BUS606: Operations and Supply Chain Management

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 $n^{*}=N+1$.