr(-r)

=

¹

-

^exp(-11.₃^{-r). (2-46)}

11.1 + 11.2 11.1 + 11.3

'l'he treatment of the case L= { .t 1 } ' .e.,={c2} is trivial. Only the in- dices 1 and 2 of case (2-36) have tobe interchanged.

1-:iow, let

L

=

{ l 1 } ,

.e.

¹

=

{c3}· (2-47)

20

This case is practically the case (2-2) for the 1-out-of-2:G system with c

1 replaced by c

3 and "failure of c

2" replaced by "failure of

c1 or c

2" and "survival of c

2

replaced by "survival of c

1 and c 2".

aence, under the condition that c

3 fails first, from (2-7), using (2-21)

CX) CX)

Fr(1) = f f

3(t) [F

1 (t)F

2 (t)

-

^F₁ ^(t+1)F₂ (t+1) Jat/Jf3 (t)F

1 (t)F

2 (t)dt

0 0

CX) CX)

= 1- ^r

r

₃ (t)F1 (t+1)F2(t+1)dt/Jf

3(t)F

1 (t)F

2(t)dt. (2-48)

0 0

In the exponential case Fr ( 1) = 1 - exp [ ( 11.

1 + 11.

2 ) T ] . (2-49)

This is most plausible, because, after the failure of c

3 (as the first failure) ,the residual system is a 2-out-of-2:G system of o

1 and c2. This finishes the conditional case.

If c

3 does not fail first, then Tr=O. Hence, from (2-21)

CX)

- -

Fr(T) = 1- f f

3 (t)F

1 (t)F

2 (t)dt

0

CX)

+ f

0

t₃ (tl [F

1 (t)F

2 (t)

-

F1 (t+1)F2(t+1) ]dt

ex,

= 1- f f3(t)F1 (t+T)F2(t+,)dt, (2-50)

0

which allows for the following simple intepretation : Tr>,, if (after c3 fails) ,both c

1 and c

2 live for at l e a s t , more units of time. In

21

the exponential case, here

{1- exp[-(A1 + A2)T]}.

A1 + A2 + A3

( 2-51)

In the case

(2-52)

since the events "c. (i=1,2,3) fails first" exclude each other, here

l.

for hot standby and s-independent T 1,T

2,T 3

00

F ( _r T)

=

f

0

00

+ f f2(t)F1(t)[F3(t) - i\(t+T)]dt

0

00

+ f f3(t)[F\(t)F2(t) - F,(t+T)F2(t+T)]dt.

0

(The check for Fr(⁰⁰)=1 is given in the appendix; see (A1-4).

Specifically, for exponential component lives :

+ - - - -A3

(2-53)

(2-54)

22

Comments: The case L={l

1}, l 1={c

1,c

2} is not discussed, since i t appears that i t has little practical meaning On the failure of c

1 er c

2 i t is known that the future operation of the system depends exclusively on c

3. The cases L={l

1} with

t

₁={c

1 ,~} and l

1

^={c2,c 3}, respectively, are trivial because always Tr=O, so that Fr(T)=1 (T).

23

2.4 The 2-out-of-3:G/F system First, let

(2-55) Then for hot standby and s-independent T1 ,T₂,T

3 the remaining sub- system is a series system of c2 and c

3. Hence, under the condition of c1 failing first, Fr is that of (2-23). Without this condition S can again fail prior to or together with c

1 , Fr is given by (2-26). For cold standby of c

3 for either c

1 or c

2, under the condition of c 1 failing first :

(2-56)

where the numerator is Pr{c2 or c

3 fail in the f i r s t , units of time after the failure of c1,and c 2 and c 3 fail after c 1 does}.

00

= f f

1 (t) [F

2(t)- F

2(t+•)F

3 (,) Jdt

0

which is (2-21) with T

3 replaced by T

3-T1. With Pr{T

1 < T

2} of (2-6), here (2-56) yields

00 00

Fr(T)

=

^f f1 (t) [F

2 (t)-F

2 (t+,)F

3 (,) ]dt / J f

1 (t)F2 (t)dt.

0 0

(Check 0bviously F (0)=0 and F (⁰⁰)=1.)

r r

(2-57)

(2-58)

24

Solutions to the cases L={t

1} with t 1={c

2} and t 1={c

3}, respectively,

~~e found frorn the case of (2-55) by trivial perrnutations of indices.

Now, let L be that of (2-52), i.e. T begins with the first cornponent r

failure. Then, for hot standby and s-independent T 1,T

2,T

3, since each cornponent can be the first to fail, by (2-21) :

00

Fr(1) = J f

1 (t) [F

2(t)F

3 (t)-

F

₂(t+1)F

3 (t+1) ]dt

0

00

+ J t ₂(t)[F

1(t)F

3 (t)-

F

₁(t+1)i\(t+1)Jdt

0

00

+ J f3 (t) [F, (t)F2 (t)-

F,

(t+1)i\ (t+1) ]dt.

0

(The check of F

4 (~)=1 is simple via (A1-4).) In the exponential case

+ 11.

3 (1 - exp [ - ( 11.

1 + 11.

2 ) 1

y } .

Cornrnent Again,cases L={l,1}, .t

1={c. ,c.} lead to F (1)=1 (1).

· i J r

(2-59)

( 2-60)

25

2.5 The 1-out-of-3:G system Again, we start with

L = { .t

1 } ,

t

1 = { c1 } . ( 2-61)

Under the condition of c

1 failing first, for hot standby and s-inde- pendent T1,T2,T

3 :

CX)

Fr(1) = J f 1(t)[i\(t)-F2(t+1)][F3 (t)-i\(t+1)]dt

0

For exponentially distributed T 1,T

2,T 3

(2-62)

(2-63)

If the condition of c

1 failing first is removed, in case that c

2 and c3 fail before c1, we will have Tr=O. Hence,

CX)

Fr(T) = J f

1 (t)F

2(t)F

3 (t)dt

0

CX)

+ J f

1 (t) {F

2 (t) [F

3 (t+1)-F

3 (t) ]+F

3 (t) [F

2 (t+1)-F

2 (t)]

0

where the last integral is

(2-64)

Pr{c2 or c3 fail as the last component at most 1 units of time after c1}.

(2-64) is readily simplified to

26

<X)

Fr(1)

=

f f

1 (t)F

2(t+,)F3(t+,)dt.

0

(2-65)

Clearly, if c

1 fails at t, both c

2 and c

3 "must" fail prior tot+,.

(The check result Fr(⁰⁰)=1 is easily verfied using (A2-1 ,2).

In the exponential case

"- 1

+ - - - ^(2-66)

In case of cold standby (of c2 for c1, and c3 for c2) Fr is the Cdf of T2 + T

3. Hence, for s-independent T2,T 3

1

Fr(,) = f f

2(t)F

3(,-t)dt.

0

For exponentially distributed T

2 and T

3, as is well-known:

since 1/11.i is the mean value of Ti.

Now, let

Under the condition of c

3 failing last, for hot standby

( 2-6 7)

(2-68)

(2-69)

27

Hence, for s-independent T1,T2,T 3 :

00

Fr(T) = J[f

1(t)F2 (t) + f2(t)F 1 (t)][F\(t) - i\(t+T)]dt

0

(Check : F (oo)=1 is shown in (A1-10,11) .) r

In the exponential case

00

Fr(T) = l1-exp(-,._

3T) ]J{,._

1exp(-,._

1t) [1-exp(-,._

2t]

0

00

/~

^{,._3exp(-,._3}t) [1- exp(-,._

1t) ][1-exp(,._

2t) ]dt

= [ 1- exp (-,._

3 T ) ] ( ,._ 1

+ ^{,._ 2} ,._ 1 + "-2

,._ 1 +

"-3 "-2 + ,._ 3 A. 1 + "-2 +

/(1 - "-3 "-3

+ ,._ 3

)

.

,._ 1 + "-3 "-2 + ^{,._ 2} ,._ 1 + "2 + ,._ 3

(2-70)

"-3

(2-71)

If the condition of c

3 failing last is removed, then Tr=O, since after the failure of

t

₁ all the components have failed. Hence, from

(2-70)

00

Fr(T) = 1- J f 3 (t)F 1 (t)F 2 (t)dt

0

00

+ J [f1 (t)F2(t)+f2(t)F1 (t) ][F3(t)-F3(t+T) ]dt.

0

(2-72)

(Check : By ·(A1-11)

28

F (00)=1.) r

In the exponential case, from (2-71)

Fr(T) ⁽ 1

+ 1

= /\ 3

/\ 1 + /\ 3 /\ 2 + /\3 /\ 1

[1- exp(-/\

3T)] ( /\ 1 + +

/\ 1 + /\3

1- ( /\ 1

+ /\ 2

=

/\ 1 + /\ 3 ^/\2 + /\ 3 /\ 1

1

+ /\ 2 + /\ 3

/\ 2 /\ 1 + "-2 /\ 2 + /\3 /\ 1 + /\ 2 + /\ 3 /\ 1 + /\2

) exp (-/\

3 T ) . + _/\2 + /\ 3

(2-73)

Finally, let Tr be the time after the second component failure with (2-74) Then, since the events "c. fails last" exclude each other, for hot

l.

standby and s-independent T 1,T

2,T 3

00

Fr(T) = J [f1(t)F₂ (t) + f 2(t)F1 (t)Hi\(t) - i\(t+T)]dt

0

00

+ J [t1 (t)F

3 (t) + t ₃ ^(t)F₁ ^{(t) J [F}₂ ^{(t) - F}₂ (t+T) Jat

0

00

+ ^f [f2(t)F3(t) + f3(t)F2(t)HF1(t) - F1(t+-r)]dt.

0

(The check for Fr(⁰⁰)=1 is given in (A1-4) ).

(2-75)

29

In the exponential case, by (2-73)

Fr ( ·r)

=

^{A 1 + A2} A1 ₊ A.3 A2 + A.3

A A.3

+( 1 ₊

A.1 ⁺ "2 "2 + A. 3

+ ( ^{A. 2}

+ - - - -

^{A. 3}

"1 + "3

F ( ^{T )} = 1 - exp ( -t... T ) •

r

A1 ⁺ A2

) [ 1 - exp ( -11. 3 ^{T) ]} A1 ⁺ 11.2 + A.3

A. 1

A.1 + A3

) [ 1- exp ( -A 2 ^{T) ]} + "2 + A3

) [ 1 - exp ( - ,...

1 ^{T ) ] • (} 2- 7 6)

"1 + A.2 + "-3

(2-76a)

30

3 THE CASE OF S-DEPENDENT COMPONENTS

In this report s-dependence of components has been limited t i l l now to the case of cold standby. Now, we will present an approach to overcome this limitation. In section 2, typically Fr(T) was found

as an integral (from O to ^{00 )} of a certain probability density, which was not a pdf. In fact, using A(t' ,t",T,S,L) as explained in the

notation list

F (T) = r

00

J ^O Pr { A ( t ' , t" , ^{T ,} s , L) }/ d t •

o at' t'=t"=t

( 3-1 )

Note that Pr{A} has partially characteristics of a Cdf of T.'s and

J

of a survivor function of TS. Since a density is desired only

for the time of failure of the last c. of an f. i t is necessary to

1 J

introduce t", even though afterwards the (partially) differentiatec function is needed only for the special case t"=t'.

Two examples will illustrate the approach of (3-1).

31

3.1 The general 1-out-of-2:G system

For L of (2-15), by the general definition of A:A(t' ,t",T,S,L) A [(T < t ')

u

(T ,,..t')Jn (T > t " + T ) , t " - ~ t • .

= 1 = 2 ² S

Hence

Pr{A} = Pr{(T1 :;; t') n (Ts > t" + T)}

+ Pr{(T2 ~ t') n (Ts > t" + T)}

- Pr{(T1:;; t') n (T2 ~ t') n (Ts > t" + T)}.

(3-2)

( 3-3) For hot standby T

5 equals T

1 or T

2, and the last event cannot happer., i.e. its probability becomes zero. So (3-3) becomes

Pr{A}

=

Pr{ (T1 :;; t') n (T2 > t" + T)}

+ Pr{(T2 ~ t') n (T1 > t" + T)}.

Check: For s-independent T

1 and T

2, from (3-4) Pr { A}

=

F

1 ( t ' ) F

2 ( t" + T) + F 2 ( t ' ) F

1 ( t" + T) ,

and

a a

at'

a

t '

Pr{A} = f

1 (t'

)F

₂ (t" + T) + f 2 (t'

)F

₁ (t" + -r),

Pr{A}

=

^f ₁(t)F

2(t+T) + f

2(t)F

1(t+,).

lt"=t'=t

1

Hence, by (3-1)

00

Fr(T) = 1- f [f

1 (t)F

2(t+T) + f

2(t)F

1 (t+,) ]dt,

0

which equals (2-16) due to (A1-1,2,3).

(3-4)

(3-5)

( 3-6)

( 3-7)

( 3-8)

32

3.2 The general 2-out-of-3:G/F system For L of (2-52)

A

= [

(T1 :;; t') U (T2 ~ t') U (T3:;; t ' ) ] n (Ts > t" + T) ,t".::t',(3-9) so that

Pr{A}

=

^{Pr{ (T}₁ ^{~ t 1 )} n (TS > ^{t" + T) }} + Pr { (T 2 ^~ t ^{1 )} n (TS > ^{t" + T) }} + Pr { (T 3 ^~ t 1)

n

(TS > t" + ^{T) }}

- Pr { (T 1 ^~ t 1)

n

(T2 ^~ t ^{1 )}

n

(TS > ^{t" + T) }} - Pr { (T 1 ^~ t 1)

n

(T3 ^~ t 1)

n

(TS > t" + T) }

Pr { (T2 ^~ t 1) n (T3 ^~ t 1)

n

(TS > ^{t" + T) }}

+ Pr { (T 1 ^~ t 1 )

n

(T2 ^~ t ^{1 )}

n

(T3 ^~ t 1)

n

(TS > ^{t" + T ) } •} (3-10) Since T

5 is always the Ti of the second ci to fail, only the first 3 probability of (3-10) do not vanish. (The other 4 are probabilities of impossible events.) Hence, since at least 2 components must be good for the system tobe goodJ for t ¹ ~t" :

Pr{A}

=

Pr{(T1 ~ t') n (T2 > t" + T) n (T3 > t" + -r)}

+ Pr { ( T

2 ^~ t ' ) n ( T

1 > t" + -r) n ( T

3 > t" + -r) }

+ Pr{(T3 ;';; t') n (T1 > t" + T) n (T2 > t" + -r)}. (3-11)

33

Check : For s-independent T 1,T

2,T

3, frorn (3-11) Pr{A}

=

F

1(t')I\(t"+T)I\(t"+T)

and

a

at' _{+ f}

- -

2 (t)F

1 (t+T)F

3 (t+T)

- -

+ F3 (t)F

1 (t+T)F

2 (t+T),

which leads to (2.5-9) by (3-1) and (A1-4).

( 3-1 2)

(3-13)

APPENDIX

A.1 Checks for diverse F 's tobe Cdf's

- - - r - - - -

In this appendix several of the Fr(r) 's of this report are checked if they are O or between O and 1 at r=+O and become 1 for r- 00 . In (2-11)

00

Fr(00)

=

J [f,(t)F2(t) +f2(t)F,(t)]dt.

0

Since, by the product rule of differentiation and by (A2-5)

,..,1e find

00

Fr( 00 ) = - [F,(t)F2(t)] = 1.

0

For r- 00 (2-16) equals (A1-1) with (A1-3) as a consequence.

(A 1 -1 )

(A1-2)

(A1-3)

In (2-26), (2-33), (2-53) and other formulas, similar to the steps from (A1-1) to (A1-3):

00

Fr(^{00 )}

=

f [f 1 (t)F 2 (t)i\(t) + f 2 (t)F 1 (t)F\(t) + f 3 (t)F 1 (t)F2 (t) Jdt

0

00

=

! {- ~t [F 1 (tli\(t)F 3 (tl J}dt

=

^1.

0

In (2-33)

00

Fr(+O) = f f3 (t)F

1 (t)F

2 (t)dt,

0

so that for equal components (with pdf and Cdf without index)

(A1-4)

(A 1-5)

35

00

F (+O) = J f(t) [F(t) J²dt = 1/3

r 0

t

see (A2-6) ] .

Frorn (2-45a), since all F. (⁰⁰)=1,

1

0

Obviously,

00

J f

1 (t+t')dt

0

0

= 1- F,(t) = F1(t).

Hence, by (A1-1,2,3), F (⁰⁰)=1.

r

0

In the exponential case (2-45a) becornes

Now,

00

A1 ^oo

=

^{[1- exp(-A}₃,)] - - - + A1A

2 J exp[-(~

1 + A 2)t]

A1 + A

2 o

00

• J exp(-A

1t') {1- exp[-A

3(t'+,) ]dt'dt.

0

J exp(~A

1t ¹ )exp[-A

3(t'+,) ]dt' =

0

Hence,

Fr(,) A1 A1 exp(-A 3 ,) + A2

=

A1 + A2 A1 + A2 A1 + A2

A1A2 exp(-A 3,) A. 1 + A.2 A1 + A3

(A 1 -6)

(A 1-7)

(A 1 - 8)

(A1-9)

36

= 1-

For 1~⁰⁰ (2-70) becornes

Fr(oo) = J [ f

1(t)F

2(t) + f

2(t)FA(t)]F

3(t)dt

0

Now, by partial integration, using (A2-5):

J [ f

1 (t)F

2 (t) + f

2 (t)F

1 (t) ]F

3 (t)dt

=

^[F₁ ^(t)F₂ (t)F\ (t)]

0 0

+ J F

2(t)F

3 ( t ) f

3 (t)dt

=

J F

1 (t)F

2( t ) f

3(t)dt.

0 0

Hence,

(A1-10)

(A1-11)

(A1-12)

37

A.2 Sorne auxiliary integrals

For non-negative randorn variables, such as Ti or Tr,with pdf fand Cdf F

00

Jf ( -r) d -r = 1

0

and

F ( - 0)

=

0, 0 ~ F ( +o) ~ 1 , F ( oo)

=

1 •

For any real function g with O~g(1)~1 (A2-1) obviously yields

00

J f(-r) g(-r)d-r ~ 1.

0

Si n ce F . ( + 0) =O ,

l

00

J f.(t)[F.(t){-¹at

l l

0

By definition

Hence,

- f . ( t ) .

l

1 ⁰⁰ d k 1

= k

J dt [Fi(t)] dt

= k.

0

1 ⁰⁰ d k

k f dt {-[F. (t)] }dt

=

0 l

k .

1

As is well known,for a>O

00

J ^{exp ( -a t) d t = 1 / a •}

0

(A2-1)

(A2-2)

(A2-3)

(A2-4)

(A2-5)

(A2-6)

(A2-7)

:: ~ . · - / '