Accepted Manuscript
Bounds for the solution to the single-period inventory model with compound renewal process input: an application to setting credit card limits J.K. Budd, P.G. Taylor PII: DOI: Reference:
S0377-2217(18)30947-0 https://doi.org/10.1016/j.ejor.2018.11.022 EOR 15471
To appear in:
European Journal of Operational Research
Received date: Revised date: Accepted date:
5 September 2017 25 September 2018 7 November 2018
Please cite this article as: J.K. Budd, P.G. Taylor, Bounds for the solution to the single-period inventory model with compound renewal process input: an application to setting credit card limits, European Journal of Operational Research (2018), doi: https://doi.org/10.1016/j.ejor.2018.11.022
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Highlights
renewal input. • These bounds are easily calculated numerically.
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• We bound the solution of the newsvendor problem with compound
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• The bounds are close for reasonable distribution choices.
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J.K. Budd∗, P.G. Taylor
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Bounds for the solution to the single-period inventory model with compound renewal process input: an application to setting credit card limits
School of Mathematics & Statistics, The University of Melbourne, Parkville 3010, Australia
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Abstract
Motivated by a desire to calculate the optimal credit limit for a credit card account in terms of the card-holder’s purchasing behaviour, we consider a single-period inventory model in which the total value of all attempted
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purchases increases in jumps throughout the period. If a purchase does not cause the limit to be exceeded, then that purchase is approved and the total value of approved purchases is increased by its value. On the other hand,
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if the limit is exceeded, then the purchase is rejected and the total value of approved purchases remains at its previous level.
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We derive an equation for the Laplace-Stieltjes transform of the expected total value of approved purchases at the end of the period. Unfortunately
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this equation appears to be intractable, a situation that we address by providing upper and lower bounds. We provide numerical examples that show that these upper and lower bounds can be close, which leads to good upper
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and lower approximations for the optimal limit. Keywords: Applied probability; Compound renewal process; Newsvendor ∗
Corresponding author Email addresses:
[email protected] (J.K. Budd),
[email protected] (P.G. Taylor)
Preprint submitted to Elsevier
Tuesday 13th November, 2018
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model; Credit card limits.
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1. Introduction In the newsvendor or single-period inventory (SPI) model, a seller holds
a quantity ` > 0 of inventory and in a single period of length T will receive total demand D, a non-negative random variable with distribution function FD . It costs the seller ν`, ν > 0 to acquire an amount ` of inventory and
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each unit sold earns a profit γ > 0. We assume that γ > ν: if that were
not the case, the vendor would make a loss on every item of inventory that he/she sells.
There are now two possibilities: either the total demand is less than or equal to the inventory held, or it exceeds the inventory. In the former
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case, the seller earns a random profit R(`) = γD − ν`. In the latter case, it is typical to assume that the entire inventory is sold, and so the profit is
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R(`) = (γ − ν)`. The expected profit as a function of the inventory held is thus
E [R(`)] = γ E [D ∧ `] − ν`
(1)
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and the associated optimisation problem is now to determine n o `ˆ := arg max γ E [D ∧ `] − ν` ,
(2)
`∈R+
the level of inventory that maximises profit. The solution to the optimisation
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problem described by Equation (2) is well-known to be `ˆ = FD−1 (γ − ν)/γ = inf{x ∈ R+ : FD (x) ≥ (γ − ν)/γ}.
(3)
This is known as the critical fractile solution. For details of its derivation, see Porteus (2002). 3
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However, there are many practical situations in which the assumption that the entire inventory is sold when the total demand exceeds the inventory
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` is not reasonable. The example that motivated this study concerns a model for a credit card account with a credit limit ` (Budd (2017)). In this context, the parameter γ is the revenue per dollar that the card issuer earns from
each approved purchase, and ν is the cost per dollar to the card issuer of providing the limit `. With the value of attempted purchases up to time
t ∈ [0, T ] given by D(t) and the value of approved purchases up to time
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t ∈ [0, T ] denoted by h(D(t), `), the profit earned by the credit card issuer at the end of the period is
R(`) = γh(D(T ), `) − ν`
(4)
which bears a similarity to (1). However, in the credit card example, the
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mechanism for keeping h(D(t), `) below ` differs from that used in the classical SPI model. At any point in t ∈ [0, T ], the credit available to the
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account holder is equal to the difference between ` and h(D(t), `). The classical SPI model corresponds to the situation where h(D(t), `) is taken to be
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hS (D(t), `) ≡ D(t) ∧ `. To implement a credit card account with such a function, the credit card issuer would have to approve an amount equal to
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` − D(t) when the card holder attempts to make a purchase with a value that exceeds ` − D(t). This does not make a lot of sense in the context – it
would require partial payment for a purchase. More realistically, the card
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issuer would employ a demand control that rejects the attempted purchase, leaving the total value of approved purchases at its previous value. There are many other situations in which the rejection policy described
above is a more suitable model than the classical SPI model. Consider a model for shipping consolidation, where items arrive randomly at a depot to 4
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be shipped to a destination and must be packed into a container with finite capacity `. If the random volume of an arriving item exceeds the available
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capacity in the container, then it may be rejected from the current container in favour of smaller items that are yet to arrive. In computing, files may
be randomly loaded into a buffer of memory of capacity `. If a file’s size
exceeds the available capacity of memory, then it will not be loaded, but subsequent attempts using smaller files may be accepted.
Suppose now that the card-issuer has some knowledge of the process
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D(t) that captures the card holder’s demand for credit, perhaps obtained by analysis of past purchasing behaviour. It would be of interest to the card-
ˆ the profit-maximising credit limit which could then be issuer to calculate `, assigned or offered to the card-holder. To do so, we would optimise the expected value of (4) with respect to `.
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Intuitively, it is clear that E [D ∧ `] ≥ E [h(D, `)], and that calculating `ˆ using the newsvendor model will underestimate the true optimal limit. In
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this paper, we attempt to find an expression for the Laplace transform of the tail function
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S(y; D(T ), `) ≡ P (h(D(T ), `) ∈ [y, `])
(5)
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of h(D(t), `) when the cumulative demand D(t) is a compound renewal process and where purchases that cause h(D(t), `) to exceed ` are rejected. We use an approach similar to the method used in Chiera and Taylor (2002),
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which exploits the Markovian nature of the process embedded at purchase instants, conditioning on the time and value of the first jump in the process. Ultimately, we show that the equation that this approach leads to is intractable, but using the same method, we derive the Laplace transform of the tail function of the total value of approved purchases in a related system 5
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in which all subsequent purchases are rejected after the first time the total value of attempted purchases exceeds `. This is a random variable which
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can be expressed in the form hR (D(t), `) ≡ sup {D(t) : D(t) ≤ `}, 0≤t≤T
(6)
which is the difference between ` and the undershoot of a compound renewal process at the first time it exceeds a level `.
We derive expressions for the Laplace transform of the expectation of
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(6) and its derivative. We then show that the expectation of h(D(t), `) is
bounded below by the expectation of (6) and above by the expectation of D(T ) ∧ `. These expressions allow us to solve two optimisation problems which provide upper and lower bounds on the optimal credit limit when demand can be rejected, effectively solving the problem of calculating `ˆ
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under the demand control with random rejection.
The newsvendor model has been studied extensively since its modern
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formulation in Arrow et al. (1951). The model has found myriad applications in inventory management and other fields and has a multitude of
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extensions and modifications. Khouja (1999) provides an excellent survey of the existing extensions of the model. A more recent survey is given in Choi
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(2012).
Related to our current paper is the analysis of Grubbstr¨om (2010). In
this paper he showed how to derive the optimal order quantity when demand
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arrives according to a compound renewal process with inter-event density x and purchase value density y and a discount factor is applied to the value of
sales as time elapses. The key to the analysis of Grubbstr¨om (2010) is the observation that the optimal order quantity can be calculated by solving a classical newsvendor problem with a particular form of demand distribution 6
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expressed as an inverse Laplace transform of a function involving the Laplace transforms of x and y. We also model the arrival of demand as a compound
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renewal process, but we stop the process at a fixed time T and do not employ a discount factor.
The undershoot and overshoot of a process with independent and identically distributed jumps is well-studied. Classical results for the joint dis-
tribution of the undershoot and overshoot using renewal theory are given in (Feller, 1971, Chap. 11). The undershoot and overshoot are related to the
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backward and forward recurrence times in a renewal process, and the joint distribution of these times as t → ∞ was studied in Winter (1989). Expres-
sions for the limiting joint distributions as ` → ∞ of the undershoot and overshoot of a subordinator (a non-decreasing L´evy process) at first passage were derived in Bertoin et al. (1999). For a concise and modern treatment
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of the first passage of subordinators, see (Kyprianou, 2014, Chap. 5).
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A similar quantity to (6) is the supremum functional sup D(u),
t > 0,
(7)
0≤u≤t
which was first studied in Kac and Pollard (1950) in relation to the distri-
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bution of the maximum of partial sums of independent random variables. The case where D(t) is a separable stochastic processes with stationary and
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independent increments was studied in Baxter and Donsker (1957), where the authors determined the double Laplace-Stieltjes transform of the distri-
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bution function σ(x, t) = P
sup D(u) ≤ x .
0≤u≤t
(8)
This result was investigated further and extended in several papers, most notably Pyke (1959), Tak´ acs (1965), Bingham (1973), and Heyde (1969). Though the random variable (7) is well-studied, the authors are not aware 7
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of a place where expressions for Laplace transform of the related random variable (6) have appeared. Furthermore, while rejection policies have also
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been investigated previously in the context of queueing theory (see, for example Perry and Asmussen (1995)), we believe that the bounds on the optimal limit that we derive are novel in the inventory-model setting.
2. Laplace transforms for the distribution of fulfilled demand
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In this section, we study the tail function defined in Equation (5). 2.1. Assumptions on the demand process
Let A(t), 0 ≤ t ≤ T denote the process of cumulative demand for some resource in a single period of length T . We assume that A(t) is a compound renewal process
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N (t)
A(t) =
X
ξj ,
(9)
j=1
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where {ξj }, i ≥ 1 is a sequence of non-negative, independent random variables with common distribution function F and N (t) is a random variable,
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independent of ξj , describing the number of events in (0, t] in a renewal process with inter-event time distribution G. For k = 1, 2, . . ., we write tk = inf{t : N (t) = k} and τk = tk − tk−1 , with t0 = 0.
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We further assume that both F and G are of exponential order, which
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is sufficient to ensure the existence of the Laplace-Stieltjes transforms f˜(θ) =
Z
∞
−θz
e
F (dz)
and g˜(ω) =
0
Z
∞
e−ωu G(du)
(10)
0
for Re(θ) > σF and Re(ω) > σG , where the respective abscissae of conver-
gence σF and σG of f˜ and g˜ are strictly less than zero. Further details on the sufficiency of this condition can be found in Doetsch (1974). 8
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2.2. Demand control 1: the SPI model
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We first consider the SPI model when the cumulative demand process is described by Equation (9). Let hS A(t), ` ≡ hS (t, `) be the demand fulfilled up to time t, 0 < t ≤ T and set hS (0, `) = 0. Set A(0) = 0 and
suppose the demand for inventory arrives at the random times 0 < t1 < · · · < tj < . . . tN (T ) < T , N (T ) ≥ 1. At the instants tj , 1 ≤ j ≤ N (T ), the seller applies a demand control which may be described as follows:
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Demand Control 1 (DC1). If the sum of the fulfilled demand hS (tj , `) and
the amount of demand ξj arriving at time tj exceeds the inventory `, then − accept an amount x = ` − hS (t− j , `) ≤ ξj , where, t = lim↓0 t − .
Under this demand control, we know that hS (t, `) = A(t) ∧ ` and that
we only need consider the distribution function of hS (T, `) in order to solve
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the optimisation problem (2). Indeed, the solution may be found by differentiating Equation (1), setting the left-hand side to 0 and re-arranging for
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`. Since
∂ ∂ E [A(T ) ∧ `] = ∂` ∂`
Z
`
P (A(T ) > y) dy = P (A(T ) > `) ,
(11)
0
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we only require an expression for the tail function of the random variable A(T ). In the case that A(t), 0 ≤ t ≤ T is a compound renewal process, the
P (A(T ) > y) =
∞ X
n=1
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tail function is given by pn (T ) 1 − F ∗n (y) ,
y ≥ 0,
(12)
where pn (T ) = P (N (T ) = n) = G∗(n+1) (T ) − G∗n (T ). Since evaluation of Equation (12) requires the evaluation of convolutions of F and G, it may be
easier to work with the Laplace transform Z ∞Z ∞ S ˜ H (ω, θ) ≡ e−(ωt+θ`) P (A(T ) > `) d` dt. 0
0
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(13)
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Multiplying Equation (12) by e−(ωt+θ`) and integrating over [0, ∞) × [0, ∞) yields (after some rearrangement)
Observe that since E hS (T, `) =
Z
∞
P (A(T ) > `) d`,
0
(14)
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1 g˜(ω) 1 − f˜(θ) S ˜ H (ω, θ) = . θω 1 − f˜(θ)˜ g (ω)
(15)
we may obtain the two-dimensional Laplace transform of the expectation of
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hS (T, `) by simply dividing Equation (14) by θ to yield 1 ˜S 1 g˜(ω) 1 − f˜(θ) H (ω, θ) = 2 . θ θ ω 1 − f˜(θ)˜ g (ω)
(16)
Equations (14) and (16) can be inverted analytically (if possible) or numeri-
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cally using, for example, the algorithms detailed in Abate and Whitt (1992), Abate and Whitt (1995) or Abate et al. (1998).
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2.3. Demand control 2: rejection with retrial In the previous section, we showed that when the demand was described
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by a compound renewal process with mark distribution F and inter-event time distribution G, an expression for the (two-dimensional) Laplace trans-
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form of the tail function of A(T ) could be found in terms of the Laplace transforms f˜ and g˜. In this section, we attempt to find a similar expression for the Laplace transform of the tail function of the fulfilled demand when
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the demand control is described by Demand Control 2 (DC2). The demand arriving at time tj , 1 ≤ j ≤ N (T ) will be fulfilled if and only if the sum of the demand already satisfied and the size of the arriving demand ξj is less than the total inventory `. 10
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Let h A(t), ` ≡ h(t, `) denote the cumulative demand already fulfilled
by time t when the total available inventory is `. We proceed using a similar
` and t > 0, we have four possibilities to consider i. No jumps occur in the interval [0, t);
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approach detailed in Chiera and Taylor (2002). Set h(0, `) = 0. For 0 < y ≤
ii. The process jumps to some z ∈ (0, y] at time u ∈ (0, t) and then re-
generates itself at this point. That is, a new process starts at z that
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behaves like the original one, but shifted by ` − z in space and t − u in time.
iii. The process jumps to some z ∈ (y, `] at time u ∈ (0, t). Note that any subsequent jumps that happen in the remaining time interval (u, t)
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cannot take the process out of the interval (y, `].
iv. The process attempts to jump to some z ∈ (`, ∞) at time u ∈ (0, t). If
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this occurs, the process regenerates from 0 with time t − u remaining for further jumps to occur.
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With τ = τ1 and ξ = ξ1 we combine the cases above to derive 0, τ >t E 1{h(t,`)−h(τ,`)∈(y−ξ,`−ξ]} , τ ≤ t, ξ ≤ y E 1{h(t,`)∈(y,`]} | τ, ξ = 1, τ ≤ t, y < ξ ≤ ` E 1{h(t,`)−h(τ,`)∈(y,`]} , τ ≤ t, ξ > `.
(17)
For 0 < ξ ≤ y, the distribution of h(t, `) − h(τ, `) is the same as that of h(t − τ, ` − ξ) by the regenerative property mentioned in the second 11
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case above. For ξ > `, the distribution of h(t, `) − h(τ, `) is equal to the
>t
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distribution of h(t − τ, `). So we have 0, τ E 1{h(t−τ,`−ξ)∈(y−ξ,`−ξ]} , τ E 1{h(t,`)∈(y,`]} | τ, ξ = 1, τ E 1{h(t−τ,`)∈(y,`]} , τ
≤ t, ξ ≤ y
≤ t, y < ξ ≤ ` ≤ t, ξ > `.
(18)
S(y, t, `) =
Z tZ
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Set S(y, t, `) ≡ P h(t, `) ∈ (y, `] . By the law of total probability, y
H(y − z, t − u, ` − z)F (dz) G(du) Z t + G(t) F (`) − F (y) + 1 − F (`) S(y, t − u, `) G(du). (19) 0
0
0
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It follows that |S(y, t, `)| ≤ G(t), which shows that S(y, t, `) is of exponential order with respect to t when G is of exponential order. Thus, the one-
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dimensional Laplace transform
˜ H(y, ω, `) ≡
Z
∞
e−ωt S(y, t, `) dt
(20)
0
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exists for all ω with Re(ω) > σG . Furthermore,
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˜ H(ψ, ω, θ) ≡
Z
0
∞Z ∞Z ` 0
e−(ωt+θ`+ψy) S(y, t, `) dy d` dt
(21)
0
exists for all ω with Re(ω) > σG , and θ and ψ with Re(θ) > 0 and Re(ψ) > 0.
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So, by multiplying the right-hand side of Equation (19) by e−(ωt+θ`+ψy) and
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integrating over (0, `] for y and (0, ∞) for ` and t, we have Z ∞Z ∞Z ` ˜ e−(ωt+θ`+ψy) H(ψ, ω, θ) = 0 "0Z Z0 0
y
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t
S(y − z, t − u, ` − z)F (dz) G(du)
0
+ G(t) F (`) − F (y) # Z t + 1 − F (`) S(y, t − u, `) G(du) dy d` dt. 0
(22)
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When ψ, ω and θ all have a strictly positive real part, we may integrate Equation (22) term by term and change the order of integration to arrive at 1 ˜ ˜ H(ψ, ω, θ) = g˜(ω)f˜(θ + ψ)H(ψ, ω, θ) + g˜(ω) f˜(θ) − f˜(θ + ψ) θωψ Z ∞Z ∞Z ` −(ωs+θ`+ψy) + g˜(ω) e S(y, s, `) 1 − F (`) dy d` ds . 0
0
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0
(23)
Note that the integral in the third term is “almost” the Laplace transform
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˜ H(ψ, ω, θ), but the term 1 − F (`) complicates the expression. We can proceed further by examining a special case. If we assume that
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ξ is exponentially distributed with parameter µ, then 1 − F (`) = e−µ` , so Equation (23) becomes
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˜ ˜ H(ψ, ω, θ) = g˜(ω)f˜(θ + ψ)H(ψ, ω, θ) + ˜ + g˜(ω)H(ψ, ω, θ + µ).
1 g˜(ω) f˜(θ) − f˜(θ + ψ) θωψ
(24)
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˜ Rearranging leads to a functional equation for H, ˜(θ) − f˜(θ + ψ) ˜ g ˜ (ω) f g ˜ (ω) H(ψ, ω, θ + µ) 1 ˜ H(ψ, ω, θ) − = , θωψ 1 − g˜(ω)f˜(θ + ψ) 1 − g˜(ω)f˜(θ + ψ)
(25)
˜ occurs on the left hand side with two where f˜(θ) = µ/(µ + θ). Note that H different values of its third argument. It is not clear to the authors how to 13
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˜ proceed further from here to obtain an expression for H(ψ, ω, θ) in terms of
2.4. Demand control 3: rejection without retrial
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f˜ and g˜.
The third term in Equation (23) prevents us from proceeding any further.
We may remove it by changing the fourth case in Equation (17) and (18) from E 1{h2 (t−τ,`)∈(y,`]} to 0. This results in a new demand control: Demand Control 3 (DC3). If the sum of the fulfilled demand hR (tj , `)
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and the amount of demand ξj arriving at time tj exceeds the inventory `, reject the demand and any further demand for the remainder of the period. Setting hR (t0 , `) = hR (0, `) = 0, the total demand fulfilled under this control can be expressed as
hR (t, `) = sup A(u) : A(u) ≤ ` .
(26)
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0≤u≤t
In order to find the Laplace transform of the tail function (27)
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S R (y, t, `) = P hR (t, `) ∈ (y, `] ,
we can proceed as we did with DC2 and obtain the three-dimensional trans-
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form of the tail function of the fulfilled demand under DC3, ˜(θ) − f˜(θ + ψ) g ˜ (ω) f 1 R ˜ (ψ, ω, θ) = H , θωψ 1 − g˜(ω)f˜(θ + ψ)
(28)
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for ω, θ and ψ with strictly positive real part.
To calculate the (two-dimensional) Laplace transform of the expectation
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of hR (t, `), we note that Z ∞Z ∞ e−(ωt+θ`) E hR (t, `) d` dt 0 0 Z ∞Z ∞ Z ` = e−(ωt+θ`) S R (y, t, `)dy d` dt 0
0
0
˜R
= H (0, ω, θ). 14
(29)
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We apply L’Hospital’s rule to Equation (28) to obtain 1 g˜(ω) f˜(θ) − f˜(θ + ψ) lim H (ψ, ω, θ) = lim ψ→0 θωψ ψ→0 1 − g˜(ω)f˜(θ + ψ) ˜R
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d ˜ g˜(ω) f (θ). (30) θω 1 − g˜(ω)f˜(θ) dθ It should be noted that the derivative of E hR (t, `) with respect to ` may not exist. Indeed, if F is lattice, then E hR (t, `) will be a step function. =−
In the case where the derivative does exist, we obtain its Laplace transform
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by multiplying Equation (30) by θ to yield ˜ R (ψ, ω, θ) = − θ lim H ψ→0
g˜(ω) d f˜(θ). ω 1 − g˜(ω)f˜(θ) dθ
(31)
2.5. Discussion of the demand controls
Figure 1 shows that the paths of the processes hS (t, `) and hR (t, `) bound
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the paths of the process h(t, `), where 0 ≤ t ≤ T . Indeed, for 0 < ` < ∞
Figure 1: Realisations of the attempted purchase process A(t) and the processes induced
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by the demand controls DC1 — DC3. The process h(t, `) is bounded above and below by
hS (t, `) and hR (t, `) respectively.
and 0 ≤ t ≤ T , if the process A(t) < `, then A(t) = h(t, `) = hS (t, `) = hR (t, `). 15
(32)
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On the other hand, if A(t) ≥ `, we will have A(t) ≥ hS (t, `) ≥ h(t, `) ≥ hR (t, `),
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from which it follows that
E [A(t)] ≥ E hS (t, `) ≥ E [h(t, `)] ≥ E hR (t, `) .
Let
(33)
n o `ˆ := arg max γ E [h(t, `)] − ν` , `∈R+
(34)
(35)
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with analogous definitions for `ˆS and `ˆR . Then it follows from the inequality (34) that
`ˆS ≤ `ˆ ≤ `ˆR ,
(36)
which implies that if we are able to solve the optimisation problem (35) with h(t, `) replaced by hS (t, `) and hR (t, `) respectively, then we will obtain
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upper and lower bounds to the solution of (35).
For unusual purchase size distributions, the gap between the bounds
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on h(t, `) given by DC1 and DC3 can be arbitrarily large. For example, if the purchase size distribution concentrates its mass above the limit `
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then the balance given by DC1 is ` while the balance given by DC3 is 0. Indeed, the difference between the upper and lower bounds is 0 if the
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underlying compound process remains under the limit, or it is equal to the undershoot of the process if the input compound process exceeds the limit. Determining the tightness of the bounds for E [h(t, `)] is equivalent to
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computing the backward recurrence time for a renewal process with interarrival distribution identical to the mark distribution. In general, this is a difficult task, but asymptotic results are well-known. In solving the optimisation problem in Equation (35) under either DC1 or DC3, we take the distributions F and G as given, calculate E [h(t, `)] 16
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and then optimise with respect to `. As we observe in Section 3 below, this optimisation appears to be analytically intractable even when F and G are
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simple distributions, although we can solve it numerically. This makes it difficult to derive a priori estimates of the tightness of the bounds analyti-
cally. However, as we show with the examples in Section 3, with ‘reasonable’ purchase size distributions, the bounds tend to be close.
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3. Numerical examples
In this section, we used the results from the previous section to calculate bounds on the optimal limit under DC2. We provide two examples in which the demand is described by a compound Poisson process with rate λ and marks that are 1) exponentially distributed with parameter µ and, 2)
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uniformly distributed on the interval [0, b]. 3.1. Exponentially distributed marks
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For a compound Poisson process with rate λ and exponentially distributed marks, the Laplace transforms of the inter-arrival time distribution
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and the mark size distribution are λ µ , Re(ω) > −λ and f˜(θ) = , Re(θ) > −µ, λ+ω µ+θ
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g˜(ω) =
(37)
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respectively. Equation (16) becomes 1 ˜S λ H (ω, θ) = θ ω θ(λ + ω) + µω
(38)
which can be inverted once to yield
! 1 ˜S 1 µ H (t, θ) = 2 1 − exp λt −1 . θ θ µ+θ
17
(39)
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Similarly Equation (31) now becomes ˜ R (ψ, ω, θ) = θ lim H ψ→0
λµ , ω(θ + µ) µω + θ(λ + ω)
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which can also be inverted analytically to yield ! µ µ 1 R 1 − exp λt θ lim H (ψ, t, θ) = −1 . ψ→0 θ µ+θ µ+θ
(40)
(41)
It does not appear to be easy to perform further analytical inversion of either Equations (39) or (41) with respect to θ, since the inverse transform of the
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exponential of a rational function is not a standard transform. As such,
we resorted to numerical inversion using the EULER algorithm as detailed in Abate and Whitt (1995).
We calculated the upper and lower bounds on the optimal limit using γ = 0.0054, ν = 0.0007 and T = 30. We took Λ = (0, 20 000] and used a
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bisection search to solve Equation (35). Table 1 shows the optimal limits calculated for a range of values of λ and µ. An important factor in this analysis is the probability that the demand
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ˆ The probability of demand being rejected will exceed the optimal limit `. is given by the tail function of A(T ). We obtained its Laplace transform by
(42)
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multiplying Equation (39) by θ to find ! 1 µ ˜ A(θ) = 1 − exp λT −1 . ψ µ+θ
The results of numerically inverting Equation (42) at the optimal limit bound corresponding to DC1 is constant given by the ratio of ν to γ. The
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corresponding results for the optimal limits calculated under DC3 are presented in Table 2. They show that the probability of demand being rejected remains constant when the rate parameter of the mark size distribution changes. Indeed, by scaling the mark size distribution by some α > 0, we scale the input process A(t) and, as evidenced by Table 1, the optimal limit. 18
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1/µ 20
λ
3
4
5
80
100
[777, 798]
[1 554, 1 596]
[2 330, 2 395]
[3 107, 3 193]
[3 884, 3 991]
600
1 200
1 800
2 400
3 000
[1 449, 1 470]
[2 899, 2 941]
[4 348, 4 411]
[5 798, 5 882]
[7 247, 7 352]
1 200
2 400
3 600
4 800
6 000
[2 105, 2 126]
[4 210, 4 252]
[6 315, 6 378]
[8 420, 8 503]
[10 525, 10 629]
1 800
3 600
5 400
7 200
9 000
[2 752, 2 773]
[5 504, 5 545]
[8 256, 8 318]
[11 008, 11 091]
[13 760, 13 863]
2 400
4 800
[3 393, 3 414]
[6 786, 6 828]
3 000
6 000
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2
60
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1
40
7 200
9 600
12 000
[10 180, 10 242]
[13 573, 13 655]
[16 966, 17 069]
9 000
12 000
15 000
Table 1: Upper and lower bounds on the optimal limit when A(t) is a compound Poisson process with rate λ and exponentially distributed marks with parameter µ. The values
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were calculated using T = 30, γ = 0.0054 and ν = 0.0007. The lower and upper bounds result from calculating the optimal limit under DC1 and DC3 respectively. The expected
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value E [A(T )] is given immediately below each set of bounds.
λ
1/µ
20
40
60
80
100
1
0.105 9
0.105 9
0.105 9
0.105 9
0.105 9
2
0.112 2
0.112 2
0.112 2
0.112 2
0.112 2
3
0.115 1
0.115 1
0.115 1
0.115 1
0.115 1
4
0.116 9
0.116 9
0.116 9
0.116 9
0.116 9
5
0.118 2
0.118 2
0.118 2
0.118 2
0.118 2
Table 2: Probability that demand will exceed the assigned optimal limit under DC3. These results were obtained using T = 30, γ = 0.0054, and ν = 0.0007. Under DC1, the decline probabilities are constant at ν/γ = 0.129 6
19
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3.2. Uniformly distributed marks For a compound Poisson process with rate λ and marks uniformly and
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continuously distributed on the interval (a, b), the Laplace transforms of the inter-arrival time distribution and mark size distribution are g˜(ω) =
λ , Re(ω) > −λ λ+ω
e−aθ − e−bθ and f˜(θ) = , Re(θ) 6= 0, θ(b − a)
(43)
while Equation (31) becomes
˜ R (ψ, ω, θ) = θ lim H ψ→0
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respectively. Note that f˜(0) ≡ 1. Setting a = 0, Equation (16) now becomes λ ebθ (bθ − 1) + 1 1 1 ˜S H (ω, θ) = 2 (44) θ θ ω λ + ebθ λ(bθ − 1) + bθω λ 1 − ebθ + bθ . ω λ(ebθ − 1) − bθebθ (λ + ω)
(45)
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Equations (45) are considerably harder to invert and so we used a twodimensional version of the EULER algorithm to obtain results for varying
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values of λ and b. We calculated the upper and lower bounds using the same values for γ, ν and T that were used in the previous example. The results are presented in Table 3. As in the previous example, we also calculated
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the probability that demand will exceed the optimal limit. These results are shown in Table 4.
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We also calculated the optimal limits and probability of exceeding the
optimal limit when the purchase size is discretely distributed on the integers
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between [0, b] for b = 5, 10, 20, 50, 100. The results are shown in Tables 5
and 6. In contrast to the results obtained in the previous example, the probability of demand exceeding the optimal limit in both the continuous and discrete case is not independent of b (and therefore the mean purchase size).
20
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b 5
λ
3
4
5
50
100
[93, 95]
[186, 189]
[372, 379]
[930, 947]
[1 860, 1 893]
75
150
300
750
1 500
[175, 177]
[351, 354]
[702, 708]
[1 754, 1 771]
[3 508, 3 541]
150
300
600
1 500
3 000
[256, 258]
[512, 515]
[1 024, 1 031]
[2 561, 2 561]
[5 121, 5 155]
225
450
900
2 250
4 500
[336, 337]
[672, 675]
[1 343, 1 350]
[3 358, 3 375]
[6 717, 6 750]
300
600
[415, 417] 375
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2
20
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1
10
1 200
3 000
6 000
[830, 833]
[1 660, 1 667]
[4 151, 4 167]
[8 301, 8 334]
750
1 500
3 750
7 500
Table 3: Upper and lower bounds on the optimal limit when A(t) is a compound Poisson process with rate λ with marks distributed uniformly and continuously on the interval
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(0, b). The values were calculated using T = 30, γ = 0.0054 and ν = 0.0007. The lower and upper bounds result from calculating the optimal limit under DC1 and DC3
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respectively. The expected value E [A(T )] is given immediately below each set of bounds.
λ
b
5
10
20
50
100
1
0.110 7
0.110 6
0.110 4
0.110 4
0.110 3
2
0.115 5
0.115 4
0.115 5
0.115 5
0.115 4
3
0.118 0
0.117 8
0.117 9
0.117 8
0.117 8
4
0.119 4
0.119 4
0.119 4
0.119 3
0.119 3
5
0.120 4
0.120 4
0.120 4
0.120 3
0.120 3
Table 4: Probability that demand will exceed the assigned optimal limit under DC3. These results were obtained using T = 30, γ = 0.0054, and ν = 0.0007. Under DC1, the decline probabilities are constant at ν/γ = 0.129 6
21
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b
λ
3
4
5
20
50
100
[111, 112]
[204, 207]
[390, 397]
[948, 964]
[1 878, 1 911]
75
150
300
750
1 500
[209, 211]
[385, 388]
[735, 742]
[1 788, 1 804]
[3 542, 3 575]
150
300
600
1 500
3 000
[306, 307]
[562, 565]
[1 074, 1 081]
[2 610, 2 627]
[5 171, 5 204]
225
450
900
2 250
4 500
[401, 403]
[737, 740]
[1 409, 1 416]
[3 424, 3 441]
[6 782, 6 816]
300
600
[496, 498] 375
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2
10
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1
5
1 200
3 000
6 000
[911, 915]
[1 741, 1 748]
[4 232, 4 248]
[8 382, 8 416]
750
1 500
3 750
7 500
Table 5: Upper and lower bounds on the optimal limit when A(t) is a compound Poisson process with rate λ with marks distributed uniformly and discretely on the interval (0, b).
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The values were calculated using T = 30, γ = 0.0054 and ν = 0.0007. The lower and upper bounds result from calculating the optimal limit under DC1 and DC3 respectively.
AC
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PT
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The expected value E [A(T )] is given immediately below each set of bounds.
λ
b 5
10
20
50
100
1
0.111 1
0.110 9
0.110 6
0.110 5
0.110 4
2
0.116 2
0.115 9
0.115 6
0.115 5
0.115 5
3
0.118 5
0.118 1
0.118 0
0.117 9
0.117 9
4
0.119 8
0.119 6
0.119 4
0.119 3
0.119 3
5
0.120 8
0.120 6
0.120 4
0.120 4
0.120 3
Table 6: Probability that demand will exceed the assigned optimal limit under DC3. These results were obtained using T = 30, γ = 0.0054, and ν = 0.0007.
22
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4. Conclusion In this paper, we have shown that bounds on the solution to the SPI
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model with random rejection can be obtained by solving two related optimisation problems when the demand process is a compound renewal process. The first is the classic SPI model, while the second is a system in which all
subsequent requests for demand are rejected after the first time the cumulative value of demand exceeds the inventory level `. The solution to the
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former model is well-known, and we derived expressions for the solution to
the latter model in terms of the Laplace-Stieltjes transforms of the interarrival time distribution and the mark-size distribution. Both solutions can easily be found numerically.
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Acknowledgments
P. G. Taylor’s research is supported by the Australian Research Council
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(ARC) Laureate Fellowship FL130100039 and the ARC Centre of Excellence for Mathematical and Statistical Frontiers (ACEMS). The authors are also grateful to two anonymous referees, whose comments significantly improved
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the paper.
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