# Optimal Maintenance of a Production System withL Intermediate Buffers

Authors: Constantinos C. Karamatsoukis; Epaminondas G. Kyriakidis
Journal: Mathematical Problems in Engineering
 (2012)

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## Abstract

We consider a production-inventory system that consists of an input-generating installation, a production unit andL intermediate buffers. It is assumed that the installation transfers the raw material j∈{1,…,L} to buffer Bj, and the production unit pulls the raw material j∈{1,...,L} from buffer Bj.
We consider the problem of the optimal preventive maintenance of the installation if the installation deteriorates stochastically with usage and the production unit is always in operative condition. We also consider the problem of the optimal preventive maintenance of the production unit if the production unit deteriorates stochastically with usage and the installation is always in operative condition. Under a suitable cost structure and for given contents of the buffers, it is proved that the average-cost optimal policy for the first (second) problem initiates a preventive maintenance of the installation (production unit) if and only if the degree of deterioration of the installation (production unit) exceeds some critical level. Numerical results are presented for both problems.


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## Body

## 1. Introduction

In the present paper, we study two problems, and we generalize the results obtained in two previous papers by Kyriakidis and Dimitrakos [1] and Pavitsos and Kyriakidis [2] that are concerned with the preventive maintenance of a production-inventory system. We consider a manufacturing system (see Figure 1) in which an input-generating installation (I) transfers L raw materials to a subsequent production unit (P). We assume that L buffers B1,…,BL have been built between the installation and the production unit. The installation transfers the raw material j∈{1,…,L} to the buffer Bj, and the production unit pulls this raw material from the buffer Bj. The buffers have finite capacities.Figure 1
The production-inventory system.In the first problem it is assumed that the installation deteriorates stochastically over time, and the production unit is always in operative condition. The deteriorating process for the installation is described by some known transition probabilities between different degrees of deterioration. A discrete-time Markov decision model is considered for the optimal preventive maintenance of the installation. The maintenance times are assumed to be geometrically distributed, and the cost structure includes operating costs of the installation, costs for storing the raw materials in the buffers, maintenance costs and costs due to production delay when the installation does not operate or operate partially and the contents of some or all buffers are below some specific levels. It is proved that for fixed contents of the buffers the policy that minimizes the long-run expected average cost per unit time is of control-limit type, that is, it initiates a preventive maintenance of the installation if and only if its degree of deterioration exceeds some critical level. This result generalizes the structural result that was obtained by Kyriakidis and Dimitrakos [1] for the case in which L=1.  In the second problem it is assumed that the production unit deteriorates stochastically over time and the installation is always in operative condition. The deteriorating process for the production unit is described by some known transition probabilities between different degrees of deterioration. A discrete-time Markov decision model is formulated for the optimal preventive maintenance of the production unit. The maintenance times are assumed to be geometrically distributed, and the cost structure includes operating costs of the production unit, costs for the maintenance of the production unit, storage costs, penalty costs, and costs due to the lost production. It is proved that for fixed contents of the buffers the average-cost optimal policy is again of control-limit type, that is, it initiates a preventive maintenance of the production unit if and only if its degree of deterioration exceeds some critical level. This result generalizes the structural result that was obtained by Pavitsos and Kyriakidis [2] for the case in which L=1.An example of this system could be a production machine that pullsL different parts from L buffers and assembles them in order to produce the final product. These parts are transferred by a feeder to the buffers. Note that in the last twenty years a great number of maintenance models for production-inventory systems have been studied (see Van Der Duyn Schouten and Vanneste [3], Meller and Kim [4], Iravani and Duenyas [5], Sloan [6], Yao et al. [7], Rezg et al. [8], Dimitrakos and Kyriakidis [9], Karamatsoukis and Kyriakidis [10], and Hadidi et al. [11]). In these models, the preventive maintenance depends on the working condition of a machine and the level of a subsequent buffer. The first problem that we study in the present paper has its origin in a model introduced by Van Der Duyn Schouten and Vanneste [3]. The states of that model consist of the age of a machine and the content of a subsequent buffer that is fed by the machine. The cost structure included costs due to lost production that were incurred when a repair was performed on the machine and the buffer was empty. The repair times of the machine were assumed to be geometrically distributed. It was proved that, for fixed buffer content, the average-cost optimal policy initiates a preventive maintenance of the machine if and only if its age is greater than or equal to a critical value.The rest of the paper is organized as follows. In the next section, we describe the problem in which only the installation deteriorates with usage, and we derive the structure of the average-cost optimal policy. In Section3, we study the case in which only the production unit deteriorates over time, and the structure of the average-cost optimal policy is derived. Numerical results are presented for both problems. In the final section, the main conclusions of the paper are summarized, and we propose topics for future research.


## 2. The Problem when the Installation Deteriorates Stochastically

We consider a production-inventory system (see Figure1) which consists of an installation (I) that supplies the buffer Bj with the raw material j∈{1,…,L} and a production unit (P) which pulls dj units of the raw material j∈{1,…,L} from buffer Bj during one unit of time. It is assumed that the production unit is always in operative condition, and that the installation may fail as time evolves. The buffer Bj, j=1,2,…,L has finite capacity which is equal to Kj units of raw material j. As long as the buffer Bj, j=1,…,L is not full and the installation is in operative condition, the installation may transfer pj(>dj) units of the raw material j∈{1,…,L} to buffer Bj during one unit of time and the difference pj-dj is stored in buffer Bj. As soon as buffer Bj, j=1,…,L is filled up the installation reduces its speed from pj to dj. The numbers pj,dj,Kj, j=1,…,L are assumed to be integers.We suppose that the installation is inspected at discrete, equidistant time epochsτ=0,1,… (say every day), and is classified into one of m+2 working conditions 0,1,…,m+1 which describe increasing levels of deterioration. Working condition 0 denotes a new installation (or functioning as good as new), while working condition m+1 means that the installation is in failed (inoperative) condition and it cannot transfer the raw materials to the buffers. The intermediate working conditions 1,…,m are operative and are ordered ascendingly to reflect their relative degree of deterioration. The transition probability of moving from working condition i at time epoch τ to working condition r at time epoch τ+1 is equal to pir. We assume that the probability of eventually reaching the working condition m+1 from any initial working condition i is nonzero. If at a time epoch τ the installation is found to be at failure state m+1, a corrective maintenance is mandatory. If at a time epoch τ the installation is found to be at any working condition i≤m, a preventive maintenance may be initiated. When a preventive maintenance is performed, the installation does not operate and it does not transfer any raw material to its buffer. It is assumed that the preventive and corrective repair times (expressed in time units) are geometrically distributed with probability of success aI and bI, that is, the probability that they will last t≥1 time units are equal to (1-aI)t-1aI and (1-bI)t-1bI, respectively. When a preventive or a corrective maintenance is performed and the buffer Bj, j∈{1,…,L} contains xj units of raw material j, the production unit pulls from buffer Bj during one unit of time min(xj,dj) units of raw material j. Both maintenances bring the installation to its perfect condition 0.We introduce the statePMto denote the situation that a preventive maintenance is performed on the installation. Then the state space of the system is the set S = {0,…,m+1,PM} × {0,…,K1} × ⋯ × {0,…,KL}, where (i,x1,…,xL)∈S is the state in which i is the working condition of the installation and xj∈{0,…,Kj},j=1,…,L, is the content of buffer Bj. A policy is any rule for choosing actions at each time epoch τ=0,1,… The possible actions are: action 1 (initiate a preventive maintenance), action 2 (initiate a corrective maintenance), action J⊆{1,…,L} (transfer raw materials only to those buffers that belong to the nonempty subset J of the set {1,…,L}). If at a time epoch τ the installation is found to be at state PM or state m+1, the action 1 or the action 2 is compulsory, respectively. If at a time epoch τ the installation is found to be at working condition i∈{0,…,m}, then we may either choose either action 1 or action J⊆{1,…,L}. Hence, the number of possible actions in this case is 2L since the number of nonempty subsets of {1,…,L} is ∑i=1L(Li)=2L-1. If at a time epoch τ action J is chosen and j belongs to J with xj<Kj, then the content of buffer Bj at next time epoch τ+1 will be min(xj+pj-dj,Kj). This increase of the buffer content will happen even if the working condition of the installation at next time epoch τ+1 is the failure state m+1. A policy is said to be stationary if at each time epoch τ=0,1,…, it chooses one action which depends only on the current state of the system.The cost structure of the problem includes operating costs of the installation, storage costs, costs due to the lost production, and maintenance costs. If the working condition of the installation isi∈{0,…,m} and the buffer Bj, j∈{1,…,L} is not full (or full) the cost of transferring pj (or dj) units of raw material j to buffer Bj during one unit of time is equal to cj(i) (or c~j(i)). Therefore, if at a time epoch τ the working condition of the installation is found to be i∈{0,…,m} and action J⊆{1,…,L} is chosen, then the operating cost until the next time epoch τ+1 is equal to ∑j∈J1cj(i)+∑j∈J2c~j(i), where J1∪J2=J,J1 corresponds to the buffers that are not full and J2 corresponds to the buffers that are full. We assume that the cost of holding a unit of raw material j∈{1,…,L} in buffer Bj for one unit of time is equal to hj. The cost rates during a preventive and a corrective maintenance of the installation are equal to cp and cf, respectively. When a preventive or a corrective maintenance is performed on the installation and all buffers B1,…,BL are empty (i.e., xj=0,j=1,…,L), the production unit does not pull any raw material j∈{1,…,L} from the buffers. In this case we incur a cost due to production delay that is equal to C>0 per unit of time. When xj≥dj,j=1,…,L, we do not incur any such cost since all buffers contain enough raw materials to satisfy the demands of the production unit for one unit of time. When the inequality xj≥dj is satisfied for some (but not for all) j=1,…,L, the demands for raw materials of the production unit for one unit of time are partially satisfied. Therefore, the productivity of the production unit is reduced in the sense that the time for the production of the final products increases, since for one unit of time, some of the raw materials that are needed for the production of the final products are not available. In this case it seems reasonable to assume that the cost rate due to production delay is equal to C∑j=1L(dj-xj)+/∑j=1Ldj, where (dj-xj)+=max(dj-xj,0) is the unavailable quantity of the raw material j during one unit of time. Similarly, if at a time epoch τ the working condition of the installation is found to be i∈{1,…,m} and the action J⊆{1,…,L} is chosen, then the cost due production delay until the time epoch τ+1 is equal to C∑j∉J(dj-xj)+/∑j=1Ldj. The following conditions on the cost structure and on the transition probabilities are assumed to be valid.Condition 1.
Forj∈{1,…,L} the sequences {cj(i)} and {c~j(i)}, 0≤i≤m, are nondecreasing with respect to i.  That is, as the working condition of the installation deteriorates, the cost of transferring the raw material j∈{1,…,L} to buffer Bj increases.Condition 2.
Forj∈{1,…,L}, c~j(i)≤cj(i), 0≤i≤m. That is, the cost of transferring pj units of raw material j∈{1,…,L} to buffer Bj during one unit of time is greater than or equal to the cost of transferring dj units of raw material j to buffer Bj during one unit of time.Condition 3.
0
<
b
I
<
a
I
≤
1. That is, the expected time required for a preventive maintenance is smaller than the expected time required for a corrective maintenance.Condition 4.
c
p
≤
c
f. That is, the cost rate of a preventive maintenance does not exceed the cost rate of a corrective maintenance.Condition 5 (An Increasing Failure Rate Assumption).
For eachk=0,…,m, the function Dk(i)=∑r=km+1pir is nondecreasing in i, 0≤i≤m.
A consequence of this condition is thatIi≤stIi+1, 0≤i≤m, where Ii is a random variable representing the next working condition of the installation if its present working condition is i.  It can be shown (see pages 122-123 in Derman [12]) that this condition is equivalent to the following one:Condition 6.
For any nondecreasing functionh(r),0≤r≤m+1, the quantity ∑r=0m+1pirh(r), 0≤i≤m, is nondecreasing in i.
We consider a discrete-time Markov decision process in which we aim to find a stationary policy which minimizes the long-run expected average cost per unit time. Note that forL=1 this problem was studied in Kyriakidis and Dimitrakos [1].


### 2.1. The Structure of the Optimal Policy

Letα(0<α<1) be a discount factor. The minimum expected n-step discounted cost Vnα(i,x-), where (i,x-)=(i,x1,…,xL) is the initial state, can be found for all n=1,2,…, recursively, from the following equations (see chapter 1 in Ross [13]):
(2.1)Vnα(i,x¯)=min{minJ⊆{1,…,L}[∑j∈Jcj(i)+∑j=1Lhjxj+∑j∉J(dj-xj)+∑j=1LdjC+α∑r=0m+1pirVn-1α(r,x-′)],Vnα(PM,x¯){∑j=1Lhjxj}},0≤i≤m,0≤xj<Kj,j=1,…,L,(2.2)Vnα(PM,x-)=cp+∑j=1Lhjxj+∑j=1L(dj-xj)+∑j=1LdjC+αaIVn-1α(0,(x--d-)+)+α(1-aI)Vn-1α(PM,(x--d-)+),0≤xj≤Kj,j=1,…,L,(2.3)Vnα(m+1,x-)=cf+∑j=1Lhjxj+∑j=1L(dj-xj)+∑j=1LdjC+αbIVn-1α(0,(x--d-)+)+α(1-bI)Vn-1α(m+1,(x--d-)+),0≤xj≤Kj,j=1,…,L,
where x-′ in (2.1) is a vector with L components in which the jth component equals to min(xj+pj-dj,Kj), if j∈J, while, if j∉J, it is equal to (xj-dj)+=max(xj-dj,0) and (x--d-)+ in (2.2), and (2.3) is the vector ((x1-d1)+,…,(xL-dL)+). The initial condition is V0α(s)=0, s∈S. Note that, if xj=Kj for some values of j∈{1,…,L}, (2.1) is valid if cj(i) is changed to c~j(i) for these values of j. Note that the first term in the curly brackets in (2.1) corresponds to the best action among all actions J⊆{1,…,L}, while the second term corresponds to action 1 (i.e., initiate a preventive maintenance of the installation). The first part of the following lemma is needed to prove that the average-cost optimal policy is of control-limit type for fixed levels of the buffers.Lemma 2.1.
For eachn=0,1,…, we have that (i)
Vnα(i,x-)≤Vnα(i+1,x-),0≤i≤m,0≤xj≤Kj,j=1,…,L,(ii)
Vnα(PM,x-)≤Vnα(m+1,x-),0≤xj≤Kj,j=1,…,L.Proof.
We will prove the lemma by induction onn. The lemma is valid for n=0, since V0α(s)=0,s∈S. We assume that it is valid for n-1(≥0). We will show that it is also valid for n. First, we prove part (ii) and then part (i).
Part (ii): LetD=Vn-1α(m+1,(x--d-)+-Vn-1α(0,(x--d-)+).
For0≤xj≤Kj,j=1,…,L, we have that
(2.4)Vnα(PM,x-)=cp+∑j=1Lhjxj+∑j=1L(dj-xj)+∑j=1LdjC+αaIVn-1α(0,(x--d-)+)+α(1-aI)Vn-1α(PM,(d--x-)+)≤cf+∑j=1Lhjxj+∑j=1L(dj-xj)+∑j=1LdjC+αaIVn-1α(0,(x--d-)+)+α(1-aI)Vn-1α(m+1,(x--d-)+)=cf+∑j=1Lhjxj+∑j=1L(dj-xj)+∑j=1LdjC+αVn-1α(m+1,(x--d-)+)-αaID≤cf+∑j=1Lhjxj+∑j=1L(dj-xj)+∑j=1LdjC+αVn-1α(m+1,(x--d-)+)-αbID=Vnα(m+1,x-).
The first inequality follows from Condition 4 and from part (ii) of the induction hypothesis. The second inequality follows from Condition 3 and the inequality D≥0 which is a consequence of part (i) of the induction hypothesis.
Part (i): We have to show that(2.5)Vnα(m,x-)≤Vnα(m+1,x-),0≤xj≤Kj,j=1,…,L,(2.6)Vnα(i,x-)≤Vnα(i+1,x-),0≤i≤m-1,0≤xj≤Kj,j=1,…,L.
Inequality (2.5) is easily verified using (2.1) with i=m and part (ii) above
(2.7)Vnα(m,x-)≤Vnα(PM,x-)≤Vnα(m+1,x-).
For 0≤i≤m-1 and 0≤xj<Kj,j=1,…,L, we have that
(2.8)Vnα(i,x¯)=min{minJ⊆{1,…,L}[∑j∈Jcj(i)+∑j=1Lhjxj+∑j∉J(dj-xj)+∑j=1LdjC+α∑r=0m+1pirVn-1α(r,x-′)],Vnα(PM,x¯)}≤min{minJ⊆{1,…,L}[∑j∈Jcj(i+1)+∑j=1Lhjxj+∑j∉J(dj-xj)+∑j=1LdjC+α∑r=0m+1pi+1,rVn-1α(r,x-′)],Vnα(PM,x¯){∑j=1Lhjxj}}=Vnα(i+1,x-).
The above inequality follows from Condition 1 and the inequality
(2.9)∑r=0m+1pirVn-1α(r,x-′)≤∑r=0m+1pi+1,rVn-1α(r,x-′)
which is implied by part (i) of the induction hypothesis and Condition 6. Hence (2.6) has been proved for xj∈{0,…,Kj-1},j=1,…,L. Similarly, we obtain (2.6) if xj=Kj for some values of j∈{1,…,L}.Since the state spaceS is finite, and the state (0,0-) is accessible from every other state under any stationary policy, it follows that there exist numbers v(s), s∈S and a constant g that satisfy the average-cost optimality equations (see Corollary 2.5 in [13], page 98). For the states (i,x-),0≤i≤m,0≤xj<Kj,j=1,…,L, the optimality equations take the following form:
(2.10)v(i,x¯)=min{minJ⊆{1,…,L}[∑j∈Jcj(i)+∑j=1Lhjxj+∑j∉J(dj-xj)+∑j=1LdjC-g+∑r=0m+1pirv(r,x-′)],v(PM,x¯){[∑j∈Jcj(i)+∑j=1Lhjxj+∑j∉J(dj-xj)+∑j=1LdjC-g+∑r=0m+1pirv(r,x-′)]}}.
If xj=Kj for some values of j∈{1,…,L}, we must change in the above equations cj(i) to c~j(i) for these values of  j. In view of part (i) of the above lemma we have the following result.Corollary 2.2.
v
(
i
,
x
-
)
≤
v
(
i
+
1
,
x
-
)
,
0
≤
i
≤
m
,
0
≤
x
j
≤
K
j
,
j
=
1
,
…
,
L.The following proposition gives a characterization of the form of the optimal policy.Proposition 2.3.
For fixed contentsx1,…,xL (0≤xj≤Kj,j=1,…,L) of buffers B1,…,BL, there exists a critical working condition i*(x1,…,xL) such that the policy that minimizes the expected long-run average cost per unit time initiates a preventive maintenance of the installation if and only if its working condition i∈{0,…,m} is greater than or equal to i*(x1,…,xL).Proof.
Suppose that for some fixedx-=(x1,…,xL) such that 0≤xj<Kj, 1≤j≤L, the optimal policy initiates a preventive maintenance of the installation at state (i,x-) where i∈{0,…,m-1}. This implies that
(2.11)v(PM,x¯)≤minJ⊆{1,…,L}[∑j∈Jcj(i)+∑j=1Lhjxj+∑j∉J(dj-xj)+∑j=1LdjC-g+∑r=0m+1pirv(r,x-′)].
To show that the optimal policy prescribes a preventive maintenance on the installation at state(i+1,x-) it is enough to show that
(2.12)v(PM,x¯)≤minJ⊆{1,…,L}[∑j∈Jcj(i+1)+∑j=1Lhjxj+∑j∉J(dj-xj)+∑j=1LdjC-g+∑r=0m+1pi+1,rv(r,x-′)].
From Conditions 1 and 6 and Corollary 2.2, it follows that the right-hand side of (2.12) is greater than or equal to the right-hand side of (2.11). Hence (2.11) implies (2.12). The same result is obtained similarly if xj=Kj for some values of j∈{1,…,L}.Remark 2.4.
In the above proposition, if for fixed contentsx1,…,xL of the buffers i*(x1,…,xL)=m+1, then the optimal policy never initiates a preventive maintenance of the installation whenever the buffer Bj, j=1,…,L, contains xj units or raw material j.


### 2.2. Numerical Results

Example 2.5.
Suppose thatL=2,  m=5,  aI=0.6,  bI=0.4,  cp=10,  cf=15,  K1=5,  K2=20,  h1=1,  h2=1,  p1=2,  d1=1,  p2=2,  d2=1,  c1(i)=0.8(i+1),  c~1(i) = 0.5(i+1),  c2(i) = 0.7(i+1),  c~2(i) = 0.5(i+1),  0≤i≤m,  and  pir=1/(m+2-i),  0≤i≤m,  i≤r≤m+1. It can be readily checked that these probabilities satisfy Condition 5. We computed the optimal policy if  C is equal to 0.5 or 15.5 by implementing the value-iteration algorithm (see Chapter 3 of Tijms [14]). Our numerical results verify the result of Proposition 2.3. In Table 1 we present the critical numbers i*(x1,x2),0≤x1≤5,0≤x2≤20. In each cell of this table the first number corresponds to C=0.5 and the second number corresponds to C=15.5. The minimum average cost was found to be 7.49 if C=0.5 which is, as expected, considerably smaller than the minimum average cost if C=15.5, which was found to be 11.63.Table 1
The critical numbersi*(x1,x2) for C=0.5,15.5.
x
2
∖
x
1
0
1
2
3
4
5
0
3, 6
3, 5
3, 5
4, 6
4, 6
4, 6
1
3, 5
2, 4
2, 4
1, 6
1, 4
1, 4
2
3, 5
2, 3
1, 2
0, 3
0, 3
2, 3
3
4, 6
1, 4
0, 3
0, 2
0, 2
3, 3
4
3, 6
1, 1
0, 0
0, 0
0, 0
3, 3
5
4, 6
1, 4
0, 0
0, 2
0, 1
3, 3
6
4, 6
1, 4
0, 2
0, 1
0, 1
2, 4
7
4, 6
0, 4
0, 2
0, 1
0, 1
2, 3
8
4, 6
0, 4
0, 2
0, 0
0, 0
1, 3
9
4, 6
0, 4
0, 2
0, 0
0, 0
2, 3
10
4, 6
0, 4
0, 2
0, 0
0, 0
1, 3
11
4, 6
0, 4
0, 2
0, 0
0, 0
1, 3
12
4, 6
0, 4
0, 2
0, 0
0, 0
1, 3
13
4, 6
0, 4
0, 1
0, 0
0, 0
1, 3
14
4, 6
0, 4
0, 1
0, 0
0, 0
1, 3
15
4, 6
0, 4
0, 1
0, 0
0, 0
1, 3
16
4, 6
0, 4
0, 1
0, 0
0, 0
1, 2
17
4, 6
0, 4
0, 1
0, 0
0, 0
1, 3
18
4, 6
0, 4
0, 1
0, 0
0, 0
1, 2
19
4, 6
0, 4
0, 1
0, 0
0, 0
1, 3
20
4, 6
0, 4
0, 1
0, 0
0, 0
1, 2We observe that forx1∈{0,…,5} and x2∈{0,…,20} the critical number that corresponds to C=0.5 is smaller than or equal to the critical number that corresponds to C=15.5. This is intuitively reasonable since if the cost due to production delay takes large values it seems disadvantageous to have all or most of the buffers empty when a maintenance is performed on the installation. Therefore in this case it seems preferable to initiate a preventive maintenance of the installation only if its degree of deterioration is relatively high. We also observe that when C=15.5 and buffer B1 or buffer B2 is empty, the optimal policy in most cases never initiates a preventive maintenance of the installation. For example i*(0,3)=6=m+1. It can be also seen from the Table 1 that i*(x1,x2) is not a monotone function with respect to x1∈{0,…,5} and with respect to x2∈{0,…,20} for constant x2 and x1, respectively. Note that, when i<i*(x1,x2),0≤x1≤5,0≤x2≤20, the value iteration algorithm gives the optimal action for the operation of the installation. For example, if C=0.5, x1=0, x2=18, the optimal action when the working condition of the installation is 3 is to transfer raw material 1 to buffer 1 (i.e.,  J={1}). If C=15.5, x1=1, x2=1, the optimal action when the working condition of the installation is 2 is to transfer raw material 1 to buffer 1 and raw material 2 to buffer 2 (i.e.,  J={1,2}).Example 2.6.
Suppose thatL=2, m=15, aI=0.3, bI=0.2, cp=10, cf=15, h1=1, h2=1, C=80, p1=4, d1=2, p2=3, d2=2, c1(i)=1.5(i+1), c~1(i)=0.75(i+1), c2(i)=2(i+1), c~2(i)=i+1, 0≤i≤m, pir=1/(m+2-i), 0≤i≤m, i≤r≤m+1. In Table 2 below we present the minimum average cost g obtained by the value-iteration algorithm for K1∈{1,…,10} and K2∈{5,10}.Table 2
The minimum average cost asK1 or K2 varies.
K
1
K
2
=
5
K
2
=
10
1
51.20
51.17
2
48.55
48.49
3
47.55
47.50
4
45.94
45.88
5
45.60
45.56
6
44.87
44.83
7
44.78
44.75
8
44.49
44.45
9
44.46
44.43
10
44.39
44.37From the above table we see that asK1 or K2 increases, the minimum average cost decreases. This is intuitively reasonable because in this example it seems favourable to have buffers with large, capacities since the cost rate C due to production delay is relatively large while the probabilities aI, bI of successful maintenances and the storage cost rates h1,h2 are relatively small.


## 2.1. The Structure of the Optimal Policy

Letα(0<α<1) be a discount factor. The minimum expected n-step discounted cost Vnα(i,x-), where (i,x-)=(i,x1,…,xL) is the initial state, can be found for all n=1,2,…, recursively, from the following equations (see chapter 1 in Ross [13]):
(2.1)Vnα(i,x¯)=min{minJ⊆{1,…,L}[∑j∈Jcj(i)+∑j=1Lhjxj+∑j∉J(dj-xj)+∑j=1LdjC+α∑r=0m+1pirVn-1α(r,x-′)],Vnα(PM,x¯){∑j=1Lhjxj}},0≤i≤m,0≤xj<Kj,j=1,…,L,(2.2)Vnα(PM,x-)=cp+∑j=1Lhjxj+∑j=1L(dj-xj)+∑j=1LdjC+αaIVn-1α(0,(x--d-)+)+α(1-aI)Vn-1α(PM,(x--d-)+),0≤xj≤Kj,j=1,…,L,(2.3)Vnα(m+1,x-)=cf+∑j=1Lhjxj+∑j=1L(dj-xj)+∑j=1LdjC+αbIVn-1α(0,(x--d-)+)+α(1-bI)Vn-1α(m+1,(x--d-)+),0≤xj≤Kj,j=1,…,L,
where x-′ in (2.1) is a vector with L components in which the jth component equals to min(xj+pj-dj,Kj), if j∈J, while, if j∉J, it is equal to (xj-dj)+=max(xj-dj,0) and (x--d-)+ in (2.2), and (2.3) is the vector ((x1-d1)+,…,(xL-dL)+). The initial condition is V0α(s)=0, s∈S. Note that, if xj=Kj for some values of j∈{1,…,L}, (2.1) is valid if cj(i) is changed to c~j(i) for these values of j. Note that the first term in the curly brackets in (2.1) corresponds to the best action among all actions J⊆{1,…,L}, while the second term corresponds to action 1 (i.e., initiate a preventive maintenance of the installation). The first part of the following lemma is needed to prove that the average-cost optimal policy is of control-limit type for fixed levels of the buffers.Lemma 2.1.
For eachn=0,1,…, we have that (i)
Vnα(i,x-)≤Vnα(i+1,x-),0≤i≤m,0≤xj≤Kj,j=1,…,L,(ii)
Vnα(PM,x-)≤Vnα(m+1,x-),0≤xj≤Kj,j=1,…,L.Proof.
We will prove the lemma by induction onn. The lemma is valid for n=0, since V0α(s)=0,s∈S. We assume that it is valid for n-1(≥0). We will show that it is also valid for n. First, we prove part (ii) and then part (i).
Part (ii): LetD=Vn-1α(m+1,(x--d-)+-Vn-1α(0,(x--d-)+).
For0≤xj≤Kj,j=1,…,L, we have that
(2.4)Vnα(PM,x-)=cp+∑j=1Lhjxj+∑j=1L(dj-xj)+∑j=1LdjC+αaIVn-1α(0,(x--d-)+)+α(1-aI)Vn-1α(PM,(d--x-)+)≤cf+∑j=1Lhjxj+∑j=1L(dj-xj)+∑j=1LdjC+αaIVn-1α(0,(x--d-)+)+α(1-aI)Vn-1α(m+1,(x--d-)+)=cf+∑j=1Lhjxj+∑j=1L(dj-xj)+∑j=1LdjC+αVn-1α(m+1,(x--d-)+)-αaID≤cf+∑j=1Lhjxj+∑j=1L(dj-xj)+∑j=1LdjC+αVn-1α(m+1,(x--d-)+)-αbID=Vnα(m+1,x-).
The first inequality follows from Condition 4 and from part (ii) of the induction hypothesis. The second inequality follows from Condition 3 and the inequality D≥0 which is a consequence of part (i) of the induction hypothesis.
Part (i): We have to show that(2.5)Vnα(m,x-)≤Vnα(m+1,x-),0≤xj≤Kj,j=1,…,L,(2.6)Vnα(i,x-)≤Vnα(i+1,x-),0≤i≤m-1,0≤xj≤Kj,j=1,…,L.
Inequality (2.5) is easily verified using (2.1) with i=m and part (ii) above
(2.7)Vnα(m,x-)≤Vnα(PM,x-)≤Vnα(m+1,x-).
For 0≤i≤m-1 and 0≤xj<Kj,j=1,…,L, we have that
(2.8)Vnα(i,x¯)=min{minJ⊆{1,…,L}[∑j∈Jcj(i)+∑j=1Lhjxj+∑j∉J(dj-xj)+∑j=1LdjC+α∑r=0m+1pirVn-1α(r,x-′)],Vnα(PM,x¯)}≤min{minJ⊆{1,…,L}[∑j∈Jcj(i+1)+∑j=1Lhjxj+∑j∉J(dj-xj)+∑j=1LdjC+α∑r=0m+1pi+1,rVn-1α(r,x-′)],Vnα(PM,x¯){∑j=1Lhjxj}}=Vnα(i+1,x-).
The above inequality follows from Condition 1 and the inequality
(2.9)∑r=0m+1pirVn-1α(r,x-′)≤∑r=0m+1pi+1,rVn-1α(r,x-′)
which is implied by part (i) of the induction hypothesis and Condition 6. Hence (2.6) has been proved for xj∈{0,…,Kj-1},j=1,…,L. Similarly, we obtain (2.6) if xj=Kj for some values of j∈{1,…,L}.Since the state spaceS is finite, and the state (0,0-) is accessible from every other state under any stationary policy, it follows that there exist numbers v(s), s∈S and a constant g that satisfy the average-cost optimality equations (see Corollary 2.5 in [13], page 98). For the states (i,x-),0≤i≤m,0≤xj<Kj,j=1,…,L, the optimality equations take the following form:
(2.10)v(i,x¯)=min{minJ⊆{1,…,L}[∑j∈Jcj(i)+∑j=1Lhjxj+∑j∉J(dj-xj)+∑j=1LdjC-g+∑r=0m+1pirv(r,x-′)],v(PM,x¯){[∑j∈Jcj(i)+∑j=1Lhjxj+∑j∉J(dj-xj)+∑j=1LdjC-g+∑r=0m+1pirv(r,x-′)]}}.
If xj=Kj for some values of j∈{1,…,L}, we must change in the above equations cj(i) to c~j(i) for these values of  j. In view of part (i) of the above lemma we have the following result.Corollary 2.2.
v
(
i
,
x
-
)
≤
v
(
i
+
1
,
x
-
)
,
0
≤
i
≤
m
,
0
≤
x
j
≤
K
j
,
j
=
1
,
…
,
L.The following proposition gives a characterization of the form of the optimal policy.Proposition 2.3.
For fixed contentsx1,…,xL (0≤xj≤Kj,j=1,…,L) of buffers B1,…,BL, there exists a critical working condition i*(x1,…,xL) such that the policy that minimizes the expected long-run average cost per unit time initiates a preventive maintenance of the installation if and only if its working condition i∈{0,…,m} is greater than or equal to i*(x1,…,xL).Proof.
Suppose that for some fixedx-=(x1,…,xL) such that 0≤xj<Kj, 1≤j≤L, the optimal policy initiates a preventive maintenance of the installation at state (i,x-) where i∈{0,…,m-1}. This implies that
(2.11)v(PM,x¯)≤minJ⊆{1,…,L}[∑j∈Jcj(i)+∑j=1Lhjxj+∑j∉J(dj-xj)+∑j=1LdjC-g+∑r=0m+1pirv(r,x-′)].
To show that the optimal policy prescribes a preventive maintenance on the installation at state(i+1,x-) it is enough to show that
(2.12)v(PM,x¯)≤minJ⊆{1,…,L}[∑j∈Jcj(i+1)+∑j=1Lhjxj+∑j∉J(dj-xj)+∑j=1LdjC-g+∑r=0m+1pi+1,rv(r,x-′)].
From Conditions 1 and 6 and Corollary 2.2, it follows that the right-hand side of (2.12) is greater than or equal to the right-hand side of (2.11). Hence (2.11) implies (2.12). The same result is obtained similarly if xj=Kj for some values of j∈{1,…,L}.Remark 2.4.
In the above proposition, if for fixed contentsx1,…,xL of the buffers i*(x1,…,xL)=m+1, then the optimal policy never initiates a preventive maintenance of the installation whenever the buffer Bj, j=1,…,L, contains xj units or raw material j.


## 2.2. Numerical Results

Example 2.5.
Suppose thatL=2,  m=5,  aI=0.6,  bI=0.4,  cp=10,  cf=15,  K1=5,  K2=20,  h1=1,  h2=1,  p1=2,  d1=1,  p2=2,  d2=1,  c1(i)=0.8(i+1),  c~1(i) = 0.5(i+1),  c2(i) = 0.7(i+1),  c~2(i) = 0.5(i+1),  0≤i≤m,  and  pir=1/(m+2-i),  0≤i≤m,  i≤r≤m+1. It can be readily checked that these probabilities satisfy Condition 5. We computed the optimal policy if  C is equal to 0.5 or 15.5 by implementing the value-iteration algorithm (see Chapter 3 of Tijms [14]). Our numerical results verify the result of Proposition 2.3. In Table 1 we present the critical numbers i*(x1,x2),0≤x1≤5,0≤x2≤20. In each cell of this table the first number corresponds to C=0.5 and the second number corresponds to C=15.5. The minimum average cost was found to be 7.49 if C=0.5 which is, as expected, considerably smaller than the minimum average cost if C=15.5, which was found to be 11.63.Table 1
The critical numbersi*(x1,x2) for C=0.5,15.5.
x
2
∖
x
1
0
1
2
3
4
5
0
3, 6
3, 5
3, 5
4, 6
4, 6
4, 6
1
3, 5
2, 4
2, 4
1, 6
1, 4
1, 4
2
3, 5
2, 3
1, 2
0, 3
0, 3
2, 3
3
4, 6
1, 4
0, 3
0, 2
0, 2
3, 3
4
3, 6
1, 1
0, 0
0, 0
0, 0
3, 3
5
4, 6
1, 4
0, 0
0, 2
0, 1
3, 3
6
4, 6
1, 4
0, 2
0, 1
0, 1
2, 4
7
4, 6
0, 4
0, 2
0, 1
0, 1
2, 3
8
4, 6
0, 4
0, 2
0, 0
0, 0
1, 3
9
4, 6
0, 4
0, 2
0, 0
0, 0
2, 3
10
4, 6
0, 4
0, 2
0, 0
0, 0
1, 3
11
4, 6
0, 4
0, 2
0, 0
0, 0
1, 3
12
4, 6
0, 4
0, 2
0, 0
0, 0
1, 3
13
4, 6
0, 4
0, 1
0, 0
0, 0
1, 3
14
4, 6
0, 4
0, 1
0, 0
0, 0
1, 3
15
4, 6
0, 4
0, 1
0, 0
0, 0
1, 3
16
4, 6
0, 4
0, 1
0, 0
0, 0
1, 2
17
4, 6
0, 4
0, 1
0, 0
0, 0
1, 3
18
4, 6
0, 4
0, 1
0, 0
0, 0
1, 2
19
4, 6
0, 4
0, 1
0, 0
0, 0
1, 3
20
4, 6
0, 4
0, 1
0, 0
0, 0
1, 2We observe that forx1∈{0,…,5} and x2∈{0,…,20} the critical number that corresponds to C=0.5 is smaller than or equal to the critical number that corresponds to C=15.5. This is intuitively reasonable since if the cost due to production delay takes large values it seems disadvantageous to have all or most of the buffers empty when a maintenance is performed on the installation. Therefore in this case it seems preferable to initiate a preventive maintenance of the installation only if its degree of deterioration is relatively high. We also observe that when C=15.5 and buffer B1 or buffer B2 is empty, the optimal policy in most cases never initiates a preventive maintenance of the installation. For example i*(0,3)=6=m+1. It can be also seen from the Table 1 that i*(x1,x2) is not a monotone function with respect to x1∈{0,…,5} and with respect to x2∈{0,…,20} for constant x2 and x1, respectively. Note that, when i<i*(x1,x2),0≤x1≤5,0≤x2≤20, the value iteration algorithm gives the optimal action for the operation of the installation. For example, if C=0.5, x1=0, x2=18, the optimal action when the working condition of the installation is 3 is to transfer raw material 1 to buffer 1 (i.e.,  J={1}). If C=15.5, x1=1, x2=1, the optimal action when the working condition of the installation is 2 is to transfer raw material 1 to buffer 1 and raw material 2 to buffer 2 (i.e.,  J={1,2}).Example 2.6.
Suppose thatL=2, m=15, aI=0.3, bI=0.2, cp=10, cf=15, h1=1, h2=1, C=80, p1=4, d1=2, p2=3, d2=2, c1(i)=1.5(i+1), c~1(i)=0.75(i+1), c2(i)=2(i+1), c~2(i)=i+1, 0≤i≤m, pir=1/(m+2-i), 0≤i≤m, i≤r≤m+1. In Table 2 below we present the minimum average cost g obtained by the value-iteration algorithm for K1∈{1,…,10} and K2∈{5,10}.Table 2
The minimum average cost asK1 or K2 varies.
K
1
K
2
=
5
K
2
=
10
1
51.20
51.17
2
48.55
48.49
3
47.55
47.50
4
45.94
45.88
5
45.60
45.56
6
44.87
44.83
7
44.78
44.75
8
44.49
44.45
9
44.46
44.43
10
44.39
44.37From the above table we see that asK1 or K2 increases, the minimum average cost decreases. This is intuitively reasonable because in this example it seems favourable to have buffers with large, capacities since the cost rate C due to production delay is relatively large while the probabilities aI, bI of successful maintenances and the storage cost rates h1,h2 are relatively small.


## 3. The Problem when the Production Unit Deteriorates Stochastically

We consider the same production-inventory system (see Figure1) as the one introduced in the previous section with the following modifications: (i) the installation is always in operative condition while the production unit may experience a failure as time evolves and (ii) as long as the buffer Bj, j=1,…,L, is not empty and the production unit is in operative condition, the production unit may pull the raw material j from buffer Bj at a constant rate of dj(>pj) units of raw material j per unit of time. When the buffer Bj is empty and the production unit is in operative condition, the production unit reduces its pull rate from dj to pj.We assume that the production unit is monitored at discrete equidistant time epochsτ=0,1,… (say every day), and is classified into one of n+2 working conditions 0,…,n+1. We suppose that working condition i is better than working condition i+1. Working condition 0 means that the production unit is new (or functioning as good as new), while working condition n+1 means that the production unit does not function, and it cannot pull the materials from the buffers. The intermediate working conditions 1,…,n are operative. If the working condition at time epoch τ is i then the working condition at time epoch τ+1 will be r with probability qir. The probability that the deterioration process of the production unit reaches eventually the failure state n+1 from any initial working condition i is assumed to be nonzero. If at a time epoch τ the production unit is found to be at the failure state n+1, a corrective maintenance is compulsory. If it is found to be at any working condition i≤n, a preventive maintenance is optional. The production unit does not operate when it is under preventive maintenance, and it does not pull any raw material from its buffer. When a preventive or a corrective maintenance is performed and the buffer Bj, j=1,…,L, contains xj units of raw material j, the installation transfers min(pj,Kj-xj) units of raw material j to buffer Bj during one unit of time. The preventive and corrective maintenance times (expressed in time units) are geometrically distributed with probability of success aP and bP, respectively. Both maintenances bring the production unit to its perfect condition 0. The state space of the system is the set S~={0,…,n+1,PM}×{0,…,K1}×⋯×{0,…,KL}, where PMrepresents the situation that the production unit is under a preventive repair. The possible actions are the same as the ones considered in the problem studied in Section 2 with the following modification: action J⊆{1,…,L} is the action of pulling raw materials only from buffers Bj, j∈J. If at a time epoch τ the production unit is found to be at working condition i∈{0,…,n}, then we may choose either the action of initiating a preventive maintenance or action J⊆{1,…,L}.The cost structure includes operating costs of the production unit, storage costs, maintenance costs, costs due to production delay, and penalty costs. The storage costshj, j∈{1,…,L}, and the maintenance costs cp,cf are defined exactly in the same way as in the problem studied in the previous section. If the working condition of the production unit is i∈{0,…,n} and the buffer Bj, j∈{1,…,L}, is nonempty (or empty), the cost of pulling dj (or pj) units of raw material j from buffer Bj during one unit of time is equal to cj(i) (or c~j(i)). We assume that the cost rate due to production delay as long as a maintenance of the production unit lasts is equal to C>0. Therefore, if at a time epoch τ the action J⊆{1,…,L} is selected and the content of buffer Bj, j=1,…,L, is xj∈{0,…,Kj}, then the cost due to production delay until time epoch τ+1 is equal to C(∑j∉Jdj+∑j∈J(dj-pj-xj)+)/∑j=1Ldj, where (dj-pj-xj)+=max(dj-pj-xj,0) is the unavailable quantity of raw material j∈J during one unit of time. A penalty cost per unit time which is equal to Pj>0, j=1,…,L, is also imposed for each unit or raw material j that is not stored in buffer Bj during a corrective or a preventive maintenance of the production unit when the buffer Bj is full. This cost is due to the necessary labor for transferring and storing the raw material in another place until the completion of the maintenance. We assume that Conditions 1–5 on the cost structure that we introduced for the problem studied in the previous section are valid if we replace bI with bP,aI with aP, and m with n. We consider a discrete-time Markov decision process in which we aim to find a stationary policy which minimizes the expected long-run average cost per unit of time. Note that for L=1, the problem was studied in Pavitsos and Kyriakidis [2].Since the state spaceS~ is finite and the state (0,K1,…,KL) is accessible from every other state under any stationary policy, it follows that there exist numbers w(s), s∈S and a constant g that satisfy the average-cost optimality equations. For the states (i,x-), 0≤i≤n, 0<xj≤Kj, j=1,…,L, the optimality equations have the following form:
(3.1)w(i,x-)=min{minJ⊆{1,…,L}A(J),w(PM,x-)},
where
(3.2)A(J)=∑j∈Jcj(i)+∑j=1Lhjxj+∑j∉Jdj+∑j∈J(dj-pj-xj)+∑j=1LdjC+∑j∉JPj(pj+xj-Kj)+-g+∑r=0m+1qirw(r,x-′),(3.3)w(PM,x-)=∑j=1Lhjxj+C+∑j=1LPj(pj+xj-Kj)+-g+aPw(0,min(x1+p1,K1),…,min(xL+pL,KL))+(1-aP)w(PM,min(x1+p1,K1),…,min(xL+pL,KL)).
If xj=0 for some values of j∈{1,…,L}, we must change in (3.2) cj(i) to cj~(i) for these values. Note that x-′ in (3.2) is a vector with L components in which the jth component is equal to (xj+pj-dj)+ if j∈J, while, if j∉J, it is equal to (xj-dj)+. It is possible to prove that w(i,x-) ≤ w(i+1,x-),  0≤i≤n,  0≤xj≤Kj,  j=1,…,L, using the dynamic programming equation for the corresponding finite-horizon problem. The method is exactly the same as that used for the proof of Corollary 2.2 in the previous section and, therefore, we omit the details. An immediate consequence of the above monotonicity result is that the result of Proposition 2.3 is valid for the problem of the optimal preventive maintenance of the production unit.


### 3.1. Numerical Examples

Example 3.1.
Suppose thatL=2, K1=7, K2=10, n=10, aP=0.6, bP=0.4, cp=0.4, cf=0.8, h1=1, h2=2, C=0.5, P1=1, P2=1, p1=1, d1=2, p2=1, d2=2, c1(i)=2(i+1), c~1(i)=1.5(i+1), c2(i)=3(i+1), c~2(i)=2.5(i+1), 0≤i≤n, and pir=1/(n+2-i), 0≤i≤n,  i≤r≤n+1. We computed the optimal policy by implementing the value-iteration algorithm. The minimum average cost was found to be 15.67. In Table 3, we present the critical numbers  i*(x1,x2),  0 ≤ x1 ≤ 7,  0 ≤ x2 ≤ 10.
We can see from Table3 that i*(x1,x2) is not a monotone function with respect to x1∈{0,…,7} or with respect to x2∈{0,…,10}, respectively. Note that, when i<i*(x1,x2),  0≤x1≤7,  0≤x2≤10, the value-iteration algorithm gives the optimal action for the operation of the production unit. For example, if x1=0,x2=9, the optimal action when the working condition of the installation is i∈{0,1,2,3,4} is to pull raw material 1 from buffer B1 and raw material 2 from buffer B2 (i.e., J={1,2}), while, when the working condition is i∈{5,6,7} the optimal condition is to pull only raw material 2 from buffer B2 (i.e., J={2}).Table 3
The critical numbersi*(x1,x2),0≤x1≤7,  0≤x2≤10.
x
2
∖
x
1
0
1
2
3
4
5
6
7
0
2
6
7
7
7
6
4
2
1
8
9
9
9
9
9
8
8
2
9
10
10
10
10
10
9
9
3
10
10
10
10
10
10
10
1
4
10
10
10
10
10
10
10
10
5
10
10
10
10
10
10
10
10
6
10
10
10
10
10
10
10
10
7
10
10
10
10
10
10
10
10
8
10
10
10
10
10
10
10
10
9
8
8
9
9
9
9
9
9
10
3
4
4
5
5
5
4
3Example 3.2.
Suppose thatL=2, K1=5, K2=15, n=10, a=0.6, b=0.4, cp=0.5, cf=0.8, C=0.5, P1=1, P2=1, p1=1, d1=2, p2=1, d2=2, c1(i)=2(i+1), c~1(i)=1.5(i+1), c2(i)=3(i+1),c~2(i)=2.5(i+1), 0≤i≤n, and pir=1/(n+2-i), 0≤i≤n, i≤r≤n+1. In Figure 2 we present the graph of the minimum average cost g(h1) as a function of h1∈{1,…,10}, if h2=1, and the graph of the minimum average cost g(h2) as a function of h2∈{1,…,10}, if h1=1. We observe that g(hi), i=1,2, increases as hi increases. The increase of the minimum average cost is more intense when h2 increases. This can be explained by the fact that the capacity of buffer B2 is considerably greater than the capacity of buffer B1.Figure 2
The minimum average costg(h1) and g(h2).


## 3.1. Numerical Examples

Example 3.1.
Suppose thatL=2, K1=7, K2=10, n=10, aP=0.6, bP=0.4, cp=0.4, cf=0.8, h1=1, h2=2, C=0.5, P1=1, P2=1, p1=1, d1=2, p2=1, d2=2, c1(i)=2(i+1), c~1(i)=1.5(i+1), c2(i)=3(i+1), c~2(i)=2.5(i+1), 0≤i≤n, and pir=1/(n+2-i), 0≤i≤n,  i≤r≤n+1. We computed the optimal policy by implementing the value-iteration algorithm. The minimum average cost was found to be 15.67. In Table 3, we present the critical numbers  i*(x1,x2),  0 ≤ x1 ≤ 7,  0 ≤ x2 ≤ 10.
We can see from Table3 that i*(x1,x2) is not a monotone function with respect to x1∈{0,…,7} or with respect to x2∈{0,…,10}, respectively. Note that, when i<i*(x1,x2),  0≤x1≤7,  0≤x2≤10, the value-iteration algorithm gives the optimal action for the operation of the production unit. For example, if x1=0,x2=9, the optimal action when the working condition of the installation is i∈{0,1,2,3,4} is to pull raw material 1 from buffer B1 and raw material 2 from buffer B2 (i.e., J={1,2}), while, when the working condition is i∈{5,6,7} the optimal condition is to pull only raw material 2 from buffer B2 (i.e., J={2}).Table 3
The critical numbersi*(x1,x2),0≤x1≤7,  0≤x2≤10.
x
2
∖
x
1
0
1
2
3
4
5
6
7
0
2
6
7
7
7
6
4
2
1
8
9
9
9
9
9
8
8
2
9
10
10
10
10
10
9
9
3
10
10
10
10
10
10
10
1
4
10
10
10
10
10
10
10
10
5
10
10
10
10
10
10
10
10
6
10
10
10
10
10
10
10
10
7
10
10
10
10
10
10
10
10
8
10
10
10
10
10
10
10
10
9
8
8
9
9
9
9
9
9
10
3
4
4
5
5
5
4
3Example 3.2.
Suppose thatL=2, K1=5, K2=15, n=10, a=0.6, b=0.4, cp=0.5, cf=0.8, C=0.5, P1=1, P2=1, p1=1, d1=2, p2=1, d2=2, c1(i)=2(i+1), c~1(i)=1.5(i+1), c2(i)=3(i+1),c~2(i)=2.5(i+1), 0≤i≤n, and pir=1/(n+2-i), 0≤i≤n, i≤r≤n+1. In Figure 2 we present the graph of the minimum average cost g(h1) as a function of h1∈{1,…,10}, if h2=1, and the graph of the minimum average cost g(h2) as a function of h2∈{1,…,10}, if h1=1. We observe that g(hi), i=1,2, increases as hi increases. The increase of the minimum average cost is more intense when h2 increases. This can be explained by the fact that the capacity of buffer B2 is considerably greater than the capacity of buffer B1.Figure 2
The minimum average costg(h1) and g(h2).


## 4. Conclusions and Future Research

We presented two discrete-time Markov decision models for the optimal condition-based preventive maintenance of a production system which consists of two machines andL intermediate buffers. The first machine transfers L different raw materials to the buffers, and the second machine draws the raw materials from the buffers. The second machine is considered to be a production unit that assembles the raw materials in order to produce the final product. In the first model, it is assumed that only the first machine deteriorates stochastically over time while the production unit is always in operative condition. It is possible to monitor the first machine at discrete equidistant time epochs and to classify it into one working condition that describes its level of deterioration. If the first machine is in failed condition a corrective maintenance must be commenced; otherwise a preventive maintenance may be performed or the action of transferring raw materials to any subset of the set of L buffers may be selected. The maintenances bring the first machine to its perfect condition. In the second model it is assumed, that only the production unit deteriorates over time, and the first machine is always in operative machine. It is possible to determine the level of deterioration of the production unit after inspecting it at discrete equidistant time epochs. If the production unit is in failed condition a corrective maintenance must be started; otherwise a preventive maintenance may be initiated or the action of pulling the raw materials from any subset of the set of L buffers may be selected. Both maintenances bring the production unit to its perfect condition.In both models we considered the problem of determining the policy that minimizes the expected long-run average cost per unit time. If the maintenance times are geometrically distributed we proved that, in both models, the optimal policy is of control-limit type, that is, for fixed contents of the buffers it prescribes a preventive maintenance of the first machine or the production unit if and only if its degree of deterioration exceeds some critical level. The proof was achieved through the corresponding finite-horizon problem.A topic for future research could be a more complicated problem in which the first machine transfers the raw materials to the buffers and the production unit draws them from the buffers in a random manner. Another topic for future research could be the study of the maintenance problems that we would have if the maintenance times are not geometrically distributed but follow some general distributions with suitable conditions.



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*Source: 673864-2012-10-09.xml*