Optimal Policies for a Finite-Horizon Production Inventory Model

The model treated in this paper is based on the following assumptions:

1. the planning horizon is finite;
2. a single item is considered;
3. products are assumed to experience deterioration while in stock;
4. shortages are not permitted;
5. initial inventory at the beginning of planning horizon is zero, also the inventory depletes to zero at the end of the planning horizon;
6. the demand function is strictly positive;

We will initially look at a single period $i$ (production cycle), say, starting at time $t_{i-1}$ and ending at time $t_{i}$, $i = 1,2...$. Some of the notations used in the model are as follows:

$H$: the total planning horizon,
$p$: the constant production rate,
$D(t)$: the demand rate at time $t, 0 \text { < } D(t) \text { < } p$,
$\alpha$: constant deteriorating rate of inventory items with $\alpha \text { > } 0$,
$t_{i}^{p}$: the time at which the inventory level reaches it is maximum in $i$th production cycle,
$K$: set up cost for the inventory model,
$c_{1}$: the cost of one unit of the item with $c_{1} \text { > }  0$,
$c_{h}$: carrying cost per inventory unit held in the model per unit time,
$TC$: total system cost during $H$.

Figure 1 shows the changes of the level of stock for a typical production period.

Figure 1 The changes of inventory levels of various components of the model for a typical production batch.

Let $I(t)$ be the level of stock at time $t$. The change in period $i$, the level of inventory, may be described by the following differential equation $i, i=1,2, \ldots$,

$I^{\prime}(t)=p-D(t)-\alpha I(t), \quad t_{i-1} \leq t \text { < } t_{i}^{p}, \quad I(t) \text { ↓ } 0 \quad \text { as } t \text { ↓ } t_{i-1}$.   (2.1)

The solution to (2.1) is given by

$I(t)=e^{-\alpha t} \int_{t_{i-1}}^{t} e^{\alpha u}\{p-D(u)\} d u, \quad t_{i-1} \leq t \text { < } t_{i}^{p}$,   (2.2)

$I^{\prime}(t)=-D(t)-\alpha I(t), \quad t_{i}^{p} \leq t \text { < } t_{i}, \quad I(t) \uparrow 0 \text { as } t \uparrow t_{i}$.   (2.3)

The solution to (2.3) is given by 

$I(t)=e^{-\alpha t} \int_{t}^{t_{i}} e^{\alpha u} D(u) d u, \quad t_{i}^{p} \leq t \text { < } t_{i}$.   (2.4)

The total costs, (excluding the setup cost) for period $i$, which consist of holding cost and deterioration cost are given by

$\begin{gathered} c_{h} \int_{t_{i-1}}^{t_{i}^{p}} e^{-\alpha t}\left[\int_{t_{i-1}}^{t} e^{\alpha u}\{p-D(u)\} d u\right] d t \\ \quad+c_{h} \int_{t_{i}^{p}}^{t_{i}} e^{-\alpha t}\left\{\int_{t}^{t_{i}} e^{\alpha u} D(u) d u\right\} d t \\ \quad+c_{1}\left\{\int_{t_{i-1}}^{t_{i}^{p}} p d u-\int_{t_{i-1}}^{t_{i}} D(u) d u\right\} \end{gathered}$ (2.5)

We will call models with this cost OHD models. It is possible to consider instead of (2.5) the form

$\begin{aligned} &c_{h} \int_{t_{i-1}}^{t_{i}^{p}} e^{-\alpha t}\left[\int_{t_{i-1}}^{t} e^{\alpha u}\{p-D(u)\} d u\right] d t \\ &\quad+c_{h} \int_{t_{i}^{p}}^{t_{i}} e^{-\alpha t}\left\{\int_{t}^{t_{i}} e^{\alpha u} D(u) d u\right\} d t+c_{1} \int_{t_{i-1}}^{t_{i}^{p}} p d u \end{aligned}$  (2.6)

which considers only holding and purchasing costs, where the expression $c_{1} \int_{t_{i-1}}^{t_{i}^{p}} p d u$ represents the purchasing cost. We call this OHP models.

Note that since the function $I$ is continuous at $t_{i}^{p}$, we have

$e^{-\alpha t_{i}^{p}} \int_{t_{i-1}}^{t_{i}^{p}} e^{\alpha u}\{p-D(u)\} d u=e^{-\alpha t_{i}^{p}} \int_{t_{i}^{p}}^{t_{i}} e^{\alpha u} D(u) d u$,   (2.7)

or

$\int_{t_{i-1}}^{t_{i}^{p}} e^{\alpha u}\{p-D(u)\} d u=\int_{t_{i}^{p}}^{t_{i}} e^{\alpha u} D(u) d u$,   (2.8)

then

$p \int_{t_{i-1}}^{t_{i}^{p}} e^{\alpha u} d u=\int_{t_{i-1}}^{t_{i}} e^{\alpha u} D(u) d u$,   (2.9)

or

$\frac{p}{\alpha}\left(e^{\alpha t_{i}^{p}}-e^{\alpha t_{i-1}}\right)=\frac{p}{\alpha} e^{\alpha t_{i-1}}\left\{e^{\alpha\left(t_{i}^{p}-t_{i-1}\right)}-1\right\}=\int_{t_{j-1}}^{t_{i}} e^{\alpha u} D(u) d u$,   (2.10)

$e^{\alpha\left(t_{i}^{p}-t_{i-1}\right)}-1=\frac{\alpha}{p} e^{-\alpha t_{i-1}} \int_{t_{i-1}}^{t_{i}} e^{\alpha u} D(u) d u$ ).  (2.11)

Therefore,

$t_{i}^{p}-t_{i-1}=\frac{1}{\alpha} \log \left\{1+\frac{\alpha}{p} e^{-\alpha t_{i-1}} \int_{t_{i-1}}^{t_{i}} e^{\alpha u} D(u) d u\right\}$.   (2.12)

Lemma 2.1. The expression of the cost in (2.5) is equal to

$\left(c_{h}+\alpha c_{1}\right)\left[\frac{p}{\alpha^{2}} \log \left\{1+\frac{\alpha}{p} e^{-\alpha t_{i-1}} \int_{t_{i-1}}^{t_{i}} e^{\alpha u} D(u) d u\right\}-\frac{1}{\alpha} \int_{t_{i-1}}^{t_{i}} D(u) d u\right]$.   (2.13)

Proof. Applying integration by parts, we get that (2.5) reduces to

$\begin{aligned} &\frac{c_{h}}{\alpha} \int_{t_{i-1}}^{t_{i}^{p}}\left\{1-e^{\alpha\left(u-t_{i}^{p}\right)}\right\}\{p-D(u)\} d u \\ &\quad+\frac{c_{h}}{\alpha} \int_{t_{i}^{p}}^{t_{i}}\left\{e^{\alpha\left(u-t_{i}^{p}\right)}-1\right\} D(u) d u \\ &\quad+c_{1} \int_{t_{i-1}}^{t_{i}^{p}} p d u-c_{1} \int_{t_{i-1}}^{t_{i}} D(u) d u \end{aligned}$   (2.14)

or

$\begin{aligned} &\frac{c_{h}}{\alpha}\left\{p \int_{t_{i-1}}^{t_{i}^{p}} d u-\int_{t_{i-1}}^{t_{i}^{p}} D(u) d u-p \int_{t_{i-1}}^{t_{i}^{p}} e^{\alpha\left(u-t_{i}^{p}\right)} d u+\int_{t_{i-1}}^{t_{i}^{p}} e^{\alpha\left(u-t_{i}^{p}\right)} D(u) d u\right\} \\ &\quad+\frac{c_{h}}{\alpha}\left\{\int_{t_{i}^{p}}^{t_{i}} e^{\alpha\left(u-t_{i}^{p}\right)} D(u) d u-\int_{t_{i}^{p}}^{t_{i}} D(u) d u\right\} \\ &\quad+c_{1} \int_{t_{i-1}}^{t_{i}^{p}} p d u-c_{1} \int_{t_{i-1}}^{t_{i}} D(u) d u \end{aligned}$   (2.15)

$\begin{aligned} =& p\left(\frac{c_{h}}{\alpha}+c_{1}\right)\left(t_{i}^{p}-t_{i-1}\right)-\frac{c_{h}}{\alpha} e^{-\alpha t_{i}^{p}} \int_{t_{i-1}}^{t_{i}^{p}} e^{\alpha u u}\{p-D(u)\} d u \\ &+\frac{c_{h}}{\alpha} e^{-\alpha t_{i}^{p}} \int_{t_{i}^{p}}^{t_{i}} e^{\alpha u} D(u) d u-\frac{c_{h}}{\alpha} \int_{t_{i-1}}^{t_{i}} D(u) d u-c_{1} \int_{t_{i-1}}^{t_{i}} D(u) d u . \end{aligned}$   (2.16)

This is equal, using (2.7), to

$p\left(\frac{c_{h}}{\alpha}+c_{1}\right)\left(t_{i}^{p}-t_{i-1}\right)-\left(\frac{c_{h}}{\alpha}+c_{1}\right) \int_{t_{i-1}}^{t_{i}} D(u) d u$.   (2.17)

Now, the lemma follows from (2.12) and (2.17).

Note that Lemma 2.1 reduced the dependence of the inventory cost in period $i$ from three variables to two variables. This reduction can be significant for an $n$-period model. Let

$R(x, y):=\left(c_{h}+\alpha c_{1}\right)\left[\frac{p}{\alpha^{2}} \log \left\{1+\frac{\alpha}{p} e^{-\alpha x} \int_{x}^{y} e^{\alpha t} D(t) d t\right\}-\frac{1}{\alpha} \int_{x}^{y} D(t) d t\right]$.   (2.18)

Remark 2.2. Let $\alpha \rightarrow 0$, in (2.18), and recall that $\log (1+x)$ may be expanded as

$x-\frac{1}{2} x^{2}+\frac{1}{3} x^{3}-\frac{1}{4} x^{4}+\frac{1}{5} x^{5}+O\left(x^{6}\right)$,   (2.19)

to get that as $\alpha \rightarrow 0$, $R(x, y)$ is equivalent to

$\left(c_{h}+\alpha c_{1}\right)\left[\int_{x}^{y}\left\{\frac{e^{\alpha(t-x)}-1}{\alpha}\right\} D(t) d t-\frac{1}{2 p}\left\{\int_{x}^{y} e^{\alpha(t-x)} D(t) d t\right\}^{2}\right]$,   (2.20)

which leads to the expression

$c_{h}\left[\int_{x}^{y}(t-x) D(t) d t-\frac{1}{2 p}\left\{\int_{x}^{y} D(t) d t\right\}^{2}\right]$.    (2.21)

This expression may be found in Hill, Omar and Smith, and Rau and Ouyang. However, their interest in finding the optimal inventory policy for their model centered around treating special cases for demand rate functions or devising heuristics.

The total inventory costs where $n$ ordered are made may be written as follows:

$\mathrm{TC}\left(t_{1}, \ldots, t_{n}\right)=n K+\sum_{i=1}^{n} R\left(t_{i-1}, t_{i}\right)$,   (2.22)

which is given by (2.18).

The objective now is to find $n$ and $t_{1}, \ldots, t_{n}$ which minimizes $TC$ subject to $t_{0}=0 \text { < } t_{1} \text { < } \cdots \text { < } t_{n}=H$. The problem becomes a mixed integer programming problem. The approach that we will use to solve it is based on a procedure developed by Benkherouf and Gilding. The next section contains the ingredients of the approach.

This section contains a summary of the work of Benkherouf and Gilding needed to tackle the problem of this paper. Proofs of the results are omitted. Interested readers may consult Benkherouf and Gilding.

Consider the problem

$\mathbf{P}: \operatorname{TC}\left(t_{1}, \ldots, t_{n} ; n\right)=n K+\sum_{i=1}^{n} R_{i}\left(t_{i-1}, t_{\mathrm{i}}\right)$,   (3.1)

subject to

$0=t_{0} \text { < } t_{1} \text { < } \cdots \text { < } t_{n}=H$.   (3.2)

It was shown in Benkherouf and Gilding that, under some technical conditions, the optimization problem (P) has a unique optimal solution which can be found from solving a system of nonlinear equations derived from the first-order optimality condition. To be precise, $t_{0}=0$ and $t_{n}=H$ and ignore the rest of the constraints (3.2).

Write

$S_{n}:=\sum_{i=1}^{n} R_{i}\left(t_{i-1}, t_{i}\right)$.   (3.3)

Assuming that $R_{i}^{\prime} s$  are twice differentiable, then, for fixed $n$, the optimal solution in (P) subject to (3.2) reduces to minimizing $S_{n}$.

Use the notation $\nabla$ for the gradient, then setting $\nabla \mathrm{TC}\left(t_{1}, \ldots, t_{n} ; n\right)=0$ gives

$(\nabla \mathrm{TC})_{i}=\left(\partial R_{i}\right)_{y}\left(t_{i-1}, t_{i}\right)+\left(\partial R_{i+1}\right)_{x}\left(t_{i}, t_{i+1}\right)=0, \quad i=1, \ldots, n-1$.   (3.4)

Two sets of hypotheses were put forward.

Hypothesis 1. The functions $R_{i}$ satisfy, for $i=1, \ldots, n \text { and } y \text { > } x$,

(1) $R_{i}(x, y) \text { > } 0$

(2) $R_{i}(x, x) \text { = } 0$

(3) $\left(\partial R_{i}\right)_{x}(x, y) \text { < } 0 \text { < } \left(\partial R_{i}\right)_{y}(x, y)$

(4) $\left(\partial_{x} \partial_{y} R\right)_{i}(x, y) \text { < } 0$

Hypothesis 2. Define

$\begin{aligned} &\mathscr{L}_{x} z=\partial_{x}^{2} z+\partial_{x} \partial_{y} z+f(x) \partial_{x} z \\ &\mathscr{L}_{y} z=\partial_{y}^{2} z+\partial_{x} \partial_{y} z+f(y) \partial_{y} z \end{aligned}$   (3.5)

then there is a continuous function $f$ such that $\mathscr{L}_{x} R_{i} \geq 0, \mathscr{L}_{y} R_{i} \geq 0$ for all $i=1, \ldots, n$, and $\left(\partial R_{i}\right)_{y}+\left(\partial R_{i+1}\right)_{x}=0$ on the boundary of the feasible set.

The next theorem shows that under assumptions in Hypotheses 1 and 2, the function $S_{n}$ has a unique minimum.

Theorem 3.1.The system (3.4) has a unique solution subject to (3.2). Furthermore, this solution is the solution of (3.1) subject to (3.2).  Recall that a function $S_{n}$ is convex in $n$ if

$S_{n+2}-S_{n+1} \geq S_{n+1}-S_{n}$.   (3.6)

This is equivalent to

$\frac{1}{2}\left(S_{n}+S_{n+2}\right) \geq S_{n+1}$.   (3.7)

Theorem 3.2. If $s_{n}$ denotes the minimum objective value of (3.1) subject to (3.2) and $R_{i}(\mathrm{x}, y)=R(x, y)$ then $s_{n}$ is convex in $n$.

Based on the convexity property of $s_{n}$, the optimal number of cycles $n*$ is given by

$n^{*}=\min \left\{n \geq 1: s_{n+1}-s_{n} \text { > } 0\right\}$.   (3.8)

Now to solve (3.4) at $i=n-1$,

$\partial R_{n-1 y}\left(t_{n-2}, t_{n-1}\right)+\left(\partial R_{n}\right)_{x}\left(t_{n-1}, H\right)=0$.   (3.9)

Assume that $t_{n-1}$ is known, $t_{n-2}$ can be found uniquely as a function of $t_{n-1}$. Repeating this process for $i=n-2$, $i=1, t_{n-3}, \ldots, t_{n}$ are a function of $t_{n-1}$. So, the search for the optimal solution of (3.5) can be conducted using a univariate search method.

This section is concerned with the optimal inventory policy for the production inventory model. The model has been introduced in Section 2. This section will investigate the extent to which the function $R$ given by (2.18) satisfies Hypotheses 1 and 2,

$R(x, y)=\left(c_{h}+\alpha c_{1}\right)\left[\frac{p}{\alpha^{2}} \log \left\{1+\frac{\alpha}{p} \int_{x}^{y} e^{\alpha(t-x)} D(t) d t\right\}-\frac{1}{\alpha} \int_{x}^{y} D(t) d t\right]$.   (4.1)

Without loss of generality, we will set $c_{h}+\alpha c_{1}$ to 1. As this will have no effect on the solution of the optimization problem where $K$ needs to be replaced by $K /\left(c_{h}+\alpha c_{1}\right)$, therefore, we set

$R(x, y)=\frac{p}{\alpha^{2}} \log \left\{1+\frac{\alpha}{p} \int_{x}^{y} e^{\alpha(t-x)} D(t) d t\right\}-\frac{1}{\alpha} \int_{x}^{y} D(\mathrm{t}) d t$.   (4.2)

Write

$G(x, y)=\frac{\alpha}{p} \int_{x}^{y} e^{\alpha(t-x)} D(t) d t$.   (4.3)

Direct computations then lead to

$\begin{aligned} \partial_{x} R &=\frac{p}{\alpha^{2}} \frac{\partial_{x} G}{1+G}+\frac{1}{\alpha} D(x), \\ \partial_{y} R &=\frac{p}{\alpha^{2}} \frac{\partial_{y} G}{1+G}-\frac{1}{\alpha} D(y), \\ \partial_{x} \partial_{y} R &=\frac{p}{\alpha^{2}} \frac{\partial_{x} \partial_{y} G(1+G)-\left(\partial_{x} G\right)\left(\partial_{y} G\right)}{(1+G)^{2}}, \\ \partial_{x}^{2} R &=\frac{p}{\alpha^{2}} \frac{\partial_{x}^{2} G(1+G)-\left(\partial_{x} G\right)^{2}}{(1+G)^{2}}+\frac{1}{\alpha} D^{\prime}(x), \\ \partial_{y}^{2} R &=\frac{p}{\alpha^{2}} \frac{\partial_{y}^{2} G(1+G)-\left(\partial_{x} G\right)^{2}}{(1+G)^{2}}-\frac{1}{\alpha} D^{\prime}(y), \\ \partial_{x} G &=-\alpha\left\{G+\frac{D(x)}{p}\right\}, \\ \partial_{y} G &=\frac{\alpha}{p} e^{\alpha(y-x)} D(y), \\ \partial_{x} \partial_{y} G &=-\alpha \partial_{y} G, \\ \partial_{x}^{2} G &=-\alpha\left\{-\alpha G-\alpha \frac{D(x)}{p}+\frac{D^{\prime}(x)}{p}\right\}, \end{aligned}$

$\partial_{y}^{2} G=\alpha \partial_{y} G+\frac{\alpha}{p} e^{\alpha(y-x)} D^{\prime}(y)$.   (4.4)

The following result indicates that $R$ obtained in (4.2) satisfies Hypothesis 1.

Lemma 4.1. The function $R$ satisfies Hypothesis 1.

Proof. It is clear that for any $x \in[0, H]$,

$R(x, x)=0$.   (4.5)

Now, direct computations show that

$\partial_{x} R(x, y)=\frac{G(x, y)}{\alpha\{1+G(x, y)\}}\{-p+D(x)\}$.   (4.6)

But $p \text { > } D(x)$, therefore $\partial_{x} R(x, y) \text { < } 0$ since $G(x, y \text { > } 0$ for $y \text { > } x$. Also, it can be shown that

$\partial_{y} R(x, y)=\frac{D(y)}{\alpha}\left\{\frac{e^{\alpha(y-x)}}{1+G(x, y)}-1\right\}$.   (4.7)

We claim that $\partial_{y} R(x, y) \text { > } 0$. Indeed, the claim is equivalent to

$F_{x}(y)=1+G(x, y)-e^{\alpha(y-x)} \text { < } 0 \quad \text { for } y \text { > } x$.   (4.8)

The function $F_{x}(y)$ ;is decreasing since

$F_{x}^{\prime}(y)=-\alpha e^{\alpha(y-x)}\left\{1-\frac{D(y)}{p}\right\}$,   (4.9)

with $F_{x}(x)=0$. Hence, the claim is true. To complete the proof, we need to examine the sign of $\partial_{x} \partial_{y} R$. Again, some algebra leads to

$\partial_{x} \partial_{y} R(x, y)=-\alpha \partial_{y} G(x, y)\left\{1-\frac{D(y)}{p}\right\}$.    (4.10)

But $\partial_{y} G(x, y) \text { > } 0$ and $p \text { > } D(x)$. Therefore,  $\partial_{x} \partial_{y} R(x, y) \text { < } 0$  for $y \text { > } x$, and the proof is complete.

Before we proceed further, we set

$Z_{x}(u)=\frac{D^{\prime}(u)}{D(u)\{1+2 G(x, u)\}}, \quad \text { with } 0 \leq x \leq u \leq H$.   (4.11)

We assume the following.

(A1)The function $Z_{x}$ is nonincreasing.

Note that as $\alpha \ text { → } 0, G(x, u) \ text { → } 0$, and consequently $Z_{x}(u)$ reduces to $D^{\prime}(u) / D(u)$. In other words, assumption (A1) implies that $D$ is logconcave. This property of the demand rate function may be found in [4, 6], when considering models with infinite production rates. As a matter of fact, this property of $D$ can also be obtained if we let $p \text { → } \infty$.

Example 4.2. Let $D(u)=\alpha e^{\beta u}$ where $\alpha \text { > } 0$ and is known and $\beta \text { > } 0$ and is known, then $Z_{x}(u)$ is is nonincreasing.

Note that $Z_{x}$ is nonincreasing which is equivalent to $1 / Z_{x}$ non-decreasing. We have

$g_{x}(u):=\frac{1}{Z_{x}(u)}=\frac{D(u)\{1+2 G(x, u)\}}{D^{\prime}(u)}=\frac{1}{\beta}\{1+2 G(x, u)\}$,   (4.12)

with $g_{x}^{\prime}(u)=(2 / \beta) \partial_{y} G(x, y) \text { > } 0$, which implies the result.

Example 4.3. Let $D(u)=a+b u$, where $b \text { > } 0$, then it is an easy exercise to check that assumption (A1) is satisfied.

Lemma 4.4. If $Z_{x}$satisfies (A1) for all $0 \leq x \leq H$ then $\mathscr{L}_{x} R \geq 0$ where $\mathscr{L}_{x}$ is defined in 3.5.

Proof. Tedious but direct algebra using the definition of $\mathscr{L}_{x}R$ leads to

$\begin{aligned} \mathscr{L}_{x} R=& \frac{\{1-(D(x) / p)\}}{\{1+G(x, y)\}^{2}}\left\{\alpha \int_{x}^{y} e^{\alpha(t-x)} D(t) d t+D(x)-e^{\alpha(y-x)} D(y)\right\} \\ &+\frac{1}{\alpha} \frac{D^{\prime}(x) G(x, y)}{1+G(x, y)}-\frac{p}{\alpha} \frac{G(x, y)}{1+G(x, y)}\left\{1-\frac{D(x)}{p}\right\} f(x). \end{aligned}$   (4.13)

$\mathscr{L}_{x} R \geq 0$ is equivalent to

$\begin{gathered} \frac{\{1-(D(x) / p)\}}{\{1+G(x, y)\}}\left\{\alpha \int_{x}^{y} e^{\alpha(t-x)} D(t) d t+D(x)-e^{\alpha(y-x)} D(y)\right\} \\ \quad+\frac{1}{\alpha} D^{\prime}(x) G(x, y)-\frac{p}{\alpha} G(x, y)\left\{1-\frac{D(x)}{p}\right\} f(x) \geq 0. \end{gathered}$   (4.14)

Let

$f(x)=-\frac{D^{\prime}(x)}{D^{\prime}(x)}+\frac{D^{\prime}(x)}{p\{1-(D(x) / p)\}}$.   (4.15)

It can be shown that (4.14) is true if

$\frac{\alpha}{p} \frac{e^{\alpha(y-x)} D(y)-D(x)-\alpha \int_{x}^{y} e^{\alpha(t-x)} D(t) d t}{G(x, y)\{1+G(x, y)\}} \leq \frac{D^{\prime}(x)}{D(x)}$.   (4.16)

Define, for $u \geq x$.

$\begin{aligned} &F_{x}(u)=e^{\alpha(u-x)} D(u)-\alpha \int_{x}^{u} e^{\alpha(t-x)} D(t) d t \\ &H_{x}(u)=\frac{\alpha}{p} \int_{x}^{u} e^{\alpha(t-x)} D(t) d t\left\{1+\frac{\alpha}{p} \int_{x}^{u} e^{\alpha(t-x)} D(t) d t\right\}. \end{aligned}$    (4.17)

The left hand side of (4.16) may be written as

$\frac{\alpha}{p} \frac{F_{x}(y)-F_{x}(x)}{H_{x}(y)-H_{x}(x)}$.   (4.18)

This is equal by extended-mean value theorem to $(\alpha / p)\left(\left(F_{x}^{\prime}(\xi)\right) /\left(H_{x}^{\prime}(\xi)\right)\right)$ for some $x \text { < } \xi \leq y$. However,

$\begin{aligned} F_{x}^{\prime}(u) &=\alpha e^{\alpha(u-x)} D^{\prime}(u), \\ H_{x}^{\prime}(u) &=\partial_{y} G(x, u)\{1+2 G(x, u)\}. \end{aligned}$   (4.19)

Therefore,

$\frac{\alpha}{p} \frac{F_{x}^{\prime}(\xi)}{H_{x}^{\prime}(\xi)}=\frac{D^{\prime}(\xi)}{D(\xi)\{1+2 G(x, \xi)\}} \leq \frac{D^{\prime}(x)}{D(x)\{1+2 G(x, x)\}}=\frac{D^{\prime}(x)}{D(x)}$.   (4.20)

The last inequality follows from assumption (A1). This completes the proof.

Now, set for $0 \leq u \leq y \leq H$,

$V_{y}(u)=-\alpha e^{\alpha(y-u)}\left\{e^{\alpha(y-u)} D(y)-D(u)-\alpha \int_{u}^{y} e^{\alpha(t-u)} D(t) d t\right\}$,   (4.21)

$W_{y}(u)=-\{1+G(u, y)\}\left\{e^{\alpha(y-u)}-1-G(u, y)\right\}$.   (4.22)

The next assumption is needed for $\mathscr{L}_{y} R \geq 0$ of Hypothesis 2 to hold.

$\text { (A2) } V_{y}^{\prime} / W_{y}^{\prime}$ is non-decreasing.

Assumption (A2) is technical and is needed to complete the result of the paper. This assumption may seem complicated but, it is not difficult to check it numerically using MATLAB or Mathematica, say, once the demand rate function is known. Moreover, it can be shown that as $\alpha \rightarrow 0$, reduces to the condition that the function

$\mathscr{F}=\frac{D^{\prime}(u)}{1-(D(u) / p)}$   (4.23)

is non-decreasing. This property is satisfied by linear and exponential demand rate functions. In fact, assumption (A1) is also, in this case, satisfied when $D$ is linear or exponential.

Lemma 4.5. If assumption (A2) is satisfied, then $\mathscr{L}_{y} R \geq 0$.

Proof. Recall that

$\mathscr{L}_{y} R=\partial_{y}^{2} R+\partial_{x} \partial_{y} R+f(y) \partial_{y} R$.   (4.24)

Direct and tedious computation leads to

$\begin{aligned} \mathscr{L}_{y} R(x, y)=& \frac{1}{\alpha} \frac{\partial_{y} G(x, y)}{\{1+G(x, y)\}^{2}} \\ & \times\left[\alpha \int_{x}^{y} e^{\alpha(t-x)} D(t) d t-\left\{e^{\alpha(y-x)} D(y)-D(x)\right\}\right] \\ &-\frac{1}{\alpha} D^{\prime}(y)+\frac{1}{\alpha} \frac{e^{\alpha(y-x)} D^{\prime}(y)}{\{1+G(x, y)\}} \\ &+\frac{1}{\alpha} f(y) D(y)\left\{\frac{e^{\alpha(y-x)}}{1+G(x, y)}-1\right\}. \end{aligned}$   (4.25)

Recall the definition of the function $f$ in (4.15). Then $\mathscr{L}_{y} R \geq 0$ is equivalent to

$\begin{gathered} \frac{\partial_{y} G(x, y)}{1+G(x, y)}\left\{e^{\alpha(y-x)} D(y)-D(x)-\alpha \int_{x}^{y} e^{\alpha(t-x)} D(t) d t\right\} \\ \leq \frac{D^{\prime}(y) D(y)}{p\{1-(D(y) / p)\}}\left\{e^{\alpha(y-x)}-1-G(x, y)\right\}, \end{gathered}$   (4/26)

or, by $\left(\partial_{y} G\right)$ and (4.8), we get that the requirement $\mathscr{L}_{y} R \geq 0$ leads to

$\alpha \frac{e^{\alpha(y-x)}\left\{e^{\alpha(y-x)} D(y)-D(x)-\alpha \int_{x}^{y} e^{\alpha(t-x)} D(t) d t\right\}}{\{1+G(x, y)\}\left\{e^{\alpha(y-x)}-1-G(x, y)\right\}} \leq \frac{D^{\prime}(y)}{\{1-(D(y) / p)\}}$.   (4.27)

The left hand side of (4.27) is equal to $\left(V_{y}(y)-V_{y}(x)\right) /\left(W_{y}(y)-W_{y}(x)\right)$, where $V_{y}$ and $W_{y}$ are given by (4.21), and (4.22) respectively.

Computations show that

$\begin{aligned} V_{y}^{\prime}(u)=& 2 \alpha^{2} e^{\alpha(y-u)}\left\{e^{\alpha(y-u)} D(y)-D(u)-\alpha \int_{u}^{y} e^{\alpha(t-u)} D(t) d t\right\} \\ &+\alpha e^{\alpha(y-u)} D^{\prime}(y) \\ W_{y}^{\prime}(u)=&-\partial_{x} G(u, y)\left\{e^{\alpha(y-u)}-1-G(u, y)\right\} \\ &-\{1+G(u, y)\}\left\{-\alpha e^{\alpha(y-u)}-\partial_{x} G(u, y)\right\}. \end{aligned}$   (4.28)

Now, the extended-mean value theorem gives that

$\frac{V_{y}(y)-V_{y}(x)}{W_{y}(y)-W_{y}(x)}=\frac{V_{y}^{\prime}(\xi)}{W_{y}^{\prime}(\xi)}, \quad$ for some $x \text { < } \xi \text { < } y$.   (4.29)

But assumption (A2) implies that $\left(V_{y}^{\prime}(\xi) / W_{y}^{\prime}(\xi)\right) \leq\left(V_{y}^{\prime}(y) / W_{y}^{\prime}(y)\right)$, where the right hand side of the above inequality is equal to

$\frac{\alpha D^{\prime}(y)}{-\{-\alpha+\alpha(D(y) / p)\}}=\frac{D^{\prime}(y)}{\{1-\alpha(D(y) / p)\}}$.   (4.30)

This is the right hand side of (4.27). Hence, $\mathscr{L}_{y} R \geq 0$.

As a consequence of Lemmas 4.4 and 4.5 and Theorem 3.1, we have the following result.

Theorem 4.6. Under the requirements that assumptions (A1) and (A2) hold the function $S_{n}=\sum_{i=1}^{N} R\left(t_{i-1}, t_{i}\right)$ with $t_{0}=0 \text { < } t_{1} \text { < } \cdots \text { < } t_{n}$, has a unique minimum, this minimum can be found using the iterative procedure.

Let $s_{n}$ be the minimal value of $S_{n}$, then the next theorem follows from Theorem 3.2.

Theorem 4.7. The function $s_{n}$ is convex in $n$.

As a consequence of Theorem 4.7, the search for the optimal inventory policy can be conducted in two grids: the integer grid and $\mathbb{R}_{+}^{n}$. That is, for fixed integer $n$, the corresponding optimal times are found from the solution of the system of nonlinear equations (3.4) with corresponding objective value $s_{n}$. Then, the optimal value of $n$ can be obtained using the following corollary.

Corollary 4.8. The optimal number of production period $n*$ is such that

(1) if $K \text { > } s_{1}-s_{2}$, $n^{*}=1$

(2) if there exists an $N \geq 2$ such that $s_{N-1}-s_{N} \text { > } K \text { > } s_{N}-s_{N+1}$, then $n^{*}=N$,

(3) if there exists an $N \geq 1$ such that $K=s_{N}-s_{N+1}$, then $n^{*}=N$ and $n^{*}=N+1$.

This paper was concerned with finding the economic-production-lot-size policy for an inventory model with deteriorating items. An optimal inventory policy was proposed for a class of cost functions named OHD models. The proposed optimality approach was based on an earlier work in Benkherouf and Gilding. The extension to OHD models should not pose any difficulty. Indeed, note that by comparing (2.5) and (2.6), the OHP and OHD models differ in the expression

$-c_{1} \int_{t_{i-1}}^{t_{i}} D(u) d u$   (5.1)

Now, consider the optimization problem (2.22) with $R$ given by (2.6). It is clear that adding $-c_{1} \int_{0}^{H} D(u) d u=-c_{1} \sum_{i=1}^{n} \int_{t_{i-1}}^{t_{i}} D(u) d u$ will have no effect on the optimization problem. Consequently, the results obtained for the OHP model apply to the OHD model.

Before we close, we revisit paper Teng et al. and note that the model in Teng et al. allows for the purchasing cost to vary with time, and therefore with fixed unit cost and no deterioration, the model in Teng et al. is a special case of the model of the present paper. The reduction (2.18) in the present paper allows a direct approach as though the problem on hand is an unconstrained optimization problem. The approach adopted in Teng et al. is the standard approach for constrained nonlinear programming problem. The key result in Teng et al. is Theorem  1 which adapted to the model of this paper with $\theta=0$ requires that $c_{h}(1-(D(t) / p)) \text { > } 0$ to hold. This is satisfied since $D(t) \text { < } p$. Theorem  1 in Teng et al. states, with no conditions imposed on $D$, that for fixed $n$ the optimal inventory policy is uniquely determined as a solution of the first order condition of the optimization problem on hand. A result similar to Theorem 3.2 related to convexity of the corresponding objective value with respect to $n$ is also presented. The following counterexample shows that Theorem  1 in Teng et al. cannot be entirely correct in its present form. Indeed, for simplicity let $n=2$ then the problem treated in Teng et al. reduces (equivalently) to minimizing (2.22) with $R$ given by (2.21). The objective function in this case is a function of a single variable. Take $D(t)=2 \sin (10 t)+2 \cos (10 t)+4, c_{h}=1, \text { and } H \simeq 4.27$, and ignore the setup cost. Figure 2 shows the plot of the objective function. It is clear that multiple critical points can be observed as well as multiple optima. The remark on Teng et al also applies to part of Balkhi.

Figure 2 Behaviour of the objective function when $D(t)=2 \sin (10 t)+2 \cos (10 t)+4$.

It is worth noting that the keys to success in applying the approach in Benkherouf and Gilding are the separability of the cost functions between periods and Hypotheses 1 and 2. With this in mind, we believe that the approach of this paper to models with shortages and possibly with costs that are a function of time are possible. The technical requirement needed to generalize the results will be slightly more involved but essentially similar.