# Optimal Replenishment and Lot Sizing Strategy for Inventory Mechanism with Step-Shaped Demand and Backordering

Authors: Lixia Zhang; Lili Han
Journal: Mathematical Problems in Engineering
 (2019)

---

## 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.


---

## Body

## 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], Wanget 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, Reshet 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. Deyet al. [13] and Murianaet al. [14], respectively, considered the inventory model with dynamic demand and stochastic demand. Different from the above, Wuet 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[0,T] and [T,∞], 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.Figure 1
Two-stage demand inventory system withλ′>λ.The remainder of the paper is organized as follows. Section2 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 (Table1) and then give the assumptions on the model.Table 1
Notation.
symbol description λ ′ demand rate in the first stage λ demand rate in the second stage K retailer’s fixed ordering cost h holding cost per unit item per unit time p backorder cost per unit item per unit time T length of the first stageAssumption 1.
For the inventory system, we assume that
( 1 ) the time horizon is infinite;
( 2 ) the leading time for each order is zero;
( 3 ) shortage is allowed and will be backordered;
( 4 ) unit purchasing cost for both stages is the same;
( 5 ) 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[0,T] and [T,∞], and the demand rate λ′ in the first stage is strictly greater than λ in the second stage; i.e., λ′>λ. 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 QBEOQ=2λK(h+p)/hp with maximal backorder BBEOQ=2λhK/p(h+p) 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 sizeQλ′BEOQ=2λ′K(h+p)/hp are made in the first stage and these ordering cycles are exactly ended at T, and orders with size QλBEOQ=2λK(h+p)/hp are made in the second stage. Under this policy, the long-run average operating cost is 2λ′Khp/h+p in the first stage and 2λKhp/h+p 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. Supposem orders with sizes Q1,…,Qm are made in the first stage, and the maximal shortage levels are, respectively, B1,…,Bm in each replenishment cycle. According to the inventory cost minimization principle, the first (m-1) replenishment cycles would be ended before T and the m-th replenishment cycle would be ended at or after T, i.e., (1)Q1+Q2+⋯+Qm-1+Bm-1<λ′T,Q1+Q2+⋯+Qm+Bm≥λ′T.Further, once the m replenishment cycles are ended, the system would resort to the BEOQ ordering policy with size QλBEOQ=2λK(h+p)/hp. Thus, to minimize the retailer’s inventory cost, we only need to determine the order sizes Q1,…,Qm, and the maximum shortage levels B1,…,Bm. In this sense, we denote the operating cost for these replenishment cycles by f(Q1,…,Qm,B1,…,Bm), and denote the time horizon of these replenishment cycles by Tm. Since shortage will be fully backordered, it holds that (2)Tm=T+Bm+∑i=1mQi-λ′Tλ.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 are2λ′Khp/h+p in the first stage and 2λKhp/h+p in the second stage, we obtain the operating cost under the IBEOQ ordering policy over the time horizon Tm,(3)fIBEOQ=T2λ′Khph+p+Tm-T2λKhph+p=T2λ′Khph+p+2Khpλh+pBm+∑i=1mQi-λ′T.Accordingly, the operating cost under the ordering policy relative to the IBEOQ ordering policy over the time horizon Tm is(4)FQ1,…,Qm,B1,…,Bm=fQ1,…,Qm,B1,…,Bm-fIBEOQ.Then, the problem of determining an optimal policy can be formulated as the following optimization problem:(5)minm,Qi,BiFQ1,…,Qm,B1,…,Bms.t.Q1+Q2+⋯+Qm-1+Bm-1<λ′TQ1+Q2+⋯+Qm+Bm≥λ′TQi,Bi>0,i=1,…,m.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, ifm orders are made in the first stage with sizes Q1,…,Qm, since shortages are allowed, according to the the size of the remaining stock ∑i=1mQi-λ′T at T, we have three scenarios: ∑i=1mQi-λ′T<0, ∑i=1mQi-λ′T=0, and ∑i=1mQi-λ′T>0. See Figures 2–4.Figure 2
∑ i = 1 m Q i < λ ′ T.Figure 3
∑ i = 1 m Q i = λ ′ T.Figure 4
∑ i = 1 m Q i > λ ′ T.


### 4.1. Scenario∑i=1mQi<λ′T

This scenario is depicted in Figure2, the stock position is exhausted and the shortage level is λ′T-∑i=1mQi at T. From the knowledge of inventory, the total operating cost over the time length Tm is(6)fQ1,…,Qm,B1,…,Bm=mK+h2λ′Q12+h2λ′∑i=2mQi-Bi-12+p2λ′∑i=1m-1Bi2+p2λBm2+p2λ′-p2λλ′T-∑i=1mQi2.Subsequently, substituting (6) and (3) into (4) yields that(7)FQ1,…,Qm,B1,…,Bm=mK+h2λ′Q12+h2λ′∑i=2mQi-Bi-12+p2λ′∑i=1m-1Bi2+p2λBm2+p2λ′-p2λλ′T-∑i=1mQi2-2Khpλh+pBm+∑i=1mQi-T2λ′Khph+p+λ′T2Khpλh+p.Then, the problem (5) can be formulated as the following optimization problem:(8)minm,Qi,BiFQ1,…,Qm,B1,…,Bms.t.Q1+Q2+⋯+Qm-1+Bm-1<λ′TQ1+Q2+⋯+Qm<λ′TQ1+Q2+⋯+Qm+Bm≥λ′TQi,Bi>0,i=1,…,mFor this problem, we have the following conclusion.Theorem 1.
For the inventory model (8), suppose m orders are made in the first stage. If m≤min⁡u1,u2, then the optimal ordering sizes and the optimal maximum shortage level in the first stage are, respectively, (9)Q1=pλ′Th+pm,Qi=λ′Tm,i=2,3,…,m,Bi=hλ′Th+pm,i=1,2,…,mand if min⁡{u1,u2}<m<max⁡{u1,u2}, then the optimal ordering sizes and the optimal maximum shortage level in the first stage are, respectively, (10)Q1=λ′β-αpTh+phλ′-αm,Qi=λ′β-αpTphλ′-αm,i=2,3,…,m,Bi=hλ′β-αpTph+phλ′-αm,i=1,2,…,m-1Bm=2λKhph+pwhere u1=λ′Thp/2λK(h+p), u2=Tλhp/2K(h+p)+h/h+p,α=(h+p)(λ′-λ), β=2λKhp(h+p).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 Wanget al. [21, 22].
( 1 ) System 1:(11)∂F∂Q1=hλ′Q1-pλ′-pλλ′T-∑i=1mQi-2Khpλh+p=μ,∂F∂Qi=hλ′Qi-Bi-1-pλ′-pλλ′T-∑i=1mQi-2Khpλh+p=μ,i=2,…,m,∂F∂Bi=-hλ′Qi+1-Bi+pλ′Bi=0,i=1,…,m-1,∂F∂Bm=pλBm-2Khpλh+p=μ,Q1+Q2+⋯+Qm-1+Bm-1<λ′T,Q1+Q2+⋯+Qm<λ′T,Q1+Q2+⋯+Qm+Bm=λ′T,Qi,Bi>0,i=1,…,m,where μ≥0 is the Lagrange multiplier corresponding with the equality constraint.
( 2 ) System 2: (12)∂F∂Q1=hλ′Q1-pλ′-pλλ′T-∑i=1mQi-2Khpλh+p=0,∂F∂Qi=hλ′Qi-Bi-1-pλ′-pλλ′T-∑i=1mQi-2Khpλh+p=0,i=2,…,m,∂F∂Bi=-hλ′Qi+1-Bi+pλ′Bi=0,i=1,…,m-2,∂F∂Bm=pλBm-2Khpλh+p=0,Q1+Q2+⋯+Qm-1+Bm-1<λ′T,Q1+Q2+⋯+Qm<λ′T,Q1+Q2+⋯+Qm+Bm>λ′T,Qi,Bi>0,i=1,…,m
For these two systems, using their former(2m-1) equations, respectively, we can all conclude that(13)Q2=⋯=Qm,B1=⋯=Bm-1=hh+pQi,Q1=ph+pQi,i=2,…,m.
For the former system, substituting (13) into the last equation and combining the other equations yield(14)Q1=pλ′Th+pm,Qi=λ′Tm,i=2,3,…,m,Bi=hλ′Th+pm,i=1,2,…,mwith μ=phλ′T/mλ(h+p)-2Khp/λ(h+p). It is easy to verify that all the inequalities of the former system hold for Qi, and Bi given above. In addition, to guarantee Qi and Bi given above constitute a KKT point of problem (8), μ≥0 is required; that is, (15)m≤λ′Tph2λKh+p=u1.So the equation set by (14) is the solution to the problem (8) provided that (16)m≤u1.
For the latter system, after some substitutions and algebra, we have: ifm(h+p)(λ′-λ)≠λ′h, then(17)Q1=ph+p·λ′2λKhph+p-pTh+pλ′-λphλ′-mh+pλ′-λ,Qi=λ′2λKhph+p-pTh+pλ′-λphλ′-mh+pλ′-λ,i=2,…,m,Bi=hh+p·λ′2λKhph+p-pTh+pλ′-λphλ′-mh+pλ′-λ,i=1,…,m-1Bm=2λKhph+p.otherwise, the equations of the latter system have no solution.
Denote(18)u1=λ′Thp2λKh+p,u2=Tλhp2Kh+p+hh+p,α=h+pλ′-λ,β=2λKhph+p.To guarantee Qi and Bi given by (17) is a KKT point of problem (8), we break up the discussion into two cases.
C a s e   1  (2λKhp(h+p)≥pT(h+p)(λ′-λ), i.e., β≥αpT). In this case, to guarantee the nonnegativity of the Qi and Bi, it needs that m(h+p)(λ′-λ)<λ′h, i.e., m<λ′h/(h+p)(λ′-λ). Substituting (17) into the second inequality and the last inequality in the latter system yields that (19)λ′Tph2λKh+p<m<Tλhp2Kh+p+hh+p.Moreover, under this case, we have (20)λ′Tph2λKh+p<Tλhp2Kh+p+hh+p<λ′hh+pλ′-λ,so Qi and Bi given by (17) are a solution to the latter system provided that (21)β≥αpT,u1<m<u2.
C a s e   2 ( β < α p T ). By a similar discussion to that for Case 1, we know that Qi and Bi given by (17) are a solution to the latter system provided that (22)β<αpT,u2<m<u1.Substituting (14) and (17) into (8), respectively, yields that the operating cost over the interval of length Tm under the two policies relative to the IBEOQ ordering policy are FI and FII, respectively.(23)FI=mK+hpλ′T22mh+p-T2λ′Khph+p,(24)FII=mK+hλ′mh+p-hβ-αpT22ph+p2λ′h-mα2+λ′λ-λ′h+pβm-hβ+γ22λph+p2λ′h-mα2-λ′βmh+p-hβ-αpTλph+p2λ′h-mα-Khh+p+λ′T2Khpλh+p-T2λ′Khph+p,where γ=λhpT(h+p).
Under the Case 2, ifu2<m<u1, after some substitutions and algebra, we can prove that FI(m)>FII(m).
Combining the discussions above, we conclude thatQ, B, and Bm given by (14) are a solution of problem (8) provided that (25)m≤minu1,u2,and (17) is a solution of problem (8) provided that (26)min⁡u1,u2<m<max⁡u1,u2.
The proof is completed by combining the discussions above.By Theorem1, 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., (27)Q2=⋯=Qm≜Q,B1=⋯=Bm-1≜B.also (28)B=hh+pQ,Q1=ph+pQ.So we can obtain the following two ordering policies for problem (5) with m orders in the first stage.Policy π I (29) Q = λ ′ T m , Q 1 = p h + p Q , B = B m = h h + p Q ,which applies to the case thatm≤min⁡{u1,u2}.Policy π I I (30) Q = λ ′ β - α p T p h λ ′ - m α , Q 1 = p h + p Q , B = h h + p Q , B m = 2 λ K h p h + p ,which applies to the case thatmin⁡{u1,u2}<m<max⁡{u1,u2}.Now, we consider the optimal ordering times in the first stage under PolicyπI and Policy πII.By the proof process of Theorem1, we know that the operating costs over the interval of length Tm under the two policies relative to the IBEOQ ordering policy are FI and FII, respectively.For functionFI, by (31)FI′′m=hpλ′T2h+pm3>0,we know that FI(m) is a convex function in m, and its minimum can be obtained at the stationary point of the function; i.e., (32)m¯I=Tλ′hp2Kh+p.SetmI1=⌊m¯I⌋,mI2=⌈m¯I⌉,mI3=⌊min⁡{u1,u2}⌋. Then due to the fact that m is a positive integer, and requires m≤min⁡{u1,u2}. So the optimal ordering times in the first stage are mI=mI1 provided that mI1≤min⁡{u1,u2} and FI(mI1)≤FI(mI2) or mI2>min⁡{u1,u2}, or the optimal ordering times are mI=mI2 provided that mI2≤min⁡{u1,u2},andFI(mI1)>FI(mI2). Moreover, if mI1>min⁡{u1,u2}, the optimal ordering times in the first stage are mI=mI3. That is (33)m∗=mI1ifmI1≤min⁡u1,u2andFImI1≤FImI2ormI2>min⁡u1,u2,mI2ifmI2≤min⁡u1,u2andFImI1>FImI2,mI3ifmI1>min⁡u1,u2.For functionFII given by (24), by (34)FII′′m=-αλ′hβ-αpT2ph+pλ′h-mα3,we know that if λ′h-mα<0, i.e., m>λ′h/α, then FII′′(m)>0 and FII(m) is a convex function in m, and its minimum can be obtained at the stationary point of the function; i.e., (35)m¯II=hλ′α+λ′h2Kph+pβα-pT>hλ′α.On the other hand, according to the proof of Theorem1, if m>λ′h/α, to guarantee Qi, Bi, and Bm given by (17) is a solution to the model (8)), it needs that (36)β<αpT,u2<m<u1.In this case, the optimal ordering times in the first stage are (37)m¯II=hλ′α+λ′h2Kph+ppT-βα=Tλ′hp2Kh+p+hλ′-λλ′λ′-λh+p,which satisfies that u2<mII¯<u1.However, ifλ′h-mα>0, i.e., m<λ′h/α, then FII′′(m)<0 and FII(m) is a concave function in m, and its minimum can be obtained at the terminal point (u1andu2) of the function, and, according to the proof of Theorem 1, under the case that (38)β≥αpT,u1<m<u2.DenotemII1=⌊m¯II⌋,mII2=⌈m¯II⌉, mII3=⌈u1⌉. mII4=⌊u2⌋. Obviously, mII1<u1, mII2>u2, mII3>u1, mII4<u2. Based on the discussion above, we obtain the optimal ordering times in the first stage under Policy πII(39)m∗=mII1,ifβ<αpT,u2<mII1≤mII2<u1,FIImII1≤FIImII2,orβ<αpT,u2<mII1<u1≤mII2;mII2,ifβ<αpT,u2<mII1≤mII2<u1,FIImII1>FIImII2orβ<αpT,mII1≤u2<mII2<u1;mII3,ifβ≥αpT,mII3<u2,mII4>u1,FIImII3≤FIImII4,β≥αpT,mII4≤u1<mII3<u2;mII4,ifβ≥αpT,mII3<u2,mII4>u1,FIImII3>FIImII4,orβ≥αpT,u1<mII4<u2≤mII3.


### 4.2. Scenario∑i=1mQi=λ′T

In this scenario,m orders are made in the first stage and they can exactly meet the demand of the first stage; see Figure 3. Obviously, Tm=T+Bm/λ. From the knowledge of inventory, we have(40)fQ1,…,Qm,B1,…,Bm=mK+h2λ′Q12+h2λ′∑i=2mQi-Bi-12+p2λ′∑i=1m-1Bi2+p2λBm2,(41)fIBEOQ=T2λ′Khph+p+Bm2Khpλh+p.Substituting (40) and (41) into formula (4) yields that(42)FQ1,…,Qm,B1,…,Bm=mK+h2λ′Q12+h2λ′∑i=2mQi-Bi-12+p2λ′∑i=1m-1Bi2+p2λBm2-Bm2Khpλh+p-T2λ′Khph+p.Then problem (5) can be formulated as the following optimization problem:(43)minm,Qi,BiFQ1,…,Qm,B1,…,Bms.t.Q1+Q2+⋯+Qm=λ′TQi,Bi>0,i=1,…,m.For this problem, we have the following conclusion.Theorem 2.
For inventory model (43), suppose m orders are made in the first stage; then the optimal ordering sizes and the optimal maximum shortage level in the first stage are, respectively, (44)Q1=pλ′Tmh+p-h,Qi=h+pλ′Tmh+p-h,i=2,3,…,mBi=hλ′Tmh+p-h,i=1,2⋯,m-1Bm=2λKhph+p.Proof.
For problem (43), its KKT point satisfies the following system: (45)∂F∂Q1=hλ′Q1=μ,∂F∂Qi=hλ′Qi-Bi-1=μ,i=2,…,m,∂F∂Bi=-hλ′Qi+1-Bi+pλ′Bi=0,i=1,…,m-1,∂F∂Bm=pλBm-2Khpλh+p=0,Q1+Q2+⋯+Qm=λ′T,Qi,Bi≥0,i=1,…,m.where μ≥0 is the optimal Lagrange multiplier.
By solving the equations in the system, we can obtain the desired conclusion.For this scenario, by Theorem2, the same as Scenario 1, the optimal ordering sizes {denoted as Q} from second order to m-th order in the first stage are the same, and the optimal maximum shortage levels {denoted as B} 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 π I I I (46) Q = h + p λ ′ T m h + p - h , Q 1 = p h + p Q , B = h h + p Q , B m = 2 λ K h p h + p .Under this policy, the operating cost over the interval of lengthTm relative to the IBEOQ ordering policy is(47)FIII=mK+hpλ′T22mh+p-h-Khh+p-T2λ′Khph+p.Now, we consider the optimal ordering times in the first stage under PoliciesπIII.For functionFIII, by (48)FIII′′m=hph+p2mh+p-h3λ′T2>0,we know that FIII(m) is a convex function in m, and its minimum can be obtained at the stationary point of the function, i.e., (49)FIII′m=K-hph+p2mh+p-hλ′T2=0.Set(50)m¯III=Tλ′hp2Kh+p+hh+p,mIII1=m¯III,mIII2=m¯III.Then, the optimal ordering times in the first stage are mIII=mIII1 provided that FIII(mIII1)≤FIII(mIII2), and mIII=mIII2 otherwise. Then, the optimal order times under Policy πIII are that (51)m∗=mIII1ifFIIImIII1≤FIIImIII2,mIII2ifFIIImIII1>FIIImIII2.


### 4.3. Scenario∑i=1mQi>λ′T

For this scenario, see Figure4; a similar discussions to that for Scenario 1 yields (52)fQ1,…,Qm,B1,…,Bm=mK+h2λ′Q12+h2λ′∑i=2mQi-Bi-12+p2λ′∑i=1m-1Bi2+p2λBm2+h2λ-h2λ′∑i=1mQi-λ′T2.So(53)FQ1,…,Qm,B1,…,Bm=mK+h2λ′Q12+h2λ′∑i=2mQi-Bi-12+p2λ′∑i=1m-1Bi2+p2λBm2+h2λ-h2λ′∑i=1mQi-λ′T2-2Khpλh+pBm+∑i=1mQi-T2λ′Khph+p+λ′T2Khpλh+p.Therefore, the problem (5) can be formulated as the following form:(54)minm,Qi,BiFQ1,…,Qm,B1,…,Bms.t.Q1+Q2+⋯+Qm-1+Bm-1<λ′TQ1+Q2+⋯+Qm>λ′TQi,Bi≥0,i=1,…,m.For this problem, we have the following conclusion.Theorem 3.
For inventory model (54), suppose m orders are made in the first stage. If μ2<m<u1+1, then the optimal ordering sizes and the optimal maximum shortage level in the first stage are, respectively, (55)Q1=pλ′αhT+βhh+pm-1α+pλ′Qi=λ′αhT+βhm-1α+pλ′,i=2,3,…,mBi=λ′αhT+βh+pm-1α+pλ′,i=1,2,…,m-1,Bm=2λKhph+p,where α=(h+p)(λ′-λ), β=2λKhp(h+p), u1=λ′Thp/2λK(h+p), u2=Tλhp/2K(h+p)+h/h+p.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: (56)∂F∂Q1=hλ′Q1+hλ-hλ′∑i=1mQi-λ′T-2Khpλh+p=0,∂F∂Qi=hλ′Qi-Bi-1+hλ-hλ′∑i=1mQi-λ′T-2Khpλh+p=0,i=2,…,m,∂F∂Bi=-hλ′Qi+1-Bi+pλ′Bi=0,i=1,…,m-1,∂F∂Bm=pλBm-2Khpλh+p=0,Q1+Q2+⋯+Qm-1+Bm-1<λ′T,Q1+Q2+⋯+Qm>λ′T,Qi,Bi≥0,i=1,…,m,
For the systems of equations, we can obtain that(57)Q1=ph+p·λ′hTh+pλ′-λ+2λKhph+phm-1h+pλ′-λ+pλ′,Qi=λ′hTh+pλ′-λ+2λKhph+phm-1h+pλ′-λ+pλ′,i=2,…,m,Bi=hh+p·λ′hTh+pλ′-λ+2λKhph+phm-1h+pλ′-λ+pλ′,i=1,…,m-1Bm=2λKhph+p.
Taking the inequality constraints into considerations, we have that(58)Tλhp2Kh+p+hh+p<m<λ′Thp2λKh+p+1,i.e., u2<m<u1+1. The proof is completed.By Theorem3, under the Scenario ∑i=1mQi>λ′T, we can obtain the following optimal ordering policy for problem (5).Policy π I V (59) Q = λ ′ α h T + β h m - 1 α + p λ ′ , Q 1 = p h + p Q , B = h h + p Q , B m = 2 λ K h p h + p ,which applies to the case thatu2<m<u1+1.The operating cost over the interval of lengthTm under this policy relative to the IBEOQ ordering policy is(60)FIV=mK+pλ′mh+p-hαhT+β22hh+p2m-1α+pλ′2+λ′λ′-λh+pβm-hβ+γ22λhh+p2m-1α+pλ′2-λ′βmh+p-hαhT+βλhh+p2m-1α+pλ′-Khh+p+λ′T2Khpλh+p-T2λ′Khph+p,Now, we consider the optimal ordering times in the first stage under PoliciesπIV.Similarly, for functionFIV given by (60), by (61)FIV′′m=αλ′pβ+αhT2hh+pλ′p+m-1α3>0,we know that FIV(m) is a convex function in m, and its minimum can be obtained at the stationary point of the function; i.e., (62)m¯IV=Tλ′hp2Kh+p+1-pλ′-λλ′α>u2.DenotemIV1=⌊m¯IV⌋,mIV2=⌈m¯IV⌉. A similar discussion yields the optimal ordering times in the first stage (63)m∗=mIV1ifu2<mIV1≤mIV2<u1+1andFIVmIV1≤FIVmIV2,oru2<mIV1<u1+1≤mIV2,mIV2ifmIV1≤u2<mIV2<u1+1andFIVmIV1>FIVmIV2,ormIV1<u2≤mIV2<u1+1.Based on the discussion about three Scenarios above, we can present our algorithm for problem (5).


### 4.4. Algorithm

Step 1.
Input value of parametersK,λ,λ′,h,p,T.Step 6.
Computeu1=λ′Thp/2λK(h+p), u2=Tλhp/2K(h+p)+h/h+p, α=h+p(λ′-λ), β=2λKhp(h+p),  mI¯=Tλ′hp/2K(h+p),mII¯=hλ′/α+λ′h/2Kp(h+p)(pT-β/α).(64)mIII¯=Tλ′hp2Kh+p+hh+p,mIV¯=Tλ′hp2Kh+p+1-pλ′-λλ′α.
Let(65)mI1=m¯I,mI2=m¯ImI3=min⁡u1,u2,,mII1=m¯II,mII2=m¯II,mIII1=m¯III,mIII2=m¯III,mII3=u1,mII4=u2,mIV1=m¯IV,mIV2=m¯IV.ComputeFI(mI1), FI(mI2), FII(mII1), FII(mII2), FII(mII3), FII(mII4), FIII(mIII1), FIII(mIII2), FIV(mIV1), and FIV(mIV2).Step 3.
IfFIII(mIII1)≤FIII(mIII2), then (66)mIII∗=mIII1,FIII∗=FIIImIII1,
otherwise(67)mIII∗=mIII2,FIII∗=FIIImIII2.Step 4.
IfmI2≤min⁡{u1,u2}, and FI(mI2)≤FI(mI1), then (68)mI∗=mI2,(69)FI∗=FImI2,
ifmI1>min⁡{u1,u2}(70)mI∗=mI3,FI∗=FImI3,
otherwise,mI∗=mI1,FI∗=FI(mI1).Step 5.
( 1 ) If β<αpT,
(a) ifu2<mII1≤mII2<u1 and FII(mII1)≤FII(mII2), or u2<mII1<u1≤mII2, then (71)mII∗=mII1,FII∗=FIImII1;
(b) ifu2<mII1≤mII2<u1 and FII(mII1)>FII(mII2), or mII1≤u2<mII2<u1, then (72)mII∗=mII2,FII∗=FIImII2,
otherwise,FII∗=∞.
(2) Ifβ≥αpT,
(a) ifmII3<u2 and mII4>u1 and FII(mII3)≤FII(mII4), or mII4≤u1<mII3<u2, then (73)mII∗=mII3,FII∗=FIImII3;
(b) ifu2>mII3, and mII4>u1 and FII(mII3)>FII(mII4), or u1≤mII4<u2<mII3, then (74)mII∗=mII4,FII∗=FIImII4,
0therwise,FII∗=∞.Step 6.
Ifu2<mIV1≤mII2<u1+1 and FIV(mIV1)≤FIV(mIV1), or u2<mIV1<u1+1≤mIV2, then (75)mIV∗=mIV1,FIV∗=FIVmIV1,
Ifu2<mIV1≤mII2<u1+1 and FIV(mIV1)>FIV(mIV1), or mIV1≤u2<mIV2<u1+1, then (76)mIV∗=mIV2,FIV∗=FIVmIV2,
otherwise,FIV∗=∞.Step 7.
CompareFI∗, FII∗, FIII∗, and FIV∗.
Ifmin⁡{FI∗,FII∗,FIII∗,andFIV∗}=FI∗, then, the optimal order times, the optimal order size from the second replenishment cycle to m-th replenishment cycle, and the optimal maximum shortage level of m-th replenishment cycle in the first stage are, respectively, (77)m∗=mI∗,Q∗=λ′Th+pm∗h+p-h,Bm∗=hh+pQ∗,
go to Step8.
Ifmin⁡FI∗,FII∗,FIII∗,andFIV∗=FII∗, then, the optimal order times, the optimal order size from the second replenishment cycle to m-th replenishment cycle, and the optimal maximum shortage levels of m-th replenishment cycle in the first stage are, respectively, (78)m∗=mIII∗,Q∗=λ′β-αpTphλ′-αm∗,Bm∗=2λKhph+p,
go to Step8.
Ifmin⁡FI∗,FII∗,FIII∗,andFIV∗=FIII∗, the optimal order times, the optimal order size from the second replenishment cycle to m-th replenishment cycle, and the optimal maximum shortage level of m-th replenishment cycle in the first stage are, respectively, (79)m∗=mI∗,Q∗=λ′Th+pm∗h+p-h,Bm∗=2λKhph+p,
go to Step8.
Ifmin⁡FI∗,FII∗,FIII∗,andFIV∗=FIV∗, the optimal order times, the optimal order size from the second replenishment cycle to m-th replenishment cycle, and the optimal maximum shortage level of m-th replenishment cycle in the first stage are, respectively, (80)m∗=mIV∗,Q∗=λ′β+αhThpλ′+αm∗-1,Bm∗=2λKhph+p,
go to Step8.Step 8.
The optimal order size of the first replenishment cycle and the optimal maximum shortage level of the first replenishment cycle (m-1)-th replenishment cycle in the first stage are, respectively,(81)Q1∗=ph+pQ∗,B∗=hh+pQ∗,
and terminate.


## 4.1. Scenario∑i=1mQi<λ′T

This scenario is depicted in Figure2, the stock position is exhausted and the shortage level is λ′T-∑i=1mQi at T. From the knowledge of inventory, the total operating cost over the time length Tm is(6)fQ1,…,Qm,B1,…,Bm=mK+h2λ′Q12+h2λ′∑i=2mQi-Bi-12+p2λ′∑i=1m-1Bi2+p2λBm2+p2λ′-p2λλ′T-∑i=1mQi2.Subsequently, substituting (6) and (3) into (4) yields that(7)FQ1,…,Qm,B1,…,Bm=mK+h2λ′Q12+h2λ′∑i=2mQi-Bi-12+p2λ′∑i=1m-1Bi2+p2λBm2+p2λ′-p2λλ′T-∑i=1mQi2-2Khpλh+pBm+∑i=1mQi-T2λ′Khph+p+λ′T2Khpλh+p.Then, the problem (5) can be formulated as the following optimization problem:(8)minm,Qi,BiFQ1,…,Qm,B1,…,Bms.t.Q1+Q2+⋯+Qm-1+Bm-1<λ′TQ1+Q2+⋯+Qm<λ′TQ1+Q2+⋯+Qm+Bm≥λ′TQi,Bi>0,i=1,…,mFor this problem, we have the following conclusion.Theorem 1.
For the inventory model (8), suppose m orders are made in the first stage. If m≤min⁡u1,u2, then the optimal ordering sizes and the optimal maximum shortage level in the first stage are, respectively, (9)Q1=pλ′Th+pm,Qi=λ′Tm,i=2,3,…,m,Bi=hλ′Th+pm,i=1,2,…,mand if min⁡{u1,u2}<m<max⁡{u1,u2}, then the optimal ordering sizes and the optimal maximum shortage level in the first stage are, respectively, (10)Q1=λ′β-αpTh+phλ′-αm,Qi=λ′β-αpTphλ′-αm,i=2,3,…,m,Bi=hλ′β-αpTph+phλ′-αm,i=1,2,…,m-1Bm=2λKhph+pwhere u1=λ′Thp/2λK(h+p), u2=Tλhp/2K(h+p)+h/h+p,α=(h+p)(λ′-λ), β=2λKhp(h+p).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 Wanget al. [21, 22].
( 1 ) System 1:(11)∂F∂Q1=hλ′Q1-pλ′-pλλ′T-∑i=1mQi-2Khpλh+p=μ,∂F∂Qi=hλ′Qi-Bi-1-pλ′-pλλ′T-∑i=1mQi-2Khpλh+p=μ,i=2,…,m,∂F∂Bi=-hλ′Qi+1-Bi+pλ′Bi=0,i=1,…,m-1,∂F∂Bm=pλBm-2Khpλh+p=μ,Q1+Q2+⋯+Qm-1+Bm-1<λ′T,Q1+Q2+⋯+Qm<λ′T,Q1+Q2+⋯+Qm+Bm=λ′T,Qi,Bi>0,i=1,…,m,where μ≥0 is the Lagrange multiplier corresponding with the equality constraint.
( 2 ) System 2: (12)∂F∂Q1=hλ′Q1-pλ′-pλλ′T-∑i=1mQi-2Khpλh+p=0,∂F∂Qi=hλ′Qi-Bi-1-pλ′-pλλ′T-∑i=1mQi-2Khpλh+p=0,i=2,…,m,∂F∂Bi=-hλ′Qi+1-Bi+pλ′Bi=0,i=1,…,m-2,∂F∂Bm=pλBm-2Khpλh+p=0,Q1+Q2+⋯+Qm-1+Bm-1<λ′T,Q1+Q2+⋯+Qm<λ′T,Q1+Q2+⋯+Qm+Bm>λ′T,Qi,Bi>0,i=1,…,m
For these two systems, using their former(2m-1) equations, respectively, we can all conclude that(13)Q2=⋯=Qm,B1=⋯=Bm-1=hh+pQi,Q1=ph+pQi,i=2,…,m.
For the former system, substituting (13) into the last equation and combining the other equations yield(14)Q1=pλ′Th+pm,Qi=λ′Tm,i=2,3,…,m,Bi=hλ′Th+pm,i=1,2,…,mwith μ=phλ′T/mλ(h+p)-2Khp/λ(h+p). It is easy to verify that all the inequalities of the former system hold for Qi, and Bi given above. In addition, to guarantee Qi and Bi given above constitute a KKT point of problem (8), μ≥0 is required; that is, (15)m≤λ′Tph2λKh+p=u1.So the equation set by (14) is the solution to the problem (8) provided that (16)m≤u1.
For the latter system, after some substitutions and algebra, we have: ifm(h+p)(λ′-λ)≠λ′h, then(17)Q1=ph+p·λ′2λKhph+p-pTh+pλ′-λphλ′-mh+pλ′-λ,Qi=λ′2λKhph+p-pTh+pλ′-λphλ′-mh+pλ′-λ,i=2,…,m,Bi=hh+p·λ′2λKhph+p-pTh+pλ′-λphλ′-mh+pλ′-λ,i=1,…,m-1Bm=2λKhph+p.otherwise, the equations of the latter system have no solution.
Denote(18)u1=λ′Thp2λKh+p,u2=Tλhp2Kh+p+hh+p,α=h+pλ′-λ,β=2λKhph+p.To guarantee Qi and Bi given by (17) is a KKT point of problem (8), we break up the discussion into two cases.
C a s e   1  (2λKhp(h+p)≥pT(h+p)(λ′-λ), i.e., β≥αpT). In this case, to guarantee the nonnegativity of the Qi and Bi, it needs that m(h+p)(λ′-λ)<λ′h, i.e., m<λ′h/(h+p)(λ′-λ). Substituting (17) into the second inequality and the last inequality in the latter system yields that (19)λ′Tph2λKh+p<m<Tλhp2Kh+p+hh+p.Moreover, under this case, we have (20)λ′Tph2λKh+p<Tλhp2Kh+p+hh+p<λ′hh+pλ′-λ,so Qi and Bi given by (17) are a solution to the latter system provided that (21)β≥αpT,u1<m<u2.
C a s e   2 ( β < α p T ). By a similar discussion to that for Case 1, we know that Qi and Bi given by (17) are a solution to the latter system provided that (22)β<αpT,u2<m<u1.Substituting (14) and (17) into (8), respectively, yields that the operating cost over the interval of length Tm under the two policies relative to the IBEOQ ordering policy are FI and FII, respectively.(23)FI=mK+hpλ′T22mh+p-T2λ′Khph+p,(24)FII=mK+hλ′mh+p-hβ-αpT22ph+p2λ′h-mα2+λ′λ-λ′h+pβm-hβ+γ22λph+p2λ′h-mα2-λ′βmh+p-hβ-αpTλph+p2λ′h-mα-Khh+p+λ′T2Khpλh+p-T2λ′Khph+p,where γ=λhpT(h+p).
Under the Case 2, ifu2<m<u1, after some substitutions and algebra, we can prove that FI(m)>FII(m).
Combining the discussions above, we conclude thatQ, B, and Bm given by (14) are a solution of problem (8) provided that (25)m≤minu1,u2,and (17) is a solution of problem (8) provided that (26)min⁡u1,u2<m<max⁡u1,u2.
The proof is completed by combining the discussions above.By Theorem1, 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., (27)Q2=⋯=Qm≜Q,B1=⋯=Bm-1≜B.also (28)B=hh+pQ,Q1=ph+pQ.So we can obtain the following two ordering policies for problem (5) with m orders in the first stage.Policy π I (29) Q = λ ′ T m , Q 1 = p h + p Q , B = B m = h h + p Q ,which applies to the case thatm≤min⁡{u1,u2}.Policy π I I (30) Q = λ ′ β - α p T p h λ ′ - m α , Q 1 = p h + p Q , B = h h + p Q , B m = 2 λ K h p h + p ,which applies to the case thatmin⁡{u1,u2}<m<max⁡{u1,u2}.Now, we consider the optimal ordering times in the first stage under PolicyπI and Policy πII.By the proof process of Theorem1, we know that the operating costs over the interval of length Tm under the two policies relative to the IBEOQ ordering policy are FI and FII, respectively.For functionFI, by (31)FI′′m=hpλ′T2h+pm3>0,we know that FI(m) is a convex function in m, and its minimum can be obtained at the stationary point of the function; i.e., (32)m¯I=Tλ′hp2Kh+p.SetmI1=⌊m¯I⌋,mI2=⌈m¯I⌉,mI3=⌊min⁡{u1,u2}⌋. Then due to the fact that m is a positive integer, and requires m≤min⁡{u1,u2}. So the optimal ordering times in the first stage are mI=mI1 provided that mI1≤min⁡{u1,u2} and FI(mI1)≤FI(mI2) or mI2>min⁡{u1,u2}, or the optimal ordering times are mI=mI2 provided that mI2≤min⁡{u1,u2},andFI(mI1)>FI(mI2). Moreover, if mI1>min⁡{u1,u2}, the optimal ordering times in the first stage are mI=mI3. That is (33)m∗=mI1ifmI1≤min⁡u1,u2andFImI1≤FImI2ormI2>min⁡u1,u2,mI2ifmI2≤min⁡u1,u2andFImI1>FImI2,mI3ifmI1>min⁡u1,u2.For functionFII given by (24), by (34)FII′′m=-αλ′hβ-αpT2ph+pλ′h-mα3,we know that if λ′h-mα<0, i.e., m>λ′h/α, then FII′′(m)>0 and FII(m) is a convex function in m, and its minimum can be obtained at the stationary point of the function; i.e., (35)m¯II=hλ′α+λ′h2Kph+pβα-pT>hλ′α.On the other hand, according to the proof of Theorem1, if m>λ′h/α, to guarantee Qi, Bi, and Bm given by (17) is a solution to the model (8)), it needs that (36)β<αpT,u2<m<u1.In this case, the optimal ordering times in the first stage are (37)m¯II=hλ′α+λ′h2Kph+ppT-βα=Tλ′hp2Kh+p+hλ′-λλ′λ′-λh+p,which satisfies that u2<mII¯<u1.However, ifλ′h-mα>0, i.e., m<λ′h/α, then FII′′(m)<0 and FII(m) is a concave function in m, and its minimum can be obtained at the terminal point (u1andu2) of the function, and, according to the proof of Theorem 1, under the case that (38)β≥αpT,u1<m<u2.DenotemII1=⌊m¯II⌋,mII2=⌈m¯II⌉, mII3=⌈u1⌉. mII4=⌊u2⌋. Obviously, mII1<u1, mII2>u2, mII3>u1, mII4<u2. Based on the discussion above, we obtain the optimal ordering times in the first stage under Policy πII(39)m∗=mII1,ifβ<αpT,u2<mII1≤mII2<u1,FIImII1≤FIImII2,orβ<αpT,u2<mII1<u1≤mII2;mII2,ifβ<αpT,u2<mII1≤mII2<u1,FIImII1>FIImII2orβ<αpT,mII1≤u2<mII2<u1;mII3,ifβ≥αpT,mII3<u2,mII4>u1,FIImII3≤FIImII4,β≥αpT,mII4≤u1<mII3<u2;mII4,ifβ≥αpT,mII3<u2,mII4>u1,FIImII3>FIImII4,orβ≥αpT,u1<mII4<u2≤mII3.


## 4.2. Scenario∑i=1mQi=λ′T

In this scenario,m orders are made in the first stage and they can exactly meet the demand of the first stage; see Figure 3. Obviously, Tm=T+Bm/λ. From the knowledge of inventory, we have(40)fQ1,…,Qm,B1,…,Bm=mK+h2λ′Q12+h2λ′∑i=2mQi-Bi-12+p2λ′∑i=1m-1Bi2+p2λBm2,(41)fIBEOQ=T2λ′Khph+p+Bm2Khpλh+p.Substituting (40) and (41) into formula (4) yields that(42)FQ1,…,Qm,B1,…,Bm=mK+h2λ′Q12+h2λ′∑i=2mQi-Bi-12+p2λ′∑i=1m-1Bi2+p2λBm2-Bm2Khpλh+p-T2λ′Khph+p.Then problem (5) can be formulated as the following optimization problem:(43)minm,Qi,BiFQ1,…,Qm,B1,…,Bms.t.Q1+Q2+⋯+Qm=λ′TQi,Bi>0,i=1,…,m.For this problem, we have the following conclusion.Theorem 2.
For inventory model (43), suppose m orders are made in the first stage; then the optimal ordering sizes and the optimal maximum shortage level in the first stage are, respectively, (44)Q1=pλ′Tmh+p-h,Qi=h+pλ′Tmh+p-h,i=2,3,…,mBi=hλ′Tmh+p-h,i=1,2⋯,m-1Bm=2λKhph+p.Proof.
For problem (43), its KKT point satisfies the following system: (45)∂F∂Q1=hλ′Q1=μ,∂F∂Qi=hλ′Qi-Bi-1=μ,i=2,…,m,∂F∂Bi=-hλ′Qi+1-Bi+pλ′Bi=0,i=1,…,m-1,∂F∂Bm=pλBm-2Khpλh+p=0,Q1+Q2+⋯+Qm=λ′T,Qi,Bi≥0,i=1,…,m.where μ≥0 is the optimal Lagrange multiplier.
By solving the equations in the system, we can obtain the desired conclusion.For this scenario, by Theorem2, the same as Scenario 1, the optimal ordering sizes {denoted as Q} from second order to m-th order in the first stage are the same, and the optimal maximum shortage levels {denoted as B} 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 π I I I (46) Q = h + p λ ′ T m h + p - h , Q 1 = p h + p Q , B = h h + p Q , B m = 2 λ K h p h + p .Under this policy, the operating cost over the interval of lengthTm relative to the IBEOQ ordering policy is(47)FIII=mK+hpλ′T22mh+p-h-Khh+p-T2λ′Khph+p.Now, we consider the optimal ordering times in the first stage under PoliciesπIII.For functionFIII, by (48)FIII′′m=hph+p2mh+p-h3λ′T2>0,we know that FIII(m) is a convex function in m, and its minimum can be obtained at the stationary point of the function, i.e., (49)FIII′m=K-hph+p2mh+p-hλ′T2=0.Set(50)m¯III=Tλ′hp2Kh+p+hh+p,mIII1=m¯III,mIII2=m¯III.Then, the optimal ordering times in the first stage are mIII=mIII1 provided that FIII(mIII1)≤FIII(mIII2), and mIII=mIII2 otherwise. Then, the optimal order times under Policy πIII are that (51)m∗=mIII1ifFIIImIII1≤FIIImIII2,mIII2ifFIIImIII1>FIIImIII2.


## 4.3. Scenario∑i=1mQi>λ′T

For this scenario, see Figure4; a similar discussions to that for Scenario 1 yields (52)fQ1,…,Qm,B1,…,Bm=mK+h2λ′Q12+h2λ′∑i=2mQi-Bi-12+p2λ′∑i=1m-1Bi2+p2λBm2+h2λ-h2λ′∑i=1mQi-λ′T2.So(53)FQ1,…,Qm,B1,…,Bm=mK+h2λ′Q12+h2λ′∑i=2mQi-Bi-12+p2λ′∑i=1m-1Bi2+p2λBm2+h2λ-h2λ′∑i=1mQi-λ′T2-2Khpλh+pBm+∑i=1mQi-T2λ′Khph+p+λ′T2Khpλh+p.Therefore, the problem (5) can be formulated as the following form:(54)minm,Qi,BiFQ1,…,Qm,B1,…,Bms.t.Q1+Q2+⋯+Qm-1+Bm-1<λ′TQ1+Q2+⋯+Qm>λ′TQi,Bi≥0,i=1,…,m.For this problem, we have the following conclusion.Theorem 3.
For inventory model (54), suppose m orders are made in the first stage. If μ2<m<u1+1, then the optimal ordering sizes and the optimal maximum shortage level in the first stage are, respectively, (55)Q1=pλ′αhT+βhh+pm-1α+pλ′Qi=λ′αhT+βhm-1α+pλ′,i=2,3,…,mBi=λ′αhT+βh+pm-1α+pλ′,i=1,2,…,m-1,Bm=2λKhph+p,where α=(h+p)(λ′-λ), β=2λKhp(h+p), u1=λ′Thp/2λK(h+p), u2=Tλhp/2K(h+p)+h/h+p.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: (56)∂F∂Q1=hλ′Q1+hλ-hλ′∑i=1mQi-λ′T-2Khpλh+p=0,∂F∂Qi=hλ′Qi-Bi-1+hλ-hλ′∑i=1mQi-λ′T-2Khpλh+p=0,i=2,…,m,∂F∂Bi=-hλ′Qi+1-Bi+pλ′Bi=0,i=1,…,m-1,∂F∂Bm=pλBm-2Khpλh+p=0,Q1+Q2+⋯+Qm-1+Bm-1<λ′T,Q1+Q2+⋯+Qm>λ′T,Qi,Bi≥0,i=1,…,m,
For the systems of equations, we can obtain that(57)Q1=ph+p·λ′hTh+pλ′-λ+2λKhph+phm-1h+pλ′-λ+pλ′,Qi=λ′hTh+pλ′-λ+2λKhph+phm-1h+pλ′-λ+pλ′,i=2,…,m,Bi=hh+p·λ′hTh+pλ′-λ+2λKhph+phm-1h+pλ′-λ+pλ′,i=1,…,m-1Bm=2λKhph+p.
Taking the inequality constraints into considerations, we have that(58)Tλhp2Kh+p+hh+p<m<λ′Thp2λKh+p+1,i.e., u2<m<u1+1. The proof is completed.By Theorem3, under the Scenario ∑i=1mQi>λ′T, we can obtain the following optimal ordering policy for problem (5).Policy π I V (59) Q = λ ′ α h T + β h m - 1 α + p λ ′ , Q 1 = p h + p Q , B = h h + p Q , B m = 2 λ K h p h + p ,which applies to the case thatu2<m<u1+1.The operating cost over the interval of lengthTm under this policy relative to the IBEOQ ordering policy is(60)FIV=mK+pλ′mh+p-hαhT+β22hh+p2m-1α+pλ′2+λ′λ′-λh+pβm-hβ+γ22λhh+p2m-1α+pλ′2-λ′βmh+p-hαhT+βλhh+p2m-1α+pλ′-Khh+p+λ′T2Khpλh+p-T2λ′Khph+p,Now, we consider the optimal ordering times in the first stage under PoliciesπIV.Similarly, for functionFIV given by (60), by (61)FIV′′m=αλ′pβ+αhT2hh+pλ′p+m-1α3>0,we know that FIV(m) is a convex function in m, and its minimum can be obtained at the stationary point of the function; i.e., (62)m¯IV=Tλ′hp2Kh+p+1-pλ′-λλ′α>u2.DenotemIV1=⌊m¯IV⌋,mIV2=⌈m¯IV⌉. A similar discussion yields the optimal ordering times in the first stage (63)m∗=mIV1ifu2<mIV1≤mIV2<u1+1andFIVmIV1≤FIVmIV2,oru2<mIV1<u1+1≤mIV2,mIV2ifmIV1≤u2<mIV2<u1+1andFIVmIV1>FIVmIV2,ormIV1<u2≤mIV2<u1+1.Based on the discussion about three Scenarios above, we can present our algorithm for problem (5).


## 4.4. Algorithm

Step 1.
Input value of parametersK,λ,λ′,h,p,T.Step 6.
Computeu1=λ′Thp/2λK(h+p), u2=Tλhp/2K(h+p)+h/h+p, α=h+p(λ′-λ), β=2λKhp(h+p),  mI¯=Tλ′hp/2K(h+p),mII¯=hλ′/α+λ′h/2Kp(h+p)(pT-β/α).(64)mIII¯=Tλ′hp2Kh+p+hh+p,mIV¯=Tλ′hp2Kh+p+1-pλ′-λλ′α.
Let(65)mI1=m¯I,mI2=m¯ImI3=min⁡u1,u2,,mII1=m¯II,mII2=m¯II,mIII1=m¯III,mIII2=m¯III,mII3=u1,mII4=u2,mIV1=m¯IV,mIV2=m¯IV.ComputeFI(mI1), FI(mI2), FII(mII1), FII(mII2), FII(mII3), FII(mII4), FIII(mIII1), FIII(mIII2), FIV(mIV1), and FIV(mIV2).Step 3.
IfFIII(mIII1)≤FIII(mIII2), then (66)mIII∗=mIII1,FIII∗=FIIImIII1,
otherwise(67)mIII∗=mIII2,FIII∗=FIIImIII2.Step 4.
IfmI2≤min⁡{u1,u2}, and FI(mI2)≤FI(mI1), then (68)mI∗=mI2,(69)FI∗=FImI2,
ifmI1>min⁡{u1,u2}(70)mI∗=mI3,FI∗=FImI3,
otherwise,mI∗=mI1,FI∗=FI(mI1).Step 5.
( 1 ) If β<αpT,
(a) ifu2<mII1≤mII2<u1 and FII(mII1)≤FII(mII2), or u2<mII1<u1≤mII2, then (71)mII∗=mII1,FII∗=FIImII1;
(b) ifu2<mII1≤mII2<u1 and FII(mII1)>FII(mII2), or mII1≤u2<mII2<u1, then (72)mII∗=mII2,FII∗=FIImII2,
otherwise,FII∗=∞.
(2) Ifβ≥αpT,
(a) ifmII3<u2 and mII4>u1 and FII(mII3)≤FII(mII4), or mII4≤u1<mII3<u2, then (73)mII∗=mII3,FII∗=FIImII3;
(b) ifu2>mII3, and mII4>u1 and FII(mII3)>FII(mII4), or u1≤mII4<u2<mII3, then (74)mII∗=mII4,FII∗=FIImII4,
0therwise,FII∗=∞.Step 6.
Ifu2<mIV1≤mII2<u1+1 and FIV(mIV1)≤FIV(mIV1), or u2<mIV1<u1+1≤mIV2, then (75)mIV∗=mIV1,FIV∗=FIVmIV1,
Ifu2<mIV1≤mII2<u1+1 and FIV(mIV1)>FIV(mIV1), or mIV1≤u2<mIV2<u1+1, then (76)mIV∗=mIV2,FIV∗=FIVmIV2,
otherwise,FIV∗=∞.Step 7.
CompareFI∗, FII∗, FIII∗, and FIV∗.
Ifmin⁡{FI∗,FII∗,FIII∗,andFIV∗}=FI∗, then, the optimal order times, the optimal order size from the second replenishment cycle to m-th replenishment cycle, and the optimal maximum shortage level of m-th replenishment cycle in the first stage are, respectively, (77)m∗=mI∗,Q∗=λ′Th+pm∗h+p-h,Bm∗=hh+pQ∗,
go to Step8.
Ifmin⁡FI∗,FII∗,FIII∗,andFIV∗=FII∗, then, the optimal order times, the optimal order size from the second replenishment cycle to m-th replenishment cycle, and the optimal maximum shortage levels of m-th replenishment cycle in the first stage are, respectively, (78)m∗=mIII∗,Q∗=λ′β-αpTphλ′-αm∗,Bm∗=2λKhph+p,
go to Step8.
Ifmin⁡FI∗,FII∗,FIII∗,andFIV∗=FIII∗, the optimal order times, the optimal order size from the second replenishment cycle to m-th replenishment cycle, and the optimal maximum shortage level of m-th replenishment cycle in the first stage are, respectively, (79)m∗=mI∗,Q∗=λ′Th+pm∗h+p-h,Bm∗=2λKhph+p,
go to Step8.
Ifmin⁡FI∗,FII∗,FIII∗,andFIV∗=FIV∗, the optimal order times, the optimal order size from the second replenishment cycle to m-th replenishment cycle, and the optimal maximum shortage level of m-th replenishment cycle in the first stage are, respectively, (80)m∗=mIV∗,Q∗=λ′β+αhThpλ′+αm∗-1,Bm∗=2λKhph+p,
go to Step8.Step 8.
The optimal order size of the first replenishment cycle and the optimal maximum shortage level of the first replenishment cycle (m-1)-th replenishment cycle in the first stage are, respectively,(81)Q1∗=ph+pQ∗,B∗=hh+pQ∗,
and terminate.


## 5. Computational Experiments and Sensitivity Analysis

In this section, we will perform some numerical experiments to test the validity of model (5) and the efficiency of the proposed algorithm.Example 1.
Consider the inventory system with the following parameters:λ′=15,λ=10,K=10,T=4,h=3,p=1.
For this inventory system, direct computing gives that(82)α=20,β=48.9898,u1=3.67,u2=3.20,mI¯=3.75,mI¯=3,mII¯=3.41,mIV¯=3.86,Obviously, β<αpT, and ⌊mII¯⌋<u2,⌈mII¯⌉>u1, by Algorithm, Policy πII does not apply and the operating cost under the other three policies relative to the IEOQ ordering policy are, respectively, {0.1923,0,-0.202}, so the optimal policy is Policy πIV2 and the inventory cost saving over the IEOQ ordering policy by the Policy πIV is 0.202; the optimal ordering times, the ordering size, and the optimal maximum shortage level in the first stage are, respectively, (83)m∗=4,Q∗=19.2660,Q1∗=4.8165,B∗=14.4495,Bm∗=12.2474.
To show the lengthT of the first stage on ordering policy, we reconsider the variation of the inventory cost saving under the optimal ordering policies when T varies from 0 to 30 on the problem. The numerical results are shown in Figure 5, from which we can see that the inventory cost saving under the optimal ordering policy over the IEOQ ordering policy increases cyclically to a constant and is nonnegative, which means that the ordering policy given by Algorithm is superior to the IEOQ ordering policy. Further, the numerical experiments show that the ordering Policies πI, πII, and πIII almost never apply for the problem and therefore the remaining stock at T is always positive for T∈(0,30]. See Figure 6.Figure 5
Variation of cost saving w.r.t.T in Example 1.Figure 6
Variation of remaining stock atT in Example 1.Example 2.
Consider the inventory system in whichλ′=15,K=10,T=15,h=3,p=1.For this inventory system, we test the effect of ratioλ′/λ on the inventory cost saving using Algorithm. To this end, we allow λ′ to vary from λ to 3λ. The numerical results are shown in Figure 7, from which we can see that the inventory cost saving is monotonically increasing with the ratio λ′/λ. Combining with the numerical results for Example 1 that the remaining stock at T under the optimal ordering policies is always positive, as shown in Figure 6, we can see that the remaining stock at T is cyclical w.r.t. the length T of the first stage and the ratio λ′/λ.Figure 7
Variation of cost saving w.r.t. ratioλ′/λ in Example 2.For the inventory model (5) with λ′>λ, the numerical results show that almost all optimal ordering policies are Policy πIV, whereas Policy πII never applies. Further, the optimal ordering policy given by Algorithm is superior to the IEOQ ordering policy.


## 6. Conclusion and Extensions

This study investigated an optimization model for the inventory mechanism with step-shaped demand and backordering. On the basis of the inventory cost analysis under different ordering policies, we presented a closed-form solution to the model. From the obtained results, we demonstrate that the economic order quantity is not necessarily an optimal ordering policy for the concerned inventory model. For the model, there are four possible optimal ordering policies, namely, PoliciesπI, πII, πIII, and πIV. Our numerical results show that Policy πII never applies. If this can be proved or confirmed in theory, then the given solution method can be simplified.



---

*Source: 7452745-2019-08-04.xml*