#### Abstract

This paper discusses the inventory mechanism with backordering and with the infinite planning horizon consisting of two stages wherein the demand rate in the first stage is strictly greater than that in the second stage. To minimize the retailer’s inventory cost, we establish a lot sizing decision model. On the basis of the inventory cost analysis, we present a closed-form solution to the model and provide an optimal replenishment and stocking strategy to the retailer. The given numerical experiments show the validity of the model.

#### 1. Introduction

The economic order quantity (EOQ) is a seminal inventory model which was originally introduced by Harris [1] and further developed by Wilson [2]. Almost 50 years later, many variations were developed that differ in assumptions on not only the life time of an item, but also the type of demand, the presence of price discounts, allowing shortages and backordering, single or multiple items, one or two warehouses, single or multiechelon modeling, average cost or discounted cash flows, and whether a delay in payment is permissible in Bakker [3]. An elaborate review of the model and its well-known extensions can be found in Nahmias [4]. To make the model more realistic, various relaxations on demand rate are made thereafter by Waters [5]. The extensions of the basic EOQ model include inventory models with a limited-time price incentives such as a one-time-only or temporary price reduction and a limited-time interest-free delayed payment privilege; see, e.g., Khouja [6], Wang* et al*. [7], and references therein.

In modern consumer products market, numerous factors depending upon natural or business conditions affect various determinants of market demand. For example, the demand rate of a new type of air-conditioner, say air-conditioner A, is high in summer, which can be taken as a constant. However, because of the seasons, the demand rate of air-conditioner A will be lower after summer and a new type of air-conditioner will be launched into market the next year. Thus, the inventory system of air-conditioner A will have two-stage demand; that is, the demand rate in the second stage is less than that in the first stage and which can be taken as a constant in the second stage.

Research considering the change in demand rate was widely studied. For a discrete case of time-varying demand pattern, Wagner and Whitin [8] established a dynamic programmimg model, and, for a time-proportional varying demand pattern, Resh* et al. *[9] proposed an optimal replenishment number and time scheduling stocking strategy. Donaldson [10] considered the inventory model in which the demand varies linearly with the time, and the model was extended by Dave [11, 12] to the case that the shortages are allowed. Dey* et al*. [13] and Muriana* et al*. [14], respectively, considered the inventory model with dynamic demand and stochastic demand. Different from the above, Wu* et al*. [15] derive the EOQ model for inventory of an item that deteriorates at a Weibull-distributed rate and assuming the demand rate with continuous function of time. Yan [16] considered the EOQ model with freshness-dependent demand for perishable items. Baker and Urban [17], Datta and Pal [18], and Urban [19] considered the inventory model wherein the demand depends on stock level, whereas Mandal and Phaujdar [20] considered the inventory model wherein the demand rate is linearly dependent on the inventory level and items are allowed to deteriorate over time.

In this paper, we consider the following extended version of the EOQ inventory model with step-shaped demand and backordering: the time horizon of the concerned inventory mechanism consists of two stages and , and the demand rate in the first stage differs from that in the second stage; see Figure 1. Further, shortage is allowed and will be backordered. The other assumptions are the same as those on the EOQ model. For this mechanism, to minimize the retailer’s inventory cost, we establish an optimization inventory decision model. By an inventory cost analysis, we present a closed-form solution to the model and provide an optimal replenishment and lot sizing strategy to the retailer. The given numerical experiments show the validity of the model.

The remainder of the paper is organized as follows. Section 2 presents the assumptions on the inventory model and notations used in the subsequent analysis. A mathematical formulation of the problem is also established in Section 3. In Section 4, we derive the optimal solution for the inventory model and provide an optimal replenishment policy to the retailer. Numerical experiments on sensitivity analysis to show the validity of the model are made in Section 5. The conclusion and some extensions are given in the last section.

#### 2. Notations and Assumptions

First, we give the notations used in the subsequent analysis (Table 1) and then give the assumptions on the model.

*Assumption 1. *For the inventory system, we assume that

the time horizon is infinite;

the leading time for each order is zero;

shortage is allowed and will be backordered;

unit purchasing cost for both stages is the same;

the fixed ordering and unit backordering costs for both stages are the same.

Without loss of generality, we also assume that the stock level is zero at the beginning inventory operation system.

#### 3. Problem Description and Model Formulation

Consider the following inventory mechanism in which the time horizon of the concerned inventory mechanism consists of two stages and , and the demand rate in the first stage is strictly greater than in the second stage; i.e., . Further, shortage is allowed and will be backordered. The other assumptions are the same as those in the EOQ model; see Figure 1. For this inventory system, if the demand rate is constant, then the optimal ordering size at each replenishment circle is with maximal backorder from the knowledge of inventory by Waters [5].

In our concerned inventory model, due to the skipping of the demand rate, to deduce the inventory cost, we should determine optimal order size and the maximal shortage level in the first stage so that the inventory cost is as lower as possible. As the inventory system would revert to the regular EOQ model with backordering, i.e., BEOQ model, once the replenishment cycles in the first stage are finished. To derive the optimal policy, we need the following “ideal” BEOQ ordering policy, abbreviated “IBEOQ”: orders with size are made in the first stage and these ordering cycles are exactly ended at , and orders with size are made in the second stage. Under this policy, the long-run average operating cost is in the first stage and in the second stage, respectively. On account of this, we may establish an optimal replenishment and stocking strategy model by minimizing the operating cost relative to the IBEOQ ordering policy.

For such an inventory system, as the demand rates of two stages are different, according to the inventory cost analysis, the retailer should make orders with different sizes in different stages. Suppose orders with sizes are made in the first stage, and the maximal shortage levels are, respectively, in each replenishment cycle. According to the inventory cost minimization principle, the first replenishment cycles would be ended before and the -th replenishment cycle would be ended at or after , i.e., Further, once the replenishment cycles are ended, the system would resort to the BEOQ ordering policy with size . Thus, to minimize the retailer’s inventory cost, we only need to determine the order sizes , and the maximum shortage levels . In this sense, we denote the operating cost for these replenishment cycles by , and denote the time horizon of these replenishment cycles by . Since shortage will be fully backordered, it holds that

In the other way, since shortage is allowed and will be backordered, using the fact that the long-run average operating cost under the IBEOQ policy are in the first stage and in the second stage, we obtain the operating cost under the IBEOQ ordering policy over the time horizon ,Accordingly, the operating cost under the ordering policy relative to the IBEOQ ordering policy over the time horizon isThen, the problem of determining an optimal policy can be formulated as the following optimization problem:

In the subsequent sections, we will first compute the optimal ordering sizes and the maximum shortage levels for fixed ordering times in the first stage and then calculate the optimal ordering times in the first stage based on the inventory cost analysis. At the end, we present the algorithm for problem (5).

#### 4. Model Solution

For the inventory system, if orders are made in the first stage with sizes , since shortages are allowed, according to the the size of the remaining stock at , we have three scenarios: , , and . See Figures 2–4.

##### 4.1. Scenario

This scenario is depicted in Figure 2, the stock position is exhausted and the shortage level is at . From the knowledge of inventory, the total operating cost over the time length is

Subsequently, substituting (6) and (3) into (4) yields thatThen, the problem (5) can be formulated as the following optimization problem:

For this problem, we have the following conclusion.

Theorem 1. *For the inventory model (8), suppose orders are made in the first stage. If , then the optimal ordering sizes and the optimal maximum shortage level in the first stage are, respectively, and if , then the optimal ordering sizes and the optimal maximum shortage level in the first stage are, respectively, where , ,, *

*Proof. *For problem (8), since the constraints are linear, any optimal solution is a KKT point which satisfies one of the following two systems. See Wang* et al*. [21, 22].

System 1:where is the Lagrange multiplier corresponding with the equality constraint.

System 2: For these two systems, using their former equations, respectively, we can all conclude thatFor the former system, substituting (13) into the last equation and combining the other equations yieldwith . It is easy to verify that all the inequalities of the former system hold for , and given above. In addition, to guarantee and given above constitute a KKT point of problem (8), is required; that is, So the equation set by (14) is the solution to the problem (8) provided that For the latter system, after some substitutions and algebra, we have: if , thenotherwise, the equations of the latter system have no solution.

Denote To guarantee and given by (17) is a KKT point of problem (8), we break up the discussion into two cases.

, i.e., . In this case, to guarantee the nonnegativity of the and , it needs that , i.e., . Substituting (17) into the second inequality and the last inequality in the latter system yields that Moreover, under this case, we have so and given by (17) are a solution to the latter system provided that . By a similar discussion to that for Case 1, we know that and given by (17) are a solution to the latter system provided that Substituting (14) and (17) into (8), respectively, yields that the operating cost over the interval of length under the two policies relative to the IBEOQ ordering policy are and , respectively.where

Under the Case 2, if , after some substitutions and algebra, we can prove that

Combining the discussions above, we conclude that , , and given by (14) are a solution of problem (8) provided that and (17) is a solution of problem (8) provided that The proof is completed by combining the discussions above.

By Theorem 1, we know that the optimal ordering sizes from second order to m-th order in the first stage are the same, and the optimal maximum shortage levels from the first order to (m-1)-th order in the first stage are the same too; i.e., also So we can obtain the following two ordering policies for problem (5) with orders in the first stage.

*Policy*

which applies to the case that

*Policy*

which applies to the case that

Now, we consider the optimal ordering times in the first stage under Policy and Policy

By the proof process of Theorem 1, we know that the operating costs over the interval of length under the two policies relative to the IBEOQ ordering policy are and , respectively.

For function , by we know that is a convex function in , and its minimum can be obtained at the stationary point of the function; i.e.,

Set . Then due to the fact that is a positive integer, and requires . So the optimal ordering times in the first stage are provided that and or , or the optimal ordering times are provided that . Moreover, if , the optimal ordering times in the first stage are . That is

For function given by (24), by we know that if , i.e., , then and is a convex function in , and its minimum can be obtained at the stationary point of the function; i.e.,

On the other hand, according to the proof of Theorem 1, if , to guarantee , , and given by (17) is a solution to the model (8)), it needs that In this case, the optimal ordering times in the first stage are which satisfies that

However, if , i.e., , then and is a concave function in , and its minimum can be obtained at the terminal point of the function, and, according to the proof of Theorem 1, under the case that

Denote , . Obviously, , , , . Based on the discussion above, we obtain the optimal ordering times in the first stage under Policy

##### 4.2. Scenario

In this scenario, orders are made in the first stage and they can exactly meet the demand of the first stage; see Figure 3. Obviously, . From the knowledge of inventory, we have

Substituting (40) and (41) into formula (4) yields that

Then problem (5) can be formulated as the following optimization problem:

For this problem, we have the following conclusion.

Theorem 2. *For inventory model (43), suppose orders are made in the first stage; then the optimal ordering sizes and the optimal maximum shortage level in the first stage are, respectively, *

*Proof. *For problem (43), its KKT point satisfies the following system: where is the optimal Lagrange multiplier.

By solving the equations in the system, we can obtain the desired conclusion.

For this scenario, by Theorem 2, the same as Scenario 1, the optimal ordering sizes d as from second order to m-th order in the first stage are the same, and the optimal maximum shortage levels d as from the first order to (m-1)-th order in the first stage are the same too, so we can obtain the following optimal ordering policy for problem (5):

*Policy*

Under this policy, the operating cost over the interval of length relative to the IBEOQ ordering policy is

Now, we consider the optimal ordering times in the first stage under Policies

For function , by we know that is a convex function in , and its minimum can be obtained at the stationary point of the function, i.e., SetThen, the optimal ordering times in the first stage are provided that , and otherwise. Then, the optimal order times under Policy are that

##### 4.3. Scenario

For this scenario, see Figure 4; a similar discussions to that for Scenario 1 yields SoTherefore, the problem (5) can be formulated as the following form:

For this problem, we have the following conclusion.

Theorem 3. *For inventory model (54), suppose orders are made in the first stage. If , then the optimal ordering sizes and the optimal maximum shortage level in the first stage are, respectively, where , , , *

*Proof. *For problem (54), since all the constraints are linear, if the problem has an optimal solution, then it is a KKT point which satisfies the following system: For the systems of equations, we can obtain thatTaking the inequality constraints into considerations, we have that i.e., . The proof is completed.

By Theorem 3, under the Scenario , we can obtain the following optimal ordering policy for problem (5).

*Policy*

which applies to the case that

The operating cost over the interval of length under this policy relative to the IBEOQ ordering policy is

Now, we consider the optimal ordering times in the first stage under Policies

Similarly, for function given by (60), by we know that is a convex function in , and its minimum can be obtained at the stationary point of the function; i.e.,

Denote . A similar discussion yields the optimal ordering times in the first stage

Based on the discussion about three Scenarios above, we can present our algorithm for problem (5).

##### 4.4. Algorithm

*Step 1. *Input value of parameters .

*Step 6. *Compute , , , , Let

Compute , , , , , , , , , and

*Step 3. *If , then otherwise

*Step 4. *If , and , then if otherwise,

*Step 5. * If ,

(a) if and , or , then (b) if and , or , then otherwise,

(2) If ,

(a) if and and , or , then (b) if , and and , or , then 0therwise,

*Step 6. *If and , or , then If and , or , then otherwise,