- Email: [email protected]
Optimal inventory policies with two modes of freight transportation
Available online at www.sciencedirect.com
European Journal of Operational Research 186 (2008) 576–585 www.elsevier.com/locate/ejor
Production, Manufacturing and Logistics
Optimal inventory policies with two modes of freight transportation Brian Q. Rieksts a
a,1
, Jose´ A. Ventura
b,*
Institute for Defense Analyses, Cost Analysis and Research Division, 4850 Mark Center Drive, Alexandria, VA 22302, USA b Harold and Inge Marcus Department of Industrial and Manufacturing Engineering, The Pennsylvania State University, University Park, PA 16802, USA Received 6 August 2004; accepted 8 January 2007 Available online 1 March 2007
Abstract Theoretical inventory models with constant demand rate and two transportation modes are analyzed in this paper. The transportation options are truckloads with fixed costs, a package delivery carrier with a constant cost per unit, or using a combination of both modes simultaneously. Exact algorithms for computing the optimal policies are derived for single stage models over both an infinite and a finite planning horizon. Ó 2007 Elsevier B.V. All rights reserved. Keywords: Supply chain management; Inventory; Transportation costs
1. Introduction Transportation and inventory are two of the major drivers in a supply chain. This paper derives policies for inventory models that include transportation costs. Different modes of shipping freight are typically categorized as either truckload (TL) transportation or less than truckload (LTL) transportation. In TL transportation, there is a fixed cost per load up to a given capacity. With TL transportation, a company may over-declare a quantity that uses less than the capacity available and transport this freight at the cost of a full load. One example of TL transportation is a company that uses a truck to transport freight. Any quantity up to the capacity of the truck may be shipped for the fixed cost due to the truck, driver, and operating expenses. Another example of TL transportation is shipping freight in an overseas container. In some instances, the quantity of freight may not be large enough to justify the cost associated with a TL shipment. For these small quantities, a LTL carrier with a cost based on the size of the shipment is the most efficient method for transporting freight. An example of LTL transportation is a company that uses a third party carrier such the United Parcel Service to ship freight. In practice, LTL cost structures are complex. *
1
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). Tel.: +1 814 845 2528; fax: +1 703 845 2211.
0377-2217/$ - see front matter Ó 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2007.01.042
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
577
Ballou [5] studies the error in approximating transportation costs as a linear function of distance. For a given distance, there are several breakpoints in the weight of the shipment at which 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. [8] consider changes in the cost function due to over-declaring an LTL shipment. Russell and Karjewski [15] also discuss LTL transportation with multiple weight breakpoints in the context of an EOQ environment. Although a realistic LTL cost structure has several breakpoints with discounts on the cost per unit, we will simplify the LTL cost by assuming a constant cost per unit. Swenseth and Godfrey [19] analyze the error associated with this constant cost per unit approximation. Our research will provide a theoretical foundation for future work on the complex cost structure used in practice. This paper proposes theoretical inventory policies with deliveries that use TL shipments, a package delivery carrier, or a combination of the two options. In some instances, the inventory and setup costs are such that the optimal order quantity is the entire capacity of a truckload and another partial load. If the partial load is not sufficiently large to over-declare an extra truckload, it is optimal to use both TL and LTL transportation to fill this order. We consider such scenarios for inventory models that assume a constant demand rate that must be met without shortages. Each order has a constant holding cost per unit and a setup cost. For transportation costs, there is a fixed cost for each load up to the capacity of a TL shipment. For LTL transportation, we assume a constant cost per unit of freight shipped that is associated with a package delivery carrier. These models are considered for one stage over both an infinite and finite horizon. Authors have researched inventory models without transportation costs that are similar to the model proposed in this paper. In 1915, Harris [9] presents the popular economic order quantity (EOQ) model. A policy for the finite horizon EOQ model is developed by Schwarz [17]. For inventory models with transportation, TL transportation is explored by Aucamp [4] and Lippman [14]. In these models, the cost per load does not change with the number of truckloads. Lee [13] also considers TL shipments for a one stage model with discounts on the per load cost as the number of truckloads increases. Hwang et al. [10] extends Lee’s freight discount model to include an all unit quantity discount on the purchasing cost. In another inventory model with transportation costs, Burns et al. [6] examines peddling shipments to several customers as opposed to shipping to each customer directly. Larson [12] develops models with inventory levels at the origin, in-transit, and at the destination. A few different functions to approximate realistic transportation costs are proposed by Langley [11]. Abdelwahab and Saregious [2] and Swenseth and Godfrey [18] consider TL and LTL inventory models as disjoint problems and compare the policies for each scenario. For maximizing profit, Abad and Aggarwal [1] develop a model with a transportation cost structure similar to [15,18]. Arcelus and Rowcroft [3] propose models with incremental quantity discounts and three different freight structures. Some of these models have similarities to the one stage model over the infinite horizon discussed in this paper. Since the one stage infinite horizon model must be solved in computing an optimal policy for the finite horizon, an algorithm is developed that is efficient for our specific model. The remainder of this paper derives policies for inventory models with two options for transportation. Section 2 presents an optimal policy for a one stage system over the infinite horizon. An optimal policy for the single stage model over the finite horizon is discussed in Section 3. Through an example, Section 4 demonstrates the cost improvement of using both modes of transportation simultaneously. Concluding remarks are provided in Section 5. 2. Infinite horizon single stage model This section considers a single stage system with an infinite horizon. An example of this system is a retailer that incurs transportation costs for incoming shipments. Except for transportation costs, the assumptions are the same as for the basic EOQ model. We assume the holding cost is independent of the purchase price and any capital invested in transportation. The costs are a setup cost K for each order and a positive unit holding cost per time unit denoted as h. Demand occurs at a constant rate, d, that must be satisfied without shortages. For transportation, a full truckload may be used with a cost of CT for each truckload. Each truckload has sufficient capacity to ship a quantity that satisfies TQ time units of demand. Note that if QT is the capacity of the truck in terms of the number of units of product, then T Q ¼ QT =d. Another option of transportation is
578
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
available for quantities that are less than a full truckload with a cost, s, for each unit shipped. An assumption is given that sdT Q > C T . Otherwise, only LTL transportation is used in the optimal policy. Similar to the EOQ model, an optimal policy may be defined by the order interval T. Each order placed at an inventory level of zero corresponds to the quantity that is exactly sufficient to satisfy demand that occurs over the interval T. Policies that only order if the inventory is zero are said to satisfy the zero-inventory ordering property. In Section 3, it is proven that even for the finite horizon an optimal policy satisfies the zero-inventory ordering property. Since the arguments of the proof hold as the finite horizon goes to infinity, an optimal policy for the single stage model over an infinite horizon satisfies the zero-inventory ordering property. The shipping cost for a quantity corresponding to an order interval T within the range jT Q 6 T 6 ðj þ 1ÞT Q is CðT Þ ¼ minfsdðT jT Q Þ þ jC T ; ðj þ 1ÞC T g: By solving for the quantity at which both modes of transportation have the same cost, the transportation cost is reduced to ( sdðT jT Q Þ þ jC T for jT Q 6 T < CsdT þ jT Q ; CðT Þ ¼ for CsdT þ jT Q 6 T < ðj þ 1ÞT Q : ðj þ 1ÞC T A graph of the total transportation cost as a function of the order interval is given in Fig. 1. The data for Figs. 1 and 2 corresponds to the example in Section 3. The average holding, setup, and transportation cost for the system is ZðT Þ ¼
K hdT CðT Þ þ þ : T 2 T
As shown by Fig. 2, Z(T) is neither concave nor convex. An algorithm for obtaining the optimal order interval may be derived from the bounds that Lemma 1 provides. Aucamp [4] and Lippman [14] prove similar bounds for their TL models. Lemma p1.ffiffiffiffiffiffiffiffiffiffiffiffiffi The ffi optimal order interval T* is in the range ½j T Q ; ðj þ 1ÞT Q where j* corresponds to * j T Q 6 2K=hd 6 ðj þ 1ÞT Q . Proof (By contradiction). Suppose that the optimal order interval T* is not on the interval ½j T Q ; ðj þ 1ÞT Q . This implies that T > ðj þ 1ÞT Q or T < j T Q . First, assume that T > ðj þ 1ÞT Q . Since the average holding and setup costs represent a convex function with a minimum on the interval ½j T Q ; ðj þ 1ÞT Q , the average holding cost and setup costs for T* are more than for ðj þ 1ÞT Q . The average transportation cost at
40 35 30 25
C(T)
20 15 10 5 0 0
2
4
6
8
T
Fig. 1. Total transportation cost vs. order interval.
10
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
579
17
15
Z(T) 13 11
9
7 1
2
3
4
5
6
7
8
9
10
T
Fig. 2. Average setup, holding, and transportation cost vs. order interval.
ðj þ 1ÞT Q is at a minimum since full truckloads are used. This implies that the average cost at T* is greater than the cost at ðj þ 1ÞT Q . If T < j T Q , a similar argument shows that the cost at T* is greater than the cost at j T Q contradicting the optimality of T*. Therefore, the optimal order interval T* is on the interval ½j T Q ; ðj þ 1ÞT Q . h In computing the optimal order interval T*, only the interval ½j T Q ; ðj þ 1ÞT Q needs to be considered. Let the average cost function be separated into the functions K jsdT Q þ jC T hdT þ sd þ Z 1j ðT Þ ¼ 2 T for the range jT Q 6 T < CsdT þ jT Q , and K þ ðj þ 1ÞC T hdT ; þ Z 2j ðT Þ ¼ 2 T for the range CsdT þ jT Q 6 T < ðj þ 1ÞT Q . In Algorithm 1, these functions are used to compute an optimal policy for the single stage infinite horizon model. Algorithm 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi Step 0: Let j* be an integer such that j T Q 6 2K=hd 6 ðj þ 1ÞT Q . pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Step 1: If K j sdT Q þ j C T > 0 and j T Q 6 2ðK j sdT Q þ j C T Þ=hd < CsdT þ j T Q , then go to Step 2. Otherwise, go to Step 3. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Step 2: Set T^ 1 ¼ 2ðK j sdT Q þ j C T Þ=hd and go to Step 4. n o Step 3: Set T^ 1 ¼ arg min Z 1j ðj T Q Þ; Z 1j CsdT þ j T Q and go to Step 4. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Step 4: If CsdT þ j T Q 6 2ðK þ ðj þ 1ÞC T Þ=hd < ðj þ 1ÞT Q , then set T^ 2 ¼ 2ðK þ ðj þ 1ÞC T Þ=hd and go to Step 6. Otherwise,n continue to Step 5. o Step 5: Set T^ 2 ¼ arg min Z 2j CsdT þ j T Q ; Z 2j ððj þ 1ÞT Q Þ and go to Step 6. Step 6: Let the optimal order interval be T 2 arg minfZ 1j ðT^ 1 Þ; Z 2j ðT^ 2 Þg. Theorem 1 proves that T* is the optimal order interval for the single stage model over the infinite horizon. Theorem 1. Algorithm 1 yields an optimal policy over the infinite horizon. Proof. By Lemma 1, only the range ½j T Qp ; ðjffiffiffiffiffiffiffiffiffiffiffiffiffi þ 1ÞT ffi Q must be considered in computing the optimal order * interval, where j corresponds to j T Q 6 2K=hd 6 ðj þ 1ÞT Q . This range is separated into the intervals
580
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
j T Q ; CsdT þ j T Q and CsdT þ j T Q ; ðj þ 1ÞT Q . First, Z 1j ðT Þ is considered over the interval j T Q ; CsdT þ j T Q . Note that if K jsdT Q þ jC T > 0, then Z 1j ðT Þ is strictly convex. This implies that if j T Q 6 ffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðK j sdT Q þ j C T Þ=hd < CsdT þ j T Q , the minimum of Z 1j ðT Þ occurs at 2ðK j sdT Q þ j C T Þ=hd . Howffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ever, if 2ðK j sdT Q þ j C T Þ=hd is not within the range or K jsT Q þ jC T 6 0, then Z 1j ðT Þ attains a min C imum at an endpoint of the range j T Q ; sdT þ j T Q . Therefore, T^ 1 is the optimal order interval for the function Z 1j ðT Þ. 2 ForpZffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi the minimum of the ffifunction will occur j ðT Þ, the function ffiis always convex. This implies thatp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi CT at 2ðK þ ðj þ 1ÞC T Þ=hd provided that þ j T 6 2ðK þ ðj þ 1ÞC T Þ=hd < ðj þ 1ÞT Q . If Q sd ffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðj þ 1ÞC T Þ=hd of the range C 2ðK þ is not in this interval, the minimum occurs at an endpoint 2 T ^ . Thus, T þ j T ; ðj þ 1ÞT is the optimal order interval for the function Z ðT Þ. Therefore, T* is Q Q 2 j sd optimal for the range ½j T Q ; ðj þ 1ÞT Q . h 3. Finite horizon single stage model For the finite horizon model, the assumptions are identical to Section 2 except that the average cost is minimized over a finite planning horizon of TH time units. In Section 2, an assertion is given without proof that an optimal policy only orders at an inventory level of zero. Lemma 2 shows this property is true for the more general finite horizon model. Lemma 2. Every optimal policy for over a finite horizon must satisfy the zero-inventory ordering property. Proof (By contradiction). Suppose there is an optimal policy that does not satisfy the zero-inventory ordering property. To define this policy, let IðtÞ denote the inventory level and QðtÞ denote the quantity ordered at time t. Since the optimal policy does not satisfy the zero-inventory ordering property, there exists a time t 0 at which an order is placed with Iðt0 Þ > 0. Suppose that the order at time t 0 is the order closest to time TH that occurs at a positive inventory level. Consider a policy that is identical to the optimal policy except that the order at t 0 occurs at time t0 þ Iðt0 Þ=d. The modified policy is feasible and reduces the total cost by hQðt0 ÞIðt0 Þ=d. This contradicts the optimality of the original policy. Therefore, an optimal policy over a finite horizon must satisfy the zero-inventory ordering property. h Since an optimal policy satisfies the zero-inventory ordering property, an optimal policy may be defined in terms of its order intervals. Let ti represent the time interval over which the ith order quantity satisfies demand. Lemma 3 provides a bound on the number of order intervals in the optimal policy for the finite horizon model. From this bound, an optimal policy is computed by solving a formulation for each of the possible number of setups in the optimal policy. In Lemma 3, the notation dxe denotes the smallest integer greater than or equal to x. l n pffiffiffiffiffiffiffiffiffiffiffiffiffiffiom Lemma 3. The optimal number of setups n* for the finite horizon model is at most 2T H = min T Q ; 2K=hd . l n pffiffiffiffiffiffiffiffiffiffiffiffiffiffiom Proof (By contradiction). Suppose that there are 2T H = min T Q ; 2K=hd þ 1 orders in the optimal policy. This that there are at least two order intervals, ti and tj, such that 0 < ti 6 n implies pffiffiffiffiffiffiffiffiffiffiffiffiffiffio tj 6 min T Q ; 2K=hd =2. Consider a policy that is identical to the optimal policy except that the order intervals ti and tj are combined into a single order interval. The average holding and setup cost is lower for the modified policy since p the average holding and setup costs for interval are represented by a convex function ffiffiffiffiffiffiffiffiffiffiffiffiffi ffi pan ffiffiffiffiffiffiffiffiffiffiffiffiffi ffi with a minimum at 2K=hd such that ti 6 tj < ti þ tj 6 2K=hd . The average transportation cost is not increased by modifying the policy since ti 6 tj < ti þ tj 6 T Q and the average transportation costs are nonincreasing on the interval ½0; T Q . However, this cost reduction contradicts the optimality of the original policy. Therefore, the optimal of setups at the retailer between consecutive warehouse orders is at most l n pffiffiffiffiffiffiffiffiffiffiffiffiffinumber ffiom 2T H = min T Q ; 2K=hd . h A formulation for the optimal policy over a finite planning horizon may be defined given that n orders occur. Since the number of orders is predetermined as n in the formulation, only the holding and transporta-
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
581
tion costs are considered in the optimization. Over the finite horizon, minimizing the total cost is equivalent to minimizing the average cost as well. The formulation is defined as n X 1 2 hdti þ Cðti Þ ðPÞ minimize 2 i¼1 n X subject to ti ¼ T H ; i¼1
ti P 0: For a single stage model over the finite horizon, a policy with stationary order intervals is not always optimal. To demonstrate this principle, consider an example with K ¼ 12; h ¼ 1; d ¼ 1; T Q ¼ 3; C T ¼ 10; s ¼ 5, and T H ¼ 10. The optimal number of orders is n ¼ 2 as the algorithm developed later in this section verifies. Given n ¼ 2 in the optimal policy, the total setup cost is 24. If both order intervals have equal length with t1 ¼ t2 ¼ 5, the holding cost is 25, the transportation cost is 40, and the total cost is 89. Consider a policy with t1 ¼ 6 and t2 ¼ 4. In this policy, the holding cost is 26, the transportation cost is 35, and the total cost is 85. Since there is a non-stationary policy with a cost less than the policy with equal order intervals, the optimal policy is non-stationary in this example. Since an optimal policy for the finite horizon model does not necessarily have order intervals that are the same, solving this problem is more difficult than the infinite horizon model. Although the order intervals in the optimal policy may be non-stationary, Lemmas 4–6 provide some properties on the relationships between these order intervals in the optimal policy. Lippman [14] presents some properties similar to these lemmas for a model with only TL transportation. Lemma 4 shows that any two orders do not differ by more than the capacity of an entire truckload. Lemma 4. If ti and tj are any two order intervals in an optimal policy, then jti tj j 6 T Q : Proof (By contradiction). Suppose that the optimal policy has at least two order intervals ti and tj such that jti tj j > T Q . Without loss of generality, let ti and tj be two arbitrary order intervals such that ti tj > T Q . Consider a modified policy that is identical to the optimal policy except that the order intervals ti and tj are replaced by the order intervals t0i ¼ ti T Q and t0j ¼ tj þ T Q . In the modified policy, the transportation cost remains the same since the changes correspond to the capacity of an entire truckload. However, the holding cost is reduced by hdT Q ðti T Q tj Þ. This contradicts the optimality of the original policy. Therefore, jti tj j 6 T Q for any order intervals ti and tj in the optimal policy. h In Lemma 5, the results of Lemma 4 are used to derive specific bounds on the order intervals given that n orders occur over a planning horizon TH. Lemma 5. If ti is an order interval in an optimal policy with n orders, then tm ðnÞ 6 ti 6 tm ðnÞ þ T Q , where tm ðnÞ ¼ jT Q and j is the integer corresponding to jT Q 6 TnH < ðj þ 1ÞT Q . Proof (By contradiction). To show that tm ðnÞ is a lower bound on each order interval, suppose that tm ðnÞ is not a lower bound in some optimal policy. That is, there exists an optimal policy such that there is an order interval ti with ti < tm ðnÞ. Let M be the set of indices j such that tj > TnH with m denoting the cardinality of M. By Lemma 4, tm ðnÞ T Q 6 ti since TnH T Q 6 ti . Transformations for two cases are provided to show a modified policy may be constructed with a lower cost. These cases are CsdT < ti ðtm ðnÞ T Q Þ < T Q and 0 6 ti ðtm ðnÞ T Q Þ 6 CsdT . First, suppose that CsdT < ti ðtm ðnÞ T Q Þ. Consider a modified policy that is identical to the optimal policy except that t0i ¼ tm ðnÞ and t0j ¼ tj ðtm ðnÞ ti Þ=m for all j 2 M. Since the holding cost for a particular order interval is a convex function of the order interval length and ti < t0i 6 t0j < tj for all j 2 M, the holding cost is reduced in the modified policy. The transportation cost does not increase by modifying ti since ti ðtm ðnÞ T Q Þ > CsdT implies that interval ti uses enough full trucks to satisfy an interval of length tm ðnÞ. Since the remaining order intervals are not increased in the modified policy, the total transportation cost does not increase. For the remaining case, suppose that ti ðtm ðnÞ T Q Þ 6 CsdT . Consider a modified
582
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
P
policy that is identical to the optimal policy except that t0i ¼ tm ðnÞ and t0j ¼ t =m ðtm ðnÞ ti Þ=m for all j j2M P
j 2 M. By transforming the order intervals from tj to j2M t j =m, the holding cost is reduced since holding
P tj =m cost is minimize with equal order intervals. The transportation cost does not change since 0 6 j2M P
tj =m. tm ðnÞ 6 CsdT . In the remaining order interval changes, holding cost is reduced since ti < t0i 6 t0j < j2M P
t =m for j 2 M yields a savings of s per unit, the overall transporSince the decrease in the intervals j j2M tation cost does not increase with the increase for the interval ti. Therefore, the transportation cost does not increase in any case. This implies that the cost of the modified policy is less than that of the optimal policy. By this contradiction, ti P tm ðnÞ for all order intervals ti. A similar argument shows that ti 6 tm ðnÞ þ T Q . h By Lemmas 4 and 5, each order interval ti in the optimal policy may be expressed as tm ðnÞ þ t0i where t0i is the additional length of the order interval over the lower bound. Note that t0i is in the range ½0; T Q . Lemma 6 shows that there are only two different lengths of order intervals in the optimal policy. Lemma 6. In the optimal policy, if ti and tj are any order intervals such that ti tm ðnÞ < CsdT and tj tm ðnÞ < CsdT , then ti ¼ tj . Similarly, if ti and tj are any order intervals in the optimal policy such that ti tm ðnÞ P CsdT and tj tm ðnÞ P CsdT , then ti ¼ tj .
Proof (By contradiction). Suppose that the optimal policy has two order intervals ti and tj such that ti tm ðnÞ < tj tm ðnÞ < CsdT . Consider a policy that is identical to the optimal policy except the order intervals ti and tj are replaced by two intervals of length ðti þ tj Þ=2. Since ti tm ðnÞ < tj tm ðnÞ < CsdT , it follows that ðti þ tj Þ=2 tm ðnÞ < CsdT . This implies that the transportation and setup costs are the same for both policies. However, the holding cost is reduced if the order intervals are equal as discussed by Schwarz [16]. The reason for this is that the holding cost is a convex function of the order interval and ti < ðti þ tj Þ=2 < tj . A contradiction occurs since the cost of the modified policy is less than the optimal policy. Therefore, if ti and tj are any order intervals such that ti tm ðnÞ < CsdT and tj tm ðnÞ < CsdT , then ti ¼ tj . A similar argument can be given to prove the case of ti tm ðnÞ P CsdT and tj tm ðnÞ P CsdT . h Since Lemma 6 guarantees that there are only two different lengths for order intervals in the optimal policy, there are only two possible values for ti in the representation of the order intervals as tm ðnÞ þ ti . Let tL correspond to the additional length of the order interval that uses less than truckload transportation and tF represent the additional length transported in a truckload. The bounds on these variables are 0 6 tL 6 CsdT and CT 6 tF 6 T Q . If the number of orders of each length is known, an alternative formulation for P with only sd a single variable may be defined. Assume that there are n total orders and j orders of length tm ðnÞ þ tF . For notational convenience, the planning horizon TH is represented as ntm ðnÞ þ T R where TR is called the reduced planning horizon. The formulation for the optimal policy in terms of tF is 2 jhd ðn jÞhd T R jtF 2 ðtF þ tm ðnÞÞ þ ðP Þ minimize ZðtF Þ ¼ þ tm ðnÞ 2 2 nj ntm ðnÞ þ sdðT R jtF Þ þ C T þ j þ nK; TQ CT 6 tF 6 T Q ; subject to sd T R ðn jÞC T TR 6 tF 6 : j jsd j 0
To compute the optimal policy, the formulation P 0 is solved assuming n setups for each n ¼ 1; . . . ; nmax , where nmax is defined as the upper bound on n according to Lemma 2. For a given number of setups n, the formulation P 0 is solved assuming j order intervals corresponding to tF. The policy that has the lowest cost among all possible combinations of n and j is the optimal policy for the finite planning horizon problem. Algorithm 2 defines these computations for an optimal policy.
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
583
Algorithm 2 l n pffiffiffiffiffiffiffiffiffiffiffiffiffiffiom Step 0: Let nmax ¼ 2T H = min T Q ; 2K=hd . Set j ¼ 0; n ¼ 1, and Z ¼ 1. . Step 1: Set L ¼ max T R ðn jÞ CsdT ; j CsdT ; U ¼ min T R ; jT Q , and ^tF ¼ TnR þ ðnjÞs nh Step 2: If L > U , go to Step 3. If j^tF < L, set ^tF ¼ L=j. Similarly, if j^tF > U , let ^tF ¼ U =j. If Z > Zð^tF Þ, let j^tF Z ¼ Zð^tF Þ; tF ¼ ^tF , tL ¼ T Rnj ; j ¼ j and n ¼ n. Step 3: If j ¼ n, go to Step 4. Otherwise, set j ¼ j þ 1 and return to Step 1. Step 4: If n < nmax , set j ¼ 0; n ¼ n þ 1, and return to Step 1. Otherwise, stop with the optimal policy that has j* order intervals of length tm ðn Þ þ tF , ðn j Þ order intervals of length tm ðn Þ þ tL , and an optimal cost of Z*. In Theorem 2, the optimality of this algorithm is proven. Theorem 2. Algorithm 2 for the finite planning horizon model yields an optimal policy. Proof. To prove that the policy is optimal, it must be verified that Steps 1 and 2 of the algorithm compute an optimal policy for the given n and j. If j ¼ 0 or j ¼ n, the solution is trivial. The algorithm assigns equal order intervals to minimize holding cost if the formulation is feasible. Now consider the remaining possibilities with 0 < j < n. To show the objective function is convex in tF given that n and j are constant, the first derivative is taken with respect to tF Z 0 ðtF Þ ¼ jhdtF
jhdT R j2 hdtF þ jsd: nj nj 2
hd Since the second derivative is positive, Z 00 ðtF Þ ¼ jhd þ jnj , the objective function is strictly convex. This implies that ZðtF Þ achieves a minimum at the value of tF where Z 0 ðtF Þ ¼ 0. The optimal value of tF obtained by solvingZ 0 ðtF Þ ¼ 0 is tF ¼ TnR þ ðnjÞs . Recall that the bounds on tF and tL define a feasible interval for tF. If the minnh imum value for ZðtF Þ is not on this feasible interval, the optimal value occurs at an endpoint of the interval. By Lemma 3, there are at most nmax setups in the optimal policy. Since Steps 0, 3, and 4 iterate through all combinations of n and j for the formulation P 0 that may be in an optimal policy, Algorithm 2 produces an optimal policy over a finite horizon. h
4. Examples In this section, two examples are presented to illustrate the cost improvement by allowing for a combination of two modes of transportation. As a benchmark, we compare using both modes to a model with only TL transportation [4] or only having the option of a constant transportation cost per unit. The case of having a constant transportation cost per unit reduces to the EOQ model [9]. For the maximum cost improvement from allowing both modes of transportation, the optimal order quantity is between one and two truckloads. If the order quantity is less than one truckload, the costs are the same as using a single mode exclusively. If the order quantity is greater than two truckloads, the relative improvement is smaller since more units are shipped with TL transportation filled to truckload capacity. For the infinite horizon case, if the optimal EOQ policy is within the range ½T Q ; 2T Q , the holding and setup cost for an order interval of length TQ or 2T Q is at most 6% more than the optimal EOQ cost. This is true by the convexity of the average holding andhsetup cost function. i Suppose TEOQ is the optimal EOQ order interpffiffiffi 1ffiffi p val. The root-two interval is defined as T EOQ ; 2T EOQ . The holding and setup costs for order intervals 2
equal to these endpoints are 6% more than the optimal EOQ cost. Since the root-two interval cannot be contained strictly within ½T Q ; 2T Q , an order interval equal to TQ or 2TQ has minimum possible average transportation cost and at most a 6% increase in holding and setup cost. Thus, the improvement can be no greater than 6% for holding and setup costs with the transportation cost also contributing to a reduction of the maximum possible savings.
584
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
Average Cost
50
45
40
LTL
35
TL Both 30 5
10
15
20
25
30
T
Fig. 3. Average overall cost for different alternatives.
To show the potential for improvement, consider the example with K ¼ 200; h ¼ 1; d ¼ 2; C T ¼ 60:2; s ¼ 3:75, and T Q ¼ 10. In this case, the optimal EOQ policy has an order interval of T ¼ 14:1. This is the optimal order interval for the case of a constant transportation cost per unit with an optimal average cost per time unit of 35.8. Fig. 3 shows the average holding, setup, and transportation cost for only TL shipments, a carrier with a constant transportation cost per unit, and a combination of the two options. Note that the scale of the y-axis is set to the range [30, 50] to magnify the minimum. For using TL transportation exclusively, the optimal order interval is T ¼ 17:9 with an average cost of 35.8. By Algorithm 1, the optimal order interval for using both modes is 13.6 with a cost of 34.7. This is 97% of the lowest cost from using one mode of transportation exclusively. Although this is a small improvement, this reduction in logistics costs could impact profit for a large company. Carter and Ferrin [7] examine the importance of transportation costs. They discuss an example of GTE, a $27 billion company, with an $84 million transportation bill in 1993. A small improvement in holding, setup, and transportation costs could be significant for a company such as GTE. For the finite horizon model, suppose we have an example with K ¼ 25, h ¼ 2; d ¼ 1; C T ¼ 1000; s ¼ 500; T Q ¼ 4, and T H ¼ 5. If the constant transportation cost per unit s is used exclusively, the optimal policy is one order for the entire horizon with a total cost of 2550. For only TL transportation, the optimal policy is a single order with a total cost of 2050. From Algorithm 2, the optimal cost is also a single order. However, the total cost is only 1550. Thus, there was a cost improvement of 24% over only using one mode of transportation. Although this example is not realistic and was designed to show a large improvement, it does show there is a much higher potential for improvement in the finite horizon model than the infinite case. 5. Conclusions In some inventory systems, it may be beneficial to use two different methods of transportation simultaneously. Although the total costs of the models in this paper are neither concave nor convex functions of the order intervals, the cost functions are either concave or convex within certain ranges. By taking advantage of this structure, algorithms to compute optimal policies are presented for single stage models over both an infinite and finite planning horizon. In the past, many authors have optimized similar models with only one mode of transportation. In Section 4 for an infinite horizon, we showed that models considering only one mode of transportation can have as much as a 3% cost reduction by allowing two modes simultaneously. For the finite horizon model, an example is provided that illustrates a 24% cost savings. This finite horizon result for a single stage is a useful result for constructing algorithms for multiple stages. In a nested system, the downstream stages reduce to a finite horizon problem given a fixed order interval at the prior stage. If the large transportation cost is sufficiently large, it would also be useful to explore models with holding costs dependent on the opportunity costs of capital invested in transportation. In other future work, these models could be generalized across the full spectrum of TL and LTL costs with several breakpoints.
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
585
Acknowledgements The authors would like to thank Robin Roundy for suggesting inventory cost models that include both truckload and less than truckload modes of transportation. References [1] P. Abad, V. Aggarwal, Incorporating transportation cost in the lot sizing and pricing decisions with downward sloping demand, International Journal of Production Economics 95 (2005) 297–305. [2] W. Adelwahab, M. Sargious, Freight rate structure and optimal shipment size in freight transportation, Logistics and Transportation Review 26 (1992) 271–292. [3] F. Arcelus, J. Rowcroft, Inventory policies with freight and incremental quantity discounts, International Journal of Systems Science 22 (1991) 2025–2037. [4] D. Aucamp, Nonlinear freight costs in the EOQ problem, European Journal of Operational Research 9 (1982) 61–63. [5] R. Ballou, The accuracy for estimating truck class rates for logistical planning, Transportation Research A 25A (1991) 327–337. [6] L. Burns, R. Hall, D. Blumenfeld, Distribution strategies that minimize transportation and inventory costs, Operations Research 33 (1985) 469–490. [7] J. Carter, B. Ferrin, Transportation costs and inventory management: Why transportation costs matter, Production and Inventory Management Journal 37 (1996) 58–62. [8] J. Carter, B. Ferrin, C. Carter, The effect of less-than-truckload rates on the purchase order lot size decision, Transportation Journal 34 (1995) 35–44. [9] F. Harris, How many parts to make at once, Operations Research 38 (1990) 947–950, reprint from Factory – The Magazine of Management 10 (1913) 135–136. [10] H. Hwang, D. Moon, S. Shinn, An EOQ model with quantity discounts for both purchasing price and freight cost, Computers and Operations Research 17 (1990) 73–78. [11] J. Langley, The inclusion of transportation costs in inventory models: Some considerations, Journal of Business Logistics 2 (1980) 106–125. [12] P. Larson, The economic transportation quantity, Transportation Journal 28 (1988) 43–48. [13] C.Y. Lee, The economic order quantity for freight discount costs, IIE Transactions 18 (1986) 318–320. [14] S. Lippman, Economic order quantities and multiple set-up costs, Management Science 18 (1971) 39–47. [15] R. Russell, J. Karjewski, Optimal purchase and transportation cost lot sizing for a single item, Decision Sciences 22 (1991) 940–952. [16] L. Schwarz, A simple continuous review deterministic one-warehouse N-retailer inventory problem, Management Science 19 (1973) 555–566. [17] L. Schwarz, Economic order quantities for products with finite demand horizons, AIIE Transactions 4 (1972) 234–237. [18] S. Swenseth, M. Godfrey, Incorporating transportation costs into inventory replenishment decisions, International Journal of Production Economics 77 (2002) 113–130. [19] S. Swenseth, M. Godfrey, Estimating freight rates for logistics decisions, Journal of Business Logistics 17 (1996) 213–231.
European Journal of Operational Research 186 (2008) 576–585 www.elsevier.com/locate/ejor
Production, Manufacturing and Logistics
Optimal inventory policies with two modes of freight transportation Brian Q. Rieksts a
a,1
, Jose´ A. Ventura
b,*
Institute for Defense Analyses, Cost Analysis and Research Division, 4850 Mark Center Drive, Alexandria, VA 22302, USA b Harold and Inge Marcus Department of Industrial and Manufacturing Engineering, The Pennsylvania State University, University Park, PA 16802, USA Received 6 August 2004; accepted 8 January 2007 Available online 1 March 2007
Abstract Theoretical inventory models with constant demand rate and two transportation modes are analyzed in this paper. The transportation options are truckloads with fixed costs, a package delivery carrier with a constant cost per unit, or using a combination of both modes simultaneously. Exact algorithms for computing the optimal policies are derived for single stage models over both an infinite and a finite planning horizon. Ó 2007 Elsevier B.V. All rights reserved. Keywords: Supply chain management; Inventory; Transportation costs
1. Introduction Transportation and inventory are two of the major drivers in a supply chain. This paper derives policies for inventory models that include transportation costs. Different modes of shipping freight are typically categorized as either truckload (TL) transportation or less than truckload (LTL) transportation. In TL transportation, there is a fixed cost per load up to a given capacity. With TL transportation, a company may over-declare a quantity that uses less than the capacity available and transport this freight at the cost of a full load. One example of TL transportation is a company that uses a truck to transport freight. Any quantity up to the capacity of the truck may be shipped for the fixed cost due to the truck, driver, and operating expenses. Another example of TL transportation is shipping freight in an overseas container. In some instances, the quantity of freight may not be large enough to justify the cost associated with a TL shipment. For these small quantities, a LTL carrier with a cost based on the size of the shipment is the most efficient method for transporting freight. An example of LTL transportation is a company that uses a third party carrier such the United Parcel Service to ship freight. In practice, LTL cost structures are complex. *
1
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). Tel.: +1 814 845 2528; fax: +1 703 845 2211.
0377-2217/$ - see front matter Ó 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2007.01.042
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
577
Ballou [5] studies the error in approximating transportation costs as a linear function of distance. For a given distance, there are several breakpoints in the weight of the shipment at which 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. [8] consider changes in the cost function due to over-declaring an LTL shipment. Russell and Karjewski [15] also discuss LTL transportation with multiple weight breakpoints in the context of an EOQ environment. Although a realistic LTL cost structure has several breakpoints with discounts on the cost per unit, we will simplify the LTL cost by assuming a constant cost per unit. Swenseth and Godfrey [19] analyze the error associated with this constant cost per unit approximation. Our research will provide a theoretical foundation for future work on the complex cost structure used in practice. This paper proposes theoretical inventory policies with deliveries that use TL shipments, a package delivery carrier, or a combination of the two options. In some instances, the inventory and setup costs are such that the optimal order quantity is the entire capacity of a truckload and another partial load. If the partial load is not sufficiently large to over-declare an extra truckload, it is optimal to use both TL and LTL transportation to fill this order. We consider such scenarios for inventory models that assume a constant demand rate that must be met without shortages. Each order has a constant holding cost per unit and a setup cost. For transportation costs, there is a fixed cost for each load up to the capacity of a TL shipment. For LTL transportation, we assume a constant cost per unit of freight shipped that is associated with a package delivery carrier. These models are considered for one stage over both an infinite and finite horizon. Authors have researched inventory models without transportation costs that are similar to the model proposed in this paper. In 1915, Harris [9] presents the popular economic order quantity (EOQ) model. A policy for the finite horizon EOQ model is developed by Schwarz [17]. For inventory models with transportation, TL transportation is explored by Aucamp [4] and Lippman [14]. In these models, the cost per load does not change with the number of truckloads. Lee [13] also considers TL shipments for a one stage model with discounts on the per load cost as the number of truckloads increases. Hwang et al. [10] extends Lee’s freight discount model to include an all unit quantity discount on the purchasing cost. In another inventory model with transportation costs, Burns et al. [6] examines peddling shipments to several customers as opposed to shipping to each customer directly. Larson [12] develops models with inventory levels at the origin, in-transit, and at the destination. A few different functions to approximate realistic transportation costs are proposed by Langley [11]. Abdelwahab and Saregious [2] and Swenseth and Godfrey [18] consider TL and LTL inventory models as disjoint problems and compare the policies for each scenario. For maximizing profit, Abad and Aggarwal [1] develop a model with a transportation cost structure similar to [15,18]. Arcelus and Rowcroft [3] propose models with incremental quantity discounts and three different freight structures. Some of these models have similarities to the one stage model over the infinite horizon discussed in this paper. Since the one stage infinite horizon model must be solved in computing an optimal policy for the finite horizon, an algorithm is developed that is efficient for our specific model. The remainder of this paper derives policies for inventory models with two options for transportation. Section 2 presents an optimal policy for a one stage system over the infinite horizon. An optimal policy for the single stage model over the finite horizon is discussed in Section 3. Through an example, Section 4 demonstrates the cost improvement of using both modes of transportation simultaneously. Concluding remarks are provided in Section 5. 2. Infinite horizon single stage model This section considers a single stage system with an infinite horizon. An example of this system is a retailer that incurs transportation costs for incoming shipments. Except for transportation costs, the assumptions are the same as for the basic EOQ model. We assume the holding cost is independent of the purchase price and any capital invested in transportation. The costs are a setup cost K for each order and a positive unit holding cost per time unit denoted as h. Demand occurs at a constant rate, d, that must be satisfied without shortages. For transportation, a full truckload may be used with a cost of CT for each truckload. Each truckload has sufficient capacity to ship a quantity that satisfies TQ time units of demand. Note that if QT is the capacity of the truck in terms of the number of units of product, then T Q ¼ QT =d. Another option of transportation is
578
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
available for quantities that are less than a full truckload with a cost, s, for each unit shipped. An assumption is given that sdT Q > C T . Otherwise, only LTL transportation is used in the optimal policy. Similar to the EOQ model, an optimal policy may be defined by the order interval T. Each order placed at an inventory level of zero corresponds to the quantity that is exactly sufficient to satisfy demand that occurs over the interval T. Policies that only order if the inventory is zero are said to satisfy the zero-inventory ordering property. In Section 3, it is proven that even for the finite horizon an optimal policy satisfies the zero-inventory ordering property. Since the arguments of the proof hold as the finite horizon goes to infinity, an optimal policy for the single stage model over an infinite horizon satisfies the zero-inventory ordering property. The shipping cost for a quantity corresponding to an order interval T within the range jT Q 6 T 6 ðj þ 1ÞT Q is CðT Þ ¼ minfsdðT jT Q Þ þ jC T ; ðj þ 1ÞC T g: By solving for the quantity at which both modes of transportation have the same cost, the transportation cost is reduced to ( sdðT jT Q Þ þ jC T for jT Q 6 T < CsdT þ jT Q ; CðT Þ ¼ for CsdT þ jT Q 6 T < ðj þ 1ÞT Q : ðj þ 1ÞC T A graph of the total transportation cost as a function of the order interval is given in Fig. 1. The data for Figs. 1 and 2 corresponds to the example in Section 3. The average holding, setup, and transportation cost for the system is ZðT Þ ¼
K hdT CðT Þ þ þ : T 2 T
As shown by Fig. 2, Z(T) is neither concave nor convex. An algorithm for obtaining the optimal order interval may be derived from the bounds that Lemma 1 provides. Aucamp [4] and Lippman [14] prove similar bounds for their TL models. Lemma p1.ffiffiffiffiffiffiffiffiffiffiffiffiffi The ffi optimal order interval T* is in the range ½j T Q ; ðj þ 1ÞT Q where j* corresponds to * j T Q 6 2K=hd 6 ðj þ 1ÞT Q . Proof (By contradiction). Suppose that the optimal order interval T* is not on the interval ½j T Q ; ðj þ 1ÞT Q . This implies that T > ðj þ 1ÞT Q or T < j T Q . First, assume that T > ðj þ 1ÞT Q . Since the average holding and setup costs represent a convex function with a minimum on the interval ½j T Q ; ðj þ 1ÞT Q , the average holding cost and setup costs for T* are more than for ðj þ 1ÞT Q . The average transportation cost at
40 35 30 25
C(T)
20 15 10 5 0 0
2
4
6
8
T
Fig. 1. Total transportation cost vs. order interval.
10
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
579
17
15
Z(T) 13 11
9
7 1
2
3
4
5
6
7
8
9
10
T
Fig. 2. Average setup, holding, and transportation cost vs. order interval.
ðj þ 1ÞT Q is at a minimum since full truckloads are used. This implies that the average cost at T* is greater than the cost at ðj þ 1ÞT Q . If T < j T Q , a similar argument shows that the cost at T* is greater than the cost at j T Q contradicting the optimality of T*. Therefore, the optimal order interval T* is on the interval ½j T Q ; ðj þ 1ÞT Q . h In computing the optimal order interval T*, only the interval ½j T Q ; ðj þ 1ÞT Q needs to be considered. Let the average cost function be separated into the functions K jsdT Q þ jC T hdT þ sd þ Z 1j ðT Þ ¼ 2 T for the range jT Q 6 T < CsdT þ jT Q , and K þ ðj þ 1ÞC T hdT ; þ Z 2j ðT Þ ¼ 2 T for the range CsdT þ jT Q 6 T < ðj þ 1ÞT Q . In Algorithm 1, these functions are used to compute an optimal policy for the single stage infinite horizon model. Algorithm 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi Step 0: Let j* be an integer such that j T Q 6 2K=hd 6 ðj þ 1ÞT Q . pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Step 1: If K j sdT Q þ j C T > 0 and j T Q 6 2ðK j sdT Q þ j C T Þ=hd < CsdT þ j T Q , then go to Step 2. Otherwise, go to Step 3. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Step 2: Set T^ 1 ¼ 2ðK j sdT Q þ j C T Þ=hd and go to Step 4. n o Step 3: Set T^ 1 ¼ arg min Z 1j ðj T Q Þ; Z 1j CsdT þ j T Q and go to Step 4. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Step 4: If CsdT þ j T Q 6 2ðK þ ðj þ 1ÞC T Þ=hd < ðj þ 1ÞT Q , then set T^ 2 ¼ 2ðK þ ðj þ 1ÞC T Þ=hd and go to Step 6. Otherwise,n continue to Step 5. o Step 5: Set T^ 2 ¼ arg min Z 2j CsdT þ j T Q ; Z 2j ððj þ 1ÞT Q Þ and go to Step 6. Step 6: Let the optimal order interval be T 2 arg minfZ 1j ðT^ 1 Þ; Z 2j ðT^ 2 Þg. Theorem 1 proves that T* is the optimal order interval for the single stage model over the infinite horizon. Theorem 1. Algorithm 1 yields an optimal policy over the infinite horizon. Proof. By Lemma 1, only the range ½j T Qp ; ðjffiffiffiffiffiffiffiffiffiffiffiffiffi þ 1ÞT ffi Q must be considered in computing the optimal order * interval, where j corresponds to j T Q 6 2K=hd 6 ðj þ 1ÞT Q . This range is separated into the intervals
580
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
j T Q ; CsdT þ j T Q and CsdT þ j T Q ; ðj þ 1ÞT Q . First, Z 1j ðT Þ is considered over the interval j T Q ; CsdT þ j T Q . Note that if K jsdT Q þ jC T > 0, then Z 1j ðT Þ is strictly convex. This implies that if j T Q 6 ffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðK j sdT Q þ j C T Þ=hd < CsdT þ j T Q , the minimum of Z 1j ðT Þ occurs at 2ðK j sdT Q þ j C T Þ=hd . Howffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ever, if 2ðK j sdT Q þ j C T Þ=hd is not within the range or K jsT Q þ jC T 6 0, then Z 1j ðT Þ attains a min C imum at an endpoint of the range j T Q ; sdT þ j T Q . Therefore, T^ 1 is the optimal order interval for the function Z 1j ðT Þ. 2 ForpZffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi the minimum of the ffifunction will occur j ðT Þ, the function ffiis always convex. This implies thatp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi CT at 2ðK þ ðj þ 1ÞC T Þ=hd provided that þ j T 6 2ðK þ ðj þ 1ÞC T Þ=hd < ðj þ 1ÞT Q . If Q sd ffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðj þ 1ÞC T Þ=hd of the range C 2ðK þ is not in this interval, the minimum occurs at an endpoint 2 T ^ . Thus, T þ j T ; ðj þ 1ÞT is the optimal order interval for the function Z ðT Þ. Therefore, T* is Q Q 2 j sd optimal for the range ½j T Q ; ðj þ 1ÞT Q . h 3. Finite horizon single stage model For the finite horizon model, the assumptions are identical to Section 2 except that the average cost is minimized over a finite planning horizon of TH time units. In Section 2, an assertion is given without proof that an optimal policy only orders at an inventory level of zero. Lemma 2 shows this property is true for the more general finite horizon model. Lemma 2. Every optimal policy for over a finite horizon must satisfy the zero-inventory ordering property. Proof (By contradiction). Suppose there is an optimal policy that does not satisfy the zero-inventory ordering property. To define this policy, let IðtÞ denote the inventory level and QðtÞ denote the quantity ordered at time t. Since the optimal policy does not satisfy the zero-inventory ordering property, there exists a time t 0 at which an order is placed with Iðt0 Þ > 0. Suppose that the order at time t 0 is the order closest to time TH that occurs at a positive inventory level. Consider a policy that is identical to the optimal policy except that the order at t 0 occurs at time t0 þ Iðt0 Þ=d. The modified policy is feasible and reduces the total cost by hQðt0 ÞIðt0 Þ=d. This contradicts the optimality of the original policy. Therefore, an optimal policy over a finite horizon must satisfy the zero-inventory ordering property. h Since an optimal policy satisfies the zero-inventory ordering property, an optimal policy may be defined in terms of its order intervals. Let ti represent the time interval over which the ith order quantity satisfies demand. Lemma 3 provides a bound on the number of order intervals in the optimal policy for the finite horizon model. From this bound, an optimal policy is computed by solving a formulation for each of the possible number of setups in the optimal policy. In Lemma 3, the notation dxe denotes the smallest integer greater than or equal to x. l n pffiffiffiffiffiffiffiffiffiffiffiffiffiffiom Lemma 3. The optimal number of setups n* for the finite horizon model is at most 2T H = min T Q ; 2K=hd . l n pffiffiffiffiffiffiffiffiffiffiffiffiffiffiom Proof (By contradiction). Suppose that there are 2T H = min T Q ; 2K=hd þ 1 orders in the optimal policy. This that there are at least two order intervals, ti and tj, such that 0 < ti 6 n implies pffiffiffiffiffiffiffiffiffiffiffiffiffiffio tj 6 min T Q ; 2K=hd =2. Consider a policy that is identical to the optimal policy except that the order intervals ti and tj are combined into a single order interval. The average holding and setup cost is lower for the modified policy since p the average holding and setup costs for interval are represented by a convex function ffiffiffiffiffiffiffiffiffiffiffiffiffi ffi pan ffiffiffiffiffiffiffiffiffiffiffiffiffi ffi with a minimum at 2K=hd such that ti 6 tj < ti þ tj 6 2K=hd . The average transportation cost is not increased by modifying the policy since ti 6 tj < ti þ tj 6 T Q and the average transportation costs are nonincreasing on the interval ½0; T Q . However, this cost reduction contradicts the optimality of the original policy. Therefore, the optimal of setups at the retailer between consecutive warehouse orders is at most l n pffiffiffiffiffiffiffiffiffiffiffiffiffinumber ffiom 2T H = min T Q ; 2K=hd . h A formulation for the optimal policy over a finite planning horizon may be defined given that n orders occur. Since the number of orders is predetermined as n in the formulation, only the holding and transporta-
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
581
tion costs are considered in the optimization. Over the finite horizon, minimizing the total cost is equivalent to minimizing the average cost as well. The formulation is defined as n X 1 2 hdti þ Cðti Þ ðPÞ minimize 2 i¼1 n X subject to ti ¼ T H ; i¼1
ti P 0: For a single stage model over the finite horizon, a policy with stationary order intervals is not always optimal. To demonstrate this principle, consider an example with K ¼ 12; h ¼ 1; d ¼ 1; T Q ¼ 3; C T ¼ 10; s ¼ 5, and T H ¼ 10. The optimal number of orders is n ¼ 2 as the algorithm developed later in this section verifies. Given n ¼ 2 in the optimal policy, the total setup cost is 24. If both order intervals have equal length with t1 ¼ t2 ¼ 5, the holding cost is 25, the transportation cost is 40, and the total cost is 89. Consider a policy with t1 ¼ 6 and t2 ¼ 4. In this policy, the holding cost is 26, the transportation cost is 35, and the total cost is 85. Since there is a non-stationary policy with a cost less than the policy with equal order intervals, the optimal policy is non-stationary in this example. Since an optimal policy for the finite horizon model does not necessarily have order intervals that are the same, solving this problem is more difficult than the infinite horizon model. Although the order intervals in the optimal policy may be non-stationary, Lemmas 4–6 provide some properties on the relationships between these order intervals in the optimal policy. Lippman [14] presents some properties similar to these lemmas for a model with only TL transportation. Lemma 4 shows that any two orders do not differ by more than the capacity of an entire truckload. Lemma 4. If ti and tj are any two order intervals in an optimal policy, then jti tj j 6 T Q : Proof (By contradiction). Suppose that the optimal policy has at least two order intervals ti and tj such that jti tj j > T Q . Without loss of generality, let ti and tj be two arbitrary order intervals such that ti tj > T Q . Consider a modified policy that is identical to the optimal policy except that the order intervals ti and tj are replaced by the order intervals t0i ¼ ti T Q and t0j ¼ tj þ T Q . In the modified policy, the transportation cost remains the same since the changes correspond to the capacity of an entire truckload. However, the holding cost is reduced by hdT Q ðti T Q tj Þ. This contradicts the optimality of the original policy. Therefore, jti tj j 6 T Q for any order intervals ti and tj in the optimal policy. h In Lemma 5, the results of Lemma 4 are used to derive specific bounds on the order intervals given that n orders occur over a planning horizon TH. Lemma 5. If ti is an order interval in an optimal policy with n orders, then tm ðnÞ 6 ti 6 tm ðnÞ þ T Q , where tm ðnÞ ¼ jT Q and j is the integer corresponding to jT Q 6 TnH < ðj þ 1ÞT Q . Proof (By contradiction). To show that tm ðnÞ is a lower bound on each order interval, suppose that tm ðnÞ is not a lower bound in some optimal policy. That is, there exists an optimal policy such that there is an order interval ti with ti < tm ðnÞ. Let M be the set of indices j such that tj > TnH with m denoting the cardinality of M. By Lemma 4, tm ðnÞ T Q 6 ti since TnH T Q 6 ti . Transformations for two cases are provided to show a modified policy may be constructed with a lower cost. These cases are CsdT < ti ðtm ðnÞ T Q Þ < T Q and 0 6 ti ðtm ðnÞ T Q Þ 6 CsdT . First, suppose that CsdT < ti ðtm ðnÞ T Q Þ. Consider a modified policy that is identical to the optimal policy except that t0i ¼ tm ðnÞ and t0j ¼ tj ðtm ðnÞ ti Þ=m for all j 2 M. Since the holding cost for a particular order interval is a convex function of the order interval length and ti < t0i 6 t0j < tj for all j 2 M, the holding cost is reduced in the modified policy. The transportation cost does not increase by modifying ti since ti ðtm ðnÞ T Q Þ > CsdT implies that interval ti uses enough full trucks to satisfy an interval of length tm ðnÞ. Since the remaining order intervals are not increased in the modified policy, the total transportation cost does not increase. For the remaining case, suppose that ti ðtm ðnÞ T Q Þ 6 CsdT . Consider a modified
582
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
P
policy that is identical to the optimal policy except that t0i ¼ tm ðnÞ and t0j ¼ t =m ðtm ðnÞ ti Þ=m for all j j2M P
j 2 M. By transforming the order intervals from tj to j2M t j =m, the holding cost is reduced since holding
P tj =m cost is minimize with equal order intervals. The transportation cost does not change since 0 6 j2M P
tj =m. tm ðnÞ 6 CsdT . In the remaining order interval changes, holding cost is reduced since ti < t0i 6 t0j < j2M P
t =m for j 2 M yields a savings of s per unit, the overall transporSince the decrease in the intervals j j2M tation cost does not increase with the increase for the interval ti. Therefore, the transportation cost does not increase in any case. This implies that the cost of the modified policy is less than that of the optimal policy. By this contradiction, ti P tm ðnÞ for all order intervals ti. A similar argument shows that ti 6 tm ðnÞ þ T Q . h By Lemmas 4 and 5, each order interval ti in the optimal policy may be expressed as tm ðnÞ þ t0i where t0i is the additional length of the order interval over the lower bound. Note that t0i is in the range ½0; T Q . Lemma 6 shows that there are only two different lengths of order intervals in the optimal policy. Lemma 6. In the optimal policy, if ti and tj are any order intervals such that ti tm ðnÞ < CsdT and tj tm ðnÞ < CsdT , then ti ¼ tj . Similarly, if ti and tj are any order intervals in the optimal policy such that ti tm ðnÞ P CsdT and tj tm ðnÞ P CsdT , then ti ¼ tj .
Proof (By contradiction). Suppose that the optimal policy has two order intervals ti and tj such that ti tm ðnÞ < tj tm ðnÞ < CsdT . Consider a policy that is identical to the optimal policy except the order intervals ti and tj are replaced by two intervals of length ðti þ tj Þ=2. Since ti tm ðnÞ < tj tm ðnÞ < CsdT , it follows that ðti þ tj Þ=2 tm ðnÞ < CsdT . This implies that the transportation and setup costs are the same for both policies. However, the holding cost is reduced if the order intervals are equal as discussed by Schwarz [16]. The reason for this is that the holding cost is a convex function of the order interval and ti < ðti þ tj Þ=2 < tj . A contradiction occurs since the cost of the modified policy is less than the optimal policy. Therefore, if ti and tj are any order intervals such that ti tm ðnÞ < CsdT and tj tm ðnÞ < CsdT , then ti ¼ tj . A similar argument can be given to prove the case of ti tm ðnÞ P CsdT and tj tm ðnÞ P CsdT . h Since Lemma 6 guarantees that there are only two different lengths for order intervals in the optimal policy, there are only two possible values for ti in the representation of the order intervals as tm ðnÞ þ ti . Let tL correspond to the additional length of the order interval that uses less than truckload transportation and tF represent the additional length transported in a truckload. The bounds on these variables are 0 6 tL 6 CsdT and CT 6 tF 6 T Q . If the number of orders of each length is known, an alternative formulation for P with only sd a single variable may be defined. Assume that there are n total orders and j orders of length tm ðnÞ þ tF . For notational convenience, the planning horizon TH is represented as ntm ðnÞ þ T R where TR is called the reduced planning horizon. The formulation for the optimal policy in terms of tF is 2 jhd ðn jÞhd T R jtF 2 ðtF þ tm ðnÞÞ þ ðP Þ minimize ZðtF Þ ¼ þ tm ðnÞ 2 2 nj ntm ðnÞ þ sdðT R jtF Þ þ C T þ j þ nK; TQ CT 6 tF 6 T Q ; subject to sd T R ðn jÞC T TR 6 tF 6 : j jsd j 0
To compute the optimal policy, the formulation P 0 is solved assuming n setups for each n ¼ 1; . . . ; nmax , where nmax is defined as the upper bound on n according to Lemma 2. For a given number of setups n, the formulation P 0 is solved assuming j order intervals corresponding to tF. The policy that has the lowest cost among all possible combinations of n and j is the optimal policy for the finite planning horizon problem. Algorithm 2 defines these computations for an optimal policy.
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
583
Algorithm 2 l n pffiffiffiffiffiffiffiffiffiffiffiffiffiffiom Step 0: Let nmax ¼ 2T H = min T Q ; 2K=hd . Set j ¼ 0; n ¼ 1, and Z ¼ 1. . Step 1: Set L ¼ max T R ðn jÞ CsdT ; j CsdT ; U ¼ min T R ; jT Q , and ^tF ¼ TnR þ ðnjÞs nh Step 2: If L > U , go to Step 3. If j^tF < L, set ^tF ¼ L=j. Similarly, if j^tF > U , let ^tF ¼ U =j. If Z > Zð^tF Þ, let j^tF Z ¼ Zð^tF Þ; tF ¼ ^tF , tL ¼ T Rnj ; j ¼ j and n ¼ n. Step 3: If j ¼ n, go to Step 4. Otherwise, set j ¼ j þ 1 and return to Step 1. Step 4: If n < nmax , set j ¼ 0; n ¼ n þ 1, and return to Step 1. Otherwise, stop with the optimal policy that has j* order intervals of length tm ðn Þ þ tF , ðn j Þ order intervals of length tm ðn Þ þ tL , and an optimal cost of Z*. In Theorem 2, the optimality of this algorithm is proven. Theorem 2. Algorithm 2 for the finite planning horizon model yields an optimal policy. Proof. To prove that the policy is optimal, it must be verified that Steps 1 and 2 of the algorithm compute an optimal policy for the given n and j. If j ¼ 0 or j ¼ n, the solution is trivial. The algorithm assigns equal order intervals to minimize holding cost if the formulation is feasible. Now consider the remaining possibilities with 0 < j < n. To show the objective function is convex in tF given that n and j are constant, the first derivative is taken with respect to tF Z 0 ðtF Þ ¼ jhdtF
jhdT R j2 hdtF þ jsd: nj nj 2
hd Since the second derivative is positive, Z 00 ðtF Þ ¼ jhd þ jnj , the objective function is strictly convex. This implies that ZðtF Þ achieves a minimum at the value of tF where Z 0 ðtF Þ ¼ 0. The optimal value of tF obtained by solvingZ 0 ðtF Þ ¼ 0 is tF ¼ TnR þ ðnjÞs . Recall that the bounds on tF and tL define a feasible interval for tF. If the minnh imum value for ZðtF Þ is not on this feasible interval, the optimal value occurs at an endpoint of the interval. By Lemma 3, there are at most nmax setups in the optimal policy. Since Steps 0, 3, and 4 iterate through all combinations of n and j for the formulation P 0 that may be in an optimal policy, Algorithm 2 produces an optimal policy over a finite horizon. h
4. Examples In this section, two examples are presented to illustrate the cost improvement by allowing for a combination of two modes of transportation. As a benchmark, we compare using both modes to a model with only TL transportation [4] or only having the option of a constant transportation cost per unit. The case of having a constant transportation cost per unit reduces to the EOQ model [9]. For the maximum cost improvement from allowing both modes of transportation, the optimal order quantity is between one and two truckloads. If the order quantity is less than one truckload, the costs are the same as using a single mode exclusively. If the order quantity is greater than two truckloads, the relative improvement is smaller since more units are shipped with TL transportation filled to truckload capacity. For the infinite horizon case, if the optimal EOQ policy is within the range ½T Q ; 2T Q , the holding and setup cost for an order interval of length TQ or 2T Q is at most 6% more than the optimal EOQ cost. This is true by the convexity of the average holding andhsetup cost function. i Suppose TEOQ is the optimal EOQ order interpffiffiffi 1ffiffi p val. The root-two interval is defined as T EOQ ; 2T EOQ . The holding and setup costs for order intervals 2
equal to these endpoints are 6% more than the optimal EOQ cost. Since the root-two interval cannot be contained strictly within ½T Q ; 2T Q , an order interval equal to TQ or 2TQ has minimum possible average transportation cost and at most a 6% increase in holding and setup cost. Thus, the improvement can be no greater than 6% for holding and setup costs with the transportation cost also contributing to a reduction of the maximum possible savings.
584
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
Average Cost
50
45
40
LTL
35
TL Both 30 5
10
15
20
25
30
T
Fig. 3. Average overall cost for different alternatives.
To show the potential for improvement, consider the example with K ¼ 200; h ¼ 1; d ¼ 2; C T ¼ 60:2; s ¼ 3:75, and T Q ¼ 10. In this case, the optimal EOQ policy has an order interval of T ¼ 14:1. This is the optimal order interval for the case of a constant transportation cost per unit with an optimal average cost per time unit of 35.8. Fig. 3 shows the average holding, setup, and transportation cost for only TL shipments, a carrier with a constant transportation cost per unit, and a combination of the two options. Note that the scale of the y-axis is set to the range [30, 50] to magnify the minimum. For using TL transportation exclusively, the optimal order interval is T ¼ 17:9 with an average cost of 35.8. By Algorithm 1, the optimal order interval for using both modes is 13.6 with a cost of 34.7. This is 97% of the lowest cost from using one mode of transportation exclusively. Although this is a small improvement, this reduction in logistics costs could impact profit for a large company. Carter and Ferrin [7] examine the importance of transportation costs. They discuss an example of GTE, a $27 billion company, with an $84 million transportation bill in 1993. A small improvement in holding, setup, and transportation costs could be significant for a company such as GTE. For the finite horizon model, suppose we have an example with K ¼ 25, h ¼ 2; d ¼ 1; C T ¼ 1000; s ¼ 500; T Q ¼ 4, and T H ¼ 5. If the constant transportation cost per unit s is used exclusively, the optimal policy is one order for the entire horizon with a total cost of 2550. For only TL transportation, the optimal policy is a single order with a total cost of 2050. From Algorithm 2, the optimal cost is also a single order. However, the total cost is only 1550. Thus, there was a cost improvement of 24% over only using one mode of transportation. Although this example is not realistic and was designed to show a large improvement, it does show there is a much higher potential for improvement in the finite horizon model than the infinite case. 5. Conclusions In some inventory systems, it may be beneficial to use two different methods of transportation simultaneously. Although the total costs of the models in this paper are neither concave nor convex functions of the order intervals, the cost functions are either concave or convex within certain ranges. By taking advantage of this structure, algorithms to compute optimal policies are presented for single stage models over both an infinite and finite planning horizon. In the past, many authors have optimized similar models with only one mode of transportation. In Section 4 for an infinite horizon, we showed that models considering only one mode of transportation can have as much as a 3% cost reduction by allowing two modes simultaneously. For the finite horizon model, an example is provided that illustrates a 24% cost savings. This finite horizon result for a single stage is a useful result for constructing algorithms for multiple stages. In a nested system, the downstream stages reduce to a finite horizon problem given a fixed order interval at the prior stage. If the large transportation cost is sufficiently large, it would also be useful to explore models with holding costs dependent on the opportunity costs of capital invested in transportation. In other future work, these models could be generalized across the full spectrum of TL and LTL costs with several breakpoints.
B.Q. Rieksts, J.A. Ventura / European Journal of Operational Research 186 (2008) 576–585
585
Acknowledgements The authors would like to thank Robin Roundy for suggesting inventory cost models that include both truckload and less than truckload modes of transportation. References [1] P. Abad, V. Aggarwal, Incorporating transportation cost in the lot sizing and pricing decisions with downward sloping demand, International Journal of Production Economics 95 (2005) 297–305. [2] W. Adelwahab, M. Sargious, Freight rate structure and optimal shipment size in freight transportation, Logistics and Transportation Review 26 (1992) 271–292. [3] F. Arcelus, J. Rowcroft, Inventory policies with freight and incremental quantity discounts, International Journal of Systems Science 22 (1991) 2025–2037. [4] D. Aucamp, Nonlinear freight costs in the EOQ problem, European Journal of Operational Research 9 (1982) 61–63. [5] R. Ballou, The accuracy for estimating truck class rates for logistical planning, Transportation Research A 25A (1991) 327–337. [6] L. Burns, R. Hall, D. Blumenfeld, Distribution strategies that minimize transportation and inventory costs, Operations Research 33 (1985) 469–490. [7] J. Carter, B. Ferrin, Transportation costs and inventory management: Why transportation costs matter, Production and Inventory Management Journal 37 (1996) 58–62. [8] J. Carter, B. Ferrin, C. Carter, The effect of less-than-truckload rates on the purchase order lot size decision, Transportation Journal 34 (1995) 35–44. [9] F. Harris, How many parts to make at once, Operations Research 38 (1990) 947–950, reprint from Factory – The Magazine of Management 10 (1913) 135–136. [10] H. Hwang, D. Moon, S. Shinn, An EOQ model with quantity discounts for both purchasing price and freight cost, Computers and Operations Research 17 (1990) 73–78. [11] J. Langley, The inclusion of transportation costs in inventory models: Some considerations, Journal of Business Logistics 2 (1980) 106–125. [12] P. Larson, The economic transportation quantity, Transportation Journal 28 (1988) 43–48. [13] C.Y. Lee, The economic order quantity for freight discount costs, IIE Transactions 18 (1986) 318–320. [14] S. Lippman, Economic order quantities and multiple set-up costs, Management Science 18 (1971) 39–47. [15] R. Russell, J. Karjewski, Optimal purchase and transportation cost lot sizing for a single item, Decision Sciences 22 (1991) 940–952. [16] L. Schwarz, A simple continuous review deterministic one-warehouse N-retailer inventory problem, Management Science 19 (1973) 555–566. [17] L. Schwarz, Economic order quantities for products with finite demand horizons, AIIE Transactions 4 (1972) 234–237. [18] S. Swenseth, M. Godfrey, Incorporating transportation costs into inventory replenishment decisions, International Journal of Production Economics 77 (2002) 113–130. [19] S. Swenseth, M. Godfrey, Estimating freight rates for logistics decisions, Journal of Business Logistics 17 (1996) 213–231.