# Sensitivity Analysis for a Redundant System with Random Inspection andR Repairmen Under Repair Pressure

Authors: Shan Gao; Zongbin Wang
Journal: Scientific Programming
 (2023)

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## Abstract

This study examines a repairableK-out-of-N+W: G system with warm standby components and R repairmen, where common cause failure (CCF), random inspection, and a repair pressure coefficient are involved. Due to the typical non-self-announcement of the standby component’s failure, performing a random inspection on warm standby components is necessary to find whether there are failed standby components. When the number of failed components is larger than R (heavy loading occurs), all the repairmen increase their repair rates under a repair pressure coefficient to reduce the waiting queue and enhance the reliability and availability of the system. By adopting a matrix-analytic method and Markov renewal theory, this study provides the steady-state distribution and some reliability indices mean time to first failure time (MTTFF) and reliability function from the arbitrary initial state. By some numerical experiments, sensitivity analysis is finally given to show the effect of the system parameters on some performance measures.


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## Body

## 1. Introduction

With the development of science and technology, the function of a repairable system is increasingly powerful and the system is becoming more complicated, which makes the problem of system reliability more prominent. In view of this, standby redundancy techniques are generally adopted to improve and ensure the reliability of the complicated system. Recently, researchers have shown more interest in the studies of complicated repairable systems in various fields, such as computing networks, the manufacturing industry, and power plants (cf. Kumar and Jain [1], Shekhar et al. [2], Wang et al. [3], Wang et al. [4], Peiravi et al. [5, 6], and references therein). However, the repairable warm standby component system with random inspection and multirepairmen under repair pressure has never been developed. In real life, the failure of an operating component can typically self-announce, but the warm standby’s failure always cannot self-announce. For example, in a power plant with multiple generators, say N operating generators and W standby generators. When an operating generator fails, it will stop working, i.e., the failure is self-announced. However, a standby generator may fail due to the degradation of its machine parts or due to ambient atmospheric pressure, temperature, and relative humidity, etc. Because the standby generator is idle while in the standby mode and its failure cannot be found in time, in this case, the failure of a standby is not self-announced. To make sure better warm standbys are available at the epoch when operating components fail, performing random inspection and timely maintenance for failed warm standbys is essential. Accordingly, this study considers random inspection. Furthermore, to reduce the number of failed components waiting in the service area and to keep the efficiency, reliability, and availability of the system, putting some appropriate repair pressure on repairmen to improve their repair rates can help solve such problems. Therefore, this work also takes the repair pressure coefficient into consideration.Past studies on repairable systems with warm standby components conventionally assumed that if an operating or a spare component fails, it is immediately sent to the repair service area according to some repair schedule. For example, Wang et al. [7] dealt with a repairable 1-out-of-M+S: G system with R unreliable service stations. Yen et al. [8] developed some reliability theory of a repairable 1-out-of-M+S: G system with warm standbys, where the repair schedule of N-policy and working breakdowns is considered. Yang and Tsao [9] introduced working vacations and retrial failed components into a warm standby system, in which steady-state availability, reliability function, and MTTFF are developed, respectively, by matrix-analytic (MA) method and Laplace transform (LS) technique. Recently, Gao and Wang [10] studied a new repairable retrial system with an unreliable server, in which warm standbys and cold standbys are simultaneously considered.However, in the above literature, assumption that when a warm standby fails, it can be immediately found and sent to repair is not realistic. In practice, due to not being fully operational and lack of monitoring, the failure of a warm standby always cannot self-announce. Hence, inspection on warm standbys is a necessary and effective method to detect the failed warm ones and sent them to be repaired. Adachi and Kodama [11] studied the availability of a 2-unit warm standby system, where the joint lifetime distribution of the two components follows a bivariate exponential distribution, the optimal problem with the inspection period as a decision variable is discussed to maximize the steady state availability. In Naidu and Gopalan [12], the cost benefit was analyzed for a repairable one-server 2-unit warm standby system, where the lifetimes of the two components are exponentially distributed. Similarly, Pandey et al. [13] presented a cost-benefit analysis of a one-server 2-unit warm standby system, in which the system is subject to partial failure mode and different inspection strategies. Levitin et al. [14] considered a nonrepairable 1-out-of-n: G warm standby system, in which periodic inspections are performed to detect failures of all the system components. Mo et al. [15] presented a scheduled checkpointing policy to make an efficient analysis for repairable computing systems. Recently, Wu et al. [16] extended the above studies into a repairable big number of components k-out-of-n: G warm standby systems, where the inspection and maintenance optimization are considered. Lately, Xiao et al. [17] studied the optimal inspection policy for a single-unit system by considering two failure modes. By considering the mission time as a renewal point for inspection policies, Zhao et al. [18] modelled periodic and sequential inspection policies with mission failure probabilities and obtained optimizations analytically.Different from existing literature, this paper studies the random inspection for a general warm redundantK-out-of-N+W: G system with R repairmen under repair pressure coefficient, in which errors may happen with a probability p¯ when an inspection arrives to inspect each failed warm standby, and the repairmen can adjust their repair rates under larger waiting load. The concept of the service pressure coefficient was first proposed by Hiller and Lieberman [19]. In particular, in a repairable system, the repair rate can be increased when a certain number of failed components are waiting in the repair area. Wang et al. [20] provided a comparative analysis of a warm-standby M/M/R system with multiple imperfect coverages under the service pressure condition. Hsu et al. [21] examined a warm-standby system with a switching failure, reboot, and repair pressure coefficient, in which the expected profit per unit time in steady-state is developed.Recently, much research on reliability theory has considered warm-standby systems with CCFs. In engineering practice, CCFs play the most lethal role in the dependent failure of each component of the system. A CCF is referred to the failure of some or all components of the system at the same time or successively within a short time interval caused by a single cause, such as design deficiency, energetic events, temperature change, humidity, or operator errors. Jain [22] introduced CCFs into a multicomponent mixed standby machining system. Cheng et al. [23] obtained the transient indices for any arbitrary initial state and the steady-state availability for a series system with CCFs. Shekhar et al. [2] studied a computing network with CCF, working vacation, and vacation interruption. By using binary indicator universal generating function (BIUGF) and implicit analysis methods, Cheng et al. [24] provided the reliability evaluation of new k-out-of-n (G) subsystems with a multistate phased mission and performance sharing subject to CCFs. Guo et al. [25] established the β-factor model to introduce CCF into a discrete-time Bayes network, in which there are two types of component failure states, independent failure and CCF states. Shekhar et al. [26] presented the steady-state analysis for a load-sharing redundant repairable system with CCFs, switching, and reboot delay, in which various performance measures are obtained by using C-K differential-difference equations, and the optimal analysis is provided by Newton–Quasi method. Recently, Shao et al. [27] established a hierarchical formal reliability analysis model of repairable PMSs with CCFs. Using the Markov process and linear dynamical systems, Mathebula and Saha [28] investigated the influence of CCFs and the quality of repairs on the reliability performance of IEC-61850 based architecture. Zeng and Sun [29] proposed an integrated model based on generalized stochastic Petri nets by considering the effect of competing failures and CCFs and gave a simplified method to compute the reliability of systems.Through the depth literature survey, we find the following research gap that none of the past works has studied, the effect of random inspection and repair pressure coefficient on the multirepairman repairable system with warm standbys and CCFs. It should be noted that our work is different from the existing work [2, 16, 20, 21, 24]. First, Wu et al. [16] considered periodic inspection intervals; however, we take random inspection intervals into consideration. And in [16], the authors considered the repair facility with unlimited maintenance capability, while in this paper, we consider R repairman under repair pressure. Second, our work is different from [2, 20, 21] in the following aspects: (1) in [20], when a warm standby fails, it can be sent to repair only when it is immediately detected and located with a coverage probability c, otherwise, it can be sent to repair after a reboot. However in our work, a failed warm standby can be sent to repair facility at the epoch when it is detected successfully by random inspections or when an operating component fails and needs a warm spare to be activated and replace it, but the activated warm standby already fails; and in [2, 21], failed warm standbys no matter caused by their limited lifetime or by switching failures can be immediately sent to the repair facility repaired; (2) in [2], the authors only studied the transient probabilities by solving the matrix equation and gave the reliability function RYt from initial condition P0,00=1; and in [20, 21], the authors only performed the steady-state analysis for the considered repairable systems and gave system performance measures, while in our paper, we not only provide the steady-state analysis of the system and some important performance indices but also present the reliability analysis and transient availability analysis from any initial working state by using Semi-Markov theory. In addition, Cheng et al. [24] studied a redundant series system with cold spares and CCFs, in which the system is equipped with only one repairman. In this paper, we extend the work of Cheng et al. [24] to the redundant parallel system with warm spares and CCFs, in which multirepairman under repair pressure and random inspection is considered. In summary, the novelty and contribution in this present work are that the pioneering study is given to the model and it analyzes the effect of random inspection and repair pressure coefficient on the repairable K-out-of-N+W: G warm standbys system with CCFs and R repairmen. The specification is listed as follows: (1) we develop the mathematical model of the multirepairman K-out-of-N+W: G system with random inspection, repair pressure coefficient, and CCF, which aims to model the operation schedule of the cloud computing model given in Section 2.2. (2) The steady-state distribution of the system, reliability functions, and transient availability analysis from any arbitrary initial state are obtained, respectively by using the matrix-analytic technique and Markov renewal theory.The rest of this paper is organized as follows. In Section2, we describe the model in more detail and provide a practical application. In Section 3, we present the steady-state distribution of the system and some important performance measures. By adopting Markov renewal theory and LST, we provide the reliability analysis and transient availability analysis respectively in Sections 4 and 5. In Section 6, we give some sensitivity numerical examples to examine the effect of the system parameters on performance measures.


## 2. Model Description and Its Practical Application

This section first describes the system in detail and establishes a two-dimensional Markov process to model the system dynamics. Secondly, a practical application example of the model is presented.


### 2.1. Model Description

In this section, a redundant repairableK-out-of-N+W: G system with R repairmen is provided, in which CCF, random inspection, and repair pressure coefficient are taken into consideration. The system involves N indistinguishable operating components in parallel with the provision of W warm-spare components. At least K operating components of the system are required for the system to work. The system is subject to single-component failures and common cause failures of more than one component. Hence, if the system breaks down caused by the shortage of operating components instead of the occurrence of a CCF, the allowed maximum number of failed components is L=N+W−K+1. The assumptions and notations are delineated as follows:(i)
Failure process: each operating (or warm standby) component has an exponentially distributed lifetime with the rateλ (or α) 0<α<λ. The failure of an operating component is self-announcing and can be detected as soon as it occurs. But the failure of a warm standby is nonself-announcing and is inspected once every random time interval. All the components may fail simultaneously when the system is up due to the effect of environmental factors, which is called CCF. Assume that the time to CCF follows an exponential distribution with rate δ. When a CCF occurs, the system fails and all the components are sent to repair immediately. The system is functional if and only if at least K components are operating.(ii)
Inspection arrival process: when the system is up and there are standby components in the system, the inspection is performed once every time interval with lengthX, where the random variable X follows an exponential distribution with rate β. The inspection time is negligible and every inspection is perfect for good standby components but imperfect for failed standby components. Namely, the inspection can always correctly report the state of good standby, but it may successfully report the state of a failed standby with probability p (called successful inspection) and may wrongly report a failed standby as operational one with probability p¯ (called unsuccessful inspection). If a failed warm standby is detected successfully, it is immediately sent to repair, otherwise, it waits for the next inspection.(iii)
Replacement of failed operating components: when an operating component fails, it is sent to be repaired immediately and a warm standby component will be activated randomly to replace it instantaneously. If the activated standby is a failed one, it is sent to be repaired, and another standby will be chosen randomly to activate. The replacement time is negligible.(iv)
Repair mode: all failed components are repairable and as good as new. Assume that the number of failed components in the waiting line (including those being repaired) isn. If n≤R, the repair time of each failed component is exponentially distributed with the base repair rate μ, in this case, the total repair rate is μn=nμ. If R<n≤L, due to too many failed components, they are waiting to be repaired and to enhance the reliability and availability of the system, the R repairmen may perform a better repair service under work pressure with a reasonable repair pressure coefficient θ>0. In this case, each repairman’s repair rate is changed from a lower rate μ to a higher one nR+1/n+1Rθμ, i.e., the total repair rate is μn=nR+1/n+1RθRμn>R. Regarding work pressure and repair pressure coefficient, interested readers are referred to Hsu et al. [21] and references therein. When the system is failed due to an occurrence of a CCF, it is repaired by an exponential law with rate ν.(v)
All stochastic processes involved in this system are mutually independent.


### 2.2. Practical Example

With the increasing development of the internet of things (IoT), cloud computing has become a new IT service model. As a cloud computing model, platform-as-a-service (PaaS) provides customers with a complete cloud platform service, seehttps://www.ibm.com/cloud/learn/paas. Every leading cloud service provider, including Amazon Web Services (AWS), Google Cloud, IBM Cloud, and Microsoft Azure, has its own PaaS offering. In cloud computing, virtualization is the key technology. Virtual machines (VMs) are software facilities, which are installed in virtual storage and are very appropriate for system management, system testing, and application development practice. As an example, a private cloud data center is considered (see Yen et al. [8]). Assume that the cloud data center consists of N operating VMs, W warm standby VMs, and R VM repair servers. The operating VMs and warm standby VMs fail independently. Failures of operating VMs are self-announced and can be reported immediately after they occur, but failures of warm standby VMs are nonself-announced and can be detected by an inspection by the Pegasus manager. In the case that an operating VM fails, a replacement procedure is initiated by activating randomly a warm VM to replace the failed one instantaneously. If the activated standby is a failed one, it will be sent to the service area to be repaired, and another standby is randomly chosen to be activated. To enhance and maintain the quality of service (QoS) of the system, when a load of repair pressure becomes heavy, the Pegasus manager would increase the repair rates of the R VM repair servers, according to a repair pressure coefficient. Due to some common causes, for example, heat up, jammer, and power failure, the virtual network of VMs of the system maybe inoperable and need a quick repair.In view of the above characteristics of the PaaS cloud service, our work is very suitable for modelling it. Our findings can provide a sound theoretical basis for the administrator of the cloud data center for analyzing reliability problems.


## 2.1. Model Description

In this section, a redundant repairableK-out-of-N+W: G system with R repairmen is provided, in which CCF, random inspection, and repair pressure coefficient are taken into consideration. The system involves N indistinguishable operating components in parallel with the provision of W warm-spare components. At least K operating components of the system are required for the system to work. The system is subject to single-component failures and common cause failures of more than one component. Hence, if the system breaks down caused by the shortage of operating components instead of the occurrence of a CCF, the allowed maximum number of failed components is L=N+W−K+1. The assumptions and notations are delineated as follows:(i)
Failure process: each operating (or warm standby) component has an exponentially distributed lifetime with the rateλ (or α) 0<α<λ. The failure of an operating component is self-announcing and can be detected as soon as it occurs. But the failure of a warm standby is nonself-announcing and is inspected once every random time interval. All the components may fail simultaneously when the system is up due to the effect of environmental factors, which is called CCF. Assume that the time to CCF follows an exponential distribution with rate δ. When a CCF occurs, the system fails and all the components are sent to repair immediately. The system is functional if and only if at least K components are operating.(ii)
Inspection arrival process: when the system is up and there are standby components in the system, the inspection is performed once every time interval with lengthX, where the random variable X follows an exponential distribution with rate β. The inspection time is negligible and every inspection is perfect for good standby components but imperfect for failed standby components. Namely, the inspection can always correctly report the state of good standby, but it may successfully report the state of a failed standby with probability p (called successful inspection) and may wrongly report a failed standby as operational one with probability p¯ (called unsuccessful inspection). If a failed warm standby is detected successfully, it is immediately sent to repair, otherwise, it waits for the next inspection.(iii)
Replacement of failed operating components: when an operating component fails, it is sent to be repaired immediately and a warm standby component will be activated randomly to replace it instantaneously. If the activated standby is a failed one, it is sent to be repaired, and another standby will be chosen randomly to activate. The replacement time is negligible.(iv)
Repair mode: all failed components are repairable and as good as new. Assume that the number of failed components in the waiting line (including those being repaired) isn. If n≤R, the repair time of each failed component is exponentially distributed with the base repair rate μ, in this case, the total repair rate is μn=nμ. If R<n≤L, due to too many failed components, they are waiting to be repaired and to enhance the reliability and availability of the system, the R repairmen may perform a better repair service under work pressure with a reasonable repair pressure coefficient θ>0. In this case, each repairman’s repair rate is changed from a lower rate μ to a higher one nR+1/n+1Rθμ, i.e., the total repair rate is μn=nR+1/n+1RθRμn>R. Regarding work pressure and repair pressure coefficient, interested readers are referred to Hsu et al. [21] and references therein. When the system is failed due to an occurrence of a CCF, it is repaired by an exponential law with rate ν.(v)
All stochastic processes involved in this system are mutually independent.


## 2.2. Practical Example

With the increasing development of the internet of things (IoT), cloud computing has become a new IT service model. As a cloud computing model, platform-as-a-service (PaaS) provides customers with a complete cloud platform service, seehttps://www.ibm.com/cloud/learn/paas. Every leading cloud service provider, including Amazon Web Services (AWS), Google Cloud, IBM Cloud, and Microsoft Azure, has its own PaaS offering. In cloud computing, virtualization is the key technology. Virtual machines (VMs) are software facilities, which are installed in virtual storage and are very appropriate for system management, system testing, and application development practice. As an example, a private cloud data center is considered (see Yen et al. [8]). Assume that the cloud data center consists of N operating VMs, W warm standby VMs, and R VM repair servers. The operating VMs and warm standby VMs fail independently. Failures of operating VMs are self-announced and can be reported immediately after they occur, but failures of warm standby VMs are nonself-announced and can be detected by an inspection by the Pegasus manager. In the case that an operating VM fails, a replacement procedure is initiated by activating randomly a warm VM to replace the failed one instantaneously. If the activated standby is a failed one, it will be sent to the service area to be repaired, and another standby is randomly chosen to be activated. To enhance and maintain the quality of service (QoS) of the system, when a load of repair pressure becomes heavy, the Pegasus manager would increase the repair rates of the R VM repair servers, according to a repair pressure coefficient. Due to some common causes, for example, heat up, jammer, and power failure, the virtual network of VMs of the system maybe inoperable and need a quick repair.In view of the above characteristics of the PaaS cloud service, our work is very suitable for modelling it. Our findings can provide a sound theoretical basis for the administrator of the cloud data center for analyzing reliability problems.


## 3. Steady-State Distribution and Performance Measures

The steady-state distribution and performance measures are explained in the following sections.


### 3.1. Steady-State Distribution

For the consideredK-out-of-N+W: G system, the state of the system at the time t is denoted by St,t≥0 with state space S=m,n,0≤m+n≤W∪0,n,W+1≤n≤L+1, where the state m,n,0≤m+n≤W denotes that there are m failed warm standbys waiting for inspection and n failed components which are being under repair or waiting for repair; the state 0,nW+1≤n≤L denotes that there are n failed components in the service area which are being under repair or waiting for repair and no warm standby exists in the system. Note that the state 0,L denotes that the system is down because the number of operating components becomes K−1 due to the single component failure. The state 0,L+1 denotes that the system is under repair due to the occurrence of a CCF. From the assumptions of the system, the process St,t≥0 is obviously a continuous Markov chain. Thus, S can be divided to two substate spaces as S=W∪F, where W=△m,n,0≤m+n≤W∪0,L−1 is the set of working states of the system, and F=△0,L,0,L+1 is the set of failure states of the system.LetQ=qs1,s2s1,s2∈S be the infinitesimal generator of the Markov process St,t≥0. To find the expression of Q, we first define some notations for a given state s=m,n:(i)
The total failure rateΛm,n of the operating components, where Λm,n=Nλ if 0≤m+n≤W and Λm,n=N+W−nλ if m=0,W+1≤n≤L−1.(ii)
The total failure rateΔm,n of the warm standbys, where Δm,n=W−m−nα if 0≤m+n≤W.(iii)
The probabilityam,k that k failed warm standbys are detected from m failed warm ones by an arriving inspection am,k=mkpkp¯m−k,k=1,2,…,m,1≤m≤W.(iv)
The probabilitybm,k that k failed warm standbys are chosen to replace a failed operating component:(1)bm,k=AmkAW−nkW−m−nW−n−k,0≤k≤m,0≤m≤W−1,1≤m+n≤W−1.Based on the above assumptions and notations, we can obtain the elementsqs1,s2,s1,s2∈S of the matrix Q in different cases as follows:(i)
Case 1:s1=m,n, 0≤m+n≤W−1:(2)qs1,s2=W−m−nα,s2=m+1,n,βam,k,s2=m−k,n+k,1≤k≤m,−qs1,s2=s1,Nλbm,k,s2=m−k,n+k+1,0≤k≤m,μn,s2=m,n−1,δ,s2=0,L+1,0,otherwise,whereqs1=W−m−nα+β1−p¯m+Nλ+μn+δ.(ii)
Case 2:s1=0,W:(3)qs1,s2=Nλ,s2=0,W+1,μW,s2=0,W−1,δ,s2=0,L+1,−qs1,s2=s1,0,otherwise,whereqs1=Nλ+μW+δ.(iii)
Case 3:s1=m,W−m,1≤m≤W:(4)qs1,s2=Nλ,s2=0,W+1,βam,k,s2=m−k,W−m+k,1≤k≤m,μW−m,s2=m,W−m−1,δ,s2=0,L+1,−qs1,s2=s1,0,otherwise,whereqs1=Nλ+β∑k=1mam,k+μW−m+δ=Nλ+β1−p¯m+μW−m+δ.(iv)
Case 4:s1=0,n,W+1≤n≤L−1:(5)qs1,s2=N+W−nλ,s2=0,n+1,μn,s2=0,n−1,δ,s2=0,L+1,−qs1,s2=s1,0,otherwise,whereqs1=N+W−nλ+μn+δ.(v)
Case 5:s1=0,L:(6)qs1,s2=μL,s2=0,L−1,−qs1,s2=s1,0,otherwise,whereqs1=μL.(vi)
Case 6:s1=0,L+1:(7)qs1,s2=ν,s2=0,0,−qs1,s2=s1,0,otherwise,whereqs1=ν.Using the lexicographical ordering for the states,Q can be written as the Block–Jacobi matrix which is given as follows:(8)Q=Q0,0Q0,1Q1,0Q1,1Q1,2Q2,0Q2,1Q2,2Q2,3⋮⋮⋮⋱⋱QW−2,0QW−2,1QW−2,2QW−2,3⋯QW−2,W−1QW−1,0QW−1,1QW−1,2QW−1,3⋯QW−1,W−1QW−1,WQW,0QW,1QW,2QW,3⋯QW,W−1QW,W,where the block Qm1,m2 denotes the transition rate from m1 failed warm standbys to m2 failed warm standbys, 0≤m1,m2≤W. Q0,0 is a L+2×L+2 square matrix and its element Q0,0i,j=q0,i,0,j,i,j=0,1,…,L+1. Q0,1 is a L+2×W matrix and its element Q0,1i,j=q0,i,1,j,i=0,1,…,L+1,j=0,1,…,W−1. Qm,01≤m≤W is a W−m+1×L+2 matrix and its element Qm,0i,j=qm,i,0,j,i=0,1,…,W−m,j=0,1,…,L+1. Qm,n1≤m,n≤W is a W−m+1×W−n+1 matrix, and its element Qm,ni,j=qm,i,n,j,i=0,1,…,W−m,j=0,1,…,W−n.Remark 1.
Assuming thatN=5,W=3,K=3, we take the element Q2,21,0 of the submatrix Q2,2 as an example to give an illustration and other elements of all submatrixes can be explained by following similar methods: the element Q2,21,0=q2,1,2,0 denotes the transition rate from the state (2, 1) to the state (2, 0) and is equal to μ1 because state (2, 1) denotes that two failed warm standbys are waiting for inspection and one failed component is being under repair, when the repairman completes the repair of the failed component at state (2, 1) with rate μ1 and then the system will immediately enter the state (2, 0).
DefinePm,nt=PSt=m,n as the transient probability that the system is in state m,n at time t, m,n∈S, and Pm,n=limt⟶∞PSt=m,n to be the corresponding steady-state probability. Let P0=P0,0,P0,1,P0,2,…,P0,L+1 and Pm=Pm,0,Pm,1,Pm,2,…,Pm,W−m,1≤m≤W. Then, P=P0,P1,…,PL,PL+1 be the steady-state distribution vector, which can be obtained by solving the matrix equations.(9)PQ=0τ,(10)PeτT=1,where 0τ is a row zero-vector of τ dimension and eτ is a column one-vector of τ dimension, τ=W+1W+2/2+N−K+2 is the total number of the states. The system of equations (9) and (10) can be solved by using the available software MATLAB.


### 3.2. Performance Measures

Based on the steady-state distributionP, we can derive some important performance measures of the system, which are classified in the following:(1)
Mean number of failed warm standby components waiting for inspection is given by the following:(11)ENw=∑m=1Wm∑n=0W−mPm,n.(2)
Mean number of failed components in the service area which are being under repair or waiting for repair is given by the following:(12)ENr=N+WP0,L+1+∑n=1LnP0,n+∑n=1W−1n∑m=1W−nPm,n.(3)
Mean number of operating components in the system is given by the following:(13)ENo=N1−P0,L+1−∑n=1N−K+1nP0,W+n.(4)
Mean number of operational warm standby components is given by the following:(14)ENow=∑n=0WW−nP0,n+∑m=1W∑n=0W−mW−m−nPm,n.(5)
The steady-state availabilityA of the system is given by A=1−P0,L−P0,L+1.(6)
Steady-state failure frequencyFF is given by the following:(15)FF=δA+KλP0,L−1.


## 3.1. Steady-State Distribution

For the consideredK-out-of-N+W: G system, the state of the system at the time t is denoted by St,t≥0 with state space S=m,n,0≤m+n≤W∪0,n,W+1≤n≤L+1, where the state m,n,0≤m+n≤W denotes that there are m failed warm standbys waiting for inspection and n failed components which are being under repair or waiting for repair; the state 0,nW+1≤n≤L denotes that there are n failed components in the service area which are being under repair or waiting for repair and no warm standby exists in the system. Note that the state 0,L denotes that the system is down because the number of operating components becomes K−1 due to the single component failure. The state 0,L+1 denotes that the system is under repair due to the occurrence of a CCF. From the assumptions of the system, the process St,t≥0 is obviously a continuous Markov chain. Thus, S can be divided to two substate spaces as S=W∪F, where W=△m,n,0≤m+n≤W∪0,L−1 is the set of working states of the system, and F=△0,L,0,L+1 is the set of failure states of the system.LetQ=qs1,s2s1,s2∈S be the infinitesimal generator of the Markov process St,t≥0. To find the expression of Q, we first define some notations for a given state s=m,n:(i)
The total failure rateΛm,n of the operating components, where Λm,n=Nλ if 0≤m+n≤W and Λm,n=N+W−nλ if m=0,W+1≤n≤L−1.(ii)
The total failure rateΔm,n of the warm standbys, where Δm,n=W−m−nα if 0≤m+n≤W.(iii)
The probabilityam,k that k failed warm standbys are detected from m failed warm ones by an arriving inspection am,k=mkpkp¯m−k,k=1,2,…,m,1≤m≤W.(iv)
The probabilitybm,k that k failed warm standbys are chosen to replace a failed operating component:(1)bm,k=AmkAW−nkW−m−nW−n−k,0≤k≤m,0≤m≤W−1,1≤m+n≤W−1.Based on the above assumptions and notations, we can obtain the elementsqs1,s2,s1,s2∈S of the matrix Q in different cases as follows:(i)
Case 1:s1=m,n, 0≤m+n≤W−1:(2)qs1,s2=W−m−nα,s2=m+1,n,βam,k,s2=m−k,n+k,1≤k≤m,−qs1,s2=s1,Nλbm,k,s2=m−k,n+k+1,0≤k≤m,μn,s2=m,n−1,δ,s2=0,L+1,0,otherwise,whereqs1=W−m−nα+β1−p¯m+Nλ+μn+δ.(ii)
Case 2:s1=0,W:(3)qs1,s2=Nλ,s2=0,W+1,μW,s2=0,W−1,δ,s2=0,L+1,−qs1,s2=s1,0,otherwise,whereqs1=Nλ+μW+δ.(iii)
Case 3:s1=m,W−m,1≤m≤W:(4)qs1,s2=Nλ,s2=0,W+1,βam,k,s2=m−k,W−m+k,1≤k≤m,μW−m,s2=m,W−m−1,δ,s2=0,L+1,−qs1,s2=s1,0,otherwise,whereqs1=Nλ+β∑k=1mam,k+μW−m+δ=Nλ+β1−p¯m+μW−m+δ.(iv)
Case 4:s1=0,n,W+1≤n≤L−1:(5)qs1,s2=N+W−nλ,s2=0,n+1,μn,s2=0,n−1,δ,s2=0,L+1,−qs1,s2=s1,0,otherwise,whereqs1=N+W−nλ+μn+δ.(v)
Case 5:s1=0,L:(6)qs1,s2=μL,s2=0,L−1,−qs1,s2=s1,0,otherwise,whereqs1=μL.(vi)
Case 6:s1=0,L+1:(7)qs1,s2=ν,s2=0,0,−qs1,s2=s1,0,otherwise,whereqs1=ν.Using the lexicographical ordering for the states,Q can be written as the Block–Jacobi matrix which is given as follows:(8)Q=Q0,0Q0,1Q1,0Q1,1Q1,2Q2,0Q2,1Q2,2Q2,3⋮⋮⋮⋱⋱QW−2,0QW−2,1QW−2,2QW−2,3⋯QW−2,W−1QW−1,0QW−1,1QW−1,2QW−1,3⋯QW−1,W−1QW−1,WQW,0QW,1QW,2QW,3⋯QW,W−1QW,W,where the block Qm1,m2 denotes the transition rate from m1 failed warm standbys to m2 failed warm standbys, 0≤m1,m2≤W. Q0,0 is a L+2×L+2 square matrix and its element Q0,0i,j=q0,i,0,j,i,j=0,1,…,L+1. Q0,1 is a L+2×W matrix and its element Q0,1i,j=q0,i,1,j,i=0,1,…,L+1,j=0,1,…,W−1. Qm,01≤m≤W is a W−m+1×L+2 matrix and its element Qm,0i,j=qm,i,0,j,i=0,1,…,W−m,j=0,1,…,L+1. Qm,n1≤m,n≤W is a W−m+1×W−n+1 matrix, and its element Qm,ni,j=qm,i,n,j,i=0,1,…,W−m,j=0,1,…,W−n.Remark 1.
Assuming thatN=5,W=3,K=3, we take the element Q2,21,0 of the submatrix Q2,2 as an example to give an illustration and other elements of all submatrixes can be explained by following similar methods: the element Q2,21,0=q2,1,2,0 denotes the transition rate from the state (2, 1) to the state (2, 0) and is equal to μ1 because state (2, 1) denotes that two failed warm standbys are waiting for inspection and one failed component is being under repair, when the repairman completes the repair of the failed component at state (2, 1) with rate μ1 and then the system will immediately enter the state (2, 0).
DefinePm,nt=PSt=m,n as the transient probability that the system is in state m,n at time t, m,n∈S, and Pm,n=limt⟶∞PSt=m,n to be the corresponding steady-state probability. Let P0=P0,0,P0,1,P0,2,…,P0,L+1 and Pm=Pm,0,Pm,1,Pm,2,…,Pm,W−m,1≤m≤W. Then, P=P0,P1,…,PL,PL+1 be the steady-state distribution vector, which can be obtained by solving the matrix equations.(9)PQ=0τ,(10)PeτT=1,where 0τ is a row zero-vector of τ dimension and eτ is a column one-vector of τ dimension, τ=W+1W+2/2+N−K+2 is the total number of the states. The system of equations (9) and (10) can be solved by using the available software MATLAB.


## 3.2. Performance Measures

Based on the steady-state distributionP, we can derive some important performance measures of the system, which are classified in the following:(1)
Mean number of failed warm standby components waiting for inspection is given by the following:(11)ENw=∑m=1Wm∑n=0W−mPm,n.(2)
Mean number of failed components in the service area which are being under repair or waiting for repair is given by the following:(12)ENr=N+WP0,L+1+∑n=1LnP0,n+∑n=1W−1n∑m=1W−nPm,n.(3)
Mean number of operating components in the system is given by the following:(13)ENo=N1−P0,L+1−∑n=1N−K+1nP0,W+n.(4)
Mean number of operational warm standby components is given by the following:(14)ENow=∑n=0WW−nP0,n+∑m=1W∑n=0W−mW−m−nPm,n.(5)
The steady-state availabilityA of the system is given by A=1−P0,L−P0,L+1.(6)
Steady-state failure frequencyFF is given by the following:(15)FF=δA+KλP0,L−1.


## 4. Reliability Analysis of the System

Let random variableYs1 be the time to first failure of the system with initial working state s1s1∈W. LST of Ys1 is denoted as Ys1∗s=Ee−sYs1|S0=s1. Let Rs1t and MTTFFs1, respectively, be the reliability function and MTTFF of the system from the initial working state i, then Rs1t=PYs1>t and MTTFFs1=EYs1=∫0+∞Rs1tdt=−d/dsYs1∗ss=0, where Rs1t is the probability that the system will perform operation continuously without failure before time t. Let Rs1∗s=∫0∞e−stRs1tdt be the LT of the reliability function Rs1t, we readily have that Rs1∗s=1−Ys1∗s/s. Thus, once we obtain the expressions of Ys1∗s,i∈W, we can get the expressions of MTTFFs1 and Rs1t.In this section, we will use Markov renewal theory and LST technique to obtainYs1∗s. First, we present the Semi-Markov kernels of the system.


### 4.1. Markov Renewal Process and Semi-Markov Kernels of the System

LetTnT0=0 be the time epoch of the n-th state transition of the Markov process St,t≥0, i.e., Tn=inft≥Tn−1,St≠STn−1. Then, STn,n≥0 gives the successive states of St,t≥0 in the intervals between its jump epochs and τn=Tn−Tn−1n≥1 denotes the sojourn time on state STn−1. The two-dimensional process STn,Tn,n≥0 is called a homogenous Markov renewal process (or homogenous semi-Markov process) (see Chapter 8, PP 318 in Ref. Rolski et al. [30]).For any initial states1∈S, from the infinitesimal generator matrix Q given in Section 3.1, we have that(16)qs1,s2x=△PNTn+1=s2,τn+1≤x|NTn=s1=qs1,s2qs11−e−qs1x,x>0,s1≠s2,s1,s2∈S,where, Qx=qs1,s2x is called as Semi-Markov Kernel.To findYs1∗s, let qs1,s2∗s=Ee−sT1IST1=s2|S0=s1=∫0+∞e−sxdqs1,s2x,s1≠s2,s1∈W,s2∈S, which denotes the LST of the length of the sojourn time T1 in the initial state s1 and the state ST1=s2 at time T1. Then,(17)qs1,s2∗s=qs1,s2s+qs1,which yields that(18)Ee−sT1|N0=s1=∑s2≠s1qs1,s2∗s=qs1s+qs1.That is to say, given that the initial states1, T1 follows exponential distribution with rate qs1.Now, we can derive the expression ofYs1∗s.


### 4.2. LST of the Time to First Failure

Considering that the first failure of the system from any initial working states1∈W can occur only when the state of the system changes, by conditioning on the length of the sojourn time T1 in initial state s1 and the state at time T1, we obtain the following:(19)Ys1∗s=Ee−sYs1|S0=s1=∑s2≠s1,s2∈S∫0+∞Ee−sYs1|T1=x,ST1=s2,S0=s1dqs1,s2x.IfST1=s2∈W, under the condition T1=x,ST1=s2,S0=s1, Ys1 is equal to x+Ys2, in this case(20)∫0+∞Ee−sYs1|T1=x,ST1=s2,S0=s1dqs1,s2x=qs1,s2∗sYs2∗s.IfST1=s2∈F, under the condition T1=x,ST1=s2,S0=s1, Ys1 is equal to x, which leads to the following:(21)∫0+∞Ee−sYs1|T1=x,ST1=s2,S0=s1dqs1,s2x=qs1,s2∗s.Then, from (2)–(7) and (17), we have the following results:(1)
Ifs1=m,n,m+n≤W−1(22)Ym,n∗s=qm,n,m+1,n∗sYm+1,n∗s+1−δm,0∑v=1mqm,n,m−v,n+v∗sYm−v,n+v∗s+∑v=0mqm,n,m−v,n+v+1∗sYm−v,n+v+1∗s+1−δn,0qm,n,m,n−1∗sYm,n−1∗s+qm,n,0,L+1∗s,whereδm,0=1 if m=0, and δm,0=0 if m≠0.(2)
Ifs1=m,n,m+n=W,(23)Ym,n∗s=qm,n,0,W+1∗sY0,W+1∗s+1−δm,0∑v=1mqm,n,m−v,n+v∗sYm−v,n+v∗s+1−δn,0qm,n,m,n−1∗sYm,n−1∗s+qm,n,0,L+1∗s.(3)
Ifs1=0,n,W+1≤n≤L−2,(24)Y0,n∗s=q0,n,0,n+1∗sY0,n+1∗s+q0,n,0,n−1∗sY0,n−1∗s+q0,n,0,L+1∗s.(4)
Ifs1=0,L−1,(25)Y0,L−1∗s=q0,L−1,0,L−2∗sY0,L−2∗s+q0,L−1,0,L∗s+q0,L−1,0,L+1∗s.LetY0∗s=Y0,0∗s,Y0,1∗s,…,Y0,L−1∗s, Ym∗s=Ym,0∗s,Ym,1∗s,…,Ym,W−m∗s,1≤m≤W, and Y∗s=Y0∗s,Y1∗s,…,YW∗sT. Then, equations (22)–(25) can be rewritten as the block matrix equation:(26)BsY∗s=bs,where(27)bs=b0s,b1s,…,bWsT,with(28)b0s=q0,0,0,L+1∗s,q0,1,0,L+1∗s,…,q0,L−2,0,L+1∗s,q0,L−1,0,L∗s+q0,L−1,0,L+1∗s,bms=qm,0,0,L+1∗s,qm,1,0,L+1∗s,…,qm,W−m,0,L+1∗s,1≤m≤W.The coefficient matrixBs is given by block as follows:(29)Bs=B0,0sB0,1sB1,0sB1,1sB1,2sB2,0sB2,1sB2,2sB2,3s⋮⋮⋮⋱⋱BW−2,0sBW−2,1sBW−2,2sBW−2,3s⋯BW−2,W−1sBW−1,0sBW−1,1sBW−1,2sBW−1,3s⋯BW−1,W−1sBW−1,WsBW,0sBW,1sBW,2sBW,3s⋯BW,W−1sBW,Ws,where B0,0s is a square matrix with order L×L and is given by the following:(30)B0,0s=−−1q0,0,0,1∗sq0,1,0,0∗s−1q0,1,0,2∗sq0,2,0,1∗s−1q0,2,0,3∗s⋱⋱⋱q0,L−2,0,L−3∗s−1q0,L−2,0,L−1∗sq0,L−1,0,L−2∗s−1.The submatrixB0,1s is a matrix with order L×W and is given by the following:(31)B0,1s=MW×WsOL−W×W,where OL−W×W is a zero-matrix with order L−W×W, MW×Ws is a W×W square matrix and is given as follows:(32)MW×Ws=−q0,0,1,0∗s−q0,1,1,1∗s−q0,2,1,2∗s⋱−q0,W−1,1,W−1∗s.For1≤m≤W−1, the submatrix Bm,m+1s is a matrix with order W−m+1×W−m and is given by the following:(33)Bm,m+1s=−qm,0,m+1,0∗sqm,1,m+1,1∗sqm,2,m+1,2∗s⋱qm,W−m−1,m+1,W−m−1∗s0.For1≤m≤W, the submatrix Bm,ms is a matrix with order W−m+1×W−m+1 and is given by the following:(34)Bm,ms=−−1qm,1,m,0∗s−1qm,2,m,1∗s−1⋱⋱qm,W−m−1,m,W−m−2∗s−1qm,W−m,m,W−m−1∗s−1.For1≤m≤W, the submatrix Bm,0s is a matrix with order W−m+1×L and is given by the following:(35)Bm,0s=OW−m+1×mDW−m+1×W−m+2sOW−m+1×L−W−2,where(36)DW−m+1×W−m+2s=−qm,0,0,m∗sqm,0,0,m+1∗sqm,1,0,m+1∗sqm,1,0,m+2∗sqm,2,0,m+2∗sqm,2,0,m+3∗s⋱⋱qm,W−m,0,W∗sqm,W−m,0,W+1∗s.For2≤m≤W,1≤n≤m−1, the submatrix Bm,ns is a matrix with order W−m+1×W−n+1 and is given by the following:(37)Bm,ns=OW−m+1×m−nEW−m+1×W−m+1s,where(38)EW−m+1×W−m+1s=−qm,0,n,m−n∗sqm,0,0,m−n+1∗sqm,1,n,m−n+1∗sqm,1,n,m−n+2∗s⋱⋱qm,W−m−1,n,W−n−1∗sqm,W−m−1,n,W−n∗sqm,W−m,n,W−n∗s.Using (26) and applying Cramer’s rule, we can solve for Ys1∗s,s1∈W as follows:(39)Ys1∗s=detBn+1sdetBs,s1=0,n,0≤n≤L−1,detBL+m−1W−m−2/2+n+1sdetBs,s1=m,n,1≤m≤W,0≤n≤W−m,where detBs denotes the determinant of Bs, detBis is obtained by replacing the i-th column of Bs by bs.UsingMTTFFs1=EYs1=−d/dsYs1∗s|s=0, we can obtain MTTFFs1,s1∈W. Furthermore, we can obtain the reliability function Rs1t by adopting the property of inverse Laplace transform(40)Rs1t=L−1Rs1∗s=1−L−1Ys1∗ss.The numerical solutions forMTTFFs1 and Rs1t,s1∈W will be given in Section 6 by employing mathematical software Matlab 2016a to solve the Matrix (26).


## 4.1. Markov Renewal Process and Semi-Markov Kernels of the System

LetTnT0=0 be the time epoch of the n-th state transition of the Markov process St,t≥0, i.e., Tn=inft≥Tn−1,St≠STn−1. Then, STn,n≥0 gives the successive states of St,t≥0 in the intervals between its jump epochs and τn=Tn−Tn−1n≥1 denotes the sojourn time on state STn−1. The two-dimensional process STn,Tn,n≥0 is called a homogenous Markov renewal process (or homogenous semi-Markov process) (see Chapter 8, PP 318 in Ref. Rolski et al. [30]).For any initial states1∈S, from the infinitesimal generator matrix Q given in Section 3.1, we have that(16)qs1,s2x=△PNTn+1=s2,τn+1≤x|NTn=s1=qs1,s2qs11−e−qs1x,x>0,s1≠s2,s1,s2∈S,where, Qx=qs1,s2x is called as Semi-Markov Kernel.To findYs1∗s, let qs1,s2∗s=Ee−sT1IST1=s2|S0=s1=∫0+∞e−sxdqs1,s2x,s1≠s2,s1∈W,s2∈S, which denotes the LST of the length of the sojourn time T1 in the initial state s1 and the state ST1=s2 at time T1. Then,(17)qs1,s2∗s=qs1,s2s+qs1,which yields that(18)Ee−sT1|N0=s1=∑s2≠s1qs1,s2∗s=qs1s+qs1.That is to say, given that the initial states1, T1 follows exponential distribution with rate qs1.Now, we can derive the expression ofYs1∗s.


## 4.2. LST of the Time to First Failure

Considering that the first failure of the system from any initial working states1∈W can occur only when the state of the system changes, by conditioning on the length of the sojourn time T1 in initial state s1 and the state at time T1, we obtain the following:(19)Ys1∗s=Ee−sYs1|S0=s1=∑s2≠s1,s2∈S∫0+∞Ee−sYs1|T1=x,ST1=s2,S0=s1dqs1,s2x.IfST1=s2∈W, under the condition T1=x,ST1=s2,S0=s1, Ys1 is equal to x+Ys2, in this case(20)∫0+∞Ee−sYs1|T1=x,ST1=s2,S0=s1dqs1,s2x=qs1,s2∗sYs2∗s.IfST1=s2∈F, under the condition T1=x,ST1=s2,S0=s1, Ys1 is equal to x, which leads to the following:(21)∫0+∞Ee−sYs1|T1=x,ST1=s2,S0=s1dqs1,s2x=qs1,s2∗s.Then, from (2)–(7) and (17), we have the following results:(1)
Ifs1=m,n,m+n≤W−1(22)Ym,n∗s=qm,n,m+1,n∗sYm+1,n∗s+1−δm,0∑v=1mqm,n,m−v,n+v∗sYm−v,n+v∗s+∑v=0mqm,n,m−v,n+v+1∗sYm−v,n+v+1∗s+1−δn,0qm,n,m,n−1∗sYm,n−1∗s+qm,n,0,L+1∗s,whereδm,0=1 if m=0, and δm,0=0 if m≠0.(2)
Ifs1=m,n,m+n=W,(23)Ym,n∗s=qm,n,0,W+1∗sY0,W+1∗s+1−δm,0∑v=1mqm,n,m−v,n+v∗sYm−v,n+v∗s+1−δn,0qm,n,m,n−1∗sYm,n−1∗s+qm,n,0,L+1∗s.(3)
Ifs1=0,n,W+1≤n≤L−2,(24)Y0,n∗s=q0,n,0,n+1∗sY0,n+1∗s+q0,n,0,n−1∗sY0,n−1∗s+q0,n,0,L+1∗s.(4)
Ifs1=0,L−1,(25)Y0,L−1∗s=q0,L−1,0,L−2∗sY0,L−2∗s+q0,L−1,0,L∗s+q0,L−1,0,L+1∗s.LetY0∗s=Y0,0∗s,Y0,1∗s,…,Y0,L−1∗s, Ym∗s=Ym,0∗s,Ym,1∗s,…,Ym,W−m∗s,1≤m≤W, and Y∗s=Y0∗s,Y1∗s,…,YW∗sT. Then, equations (22)–(25) can be rewritten as the block matrix equation:(26)BsY∗s=bs,where(27)bs=b0s,b1s,…,bWsT,with(28)b0s=q0,0,0,L+1∗s,q0,1,0,L+1∗s,…,q0,L−2,0,L+1∗s,q0,L−1,0,L∗s+q0,L−1,0,L+1∗s,bms=qm,0,0,L+1∗s,qm,1,0,L+1∗s,…,qm,W−m,0,L+1∗s,1≤m≤W.The coefficient matrixBs is given by block as follows:(29)Bs=B0,0sB0,1sB1,0sB1,1sB1,2sB2,0sB2,1sB2,2sB2,3s⋮⋮⋮⋱⋱BW−2,0sBW−2,1sBW−2,2sBW−2,3s⋯BW−2,W−1sBW−1,0sBW−1,1sBW−1,2sBW−1,3s⋯BW−1,W−1sBW−1,WsBW,0sBW,1sBW,2sBW,3s⋯BW,W−1sBW,Ws,where B0,0s is a square matrix with order L×L and is given by the following:(30)B0,0s=−−1q0,0,0,1∗sq0,1,0,0∗s−1q0,1,0,2∗sq0,2,0,1∗s−1q0,2,0,3∗s⋱⋱⋱q0,L−2,0,L−3∗s−1q0,L−2,0,L−1∗sq0,L−1,0,L−2∗s−1.The submatrixB0,1s is a matrix with order L×W and is given by the following:(31)B0,1s=MW×WsOL−W×W,where OL−W×W is a zero-matrix with order L−W×W, MW×Ws is a W×W square matrix and is given as follows:(32)MW×Ws=−q0,0,1,0∗s−q0,1,1,1∗s−q0,2,1,2∗s⋱−q0,W−1,1,W−1∗s.For1≤m≤W−1, the submatrix Bm,m+1s is a matrix with order W−m+1×W−m and is given by the following:(33)Bm,m+1s=−qm,0,m+1,0∗sqm,1,m+1,1∗sqm,2,m+1,2∗s⋱qm,W−m−1,m+1,W−m−1∗s0.For1≤m≤W, the submatrix Bm,ms is a matrix with order W−m+1×W−m+1 and is given by the following:(34)Bm,ms=−−1qm,1,m,0∗s−1qm,2,m,1∗s−1⋱⋱qm,W−m−1,m,W−m−2∗s−1qm,W−m,m,W−m−1∗s−1.For1≤m≤W, the submatrix Bm,0s is a matrix with order W−m+1×L and is given by the following:(35)Bm,0s=OW−m+1×mDW−m+1×W−m+2sOW−m+1×L−W−2,where(36)DW−m+1×W−m+2s=−qm,0,0,m∗sqm,0,0,m+1∗sqm,1,0,m+1∗sqm,1,0,m+2∗sqm,2,0,m+2∗sqm,2,0,m+3∗s⋱⋱qm,W−m,0,W∗sqm,W−m,0,W+1∗s.For2≤m≤W,1≤n≤m−1, the submatrix Bm,ns is a matrix with order W−m+1×W−n+1 and is given by the following:(37)Bm,ns=OW−m+1×m−nEW−m+1×W−m+1s,where(38)EW−m+1×W−m+1s=−qm,0,n,m−n∗sqm,0,0,m−n+1∗sqm,1,n,m−n+1∗sqm,1,n,m−n+2∗s⋱⋱qm,W−m−1,n,W−n−1∗sqm,W−m−1,n,W−n∗sqm,W−m,n,W−n∗s.Using (26) and applying Cramer’s rule, we can solve for Ys1∗s,s1∈W as follows:(39)Ys1∗s=detBn+1sdetBs,s1=0,n,0≤n≤L−1,detBL+m−1W−m−2/2+n+1sdetBs,s1=m,n,1≤m≤W,0≤n≤W−m,where detBs denotes the determinant of Bs, detBis is obtained by replacing the i-th column of Bs by bs.UsingMTTFFs1=EYs1=−d/dsYs1∗s|s=0, we can obtain MTTFFs1,s1∈W. Furthermore, we can obtain the reliability function Rs1t by adopting the property of inverse Laplace transform(40)Rs1t=L−1Rs1∗s=1−L−1Ys1∗ss.The numerical solutions forMTTFFs1 and Rs1t,s1∈W will be given in Section 6 by employing mathematical software Matlab 2016a to solve the Matrix (26).


## 5. Transient Availability Analysis

DefineAs1t to be the transient availability at time t from any arbitrary initial state s1∈S, i.e.,(41)As1t=PSt∈W|S0=s1,s1∈S.The LT ofAs1t is denoted as As1∗s=∫0+∞e−stAs1tdt. In the following, we first derive the expression of As1∗s based on the Markov renewal theory.If the initial states1∈W, conditioning on the time of the first state transition T1≤t or T1>t and using (18) leads to the following:(42)As1t=PSt∈W,T1≤t|S0=s1+PSt∈W,T1>t|S0=s1=PSt∈W,T1≤t,ST1∈S,ST1≠s1|S0=s1+PT1>t|N0=i=∑s2≠s1,s2∈S∫0tAs2t−xdqs1,s2x+e−qs1t.If initial states1=0,L or 0,L+1, we have(43)As1t=PSt∈W,T1≤t|S0=s1=∑s2∈W∫0tAs2t−xdqs1,s2x.Taking LTs on both sides of (42) and (43) and from (2)–(7), we can obtain the following:(1)
Ifs1=m,n,m+n≤W−1(44)Am,n∗s=qm,n,m+1,n∗sAm+1,n∗s+1−δm,0∑v=1mqm,n,m−v,n+v∗sAm−v,n+v∗s+∑v=0mqm,n,m−v,n+v+1∗sAm−v,n+v+1∗s+1−δn,0qm,n,m,n−1∗sAm,n−1∗s+qm,n,0,L+1∗sA0,L+1∗s+1s+qm,n.(2)
Ifs1=m,n,m+n=W,(45)Am,n∗s=qm,n,0,W+1∗sA0,W+1∗s+1−δm,0∑v=1mqm,n,m−v,n+v∗sAm−v,n+v∗s+1−δn,0qm,n,m,n−1∗sAm,n−1∗s+qm,n,0,L+1∗sA0,L+1∗s+1s+qm,n.(3)
Ifs1=0,n,W+1≤n≤L−2,(46)A0,n∗s=q0,n,0,n+1∗sA0,n+1∗s+q0,n,0,n−1∗sA0,n−1∗s+q0,n,0,L+1∗sA0,L+1∗s+1s+qm,n.(4)
Ifs1=0,L−1,(47)A0,L−1∗s=q0,L−1,0,L−2∗sA0,L−2∗s+q0,L−1,0,L∗sA0,L∗s+q0,L−1,0,L+1∗sA0,L+1∗s+1s+q0,L−1.(5)
Ifs1=0,L,(48)A0,L∗s=q0,L,0,L−1∗sA0,L−1∗s.(6)
Ifs1=0,L+1,(49)A0,L+1∗s=q0,L+1,0,0∗sA0,0∗s.To obtain the solutions ofAs1s,s1∈S, we denote the set of equations (44)–(49) in matrix equation form. Let(50)A0∗s=A0,0∗s,A0,1∗s,…,A0,L+1∗s,Am∗s=Am,0∗s,Am,1∗s,…,Am,W−m∗s,1≤m≤W,ς0s=1s+q0,0,1s+q0,1,…,1s+q0,L−1,0,0,ςms=1s+qm,0,1s+qm,1,…,1s+qm,W−m,1≤m≤W.Then,A∗s=A0∗s,A1∗s,…,AW∗sT and ςs=ς0s,ς1s,…,ςWsT are two column vectors with order τ×1. Thus, the equations (44)–(49) can be rewritten as the block matrix equation.(51)ΥsA∗s=ςs.Comparing (22)–(25) and (44)–(49), we find that the coefficient matrix Υs has the following form:(52)Υs=Υ0,0sΥ0,1sΥ1,0sB1,1sB1,2sΥ2,0sB2,1sB2,2sB2,3s⋮⋮⋮⋱⋱ΥW−2,0sBW−2,1sBW−2,2sBW−2,3s⋯BW−2,W−1sΥW−1,0sBW−1,1sBW−1,2sBW−1,3s⋯BW−1,W−1sBW−1,WsΥW,0sBW,1sBW,2sBW,3s⋯BW,W−1sBW,Ws,where, the submatrix Υ0,1s is a matrix with order L+2×W and is given by the following:(53)Υ0,1s=B0,1sOL×2O2×LO2×2.The submatrixΥ0,0s is a square matrix with order L+2×L+2 and is given by the following:(54)Υ0,0s=B0,0sΓsΔsI2,where, I2 is a second-order identity matrix, Γs is a matrix with order L×2 and is given by the following:(55)Γs=0−q0,0,0,L+1∗s0−q0,1,0,L+1∗s⋮⋮0−q0,L−2,0,L+1∗s−q0,L−1,0,L∗s−q0,L−1,0,L+1∗s,and Δs is a matrix with order 2×L given as follows:(56)Δs=00⋯0−q0,L,0,L−1∗s−q0,L+1,0,0∗s0⋯00.For1≤m≤W, the submatrix Υm,0s is a matrix with order W−m+1×L+2 and is given by the following:(57)Υm,0s=Bm,0sΓms,where, Γms is a matrix with order W−m+1×2 and is given by the following:(58)Γms=0−qm,0,0,L+1∗s0−qm,1,0,L+1∗s⋮⋮0−qm,W−m,0,L+1∗s.Cramer’s rule is used on equation (51) to find the expression of A∗s. Using the available software Matlab, we can obtain their numerical solutions, which are given in Section 6. By inverse Laplace transform, we can get the transient availability As1t at time t from any arbitrary initial state s1∈S, i.e., As1t=L−1As1∗s and the steady-state availability A=limt⟶+∞As1t=lims⟶0sAs1∗s,s1∈S.


## 6. Numerical Results

To illustrate the application of the results in this paper, sensitivity analysis of performance measures is performed numerically by using MATLAB. Firstly, we examine the effects of the initial state (i.e., the number of failed warm standby components and the number of failed components in the waiting line) on the system reliability in Figure1. And in Table 1, we listed the effect of different working pressure coefficients θ=0.1,0.7,4.9 on MTTFF. In Figure 1 and Table 1, we take N=5,W=3,K=3,λ=0.3;α=0.03,δ=0.0001,β=0.01,θ=0.7,μ=1.5,ν=2,p=0.95 as general case.Figure 1
Curves of reliability functions from different initial states.Table 1
MTTFF for various pressure coefficients and initial states.
θ=0.7c=0.9c=1.2MTTFF0,0107.85116.85132.00MTTFF0,1107.32116.32131.48MTTFF0,2106.24115.25130.41MTTFF0,3103.51112.52127.69MTTFF0,496.91105.80120.78MTTFF0,577.3485.1798.44MTTFF1,0105.66114.65129.79MTTFF1,1104.55113.54128.67MTTFF1,2102.23111.18126.26MTTFF2,0102.51111.48126.57MTTFF2,1100.05108.98124.01MTTFF3,097.61106.49121.48To show clearly the effect of different initial states on reliability function, we plot the curves grouped by the numbers of failed components, group 1 includes states0,0,0,1,0,2,0,3,0,4,and0,5, group 2 consists of states 1,0,1,1,and1,2, and group 3 contains the states 2,0,2,1,and3,0. From Figure 1, we have the following results:(i)
For any initial working states1, as function of time t, Rs1t always decreases and tends to zero. Such phenomenon agrees with the theoretic result.(ii)
With the same number of failed warm standbys, the reliability is decreasing with the increasing number of failed components in the repair service area, which can see more clearly from the magnification of one segment of the cures in each subplot. The reason is that when the number of failed components in the repair service is bigger, the availability of the good warm standby is smaller, then when an operating component fails, it has a higher chance to be replaced by a good warm standby.From Table1, we see that(i)
With the increase of the repair pressure coefficient, the mean time to first failure becomes longer no matter what the initial working state is. This result just illustrates that increasing the repair pressure can help prolonging the MTTFF, which is very important for the system because MTTFF is an important characteristic index for evaluating system reliability.(ii)
For the initial working statem,n, given that m and other system parameters are fixed, we have that with the value of n increasing, MTTFF is decreasing, that is MTTFF0,0>MTTFF0,1>MTTFF0,2>MTTFF0,3>MTTFF0,4>MTTFF0,5, MTTFF1,0>MTTFF1,1>MTTFF1,2, and MTTFF2,0>MTTFF2,1. On the other hand, we also observe that when m+n is fixed, the bigger the value of m, the smaller the value of MTTFFm,n, i.e., we have the relations MTTFF1,0<MTTFF0,1 and MTTFF2,0<MTTFF1,1<MTTFF0,2. That is because when m+n is given, the value of m is bigger, which leads to more failed warm standbys waiting for the arrival of a random inspection or a new failed operating component, only in this case it can obtain their repair. Such result also indicates the necessity of the random inspection.Secondly, we perform the sensitivity analysis for the performance measuresENw,ENr,ENo,A and FF by taking N=5,W=3,K=3,λ=0.3;α=0.03,δ=0.0001,β=0.01,θ=0.7,μ=3,ν=8,p=0.95 as general default values, the numerical results are shown in Figures 2–5.Figure 2
The effect ofλ on performance measures ENw,ENr,ENo,A and FF for different values of μ. (a) Mean number of failed warm standby components. (b) Mean number of failed components in the service area. (c) Mean number of operating components. (d) Steady-state availability. (e) Steady-state failure frequency.
(a)(b)(c)(d)(e)Figure 3
The effect ofα on performance measures ENw,ENr,ENo,A and FF for different values of β. (a) Mean number of failed warm standby components. (b) Mean number of failed components in the service area. (c) Mean number of operating components. (d) Steady-state availability. (e) Steady-state failure frequency.
(a)(b)(c)(d)(e)Figure 4
The effect ofδ on performance measures ENw,ENr,ENo,A and FF for different values of ν. (a) Mean number of failed warm standby components. (b) Mean number of failed components in the service area. (c) Mean number of operating components. (d) Steady-state availability. (e) Steady-state failure frequency.
(a)(b)(c)(d)(e)Figure 5
The effect ofp on performance measures ENw,ENr,ENo,A and FF for different values of θ. (a) Mean number of failed warm standby components. (b) Mean number of failed components in the service area. (c) Mean number of operating components. (d) Steady-state availability. (e) Steady-state failure frequency.
(a)(b)(c)(d)(e)From Figures2–5, we can observe that(i)
Figure2 shows that, with increasing of the value λ, the values of the mean failed warm standbys ENw, the mean operating components ENo, and the availability A are decreasing, but the mean failed components ENr and the failure frequency FF are increasing. The reason is due to that as the operating component’s failure rate λ increases, it can promote failed warm standbys to enter the repair service and decrease the number of operating components, thus ENw and ENo are decreasing functions of λ, while as ENr is increasing in function. Furthermore, the increasing of λ leads to the increasing of failure frequency and decreasing of the availability. From Figure 4, we can see the same effect of δ on these performance measures. The parameter δ is the arriving rate of CCF, when it becomes bigger, the time that the system enters failure becomes shorter.As our expectations, Figures2 and 4 display that the increasing of the repair rates μ and ν makes the mean number of operating components ENw,ENo and availability A both increase and makes the mean number of failed components in the repair service area ENr decrease. It should be noted that there is some differentiation in Figures 2 and 4, with the increasing of the repair rate μ, the failure frequency FF decreases, which agrees with our expectation, but FF increases with the increasing of the repair rate ν. This recommends that the decision maker and managers can considering to solve QoS by choosing appropriate service level and suggests that we should not make judgement based on subjective conjecture.(ii)
Figure3 reveals that when the failure rate of warm standbys increases, the mean number of failed warm standbys ENw and failed components in the repair service area ENr increase, which leads to the availability of good warm standbys decrease when an operating component fails and needs a well warm standby to replace it, and then the mean number of operating components ENo and the availability A decrease, and the failure frequency FF increases. Figure 3 also shows that with the increasing of the value of β, the mean number of failed warm standbys and the failure frequency are decreasing, the operating components and the availability are increasing. Furthermore, from Figure 5, we see the alike impact of the successful probability p of the random inspection on these performance measures. Such results illustrate the importance of the random inspection for the failed warm standbys.(iii)
Figure5 also depicts the effect of the pressure coefficient on these performance measures. From Figure 5, we can see that enhancing the repair rate by the appropriate increasing of the pressure coefficient can help reducing workload and improving the system availability, which can be justified by the trend of the curves given in Figure 5.


## 7. Conclusions

Common cause failure is a common type of failure dependency in repairable systems. The present study focused on the sensitivity analysis for a redundant repairable system with common cause failure, random inspection, and variable repair rates. By using Markov theory and Markov-renewal theory, we obtained the performance measures, reliability functions, and transient availability from any arbitrary initial state. This model can be used for modelling various repairable systems in engineering systems, industrial systems, and computing systems. The results and methods given in this paper have great importance in valuating the performance and designing the system parameters for the advancement of the respective modern engineering industries.However, the proposed method is limited to the assumption on exponential distributions. As a possible direction of future research, one could consider redundant systems with random inspection and repair pressure coefficient, in which the lifetime of operating components and warm standbys follow phase-type distributions. In this paper, we assume that theK-out-of-N+W: G repairable system only consists of identical components and the inspection time is negligible. Actually, the repairable systems can consist of multiple types of components and inspection takes time. So, as a second direction of future research, one could consider redundant series systems with multitype repairable system with non-negligible random inspection and variable repair rates. In this paper, we only consider the inspection imperfect for failed warm components, but a perfect warm standby may be reported as a failed standby. Therefore, as a third interesting research direction, one can consider the case that inspection is imperfect for all standbys instead of only for failed standbys.



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*Source: 5265698-2023-01-06.xml*