a power-of-two ordering policy for one warehouse, multi-retailer

8/2/2019 A Power-Of-two Ordering Policy for One Warehouse, Multi-Retailer

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A Power-of-two Ordering Policy for One Warehouse, Multi-Retailer

Systems with Stochastic Demand

Leon Yang Chu Zuo-Jun Max ShenDepartment of Industrial & Systems Engineering, University of Florida, Gainesville, FL 32611

Department of Industrial Engineering & Operations Research, University of California, Berkeley, CA 94720

May 15, 2005

Abstract

We study a two-echelon supply chain with one warehouse and N retailers facing stochastic

demand. An easy-to-implement inventory policy, the so-called power-of-two policy, is proposed

to manage inventory for the system. To maintain a certain service level, safety stocks are kept

at the warehouse and each retailer outlet to buffer random demand. Our analysis highlights the

important role of warehouse safety stock level, which, in addition to the length of the warehouse

order interval, signicantly affect the length of the retailers order interval. We develop a

polynomial algorithm to nd a power-of-two ordering policy for arbitrary target service level,

whose cost is guaranteed to be no more than 32 1.26 times the optimal cost. Such a policycan be computed in time O(N 3 ).

1 Introduction

We consider a distribution system with one warehouse and multiple retailers. Random demand

occurs at the retailers only. The warehouse orders from an outside supplier with unlimited supply

and replenishes the retailers inventories. To maintain appropriate service levels, safety stock is

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maintained at the supplier and the retailers. The warehouse and each retailer follow a periodic

review inventory policy with certain service requirements. The planning horizon is innite and the

objective is to minimize the long-run average system-wide cost.

Multi-echelon stochastic inventory models have received much attention in the research com-

munity. In terms of exact results, most papers focus on continuous review policy. For example,

Axsater (2000) provides exact results for compound Poisson demand and nonidentical retailers

when the warehouse and the retailers use the ( R,nQ ) policy. Cheung and Hausman (2000) analyze

exact results for the central warehouse serving nonidentical retailers. Chen and Zheng (1997) pro-

vide exact results when the central warehouse uses continuous review reorder point policy, while

the reorder point is calculated according to echelon stock inventory. For periodic review models,

Liljenberg (1996) and Cachon (2001) provide exact valuation methods for identical retailers under

batch ordering policy.

It is recognized that the structure of an optimal policy for most multi-echelon stochastic inven-

tory systems is either unknown or extremely complex, thus, it is difficult to implement them in

practice. As a result, much research has focused on heuristic policies that are easier to implement.

For excellent reviews of this literature, we refer the readers to Axsater (1993c) and Federgruen

(1993). Recent developments include Lee and Billington (1993), Tempelmeier (1993), Hausman

and Erkip (1994), Chen and Samroengraja (2000).

However, since these policies are only heuristic, a natural question to ask is how much is the gap

between the heuristic solution and the optimal one. The literature only provides some numerical

evidence (e.g., Federgruen and Zipkin 1984a,b and Chen and Zheng 1993b,1994a,b). But no matter

how extensively the numerical experiments have been conducted, there are no guarantees with

respect to the heuristics performance on other problem instances. Therefore, it is much more

desirable to have a heuristic policy with a good worst-case performance, i.e., a good a priori

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guarantee for its performance on any problem instance.

For a single-echelon continuous-review inventory system with stationary stochastic demand,

Zheng (1992) shows that the maximum relative error is bounded by 0.125 when using the deter-

ministic EOQ formula as a heuristic solution in the ( Q, r ) policy. Axsater (1996) gives a slightly

better bound of 0.118. Levi, Roundy, and Shmoys (2004b) analyze stochastic inventory prob-

lems with correlated and non-stationary demands. They provide a 2-approximation algorithm for

the periodic-review stochastic inventory control problem and a 3-approximation algorithm for the

stochastic lot-sizing problem.

Unfortunately, for multi-echelon stochastic inventory systems, we have not seen any work in

this area. In contrast, for a wide range of multi-echelon, deterministic inventory systems, it is well

known that a class of easily implementable policies, the so-called power-of-two policies, perform

surprisingly well (see, e.g., Maxwell and Muckstadt 1985, and Roundy 1985,1986). Under a power-

of-two policy, each item is replenished at constant reorder intervals which are power-of-two multiples

of some xed or variable base planning period. An optimal power-of-two policy can be found very

easily and its cost is guaranteed to be within 2% of optimality. Recent works that analyze worst-case

bound performance of other easy-to-implement policies include Chan, Muriel, Shen, Simchi-Levi,

and Teo (2002), Chen (2000), Levi, Roundy, and Shmoys (2004a), and Shen, Simchi-Levi, and Teo

(2002).

In this paper, we apply the power-of-two policy to OWMR models with stochastic demand. To

ensure a certain service level in this random demand environment, safety stocks are needed at the

warehouse and each retailer site. Inventory replenishment decisions are then based on the average

demand rates, similar to the spirits of Zheng (1992) and Axsater (1996). We develop a polynomial

algorithm to nd a close-to-optimal power-of-two ordering policy for arbitrary target service level.

Our approach highlights the important role of warehouse safety stock level, which, in addition to

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the length of the warehouse order interval, signicantly affect the length of the retailers order

interval.

Since it is often very difficult in practice to assess shortage costs for an external customer, we

adopt a common service denition that species the probability that a demand or a collection of

demands is met. We assume that the demand at each retailer during the lead time is a continuous

random variable D i . Let i = var (D i ) be the standard deviation of demand during the lead time.Then the amount of safety stocks required can be calculated as z i , where the value of z depends

on the service requirement and the type of service measurement. For instance, if we specify a 90%

service level, then z = 1 .28 for Type 1 service level (Nahmias, 1997). We adopt Type 1 service level

measurement in this paper, and we also discuss how to extend the results to systems with Type 2

service level.

The rest of this paper is organized as follows: Section 2 describes the power-of-two inventory

model. Section 3 discusses the OWMR model without power-of-two constraints. Based on this

relaxation model, we discuss how to construct a close-to-optimal power-of-two ordering policy in

Section 4. Section 5 describes briey how our approach can be extended to systems with Type 2

service level requirements. We conclude in Section 6.

2 Model

We use term facility to refer to either the warehouse or a retailer. Facility 0 is the warehouse and

retailers are facilities 1 through N . Let the holding cost rate and the xed setup cost at retailer

n , n = 1 , . . . , N , be hn , K n , and at the warehouse be h0, K 0. Let the average demand at retailer n

be n and the variance 2n per unit time. We dene echelon holding cost rate hn hn h0, whichcan also take negative values (salvage values). Order lead time is L0 at the warehouse and Ln for

each retailer n, n = 1 , . . . , N .

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A power-of-two (POT) policy is one in which each facility n places an order once every T n > 0

units of time beginning at t = 0, and the ratio T n /T 0 is an integer power of two. In this model, the

retailers orders are synchronized , i.e., all N retailers order in the same periods. Lee et al. (1997)

and Cachon (1999) study the scheduled ordering policies in which the retailers orders are balanced ,

i.e., the same number of retailers order at each period. These researches demonstrate that the

scheduled ordering policies reduce supplier demand variance and may lead to lower total supply

chain cost. However, as pointed out by Cachon, it is not clear that balancing the retailer order

intervals will lower costs. With synchronized ordering the supplier can anticipate a large demand

every T 0 periods, so the supplier can arrange to have inventory arrive just before the retailers order.

Therefore, we rst focus on the synchronized ordering policy in this paper, and leave the scheduled

ordering policy as one important future research direction.

An POT inventory control policy for the system can be characterized by an ( N + 1)-tuple,

T = ( T 0, T 1, , T N ), and the corresponding average cost can be written as

c(T ) K 0/T 0 + h0s0 +1nN

(cn (T 0, T n ) + hn sn )

where cn (T 0, T n ) includes the setup costs and cycle inventory costs associated with retailer n, and

can be approximated by cn (T 0, T n ) = K n /T n + n2 (hn T n + h0(T 0T n )) where xy denotes the

larger of x and y. (Note for the deterministic demand case, setup costs and cycle inventory costs

associated with retailer n is K n /T n + n2 (hn T n + h0(T 0

T n )), and we can continue to use this

formula under stochastic demand when we have a high service level and the probability of out-of-

stock can be ignored in cost computation.) The safety stock level sn at retailer n can be written

as n (Ln + T n )2n = n nLn + T n , where n depends on the specied Type 1 service level. Tocalculate the safety stock level at the warehouse, let us rst examine how each retailer contributes

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to the variance of order at the warehouse.

Case 1. T n T 0: Retailer n orders less frequently than the warehouse does. Each time retailern orders, the order introduces variance L02n to the warehouse.

Case 2. T n < T 0: Retailer n orders more frequently than the warehouse does, and whenever

the warehouse orders, retailer n also orders. After each warehouses order, retailer n would order

T 0/T n 1 more times until warehouses next order. In this case, retailer ns orders introducevariance ( L0 + T 0 T n )2n to the warehouse.

The maximum safety stock level for warehouse occurs when all the retailers order at the same

time, and the corresponding safety stock level s0 is 0 L0 n

2n + n (0(T

0 T n ))2n , where 0

depends on the specied service level. This square root formula can be used for any normalized

demand distribution (Eppen, 1981), or it can be used as a good pooling approximation for demand

distributions that are not normalized. For single-parameter demand distributions (e.g. Poisson)

or additive two-parameter demand distributions that have a constant variance-to-mean ratio, it is

possible to construct a appropriate safety stock pooling function numerically by plotting the total

amount of stock required to achieve the given customer service level (as a function of the average

lead time demand) then subtracting the average lead time demand. See Caggiano et al. (2002) for

more discussions.

Our problem can be formulated as follows:

Minimize c(T )

K 0/T 0 + h0s0 +

1nN (cn (T 0, T n ) + hn sn )

Subject to T n > 0 for n = 0 , 1, , N

T n /T 0 is an integer power of two

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where cn (T 0, T n ) := K n /T n + n2 (hn T n + h0(T 0T n )) ,

sn := n nLn + T n , ands0 := 0

L0 n 2n + n (0

(T 0

T n ))2n ,

3 Relaxation of the OWMR Model

Now let us rst drop the power-of two ratio constraints that specify T n /T 0 must be an integer

power of two, and minimize the corresponding relaxation problem. Let Gdenote the retailer setwith T n > T 0, E the retailer set with T n = T 0, and Lthe retailer set with T n < T 0. We solve thisrelaxation problem by discomposing it into smaller subproblems.

3.1 Determining G, E , and LNote if n = 0, for all n = 1 , . . . , N , our problem reduces to the original problem in Roundy (1985),

and we can follow his approach to decide whether a retailer belongs to G, E , or L. In the following,we assume n > 0 for n = 1 , . . . , N .

Let f n (x) = K n /x + n2 hn x + hn n nLn + x. For T n T 0, f n (T n ) = cn (T 0, T n ) + hn sncharacterizes the cost at retailer n . By Corollary 1 in Appendix, the rst order condition of

f n (x) has unique solution, which minimizes f n (x). Denote this minimum solution n . Dene

gn (x) = K n /x + n2 hn x + hn n nLn + x For T n T 0, gn (T n ) = cn (T 0, T n ) + hn sn 2 h0T 0characterizes the cost at retailer n . By Lemma 5 in Appendix, gn (x) = 0 has at most two solutions,

and gn (x) has at most one local minimum. If gn (x) has one local minimum, denote it as n ; if

g(x) has no local minimum, then gn (x) is an decreasing function and we dene n = . f n (x) isnegative in (0 , n ) and positive in ( n , ), while gn (x) is negative on (0 , n ). gn ( n ) = 0 if n < .

Lemma 1 n > n for all n .

Proof: Since gn (x) = f n (x) n2 h0 < f n (x), and f n (x) < 0 on (0, n ), we know that gn is negative

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on (0, n ), and gn ( n ) < 0. As either n = or gn ( n ) = 0, we conclude that n > n .

To determine whether n belongs to G, E , or L, that is, whether T n is greater, equal to, or lessthan T 0, we rst need to know how T n inuences the total cost. Note there are four terms in the

cost function that contain T n : K n /T n , n2 (hn T n + h0(T 0T n )), hn n nLn + T n , and h0s0, where

s0 = 0 L0 n 2n + n (0(T 0 T n ))2n .We observe that

If T n T 0, the warehouse stock level s0 is independent of T n , and the total cost is a functionof T n that varies according to f n (T n ), that is, c ( T )T n =

f n (T n )T n .

If T n T 0, s0 is a decreasing function of the order interval T n . We may explicitly expressthe relationship by writing s0(T n ) instead of s0. The total cost varies according to gn (T n ) +

h0s0(T n ), that is, c (T )

T n = (gn (T n )+ h 0 s 0 (T n ))

T n .

Proposition 1 n

Gif and only if T 0 < n .

Proof: Let T n be the optimal T n for given T 0.

only if =

: suppose T 0 n , f n (T n ) is a strictly increasing function on ( T 0, ). Since the total costchanges according to f n (T n ), we can decrease the total cost by setting T n equal to T 0 if T n > T 0.

Thus, T n T 0, and n LE .if

= : suppose T 0 < n , we show that T n = n > T 0 and n G. Recall that when T n T 0,the total cost varies according to gn (T n ) + h0s0(T n ). The derivative of the total cost is ( gn (T n ) +

h0s0(T n )) = f n (T n ) n2 h0 h020

2n

2s 0 . Since f n (T n ) is a decreasing function on (0 , n ) and 0 < T 0 T 0, and n G.

Proposition 2 n E if n T 0 n .

Proof: Since n T 0, n LE . T n (0, T 0] minimizes the total cost, which is equivalent to min-imizing (gn (T n ) + h0s0(T n )) on (0 , T 0]. Since, T 0 n , gn (T n ) is a decreasing function on (0 , T 0].Note s0(T n ) is also a decreasing function. Thus, to minimize cost, T n equals T 0, that is, n E .

If the demand is deterministic, our denitions of n and n are consistent with the corresponding

denitions in Roundy (1985). In this case, n

Gif and only if T

0<

n; n

E if and only if

n T 0 n ; and n Lif and only if n < T 0. However, with stochastic demand, if T 0 > n , nmay belong to E or Ldepending on the value of T 0, and the level of warehouse safety stock s0. Weillustrate this observation graphically in Figure 1 where the boundary function r n (T 0) is used to

determine whether n Lor E . We explain how to dene r n (T 0) and efficiently enumerate all thepossible partitions in the remainder of this section.

First note that if T 0 > n , then T 0 > n , T n T 0 by Proposition 1, and the total costvaries according to gn (T n ) + h0s0(T n ), where s0 = 0 L0 n 2n + n (0(T 0 T n ))2n . De-ne C n = ( L0 i

2i + i= n (0(T 0 T i ))2i ), then we can also express s0 as follows: s0 =

0 C n + (0 (T 0 T n ))2n .To determine whether n E or L, we need to compare gn (T 0)+ h0s0(T 0) with gn (T n )+ h0s0(T n )

for T n (0, T 0). Dene gn (T n ) h0s0(T n ) + h0s0(T 0) + gn (T 0), it suffices to compare gn (T 0) and

gn (T n ) on (0, T 0). Note gn (T n ) depends not only on T n , but also on T 0 and C n .

In Figure 2, we draw gn (T n ) and gn (T n ) as functions of T n on (0, T 0]. Note that gn (T n ) ap-

proaches positive innity when T n goes to zero, while gn (T n ) gn (T 0) for all T n (0, T 0). Thus,gn (T n ) is either below, or intersecting, or tangent with gn (T n ) on (0, T 0).

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0T0

s 0

n

n

G

E

L

r n(T0)

Figure 1: The relationship of n G, E or Lwith given T 0 and s0

If gn (T n ) is below gn (T n ) on (0, T 0), that is, gn (T n ) > gn (T n ) gn (T n ) + h0s0(T n ) >

gn (T 0) + h0s0(T 0) for T n (0, T 0). This means that T 0 minimizes gn (.) + h0s0(.) on (0, T n ].

Furthermore, T n (0, T 0] minimizes the total cost, which is equivalent to minimize ( gn (T n ) +

h0s0(T n )) on (0 , T 0]. Thus, T n = T 0, and n E .

If gn (T n ) intersects with gn (T n ), that is, gn (T n ) + h0s0(T n ) < g n (T 0) + h0s0(T 0) for someT n (0, T 0). T n lies in (0, T 0), and n L.

If gn (T n ) is tangent with gn (T n ) on (0, T 0) without intersecting with gn (T n ), that is, gn (T n ) +h0s0(T n ) has multiple minimum solutions on (0 , T 0], and n can belong to either E or L.

The above discussion sheds light on how to determine whether n E or L. However, it is stillnot easy to implement since gn (T n ) depends not only on T 0, but also on C n , the variance of order

at the warehouse contributed by other retailers. We hope to nd a better way to determine n E 10


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Tn

(T0, g

n(T

0))

gn(T

n)

gn(T

n)

Figure 2: gn (T n ) and gn (T n ) at C n = C n

or Lwithout calculating C n . To do this, let us rst investigate how C n affects gn (T n ). Note that

for a xed T 0, if we decrease the value of C n , gn (T n ) will shift downward; if we increase the value of

C n , gn (T n ) will shift upward; and if we keep on increasing C n till it approaches positive innity,

gn (T n ) will converge to gn (T 0) for each T n [0, T 0], that is, gn (T n ) will become a horizontal line.

Since gn (T n ) approaches positive innity as T n approaches zero and gn ( n ) < g n (T 0) with n < T 0,

gn (T n ) intersects with gn (T n ) on [0, T 0]. Therefore, when C n approaches positive innity, T n has

little impact on the the warehouse safety stock level s0; and n

Lif and only if T 0 > n as in

Roundy (1985) under this scenario.

For given T 0, we are interested in the value of C n at which n shifts from Lto E . If we decreasethe value of C n to zero, there are only two possible scenarios:

I) gn (T n ) intersects with gn (T n ) on (0, T 0).

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II) gn (T n ) is below or tangent with gn (T n ) on (0, T 0).

In this case, there exists unique C n such that gn (T n ) is tangent with gn (T n ) when C n = C n .

According to the different positions of the tangent points, we have the following two sub-cases:

IIa) T 0, the end point, is the unique tangent point when we set C n = C n .

IIb) There exists internal tangent point T n (0, T 0) when we set C n = C n . Let T n be the

rightmost tangent point on (0 , T 0) if there are more than one internal tangent point.

Let us rst focus on Scenario IIb) when the tangent point is T n (0, T 0), and see how we can

determine whether n E or Laccording to the values of T 0 and C n . Based on the relative valuesof C n and C n , we have the following three cases:

C n < C n : gn (T n ) is below gn (T n ) on (0, T 0). Recall that gn (T n ) > gn (T n ) gn (T n ) +h0s0(T n ) > g n (T 0) + h0s0(T 0) for T n (0, T 0), this means that T 0 minimizes gn (.) + h0s0(.)

on (0, T n ]. We also know that T n minimizes gn (T n ) + h0s0(T n ) on (0, T 0], thus T n = T 0 and

n E . The corresponding warehouse safety stock level is s0 = 0C n < 0 C n .

C n = C n : gn (T n ) is tangent with gn (T n ) on T n (0, T 0). gn (T n ) + h0s0(T n ) has multiple

minimum solutions on (0 , T 0]. Since T n minimizes gn (T n ) + h0s0(T n ) on (0, T 0], one possi-

ble value of T n is T n = T 0 and n E , then s0 = 0 C n ; another possible value is T n = T n (orother tangent points) and n L, then s0 = 0 C n + ( T 0 T n )2n or s0 0 C n + ( T 0 T n )2n .C n > C n : gn (T n ) intersects with gn (T n ) on T n (0, T 0). Since T n minimizes gn (T n ) +

h0s0(T n ) on (0, T 0], T n < T 0, and n L. Furthermore, by Lemma 10 in Appendix, theoptimal T n is no more than T n , then s0 = 0 C n + ( T 0 T n )2n > 0 C n + ( T 0 T n )2n .

Note that the optimal warehouse safety stock level s0 is either smaller or equal to 0 C n whenn E , or larger or equal to 0 C n + ( T 0

T n )2n when n L, and s0 will never pick a value

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within 0 C n , 0 C n + ( T 0 T n )2n . Thus, if we can dene a function r n (T 0) whose valueis always between 0 C n and 0 C n + ( T 0 T n )2n for each retailer n, then for retailer n , if T 0 > n , n

E if and only if s0

r n (T 0 ) and n

Lif and only if s0 > r n (T 0 ). This way, we can

use T 0 and s0 to nd out whether n E or Lwithout calculating C n , whose value varies with n.

TnTn

slope: h 0 0 n2 /2/(C n

* +(T0Tn) n2)1/2

slope: h 0 0 n2 /2/(C n

* )1/2

(T0,gn(T0))

gn(Tn)

gn(Tn)

slope: h 0 02

n2 /2/r n(T0)

Figure 3: The slopes of the tangent lines at T n = T 0 and T n = T n

Next we show such function r n (T 0) does exist. In Figure 3, we draw a line through ( T 0, gn (T 0))

such that this line is tangent with gn (T n ) and lies below gn (T n ) on (0, T 0). We dene r n (T 0) such

that h020

2n

2r n (T 0 ) is the slope of the tangent line.

Lemma 2 0 C n r n (T 0) < 0 C n + ( T 0 T n )2nProof: We x C n = C n . h0s0(T n ) is a convex function of T n ; the slope (derivative with regardto T n ) is (h0s0(T n )) = h0

20 2n

2s 0 (T n ) , which is an increasing function of T n .

Because gn (T n ) is convex and below gn (T n ) on (0, T 0] at C n = C n , the slope at T 0, h020

2n

2s 0 (T 0 ) =

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h020

2n

20 C n , is no less than the slope of the tangent line h020

2n

2r n (T 0 ) . That is, 0 C n r n (T 0).Meanwhile, the slope at T n , h0

20 2n

2s 0 ( T n )= h0

20 2n

20 C n +( T 0

T n ) 2n, is less than the slope of the tangent

line. That is, r n (T 0) < 0

C

n + ( T 0

T n )2n .

Thus, 0 C n r n (T 0) < 0 C n + ( T 0 T n )2n . Figure 3 shows the graphic proof and howto represent r n (.) as a function of T 0.

Thus if T 0 > n , we can claim that under Scenario IIb), if s0 r n (T 0), then n E , and if s0 > r n (T 0), then n L.

This statement is also true for Scenario IIa), where gn (T n ) is tangent with gn (T n ) at T 0 when

C n = C n . Under this scenario, r n = 0 C n . When n E , C n C n , and we haves0 = 0C n , which is indeed no more than r n = 0 C n . On the other hand, when n L, thatis C n > C n , we have s0 > 0

C n , which is larger than r n = 0 C n . Thus, if s0 r n (T 0),then n E , and if s0 > r n (T 0), then n L.

Furthermore, this statement is also true for Scenario I) where gn (T n ) intersects with gn (T n )

when we decreases C n to zero. Under this scenario, n L, T n minimizes gn (T n ) + h0s0(T n ) on(0, T 0). Since gn (T n ) = h0s0(T n ) + h0s0(T 0) + gn (T 0), T n maximizes gn (T n ) gn (T n ) on (0, T 0),(T n , gn (T n )) lies above gn (T n ) as gn (T n ) intersects with gn (T n ). Since the tangent line is either

below or tangent with gn (T n ), (T n , gn (T n )) lies above the tangent line. In the following, we show

that the optimal warehouse safety stock level s0 is greater than r n (T 0). Suppose s0 is no more than

r n (T 0), then h0 20 2n2s 0 , the slope of gn (T n ) at T n , is no less than the slope of the tangent line, h0

20 2n2r n .

Since s0(T n ) is concave, then gn (T n ) is convex, and the slope of gn (T n ) is no less than the slope of

the tangent line on [ T n , T 0]. Therefore, gn (T n ) is above the tangent line on [ T n , T 0]. However, both

gn (T n ) and the tangent line pass through ( T 0, gn (T 0)). Contradictory! Thus, the optimal warehouse

safety stock level s0 > r n , and we can still use r n to determine whether n E or L.

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Finally, we can claim that if T 0 > n , then n E if s0 r n (T 0), and n Lif s0 > r n (T 0) underall possible scenarios.

The curve s0 = r n (T 0) and the line T 0 = n for retailer n cut the place into three areas. A total

of N curves and N lines cut the plane into many small areas. Each small area corresponds to one

possible partition of G, E , and L. In the next, we show that there are at most O(n2) areas.

Theorem 1 There exists a constant k, such that for any two retailers i and j , r i (T 0) and r j (T 0)

do not intersect for more than k times.

Proof: See Appendix.

By Theorem 1, a curve r i(T 0) intersects no more than k times with another curve r j (T 0), and

a curve for a particular retailer may have at most O(n) intersections. Thus, there will be at most

O(n2) intersections, and we can nd the T 0 value for each intersection since each curve r n (T 0) can

be represented by a polynomial by Lemma 11 in the Appendix. By ranking these O(n2) values of

n s and n s, we can divide the T 0 axis into O(n2) segments. Starting from the leftmost interval,

which corresponds to all n G, as we increase the value of T 0 and move from left to right, we mayintroduce new partitions as T 0 crosses n , n or an intersection point.

The following example illustrates how we can enumerate all the possible partitions efficiently.

To keep the example simple, we assume there are only two retailers and r 1 and r 2 intersects only

once at T 12

as shown in Figure 4. 1,

1,

2,

2, and T

12divide the T axis into six intervals. We

starts at T 0 = 0 and continuously increase the value of T 0 till it passes T 12 , and we discuss the

possible partition changes during this process.

(0, 1): the only possible partition is ( G, E , L)= ( {1, 2},,).

( 1, 1): Retailer 1 is moved from Gto E , and we get a new partition ( G, E , L)= ( {2}, {1},).

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T0

s 0

1 1 2 2 T12

{1,2},, {2},{1},

{2},{1},

r1(0)

r2(0)

{2},,{1} ,{2},{1}

,{1,2},

,,{1,2}

,{1},{2}

,{1,2},

Figure 4: Example of Enumerating Possible Partitions

( 1, 2): A new partition, ( {2},, {1}), is added for the case s0 > r 1(T 0), in which 1 Landthe rest of retailers remains unchanged. Now we have two possible partitions, (

{2

},

,

{1

})

and ( {2}, {1},).

( 2, 2): Retailer 2 is removed from Gto E for all the partitions in the previous interval. Thus,we have two possible partitions, ( , {2}, {1}) and (, {1, 2},).

( 2, T 12): A new partition, ( ,, {1, 2}), is added for the case s0 > r 2(T n ) in which 2 Land

the rest of retailers remains unchanged. Now we have three possible partitions, ( ,, {1, 2}),(, {2}, {1}), and (, {1, 2},).

(T 12 , ): A new partition ( , {1}, {2}) is added and an old partition ( , {2}, {1}) is removed.We have three possible partitions, ( ,, {1, 2}), (, {1}, {2}), and (, {1, 2},).

In this example, observe that for each of the intervals, we have at most 3 possible partitions as

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there are at most two r i (T 0) functions separating the partitions. We list the partitions corresponding

to large s0 (s0 > r i(T 0)) rst, then the partitions corresponding to small s0 (s0 < r i (T 0)). Note no

sorting is needed since the newly introduced partition when T 0 crosses n always have large s0.

In general, the partition enumeration can be carried out in O(N 2 log N ), and the complexity is

determined by sorting the n s, n s, and the O(N 2) intersection positions.

3.2 Solution algorithm with known G, E , and LRecall the average cost of POT policy can be written as:

c(T ) = K 0/T 0 + h0s0 +1nN

(cn (T 0, T n ) + hn sn )

For given G, E , and L, we regroup the terms as

K 0/T 0 + h0s0 +nE

(cn (T 0, T n ) + hn sn ) +nG

(cn (T 0, T n ) + hn sn ) +nL

(cn (T 0, T n ) + hn sn )

where

nE (cn (T 0, T n ) + hn sn ) = nE (K n /T 0 + n2 hn T 0 + hn n nLn + T 0),

nG(cn (T 0, T n ) + hn sn ) = nG(K n /T n + n2 hn T n + hn n nLn + T n ), and

nL(cn (T 0, T n ) + hn sn ) = nL(K n /T n + n2 hn T n + hn n nLn + T n ) + nL n2 h0T 0

For n G, we can solve the optimal T n by minimizing K n /T n + n2 hn T n + hn n nLn + T n , asthe rst order condition characterizes the optimal solution. The solution is exactly n we dened in

previous subsection. However, we can not use the same approach to solve T n with n L, because

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s0 is also a function of these T n s. The rst order condition with respect to T n is

K nT 2n

+n2

hn + hn n n1

2Ln + T n h0202n2s0

= 0 (1)

The rst order condition of T 0 gives us

nL

n2

h0 +nE

(K nT 20

+n2

hn + hn n n1

2Ln + T 0 ) K 0T 20

+ h020 nL

2n

2s0= 0 (2)

Equations (1) and (2) may have multiple solutions. According to (1), we can write T n as a

function of the warehouse stock level s0 by Corollary 2 in the Appendix. Furthermore, by Lemma 8

in the Appendix, each T n (s0) is a strictly convex function of s0. According to (2), we can write T 0

as a function of warehouse stock level s0 by Corollary 3, and T 0(s0) is a strictly concave function of

s0 by Lemma 9 in the Appendix. Thus, if we know the optimal s0, we can determine the optimal

T 0 and T n s.

Note that if we know the values of T 0 and T n s, we can calculate s0 by

0 L0 n 2n + ( nL

2n )T 0 nL

2n T n

That is, the optimal solution ( s0, T 0, T n ) must also satisfy the equation that denes s0. Dene

(s0) = s20 = 20 (L0 n 2n +( nL2n )T 0(s0) nL2n T n (s0)). ( s0) is a strictly concave function

of s0. At optimality, we must have ( s

0) = s2

0, or

(s

0) = s

0, according to the denition of s

0.

Since ( (s0) s0)s 0 = 14 2 (s 0 )( (s 0 )) 2 3 / 2 < 0 for (s0) < 0, (s0) s0 is also a strictlyconcave function, and (s0) s0 = 0 has at most two solutions. Via a binary search or a goldensearch, we can nd these potential optimal s0 values, and determine the corresponding T n s and T 0

in O(n). By comparing the two solutions, we can select the solution with lower total cost. Thus,

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given G, E , and L, we can solve the minimization problem efficiently. As there are only O(n2)possible partition of G, E , and L, we can solve the relaxation problem in O(n3).

4 A Close-to-optimal Power-of-two Ordering Policy

In this section, we will use the optimal solution of the relaxation problem to build a close-to-optimal

power-of-two ordering policy.

We investigate two scenarios: i) the base interval length T L is given and xed, nd a power-of-

two ordering policy satisfying T n = 2 pn T L ( pn Z ); ii) we have the exibility to choose the base

interval length T L and the corresponding power-of-two ordering policy.

4.1 A Close-to-optimal Power-of-two Ordering Policy with Given T L

For arbitrary target service level, we can construct an ordering policy satisfying T n = 2 pn T L ( pn

Z ), which is no more than 5+1

2 1.618 times the cost of the relaxation problem by thefollowing procedure:

i) Let T n denote the optimal solution of the relaxation problem. Dene qn = log2(T n /T L ) ,

that is, qn is the integer part of log 2(T n /T L ), and dene r n = log 2(T n /T L ) qn .

ii) For n {0}E , if r n < log2(), set T n = 2 qn T L , otherwise, set T n = 2 qn +1 T L .

iii) For n G, if r n 0.5, set T n = 2 qn T L , otherwise, set T n = 2 qn +1 T L .

iv) For n L, we consider two cases: qn < q 0 and qn = q0.

qn < q 0: we set T n = 2 qn T L if r n < 0.5 and set T n = 2 qn +1 T L otherwise.

qn = q0:

if r 0 < log2(), we set T n = 2qn T L = T 0;

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if r 0 log2(), we set T n = 2 qn T L if r n 1 log2()and set T n = 2 qn +1 T L if r n > 1 log2().

Following the above procedure, if n G, then T n T 0; if n E , then T n = T 0; and if n L,then T n T 0. Now we show that each term in the objective function is positive, and the cost ratiobetween the constructed ordering policy and the optimal solution for the relaxation problem is no

more than .

The objective function has the following terms: K nT n for all n , n2 hn T n for n L, n2 h0T 0 for n

L, n

2h

nT n for n

GE , h

nn nLn + T n for n

GEL, and h00

L0

n2

n+

nL(T 0

T n )2

n.

To prove that each term in the objective function is positive, it suffices to show hn > 0 for

n L. Now suppose this is not ture, and hn 0, then both gn (T n ) and h0s0(T n ) are decreasingfunctions on (0 , T 0); thus, n G E . This is a contradiction. Therefore, hn > 0 for n L.

Next we show that the cost ratio between the constructed ordering policy and the optimal

solution for the relaxation problem is no more than . The proof of this result replies on the

following two lemmas.

Lemma 3 max{T nT n , T

nT n }< for all n .

Lemma 4 max nL{T 0 T nT 0 T n } 2.

The detailed proofs are provided in the Appendix.

Theorem 4.1 The cost ratio between the constructed ordering policy and the optimal solution for

the relaxation problem is no more than

Proof: Lemma 3 says that T

nT n < for all n . Thus, the cost ratios between

K nT n and

K nT n

are less than

. Similarly, the ratios between n2 hn T n and n2 hn T n , between

n2 h0T 0 and

n2 h0T 0 , and between

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n2 hn T n and

n2 hn T n are all less than . Furthermore, the ratios between hn n nLn + T n and

hn n n Ln + T n are less than < for all n G E L.So far, we have compared all the terms except the warehouse safety stock cost. According

to Lemma 4, it is easy to show that the ratios between h00 L0 n 2n + nL(T 0 T n )2n andh00 L0 n 2n + nL(T 0 T n )2n are no more than 2 = .

We have shown that each term in the objective function of the power-of-two ordering policy is

no more than 1.618 times of the corresponding term in the optimal solution for the relaxation

problem. Since the relaxation problem is a lower bound of optimal periodic ordering policy, we

claim that the cost of the resulting power-of-two ordering policy is no more than 1.618 times the

optimal cost.

4.2 A Close-to-optimal Power-of-two Ordering Policy with Unrestricted T L

For arbitrary target service level, if the base period T L is not xed in advance, we can construct a

power-of-two ordering policy whose cost is no more than 32 1.260 times the cost of the relaxationproblem by the following procedure:

i) Let T n denote the optimal solution of the relaxation problem. Dene T L = T 0 /32, qn =

log2(T n /T L ) , that is, qn is the integer part of log 2(T n /T L ), and dene r n = log 2(T n /T L )qn ;

ii) For n {0}E , set T n = T L .

iii) For n G,

K n /T n n2 hn T n : if r n 1/ 3, set T n = 2 qn T L , otherwise, T n = 2 qn +1 T L .

K n /T n < n2 hn T n : if r n 2/ 3, set T n = 2 qn T L , otherwise, T n = 2 qn +1 T L .

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iii) For n L,

K n /T n n2 hn T n : if r n 1/ 3, set T n = 2 qn T L , otherwise, T n = 2 qn +1 T L .

K n /T n 0, and di 0, there existsa critical value x, such that f (x) is a convex function of x on (0, x), and a concave function of x

on (x, ).

Proof: f (x) = 2ax 3 i ci4 1(di + x)3 / 2 . f (x) approaches positive innity as x approaches 0 from above,and f (x) approaches zero from below as x approaches positive innity. Thus, there exists xsuch

that f (x) = 0. We prove that there is a unique xsuch that f (x) = 0 by contradiction. Assume

there exist x1 and x2 = x 1 ( > 1) such that f (x1) = f (x2) = 0. It follows that di + x 2di + x 1 andci4

1(di + x 1 )3 / 2 3/ 2 ci4 1(di + x 2 )3 / 2 . Thus, f (x1) 3 2ax 32 3/ 2 i ci4 1(di + x 2 )3 / 2 = ( 3 3/ 2) 2ax 32 > 0 for

> 1, which is a contradiction. That is, there exists unique xsuch that f (x) = 0. Furthermore,

f (x) > 0 on (0, x), and f (x) < 0 on (x, ), i.e., f (x) is a convex function of x on (0, x), andf (x) is a concave function of x on (x, ).

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Corollary 1 For f (x) = ax + bx + cd + x with a > 0, b > 0, c 0 and d 0, the rst order condition characterizes the optimal solution of the minimization problem.

Proof: If c = 0, f (x) is convex, and the rst order condition characterizes the optimal solution of

the minimization problem. Otherwise, by Lemma 5, we can have at most two values satisfying the

rst order condition: one between 0 and the critical x, which corresponds to a local minimum, and

one greater than x, which corresponds to a local maximum. Also observe that f (x) approaches

positive innity as x approaches positive innity, so there is no local maximum on ( x, ). Thus,the rst order condition has a unique solution, which corresponds to a local minimum. The unique

solution is also global minimum.

Lemma 6 For f (x) = ax + bx + cd + x with a > 0, c > 0 and d 0, let xdenote the uniquesolution for f (x) = 0 as in Lemma 5. Consider g(x) f (x) + eh x on [0, h] with e > 0 and h > 0. If the optimal solution of minimizing g(x) is an internal point, then it lies between 0 and

x.

Proof: It suffices to prove the case with h > x . Since eh x is a concave function, by Lemma 5,g(x) is concave on ( x, h), and no internal point of ( x, h) would correspond to an optimal solution

of minimizing g(x). Thus, the internal optimal solution must lie between 0 and x.

Corollary 2 Let f (x) = ax + bx + cd + x with a > 0, c 0, and d 0. Assume the optimal solution of minimizing g(x)

f (x) + eh

x with e > 0 and h > 0 is an internal point, then the

optimal x is a function of the derivative of eh x.

Proof: The optimal solution satises the rst order condition g (x) = 0. Thus, f (x) = (eh x)is satised at the optimal solution. The result is obvious when c = 0. If c > 0, by Lemma 6, the

solution lies between 0 and x. f (x) is positive on (0 , x), thus f (x) is monotone, and if we know

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the derivative of eh x, the optimal solution x can be uniquely determined. The optimal value xis a monotone decreasing function of this derivative ( eh x) , since f (x) is monotone increasing.

Lemma 7 For f (x) = ax

+ bx + i cidi + x with a > 0, ci

0, ci > 0, and di

0, let x

denote the unique solution for f (x) = 0 as in Lemma 5. Consider g(x) f (x) + eh + x with e > 0. If the optimal solution of minimizing g is an internal point, then it lies between 0 and x.

Proof: Due to the concavity, proof is similar to the proof of Lemma 6.

Corollary 3 Let f (x) = ax + bx+ i cidi + x with a > 0, ci 0, and di 0. Assume the optimal

solution of minimizing g(x) f (x) + e

h + x with e > 0 is an internal point, then the optimal xis a function of the derivative of eh + x.

Proof: Proof is similar to the proof of Corollary 2.

Lemma 8 For f (x) = ax + bx + cd + x with a > 0, c 0, and d 0, let xbe the unique solution for f (x) = 0 (as in Lemma 5) if c > 0 and let xbe if c = 0 . Let T (s) denote the unique

solution in (0, x) such that f (T (s)) C s = 0 , where C > 0 is a constant, then T (s) is a strictly

convex function of s.

Proof: f (T )|T = T (s) = C s > 0, and f (T ) > 0 as T (0, x).f (T ) > 0

2aT 3 >

14

c(d+ T ) 3 / 2

6aT 4 >

34

1T

c(d+ T )3 / 2 34 c(d+ T )5 / 2 38 c(d+ T )5 / 2

f (T ) =

6aT 4 +

38

c(d+ T )5 / 2 < 0

Thus, dT ds = 1C (f (T ))2

f (T ) < 0. Furthermore, since f (T ) > 0 and f (T ) < 0,

d2T ds2

= 1C

2f (T )( f (T ))2 (f (T ))2f (T )(f (T ))2

dT ds

> 0

That is, T (s) is a strictly convex function of s.

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Lemma 9 For f (x) = ax + bx + i cidi + x with a > 0, b > 0, ci 0 and di 0, let xbe theunique solution for f (x) = 0 (as in Lemma 5) if i ci > 0 and let xbe if i ci = 0 . Let T (s)be the unique solution in (0, x) such that f (T (s)) + C s = 0 , where C > 0 is a constant, then T (s)

is a strictly concave function of s.

Proof: f (T )|T = T (s) = C s < 0, and f (T ) > 0 as T (0, x). By the same arguments in Lemma 8,f (T ) > 0f (T ) < 0.

dT ds =

1C

(f (T )) 2

f (T ) > 0,d2 T ds 2 =

1C

2f (T )( f (T )) 2 (f (T ))2 f (T )

(f (T )) 2dT ds . Since

dT ds > 0

and f (T ) < 0, in order to show d2 T

ds 2 < 0, it suffices to show that 2( f (T ))2 f (T )f (T ) > 0.

It is easy to show that

2(f (T ))2 = 2(2aT 3

14 i

ci(di + T )3/ 2

)2 =8a2

T 6 2aT 3 i

ci(di + T )3/ 2

+18

(i

ci(di + T )3/ 2

)2

Furthermore,

0 < f (T ) =6aT 4

38 i

ci(di + T )5/ 2

6aT 4

and

0 < f (T ) =a

T 2 b12 i

ci(di + T )1/ 2

0. Thus

f (T )f (T ) = ( f (T ))(f (T )) < (a

T 2 12 i

ci(di + T )1/ 2

)6aT 4

=6a2

T 6 3aT 4 i

ci(di + T )1/ 2

and

2(f (T ))2 f (T )f (T ) >2a2

T 6 2aT 3 i

ci(di + T )3/ 2

+18

(i

ci(di + T )3/ 2

)2 +3aT 4 i

ci(di + T )1/ 2

We know 3aT 4 >38 i

ci(di + T )5 / 2

for f (T ) = 2aT 3 14 i ci(di + T )3 / 2 > 0, and ( i ci(di + T )5 / 2 )( i ci(di + T )1 / 2 )

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( ici

(di + T )3 / 2)2 by Cauchy inequality, thus we have

2(f (T ))2 f (T )f (T ) >2a2

T 6 2aT 3 i

ci(di + T )3/ 2

+18

(i

ci(di + T )3/ 2

)2 +38

(i

ci(di + T )3/ 2

)2

= 2( aT 3 12 i ci(di + T )3 / 2 )2 0We conclude that d

2 T ds 2 < 0, and T (s) is a strictly concave function of s.

Lemma 10 If C n > C n , the optimal T n that minimizes gn (T n ) + h00 C n + ( T 0 T n )2n on (0, T 0) is no more than T n .

Proof: Recall that T n is the rightmost tangent point on (0 , T 0). Therefore, T n minimizes gn (T n ) gn (T n ) on (0, T 0). By denition, gn (T n ) = h0s0(T n ) + h0s0(T 0) + gn (T 0), T n minimizes ( gn (T n ) +h00 C n + ( T 0 T n )2n ) on (0, T 0). For T n < T n < T 0, we have

gn (

T n ) + h00

C n

+ ( T 0

T n )2n

< g n (T n ) + h00

C n

+ ( T 0

T n )2n

That is, for T n < T n < T 0,

gn (T n ) gn ( T n ) > h 00 C n + ( T 0 T n )2n h00 C n + ( T 0 T n )2nAlso, for T n < T n < T 0,

h00 C n + ( T 0 T n )2n h00 C n + ( T 0 T n )2n> h 00 C n + ( T 0 T n )2n h00 C n + ( T 0 T n )2n

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because C n > C n . Thus, for T n < T n < T 0,

gn (T n ) gn (

T n ) > h 00 C n + ( T 0

T n )2n h00

C n + ( T 0 T n )2n

or

gn ( T n ) + h00 C n + ( T 0 T n )2n < g n (T n ) + h00 C n + ( T 0 T n )2nThat is, the optimal solution T n is no more than T n .

In the next, we show that the curve ( r i (T 0)s separate the plane into O(n2) areas. Let us dene

an K n , bn n2 hn , cn hn n n , dn Ln , and en h0202n / 2, then gn (x) = an /x + bn x +cndn + x. Let ( T , gn ( T )) be the tangent point. According to the denition of r n (T 0), we have

gn ( T ) =en

r n (T 0)(3)

gn (T 0)

gn (

T ) = ( T 0

T )gn (

T ) (4)

We rst show the curve r n (T 0) satises some polynomial function.

Lemma 11 There exists a constant C 1, and a nonzero polynomial P with degree less than C 1 such

that P (r n (T 0), T 0, a n , bn , cn , dn , en ) = 0 .

Proof: By (3) and (4),

gn (T 0) T 0en

r n (T 0)= gn ( T ) T gn ( T )=

an

T + bn T + cn dn + T +

an

T bn T

cn T 2 dn + T =

2an

T

+cn2

2dn + T

dn + T 30


31/37


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equation R = 0 as R1 + R2 3q + R3 3 q2 = 0, where R1, R2, R3 are functions of r n (T 0), T 0,an ,bn ,cn ,dn ,en . If R3 = 0, then we can move R1 to the right-hand-side of the equation, and

cubing both sides to remove the cubic root operations. If R3 = 0, multiply both sides by R3,

and we have R22 R1R3 = R233 q2 + R3R2 3q + R22 . Multiplying both sides by R3 3q R2,

we get (R22 R1R3)R3 3q = R23q R1R2R3. By cubing both sides, we no longer take cubicroot of term q.

Thus, we can generate a nonzero polynomial P such that P (r n (T 0), T 0, an , bn , cn , dn , en ) = 0.

The constructive procedure shows that there exists a constant C 1 such that the degree of P is less

than C 1.

Proof of Theorem 1.

By the proof of Lemma 11, we know that for any retailer n, we can get a nonzero polynomial

P n such that P n (r n (T 0), T 0) = 0 and the degree of P n is no more than some constant C 1.

To prove Theorem 1, it suffices to show that P i (x, y) = 0 and P j (x, y) = 0 will not intercept

with each other for more than some constant times when both P i and P j are nonzero polynomials

with degree no more than C 1. Without loss of generality, we can assume P i (x, y) and ( P i (x, y ))y ,

the derivative of P i (x, y ) with respect to y, have no factor in common, and P j (x, y) and ( P j (x, y ))y ,

the derivative of P j (x, y) with respect to y, have no factor in common, because factoring out the

common factors does not change the shape of P i (x, y ) = 0 or P j (x, y) = 0.

By treating P i(x, y ) and P j (x, y ) as polynomials of variable y, we can factor the polynomials

P i and P j by iterated division. Thus, P i(x, y) = u(x, y ) p(x, y) and P j (x, y ) = v(x, y) p(x, y ), where

p(x, y ) is a common factor of P i (x, y ) and P j (x, y) while u(x, y ) and v(x, y ) have no common

polynomial factor.

P i(x, y) = 0 and P j (x, y ) = 0 will intercept with each other if and only if i) u(x, y ) and v(x, y)

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intercept with each other; ii) u(x, y) or v(x, y) intercept with p(x, y ); iii) p(x, y) intercepts itself.

i) u(x, y ) and v(x, y) may intercept with each other at ( x, y ) only if u(x, y) = v(x, y ) = 0. Since

u(x, y) and v(x, y) have no common factor involving y, by iterated division, ( x, y ) must satisfy a

series of polynomial equations, the last of which has degree of y equal to zero, that is x must be a

root of a polynomial equation. Since the degrees of u(x, y), v(x, y ) are no more than C 1, the degree

of the resulting polynomial is bounded by some constant number C 2. Thus, we have at most C 2

possible values of x, each of which corresponds to at most C 1 possible values of y, and there are at

most C 1C 2 interceptions between u and v.

ii) Since P i (x, y) and ( P i(x, y))y have no factor in common, and P j (x, y) and ( P j (x, y ))y have

no factor in common, and there are at most C 1C 2 interceptions between u(x, y ) and p(x, y ) by the

same argument. Similarly, there are at most C 1C 2 interceptions between v(x, y) and p(x, y ).

iii) For the case where p(x, y) = 0 intercepts itself at ( x, y), the interception can only happen

if y is a multiple root for the given x. Thus, p(x, y) may intercept with itself only if p(x, y) =

( p(x, y ))y = 0. Note that p(x, y ) and ( p(x, y)) y = 0 have no factor in common, because P i (x, y) and

(P i (x, y ))y have no factor in common. Since the degrees of p(x, y ), ( p(x, y))y are no more than C 1,

and p(x, y ), ( p(x, y ))y have no factor in common, this can only happen when x satises a polynomial

with degree bounded by the constant number C 2. Thus, we have at most C 2 possibility of x, each

of which corresponds to at most C 1 possibility of y, and there are at most C 1C 2 interceptions of

p(x, y ).

Thus, by setting k equal to 4 C 1C 2, we can guarantee that the difference of the curves ( r i(T 0) r j (T 0)) does not change sign for more than k times for any two retailers i and j .

Proof of Lemma 3 : max{T nT n , T

nT n }< for all n .

We discuss the following three cases:

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For n {0}E : T

nT n = 2

r n if r n < log2(), andT nT n = 2

r n 1 if r n log2(). Since 0 r n < 1,2 T

nT n < , max{T nT n , T

nT n }< for n {0}E .

For n G,T nT n = 2

r n

if r n < 0.5, andT nT n = 2

r n

1

if r n 0.5, max{T nT

n ,T nT n }

2 < for

n G.

For n L, when qn < q 0, same as the case n G, max{T nT n , T

nT n }< ; when qn = q0, if

r 0 < log2(), we haveT nT n = 2

r n . Since n Land T n < T 0 , r n < r 0 < log2(), and1 T

nT n < ; if r 0 log2(), we have T

nT n = 2

r n if r n 1 log2() and T

nT n = 2

r n 1 if

r n > 1

log2(); and 1

a power-of-two ordering policy for one warehouse, multi-retailer

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