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Redundancy Allocation usingComponent-ImportanceMeasures for MaximizingSystem ReliabilityDebasis Bhattacharyaa & Soma Roychowdhuryb

a Visva-Bharati University, Santiniketan, 731235,Indiab Indian Institute of Social Welfare and BusinessManagement, Calcutta, 700073, IndiaPublished online: 11 Mar 2014.

To cite this article: Debasis Bhattacharya & Soma Roychowdhury (2014) RedundancyAllocation using Component-Importance Measures for Maximizing System Reliability,American Journal of Mathematical and Management Sciences, 33:1, 36-54, DOI:10.1080/01966324.2013.877363

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American Journal of Mathematical and Management Sciences, 33:36–54, 2014Copyright C© Taylor & Francis Group, LLCISSN: 0196-6324 print / 2325-8454 onlineDOI: 10.1080/01966324.2013.877363

REDUNDANCY ALLOCATION USINGCOMPONENT-IMPORTANCE MEASURES FOR MAXIMIZING

SYSTEM RELIABILITY

DEBASIS BHATTACHARYA1 and SOMA ROYCHOWDHURY2

1Visva-Bharati University, Santiniketan, 731235, India2Indian Institute of Social Welfare and Business Management, Calcutta,

700073, India

SYNOPTIC ABSTRACT

Redundancy allocation, being an important and effective way to improve sys-tem reliability, has been discussed by many authors. The main idea, which hasbeen advocated in this article, lies in the fact that some components have moresignificance in the functioning of the system than the others; because of this,it is expected that if allocation of the redundant component is made accordingto some component-importance measure, optimality can be achieved easily. Thenovelty of this study is that the problem of redundancy allocation has been solvedby allocating redundant components according to some component-importancemeasures that are not very difficult to obtain and comprehend for an engineeredsystem, even when there is no information about the reliability of the compo-nents. The component structural-importance measure and reliability-importancemeasure have been used in this work for the purpose of allocating redundancy.The results derived can be used to maximize system reliability using redundancyallocation for a general n-component coherent system. Applications of the resultshave been illustrated with examples of coherent systems.

Key Words and Phrases: active redundancy, component-importancemeasure, optimization, structure function, system reliability.

1. Introduction

Improving system reliability is an essential area of concern to relia-bility practitioners. Reliability enhancement or improvement canbe achieved in various ways; viz., by adopting a “preventive main-tenance policy,” which prevents premature failure or degradation

Address correspondence to Soma Roychowdhury, Indian Institute of Social Welfareand Business Management, Calcutta, 700073, India. E-mail: srcdb@yahoo.com

This work was done when the authors were visiting the Department of Statistics,University of California, Davis, CA, USA, as visiting professors.

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Redundancy Allocation Using Component-Importance Measures 37

of components; by using “quality control technique,” which en-sures that the product used to develop the system is of approvedquality or standards and the system operates according to the ap-proved procedures; by introducing “monitoring, surveillance test-ing, and inspection” so that detectable causes cannot accumulatewithout attention and thus cause a major failure of the system; bybringing “equipment diversity,” which ensures that componentsof different manufacturers are used interchangeably while the sys-tem is developed; and so on. Another important and effective wayto achieve higher reliability is to add some additional function-ally identical components to the system in order to increase thelikelihood that a sufficient number of components will survive af-ter absorbing the shock from a failure and that the system willstill continue to function. These components are known as re-dundant components. Generally, there are two commonly usedtypes of redundancies—standby redundancy and active or paral-lel redundancy. In standby redundancy, a redundant componentis attached in such a way that it starts functioning immediately af-ter the failure of the component to which it is attached. Activeredundancy is used when it is difficult or not possible to replacethe failed components during operation of the system. The ac-tive redundant components are connected in parallel with thecomponents of the system, and they function simultaneously withthe original components. In this article, the active, or parallel, re-dundancy at component level is considered in order to increasethe system reliability. It has been noted in Barlow and Proschan(1981) that in the case of active redundancy, component-wise re-dundancy works much better than system-wise redundancy. At thispoint, it is worthwhile to mention two points: (1) this tactic to im-prove system reliability loses its positive impact sometimes becauseof failure of the system due to a common cause, and (2) increas-ing system redundancy generally involves extra cost.

System reliability is improved whenever a redundant com-ponent is added to a component of the coherent system, butthe amount of increase depends on the choice of componentto which redundancy is allocated. This phenomenon is veryexpected, because some components may play more importantroles in the functioning of the system than the others. We needto allocate the redundant components in such a way that the in-crease in system reliability is maximized. We expect that the

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38 D. Bhattacharya and S. Roychowdhury

allocation of redundant components according to somecomponent-importance measure may lead us to achieve theoptimality. There are various component-importance measures ofwhich structural-importance measures and reliability-importancemeasures are very popular. In this study, the reliability of any givencoherent system has been optimized by allocating redundancyaccording to those component-importance measures.

Redundancy allocation under parallel redundancy used atthe component level for a coherent system leads to an increase insystem lifetime (Barlow & Proschan, 1981). Boland, El-Neweihi,and Proschan (1988) discussed an optimal allocation strategy ina k-out-of-n system, which allocates the redundant componentsequally among n components using a stochastic ordering crite-rion. Shaked and Shanthikumar (1992) considered the lifetimesof the original components and redundant components to be in-dependent and identically distributed (iid) and the lifetimes ofthe original components and the residual components to be in-dependently distributed for optimally allocating m active redun-dancies to an n-component series system. Singh and Singh (1997)preferred the strategy of balanced allocation to the strategy of al-locating larger number of components to stronger components.The problem of allocating m active redundancies to a k-out-of-nsystem has been considered in recent works of Li and Hu (2008)and Hu and Wang (2009), wherein the lifetimes of all workingcomponents and active redundancies are considered to be iid. Allof the above works considered a specific system design, not a gen-eral coherent system.

The present work finds a solution toward achieving max-imum reliability improvement of any coherent system whileallocating the redundant component, using an appropriatecomponent-importance measure. The novelty of this study is thatit solves the redundancy allocation problem for any system in gen-eral. No particular form of system design has been assumed tosolve the problem. Moreover, the proposed method is capable ofsolving the problem even when there is no information about thereliability of the components. It allocates redundant componentsaccording to some component-importance measures that are notvery difficult to obtain or comprehend for any coherent system.Examples of some important engineered systems have been dis-cussed to show how the rule works. The situation in which we are

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Redundancy Allocation Using Component-Importance Measures 39

allowed to add only one redundant component at a time is con-sidered. The problem is to select the best system component towhich the redundant component is to be added in order to maxi-mize the system reliability. The same method can be used repeat-edly for adding more than one redundant component.

This article is organized as follows: the present section intro-duces the work and discusses earlier works in this area of research.Section 2 presents the notations and definitions. The main re-sults are discussed in Section 3, and Section 4 shows examplesof how the results derived in Section 3 can be used in making adecision.

2. Notation and Definitions

Let us consider a binary coherent system (I , ϕ), composed ofn components, whose lifetimes are independently distributedamong themselves and are independent of the lifetimes of theredundant components, where I is the index set {1, 2, . . . , n}of n components, and ϕ is the structure function of the system,which takes the value 0 if the system is in failing state, and 1 ifthe system is in functioning state. Basically, ϕ describes a mappingfrom the space {0, 1}n to {0,1}. In a coherent system, every com-ponent has some contribution toward the system performance,in other words, the system contains no component whose func-tioning or failure has absolutely no effect on the system life, irre-spective of the condition of the state in which the remaining com-ponents might be; and the performance of the system improvesby improving any component or subsystem (Barlow & Proschan,1981). Let the state (functioning or failing) of the ith componentbe denoted by a binary variable, xi(t), where xi(t) is 0 if the ithcomponent is in a failing state, and 1 if the ith component is ina functioning state at time t. Here, xi (t) ∼ binomial (1, pi(t)),where pi(t) = P(Yi >t), the reliability of the ith component attime t, where Yi is the life of the ith component of the system,i = 1, 2, . . . , n. Hence, E(xi(t)) = pi(t). Thus, the system reli-ability R(t) at time t, defined as the probability that the systemsurvives at least up to time t, can be written as

R(t) = P(T > t) = h(p(t)) = E [ϕ(x, t)], (1)

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40 D. Bhattacharya and S. Roychowdhury

where ϕ = ϕ(x, t) is the structure function of the system, x = x(t)= (x1(t), x2(t), . . . , xn(t)) is the state vector, and p(t) = (p1(t),p2(t), . . . , pn(t)) is the vector of component reliabilities. The statevector x is a vector of n-tuple consisting of zeros and ones and canassume any of 2n possible values represented by the vertices of theunit cube in n-dimensional space, viz., (0, 0, . . . , 0), (1, 0, . . . ,0),(0, 1, . . . , 0), . . . , (1, 1, 0, . . . , 0), . . . , (1, 1, . . . , 1). From nowon, we will suppress the time variable t in the expressions of xi(t),pi(t), x(t), p(t), ϕ(x, t) for the sake of notational convenience.

A very well-known concept of comparing the relative impor-tance of components in a coherent system with structure functionϕ is the structural importance of a component, denoted by Iϕ(i) forthe ith component and defined by

Iϕ(i) =∑

[x:xi =1]

[ϕ

(1i , x(n−1)×1

) − ϕ(0i , x(n−1)×1

)]2n−1

, (2)

where (1i , x(n−1)×1) and (0i , x(n−1)×1) denote the vectors (x1(t),x2(t), . . . , xi−1(t), 1, xi+1(t), . . . , xn(t)) and (x1(t), x2(t), . . . ,xi−1(t), 0, xi+1(t), . . . , xn(t)), respectively. For two components iand j, we say that the component i is structurally more importantthan the component j if Iϕ(i) > Iϕ( j). Note that the structural im-portance of a component is related to the structure of the systembut does not depend on the component reliabilities. In situationswhen the system design is known, that is, ϕ(x) is known, but noinformation about component reliabilities is available, structuralimportance is used to measure the relative importance of differ-ent components.

If both the structure function ϕ(x) and the vector of compo-nent reliabilities p are known, then reliability importance is usedto measure the relative importance of components. The reliabilityimportance of the ith component is defined as the rate of changein system reliability with respect to the change in the reliability, pi,of the ith component, and is denoted by RI (i).

From the expression of h(p) as given in (1), it is clear that

h(p) ≡ h(pn×1)

= P (ϕ(x) = 1)

= P(system works)

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Redundancy Allocation Using Component-Importance Measures 41

= P(system works and component i works)

+ P(system works and component i fails)

= p i h(1i , pn−1×1) + (1 − p i )h(0i , pn−1×1),

where

(1i , pn−1×1) = (p 1, p 2, . . . , p i−1, 1, p i+1, . . . , p n),

and

(0i , pn−1×1) = (p 1, p 2, . . . , p i−1, 0, p i+1, . . . , p n).

Now, finally,

RI (i) = ∂h(pn×1)∂p i

= h(1i , pn−1×1) − h(0i , pn−1×1). (3)

Note that in the case p i = 12 , i = 1, 2, . . . , n, the Birnbaum

reliability importance reduces to Birnbaum structural importance(Birnbaum, 1968). By notation, (i) �b ( j), we mean that compo-nent i is a better choice than component j for the addition of aredundant component. It is better in the sense that componenti gives more rise in system reliability than component j when aredundant component is added to it.

3. Main Results

Here we consider the following situations:

1. Only system design is known (i.e., the structure function ϕ isknown, but no information about component reliabilities isavailable). The reliabilities of the spares of all system compo-nents are the same.

2. The system design is known, and the reliabilities of all systemcomponents are the same. The reliabilities of the spares of allsystem components are the same.

3. The system design is known, and all component reliabilitiesmay not be the same. The reliabilities of the spares of all systemcomponents are the same.

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42 D. Bhattacharya and S. Roychowdhury

4. The system design is known, and all component reliabilitiesare not necessarily the same. The reliabilities of the spares ofall system components may be different.

Let us first consider a situation for which we have no knowl-edge about the component reliabilities. Suppose we let r be the re-liability of the redundant component to be added and h(i)(pn×1)be the reliability of the corresponding augmented system whenthe redundant component is added to the ith component of thesystem. Now we prove the following result, which helps us choosethe system component to which a redundant component is to beadded in order to maximize the gain in system reliability.

Result 1. In the absence of knowledge about the component reliabilities,the component i gives more rise in system reliability than component j; thatis, (i) �b ( j)when a redundant component is added to it if

Iϕ(i) > Iϕ( j).

Proof. By pivotal decomposition, we can write the system reliabil-ity h(pn×1) as

h(pn×1) = p i × h(0i , p(n−1)×1) + (1 − p i ) × h

(1i , p(n−1)×1)

= h(0i , p(n−1)×1) + p i × (

h(1i , p(n−1)×1))

− h(0i , p(n−1)×1)

= h(0i , p(n−1)×1) + p i × E

[ϕ

(1i , x(n−1)×1)

− ϕ(0i , x(n−1)×1)]

= h(0i , p(n−1)×1) + p i ×

∑x

[ϕ

(1i , x(n−1)×1)

− ϕ(0i , x(n−1)×1) ]× P(x : ϕ(1i , x(n−1)×1)

− ϕ(0i , x(n−1)×1) = 1).

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Redundancy Allocation Using Component-Importance Measures 43

Note that ϕ(1i , x(n−1)×1) − ϕ(0i , x(n−1)×1) = 1 only ifϕ(1i , x(n−1)×1) = 1 and ϕ(0i , x(n−1)×1) = 0, that is,ϕ(1i , x(n−1)×1) > ϕ(0i , x(n−1)×1).Hence,

∑x

[ϕ(1i , x(n−1)×1) − ϕ

(0i , x(n−1)×1)] × P(x : ϕ

(1i , x(n−1)×1)

− ϕ(0i , x(n−1)×1) = 1

)=

∑x

1 × P(x :ϕ

(1i , x(n−1)×1) > ϕ

(0i , x(n−1)×1))

= 12n−1

× |x : ϕ(1i , x(n−1)×1) > ϕ

(0i , x(n−1)×1)| (4)

In the case of unavailability of any knowledge about pi’s, usingp i = 1

2 , i = 1, 2, . . ., n, and denoting cardinality of the set by | · |,(4) reduces to

12n−1

× |x : ϕ(1i , x(n−1)×1) > ϕ(0i , x(n−1)×1)| = Iϕ(i).

Note that in xn×1, excepting the ith element, all (n −1) el-ements can take one of the two values, viz., 0 or 1, which ispossible in 2n−1 ways. Out of these 2n−1 possibilities, m = |x :ϕ(1i , x(n−1)×1) > ϕ(0i , x(n−1)×1)| is the number of x’s for whichϕ(1i , x(n−1)×1) > ϕ(0i , x(n−1)×1). Thus, m

2n−1 (= Iϕ(i)) gives the to-tal of probabilities P(x : ϕ(1i , x(n−1)×1) > ϕ(0i , x(n−1)×1)) over allsuch x’s for which ϕ(1i , x(n−1)×1) > ϕ(0i , x(n−1)×1), or, Iϕ(i) =

m2n−1 = ∑

x P(x : ϕ(1i , x(n−1)×1) > ϕ(0i , x(n−1)×1)).Hence,

h(pn×1)=h(0i , p(n−1)×1)+ 1

2× Iϕ(i), since p i = 1

2, i = 1, 2, . . . , n.

(5)

If a redundant component with reliability r is added to compo-nent i, the reliability of the new augmented ith component, p ′

i ,

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44 D. Bhattacharya and S. Roychowdhury

will be

p ′i = p i + r (1 − p i ) = 1

2+ r

2,

and hence, from (5), the reliability of the corresponding aug-mented system will be

h(i)(pn×1) = h(0i , p(n−1)×1) +

(12

+ r2

)× Iϕ(i)

= h(pn×1) + r2

× Iϕ(i). (6)

Similarly, if, instead, a redundant component with reliability r isadded to component j, the reliability of the augmented system willbe

h( j)(pn×1) = h(0 j , p(n−1)×1) +(

12

+ r2

)

× Iϕ( j) = h(pn×1) + r2

× Iϕ( j). (7)

If Iϕ(i) > Iϕ( j), then comparing (6) and (7), we get

h(i)(pn×1) > h( j)(pn×1).

Hence, the result. �

Let us now consider the case wherein all component relia-bilities are the same (=p). We prove the following result, whichguides us to choose the best system component to accommodatethe addition of the redundant component.

Result 2. If reliabilities of all system components are the same, the compo-nent i gives more improvement in system reliability than component j does;that is, (i) �b ( j) when a redundant component is added to it if

RI (i) > RI ( j).

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Redundancy Allocation Using Component-Importance Measures 45

Proof. By pivotal decomposition, the system reliability h(pn×1)can be written as

h(pn×1) = p i × h(1i , p(n−1)×1) + (1 − p i ) × h(0i , p(n−1)×1)

= h(0i , p(n−1)×1) + p i × RI (i). (8)

For pi = p, i = 1, 2, . . ., n, (8) becomes

h(pn×1) = h(0i , p(n−1)×1) + p × RI (i).

Now, if a redundant component is added to component i, thereliability of the augmented component i will be {p + r (1 − p )},and the reliability of the new augmented system becomes

h(i)(pn×1) = h(0i , p(n−1)×1) + {p + r (1 − p )} × RI (i)

= h(pn×1) + {r (1 − p )} × RI (i). (9)

Similarly,

h( j)(pn×1) = h(pn×1) + {r (1 − p )} × RI ( j). (10)

Thus, it is evident from (9) and (10) that

h(i)(pn×1) > h( j)(pn×1) if RI (i) > RI ( j).�

Now, suppose that all component reliabilities are not thesame. The following result gives the condition for component ito be a better choice than component j for the addition of a re-dundant component:

Result 3. System reliability improvement is higher in the case when aredundant component is added to component i than to component j; thatis, (i) �b ( j) if

(1 − p i ) × RI (i) > (1 − p j ) × RI ( j).

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46 D. Bhattacharya and S. Roychowdhury

Proof. The system reliability h(pn×1) can be written as

h(pn×1) = p i × h(1i , p(n−1)×1) + (1 − p i ) × h(0i , p(n−1)×1)

= h(0i , p(n−1)×1)+p i × (h(1i , p(n−1)×1) − h(0i , p(n−1)×1))

= h(0i , p(n−1)×1) + p i × RI (i).

If the redundant component is added to the ith system compo-nent, the system reliability becomes

h(i)(pn×1) = h(0i , p(n−1)×1) + {p i + r (1 − p i )} × RI (i)

= h(pn×1) + r (1 − p i ) × RI (i), i = 1, 2, . . . , n. (11)

Similarly, if the redundant component is added to the jth systemcomponent, j �= i, the system reliability becomes

h( j)(pn×1) = h(pn×1) + r (1 − p j ) × RI ( j). (12)

Comparing (11) and (12), we can write,

h(i)(pn×1) > h( j)(pn×1)

if (1 − p i ) × RI (i) > (1 − p j ) × RI ( j).�

The reliabilities of the spares (redundant components) of dif-ferent components of the system may not necessarily be the same.In the case in which they are different, the above result needs tobe slightly modified. Suppose the reliability of the spare of the ithcomponent to be ri , i = 1, 2, . . ., n. Then we have the followingresult:

Result 4. System reliability improvement is higher in the case in which aredundant component is added to component i than to component j; thatis, (i) �b ( j) if

ri (1 − p i ) × RI (i) > r j (1 − p j ) × RI ( j).

Proof. Similar to the proof of Result 3. �

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Redundancy Allocation Using Component-Importance Measures 47

FIGURE 1 A series-parallel system.

4. Application of the Results

4.1. A Series-Parallel System

Let us consider an example of the system, as shown inFigure 1.

The reliability of the above system is

h(p) = p 1 × (p 2 + p 3 − p 2 × p 3).

Let us determine the structural importance of the systemcomponents. For computing the structural importance of com-ponent 1 of System 1, we construct Table 1.

From Table 1, by (2), we get

Iϕ(1) =∑

[x:xi =1] [ϕ(1i , x) − ϕ(0i , x)]

2n−1= 3

4. (13)

TABLE 1 Computation of structural importance ofcomponent 1

ϕ(x1 = 1, x2, x3) ϕ(x1 = 0, x2, x3) ϕ(1, x) – ϕ(0, x)

ϕ(1, 1, 1) = 1 ϕ(0, 1, 1) = 0 1ϕ(1, 1, 0) = 1 ϕ(0, 1, 0) = 0 1ϕ(1, 0, 1) = 1 ϕ(0, 0, 1) = 0 1ϕ(1, 0, 0) = 0 ϕ(0, 0, 0) = 0 0

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48 D. Bhattacharya and S. Roychowdhury

Similarly, we can obtain Iϕ(2) = Iϕ(3) = 1/4. Now let us deter-mine the reliability importance of the components.

RI (1) = ∂h(p)∂p 1

= p 2 + p 3 − p 2 × p 3, (14)

RI (2) = ∂h(p)∂p 2

= p 1 × (1 − p 3), (15)

and RI (3) = ∂h(p)∂p 3

= p 1 × (1 − p 2). (16)

If a redundant component with reliability r is added to compo-nent 1, the reliability of the augmented system will be

h(1)(p) = (p 1 + r − rp 1) × (p 2 + p 3 − p 2 × p 3)

= h(p) + r × (1 − p 1) × (p 2 + p 3 − p 2 × p 3). (17)

Similarly,

h(2)(p) = p 1 × {(p 2 + r − rp 2) + p 3 − (p 2 + r − rp 2) × p 3}= h(p) + r × p 1 × (1 − p 2) × (1 − p 3), (18)

and h(3)(p) = p 1 × {(p 3 + r − rp 3) + p 2 − (p 3 + r − rp 3) × p 2}= h(p) + r × p 1 × (1 − p 2) × (1 − p 3).

(19)

(i) Now suppose the component reliabilities to be unknown.By comparing Iϕ(1), the structural importance of compo-nent 1 as obtained in (13), with Iϕ(2) and Iϕ(3), we findIϕ(1) to be the maximum. Thus, by Result 1, the redun-dant component should be added to component 1 in or-der to get maximum rise in system reliability. In this case,because all pi’s are unknown, we assume that pi = 1

2 , forall i. From (17)–(19), the reliability of the augmented sys-tems would become

h(1)(p) =(

12

+ r − r × 12

)×

(12

+ 12

− 14

)

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Redundancy Allocation Using Component-Importance Measures 49

= h(p) + r ×(

1 − 12

)×

(12

+ 12

− 14

)

= h(p) + 3r8

, and

h(2)(p) = h(3)(p)

= h(p)+r × p 1 × (1−p 2) × (1 − p 3)=h(p) + r8,

which shows that the gain in system reliability is maximumif the redundant component is added to component 1.

(ii) Next suppose the component reliabilities are all equal(= p). Then, using (14)–(16), the reliability importanceof the components becomes

RI (1) = 2p − p 2, and

RI (2) = RI (3) = p − p 2 < RI (1).

Thus, by Result 2, the redundant component should beadded to component 1, whose reliability importance ismaximum. This will give maximum rise in system relia-bility. Now let us determine the reliabilities of the aug-mented systems.

h(1)(p) = h(p) + r × (1 − p ) × (2p − p 2)

= h(p) + r × p × (1 − p ) + r × p × (1 − p )2,

h(2)(p) = h(3)(p) = h(p) + r × p × (1 − p )2,

which shows that the rise in system reliability will be max-imum when the redundant component is added to com-ponent 1, which has the highest reliability importance.

(iii) Now suppose the component reliabilities to be different.Let p 1 = 0.9, p 2 = 0.6, p 3 = 0.7. Here,

RI (1) = 0.6 + 0.7 × (1 − 0.6) = 0.88,

RI (2) = 0.9 × (1 − 0.7) = 0.27, and

RI (3) = 0.9 × (1 − 0.6) = 0.36.

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50 D. Bhattacharya and S. Roychowdhury

Hence,

(1 − p 1) × RI (1) = 0.1 × 0.88 = 0.088,

(1 − p 2) × RI (2) = 0.4 × 0.27 = 0.108, and

(1 − p 3) × RI (3) = 0.3 × 0.36 = 0.108.

Thus,

(1 − p 2) × RI (2) = (1 − p 3) × RI (3) > (1 − p 1) × RI (1).

Hence, by Result 3, the redundant component should beadded to component 2 or 3 in order to maximize the gainin system reliability. Let us now determine the gain in sys-tem reliability using (17)–(19).

h(1)(p) = h(p) + r × (1 − p 1) × (p 2 + p 3 − p 2 × p 3)

= h(p) + r × 0.1 × 0.88 = h(p) + 0.088r

h(2)(p) = h(3)(p) = h(p) + r × p 1 × (1 − p 2) × (1 − p 3)

= h(p) + 0.108r > h(1)(p).

(iv) Now, in addition, suppose the reliabilities of the sparesof different components to be not the same. Let p 1 =0.9, p 2 = 0.6, p 3 = 0.7, r1 = 0.8, r2 = r3 = 0.6.Let us compute the following:

r1(1 − p 1) × RI (1) = 0.8 × 0.108 = 0.0704, and

r2(1 − p 2) × RI (2) = r3(1 − p 3) × RI (3) = 0.6 × 0.108

= 0.0648.

Because r1(1−p 1)×RI (1) > r2(1−p 2)×RI (2) = r3(1−p 3) × RI (3), the redundant component should be addedto component 1 to get maximum increase in system relia-bility, by Result 4.

Now let us find the reliability of the augmented system,when the redundant component is added to different system

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Redundancy Allocation Using Component-Importance Measures 51

FIGURE 2 A bridge system.

components. Using (17)–(19) we get

h(1)(p) = h(p) + r1 × (1 − p 1) × (p 2 + p 3 − p 2 × p 3)

= h(p) + 0.0704,

h(2)(p) = h(p) + r2 × p 1 × (1 − p 2) × (1 − p 3)

= h(p) + 0.0648, and

h(3)(p) = h(p) + r3 × p 1 × (1 − p 2) × (1 − p 3)

= h(p) + 0.0648,

which shows that h(2)(p)and h(3)(p) both are less than h(1)(p).

4.2. A Bridge System

Let us consider the bridge system as shown in Figure 2.The system reliability is

h(p) = h(p5×1) = p 1p 4 + p 2p 5 + p 2p 3p 4 + p 1p 3p 5

− p 1p 3p 4p 5 − p 1p 2p 3p 5 − p 1p 2p 3p 4 − p 2p 3p 4p 5

− p 1p 2p 4p 5 + 2p 1p 2p 3p 4p 5.

By (2), the structural importance of each of the components is

Iϕ(1) = Iϕ(2) = Iϕ(4) = Iϕ(5) = 3/8, Iϕ(3) = 1/8. (20)

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52 D. Bhattacharya and S. Roychowdhury

The reliability importance of each of the different components isas follows:

RI (1) = ∂h(p)∂p 1

= p 4 + p 3p 5 − p 3p 4p 5 − p 2p 3p 5

−p 2p 3p 4 − p 2p 4p 5 + 2p 2p 3p 4p 5, (21)

RI (2) = ∂h(p)∂p 2

= p 5 + p 3p 4 − p 1p 3p 5 − p 1p 3p 4 − p 3p 4p 5

−p 1p 4p 5 + 2p 1p 3p 4p 5, (22)

RI (3) = ∂h(p)∂p 3

= p 2p 4 + p 1p 5 − p 1p 4p 5 − p 1p 2p 5

−p 1p 2p 4 − p 2p 4p 5 + 2p 1p 2p 4p 5, (23)

RI (4) = ∂h(p)∂p 4

= p 1 + p 2p 3 − p 1p 3p 5 − p 1p 2p 3 − p 2p 3p 5

−p 1p 2p 5 + 2p 1p 2p 3p 5, (24)

RI (5) = ∂h(p)∂p 5

= p 2 + p 1p 3 − p 1p 3p 4 − p 1p 2p 3 − p 2p 3p 4

−p 1p 2p 4 + 2p 1p 2p 3p 4. (25)

Now consider the following situations:

(i) We have no idea about the component reliabilities.(ii) All component reliabilities are known to be equal.

(iii) All component reliabilities are different.(iv) All component reliabilities may not be the same, and the

reliabilities of the spares of all components are also differ-ent.

Now, using the results proved in a previous section, we choose thebest system component to accommodate the addition of a redun-dant component and thus maximize the system reliability.

(i) In the case in which we have no idea about componentreliabilities, by Result 1 we choose to add the redundantcomponent to the component that has maximum struc-tural importance. Here, by (20), components 1, 2, 4, and

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Redundancy Allocation Using Component-Importance Measures 53

5 have the same structural importance, which is higherthan that of component 3. Hence, the redundant compo-nent can be added to any of the four components 1, 2, 4,or 5. In each of these cases, the gain in system reliabilitywill be maximum.

(ii) If the component reliabilities are known to be equal,then the reliability importance of each of the componentsshould be compared according to Result 2. Here, using(21)–(25), we get the reliability importance of each com-ponent as

RI (1) = RI (2) = RI (4) = RI (5)

= p + p 2 − 4p 3 + 2p 4, and

RI (3) = 2p 2 − 4p 3 + 2p 4 < p + p 2 − 4p 3 + 2p 4

= RI (1) = RI (2) = RI (4) = RI (5).

Hence, by Result 2, the redundant component can beadded to any of the components 1, 2, 4, or 5, and in allcases, the gain in system reliability will be the same.

(iii) Suppose the component reliabilities to be: p 1 = 0.9, p 2 =0.6, p 3 = 0.7, p 4 = 0.8, p 5 = 0.75. By (21)–(24), the relia-bility importance of each of the components is as follows:RI (1) = 0.398, RI (2) = 0.1295, RI (3) = 0.066, RI (4) =0.3165, and RI (5) = 0.1848. Then compute

(1 − p 1) × RI (1) = 0.1 × 0.398 = 0.0398,

(1 − p 2) × RI (2) = 0.4 × 0.1295 = 0.0518,

(1 − p 3) × RI (3) = 0.3 × 0.066 = 0.0198,

(1 − p 4) × RI (4) = 0.2 × 0.3165 = 0.0633, and

(1 − p 5) × RI (5) = 0.25 × 0.1848 = 0.0462.

Here, (1 − p 4) × RI (4) is maximum. Hence, byResult 3, the redundant component should be added tocomponent 4 in order to maximize the system reliability.

(iv) Suppose the reliabilities of the spares of all the com-ponents are not the same. The component reliabilities

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54 D. Bhattacharya and S. Roychowdhury

are p 1 = 0.9, p 2 = 0.6, p 3 = 0.7, p 4 = 0.8, p 5 = 0.75, andthe reliabilities of the redundant components are r1 =0.75, r2 = 0.8, r3 = 0.65, r4 = 0.6, r5 = 0.7. Now let uscompute the following:

r1(1 − p 1) × RI (1) = 0.75 × 0.0398 = 0.02985,

r2(1 − p 2) × RI (2) = 0.8 × 0.0518 = 0.04144,

r3(1 − p 3) × RI (3) = 0.65 × 0.0198 = 0.01287,

r4(1 − p 4) × RI (4) = 0.6 × 0.0633 = 0.03798, and

r5(1 − p 5) × RI (5) = 0.7 × 0.0462 = 0.03234.

Here, r2(1 − p 2) × RI (2) is maximum, and hence, by Re-sult 4, component 2 is to be chosen to have the additionof a spare component in order to maximize the systemreliability.

References

Barlow R. E., & Proschan, F. (1981). Statistical theory of reliability and life testing:Probability models. Silver Spring, MD: To Begin With.

Birnbaum, Z. W. (1968). On the importance of different components in a multi-component system (Technical Report No. 54, pp. 1–21). Seattle, WA: Depart-ment of Mathematics, University of Washington.

Boland, P. J., El-Neweihi, E., & Proschan, F. (1988). Active redundancy allocationin coherent systems. Probability in the Engineering and Informational Sciences, 2,343–353.

Hu, T., & Wang, Y. (2009). Optimal allocation of active redundancies in r -out-of-n systems. Journal of Statistical Planning and Inference, 139(10), 3733–3737.

Li, X., & Hu, X. (2008). Some new stochastic comparisons for redundancy allo-cations in series and parallel systems. Statistics and Probability Letters, 78(18),3388–3394.

Shaked, M., & Shanthikumar, J. G. (1992). Optimal allocation of resourcesto nodes of parallel and series systems. Advances in Applied Probability, 24,894–914.

Singh, H., & Singh, R. S. (1997). Optimal allocation of resources to nodes ofseries systems with respect to failure rate ordering. Naval Research Logistics,44(1), 147–152.

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