Warranty and redundancy design with uncertain quality.

1. Introduction

The role and importance of warranty has changed significantly over the last two decades. Currently, almost all products are sold with some form of warranty attached. Warranty is of importance to both manufacturers and consumers. From the consumers' point of view, warranty

There are many different aspects to warranty and these have been studied by researchers from diverse disciplines. The three part survey of the topic conducted by Blischke and Murthy [1] and Murthy and Blischke [2,3] developed a framework, which enabled the issues to be viewed in a proper perspective. The Product Warranty Handbook [4] deals with several of these issues in greater depth. One of the key issues is the warranty cost to the manufacturer.

Selling a product with a warranty results in additional costs to the manufacturer due to servicing of the warranty. Cost analysis for a variety of warranty policies can be found in the paper of Blischke and Murthy [5]. Most cost analysis models assume that all items have the same failure time distribution. In real life, due to manufacturing-based quality variation, a fraction of the components fail to meet the design specification. The expected warranty cost for a nonconforming item is bigger than that for a conforming one. One way of reducing warranty cost in this case is through effective quality control. The literature on quality control is vast. Murthy and Djamaludin [6] have presented a brief review of models that deal with quality variations and control in manufacturing. Optimal quality control strategies such as lot sizing (to reduce the likelihood of nonconforming items being produced) and inspection (to weed out nonconforming items) are considered for products sold with a warranty [7-9].

The expected warranty cost depends on the reliability of the product. The higher the product reliability then the smaller is the expected warranty cost. Product reliability can be improved in two ways. The first way is through the use of redundancy, in which one or more components are replicated to create a module of similar components rather than a single component. Three different types of redundancy have been used: these are described as hot, cold and warm. A brief review of the literature on this topic can be found in Murthy and Hussain [10] who have examined the use of these redundancy types in the context of warranties when there are no quality variations (i.e., the items are statistically similar). The second approach to improve product reliability is through a reliability development program where the improvement is achieved through a series of test-fix-test cycles. Optimal development strategies for the case when products are sold with a warranty can be found in Murthy and Nguyen [11] and also in Hussain [12].

In this paper, we consider the case where items are sold with a warranty and some of the items are nonconforming due to quality variations. The expected warranty cost can be reduced by quality control (testing to weed out nonconforming items) and/or building in redundancy. However these approaches increase the manufacturing cost due either to the cost of replicated components if redundancy is used or the cost of inspection (or testing) if quality control is used. Hence, the optimal decision must take into account these tradeoffs. We develop a model to answer the following questions:

(1) Is it worthwhile building in redundancy?

(2) Is it worthwhile testing to weed out nonconforming components?

(3) If redundancy and testing are to be used, should the testing be done at component or module level?

As a result, we need to consider the following five options:

Option 1: No redundancy and no testing.

Option 2: No redundancy and component level testing.

Option 3: Redundancy and no testing.

Option 4: Redundancy and component level testing.

Option 5: Redundancy and module level testing.

The optimal choice depends on various factors such as the reliability of the product, the type of redundancy, the warranty policy and the warranty period.

We confine our attention to a critical component of a product, which is sold with a Free Replacement Warranty (under this policy, the manufacturer provides a free replacement for failures within the warranty period, T). The component is assumed to be nonrepairable and needs to be replaced on failure. Typical examples are a bulb in a projector, a solenoid valve in a gas burner, a transistor power amplifier used to drive a control motor, and an integrated circuit of an electronic device. Due to quality variation, some components are nonconforming. We focus our attention on the warranty cost associated with this component. We note that the total warranty cost for the product can be obtained through summation of the warranty costs for its various components. We consider all three types of redundancy but restrict our analysis to hot standby redundancy with a single replication of the component. As a result, the module contains two components with redundancy and one without redundancy.

Possible areas of application of the model include:

(1) Impact of reliability and/or quality improvement on warranty cost and total cost.

(2) Analysis of different options in order to determine which option yields the minimum cost.

The outline of the paper is as follows. In Section 2 we give details of the model formulation. The analysis of the five options with hot standby redundancy is performed in Section 3. Section 4 deals with some numerical examples. In Section 5, we discuss the analysis of cold and warm standby redundancy. Finally, in Section 6 we conclude with some topics for further research.

2. Model formulation

In this section, we give details of the model formulation.

2.1. Component quality and failure distribution

It is assumed that the production process is in a steady state. The components produced are nonconforming with a probability p and conforming with probability 1 - p. The case p = 0 corresponds to perfect quality.

The failure distribution of a conforming component is described by the probability that it fails at or before time x, F(x) with F(0) = 0. One can define two types of non-conformance. The first type of nonconforming component is nonoperational when put into use. Such nonconforming components are easily detected by testing for a very short time. The second type of nonconforming component is operational when put into use but its failure characteristics are poor (e.g., smaller mean time to failure) as compared with those of a conforming component. In this paper, we confine our attention to the first type of nonconformance. Hence, the distribution function for a nonconforming item is given by the Heaviside function U(x), with U(x) = 1 for x [greater than or equal to] 0, since the time to failure is zero at a probability of one. The nonconforming items detected under testing are scrapped. Similarly, failed items are nonrepairable and need to be replaced by new ones over the warranty period.

2.2. Redundancy

We consider all three type of redundancy with duplication of the critical component. We use the term "module" so that with redundancy, the module contains two components and without redundancy it has only one component.

The failure distribution for the module depends on: (1) F(x); (2) p; (3) the type of redundancy; and (4) the option used. Let [G.sub.i](x) denote the distribution function for module failure under option i, 1 [less than or equal to] i [less than or equal to] 5.

We first consider the hot standby case and discuss this analysis in detail. Under hot standby redundancy conditions the time to failure for the module is given by the longer of the two individual lifetimes since both components are put in use at the same time. Later, we consider the cold and warm standby cases.

2.3. Testing

We consider two types of testing. In the first option 100% testing at component level is carried out to weed out nonconforming components. The testing is assumed to take a negligible time and to be perfect (error free). Hence all nonconforming components are detected and scrapped. As a result, the module contains only conforming components.

In the second type, which is applicable only when redundancy is used, the testing is done at module level. Modules which are nonoperational are detected and scrapped. This situation arises only when both of the components of a module are nonconforming. In other cases, there is at least one conforming component in the module and hence the module is operational. As before, the testing is assumed to be 100% and error free.

2.4. Warranty

The items are sold with a Free Replacement Warranty (FRW) policy. Under this policy, the manufacturer provides a free replacement for failures within the warranty period T.

We assume the following:

(1) During the warranty period, module failures are detected instantaneously and result in immediate warranty claims.

(2) All claims are valid and the manufacturer provides a replacement module.

(3) The time needed to execute a claim and obtain a replacement module is relatively short compared with the mean time between failures so that it can be treated as being nearly zero.

These are reasonable assumptions and are used in most warranty cost analyses. They can be relaxed, but at the cost of the analysis becoming more complex.

2.5. Modeling costs

The following costs are relevant to the modeling.

[C.sub.m] = Cost of manufacturing a component.

[C.sub.t][[C.sub.T]] = Cost of testing a component [module] ([C.sub.T] [greater than] [C.sub.t]).

[C.sub.s][[C.sub.s]] = Cost of scrapping a nonconforming component [module] ([C.sub.S] [greater than] [C.sub.s]).

[C.sub.h] = Cost of handling each warranty claim.

[Phi][C.sub.m] = Cost of switch (for cold and warm standby).

With redundancy and/or testing, the total manufacturing cost increases. We assume that the number of items produced is sufficiently large so that we can use asymptotic results in computing the total manufacturing ([C.sub.m] + testing + scrapping) cost per module. Let [C.sub.mi] denote this cost for option i, 1 [less than or equal to] i [less than or equal to] 5. Expressions for these costs will be presented in a later section of the paper.

Let [N.sub.i](T) denote the number of warranty claims over the warranty period under option i, 1 [less than or equal to] i [less than or equal to] 5. The claims occur according to a renewal process with the time between renewals being distributed according to [G.sub.i](x). Let [M.sub.i](T) denote the expected number of claims. The total expected cost (associated with the critical component) to the manufacturer under option i is given by

[J.sub.i](T) = [C.sub.mi] + ([C.sub.mi] + [C.sub.h]) [M.sub.i](T), (1)

where the first term represents the cost of the item sold and second represents the expected cost of servicing the module over the warranty period.

2.6. Decision problem

The decision problem (for a specified redundancy) is to choose the optimal option [i.sup.*]. It is given by the value of i, which yields a minimum value for [J.sub.i], 1 [less than or equal to] i [less than or equal to] 5.

3. Model analysis for the hot standby case

We first derive expressions for the total expected cost to the manufacturer, for each of the five options with hot standby being considered for Options 3-5. Later on, we present some comparative results.

3.1. Option 1: No redundancy and no testing

Since components are untested and the module consists of a single component, the failure distribution of a module is given by a mixture of the distribution functions for conforming and nonconforming components. As a result, [G.sub.1] (x), is given by

[G.sub.1](x) = pU(x) + (1 - p)F(x). (2)

Note that [G.sub.1] (0) [greater than] 0 but F(0) = 0. The expected number of replacements under warranty is given by

[M.sub.1](T) = ([M.sub.F](T) + p)/(1 - p), (3)

where [M.sub.F](x) is the renewal integral equation associated with the distribution function F(x) and is given by

[M.sub.F](T) = F(T) + [integral of] [M.sub.F](T - x)dF(x) between limits, T and 0. (4)

Details of the derivation can be found in the Appendix. Also,

[C.sub.m1] = [C.sub.m]. (5)

As a result, we have

[J.sub.1](T) = [C.sub.m] + ([C.sub.m] + [C.sub.h])[M.sub.1](T), (6)

where [M.sub.1](T) is given by (3).

3.2. Option 2: No redundancy and component level testing

In this case, every component is tested so each module consists of a single conforming component. This implies

[G.sub.2](x) = F(x), (7)

and

[M.sub.2](T) = [M.sub.F](T). (8)

It is easily shown that

[C.sub.m2] = ([C.sub.m] + [C.sub.t] + p[C.sub.s])/(1 - p). (9)

From (1) we have

[J.sub.2](T) = [C.sub.m2] + ([C.sub.m2] + [C.sub.h])[M.sub.F](T), (10)

with [C.sub.m2] being given by (9).

3.3. Option 3: Redundancy and no testing

Since no testing is carried out, a module can contain 0, 1 or 2 nonconforming components. There is a probability [p.sup.2] that both components are nonconforming and thus the module is nonoperational. In this case the module failure distribution is given by U(x). There is a probability of 2p(1 - p) that one of the components is nonconforming, and in this case the module failure distribution is given by F(x). Finally, there is a probability [(1 - p).sup.2] that both components are conforming and in this case the module failure distribution is given by [[F(x)].sup.2]. As a result, [G.sub.3](x) is given by

[G.sub.3](x) = [p.sup.2]U(x) + (1 - [p.sup.2])H(x), (11)

where

H(x) = {[(1 - p).sup.2][[F(x)].sup.2] + 2p(1 - p)F(x)}/(1 - [p.sup.2]). (12)

Note that [G.sub.3](0) [greater than] 0 and H(0) = 0. Following the approach used in the Appendix, the expected number of claims under a warranty is given by

[M.sub.3](T) = [[M.sub.H](T) + [p.sup.2]]/(1 - [p.sup.2]), (13)

with [M.sub.H](T) being given by

[M.sub.H](T) = H(T) + [integral of] [M.sub.H](T - x)dH(x) between limits T and 0. (14)

Since each module has two components

[C.sub.m3] = 2[C.sub.m]. (15)

From (1) we have

[J.sub.3](T) = 2[C.sub.m] + (2[C.sub.m] + [C.sub.h])[M.sub.3](T), (16)

where [M.sub.3](T) is given by (13).

3.4. Option 4: Redundancy and component level testing

Since components are tested at component level and nonconforming ones are scrapped, each module consists of two conforming components at the time of purchase. In this case,

[G.sub.4](x) = [[F(x)].sup.2]. (17)

Also,

[C.sub.m4] = 2[C.sub.m2], (18)

with [C.sub.m2] being given by (9). From (1) we have

[J.sub.4](T) = 2[C.sub.m2] + (2[C.sub.m2] + [C.sub.h])[M.sub.4](T), (19)

where [M.sub.4](T) is the renewal function associated with the [G.sub.4](x) given by (17).

3.5. Option 5: Redundancy and module level testing

In this case we have

[G.sub.5](x) = H(x), (20)

with H(x) being given by (12) since all nonoperational modules are scrapped after testing. The expected number of replacements under warranty is given by

[M.sub.5](T) = [M.sub.H](T), (21)

with [M.sub.H] (T) being given by (14).

The cost per module released is given by

[C.sub.m5] = (2[C.sub.m] + [C.sub.T] + [p.sup.2][C.sub.S])/(1 - [p.sup.2]). (22)

As a result, from (1), we have

[J.sub.5](T) = [C.sub.m5] + ([C.sub.m5] + [C.sub.h]) [M.sub.H](T), (23)

with [C.sub.m5] being given by (22) and [M.sub.H](T) being given by (14).

3.6. Comparison

For the no redundancy case [Options 1 and 2] we have the following results.

Proposition 1. (i) [C.sub.m1] [less than] [C.sub.m2] and (ii) [M.sub.1](T) [greater than] [M.sub.2](T).

Proof: Trivial and follows from (5) and (9) and from (3) and (8).

Proposition 2. Testing (Option 2) is better than no testing (Option 1) iff

[C.sub.h] [greater than] [C.sub.t]/p + [C.sub.s].

Proof: The use of testing is justified iff [J.sub.2](T) [less than] [J.sub.1](T). The result follows from (6) and (10) after some simplification.

Note that the inequality in Proposition 2 does not depend on F(x) and T but only on p and the cost parameters. When p = 0, no testing is the better option since [C.sub.h] is finite. This is to be expected since all components are conforming. If [C.sub.h] [less than] [C.sub.t] + [C.sub.s], then no testing is the better option irrespective of p. This occurs when either [C.sub.t] and/or [C.sub.s] are large relative to [C.sub.h]. Intuitively, the costs of testing and scrapping make testing uneconomical.

If [C.sub.h] [greater than] [C.sub.t] + [C.sub.s], then there exists a [p.sub.a] ([less than] 1 and given by [C.sub.h] = [C.sub.t]/[p.sub.a] + [C.sub.s]) such that for all p greater than [p.sub.a], testing is the better option and for p smaller than [p.sub.a], no testing is the better option. This is to be expected, since testing is less useful when p is small.

For the hot standby redundancy case (Options 3-5) we have the following results.

Proposition 3. [C.sub.m3] [less than] [C.sub.m4] and [C.sub.m3] [less than] [C.sub.m5].

Proof: Trivial and follows from (15), (18) and (22).

Proposition 4. [M.sub.4](T) [less than] [M.sub.5](T) [less than] [M.sub.3](T).

Proof: From (13) and (21) we have [M.sub.3](T) [greater than] [M.sub.5](T). Note that

[G.sub.5](x) = H(x)

= {2p(1 - p)F(x) + [(1 - p).sup.2][[F(x)].sup.2]}/(1 - [p.sup.2]) [greater than] {2p(1 - p) + [(1 - p).sup.2]}[[F(x)].sup.2]/(1 - [p.sup.2])

= [[F(x)].sup.2] = [G.sub.4](x).

As a result, from Stoyan [13] we have [M.sub.4](T) [less than] [M.sub.5](T) and hence the result.

Proposition 5. Module level testing (Option 5) is better than no testing (Option 3) iff

[C.sub.h] [greater than] [C.sub.T]/[p.sup.2] + [C.sub.S].

Proof: The use of module level testing is justified iff [J.sub.5](T) [less than] [J.sub.3](T). The result follows from (16) and (23) after some simplification.

Proposition 6. Component level testing (Option 4) is better than no testing (Option 3) iff

[C.sub.h]/[C.sub.m] [greater than] [2(p + [C.sub.t]/[C.sub.m] + p[C.sub.s]/[C.sub.m])(1 + [M.sub.4](T))]/[(1 - p)([M.sub.3](T) - [M.sub.4](T))] - 2.

Proof: The use of component level testing is justified iff [J.sub.4](T) [less than] [J.sub.3](T). The result follows from using (16) and (19).

Note that Proposition 6 involves [M.sub.3](T) and [M.sub.4](T). For general F(x), these cannot be expressed in an analytical form. However, when F(x) is exponentially distributed with a parameter [Mu] (= mean failure time), they can be expressed in an analytical form. [M.sub.3](T) is given by (13) with

[M.sub.H](T) = 2{(3 + p)[(1 + p).sup.2]T - [Mu](1 - [p.sup.2]) x {1 - exp[-(3 + p)T/([Mu](1 + p))]}}/[[Mu](1 + p)[(3 + p).sup.2]], (24)

and

[M.sub.4](T) = 2{3T - [Mu]{1 - exp[-(3T/[Mu])]}}/(9[Mu]). (25)

Using (24) and (25), the inequality in Proposition 6 becomes

[Mathematical Expression Omitted]

where v = T/[Mu]. This involves only the parameters of the model and hence is easier to check.

4. Numerical example (hot standby)

In this section we present some numerical results. The total expected costs and other cost parameters are normalized with respect to the component manufacturing cost ([C.sub.m]). The following nominal values are used:

[C.sub.T]/[C.sub.m] = [C.sub.t]/[C.sub.m] = 0.1 and [C.sub.S]/[C.sub.m] = [C.sub.s]/[C.sub.m] = 0.0.

We consider two different failure distributions, namely exponential and Weibull. Finally, we consider four different warranty periods (ranging from 1 to 4 years) and four different combinations of [C.sub.h]/[C.sub.m] and p as listed in Table 1.

4.1. Exponential distribution

Let [Mu] = 4 years so that the mean time to failure is 4 years. The results for no redundancy (Options 1 and 2) and for hot standby redundancy (Options 3-5) are given in Tables 2 and 3. The optimal choice is indicated by an * next to the optimal total expected cost [J.sub.i](T).

For parameter combination A1, [J.sub.1](T) = [J.sub.2](T) for all four values of T. This is not true in general, as can be seen from the results for the other parameter combinations.

For parameter combination A1, the optimal strategy is Option 1 (no redundancy and no testing) for T[less than or equal to]2 and Option 3 (redundancy and no testing) for T[greater than or equal to]3. The reason for this is that since p and [C.sub.h]/[C.sub.m] are small, the savings in the warranty cost with testing [component and module levels] is less than the increase in the cost of the module due to testing and scrapping of nonconforming items. However, for large T, building in redundancy is worthwhile.

Table 1. The parameter values used in the modeling study

Combination                  p           [C.sub.h]/[C.sub.m]

A1                          0.05                  2
A2                          0.05                  5
B1                          0.15                  2
B2                          0.15                  5

[TABULAR DATA FOR TABLE 2 OMITTED]

For parameter combination A2, the optimal strategy is Option 3 (redundancy and no testing) for all values of T. This is to be expected, since p is small (so that testing is uneconomical) and [C.sub.h]/[C.sub.m] is large (so that using redundancy reduces the number of claims over the warranty period).

For parameter combination B1, the optimal strategy is Option 2 (no redundancy and testing) for T [less than or equal to] 2 and Option 3 (redundancy and no testing) for T [greater than or equal to] 3.

For parameter combination B2, the optimal strategy is Option 5 (redundancy and module level testing) for T = 1 and Option 4 (redundancy and component level testing) for T [greater than or equal to] 2. This is to be expected since p and [C.sub.h]/[C.sub.m] are large so testing and use of redundancy are worthwhile.

4.2. Weibull distribution

Here F(x) is the Weibull distribution given by F(x) = 1 - exp[-[([Theta]x).sup.[Gamma]]] with parameter values:

[Theta] = 0.22570/year and [Gamma] = 1.50.

The mean time to failure is 4 years. The shape parameter y [greater than] 1 implies an increasing failure rate.

[TABULAR DATA FOR TABLE 3 OMITTED]

[TABULAR DATA FOR TABLE 4 OMITTED]

The results for no redundancy (Options 1 and 2) and for redundancy (Options 3-5) are given in Tables 4 and 5. As can be seen, the results are very similar to the exponential case and hence need no further discussion.

Based on these two examples, the following inference can be drawn. When p and [C.sub.h]/[C.sub.m] are both small, then no redundancy and no testing is the optimal strategy. When p is small and [C.sub.h]/[C.sub.m] is large, the strategy changes to redundancy and no testing since there are few nonconforming components. When p is large and [C.sub.h]/[C.sub.m] is small, then testing and no redundancy is the strategy for small T values and this changes to testing and no redundancy for large values of T. Finally, when both p and [C.sub.h]/[C.sub.m] are large, then the optimal strategy is to use both redundancy and testing.

5. Model analysis for cold and warm standby cases

The analyses for the cold and warm standby cases are similar to that for the hot standby case. However, the expressions for [G.sub.i](x) and [C.sub.mi], 3 [less than or equal to] i [less than or equal to] 5, are different. In this section we indicate these differences. Further details, [TABULAR DATA FOR TABLE 5 OMITTED] along with numerical examples, can be found in Hussain [12].

5.1. Cold standby

Under cold standby redundancy conditions component 1 is in an operational state and component 2 is in an idle state when the module is put into use. When component 1 fails, component 2 is switched on. We assume a perfect (or reliable) switch so that the time to module failure is the sum of the individual failure times for the two components of the module.

Option 3.

In this case

[G.sub.3](x) = [G.sub.1](x) * [G.sub.1](x), (26)

where [G.sub.1](x) is given by (2) and * is the convolution operator. As a result, [G.sub.3](x) can be written in the form given in (11) with H(x) being given by

H(x) = {[(1 - p).sup.2]F(x) * F(x) + 2p(1 - p)F(x)}/(1 - [p.sup.2]). (27)

Also

[C.sub.m3] = (2 + [Phi])[C.sub.m], (28)

since we have the additional cost of the switch.

Option 4.

In this case

[G.sub.4](x) = F(x) * F(x), (29)

and

[C.sub.m4] = (2[C.sub.m2] + [Phi][C.sub.m]), (30)

with [C.sub.m2] being given by (9).

Option 5. In this case

[G.sub.5](x) = {[(1 - p).sup.2]F(x) * F(x) + 2p(1 - p)F(x)}/(1 - [p.sup.2]), (31)

and

[C.sub.m5] = (2[C.sub.m] + [Phi][C.sub.m] + [C.sub.T] + [p.sup.2][C.sub.S])/(1 - [p.sup.2]). (32)

5.2. Warm standby

Under warm standby redundancy conditions component 1 is in operational (fully energized) mode and component 2 is in partially energized mode when the module is put into use. The deterioration in the partially energized mode is slower than that in the fully energized mode and, as a result, the mean time to failure in the partially energized mode is larger than that in the fully energized mode. Let L(x) denote the distribution function for component failure in the partially energized mode.

If component 1 fails before component 2, then component 2 is switched on after failure. If [X.sub.1] is the age of component 1 at failure then the effective age of component 2 at the instant of switching is given by [Mathematical Expression Omitted] and is related to [X.sub.1] as follows:

[Mathematical Expression Omitted]. (33)

The module failure occurs when component 2 fails. Note that the failure time for component 2 is dependent on [X.sub.1]. If component 2 fails before component 1, then module failure occurs with the failure of component 1.

The failure distribution for module failure is more complex than for other forms of standby.

Option 3.

In this case, [G.sub.3](x) is given by (11) with H(x) being given by

[Mathematical Expression Omitted] (34)

and [C.sub.m3] is given by (28).

Option 4.

In this case,

[Mathematical Expression Omitted] (35)

and [C.sub.m4] is given by (30).

Option 5.

In this case

[Mathematical Expression Omitted], (36) and [C.sub.m5] is given by (32).

6. Conclusion and extension

In this paper we have considered the case of products that have an uncertain quality that are sold with a warranty. We have discussed the use of redundancy and/or quality control techniques to minimize the expected total (manufacturing + warranty) cost in the case where products are sold with a FRW policy and the redundancy involves a single replication. The model studied in this paper can be extended in several ways which include:

(i) It is a simple extension to obtain the optimal redundancy and quality control strategies when the product is sold with a Pro-Rata Warranty policy (PRW) or combination warranty policies involving FRW and PRW at different stages of the warranty period.

(ii) Another extension is to examine the case where the redundancy involves more than one replicate. In this case, each module consists of K ([greater than] 2) components. The analysis method in this case is similar to that for K = 2, studied in this paper. However, the expressions for [G.sub.i](x), 1 [less than or equal to] i [less than or equal to] 5, are more complex.

(iii) Another useful model is that nonconforming components are operational when put into use but have inferior reliability characteristic (e.g., a smaller mean time to failure). The authors are currently studying this problem and some preliminary results can be found in Hussain and Murthy [14].

(iv) As mentioned in the Introduction, product reliability can be improved through product development. The authors have developed several models for determining the optimal development strategies that take into account the warranty and the uncertainty in the outcome of the development. These will be reported elsewhere.

References

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[2] Murthy, D.N.P. and Blischke, W.R. (1992) Product warranty management-II: an integrated framework for study. European Journal of Operations Research, 62, 261-281.

[3] Murthy, D.N.P. and Blischke, W.R. (1992) Product warranty management-III: a review of mathematical models. European Journal of Operations Research, 63, 1-34.

[4] Blischke, W.R. and Murthy, D.N.P. (eds.) (1996) Product Warranty Handbook, Marcel Dekker New York.

[5] Blischke, W.R. and Murthy, D.N.P. (1993) Warranty Cost Analysis, Marcel Dekker, New York.

[6] Murthy, D.N.P. and Djamaludin, I. (1990) Quality control in single stage production system: open and closed loop policies. International Journal of Production Research, 28, 2219-2242. [7] Murthy, D.N.P., Wilson, R.J. and Djamaludin, I. (1993) Product warranty and quality control. Quality and Reliability Engineering International, 9, 431-443.

[8] Djamaludin, I. (1993) Quality control schemes for items sold with warranty. PhD thesis, The University of Queensland, Queensland, Australia.

[9] Djamaludin, I., Wilson, R.J. and Murthy, D.N.P. (1995) Economic lotsizing and inspection strategy for items sold with warranty. Mathematical and Computer Modelling, 22(10-12), 35-44.

[10] Murthy, D.N.P. and Hussain, A.Z.M.O. (1995) Warranty and optimal redundancy design. Engineering Optimization, 23, 301314.

[11] Murthy, D.N.P. and Nguyen, D.G. (1987) Optimal development testing policies for products sold under warranty. Reliability Engineering, 19, 113-123.

[12] Hussain, A.Z.M.O. (1997) Warranty and product reliability. PhD thesis, The University of Queensland, Brisbane, Australia.

[13] Stoyan, D. (1983) Comparison Methods for Queues and Other Stochastic Models, John Wiley and Sons, New York, p. 30.

[14] Hussain, A.Z.M.O. and Murthy, D.N.P. (1996) Warranty and optimal redundancy design with uncertain quality, in Stochastic Models in Engineering, Technology and Management, Wilson, R.J., Murthy, D.N.P. and Osaki, S. (eds.)., The University of Queensland, Brisbane, Australia, pp. 226-235.

Appendix

We derive the expected value of [N.sub.1](t) conditional on, [X.sub.1], the time to failure of the first module. Then from simple probabilistic arguments, we have

[Mathematical Expression Omitted].

On removing the conditioning, and noting that [M.sub.1](t) = E[[N.sub.1](t)], we have

[M.sub.1](t) = p[1 + [M.sub.1](t)] + (1 - p) [integral of] [1 + [M.sub.1](t - x)] dF(x) between limits T and 0,

which can be rewritten as

[M.sub.1](t) = p/(1 - p) + F(t) + [integral of] [M.sub.1](t - x)dF(x) between limits T and 0.

On taking the Laplace transform, we have

[Mathematical Expression Omitted],

where [Mathematical Expression Omitted] and [Mathematical Expression Omitted] are the Laplace transforms of [M.sub.1](t) and f(t) respectively. This can be rewritten, after some simple manipulation, as follows:

[Mathematical Expression Omitted]. On taking the Laplace inverse transformation, we have

[M.sub.1](t) = [p + [M.sub.F](t)]/(1 - p)

where

[M.sub.F](t) = F(t) + [integral of] [M.sub.F](t - x)dF(x) between limits t and 0.

Biographies

A.Z.M. Obaid Hussain obtained a B.Sc. degree in Mechanical Engineering from the Bangladesh University of Engineering and Technology, an M.S. degree in Management from the Arthur D'Little Management Education Institute, Massachusetts, USA and a Ph.D. degree from the Department of Mechanical Engineering at the University of Queensland, Australia. He is currently a research scientist/engineer with the CSIRO Division of Mathematical and Information Sciences based in Melbourne. His research interests include reliability and quality engineering and warranty management. He has extensive experience in the industrial utilization of natural gas and in the management of the construction of pipelines and related facilities.

D.N.P. Murthy obtained a B.E. degree in Electrical Engineering from Jabalpur University and an M.E. degree in Applied Electronics from the Indian Institute of Science in India. He also obtained M.S. and Ph.D. (Applied Mathematics) degrees from Harvard University. He is currently Professor of Engineering and Operations Management in the Department of Mechanical Engineering at the University of Queensland. He has held visiting academic appointments at several universities in the USA and India. His current research interest include various aspects of technology management (new product development, R&D management, strategic management of technology), operations management (lot sizing, quality, maintenance) and post-sale support (warranties, service contracts). He has authored or coauthored over 110 journal papers and 90 conference papers. He is the coauthor of two books (Mathematical Modelling, and Warranty Cost Analysis) and the coeditor of the Product Warranty Handbook. Currently, he is writing a book on reliability theory to be published by John Wiley in 1998. He is a member of several professional societies and is on the Editorial Board of five international journals. He has run several courses and consulted for a variety of private and public sector organisations on reliability, quality, warranty, maintenance, manufacturing, strategic management and technology management.