# Maintenance Optimization of a 2-Component Swappable Series System Using the Delay-Time Concept

Authors: Liying wang; Wenhua Zhang
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

In the 2-component swappable series system, the two components undertake tasks with different loads and degrade at different speeds. To prolong the lifetime of the series system, these two components are swapped in the operating process of the system in practice. This is common in the maintenance of duplexing steelmaking systems, tires of vehicles, and steel rails in curves. The failure process of each component in the system is modeled based on a two-stage delay-time concept and divided into two stages: normal and defective. Inspections are carried out periodically on the system. Two components may be swapped once at an inspection time that the two components are both in the normal stage. Due to the increase or decrease of loads, normal and defective time distributions after the swap are assumed to be different from those prior to the swap. The system is subjected to failure, inspection, and age-based renewals. The number of inspections over the maximum usage time of the system and the swap time are optimized jointly by minimizing the expected cost per unit time in a long run. A numerical example is presented to demonstrate the model.


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## Body

## 1. Introduction

The motivation of the paper comes from the operating process of the duplexing steelmaking system and maintenance of vehicles and steel rails in curves. The duplexing steelmaking system usually consists of two converters. One converter is for dephosphorization, and the other is for decarburization. Since each converter can fail mainly due to one failure mode, we treat them as components. Due to the heavier load, the decarburization converter degrades more quickly than the dephosphorization one. To prolong the system’s lifetime, the decarburization converter will be used for dephosphorization at a later stage of the furnace life; that is, the two converters are swapped. The swap time is usually determined by engineers according to their experiences. The swap can also be found in the process of vehicle maintenance. For a front drive vehicle, the tire wear of front and rear wheels is different. To keep the safety, the handling, and the stability, they are swapped periodically according to the handbook of vehicle maintenance. The swap number and times are also mapped out by original equipment manufacturers. In a curve, the outside steel rails degrade more quickly than the inside ones; they are also swapped periodically in the maintenance of railway lines. Some literatures studied maintenance policies with the swap. Taking advantage of the different degradation rates of batteries, Almuhtady et al. [1] introduced a policy incorporating implementation of swapping and substitution actions. Fu et al. [2] studied a new maintenance policy, which performs periodic preventive system replacements and periodic preventive component reallocations between system replacements as well as minimal repairs for emergency failures. Sun et al. [3] quantified the benefit of incorporating the reallocation into the condition-based maintenance.The delay-time concept was proposed by Christer [4]. The failure process in the delay-time concept is divided into two/three stages, named normal and defective/normal, minor defective and major defective (Christer and Waller [5]; Wang [6]; Wang [7]; Wang et al. [8]; and Yang [9]). The defective stages are not self-announcing and they can be identified by inspections. When the component is detected to be in a defect state at the detection time, the component will be repaired. Delay-time-based maintenance models can capture the relationship between the number of failures and inspection intervals, so they are widely used in describing the degradation of plants and optimizing of inspection intervals (Baker and Wang [10]; Baker and Christer [11]; Jia and Christer [12]; Wang [13]; Aven and Castro [14]; Jones et al. [15]; and Scarf et al. [16]). Single-component and multicomponent models have been proposed in the literature (Wang et al. [17]; Lu et al. [18]; and Liu et al. [19]). An adjustable inspection and replacement strategy for systems subject to 0-1 logic failure and two-stage failure process was proposed by Ma et al. [20]. Peng et al. [21] constructed a single-unit system subject to two types of failures: a traditional catastrophic failure and a two-stage delayed failure. Maintenance strategy considering imperfect inspections is discussed in their paper. Yang et al. [22] considered the optimization of a preventive maintenance policy based on dependent two-stage deterioration and external shocks. For systems experiencing internal failures and external shocks, the effects of external shocks were discussed as two stages of random virtual age increment, and a mission abort strategy was proposed (Qiu and Cui [23]). Zhao et al. [24] investigated the joint optimization of inspection and condition-based mission abort policies for systems subject to continuous degradation. A new model whose degradation was described by dependent two-stage failure process with competing failures was built by Qiu and Cui [25]. Heydari [26] gave a periodic inspection policy for products with extended warranty period-based two-stage failure process. Based on three-stage failure process, Wang et al. [27] proposed a condition-based maintenance policy for products sold with renewal warranty contract.Due to the advantage mentioned above, we built the 2-component swappable series system using the delay-time concept in this paper. The new contributions of this paper are as follows. By using the concept of delay time, the failure process of each component is divided into normal and defect stages, which can be detected by inspections. The two components can be swapped once at a detection time. Inspection renewals are adopted at detection times depending on the states of the two components. So, inspection and swapping are incorporated into maintenance in this paper. By jointly optimizing the number of inspections and swap time, long-term expected cost per unit time is minimized. It is worth noting that the cost of swapping components is usually much lower than the cost of replacing them, so when the workload of each component is unbalanced, swap with replacement can further reduce the cost of running the system.The remaining part of the paper is organized as follows. Section2 provides the modeling assumptions and notations. Section 3 establishes the probabilities of failure renewal, inspection renewal, and the overhaul. Section 4 models the long run expected cost per unit time. Section 5 presents a numerical example. Section 6 concludes the paper and suggests some future works.


## 2. Modeling Assumptions and Notations

We consider a series system with two components, namely, components 1 and 2.(1)
The two components undertake the same tasks but with different loads. So, they degrade at different speeds, and their lifetimes are statistically different. The failure process of each component is divided into two stages, normal and defective stages. The distributions in different stages and for different components are statistically independent.(2)
The maximum usage time of the system isTage. The system is subjected to an overhaul (replace both components) when it reaches Tage; no failure occurs and no defect is found before Tage. Once a component fails, both components are repaired or replaced, which can be regarded as the failure renewal of the system. Note that both components are replaced at Tage. While on a failure renewal, a component fails and the other component may be in normal or defective states; it can be restored at a lower cost. Furthermore, the usage times of the two components on a failure renewal are shorter than those at Tage; they can also be restored at a lower cost. So, we assume that the cost of an overhaul is greater than that of a failure renewal.(3)
The system is subjected to equispaced inspections in0,Tage. M is the number of inspections in (0, Tage), and it is a decision variable. Let T=Tage/M+1. Then, T is the inspection interval and both components of the system are inspected at times kTk=1,2,…,M. Inspections are perfect. The downtime due to inspections is negligible compared with the inspection intervals or inspections can be performed online, so there is no downtime. However, the costs of inspections are considered.(4)
The two components are swapped once at timeτ=m′Tm′∈1,2,…,M provided the system has not failed before τ and the no defect is found before τ. Due to the variety of the load, the hazard function of normal time distributions and the delay time distribution will be different from those prior to the swap. Assume that extra cost is incurred for the swap.(5)
If one or two components are found to be in the defective state by an inspection, the two components are repaired or replaced, which can be regarded as the inspection renewal of the system.Assumptions (1) and (3) can be justified in previous delay time modeling applications (Wang [6]; Wang et al. [8]). According to assumption (2), Tage in this paper is the maximum usage time of the system. It is fixed, instead of a decision variable. The simultaneous optimization of M,T,τ in the case that Tage is assumed to be the threshold of age-based renewal will be conducted in the future.To reduce the cost of system startup and shutdown, we assume that the swap is carried out at inspection time in assumption (4). This is a common case in practice. Furthermore, due to the swap, the component that takes lighter loads prior to the swap will take heavier ones after the swap, so it will degrade more quickly than itself in the case that swap is not carried out. On the contrary, the component will degrade more slowly than another component in the case that swap is not carried out. So, the assumptions about hazard functions in assumption (4) are reasonable. As stated in assumption (4), the two components are swapped only once in this paper; more than one swap may happen in practice. This will be left to a separate paper.The explanations of assumption (5) are as follows. If one component is found to be in the defective state or fails, there could be an opportunity to check another component and conduct repair or replacement. This assumption can be relaxed with more mathematics involved, but now we keep in for simplicity.We will use the following notations in subsequent modeling:Xi is the random time of the normal stage of component ii=1,2λit is the hazard function of component ii=1,2 in the normal stage prior to the swapλ˜it is the hazard function of component ii=1,2 in the normal stage after the swapFit is the cumulative distribution function (cdf) of Xii=1,2fit is the probability density function (pdf) of Xii=1,2Yi is the delay time of component i prior to the swapgit is the pdf of Yii=1,2Git is the cdf of Yii=1,2Y˜i is the delay time of component ii=1,2 after the swapg˜it is the pdf of Y˜ii=1,2.G˜it is the cdf of Y˜ii=1,2T is the interval of inspectionsM is the number of inspectionsTage is the maximum usage time of the systemτ is the time of the swapTf is the random time that the system fails in the case that the two components are not swappedT1d is the random time that the system is renewed due to the defect of component 1 or component 2 in the case that the two components are not swappedT2d is the random time that the system is renewed due to the defects of components 1 and 2 in the case that the two components are not swappedTsf is the random time that the system fails in the case that the two components are swappedTs1d is the random time that the system is renewed due to the defect of component 1 or 2 in the case that the two components are swappedTs2d is the random time that the system is renewed due to the defect of component 1 and 2 in the case that the two components are swappedZ is the failure renewal timefmz is the probability density function of the time that Z exceeds the last inspection in the case that Tf∈m−1T,mT (before the swap, z∈0,T)f˜mz is the probability density function of the time that Z exceeds the last inspection in the case that Tf∈m−1T,mT (after the swap, z∈0,T)Ci is the average cost per inspectionCf is the average cost per failure renewalC1d is the average cost per inspection renewal in the case that one component of the system is identified to be in defective stateC2d is the average cost per inspection renewal in the case that two components of the system are identified to be in defective stateCs is the average cost per swapCa is the average cost of the overhaulAccording to the assumptions and notations above, the survival function of componentii=1,2 in the normal stage:(1)1−Fit=exp−∫0tλiudu,t≤τ,exp−∫0τλiudu−∫τtλ˜iudu,t>τ.So, the cdf ofXii=1,2, Fit, can be determined by λiti=1,2 and λ˜it. From equation (1), the pdf of Xii=1,2(2)fit=λitexp−∫0tλiudu,t≤τ,λ˜itexp−∫0τλiudu−∫τtλ˜iudu,t>τ.The decision variables in this paper are the number of inspectionsM and the swap time τ. The objective is to minimize the long run expected cost per unit time.


## 3. Probabilistic Models for Failure, Defect Stage Identifications

We use the long run expected cost per unit time as a measure to optimize the number of inspectionsM and the swap time τ. In order to derive the long run expected cost per unit time, the renewal probabilities prior to and after the swap need to be formulated. There are four mutually exclusive scenarios at the end of a renewal cycle, that is, a failure renewal, an overhaul, inspection renewals due to one or two components’ defects.Figure 1
The system fails before any defective stage of the two components is identified by the inspection.∘ denotes defective stage arrives; • denotes failure.Figure 2
Illustration of a renewal due to the defect of one component. (a) Component 1 is defective; component 2 is normal. (b) Component 2 is defective; component 1 is normal.
(a)(b)Figure 3
Illustration of a renewal due to defects of two components.


### 3.1. Probability of a Renewal Prior to the Swap

(1)
The system fails atTfTf∈mT,m+1T,m+1T≤τ. Since the system considered is a series one, the system will fail once one of the components fails. It leads from assumptions (3) and (5) that the defect of the failed component must have started at some time uu∈mT,m+1T, the delay time is less than the interval m+1T−u, and another component is in normal or defective stage at the same time, as shown in Figure 1.Therefore, the probability of such a failure renewal is given as(3)PmT<Tf<m+1T=PmT<X1<m+1T,0<Y1<m+1T−X1,X2>X1+Y1+PmT<X1<m+1T,0<Y1<m+1T−X1,mT<X2<X1+Y1,Y2>X1+Y1−X2+PmT<X2<m+1T,0<Y2<m+1T−X2,X1>X2+Y2+PmT<X2<m+1T,0<Y2<m+1T−X2,mT<X1<X2+Y2,Y1>X2+Y2−X1=∫mTm+1Tf1u∫0m+1T−ug1v1−F2u+v+∫mTu+vf2x∫u+v−x∞g2ydydxdvdu+∫mTm+1Tf2u∫0m+1T−ug2v1−F1u+v+∫mTu+vf1x∫u+v−x∞g1ydydxdvdu,0≤m≤m′.Note that the failure renewal occurs prior to the swap, and the swap timeτ=m′T, so 0≤m≤m′. The first term in equation (3) corresponds to the probability that Component 1 fails and the second one is that of Component 2.Similarly, the probability of failure renewal withinTm,Tm+z:(4)PmT<Tf<mT+z=∫mTmT+zf1u∫0mT+z−ug1v1−F2u+v+∫mTu+vf2x1−G2u+v−xdxdvdu+∫mTmT+zf2u∫0mT+z−ug2v1−F1u+v+∫mTu+vf1x1−G1u+v−xdxdvdu,0≤m≤m′.The failure renewal time,Z, is the time from a new system’s operating to its first failure before the maximum usage time of the system. Its probability density function is a segment one on intervals m−1T,mT1≤m≤M+1. Let fmz be the probability density function of the time that Z exceeds m−1T in the scenario that the system fails at Tf, Tf∈m−1T,mT1≤m≤M+1, before which no defective stage has been identified by inspections. Then, by differentiating equation (4) with respect to z, we have(5)fmz=∫mTmT+zf1u1−F2mT+z+∫mTmT+zf2x1−G2mT+z−xdxg1mT+z−udu+∫mTmT+zf2x1−F1mT+z+∫mTmT+zf1u1−G1mT+z−udug2mT+z−xdx,z∈0,T,0≤m≤m′.The same derivation for the pdf will be used subsequently and it is omitted in the following for simplicity.(2)
The component is renewed atTm+1Tm+1≤τ due to the defect of component 1 or 2. According to assumptions (3) and (4), in this scenario, one component’s defect arrives at time uu∈Tm,Tm+1, its defective stage must be longer than Tm+1−u and another component’s duration time in the normal stage is longer than Tm+1, as shown in Figure 2. So, the probability is(6)PT1d=m+1T=PmT<X1<m+1T,Y1>m+1T−X1,X2>m+1T+PmT<X2<m+1T,Y2>m+1T−X2,X1>m+1T=1−F2m+1T∫mTm+1Tf1u1−G1m+1T−udu+1−F1m+1T∫mTm+1Tf2x1−G2m+1T−xdx,0≤m≤m′−1.(3)
The component is renewed atm+1Tm+1T<τ due to the defects of components 1 and 2. In this scenarios, the two components’ defects occur between mT and m+1T, and their defective stages must skip over m+1T; see Figure 3.Then, we have the probability of such an event, according to the independency of the two components,(7)PT2d=m+1T=PmT<X1<m+1T,Y1>m+1T−X1,mT<X2<m+1T,Y2>m+1T−X2=∫mTm+1Tf1u1−G1m+1T−udu∫mTm+1Tf2x1−G2m+1T−xdx,0≤m≤m′−1.


### 3.2. Probability of a Renewal after the Swap

Using a similar derivation to that in Section3.1, we can obtain the following renewal probabilities after the swap. From assumption (4) and notations in Section 2, after the swap, the pdf and cdf of the duration time in defective stage for component ii=1,2 are g˜iu and G˜iu, respectively. Furthermore, the lifetime, Tage, can be represented as M+1T.(1)
The system fails atTfTf∈mT,m+1T,mT≥τ. In this scenario, the duration times in the normal stage of the two components are longer than τ, as shown in Figure 4.So, the pdf of failure atTf, Tf∈mT,m+1T,z∈0,T,mT≥τ is given as(8)f˜mz=limΔz⟶0PTf∈mT+z,mT+z+dzΔz=∫mTmT+zf1u1−F2mT+z+∫mTmT+zf2x1−G˜2mT+z−xdxg˜1mT+z−udu+∫mTmT+zf2x1−F1mT+z+∫mTmT+zf1u1−G˜1mT+z−udug˜2mT+z−xdx,m′≤m≤M.Similarly, the probability the system fails atTsfTsf∈mT,m+1T,mT≥τ is(9)PmT<Tsf<m+1T=∫mTm+1Tf1u∫0m+1T−ug˜1v1−F2u+v+∫mTu+vf2x∫u+v−x∞g˜2ydydxdvdu+∫mTm+1Tf2x∫0m+1T−xg˜2y1−F1x+y+∫mTx+yf1u∫x+y−u∞g˜1vdvdudydx,m′≤m≤M.(2)
The component is renewed atm+1Tm+1T>τ due to the defect of component 1 or 2; see Figure 5.The probability of such an event is(10)PTs1d=m+1T=1−F2m+1T∫mTm+1Tf1u1−G˜1m+1T−udu+1−F1m+1T∫mTm+1Tf2x1−G˜2m+1T−xdx,m′≤m≤M−1.(3)
The system is renewed atTm+1Tm+1>τ due to the defects of component 1 and 2; see Figure 6.The renewal probability in this scenario is(11)PTs2d=m+1T=PTm<X1<m+1T,Y1>m+1T−X1,Tm<X2<m+1T,Y2>m+1T−X2=∫mTm+1Tf1u1−G˜1m+1T−udu∫mTm+1Tf2x1−G˜2m+1T−xdx,m′≤m≤M−1.(4)
An overhaul is performed atTage. According to the assumptions and notations in Section 2, when both components being normal at TM but not failed at Tage, an overhaul will be performed at Tage; see Figure 7.Figure 4
Illustration of the system failure after the swap.So, the overhaul probability is(12)Prenewal atTage=PX1>Tage,X2>Tage+PMT<X1<Tage,Y˜1>Tage−X1,X2>Tage+PMT<X2<Tage,Y˜2>Tage−X2,X1>Tage+PMT<X1<Tage,Y˜1>Tage−X1,MT<X2<Tage,Y˜2>Tage−X2=1−F1Tage1−F2Tage+1−F2Tage∫MTTagef1u1−G˜1Tage−udu+1−F1Tage∫MTTagef2x1−G˜2Tage−xdx+∫MTTagef1u1−G˜1Tage−udu∫MTTagef2x1−G˜2Tage−xdx.Figure 5
Illustration of a renewal after swap due to the defect of one component. (a) Component 1 is defective; component 2 is normal. (b) Component 2 is defective; component 1 is normal.
(a)(b)Figure 6
Illustration of a renewal after swap due to the defects of two components.Figure 7
Illustration of an age renewal.
(a)(b)(c)(d)


## 3.1. Probability of a Renewal Prior to the Swap

(1)
The system fails atTfTf∈mT,m+1T,m+1T≤τ. Since the system considered is a series one, the system will fail once one of the components fails. It leads from assumptions (3) and (5) that the defect of the failed component must have started at some time uu∈mT,m+1T, the delay time is less than the interval m+1T−u, and another component is in normal or defective stage at the same time, as shown in Figure 1.Therefore, the probability of such a failure renewal is given as(3)PmT<Tf<m+1T=PmT<X1<m+1T,0<Y1<m+1T−X1,X2>X1+Y1+PmT<X1<m+1T,0<Y1<m+1T−X1,mT<X2<X1+Y1,Y2>X1+Y1−X2+PmT<X2<m+1T,0<Y2<m+1T−X2,X1>X2+Y2+PmT<X2<m+1T,0<Y2<m+1T−X2,mT<X1<X2+Y2,Y1>X2+Y2−X1=∫mTm+1Tf1u∫0m+1T−ug1v1−F2u+v+∫mTu+vf2x∫u+v−x∞g2ydydxdvdu+∫mTm+1Tf2u∫0m+1T−ug2v1−F1u+v+∫mTu+vf1x∫u+v−x∞g1ydydxdvdu,0≤m≤m′.Note that the failure renewal occurs prior to the swap, and the swap timeτ=m′T, so 0≤m≤m′. The first term in equation (3) corresponds to the probability that Component 1 fails and the second one is that of Component 2.Similarly, the probability of failure renewal withinTm,Tm+z:(4)PmT<Tf<mT+z=∫mTmT+zf1u∫0mT+z−ug1v1−F2u+v+∫mTu+vf2x1−G2u+v−xdxdvdu+∫mTmT+zf2u∫0mT+z−ug2v1−F1u+v+∫mTu+vf1x1−G1u+v−xdxdvdu,0≤m≤m′.The failure renewal time,Z, is the time from a new system’s operating to its first failure before the maximum usage time of the system. Its probability density function is a segment one on intervals m−1T,mT1≤m≤M+1. Let fmz be the probability density function of the time that Z exceeds m−1T in the scenario that the system fails at Tf, Tf∈m−1T,mT1≤m≤M+1, before which no defective stage has been identified by inspections. Then, by differentiating equation (4) with respect to z, we have(5)fmz=∫mTmT+zf1u1−F2mT+z+∫mTmT+zf2x1−G2mT+z−xdxg1mT+z−udu+∫mTmT+zf2x1−F1mT+z+∫mTmT+zf1u1−G1mT+z−udug2mT+z−xdx,z∈0,T,0≤m≤m′.The same derivation for the pdf will be used subsequently and it is omitted in the following for simplicity.(2)
The component is renewed atTm+1Tm+1≤τ due to the defect of component 1 or 2. According to assumptions (3) and (4), in this scenario, one component’s defect arrives at time uu∈Tm,Tm+1, its defective stage must be longer than Tm+1−u and another component’s duration time in the normal stage is longer than Tm+1, as shown in Figure 2. So, the probability is(6)PT1d=m+1T=PmT<X1<m+1T,Y1>m+1T−X1,X2>m+1T+PmT<X2<m+1T,Y2>m+1T−X2,X1>m+1T=1−F2m+1T∫mTm+1Tf1u1−G1m+1T−udu+1−F1m+1T∫mTm+1Tf2x1−G2m+1T−xdx,0≤m≤m′−1.(3)
The component is renewed atm+1Tm+1T<τ due to the defects of components 1 and 2. In this scenarios, the two components’ defects occur between mT and m+1T, and their defective stages must skip over m+1T; see Figure 3.Then, we have the probability of such an event, according to the independency of the two components,(7)PT2d=m+1T=PmT<X1<m+1T,Y1>m+1T−X1,mT<X2<m+1T,Y2>m+1T−X2=∫mTm+1Tf1u1−G1m+1T−udu∫mTm+1Tf2x1−G2m+1T−xdx,0≤m≤m′−1.


## 3.2. Probability of a Renewal after the Swap

Using a similar derivation to that in Section3.1, we can obtain the following renewal probabilities after the swap. From assumption (4) and notations in Section 2, after the swap, the pdf and cdf of the duration time in defective stage for component ii=1,2 are g˜iu and G˜iu, respectively. Furthermore, the lifetime, Tage, can be represented as M+1T.(1)
The system fails atTfTf∈mT,m+1T,mT≥τ. In this scenario, the duration times in the normal stage of the two components are longer than τ, as shown in Figure 4.So, the pdf of failure atTf, Tf∈mT,m+1T,z∈0,T,mT≥τ is given as(8)f˜mz=limΔz⟶0PTf∈mT+z,mT+z+dzΔz=∫mTmT+zf1u1−F2mT+z+∫mTmT+zf2x1−G˜2mT+z−xdxg˜1mT+z−udu+∫mTmT+zf2x1−F1mT+z+∫mTmT+zf1u1−G˜1mT+z−udug˜2mT+z−xdx,m′≤m≤M.Similarly, the probability the system fails atTsfTsf∈mT,m+1T,mT≥τ is(9)PmT<Tsf<m+1T=∫mTm+1Tf1u∫0m+1T−ug˜1v1−F2u+v+∫mTu+vf2x∫u+v−x∞g˜2ydydxdvdu+∫mTm+1Tf2x∫0m+1T−xg˜2y1−F1x+y+∫mTx+yf1u∫x+y−u∞g˜1vdvdudydx,m′≤m≤M.(2)
The component is renewed atm+1Tm+1T>τ due to the defect of component 1 or 2; see Figure 5.The probability of such an event is(10)PTs1d=m+1T=1−F2m+1T∫mTm+1Tf1u1−G˜1m+1T−udu+1−F1m+1T∫mTm+1Tf2x1−G˜2m+1T−xdx,m′≤m≤M−1.(3)
The system is renewed atTm+1Tm+1>τ due to the defects of component 1 and 2; see Figure 6.The renewal probability in this scenario is(11)PTs2d=m+1T=PTm<X1<m+1T,Y1>m+1T−X1,Tm<X2<m+1T,Y2>m+1T−X2=∫mTm+1Tf1u1−G˜1m+1T−udu∫mTm+1Tf2x1−G˜2m+1T−xdx,m′≤m≤M−1.(4)
An overhaul is performed atTage. According to the assumptions and notations in Section 2, when both components being normal at TM but not failed at Tage, an overhaul will be performed at Tage; see Figure 7.Figure 4
Illustration of the system failure after the swap.So, the overhaul probability is(12)Prenewal atTage=PX1>Tage,X2>Tage+PMT<X1<Tage,Y˜1>Tage−X1,X2>Tage+PMT<X2<Tage,Y˜2>Tage−X2,X1>Tage+PMT<X1<Tage,Y˜1>Tage−X1,MT<X2<Tage,Y˜2>Tage−X2=1−F1Tage1−F2Tage+1−F2Tage∫MTTagef1u1−G˜1Tage−udu+1−F1Tage∫MTTagef2x1−G˜2Tage−xdx+∫MTTagef1u1−G˜1Tage−udu∫MTTagef2x1−G˜2Tage−xdx.Figure 5
Illustration of a renewal after swap due to the defect of one component. (a) Component 1 is defective; component 2 is normal. (b) Component 2 is defective; component 1 is normal.
(a)(b)Figure 6
Illustration of a renewal after swap due to the defects of two components.Figure 7
Illustration of an age renewal.
(a)(b)(c)(d)


## 4. Cost Models

### 4.1. The Expected Renewal Cycle Cost

Based on overhaul and different renewal probabilities formulated earlier and the costs due to inspections and replacements, we can obtain the expected renewal cycle cost.The expected cost caused by a failure renewalECfail,M,τ is given by(13)ECfail,M,τ=∑m=0m′−1mCi+CfPmT<Tf<m+1T+∑m=m′MmCi+CfPmT<Tsf<m+1T.Using equations (4), (5), (9), and (10), the expected cost of an inspection renewal ECinsp,M,τ is given by(14)ECinsp,M,τ=∑m=0m′−1m+1CiPT1d=m+1T+PT2d=m+1T+C1dPT1d=m+1T+C2dPT2d=m+1T+∑m=m′M−1m+1CiPTs1d=m+1T+PTs2d=m+1T+C1dPTs1d=m+1T+C2dPTs2d=m+1T.The expected cost of an overhaul is(15)ECage,M,τ=MCi+CaPrenewal atTage.The expected cost of a swap is(16)ECswap,M,τ=Cs∑m=m′M−1PTs1d=m+1T+PTs2d=m+1T+Cs∑m=m′MPmT<Tsf<m+1T+CsPrenewalatTage.


### 4.2. The Expected Renewal Cycle Length

From the pdfs of a failure renewal, the expected length of a failure renewalELfail is given by(17)ELfail,M,τ=∑m=0m′−1∫0TmT+zfmzdz+∑m=m′M∫0TmT+zf˜mzdz.The expected length of an inspection renewal is given as(18)ELinsp,M,τ=∑m=0m′−1m+1TPT1d=m+1T+PT2d=m+1T+∑m=m′M−1m+1TPTs1d=m+1T+PTs2d=m+1T.According to the expected cycle cost and length with different renewals, the long run expected cost per unit time can be given using the renewal reward theorem (Ross [28], as shown in equation (18)):(19)CM,τ=ECM,τELM,τ=ECfail,M,τ+ECinsp,M,τ+ECage,M,τ+ECswap,M,τELfail,M,τ+ELinsp,M,τ+TagePrenewal atTage,where ECM,τ is the expected cycle cost and ELM,τ is the expected renewal cycle length.We seek for the optimal inspection numberM and swap time τ which minimize equation (19).


## 4.1. The Expected Renewal Cycle Cost

Based on overhaul and different renewal probabilities formulated earlier and the costs due to inspections and replacements, we can obtain the expected renewal cycle cost.The expected cost caused by a failure renewalECfail,M,τ is given by(13)ECfail,M,τ=∑m=0m′−1mCi+CfPmT<Tf<m+1T+∑m=m′MmCi+CfPmT<Tsf<m+1T.Using equations (4), (5), (9), and (10), the expected cost of an inspection renewal ECinsp,M,τ is given by(14)ECinsp,M,τ=∑m=0m′−1m+1CiPT1d=m+1T+PT2d=m+1T+C1dPT1d=m+1T+C2dPT2d=m+1T+∑m=m′M−1m+1CiPTs1d=m+1T+PTs2d=m+1T+C1dPTs1d=m+1T+C2dPTs2d=m+1T.The expected cost of an overhaul is(15)ECage,M,τ=MCi+CaPrenewal atTage.The expected cost of a swap is(16)ECswap,M,τ=Cs∑m=m′M−1PTs1d=m+1T+PTs2d=m+1T+Cs∑m=m′MPmT<Tsf<m+1T+CsPrenewalatTage.


## 4.2. The Expected Renewal Cycle Length

From the pdfs of a failure renewal, the expected length of a failure renewalELfail is given by(17)ELfail,M,τ=∑m=0m′−1∫0TmT+zfmzdz+∑m=m′M∫0TmT+zf˜mzdz.The expected length of an inspection renewal is given as(18)ELinsp,M,τ=∑m=0m′−1m+1TPT1d=m+1T+PT2d=m+1T+∑m=m′M−1m+1TPTs1d=m+1T+PTs2d=m+1T.According to the expected cycle cost and length with different renewals, the long run expected cost per unit time can be given using the renewal reward theorem (Ross [28], as shown in equation (18)):(19)CM,τ=ECM,τELM,τ=ECfail,M,τ+ECinsp,M,τ+ECage,M,τ+ECswap,M,τELfail,M,τ+ELinsp,M,τ+TagePrenewal atTage,where ECM,τ is the expected cycle cost and ELM,τ is the expected renewal cycle length.We seek for the optimal inspection numberM and swap time τ which minimize equation (19).


## 5. A Numerical Example

### 5.1. Optimisation of the Number of Inspections and the Time of Swap

To solve equation (18), the hazard functions λit and λ˜iti=1,2, the pdfs of Yi and Y˜ii=1,2, git and g˜it, must be available. From the literature in reliability, the time to failure of a component is popularly described as the Weibull distribution. So, we assume(20)λ1t=β1α1tα1β1−1,λ˜1t=β˜1α˜1tα˜1β˜1−1,λ2t=μ1ν1tν1μ1−1,λ˜2t=μ˜1ν˜1tν˜1μ˜1−1,g1t=β2α2tα2β2−1e−t/α2β2,g˜1t=β˜2α˜2tα˜2β˜2−1e−t/α˜2β˜2,g2t=μ2ν2tν2μ2−1e−t/v2μ2,g˜2t=μ˜2v˜2tv˜2μ˜2−1e−t/v˜2μ˜2,where α1,α2, α˜1, α˜2, ν1, ν˜1, ν2, and v˜2are scale parameters and β1,β2, β˜1, β˜2, μ1, μ˜1, μ2, and μ˜2 are shape parameters and they are bigger than 0. The values of these parameters are listed in Tables 1 and 2 .Table 1
The parameters of hazard functions.
Component 1Component 2λ1tλ˜1tλ2tλ˜2tα1β1α˜1β˜1ν1μ1ν˜1μ˜1260.831.27260.831.32220.611.45220.611.3Table 2
The parameters of delay times.
Component 1Component 2g1tg˜1tg2tg˜2tα2β2α˜2β˜2ν2μ2ν2μ˜2220.41.5220.41.6160.021.7160.021.25It can be shown from Table1 that component 2 degrades more quickly than component 1 prior to the swap. While after the swap, the trend is the opposite. The other parameters of the system are listed in Table 3.Table 3
The other parameters of the system.
CiCfC1dC2dCsCaTage0.1120.60.8160360We ﬁrst use the model presented in Section3 based on the parameters in Tables 1 and 2. To analysis the effect of the component swap, we plot the figures of the maximal and minimal expected cost per unit time of equation (19) in terms of the inspection numbers in the cases that the swap is carried out and the expected cost per unit time in the case that the swap is not carried out, as shown in Figure 8. Note that for a given number of inspections M, the inspection interval T=360/M+1, and the time of swap τ may be m′Tm′=1,2,…,M; the maximal and minimal expected cost per unit time are thus different. The figures indicate the joint optimization of number of inspections and swap time can decrease the expected cost per unit time. Detailed comparison analysis will be discussed in Section 5.2.Figure 8
Illustration of the expected cost per unit time in terms of the number of inspections.To obtain the optimal decision variables, the minimal expected cost per unit time in terms of the inspection numbers is shown in Figure9. The figures indicate that the expected cost per unit time is minimized when the inspection number M∗=11 or M∗=12. So, the optimal inspection interval T∗=360/11+1=30 or T∗=360/12+1=27.69. Figure 10 is about the expected cost per unit time in terms of the swap time given T∗=30 or T∗=27.69. The figures indicate that when m′=5, the expected cost per unit time is minimized. Hence, the optimal time of swap τ=5T=150 or τ=5T=138.46 with the expected cost per unit time of 0.01776.Figure 9
Illustration of the minimal expected cost per unit time in terms of the number of inspections.Figure 10
Illustration of the expected cost per unit time in terms of the time of swap givenM=12.


### 5.2. Comparison Analysis

For comparison, the long run expected cost per unit time without swap is given below.According to model assumptions, the expected cost of a failure renewal in the case of no swapECfail,M is given by(21)ECfail,M=∑m=0MMCi+CfPmT<Tf<m+1T,and the expected cost of an inspection renewal ECinsp,M is given by(22)ECinsp,M=∑m=0Mm+1Ci+C1dPT1d=m+1T+∑m=0Mm+1Ci+C2dPT2d=m+1T.The expected cost of an overhaul in the case of no swap is(23)ECage,M=MCi+CaPrenewalatTage.The expected length of a failure renewal in the case of no swapELfail,M is given by(24)ELfail,M=∑m=0M∫0TmT+zfmzdz.The expected length of an inspection renewal in the case of no swap is given as(25)ELinsp,M=∑m=0Mm+1TPT1d=m+1T+PT2d=m+1T.Consequently, the long run expected cost per unit time can be given as(26)CM=ECMELM=ECfail,M+ECinsp,M+ECage,MELfail,M+ELinsp,M+TagePrenewalatTage.Substituting parameters into the aforementioned formulas, the long run expected cost per unit time for the model without swap can be given (see Table4). The table indicates that the optimal inspection number is 15 and the corresponding cost rate is 0.02593, which is greater than 0.01776, the cost rate in the case of swap.Table 4
Cost rates for various inspection numbers (without swap).
MELMECMCM1252.166213.92840.05522211.58009.31320.04403190.17087.13970.03754177.33395.96380.03365168.86035.25630.03116162.87284.79790.02957158.42594.48480.02838154.99634.26290.02759152.27254.10160.026910150.05813.98220.026511148.22283.89300.026312146.67723.82610.026113145.35783.77630.026014144.21863.73960.025915143.22523.71340.025916142.35113.69550.026017141.57633.68440.026018140.88473.67900.026119140.26363.67820.026220139.70273.68140.026421139.19373.68790.026522138.72983.69730.026723138.30523.70920.026824137.91523.72320.027025137.55563.73910.0272As the long run expected cost rate of the swap model increases with the increasing of the extra cost for the swap, the maintenance police proposed in this paper may have lost the economic efficiency when the extra cost for the swap is large enough. We will discuss this problem by a numerical method.LetCs change from 2 to 6 by step 1 and the other system parameters be the same as those in Section 5.1. The optimal strategies M∗,τ∗ and their corresponding expected cost rates CM∗,τ∗ are presented in Table 5.Table 5
Optimum policy for variousCs.
CsM∗τ∗CM∗,τ∗215157.50.02082315157.50.02326416169.40.02527516169.40.02672617200.00.02779The table indicates the cost rate forCs=4 is smaller than 0.025927, the cost rate for the model without swap. While for Cs=5, the cost rate is greater than 0.025927. So, the threshold for the cost of the swap so that no benefit is derived is between 4 and 5.


## 5.1. Optimisation of the Number of Inspections and the Time of Swap

To solve equation (18), the hazard functions λit and λ˜iti=1,2, the pdfs of Yi and Y˜ii=1,2, git and g˜it, must be available. From the literature in reliability, the time to failure of a component is popularly described as the Weibull distribution. So, we assume(20)λ1t=β1α1tα1β1−1,λ˜1t=β˜1α˜1tα˜1β˜1−1,λ2t=μ1ν1tν1μ1−1,λ˜2t=μ˜1ν˜1tν˜1μ˜1−1,g1t=β2α2tα2β2−1e−t/α2β2,g˜1t=β˜2α˜2tα˜2β˜2−1e−t/α˜2β˜2,g2t=μ2ν2tν2μ2−1e−t/v2μ2,g˜2t=μ˜2v˜2tv˜2μ˜2−1e−t/v˜2μ˜2,where α1,α2, α˜1, α˜2, ν1, ν˜1, ν2, and v˜2are scale parameters and β1,β2, β˜1, β˜2, μ1, μ˜1, μ2, and μ˜2 are shape parameters and they are bigger than 0. The values of these parameters are listed in Tables 1 and 2 .Table 1
The parameters of hazard functions.
Component 1Component 2λ1tλ˜1tλ2tλ˜2tα1β1α˜1β˜1ν1μ1ν˜1μ˜1260.831.27260.831.32220.611.45220.611.3Table 2
The parameters of delay times.
Component 1Component 2g1tg˜1tg2tg˜2tα2β2α˜2β˜2ν2μ2ν2μ˜2220.41.5220.41.6160.021.7160.021.25It can be shown from Table1 that component 2 degrades more quickly than component 1 prior to the swap. While after the swap, the trend is the opposite. The other parameters of the system are listed in Table 3.Table 3
The other parameters of the system.
CiCfC1dC2dCsCaTage0.1120.60.8160360We ﬁrst use the model presented in Section3 based on the parameters in Tables 1 and 2. To analysis the effect of the component swap, we plot the figures of the maximal and minimal expected cost per unit time of equation (19) in terms of the inspection numbers in the cases that the swap is carried out and the expected cost per unit time in the case that the swap is not carried out, as shown in Figure 8. Note that for a given number of inspections M, the inspection interval T=360/M+1, and the time of swap τ may be m′Tm′=1,2,…,M; the maximal and minimal expected cost per unit time are thus different. The figures indicate the joint optimization of number of inspections and swap time can decrease the expected cost per unit time. Detailed comparison analysis will be discussed in Section 5.2.Figure 8
Illustration of the expected cost per unit time in terms of the number of inspections.To obtain the optimal decision variables, the minimal expected cost per unit time in terms of the inspection numbers is shown in Figure9. The figures indicate that the expected cost per unit time is minimized when the inspection number M∗=11 or M∗=12. So, the optimal inspection interval T∗=360/11+1=30 or T∗=360/12+1=27.69. Figure 10 is about the expected cost per unit time in terms of the swap time given T∗=30 or T∗=27.69. The figures indicate that when m′=5, the expected cost per unit time is minimized. Hence, the optimal time of swap τ=5T=150 or τ=5T=138.46 with the expected cost per unit time of 0.01776.Figure 9
Illustration of the minimal expected cost per unit time in terms of the number of inspections.Figure 10
Illustration of the expected cost per unit time in terms of the time of swap givenM=12.


## 5.2. Comparison Analysis

For comparison, the long run expected cost per unit time without swap is given below.According to model assumptions, the expected cost of a failure renewal in the case of no swapECfail,M is given by(21)ECfail,M=∑m=0MMCi+CfPmT<Tf<m+1T,and the expected cost of an inspection renewal ECinsp,M is given by(22)ECinsp,M=∑m=0Mm+1Ci+C1dPT1d=m+1T+∑m=0Mm+1Ci+C2dPT2d=m+1T.The expected cost of an overhaul in the case of no swap is(23)ECage,M=MCi+CaPrenewalatTage.The expected length of a failure renewal in the case of no swapELfail,M is given by(24)ELfail,M=∑m=0M∫0TmT+zfmzdz.The expected length of an inspection renewal in the case of no swap is given as(25)ELinsp,M=∑m=0Mm+1TPT1d=m+1T+PT2d=m+1T.Consequently, the long run expected cost per unit time can be given as(26)CM=ECMELM=ECfail,M+ECinsp,M+ECage,MELfail,M+ELinsp,M+TagePrenewalatTage.Substituting parameters into the aforementioned formulas, the long run expected cost per unit time for the model without swap can be given (see Table4). The table indicates that the optimal inspection number is 15 and the corresponding cost rate is 0.02593, which is greater than 0.01776, the cost rate in the case of swap.Table 4
Cost rates for various inspection numbers (without swap).
MELMECMCM1252.166213.92840.05522211.58009.31320.04403190.17087.13970.03754177.33395.96380.03365168.86035.25630.03116162.87284.79790.02957158.42594.48480.02838154.99634.26290.02759152.27254.10160.026910150.05813.98220.026511148.22283.89300.026312146.67723.82610.026113145.35783.77630.026014144.21863.73960.025915143.22523.71340.025916142.35113.69550.026017141.57633.68440.026018140.88473.67900.026119140.26363.67820.026220139.70273.68140.026421139.19373.68790.026522138.72983.69730.026723138.30523.70920.026824137.91523.72320.027025137.55563.73910.0272As the long run expected cost rate of the swap model increases with the increasing of the extra cost for the swap, the maintenance police proposed in this paper may have lost the economic efficiency when the extra cost for the swap is large enough. We will discuss this problem by a numerical method.LetCs change from 2 to 6 by step 1 and the other system parameters be the same as those in Section 5.1. The optimal strategies M∗,τ∗ and their corresponding expected cost rates CM∗,τ∗ are presented in Table 5.Table 5
Optimum policy for variousCs.
CsM∗τ∗CM∗,τ∗215157.50.02082315157.50.02326416169.40.02527516169.40.02672617200.00.02779The table indicates the cost rate forCs=4 is smaller than 0.025927, the cost rate for the model without swap. While for Cs=5, the cost rate is greater than 0.025927. So, the threshold for the cost of the swap so that no benefit is derived is between 4 and 5.


## 6. Conclusions and Further Research

In this paper, we develop a 2-component swappable series system model based on a two-stage failure process characterized by the delay-time concept. The system is subjected to periodic inspections at an interval and the two components can be swapped once at some epoch that an inspection is carried out. Our objective is to optimize the inspection interval and the time of swap jointly by minimizing the expected cost per unit time. The motivation of the system comes from the performance process of steelmaking converter system and the maintenance of vehicles and railway lines. The results obtained in this paper may be useful in improving the economic efficiency of these systems. The numerical example demonstrates the results of the proposed model. Economic efficiency of the proposed model is stated by comparison analysis.There are several further research topics which will be studied in the future. First, the number of swaps should be optimized jointly with the inspection interval and the time of swap. Second, imperfect inspection, repairs, and PMs should be discussed. Third, the parameters estimations of the Weibull distributions and how to describe relationships of two failure processes, prior to and after swap, are also important topics.



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