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Optimal inventories for repairable redundant systems with aging components
JOURN,,,.
OF OI’ERAI
IONS MANAGFMlINl
Vol. 5. No. 7. Ma) 1985
Optimal Inventories for Repairable Redundant Systems with Aging Components Charles H. Smith* Margaret K. Schaefer**
EXECUTIVE
SUMMARY
Complex systems that are required to perform very reliably are often designed to be “faultparts have failed. Often faulttolerant,” so that they can function even though some component tolerance is achieved through redundancy, involving the use of extra components. One prevalent redundant component configuration is the m-out-of-n system, where at least m of n identical and independent components must function for the system to function adequately. Often machines containing m-out-of-n systems are scheduled for periodic overhauls, during which all failed components are replaced, in order to renew the machine’s reliability. Periodic overhauls are appropriate when repair of component failures as they occur is impossible or very costly. This will often be the case for machines which are sent on “missions” during which they are unavailable for repair. Examples of such machines include computerized control systems on space vehicles, military and commercial aircraft, and submarines. An interesting inventory problem arises when periodic overhauls are scheduled. How many spare parts should be stocked at the maintenance center in order to meet demands? Complex electronic equipment is rarely scrapped when it fails. Instead, it is sent to a repair shop, from which it eventually returns to the maintenance center to be used as a spare. A Markov model of spares availability at such a maintenance center is developed in this article. Steady-state probabilities are used to determine the initial spares inventory that minimizes total shortage cost and inventory holding cost. The optimal initial spares inventory will depend upon many factors, including the values of m and n, component failure rate, repair rate, time between overhauls, and the shortage and holding costs. In a recent paper, Lawrence and Schaefer [4] determined the optimal maintenance center inventories for fault-tolerant repairable systems. They found optimal maintenance center inventories for machines containing several sets of redundant systems under a budget constraint on total inventory investment. This article extends that work in several important ways. First, we relax the assumption that the parts have constant failure rates. In this model, component failure rates increase as the parts age. Second, we determine the optimal preventive maintenance policy, calculating the optimal age at which a part should be replaced even if it has not failed because the probability of subsequent failure has become unacceptably high. Third, we relax the earlier assumption that component repair times are independent, identically distributed random variables. In this article we allow congestion to develop at the repair shop, making repair times longer when
* Virginia Commonwealth University, Richmond, Virginia. ** College of William and Mary, Williamsburg, Virginia.
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there are many items requiring repair. Fourth, we introduce a more efficient solution method, marginal analysis, as an alternative to dynamic programming, which was used in the earlier paper. Fifth, we modify the model in order to deal with an alternative objective of maximizing the jobcompletion rate. In this article, the notation and assumptions of the earlier model are reviewed. changes in the model development and solution in order to extend the model Several illustrative examples are included.
The requisite are described.
INTRODUCTION Many types of complex machines that must perform very reliably are designed so that they can function when some components have failed. Often an m-out-of-n redundant component configuration is used, in which a system functions if at least m of the n identical independent components function. Machines with such systems are often scheduled for periodic overhauls at a maintenance center, when all failed parts are replaced by good ones to renew the machine’s reliability. In this article a model to determine the optimal repairable parts inventory for a maintenance center serving machines with redundant systems is developed. The machines must leave the maintenance center in “good as new” condition in order to meet high reliability requirements. Upon arrival at the maintenance center, the machine is inspected. It is possible that none of the parts have failed. All failed parts are removed and sent immediately to a repair shop, from which they eventually return to be used in future overhauls. Stocking only one repairable part of each type may be inadequate since the maintenance center would then be subject to shortages, necessitating expensive emergency repair if demands occur during the part repair time, which could be several days or even weeks. In earlier research, Lawrence and Schaefer [4] determined optimal inventories for multiple redundant repairable systems subject to a budget constraint on total investment. The relevant costs included the inventory holding costs and the shortage costs if a part demand cannot be immediately met from available inventory. Critical to the analysis was the determination of steady-state probabilities of stockout of parts at the time of inspection. We substantially generalize that model here by 1) allowing the component failure rates to increase over time instead of being constant over time; 2) considering preventive maintenance for the aging components, and allowing the threshold replacement age to be determined by a reliability constraint; 3) generalizing the model of repair shop operation to include congestion; 4) suggesting marginal analysis as an efficient alternative solution method to dynamic programming; and 5) modifying the model to treat the alternative objective of maximizing job-completion rate. The article is organized in the following manner. In the first section we discuss related models in the literature. The next section contains the model development, including a summary of the Lawrence and Schaefer model. An alternative model for the repair center portion of the study is then described, followed by a treatment of the job-completion rate objective. We next apply our model in illustrative examples, solved using dynamic programming and marginal analysis. RELATED
MODELS
Although much has been written about fault-tolerant systems, emphasis has been placed upon reliability. There is very little literature on the problem of optimal maintenance
340
APICS
center inventories. A recent survey by Nahmias [5] of the extensive repairable inventory literature contains several multi-item models, but, in these models, failure of a single item triggers an immediate demand for a spare. None of them considers the overhaul context, where repair is not required immediately but as a result several parts may be demanded simultaneously at the overhaul. Schaefer [7] found optimal maintenance center inventories for complex machines under a job-completion criterion, where a single stockout penalty was assessed if the overhaul could not be completed because of shortage. The model handled machines with several different types of parts, but exact results were obtained only when part failure rates were low and there was only one part of each type. Lawrence and Schaefer [4] developed the model that is generalized here. Their model assessed a different stockout penalty for each type and number of parts missing, minimizing the expected losses over the parts. Redundant systems were handled explicitly, and part failure rates were not required to be low. That model used a discrete Markov process to generate the steady-state probabilities of stockouts. Other examples of this technique may be found in Tainiter [S], who found the steady-state probabilities for an inventory of rental cars, where the time of rental is analogous to repair time, and in Hadley and Whitin [3], where it was advocated for any invetory system whose condition is reviewed only at discretely space intervals of time. This article generalizes the Lawrence and Schaefer model by allowing increasing component failure rates over time. This generalization allows optimal inventories to be calculated for a much broader class of components, namely, those that wear out gradually over time. In our model we determine the steady-state age distribution of a part at overhaul using Markov chain analysis. This technique has been employed in perishable item inventory theory (see [ 1, 2, 61). The analogies with the present case are clear: in each time period items are either demanded (fail) or they age another period, to leave the system once they reach a given threshold age. Yet the results of these models cannot be used here because these models assume that demands may be filled with units chosen by the decision maker, usually using a FIFO strategy, whereas in our application, items fail randomly. In [4], dynamic programming was used to solve this problem for several independent mi-out-of-n systems subject to an overall budget constraint for parts. This article also shows that a more efficient marginal analysis procedure can be employed to find some of the solutions. MODEL
DEVELOPMENT
Our model is presented in several steps. First, we provide the notation and a summary of the basic one-system model. Next, we extend the basic case by permitting increasing component failure rates with a specified replacement age. The following section analyzes the problem when a replacement age is not given exogenously. Finally, the model is extended to multiple systems with an overall budget constraint. Basic Case: Single M-Out-of-N
System
First, we briefly review the assumptions and notation for the single m-out-of-n system with constant failure rate as analyzed in [4]. Consider a maintenance center serving a set of identical machines containing a single m-out-of-n system of independent identical parts and experiencing identical workloads. We will determine the initial inventory of
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spare parts that minimizes expected stockout costs plus holding costs. We assume that each machine arrives for overhaul in functioning condition so that at least m parts are working and n - m or fewer have failed during the cycle; that is, the machine is so reliable that the probability that more than n - m parts have failed is neglible. This is a reasonable assumption since redundancy is employed to ensure high reliability. Each failed part is immediately sent to a repair shop, from which it eventually returns to the stock of spares at the maintenance center. The total number of spares on hand and undergoing repair is a constant, s. Stockouts may occur if demand for spares at an overhaul exceeds the number of spares presently on hand. For ease of exposition we assume that one machine arrives for overhaul each day, and the cycle time between overhauls is T hours. We also assume each machine.started the cycle in good-as-new condition, and that even if the part is stocked out, the machine is released from overhaul without appreciable delay, although a shortage penalty is assessed. Thus there are no back-orders. We determine steady-state stockout probabilities by modeling the inventory process as a Markov chain, with time intervals of one day and states representing the number of spares available at the maintenance center at the end of the day. Stockouts, i.e. situations where demand for spares exceeds current supply, are denoted by negative values. Thus the possible states are m - n, . . . , s, where s is the number of spares initially stocked. We introduce the following notation: h = constant component failure rate, measured in failures/hour; vk = probability that k parts are required at an inspection, k = 0, 1, . . . , n - m a = 1 - exp(-AT) = probability that a single part has failed during the cycle time T hours p = exponential repair rate measured in repairs/day b = 1 - exp(-p) = probability that a part which is currently at the repair shop will be returned to the maintenance center during the next day R(t, u) = conditional probability that t spares return to the maintenance center from the repair shop during a day that begins with u in the repair shop, 0 I t ll_llS P(s) = matrix of transition probabilities for each value of s The probability that k parts are demanded at an inspection is n
0
V k=
k
Pk( 1 - PYk
; b’( 1 - b),-’
R(t, u) = 0
For a given initial inventory level s, the elements Pij of P(s) are probabilities of transition from state i immediately following one inspection to state j following the next inspection. They are given by min(s-j,n-m)
Pij =
C
vkR(k +j
-i,
s-
i)
for
i = 0,. . . ,s
and
j = m-n,.
. . ,s.
k=max(O,i-j)
Since there are no backorders, Pij = Poj for i = m - n, . . . , -1 and all j. As is well known, the steady-state probabilities ri of having i parts on hand solutions of the system
342
are the
APES
f:
KiPij =
j=m-n,...,s-
Tj;
c
Ti
=
1
1
i=m-n
P - j of a shortage We denote by Bj(s) for j = I, . . . , n - m the steady-state probability of j items when the initial inventory was s. Let H denote the holding cost per day for each spare part in the system and Lj the shortage cost for each day in which a shortage of j parts occurs. The shortage cost Lj may be thought of as the emergency repair or borrowing cost of j items. Lj will usually increase as j increases. We seek a choice of s to minimize the sum of holding costs and long-term expected shortage costs. This problem may be formulated as: Find a non-negative
integer
s which minimizes n-m TC(S) = HS + C LjBj(s)
(PI)
j=l
This problem
is easily solved by enumeration.
Single M-Out-of-N
System with Increasing Failure Rate
In this section we consider a single m-out-of-n system where unfailed parts are replaced at an exogenously determined age M. Later it is shown that the optimal value of M can be determined by utilizing a constraint on the probability of mission failure between inspections. Our generalization of this problem removes the requirement that X and, therefore, a must be constant over time. In fact we allow an arbitrary increasing failure rate. All we require is the capability of determining the probability of part failure for parts at each age. For example, we may have a failure rate that is dependent on the item’s age but that is constant during the interval between inspections. Thus, instead of a single constant X, there is a set {hi} corresponding to the failure rate of a part during the period ending with its inspection at age i. Increasing failure rates may be incorporated into the earlier model by changing only the determination of the vk values. In order to calculate the probability that k items are required at an overhaul, we must analyze an additional Markov chain to generate the steady-state age distribution for each of the n identical parts. The following states can be defined with respect to a given part at the time of overhaul inspection: l l l
State 0 means the part must be replaced due to failure since the last inspection. State i means the part has successfully completed i cycles for i = I, . . . , M - 1. State M means the part has successfully completed M cycles, but it must now be replaced due to age.
For the single m-out-of-n system, the transition probabilities for the associated transition matrix are denoted pij. From given failure rate information one would determine the failure probability ai (i = I, . . . , M) that a part fails during the period ending with its inspection at age i given that it was functioning at the last inspection. For example, when the failure rate is a constant Xi during an interval between inspections, ai = 1 - exp(-XiT).
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All parts that considerations. Pi,i+l
=
1 -
function until age M without failure are replaced We have the following description of the pij: ai+l
Pi.0 = ai+ PM,j
=
P0.j
for
i=O,
for
i=O,
for
j = 0,l
l,...,M-
due
to threshold
1
l,...,M-
1
pi,j = 0 elsewhere For example,
if M = 4 the transition
matrix
is the following:
0 a, 1 - al 0 0 0 l-a2 0 1 a2 0 0 0 2 a3 0 l-a3 0 0 3 a4 0 0 1 - a4 4 i a, 1 - al 0 0 1 0 Let the steady-state probabilities for the above Markov chain be denoted Gi. The steady-state probability of a demand for a replacement part at inspection for a given position is simply 7;a = Go + GM, the sum of the probabilities of replacement due to failure and due to age, respectively. Fortunately, the particular structure of the above transition matrix results in closedform expressions for the steady-state probabilities. In particular k. and G,,, are as follows, where ai = 1 - ai: 1 -a&- *aM ,. To = 1 + a, + a&
+ * * ’ + ala2’ ’ ‘aMpI _a]& ’ ’ ’ aM
A
KM = 1 + a, + a,& + - - - + ala2- - ‘i&,-I Since all n parts fail and age independently, each part has the same steady-state replacement demand probabilities. Thus, the probability that k repair parts are needed at inspection is given by n Vk = &k(l - &)“_k k
0
After the vk’s are determined, further analysis proceeds as it did in [4]. For a single m-out-of-n system the optimal level of initial inventory s to minimize TC(s) in (Pl) is easily determined by enumeration. Single M-Out-of-N System with Increasing Failure Rate and Replacement Age M Not Given In this section we show how to choose the preventive replacement age of the component requirement on the machine’s performance in order to meet a “mission reliability” between overhauls. Suppose that T, the time between overhauls is given, as are the component failure probabilities {ai}. We want to determine M, the number of cycles after which a working component has to be replaced because of deterioration. The performance standard requires that the probability that the m-out-of-n system will fail is less than a small number CY,where failure is defined as having more than n - m parts fail between overhauls.
344
APICS
To solve this problem, the steady-state probability values of M. Let this be denoted Go(M). The optimal which
so, must be calculated for many value of M is the largest value for
&,(M)‘( 1 - .TTa(M))=j < (Y Since the probability of mission failure clearly increases as M increases, a simple search procedure may be used to find the optimal M. In case the optimal M is infinite, the limit of i&(M) exists and is used to compute vk. Multiple
M,-Out-of-N,
Systems
with a Budget Constraint
In [4], the analysis was further extended to the determination of the optimal allocation of initial spares among several sets of redundant systems where the overall inventory holding cost for spares is constrained by a budget. Since the r systems are assumed to function independently, the vk and Bj(s) values can be determined as above separately for each of the systems. In particular, we suppose that each machine is composed of r different types of parts, each arranged in independent m,-out-of-n, configurations. For each q, H, denotes the (integer), holding cost per item per day and Ljq the stockout cost for each day in which a shortage of j items occurs. Each type of part has associated with it a Markov chain as in the single-item case. Let Bjq(Sq) be the steady-state probability of a shortage of p items of type q when sq were initially on hand, determined as shown above. The total holding and expected stockout costs can be minimized by considering each item separately as in (Pl) above. If, however, there is a limit B on total inventory investment (in dollars, with B integer), we are led to the following problem: Find non-negative integers sl, . . , s, to minimize total expected cost r “4-m, C(S) = C C LjqBjq(Sq) q=l j=l s.t. 2 H,s, I B. q=1
(P2)
It is sufficient to consider for each q only values of sq less than or equal to [B/H,] since this is the maximum affordable number of spares of type q. Problem (P2) can now be solved by the standard dynamic programming algorithm [4]. Thus far we have generalized Lawrence and Schaefer by allowing arbitrary increasing failure rates and permitting the optimum part replacement ages to be decision variables of the problem. We have determined new values for vk which can be used in the earlier model. Thus the dynamic programming approach discussed there is still valid. An efficient alternative to dynamic programming for solving (P2) is marginal analysis. In marginal analysis, the spares vector for the r subsystems is initialized by s = (si, s2, according to which part . . . , s,) = (0, 0, . . . ) 0). Spare parts are added to s one-by-one has the highest incremental benefit-cost ratio. The benefit-cost ratio of increasing sq by 1 is [(expected stockout costls,) - (expected stockout costIs, + 1)1/H,, which equals k--m, n4-m4 [ C LjqBjq(sq) C LjqBjq(sq + l)l/Hq j=l
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345
The allocations achieved with marginal analysis are a subset of those found by dynamic programming. The required condition for marginal analysis is that the probabilities of stockout Bj(S) be convex and decreasing. This convexity condition must hold for each type of part in order for the marginal analysis allocations to be optimal, and it should be verified by inspection of the Bj(S) values in any application. (In our example problem, this condition appeared to hold for all systems with a 5 0.05.) Using dynamic programming, optimal allocations are found for all intermediate values of the budget, B’ = 0, 1, 2, . . . , B. Using marginal analysis, not all of those allocations are produced, in general, but the ones that are produced are optimal for that particular value of B’. Only in the case where Hi = Hj, for all i and j, i.e. where holding costs are all equal, does marginal analysis produce all of the optimal solutions. Since marginal analysis only involves calculating and updating a single benefit-cost ratio vector of size r, its computational advantage over dynamic programming is enormous. Storage requirements are minimal. Another advantage that marginal analysis has over dynamic programming is that the steady-state probabilities Bjq(sq) can be calculated only as needed, rather than calculated ahead of time. This considerably reduces the number of sets of simultaneous equations that must be solved. We illustrate the methods in a simple example below where r = 4 and B = 30. Marginal analysis was many times faster than dynamic programming. The advantage would become even more pronounced for problems with larger B and r values. REPAIR
CENTER
SERVICE
ALTERNATIVE
In [4] and so far in this paper, the repair shop has functioned as if it had an infinite number of servers; that is, items arriving at the repair shop are handled independently, as if they immediately begin undergoing repair. This followed from the assumption that successive repair times were independent, indentically distributed random variables. In many situations it is more appropriate to treat the repair shop as a queuing system with a limited number of servers. In particular, we now discuss the solution of this problem when the repair shop is regarded as a one-server queuing system with a first in-first out queue discipline. R(t, u) no longer has a binomial distribution. Instead, any time a queue of failed parts exists at the repair shop, items will leave according to a Poisson distribution until no further items remain. We have R(t, u) = (e-‘p’)/t!
if
o
R(u, u) = 5
= 1 - u$ R(k, u)
and (e-‘pk)/k!
k=O
k=u
Otherwise, the analysis proceeds as before. Recalling that the states of the Markov chain are the number of items on hand, we see that the Markov assumption that the next state of the system depends only on the present state remains valid. The assumption involving the number of servers can have significant effects upon the solution of a problem, as illustrated by our example below. MAXIMIZING
JOB-COMPLETION
In this section we show how to modify rate. In Schaefer [7], optimal maintenance
346
RATE the model in order to maximize job-completion center inventories were found that maximized
APES
job-completion rate, defined as the probability that no shortage of any part would occur at an overhaul, subject to a constraint on total inventory investment. A single penalty, L, was levied if even one required part was not available. In the earlier paper, m-out-ofn systems were not considered explicitly-part failures had to be so low that the probability that more than one part of each type was required at an overhaul was negligible. The model of this research can easily be modified to solve the problem of maximizing job-completion rate subject to a constraint on holding costs. We wish to solve: Find a spares vector s = (s,, s2, . . _ , s,) to maximize fi f,(s,) subject q=l
to i H,s, YZB q=l
(P3)
where f,(s,) is the probability that there is no shortage of part q at an overhaul, given an initial inventory of s,. We can find the probability of no shortage easily; it is just nq-mq f&q) = 1 C b&J p=l (P3) may above.
be solved
by dynamic
programming
or by marginal
analysis,
as explained
EXAMPLE Consider a machine containing four kinds of m,-out-of-n, systems with the data given in Table 1. The budget constraint B is $30. We will use M = 5 as the replacement age for each type of part. The probability of emergency failure using M = 5 ranges from a high of 0.3695 X lops for System 2 to a low of 0.163 1 X 10e6 for System 4. TABLE System
1 Data
1 6 10 4
9 m n H
Lj, j=l 2 3 4 b
24 60 150 240
75 210 600
15 30 45 60
20 50 200 500
.5
.4
.5
.6
0.005 0.010 0.020 0.030 0.040
0.01 0.02 0.03
0.030 0.035 0.040 0.045 0.050
0.010 0.015 0.020 0.025 0.030
ai i= 1 2 3 4 5
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of Operations
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347
TABLE 2 Optimal Inventories B *0 1 2 3 *4 5 6 7 *s 9 10 11 12 13 *14
s*
C (s*)
TC (s*)
B
(0, 0, 0, 0)
$60.94 60.94 60.94 60.94 51.73 51.73 50.48 50.48 43.67 43.67 41.27 41.27 39.70 39.70 33.22
$60.94 60.94 60.94 60.94 55.73 55.73 56.48 56.48 51.67 5 I .67 5 1.27 51.27 51.70 51.70 47.22
16 17 *18 19 20 21 22 23 *24 25 26 27 28 29 30
(0, 0, 0, 0) (0, 0, 0, 0) (0, 0, 0, 0) (0, l,O,O) (0, 1, 0, 0) (O,O,O, 1) (O,O,O, 1) (1, l,O,O) (1, 1, 0, 0) (0, 1, 0, 1) (0, 1, 0, 1) (2, 1, 0, 0) (2, 1, 0, 0) (1, 1, 0, 1)
s* (1, (1, (2, (2, (1, (1, (1, (1, (2, (2, (2, (2. (1, (1, (2,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1. 1, 1, 1, 1. 2,
0, 0, 0, 0, 0, 0, 1, 1. 0, 0, 1, 1. 1, 1, 1,
1) 1) 1) 1) 2) 2) 1) 1) 2) 2) 1) 1) 2) 2) 1)
C (s*)
TC (s*)
$33.22 33.22 29.24 29.24 28.36 28.36 26.85 26.85 24.39 24.39 22.88 22.88 22.00 22.00 20.6 I
$47.22 47.22 47.44 47.44 48.36 48.36 48.85 48.85 48.39 48.39 48.88 48.88 50.00 50.00 50.6 1
The optimal solutions obtained from dynamic programming are listed in Table 2. Consider now the same problem except let the repair shop function as a one-server operation. Then the results for selected values of B are given in Table 3. The asterisks in the tables indicate those solutions that were also obtained by marginal analysis. Marginal analysis provides a much faster computational alternative than dynamic programming, but it will not necessarily generate all optimal solutions. The column labeled TC(s*) in the tables represents the total of stockout costs C(s*) plus holding
costs
i H,s,. In these examples the minimum total cost requires $14 in q=l holding costs. Note that the expected stockout cost C(s*) for the single-server repair shop is always at least as great as for the infinite server repair shop. While in the example the repair shop alternatives tend to have the same optimal solutions, this is not true for
TABLE 3 Optimal Inventories-Single-Server B 0 2 *4 6 *8 10 12 *14
348
S*
(030, 0, 0) (0. (0, (0, (1, (0, (2, (1,
0, 0, 0) 1, 0, 0) 0, 0, 1) 1, 0, 0) 1, 0, 1) l,O,O) 1, 0, 1)
c
(s*)
$60.94 60.94 51.73 50.48 43.67 41.27 41.02 33.22
TC (s*)
B
$60.94 60.94 55.73 56.48 51.67 51.27 53.02 47.22
16 18 20 ‘22 24 *26 28 30
Repair Shop s* (I, (2, (1, (1, (1, (2, (1, (2,
1, 1, 1, 1, 1, 1, 1, 2,
0, 1) 0, 1) 0%2) 1, 1) 1, 1) 1, 1) 1, 2) 1, 1)
C (s*)
TC (s*)
$33.22 30.56 29.92 26.85 26.85 24.20 23.55 22.27
$47.22 48.56 49.92 48.85 48.85 50.20 51.55 52.27
APES
B = 24. Note that when s* differs, the single-server model tends to spread across more of the systems in order to avoid the consequences of congestion.
the spares
CONCLUSIONS We have developed a model that determines optimal maintenance center inventories for a set of m-out-of-n systems on a machine subject to scheduled overhauls. The components may have constant or increasing failure rates, and the repair shop may be subject to congestion. Solution methods include dynamic programming and marginal analysis.
REFERENCES 1. Chazan. D. and S. Gal “A Markovian Model for a Perishable Product Inventory,” Management Science, Vol. 23, No. 5, (1977) pp. 512-514. 2. Graves, S., “The Application of Queuing Theory to Continuous Perishable Inventory Systems,” Managemenl Science, Vol. 28, No. 4, (1982) pp. 400-406. 3. Hadley, G. and T. Whitin, Analysis of Inventory Sysfems, Prentice-Hall, Inc., Englewood Cliffs, N.J., 1963. 4. Lawrence, S.H. and M.K. Schaefer, “Optimal Maintenance Center Inventories for Fault-Tolerant Repairable Systems,” Journal qf Operations Management, Vol. 4, No. 2, (February 1984) pp. 175181.
Journal of Operations
Management
5. Nahmias, Systems:
S., “Managing Reparable Item Inventory A Review,” Chapter 12 in Multi-Level Production/Inventory Control Syslems, edited by L.B. Schwarz, North-Holland, Amsterdam, 198 1. 6. Pegels, C. and A. Jelmert, “An Evaluation of Blood-Inventory Policies: A Markov Chain Application,” Operations Research, Vol. 18, No. 6, (1970) pp. 1087-1098. 7. Schaefer, M.K., “A Multi-Item Maintenance Center Inventory Model for Low-Demand Reparable Items,” Management Science, Vol. 29, No. 9, (1983), pp. 1062-1068. 8. Tainiter, M., “Some Stochastic Inventory Models for Rental Situations,” Management Science, Vol. 11, No. 2, (1964), pp. 316-326.
349
OF OI’ERAI
IONS MANAGFMlINl
Vol. 5. No. 7. Ma) 1985
Optimal Inventories for Repairable Redundant Systems with Aging Components Charles H. Smith* Margaret K. Schaefer**
EXECUTIVE
SUMMARY
Complex systems that are required to perform very reliably are often designed to be “faultparts have failed. Often faulttolerant,” so that they can function even though some component tolerance is achieved through redundancy, involving the use of extra components. One prevalent redundant component configuration is the m-out-of-n system, where at least m of n identical and independent components must function for the system to function adequately. Often machines containing m-out-of-n systems are scheduled for periodic overhauls, during which all failed components are replaced, in order to renew the machine’s reliability. Periodic overhauls are appropriate when repair of component failures as they occur is impossible or very costly. This will often be the case for machines which are sent on “missions” during which they are unavailable for repair. Examples of such machines include computerized control systems on space vehicles, military and commercial aircraft, and submarines. An interesting inventory problem arises when periodic overhauls are scheduled. How many spare parts should be stocked at the maintenance center in order to meet demands? Complex electronic equipment is rarely scrapped when it fails. Instead, it is sent to a repair shop, from which it eventually returns to the maintenance center to be used as a spare. A Markov model of spares availability at such a maintenance center is developed in this article. Steady-state probabilities are used to determine the initial spares inventory that minimizes total shortage cost and inventory holding cost. The optimal initial spares inventory will depend upon many factors, including the values of m and n, component failure rate, repair rate, time between overhauls, and the shortage and holding costs. In a recent paper, Lawrence and Schaefer [4] determined the optimal maintenance center inventories for fault-tolerant repairable systems. They found optimal maintenance center inventories for machines containing several sets of redundant systems under a budget constraint on total inventory investment. This article extends that work in several important ways. First, we relax the assumption that the parts have constant failure rates. In this model, component failure rates increase as the parts age. Second, we determine the optimal preventive maintenance policy, calculating the optimal age at which a part should be replaced even if it has not failed because the probability of subsequent failure has become unacceptably high. Third, we relax the earlier assumption that component repair times are independent, identically distributed random variables. In this article we allow congestion to develop at the repair shop, making repair times longer when
* Virginia Commonwealth University, Richmond, Virginia. ** College of William and Mary, Williamsburg, Virginia.
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there are many items requiring repair. Fourth, we introduce a more efficient solution method, marginal analysis, as an alternative to dynamic programming, which was used in the earlier paper. Fifth, we modify the model in order to deal with an alternative objective of maximizing the jobcompletion rate. In this article, the notation and assumptions of the earlier model are reviewed. changes in the model development and solution in order to extend the model Several illustrative examples are included.
The requisite are described.
INTRODUCTION Many types of complex machines that must perform very reliably are designed so that they can function when some components have failed. Often an m-out-of-n redundant component configuration is used, in which a system functions if at least m of the n identical independent components function. Machines with such systems are often scheduled for periodic overhauls at a maintenance center, when all failed parts are replaced by good ones to renew the machine’s reliability. In this article a model to determine the optimal repairable parts inventory for a maintenance center serving machines with redundant systems is developed. The machines must leave the maintenance center in “good as new” condition in order to meet high reliability requirements. Upon arrival at the maintenance center, the machine is inspected. It is possible that none of the parts have failed. All failed parts are removed and sent immediately to a repair shop, from which they eventually return to be used in future overhauls. Stocking only one repairable part of each type may be inadequate since the maintenance center would then be subject to shortages, necessitating expensive emergency repair if demands occur during the part repair time, which could be several days or even weeks. In earlier research, Lawrence and Schaefer [4] determined optimal inventories for multiple redundant repairable systems subject to a budget constraint on total investment. The relevant costs included the inventory holding costs and the shortage costs if a part demand cannot be immediately met from available inventory. Critical to the analysis was the determination of steady-state probabilities of stockout of parts at the time of inspection. We substantially generalize that model here by 1) allowing the component failure rates to increase over time instead of being constant over time; 2) considering preventive maintenance for the aging components, and allowing the threshold replacement age to be determined by a reliability constraint; 3) generalizing the model of repair shop operation to include congestion; 4) suggesting marginal analysis as an efficient alternative solution method to dynamic programming; and 5) modifying the model to treat the alternative objective of maximizing job-completion rate. The article is organized in the following manner. In the first section we discuss related models in the literature. The next section contains the model development, including a summary of the Lawrence and Schaefer model. An alternative model for the repair center portion of the study is then described, followed by a treatment of the job-completion rate objective. We next apply our model in illustrative examples, solved using dynamic programming and marginal analysis. RELATED
MODELS
Although much has been written about fault-tolerant systems, emphasis has been placed upon reliability. There is very little literature on the problem of optimal maintenance
340
APICS
center inventories. A recent survey by Nahmias [5] of the extensive repairable inventory literature contains several multi-item models, but, in these models, failure of a single item triggers an immediate demand for a spare. None of them considers the overhaul context, where repair is not required immediately but as a result several parts may be demanded simultaneously at the overhaul. Schaefer [7] found optimal maintenance center inventories for complex machines under a job-completion criterion, where a single stockout penalty was assessed if the overhaul could not be completed because of shortage. The model handled machines with several different types of parts, but exact results were obtained only when part failure rates were low and there was only one part of each type. Lawrence and Schaefer [4] developed the model that is generalized here. Their model assessed a different stockout penalty for each type and number of parts missing, minimizing the expected losses over the parts. Redundant systems were handled explicitly, and part failure rates were not required to be low. That model used a discrete Markov process to generate the steady-state probabilities of stockouts. Other examples of this technique may be found in Tainiter [S], who found the steady-state probabilities for an inventory of rental cars, where the time of rental is analogous to repair time, and in Hadley and Whitin [3], where it was advocated for any invetory system whose condition is reviewed only at discretely space intervals of time. This article generalizes the Lawrence and Schaefer model by allowing increasing component failure rates over time. This generalization allows optimal inventories to be calculated for a much broader class of components, namely, those that wear out gradually over time. In our model we determine the steady-state age distribution of a part at overhaul using Markov chain analysis. This technique has been employed in perishable item inventory theory (see [ 1, 2, 61). The analogies with the present case are clear: in each time period items are either demanded (fail) or they age another period, to leave the system once they reach a given threshold age. Yet the results of these models cannot be used here because these models assume that demands may be filled with units chosen by the decision maker, usually using a FIFO strategy, whereas in our application, items fail randomly. In [4], dynamic programming was used to solve this problem for several independent mi-out-of-n systems subject to an overall budget constraint for parts. This article also shows that a more efficient marginal analysis procedure can be employed to find some of the solutions. MODEL
DEVELOPMENT
Our model is presented in several steps. First, we provide the notation and a summary of the basic one-system model. Next, we extend the basic case by permitting increasing component failure rates with a specified replacement age. The following section analyzes the problem when a replacement age is not given exogenously. Finally, the model is extended to multiple systems with an overall budget constraint. Basic Case: Single M-Out-of-N
System
First, we briefly review the assumptions and notation for the single m-out-of-n system with constant failure rate as analyzed in [4]. Consider a maintenance center serving a set of identical machines containing a single m-out-of-n system of independent identical parts and experiencing identical workloads. We will determine the initial inventory of
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spare parts that minimizes expected stockout costs plus holding costs. We assume that each machine arrives for overhaul in functioning condition so that at least m parts are working and n - m or fewer have failed during the cycle; that is, the machine is so reliable that the probability that more than n - m parts have failed is neglible. This is a reasonable assumption since redundancy is employed to ensure high reliability. Each failed part is immediately sent to a repair shop, from which it eventually returns to the stock of spares at the maintenance center. The total number of spares on hand and undergoing repair is a constant, s. Stockouts may occur if demand for spares at an overhaul exceeds the number of spares presently on hand. For ease of exposition we assume that one machine arrives for overhaul each day, and the cycle time between overhauls is T hours. We also assume each machine.started the cycle in good-as-new condition, and that even if the part is stocked out, the machine is released from overhaul without appreciable delay, although a shortage penalty is assessed. Thus there are no back-orders. We determine steady-state stockout probabilities by modeling the inventory process as a Markov chain, with time intervals of one day and states representing the number of spares available at the maintenance center at the end of the day. Stockouts, i.e. situations where demand for spares exceeds current supply, are denoted by negative values. Thus the possible states are m - n, . . . , s, where s is the number of spares initially stocked. We introduce the following notation: h = constant component failure rate, measured in failures/hour; vk = probability that k parts are required at an inspection, k = 0, 1, . . . , n - m a = 1 - exp(-AT) = probability that a single part has failed during the cycle time T hours p = exponential repair rate measured in repairs/day b = 1 - exp(-p) = probability that a part which is currently at the repair shop will be returned to the maintenance center during the next day R(t, u) = conditional probability that t spares return to the maintenance center from the repair shop during a day that begins with u in the repair shop, 0 I t ll_llS P(s) = matrix of transition probabilities for each value of s The probability that k parts are demanded at an inspection is n
0
V k=
k
Pk( 1 - PYk
; b’( 1 - b),-’
R(t, u) = 0
For a given initial inventory level s, the elements Pij of P(s) are probabilities of transition from state i immediately following one inspection to state j following the next inspection. They are given by min(s-j,n-m)
Pij =
C
vkR(k +j
-i,
s-
i)
for
i = 0,. . . ,s
and
j = m-n,.
. . ,s.
k=max(O,i-j)
Since there are no backorders, Pij = Poj for i = m - n, . . . , -1 and all j. As is well known, the steady-state probabilities ri of having i parts on hand solutions of the system
342
are the
APES
f:
KiPij =
j=m-n,...,s-
Tj;
c
Ti
=
1
1
i=m-n
P - j of a shortage We denote by Bj(s) for j = I, . . . , n - m the steady-state probability of j items when the initial inventory was s. Let H denote the holding cost per day for each spare part in the system and Lj the shortage cost for each day in which a shortage of j parts occurs. The shortage cost Lj may be thought of as the emergency repair or borrowing cost of j items. Lj will usually increase as j increases. We seek a choice of s to minimize the sum of holding costs and long-term expected shortage costs. This problem may be formulated as: Find a non-negative
integer
s which minimizes n-m TC(S) = HS + C LjBj(s)
(PI)
j=l
This problem
is easily solved by enumeration.
Single M-Out-of-N
System with Increasing Failure Rate
In this section we consider a single m-out-of-n system where unfailed parts are replaced at an exogenously determined age M. Later it is shown that the optimal value of M can be determined by utilizing a constraint on the probability of mission failure between inspections. Our generalization of this problem removes the requirement that X and, therefore, a must be constant over time. In fact we allow an arbitrary increasing failure rate. All we require is the capability of determining the probability of part failure for parts at each age. For example, we may have a failure rate that is dependent on the item’s age but that is constant during the interval between inspections. Thus, instead of a single constant X, there is a set {hi} corresponding to the failure rate of a part during the period ending with its inspection at age i. Increasing failure rates may be incorporated into the earlier model by changing only the determination of the vk values. In order to calculate the probability that k items are required at an overhaul, we must analyze an additional Markov chain to generate the steady-state age distribution for each of the n identical parts. The following states can be defined with respect to a given part at the time of overhaul inspection: l l l
State 0 means the part must be replaced due to failure since the last inspection. State i means the part has successfully completed i cycles for i = I, . . . , M - 1. State M means the part has successfully completed M cycles, but it must now be replaced due to age.
For the single m-out-of-n system, the transition probabilities for the associated transition matrix are denoted pij. From given failure rate information one would determine the failure probability ai (i = I, . . . , M) that a part fails during the period ending with its inspection at age i given that it was functioning at the last inspection. For example, when the failure rate is a constant Xi during an interval between inspections, ai = 1 - exp(-XiT).
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All parts that considerations. Pi,i+l
=
1 -
function until age M without failure are replaced We have the following description of the pij: ai+l
Pi.0 = ai+ PM,j
=
P0.j
for
i=O,
for
i=O,
for
j = 0,l
l,...,M-
due
to threshold
1
l,...,M-
1
pi,j = 0 elsewhere For example,
if M = 4 the transition
matrix
is the following:
0 a, 1 - al 0 0 0 l-a2 0 1 a2 0 0 0 2 a3 0 l-a3 0 0 3 a4 0 0 1 - a4 4 i a, 1 - al 0 0 1 0 Let the steady-state probabilities for the above Markov chain be denoted Gi. The steady-state probability of a demand for a replacement part at inspection for a given position is simply 7;a = Go + GM, the sum of the probabilities of replacement due to failure and due to age, respectively. Fortunately, the particular structure of the above transition matrix results in closedform expressions for the steady-state probabilities. In particular k. and G,,, are as follows, where ai = 1 - ai: 1 -a&- *aM ,. To = 1 + a, + a&
+ * * ’ + ala2’ ’ ‘aMpI _a]& ’ ’ ’ aM
A
KM = 1 + a, + a,& + - - - + ala2- - ‘i&,-I Since all n parts fail and age independently, each part has the same steady-state replacement demand probabilities. Thus, the probability that k repair parts are needed at inspection is given by n Vk = &k(l - &)“_k k
0
After the vk’s are determined, further analysis proceeds as it did in [4]. For a single m-out-of-n system the optimal level of initial inventory s to minimize TC(s) in (Pl) is easily determined by enumeration. Single M-Out-of-N System with Increasing Failure Rate and Replacement Age M Not Given In this section we show how to choose the preventive replacement age of the component requirement on the machine’s performance in order to meet a “mission reliability” between overhauls. Suppose that T, the time between overhauls is given, as are the component failure probabilities {ai}. We want to determine M, the number of cycles after which a working component has to be replaced because of deterioration. The performance standard requires that the probability that the m-out-of-n system will fail is less than a small number CY,where failure is defined as having more than n - m parts fail between overhauls.
344
APICS
To solve this problem, the steady-state probability values of M. Let this be denoted Go(M). The optimal which
so, must be calculated for many value of M is the largest value for
&,(M)‘( 1 - .TTa(M))=j < (Y Since the probability of mission failure clearly increases as M increases, a simple search procedure may be used to find the optimal M. In case the optimal M is infinite, the limit of i&(M) exists and is used to compute vk. Multiple
M,-Out-of-N,
Systems
with a Budget Constraint
In [4], the analysis was further extended to the determination of the optimal allocation of initial spares among several sets of redundant systems where the overall inventory holding cost for spares is constrained by a budget. Since the r systems are assumed to function independently, the vk and Bj(s) values can be determined as above separately for each of the systems. In particular, we suppose that each machine is composed of r different types of parts, each arranged in independent m,-out-of-n, configurations. For each q, H, denotes the (integer), holding cost per item per day and Ljq the stockout cost for each day in which a shortage of j items occurs. Each type of part has associated with it a Markov chain as in the single-item case. Let Bjq(Sq) be the steady-state probability of a shortage of p items of type q when sq were initially on hand, determined as shown above. The total holding and expected stockout costs can be minimized by considering each item separately as in (Pl) above. If, however, there is a limit B on total inventory investment (in dollars, with B integer), we are led to the following problem: Find non-negative integers sl, . . , s, to minimize total expected cost r “4-m, C(S) = C C LjqBjq(Sq) q=l j=l s.t. 2 H,s, I B. q=1
(P2)
It is sufficient to consider for each q only values of sq less than or equal to [B/H,] since this is the maximum affordable number of spares of type q. Problem (P2) can now be solved by the standard dynamic programming algorithm [4]. Thus far we have generalized Lawrence and Schaefer by allowing arbitrary increasing failure rates and permitting the optimum part replacement ages to be decision variables of the problem. We have determined new values for vk which can be used in the earlier model. Thus the dynamic programming approach discussed there is still valid. An efficient alternative to dynamic programming for solving (P2) is marginal analysis. In marginal analysis, the spares vector for the r subsystems is initialized by s = (si, s2, according to which part . . . , s,) = (0, 0, . . . ) 0). Spare parts are added to s one-by-one has the highest incremental benefit-cost ratio. The benefit-cost ratio of increasing sq by 1 is [(expected stockout costls,) - (expected stockout costIs, + 1)1/H,, which equals k--m, n4-m4 [ C LjqBjq(sq) C LjqBjq(sq + l)l/Hq j=l
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345
The allocations achieved with marginal analysis are a subset of those found by dynamic programming. The required condition for marginal analysis is that the probabilities of stockout Bj(S) be convex and decreasing. This convexity condition must hold for each type of part in order for the marginal analysis allocations to be optimal, and it should be verified by inspection of the Bj(S) values in any application. (In our example problem, this condition appeared to hold for all systems with a 5 0.05.) Using dynamic programming, optimal allocations are found for all intermediate values of the budget, B’ = 0, 1, 2, . . . , B. Using marginal analysis, not all of those allocations are produced, in general, but the ones that are produced are optimal for that particular value of B’. Only in the case where Hi = Hj, for all i and j, i.e. where holding costs are all equal, does marginal analysis produce all of the optimal solutions. Since marginal analysis only involves calculating and updating a single benefit-cost ratio vector of size r, its computational advantage over dynamic programming is enormous. Storage requirements are minimal. Another advantage that marginal analysis has over dynamic programming is that the steady-state probabilities Bjq(sq) can be calculated only as needed, rather than calculated ahead of time. This considerably reduces the number of sets of simultaneous equations that must be solved. We illustrate the methods in a simple example below where r = 4 and B = 30. Marginal analysis was many times faster than dynamic programming. The advantage would become even more pronounced for problems with larger B and r values. REPAIR
CENTER
SERVICE
ALTERNATIVE
In [4] and so far in this paper, the repair shop has functioned as if it had an infinite number of servers; that is, items arriving at the repair shop are handled independently, as if they immediately begin undergoing repair. This followed from the assumption that successive repair times were independent, indentically distributed random variables. In many situations it is more appropriate to treat the repair shop as a queuing system with a limited number of servers. In particular, we now discuss the solution of this problem when the repair shop is regarded as a one-server queuing system with a first in-first out queue discipline. R(t, u) no longer has a binomial distribution. Instead, any time a queue of failed parts exists at the repair shop, items will leave according to a Poisson distribution until no further items remain. We have R(t, u) = (e-‘p’)/t!
if
o
R(u, u) = 5
= 1 - u$ R(k, u)
and (e-‘pk)/k!
k=O
k=u
Otherwise, the analysis proceeds as before. Recalling that the states of the Markov chain are the number of items on hand, we see that the Markov assumption that the next state of the system depends only on the present state remains valid. The assumption involving the number of servers can have significant effects upon the solution of a problem, as illustrated by our example below. MAXIMIZING
JOB-COMPLETION
In this section we show how to modify rate. In Schaefer [7], optimal maintenance
346
RATE the model in order to maximize job-completion center inventories were found that maximized
APES
job-completion rate, defined as the probability that no shortage of any part would occur at an overhaul, subject to a constraint on total inventory investment. A single penalty, L, was levied if even one required part was not available. In the earlier paper, m-out-ofn systems were not considered explicitly-part failures had to be so low that the probability that more than one part of each type was required at an overhaul was negligible. The model of this research can easily be modified to solve the problem of maximizing job-completion rate subject to a constraint on holding costs. We wish to solve: Find a spares vector s = (s,, s2, . . _ , s,) to maximize fi f,(s,) subject q=l
to i H,s, YZB q=l
(P3)
where f,(s,) is the probability that there is no shortage of part q at an overhaul, given an initial inventory of s,. We can find the probability of no shortage easily; it is just nq-mq f&q) = 1 C b&J p=l (P3) may above.
be solved
by dynamic
programming
or by marginal
analysis,
as explained
EXAMPLE Consider a machine containing four kinds of m,-out-of-n, systems with the data given in Table 1. The budget constraint B is $30. We will use M = 5 as the replacement age for each type of part. The probability of emergency failure using M = 5 ranges from a high of 0.3695 X lops for System 2 to a low of 0.163 1 X 10e6 for System 4. TABLE System
1 Data
1 6 10 4
9 m n H
Lj, j=l 2 3 4 b
24 60 150 240
75 210 600
15 30 45 60
20 50 200 500
.5
.4
.5
.6
0.005 0.010 0.020 0.030 0.040
0.01 0.02 0.03
0.030 0.035 0.040 0.045 0.050
0.010 0.015 0.020 0.025 0.030
ai i= 1 2 3 4 5
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of Operations
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TABLE 2 Optimal Inventories B *0 1 2 3 *4 5 6 7 *s 9 10 11 12 13 *14
s*
C (s*)
TC (s*)
B
(0, 0, 0, 0)
$60.94 60.94 60.94 60.94 51.73 51.73 50.48 50.48 43.67 43.67 41.27 41.27 39.70 39.70 33.22
$60.94 60.94 60.94 60.94 55.73 55.73 56.48 56.48 51.67 5 I .67 5 1.27 51.27 51.70 51.70 47.22
16 17 *18 19 20 21 22 23 *24 25 26 27 28 29 30
(0, 0, 0, 0) (0, 0, 0, 0) (0, 0, 0, 0) (0, l,O,O) (0, 1, 0, 0) (O,O,O, 1) (O,O,O, 1) (1, l,O,O) (1, 1, 0, 0) (0, 1, 0, 1) (0, 1, 0, 1) (2, 1, 0, 0) (2, 1, 0, 0) (1, 1, 0, 1)
s* (1, (1, (2, (2, (1, (1, (1, (1, (2, (2, (2, (2. (1, (1, (2,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1. 1, 1, 1, 1. 2,
0, 0, 0, 0, 0, 0, 1, 1. 0, 0, 1, 1. 1, 1, 1,
1) 1) 1) 1) 2) 2) 1) 1) 2) 2) 1) 1) 2) 2) 1)
C (s*)
TC (s*)
$33.22 33.22 29.24 29.24 28.36 28.36 26.85 26.85 24.39 24.39 22.88 22.88 22.00 22.00 20.6 I
$47.22 47.22 47.44 47.44 48.36 48.36 48.85 48.85 48.39 48.39 48.88 48.88 50.00 50.00 50.6 1
The optimal solutions obtained from dynamic programming are listed in Table 2. Consider now the same problem except let the repair shop function as a one-server operation. Then the results for selected values of B are given in Table 3. The asterisks in the tables indicate those solutions that were also obtained by marginal analysis. Marginal analysis provides a much faster computational alternative than dynamic programming, but it will not necessarily generate all optimal solutions. The column labeled TC(s*) in the tables represents the total of stockout costs C(s*) plus holding
costs
i H,s,. In these examples the minimum total cost requires $14 in q=l holding costs. Note that the expected stockout cost C(s*) for the single-server repair shop is always at least as great as for the infinite server repair shop. While in the example the repair shop alternatives tend to have the same optimal solutions, this is not true for
TABLE 3 Optimal Inventories-Single-Server B 0 2 *4 6 *8 10 12 *14
348
S*
(030, 0, 0) (0. (0, (0, (1, (0, (2, (1,
0, 0, 0) 1, 0, 0) 0, 0, 1) 1, 0, 0) 1, 0, 1) l,O,O) 1, 0, 1)
c
(s*)
$60.94 60.94 51.73 50.48 43.67 41.27 41.02 33.22
TC (s*)
B
$60.94 60.94 55.73 56.48 51.67 51.27 53.02 47.22
16 18 20 ‘22 24 *26 28 30
Repair Shop s* (I, (2, (1, (1, (1, (2, (1, (2,
1, 1, 1, 1, 1, 1, 1, 2,
0, 1) 0, 1) 0%2) 1, 1) 1, 1) 1, 1) 1, 2) 1, 1)
C (s*)
TC (s*)
$33.22 30.56 29.92 26.85 26.85 24.20 23.55 22.27
$47.22 48.56 49.92 48.85 48.85 50.20 51.55 52.27
APES
B = 24. Note that when s* differs, the single-server model tends to spread across more of the systems in order to avoid the consequences of congestion.
the spares
CONCLUSIONS We have developed a model that determines optimal maintenance center inventories for a set of m-out-of-n systems on a machine subject to scheduled overhauls. The components may have constant or increasing failure rates, and the repair shop may be subject to congestion. Solution methods include dynamic programming and marginal analysis.
REFERENCES 1. Chazan. D. and S. Gal “A Markovian Model for a Perishable Product Inventory,” Management Science, Vol. 23, No. 5, (1977) pp. 512-514. 2. Graves, S., “The Application of Queuing Theory to Continuous Perishable Inventory Systems,” Managemenl Science, Vol. 28, No. 4, (1982) pp. 400-406. 3. Hadley, G. and T. Whitin, Analysis of Inventory Sysfems, Prentice-Hall, Inc., Englewood Cliffs, N.J., 1963. 4. Lawrence, S.H. and M.K. Schaefer, “Optimal Maintenance Center Inventories for Fault-Tolerant Repairable Systems,” Journal qf Operations Management, Vol. 4, No. 2, (February 1984) pp. 175181.
Journal of Operations
Management
5. Nahmias, Systems:
S., “Managing Reparable Item Inventory A Review,” Chapter 12 in Multi-Level Production/Inventory Control Syslems, edited by L.B. Schwarz, North-Holland, Amsterdam, 198 1. 6. Pegels, C. and A. Jelmert, “An Evaluation of Blood-Inventory Policies: A Markov Chain Application,” Operations Research, Vol. 18, No. 6, (1970) pp. 1087-1098. 7. Schaefer, M.K., “A Multi-Item Maintenance Center Inventory Model for Low-Demand Reparable Items,” Management Science, Vol. 29, No. 9, (1983), pp. 1062-1068. 8. Tainiter, M., “Some Stochastic Inventory Models for Rental Situations,” Management Science, Vol. 11, No. 2, (1964), pp. 316-326.
349