International Journal of Production Economics 187 (2017) 216–228
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Int. J. Production Economics
journal homepage: www.elsevier.com/locate/ijpe
Optimality of (s, S) policies in EOQ models with general cost structures
a,⁎
b
Sandun Perera , Ganesh Janakiraman , Shun-Chen Niu
a
b
MARK
b
School of Management, University of Michigan–Flint, Flint, MI 48502, USA
Naveen Jindal School of Management, The University of Texas at Dallas, Richardson, TX 75080, USA
A R T I C L E I N F O
A BS T RAC T
Keywords:
EOQ models
(s, S) optimality
General ordering/procurement cost structures
The classical economic-order-quantity (EOQ) model is at the heart of supply chain optimization and the theory
of inventories. We study an extension of the original EOQ model that permits a minimal set of assumptions on
the ordering/procurement and holding/backorder costs, and establish necessary and sufficient conditions for
the existence of an optimal policy of the (s, S ) type. Our work lends theoretical credibility to the practice of using
(s, S ) policies for virtually any cost structure of practical interest. Our proof is constructive and elementary, in
the sense that it is based on first principles that do not rely on advanced mathematical machinery. We also prove
that an optimal policy, not necessarily of the (s, S ) type, always exists within our general EOQ framework.
1. Introduction
The economic-order-quantity (EOQ) model (Harris, 1913)1, along
with the Newsvendor model, is at the heart of the practice of supply
chain optimization and the theory of inventories. The EOQ model
assumes that demand arises continuously and at a constant rate.
Several versions of this model of trading off inventory-holding costs
(and, in some cases, shortage costs) with procurement/logistics costs
(e.g., fixed ordering costs, volume discounts, etc.) are presented in
basic texts and courses in Operations Research and Operations
Management. The typical analysis of the EOQ model involves assuming
a policy with the following structure: Place a constant order quantity
every time the net-inventory level decreases to a threshold. The order
quantity and the threshold are then optimized. This approach usually
leads to square-root formulas or algorithms to determine these
quantities (Porteus, 2002; Winston, 2004; Zipkin, 2000). In this paper,
we take a step back and investigate the important issue of when/why
such a policy structure—commonly referred to as (s, S )—is optimal in
the EOQ model.
We present an elementary proof for the existence of an optimal
policy of the (s, S ) type in the EOQ model under general ordering/
procurement and holding/shortage cost structures. By elementary
proof, we mean a rigorous proof from first principles that do not
require significant mathematical machinery. This is the main contribu-
tion of our work. To shed some light on this contribution relative to the
literature, we will begin with a discussion both about our work and on
the “EOQ proofs” that exist in the literature.
The assumptions in the standard textbook version of the EOQ
model are as follows. The ordering cost includes a set-up cost plus a
linear variable cost. Inventory holding costs are also linear.
Backordering is either not allowed or is allowed with a linear shortage
cost. This is the model for which convenient square-root formulas are
available for the optimal re-order point, s, and the optimal re-order
quantity, S − s . Generalizations to accommodate volume discounts in
ordering costs are also presented frequently; in such cases, the optimal
values of s and S − s do not take simple square-root forms, but
algorithms are presented for computing them. What is striking is that
the use of an (s, S ) policy is usually taken as granted; and in fact, formal
discussions of whether or not there exists an optimal policy of the (s, S )
type in the EOQ model appeared in the literature only recently (relative
to the history of this model). For example, an (s, S )-optimality proof for
a particular type of EOQ model without backordering is given in
Lippman (1971). In this paper, the ordering cost structure is that of a
set-up cost for placing an order plus a set-up cost for every new truck
used, where a truck has a finite capacity. Beyer and Sethi (1998)
appear2 to be the first authors that attempted an (s, S )-optimality proof
for the EOQ model under a more general cost structure. Other extant
optimality analyses for the standard EOQ model and a variety of
⁎
Corresponding author.
E-mail addresses: [email protected] (S. Perera), [email protected] (G. Janakiraman), [email protected] (S.-C. Niu).
1
We refer the reader to Erlenkotter (1989, 1990) for interesting historical accounts on the EOQ model.
2
Beyer and Sethi state: “Every OR textbook contains a calculus derivation of the EOQ formula. However, such a derivation does not establish formally that the EOQ formula provides
a production policy that minimizes the long-run average cost over the class of all admissible policies. Certainly there are many possible ways to prove the formula: But there does not
seem to be a publication devoted directly to providing a mathematically rigorous proof of the famous formula. It is also possible that there are a number of papers dealing with complex
inventory models, of which the simple lotsize model is a special case.”
http://dx.doi.org/10.1016/j.ijpe.2016.09.017
Received 4 July 2016; Received in revised form 5 September 2016; Accepted 21 September 2016
Available online 23 September 2016
0925-5273/ Published by Elsevier B.V.
International Journal of Production Economics 187 (2017) 216–228
S. Perera et al.
inventory-holding and backordering cost functions are monotone.
These assumptions are rather minimal, potentially the weakest reasonable assumptions within the EOQ framework. Hence, any cost structure
of practical interest is subsumed in our formulation. Some well-known
examples are all-unit discounts, incremental discounts, and truckload
discounts. These are discussed, for example, in Altintas et al. (2008),
Benton and Park (1996), Chen (2009), Federgruen and Lee (1990), Li
et al. (2004, 2012), and Zipkin (2000). Other variants of these cost
structures, such as modified all-unit discounts (cf. Chan et al., 2002)
and generalized truckload discount (cf. Li et al., 2004), are also
covered. The widely-studied multiple set-up costs structures (Alp
et al., 2014; Lippman, 1969, 1971) and the interesting quantitydependent fixed costs scheme discussed in Caliskan-Demirag et al.
(2012) are also allowed under our ordering-cost assumptions.
Moreover, when a supplier imposes stringent constraints on the size
of the orders, e.g., batch-ordering3, minimum-order-quantity (MOQ)4
and a combination of both batch-ordering and MOQ (cf. Zhu et al.,
2015) restrictions, the cost of ordering the forbidden amounts could be
(theoretically) considered to be infinite; our results also apply to these
and other scenarios where the ordering costs could be infinite in some
regions. Finally, we emphasize that some of the interesting orderingcost functions noted above are not increasing5, e.g., all-unit discounts6,
batch-ordering, and MOQ. To the best of our knowledge, an attempt to
study (s, S )-optimality for not-necessarily-increasing ordering-cost
functions does not exist in the literature.
The remainder of the paper is organized as follows. We present the
model, assumptions, and the main results formally in Section 2.
Section 3 is devoted to proving the main results. Finally, in Section
4, we present some concluding remarks.
deterministic extensions can be found in Adelman and Klabjan (2005),
Brimberg and Hurley (2006), Hassin and Meggido (1991), and Sun
(2004).
In contrast, it is interesting to note that multiple proofs of (s, S )
optimality in stochastic models exist and are celebrated (Scarf, 1960;
Iglehart, 1963a, 1963b; Veinott, 1966; Zheng, 1991, 1994). These
proofs are either for discrete-time (i.e., periodic review) models or for
continuous-time models in discrete-space (i.e., demands and order
quantities are integer-valued). The proofs in this literature do not
readily accommodate the continuous-space, continuous-time features
of the EOQ model.
The proof proposed by Beyer and Sethi (1998) is based on the
method of Quasi-Variational Inequalities (QVI). Their formulation
allows a cost structure that is weaker than that in the standard EOQ
model; but the QVI approach makes it necessary for the authors to
introduce several technical assumptions on the cost functions (e.g.,
continuity and differentiability). In a subsequent paper, Presman and
Sethi (2006) also apply the QVI method to analyze a stochastic version
of the EOQ model in which the demand process is the sum of a
deterministic process and a compound Poisson process. In this latter
paper, the ordering-cost function is restricted to have the standard
structure of a fixed set-up cost plus a linear variable cost.
Our proof of (s, S )-optimality is entirely different and is built from
first principles, which allows us to work with a rather general model
with minimal cost assumptions. The proof is based on a lowerbounding approach, and it involves two steps.
The first step is to consider a restricted class of policies that order
only when net-inventory is non-positive and always raise it to a nonnegative level upon ordering. That this restriction is without loss of
optimality requires a proof. To prove this, we develop a coupling that
takes any policy that violates the desired property and constructs a new
policy that satisfies the property while always having a lower amount of
inventory and a lower amount of backorder than the original. Thus, by
construction, the latter policy has lower cumulative holding and
shortage costs than the former at any time epoch. To complete the
proof, the ordering costs accumulated by the two policies also need to
be considered. To this end, we develop an inequality that compares the
ordering costs for these two policies over two suitably-chosen nested
time intervals, and combine it with the other cost comparison above to
show that the constructed policy has a long-run average total cost that
is no greater than the original.
In the second step, we show that the total costs for policies in the
restricted class of policies above can be naturally decomposed into
those in successive “ordering cycles”, defined as time periods that span
two consecutive zero-inventory epochs (i.e., time instants when the net
inventory is zero). We then exploit this cost decomposition to show that
the long-run average costs of policies in this restricted class are
bounded from below by the infimum of the average costs over any of
such cycles. This lower bound then yields a simple characterization that
an (s, S ) policy is optimal under our cost assumptions if and only if an
optimal solution exists for the problem of minimizing the long-run
average cost within the class of (s, S ) policies (i.e., whenever the
infimum above is attained by an (s, S ) policy).
To cover the scenario where the above infimum is not attained, that
is, to have a full closure for the (s, S )-optimality problem within the
EOQ framework, we also provide a simple sufficient condition on the
ordering-cost function (lower semi-continuity) that ensures the existence of an optimal (s, S ) policy. Furthermore, whenever this existence does not prevail, we show that our approach enables us to
constructively establish that an optimal policy, not of the (s, S ) type,
always exists.
Next, we briefly summarize the cost assumptions in our model, and
indicate, in particular, the breadth of ordering cost structures that can
be accommodated in our formulation. We only assume that the
ordering-cost function is uniformly bounded from below by a strictly
positive constant for all strictly positive order quantities, and that the
2. Model formulation and the main results
In this section, we present our cost assumptions and state our main
results. We first introduce the following notation:
λ :=the constant demand rate, λ > 0,
c(q ):=non-negative cost for ordering q units, q≥0,
g(x ):=non-negative holding/shortage cost rate when net-inventory
level is x, −∞ < x < ∞, and
I0:=net-inventory level at time 0.
In the standard EOQ model with backorder, it is assumed that c(·) is
composed of a fixed set-up cost and a linear variable cost (i.e.,
c(q ) = K 1{q >0} + νq , where K > 0 and ν ≥ 0 are constants), and g(x ) is
of the form h max(0, x ) + b max(0, − x ) for some positive constants h
and b. In contrast, we only need the following minimal assumptions on
c(·) and g(·).
Assumption 1. The function c(·) satisfies c(0) = 0 and 0 < K ≤ c(q )
for all q > 0 , where c(q) is not necessarily finite-valued.
3
This means that there is a constraint on the order size that stipulates that orders
must be integer multiples of a constant B; see Chen (2000) and Li et al. (2004). Such a
constraint can be interpreted as saying that the ordering cost, c(·), is finite at all integer
multiples of B, but is infinite elsewhere (see Fig. A.1 in Appendix A).
4
This type of order constraint is widely used. We refer the reader to Zhao and
Katehakis (2006), where the authors note that: “Minimum Order Quantity (MOQ) is
widely used in many industries (e.g., apparel, pharmaceutical, and consumer packaged
products). In these industries, customers (e.g., distributors or retailers) typically have
two choices, either not to order from their suppliers or to order at least a minimum
quantity, namely the MOQ”. Such a cost scheme can be reinterpreted as saying that the
cost of ordering q units, c(q), is infinite when q is strictly positive but is less than MOQ,
and is finite elsewhere (see Fig. A.2 in Appendix A). The reader is also referred to Hellion
et al. (2012), Kiesmüller et al. (2011), and Zhou et al. (2007) for further discussion
regarding the importance of the MOQ constraint.
5
Throughout this paper, by “increasing” and “decreasing”, we mean “non-decreasing”
and “non-increasing”, respectively.
6
The ordering-cost function for the all-unit discount scheme is illustrated in Fig. A.3
of Appendix A.
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International Journal of Production Economics 187 (2017) 216–228
S. Perera et al.
if the function α(·, ·) has a minimizer.
Assumption 2. The function g(·) is increasing on [0, ∞) and
decreasing on (−∞, 0], with lim x→±∞ g(x ) = ∞.
Our second result provides a simple sufficient condition for the
existence of a minimizer for α(·, ·).
Now, for 0 ≤ a ≤ b , let C π[a, b] denote the total cost (i.e., the sum of
ordering, holding and backordering costs) incurred over the time
interval [a,b]7 for a given policy π . Here, a policy π is a specification
of an infinite sequence of ordering time instants and order sizes. We
will measure the performance of a policy π by its long-run average cost,
which is defined as
f
π
C π[0, T ]
= lim sup
.
T
T →∞
Theorem 2. If c(·) is lower semi-continuous9, then, α(·, ·) is
guaranteed to have a minimizer; hence, there exists an (s, S ) policy
that is optimal in Π .
Since a minimizer for α(·, ·) may not exist, the remaining question is
whether or not there exists a policy in Π (not necessarily of the (s, S )
type) with long-run average cost α*. Our final result answers this
question in the affirmative.
(1)
π
Our objective is to minimize f over an admissible policy class, which
we specify next.
Let tnπ and qnπ respectively be the n-th ordering epoch and the size of
the n-th order under a given policy π . Observe that if lim n →∞ tnπ < ∞,
then f π = ∞8. Hence, we will restrict our attention to policies that
satisfy the condition that lim n →∞ tnπ = ∞; and formally define an
admissible policy as follows:
Theorem 3. An optimal policy always exists in Π .
3. Proofs of the theorems
3.1. Proof of Theorem 1
The first step in our proof is to restrict consideration from the class
of all policies Π to a subset of policies that are more amenable to
analysis. In light of Assumption 2, an immediate intuition is that any
attempts to maintain the net-inventory level “as close to zero as
possible” should result in a reduction in total cost. When backordering
is not allowed, this intuition can be formalized as: Order only when the
inventory level decreases to zero. Indeed, in that case, it is a wellknown result (Beyer and Sethi, 1998; Brimberg and Hurley, 2006;
Lippman, 1971; Porteus, 2002; Zipkin, 2000) that restricting attention
to this smaller policy class, which we denote by Π 0p , can be made
without loss of generality. When backordering is allowed, it is a natural
conjecture that the corresponding class of policies to consider is Π 0,
defined as policies that order only when net-inventory is non-positive
and always raise the inventory to a non-negative level upon ordering.
(Clearly, Π 0p ⊂ Π 0 .) The following lemma rigorously confirms this
conjecture10.
Definition 1. An admissible policy π is a sequence {(tnπ , qnπ ): n ≥ 1},
where t1π ≥ 0 , tnπ < tnπ+1, lim n →∞tnπ = ∞, and qnπ > 0 for all n.
Let Π denote the set of all admissible policies. Our goal is to identify
conditions under which an optimal policy exists in Π ; that is, we seek a
*
policy π * ∈ Π that satisfies f π = infπ ∈Π f π . In addition, of particular
interest is the question of whether or not the search for an optimal
policy in Π can be limited to policies of the (s, S ) type, where an (s, S )
policy is defined to be one that raises the net-inventory level to S every
time it decreases to s.
Next, observe that for a given (s, S ) policy with s=x and S=y, where
x < y , the long-run average cost α(x, y ) is given by
y
α (x , y ) =
c (y − x ) + ∫ g (z )
x
( y − x )/ λ
dz
λ .
(2)
Let α*≔inf{(x, y): x < y} α (x , y ); and note that, in general, α* may not be
finite. To have a clean setup, we shall assume that there exists at least
one policy π in Π with f π < ∞. Clearly, this assumption can be made
without loss of generality; and in particular, it implies that there must
exist x 0 < y0 such that c( y0 − x 0 ), the right limit of g(·) at x0, and the left
limit of g(·) at y0 are all finite. It then follows from Assumption 2 that
α(x 0 , y0 ) < ∞, which ensures the finiteness of α*. In the same spirit, we
will further assume that if I0 is positive, then the left limit of g(·) at I0 is
finite.
We are now ready to state the main results of this paper. Unless
explicitly indicated otherwise, we assume henceforth that Assumptions
1 and 2 hold. The first result establishes a lower bound for f π for all
π ∈ Π.
Lemma 1. It is sufficient to restrict attention to policies in Π 0 ; that is,
inf f π = inf f π .
π ∈Π
π ∈Π 0
Proof. Clearly, it is sufficient to show that for a given π ∈ Π⧹Π 0 , there
∼
exists a corresponding policy π∼ ∈ Π 0 that satisfies f π ≤ f π .
π
π
For n ≥ 1, denote by xn and yn , respectively, the net-inventory levels
just before and after the n-th order under a policy π in Π. Note that by
definition, qnπ = ynπ − xnπ . Observe that for any n ≥ 1, the vector (xnπ , ynπ )
associated with the n-th order must belong to one of the following three
categories: (i) 0 < xnπ < ynπ , (ii) xnπ ≤ 0 and ynπ ≥ 0 with xnπ < ynπ , or (iii)
xnπ < ynπ < 0 . Clearly, every order for a π ∈ Π 0 is by definition in
Category (ii). In contrast, any π ∈ Π⧹Π 0 must have at least one order
in either Category (i) or Category (iii).
Let π be an arbitrary policy in Π⧹Π 0 ; and let π∼ be the policy
constructed from π according to the following prescription:
Theorem 1. The long-run average cost of any policy π ∈ Π is
bounded from below by α*.
∼
tnπ = tnπ +
It follows immediately from Theorem 1 that if there exists a policy
in Π whose long-run average cost exactly equals α*, then that policy is
optimal. In particular, if a minimizer (x*, y*) for the function α(·, ·)
exists in the set {(x, y ): x < y}, then the (s, S ) policy with s = x* and
S = y* is optimal in Π . Thus, the following characterization holds.
∼
tnπ = tnπ +
∼
tnπ
= tnπ
xnπ
λ
ynπ
λ
and
∼
if xnπ > 0;
and qnπ = qnπ ,
∼
if ynπ < 0;
∼
qnπ
= qnπ ,
otherwise.
and qnπ = qnπ ,
and
π
For t ≥ 0 , denote by I (t ) the net-inventory level at time t under a
policy π in Π (with I π (0) = I0 for all π ∈ Π ). Clearly, the inventory
Corollary 1. An optimal policy of the (s, S ) type exists in Π if and only
7
Similar notation for other types of intervals, namely [a, b ), (a, b], and (a,b), will be
adopted throughout this paper.
8
Suppose that lim n →∞ tnπ = M < ∞ for a policy π. Then, we have that tnπ ≤ M for all
n ≥ 1, and this implies that for every N, the number of orders placed in the time interval
[0, M ] exceeds N. For any T > M , let N = T 2 . Then, we have C π[0, T ] ≥ KT 2 . Therefore,
C π[0, T ]/T ≥ KT for all T > M . Letting T → ∞ then yields f π is infinite for all π with
lim n →∞ tnπ < ∞.
9
A function h: n → [ − ∞, ∞] is called lower semi-continuous at x 0 if for every ε > 0
there exists a neighborhood B ⊂ n of x 0 such that h(x) ≥ h(x 0) − ε for all x ∈ B . The
function h is said to be lower semi-continuous on E ⊂ n if it is lower semi-continuous at
every point in E.
10
When backordering is allowed, the necessary argument to extend the result by the
above-cited authors is new, and not immediate.
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International Journal of Production Economics 187 (2017) 216–228
S. Perera et al.
Fig. 1. Construction of a policy π∼ ∈ Π 0 from a given π ∈ Π⧹Π 0 .
trajectory {I π (t ), t ≥ 0} is completely determined by I0 and the
sequence {(tnπ , qnπ ), n ≥ 1}. The inventory trajectories under both π
and π∼ are illustrated in Fig. 1. It is easily seen from this figure that if
∼
∼
xnπ > 0 for any n, then xnπ = 0 (see, e.g., epochs t2π and t2π ); and similarly
∼
π∼
π
that if yn < 0 for any n, then yn = 0 (see, e.g., epochs t8π and t8π ).
∼
Therefore, π ∈ Π 0 by construction.
We now compare the costs under π and π∼. Consider a given time
interval [0, T ]. Observe that if an order in π is relocated to arrive at one
in π∼, then this corresponding order in π∼ could be either a postponement or an early placement of the original. Consequently, the total
number of orders under π∼ in [0, T ] can be larger than that of π over the
same interval. To facilitate cost comparison, we will therefore consider
the total cost under π for a possibly larger interval [0, T + δ ], where δ is
an appropriately-chosen non-negative increment to T that depends on
both π and T.
Our intent is to pick δ so as to ensure that every order placed by π∼
within [0, T ] originates from an order placed by π within [0, T + δ ]. For
this purpose, let τ∼π (T ) be the epoch at which the last order under π∼ in
[0, T ] is placed; let τ π (T ) be the epoch at which the “corresponding”
order under π (the one with the same order index) is placed; and
denote by y π (T ) the net-inventory level under π immediately after the
placement of this corresponding order. Then, we will let δ = (y π (T ))− / λ ,
where (y π (T ))−≔ − min(y π (T ), 0) (the negative part of y π (T )), and claim
that the following inequality holds:
π∼
π
π
−
C [0, T ] ≤ C [0, T + ( y (T )) / λ]
for all T ≥ 0.
and (5) yields (3).
∼
We are finally ready to show that f π ≤ f π holds under our
π
construction. If f = ∞, we obviously have nothing to prove; we will
therefore assume that f π is finite.
We begin by noting that (3) can be rewritten as:
∼
C π[0, T + ( y π (T ))− / λ] T + ( y π (T ))− / λ
C π [0, T ]
≤
.
T
T + ( y π (T ))− / λ
T
Upon taking lim sup , (6) yields
∼
fπ
(y π (T ))−
1
lim sup
.
π ≤ 1 +
f
λ T →∞
T
lim sup
n →∞
(3)
( y π (Tn ))−
= 0.
Tn
(8)
In words, this means that whenever there exists a sequence
{y π (Tn ), n ≥ 1} that strictly descends to −∞ as n → ∞, the average
growth of the magnitude ( y π (Tn ))− is negligible relative to that of Tn.
Since g(·) is assumed to satisfy lim x→−∞ g(x ) = ∞ and f π is finite, it is
intuitive that (8) should hold. We next formalize this intuition and
complete the proof by establishing (8).
First, observe that whenever y π (T ) is negative (e.g., when T = t8π in
Fig. 1), we have τ∼π (T ) = τ π (T ) − ( y π (T ))− / λ . This scenario is illustrated
in Fig. 2 (T′ in the figure indicates another possible location for T).
From this figure, we see that the inventory trajectory under π in the
interval [τ∼π (T ), τ π (T )) (with the right endpoint open) is no higher than
the diagonal border of the shaded region spanning the same interval.
That is, I π (t ) ≤ − λ(t − τ∼π (T )) holds for any t in [τ∼π (T ), τ π (T )). It then
easily follows that the shortage cost under π during [τ∼π (T ), τ π (T )) is no
less than g( y π (T )/2)(( y π (T ))− /2)(1/λ ) (see corresponding interval markers in Fig. 2). Therefore,
(4)
π
always holds. Obviously, (4) is valid if τ (T ) ≤ T . Suppose on the other
hand that T < τ π (T ). Then, since τ∼π (T ) ≤ T by assumption, we neces∼
sarily have y π (T ) < 0 (e.g., when T is located in the interval [t8π , t8π ) in
π
Fig. 1), implying that I (T ) must not be positive. Hence, we have
λ(τ π (T ) − T ) ≤ − y π (T ) = ( y π (T ))−; and this is equivalent to (4).
We now prove (3). To facilitate the argument, it will be convenient
to split the total cost into two parts and then compare them separately.
For 0 ≤ a ≤ b and any given π , let R π [a, b] and H π [a, b] be the
cumulative ordering cost and, respectively, the cumulative sum of both
holding and shortage costs over the time interval [a, b]. By definition,
we have C π[a, b] = R π [a, b] + H π [a, b]. Recall that τ∼π (T ) is the last
ordering epoch under π∼ in [0, T ]. It follows that
∼
∼
R π [0, T ] = R π [0, τ∼π (T )] = R π [0, τ π (T )] ≤ R π [0, T + (y π (T ))− / λ],
(7)
Clearly, if y π (T ) is uniformly bounded from below for all T, then (7)
∼
implies11 that f π ≤ f π . (Note in particular that if backordering is not
allowed, then this conclusion is immediate.) Therefore, to complete the
proof, it is sufficient to show that for any given increasing sequence
π
−
π
−
{Tn}∞
n=1 satisfying lim n →∞Tn = ∞, lim n →∞( y (Tn )) = ∞, and ( y (Tn )) is
strictly increasing in n with ( y π (T1))− > 0 (the last requirement is
without loss of generality since we are concerned with lim sup ), we have
Before proving (3), we will first show that
τ π (T ) ≤ T + ( y π (T ))− / λ
(6)
(5)
11
In a previous version of this paper, the requirement of having a uniform lower
bound was introduced. The authors are grateful to two anonymous reviewers of that
earlier version for demonstrating that this requirement is not necessary. Their contributions led to a significantly shortened proof of Theorem 1.
where the inequality is due to (4). Clearly, we also have
∼
H π [0, T ] ≤ H π [0, T ] ≤ H π [0, T + (y π (T ))− /λ]. Summing this inequality
219
International Journal of Production Economics 187 (2017) 216–228
S. Perera et al.
decompose the total cost over a given interval under a policy into those
over successive ordering cycles, defined as intervals that span two
consecutive zero-inventory epochs; and then bound the total cost in
every ordering cycle from below by α* times the length of the same
cycle.
We begin with an analysis of the total costs in successive ordering
cycles. Consider an arbitrary policy π in Π 0. Let u1π ≔max[0, t1π + x1π / λ]
and v1π ≔t1π + y1π / λ ; and for i ≥ 2 , let uiπ ≔tiπ + xiπ / λ and viπ ≔tiπ + yiπ / λ . We
will formally refer to the closed time interval [uiπ , viπ ], i ≥ 1, as the i-th
ordering cycle. Note that, since π ∈ Π 0 , we have xiπ ≤ 0 , yiπ ≥ 0 , and
xiπ < yiπ for all i ≥ 1; therefore, these intervals are all well defined (and
have positive durations). It is also easily seen that viπ = uiπ+1 holds for all
i ≥ 1. These notations are illustrated in Fig. 3.
Next, let
κiπ ≔c(yiπ − xiπ ) +
∫x
yiπ
π
i
g (z )
dz
,
λ
for i ≥ 1.
(13)
For i ≥ 2 , we will assign (or define) κiπ as the total cost associated with
the i-th order, or alternatively with the i-th ordering cycle12. For i=1,
Fig. 2. Lower bound for shortage cost over [τ∼π (T ), τ π (T )) .
Fig. 3. Illustration of ordering cycles.
π
−
π
−
π
π
π
−
observe that t1π + x1π / λ could be strictly negative so that [u1π , v1π ] may not
be a “full” cycle; therefore, we should and will assign
δ1π ≔c(y1π − x1π ) + H π [u1π , v1π ] (≤κ1π ) as the total cost associated with the
first ordering cycle.
Given the above cost assignments, we will next establish a lower
bound for the total cost over a time interval. Surprisingly, for a given
time epoch T, instead of working with the default choice of interval
[0, T ], the key idea turns out to be the consideration of a suitablyenlarged time interval. Details are now developed.
Consider the last order in [0, T ] under π . For t ≥ 0 , denote by N π (t )
the total number of orders in [0, t ]; formally,
π
g( − (y (T )) /2)(y (T )) /(2λ ) ≤ C [τ (T ) − (y (T )) / λ , τ (T )]
≤ C π[0, τ π (T )]
≤ C π[0, T + (y π (T ))− / λ],
(9)
where the last inequality is again due to (4).
Now, for our given π , let {Tn}∞
n=1 be a sequence satisfying the
stipulations in conjunction with (8). (An example can be constructed
to show that such sequences could exist.) Then, after setting T = Tn in
(9) and dividing the resulting inequality by Tn + ( y π (Tn ))− / λ , we see, in
light of the assumed finiteness of f π , that for any ϵ > 0 ,
g( − ( y π (Tn ))− /2)(y π (Tn ))− /(2λ )
≤ fπ + ϵ
Tn + ( y π (Tn ))− / λ
N π (t ) ≔max{n ≥ 1: tnπ ≤ t}.
(10)
holds for all sufficiently large n; and furthermore, g( − (y π (Tn ))− /2) is
necessarily finite. Since (y π (Tn ))− is positive, a rearrangement of (10)
now yields
π
−
⎤
Tn
1 ⎡ g( − ( y (Tn )) /2)
− 1⎥ ≤
.
⎢
π
λ⎣
2( f + ϵ)
( y π (Tn ))−
⎦
Whenever N π (T ) ≥ 1, denote by vNπ (T ) the right endpoint of the ordering
cycle associated with the N π (T )-th order; that is, let vNπ (T )≔v Nππ (T ).
Clearly, we have vNπ (T ) = T + I π (T )/ λ .
We will next examine the total cost under π over the interval
[0, T + (I π (T ))+ / λ], where (I π (T ))+≔max(I π (T ), 0) (the positive part of
I π (T )). The motivation behind this enlargement of [0, T ] is that
whenever vNπ (T ) overshoots T, which occurs if and only if I π (T ) > 0 ,
(11)
Finally, since lim x→−∞ g(x ) = ∞, we can further assume without loss
of generality that the left-hand side of (11) is positive. It follows that
0<
⎡ g( − ( y π (Tn ))− /2)
⎤
( y π (Tn ))−
≤λ⎢
− 1⎥
⎣
⎦
Tn
2( f π + ϵ)
Letting n → ∞ in (12) then completes the proof.
12
Note that the ordering cycles are not disjoint; hence, the language here may appear
inappropriate. However, the intersection between any two consecutive cycles is a single
point. Therefore, the total holding and shortage costs over any time interval are not
affected by these intersections. If we adopted disjoint ordering cycles, then the indexing
of the successive ordering costs would be complicated by the fact that orders could
appear at either end of a cycle (see Fig. 3). Our cost assignment scheme here is intended
to overcome this notational complication, by allowing intersections in the definition of
ordering cycles.
−1
.
(12)
□
Given the reduction established in Lemma 1, the next step is to
analyze the average costs of policies in Π 0. Our approach is to first
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International Journal of Production Economics 187 (2017) 216–228
S. Perera et al.
the entire N π (T )-th ordering cycle would be included in the cost
calculation, and this facilitates the development of a “clean” lower
bound.
It is easily seen that13
C π[0, T ] ≥ H π [T − (I π (T ))− / λ , T ]
≥ g( − (I π (T ))− /2)((I π (T ))− /2)(1/λ ).
N π (T )
C π[0, T + (I π (T ))+ / λ] = H π [0, u1π ) + δ1π +
∑
κiπ + H π [min(vNπ (T ), T ), T ],
(14)
i =2
∞>
where we have assumed without loss of generality that T is sufficiently
large so that N π (T ) ≥ 2 . Observe further that, for i ≥ 2 , the average cost
incurred in the i-th ordering cycle is given by κiπ /(viπ − uiπ ). Since
viπ − uiπ = (yiπ − xiπ )/ λ , it follows from (2) and (13) that
κiπ = α(xiπ , yiπ )(viπ − uiπ ) ≥ α*(viπ − uiπ ). Noting that the first two and
the last terms on the right-hand side of (14) are non-negative, we then
have
Consider any (x, y ) with x < y . First, observe from (2) and
Assumption 2 that when x > 0 or y < 0 , we have α(x, y ) ≥ α(0, y − x )
or α(x, y ) ≥ α(x − y, 0), respectively. Hence, we can limit our search of
(x*, y*) to + 0 ≔{(x , y ): x < y and x ≤ 0 ≤ y}. We will next argue that we
can further restrict the search to a closed and bounded subset of + 0 .
The idea is to show that α(x, y ) is uniformly bounded from below when
y − x is sufficiently large.
For a given q > 0 , let (x, y ) be any vector in + 0 such that y − x = q .
Then, from (2), we have
κiπ ≥ α*(vNπ (T ) − u 2π )
i =2
⎞
⎛
I π (T )
= α*⎜T +
− u 2π ⎟ .
⎠
⎝
λ
Since π ∈ Π 0 , it is easily seen that if I0 ≤ 0
and that if I0 > 0 (i.e., u1π > 0 ), then
(i.e., u1π
= 0 ),
(15)
then u 2π
≤ q1π / λ ;
u 2π = u1π + q1π / λ=(I0 + q1π )/ λ .
Therefore,
y
u 2π ≤
(I0 )+ + q1π
λ
α (x , y ) =
.
(16)
Lemma 2. For any policy π ∈ Π 0 ,
⎛
(I0 )+ + q1π ⎞
I π (T )
⎟⎟
C [0, T + (I (T )) / λ] ≥ α*⎜⎜T +
−
λ
λ
⎠
⎝
π
c (q ) + ∫ g (z )
x
q /λ
q
.
(23)
Since g(z ) → ∞ as z → ∞, we have that for a given M > 0 , there exists a
positive z 0 such that g(z ) > M for all z > z 0 . It follows that, for all y > z 0 ,
Recall that our aim is to show that f π ≥ α*. In light of this
objective, we see that the significance of Lemma 2 is that it uncovers
the fact that the successful development of a tight lower bound for the
average cost under π hinges on the asymptotic behavior of I π (T )/ T as
T → ∞. We now take up this task.
Dividing both sides of (17) by T + (I π (T ))+ / λ and using the fact that
I π (T ) = (I π (T ))+ − (I π (T ))− lead to
∫z
∫0
0
∫z
y
M dz = M (y − z 0 ).
0
y
g(z )dz ≥
∫z
y
g(z )dz > M (y − z 0 ).
0
⎛y
z ⎞
α (x , y ) > M ⎜ − 0 ⎟ .
q⎠
⎝q
Observe that the condition y ≥ − x implies that 2y ≥ y − x = q , or
y / q ≥ 1/2 ; therefore,
Upon taking lim sup , this yields
⎛1
z ⎞
α (x , y ) > M ⎜ − 0 ⎟ .
q⎠
⎝2
(24)
(18)
Next, for any ε > 0 , there exists a q0 > 0 such that z 0 / q < ε whenever
q > q0 . Hence, from (24), we have α(x, y ) > M (1/2 − ε ) whenever q > q0
and y > z 0 . Note that, when y ≥ − x , the requirement q > max(q0 , 2z 0 )
guarantees that q > q0 and y > z 0 . Therefore, α(x, y ) > M (1/2 − ε ) holds
if q > max(q0 , 2z 0 ) and y ≥ − x .
The analysis for the other case with −x > y is parallel. Again, from
(22), we have
Finally, we claim that whenever f π is finite, we have
(I π (T ))−
= 0.
T
g(z )dz >
Dividing this inequality by q and using (23) then yield
((I0 )+ + q1π )/λ
(I π (T ))− / λ
−
.
T
T
fπ
1
(I π (T ))−
≥ 1 − lim sup
.
*
λ T →∞
T
α
y
This implies that
((I0 )+ + q1π )/ λ
1 C π[0, T + (I π (T ))+ / λ]
( I π (T ))− / λ
≥1−
−
π
+
π
+
T + (I (T )) / λ
T + (I (T )) / λ
T + (I π (T ))+ / λ
α*
T →∞
∫0 g(z )dz
(17)
holds for all sufficiently large T.
lim sup
(22)
y
α (x , y ) ≥
+
≥1−
dz
y
0
g(z )dz
∫ g(z )dz
λ ≥ ∫x
+ 0
.
q
q
Observe that with q fixed, we have either y ≥ − x or −x > y . We will
next examine these two scenarios.
Suppose first that y ≥ − x . Clearly, (22) implies that
Combining (15) and (16) then yields the following lemma.
π
(21)
3.2. Proof of Theorem 2
N (T )
∑
fπ
1
≥
lim sup g( − (I π (Tn ))− /2).
ϵ
2λ n →∞
Since the right-hand side of (21) is unbounded when n → ∞, we have
arrived at a contradiction. This establishes (19) (because ϵ is arbitrary),
which together with (18) imply f π ≥ α*; and hence the proof of
Theorem 1 is complete.
π
C π[0, T + (I π (T ))+ / λ] ≥
(20)
After setting T = Tn in (20), dividing both sides of the resulting
inequality by Tn, and then taking lim sup , we obtain
(19)
The proof of this claim is similar to that for (8). Suppose that there exist
a positive ϵ and an increasing sequence {Tn}∞
n=1 with lim n →∞Tn = ∞ that
satisfy (I π (Tn ))− / Tn > ϵ for all n. Consider the time interval
[T − (I π (T ))− / λ , T ] (which would be a single point when I π (T ) ≥ 0 ),
and observe that
0
α (x , y ) ≥
∫x g(z )dz
q
.
An argument similar to that for the previous case then shows that there
exist positive constants X1 and z1 such that α(x, y ) > M (1/2 − ε ) holds if
q > max(q1, 2z1) and −x > y . This now allows us to conclude that
If u1π = 0 , the interval [0, u1π ) is vacuous; otherwise, this interval does not contain
any order.
13
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International Journal of Production Economics 187 (2017) 216–228
S. Perera et al.
scenarios: (i) I0 > 0 , (ii) 0 > I0 ≥ x 0 , and (iii) 0 ≥ x 0 > I0 . For scenario
(i), we can always do nothing until inventory hits level 0 (recall that the
left limit of g(·) at I0 was assumed to be finite whenever I0 is
positive). For scenario (ii), we will simply follow the (x 0 , y0 ) policy
and let it run until the net-inventory level hits level 0 for the first
time. For scenario (iii), we will immediately place an order of size
nl (y0 − x 0 ), where nl is the smallest positive integer such that
nl (y0 − x 0 ) + I0 > x 0 . Note that the cost of placing this order is finite
due to the sub-additivity of c(·) and the finiteness of c(y0 − x 0 );
moreover, since this order takes the net-inventory level up to a
position higher than x0 (but no greater than y 0), we can then
duplicate the action prescribed for scenario (ii). It follows that, for
all three scenarios, we will be able to reach a zero-inventory epoch
in finite time. Since the total cost involved is guaranteed to be
finite, this establishes the claim.
Next, let ε0 ≔ α0 − α*, which is positive. Now, with
α(x, y ) > M (1/2 − ε ) holds if q = y − x > qm≔ max (q0 , 2z 0 , q1, 2z1), regardless of the relative sizes of −x and y.
Next, recall that there exists an (x 0 , y0 ) such that α(x 0 , y0 ) < ∞ (see
the paragraph following (2)). Without loss of generality, we can further
assume that (x 0 , y0 ) is in + 0 . Then, since M and ε are arbitrary, they can
be chosen to satisfy M (1/2 − ε ) ≥ α(x 0 , y0 ). Therefore, for any
(x, y ) ∈ + 0 such that y − x > qm , we have that α(x, y ) > α(x 0 , y0 ). Since
(x 0 , y0 ) ∈ + 0 , we have established that
inf α(x, y ) =
x<y
inf α(x, y ),
(x, y )∈+
(25)
where + ≔{(x, y ): 0 < y − x ≤ qm and x ≤ 0 ≤ y}.
Recall that α(x, y ) is originally defined for x < y . This original
domain can be extended to include (0, 0) with the convention that
α(0, 0) = ∞. That is, we now consider α(x, y ) on the closed and bounded
(i.e., compact) set + = + ∪ (0, 0). Since c(·) is lower semi-continuous,
and y − x is continuous on + , it can be easily shown, using the
definition of lower semi-continuity, that c(y − x ) is lower semi-continuous on + . Next, since c(y − x ) and 1/(y − x ) are both lower semicontinuous, c(y − x )/(y − x ) is also lower semi-continuous on + .
y
Observe further that ∫ g(z )dz /(y − x ) is continuous. Therefore, α(x, y )
x
is lower semi-continuous on + . Note that from Assumption 1, we have
limq→0+ c(q ) ≥ K , implying that lim (x, y)→(0−,0+) α(x , y ) = ∞ = α(0, 0). This
means that α(x, y ) is continuous at (0, 0). Therefore, α(x, y ) is lower
semi-continuous on + , and hence it attains its minimum on + .
However, since (x 0 , y0 ) ∈ + and α(x 0 , y0 ) < ∞ = α(0, 0), the point
(0, 0) cannot be a minimizer of α(x, y ) over + . Hence, α(x, y ) attains
its minimum on + . Thus, from (25), we conclude that there exists an
(x*, y*) ∈ + such that α*≔α(x*, y*) = infx < y α(x, y ).
(s0 , S0 ) ≔ (x 0 , y0 ),
the definition of infimum implies that there exists a sequence
{(sn , Sn ), n ≥ 0} in + such that αn≔α(sn , Sn ) ≤ α* + εn , where εn≔ε0 /2n ,
for all n ≥ 0 ; moreover, this sequence can be chosen so that the αn 's are
decreasing in n. Observe that, since αn ≤ α* + ε0 and Sn − sn ≤ qm for all
n ≥ 0 , we have
c(Sn − sn ) ≤ c(Sn − sn ) +
Sn
g(z )dz / λ
n
≤ (α* + ε0 )(Sn − sn )/ λ
≤ κ+ ,
(27)
where κ+ ≔α0 qm / λ , a finite constant. Note in addition that since (sn , Sn )
is in + for every n ≥ 0 , we always have sn +1 ≤ Sn .
For n ≥ 0 , define wn≔(Sn − sn )/ λ . Since I0 = 0 , the exact length of the
period in which policy π * follows the (s0 , S0 ) policy is ζ0≔k 0 w0 . We shall
refer to the time interval [0, ζ0] as the 0-th policy segment, or just the 0th segment, of π *; and from epoch ζ0 onward, we will let π * consist of a
sequence of contiguous segments, where, for n ≥ 1, the n-th segment is
a time interval of length ζn≔kn wn , where kn is a positive integer, in
n
which the (sn , Sn ) policy is employed. Let Γn≔ ∑ j =0 ζj , for n ≥ 0 ; then,
note that Γn is the epoch at which the n-th segment terminates and at
which the (n + 1)-th segment begins.
To “connect” the (s, S ) policies in successive policy segments, notice
however that there exists a complication, namely that for any n ≥ 1, it is
possible to have Sn −1 = 0 = sn . If such a scenario should occur, we will
let π * place a single order of size Sn − sn−1 at Γn−1, as opposed to the
“inadmissible” action of placing two concurrent orders of respective
sizes Sn −1 − sn −1 = − sn −1 and Sn − sn = Sn . If, on the other hand,
Sn −1 = 0 = sn does not apply, then, at most one order could exist at
Γn−1 and it is easily seen that the (sn −1, Sn −1) policy would “automatically” switch into the (sn , Sn ) policy at that epoch without any intervention; this is because Γn−1 is a zero-inventory epoch if and only if
Sn −1 = 0 = sn does not hold.
To complete our definition of policy π *, what remains is the
specification of the kn's. Recall that with our choice of εn's above, the
convergence of the αn 's to α* is at least at an exponential rate. The basic
intuition behind this last part of our construction, then, is that the
lengths of the successive policy segments in π *, which are determined
by the kn's, can be made to increase at a sufficient pace so that the longrun average cost of π * actually achieves α*.
We will now proceed by setting k 0 = 1; this means that, starting
from time 0, the (s0 , S0 ) policy is followed for a time interval of length
ζ0 = w0 . Hence, the total cost contribution (see (13)) associated with
the ordering cycle spanning [0, Γ0] equals α0 w0 , or α0 Γ0 . Next, at epoch
Γ0, observe that if S0 = 0 = s1 holds, then the “combined” order of size
S1 − s0 there creates a larger (or merged) ordering cycle of duration
w0 + w1. Moreover, since c(·) is sub-additive, the total cost contribution
associated with this combined order is no greater than α0 w0 + α1 w1,
which, interestingly, precisely equals the total of the cost contributions
3.3. Proof of Theorem 3
In light of (25), we will assume that α* is not attained by any
(x, y ) ∈ + , as otherwise this theorem is a consequence of Theorem 1.
Under this assumption, we will construct a policy in Π whose long-run
average cost exactly equals α*. Specifically, our strategy is to show that
there exists a policy π * whose cumulative cost function satisfies, for all
sufficiently large T, an inequality of the form
*
C π [0, T ]
≤ α* + β (T ),
T
∫s
(26)
where β (T ) is a (non-negative) function that converges to 0 as T → ∞.
*
*
Clearly, (26) implies that f π ≤ α* holds. Since f π ≥ α* also holds
*
(Theorem 1), we will then have the desired result that f π = α*.
Before proceeding, we will make the additional assumption that the
ordering-cost function is sub-additive, i.e.,
c(q1 + q2 ) ≤ c(q1) + c(q2 )
for all positive q1 and q2. This assumption is without loss of generality
as any cost function that is not sub-additive can be modified to one that
is, by splitting an order of size q1 + q2 into two orders of sizes q1 and q2
each and placing these orders consecutively with little time delay to
achieve a lower ordering cost14.
Our starting point is the vector (x 0 , y0 ) in + with
α0 ≔ α(x 0 , y0 ) < ∞,
defined in Section 3.2. We will begin by letting π * follow the (x 0 , y0 )
policy for k0 ordering cycles, where k0 is a positive integer. To initialize
this (x 0 , y0 ) policy, we will assume I0 = 0 and claim that this assumption
can be made without loss of generality. To see the validity of this claim,
observe that if I0 = 0 does not hold, then there are three possible
14
For discrete-time inventory models with a constraint on total production capacity,
ordering q1 and q2 units concurrently could be infeasible; and this would result in an
example of an ordering-cost function that is not sub-additive. However, in our
continuous-time EOQ setting, these two orders could be placed “back-to-back”; that is,
within very little time of each other.
222
S. Perera et al.
International Journal of Production Economics 187 (2017) 216–228
associated with the two “scheduled” orders of sizes S0 − s0 and S1 − s1 at
Γ0 if we had not been “forced” into combining these orders. Recall that
our goal is to develop an upper bound on the cumulative total cost over
time. It follows that for bounding purposes, we can and will conveniently “pretend” that the ordering cost for the single order at Γ0 is
equal to c(S0 − s0 ) + c(S1 − s1) if S0 = 0 = s1. With this device, we then
have that the total cost associated with the order(s) during the “w0 + w1
interval” that covers Γ0 always equals α0 w0 + α1 w1, regardless of
whether or not S0 = 0 = s1; in addition, we will assign c(S0 − s0 ) as a
cost increment that belongs to the interval [0, Γ0] and c(S1 − s1) as one
that belongs to [Γ0, Γ]
1 . This assignment scheme will also be adopted
without further comment for all future segments.
Next, let k1 be sufficiently large so that Γ1 − Γ0 = ζ1 ≥ ζ0 = Γ0 .
Clearly, if there is no combined order at Γ1, then the increment to
the total cost in the interval [Γ0, Γ]
1 would be exactly α1 ζ1; otherwise,
there exists an additional cost c(S2 − s2 ) = c(S2 ) at Γ1 that is from the
next segment [Γ1, Γ2]. Recall from (27) that c(Sn − sn ) ≤ κ+ for all n ≥ 0 ;
hence, we have c(S2 ) ≤ κ+ . It follows that
*
*
lθ (T ) ] + κ+ .
C π [0, T ] ≤ C π [0,Γ
*
C π [0, Γ]
1 ≤
Γθ (T ) and then invoking (31) and (30) now
Dividing this inequality by l
yield (26) with
β (T )≔ε0
Finally, observe that θ (T ) → ∞ as T → ∞; hence, we indeed have
β (T ) → 0 as T → ∞. This completes the proof of Theorem 3.
4. Concluding remarks
Our definition of the policy class Π in Section 2 only allows
deterministic policies. However, in the interest of generality, it is also
worthwhile to consider randomized and history-dependent policies.
Denote this larger class of policies by Π͠ . We now show that whenever
α* is attained by an (x*, y*), then the corresponding (x*, y*) policy is, in
fact, also optimal over Π͠ . Somewhat surprisingly, such an extension is
an easy consequence of our proof of Theorem 1.
For π ∈ Π͠ , define
1
∑ αj ζj + κ+
j =0
(28)
f π = lim sup
always holds. Since we also have α1 ≤ α0 = α* + ε0 , α1 ≤ α* + ε1, and
ζ1 ≥ ζ0 , it is easily seen that (28) further implies that15
T →∞
E(C π[0, T ])
,
T
where E denotes expectation17; and let this be our performance
measure. Observe that
*
κ
C π [0, Γ]
1
1
≤ α* + (ε0 + ε1) + + .
Γ1
2
Γ1
2κ
θ (T ) + 2
+ +.
Γθ (T )
2θ (T )+1
(29)
E(C π[0, T ])
E(C π[0, T ])
≥ lim inf
T
→∞
T
T
T →∞
π
⎛
⎞
C [0, T ]
≥ E⎜lim inf
⎟,
⎝ T →∞
⎠
T
f π = lim sup
Note that, since ε1 = ε0 /2 , the term (1/2)(ε0 + ε1) can be rewritten as
ε0(n + 2)/2n +1 with n=1.
Our intent beyond k1 should now be clear. For any n ≥ 2 , we will let
kn be sufficiently large so that the length of the n-th segment (i.e., ζn) is
no shorter than the total length of all previous segments (i.e., Γn−1).
That is, we will require that the kn's be chosen so that Γn − Γn −1 ≥ Γn −1
holds for every n ≥ 1; and this completes our construction of π *. It
should also be clear that this construction is designed to ensure that,
for all n ≥ 1, we have
(32)
holds. Next, note that the cost associated with the interval
lθ (T ), Γ
lθ (T ) + wθ (T )+1] equals αθ (T )+1 wθ (T )+1, and that, in light of (27), this
[Γ
cost is bounded from above by κ+ . Hence, we have
where the second inequality is due to Fatou's Lemma. Next, note that
our proof of Theorem 1 directly applies to every realized inventory
trajectory under π ; and moreover, the argument remains valid when
lim sup in (1) is replaced by lim inf . That is, lim infT →∞C π[0, T ]/ T ≥ α*
holds for every sample path. This implies that the right-hand side of
*
(32) is also bounded below by α*. Since f π = α*, we see that π * is
optimal over Π͠ .
Finally, we note that under stronger assumptions on the cost
structure, simple and direct proofs for our Theorem 1 exist. Since
such proofs have been desperately absent in standard textbooks, we
believe they are of pedagogical interest. We will articulate two
examples, but, in the interest of brevity, delegate the details to the
Appendix. The first example is for the special case when c(·) is
increasing and backordering is not allowed (i.e., g(x ) = ∞ for all
x < 0 ). We have uncovered an extremely short proof. The argument,
which complements the extant literature (see, e.g., Beyer and Sethi
(1998) and Lippman (1971)), is given in Appendix B. When compared
with our full proof of Theorem 1, it can be seen from Appendix B that
the shortness of the proof is primarily due to the no-backordering
constraint. In the absence of this requirement, it turns out that a direct
proof of Theorem 1 that does not rely on the reduction in Lemma 1 also
exists if the cost structure is restricted to that of the standard EOQ
model; and this is our second example. The argument for this extension
of the original EOQ model in Harris (1913) is provided in Appendix C.
15
This follows from the fact that a ≥ b ≥ 0 and y ≥ x > 0 together imply that
(ax + by )/(x + y ) ≤ (1/2)(a + b ) .
16
Note that the right-hand-side ratio is no less than αθ (T ) . This inequality then follows
from the fact that for any pair of positive x and y, a ≥ (ax + by )/(x + y ) if and only if
a ≥ b ≥ 0.
17
Conceptually, this expectation is on the probability space of all inventory trajectories generated by π .
*
κ
C π [0, Γn]
n+2
≤ α* + ε0 n +1 + + .
Γn
Γn
2
(30)
This inequality is an immediate extension of (29), obtained by
*
*
*
decomposing C π [0, Γn] into the sum of C π [0, Γn −1] and C π [Γn −1, Γn]
and then carrying out a simple induction on n.
Consider now an arbitrary T >Γ 0. Let θ (T ) be the index of the last
completed segment under π * by time T; and correspondingly, define
χ (T )≔max{j ≥ 0: Γθ (T ) + j wθ (T )+1 ≤ T}. Also, denote Γθ (T ) + χ (T )wθ (T )+1
lθ (T ) ≤ T < l
Γθ (T ) . Then, by definition, we have Γθ (T ) ≤ Γ
Γθ (T ) + wθ (T )+1.
by l
Since αθ (T )+1 is no greater than αθ (T ) (which, in turn, is no greater than
any αn for 0 ≤ n ≤ θ (T )), it immediately follows that16
*
*
lθ (T )]
C π [0, Γ
C π [0, Γθ (T )]
≤
l
Γθ (T )
Γθ (T )
(31)
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S. Perera et al.
Appendix A. Well-known examples of ordering-cost functions that are not increasing.
Fig. A.1. Ordering-cost function with batch-ordering constraint.
Fig. A.2. Ordering-cost function with MOQ constraint.
Fig. A.3. Ordering-cost function with all-unit discounts.
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S. Perera et al.
Appendix B. Proof of Theorem 1 with an increasing c(·) and no backordering
Recall that Π 0p denotes the class of policies that place an order only when the inventory level decreases to zero. Clearly, Lemma 1 remains valid
after replacing Π 0 by Π 0p . The proof for this special case is well known and much shorter, as we do not have the complications that arise from
allowing backorders.
Let π be a policy in Π⧹Π 0p . Then, π must place at least an order at some epoch t with a positive I π (t ); clearly, this is equivalent to the existence of
π
xi > 0 for some i ≥ 1. Let π∼ be the policy constructed from π by setting:
∼
tnπ = tnπ +
∼
tnπ
=
tnπ
xnπ
λ
and
∼
if xnπ > 0;
qnπ ,
otherwise.
and qnπ = qnπ ,
∼
qnπ
=
and
An illustration of the inventory trajectories under both π and π∼ is shown in Fig. B.1. It is easily seen from this figure (see the shaded region) that
the cumulative sum of holding and shortage costs under π∼ over time is dominated by that under π . Moreover, it is clear that for a given T ≥ 0 , every
order placed within [0, T ] under π∼ corresponds to an order placed under π within the same interval (and furthermore π may have additional orders
in [0, T ], e.g., the order at epoch t6π in Fig. B.1). It follows that the total ordering cost under π∼ is no greater than that under π within [0, T ]. Taken
∼
∼
together, these two observations imply that C π [0, T ]/ T ≤ C π[0, T ]/ T for all T ≥ 0 . By letting T → ∞, we then have f π ≤ f π .
π
p
Now, consider a policy π ∈ Π 0 . Let T be a time epoch that is sufficiently large so that N (T ) ≥ 2 ; then, the total cost under π in [0, T ] can be
written as:
N π (T )−1
C π[0, T ] = C π[0, t1π ) +
∑
C π[tiπ , tiπ+1) + C π[tNπ (T ), T ] .
(B.1)
i =1
This cost decomposition is illustrated in Fig. B.2.
Now, it is immediate that
C π[tiπ , tiπ+1) = α(0, yiπ )(tiπ+1 − tiπ ) ≥ α*(tiπ+1 − tiπ ).
(B.2)
Moreover, since c(·) and g(·) are increasing functions, we see from the shaded area in Fig. B.2 that if T −
similar to tNπ (T ) , we have C π[tNπ (T ), T ]≥α(0, yNπ(T ) − I π (T ))(T − tNπ (T )) and hence
C π[tNπ (T ), T ] ≥ α* (T − tNπ (T )).
Notice in addition that this lower bound continues to hold even if T −
that
tNπ (T )
is positive, then, with
yNπ(T )
defined
(B.3)
π
= 0 . Since C [0,
t1π )
≥ 0 and
Fig. B.1. Construction of a policy π∼ from a given π ∈ Π⧹Π 0p .
Fig. B.2. Cost decomposition over [0, T ] for π ∈ Π 0p .
225
tNπ (T )
t1π
= I0 / λ , we then conclude from (B.1)–(B.3)
International Journal of Production Economics 187 (2017) 216–228
S. Perera et al.
C π[0, T ] ≥ α* (T − I0 / λ ).
(B.4)
Dividing both sides of (B.4) by T and then letting T → ∞ now yield f π ≥ α*, completing the proof for this special case of Theorem 1.
Finally, we note that our proof in this section is similar in spirit to the approach in Lippman (1971). Lippman studies a single-product, singlelocation model without backordering. The holding cost is linear, and the ordering-cost structure allows multiple set-up costs. Under these specific
cost assumptions, Lippman proves (s, S )-optimality using a lower-bounding argument. Since the multiple set-up cost function is increasing and left
continuous, (B.4) and our Theorem 2 also imply that an optimal (s, S ) policy exists in Lippman's model.
Appendix C. A simple proof of the standard EOQ formula with backorder
We first recall that the cost assumptions in the standard EOQ model with backorder are: c(q ) = K 1{q >0}+ νq and g(x ) = h max(0, x ) + b max(0,−x )
for some constants K > 0 , ν ≥ 0 , h > 0 , and b > 0 . Now, for a given policy π , let Q π [0, T ] and ηπ [0, T ] be the cumulative order quantities and,
respectively, the total number of orders placed under π in the time interval [0, T ]. Then,
ν Q π [0, T ] + Kηπ [0, T ] + H π [0, T ]
T
T →∞
ν Q π [0, T ] + Kηπ [0, T ] + H π [0, T ]
≥ lim inf
T →∞
T
Kηπ [0, T ] + H π [0, T ]
Q π [0, T ]
≥ ν lim inf
+ lim inf
.
T →∞
T →∞
T
T
f π = lim sup
(C.1)
Next, we claim that we can restrict attention to policies that satisfy
lim inf
T →∞
Q π [0, T ]
≥ λ.
T
(C.2)
The intuition behind this claim is obvious: If the long-run rate of procurement is less than the demand rate, then the amount of backorder will
diverge to infinity. This is formalized in the following lemma.
Lemma 3. The long-run average cost f π for any policy π that violates inequality (C.2) is infinite.
Proof. Suppose lim infT →∞Q π [0, T ]/ T < λ for a given policy π . Then, by the definition of lim inf , there exist an ε ∈ (0, λ ), an n 0 > 0 , and an increasing
π
π
π
sequence {Tn}∞
n=1 with lim n →∞Tn = ∞ such that Q [0, Tn ]/ Tn < λ − ε for all n ≥ n 0 . Observe that I (Tn ) = I0 + Q [0, Tn ] − λTn ; and this, together with
π
π
the previous inequality, implies that I (Tn ) < I0 − εTn for all n ≥ n 0 . (Hence, I (Tn ) diverges to −∞ as n → ∞.) Suppose further that n (≥n 0 ) is
sufficiently large so that I0 − εTn is strictly negative and Tnl ≔Tn − (εTn − I0 )/ λ = (1− ε / λ )Tn + I0 / λ is non-negative. The time epochs Tnl and Tn are
illustrated in Fig. C.1. From this figure, we also see that (εTn − I0 )/ λ is the amount of time necessary for the net-inventory level to drop down to
I0 − εTn , assuming that it starts from level 0 at time Tnl and no future orders are placed. Now, consider the time interval (Tnl, Tn]. Clearly, the
cumulative shortage cost in [0, Tn] is no less than that within the subinterval (Tnl, Tn]. Moreover, the latter cost is, in turn, greater than
b(1/2)(εTn − I0 )((εTn − I0 )/ λ ), which is the total shortage cost in (Tnl, Tn] if the inventory level under π were zero at epoch Tnl and no future orders were
placed. (The diagonal border of the shaded area in Fig. C.1 always lies entirely above the inventory trajectory under π in (Tnl, Tn].) It follows that
C π[0, Tn]/ Tn > b(1/2)(ε − I0 / Tn )((εTn − I0 )/ λ ). Finally, letting n → ∞ in this inequality yields that f π = ∞.
□
From Lemma 3, we have that the first limit in (C.1) is no less than νλ . We now develop a lower bound for the second limit in (C.1). Observe that
this limit can be interpreted as the objective function in a standard EOQ model with ν = 0 . Denote by α0(x, y ) the function (2) for this special case. It
is easily shown that α0(x, y ) attains its minimum at s0*= − 2λKh /[b(b + h )] and S0*= 2λKb /[h(b + h )] . We will next show that, for any π ∈ Π (not
necessarily limited to Π 0),
Fig. C.1. Lower bound for backordering cost over [0, Tn].
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International Journal of Production Economics 187 (2017) 216–228
S. Perera et al.
Fig. C.2. Cost decomposition over [0, T ].
lim inf
T →∞
Kηπ [0, T ] + H π [0, T ]
≥ α0*≔α0(s0*, S0*).
T
(C.3)
The argument relies on a lower bound that is similar in spirit to (B.4).
For a given policy π , denote by C0π the cumulative cost function under the assumption that ν = 0 . Then, (B.1) again holds18 with C0π replacing C π ;
and this is illustrated in Fig. C.2.
It is easily seen from Fig. C.2 and (2) that, for 1 ≤ i ≤ N π (T ) − 1,
C0π[tiπ , tiπ+1) = K +
∫x
yiπ
π
i +1
g (z )
dz
= α0(xiπ+1, yiπ )(tiπ+1 − tiπ ) ≥ α0*(tiπ+1 − tiπ ).
λ
Now, consider the particular scenario in which both t1π and T − tNπ (T ) are positive. Then, similar reasoning also shows that
C0π[tNπ (T ), T ] = α0(I π (T ), yNπ (T ) )(T − tNπ (T )) ≥ α0*(T − tNπ (T )); and that C0π[0, t1π ) = α0(x1π , I0 )(t1π −0) − K ≥ α0* t1π − K . Consequently, from the modified (B.1),
we have
Kηπ [0, T ] + H π [0, T ] = C0π[0, T ] ≥ α0* T + γ ,
(C.4)
may not be positive, there are three other scenarios. It is easily seen that for all scenarios, the constant γ
where γ = − K . Since and/or T −
equals −K , 0, or K. Dividing both sides of (C.4) by T and letting T → ∞ now yield (C.3).
*
From (C.1)–(C.3), we now have f π ≥ νλ + α0*. It is easily verified that f π0 = νλ + α0*, where π0* denotes the (s0*, S0*) policy. Therefore, π0* is
optimal over Π .
t1π
tNπ (T )
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