# Joint Economic Design of CUSUM Control Chart and Age-Based Imperfect Preventive Maintenance Policy

Authors: Yaping Li; Zhen Chen; Ershun Pan
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

The close relationship between statistical process control and maintenance has attracted lots of researchers to focus on the jointly economic design of control chart (a main tool of statistical process control) and preventive maintenance policy, and much progress has been made in this field. However, in the existing literatures, theX- chart is used most, and other charts are rarely considered. In this paper, the economic design of CUSUM chart and age-based imperfect preventive maintenance policy is presented. The process is considered as a multiphase system, and a recursive algorithm is used to model each phase. Besides, a sampling policy under the non-Markovian deterioration assumption is employed, and an age-based imperfect preventive maintenance policy is used. An optimization model with the objective of minimizing the expected cost per unit time is constructed to obtain the near-optimal solution of decision variables: the age of the machine for maintenance, the number of age-based maintenances, sample size, sampling intervals, and the decision interval coefficient and reference value coefficient of CUSUM chart. The solution procedure of the model is provided. Also, sensitivity analysis is performed on the decision variables for each of the various parameters.


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## Body

## 1. Introduction

### 1.1. Prior Literature and Motivation

Control chart, as an effective tool of statistical process control, has been used extensively by practitioners to monitor production process and help to identify and eliminate assignable causes. Since Duncan [1] first proposed an economic design method of X- charts to maintain current control of a process; the economic design of control charts has been one import issue in the quality control field. Also optimizing maintenance strategy is a hot issue in the reliability field. To study the two problems separately is necessary and reasonable; however, in most cases, the two problems are relevant and interact on each other. In the course of statistical process control, planned/preventive maintenances need to be carried out to decrease the failure rate of the machine and reduce product variation. Similarly, corrective maintenances need to be performed to restore an out-of-control state back to an in-control state and thus have an impact on the failure mode of the machine which ultimately leads to a reduction in quality shift [2, 3] and further change the process control requirement. The close relationship between quality and maintenance has led lots of researchers to focus on the integrated model of control chart and maintenance which are more realistic in practice.There has been much research on the integrated optimization for control chart and maintenance. Lochner [4], Kneile, Stephens and Vasudeva [5], and Katter et al. [6] preliminary studied the relationships between quality and maintenance. Tagaras [7] first put forward the cost model for statistical process control and maintenance. Ben-Daya and Rahim [8] developed a joint optimization model to determine preventive maintenance (PM) level and the design parameters of the X- control chart, namely, inspection intervals, sample size, and control limit, in which inspection intervals are different and the failure mechanism follows a general distribution with the increasing hazard. Later, Lee and Rahim [9] considered replacement cost and the remaining value of the machine in the economically integrated model in which maintenance cost is the function of machine age. Cassady et al. [10] proposed an integrated model of X- control chart and age-limited PM strategy, which captures the costs associated with product inspection, process downtime, and poor quality. Same to Cassady et al. [10], considering X- control chart and age-limited PM strategy, Yeung et al. [11] used discrete Markov process to get the approximately optimal joint strategy. Based on the research of Linderman et al. [12], Zhou and Zhu [13] studied the economic design of the integrated model of control chart and maintenance management. The grid-search approach was used to search the optimal solution. Chan and Wu [14] used CCC (cumulative count of conforming chart) and planned maintenance policy to optimize the integrated model. Mehrafrooz and Noorossana [15] presented an integrated model which considered complete failure and planned maintenance simultaneously. Six scenarios in production process are analyzed and a procedure for calculating average cost per time unit was proposed too. Yin et al. [16] developed an integrated model which considered the delayed monitoring and ten scenarios were analyzed in the paper. Charongrattanasakul and Pongpullponsak [17] proposed integrated approach for process control and maintenance by EWMA control chart. Genetic algorithm is used to find the optimal values of variables in the model. Later, Ardakan [18] studied the economic design of multiple variable EWMA control chart and maintenance strategy. Shrivastava et al. [19] proposed jointly optimal design of PM and CUSUM control chart with consideration of minimal corrective maintenance (CM) and imperfect maintenance strategy. The aim of the model is to minimize the cost per unit time to get the values of variables. Li et al. [20] considered machine health condition in jointly optimizing predictive maintenance policy and X-bar control chart. Markov method gets popular in recent years. Ho and Quinino [21] developed a Markov model with several control zones to analyze the maintenance performance. Xiang [22] proposed a model for a production process that deteriorates according to a discrete-time Markov chain and further provided a breakthrough in designing an efficient solution algorithm in obtaining analytical results. Liu et al. [23] used a five-state continuous time Markov chain for a two-unit series system to find the optimal control chart parameters by minimizing the average maintenance costs. Zhang et al. [24] proposed a delayed maintenance policy, estimated the state probabilities during the delayed period by Bayesian theory, and used Markov method to model the monitoring-maintenance process.If defining that a new phase starts after a PM and that a cycle may contain many phases, we find that the above-mentioned literature simplified the model by discussing a single phase without considering the repetitiveness of maintenance, or a multiphase process in which a PM was performed at every sampling, or a multiphase process with equivalent sampling intervals in the cycle. Moreover, in previous literature sampling number was usually set as a decision variable, as well as sampling interval, and thus the age of the machine can be calculated by them when PM was performed. We find that the age of the machine varies with the age-based PM actions, which caused unnecessary or absent maintenance actions with a high probability due to ignoring the actual age condition of the machine. In addition, in most references discussed aboveX- control chart is used most, but other control charts like CUSUM chart and EWMA chart are studied by only few researchers. To our knowledge, so far there is only one paper [19] about the joint optimization of CUSUM chart and PM policy. However, its researchers calculated the expected cycle cost and the expected cycle time by a direct analysis on different production scenarios, which cannot work very well for a more complex production process. This is because a direct analysis simplifies the process greatly due to its limitation that it has no ability to cover all possible scenarios.The motivation of this paper is based on these observations from the literatures. We consider a multiphase production system in which sampling for quality control inspection is carried out with unequal intervals and age-based PMs are performed in parallel. To guarantee the rationality of PM actions, we propose an age-based imperfect PM policy by setting the age of the machine for maintenance as a decision variable. We use a CUSUM chart to jointly model with PM policy by a recursive algorithm ([8, 9] Nourelfath, Nahas, and Ben-Daya, 2016)). The objective of the paper is to minimize the expected cost per unit time to obtain the near-optimal solutions of the age of the machine for maintenance, the number of age-based maintenances, sample size, sampling intervals, and the decision interval coefficient, and reference value coefficient of CUSUM control chart.


### 1.2. Contributions and Outline

The paper develops an integrated model to simultaneously optimize the design parameters of CUSUM chart and PM policy and presents the following important characteristic.First, a jointly economic design of CUSUM control chart and age-based imperfect PM policy is studied. To make up the deficiency of the direct analysis method in the literature [19], a recursive algorithm, modified from previously mentioned literatures, is proposed to model the problem. Using recursive algorithm can consider much more scenarios than using the direct analysis and thus obtain a relatively accurate solution for decision variables. For the characteristic of the objective function that it behaves similar to a convex function of N (the number of age-based maintenances), a solution procedure is developed to find a near-optimal policy which is also effective on sensitivity analysis.Second, a multiphase process constituted by multiple PMs and multiple samplings in each PM interval is studied. Inequivalent sampling intervals are used, and an age-based imperfect preventive maintenance policy is employed. If the machine functions without a detected failure forT0 time units, then PM is performed, and each PM restores the machine to a state between as good as new and as bad as old. The age of the machine after a PM will be reduced to a certain value, not zero because of the imperfectness of maintenance. T0, as the age of the machine for maintenance, is designed to be a decision variable. As a result, our policy will reduce the probability of unnecessary or absent maintenances.The remainder of this paper is organized as follows. Section2 provides problem statement and model assumptions, in which quality control, imperfect PM, and sampling interval are discussed. Section 3 introduces how to obtain the expected phase cost and expected phase time using the recursive method. On the basis of this, the expected cycle cost and expected cycle time are calculated by taking a consideration of the repetitiveness of the sampling and maintenance. Then to minimize the expected cost per unit time, the joint model is built; further model features and solution algorithm are reported. Next, in Section 4 a numerical case is used to verify the effectiveness of the model and perform the sensitivity analysis on the decision variables for each of the various parameters. At last, conclusions and possible research in the future are presented in Section 5.


## 1.1. Prior Literature and Motivation

Control chart, as an effective tool of statistical process control, has been used extensively by practitioners to monitor production process and help to identify and eliminate assignable causes. Since Duncan [1] first proposed an economic design method of X- charts to maintain current control of a process; the economic design of control charts has been one import issue in the quality control field. Also optimizing maintenance strategy is a hot issue in the reliability field. To study the two problems separately is necessary and reasonable; however, in most cases, the two problems are relevant and interact on each other. In the course of statistical process control, planned/preventive maintenances need to be carried out to decrease the failure rate of the machine and reduce product variation. Similarly, corrective maintenances need to be performed to restore an out-of-control state back to an in-control state and thus have an impact on the failure mode of the machine which ultimately leads to a reduction in quality shift [2, 3] and further change the process control requirement. The close relationship between quality and maintenance has led lots of researchers to focus on the integrated model of control chart and maintenance which are more realistic in practice.There has been much research on the integrated optimization for control chart and maintenance. Lochner [4], Kneile, Stephens and Vasudeva [5], and Katter et al. [6] preliminary studied the relationships between quality and maintenance. Tagaras [7] first put forward the cost model for statistical process control and maintenance. Ben-Daya and Rahim [8] developed a joint optimization model to determine preventive maintenance (PM) level and the design parameters of the X- control chart, namely, inspection intervals, sample size, and control limit, in which inspection intervals are different and the failure mechanism follows a general distribution with the increasing hazard. Later, Lee and Rahim [9] considered replacement cost and the remaining value of the machine in the economically integrated model in which maintenance cost is the function of machine age. Cassady et al. [10] proposed an integrated model of X- control chart and age-limited PM strategy, which captures the costs associated with product inspection, process downtime, and poor quality. Same to Cassady et al. [10], considering X- control chart and age-limited PM strategy, Yeung et al. [11] used discrete Markov process to get the approximately optimal joint strategy. Based on the research of Linderman et al. [12], Zhou and Zhu [13] studied the economic design of the integrated model of control chart and maintenance management. The grid-search approach was used to search the optimal solution. Chan and Wu [14] used CCC (cumulative count of conforming chart) and planned maintenance policy to optimize the integrated model. Mehrafrooz and Noorossana [15] presented an integrated model which considered complete failure and planned maintenance simultaneously. Six scenarios in production process are analyzed and a procedure for calculating average cost per time unit was proposed too. Yin et al. [16] developed an integrated model which considered the delayed monitoring and ten scenarios were analyzed in the paper. Charongrattanasakul and Pongpullponsak [17] proposed integrated approach for process control and maintenance by EWMA control chart. Genetic algorithm is used to find the optimal values of variables in the model. Later, Ardakan [18] studied the economic design of multiple variable EWMA control chart and maintenance strategy. Shrivastava et al. [19] proposed jointly optimal design of PM and CUSUM control chart with consideration of minimal corrective maintenance (CM) and imperfect maintenance strategy. The aim of the model is to minimize the cost per unit time to get the values of variables. Li et al. [20] considered machine health condition in jointly optimizing predictive maintenance policy and X-bar control chart. Markov method gets popular in recent years. Ho and Quinino [21] developed a Markov model with several control zones to analyze the maintenance performance. Xiang [22] proposed a model for a production process that deteriorates according to a discrete-time Markov chain and further provided a breakthrough in designing an efficient solution algorithm in obtaining analytical results. Liu et al. [23] used a five-state continuous time Markov chain for a two-unit series system to find the optimal control chart parameters by minimizing the average maintenance costs. Zhang et al. [24] proposed a delayed maintenance policy, estimated the state probabilities during the delayed period by Bayesian theory, and used Markov method to model the monitoring-maintenance process.If defining that a new phase starts after a PM and that a cycle may contain many phases, we find that the above-mentioned literature simplified the model by discussing a single phase without considering the repetitiveness of maintenance, or a multiphase process in which a PM was performed at every sampling, or a multiphase process with equivalent sampling intervals in the cycle. Moreover, in previous literature sampling number was usually set as a decision variable, as well as sampling interval, and thus the age of the machine can be calculated by them when PM was performed. We find that the age of the machine varies with the age-based PM actions, which caused unnecessary or absent maintenance actions with a high probability due to ignoring the actual age condition of the machine. In addition, in most references discussed aboveX- control chart is used most, but other control charts like CUSUM chart and EWMA chart are studied by only few researchers. To our knowledge, so far there is only one paper [19] about the joint optimization of CUSUM chart and PM policy. However, its researchers calculated the expected cycle cost and the expected cycle time by a direct analysis on different production scenarios, which cannot work very well for a more complex production process. This is because a direct analysis simplifies the process greatly due to its limitation that it has no ability to cover all possible scenarios.The motivation of this paper is based on these observations from the literatures. We consider a multiphase production system in which sampling for quality control inspection is carried out with unequal intervals and age-based PMs are performed in parallel. To guarantee the rationality of PM actions, we propose an age-based imperfect PM policy by setting the age of the machine for maintenance as a decision variable. We use a CUSUM chart to jointly model with PM policy by a recursive algorithm ([8, 9] Nourelfath, Nahas, and Ben-Daya, 2016)). The objective of the paper is to minimize the expected cost per unit time to obtain the near-optimal solutions of the age of the machine for maintenance, the number of age-based maintenances, sample size, sampling intervals, and the decision interval coefficient, and reference value coefficient of CUSUM control chart.


## 1.2. Contributions and Outline

The paper develops an integrated model to simultaneously optimize the design parameters of CUSUM chart and PM policy and presents the following important characteristic.First, a jointly economic design of CUSUM control chart and age-based imperfect PM policy is studied. To make up the deficiency of the direct analysis method in the literature [19], a recursive algorithm, modified from previously mentioned literatures, is proposed to model the problem. Using recursive algorithm can consider much more scenarios than using the direct analysis and thus obtain a relatively accurate solution for decision variables. For the characteristic of the objective function that it behaves similar to a convex function of N (the number of age-based maintenances), a solution procedure is developed to find a near-optimal policy which is also effective on sensitivity analysis.Second, a multiphase process constituted by multiple PMs and multiple samplings in each PM interval is studied. Inequivalent sampling intervals are used, and an age-based imperfect preventive maintenance policy is employed. If the machine functions without a detected failure forT0 time units, then PM is performed, and each PM restores the machine to a state between as good as new and as bad as old. The age of the machine after a PM will be reduced to a certain value, not zero because of the imperfectness of maintenance. T0, as the age of the machine for maintenance, is designed to be a decision variable. As a result, our policy will reduce the probability of unnecessary or absent maintenances.The remainder of this paper is organized as follows. Section2 provides problem statement and model assumptions, in which quality control, imperfect PM, and sampling interval are discussed. Section 3 introduces how to obtain the expected phase cost and expected phase time using the recursive method. On the basis of this, the expected cycle cost and expected cycle time are calculated by taking a consideration of the repetitiveness of the sampling and maintenance. Then to minimize the expected cost per unit time, the joint model is built; further model features and solution algorithm are reported. Next, in Section 4 a numerical case is used to verify the effectiveness of the model and perform the sensitivity analysis on the decision variables for each of the various parameters. At last, conclusions and possible research in the future are presented in Section 5.


## 2. Problem Statement and Assumptions

The nomenclature is defined as Table1.Table 1
The nomenclature.
Nomenclature T 0 The age of the machine for maintenance N The number of age-based maintenances n Sample size h k j Thejth sampling interval in the kth phase h H The decision interval coefficient of CUSUM chart L The reference value coefficient of CUSUM chart α Probability of type I error β Probability of type II error Z 1 Average time of an PM Z 2 Average time of a replacement c f Fixed cost of each sampling c v Average variable cost per unit c C p M Average cost of a compensatory maintenance c C M Average cost of a CM c P M Average cost of a PM c R P Average cost of a replacement E Q C The expected quality control cost in the cycle E P M C The expected PM cost in the cycle E T The expected cycle time E Q C k The expected quality control cost in thekth phase E P M C k The expected PM cost in thekth phase E T k The expected phase time of thekth phase E c The expected cost per unit time in the cycleWe consider a manufacturing system with a single deteriorating machine which produces a type of products in a finite time horizon. The finite time horizon is usually called a cycle time. In the cycle, a CUSUM control chart is used to monitor the state of the production system and meanwhile maintenance policy is performed as scheduled. A production cycle begins with a system which is assumed to be in an in-control state, producing items of acceptable quality. The process is stopped at timest1,t2,⋯,tN for planned PM activities. Before the process stops, sampling for inspection is performed with unequal sampling intervals. If a failure occurs before PM action, a corrective maintenance (CM) is performed first to restore the process to an as bad as old state, then PM. Let the number of age-based maintenances,N, be a decision variable. Thus, the production cycle is divided intoN phases. A phase cycle ends either with a true alarm signaling that the process is out of control or when the time to perform the PM activity is up, whichever occurs first. The process is then restored to the in-control state by maintenance. The production cycle ends when theNth PM time arrives and a replacement is carried out instead of the PM action. All PM activities bring the machine to a state between as good as new and as bad as old, called imperfect maintenance too. The replacement will bring the process to as good as new state. Production process withN-1 PM activities and a replacement is shown as Figure 1.Figure 1
Production process in a cycle.The production process of thekth phase with both quality control inspection and PM activities is shown in Figure 2, in which s is the abbreviation of sampling for quality control inspection; other mathematical notations will be explained in the following parts.Figure 2
Production process in thekth phase.


### 2.1. Quality Control

Suppose the variation of quality characteristicx follows a normal distribution with mean μ and standard deviation σ in the in-control sate, i.e., x~Nμ,σ2, where both μ and σ are known. But after a while, the process may shift to an out-of-control state with the mean of the quality characteristic changing from μ0μ0=μ to μ0±δμσ0, where μ0,σ0 are the mean and standard deviation of samples, respectively, δμ is the magnitude of quality shift, and the standard deviation is assumed to remain the same. This particular type of shift can be attributed to an equipment failure which is subtle and cannot be recognized without shutting down the process and performing close inspection of the equipment.We apply Table CUSUM control chart to monitor the production process. Let statisticC+ denote the accumulating derivations from μ0 that are above target; let another statistic C- denote the accumulating derivations from μ0 that are below target. The statistics C+,C- are called one-sided upper and lower CUSUMs, respectively. They are computed as follows:(1)Ci+=max⁡0,xi-μ0+K+Ci-1+Ci-=max⁡0,μ0-K-xi+Ci-1-,where i denotes the ith sample, the starting values are C0+=C0-=0, K=Lσ0 is the reference value, and H=hHσ0 is the decision interval to determine if the process has been out of control.Siegmund’s average running length (ARL) approximation [25] is recommended mostly because of its simplicity. For a one-sided CUSUM (that is, Ci+ or Ci-) with parameters hH andL, Siegmund’s approximation is(2)ARL=exp⁡-2Δb+2Δb-12Δ2.For Δ≠0, where Δ=δμ-L for the upper one-sided CUSUM Ci+, Δ=-δμ-L for the lower one-sided CUSUM Ci-, and b=hH+1.166. For Δ=0, one can use ARL=b2.The quantityδμ represents the magnitude of quality shift, for which theARL is to be calculated. Therefore, if δμ=0, we would calculate ARL0+,ARL0- from (2),(3)ARL0+=ARL0-=exp⁡2Lb-2Lb-12L2Whereas ifδμ≠0, we would calculate ARL1+,ARL1- from (2),(4)ARL1+=exp-2δμ-Lb+2δμ-Lb-12δμ-L2(5)ARL1-=exp-2-δμ-Lb+2-δμ-Lb-12-δμ-L2To obtain theARL of the two-sided CUSUM from theARLs of the two-sided statistics, ARL+,ARL-, we use(6)1ARL=1ARL++1ARL-Since a sample is taken from the process, the control chart can lead to both Type I errors and Type II error. Letα denote the probability of a Type I error and β denote the probability of a Type II error. Then, α and β are, respectively, calculated corresponding to ARL0,ARL1 by(7)α=1ARL0(8)β=1-1ARL1


### 2.2. Imperfect Preventive Maintenance

A more realistic situation is one in which the failure mechanism of a preventively maintained system changes. The imperfect PM will bring that the failure rate of the system is somewhere between as good as new and as bad as old after maintenance. One way to model this is to assume a reduction in the age of the machine. Letwk be the age of the machine after thekth PM, η be the imperfectness factor, and 0<η≤1. As scheduled, a PM action is performed only when the age of the machine has arrived T0. The following equation is assumed:(9)wk=1-ηkT0.η is a measurement for the degradation of the age of the machine under the effect of a PM activity. In particular, η=1 represents a PM (a replacement is performed instead of a scheduled PM) which restores the system to as good as new state.Letmk be the sampling number in the kth phase. If wk-1+∑j=1Mhkj≤T0 and wk-1+∑j=1M+1hkj>T0, then mk=M. Note that if mk=0, it means that there is no time to sample for quality inspection and a failure would occur quickly after maintenance. So we assume that the machine is not worth being maintained again and a replacement should be performed when mk=0.LetΔk be the residual time in the phase after the mkth sampling. It is obvious that wk-1,hkj,T0,Δk satisfy(10)wk-1+∑j=1mkhkj+Δk=T0.


### 2.3. Sampling Interval

The time for the process to be in the in-control state, before the shift occurs, is assumed to be an arbitrary probability distribution, but we used the Weibull distribution for it is often used to model the time to failure of many different physical systems, especially in electrical and mechanical components and systems. If the time that the process remains in-control obeys a two-parameter Weibull distribution, let the probability density function be(11)ft=γνtν-1e-γtν,t>0,ν≥1,γ>0,where γ is the scale parameter and ν is the shape parameter of the distribution.Letλt be the hazard function, defined by(12)λt=ftF-t.From (11) and (12), the hazard function λt is given by(13)λt=γνtν-1Similar to Banerjee and Rahim [26] and Ben-Daya and Rahim [8], we choose the length of sampling intervals such that the integrated hazard over each interval is the same for all intervals; that is,(14)∫tki-1tkiλtdt=∫wk-1tk1λtdt,i=2,⋯,mk.Suppose sampling intervals arehk1,hk2,⋯,hkmk; then w0=0,tk0=wk-1,tki=wk-1+∑j=1ihkj,i=1,⋯,mk, so condition (14) becomes (15)∫tki-1tki-1+hkiλtdt=∫wk-1wk-1+hk1λtdt,i=1,2,⋯,mk.So the length of the sampling interval, hki, can be given by the following equation:(16)wk-1+hk1ν-wk-1v=tki-1+hkiν-tki-1ν,i=2,⋯,mkInstitutetki=wk-1+∑j=1ihkj,i=1,⋯,mk  to (16); we get hki by(17)hki=wk-1+∑j=1i-1hkjν+wk-1+hk1v-wk-1vv-wk-1+∑j=1i-1hkj,i=2,⋯,mkThe integrated hazard over each interval is the same for all the first interval in each phase; that is,(18)∫wk-1tk1λtdt=∫0t11λtdt,k=1,2,⋯,NThe first sampling intervals areh11,h21,⋯,hN1, and w0=0,tk1=wk-1+hk1,k=1,⋯,N, so condition (18) becomes (19)∫wk-1wk-1+hk1λtdt=∫0h11λtdt,k=1,2,⋯,NThus, hk1 can be calculated by the following equation:(20)hk1=h11ν+wk-1vv-wk-1,k=1,⋯,NFurther, by substituting (20) into (17), the length of the sampling interval hki,i=2,⋯,mk can be given by(21)hki=wk-1+∑j=1i-1hkiν+h11vv-wk-1+∑j=1i-1hki,i=2,⋯,mkBy (21), it is conceivable to reduce the problem of finding ∑k=1Nmk distinct sampling intervals to the much more manageable problem of finding h11.


## 2.1. Quality Control

Suppose the variation of quality characteristicx follows a normal distribution with mean μ and standard deviation σ in the in-control sate, i.e., x~Nμ,σ2, where both μ and σ are known. But after a while, the process may shift to an out-of-control state with the mean of the quality characteristic changing from μ0μ0=μ to μ0±δμσ0, where μ0,σ0 are the mean and standard deviation of samples, respectively, δμ is the magnitude of quality shift, and the standard deviation is assumed to remain the same. This particular type of shift can be attributed to an equipment failure which is subtle and cannot be recognized without shutting down the process and performing close inspection of the equipment.We apply Table CUSUM control chart to monitor the production process. Let statisticC+ denote the accumulating derivations from μ0 that are above target; let another statistic C- denote the accumulating derivations from μ0 that are below target. The statistics C+,C- are called one-sided upper and lower CUSUMs, respectively. They are computed as follows:(1)Ci+=max⁡0,xi-μ0+K+Ci-1+Ci-=max⁡0,μ0-K-xi+Ci-1-,where i denotes the ith sample, the starting values are C0+=C0-=0, K=Lσ0 is the reference value, and H=hHσ0 is the decision interval to determine if the process has been out of control.Siegmund’s average running length (ARL) approximation [25] is recommended mostly because of its simplicity. For a one-sided CUSUM (that is, Ci+ or Ci-) with parameters hH andL, Siegmund’s approximation is(2)ARL=exp⁡-2Δb+2Δb-12Δ2.For Δ≠0, where Δ=δμ-L for the upper one-sided CUSUM Ci+, Δ=-δμ-L for the lower one-sided CUSUM Ci-, and b=hH+1.166. For Δ=0, one can use ARL=b2.The quantityδμ represents the magnitude of quality shift, for which theARL is to be calculated. Therefore, if δμ=0, we would calculate ARL0+,ARL0- from (2),(3)ARL0+=ARL0-=exp⁡2Lb-2Lb-12L2Whereas ifδμ≠0, we would calculate ARL1+,ARL1- from (2),(4)ARL1+=exp-2δμ-Lb+2δμ-Lb-12δμ-L2(5)ARL1-=exp-2-δμ-Lb+2-δμ-Lb-12-δμ-L2To obtain theARL of the two-sided CUSUM from theARLs of the two-sided statistics, ARL+,ARL-, we use(6)1ARL=1ARL++1ARL-Since a sample is taken from the process, the control chart can lead to both Type I errors and Type II error. Letα denote the probability of a Type I error and β denote the probability of a Type II error. Then, α and β are, respectively, calculated corresponding to ARL0,ARL1 by(7)α=1ARL0(8)β=1-1ARL1


## 2.2. Imperfect Preventive Maintenance

A more realistic situation is one in which the failure mechanism of a preventively maintained system changes. The imperfect PM will bring that the failure rate of the system is somewhere between as good as new and as bad as old after maintenance. One way to model this is to assume a reduction in the age of the machine. Letwk be the age of the machine after thekth PM, η be the imperfectness factor, and 0<η≤1. As scheduled, a PM action is performed only when the age of the machine has arrived T0. The following equation is assumed:(9)wk=1-ηkT0.η is a measurement for the degradation of the age of the machine under the effect of a PM activity. In particular, η=1 represents a PM (a replacement is performed instead of a scheduled PM) which restores the system to as good as new state.Letmk be the sampling number in the kth phase. If wk-1+∑j=1Mhkj≤T0 and wk-1+∑j=1M+1hkj>T0, then mk=M. Note that if mk=0, it means that there is no time to sample for quality inspection and a failure would occur quickly after maintenance. So we assume that the machine is not worth being maintained again and a replacement should be performed when mk=0.LetΔk be the residual time in the phase after the mkth sampling. It is obvious that wk-1,hkj,T0,Δk satisfy(10)wk-1+∑j=1mkhkj+Δk=T0.


## 2.3. Sampling Interval

The time for the process to be in the in-control state, before the shift occurs, is assumed to be an arbitrary probability distribution, but we used the Weibull distribution for it is often used to model the time to failure of many different physical systems, especially in electrical and mechanical components and systems. If the time that the process remains in-control obeys a two-parameter Weibull distribution, let the probability density function be(11)ft=γνtν-1e-γtν,t>0,ν≥1,γ>0,where γ is the scale parameter and ν is the shape parameter of the distribution.Letλt be the hazard function, defined by(12)λt=ftF-t.From (11) and (12), the hazard function λt is given by(13)λt=γνtν-1Similar to Banerjee and Rahim [26] and Ben-Daya and Rahim [8], we choose the length of sampling intervals such that the integrated hazard over each interval is the same for all intervals; that is,(14)∫tki-1tkiλtdt=∫wk-1tk1λtdt,i=2,⋯,mk.Suppose sampling intervals arehk1,hk2,⋯,hkmk; then w0=0,tk0=wk-1,tki=wk-1+∑j=1ihkj,i=1,⋯,mk, so condition (14) becomes (15)∫tki-1tki-1+hkiλtdt=∫wk-1wk-1+hk1λtdt,i=1,2,⋯,mk.So the length of the sampling interval, hki, can be given by the following equation:(16)wk-1+hk1ν-wk-1v=tki-1+hkiν-tki-1ν,i=2,⋯,mkInstitutetki=wk-1+∑j=1ihkj,i=1,⋯,mk  to (16); we get hki by(17)hki=wk-1+∑j=1i-1hkjν+wk-1+hk1v-wk-1vv-wk-1+∑j=1i-1hkj,i=2,⋯,mkThe integrated hazard over each interval is the same for all the first interval in each phase; that is,(18)∫wk-1tk1λtdt=∫0t11λtdt,k=1,2,⋯,NThe first sampling intervals areh11,h21,⋯,hN1, and w0=0,tk1=wk-1+hk1,k=1,⋯,N, so condition (18) becomes (19)∫wk-1wk-1+hk1λtdt=∫0h11λtdt,k=1,2,⋯,NThus, hk1 can be calculated by the following equation:(20)hk1=h11ν+wk-1vv-wk-1,k=1,⋯,NFurther, by substituting (20) into (17), the length of the sampling interval hki,i=2,⋯,mk can be given by(21)hki=wk-1+∑j=1i-1hkiν+h11vv-wk-1+∑j=1i-1hki,i=2,⋯,mkBy (21), it is conceivable to reduce the problem of finding ∑k=1Nmk distinct sampling intervals to the much more manageable problem of finding h11.


## 3. Model Development

Since the production cycle is divided intoN phases, and each phase has similar scenarios, we first take thekth phase for example to analyse the problem and then solve the problem for the whole cycle. The approach used to derive the expected phase time ETk and the expected quality control cost EQCk is very similar to one that was earlier used by Ben-Day and Rahim [8]. The difference is that a sampling is carried out at the end of the last inspection interval because of the existence of residual time Δk after mk samplings and that CM is a minor repair, after which the PM will be carried out immediately; thus, there areN PMs (particularly the last PM is replaced by a replacement) in the cycle. Furthermore, we assume that(22)cCpM≤cCM≤cPM≤cRP.


### 3.1. Expected Cycle Time Analysis

Taking the time of thekth phase, for example, we first compute the expected phase time ETk and then consider the expected cycle time ET.Letpkj,j=1,2,⋯,mk be the conditional probability that the process shifts to an out-of-control state during the time interval tkj-1,tkj, given that the process was in in-control state at time tkj-1. In addition, we assume that the process starts in the in-control state in each phase, so pk0=0. The theorem about expected phase time is given.Theorem 1.
In thekth phase, the expected phase time ETk is(23)ETk=∑j=1mkhkj∏i=0j-11-pki+β∑j=1mkpkj∑i=j+1mkβi-j-1hki+βmk-jΔk∏i=0j-11-pki+∏j=1mk1-pkjΔk.Proof.
LetETkj,j=0,1,2,⋯,mk be the expected residual time in the phase beyond time tkj excluding Z1 or Z2, given that the process is in control at time tkj. Obviously ETk=ETk0. Meanwhile, let ERkj,j=1,2,⋯,m be the expected residual time in the phase beyond time tkj excluding Z1 or Z2, given that the process is out of control at time tkj and a true alarm has not emitted yet.
Further, from the definition ofpkj, we get pkj=(Fktkj-Fktkj-1)/(1-Fktkj-1),j=1,2,⋯,mk, where Fk· is the cumulative function of the Weibull distribution in thekth phase.
Consider the possible states of the machine, at the end of the first sampling interval during thekth phase, i.e., at time tk1. In order to find an expression for ETk, the expected residual time beyond time tk1 in the phase and the associated probabilities are presented in Table 2.
From the information shown in Table2, we get(24)ERkmk=Δk,ERkj=hj+1+βERkj+1=∑i=j+1mkβi-j-1hki+βmk-jERkmk,j=1,2,⋯,mk-1.Then we get(25)ETk=hk1+βpk1ERk1+1-pk1ETk1,ETkmk=Δk,ETkj=hkj+1+βpkj+1ERkj+1+1-pkj+1ETkj+1,j=1,2,⋯,mk-1,ETk=∑j=1mkhkj∏i=0j-11-pki+β∑j=1mkpkjERkj∏i=0j-11-pki+∏j=1mk1-pkjΔk.Substituting (24) into (25), we get (23).
Considering PM time, the expected cycle time is(26)ET=∑k=1NETk+N-1Z1+Z2.Table 2
The expected residual times and associated probabilities.
State Probability Expected residual time In control and no alarm 1 - p k 1 1 - α E T k 1 In control but a false alarm 1 - p k 1 α E T k 1 Out of control but no alarm p k 1 β E R k 1 Out of control and a true alarm p k 1 1 - β 0


### 3.2. Cost Analysis

#### 3.2.1. Quality Control Cost

In this paper, the expected quality control costEQC is used to express all cost happening for quality control. Quality control cost mainly includes quality loss costs respective for conforming items and nonconforming items, the cost of false alarm, the cost of finding assignable causes and restoring the process, and sampling and inspection cost. Similar to the analysis of the expected cycle time, we first compute the expected quality control cost of thekth phase EQCk and then consider the expected quality control cost of the cycle EQC.To find an expression forEQCk, we need to search expressions for nonconforming items.LetNkj be the expected sampling number conducted after time tkj given that the process is in the out-of-control state at time tkj. Nkj can be given by(27)Nmk=0,Nmk-1=1,Nkj=∑i=1mk-ji1-ββi-1+mk-jβmk-j,j=1,2,⋯,mk-2.Letτkj be the conditional expected in-control time within the time interval tk,j-1,tkj, given that the process was in in-control state at time tk,j-1; then τkj can be expressed by(28)τkj=∫tkj-1tkjt-tkj-1fktdtFktkj-Fktkj-1,j=1,2,⋯,mk.In particular, τkΔ is the conditional expected in-control time within the time interval tkmk,T0, given that the process was in in-control state at time tkmk, τkΔ=∫tkmkT0t-tkmkfktdt/(FkT0-Fktkmk).Theorem 2.
In thekth phase, the expected quality control cost EQCk is(29)EQCk=cf+cvn1+∑j=1mk-1∏i=1j1-pki+∑j=2mkpkjQ0τkj+Q1hkj-τkj∏i=1j-11-pki+cCM∑j=2mkpkj∏i=1j-11-pki+β∑j=2mkpkjNkjcf+cvn+Q1ERkj∏i=1j-11-pki+αcCpM∑j=1mk∏i=1j1-pki+Q0∑j=1mkhkj∏i=1j1-pki+∏i=1mk1-pki1-pkΔQ0Δk+pkΔQ0Δk+Q1Δk-τkΔ+cCM+pk1Q0τk1+Q1hk1-τk1+pk1cCM+βpk1Nk1cf+cvn+Q1ERk1where pkΔ is the conditional probability that the process shifts to the out-of-control state during the time interval tkmk,T0, given that the process was in in-control state at time tkmk, pkΔ=(FkT0-Fktkmk)/(1-Fktkmk).Proof.
LetEQCkj,j=1,2,⋯,mk-1 be the expected residual quality control cost in the phase beyond time tkj, given that the process is in control at time tkj.
Consider the possible states of the machine at the end of the first sample interval, i.e., at timetk1, during thekth phase. In order to find an expression for EQCk, the expected residual quality control cost in the phase and the associated probabilities are presented in  Table 3.
Similar to the way of developingETk, the following equations are presented by(30)EQCk=cf+cvn+pk1Q0τk1+Q1hk1-τk1+pk1cCM+βpk1Nk1cf+cvn+Q1ERk1+1-pk1EQCk1+1-pk1αcCPM+1-pk1Q0hk1
Forj=1,2,⋯,mk-1,(31)EQCkj=cf+cvn+pkj+1Q0τkj+1+Q1hkj+1-τkj+1+pkj+1cCM+βpkj+1Nkj+1cf+cvn+Q1ERkj+1+1-pkj+1EQCkj+1+1-pkj+1αcCPM+1-pkj+1Q0hkj+1EQCkmk=1-pkΔQ0Δk+pkΔQ0Δk+Q1Δk-τkΔ+cCMSo (32)EQCk=cf+cvn1+∑j=1mk-1∏i=1j1-pki+∑j=2mkpkjQ0τkj+Q1hkj-τkj∏i=1j-11-pki+cCM∑j=2mkpkj∏i=1j-11-pki+β∑j=2mkpkjNkjcf+cvn+Q1ERkj∏i=1j-11-pki+αcCPM∑j=1mk∏i=1j1-pki+Q0∑j=1mkhkj∏i=1j1-pki+∏i=1mk1-pkiEQCkmk+pk1Q0τk1+Q1hk1-τk1+pk1cCM+βpk1Nk1cf+cvn+Q1ERk1By substituting (31) into (32), we get (29).
The expected quality control cost is the sum ofN phase quality control costs; that is, (33)EQC=∑k=1NEQCk.Table 3
The expected residual costs and associated probabilities.
State Probability Current cost Expected residual cost In control and no alarm 1 - p k 1 1 - α c f + c v n + Q 0 h k 1 E Q C k 1 In control but a false alarm 1 - p k 1 α c f + c v n + c C p M + Q 0 h k 1 E Q C k 1 Out of control but no alarm p k 1 β c f + c v n + Q 0 τ k 1 + Q 1 h k 1 - τ k 1 c C M + N k 1 a + b n + Q 1 E R k 1 Out of control and a true alarm p k 1 1 - β c f + c v n + Q 0 τ k 1 + Q 1 h k 1 - τ k 1 c C M


#### 3.2.2. PM Cost

The expected PM cost is the sum ofN -1 PM costs and a replacement cost; that is, (34)EPMC=N-1cPM+cRP.


### 3.3. Modelling

Obviously, the expected total cost in the cycle time is the sum of the expected quality control cost and the expected PM cost, both discussed in Section3.2, expressed by(35)EC=EQC+EPMCThe expected cost per unit time can be considered as the ratio of the expected total cost in the cycle to the expected cycle time. As a result, the expected cost per unit time is expressed by(36)Ec=ECETBy minimizing the cost per unit time for the cycle, the integrated model of preventive maintenance policy and statistical process control can be expressed by(37)minEcs.t.ARL0≥C1,ARL1≤C2,a1≤T0≤b1,a2≤N≤b2,a3≤n≤b3,a4≤h1≤b4,a5≤hH0≤b5,a6≤L≤b6,in which T0,N,n,h1,hH,L are six decision variables, C1,C2 represent constraints on the average running lengths of the in-control state and out-of-control state, respectively, and ai,bi,i=1,2,3,4,5,6 represent constraints on decision variables.


### 3.4. Solution Procedure

Due to the cost relationship assumed in (22), performing no PM will never be optimal, and performing too many PMs will cause rising cost. With increasing PM number, the objective value will fall first until the optimal policy is reached and then rise. To solve the problem, theN must be iterated over until the cost objective begins to increase. Therefore, our solution algorithm is summarized as follows.Step 1.
Begin withN=1, iterate over specified values of T0,n,h1,hH,L, and record best Ec.Step 2.
LetN=N+1, iterate over specified values of T0,n,h1,hH,L, and record best Ec′.Step 3.
CompareEc and Ec′. If Ec′<Ec, return to Step 2; else the best solution is Ec.Because the objective model contains implicit function, differentiation, and nonlinear constraints, it is hard to solve the problem by Hessian matrix. We use a total enumeration method for specified ranges ofT0,n,h1,hH,L to get the solution. The decision variables T0,n,h1,hH,L are not tested for all possible values, but rather the most probable and practical values. For eachN, the T0 values are tested from Δh to the mean time to failure (MTTF) in increments of Δh, because we think the PM interval should be not greater than MTTF; Δh hours is the requirement of accuracy in practice. Then values are tested from 1 to 7, 10,15,30. This is drawn on the experience of Cassady et al. [10] and Yeung et al. [11]. Same to the requirement of accuracy of T0, the h1 values are tested in increments of Δh in a certain range depending on circumstances. The hH values are tested from 0.1 to 10 in increments of 0.1, and the L values are tested from 0.25 to 2 in increments of 0.25. Both of hH and L values are determined on the basis of the table given by Hawkins (1993) that concluded L values and the corresponding hH values that will achieve ARL0=370. Due to the fact that we do not test every possible value of six decision variables, we define the solution of our algorithm to be near-optimal, rather than optimal.


## 3.1. Expected Cycle Time Analysis

Taking the time of thekth phase, for example, we first compute the expected phase time ETk and then consider the expected cycle time ET.Letpkj,j=1,2,⋯,mk be the conditional probability that the process shifts to an out-of-control state during the time interval tkj-1,tkj, given that the process was in in-control state at time tkj-1. In addition, we assume that the process starts in the in-control state in each phase, so pk0=0. The theorem about expected phase time is given.Theorem 1.
In thekth phase, the expected phase time ETk is(23)ETk=∑j=1mkhkj∏i=0j-11-pki+β∑j=1mkpkj∑i=j+1mkβi-j-1hki+βmk-jΔk∏i=0j-11-pki+∏j=1mk1-pkjΔk.Proof.
LetETkj,j=0,1,2,⋯,mk be the expected residual time in the phase beyond time tkj excluding Z1 or Z2, given that the process is in control at time tkj. Obviously ETk=ETk0. Meanwhile, let ERkj,j=1,2,⋯,m be the expected residual time in the phase beyond time tkj excluding Z1 or Z2, given that the process is out of control at time tkj and a true alarm has not emitted yet.
Further, from the definition ofpkj, we get pkj=(Fktkj-Fktkj-1)/(1-Fktkj-1),j=1,2,⋯,mk, where Fk· is the cumulative function of the Weibull distribution in thekth phase.
Consider the possible states of the machine, at the end of the first sampling interval during thekth phase, i.e., at time tk1. In order to find an expression for ETk, the expected residual time beyond time tk1 in the phase and the associated probabilities are presented in Table 2.
From the information shown in Table2, we get(24)ERkmk=Δk,ERkj=hj+1+βERkj+1=∑i=j+1mkβi-j-1hki+βmk-jERkmk,j=1,2,⋯,mk-1.Then we get(25)ETk=hk1+βpk1ERk1+1-pk1ETk1,ETkmk=Δk,ETkj=hkj+1+βpkj+1ERkj+1+1-pkj+1ETkj+1,j=1,2,⋯,mk-1,ETk=∑j=1mkhkj∏i=0j-11-pki+β∑j=1mkpkjERkj∏i=0j-11-pki+∏j=1mk1-pkjΔk.Substituting (24) into (25), we get (23).
Considering PM time, the expected cycle time is(26)ET=∑k=1NETk+N-1Z1+Z2.Table 2
The expected residual times and associated probabilities.
State Probability Expected residual time In control and no alarm 1 - p k 1 1 - α E T k 1 In control but a false alarm 1 - p k 1 α E T k 1 Out of control but no alarm p k 1 β E R k 1 Out of control and a true alarm p k 1 1 - β 0


## 3.2. Cost Analysis

### 3.2.1. Quality Control Cost

In this paper, the expected quality control costEQC is used to express all cost happening for quality control. Quality control cost mainly includes quality loss costs respective for conforming items and nonconforming items, the cost of false alarm, the cost of finding assignable causes and restoring the process, and sampling and inspection cost. Similar to the analysis of the expected cycle time, we first compute the expected quality control cost of thekth phase EQCk and then consider the expected quality control cost of the cycle EQC.To find an expression forEQCk, we need to search expressions for nonconforming items.LetNkj be the expected sampling number conducted after time tkj given that the process is in the out-of-control state at time tkj. Nkj can be given by(27)Nmk=0,Nmk-1=1,Nkj=∑i=1mk-ji1-ββi-1+mk-jβmk-j,j=1,2,⋯,mk-2.Letτkj be the conditional expected in-control time within the time interval tk,j-1,tkj, given that the process was in in-control state at time tk,j-1; then τkj can be expressed by(28)τkj=∫tkj-1tkjt-tkj-1fktdtFktkj-Fktkj-1,j=1,2,⋯,mk.In particular, τkΔ is the conditional expected in-control time within the time interval tkmk,T0, given that the process was in in-control state at time tkmk, τkΔ=∫tkmkT0t-tkmkfktdt/(FkT0-Fktkmk).Theorem 2.
In thekth phase, the expected quality control cost EQCk is(29)EQCk=cf+cvn1+∑j=1mk-1∏i=1j1-pki+∑j=2mkpkjQ0τkj+Q1hkj-τkj∏i=1j-11-pki+cCM∑j=2mkpkj∏i=1j-11-pki+β∑j=2mkpkjNkjcf+cvn+Q1ERkj∏i=1j-11-pki+αcCpM∑j=1mk∏i=1j1-pki+Q0∑j=1mkhkj∏i=1j1-pki+∏i=1mk1-pki1-pkΔQ0Δk+pkΔQ0Δk+Q1Δk-τkΔ+cCM+pk1Q0τk1+Q1hk1-τk1+pk1cCM+βpk1Nk1cf+cvn+Q1ERk1where pkΔ is the conditional probability that the process shifts to the out-of-control state during the time interval tkmk,T0, given that the process was in in-control state at time tkmk, pkΔ=(FkT0-Fktkmk)/(1-Fktkmk).Proof.
LetEQCkj,j=1,2,⋯,mk-1 be the expected residual quality control cost in the phase beyond time tkj, given that the process is in control at time tkj.
Consider the possible states of the machine at the end of the first sample interval, i.e., at timetk1, during thekth phase. In order to find an expression for EQCk, the expected residual quality control cost in the phase and the associated probabilities are presented in  Table 3.
Similar to the way of developingETk, the following equations are presented by(30)EQCk=cf+cvn+pk1Q0τk1+Q1hk1-τk1+pk1cCM+βpk1Nk1cf+cvn+Q1ERk1+1-pk1EQCk1+1-pk1αcCPM+1-pk1Q0hk1
Forj=1,2,⋯,mk-1,(31)EQCkj=cf+cvn+pkj+1Q0τkj+1+Q1hkj+1-τkj+1+pkj+1cCM+βpkj+1Nkj+1cf+cvn+Q1ERkj+1+1-pkj+1EQCkj+1+1-pkj+1αcCPM+1-pkj+1Q0hkj+1EQCkmk=1-pkΔQ0Δk+pkΔQ0Δk+Q1Δk-τkΔ+cCMSo (32)EQCk=cf+cvn1+∑j=1mk-1∏i=1j1-pki+∑j=2mkpkjQ0τkj+Q1hkj-τkj∏i=1j-11-pki+cCM∑j=2mkpkj∏i=1j-11-pki+β∑j=2mkpkjNkjcf+cvn+Q1ERkj∏i=1j-11-pki+αcCPM∑j=1mk∏i=1j1-pki+Q0∑j=1mkhkj∏i=1j1-pki+∏i=1mk1-pkiEQCkmk+pk1Q0τk1+Q1hk1-τk1+pk1cCM+βpk1Nk1cf+cvn+Q1ERk1By substituting (31) into (32), we get (29).
The expected quality control cost is the sum ofN phase quality control costs; that is, (33)EQC=∑k=1NEQCk.Table 3
The expected residual costs and associated probabilities.
State Probability Current cost Expected residual cost In control and no alarm 1 - p k 1 1 - α c f + c v n + Q 0 h k 1 E Q C k 1 In control but a false alarm 1 - p k 1 α c f + c v n + c C p M + Q 0 h k 1 E Q C k 1 Out of control but no alarm p k 1 β c f + c v n + Q 0 τ k 1 + Q 1 h k 1 - τ k 1 c C M + N k 1 a + b n + Q 1 E R k 1 Out of control and a true alarm p k 1 1 - β c f + c v n + Q 0 τ k 1 + Q 1 h k 1 - τ k 1 c C M


### 3.2.2. PM Cost

The expected PM cost is the sum ofN -1 PM costs and a replacement cost; that is, (34)EPMC=N-1cPM+cRP.


## 3.2.1. Quality Control Cost

In this paper, the expected quality control costEQC is used to express all cost happening for quality control. Quality control cost mainly includes quality loss costs respective for conforming items and nonconforming items, the cost of false alarm, the cost of finding assignable causes and restoring the process, and sampling and inspection cost. Similar to the analysis of the expected cycle time, we first compute the expected quality control cost of thekth phase EQCk and then consider the expected quality control cost of the cycle EQC.To find an expression forEQCk, we need to search expressions for nonconforming items.LetNkj be the expected sampling number conducted after time tkj given that the process is in the out-of-control state at time tkj. Nkj can be given by(27)Nmk=0,Nmk-1=1,Nkj=∑i=1mk-ji1-ββi-1+mk-jβmk-j,j=1,2,⋯,mk-2.Letτkj be the conditional expected in-control time within the time interval tk,j-1,tkj, given that the process was in in-control state at time tk,j-1; then τkj can be expressed by(28)τkj=∫tkj-1tkjt-tkj-1fktdtFktkj-Fktkj-1,j=1,2,⋯,mk.In particular, τkΔ is the conditional expected in-control time within the time interval tkmk,T0, given that the process was in in-control state at time tkmk, τkΔ=∫tkmkT0t-tkmkfktdt/(FkT0-Fktkmk).Theorem 2.
In thekth phase, the expected quality control cost EQCk is(29)EQCk=cf+cvn1+∑j=1mk-1∏i=1j1-pki+∑j=2mkpkjQ0τkj+Q1hkj-τkj∏i=1j-11-pki+cCM∑j=2mkpkj∏i=1j-11-pki+β∑j=2mkpkjNkjcf+cvn+Q1ERkj∏i=1j-11-pki+αcCpM∑j=1mk∏i=1j1-pki+Q0∑j=1mkhkj∏i=1j1-pki+∏i=1mk1-pki1-pkΔQ0Δk+pkΔQ0Δk+Q1Δk-τkΔ+cCM+pk1Q0τk1+Q1hk1-τk1+pk1cCM+βpk1Nk1cf+cvn+Q1ERk1where pkΔ is the conditional probability that the process shifts to the out-of-control state during the time interval tkmk,T0, given that the process was in in-control state at time tkmk, pkΔ=(FkT0-Fktkmk)/(1-Fktkmk).Proof.
LetEQCkj,j=1,2,⋯,mk-1 be the expected residual quality control cost in the phase beyond time tkj, given that the process is in control at time tkj.
Consider the possible states of the machine at the end of the first sample interval, i.e., at timetk1, during thekth phase. In order to find an expression for EQCk, the expected residual quality control cost in the phase and the associated probabilities are presented in  Table 3.
Similar to the way of developingETk, the following equations are presented by(30)EQCk=cf+cvn+pk1Q0τk1+Q1hk1-τk1+pk1cCM+βpk1Nk1cf+cvn+Q1ERk1+1-pk1EQCk1+1-pk1αcCPM+1-pk1Q0hk1
Forj=1,2,⋯,mk-1,(31)EQCkj=cf+cvn+pkj+1Q0τkj+1+Q1hkj+1-τkj+1+pkj+1cCM+βpkj+1Nkj+1cf+cvn+Q1ERkj+1+1-pkj+1EQCkj+1+1-pkj+1αcCPM+1-pkj+1Q0hkj+1EQCkmk=1-pkΔQ0Δk+pkΔQ0Δk+Q1Δk-τkΔ+cCMSo (32)EQCk=cf+cvn1+∑j=1mk-1∏i=1j1-pki+∑j=2mkpkjQ0τkj+Q1hkj-τkj∏i=1j-11-pki+cCM∑j=2mkpkj∏i=1j-11-pki+β∑j=2mkpkjNkjcf+cvn+Q1ERkj∏i=1j-11-pki+αcCPM∑j=1mk∏i=1j1-pki+Q0∑j=1mkhkj∏i=1j1-pki+∏i=1mk1-pkiEQCkmk+pk1Q0τk1+Q1hk1-τk1+pk1cCM+βpk1Nk1cf+cvn+Q1ERk1By substituting (31) into (32), we get (29).
The expected quality control cost is the sum ofN phase quality control costs; that is, (33)EQC=∑k=1NEQCk.Table 3
The expected residual costs and associated probabilities.
State Probability Current cost Expected residual cost In control and no alarm 1 - p k 1 1 - α c f + c v n + Q 0 h k 1 E Q C k 1 In control but a false alarm 1 - p k 1 α c f + c v n + c C p M + Q 0 h k 1 E Q C k 1 Out of control but no alarm p k 1 β c f + c v n + Q 0 τ k 1 + Q 1 h k 1 - τ k 1 c C M + N k 1 a + b n + Q 1 E R k 1 Out of control and a true alarm p k 1 1 - β c f + c v n + Q 0 τ k 1 + Q 1 h k 1 - τ k 1 c C M


## 3.2.2. PM Cost

The expected PM cost is the sum ofN -1 PM costs and a replacement cost; that is, (34)EPMC=N-1cPM+cRP.


## 3.3. Modelling

Obviously, the expected total cost in the cycle time is the sum of the expected quality control cost and the expected PM cost, both discussed in Section3.2, expressed by(35)EC=EQC+EPMCThe expected cost per unit time can be considered as the ratio of the expected total cost in the cycle to the expected cycle time. As a result, the expected cost per unit time is expressed by(36)Ec=ECETBy minimizing the cost per unit time for the cycle, the integrated model of preventive maintenance policy and statistical process control can be expressed by(37)minEcs.t.ARL0≥C1,ARL1≤C2,a1≤T0≤b1,a2≤N≤b2,a3≤n≤b3,a4≤h1≤b4,a5≤hH0≤b5,a6≤L≤b6,in which T0,N,n,h1,hH,L are six decision variables, C1,C2 represent constraints on the average running lengths of the in-control state and out-of-control state, respectively, and ai,bi,i=1,2,3,4,5,6 represent constraints on decision variables.


## 3.4. Solution Procedure

Due to the cost relationship assumed in (22), performing no PM will never be optimal, and performing too many PMs will cause rising cost. With increasing PM number, the objective value will fall first until the optimal policy is reached and then rise. To solve the problem, theN must be iterated over until the cost objective begins to increase. Therefore, our solution algorithm is summarized as follows.Step 1.
Begin withN=1, iterate over specified values of T0,n,h1,hH,L, and record best Ec.Step 2.
LetN=N+1, iterate over specified values of T0,n,h1,hH,L, and record best Ec′.Step 3.
CompareEc and Ec′. If Ec′<Ec, return to Step 2; else the best solution is Ec.Because the objective model contains implicit function, differentiation, and nonlinear constraints, it is hard to solve the problem by Hessian matrix. We use a total enumeration method for specified ranges ofT0,n,h1,hH,L to get the solution. The decision variables T0,n,h1,hH,L are not tested for all possible values, but rather the most probable and practical values. For eachN, the T0 values are tested from Δh to the mean time to failure (MTTF) in increments of Δh, because we think the PM interval should be not greater than MTTF; Δh hours is the requirement of accuracy in practice. Then values are tested from 1 to 7, 10,15,30. This is drawn on the experience of Cassady et al. [10] and Yeung et al. [11]. Same to the requirement of accuracy of T0, the h1 values are tested in increments of Δh in a certain range depending on circumstances. The hH values are tested from 0.1 to 10 in increments of 0.1, and the L values are tested from 0.25 to 2 in increments of 0.25. Both of hH and L values are determined on the basis of the table given by Hawkins (1993) that concluded L values and the corresponding hH values that will achieve ARL0=370. Due to the fact that we do not test every possible value of six decision variables, we define the solution of our algorithm to be near-optimal, rather than optimal.


## 4. Numerical Example

### 4.1. Example

In this section, we present a numerical example to illustrate the implementation of our methodology.The parameters of the example are similar to the data of the model presented in [7] with some modified data for maintenance. Modifications are due to the fact that the maintenance mechanism of this paper is more suitable for the situation cCpM≤cCM≤cPM≤cRP, but the data given in the reference does not satisfy this requirement. Assume that the time for the process to be in the in-control state follows a Weibull distribution with parameters γ=0.05 and ν=2. Let cf=2.0$, cv=0.5$, Q0=50$, Q1=950$, and δμ=1. Modifications are considered withη=0.9, Z1=2, Z2=6, cCpM=50$, cCM=200$, cPM=500$, and cRP=2000$. With these data, we try to find the near-optimal values of T0,N,n,h1,hH,L.In order to limit the number of false alarms without compromising performance,ARL0 is ruled to be greater than 370. A shift in the out-of-control process should be checked out as soon as possible once assignable cause occurs and ARL1 is set to be less than 10. Because the time to failure follows the Weibull distribution with parameters γ=0.05 and ν=2, MTTF is 4 hours, so the T0 values are tested from 0.1 to 4 in increments of 0.1, where 0.1 hours can meet the requirement of accuracy. The h1 values are assumed to be tested from 1 to 3 in increments of 0.1. Using the proposed method, the near-optimal solution is T0,N,n,h1,hH,L∗=1.8,12,1,1.3,4.8,0.5 under which Ec∗=221.7121$.With the change of the number of age-based maintenances, the variation of objective value is shown in Figure1. With no PM (N=1), the expected cost per unit time is 279.8675$ which is higher than all those with PMs; that is, N≥2. The expected cost per unit time decreases until the optimal policy (N=12) is reached, and the subsequent values of N cause the objective to increase until the sampling number after a PM is zero (N=16) which means the machine should be replaced. The variation of objective value following the rising number of age-based maintenances is shown in Figure 3.Figure 3
The variation of objective value following the rising number of age-based maintenances.


### 4.2. Sensitivity Analysis

Sensitivity analyses on both the decision variables and the objective value for each of several various parameters are reported. When a parameter is varied, other parameters are kept unchanged. The results are listed in Tables4–6.Table 4
Sensitivity analysis forδμ.
δ μ T 0 N n h 1 h H L E ( c ) 1 1.8 12 1 1.3 4.8 0.5 221.7121 1.5 1.8 11 1 1.3 3.4 0.75 221.1518 2 2.2 10 1 1.2 2.5 1 219.5052 2.5 2.5 9 1 1.2 2 1.25 216.8554Table 5
Sensitivity analysis forη.
η T 0 N n h 1 h H L E ( c ) 0.99 1.6 30 1 1.3 4.8 0.5 198.5900 0.9 1.8 12 1 1.3 4.8 0.5 221.7121 0.81 1.9 8 1 1.2 4.8 0.5 232.4033 0.72 1.9 6 1 1.1 4.8 0.5 240.4450Table 6
Sensitivity analysis forcPM.
c P M T 0 N n h 1 h H L E(c) 400 1.6 15 1 1.0 4.8 0.5 193.6791 500 1.8 12 1 1.3 4.8 0.5 221.7121 600 1.9 7 1 1.5 4.8 0.5 245.5956 700 2.0 4 1 1.6 4.8 0.5 264.8867In general, we notice that the model always achieves the best economic benefit with the choice ofn=1. This phenomenon reflects the conclusion that the CUSUM chart often works best with n=1 [27] which is also applicable in the integrated model of control chart and maintenance. The hH value andL value are closely related to the magnitude of quality shift, and the former decreases and the later increases as δμ increases. However, both of the two parameters do not vary with changes of other parameters as long as the MTTF maintains unchanged, which also tremendously speeds up the solution procedure algorithm. Another phenomenon is that the change directions ofN and T0 are always opposite; that is, one decreases and another must increase, no matter which parameter causes the changes.In Table4, we can conclude that T0 gets larger as δμ increases. The increments in δμ imply the demand on quality in the system which is becoming less strict, and the machine condition has a direct impact on quality, which has a close relationship with the age. So the increasing shift must cause a rising age of the machine for maintenance. As δμ increases, the number for maintenance decreases and further leads to a reduction in the cost objective, but this reduction is very small. Due to the rising quality shift, the sampling interval becomes smaller to guarantee the failure once it occurs to be detected as soon as possible.In Table5, the PM number decreases as the η increases that is because η is a measurement for the degradation of the age of the machine under the effect of a PM activity, and the smaller the η, the worse the restoration effect. As η decreases, the number for maintenance decreases and further leads to a reduction in the cost objective. Due to the gradually worse manufacturing conditions, the quality will get worse, so the sampling interval becomes smaller to guarantee the failure to be detected as soon as possible.In Table6, the increments of the cPM lead to a large reduction in maintenance number; however, the cost objective still becomes larger. The reason is that the cost reduced by the lower PM number, as well as the lower frequency of sampling for inspection, cannot balance that raised by a higher PM cost.


## 4.1. Example

In this section, we present a numerical example to illustrate the implementation of our methodology.The parameters of the example are similar to the data of the model presented in [7] with some modified data for maintenance. Modifications are due to the fact that the maintenance mechanism of this paper is more suitable for the situation cCpM≤cCM≤cPM≤cRP, but the data given in the reference does not satisfy this requirement. Assume that the time for the process to be in the in-control state follows a Weibull distribution with parameters γ=0.05 and ν=2. Let cf=2.0$, cv=0.5$, Q0=50$, Q1=950$, and δμ=1. Modifications are considered withη=0.9, Z1=2, Z2=6, cCpM=50$, cCM=200$, cPM=500$, and cRP=2000$. With these data, we try to find the near-optimal values of T0,N,n,h1,hH,L.In order to limit the number of false alarms without compromising performance,ARL0 is ruled to be greater than 370. A shift in the out-of-control process should be checked out as soon as possible once assignable cause occurs and ARL1 is set to be less than 10. Because the time to failure follows the Weibull distribution with parameters γ=0.05 and ν=2, MTTF is 4 hours, so the T0 values are tested from 0.1 to 4 in increments of 0.1, where 0.1 hours can meet the requirement of accuracy. The h1 values are assumed to be tested from 1 to 3 in increments of 0.1. Using the proposed method, the near-optimal solution is T0,N,n,h1,hH,L∗=1.8,12,1,1.3,4.8,0.5 under which Ec∗=221.7121$.With the change of the number of age-based maintenances, the variation of objective value is shown in Figure1. With no PM (N=1), the expected cost per unit time is 279.8675$ which is higher than all those with PMs; that is, N≥2. The expected cost per unit time decreases until the optimal policy (N=12) is reached, and the subsequent values of N cause the objective to increase until the sampling number after a PM is zero (N=16) which means the machine should be replaced. The variation of objective value following the rising number of age-based maintenances is shown in Figure 3.Figure 3
The variation of objective value following the rising number of age-based maintenances.


## 4.2. Sensitivity Analysis

Sensitivity analyses on both the decision variables and the objective value for each of several various parameters are reported. When a parameter is varied, other parameters are kept unchanged. The results are listed in Tables4–6.Table 4
Sensitivity analysis forδμ.
δ μ T 0 N n h 1 h H L E ( c ) 1 1.8 12 1 1.3 4.8 0.5 221.7121 1.5 1.8 11 1 1.3 3.4 0.75 221.1518 2 2.2 10 1 1.2 2.5 1 219.5052 2.5 2.5 9 1 1.2 2 1.25 216.8554Table 5
Sensitivity analysis forη.
η T 0 N n h 1 h H L E ( c ) 0.99 1.6 30 1 1.3 4.8 0.5 198.5900 0.9 1.8 12 1 1.3 4.8 0.5 221.7121 0.81 1.9 8 1 1.2 4.8 0.5 232.4033 0.72 1.9 6 1 1.1 4.8 0.5 240.4450Table 6
Sensitivity analysis forcPM.
c P M T 0 N n h 1 h H L E(c) 400 1.6 15 1 1.0 4.8 0.5 193.6791 500 1.8 12 1 1.3 4.8 0.5 221.7121 600 1.9 7 1 1.5 4.8 0.5 245.5956 700 2.0 4 1 1.6 4.8 0.5 264.8867In general, we notice that the model always achieves the best economic benefit with the choice ofn=1. This phenomenon reflects the conclusion that the CUSUM chart often works best with n=1 [27] which is also applicable in the integrated model of control chart and maintenance. The hH value andL value are closely related to the magnitude of quality shift, and the former decreases and the later increases as δμ increases. However, both of the two parameters do not vary with changes of other parameters as long as the MTTF maintains unchanged, which also tremendously speeds up the solution procedure algorithm. Another phenomenon is that the change directions ofN and T0 are always opposite; that is, one decreases and another must increase, no matter which parameter causes the changes.In Table4, we can conclude that T0 gets larger as δμ increases. The increments in δμ imply the demand on quality in the system which is becoming less strict, and the machine condition has a direct impact on quality, which has a close relationship with the age. So the increasing shift must cause a rising age of the machine for maintenance. As δμ increases, the number for maintenance decreases and further leads to a reduction in the cost objective, but this reduction is very small. Due to the rising quality shift, the sampling interval becomes smaller to guarantee the failure once it occurs to be detected as soon as possible.In Table5, the PM number decreases as the η increases that is because η is a measurement for the degradation of the age of the machine under the effect of a PM activity, and the smaller the η, the worse the restoration effect. As η decreases, the number for maintenance decreases and further leads to a reduction in the cost objective. Due to the gradually worse manufacturing conditions, the quality will get worse, so the sampling interval becomes smaller to guarantee the failure to be detected as soon as possible.In Table6, the increments of the cPM lead to a large reduction in maintenance number; however, the cost objective still becomes larger. The reason is that the cost reduced by the lower PM number, as well as the lower frequency of sampling for inspection, cannot balance that raised by a higher PM cost.


## 5. Conclusions and Future Research

We have developed an integrated model to simultaneously optimize the parameters of CUSUM control chart and age-based PM policy, in which, a more complex multiphase process of multiple PMs and multiple samplings in each PM interval is considered. The imperfect PM policy is used in the model in which each PM restores the process to a state between as good as new and as bad as old. Different from many other researchers, we use CUSUM chart to jointly optimize the maintenance policy and we find some useful conclusions. Different from Shrivastava et al. [19], a recursive algorithm modified from previous research is used to establish the model. Furthermore, we analyze the characteristic of the cost objective function and find that it behaves similar to a convex function of maintenance number; on the basis of this, the solution algorithm is suggested. A numerical case is utilized to illustrate the effectiveness of the model and its solution algorithm. Sensitivity analysis on decision variables for each various parameters is reported too.Managers can use our methodology to control the production system for the requirement of small quality shift. By our model, the parameters of CUSUM chart and maintenance decisions can be determined. Our analysis shows that managers would be best served by focusing their attention on reducing preventive maintenance cost and improving the effectiveness of each PM, that is to make imperfectness factor as larger as possible. It is also shown that the increments of the magnitude of quality shift lead to a very limited reduction in the cost objective, so we think it is better to produce high quality products if the conditions permit.For Shewhart control chart is most in use in the existing integrated models of SPC and maintenance policy but other charts are few, it is of interest to combine CUSUM chart with various maintenance policies to find more useful information for managers.



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*Source: 9246372-2018-06-12.xml*