Managing premium wines using an $(s - 1,s)$ inventory policy: a heuristic solution approach

Abstract

Operations research models are increasingly being used to support decision making in the wine industry. However, they have not yet been used to support inventory management decisions. In this paper, we develop a heuristic procedure for managing the stock of premium wines motivated by the operations of a small export-focused winery we worked with. Following an $(s-1,s)$ inventory policy, we assume that the decision maker aims to minimize the steady-state expected values of work in process, overage, and underage costs. The developed heuristic is as follows. First, we approximate the dynamics of the labeling process by a group scheduling policy to obtain the mean delays for each labeled product. Then, we address the problem of setting the inventory positions for the whole product portfolio by solving one newsvendor-type problem for each end-product. We provide some theoretical insights, a numerical example, and we analyze the accuracy of our procedure.

Appendix A

Proof of Theorem 4, part 1

Let consider the concept of stochastic ordering. See Ross et al. (1996, Chapter 9), for further reference. It is said that a random variable X is stochastically larger than a random variable Y, denoted by $X\ge _{st}Y$ if (20) holds.

$$\begin{aligned} P\{X\ge x\} \ge P\{Y\ge x\}, \quad \forall x \in \text {Support}\left( X\right) \cap \text {Support}\left( Y\right) \end{aligned}$$

(20)

For the newsvendor problem, Song (1994) proves that a stochastically larger demand implies a non-lower optimal stock level. Moreover, Ross et al. (1996) show that a Poisson random variable is stochastically increasing in its mean. Therefore, if $\lambda _{i}\tau _i < \lambda _{j}\tau _j$, then, $X_{j} \ge _{st} X_{i}$, and thus $s_{i}^{*} \le s_{j}^{*}$. $\square$

Proof of Theorem 4, part 2

First, we need to state some properties of the newsvendor problem under Poisson demand (Lemmas 1, 2 and 3). Let $h^*(\lambda )$: ${\mathbb {R}}^+$$\rightarrow$${\mathbb {R}}^+$ be defined as,

$$\begin{aligned} h^*(\lambda ) = \min _{s \in {\mathbb {Z}}^+} f(s,\lambda ) = \min _{s \in {\mathbb {Z}}^+} C_{I}\sum _{x=0}^{s}(s-x)p(x,\lambda ) + C_{B}\sum _{x=s}^{\infty }(x-s)p(x,\lambda ) \end{aligned}$$

(21)

where $p(x,\lambda )=e^{-\lambda }\frac{{\lambda ^x }}{{x!}}$. Hence, $h^*(\lambda )$ is the minimum expected cost, in the newsvendor problem context, if the demand follows a Poisson distribution with parameter $\lambda$. We first restate the following lemma by Jain et al. (2012). $\square$

Lemma 1

(Jain et al. (2012)) Let $s^{*}(\lambda )=\arg \min _{s \in {\mathbb {Z}}^+}f(s,\lambda )$. Then, for each compact and not empty subset of ${\mathbb {R}}^+$ of the form ${\mathbb {S}}({\bar{\lambda }}) = \{x \in {\mathbb {R}}^+ : 0 \le x \le {\bar{\lambda }}\}$, with ${\bar{\lambda }} > -\ln (\epsilon )$, there exists a partition of ${\mathbb {S}}\left( \bar{\lambda } \right) = \displaystyle \cup _{l=0}^{{\bar{s}}-1}[\lambda _{l-1},\lambda _{l}[ \ \cup \ [\lambda _{{\bar{s}} -1},\lambda _{{\bar{s}}}]$, with $\lambda _{-1} = 0$ and $\lambda _{{\bar{s}}} = {\bar{\lambda }}$, such that, $s^{*}(\lambda )=l$ on $[\lambda _{l-1},\lambda _{l}[$, $\forall \ 0 \le l <{\bar{s}}$, $s^{*}(\lambda ) = {\bar{s}}$ on $[\lambda _{{\bar{s}}-1},\lambda _{{\bar{s}}}]$, where $\lambda _{l}$ is the unique root of the transcendental equation,

$$\begin{aligned} t_{l}(\lambda _{l}) = \sum _{i=0}^{l}e^{ - \lambda _{l}}\frac{{\lambda _{l}^i }}{{i!}}-\epsilon =0 \end{aligned}$$

(22)

over $[0,{\bar{\lambda }}]$, with $0< \epsilon <1$.

Proof of Lemma 1:

The existence of parameters $\left\{ \lambda _{l}\right\} _{l=0}^{{\bar{s}} -1}$, and thus the partition of ${\mathbb {S}}\left( \bar{\lambda } \right)$, was proven by Jain et al. (2012). In their proof, replace $\lambda$ by 1, and suppose ${\bar{\lambda }} = T$. On the other hand, uniqueness of these parameters can be proved by contradiction. For $l = \{0..{\bar{s}} -1\}$, suppose that $\xi _{1}^{l}$ and $\xi _{2}^{l}$ are two roots of $t_{l}(\xi )$ in $[0,{\bar{\lambda }}]$, with $\xi _{1}^{l} \ne \xi _{2}^{l}$, and by construction, both are greater than 0. In this context, note that $t_{l}(\xi )$ is differentiable and hence, continuous on $[\xi _{1}^{l}, \xi _{2}^{l}]$. Thus, by the Mean Value Theorem, we can find a number $\nu$ in $(\xi _{1}^{l}, \xi _{2}^{l})$ such that (23) holds.

$$\begin{aligned} \frac{t_{l}(\xi _{2}^{l}) - t_{l}(\xi _{1}^{l})}{\xi _{2}^{l} - \xi _{1}^{l}}=t_{l}^{\prime }(\nu ) \end{aligned}$$

(23)

Since $t_{l}(\xi _{2}^{l}) = t_{l}(\xi _{1}^{l}) =0$, because they are roots of $t_{l}(\xi )$, we get,

$$\begin{aligned} 0=t_{l}^{\prime }(\nu ) \end{aligned}$$

(24)

Recall that $t_{l}^{\prime }(\xi ) = -e^{-\xi }\frac{\xi ^{l}}{l!}$ is negative for all $\xi > 0$. So, there is not such $\nu$ in $(\xi _{1}^{l}, \xi _{2}^{l})$, for which $0=t_{l}^{\prime }(\nu )$, implying that $\xi _{1}^{l} = \xi _{2}^{l}$, and thus the roots of $t_{l}\left( \lambda \right)$ are unique in $\lambda$ for each valid value of l. $\square$

Due to Lemma 1, $s^{*}(\lambda )$ is an increasing step function in $\lambda$. Besides, close form expressions are available only for $\lambda _{0} = -\ln (\epsilon )$ and $\lambda _{1}=-1-W_{-1}(-\frac{\epsilon }{e})$, where $W_{-1}(x)$ is the lower branch of the Lambert W function. We now prove continuity, convexity and increasing behaviours of $h^*\left( \lambda \right)$ in $\lambda$.

Lemma 2

For any interval of ${\mathbb {S}}({\bar{\lambda }})$, $h^*(\lambda )$ is continuous, convex and increasing in $\lambda$.

Proof of Lemma 2

Convexity was proved by Rossi et al. (2014) (see their Appendix B). The increasing behavior was proved by Jain et al. (2012, Lemma 2). For the sake of the explanation, we briefly describe both proofs.

First, note that $p(x,\lambda )$ is a continuous function of $\lambda$. Thus $f(s,\lambda )$ is also continuous in $\lambda$.

Second, note that $f(s,\lambda )$ can be expressed as $C_{I}E[s - X]^{+} + C_{B}E[X - s]^{+}$, where $x^{+} = max\{x,0\}$, E[X] denotes expectation, and $X\sim \text {Poisson}\left( \lambda \right)$. Recall that $(X - s)^{+} = (s - X)^{+} + (X - s)$. Due to $E(s - X)^{+}$ can be written as $\sum _{j=0}^{s-1}P(X \le j)$, then $f(s,\lambda )$ can be re-expressed as (25).

$$\begin{aligned} f(s,\lambda )= & {} (C_{I}+C_{B})\sum _{j=0}^{s-1}\sum _{x=0}^{j}p(x,\lambda ) + C_{B}(\lambda - s) \end{aligned}$$

(25)

$$\begin{aligned} \frac{\partial f\left( s,\lambda \right) }{\partial \lambda }= & {} -(C_{I}+C_{B})\sum _{j=0}^{s-1}p(j,\lambda ) + C_{B} \end{aligned}$$

(26)

$$\begin{aligned} \frac{\partial ^{2}f\left( s,\lambda \right) }{\partial \lambda ^{2}}= & {} (C_{I}+C_{B})p(s-1,\lambda ) \end{aligned}$$

(27)

Note that $\frac{\partial ^{2}f(s,\lambda )}{\partial \lambda ^{2}} \ge 0, \ \forall \lambda >0$, and given $s \in {\mathbb {Z}}^{+}$, so $f(s,\lambda )$ is convex in $\lambda$.

And third, let $\lambda ^{*} = \arg \min _{\lambda >0} f\left( s,\lambda \right)$ be the unique root of $\sum _{x=0}^{s-1}e^{ - \lambda ^{*}}\frac{{(\lambda ^{*})^x }}{{x!}}=\epsilon$. Hence, $f(s,\lambda ) \ge f(s, \lambda ^{*})$, $\forall \lambda \ge \lambda ^{*}$. Now consider the interval $[\lambda _{l-1},\lambda _{l}[$. If $\lambda \in [\lambda _{l-1},\lambda _{l}[$, then by Lemma 1, $s^{*}(\lambda )$=$\hbox {argmin}_{s \in {\mathbb {Z}}^+}f(s,\lambda ) = l$. But recall that if $s^{*}(\lambda )=l$, $f(l,\lambda )$ reach its minimum at $\lambda _{l-1}$. Thus, $f(l,\lambda ) \ge f(l,\lambda _{l-1})$. Therefore, $h^*(\lambda ) = f(l,\lambda )$ is increasing in $[\lambda _{l-1},\lambda _{l}[, \ \forall l=0..{\bar{s}}-1$, and also for $[\lambda _{{\bar{s}} -1},\lambda _{{\bar{s}}}]$. $\square$

Finally, let us consider Lemma 3.

Lemma 3

For any interval $[\lambda _{l-1},\lambda _{l}[$, $0 \le l <{\bar{s}}$, it holds that $h^*(\lambda _{l}) = f(l,\lambda _{l}) = f(l+1,\lambda _{l})$, which implies $s^{*}(\lambda _{l})$=$\hbox {argmin}_{s \in {\mathbb {Z}}^+}f(s,\lambda _{l})$={l, l+1}.

Proof of Lemma 3:

We need to prove that,

$$\begin{aligned} f(l,\lambda _{l}) = f(l+1,\lambda _{l}) \end{aligned}$$

(28)

Consider (25),

$$\begin{aligned}&(C_{I}+C_{B})\sum _{j=0}^{l-1}\sum _{x=0}^{j}p(x,\lambda _l) + C_{B}(\lambda _{l} - l) \nonumber \\&\quad = (C_{I}+C_{B})\sum _{j=0}^{l}\sum _{x=0}^{j}p(x,\lambda _l) + C_{B}(\lambda _{l} - l -1) \end{aligned}$$

(29)

Rearranging terms, it is clear that,

$$\begin{aligned} C_{B} = (C_{I} + C_{B})\sum _{x=0}^{j}p(x,\lambda _l) \end{aligned}$$

(30)

Which is equivalent to,

$$\begin{aligned} \epsilon = \sum _{x=0}^{j}p(x,\lambda _l) \end{aligned}$$

(31)

and this last equality holds due to Lemma 1.

Due to Lemmas 2 and 3, $h^*(\lambda )$ is both continuous and increasing in ${\mathbb {S}}({\bar{\lambda }})$. Now suppose that $0< \lambda _{i} < \lambda _{j} \le {\bar{\lambda }}$. Then $h^*(\lambda _{i}) < h^*(\lambda _{j})$. Let $\lambda _{i} = \lambda _{i}\tau _i$ and $\lambda _{j} = \lambda _{j}\tau _j$. Thus the following holds: $h^*(\lambda _{i}\tau _i) = f_{i}(s_{i}^{*}) < f_{j}(s_{j}^{*}) = h^*(\lambda _{j}\tau _j)$, which proves the second part of Theorem 4. $\square$