A fast algorithm for a redundancy optimization problem

Microelectron. Reliab., Vol. 25, No. 3, pp. 511 523, 1985. Printed in Great Britain.

0026-2714/8553.00+ .00 © 1985 Pergamon Press Ltd.

A FAST ALGORITHM FOR A REDUNDANCY OPTIMIZATION PROBLEM V. DAKSHINA MURTY and K.B. MISRA Reliability

Engineering Centre,

Indian I n s t i t u t e of Technology, Kharagpur, 721302 (W.B.),

India

(Received f o r p u b l i c a t i o n 30 January 1985) ABSTRACT: This

paper

optimal

presents

a simple

heuristic

redundancy a l l o c a t i o n .

technique

There is

number of steps because of the c r i t e r i o n

a drastic

to

arrive

at

the

r e d u c t i o n in

the

evolved, which allows f o r the

a d d i t i o n of one or more u n i t s , s i m u l t a n e o u s l y , at more than one stage. Examples have been provided to

demonstrate the

e f f e c t i v e n e s s of

the

technique. 1.0

NOTATION: n

=

number of stages/subsystems in the system

m

=

number of c o n s t r a i n t s

rj

=

reliability

=

(

i

-

qj

of a s i n g l e component at stage j

)

xj

=

number of u n i t s at stage j

X, X

=

a l l o c a t i o n [ X l , x 2 , . . . , xn ]

, and the optimal

allocation, respectively Rj

reliability

--

bi

=

( I - Qj)

=

total

of subsystem j

resource

available

corresponding

to

the

constraint gi

=

total

fi

=

slackness f a c t o r corresponding to the i th c o n s t r a i n t

=

amount consumed by the j t h

gij(xj)

resource consumed from the i th c o n s t r a i n t

stage from the i th

constraint AGij(x j )

=

gij(xj

+ i

)

gij(xj)

* Sponsored by the dawaharlal Nehru Technological U n i v e r s i t y , ( A . P . ) , Hyderabad under the Q u a l i t y Improvement Programme.

51l

i th

512

V. DAKSHINAMURTYand K. B. MISRA

!

dlj

-

desirability

factor of the jth

stage based on the

I th constraint II

dlj

=

r e l a t i v e f i g u r e of merit of the j t h stage based on i th constraint

Dij

=

s e l e c t i v i t y factor of the j t h

stage based on the

i th constraint =ij

=

u n i t r e l i a b i l i t y - c o s t r a t i o of the j t h stage based on the i th constraint

6ij

=

a constant dependent upon

xj

and a i j

gij

=

a constant dependent upon xj and AGij(x j )

Cst

=

t o t a l resource consumed by the system, [X], the t th constraint being the t l g h t one

ACst

=

incremental resource consumption consequent on the addition of a u n i t

2.0

INTRODUCTION: Several

h e u r i s t i c approaches have been reported

in

the

l i t e r a t u r e and a majority of them have been surveyed in [ 2-3 ] . Every heuristic

approach makes use

of

one

criterion

or

the

other

for

judicious choice of a stage, to which an addition of a unit causes the highest

improvement in

the

c o s t - b e n e f l t considerations.

system r e l i a b i l i t y ,

v i e w e d f r o m the

While in some approaches, the i n i t i a l

a l l o c a t i o n i s one u n i t at each stage, in some o t h e r s , ( f o r example,[1]) a better s t a r t i n g a l l o c a t i o n is a r r i v e d at based on some other considerations.

However, a f t e r having obtained an i n i t i a l

allocation, all

these methods share one thing in common,viz., to continue to add a single u n i t at the most desirable stage, in so f a r as the resources permit and the constrained optimization problem is solved. cedure

is

undoubtedly

simple,

but

may

not

be

This pro-

very

efficient

computationally, e s p e c i a l l i y , when problems involve a large number of subsystems. Organized number required optimal) adding

of

efforts

steps

should,

(iterations)

and

f o r o b t a i n i n g the o p t i m a l result. several

therefore, hence (or,

the

s|multaneously

at

one

made to

reduce

computational

atleast

One of the ways in which t h i s units

be

a fairly

the

effort

good near-

can be e f f e c t e d i s by or

more

stages

at

any

Algorithm

iteration.

It

technique

p r o p o s e d accomplishes

513

is demonstrated through t h i s paper that the solution this

task

in

a

simple

and

s t r a i g h t - f o r w a r d manner. Further, constrants, therefore

in any design problem, though there may be very many

only the

c r i t e r i o n is

one of

them is

algorithm

shall

introduced in

generally the most active one and be

found to

be

useful.

A simple

the algorithm to pickout the most active

constriant. 3.0

THE PROBLEM: The problem that is intended

to be solved, can be stated

as

determine the vector,

X* = [ X l * , X 2 * , . . . X n * ] so as to

maximi ze n

Rs = ~ j=l

[ 1 - (1 - r~)Xj ]

(1)

subject to constraints of a general nature

of the form

n

gi = E g i j ( x j ) j=l

~< bi

(2) for i = 1 , 2 , . . . , m

4.0

THE PROPOSED SOLUTION TECHNIQUE:

4.1

A l l Linear-constraints Problem To

consider an a l l as

highlight

the

salient

features

of

the

method, we

l i n e a r - c o n s t r a i n t problem, with the constraints given n

~ glj(Xj) gl =j=l

4

bl

4

b2

4

A

(3)

n

g2

j~ig2j(xj )

or, e q u i v a l e n t l y , n

As= ~ aj

xj

(3 a) n

Ls = E l j j=1

xj

~< L

where AS and L s represent the t o t a l consumption of the resources by the system, f r o m out of the maximum a v a i l a b l e resources of A and L, r e s p e c t i v e l y ; while

aj

and l j

represent the

consumption of

the

resources by a single u n i t at stage j . We begin to solve the problem by f i r s t

assuming that the t th

constraint is the t i g h t one and accordingly defining the d e s i r a b i l i t y

514

V. DAKSHINAMURTY and K. B. MISRA

f a c t o r s [ 1 ] as !

dtj

(4)

= (ARs/Rs)/(ACst/Cst)

f o r j = 1,2 . . . . ,n = (ARj/Rj)/[AGtj(xj)/Cst) where aGtj(x j )

= In

factors

are

merely to

not

gtj(xj

+1 ) - g t j ( x j

as much as the required

determine the

at

(5)

)

absolute values of

any

stage

of

the

stage at

which

a unit

the d e s i r a b i l i t y

solution is

to

other

than

be added, and

f u r t h e r , since Cst remains the same, we may replace the d e s i r a b i l i t y f a c t o r s by the r e l a t i v e f i g u r e s of m e r i t , defined as dtj'' In (6), AGtj(x j )

= (rj

Qj/Rj)/AGtj(x j)

(6)

i s a constant and is equal to the amount of resource

required to add a single u n i t (or,

in otherwords, u n i t - c o s t ) at stage j ,

so t h a t dtj''

= [ rj/aGtj(xj)]

or, simply,

Dtj

=

where

Dtj

=

:tj

Qj /Rj

Qj / Rj

(7)

s e l e c t i v i t y f a c t o r of the j t h

stage based on

the t th c o n s t r a i n t mtj

=

unit r e l i a b i l i t y - c o s t rj/aj,

rj/lj,

r a t i o ( f o r example,

etc.)

The usual procedure would have been to c a l c u l a t e Dtj f o r the various stages and to add an a d d i t i o n a l u n i t to the stage t '

such

that Dtt' if

= max ( D t j ) , J

the current balance of the resources so permits and so on.

However,

our method deviates from t h i s point onwards. Let us a r b i t r a r i l y

choose a stage, say, stage k, as the

reference stage. Let the current a l l o c a t i o n at stage k be xk, with the s e l e c t i v i t y f a c t o r Dtk. k,[i.e.,

We can add an a d d i t i o n a l u n i t at stage

we can replace xk < - - - ( X k + l ) ] i f Dtk =

max J

and only i f ,

(Dtj)

(8)

However, in the general case, i t units

may have to

be added at

several

other

may not be so. stages

Several

( based on the

Algorithm

r e l a t i v e values of

the

515

s e l e c t i v i t y factors)

e l i g i b l e to have an additional u n i t .

before stage k becomes

Or, a l t e r n a t i v e l y , i f we intend

to add additional units in succession at stage k,

in view of (8), we

have to modify the s e l e c t i v i t y factors of the other stages such that Dtk

~

Dtj for j=1,2 . . . . ,n j~k

i.e.,

atk Qk/Rk Further

a t j Qj/Rj

(9)

s i m p l i f i c a t i o n of

xj where

~

(9)

yields

the

~ log( 1+ a t j . B t k ) - l / l o g (qj)

Btk

=

relation

(10)

Rk/atk Qk , and xj is the smallest

integer which s a t i s f i e s (10) and consequently, represents the current allocation continued

at

stage

till

j.

Now, xk

such time

<---

either

(Xk+ I )

a constraint

and the is

process is

exactly

satisfied

(which may happen only in a r e l a t i v e l y few cases) or i s v i o l a t e d . In the former case, we obtain the optimal redundancy a l l o c a t i o n . latter

case,

however, we reset xk <---

a l l o c a t i o n is

taken as the i n i t i a l

(Xk-1)

In the

and the corresponding

a l l o c a t i o n , and the step by step

procedure of adding one unit at a time, based on the current highest value of the s e l e c t i v i t y f a c t o r ,

is resorted to t i l l

optimization problem i s

The following example indicates the

solved.

the constrained

various steps. 4.1.1

Example: Let us consider the redundancy a l l o c a t i o n problem with the

following data Stage,j

1

2

0.8

0.85

aj

5

4

6

5

3

lj

1.2

2.3

3.4

4.5

5.6

Comp. r e l , r j

3

4

0.90

0.65

5 0.75

Resource/unit

5

gl = T

j=1

g2

5 = E

j=l

aj xj

1j

xj

,,<

..<

60

75

516

V. DAKSHINAMURTY and K. B. MISRA

Step

1:

Let

us

arbitrarily

choose the f i r s t t h a t the f i r s t

(

and w i t h o u t any loss

stage as the reference stage ( i . e . ,

c o n s t r a i n t is the t i g h t one ( i . e . ,

the various ~ l j

of

k = 1) and assume

t = 1) and c a l c u l a t e

as

j

1

2

3

4

alj

0.1600

0.2125

0.1500

0.1300

Step 2:

generality)

Initially,

5 0.2500

l e t us set Xk=Xl=2, so t h a t Q1 = 0.04 and Bl1=150,

so t h a t x2 ~ log( 1 + ~ 1 2 . B 1 1 ) ' 1 / l o g (q2) = 1.8411, or

x2 = 2.

Likewise, we obtain x3 = 2, x4=3 and x5 = 3, so t h a t

the a l l o c a t i o n X = [ 2 , 2 , 2 , 3 , 3 ] . Step 3:

We now compute the resources consumed as

and since none of the c o n s t r a i n t s is

g1=54 and g2=44.1

v i o l a t e d , we also c a l c u l a t e the

slackness f a c t o r s , fl

= (

bl

-

gl

)/bl

= (

60 -

54 )/60

=

0.1000,

and

s i m i l a r l y f2 = 0.4120. Since f l the t i g h t one. the

tight

one.

begin w i t h , i t

is

the smaller of

the f i r s t

c o n s t r a i n t is

(Ofcourse, we had assumed t h a t the f i r s t If

c o n s t r a i n t is

any other one were to be assumed to be t i g h t

to

would have been replaced now.)

Step 4:Since none of i.e.,

the two,

the c o n s t r a i n t s is

v i o l a t e d , we set Xk<--(Xk+l)

x I < - - - 3 and the procedure is repeated to obtain the a l l o c a t i o n

X = [

3 , 3 , 3 , 5 , 4 ] , with

65.6 reset xI

Since one of <---

the c o n s t r a i n t s ( c o n s t r a i n t 1)

v i o l a t e d , we

2 and the corrresponding a l l o c a t i o n , v i z . ,

[2,2,2,3,3],

Since the f i r s t

s e l e c t i v i t y factors, Dlj j D l j ( X 10-3 )

= 82 and g2 =

is

is taken as the i n i t i a l Step 5:

the corresponding values of gl

allocation. c o n s t r a i n t is the t i g h t one, we c a l c u l a t e the , as

1 6.6667

2 4.8913

3

4

5

1.5152

5.8234

3.9683

Since D11 is the highest among the s e l e c t i v i t y f a c t o r s , and since the balance of resources, v i z . ,

( 60 - 54 ) = 6 allows f o r the a d d i t i o n of

a u n i t at stage 1, we add an a d d i t i o n a l u n i t at stage 1 to o b t a i n the a l l o c a t i o n X = [ 3 , 2 , 2 , 3 , 3 , ] , with the corresponding values of gl = 59 and g2 = 4 5 . 3

Thus X* = [

3,2,2,3,3,],

since

no u n i t

can be a d d e d

Algorithm

517

at any stage without v i o l a t i n g the f i r s t Instead of

the f i r s t

stage,

chosen as the reference stage,

constraint.

if

(i.e.,

the fourth

stage were to be

k = 4 ), wlth x4 = 2, we obtain

the a l l o c a t i o n X = [ 2 , 2 , 1 , 2 , 2 ]

and on s e t t i n g x4 = 3, we obtain the

a l l o c a t i o n as X = [ 3,2,2,3,3 ],

which also happens to be the optimal

a l l o c a t i o n , X* 4.2

All Non-linear Constraints Problem: i

Again, we

consider

an

to h i g h l i g h t the s a l i e n t features of the method, all

non-linear

constraints

problem,

with

the

constraints given as n

gi =

where a l l

z

j=l

gij

(xj)

~< bi

(II)

the constraints are non-linear. The

desirability

factors

may o n c e again

be

defined,

assuming the t th c o n s t r i a n t to be the t i g h t one, as I

dtj

=( a Rs/Rs)/( a Cst/Cst ) =( A R j / R j ) / [ A Gtj ( x j ) / C s t ) ] = ( r j Q j ) / [ a Gtj ( x j ) / C s t )

where A Gtj ( x j )

= gtj

(xj + 1 ) - g t j

(12)

(xj)

As in the case of the l i n e a r - c o n s t r a i n t s problem, we can define the s e l e c t i v i t y factors, as

Dtj

:

( r j Q j ) / [ A Gtj ( x j ) Rj]

(13)

f o r j = 1,2 . . . . . n By section

any

4.1,

adopting

we may s t a t e

(arbitrarily)

a

reasoning

that

if

similar

an a d d i t i o n a l

c h o s e n stage,

say,

to

the

unit

is

stage

k,

one

given

t o be added a t

the

following

r e l a t i o n s h i p should hold, v i z . , Dtk

)

Dtj for

j

- 1,2 . . . . .

J + k

t.e.,

rk

Qk/[ AGtk(Xk)R ~ ~ rj

qj/[ AGtj(x J) Rj]

,

tn

n

518

V. DAKSHINAMURTY and K. B. MISRA

which leads to the r e l a t i o n s h i p

xj where

log [ 1 +

rj/(aGtj(xj).~tk)]'l/log(qj) (14) and xj i s the smallest

Qtk = rk Qk/ [ aGtk (Xk) Rk ]

i n t e g e r s a t i s f y i n g (14) and hence represents the current a l l o c a t i o n at stage j . linear

The r e s t of the procedure is constraints

problem.

the same as in the case of the

The

following

example

illustrates

the

the

redundancy a l l o c a t i o n problem with

various steps. 4.2.1

Example: Let

us consider

the f o l l o w i n g data [ 3 ] Stage,j

I

2

3

4

5

Comp.rel, r j

0.80

0.85

0.90

0.65

0.75

Constant, pj

1

2

3

4

2

Constant, cj

7

7

5

9

4

Constant, wj

7

8

8

6

9

n

5

gl : j=1~g l j

(xj) =

g2 = T g2j (x~)~ j =1

=

J~'=PJ xj 2 ~<

5 T c j ( x j + e X j / 4 ) ~< 175 j =1

n

5

g3 =j=Tlg3j ( x j )

Step 1: the

Let us a r b i t r a r i l y

first

stage

(arbitrarily) (i.e., Step 2:

100

as the

= j=1~wj xj eXj/4

(and without

reference

assume t h a t the f i r s t

~< 200

loss of

stage,

(i.e.,

generality) select k = I)

and l e t

us

c o n s t r a i n t is the t i g h t c o n s t r a i n t ,

t = 17. Initially,

let

AGl1(2)= g11(3) - g11(2)

us set xk = x I = 5 and

= 2,

so t h a t Q1 = 0.04 and

QI1 = 0.00667.

Now, l e t us assume t h a t x2 = 1, so t h a t AG12(1) = g12(2)- g12 (1)=6 and the R.H.S. x2=

2

and

of

(14) i s equal to 1.6353.

calculate

the

R.H.S.

of

(14)

Since 1 ~ 1.6353, as

1.3816.

smallest integer s a t i s f y i n g (14), we assign x2 = 2.

As 2

we set is

the

S i m i l a r l y , x3 ,

x4, and x5 are found to be 2 each, so t h a t the a l l o c a t i o n becomes X = [ 2,2,2,2,2]. Step 3:

Now, we compute the consumption of the resources as gl = 48,

g2 = 116.759 and g3 = 125.303.

Algorithm

Since

none of

the

constraints

slackness f a c t o r s as f l Step 4: i.e.,

is

519

violated,

we f u r t h e r c a l c u a l t e the

= 0.5200, f2 = 0.3328, and f3 = 0.3735.

Since f2 i s the l e a s t among the various s e l e c t i v i t y

since i t

we replace t

signifies

that

the second c o n s t r a n t i s

< - - - 2 and set x I

<---

an a l l o c a t i o n of [ 3 , 3 , 3 , 5 , 4 ] . resources

is

constraints

gl

= 186,

are

the t i g h t

one,

3 and repeat the procedure to get

The corresponding consumption of the

g2 = 200.509

violated,

factors,

we reset

and g3 = 348.641.

xI

o b t a i n the a l l o c a t i o n as [ 2 , 2 , 2 , 3 , 3

<---

],

2 and go to

Since

the

step 2

and

which is taken as the i n i t i a l

allocation. Step 5:

Now, we c a l c u l a t e the s e l e c t i v i t y j

1

D2j(x 10-3 )

2

3.2432

f a c t o r s , D2j , as

3

1.9036

4

1.2383

5

2.0204

1.8586

Since D21 i s the h i g h e s t among the s e l e c t i v i t y

factors,

addition

by

of

resources

a unit

at

available,

stage we

1

add

is

permitted

one

unit

c o n f i g u r a t i o n is X = [ 3,2,2,3,3 ] , 192.481.

Also,

since

the

at

the

stage

and since the balance

1,

so

of

the

that

the

w i t h gl = 83, g2 = 146.125 and g3=

addition

of

a unit

at

any

stage

is

not

p o s s i b l e , the c u r r e n t a l l o c a t i o n i s the optimal redundancy a l l o c a t i o n , viz.,

X*

= [

3,2,2,3,3

].

The various

results

are

summarized and

given in Table I . Table 1 .

xk

2 3

*

MR 25:3-H

.

.

.

.

.

t

.

.

.

.

.

.

.

.

Computational Results w i t h .

.

.

.

Allocation,X

1 [2,2,2,2,2] (assumed) 2

[3,3,3,5,4]

2

[2,2,2,3,3]

2

[3,2,2,3,3]*

.

.

.

.

.

.

.

.

.

.

fl

0.5200

Optimal redundancy a l l o c a t i o n

.

.

.

.

.

.

.

f2

0.3328

.

.

.

k .

= .

.

1 .

.

f3

0.3735

.

.

.

.

.

.

.

.

.

.

.

.

. o

..

..

Remarks

t <--- 2 Constraint violation. Xk < - - - 2 Initial allocation Selectivity f a c t o r consideration

.. . .

.

.

.

.

.

520

V, DAKSHIN'AMURTY and K. B. MISRA

4.3

Mixed ( l i n e a r and n o n - l i n e a r ) C o n s t r a i n t s Problem: We now present a u n i f i e d a l g o r i t h m to t a c k l e the problems

with

both

linear

and n o n - l i n e a r c o n s t r a i n t s .

c o n s i s t s of s e l e c t i n g a r b i t r a r i l y one of

the

followed

constraints

to

obtain

assumed t i g h t

to

the

be t i g h t .

is

linear

resources consumed are computed. violated,

slackness f a c t o r s

brief,

the

method

a reference stage k and assuming any One or

c o n f i g u r a t i o n X,

constraint

In

or

the

other

procedure

is

depending upon whether the non-linear,

and the

various

In case none of the c o n s t r a i n t s is

are c a l c u l a t e d and are used to a s c e r t a i n

which one among the various c o n s t r a i n t s is the most a c t i v e one and the rest

of

the

calculations

are

based

on

this

constraint

and

constrained o p t i m i z a t i o n problem is solved. 4.3.1

The A l g o r i t h m (i)

Determine the sets M1 and M2 , where MI= [ i / i th c o n s t r a i n t is a l i n e a r one] M2= [ i / i th c o n s t r a i n t is a n o n - l i n e a r one]

(2)

Choose ( a r b i t r a r i l y )

a reference stage, say, stage k.

Set L < - - ° 0 ; M < - - - 2 ; t < - - - 1 and ITER < - - - 0 (3)

Set xk < - - - M. If

(4)

Let the t th be the t i g h t c o n s t r a i n t .

t E M1, go to step 4; else, go to step 5.

(i)

If

ITER = 1, go to ( i i ) ; :ij

: rj/AGij

(I),

else, c a l c u l a t e for a l l

i E M1

Set ITER < - - - 1. (ii)

Calculate

Btk :

( 1 - Qk)/Qk mtk,

where Qk = (qk)xk and obtain the X = [ xj

vector

I xj ~ l o g ( l + ~ t j B t k ) ' I / l o g ( q j ) ;

xj i s an i n t e g e r ] Go to step 6 (5)

C a l c u l a t e ~tk = r k Qk/[AGtk(Xk)(1 - Qk )] where

Qk = (qk)xk and

aGtk (Xk) = gtk (Xk + 1) - gtk ( X k ) Determine the v e c t o r , X = ~xj

I xj ~ log [ 1+

log ( q j )

; xj

rj/(aGtj(xj)Qtk) ]/

i s an i n t e g e r }

the

Algorithm

521

n

(6)

C a l c u l a t e the resources consumed, gi =j =El g i j ( x j ) One o f the f o l l o w i n g cases a r i s e s : (i)

gi If

< bi

, for all

L

1,

=

go

i

= 1,2 . . . . ,m

to

step

7;

else,

calcualte

the

slackness f a c t o r s , fi

= ( bi - gi

)/

bi

, for i

= 1,2,...,m

J

and determine [ t

I ft'

= min J

(fj)

!

Set t (ii)

<--- t

, M <---

gv = bv and gi

~

(M + 1 ) and go t o step 3.

bi

, for

i = 1 , 2 . . . . ,m i @v

Go t o step 8 (iii)

gi > bi

, f o r any one o f i

Set L < - - (7)

Calculate Dtj with

the

1, M < - - -

(M - 1) and go to step 3

for j=l,2,...,n

current

highest

eliminate it

If

all

and add a u n i t to the stage

value,

resources c o r r e s p o n d i n g to or,

= 1,2 . . . . ,m

if

the balanace o f

the

every c o n s t r a i n t so p e r m i t s ;

from f u r t h e r c o n s i d e r a t i o n , o t h e r w i s e .

the

stages

are

eliminated

from

further

c o n s i d e r a t i o n , go t o step 8. (8)

The c u r r e n t a l l o c a t i o n is X*

(9)

4.3.2

=

[ X l * , x2* , . . . ,x n*

]

C a l c u l a t e the system r e l i a b i l i t y , n Rs = ~ [ 1 - ( 1 - r j ) X j j=l

]

Example: To demonstrate the

let

the o p t i m a l a l l o c a t i o n ,

us c o n s i d e r the

as in 4 . 2 . 1

a p p l i c a t i o n of

the above a l g o r i t h m ,

redundancy a l l o c a t i o n problem w i t h

the data same

t o g e t h e r w i t h the a d d i t i o n a l c o n s t r a i n t s given as n 5 g4 : E (x j ) : ~ aj xj ~ 60 j = l g4j j=l n 5 = E gsj ( x j ) = ~ l xj ~ 75 g5 j =1 j =1 J '

and the v a r i o u s values of aj The step w i t h the f i r s t

and l j

by step

being the same as given in 4 . 1 . 1 .

implementation of

c o n s t r a i n t assumed to be t i g h t

the

above a l g o r i t h m

and w i t h the f i r s t

stage

522

V. DAKSHINAMURTY and K. B. MISRA

as the reference stage is given below.

Step

No.

Result

of

implementation.

(1)

MI:

(2)

k = i;

(3)

xI

(5)

~iI = 0.00667,

(6)

gl

= 48,

92

94

= 46,

and

gi

< bi

(i)

[

4,5

(4)

(i)

,

= 0.2333,

t'

<---

4

= 3

ITER

;

I

X = [

95

= 0

L=o

go t o

step

93 = 1 2 5 . 3 0 3

= 34 all

f2

i

= 1,2,...,5;

= 0.3328,

and

f5

t

<---

= 4

;

f3

L~I

= 0.3375

= 0.5467 4

, M <---

3

t

E MI

; Go t o

step

3

4

5

0.2125

0.1500

0.1300

0.2500

:5j0.6667

0.3696

0.2647

0.1444

0.1339

B41

<---

I

= 775;

X = [

step

and

Some g i

> bi"

step

(ii)

B41

:

91

:

step

<

L :

I.

L=I;

M <---2

t ~M I

Go t o

Go t o

(ii)

X = [

2,2,2,3,3

and

bi

step

]

6 92 = 1 3 5 . 8 4 7 ,

94 = 5 4 , gi

g5 = 6 5 . 6

= 4;

i.

78,

93:348.641

3

= 150,

Go t o

;

6

g4 = 8 2 ,

ITER

3,3,3,5,4]

g2 = 2 0 0 . 5 0 9 ,

t

5

2,2,2,2,2]

=4j0.1600

(i)

(i)

t EM2;

2

×1 = 2 ;

(6)

ITER

@ i

Go t o

(3)

= i;

]

3

t

gl = 186,

(iii)

1,2,3

= 116.759,

,

step

[

= i;

for

f4

Go t o

(4)

t

t

0.5200,

ITER

(6)

2.

:

j

(ii)

2;

fl

xI

; M2 :

M :

<---

Go t o

(3)

]

,

g5 : for

Go t o

= 171.106

44.1

all step

g3

i 7

:

1,2,...,5

4

4

Algorithm

(7)

j

1

523

2

3

4

5

D4j (xlO -3 )

4.8913 1.515~ 5.8234 3.9683

6.6667

D41 is the highest among a l l the s e l e c t i v i t y factors.

So, xI <--- 3.

A l l other stages are eliminated from f u r t h e r consideration. (8)

Go to step 8.

The current a l l o c a t i o n is the optimal redundancy

allocation,

X* = [ 3,2,2,3,3 ] . (9)

5.0

System r e l i a b i l i t y ,

Rs = 0.90447

CONCLUSION: A

redundancy

simple

heuristic

p r o b l e m s with

constraints,

is

presented.

algorithm,

multiple There is

capable

linear

and/or

tackling non-linear

a considerable saving in

computational e f f o r t and the success rate is high. is

of

W h i l e the choice

a r b i t r a r y , in the case of mixed constraints problems, i t

is found

advantageous to assume that one among the l i n e a r constraints is t i g h t one to begin with. reliability 6.0

The algorithm w i l l

the

the

be of immense use to the

designer, e s p e c i a l l y , in the e a r l i e r stages of design.

ACKNOWLEDGEMENTS: The f i r s t

Jawaharlal Indian

Nehru Technological

Institute

opportunity

author is

of

and the

grateful

U n i v e r s i t y , ( A . P . ) , Hyderabad and the

Technology, necessary

to the a u t h o r i t i e s of the

Kharagpur,

facilities

to

for

providing

carry

out

the

him

the

present

work. 7.0

REFERENCES: 1.

K.B. Misra, A simple approach for constrained redundancy optimization

problem,

IEEE Trans.

Reliab.

R - 2 1 , 30-34

(1972) 2.

W.Kuo, C.L. Hwang and F.A. Tillman, A note on h e u r i s t i c methods in optimal system r e l i a b i l i t y ,

IEEE Trans. Reliab.

R-27, 320-324 (1978) 3.

F.A.

Tillman,

C.L.

Hwang and W. Kuo, Optimization

Systems R e l i a b i l i t y , Marcel Dekker (1980)

of