Determining the Optimum Ordering Policy in Multi-Item Joint Replenishment Problem Using a Novel Method

Abstract

This work investigates the joint replenishment problem (JRP) involving multiple items where economies exist for replenishing several items simultaneously. The demand rate for each item is known and constant. Shortages are not permitted and lead times are negligible. Many heuristic algorithms have been proposed to find quality solutions for the JRP. In this paper, cycle time division and recursive tightening methods are developed to calculate an efficient and optimal replenishment policy for JRP. Two theorems are demonstrated to guarantee that an optimal solution to the problem can be derived using cycle time division and recursive tightening methods. Restated, cycle time division and recursive tightening methods theoretically yield the optimal solution in 100% of instances. The complexity of cycle time division and recursive tightening methods is just O(NlogN), where N represents the number of items involved in the problem. Numerical examples are included to demonstrate the algorithmic procedures.

1. Introduction

The joint replenishment problem (JRP) is an inventory problem involving multiple items where economies exist for replenishing several items simultaneously. These cost savings can be achieved when the replenishment of several items is coordinated.

The JRP has received considerable attention over recent decades. Detailed surveys of the JRP are available in [1–3]. Numerous heuristic algorithms have been developed for the JRP [4–8], but few optimal algorithms have been proposed. Balintfy [9] was among the first to consider the case in which joint ordering of several items may result in savings in set-up costs. A general formulation to the JRP and a graphical solution for the two-item case were provided in [10]. Furthermore, the author of [11] was the first to present an optimal algorithm for the inventory problem involving multiple items supplied by a single supplier with the assumption of purchase orders placed at equal time intervals. A branch and bound algorithm for devising an optimal policy was proposed in [12]. Furthermore, in [13], the authors demonstrated that the assumption of [11] of equally spaced manufacturing runs does not necessarily yield an optimal solution in situations involving more than two items. Additionally, a heuristic method was proposed for determining order quantities in joint replenishments with deterministic demand in [14]. Then, the algorithm of [14] was improved by [15, 16]. Moreover, an effective heuristic algorithm was presented in [17]. Finally, an optimal production-packaging policy for jointly replenished items was demonstrated in [18].

Additionally, a simple heuristic procedure, called RAND, was proposed based on the comparison of several values of the cycle time 𝑇 [19]. Furthermore, the range of cycle time was divided into 𝑚 equally spaced values of 𝑇. Subsequently, for each value of 𝑇, the improved Silver algorithm [15] was applied to calculate the total costs and select the lowest cost as the optimal solution. In [20], the authors modified the RAND using a better estimate of the minimal value of the cycle time 𝑇. Simulation experiments were performed to demonstrate the superiority of the modified RAND procedure. The author of [21] pointed out that the lower bound of the Goyal method [11] cannot guarantee to yield an optimal solution. He thus recommended the correct lower bound on 𝑇 and demonstrated a heuristic algorithm for deriving the optimal among the various cyclic policies. In [22], a heuristic method based on relaxing the nonlinear mixed integer programming (NLP) formulation of JRP was proposed. Beginning with the ordering frequencies derived from the solution of NLP relaxation, the near-optimal replenishment policies were determined. Subsequently, the authors of [23] performed a simulation experiment to compare the efficient exhaustive search heuristic for JRP proposed by [20] with that proposed by [22]. The problem in discrete time was considered and the lower and upper bounds of the optimal cycle time were derived in [24]. Additionally, a heuristic algorithm was presented to reduce the computational effort by improving the bounds on the basic cycle time 𝑇 in [25]. Furthermore, an alternative optimal approach based on global optimization theory was proposed in [26]. The problem can efficiently be solved via Lipschitz optimization. In [27], the authors proposed heuristic methods for optimizing the cyclic policy and the strict cyclic policy for the JRP. The method of [27] for determining the optimal strict cyclic policy that does not guarantee an optimal solution was demonstrated in [28]. A modified method was thus devised for obtaining the optimal strict cyclic policy. The author of [29] considered the deterministic multiproduct inventory model. The problem is to determine a production schedule to meet the demands and minimize the sum of production and holding costs. Also, some qualitative properties of the multi-item joint replenishment inventory problem are obtained. The JRP for 𝑀 items under deterministic demand, with a minimum order, quantity constraint for each item in the replenishment order was studied in [30]. The bounds on the basic cycle time are derived and a global optimization procedure is proposed to solve the JRP with constraints. A heuristic method based on a spreadsheet technique for solving joint replenishment problems with strict cycle policies was proposed in [31]. Based on the test data, the modified RAND algorithm of [20] and Nilsson et al. algorithm of [31] achieve the optimal solutions in 81.7% and 92.9% of 48,000 instances, respectively. In [32], the author presented an analysis and solution method to solve the JRP with correction factor, which is a generalization of the method presented in [30]. Also, genetic algorithms (GAs) were developed for solving the JRP and we compared their performance to the RAND method in the literature [4, 33–35]. The algorithm first decides grouping, that is, which items are replenished together, and then assigns a fixed replenishment cycle for the items in the same group.

This study proposes the cycle time division (CTD) and recursive tightening (RT) methods to calculate an efficient and optimal strict cyclic policy for JRP. Two theorems are demonstrated to guarantee that the problem can be solved optimally using the CTD and RT methods. Restated, CTD and RT methods theoretically yield the optimal solution in every instance. The complexity of the CTD and RT methods is just O(NlogN).

The remainder of this paper is organized as follows. Section 2 introduces a statement of the problem and the required notations. Sections 3 and 4 then describe the cycle time division and recursive tightening methods. Additionally, Section 4 demonstrates the validity of two theorems. Subsequently, Section 5 proposes an optimization algorithm. Section 6 then presents numerical examples. Finally, conclusions are drawn in Section 7.

2. Assumptions, Notations, and Problem Statement

In the formulation of the joint replenishment problem, the following assumptions are made:(1)the demand rate for each item is known and constant;(2)shortages are not allowed;(3)the replenishment rate is infinite (zero lead time);(4)there is an infinite horizon.Moreover, to discuss the joint replenishment problem, we introduce the following notation: 𝑁: number of items; 𝐴: major ordering cost associated with each replenishment ($/order); 𝑖: 1,2,…,𝑁, an item index; 𝑑_𝑖: annual demand for item 𝑖 (units/year); 𝑎_𝑖: the minor ordering cost incurred if item 𝑖 is ordered in a replenishment ($/order); ℎ_𝑖: holding cost per unit per year of item 𝑖 ($/unit/year); 𝑇: time between successive replenishments (years), ( basic cycle time, decision variable); 𝑄_𝑖: order quantity of item 𝑖; 𝑇_𝑖: time interval between successive replenishments of item 𝑖 (years); 𝑚_𝑖: frequency of ordering item 𝑖 (an integer multiple, decision variable).The JRP is an inventory problem concerning multiple items where economies exist for replenishing several items together. There are 𝑁 items, and for each item 𝑖, demand occurs at a constant annual rate of 𝑑_𝑖. Holding costs are incurred at the rate of ℎ_𝑖 per unit of item 𝑖 held. A major ordering cost 𝐴 is incurred for every replenishment (independent of the items that are replenished) and a minor ordering cost 𝑎_𝑖 for each item 𝑖 included in the replenishment. The cycle time for every item is an integer multiple 𝑚_𝑖 of the replenishment basic cycle time 𝑇. Thus, the cycle time for item 𝑖 is 𝑇_𝑖=𝑚_𝑖𝑇, and the order quantity for item 𝑖 is 𝑄_𝑖=𝑇_𝑖𝑑_𝑖=𝑇𝑚_𝑖𝑑_𝑖. The objective of the JRP is to develop inventory policies so as to minimize the total annual costs. The problem involves determining the basic cycle time 𝑇 and the integer multiples 𝑚_𝑖 for each item 𝑖. Let 𝐌=(𝑚₁,𝑚₂,…,𝑚_𝑁) be the replenishment frequency vector and let 𝐶(𝑇,𝐌) be the objective annual cost function, including the holding and ordering costs, which can be expressed as 𝐶()=1𝑇,𝐌𝑇𝐴+^𝑁_𝑖=1𝑎_𝑖𝑚_𝑖+^𝑁_𝑖=1𝑚_𝑖𝑑_𝑖ℎ_𝑖𝑇2. (1)

3. Cycle Time Division Method

For any fixed vector 𝐌,𝐶(𝑇,𝐌) is convex in 𝑇, since 𝜕²𝐶(𝑇,𝐌)𝜕𝑇²=2𝑇³𝐴+^𝑁_𝑖=1𝑎_𝑖𝑚_𝑖>0. (2) Setting 𝜕𝐶(𝑇,𝐌)/𝜕𝑇 to zero and solving it for 𝑇(𝐌) with fixed vector 𝐌, it follows that 𝑇(𝐌)=2𝐴+∑^𝑁_𝑖=1𝑎_𝑖/𝑚_𝑖∑^𝑁_𝑖=1𝑚_𝑖𝑑_𝑖ℎ_𝑖. (3) Substituting back optimal 𝑇(𝐌) into (1), the total cost function can be expressed as 𝐶(𝐌)=2𝐴+^𝑁_𝑖=1𝑎_𝑖𝑚_𝑖^𝑁_𝑖=1𝑚_𝑖𝑑_𝑖ℎ_𝑖. (4) For a given basic cycle time 𝑇, the associated ordering costs decrease with increasing the frequency of ordering items. However, the inventory level and holding cost inevitably increase accordingly. Therefore, a decision regarding the integer multiples involves a tradeoff between the holding and ordering costs. Consequently (for item 𝑗), for each pair of 𝑚_𝑗 and 𝑚_𝑗+1, define 𝑇=𝑇^𝑠_{𝑗,𝑚𝑗} such that 𝐶𝑇^𝑠_{𝑗,𝑚𝑗},𝑚₁,𝑚₂,…,𝑚_𝑗,…,𝑚_𝑁=𝐶𝑇^𝑠_{𝑗,𝑚𝑗},𝑚₁,𝑚₂,…,𝑚_𝑗+1,…,𝑚_𝑁 (5) or 𝑎_𝑗𝑇^𝑠_{𝑗,𝑚𝑗}𝑚_𝑗+𝑇^𝑠_{𝑗,𝑚𝑗}𝑚_𝑗𝑑_𝑗ℎ_𝑗2=𝑎_𝑗𝑇^𝑠_{𝑗,𝑚𝑗}𝑚_𝑗+𝑇+1_{𝑠𝑗,𝑚𝑗}𝑚_𝑗+1𝑑_𝑗ℎ_𝑗2 (6) as a critical cycle time. A given item 𝑗 may have multiple critical cycle times.

Solve for 𝑇^𝑠_{𝑗,𝑚𝑗} then; 𝑇^𝑠_{𝑗,𝑚𝑗}=2𝑎_𝑗1/𝑚_𝑗−1/𝑚_𝑗+1ℎ_𝑗𝑑_𝑗. (7) Additionally, the difference in the objective cost function between 𝑚_𝑗 and (𝑚_𝑗+1) is 𝐶𝑇,𝑚₁,𝑚₂,…,𝑚_𝑗,…,𝑚_𝑁−𝐶𝑇,𝑚₁,𝑚₂,…,𝑚_𝑗+1,…,𝑚_𝑁=𝑎_𝑗𝑇1𝑚_𝑗−1𝑚_𝑗−+1𝑇𝑑_𝑗ℎ_𝑗2. (8) Taking partial derivative of this difference with respect to 𝑇, then 𝜕𝐶𝑇,𝑚₁,…,𝑚_𝑗,…,𝑚_𝑁−𝐶𝑇,𝑚₁,…,𝑚_𝑗+1,…,𝑚_𝑁×(𝜕𝑇)⁻¹=−𝑎_𝑗𝑇²1𝑚_𝑗−1𝑚_𝑗−𝑑+1^𝑗ℎ_𝑗2<0. (9) Since 𝐶(𝑇^𝑠_{𝑗,𝑚𝑗},𝑚₁,…,𝑚_𝑗,…,𝑚_𝑁)=𝐶(𝑇^𝑠_{𝑗,𝑚𝑗},𝑚₁,…,𝑚_𝑗+1,…,𝑚_𝑁) and 𝜕[𝐶(𝑇,𝑚₁,…,𝑚_𝑗,…,𝑚_𝑁)−𝐶(𝑇,𝑚₁,…,𝑚_𝑗+1,…,𝑚_𝑁)]/𝜕𝑇<0, it follows that(a)𝐶(𝑇,𝑚₁,…,𝑚_𝑗,…,𝑚_𝑁)>𝐶(𝑇,𝑚₁,…,𝑚_𝑗+1,…,𝑚_𝑁) for 𝑇<𝑇^𝑠_{𝑗,𝑚𝑗};(b)𝐶(𝑇,𝑚₁,…,𝑚_𝑗,…,𝑚_𝑁)<𝐶(𝑇,𝑚₁,…,𝑚_𝑗+1,…,𝑚_𝑁) for 𝑇>𝑇^𝑠_{𝑗,𝑚𝑗}.

Hence, the following lemma is obtained.

Lemma 1. For multiple items jointly replenished, if basic cycle time 𝑇 is greater than the critical cycle time 𝑇^𝑠_{𝑗,𝑚𝑗}, then 𝑚_𝑗 is chosen as the frequency of ordering item 𝑗; otherwise, (𝑚_𝑗+1) is selected as the frequency of ordering item 𝑗.

Applying results from the study of [13], the lower bound for cycle time is 𝑇_𝐿=min{𝑎_𝑖/𝑑_𝑖ℎ_𝑖,𝑖=1,2,…,𝑁}. Therefore, for 𝑗=1,2,…,𝑁, use (7) to find all critical cycle is times 𝑇^𝑠_{𝑗,𝑚𝑗}, namely, 𝑇^𝑠_𝑗,1, 𝑇^𝑠_𝑗,2, and so on, until the resulting critical cycle time less than 𝑇_𝐿 for each item. Let 𝑇^𝑠_{𝑗,𝑚𝑗(max)} be the smallest critical cycle time for item 𝑗 then there are decreasing sequential critical cycle times 𝑇^𝑠_𝑗,1, and 𝑇^𝑠_𝑗,2,…, 𝑇^𝑠_{𝑗,𝑚𝑗(max)−1}, 𝑇^𝑠_{𝑗,𝑚𝑗(max)}, where 𝑇^𝑠_{𝑗,𝑚𝑗(max)−1}≥𝑇_𝐿 and 𝑇^𝑠_{𝑗,𝑚𝑗(max)}<𝑇_𝐿 for item 𝑗. Merge the critical cycle times for all items to establish an ordered critical cycle time vector 𝐒=(𝑠₁,𝑠₂,…,𝑠_𝑁𝑐) such that 𝑠₁≥𝑠₂≥⋯≥𝑠_𝑁𝑐, where 𝑠_𝑁𝑐=max{𝑇^𝑠_{𝑗,𝑚𝑗(max)},𝑗=1,2,…,𝑁}. Let RG₁=(𝑟₁,1) be the first cycle time range, where 𝑟₁=𝑠₁. If the basic cycle time 𝑇 falls in RG₁, then 𝑇>𝑇^𝑠_𝑖,1, for 𝑖=1,2,3,…,𝑁. Moreover, since 𝑟₁ is the largest among the critical cycle time for all items, the local optimal integer multiples for every item is 1; that is, 𝐌₁=(1,1,…,1) is the local optimal replenishment frequency vector for this case according to Lemma 1.

Next, construct RG₂=[𝑟₂,𝑟₁] where 𝑟₂=𝑠₂. If 𝑟₂ (or 𝑠₂) and 𝑟₁ are the respective critical cycle times for items 𝑗 and 𝑘, define 𝐌^′=(0,…,0,1,0,…,0), where the only non-zero element of 𝐌^′ occurs in the 𝑗th position. Suppose the basic cycle time 𝑇 lies in RG₂, then 𝑇^𝑠_𝑗,1≥𝑇 and 𝑇≥𝑇^𝑠_𝑖,1, for 𝑖=1,2,…,𝑁, and 𝑖≠𝑗. In this case, the local optimal integer multiples for item 𝑗 is increased to two, while the local optimal integer multiples for the other items remain at one; thus 𝐌₂=𝐌₁+𝐌^′, and the corresponding local optimal replenishment frequency vector is 𝐌₂. This process is continued and the local optimal replenishment frequency vector 𝐌_𝑖 is obtained for the cycle time range RG_𝑖, for 𝑖=1,2,…,𝑁𝑐, provided the critical order intervals differ for all items.

Given 𝑠_𝑖+1=𝑠_𝑖 and 𝑠_𝑖+1>𝑠_𝑖+2,where 𝑠_𝑖 and 𝑠_𝑖+1 are the critical cycle times for items 𝑒 and 𝑓, respectively, the aforementioned procedure is modified slightly as follows. Let RG_𝑖+1=[𝑟_𝑖+1,𝑟_𝑖], 𝐌^′=(0,…,0,1,0,…,0,1,0,…,0), and 𝐌_𝑖+1=𝐌_𝑖+𝐌^′, where 𝑟_𝑖+1=𝑠_𝑖+2, 𝑟_𝑖=𝑠_𝑖, and the two 1’s of 𝐌^′occupy the 𝑒th and 𝑓th positions, respectively. Thus, the local optimal replenishment frequency vector is 𝐌_𝑖+1 for the situation where the basic cycle time 𝑇 falls within the range RG_𝑖+1. A similar modification is performed if more than two critical cycle times are equal.

Upon the completion of the procedure, there exists 𝑈 resulting cycle time ranges RG₁,RG₂,…,RG_𝑈  with respective associated replenishment frequency vectors 𝐌₁,𝐌₂,…,𝐌_𝑈, where 𝑈≤𝑁𝑐.

4. Recursive Tightening Method

Clearly, at least one element in 𝐌_𝑖+1 is greater than that in 𝐌_𝑖 at the same position. An application of (3) yields 𝑇(𝐌_𝑖)>𝑇(𝐌_𝑖+1). Consequently, the following lemma is obtained.

Lemma 2. For multiple items jointly replenished, 𝑇(𝐌_𝑖)>𝑇(𝐌_𝑖+1).

Next, two properties are established to identify the candidate replenishment frequency vectors for the problem being investigated.

Theorem 3. For multiple items jointly replenished, if the basic cycle time 𝑇(𝐌_𝑖) obtained by (3) does not lie within its associated cycle time range 𝑅𝐺_𝑖 but rather lies in the range 𝑅𝐺_𝑗=[𝑟_𝑗,𝑟_𝑗−1], where 𝑖<𝑗, then 𝐶(𝑇(𝐌_𝑖),𝐌_𝑖)>𝐶(𝑇(𝐌_𝑖),𝐌_𝑗). Consequently, none of the replenishment frequency vector 𝐌_𝑖,𝐌_𝑖+1,…, or 𝐌_𝑗−1 is the global optimal replenishment frequency vector, and none of 𝑇(𝐌_𝑖),𝑇(𝐌_𝑖+1),…, or 𝑇(𝐌_𝑗−1) is the global optimal cycle time.

Proof. Let the corresponding replenishment frequency vector of RG_𝑖 be 𝐌_𝑖=(𝑚_(𝑖)1,𝑚_(𝑖)2,…,𝑚_(𝑖)𝑁), and let that of RG_𝑗 be 𝐌_𝑗=(𝑚_(𝑗)1,𝑚_(𝑗)2,…,𝑚_(𝑗)𝑁), where 𝑖<𝑗. From the construction of 𝐌_𝑖 and 𝐌_𝑗, it is evident that 𝑚_(𝑖)1≤𝑚_(𝑗)1,𝑚_(𝑖)2≤𝑚_(𝑗)2,…, and 𝑚_(𝑖)𝑛≤𝑚_(𝑗)𝑁. Consider RG_𝑖=[𝑟_𝑖,𝑟_𝑖−1] and its adjacent cycle time range RG_𝑖+1=[𝑟_𝑖+1,𝑟_𝑖]. It is known that 𝑇(𝐌_𝑖) does not lie within RG_𝑖 but rather lies in RG_𝑗, where 𝑖<𝑗, so we have 𝑇(𝐌_𝑖)<𝑟_𝑖. With very few distinctions (at least one), most of the elements of 𝐌_𝑖 and 𝐌_𝑖+1 are the same. For the sake of simplicity, assume that there is only one element different from 𝐌_𝑖 and 𝐌_𝑖+1, and let it be the 𝑘th element or 𝑚_(𝑖)𝑘+1=𝑚_(𝑖+1)𝑘. Since 𝑇(𝐌_𝑖)<𝑟_𝑖, according to Lemma 1, 𝑚_(𝑖+1)𝑘 is chosen as the integer multiples for item 𝑘; consequently, 𝐶(𝑇(𝐌_𝑖),𝐌_𝑖)>𝐶(𝑇(𝐌_𝑖),𝐌_𝑖+1). Similarly, 𝐶(𝑇(𝐌_𝑖),𝐌_𝑖+1)>𝐶(𝑇(𝐌_𝑖),𝐌_𝑖+2), and 𝐶(𝑇(𝐌_𝑖),𝐌_𝑖+2)>𝐶(𝑇(𝐌_𝑖),𝐌_𝑖+3),…, 𝐶(𝑇(𝐌_𝑖),𝐌_𝑗−1)>𝐶(𝑇(𝐌_𝑖),𝐌_𝑗). Hence, it follows that 𝐶(𝑇(𝐌_𝑖),𝐌_𝑖)>𝐶(𝑇(𝐌_𝑖),𝐌_𝑗).
From Lemma 2, it is known that 𝑇(𝐌_𝑖)>𝑇(𝐌_𝑖+1)>⋯>𝑇(𝐌_𝑗−1)>𝑇(𝐌_𝑗), where 𝑖<𝑗. Suppose that 𝑇(𝐌_𝑖+1),…,𝑇(𝐌_𝑗−1), and 𝑇(𝐌_𝑗) lie in the cycle time ranges RG_𝑤=[𝑟_𝑤,𝑟_𝑤−1],…,RG_𝑦=[𝑟_𝑦,𝑟_𝑦−1], and RG_𝑧=[𝑟_𝑧,𝑟_𝑧−1], respectively, where 𝑖≤𝑗−1<𝑗≤𝑤≤⋯≤𝑦≤𝑧. From Lemma 1, it has been shown that 𝐶(𝑇(𝐌_𝑖),𝐌_𝑖)>𝐶(𝑇(𝐌_𝑖),𝐌_𝑗), 𝐶(𝑇(𝐌_𝑖+1),𝐌_𝑖+1)>𝐶(𝑇(𝐌_𝑖+1),𝐌_𝑤),…, 𝐶(𝑇(𝐌_𝑗−1),𝐌_𝑗−1)>𝐶(𝑇(𝐌_𝑗−1),𝐌_𝑦) and 𝐶(𝑇(𝐌_𝑗),𝐌_𝑗)≥𝐶(𝑇(𝐌_𝑗),𝐌_𝑧). Hence, none of 𝐌_𝑖,𝐌_𝑖+1,…,𝐌_𝑗−2, and 𝐌_𝑗−1 is the global optimal replenishment frequency vector.
Again by Lemma 2, we have (𝐌_𝑧)≤𝑇(𝐌_𝑦)≤𝑇(𝐌_𝑤)≤𝑇(𝐌_𝑗)<𝑇(𝐌_𝑗−1)<⋯<𝑇(𝐌_𝑖+1)<𝑇(𝐌_𝑖). Therefore, it follows that 𝐶(𝑇(𝐌_𝑖),𝐌_𝑗)>𝐶(𝑇(𝐌_𝑗),𝐌_𝑗), 𝐶(𝑇(𝐌_𝑖+1),𝐌_𝑤)>𝐶(𝑇(𝐌_𝑤),𝐌_𝑤),…, 𝐶(𝑇(𝐌_𝑗−1),𝐌_𝑦)>𝐶(𝑇(𝐌_𝑦),𝐌_𝑦), and 𝐶(𝑇(𝐌_𝑗),𝐌_𝑧)≥𝐶(𝑇(𝐌_𝑧),𝐌_𝑧). Hence, none of 𝑇(𝐌_𝑖),𝑇(𝐌_𝑖+1),…,𝑇(𝐌_𝑗−2) and 𝑇(𝐌_𝑗−1) is the global optimal cycle time.

Theorem 4. For multiple items jointly replenished, if the basic cycle time 𝑇(𝐌_𝑖) calculated by (3) does not lie within its associated cycle time range 𝑅𝐺_𝑖 but rather lies in the range 𝑅𝐺_𝑗=[𝑟_𝑗,𝑟_𝑗−1], where 𝑗<𝑖, then 𝐶(𝑇(𝐌_𝑖),𝐌_𝑖)>𝐶(𝑇(𝐌_𝑖),𝐌_𝑗). Consequently, none of the vectors 𝐌_𝑗+1,𝐌_𝑗+2,…,𝐌_𝑖−1, or 𝐌_𝑖 is the global optimal replenishment frequency vector and none of 𝑇(𝐌_𝑗+1),𝑇(𝐌_𝑗+2),…,𝑇(𝐌_𝑖−1), or 𝑇(𝐌_𝑖) is the global optimal cycle time.

Proof. Similar to the proof of Theorem 3.

Next, we use Theorems 3 and 4 to explain how to find the optimal 𝑇 and 𝐌_𝑖. Since the minimal integer multiples for each item are one, 𝐌₁ is used in (3) to calculate the maximal cycle time as follows: 𝑇_max𝐌₁=2𝐴+∑^𝑁_𝑖=1𝑎_𝑖∑^𝑁_𝑖=1𝑑_𝑖ℎ_𝑖. (10)

Suppose that the maximal cycle time 𝑇_max(𝐌₁) falls in RG_𝑖=[𝑟_𝑖,𝑟_𝑖−1]. If 𝑖=1, then set 𝐌_min=𝐌₁,𝑇_max=𝑇(𝐌_min) and stop. Otherwise, consider the situation where 𝑖>1. From Theorem 3, none of the replenishment frequency vectors 𝐌₁,𝐌₂,…, and 𝐌_𝑖−1 provides a solution for the global optimal integer multiples, and none of 𝑇(𝐌₁),𝑇(𝐌₂),…,𝑇(𝐌_𝑖−2), and 𝑇(𝐌_𝑖−1) is the global optimal cycle time. The resulting minimal replenishment frequency vector 𝐌_𝑖 is designated as 𝐌_min. Next, 𝐌_min is used in (3) to calculate the cycle time 𝑇(𝐌_min), while RG_𝑗=[𝑟_𝑗,𝑟_𝑗−1] and 𝐌_𝑗 denote its associated cycle time range and replenishment frequency vector. If 𝐌_min=𝐌_𝑗, then stop; otherwise, set 𝐌_min equal to 𝐌_𝑗 and repeat the procedure until the aforementioned stopping criterion is reached. This process is termed the downward recursive tightening procedure while the resulting 𝐌_min is an upper bound of the candidate vector set.

The same approach can be applied in the opposite direction to determine the lower bound of the candidate vector set.  𝐌_𝑈  is used in (3) to calculate the minimal cycle time 𝑇_min(𝐌_𝑈). Suppose the resulting 𝑇_min(𝐌_𝑈) falls in RG_𝑖=[𝑟_𝑖,𝑟_𝑖−1], where 𝑖<𝑈. From Theorem 4, none of the replenishment frequency vectors 𝐌_𝑈,𝐌_𝑈−1,…,𝐌_𝑖+2, and 𝐌_𝑖+1 provides a solution for the global optimal integer multiples, and none of 𝑇(𝐌_𝑈),𝑇(𝐌_𝑈−1),…,𝑇(𝐌_𝑖+2) and 𝑇(𝐌_𝑖+1) is the global optimal cycle time. The associated replenishment frequency vector 𝐌_max is set equal to 𝐌_𝑖 and 𝐌_max is used in (3) to compute 𝑇(𝐌_max), while RG_𝑗=[𝑟_𝑗,𝑟_𝑗−1] and 𝐌_𝑗 denote the associated cycle time range and replenishment frequency vector. If 𝐌_max=𝐌_𝑗, then stop; otherwise, set 𝐌_max equal to 𝐌_𝑗 and repeat the procedure until the stopping criterion is satisfied and the resulting 𝐌_max yielded is the lower bound of the candidate vector set. This approach is termed the upward recursive tightening procedure.

Finally, the replenishment frequency vectors between the lower and the upper bounds of the candidate vector set are compared for optimality.

5. Optimization Algorithm

The following procedure integrates the cycle time division and the recursive tightening methods to obtain the optimal values of 𝑇 and 𝐌 for the problem being studied.

Step 1. Set 𝑇_𝐿=min{𝑎_𝑖/𝑑_𝑖ℎ_𝑖,𝑖=1,2,…,𝑁}.

Step 2. For 𝑗=1,2,…,𝑁, do(a)set 𝑚_𝑗=1.(b)compute the critical cycle time 𝑇^𝑠_{𝑗,𝑚𝑗} using (7);(c)if 𝑇^𝑠_{𝑗,𝑚𝑗}≥𝑇_𝐿, then set 𝑚_𝑗=𝑚_𝑗+1; go to Step 2(b); else set 𝑚_𝑗(max)=𝑚_𝑗.

Step 3. Merge all the critical cycle times to establish a critical cycle time vector 𝐒=(𝑠₁,𝑠₂,…,𝑠_Nc) such that 𝑠₁≥𝑠₂≥⋯≥𝑠_𝑁𝑐, where 𝑠_𝑁𝑐=max{𝑇^𝑠_{𝑗,𝑚𝑗(max)},𝑗=1,2,…,𝑁} and 𝑁𝑐≤∑^𝑁_𝑗=1𝑚_𝑗(max). Construct vector 𝐁=(𝑏_𝑘) such that 𝑏_𝑘 is the item number with which 𝑟_𝑘 is associated for 𝑘 = 1, 2, …, Nc.

Step 4. Set 𝐌₁=(1,1,…,1), 𝐌^′=𝟎, 𝑘=1, 𝑟₁=𝑠₁, and RG₁=(𝑟₁,1).

Step 5. For 𝑗=2,3,…,𝑁𝑐, (a)set 𝐌^′′=(0,…,0,1,0,…,0), where the only non-zero element of 𝐌^′′ is in the (𝑏_𝑗−1)th position.(b)if 𝑠_𝑗−1>𝑠_𝑗, then set 𝑟_𝑘+1=𝑠_𝑗, RG_𝑘+1=[𝑟_𝑘+1,𝑟_𝑘], and 𝐌_𝑘+1=𝐌_𝑘+𝐌^′+𝐌^′′, 𝑘=𝑘+1,𝐌^′=𝟎;
else set 𝐌^′=𝐌^′+𝐌^′′.

Step 6. Set 𝐌_𝑈=𝐌_𝑘.

Step 7. Use M₁ in (10) to calculate 𝑇_max(M₁) and obtain its associated replenishment frequency vector 𝐌_min.

Step 8. If 𝐌_min=𝐌₁, then set 𝑇_max=𝑇_max(𝐌₁), 𝑒=1, and go to Step 11.

Step 9. Use 𝐌_min and (3) to calculate 𝑇(𝐌_min).

Step 10. If 𝐌^′=𝐌_min, then set 𝑇_max=𝑇(𝐌_min), 𝑒 = the index of the cycle time range in which 𝑇(𝐌_min) falls, and go to Step 11;
else set 𝐌_min=𝐌^′ and go to Step 9.

Step 11. Use 𝐌_𝑈 in (3) to find 𝑇_min(𝐌_𝑈) and identify its associated replenishment frequency vector 𝐌_max.

Step 12. If 𝐌_max=𝐌_𝑈, then set 𝑇_min = 𝑇_min(𝐌_𝑈), 𝑓=𝑈, and go to Step 15.

Step 13. Use 𝐌_max in (3) to calculate 𝑇(𝐌_max) and its associated replenishment frequency vector 𝐌^′.

Step 14. If 𝐌^′=𝐌_max, then set 𝑇_min=𝑇(𝐌_max), 𝑓 = the index of the cycle time range in which 𝑇(𝐌_max) falls, and go to Step 15;
else set 𝐌_max=𝐌^′,and go to Step 13.

Step 15. Find 𝐶(𝑇^∗,𝐌^∗)=Min{𝐶(𝑇_𝑖,𝐌_𝑖), 𝑖=𝑒,𝑒+1,…,𝑓−1,𝑓, where 𝐌_𝑖 is the candidate replenishment frequency vector}.

Step 16. 𝑇^∗ and 𝐌^∗ are the optimal solutions with the optimal objective function of 𝐶(𝑇^∗,𝐌^∗).

The complexity of the algorithm is analyzed as follows. Suppose 𝑁 items in the jointly replenished system, where item 𝑗 has 𝑚_𝑗 critical cycle times, and 𝑚=(1/𝑁)∑^𝑁_𝑗=1𝑚_𝑗. Thus the complexity involved in finding the critical cycle times for all items is O(𝑁𝑚), or simply O(N). Forming the critical cycle time vector 𝐑 requires 𝐎(𝑁log𝑁) operations, and the complexity involved in generating the cycle time ranges is O(N). The recursive tightening procedure uses O(N) operations to identify the lower and upper bounds of the candidate vector set and O(N) operations to identify the optimum in the candidate vector set. Thus, the complexity of the overall solution procedure is 𝐎(𝑁log𝑁).

6. Numerical Examples

Example 1. Consider a multi-item inventory system with seven items involving the data summarized in Table 1. The proposed algorithm is applied. In Step 1, 𝑇_𝐿=0.0516. The resulting critical cycle times of each item are listed in Table 2. Furthermore, Table 3 lists the cycle time ranges established by merging all the critical cycle times and the associated replenishment frequency vectors found.

Table 1 

Data of Example 1 (the parameter’s values are adopted from [31]).

Table 2 

The critical cycle times of Example 1.

Table 3 

The cycle time ranges and associated replenishment frequency vectors for Example 1.

Starting with 𝐌₁ = (1, 1, 1, 1, 1, 1, 1), the upper bound candidate replenishment frequency vector identified via the downward recursive tightening procedure is 𝐌₇= (1, 1, 1, 2, 2, 3, 3) as illustrated in Table 4. Similarly, the lower bound candidate replenishment frequency vector identified via the upward recursive tightening procedure is also 𝐌₇ = (1, 1, 1, 2, 2, 3, 3). This indicates that the optimal replenishment frequency vector 𝐌^∗ is 𝐌⁷ = (1, 1, 1, 2, 2, 3, 3) with optimal basic cycle time of 𝑇^∗=0.1148 time units and the corresponding optimal objective function of 𝐶(𝑇^∗,𝐌^∗) = $1669.96/time unit. This solution is the best solution our procedure can find and also in this case the optimal solution of the problem (can be verified using for instance Goyal [18] and Nilsson et al. [31]).

Table 4 

The results of the recursive tightening procedures for Example 1.

Example 2. Consider a multi-item inventory system with fourteen items involving the data listed in Table 5. The proposed algorithm is performed. In Step 1, 𝑇_𝐿 = 0.0516. The resulting critical cycle times of each item are listed in Table 6. The cycle time ranges formed by merging all the critical cycle times and the associated replenishment frequency vectors found are listed in Table 7.

Table 5 

The data of Example 2.

Table 6 

The critical cycle times of Example 2.

Table 7 

The cycle time ranges and associated replenishment frequency vectors for Example 2.

Starting with M₁ = (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1), the upper bound candidate replenishment frequency vector identified via the downward recursive tightening procedure is 𝐌₉= (1, 1, 1, 2, 2, 3, 3, 1, 2, 1, 1, 1, 1, 2) as shown in Table 8. Similarly, the lower bound candidate replenishment frequency vector identified via upward the recursive tightening procedure is 𝐌₁₅ = (1, 1, 2, 2, 2, 3, 4, 2, 2, 2, 2, 1, 2, 2). Between the lower and upper bounds, 𝐌₁₀, 𝐌₁₁, 𝐌₁₂,𝐌₁₃, and 𝐌₁₄ are not the candidate replenishment frequency vectors (it is becuase the corresponding cycle time is not falling in the range that determines 𝐌_𝑖). Comparison of the objective function of the replenishment frequency vectors in the candidate vector set of {𝐌₉,𝐌₁₅} indicates that the optimal replenishment frequency vector 𝐌^∗ is 𝐌₁₅ = (1, 1, 2, 2, 2, 3, 4, 2, 2, 2, 2, 1, 2, 2) with optimal basic cycle time of 𝑇^∗=0.0966 time units, and the corresponding optimal objective function of 𝐶(𝑇^∗,𝐌^∗)= $2713.77/time unit.

Table 8 

The results of the recursive tightening procedures for Example 2.

Comparing Examples 1 and 2 reveals that the number of critical cycle times increases with increasing number of items. Again, the results from Tables 3 and 7 indicate that the number of cycle time regions generated increases with increasing number of critical cycle times; however, the span of each cycle time region is inevitably reduced. A reduction in the span of a cycle time region may eventually lessen its probability of containing the 𝑇(𝐌) computed via (3). As a result, the RT procedures generate 1 and 2 candidate replenishment frequency vectors in Examples 1 and 2, respectively. Restated, the larger the number of items involved in a distribution system, the better the efficiency of the proposed algorithm.

7. Concluding Remarks

This work studies the joint replenishment problem involving multiple items where economies can be achieved by replenishing several items simultaneously. To solve the problem being studied, this study devises a cycle time division method for establishing a set of ranges that contain the optimal cycle time and corresponding optimal replenishment frequency vector of the problem and subsequently develops a recursive tightening procedure to further reduce the set of replenishment frequency vectors to facilitate the search for the optimal cycle time and associated replenishment frequency for each item. The proposed solution procedure has a polynomial complexity of O(NlogN), where 𝑁 denotes the number of items involved in the problem. Two theorems are proven, thus guaranteeing that an optimal solution to the problem can be obtained using cycle time division and recursive tightening. Restated, CTD and RT methods theoretically can achieve the optimal solution in all cases. In addition, in comparison with the heuristic method proposed by Nilsson et al. [31], our new method performs uninterestingly and complexly. However, comparing with many newer heuristic methods, we argue that the optimal solution searching is guaranteed to this new method, and the larger the number of items the system involved, the better the efficiency this novel method had.

Acknowledgments

The authors are grateful for the valuable comments and suggestions made by the anonymous referees. This research was supported by the National Science Council, Taiwan, R.O.C., under Grant no. NSC- 101-2410-H-147-002.