# Planning Horizon for Production Inventory Models with Production Rate Dependent on Demand and Inventory Level

Authors: Jennifer Lin; Henry C. J. Chao; Peterson Julian
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

This paper discusses why the selection of a finite planning horizon is preferable to an infinite one for a replenishment policy of production inventory models. In a production inventory model, the production rate is dependent on both the demand rate and the inventory level. When there is an exponentially decreasing demand, the application of an infinite planning horizon model is not suitable. The emphasis of this paper is threefold. First, while pointing out questionable results from a previous study, we propose a corrected infinite planning horizon inventory model for the first replenishment cycle. Second, while investigating the optimal solution for the minimization problem, we found that the infinite planning horizon should not be applied when dealing with an exponentially decreasing demand. Third, we developed a new production inventory model under a finite planning horizon for practitioners. Numerical examples are provided to support our findings.


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## Body

## 1. Introduction

Inventory models, in general, can be classified into two categories: infinite and finite planning horizon. For inventory models with the finite planning horizon, the goal is to minimize the total cost. On the other hand, without the present value, that is, not considering the time value of money, the total cost for the entire infinite planning horizon will go to infinity such that researchers are not able to compare the total cost for different inventory policies. The prevailing solution to this dilemma is to minimize the average cost of the first replenishment cycle because of a constant demand that implies an identical replenishment policy for the second replenishment cycle. As a result, the minimization of the average cost for the first replenishment cycle will lead to the optimal solution. The original paper of Wilson’s EOQ model [1] is an example of an infinite planning horizon problem.It should be noted that practitioners in previous studies seemed to randomly decide whether to use an infinite planning horizon or a finite one. That is, they make their choice either by routine experience or by referencing other studies without explaining or considering the choice that fits the characteristics of date on hand. For examples, under the assumptions of time-vary demand, production, and deterioration rate, Goyal and Giri [2] developed two models by employing different modeling approaches over an infinite planning horizon. On the other hand, Goyal’s model [3] was considered as a finite planning horizon problem over time period [0,T], where the replenishment cycle did not repeat itself in the same manner. It infers that each replenishment cycle within the planning horizon [0,T] has different optimal solution such that the solution finding process requires the minimization of the total cost over the entire time period. For both studies, the reasoning behind the selection of either planning horizons was not explained. The purpose of this paper is to point out that in practice, some inventory models work sensibly over an infinite planning horizon. Managers under a highly competitive environment should be making correct and coherent decisions toward the development of inventory models that fit the pursuit of effective cost control.Many papers have also discussed production inventory models under different conditions. By viewing the production rate as a variable, Bhunia and Maiti [4] developed two inventory systems. In the first system, the production rate was dependent on the inventory level, while the production rate was dependent upon the demand in the second. Su and Lin [5] combined the two models creating a model where production rate is dependent on both inventory level and demand. Moreover, Su and Lin [5] assumed that shortages were allowed with complete backlog and an exponentially decreasing demand.We will show that finding the minimum value of the first replenishment cycle is not reasonable with an exponentially decreasing demand since the optimal solution for the production period will go to infinity, implying that the average cost is decreasing to zero. In response, we have developed a finite planning horizon production inventory model.There are two primary reasons that justify assuming that the demand will decrease exponentially. First, the numerous innovations in the field of technology contribute to the expedited release of new merchandises, tremendously decreasing demand for the existing products in the market. Second, rapid changes in consumer preferences also greatly impact the sales of current merchandise. For instance, less than a year after a new camera cellular phone is introduced, an even newer generation will hit the market, with higher dpi than the previous generation. As a result, the demand for the old cell phone will plunge drastically.Su and Lin [5] tried to extend the findings of Bhunia and Maiti [4], but their derivation for the differential equations with boundary conditions contained questionable results. Moreover, they could not analyze how many local minimum points exist. Up to now, there have been four published papers that have referred to Su and Lin [5] in their studies, Chu and Chung [6], Alfares et al. [7], Feng and Yamashiro [8], and Kang [9]. However, none of these papers have been made aware of the fundamental flaw in Su and Lin [5].The derivation of Su and Lin [5] for the inventory level of the third phase contained questionable results such that their findings for relations among decision variables and their objective function also had questionable results. Moreover, we showed that their model is not suitable for infinite planning horizon, and then we studied the inventory model with finite planning horizon. There are two closely related papers, Yang et al. [10] and Lin et al. [11], that are considered for the finite planning horizon. There two models are developed for the EOQ with one decision variable to show that the optimal replenishment policy is independent of the demand type. However, there are four decision variables in our EPQ inventory model. We find two relations among these decision variables of the optimal solution for the infinite planning horizon, then two independent decision variables are left. For the finite planning horizon, we proved that there is only one decision variable left. These two papers have significant contribution for the theoretical development of EOQ inventory models, but their findings cannot be applied to our EPQ inventory model.There are four phases for an EPQ inventory model, and then we proved that there is an upper bound for the elapse time for the first phase. It is an important finding when we applied a program to locate the optimal solution. For the infinite planning horizon, we showed that four decision variables are related, so only two independent decision variables are left, and we find the relations among decision variables that reduced the tedious computation for the minimum value. For the finite planning horizon, in each replenishment cycle, we proved that there is only one independent decision variable that achieves the efficiency for computation. Our first main contribution is providing an analytical approach to solve the optimal solution such that the result from computer programs is supported by the mathematical theorem. Our second main contribution is to reduce the number of independent decision variables to its minimum such that for obtaining the optimal solution, computer programs can be executed effectively.


## 2. Notation and Assumptions

To avoid confusion, we will use the same assumptions and notation as Su and Lin [5]:θ:
deterioration rate,Im:
maximum inventory level,Ib:
unfilled order backlog,C:
setup cost for each new cycle,Cd:
the cost of a deteriorated item,Ci:
inventory carrying cost per unit time,Cs:
shortage cost per unit,K:
total average cost of the system.The assumptions below are used.(1)
A single item is considered over (a) an infinite planning horizon for the first model and (b) a finite planning horizon ofT units of time for the second model which is subject to a constant deterioration rate.(2)
Demand rate,D(t), is known and decreases exponentially so that D(t)=Aexp(-λt), where A is the initial demand rate and λ is the decreasing rate of demand, 0≤λ≤1.(3)
I(t) is the inventory level.(4)
Production rate,P(t), depends on both the demand and the inventory level with P(t)=a+bD(t)-cI(t), a>0,0≤b<1, and 0≤c<1.(5)
Deterioration of the units is considered only after those units are received and put in inventory.(6)
There is no replacement or repair of deteriorated items.(7)
Shortages are allowed and fully backordered.(8)
Two extra conditions,λ>c+θ and a≥A, are added (explained in Section 4).Remark 1.
Su and Lin [5] assumed that T is a prescribed period of time and denoted by t4=T. Note that in the beginning, Su and Lin [5] tried to develop a production inventory model for a finite planning horizon, say [0,T]. However, during their derivation, they considered the problem of minimizing the average cost for the first replenishment cycle in the infinite planning horizon.  To clearly distinguish the difference between infinite and finite planning horizon, we will separate the problem into two cases.Case (a).
We minimize the average cost of the first replenishment cycle. It is a minimization problem with an infinite planning horizon.Case (b). We minimize the total cost over a finite planning horizon of [0,T].


## 3. A Review of Su and Lin [5]

In Su and Lin [5], the first replenishment cycle can be divided into four phases based on the time interval:(a)
the first phase[0,t1]: the production dominates demand and deterioration, and the inventory level accumulates,(b)
the second phase[t1,t2]: no production activity takes place. Demand and deterioration dominate, and so the inventory level gradually drops to zero at t2,(c)
the third phase[t2,t3]: no production and no deterioration take place. The shortage accumulates to Ib at t3,(d)
the fourth phase[t3,t4]: the production is resumed, shortages accumulated during the third phase are fully backordered, and the inventory level returns to zero at t4.The differential equations developed by Su and Lin [5] for governing stock levels over the four different phases during the first replenishment cycle, [0,t4], can be expressed as follows:
(1)ddtI(t)=P(t)-D(t)-θI(t)=a+(b-1)Aexp(-λt)-(c+θ)I(t),0<t<t1,ddtI(t)=-D(t)-θI(t)=-Aexp(-λt)-θI(t),t1<t<t2,ddtI(t)=-D(t)=-Aexp(-λt),t2<t<t3,ddtI(t)=P(t)-D(t)=a+(b-1)Aexp(-λt)-cI(t),t3<t<t4.
Under the boundary conditions,
(2)I(0)=0,I(t1)=Im,I(t2)=0,I(t3)=-Ib,I(t4)=0,
Su and Lin [5] found that
(3)I(t)=ac+θ(1-exp(-(c+θ)t))+A(1-b)λ-c-θ(exp(-λt)-exp(-(c+θ)t)),0≤t≤t1,(4)I(t)=Aexp(-λt)λ-θ(1-exp(-(λ-θ)(t2-t))),t1≤t≤t2,(5)I(t)=(Aλ)(exp(-λt)-1),t2≤t≤t3,(6)I(t)=-ac(exp(c(t4-t))-1)-A(1-b)λ-cexp(-λt)(exp(-(λ-c)(t4-t))-1),t3≤t≤t4.
However, the result Su and Lin [5] derived in (5) is false. The expression should be revised as
(7)I(t)=(Aλ)(exp(-λt)-exp(-λt2)),t2≤t≤t3.
Owing to an error in (5) of their derivations for Im and Ib, the relation between t1 and t2, say t2=R(t1), and the relation between t3 and t4, say t3=R(t4), all contain questionable results. It implies that their objective function, K(t1,t2,t3,t4), is also false.Su and Lin [5] derived the expression, K(t1,t2,t3,t4)=K(t1,t2(t1),t3(t4),t4), so that the objective function has two independent variables, t1 and t4. They computed  ∂K/∂t1=0 and ∂K/∂t4=0. However, they could not analyze whether a system that is comprised of ∂K/∂t1=0 and ∂K/∂t4=0 has solution.


## 4. Our Improvement for Infinite Planning Horizon Model

It should be pointed out that the result of (3) is based on the condition λ≠c+θ. On the other hand, if λ=c+θ, (3) should be revised as
(8)I(t)=(aλ)(1-exp(-λt))-At(1-b)exp(-λt),0≤t≤t1.
Hence, if we try to provide a complete study for the production inventory model of Su and Lin [5], then our model should be divided into seven cases: case (1): c+θ<λ, case (2): c+θ=λ, case (3): c<λ<c+θ, case (4): c=λ, case (5): θ<λ<c, case (6): θ=λ, and case (7): θ>λ.To focus on the investigation of a production inventory model where the production rate is dependent both on demand and inventory level, demand is exponentially decreasing, and shortages are fully backordered, we add two extra conditions:λ>c+θ and a≥A.The reasoning behind the addition of an extra condition,a≥A, is as follows: when t=0, the demand rate D(0)=A, the inventory level I(0)=0, and the production rate P(0)=a+bA. For the accumulation of inventory during the first phase [0,t1], it implies that a+bA≥A for 0≤b<1. For the special case of b=0, we know that a≥A is valid.  Therefore, we derive that a≥(1-b)A when 0<b<1.Ifa<A, then the domain of b has a lower bound satisfying the expression, b≥1-(a/A), such that the domain of b should be changed from [0,1) to [1-(a/A),1).Moreover, Su and Lin [5] assumed in their numerical example that A=200 and a=200.  Their assumption provides support for our extra condition of a≥A. They also assumed that λ=0.3, θ=0.05, and c=0.2 which provides evidence that our condition, λ>c+θ, is reasonable. Moreover, the condition of λ>c+θ will focus on the development of a production inventory that is compatible with the numerical examples in Su and Lin [5] and to avoid tedious discussion for different inventory models with different relations among λ,c, and θ.Based on (3), (4), (6), and (7) and the boundary conditions of (2), we derive that
(9)Im=ac+θ(1-exp(-(c+θ)t1))+A(1-b)λ-c-θ(exp(-λt1)-exp(-(c+θ)t1))=Aexp(-λt1)λ-θ(1-exp(-(λ-θ)(t2-t1))),Ib=(Aλ)(exp(-λt2)-exp(-λt3))=(ac)(exp(c(t4-t3))-1)+A(1-b)λ-c(exp((c-λ)(t4-t3))-1)exp(-λt3).
From (9), we find the relation between t1 and t2 and then the relation among t2, t3, and t4:
(10)t2=1θ-λln[(exp((θ-λ)t1)-exp(-ct1))exp((θ-λ)t1)1θ-λlncc-a(λ-θ)A(c+θ)(exp(θt1)-exp(-ct1))1θ-λlncc-(λ-θ)(1-b)λ-c-θ1θ-λlncc×(exp((θ-λ)t1)-exp(-ct1))((θ-λ)t1)],(11)Aλ(exp(-λt2)-exp(-λt3))+ac+A(1-b)λ-cexp(-λt3)=(ac+A(1-b)λ-cexp(-λt4))exp(c(t4-t3)).
We will simplify a four-variable problem, t1, t2, t3, and t4, to a two-variable problem of t1 and t3. During [0,t1], production, demand, and deterioration interact with each other to accumulate items that will be consumed and thus deteriorate during [t1,t2] such that, trivially, t2-t1 is dependent on t1. We will derive the detailed relation between t2-t1 and t1, that is, t2 and t1. During [t2,t3], the shortages will accumulate to be backlogged during [t3,t4] so that naturally t4-t3 is dependent on t3-t2. We will derive the detailed relation between t4-t3 and t3-t2, which is the relation of (i) t4 and (ii) t3 with t2. Due to the fact that demand is varied, t2 will influence the shortage during [t2,t3].In the following, we will prove thatt2 can be uniquely decided if t1 is given. Whent1 and t2 are given, by using the relation in (11), the unique value of t4 can be derived if t3 is also given. Hence, we will simplify a four-variable problem to a two-variable problem. Let us rewrite (10) as
(12)1-exp(-(λ-θ)(t2-t1))=a(λ-θ)A(c+θ)[exp(λt1)-exp((λ-c-θ)t1)]+(λ-θ)(1-b)λ-c-θ[1-exp((λ-c-θ)t1)].
We tried to find the condition of t1 under which there is a solution to t2 with t2≥t1, satisfying (12). For the later discussion, given that t1, we denote the unique solution of t2 that satisfies (12) as t2(t1). We will prove that the feasible domain of t1 is bounded, guaranteeing the existence of t2.Motivated by (12), we assume the following auxiliary function, g(t1), to be
(13)g(t1)=a(λ-θ)A(c+θ)[exp(λt1)-exp((λ-c-θ)t1)]+(λ-θ)(1-b)λ-c-θ[1-exp((λ-c-θ)t1)].
Taking the derivative of g(t1) with respect to t1 yields
(14)ddt1g(t1)=a(λ-θ)A(c+θ)[λexp(λt1)-(λ-c-θ)exp((λ-c-θ)t1)]-(λ-θ)(1-b)exp((λ-c-θ)t1).
Under the conditions a≥A and λ>c+θ, it follows that
(15)ddt1g(t1)≥(λ-θ)(c+θ)[λexp(λt1)-(λ-c-θ)exp((λ-c-θ)t1)]-(λ-θ)exp((λ-c-θ)t1)=λ(λ-θ)c+θ[exp(λt1)-exp((λ-c-θ)t1)]>0,
showing that g(t1) is an increasing function from g(0)=0 to limt1→∞g(t1)=∞, since
(16)limt1→∞g(t1)exp((λ-c-θ)t1)=limt1→∞[a(λ-θ)A(c+θ)(exp((c+θ)t1)-1)limt1→∞c+(λ-θ)(1-b)λ-c-θ(exp(-(λ-c-θ)t1)-1)]=∞.
There is a unique point, say t1#, that satisfies g(t1#)=1.From (3) and (13), we have
(17)λ-θAexp(λt1#)I(t1#)=g(t1#)=1.
From λ>c+θ, the inequality, λ>θ, is held. It follows that
(18)I(t1#)=Aλ-θexp(-λt1#).
By referring to (12), we obtain
(19)g(t1#)=1=1-exp(-(λ-θ)(t2(t1#)-t1#)).
According to (19), we showed that t2(t1#)-t1# must go to ∞ so that t2(t1#) will go to ∞ as well. We will express the result as limt1→t1#t2(t1)=∞ and summarize our findings in the next lemma.Lemma 2.
I
(
t
1
#
)
=
(
A
/
(
λ
-
θ
)
)
e
-
λ
t
1
# and  limt1→t1#t2(t1)=∞.From Lemma2, we know that the feasible domain of t1 should be set as
(20)0≤t1<t1#.
Given t1, with t1<t1#, then g(t1)<1 so that there is a unique t2, say t2(t1), that satisfies
(21)1-exp(-(λ-θ)(t2(t1)-t1))=g(t1).
We may explicitly express t2(t1) as
(22)t2(t1)=t1-1λ-θln[1-a(λ-θ)A(c+θ)-1λ-θlncc×[exp(λt1)-exp((λ-c-θ)t1)]-1λ-θlncc-(λ-θ)(1-b)λ-c-θ[1-exp((λ-c-θ)t1)]a(λ-θ)A(c+θ)].
We will summarize our findings in the following lemma.Lemma 3.
Ift1<t1#, then there is a unique t2, say t2(t1), as in (22) so that (10) is satisfied.Next, we consider the relation amongt2, t3, and t4 by rewriting (11) as
(23)[Aλ(exp(-λt2)-exp(-λt3))+ac+A(1-b)λ-cexp(-λt3)]×exp(ct3)=(ac+A(1-b)λ-cexp(-λt4))exp(ct4).
Motivated by (23), we assume the following auxiliary function:
(24)f(t4)=(ac+A(1-b)λ-cexp(-λt4))exp(ct4).
We find that
(25)f′(t4)=exp(ct4)[a-A(1-b)exp(-λt4)]>a-A(1-b)≥0.
Under our assumptions of a≥A and 0≤b<1, it can be inferred that f(t4) increases from f(t3)=(a/c)exp(ct3)+(A(1-b)/(λ-c))exp((c-λ)t3) to limt4→∞f(t4)=∞.The relation below,(26)[Aλ(exp(-λt2)-exp(-λt3))+ac+A(1-b)λ-cexp(-λt3)]×exp(ct3)≥acexp(ct3)+A(1-b)λ-cexp((c-λ)t3)=f(t3),
holds since t2≤t3. Therefore, if t1 and t3 are decided, with the restriction t1<t1#, then there is a unique t2(t1) that satisfies (10). Also, from (26) and the increasing function f(t4), we know that for a given t3 under the condition, t2(t1)≤t3, there is a unique point explicitly denoted as t4(t1,t2(t1),t3), simply say t4, that satisfies the condition, t4≥t3, such that the following expression,
(27)[Aλ(exp(-λt2)-exp(-λt3))+ac+A(1-b)λ-cexp(-λt3)]×exp(ct3)=f(t4),
satisfies (23). We will summarize our results in the next lemma.Lemma 4.
Ift1 and t3 are given with t1<t1# and t2(t1)≤t3, then there is a unique t4, denoted as t4(t1,t2(t1),t3) that satisfies (11).Up to this point, the corrected objective function below can be provided as(28)K(t1,t2,t3,t4)={ac+θ(t1+exp(-(c+θ)t1)-1c+θ)+A(1-b)λ-c-θcc×(exp(-(c+θ)t1)-1c+θ-exp(-λt1)-1λ)cc+Aλ-θ(exp(-λt1)-exp(-λt2)λ+exp(-(λ-θ)t2)c+Aλ-θcc×exp(-θt2)-exp(-θt1)θ)}θCd+Cit4+Cst4{exp(-λt4)-exp(-λt3)λAλ2(exp(-λt3)-exp(-λt2))+Cst4ccc+Aλexp(-λt2)(t3-t2)-ac2(1-exp(c(t4-t3)))+Cst4ccc-ac(t4-t3)+A(1-b)λ-c+Cst4ccc×(exp(-(λ-c)t4)(exp(-ct3)-exp(-ct4)c)+Cst4cccccc+exp(-λt4)-exp(-λt3)λ)}+Ct4,
with the conditions 0<t1≤t2≤t3≤t4.Givent1, with t1<t1# and (22), t2(t1) can be derived. Given a t3 that satisfies t2(t1)≤t3, then, by (23), t4 can also be obtained. We have learned from above discussion that only t1 and t3 are independent variables.Hence, the problem becomes the minimization ofK(t1,t2(t1),t3,t4(t1,t2(t1),t3)) under two restrictions:
(29)t1<t1#,t2(t1)≤t3.
We have derived a two-variable minimum problem of t1 and t3 under the conditions of (29), for the infinite horizon minimum cost inventory model. The findings are concluded in the next theorem.Theorem 5.
For the production inventory model with infinite planning horizon, if one minimizes the average cost for the first replenishment cycle, then there are two necessary conditions,t1<t1# and t2(t1)≤t3, for the production inventory model of Su and Lin [5].


## 5. Numerical Examples for Infinite Planning Horizon Inventory Model

We will employ the same numerical examples as Su and Lin [5] for comparison purposes where A=200, λ=0.3, θ=0.05, C=100, Cd=3, Cs=10, Ci=1, a=200, b=0.2, and c=0.2.  Some computation results are showed in Table 1 with t1#=3.6699 arranged according to a sequence of different values of t3.Table 1
The results for average cost of the first cycle for infinite planning horizon.
t
1
t
2
t
3
t
4
K
(
t
1
,
t
2
,
t
3
,
t
4
)
1.4683
2.2261
2.3148
2.3885
88.8785
2.4280
4.6512
4.9220
4.9876
102.8907
3.2554
9.1627
9.9433
9.9794
108.9192
3.6192
17.6260
19.9899
19.9985
77.4016
3.6626
25.4135
29.9900
29.9912
54.0973
3.6695
37.6621
477.5669
477.5670
3.4965
3.6696
39.0597
1745.8406
1745.8407
0.9889
3.6698
46.3253
5929.4429
5929.4434
0.2818
3.6698
46.3253
14903.3271
14903.3281
0.1158
3.6698
46.3253
19998.2832
19998.2852
0.0879From the numerical examples in Table1, it reveals that if we prolong the replenishment cycle, then the average cost will eventually decrease. The rationale is that with a negatively exponential demand function, the market demand will dramatically decrease, especially in a longer inventory horizon, which will in turn significantly bring down the corresponding average holding and shortage costs. On the other hand, when we prolong the shortage phase with t3=19998.2832 (the personal computer’s computational limit) in order to reduce the average cost, the ordinary customers may lose patience when waiting for the backorder.  Hence, a full backorder cannot be performed. In other words, it is impossible to simultaneously achieve full backorder and minimize average inventory cost. We may conclude that for the negatively exponential demand, D(t)=Aexp(-λt), the infinite planning horizon production inventory model is not adequate. Therefore, we stop the discussion of case (a) in the infinite planning horizon.


## 6. Our Proposed Production Inventory Model with Finite Planning Horizon

Next, we consider case (b) with a finite planning horizon, denoted by[0,T]. To simplify the discussion, we assume that there is one replenishment cycle during the finite planning horizon. Our results can be easily extended to several replenishment cycles. In this setting, (28) should be revised as follows:
(30)K(t1,t2,t3,t4=T)={ac+θ(t1+exp(-(c+θ)t1)-1c+θ)+A(1-b)λ-c-θcc×(exp(-(c+θ)t1)-1c+θ-exp(-λt1)-1λ)cc+Aλ-θ(exp(-λt1)-exp(-λt2)λ+exp(-(λ-θ)t2)cc+Aλ-θcc×exp(-θt2)-exp(-θt1)θ)}θCd+CiT+CsT{exp(-λT)-exp(-λt3)λ)Aλ2(exp(-λt3)-exp(-λt2))+CsTcc+Aλexp(-λt2)(t3-t2)-ac2(1-exp(c(T-t3)))+CsTcc-ac(T-t3)+A(1-b)λ-c+CsTcc×(exp(-(λ-c)T)(exp(-ct3)-exp(-cT)c)+CsT+ccc+exp(-λT)-exp(-λt3)λ)}+CT.
Here, we will derive a stronger condition than t1<t1# for the feasible domain of t1. For a given t4, from (10), since t2(t1) is an increasing function of t1, there is a unique point, say t1∧(t4) with t2(t1∧(t4))=t4. Under the condition
(31)t1<t1∧(t4),
the desired result t2(t1)<t4 is achieved, since t2(t1) is an increasing function of t1.Lemma 6.
For a givent4, the feasible domain of t1 is [0,t1∧(t4)), implying that t2(t1)<t4 with t2(t1∧(t4))=t4.In the following, whent4 is given, if we take a t1 that satisfies (31), then we will prove that there is a unique t3 that satisfies (23).Based on (23), let us assume other two auxiliary functions, k(t) and h(t3), where
(32)k(t)=acexp(ct)+A(1-b)λ-cexp(-(λ-c)t),fort2≤t≤T.
With a restricted domain, k(t) is related to our previous auxiliary function f(t4) of (24) and
(33)h(t3)=[A(1-b)λ-cAλ(exp(-λt2)-exp(-λt3))+acc+A(1-b)λ-cexp(-λt3)]exp(ct3),fort2≤t3≤T.
From (d/dt)k(t)=aexp(ct)-A(1-b)exp(-(λ-c)t)>0, under the conditions a≥A and 0≤b<1, k(t) is an increasing function which implies that
(34)k(t2)=acexp(ct2)+A(1-b)λ-cexp(-(λ-c)t2)<k(T)=acexp(cT)+A(1-b)λ-cexp(-(λ-c)T).
On the other hand, the expression
(35)ddt3h(t3)=Aλ[λexp(-(λ-c)t3)=Aλic+c(exp(-ct2)-exp(-λt3))exp(ct3)]+aexp(ct3)-A(1-b)exp((c-λ)t3)>0
shows that h(t3) is an increasing function. If we apply (33) and (34), then h(t3) increases from
(36)h(t2)=k(t2)<k(T),
to
(37)h(T)=Aλ(exp(-λt2)-exp(-λT))+k(T)>k(T).
Therefore, there is a unique point, say t3(t2), that satisfies
(38)h(t3(t2))=k(T).
When t4=T is given, based on the previous discussion, if t1 is given with t1<t1∧(t4), then we have t2(t1)<t4.  From (38), there exists a unique point, t3(t2), with h(t3(t2(t1)))=K(T) such that t2(t1), t3(t2(t1)), and t4 satisfy (23). Hence, for a finite-horizon minimum cost inventory model, we have simplified a four-variable problem to a one-variable problem.Hence, in the following, if we only consider thoset1s that satisfy the condition of (31), then
(39)t2(t1)<t4.
By (23) and (38), the relation, t3=t3(t2), implies that
(40)t4(t1,t2(t1),t3(t2(t1)))=t4,
where t4(t1,t2(t1),t3), defined in Lemma 4, satisfies (11).The objective function becomes a one-variable problem(41)K(t1)=K(t1,t2(t1),t3(t2(t1)),t4=T)=CT+θCd+CiT×{ac+θ(t1+exp(-(c+θ)t1)-1c+θ)+cc+A(1-b)λ-c-θ(exp(-(c+θ)t1)-1c+θ+cc+A(1-b)λ-c-θccc-exp(-λt1)-1λ)+cc+Aλ-θ(exp(-λt1)-exp(-λt2)λ+cc+Aλ-θcc+exp(-(λ-θ)t2)+cc+Aλ-θcc×exp(-θt2)-exp(-θt1)θ)}+C3T{exp(-λT)-exp(-λt3)λ)Aλ2(exp(-λt3)-exp(-λt2))+C3Tcc+Aλexp(-λt2)(t3-t2)+C3Tcc-ac2(1-exp(c(T-t3)))-ac(T-t3)+C3Tcc+A(1-b)λ-c((exp(-ct3)-exp(-cT)c)exp(-(λ-c)T)+C3Tcc+A(1-b)λ-ccc×(exp(-ct3)-exp(-cT)c)+C3Tcc+A(1-b)λ-ccc+exp(-λT)-exp(-λt3)λ)}.
We will summarize our findings in the next theorem.Theorem 7.
For the production inventory model with the finite planning horizon,[0,t4], if one only considers one replenishment cycle, there is a natural restriction t1<t1∧(t4), creating a one-variable minimum problem.From Theorem7, computer program as MathCAD can be adopted to locate the optimal solution. We may point out that the benefits to the simplified production inventory model that we have proposed include (a) easy to use for decision makers, (b) reduction of the solution space (computation time) in determining the parameter setting, and (c) reduction of the model complexity.


## 7. Numerical Example for the Finite Planning Horizon

For the finite planning horizon production inventory model with the same data,A=200, λ=0.3, θ=0.05, C=100, Cd=3, Cs=10, Ci=1, a=200, b=0.2, c=0.2, and T=2, we find the optimal solution, t1*=1.2742. With (12), it shows that t2*=1.8620. With (38), it shows that t3*=1.9306. Finally, with (41), we find that the minimum cost is K(t1*,t2*,t3*,t4*)=89.7151. The above discussion is based on the preset condition that there is one replenishment cycle. However, under the multiple replenishment cycles, the total setup cost will be at least 200, and then the average cost during [0,2] is more than one hundred that is larger than the result of one replenishment cycle. It specifies that the average cost for multiple replenishment cycles is much larger than that of one replenishment cycle. Hence, for this numerical example, we only consider one replenishment cycle.Particle swarm optimization is applied to check our findings. Both approaches have the same optimal solution.


## 8. Conclusion and Further Direction

We have shown that with an exponentially decreasing demand, the goals of simultaneously minimizing the average cost for the first replenishment cycle and fully backordering the shortage items cannot be applied for an infinite-horizon minimum cost inventory model.  The result of our investigation explicitly reveals that using the infinite planning horizon model is inappropriate in practice. For the finite planning horizon, we have shown that the four-phase production inventory model can be converted to a single variable problem in order to find the minimum solution. Our study not only provides a sound operational formulation but also offers a practical and efficient approach in the location of the optimal solution.The study we have carried out can probably be viewed as the first attempt to solve a finite planning horizon production inventory model. In the future, it would be interesting to show that our objective function is convex to ensure the existence of a local minimum. Moreover, the issues of how to decide the optimal solution under several replenishment cycles and how to verify the convexity of the minimum value under multiple replenishment cycles deserve further study.



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*Source: 961258-2013-03-28.xml*