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We consider a maintenance shop that is responsible for the availability of a fleet of assets; e.g., trains. Unavailability of assets may be due to active maintenance time or unavailability of spare parts. Both spare assets and spare parts may be stocked in order to ensure a certain fleet readiness, which is the probability of having sufficient assets available for the primary process (e.g., running a train schedule) at any given moment. This is different from guaranteeing a certain average availability, as is typically done in the literature on spare parts inventories. We analyze the corresponding system, assuming continuous review and base stock control. We propose an algorithm, based on a marginal analysis approach, to solve the optimization problem of minimizing holding costs for spare assets and spare parts. Since the problem is not item separable, even marginal analysis is time-consuming, but we show how to efficiently solve this problem. Using a numerical experiment, we show that our algorithm generally leads to a solution that is close to optimal and that it is much faster than an existing algorithm for a closely related problem. We further show that the additional costs that are incurred when the problem of stocking spare assets and spare parts is not solved jointly can be significant. A key managerial insight is that typically the number of spare assets to be acquired is very close to a lower bound that is determined only by the active maintenance time on the assets. It is typically not cost-effective to acquire more spare assets to cover spare parts unavailability.
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Fleet readiness: stocking spare parts and high-tech assets
Abstract
We consider a maintenance shop that is responsible for the availability of a fleet of assets,
e.g., trains. Unavailability of assets may be due to active maintenance time or unavailability
of spare parts. Both spare assets and spare parts may be stocked in order to ensure a
certain fleet readiness, which is the probability of having sufficient assets available for the
primary process (e.g., running a train schedule) at any given moment. This is different
from guaranteeing a certain average availability, as is typically done in the literature on
spare parts inventories. We analyze the corresponding system, assuming continuous review
and base stock control. We propose an algorithm, based on a marginal analysis approach,
to solve the optimization problem of minimizing holding costs for spare assets and spare
parts. Since the problem is not item separable, even marginal analysis is time consuming,
but we show how to efficiently solve this. Using a numerical experiment, we show that our
algorithm generally leads to a solution that is close to optimal, and that it is much faster
than an existing algorithm for a closely related problem. We further show that the additional
costs that are incurred when the problem of stocking spare assets and spare parts is not
solved jointly, can be significant. A key managerial insight is that typically, the number of
spare assets to be acquired is very close to a lower bound that is determined only by the
active maintenance time on the assets. It is typically not cost effective to acquire more spare
assets to cover spare parts unavailability.
Keywords: Maintenance ·Inventory ·Fleet sizing ·Spare parts
1 Introduction
Many important services and operations depend on the availability of a sufficiently large fleet
of assets. An airline, for example, depends on a fleet of aircraft to service all planned flights,
while a railway company depends on a fleet of rolling stock to make the train schedule work.
Other examples exist in the defense and maritime industries. In all such cases, the availability
of assets (the fraction of time that they are available to operate) is not the most appropriate
measure of fleet performance. For example, consider a train operator that requires 90 trains to
run its schedule as planned, while it owns 100 trains that achieve a 90% availability. The train
operator seems to do fine, but it could be that half of the time there are 95 trains available,
while the other half of the time there are 85 trains available: during the latter half of the time,
the train operator cannot run its schedule. Similarly, consider an air force that has brought a
number of airplanes to a war zone. To fly a relatively safe mission, it requires 40 airplanes to be
available. If the average availability of its 50 airplanes is 80%, the target seems to be achieved.
However, it may be that most of the time about 35 airplanes are available, while sometimes
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there are almost 50 available. Then very often, there are not enough airplanes available to fly
a safe mission. In both examples, a more accurate measure of performance is the fraction of
time that sufficient assets are available to fulfill the function of the fleet, i.e., the probability
that sufficient assets are available at an arbitrary moment in time. We refer to this performance
measure as fleet readiness.
When kassets are needed to fulfill the function of the fleet, then typically N > k assets
need to be acquired to achieve a high fleet readiness, since assets are subject to failures and
need maintenance. The maintenance time of an asset consists of two main parts: the active
maintenance time in which the actual maintenance operations occur (usually the replacement of
line replaceable units) and the maintenance delay time which is the waiting time for maintenance
resources to become available (some authors call this time to support). A major culprit for
maintenance delay is a lack of spare parts needed for replacement.
High fleet readiness can be achieved by a combination of the following: (1) Buying assets in
addition to what is necessary to run daily operations; (2) Reducing the maintenance delay time
by stocking spare parts; (3) Reducing the required number of maintenance actions by increasing
asset reliability; Or (4) improving the speed of maintenance/replacement operations. This paper
focuses on the first two options as these amount to investment decisions of a logistical nature.
The last two options can usually only be achieved by making asset engineering modifications
that are specific to the technology of the asset.
Buying as many assets as a given budget allows is a popular method to increase fleet readiness
but it is not always effective. The money needed to buy assets and spare parts usually comes
from the same budget. In the last decades of the previous century, the Dutch defense engaged
in what has come to be called ‘carcass politics’1. Under carcass politics, the available budget
to establish a fleet is spent as much as possible on buying complete assets, and the remainder
is spent on spare parts. Spare parts become short in supply soon after this and as a result,
technicians start using parts from complete assets leaving only a ‘carcass’ behind. This practice
is often referred to as cannibalization. Clearly, this practice does not necessarily lead to high
fleet readiness. There is a trade-off between investing in assets and spare parts to meet a certain
fleet readiness and this paper explores this trade-off, using a model of a single stock point with
continuous review, stochastic lead times, and base stock control.
The trade-off between investing in assets or spare parts to realize a certain fleet readiness
objective is non-trivial. In general, this problem is non-convex and the analysis cannot be
separated into an analysis per spare part type and asset: Its evaluation requires the convolution
of backorder distributions per spare part type. This is in stark contrast with many spare part
inventory problems in which the resulting optimization problems are convex and separable per
item (see, e.g., Sherbrooke, 2004; Muckstadt, 2005; Basten and Van Houtum, 2014). Assets and
spare parts achieve a certain fleet readiness jointly and so their analysis cannot be separated.
In fact, we will show that their joint analysis is mathematically a generalization of multi-
echelon inventory theory, even though we consider only a single stock point. Unfortunately,
1The Dutch word is ‘rompenpolitiek’, see, e.g., Tjepkema (2010)
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this generalization is not susceptible to standard tools such as Clark and Scarf decomposition
(Clark and Scarf, 1960) and METRIC type inventory models (Sherbrooke, 1968).
Our main contributions in this paper are the following: We consider the problem of deciding
on asset investment and spare part investment jointly, whereas previous work considers them
separately, see also Section 2, which we also often see in practice. However, both are sizeable
investments that serve a common purpose: achieving high fleet readiness. Fleet readiness is
usually not used as service measure in this setting because it is computationally demanding to
evaluate and difficult to optimize. We show that evaluation requires performing a large number
of convolutions, but we also show how the number of convolutions can be lowered exponentially.
We next develop a greedy heuristic for this problem that is computationally efficient. In a
numerical experiment, we compare our heuristic with enumeration on small instances and find
that our heuristic finds the optimal solution of 51% of our test instances and has an average
optimality gap on the other instances of 3.7%. Our algorithm is 50 times faster on medium size
instances than an existing algorithm that was developed for a related problem. (The existing
algorithm takes too much time to perform a comparison on large instances.) Our key managerial
insight is that spare assets should be acquired to cover active maintenance time on the assets;
it is typically not cost effective to acquire more spare assets to cover spare parts unavailability.
In other words, the use of ‘carcass politics’ is not cost effective.
The remainder of this paper is organized as follows. We discuss related literature in Section 2
and position our work with respect to previous work. In Section 3, we explain the model and
the optimization problem that we focus on. We analyze the model in Section 4; we show that
the problem is not convex and we prove some other properties. We use those to construct an
algorithm to solve the optimization problem in Section 5. In Section 6, we perform a numerical
experiment, and we conclude in Section 7.
2 Related literature
Since our main contributions are the combination of the fleet sizing and spare part investment
decisions subject to a service level constraint that is not often used, this literature review is
structured as follows: We discuss the fleet readiness measure in Section 2.1, fleet sizing in
Section 2.2 and spare parts optimization in Section 2.3. In Section 2.3, we specifically focus on
a closely related paper by De Smidt-Destombes et al. (2011).
2.1 Fleet readiness
Fleet readiness as a performance measure is not as common as availability. Some authors
however, already noted that in many instances readiness is a more appropriate performance
measure. Safaei et al. (2011), for instance, consider a deterministic maintenance scheduling
problem subject to a manpower constraint and a fleet readiness constraint. Jin and Wang
(2012) use the fleet readiness measure in the context of performance based contracting. They
approximate this measure by using the availability as the probability that a vehicle is available
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at an arbitrary moment in time and then use the binomial distribution to compute the fleet
readiness. This approximation is more tractable than actual fleet readiness but it assumes
that the availability of different vehicles is uncorrelated at any particular time point. A similar
approach has been followed by Costantino et al. (2013) in a multi-echelon, multi-indenture spare
parts inventory setting. Some authors use the term fleet readiness as the average number of
vehicles of a fleet that are available, e.g., Sherbrooke (1971) and Salman et al. (2007). That is,
these authors consider the availability times the size of the fleet rather than the fleet readiness
as we define it.
A closely related concept from the reliability engineering literature is the availability of a
k-out-of-Nsystem (e.g., De Smidt-Destombes et al., 2004). In this setting, a system consists
of Ncomponents and only functions if at least kout of those Ncomponents are operational.
The availability is then defined as the probability that at least kout of the Ncomponents are
operational. In our setting, we would say that a fleet is ready if at least kout of Nassets are
operational, or alternatively, if not more than Nkassets are unavailable. Thus these measures
are equivalent.
2.2 Fleet sizing
Fleet sizing for vehicles has been studied in different settings. Hoff et al. (2010) and Pantuso
et al. (2014) provide a review of these models in the general and maritime setting, respectively.
Most of these models are deterministic and are concerned with calculating the minimum fleet
size necessary to perform daily operations. Our model takes this minimum number of vehicles
needed as an input and supports the investment decision in additional vehicles (or other assets)
and spare parts to make sure that the fleet is operationally ready with a certain probability
at any moment in time. Hoff et al. (2010) already note that dealing with uncertainty is an
important aspect to incorporate when making the fleet sizing decision. Our work partially fills
this gap by providing a model that deals with the uncertainty in the number of vehicles down
for maintenance or lack of a spare part.
2.3 Spare parts optimization
The optimization of spare part inventory decisions has a long history that started with the
work of Feeney and Sherbrooke (1966) and Sherbrooke (1968). This line of research has led to
a large stream of literature that has been consolidated in the books by Sherbrooke (2004) and
Muckstadt (2005) and the review papers by Guide Jr. and Srivastava (1997), Kennedy et al.
(2002), and Basten and Van Houtum (2014). Here, we focus on the most closely related work,
which is the paper of De Smidt-Destombes et al. (2011). In that paper, the authors consider a
fleet that is taken on a mission with a package of spare parts. The objective is to minimize the
investment in this spare parts package subject to a constraint on the probability that the fleet
remains ready throughout the mission. We will show that that constraint is mathematically
equivalent to the fleet readiness constraint as used in this paper. We extend the model in two
ways: (1) We also consider the size of the fleet as a decision variable and (2) we account for
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Figure 1: Modeled system
the fact that maintenance itself requires time and renders an asset unavailable. We show that,
even for a fixed fleet size, optimizing the spare parts package is not a separable and convex
problem. Despite this, De Smidt-Destombes et al. (2011) use a marginal analysis approach and
we pursue a similar approach. As a new contribution, we benchmark this approach with respect
to the optimal solution found by enumeration. We find that our algorithm yields high quality
solutions. Furthermore, we provide results that make algorithms based on marginal analysis
more tractable by giving easy to compute bounds so that gradients do not need to be computed
for every direction of ascent. In addition, we provide an algorithm that computes the gradient
in O(log n) time instead of O(n) time, with nbeing the number of distinct spare part types.
3 Model description
The system that we analyze is shown in Figure 1. We consider a fleet of assets that are composed
of line replaceable units (LRUs). We let Idenote the set of LRUs and we reserve the index 0
for the assets; we denote the set of LRUs and assets by I0=I∪ {0}. Assets fail randomly due
to a failure in exactly one LRU iI; such failures occur according to a Poisson process with
intensity λiand the total intensity over all LRUs is denoted by λ0=PiIλi.λiindicates the
failure rate of LRU iover the complete fleet and it is constant over time, i.e., it is not influenced
by the number of assets that is operational at any point in time. The assumption of a constant
failure intensity is regularly made in the spare parts inventory literature, see, e.g., Basten and
Van Houtum (2014, p.40). It is reasonable, since the required fleet readiness is typically high.
Furthermore, when there is a shortage of operational assets, which would lead to a slightly
lower failure intensity, then often some work load is routed to the operational assets so that
their failure intensity increases.
An asset is repaired by replacement of the failed part by a functioning spare part. In the
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remainder of this paper, if we refer to (spare) parts, components, or items of LRU type i, we
say parts of LRU i. We assume that disassembly of the failed part takes negligible time (i.e.,
is instantaneous); assembly of the functioning spare part into the asset takes exactly µitime
units for LRU iIif a spare part is available immediately from stock (i.e., µiis deterministic).
After being repaired, the asset is sent to the pool of stand-by assets. We also refer to this pool
as the stock of spare assets.
The failed part of LRU iIis sent to the repair shop; its repair lead time is generally
distributed with mean Titime units. Repair times of parts of the same LRU are independent
and identically distributed (i.i.d.) and repair times of parts of different LRUs are independent
of each other. In other words, we assume that the repair shop has an infinite number of servers,
or that the repair shop is able to schedule repairs and hire capacity such that it can guarantee a
certain average repair time (we have made an analogous assumption for the maintenance shop).
After being repaired, a part is returned to stock. Repairs may be performed either at an internal
repair shop, or they may be outsourced to an external repair shop. In fact, the model can also
be used if parts are discarded and replaced by new parts. In that case, repair lead time should
be read as supply lead time or order-and-ship time.
All stock points are controlled using a continuous review (Si1, Si) base stock policy (i.e.,
one-for-one replenishment) with Sibeing the base stock level for LRU iIor the asset (i= 0).
(Notice that S0does not indicate the fleet size; if, say, 100 assets are required in the primary
process of the user and the base stock level, S0, is equal to 10, then the fleet size is 110.)
Our assumptions mean that failed assets or parts are not batched, but immediately sent into
maintenance or repair. If the setup costs for a repair are low, this is a reasonable assumption.
As a result, this assumption is commonly made in the spare parts literature (see, e.g. Basten
and Van Houtum, 2014, p.40); we come back to this assumption in Remark 3.1.
Under this policy, the dynamics of the system can be described as follows: Let Di(t0, t)
denote the demand for LRU i(or, equivalently, the number of failures in parts of LRU i)
between times t0and t. Let Xi(t) denote the number of parts of LRU iIin repair, also
called the pipeline of LRU i, at time t. Then, if the repair lead time Tiis deterministic, the
pipeline at time tconsists of all demands between times tTiand t, i.e., Xi(t) = Di(tTi, t).
Due to Palm’s theorem (Palm, 1938), this equality still holds in distribution if the repair lead
time is not deterministic. The number of backorders for LRU iIis denoted by Bi(t, Si) and
satisfies Bi(t, Si) = [Xi(t)Si]+, with [x]+= max{0, x}. In other words, there are backorders
if the number of parts in the pipeline is higher than the base stock level, with the number
of backorders being equal to the difference. We denote by Y0(t) the number of assets in the
maintenance shop that are actively being maintained at time t(i.e., the assets that are waiting
for a spare part are not included). Since the assembly time µiis assumed to be deterministic,
this is the summation over all LRUs of the demands for that LRU between times tµiand
t, i.e., Y0(t) = PiIDi(tµi, t). For notational convenience, we introduce Sas the vector of
all base stock levels Sifor iI. The pipeline X0(t, S) of assets in the maintenance shop at
time tis equal to the number of assets that came in for maintenance after time tµi, plus the
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number of assets that should have finished maintenance, but were delayed due to backordered
spare parts:
X0(t, S) = Y0(t) + X
iI
Bi(tµi, Si) = X
iI
Di(tµi, t) + X
iI
[Di(tµiTi, t µi)Si]+,(1)
while the number of assets short is denoted by B0(t, S0)=[X0(t, S)S0]+, with S0being the
vector of all base stock levels Sifor iI0. (This can also be interpreted as the number of
backordered assets.) The readiness, R(S0), is the probability of not being any assets short in
steady state: R(S0) = limt→∞ P{B0(t, S0)=0}. In other words, the readiness is the long-term
probability that the number of assets in maintenance (i.e., in the pipeline) does not exceed the
number of spare assets.
Remark 3.1. If the asset consists of one LRU only, our system simplifies to a two-echelon
serial inventory system, the Clark-Scarf model (Clark and Scarf, 1960). Specifically, when
|I|= 1, Y0(t) can be interpreted as the number of orders in transit from the upstream stock
point to the downstream stock point at time t, while B1(tµ1, S1) represents the orders from
the downstream stock point that are backordered at the upstream stock point at time tµ.
By allowing |I|>1, we are dealing with a generalization of a two-echelon serial inventory
system under base stock control. Since base stock policies are optimal in the Clark-Scarf model
and many of its generalizations, including some convergent systems, (see for an overview Van
Houtum, 2006) our assumption of base stock control seems reasonable.
Remark 3.2. When µi= 0 and Ti=Tfor all iI, then R(S0) can also be interpreted as
the probability that the fleet remains ready during a mission of length Twhen a spare parts
package of size Sis brought on the mission. (Note that Ti=Tfor all iIimplies that spare
parts cannot be repaired during the mission.) For a fixed asset base stock level S0, this is the
setting that De Smidt-Destombes et al. (2011) consider.
The costs of holding spare assets and LRUs are linear in their base stock level: ciper unit
for asset or LRU iI0. Our goal is to find the base stock levels that minimize the total costs
C(S0) = PiI0ciSi, such that the target readiness Robj is achieved. Formally, our optimization
problem, Problem (P), is:
(P) min
S0N|I0|
0
C(S0)
subject to R(S0)Robj,
with N0=N∪ {0}being the set of non-negative integers. We emphasize that Problem (P) is
not separable per item because R(S0) cannot be written as a sum of terms that depend on one
Sionly. Further note that the costs can also comprise initial investment costs instead of holding
costs.
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4 Analysis
In this section, we give results on the behavior of the fleet readiness as a function of the number
of spare parts and spare assets. We use these results to explain, in Section 5, why we make
certain choices in the algorithm that we use to solve Problem (P). Since we consider the system
in steady state, we suppress the time parameter in the state variables from now on, and we show
their distributions in Lemma 1. The results in this lemma are very similar to the derivation of
results for the standard two-echelon model for spare parts (see, e.g., Graves, 1985; Basten and
Van Houtum, 2013).
Lemma 1. In steady state, the state variables are distributed as follows:
(i) For iI, the pipeline, Xi, is Poisson distributed with mean λiTi, i.e.:
P{Xi=x}=(λiTi)x
x!eλiTi, x N0.
(ii) For iI, the distribution of the number of backorders, Bi(Si), is given by:
P{Bi(Si) = b}=
P{XiSi},if b= 0;
P{Xi=Si+b},if bN.
(iii) The number of assets that are actively being maintained, Y0, is Poisson distributed with
mean PiIλiµi, i.e.:
P{Y0=y}=PiIλiµiy
y!ePiIλiµi, y N0.
(iv) The distribution of the asset pipeline, X0(S), is given by:
P{X0(S) = x}=
x
X
y=0
P{Y0=y}PPiIBi(Si) = xy, x N0.
(v) The distribution of the number of assets short, B0(S0), is given by:
P{B0(S0) = b}=
P{X0(S)S0},if b= 0;
P{X0(S) = S0+b},if bN.
Proof. (i) Under a base stock policy, each demand for LRU iItriggers an order, and each
order is delivered after an i.i.d. amount of time with mean Ti. Therefore, the number of
outstanding orders of LRU iIbehaves as the number of customers in an M/G/-queue
with arrival rate λiand processing time with mean Ti. By Palm’s Theorem (Palm, 1938),
the number of customers in such a queue in steady state is Poisson distributed with mean
λiTi.
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(ii) The number of backorders at time tis defined as Bi(t, Si) = [Xi(t)Si]+. There-
fore, we have P{Bi(Si)=0}=P{[XiSi]+= 0}=P{XiSi}, and P{Bi(Si) = b}=
P{[XiSi]+=b}=P{Xi=Si+b}for bN.
(iii) By definition, Y0(t) = PiIDi(tµi, t). Since µiis deterministic, Di(tµi, t) denotes
the number of arrivals of a Poisson process with intensity λiin a time interval of length
µi. Therefore Di(tµi, t) has a Poisson distribution with mean λiµi. Since the Pois-
son distribution is closed under convolution, Y0(t) has a Poisson distribution with mean
PiIλiµifor each tand in particular as t→ ∞.
(iv) First observe that since µiis deterministic, [tµiTi, t µi) and [tµi, t) are disjoint
intervals. By Equation (1) and the independent increments of the Poisson process, this im-
plies that Y0and PiIBi(Si) are independent random variables. Now, using Equation (1)
and letting t→ ∞, we have:
P{X0(S) = x}=PY0+PiIBi(Si) = x
=
x
X
y=0
P{Y0=y}PY0+PiIBi(Si) = x|Y0=y
=
x
X
y=0
P{Y0=y}PPiIBi(Si) = xy,
where the final equality follows from the independence of Y0and PiIBi(Si).
(v) This is analogous to part (ii).
We use additional notation in this section: Let eibe a vector of length |I0|with all zeros,
except at the location corresponding to the base stock level of spare assets (i= 0: S0) or spare
LRUs (iI:Si). Furthermore, notice that concavity of R(S0) in Sifor iI0is equivalent to
R(S0+ei)R(S0)R(S0+ 2ei)R(S0+ei) for S0N|I0|
0, while joint concavity in i, j I0
is equivalent to R(S0+ej)R(S0)R(S0+ei+ej)R(S0+ei) for S0N|I0|
0.
R(S0) is not in general jointly concave in S0and Siwith iI. As a counter example,
consider an asset consisting of one LRU, indexed 1, with λ1= 2 and µ1=T1= 1. Evaluating
R(S0) gives the following results: R(0,0) 0.1353, R(1,0) 0.4061, R(0,1) 0.2707, and
R(1,1) 0.6090. It is easily seen that if either S0or S1is increased, the readiness increases.
However, R(0,1) R(0,0) 0.1353 < R(1,1) R(1,0) 0.2030, and R(1,0) R(0,0)
0.2707 < R(1,1) R(0,1) 0.3383. This means that R(S0) is not jointly concave. For larger
values of S0, the inequalities required for concavity do typically hold.
Because R(S0) is not jointly concave in general, our algorithm enumerates the number of
spare assets (see Section 5); Proposition 1 gives bounds on its optimal value. The lower bound,
SLB
0, is the number of spare assets that are required to achieve the target readiness if there are
unlimited spare parts available. In Section 6, we show that this lower bound is usually tight,
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implying that spare assets are mainly stocked to cover the active maintenance time, while they
are not stocked to cover spare parts unavailability.
Proposition 1. The optimal number of spare assets for Problem (P), denoted as S
0, is bounded
as follows:
(i) S
0SLB
0, with SLB
0being the smallest integer Sthat satisfies P{Y0S} ≥ Robj.
(ii) S
0SUB
0, with SUB
0being the smallest integer Sfor which there exists S00=S0
0, S0
1, . . . , S0
|I|
with SLB
0S0
0S,C(S00)< c0(S+ 1) and R(S00)Robj.
Proof. For part (i): Since X0(S)d
=Y0+PiIBi(Si) by definition ( d
= denotes equality in dis-
tribution), we have that P{X0(S)x} ≥ P{Y0x}for all xN0because Bi(Si) are non-
negative random variables. The readiness constraint in Problem (P) requires P{X0(S)S0} ≥
Robj, so a feasible S0must satisfy P{Y0S0} ≥ Robj.
For part (ii): S00represents a feasible solution, since R(S00)Robj, and the associated cost
of this solution is C(S00). Stocking S+ 1 spare assets costs c0(S+ 1). If c0(S+ 1) > C(S00),
then the optimal solution cannot contain S+1 spare assets (or more), because there is a feasible
solution with cost lower than c0(S+ 1). This means that Sis an upper bound on the number
of spare assets in the optimal solution, S
0.
Problem (P) is also not in general jointly concave in Siand Sjwith i, j Iand i6=j. As
a counter example, consider an asset consisting of two LRUs, indexed 1 and 2, with S0=µ1=
µ2= 0. Joint concavity is equivalent to R(0,S+e1)R(0,S)R(0,S+e1+e2)R(0,S+e2)
for S0N|I0|
0, which would mean that
(1 P{X1S1+ 1}P{X2S2})(1 P{X1S1}P{X2S2})
(1 P{X1S1+ 1}P{X2S2+ 1})(1 P{X1S1}P{X2S2+ 1}),
P{X1S1}P{X2S2} − P{X1S1+ 1}P{X2S2}
P{X1S1}P{X2S2+ 1} − P{X1S1+ 1}P{X2S2+ 1},
(P{X1S1} − P{X1S1+ 1})P{X2S2}
(P{X1S1} − P{X1S1+ 1})P{X2S2+ 1},and
P{X2S2} ≥ P{X2S2+ 1}.
However, P{X2=S2+ 1}>0 for all S20, so that P{X2S2+ 1}>P{X2S2}, showing
that this problem is not jointly concave.
We show other convexity results in Proposition 2, which our algorithm uses to determine
lower bounds. The proof can be found in Appendix A; it uses properties of the shape of the
Poisson distribution.
Proposition 2. The second order difference function for the number of spare LRUs iI,
2
iR(S0) = ∆iR(S0+ei)iR(S0), behaves as follows:
(i) If S0+Si<dλiTie − 2, then 2
iR(S0)>0, that is, R(S0)is strictly convex in Si.
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(ii) If Si≥ dλiTie − 2, then 2
iR(S0)0, that is, R(S0)is concave in Si.
Notice that:
A similar result as part (ii) has been shown by Rustenburg (2000, p.41).
The behavior of ∆2
iR(S0) is not clear beforehand in all cases that are not covered by
Proposition 2 (i.e., if both Si<dλiTie − 2 and S0+Si≥ dλiTie − 2).
Proposition 3 gives a result that allows our algorithm to avoid performing unnecessary
calculations: Our algorithm uses a marginal analysis approach. In each iteration, an additional
spare part is stocked of the LRU that gives the biggest ‘bang for the buck’. Proposition 3 gives
an upper bound on how much this ‘bang for the buck’ may have changed for a certain LRU
from one iteration to the next. Our algorithm can thus quickly check if the ‘bang for the buck’
of a certain LRU may be sufficiently high to perform time consuming exact calculations. Since
the proof is long and does not give insight into the problem, it is deferred to Appendix A.
Proposition 3. If Si≥ dλiTie − 2and Sj≥ dλjTje − 2, with i, j I, then:
iR(S0+ej)iR(S0)<P{Xj=Sj+ 1}P{Xi=Si+ 1}.
5 Algorithm
We give the pseudo code of our algorithm in Figure 2 and we explain the complete algorithm in
Section 5.1. Next, we focus on how to compute the convolutions in Line 11 of our algorithm in
Section 5.2. This is a very time consuming step in the algorithm and we propose a novel way
to do this efficiently.
5.1 Overview
The algorithm functions as follows. It enumerates the asset base stock level between a lower
bound (Line 1) and an upper bound (Line 3), based on parts (i) and (ii) of Proposition 1,
respectively. Although those bounds are simple, we still find in our numerical experiment
(Table 7) that typically, the number of enumerated spare asset levels is small, i.e., the difference
between the lower and upper bounds is small.
For each asset base stock level, each LRU base stock level is initialized at a lower bound
based on part (ii) of Proposition 2 (Line 4). Notice that this lower bound guarantees that
the readiness is concave in each LRU base stock level. However, notice further that it is not
guaranteed that the optimal LRU base stock level is above this lower bound: Consider a system
in which all LRUs are inexpensive, but there is one very expensive LRU iwith a failure rate λi
and repair time Tisuch that λiTi= 2 + , with being a very small number. The lower bound
then ensures that Si= 1, while it may be optimal to have Si= 0. In practice, however, such
examples are seldom encountered and each optimal LRU base stock level will typically be above
the lower bound.
11
1: S0min{SN0|P{Y0S} ≥ Robj}
2: Calculate the probability mass function of Y0
3: while c0S0Cbest do
4: Simax{0,dλiTie − 2}for all iI
5: Calculate the probability mass functions of Bifor all iI, and of Y0+PiIBi
6: Rcur R(S0); Γbest 0
7: ibest ← −1; P{X1=S1} ← 1; Γi1/cifor all iI
8: while Rcur < Robj do
9: for iIdo
10: ΓiΓi+P{Xibest =Sibest }P{Xi=Si+1}
ci
11: if ΓiΓbest or i=ibest then
12: ΓiR(S0+ei)Rcur
ci
13: Γbest max{Γi,Γbest}
14: end if
15: end for
16: ibest arg maxiIΓi;Sibest Sibest + 1; Rcur R(S0)
17: end while
18: if C(S0)< Cbest then
19: Cbest C(S0)
20: end if
21: S0S0+ 1
22: end while
Figure 2: Greedy algorithm for Problem (P)
Then, the probability mass functions are computed of Bifor iI, and of Y0+PiIBi
(Line 5). Using a smart way of ordering and storing these computations results in a reduction
of the computations that are performed per iteration of the marginal analysis approach that is
used to stock additional spare parts (Line 12). We explain this in detail in Section 5.2. Further
variables are initialized in Lines 6 and 7.
As long as the target readiness has not been reached (Line 8), for each LRU (Line 9) an
upper bound on the increase in readiness per additionally invested dollar is calculated (Line
10), using the result in Proposition 3. If the upper bound is such that the current LRU would
be the best option encountered thus far (Line 11), exact calculations are performed (Line 12)
and it is checked if it is really the best LRU encountered thus far (Line 13). In our numerical
experiment (Table 6), we find that using the upper bound for a first check, saves over 50% of
computation time for problem instances with 256 LRUs and that the relative savings increase
with an increasing problem size.
Notice that Line 7 ensures that in the first iteration of the while loop (Lines 8 to 17), the
first condition of the if-clause on Line 11 is always true. This means that the first LRU that is
checked, is by definition the best encountered thus far. The second condition of that if-clause
is required because Proposition 3, which is used in Line 10, applies only for i6=j.
In Line 16, the base stock level of the right LRU is increased and the readiness is adapted.
As soon as the target readiness is reached for the current asset base stock level (Line 8), the
12
marginal analysis approach is stopped and it is checked if a new best solution is found (Line
18). Next, the asset base stock level is increased by one (Line 21) so that a new iteration can
start if the upper bound has not been reached yet (Line 3).
5.2 Convolutions
The computationally most demanding step in Algorithm 2 is the computation of Γi= (R(S0+
ei)Rcur)/ci(Line 12), specifically the evaluation of R(S0+ei) = P{B0(S0+ei)=0}=
P{U(S+ei)S0}, with U(S) = Y0+PiIBi(Si). This requires computing the probability
mass function of U(S+ei) by convolution; we provide an algorithm to compute that using
results that have already been computed for U(S).
We require some additional notation. Let Bi(Si) be a vector containing the probability mass
function of Bi(Si) = (XiSi)+up to S0, i.e., Bi(Si) = (P{Bi(Si)=0},...,P{Bi(Si) = S0}).
Similarly, let Y0= (P{Y0= 0},...,P{Y0=S0}) and U(S) = (P{U(S)=0},...,P{U(S) = S0}).
Furthermore, let abdenote the convolution of the vectors aand bof equal length: If c=ab,
then chas the same length as both aand band the i-th element of cis given by ci=Pi
j=0 aijbj.
(Note that we number elements in a vector starting from 0.) The convolution operator satisfies
commutativity (ab=ba) and associativity ((ab)c=a(bc)). Finally, we let
Ba,b(S) = Ba(Sa) · · · ∗ Bb(Sb) for ab.
The complexity in computing U(S) lies in the computation of B1,|I|(S) and its complexity
increases with |I|. After computing Bi(Si) for all iI, the straightforward way to compute
B1,|I|(S) is to successively compute as follows: B1,2(S) = B1(S1)B2(S2), B1,3(S) = B1,2(S)
B3(S3), . . .,B1,|I|(S) = B1,|I|−1(S)B|I|(S|I|). This requires performing |I| − 1 convolutions
and this is what the algorithm of De Smidt-Destombes et al. (2011) does.
Instead, our algorithm builds up a tree starting from its leaves. An example for |I|= 8
is shown in Figure 3. Formally the procedure works as follows: Compute B1,2(S) = B1(S1)
B2(S2), . . .,B|I|−1,|I|(S) = B|I|−1(S|I|−1)B|I|(S|I|). Then, compute B1,4(S) = B1,2(S)
B3,4(S), . . .,B|I3|,|I|(S) = B|I|−3,|I|−2(S)B|I|−1,|I|(S). Continue in this manner until arriving
at the root node of the tree: B1,|I|(S). This procedure also requires |I| − 1 convolutions.
However, computing B1,|I|(S+ei), for some iI, can now be done efficiently by reusing most
results in the tree: Ba,b(S+ei) = Ba,b(S) whenever i<aor b<i. This is easily seen when
we reconsider the example where |I|= 8: Figure 4 shows the tree that results when computing
B1,8(S+e3); if B1,8(S) is already evaluated, this only requires the evaluation of 4 nodes. Of
those 4 nodes, one concerns the determination of B3(S3+ 1) and 3 = log2(8) require taking a
convolution. The same reasoning can be applied for general |I|and yields the following result.
Proposition 4. After an initial evaluation of B1,|I|(S)which requires O(|I|)convolutions, all
subsequent evaluations of B1,|I|(S+ei)with iIrequire performing only O(log |I|)convolutions.
The only thing that we have not explained yet is when to perform the convolution with Y0.
It would be straightforward to do this at the end (i.e., at the root of the tree), but that would
require an additional convolution each time that an LRU stock level is increased. Therefore,
13
Figure 3: Computation of B1,|I|(S) for |I|= 8 via a tree structure. Each non-leaf node in this
tree is obtained by convolution of its two children nodes.
Figure 4: Computation of B1,|I|(S+e3) for |I|= 8 via a tree structure. This tree is identical
to the tree for the computation of B1,|I|(S) except in the shaded nodes.
this convolution is performed in the beginning: First, B1,1(S) = Y0B1(S1) is calculated,
which is used to calculate B1,2(S) = B1,1(S)B2(S2).
6 Numerical experiment
We use the numerical experiment to answer four questions:
1. What is the quality of the solutions of our algorithm?
2. How should investments be divided between assets and spare parts?
3. What is the value of jointly optimizing spare assets and spare parts?
4. What is the computational effort required to run our algorithm?
14
Set 1 Set 2
# LRUs |I|2; 4; 8 16; 64; 256; 1,024
Set 1 & Set 2
Maximum assembly time µmax 0.001; 0.01
Maximum resupply lead time Tmax 0.01; 0.1
Average costs of an LRU cave 100; 1,000
Relative costs of an asset crel 0.5; 1; 2
Target readiness Robj 0.9; 0.95; 0.975
Table 1: Settings of the parameters that are varied in the numerical experiment
To this end, we generate two sets of problem instances. Set 1 consists of smaller problem
instances and is mainly used to answer Question 1, by comparing the solution of our algorithm
with the optimal solution, found by enumeration. Set 2 consists of larger problem instances and
is used for two reasons. First, it is used to get managerial insights, i.e., to answer Questions 2
and 3. Second, it is used to compare the computation times of our algorithm with that of De
Smidt-Destombes et al. (2011), thus answering Question 4. We explain how we generate both
sets of problem instances in Section 6.1. We answer Question 1 in Section 6.2, Question 2 and
3 in Section 6.3, and Question 4 in Section 6.4.2
6.1 Setup
Table 1 shows the settings for the parameters that are varied in our numerical experiment for
the two sets of problem instances. (When the number of LRUs is varied, also λiis varied; we
explain this below.) We use a full factorial design per set and we generate ten problem instances
per combination of parameters. As a result, Set 1 and Set 2 consist of 2,160 and 2,880 problem
instances, respectively. The way in which we generate problem instances leads to instances that
are realistic in practice, and to a wide range of combinations of parameter values in the sets.
There are |I|LRUs and a value µis drawn from a uniform distribution on the range [0, µmax].
Then, for each LRU iI(see the explanation below):
µiµ,
Tiis drawn from a uniform distribution on the range [0, T max],
λi128
|I|in Set 1 and λi1,024
|I|in Set 2, and
ciis drawn from an exponential distribution with mean 1
cave , and 10 is added. This
effectively means that there are no LRUs with costs of less than 10, and the mean costs
are cave + 10.
2The experiment is implemented in Python 3.4 and performed on an Intel Xeon E5530 @ 2.4 GHz with 8 GB
RAM, running Windows Server 2008 R2 Enterprise Service Pack 1.
15
# LRUs: |I|248
% Problem instances with optimal solution 73% 55% 26%
Average additional costs in remaining instances 2.8% 3.8% 4.0%
Maximum additional costs in remaining instances 63% 40% 93%
Table 2: Set 1: Quality of solutions found by our algorithm
The same value µcan be used for all iI, since it influences only the number of assets in
active maintenance, Y0.
λiis relevant only for calculating the distribution of the number of parts in resupply, Xifor
iI, and the distribution of the number of assets in active maintenance, Y0. Since the average
number of parts in resupply is varied by varying Tiand since the average number of assets in
the maintenance shop is varied by varying µ,λican be kept constant in each problem instance.
However, when the number of LRUs is varied, λiis also varied. Our aim is to get solutions in
which the optimal number of spare assets and spare parts is realistic and higher than zero. We
therefore show the average number of spare assets and spare parts in the solutions in the next
section. The largest problem instances of Set 2 are the most realistic ones, with 1,024 LRUs
and a demand rate per LRU of 1.
The costs of holding a spare asset are equal to crel times the summation of the costs of
holding one of each of the spare parts, i.e., c0=crel PiIci. Finally, the target readiness, Robj,
is varied.
6.2 Quality of solutions
Whenever we give results for multiple parameter settings in any of the tables in this or the
next sections, then the numbers are the averages over all problem instances with one of those
settings. We sometimes additionally give a line with a maximum, which is then the maximum
over all problem instances with one of those settings. For example, in Table 3, the first value on
the top line gives the average number of spare assets in the solutions for the problem instances
with 16 LRUs, while the second value on that line gives the maximum number of spare assets
in any of the solutions to the problem instances with 16 LRUs.
We use Table 2 to answer Question 1: It shows for the problem instances in Set 1 how our
algorithm performs compared with the optimal solution. Many problem instances, 51%, are
solved to optimality by our marginal analysis approach, and the average difference with the
optimal solution on the other instances is small: 3.7% on average. (The values vary depending
on the problem size.) The maximum difference is large, 93%, but large differences for these
small instances can be caused by small mistakes. For example, for each of the three problem
instances on which our algorithm incurs more than 50% additional costs, the number of stocked
spare parts is correct, but one spare part is of the wrong LRU type. Furthermore, for all but
one of the fourteen problem instances on which our algorithm incurs more than 25% additional
costs, the achieved readiness is at least two percentage points higher, i.e., there is a large
overshoot (remember that the readiness target is at least 90% in our problem instances). The
16
# Spare assets in solution
Maximum
# Spare assets above the LB
Maximum
# Spare parts in solution
Divided by # LRUs
Additional costs of LB algorithm (%)
Maximum (%)
# LRUs: |I|
16 5.6 18 0.59 4 67 4.2 3.9 85
64 5.2 17 0.20 1 151 2.4 2.1 76
256 5.1 17 0.09 1 396 1.5 1.0 55
1,024 5.0 17 0.04 1 1,251 1.2 0.5 43
Relative costs of an asset: crel
0.5 5.6 18 0.55 4 445 2.1 4.4 85
1 5.1 17 0.12 1 470 2.4 1.1 46
2 5.0 17 0.02 1 483 2.5 0.2 24
Maximum assembly time: µmax 0.001 1.9 6 0.27 3 487 2.4 3.0 85
0.01 8.5 18 0.19 4 445 2.2 0.8 34
Maximum resupply lead time: Tmax 0.01 5.1 18 0.11 2 300 1.1 1.3 85
0.1 5.3 18 0.35 4 632 3.5 2.4 62
Average costs of an LRU: cave 100 5.2 18 0.23 3 461 2.3 1.9 85
1000 5.2 18 0.23 4 471 2.4 1.9 84
Target readiness: Robj
0.9 4.6 16 0.20 4 449 2.2 1.1 34
0.95 5.2 17 0.22 3 467 2.3 1.6 51
0.975 5.8 18 0.27 3 482 2.4 2.9 85
Table 3: Set 2: Key results for all parameters
only exception is one problem instance with four LRUs in which our algorithm stocks one asset
too many. If we look at the problem instances for which our algorithm achieves a lower readiness
than in the optimal solution, we see that the highest additional costs are 13% (the difference
in readiness is 0.2 percentage point). All in all, we believe that our algorithm typically finds
good solutions, especially considering that the overshoot typically decreases if the problem size
increases (see, e.g. Basten and Van Houtum, 2014, p.42).
6.3 Managerial insights
To answer Question 2, Table 3 gives insight into the solutions that our algorithm finds (we
explain the lower bound algorithm below). Recall that the lower bound (LB) in part (i) of
Proposition 1, the lower bound that our algorithm uses, is such that the spare assets cover
17
# Spare assets above the lower bound 0 1 2 3 4 >4
# Problem instances 2,349 424 83 23 1 0
Table 4: Set 2: Analysis of spare asset levels
# LRUs: |I|2 4 8
# Spare assets in optimal solution 2.8 1.8 1.6
maximum 12 6 5
# Spare assets above the lower bound 1.5 0.6 0.4
maximum 11 5 3
# Spare parts in optimal solution 5.2 7.6 12.3
divided by # LRUs 2.6 1.9 1.5
Table 5: Set 1: Analysis of optimal solutions
the active maintenance time. That means that if our algorithm stocks more spare assets, then
apparently, spare assets are also used to cover spare parts unavailability. However, this is not
the case in our numerical experiment: The number of spare assets in the solution is close to the
lower bound; it is the same for 82% of the problem instances, see Table 4. Furthermore, the
gap becomes smaller when the problem size increases, to 0.04 on average for problem instances
with 1,024 LRUs. In fact, for more than 16 LRUs, the gap is never more than 1. We further
see that a gap of more than 1 only occurs if assets are relatively inexpensive, i.e., crel = 0.5.
However, this value of crel will not be realistic in most practical settings. The other parameters
have a very limited influence on the number of additional spare assets to stock. Table 5 shows
that also in the optimal solution, on Set 1, the number of spare assets is typically close to the
lower bound.
This suggests that the lower bound is useful in practice to get an idea of the fleet size to
acquire, while it is easy to calculate. For that reason, we have also implemented what we call
the lower bound algorithm. Our algorithm, as defined in Section 5 and Figure 2, stocks spare
parts for a number of spare asset levels, starting at the lower bound in part (i) of Proposition 1.
To show the value of jointly optimizing the spare asset and spare parts stock levels, we need to
compare the solutions of our algorithm with those of an algorithm in which first the spare asset
level is determined and then the spare parts stock levels. For a fair comparison, the spare asset
level should be determined in a meaningful way. Since we have seen above that our algorithm
often finds a solution in which the lower bound on the asset stock level is used, this is the asset
stock level that the lower bound algorithm uses. So, the lower bound algorithm first sets the
number of spare assets to stock equal to the lower bound from part (i) of Proposition 1, and
then stocks spare parts in the same way as our algorithm does. By comparing solutions of our
algorithm with those of the lower bound algorithm, we answer Question 3.
Table 3 shows the additional costs that the lower bound algorithm incurs for the different
parameter settings. We see that the additional costs are sometimes large. Although the average
18
# LRUs: |I|16 64 256
(1) Computation time (seconds): using the bound 0.4 3.6 22.6
(2) Computation time (seconds): not using the bound 0.5 5.5 48.9
(3) Computation time (seconds): De Smidt-Destombes et al. 1.4 41.8 1,146.8
Relative computation time (2)/(1) 1.2 1.5 2.2
Relative computation time (3)/(1) 3.3 11.6 50.7
Relative computation time (3)/(2) 2.7 7.6 23.5
Table 6: Set 2: Computation times of our algorithms and that of De Smidt-Destombes et al.
(2011), with ‘bound’ referring to the use (or not) of the results in Proposition 3
additional costs decreases when the number of LRUs increases, there is still an instance with
1,024 LRUs where using the lower bound algorithm leads to 43% additional costs. The lowest
average additional costs are achieved when crel = 2. This makes sense intuitively since assets
are relatively expensive, so it is likely that it is more cost effective to invest in spare parts than
in spare assets. However, even when crel = 2, the maximum additional costs are 24%. This
clearly shows that joint optimization of spare asset levels and spare LRU levels is necessary in
practice, since it is never certain upfront that it is safe to use the lower bound on the asset stock
levels, i.e., the lower bound algorithm.
6.4 Computational effort
In this section, we answer Question 4. Table 6 shows the computation times for three algorithms:
our algorithm that uses the bound based on Proposition 3, our algorithm that does not use that
bound, and the algorithm by De Smidt-Destombes et al. (2011). Since the algorithm of De
Smidt-Destombes et al. requires more computation time than our algorithms, we have only run
the problem instances of up to 256 LRUs using their algorithm. Note that by construction, all
three algorithms find identical solutions.
Given the number of convolutions that both algorithms perform for each LRU in each
iteration of the marginal analysis approach, we would expect that, when not using the bound
based on Proposition 3, the algorithm of De Smidt-Destombes et al. would require for 16, 64,
and 256 LRUs 4, 10.67 and 32 times as much computation time, respectively, being |I|
log2|I|. We
find that the relative performance of our algorithm is about 70% of what we expected, i.e., their
algorithm requires 2.7, 7.6 and 23.5 times as much computation time on average, respectively.
This is probably due to our algorithm requiring more storage and overhead. We further see that
using the bound that is based on Proposition 3 saves a considerable amount of computation
time, with the savings increasing with an increasing problem size: For the instances with 256
LRUs, our algorithm that does not use the bound requires 2.2 times as much computation time
on average as our algorithm that does use the bound.
Table 7 shows the computation times on Set 2 for our algorithm that uses the bound. The
computation times increase if the number of LRUs or the maximum resupply lead time increases.
In both cases, this is due to the fact that more parts need to be stocked, so more iterations of
19
# LRUs: |I|Rel. costs of Max. res.
an asset: crel lead t.: Tmax
16 64 256 1,024 0.5 1 2 0.01 0.1
Computation time (seconds) 0.4 3.6 22.6 229.9 95.4 61.5 35.5 5.2 123.1
# Spare asset levels enumerated 4.0 2.9 2.1 1.8 4.2 2.4 1.5 1.6 3.8
Maximum 12 8 5 4 12 7 4 4 12
Max. ass. Ave. costs of Target readiness:
time: µmax an LRU: cave Robj
0.001 0.01 100 1,000 0.9 0.95 0.975
Computation time (seconds) 68.2 60.1 63.0 65.3 61.6 63.9 66.9
# Spare asset levels enumerated 2.8 2.6 2.7 2.7 2.6 2.7 2.8
Maximum 12 11 12 12 10 11 12
Table 7: Set 2: Computation times for all parameters
the greedy algorithm need to be performed. In the former case, an additional reason is that
more LRUs need to be checked in each iteration, since Problem (P) is not item-separable. If the
relative costs of an asset increase, then the computation times decrease. The key reason for this
is that fewer spare asset levels need to be enumerated. The influence of the other parameters
on the computation time is limited.
7 Conclusions and recommendations
We have considered the problem of jointly optimizing the number of spare LRUs and spare
assets, i.e., the spare parts inventories and fleet size. This is a problem that needs to be solved
by companies that use a fleet of assets, e.g., railway operators, shipping companies or defense
organizations. We have found that the optimization problem is challenging since it is not item-
separable, nor jointly concave. Despite this challenge, we have been able to construct an efficient
and effective algorithm. In a numerical experiment, we have shown that our algorithm typically
finds solutions that are close to optimal and that our algorithm is relatively fast, both due to
the order in which convolutions are performed and due to the bound based on Proposition 3
that is used to avoid performing unnecessary computations.
We have also shown that the asset stock level that our algorithm finds is often close to, or
even the same as, the easily computable lower bound in Proposition 1. In fact, in all instances
in our numerical experiment with more than 16 LRUs, our algorithm stocks exactly the lower
bound, or one more. Since the lower bound is the number of spare assets that are required
to achieve the target readiness if there are unlimited spare parts available, our findings imply
that spare assets should be acquired to cover the active maintenance time, while they should
not be acquired to cover spare parts unavailability. This is the exact opposite of the ‘carcass
politics’ that we mentioned in Section 1: the policy of spending as much of the available budget
on buying complete assets, and spending only the remainder on spare parts. Still, setting the
20
asset stock level to the lower bound and then stocking spare parts can lead to large additional
costs. This means that joint optimization is necessary in practice.
It would be interesting to extend our work by modeling the maintenance processes more
realistically. We have now assumed that the repair shop that repairs failed parts has ample
servers. This may be realistic in many settings, since it can represent lead time agreements
with the repair shop, but in other settings it may not be. We have further assumed that assets
fail due to a failure in exactly one LRU and that only this LRU is replaced. In practice, often
multiple LRUs are replaced at the same time, for example since conditions of components are
monitored. It may be interesting to model this. It then becomes relevant to incorporate whether
these replacements need to be performed sequentially, or can be performed in parallel.
Another interesting extension would be to consider the optimization of the LRU level itself:
In case of a failure in a part of a certain LRU type, it may be possible to either exchange and
repair that part, or a module in which the part is contained. This influences the exchange times,
the required resources for the exchange, and the types and amounts of spare parts to stock.
Some first results on that problem, without considering spare LRUs and spare assets, can be
found in Parada Puig and Basten (2015).
Acknowledgements
The authors thank two reviewers, an associate editor, and the department editor, Sila C¸ etinkaya,
for their helpful comments, which enabled us to improve the paper considerably. The authors
further thank Bob Huisman of NedTrain for discussions on the real-life situation that led to the
model in this paper. The first author gratefully acknowledges the support of the Lloyd’s Register
Foundation (LRF). LRF helps to protect life and property by supporting engineering-related
education, public engagement and the application of research. The second author thanks the
Netherlands Foundation for Scientific Research for funding this research.
A Proofs
Proof of Proposition 2
We first derive the difference function for the number of spare LRUs iI. For notational
convenience, let Zi=Y0+PkI\{i}Bk. Then:
iR(S0) = R(S0+ei)R(S0)
=1PZi+ [XiSi1]+> S01PZi+ [XiSi]+> S0
=PZi+ [XiSi]+> S0PZi+ [XiSi1]+> S0
=
X
x=0
PZi+ [XiSi]+> S0|Xi=xP{Xi=x}
X
x=0
PZi+ [XiSi1]+> S0|Xi=xP{Xi=x}
21
=
Si+S0+1
X
x=Si+1
PZi+ [XiSi]+> S0|Xi=xP{Xi=x}
Si+S0+1
X
x=Si+1
PZi+ [XiSi1]+> S0|Xi=xP{Xi=x}
=
S0+Si+1
X
x=Si+1
[P{Zi> S0+Six} − P{Zi> S0+Si+ 1 x}]P{Xi=x}
=
S0+Si+1
X
x=Si+1
P{Zi=S0+Si+ 1 x}P{Xi=x}
=
S0+1
X
b=1
P{Xi=Si+b}P{Zi=S0+ 1 b}.
The fifth equation holds because if Xi< Si+ 1, then [XiSi]+= [XiSi1]+= 0, and if
Xi> Si+S0+ 1, then P{Zi+ [XiSi]+> S0}=P{Zi+ [XiSi1]+> S0}= 1.
We next derive the second order difference function:
2
iR(S0)=∆iR(S0+ei)iR(S0)
=
S0+1
X
b=1
P{Xi=Si+1+b}P
Y0+X
kI\{i}
Bk=S0+ 1 b
S0+1
X
b=1
P{Xi=Si+b}P
Y0+X
kI\{i}
Bk=S0+ 1 b
=
S0+1
X
b=1
[P{Xi=Si+1+b} − P{Xi=Si+b}]P
Y0+X
kI\{i}
Bk=S0+ 1 b
.
By Lemma 1, Xiis a Poisson distributed random variable with mean λiTiso that we may
express its probability mass function recursively as P{Xi=k}=λiTi
kP{Xi=k1}, for k > 0.
Now, for part (i): If S0+Si<dλiTie − 2, then P{Xi=Si+1+b}>P{Xi=Si+b}for
b∈ {1, . . . , S0+ 1}, so that ∆2
iR(S0)>0.
For part (ii): If Si≥ dλiTie − 2, then P{Xi=Si+1+b} ≤ P{Xi=Si+b}for b
{1, . . . , S0+ 1}, so that ∆2
iR(S0)0.
Proof of Proposition 3
For notational convenience, let Zij =Y0+PkI\{i,j}Bk. Then:
iR(S0+ej)iR(S0)
=
S0+1
X
b=1
P{Xi=Si+b}
22
PZij + [XjSj1]+=S0+ 1 bPZij + [XjSj]+=S0+ 1 b
=
S0+1
X
b=1
P{Xi=Si+b}
S0+1b
X
z=0
P{Zij =z}
P[XjSj1]+=S0+ 1 bzP[XjSj]+=S0+ 1 bz
=
S0+1
X
b=1
P{Xi=Si+b}
S0b
X
z=0
P{Zij =z}
P[XjSj1]+=S0+ 1 bzP[XjSj]+=S0+ 1 bz
+
S0+1
X
b=1
P{Xi=Si+b}P{Zij =S0+ 1 b}
P[XjSj1]+= 0P[XjSj]+= 0
=
S0+1
X
b=1
P{Xi=Si+b}
S0b
X
z=0
P{Zij =z}
[P{Xj=S0+Sj+ 2 bz} − P{Xj=S0+Sj+ 1 bz}]
+P{Xj=Sj+ 1}
S0+1
X
b=1
P{Xi=Si+b}P{Zij =S0+ 1 b}.(2)
After the third equation, the case that z=S0+ 1 bis considered separately. Furthermore,
we define Pc
x=bax= 0 if c < b.
We now require two results, which we prove below, that we combine to prove Proposition 3.
The first result is that the first of the two terms in Equation (2) is negative, while the second
result is that the second term in that equation is smaller than P{Xj=Sj+ 1}P{Xi=Si+ 1}.
The summation of the two terms is then also smaller than P{Xj=Sj+ 1}P{Xi=Si+ 1}.
1. Xjin P{Xj=S0+Sj+ 2 bz}and P{Xj=S0+Sj+ 1 bz}ranges from Sj+ 1
and Sj, to Sj+S0+ 1 and Sj+S0, respectively. Since Sj≥ dλjTje − 2, the first term in
Equation (2) must be negative (due to the properties of the Poisson distribution discussed
in the proof of Proposition 2).
2. Since Si≥ dλiTie − 2, it holds that:
S0+1
X
b=1
P{Xi=Si+b}P
Y0+X
kI\{i,j}
Bk=S0+ 1 b
<P{Xi=Si+ 1}
S0+1
X
b=1
P
Y0+X
kI\{i,j}
Bk=S0+ 1 b
<P{Xi=Si+ 1}.
As a result, the second term in Equation (2) is smaller than P{Xj=Sj+ 1}P{Xi=Si+ 1}.
23
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