Two-stage inventory models with a bi-modal transportation cost

Computers & Operations Research 37 (2010) 20 -- 31

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Two-stage inventory models with a bi-modal transportation cost B.Q. Rieksts a , José A. Ventura b, ∗ a b

Institute for Defense Analyses, Cost Analysis and Research Division, 4850 Mark Center Drive, Alexandria, VA 22302, USA Harold and Inge Marcus Department of Industrial and Manufacturing Engineering, The Pennsylvania State University, University Park, PA 16802, USA

A R T I C L E

I N F O

Available online 26 March 2009 Keywords: Inventory Transportation cost Truckload Less than truckload

A B S T R A C T

This paper discusses inventory models over an infinite planning horizon with constant demand rate and two modes of transportation. These transportation options include truckloads and a less than truckload carrier. An optimal algorithm is derived for a one-warehouse one-retailer system. A power-of-two heuristic algorithm is also proposed for a one-warehouse multi-retailer system. Computational results are provided to show that, on the average, the heuristic algorithm is at least 94% effective. © 2009 Elsevier Ltd. All rights reserved.

1. Introduction Transportation and inventory costs are important factors in optimizing profits in a supply chain. This paper develops policies for inventory models in supply chains that include transportation costs. Two typical modes of transporting freight are truckload (TL) and less than truckload (LTL) shipments. To transport freight with a TL shipment, a fixed cost is incurred for each load up to a given capacity. For a particular load, the cost is independent of the quantity shipped. The practice of transporting freight that is less than the full capacity at the cost of a full load is called over-declaring. Some examples of a TL shipment are an overseas container or an actual truck that is designated for a specific product. For a truck, the cost due to the truck, driver, and operating expenses is proportional to the distance traveled. Any variable cost dependent on the quantity of freight is considered negligible. In LTL transportation, the cost is based on the quantity shipped and distance traveled. For small quantities, using an entire TL may be more expensive than LTL transportation. An example of LTL transportation is a third party carrier such as the United Parcel Service (UPS). By consolidating freight from different companies into the same shipment, third party carriers can afford to transport small quantities. Our paper presents inventory models which include TL shipments, an LTL carrier, or a combination of the two options. In practice, LTL carriers have multiple breakpoints in the quantity shipped where the per unit cost decreases. Within these weight breakpoints, a company may over-declare a quantity to take advantage of the next discount. Carter et al. [9] consider changes in the cost function due to

∗ Corresponding author. Tel.: +1 814 865 3841; fax: +1 814 863 4745. E-mail addresses: [email protected] (B.Q. Rieksts), [email protected] (J.A. Ventura). 0305-0548/$ - see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.cor.2009.02.026

over-declaring an LTL shipment. A simplifying assumption is made in our paper that the LTL cost is a constant cost per unit. Swenseth and Godfrey [35] analyze the error associated with this constant cost per unit approximation. Our paper provides a theoretical foundation for future work with multiple LTL breakpoints. For some systems, it may be optimal to use both TL and LTL transportation simultaneously. If the inventory and setup costs are dominant, the order quantity may be a combination of full loads and a partial load. If the quantity of the partial load is not sufficient to justify another truckload using TL transportation, it is optimal to use both modes of freight transportation. The inventory systems in this research consider these transportation scenarios in an environment with assumptions similar to the economic ordering quantity (EOQ) model. At each stage, there is a positive holding cost per unit and a setup cost for each order. Setup costs represent administrative costs for an order as well as any manufacturing setup costs to transform raw materials into products or for a process to add value to products. In addition to this fixed component, purchasing raw materials or products from an external supplier has a variable cost that corresponds to labor or materials. We assume this variable cost is proportional to quantity, so it is independent of the inventory decisions in our research. For the transportation costs, there is a constant fixed cost for each load up to the capacity of a TL shipment and a constant cost per unit of freight shipped with LTL transportation. The goal is to minimize the average holding, setup, and transportation costs. Over the infinite horizon, we develop a heuristic policy for a one-warehouse multi-retailer system as well as an optimal policy for the special case of a one-warehouse one-retailer system. Several authors have researched continuous review inventory models without transportation costs that provide a foundation for the models proposed in our paper. In 1915, Harris [19] presents the popular EOQ model. For a finite horizon, Schwarz [33] develops an optimal policy for the EOQ model. Schwarz [32] also derives an optimal inventory policy for a one-warehouse one-retailer model

B.Q. Rieksts, J.A. Ventura / Computers & Operations Research 37 (2010) 20 -- 31

and properties for the optimal policy of a one-warehouse model with multiple retailers. Furthermore, Roundy [31] derives a 98% effective power-of-two heuristic for a one-warehouse multi-retailer system. For inventory models with transportation costs and continuous review, TL transportation is explored by Aucamp [5] and Lippman [29]. In these models, the cost per load does not change with the number of truckloads. Lee [25] considers TL shipments for a onestage model with discounts on the cost per load as the number of truckloads increase. Hwang et al. [20] incorporate an all unit quantity discount on the purchasing cost in Lee's freight discount model. In Tersine et al. [37], a composite EOQ model is also developed with both quantity and freight discounts. Abdelwahab and Sargious [2], as well as Swenseth and Godfrey [36], consider TL and LTL inventory models as disjoint problems and compare the policies for each scenario. Also, Abad and Aggarwal [1] maximize profit in a model with multiple breakpoints for LTL rates and TL shipments. In the distribution of a product to several customers, Burns et al. [8] investigate the options of shipping to each retailer directly or peddling deliveries to more than one customer per truckload. Although our research addresses continuous review inventory policies, several researchers have focused on transportation costs with periodic demand. For periodic demand, Lee [26] develops optimality properties and a polynomial algorithm to solve a model with a TL transportation cost and a minimum order quantity. Alp et al. [3] include stochastic lead-times in a model with a TL transportation cost and periodic demand. For the special case of deterministic lead-times, they show the problem reduces to a problem with at most one truckload, which is the capacitated problem with a fixed dispatching cost. In other research for periodic review models, Li et al. [28] derive a dynamic programming algorithm for a single stage model with the same TL and LTL transportation cost structure that is used in our paper. They also solve a problem where the order quantities are restricted to be a multiple of a full truckload. Chan et al. [11] also present results for discrete demand and an LTL transportation cost called a modified all-unit discount cost. A more general cost structure than the model in our paper is studied in Jaruphongsa et al. [21]. They allow for multiple modes of transportation with each mode having a fixed cost component, a unit cost component, and a fixed cost for each load defined by a cargo capacity. They develop optimality properties and dynamic programming algorithms to compute an optimal solution. For cases with only two modes of transportation, their algorithms are polynomial. In a one-warehouse one-retailer system with periodic review, Lee et al. [27] analyze a model with holding and setup costs at each stage, a TL transportation cost to the retailer, and a waiting cost at the retailer for unsatisfied demand. They also derive optimality properties and use a network approach to solve the model in polynomial time. Jaruphongsa et al. [22] solve a model similar to that of Lee et al. [27], but they include both TL and LTL transportation costs. Also, for a onewarehouse one-retailer system, Kaminsky and Simchi-Levi [23] use a dynamic programming approach for a model with a concave transportation cost and discrete demand. For multiple items, Anily and Tzur [4] develop a model with a TL transportation cost. Items have different holding costs, but are assumed to require the same volume and weight for transportation. They use dynamic programming to compute the arc costs required for a shortest path formulation of the problem. Also, for periodic review, Chan et al. [12] extend the results of Chan et al. [11] to a one-warehouse multi-retailer model with a modified all unit-discount cost for LTL transportation. In solving the one-warehouse one-retailer model in our paper, a sequence of subproblems with concave transportation costs and continuous demand is solved. For concave production costs, Dobos [14] and Bensoussan and Proth [6] explore optimal control problems for the production rates of inventory systems with continuous

21

demand that is not necessarily constant. Several authors such as Lal and Staelin [24], Dada and Srikanth [13], and Weng [40] consider concave costs in the context of incremental quantity discounts to coordinate channels between buyers and sellers. An overview of literature in quantity discounts is provided by Benton and Park [7]. Some authors consider continuous review models with two facilities and transportation costs that are particularly relevant to this research. Hahm and Yano [15] extend Schwarz's [32] results for a one-warehouse one-retailer model to consider production and delivery frequency with TL transportation. Hahm and Yano [16–18] develop extensions of their own model to multiple items. Often in the literature, the one-warehouse one-retailer problem is referred to as the buyer–vendor problem. Toptal et al. [39] solve a version of the buyer–vendor problem with TL transportation costs. However, both Hahm and Yano [15] and Toptal et al. [39] assume that the order intervals at each facility are stationary. Toptal and Cetinkaya [38] examine channel coordination for the model of Toptal et al. [39] and analyze the differences between centralized and decentralized solutions. Cetinkaya and Lee [10] consider policies with nonstationary order intervals for outbound transportation decisions with TL shipments. Their model has similarities to the one-warehouse one-retailer model discussed in this research. However, our research generalizes their results to include a setup cost at the retailer and an LTL transportation cost. Rieksts and Ventura [30] develop results for the single stage model that provide a foundation for this paper. In the remaining sections of this paper, two inventory models that consider TL and LTL transportation costs simultaneously are proposed. Section 2 summarizes the relevant work for the single stage model derived by Rieksts and Ventura [30]. These results are useful in the development of the optimal policy for a one-warehouse one-retailer system that Section 3 presents. In Section 4, a power-oftwo heuristic is presented for a one-warehouse multi-retailer model. Section 5 demonstrates the effectiveness of the heuristic through a computational study. Conclusions and directions for future research are given in Section 6. 2. Single stage models Results for single stage models over an infinite and finite planning horizon are utilized in developing an optimal policy for two stages. This section defines basic notation, optimality properties, and two algorithms from [30] for single stage models. In the single stage model, demand occurs at a constant rate d per time unit. To definite the costs, let h be the positive per unit holding cost and K be the positive setup cost incurred for each order. For transportation costs, s represents a per unit transportation cost for an LTL carrier. A TL shipment has a cost of CT per load where the quantity QT is the capacity of a load. For convenience of notation, the capacity is considered as the time (TQ ) required for consuming QT units of product at demand rate d. That is, TQ = QT /d. An assumption is made that sdT Q > CT . Otherwise, only LTL transportation would be used. For both infinite and finite horizons, an optimal policy satisfies the zeroinventory ordering property. That is, an order is only placed if the inventory level is zero. Since the zero-inventory ordering property is satisfied, the optimal policy may be defined in terms of its order intervals. A policy is called stationary if its order intervals have a constant length. Over the infinite horizon, there is an optimal policy that is stationary for the single stage model. If T denotes the order interval for a stationary policy, then the number of full truckloads used to transport the order is defined as T/TQ . The average cost for a stationary policy with order interval T is Z(T) = K/T + hdT/2 + C(T)/T,

(1)

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B.Q. Rieksts, J.A. Ventura / Computers & Operations Research 37 (2010) 20 -- 31

300 250 C(T)

200 150 100 50 0 0

5

10

15

20

25

T Fig. 1. Total transportation cost as a function of the order interval.

LTL Only TL Only Both TL and LTL

30 25 C(T)/T

20 15 10 5 0 0

5

10

15

20

25

T Fig. 2. Average transportation cost vs. order interval.

where the transportation cost function C(T) is defined as C(T) = min{sd(T − TQ T/TQ ) + CT T/TQ , (T/TQ  + 1)CT }.

(2)

To illustrate the calculations of these transportation costs, we consider an example with QT = 10, d = 2, s = 10, and CT = 50. In this example, the capacity of a truck in the time domain is TQ = QT /d = 5. Fig. 1 displays the total transportation cost as a function of the order interval T. Now we compute the transportation costs for three different order intervals to demonstrate different scenarios for shipments. First, suppose the order interval is T = 2. In this case, the cost of using LTL exclusively is less than the cost of transporting a partially filled TL shipment, so the total transportation cost is C(2) = sdT = 40. In the second scenario, let the order interval be T = 6. Both modes of transportation are used for a shipment of this size to yield a transportation cost of C(6) = sd(T − TQ T/TQ ) + CT T/TQ  = 70. For the final scenario, suppose the order interval is T = 8. Since using TL transportation exclusively has a lower cost than using both modes simultaneously, an over-declared TL shipment is used with a cost of C(8) = (T/TQ  + 1)CT = 100. The inventory models developed in this paper have two stages, but transportation costs for a facility are computed in the same manner as the single stage model. Fig. 2 compares the average transportation cost of using TL exclusively, LTL exclusively, or both modes of transportation simultaneously. The cost savings from allowing two modes of transportation simultaneously is illustrated in this figure. The maximum advantage occurs when an order requires a full TL load and a small additional amount that could be sent with an LTL carrier. As the number of full TL loads in a shipment increases, the advantage gained from using concurrent modes of transportation decreases. To quantify the savings, we can compare the optimal costs of using each mode exclusively to using both modes simultaneously. Consider the transportation costs for Figs. 1 and 2, but change the LTL rate to s = 6. Also, let K = 100 and h = 2. For these costs, the optimal policy for using LTL transportation exclusively is the EOQ solution with an order interval of 7.1 and an average cost of 40.3.

Aucamp [5] provides the optimal policy for using TL transportation exclusively with an optimal average cost of 40 for alternate optimal order intervals of 5 and 10. For allowing both modes of transportation simultaneously, the optimal order interval is T ∗ =6.7 with cost Z(T ∗ )= 38.8. Thus, this example shows a cost savings of 3% by allowing two modes of transportation. According to Rieksts and Ventura [30], the cost savings from using both modes of transportation simultaneously compared to a single mode can be no more than 6% due to convexity of holding and setup costs. The tradeoff between holding, setup, and transportation costs further reduces the savings. In the computation of an optimal policy allowing both modes of transportation, bounds on the optimal order interval are useful since the cost is neither convex nor concave. For the infinite horizon single stage  model, T ∗ must be within the range [x∗ TQ , (x∗ + 1)TQ ], where x∗ =  2K/hd/TQ . Observe in Fig. 2 that the average transportation cost for allowing both modes simultaneously has minimum at each endpoint of the range [x∗ TQ , (x∗ + 1)TQ ]. Full truckloads are being utilized with an average transportation cost of 10. By convexity, holding and setup costs are only greater outside this range than at the endpoints (see [30] for a formal proof of the bounds). If the optimal order interval T ∗ is greater than x∗ TQ + CT /sd, then x∗ + 1 loads of TL transportation are used. Otherwise, x∗ loads of TL transportation are used and the remaining freight is shipped with an LTL carrier. Over the range [x∗ TQ , x∗ TQ + CT /sd], it can be shown that the average holding, setup, and transportation cost is either concave or convex depending on the parameters of the problem. Also, the average cost is convex over the range [x∗ TQ + CT /sd, (x∗ + 1)TQ ]. In Appendix A, Algorithm RV1 provides the steps for computing the optimal order interval T ∗ [30]. In the finite horizon model for one stage, let TH correspond to the length of the finite planning horizon. Although these order intervals may be non-stationary for the finite horizon model, there are some properties that the optimal policy must satisfy. One property is that at most n max orders occur in an optimal policy, where nmax = 2TH /min{TQ , 2K/hd}. If more than nmax orders occur, then the policy would have two order intervals that could be combined  into a single order interval that is less than TQ and less than 2K/hd. By combining two order intervals, the transportation cost does not increase. The cost is reduced by the convexity of the holding and setup costs (see [30] for a formal proof). Given that there are n orders in an optimal policy, each order interval has a minimum length of tm (n), where tm (n) = TH /(nT Q )TQ . Furthermore, these order intervals may have at most two different lengths. One type of order interval only uses TL transportation and may be expressed as tF +tm (n), where CT /sd  tF  TQ . The other order intervals may use LTL transportation and are represented as tL + tm (n), where 0  tL  CT /sd. For the purpose of notation, the planning horizon TH is reduced to TR = TH − ntm (n), where TR is called the reduced planning horizon. Given j order intervals of length tF + tm (n) and n − j intervals of length tL + tm (n), the order interval tL may be expressed in terms of tF and TR . This implies that the single variable formulation for an optimal policy in terms of tF for a finite horizon is (P)

minimize ZH (tF ) =

jhd (tF + tm (n))2 2 2  (n − j)hd TR −jtF + +tm (n) +sd(TR − jtF ) 2 n−j +CT (ntm (n)/TQ +j)+nK

subject to

CT /sd  tF  TQ , TR /j − (n − j)CT /jsd  tF  TR /j.

(3) (4)

By the convexity of the objective, the formulation (P) is easily solved as shown by Rieksts and Ventura [30]. To obtain the optimal policy

B.Q. Rieksts, J.A. Ventura / Computers & Operations Research 37 (2010) 20 -- 31

over the finite horizon, the minimum cost policy is selected among the optimal policies for various cases of problem P corresponding to n = 1 to nmax and j = 0 to n. The steps for this algorithm are given in Appendix A by Algorithm RV2. These results are used in Section 3. 3. One-warehouse one-retailer model over an infinite horizon In this section, an exact algorithm is developed to compute a policy for a one-warehouse one-retailer model over an infinite planning horizon. Transportation costs are incurred for transferring a single product between Stage 0 referring to the warehouse and Stage 1 corresponding to the retailer. Applications of this model are not limited to a warehouse directly supplying a single retailer. For example, Fig. 3 provides an example with a manufacturer and a distribution center. In the system, raw materials are transferred from an external supplier to the manufacturer. The manufacturer incurs a setup cost for each production run to convert raw materials into the product. From the manufacturer, the product is shipped to the distribution center that must satisfy customer demand. Products from one production run may be held in inventory to supply several orders from the distribution center. The assumptions of the one-warehouse one-retailer model are the following.

long-run average setup, holding, and transportation cost in a system with centralized control. The policies under consideration are restricted to any feasible policy that satisfies downstream demand. These policies are defined by the number of setups, the setup times and the order quantities. Let the number of setups at Stage i through time t be denoted as ni (t) for i = 0, 1. Also, suppose i,k is the time for the kth setup and i,k is the kth order quantity at Stage i, for i = 0, 1. Let i be the sequence of setup times {i,k } and i be the sequence of order quantities {i,k } at Stage i, for i = 0, 1. The inventory levels at time u with u ∈ [0, t] are defined to be I0 [u; n0 (u), n1 (u), 0 , 1 , 0 , 1 ] and I1 [u; n1 (u), 1 , 1 ], respectively. The average cost of the optimal policy is determined by limt→∞ z(t)/t, where ⎧ ⎨  t h0 I0 [u; n0 (u), n1 (u), 0 , 1 , 0 , 1 ] du z(t) = minimize ⎩ n0 (t),n1 (t), 0 0 ,1 ,0 ,1

+ h1

Most of these assumptions are consistent with previous research by Roundy [31] and Schwarz [32] for models without transportation costs. Although the demand is assumed to be constant and deterministic (1), these algorithms could be implemented in practice by requiring safety stock at all times. The goal is to minimize the

External Supplier

Manufacturer (Stage 0)

 0

t

I1 [u; n1 (u), 1 , 1 ] du + K0 n0 (t)

+K1 n1 (t) +

n 1 (t)

⎫ ⎬

C(1,k /d)

k=1

subject to 1. Customer demand occurs at a constant deterministic rate of d products per time unit. 2. The inventory levels are under continuous review. 3. No shortages are permitted. 4. The costs for transporting raw materials or products to Stage 0 are incurred by the external supplier and not considered in this model. 5. A transportation cost is included for shipping products from Stages 0 to 1. This cost is defined in the same manner as Section 2 with both TL and LTL modes of transportation available. 6. For each shipment received by either stage, there is a positive setup cost denoted as K0 and K1 . 7. The purchasing cost of raw materials or products consists of a variable cost and fixed component. Since the demand rate is constant, the variable purchasing cost does not affect the optimization. The fixed component is considered as part of the Stage 0 setup cost K0 . 8. Holding costs are also incurred at each stage with h0 and h1 representing the unit holding cost per time unit at Stages 0 and 1, respectively. For notational convenience, holding costs are represented in terms of echelon holding costs or the additional holding costs in comparison to the previous stages. Let the echelon holding costs be h0 = h0 at the warehouse and h1 = h1 − h0 at the retailer, with hi > 0 for i = 0, 1. 9. All cost rates are constant with respect to time. 10. The initial inventory level is zero. 11. The lead-times are constant and deterministic and, thus considered negligible in the computations.

23

⎭

I0 [u; n0 (u), n1 (u), 0 , 1 , 0 , 1 ]  0 for all u ∈ [0, t], I1 [u; n1 (u), 1 , 1 ]  0

for all u ∈ [0, t].

Since this set of policies is very broad, some properties of an optimal policy are derived to develop a tighter formulation for the twostage problem. One property is that Stage 0 only has a setup at a given time if Stage 1 places an order at that time. A policy with this property is called a nested policy. Property 1 shows that an optimal policy must be nested and satisfy the zero-inventory ordering property over a finite horizon. Schwarz [32] provides proves this property with inventory transformations for a model without transportation costs. These proofs are valid for the model in this paper since transportation costs do not change in Schwarz's inventory transformations. Although all optimal policies over an infinite planning horizon may not satisfy these criteria, there does exist at least one optimal policy with these properties. Sun [34] provides arguments to show that an optimal policy over an infinite horizon with these properties may be constructed with a sequence of finite horizon policies. Property 1. In the one-warehouse one-retailer model, an optimal policy must be nested and satisfy the zero-inventory ordering property at each stage over a finite horizon. Since the optimal policy is nested and satisfies the zero-inventory ordering property, there exists an optimal policy with a stationary order interval at Stage 0. That is, consider an optimal policy with non-stationary order intervals at Stage 0. There is an alternative optimal policy by repeating one of the order intervals. Let T0 denote the stationary Stage 0 order interval. Lemma 1 provides bounds for the optimal value of T0 . Lemma 1. Let T1∗ be the optimal policy for Stage 1 produced by Algo rithm RV1, T0 = 2K0 /h0 d, and  =T0 /T1∗ . Then any optimal stationary order interval at Stage 0, T0 , is in the range [T1∗ , ( + 1)T1∗ ].

Distribution Center (Stage 1)

Fig. 3. Application of one-warehouse one-retailer model.

Customer

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B.Q. Rieksts, J.A. Ventura / Computers & Operations Research 37 (2010) 20 -- 31

Proof (By contradiction). Suppose that there exists an optimal policy with a Stage 0 order interval T0 such that T0 < T1∗ or T0 > ( + 1)T1∗ . First, assume that the case with T0 < T1∗ is true. Consider that there exists a feasible stationary policy with a Stage 0 order interval Tˆ 0 =

T1∗ and Stage 1 order interval T1∗ . Since the holding and setup costs

at Stage 0 represent a convex function with a minimum at

T0 ,

these

costs at Stage 0 for the interval Tˆ 0 are less than those for T0 . Consider that the order interval T1∗ is optimal for Stage 1 for the single stage model. This implies the cost incurred at Stage 1 for the policy for Tˆ 0 is less than or equal to the cost at Stage 1 for the policy corresponding to T0 . Thus, the cost for the policy with interval T0 is more than that for policy Tˆ 0 . Since this contradicts the optimality of T0 , the optimal Stage 0 order interval has a lower bound of T1∗ . A similar argument verifies that ( + 1)T1∗ is an upper bound on T0 . Therefore, T0 is in the range [T1∗ , ( + 1)T1∗ ].  The bounds on the Stage 0 order interval that Lemma 1 provides are important for the algorithm to compute an optimal policy. Let these bounds be denoted by T0 min = T1∗ and T0 max = ( + 1)T1∗ . By Property 1, the optimal policy for the two-stage model satisfies the zero-inventory ordering property and is nested. Given the optimal Stage 0 order interval, an optimal policy at Stage 1 can be computed by utilizing the results for the single stage model over a finite planning horizon. If there are n orders at Stage 1 between consecutive Stage 0 orders, let tk denote the length of the kth order interval at Stage 1, for k = 1, . . . , n. By assuming the upper bound T0 max and using the results for the single stage model over a finite horizon, an upper bound on the maximum number  of orders at Stage 1 is nmax = 2T0 max /min{TQ , T1 }, where T1 = 2K1 /h1 d. For n orders at ˆ for an optimal policy is Stage 1, the formulation (P) ⎛

ˆ (P)

minimize

⎞ n 2 T0 K0 /T0 +h0 dT 0 /2+ ⎝ (K1 + h1 dtk /2 + C(tk ))⎠ k=1

subject to

n

tk = T0 ,

(5)

Table 1 Possible order intervals at Stage 1 in the optimal policy.



tF  (T0R )

tL (T0R )

1 2 3 4 5

T0R /n + (n − j)s/nh1 T0R /j TQ T0R /n –

T0R /n − js/nh1 0 (T0R − jT Q )/(n − j) – T0R /n

interval, T0R . The order intervals for the th form as a function of T0R are denoted as tF  (T0R ) and tL (T0R ), for  =1, . . . , 5. Since the objective for (P) is strictly convex, one possible form of the optimal policy occurs at tF1 (T0R )=T0R /n+(n−j)s/nh1 and tL1 (T0R )=T0R /n−js/nh1 with ZH (tF ) = 0. Two of the other forms correspond to tight constraints in the formulation (P). If the upper bound from Eq. (4) is tight, the optimal policy is of the form tL2 (T0R ) = 0 and tF2 (T0R ) = T0R /j. If the upper bound from Eq. (3) is tight, the optimal policy has the form tF3 (T0R )=TQ and tL3 (T0R )=(T0R −jT Q )/(n−j). The cases where the lower bounds from Eqs. (3) and (4) are tight do not need to be considered as possible optimal policies for the two-stage model. Although these cases may be optimal for specific values of n and j, they are always dominated by an extreme case with j = 0 or j = n. The last two forms of the optimal policy correspond to these extreme values. If j = n, then tF4 (T0R ) = T0R /n and there are no intervals of length tL4 (T0R ). Similarly, if j = 0, then tL5 (T0R ) = T0R /n and there are no intervals of length tF5 (T0R ). Observe that each possible form of the optimal policy is a function of T0R and the parameters of the problem. If a specific form of ˆ the optimal policy and range of T0 is assumed, the formulation (P) may be transformed into a formulation with T0R as the only variable. Let these single variable formulations be denoted as (Pˆ  ), where  corresponds to the different options in Table 1. The general representation of these formulations is  jh1 d (Pˆ  ) minimize Z (T0R ) = (tF  (T0R ) + tm (n))2 2

k=1

T0 min  T0  T0 max ,

+

(6)

T0 , t1 , . . . , tn  0.

+ (n − j)sdtL (T0R ) + nK 1

An optimal policy for the two-stage model may be computed by ˆ for n=1, . . . , nmax and selecting solving a sequence of formulations (P) the best policy. ˆ we assume the value of T0 in difTo solve the formulation (P), ferent segments of the interval [T0 min , T0 max ]. First, suppose T0 = T0 min . By the results for the finite horizon, each of the n orders has an order interval tk in the range [tm (n), tm (n) + TQ ] where tm (n) = ˆ is valid for all T0 on the interTQ T0 min /nT Q . This formulation (P) val [ntm (n), ntm (n) + nT Q ] with ntm (n)  T0 min . After the optimal policy is computed for this segment of T0 , tm (n) is incremented by TQ and the next segment is considered. This procedure is repeated until T0 max < ntm (n). In computing the optimal policy for a particular segment, the Stage 0 order interval is expressed as T0 = T0R + ntm (n) for ntm (n)  T0  ntm (n) + nT Q . T0R is called the reduced Stage 0 order interval with bounds 0  T0R  nT Q . Results for the finite horizon single stage model show that each tk may be represented as tk = tL + tm (n) or tk = tF + tm (n). If j of the n setups at Stage 1 correˆ spond to order intervals tF + tm (n), the constraint from Eq. (5) of (P) maybe expressed as jtF + (n − j)tL = T0R .

(n − j)h1 d (tL (T0R ) + tm (n))2 2

(7)

For the finite horizon, there are five possible forms for the optimal policy given j and n. Table 1 displays these different forms of the order intervals tF and tL as a function of the reduced Stage 0 order

+ K0 + CT (ntm (n)/TQ + j)  h0 d (ntm (n) + T0R )2 (ntm (n) + T0R )−1 + 2 subject to

L  T0R  U ,

where L and U correspond to the bounds from Eqs. (3) and (4) on tF  (T0R ) and tL (T0R ), for  = 1, . . . , 5. The specific bounds L and U are given in Table B.1 in Appendix B. The optimal policy may be computed by solving each of these formulations (Pˆ  ) for every combination of j, n, and the segments containing T0 . For each (Pˆ  ) the first derivative of the cost function, Z (T0R ), is of the form 2 Z (T0R ) = (a T0R + b T0R + c )(ntm (n) + T0R )−2 ,

(8)

where ai , bi , and ci are given in Tables B.2 and B.3 in Appendix B. The second derivative is Z

(T0R ) = (2a ntm (n)T0R −b T0R +2c +b ntm (n))(ntm (n)+T0R )−3 .

(9)

Since it can be verified that 2a ntm (n)−b =0, the second derivatives are either negative or non-negative for all T0R > 0 depending upon the parameters of the model, for  = 1, . . . , 5. This implies that the objective function is either concave or convex. If the single variable objective function, Z (T0R ), is concave, its minimum occurs at an

B.Q. Rieksts, J.A. Ventura / Computers & Operations Research 37 (2010) 20 -- 31

endpoint of the interval [L , U ]. If Z (T0R ) is convex, the root of the first derivative must be considered as well. Let R denote the root of the first derivative of objective function Z (T0R ) if a real root exists. In each case, R is the root of a quadratic equation of the form 2 + b T0R + c = 0. a T0R

(10)

From the quadratic formula, if a real R exists, it is of the form R =

−b +



b2 − 4a c

2a

25

T0 > 0 depending upon the parameters of the model, the objective functions are either convex or concave. This implies that if the formulation is feasible, the minimum for each (Pˆ  ) occurs at the root R or an endpoint of the interval [L , U ]. Since Steps 5 and 6 select this appropriate minimum and Steps 1–4, 7 and 8 iterate through each possible formulation, the policy produced by the algorithm is optimal.  4. One-warehouse multi-retailer model over an infinite horizon

.

(11)

Since b > 0 and T0R must be positive, the other root does not need to be considered. If b2 − 4a c  0 and L  R  U , the minimum of Z (T0R ) occurs at R . Otherwise, the minimum occurs at an endpoint of the interval [L , U ]. Algorithm A1 presents the steps for explicitly computing an optimal policy for the two-stage model.   Algorithm A1. Step 0: Let T0 = 2K0 /h0 d, T1 = 2K1 /h1 d, T1∗ be the optimal policy for Stage 1 produced by Algorithm RV1, and  = T0 /T1∗ . Set T0 min = T1∗ , T0 max = ( + 1)T1∗ , n = 1, nmax = 2T0 max /min{TQ , T1 }, and Z ∗ = ∞. Step 1: Set tm (n) = TQ T0 min /nT Q . Step 2: Set j = 0,  = 5, and go to Step 4. Step 3: Set  = 1. Step 4: If U < L , go to Step 8. If Z

(T0R ) > 0, b2 − 4a c  0 and L  R  U , go to Step 5. Otherwise, go to Step 6. Step 5: Let ˆj = j, nˆ = n, ˆ = , Tˆ 0 = ntm (n) + R , Zˆ = Z (R ), and go to Step 7. Step 6: Set ˆj = j, nˆ = n, ˆ = , Tˆ 0 = ntm (n) + arg min{Z (L ), Z (U )}, and Zˆ = min{Z (L ), Z (U )}. Step 7: If Zˆ  Z ∗ , go to Step 8. Otherwise, update the current best ˆ ˆ ∗ = ˆ , T ∗ = Tˆ 0 , and Z ∗ = Z. policy by setting j∗ = ˆj, n∗ = n, 0

Step 8: If  < 3, set  =  + 1 and return to Step 4. If j < (n − 1), set j = j + 1 and return to Step 3. If j < n, set j = n,  = 4, and return to Step 4. If ntm (n) + nT Q < T0 max , let tm (n) = tm (n) + TQ and return to Step 2. If n < nmax , set n = n + 1 and return to Step 1. Otherwise, ∗ stop with the optimal policy defined in terms of j∗ , n∗ ,  , T0∗ , and Z ∗ . Theorem 1 proves that Algorithm A1 is optimal for the two-stage model.

Theorem 1. The policy derived with Algorithm A1 is optimal for the two-stage model over the infinite horizon. Proof. By Property 1, there exists an optimal policy for the twostage model that is nested and satisfies the zero-inventory ordering property. This implies that there is an optimal policy with a stationˆ is a valid formulation ary order interval at Stage 0. Furthermore, (P) for the optimal policy for the two-stage model, given n order intervals exist at Stage 1 between consecutive orders at Stage 0. Lemma 1 provides the bounds T0 min and T0 max on the Stage 0 order interval. By assuming T0 max and utilizing results for the single stage finite horizon model, the maximum number of order intervals at Stage 1 between consecutive Stage 0 orders is at most nmax . An optimal policy for the two-stage model can be computed by considering n = 1, . . . , nmax setups at Stage 1 for all T0 in the range [T0 min , T0 max ]. The finite horizon results guarantee that the optimal policy for Stage 1 between consecutive orders at Stage 0 is one of the five scenarios given in Table 1. For a given n, the range [T0 min , T0 max ] may be reduced into several segments [ntm (n), ntm (n) + nT Q ]. For a particular segment, the different scenarios in Table 1 are represented as the formulations (Pˆ  ) with a single variable. Since the second derivatives of the objective functions are either non-negative or negative for all

In the one-warehouse multi-retailer system, computing an optimal policy is difficult since the optimal policy is not necessarily nested. Roundy [31] shows that this is true even for the model without transportation costs. In this section, a lower bound on the minimum cost and a power-of-two heuristic policy are developed for a one-warehouse multi-retailer system with transportation costs. For this model, the warehouse, denoted as Facility 0, buys a product from an external supplier. From the warehouse, the product is transferred to satisfy the demand of each of the y retailers. Fig. 4 illustrates an example of a one-warehouse multi-retailer system. Facility i denotes the ith retailer, i = 1, . . . , y. Let Y = {1, . . . , y} denote the set of indices for the retailers. The demand at each retailer is defined as di , for i ∈ Y. Similar to the models for one-stage and one-warehouse one-retailer systems, the objective is to minimize the long-run average setup, holding, and transportation costs. The assumptions are analogous to those listed for the one-warehouse one-retailer model. An assumption is made that the external supplier incurs the transportation cost to the warehouse. The purchasing cost given to the warehouse has fixed and variable components. The variable cost does not affect the optimization and is not included in the objective function. The fixed cost is considered as part of the setup cost at the warehouse. The setup cost at the warehouse also includes any administrative costs for an order. Let Ki denote the setup cost at Facility i, for i ∈ {0} ∪ Y. For holding costs, let hi be the holding cost per unit at the ith facility, for i ∈ {0} ∪ Y. An assumption is made that the unit holding cost at the warehouse is less than at the retailers. That is, h0 > hi , i ∈ Y. In the objective function, the holding costs are represented in terms of echelon holding costs as h0 = h0 at the warehouse and hi = hi − h0 at the ith retailer, for i ∈ Y. These assumptions are also consistent with the one-warehouse multi-retailer model of Roundy [31]. Another possible application of this model is a manufacturing facility with a setup cost for each production run. After a production run, the product is transferred to a set of y distributors with constant demand rates. For transportation, both TL and LTL modes of transportation are available to transfer the product from Facilities 0 to i, for i ∈ Y. For TL transportation, the cost for transporting a truckload from Facilities 0 to i is denoted by CTi , for i ∈ Y. The capacity of a standard truck is denoted as QT units of product. TQi represents the time required for Facility i to consume QT units. That is, TQi = QT /di . For the LTL carrier, the cost of shipping a unit of product from Facilities 0 to i is denoted as si , for i ∈ Y. An assumption is made that CTi < si QT . If this is not the case, LTL transportation always has a lower cost for Facility i.

External Supplier Regional Warehouse

Retailer 1

Retailer 2

. . . Retailer y-1

Fig. 4. One-warehouse multi-retailer system.

Retailer y

26

B.Q. Rieksts, J.A. Ventura / Computers & Operations Research 37 (2010) 20 -- 31

Since this is a constant cost, the heuristic in this section is applied without a transportation cost for this facility. With CTi < si QT , the total transportation cost for shipping di T units from Facilities 0 to i to satisfy the demand at Facility i for a time interval T is  si di (T − xT Qi ) + xC Ti for xT Qi  T < CTi /(si di ) + xT Qi , (12) Ci (T) = (x+1)CTi for CTi /si di +xT Qi  T < (x+1)TQi , where the integer x = T/TQi  represents the number of full TL shipments for the order. Similar to the one-warehouse one-retailer model, all policies that satisfy downstream demand are considered feasible. As shown in Section 3, even the case with one retailer can have non-stationary order intervals at the retailer in an optimal policy. Although an optimal policy is difficult to characterize even without transportation costs, a lower bound on the optimal cost can be computed by extending the results of Roundy [31]. A power-of-two heuristic policy is developed and compared to this benchmark. This heuristic policy has a structure that is simple to implement and compares favorably to a lower bound in a computational study. In the heuristic, a power-of-two policy with stationary order intervals is computed. Since the policy may be non-nested, the order quantities at the warehouse are not necessarily stationary as discussed by Roundy [31]. Consider an example with two retailers with demand one at each retailer. The first retailer orders only one unit from the warehouse at each time unit. However, the second retailer orders two units from the warehouse every other time unit. Suppose the warehouse orders only enough to satisfy orders it receives from these retailers at each time unit. This implies the warehouse has stationary order intervals of length one time unit, but has nonstationary order quantities alternating between orders of one and three units. Although a policy with stationary order intervals may not be optimal even for a single retailer, computational results show the cost of this policy is close to the lower bound. To define a policy for the one-warehouse multi-retailer system with stationary order intervals, let T = (T0 , . . . , Ty ), where Ti denotes the order interval for Facility i, for i ∈ {0} ∪ Y. In a power-of-two policy, each order interval must be a multiple of the base planning period TB . This base planning period may be selected as a convenient time interval such as a week or optimized to compute a more efficient policy. The formulation for the optimal power-of-two formulation is given as (Ki /Ti + hi di Ti /2 (P02) minimize Z(T) = K0 /T0 + i∈Y

+ h0 di max{T0 , Ti }/2 + Ci (Ti )/Ti ) subject to Ti = 2i TB

for i ∈ {0} ∪ Y,

TB  0,

i integer for i ∈ {0} ∪ Y. In the objective function, there are two cases to consider for the average holding cost at each retailer since the policy is not nested. These cases are Ti < T0 and Ti  T0 . For the case of Ti  T0 , the product for Facility i is crossdocked at the warehouse and sent directly to the retailer. This implies the average holding cost is (h0 + hi )di Ti /2. In the other case with Ti < T0 , the inventory at the warehouse replenishes orders at the retailer every Ti time units with a holding cost of hi di Ti /2 + h0 di T0 /2. For a model without transportation costs, Roundy [31] powerof-two policy for a fixed base period is within 6% of an analytical lower bound. A counterexample shows that this analytical bound is no longer valid for the model in this section with transportation costs. For this example, we consider a one-warehouse one-retailer model since Algorithm A1 allows us to compute an optimal policy.

The parameters of the example are K0 = 100, h0 = 1, K1 = 100, h1 = 1, TQ = 10, d = 2, CT = 100, and s = 20. From Algorithm A1, an optimal ∗ policy is defined by j∗ = 0, n∗ = 1,  = 5, T0∗ = 10, and Z ∗ = 50. Now suppose the base planning period for a power-of-two policy is TB = 6. The optimal power-of-two policy for these parameters can be determined by enumerating policies within bounds similar to those for the single stage model over the infinite horizon. This solution is T0 = 12, T1 = 12, and Z(T) = 55.7. Thus, a difference of more than 10% exists between the overall optimal cost and the optimal cost of a power-of-two policy with a fixed base period. Although Roundy's analytical bounds on effectiveness are not valid for the one-warehouse multi-retailer model in this paper, we develop a lower bound on the optimal cost to use as a benchmark. The formulation (P02) may be modified to produce this lower bound as shown by Theorem 2. Theorem 2. The optimal cost for the one-warehouse multi-retailer model is bounded below by the optimal policy for the formulation (LB) minimize



K0 /T0 + +



(Ki /Ti + hi di Ti /2 + h0 di max{T0 , Ti }/2)

i∈Y

CTi /TQi

i∈Y

subject to Ti  0

for i ∈ {0} ∪ Y.

Proof. Since CTi < si QT , a lower bound on the transportation cost to Facility i is CTi /TQi . This is the scenario where TL transportation is used with a truck at full capacity. Thus, the average transportation  cost for the system is bounded below by i∈Y CTi /TQi . Since this term is constant, it does not effect the optimization. In Roundy [31], the  formulation LB without the constant i∈Y CTi /TQi is shown to be a lower bound on the optimal holding and setup cost for the onewarehouse multi-retailer system. Since the model of Roundy [31] is a relaxation of the model with transportation costs, the optimal holding and setup cost to LB is a lower bound to the optimal holding and setup cost with transportation costs. Therefore, LB is a lower bound on the optimal holding, setup, and transportation cost for the one-warehouse multi-retailer model.  The algorithm by Roundy [31] may be utilized to compute a numerical value for the lower bound. We modify Roundy's algorithm with two changes to obtain a stationary power-of-two policy for the model with transportation costs. In Roundy's algorithm for a problem without transportation costs, the power-of-two constraint is relaxed and a policy is computed for the relaxed formulation. The relaxed policy is rounded to a powerof-two policy. We develop a modified power-of-two heuristic called Algorithm H1. The first modification of Roundy's algorithm involves the computation of retailer order intervals. In Roundy's algorithm, the cost at a retailer in the relaxed  problem is minimized according  to Harris's [19] EOQ solution by 2Ki /(h0 + hi )di if T0 < Ti or 2Ki /hi di if T0 > Ti . In Algorithm H1, the EOQ solutions are replaced by solutions from Algorithm RV1 that consider transportation cost. In computing the relaxed policy, let  i be the optimal order interval for the single stage model over the infinite planning horizon by Algorithm RV1 for Facility i with a holding cost of h0 + hi , for i ∈ Y. Suppose the optimal stationary warehouse order interval T0 is less than Ti . In this case, the cost for the single stage policy  i is a lower bound on the average cost at Facility i for any policy with a stationary time between orders. Similarly, if T0 > Ti , let i be the single stage policy with holding cost hi from Algorithm RV1 for Facility i, for i ∈ Y. Observe that  i is always less than or equal to i . For a given T0 , let G = {i|T0 <  i }, E = {i| i  T0  i }, and L = {i|T0 > i }. To compute the policy with the

B.Q. Rieksts, J.A. Ventura / Computers & Operations Research 37 (2010) 20 -- 31

power-of-two constraint relaxed, the breakpoints  i and i are sorted in non-decreasing order, for i ∈ Y. Roundy's algorithm considers T0 on intervals between consecutive breakpoints. The second change from Roundy's algorithm is that each range of breakpoints must be considered. Roundy's algorithm takes advantage of the convexity of the objective to reduce the search space across breakpoints. However, Algorithm H1 must consider each pair of breakpoints since the computations of  i and i involve nonconvex transportation costs. Let LBP and RBP denote a pair of consecutive breakpoints. For each pair LBP and RBP, a policy is computed by assuming LBP  T0  RBP. In these policies, the order intervals at the retailers are set to  i for i ∈ G, i for i ∈ L, and T0 for i ∈ E. The warehouse order interval T0 is optimized over the interval [LBP, RBP] with respect the setup and holding costs at the Facility i, for each i ∈ {0} ∪ E. By convexity, this problem is easily solved. After a policy is computed for each pair of consecutive breakpoints, the order intervals for the lowest cost policy are rounded off to the nearest power-of-two of the base planning period TB . Algorithm H1 outlines the steps of the power-of-two heuristic. Algorithm H1. Step 1: For i ∈ Y, let i be an optimal order interval for the single stage model over the infinite horizon with h=hi , K =Ki , CT = CTi , s = si , TQ = QT /di , and d = di from Algorithm RV1. Similarly, let  i be an optimal order interval for the single stage model with h = hi + h0 . Step 2: Sort the breakpoints i and  i in non-decreasing order. Assume that T0 is greater than the largest breakpoint. Let G = E = ∅, L = Y and Z(T∗ ) = ∞.     Step 3: Set Tˆ 0 = 2 Ki /( di (hi + h0 ) + di h0 ), LBP = i∈E

i∈E

i∈L

max{maxi∈L {i }, maxi∈E { i }}, and RBP=min{mini∈G { i }, mini∈E {i }, ∞}. If Tˆ 0 < LBP, let Tˆ 0 = LBP. If Tˆ 0 > RBP, let Tˆ 0 = RBP. Step 4: Set Tˆ i = i for i ∈ L, Tˆ i = Tˆ 0 for i ∈ E, and Tˆ i =  for i ∈ G. i

ˆ < Z(T∗ ), let T∗ =T. ˆ If LBP corresponds to i , let L=L−{i} Step 5: If Z(T) and E = E ∪ {i}. If LBP corresponds to  i , let E = E − {i} and G = G ∪ {i}. If L = E = ∅, go to Step 6. Otherwise, return to Step 3. √ Step 6: Set Ti =2i TB where i corresponds to 2i −1 < Ti∗ / 2TB  2i , for i ∈ {0} ∪ Y.

Note that the transportation costs are not considered in computing T0 . Since different retailers may have different TQi 's for i ∈ E, the objective function may become non-convex. To optimize transportation cost for T0 , we must consider local minimums in [LBP, RBP]. These local minimums must be considered from the T0 that minimizes holding and setup cost to the closest values of T0 with minimum transportation cost in each direction. The transportation cost for T0 achieves a minimum at common multiples of all TQi 's for i ∈ E where only TL shipments are used at their capacity. Every local minimum corresponding to a multiple of a TQi for i ∈ E must be considered in this neighborhood. However, rounding this relaxed solution to a power-of-two policy can significantly increase the transportation cost. Recall the example with a 10% increase in overall cost by rounding to a power-of-two policy. Due to convexity, rounding to a power-of-two policy with minimum holding and setup costs will increase these costs by at most 6%. Furthermore, if the transportation cost for Facility i is significant, policies with i ∈ E are dominated by policies with i ∈ L ∪ G that include transportation costs. The cost reduction by optimizing this non-convex transportation cost function does not justify the complexity that would be added to the heuristic. However, we do develop an algorithm for selecting the base planning period that includes the optimization of transportation costs. If the base planning period is selected as a convenient time interval, rounding off to the nearest power-of-two may yield an increase in transportation costs. Since TQi varies between the retailers, the base period cannot be set as a multiple of the time required to

27

consume a truckload of product. In an attempt to minimize the increase in cost due to round off, we optimize the holding, setup, and transportation cost of a power-of-two √ policy B(TB ) with respect to TB √ across the root-two interval [ 0.5, 2]. The optimization problem is minimize B(TB ) = (Ki /TB 2i + hi di TB 2i /2 i∈Y

+ h0 di max{TB 20 , TB 2i }/2 + Ci (TB 2i )/TB 2i ) + K0 /TB 20 √ √ subject to 0.5  TB  2, √ √ 0.5Ti∗  TB 2i  2Ti∗

for i ∈ {0} ∪ Y,

i integer for i ∈ {0} ∪ Y. Roundy [31] presents an algorithm to compute an optimal TB that minimizes holding and setup costs. We modify Roundy's algorithm to include transportation costs in the optimization. Let i be an integer corresponding to 2i −1 < Ti∗  2i , for i ∈ {0} ∪ Y. Note that if √ TB is on the root-two interval and TB  2Ti∗ /2i , then Ti = 2i TB . If √ TB > 2Ti∗ /2i , then Ti = 2i −1 TB . This implies that there are y + 1 breakpoints on the root-two interval at which i decreases from i to √i − 1 as TB increases. Let these breakpoints be denoted as BP i = 2Ti∗ /2i , for i ∈ {0} ∪ Y. Unlike the problem Roundy solved, a segment between two consecutive breakpoints may contain local extrema due to the transportation costs. We modify Roundy's algorithm by computing transportation cost breakpoints, TBP p , between consecutive BP i 's. As TB changes, the transportation cost at each retailer has breakpoints between using some LTL transportation and use TL exclusively. These breakpoints correspond to xT Qi or xT Qi + CTi /(si di ) for some non-negative integer x. Since B(TB ) is either concave or convex between consecutive transportation cost breakpoints, the optimal value of TB is easily computed for each of these segments. Algorithm A2 provides the steps for computing the optimal base planning period, TB∗ , over the entire root-two interval. Algorithm A2. Step 1: Let i = log2 Ti∗  for i ∈ {0} ∪ Y and B(TB∗ ) = ∞. √ √ √ Set BP 0 = 0.5, BP y+2 = 2, and BP i+1 = 2Ti∗ /2i , i ∈ {0} ∪ Y. Step 2: Sort the breakpoints BP m in non-decreasing order. Let BP [m] denote the mth smallest breakpoint. Also, let I(m)=i denote the index of the facility corresponding BP i+1 = BP [m] for m = 1, . . . , y + 1. Set m = 0, A = {0} ∪ Y, and B = ∅. Step 3: Let TBP0 = BP[m] , p = 1, and i = 1.

Step 4: If i ∈ A, let  i = i . If i ∈ B, let  i = i −1. Let x=BP [m] 2i /TQi .

Step 5: If xT Qi +CTi /(si di ) > BP [m] 2i and xT Qi +CTi /(si di ) < BP [m+1] 2i ,

then TBP p = (xT Qi + CTi /(si di ))/2i and p = p + 1.



Step 6: If (x + 1)TQi < BP [m+1] 2i , then TBP p = (x + 1)TQi /2i and p = p + 1.

Step 7: Let x = x + 1. If xT Qi + CTi /(si di ) < BP [m+1] 2i , go to Step 5.

 If xT Qi + CTi /(si di )  BP [m+1] 2 i and i < y, set i = i + 1 and go to Step 4.

If xT Qi + CTi /(si di )  BP [m+1] 2i and i = y, set TBP p = BP [m+1] , pmax = p, and continue to Step 8. Step 8: Sort the transportation cost breakpoints TBP p in nondecreasing order and let TBP [p] denote the pth smallest breakpoint. Let p = 0.

Step 9: Let i = 1, KB = K0 /20 , and HB = 0.



 Step 10: If TBP [p+1] 2 i  TBP [p] 2i /TQi TQi + CTi /(si di ), then



 HB = HB + hi di 2 i /2, and KB = KB + (Ki + TBP [p] 2i /TQi CTi −



TBP [p] 2i /TQi TQi si di )/2i .



Step 11: If TBP [p] 2i  TBP [p] 2i /TQi TQi + CTi /(si di ), then HB = HB +



hi di 2i /2 and KB = KB + (Ki + (1 + TBP [p] 2i /TQi )CTi )/2i .

Step 12: If  0 <  i , then HB = HB + h0 di 2i /2. Otherwise, HB = HB +

h0 di 20 /2.



28

B.Q. Rieksts, J.A. Ventura / Computers & Operations Research 37 (2010) 20 -- 31

Step 13: If i < y, then set i = i + 1 and go to Step 10. Otherwise, continue to Step 14.  Step 14: If KB > 0 and TBP [p]  KB /HB  TBP [p+1] , then Tˆ B =  KB /HB . Otherwise, Tˆ B =arg min{B(TBP [p] ), B(TBP [p+1] )}. If B(Tˆ B ) < B(T ∗ ),

Table 3 Effectiveness of policies for one retailer.

B

then TB∗ = Tˆ B and Ti = TB∗ 2i for i ∈ {0} ∪ Y. Step 15: If p < (pmax − 1), let p = p + 1 and return to Step 9. If m  y, set m = m + 1, A = A − I(m), B = B ∪ I(m), and return to Step 3. Otherwise, stop with TB∗ as the optimal base period and T as the corresponding power-of-two policy.

Computational results show that optimizing TB with Algorithm A2 reduces the average cost for policies computed by Algorithm H1. 5. Computational study The power-of-two heuristic is tested through a computational study. We generate parameters for test problems according to a uniform distribution. Scenario A in Table 2 displays the baseline distributions. For sensitivity analysis, we shift the distribution of each parameter lower and then higher to define the additional scenarios in Table 2. The capacity of a truck is assumed to be QT = 200 for all scenarios. We assume the difference between transportation cost rates at different retailers is proportional to the distance between retailers. To implement this assumption, the ratio CTi /si is assumed to be identical for all i ∈ Y. We consider an example with two retailers to demonstrate how transportation costs are determined for different retailers. Suppose a warehouse has two retailers with Retailer 2 twice the distance from the warehouse as Retailer 1. Let the parameters for TL shipments be CT1 = 20 and CT2 = 40. Assume the ratio CTi /si = 10 is for i = 1, 2. This implies the parameters for LTL shipments are s1 = 2 and s2 = 4. By assuming CTi /si is identical for all retailers, ratios between each si are the same as the corresponding ratios between CTi . Let the demands be d1 = 20 and d2 = 40. Given QT = 200, we have TQ1 = 10 and TQ2 = 5. For each retailer, the transportation cost is computed from si , CTi , TQi , and di in the same manner as the example shown in Fig. 1. Table 3 displays the computational results for one retailer. For each Scenario A–K, 100 test problems were generated according to parameter distributions in Table 2. These problems were solved with Algorithm H1 for a fixed base period (TB = 1) and a variable base period computed with Algorithm A2. For each scenario, Table 3 displays the minimum, average, and maximum effectiveness of the costs of these polices computed with respect to the lower bound in Theorem 2. We observe a gap between the optimal cost and the lower bound as shown by the last three columns of Table 3. The effectiveness of the heuristic would increase if compared to the optimal cost instead of the lower bound. However, the optimal cost is not known for more than one retailer.

Table 2 Scenarios for parameter distributions in computational study. Scenario

A B C D E F G H I J K

Parameters hi

Ki

di

CTi

CTi /si

U(2,6) U(1,2) U(6,10) U(2,6) U(2,6) U(2,6) U(2,6) U(2,6) U(2,6) U(2,6) U(2,6)

U(100,200) U(100,200) U(100,200) U(50,100) U(200,300) U(100,200) U(100,200) U(100,200) U(100,200) U(100,200) U(100,200)

U(50,800) U(50,800) U(50,800) U(50,800) U(50,800) U(25,50) U(800,1000) U(50,800) U(50,800) U(50,800) U(50,800)

U(40,60) U(40,60) U(40,60) U(40,60) U(40,60) U(40,60) U(40,60) U(20,40) U(60,80) U(40,60) U(40,60)

U(50,150) U(50,150) U(50,150) U(50,150) U(50,150) U(50,150) U(50,150) U(50,150) U(50,150) U(25,50) U(150,200)

√ √ TB ∈ [ 0.5, 2]

Scenario TB = 1

A B C D E F G H I J K

Optimal

Min. (%)

Avg. (%)

Max. (%)

Min. (%)

Avg. (%)

Max. (%)

Min. (%)

Avg. (%)

Max. (%)

85.26 82.73 86.20 81.25 86.27 86.31 86.02 88.86 85.18 84.29 85.89

95.38 95.59 96.04 93.09 96.95 94.47 96.42 96.29 94.80 94.74 96.79

99.13 99.75 99.87 99.89 99.96 99.06 99.56 99.48 98.93 99.13 99.72

94.66 95.99 94.61 90.46 97.01 91.09 95.87 95.92 93.09 93.47 96.86

98.35 98.59 97.65 96.04 98.86 97.29 98.40 98.78 98.02 98.09 99.27

100.00 100.00 99.97 99.97 100.00 99.35 99.91 100.00 100.00 100.00 100.00

94.66 96.07 94.61 90.46 97.01 91.09 96.02 95.92 93.09 93.47 96.86

98.36 98.60 97.65 96.04 98.86 97.29 98.42 98.78 98.03 98.10 99.27

100.00 100.00 99.97 99.97 100.00 99.35 99.91 100.00 100.00 100.00 100.00

Table 4 Effectiveness of power-of-two policies. √ √ TB ∈ [ 0.5, 2]

Scenario

y

TB = 1 Min. (%)

Avg. (%)

Max. (%)

Min. (%)

Avg. (%)

Max. (%)

A B C D E F G H I J K A B C D E F G H I J K A B C D E F G H I J K A B C D E F G H I J K

10 10 10 10 10 10 10 10 10 10 10 50 50 50 50 50 50 50 50 50 50 50 100 100 100 100 100 100 100 100 100 100 100 200 200 200 200 200 200 200 200 200 200 200

92.90 88.92 89.26 86.23 91.78 90.32 92.74 94.26 91.69 90.31 94.79 92.86 92.04 90.63 86.53 94.07 90.46 91.40 94.84 91.13 92.58 95.56 93.04 92.69 90.49 86.37 94.33 90.08 92.28 94.73 91.62 92.85 95.83 93.81 92.71 91.15 87.13 94.52 90.09 93.39 95.37 92.52 93.33 95.86

95.07 95.50 94.17 91.64 95.19 95.76 96.82 96.29 93.98 94.24 97.51 95.01 94.45 93.18 90.77 95.65 95.55 96.25 96.27 93.90 94.11 97.57 94.89 94.43 92.99 90.34 95.76 95.42 95.84 96.17 93.77 94.03 97.47 94.89 94.59 93.21 90.57 95.84 95.68 95.78 96.14 93.78 93.92 97.45

97.77 98.27 97.14 95.94 98.33 98.61 99.21 98.24 97.34 96.95 99.11 96.82 96.98 95.46 94.74 96.73 98.16 98.86 97.73 96.10 95.55 99.10 96.53 96.50 95.65 94.08 96.57 98.19 98.50 97.35 95.81 94.76 98.85 96.36 96.32 95.51 94.05 96.71 97.87 98.35 97.13 95.68 94.54 98.58

93.48 93.56 92.13 87.52 95.77 91.26 97.39 95.25 92.03 92.23 96.64 93.99 93.40 91.84 87.66 95.53 90.70 97.83 95.47 92.81 93.37 96.54 93.61 93.12 91.63 87.03 95.68 90.38 97.76 95.16 92.39 93.30 96.22 93.86 93.37 92.17 87.74 95.78 90.38 97.78 95.38 92.62 93.40 96.57

96.05 96.50 94.91 91.99 97.36 96.07 98.96 97.13 95.14 95.16 98.27 95.44 95.63 94.19 91.16 96.96 95.75 98.52 96.56 94.48 94.45 97.82 95.19 95.43 93.94 90.75 96.84 95.60 98.40 96.36 94.20 94.25 97.64 95.13 95.41 93.99 90.88 96.79 95.86 98.30 96.28 94.15 94.11 97.62

98.12 98.33 97.51 96.04 98.63 98.92 99.48 98.68 97.62 97.35 99.49 97.18 97.47 96.65 94.86 98.22 98.33 99.26 97.87 96.57 95.70 99.12 96.85 97.16 96.63 94.89 97.90 98.20 99.03 97.51 96.31 94.97 98.88 96.52 97.04 96.46 94.74 97.55 98.12 98.87 97.23 95.93 94.82 98.68

Similar to Table 3, Table 4 presents the minimum, average, and maximum effectiveness across 100 test problems for Scenarios A–K and different numbers of retailers. We note that Scenario D has the lowest effectiveness, but this scenario also corresponds to the largest gap between the optimal cost and the lower bound in Table 3. As the

B.Q. Rieksts, J.A. Ventura / Computers & Operations Research 37 (2010) 20 -- 31

6. Conclusions

Effectiveness

100%

In this paper, an optimal policy is derived for a one-warehouse one-retailer inventory system that includes the option of transporting freight with two different modes. Although the optimal policy may have non-stationary order intervals, a tractable optimal algorithm is developed. The order of this algorithm is O(n2 ), where n is the upper bound on the number of setups at the retailer. Section 4 develops a heuristic algorithm to compute power-oftwo policies for a one-warehouse multi-retailer model. A counter example shows that it is impossible to construct a 94% effective power-of-two policy in every instance. However, we provide a lower bound for a decision maker to evaluate the effectiveness of a policy. Computational results demonstrate our power-of-two policies have a cost within 6% of the theoretical lower bound on average for 5500 test problems. We also observed a gap between the lower bound and the optimal cost for one retailer, so our policies could be more effective compared to the optimal cost. For a one-stage model with EOQ assumptions, we noted that using two modes of transportation compare to using one mode exclusively has at most a 6% cost savings while we have only been able to demonstrate a 3% savings. For the one-warehouse, multi-retailer model, a decision maker can quantify the cost difference between heuristic power-of-two policies using both modes of transportation simultaneously compared to those using a single mode exclusively. For TL transportation, a power-of-two policy can be computed with Algorithms H1 and A2 by assuming a sufficiently large LTL cost. For LTL transportation, since the cost of shipping freight is proportional to quantity, the transportation cost of using LTL exclusively does not affect the optimization and Roundy's power-of-two policies can be used. Although these are only heuristic solutions, a decision maker can compare policies using only TL shipments, only LTL shipments, and those using both modes developed in Section 4. In future research, our algorithms could be extended to include other factors such as quantity discounts or backlogging. Perhaps tighter lower bounds for the current one-warehouse multi-retailer model could be developed as well.

95% Min Effectiveness Avg Effectiveness Max Effectiveness

90% 85% 80% 1

10 50 100 Number of Retailers (n)

200

Fig. 5. Effectiveness across all scenarios for TB = 1.

Effectiveness

100% 95% 90% Min Effectiveness Avg Effectiveness Max Effectiveness

85% 80% 1

10 50 100 Number of Retailers (n)

200

√ √ Fig. 6. Effectiveness across all scenarios for TB ∈ [ 0.5, 2].

Table 5 Computational times across all scenarios for each number of retailers. n

1 10 50 100 200

29

√ √ TB ∈ [ 0.5, 2]

TB = 1 Min. (s)

Avg. (s)

Max. (s)

Min. (s)

Avg. (s)

Max. (s)

< 0.01 < 0.01 < 0.01 < 0.01 < 0.01

< 0.01 < 0.01 < 0.01

0.02 0.02 0.06 0.02 0.06

< 0.01 < 0.01 < 0.01 < 0.01 < 0.01

< 0.01 < 0.01

0.02 0.02 0.02 0.05 0.13

0.01 0.03

0.01 0.02 0.08

Acknowledgment The authors would like to thank Robin Roundy for suggesting inventory cost models that include two modes of transportation. Appendix A. Single stage algorithms for inventory models with a bi-modal transportation cost A.1. Algorithm RV1: infinite horizon [30]

setup costs decrease, the optimal order intervals decrease. Since the lower bound assumes a transportation cost with full TL shipments, the gap in transportation costs from the lower bound increases as fewer full TL shipments are used as is the case in Scenario D. Figs. 5 and 6 display the data in Table 4 as a function of the number of retailers. Fig. 5 shows results for a fixed base period while Fig. 6 is for a variable base period. As Fig. 6 shows, the average effectiveness achieved from optimizing the base period is at least 95% for each number of retailers. Table 5 displays the computational times required to compute these power-of-two polices. The minimum, average, and maximum times to compute polices are displayed as a function of the number of retailers across all scenarios. The algorithms were implemented in C with MS Visual C + + 2005 Express. The results were computed using a PC with a 3.0 GHz Pentium 4 processor and 384 MG of RAM. The computational time was less than 0.15 s for each of the 5500 test problems. Also, we observed that less than 0.05 s was required to compute each of the optimal polices for one retailer in Table 3.

 Step 0: Let x∗ be an integer such that x∗ TQ  2K/hd  (x∗ + 1)TQ .  Step 1: If K −x∗ sdT Q +x∗ CT > 0 and x∗ TQ  2(K−x∗ sdT Q +x∗ CT )/hd


Step 2: Set Tˆ 1 = 2(K − x∗ sdT Q + x∗ CT )/hd and go to Step 4. ∗ ∗ Step 3: Set Tˆ 1 = arg min{Z(x  TQ ), Z(CT /sd + x TQ )} and go to Step 4. Step 4: If CT /sd + x∗ TQ  2(K + (x∗ + 1)CT )/hd < (x∗ + 1)TQ , then  set Tˆ 2 = 2(K + (x∗ + 1)CT )/hd and go to Step 6. Otherwise, continue to Step 5. Step 5: Set Tˆ 2 = arg min{Z(CT /sd + x∗ TQ ), Z((x∗ + 1)TQ )} and go to Step 6. Step 6: Let the optimal order interval be T ∗ ∈ arg min{Z(Tˆ 1 ), Z(Tˆ 2 )}. A.2. Algorithm RV2: finite horizon [30]  Step 0: Let nmax = 2TH /min{TQ , 2K/hd}. Set j = 0, n = 1, and Z ∗ = ∞.

30

B.Q. Rieksts, J.A. Ventura / Computers & Operations Research 37 (2010) 20 -- 31

Step 1: Set L = max{TR − (n − j)CT /sd, jC T /sd}, U = min{TR , jT Q }, and tˆF = TR /n + (n − j)s/nh. Step 2: If L > U, go to Step 3. If jtˆF < L, set tˆF =L/j. Similarly, if jtˆF > U, let tˆF = U/j. If Z ∗ > ZH (tˆF ), let Z ∗ = ZH (tˆF ), tF∗ = tˆF , tL∗ = (TR − jtˆF )/(n − j), j∗ = j and n∗ = n. Step 3: If j = n, go to Step 4. Otherwise, set j = j + 1 and return to Step 1. Step 4: If n < nmax , set j = 0, n = n + 1, and return to Step 1. Otherwise, stop with the optimal policy that has j∗ order intervals of length tm (n∗ ) + tF∗ , (n∗ − j∗ ) order intervals of length tm (n∗ ) + tL∗ , and an optimal cost of Z ∗ .

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Appendix B. Optimal parameters for the two-stage formulations (Pˆ  ) The specific bounds L and U are given in Table B.1. The coefficients of ai , bi , and ci are given in Tables B.2 and B.3.

Table B.1 Upper and lower bounds for the formulations (Pˆ  ).

 1 2

L



js nC T (n − j)s − max , h1 sd h1 jC T sd



U



min

nC T js (n − j)s + , nT Q − sd h1 h1



jT Q

3

jT Q

jT Q +

4

nC T sd

nT Q

5

0

nC T sd

(n − j)CT sd

Table B.2 Coefficients a and b for the derivative in Eq. (8).

 1 2 3 4 5

a h1 d h0 d + 2n 2 h1 d h0 d + 2j 2 h1 d h0 d + 2(n − j) 2 h1 d h0 d + 2n 2 h1 d h0 d + 2n 2

b h1 dtm (n) + nh0 dtm (n) nh1 dtm (n) + nh0 dtm (n) j nh1 dtm (n) + nh0 dtm (n) n−j h1 dtm (n) + nh0 dtm (n) h1 dtm (n) + nh0 dtm (n)

Table B.3 Coefficient c for the derivative in Eq. (8).



c 2

1

2

3

4 5

2

nh1 dtm (n) n2 h0 dtm (n) + + (n − j)sdtm (n) − nK 1 − K0 − CT 2 2   2 2 ntm (n) nh1 dtm (n) n2 h0 dtm (n) +j + − nK 1 − K0 − CT 2 2 TQ



ntm (n) +j TQ



   2  2 2 jh1 dT Q ntm (n) nh1 dtm (n) n2 h0 dtm (n) jnh1 dT Q tm (n) j +j + − − 1+ + nsdtm (n) + jsdT Q − nK 1 − K0 − CT 2 2 n−j 2 n−j TQ   2 2 ntm (n) nh1 dtm (n) n2 h0 dtm (n) +n + − nK 1 − K0 − CT 2 2 TQ   2 2 ntm (n) nh1 dtm (n) n2 h0 dtm (n) + + nsdtm (n) − nK 1 − K0 − CT 2 2 TQ

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