Analysis of inventory policies for perishable items with fixed leadtimes and lifetimes

Abstract

This paper deals with a continuous review inventory system with perishable items and Poisson demand. Lifetimes and leadtimes are assumed to be fixed. First, a quite general base-stock model is developed where a number of combinations between backorder- and lost sales policies are evaluated and optimized. The solution technique for all these combinations is exact. Secondly, we consider the case with non-negligible ordering costs and assume that the inventory policy employed is the commonly used ($R,Q$) policy. We develop a new heuristic approach for evaluating and analyzing the proposed $(R,Q)$ model and compare our results with those obtained by related papers. This heuristic approach uses the base-stock model developed as a building block. The results reveal that our approach works reasonably well in all cases considered.

Appendix 1

1.1 Proof of Proposition 1

Let us compare the state of the process at time points $t$ and $t+\delta$, where $\delta$ is a small positive number. In the light of (1), we consider two different time intervals. Consider first the case when $0\le T_1<\tau$ and $T_{S}\not =0$. Now, if there is no demand during $[t,t+\delta )$, all units in the system will age $\delta$ time units. As a result, we have

$$\begin{aligned} \begin{aligned} f_{T_1,T_2,\ldots ,T_S}(t+\delta ,t_1,t_2,\ldots ,t_S)&= f_{T_1,T_2,\ldots ,T_S}(t,t_1-\delta ,t_2-\delta ,\ldots ,t_S-\delta )\\&\quad \quad \cdot (1-\lambda \delta +o(\delta ))+o(\delta ). \end{aligned} \end{aligned}$$

(25)

Note that (25) can be reformulated as a sum of partial derivatives of $t, t_1,\ldots ,t_S$ by using the standard trick of telescoping sums, and by taking the limit as $\delta \rightarrow 0$. That is, by adding and subtracting terms we get

$$\begin{aligned}&\frac{f_{T_1,T_2,\ldots ,T_S}(t+\delta ,t_1,t_2,\ldots ,t_S)-f_{T_1,T_2,\ldots ,T_S}(t,t_1,t_2,\ldots ,t_S)}{\delta } \nonumber \\&\quad \quad + \frac{1}{\delta }\sum _{k=1}^{S} \{f_{T_1,T_2,\ldots ,T_{S}}(t,t_1-\delta ,\ldots ,t_{k-1}-\delta ,t_k,\ldots ,t_{S}) \nonumber \\&\quad \quad -f_{T_1,T_2,\ldots ,T_{S}}(t,t_1-\delta ,\ldots ,t_{k}-\delta ,t_{k+1},\ldots ,t_{S})\} \nonumber \\&\quad = -\lambda \cdot f_{T_1,T_2,\ldots ,T_{S}}(t,t_1-\delta ,\ldots ,t_{S}-\delta ). \end{aligned}$$

(26)

As mentioned, by taking the limit as $\delta \rightarrow 0$, we obtain

$$\begin{aligned}&\frac{\partial }{\partial t} f_{T_1,T_2,\ldots ,T_{S}}(t,t_1,\ldots ,t_{S})+\sum _{k=1}^{S} \frac{\partial }{\partial t_k}f_{T_1,T_2,\ldots ,T_{S}}(t,t_1,\ldots ,t_{S})\nonumber \\&\quad \quad = -\lambda \cdot f_{T_1,T_2,\ldots ,T_{S}}(t,t_1,\ldots ,t_{S}) \end{aligned}$$

(27)

In order to obtain the limiting distribution of $f_{T_1,T_2,\ldots ,T_{S_i}}(t,t_1,\ldots ,t_{S})$, take the limit as $t\rightarrow \infty$. This implies that

$$\begin{aligned} \sum _{k=1}^{S} \frac{\partial }{\partial t_k}f_{T_1,T_2,\ldots ,T_{S}}(t_1,\ldots ,t_{S})= -\lambda \cdot f_{T_1,T_2,\ldots ,T_{S}}(t_1,\ldots ,t_{S}). \end{aligned}$$

(28)

Similarly as (28) we obtain

$$\begin{aligned} \sum _{k=1}^{S} \frac{\partial }{\partial t_k}f_{T_1,T_2,\ldots ,T_{S}}(t_1,\ldots ,t_{S})= -\mu \cdot f_{T_1,T_2,\ldots ,T_{S}}(t_1,\ldots ,t_{S}) \end{aligned}$$

(29)

for the case when $\tau \le t_1<T$ and $t_{S}\not =0$.

In order to derive the solutions of (28) and (29), note that (28) and (29) are first order linear partial differential equations. In general, it is relatively easy to solve such equations by using, e.g., the characteristic method, see e.g., Strauss (1992). By using this method, the solutions of (28) and (29) can in a compact way be stated as

$$\begin{aligned} f_{T_1,T_2,\ldots ,T_{S}}(t_1,t_2,\ldots ,t_{S}) = \varphi (t_1-t_2,t_2-t_3,\ldots ,t_{S-1}-t_{S}) \exp \left\{ -\int \limits _0^{t_1} \eta (t) dt\right\} , \end{aligned}$$

(30)

where $\varphi (t_1-t_2,t_2-t_3,\ldots ,t_{S-1}-t_{S})$ is a differentiable function determined by the boundary conditions describing the process when a demand occurs or when an item perishes. It turns out that $\varphi (t_1-t_2,t_2-t_3,\ldots ,t_{S-1}-t_{S})$, in this specific case, is a constant. To prove this fact, it is possible to use exactly the same proof as in Schmidt and Nahmias (1985) with small modifications. Hence, we refer to Schmidt and Nahmias (1985) for these details.

Using the fact that $\varphi (t_1-t_2,t_2-t_3,\ldots ,t_{S-1}-t_{S})$ is a just a constant, $C$, we can express (30) more explicitly as

$$\begin{aligned} f_{T_1,T_2,\ldots ,T_{S}}(t_1,t_2,\ldots ,t_{S}) = \left\{ \begin{array}{lll} C e^{-\lambda t_1} &{}\quad \hbox {for}\; 0\le t_1 < \tau \\ C e^{-(\lambda \tau + \mu (t_1-\tau ))} &{}\quad \hbox {for}\; \tau \le t_1 \le T. \end{array}\right. \end{aligned}$$

$\square$

1.2 Proof of Proposition 2

In order to calculate the average stock on hand we will use the well known Little’s law. Note that, if an item has been consumed by a customer demand, the average time the item has spent in stock is $\int _{L}^T (t_1-L)f_{T_1}(t_1)dt_1$. However, if an item has perished, the item has spent exactly $T-L$ units of time in stock. Hence, the average stock on hand is obtained as

$$\begin{aligned} E(IL)^+ = \mu \int \limits _{L}^T (t_1-L)f_{T_1}(t_1)dt_1 + (T-L)\pi . \end{aligned}$$

(31)

The integral in (31) can be expanded as

$$\begin{aligned} \int \limits _{L}^T (t_1-L)f_{T_1}(t_1)dt_1&= \int \limits _{L}^T (t_1-L) Ce^{-(\lambda \tau + \mu (t_1-\tau ))} \frac{t_1^{S-1}}{(S-1)!}dt_1 \nonumber \\&= e^{(\mu -\lambda )\tau }\left( \frac{C\cdot S}{\mu ^{S+1}} \int \limits _{L}^T \frac{e^{-\mu t_1} t_1^S \mu ^{S+1}}{S!}dt_1\right. \nonumber \\&\quad \left. - \frac{C\cdot L}{\mu ^{S}} \int \limits _{L}^T \frac{e^{-\mu t_1} t_1^{S-1}\mu ^{S}}{(S-1)!}dt_1\right) . \end{aligned}$$

(32)

By recognizing $P(X \le x)=\int _0^{x} e^{-\mu t_1}\mu ^k {t_1}^{k-1}/(k-1)!dt_1=1-\sum _{n=0}^{k-1} e^{-\mu x} (\mu x)^n/n!$ as the distribution function of a r.v. $X\in \hbox {Erlang}(\mu ,k)$, the integrals in (32) can be simplified as

$$\begin{aligned} \int \limits _{L}^T \frac{e^{-\mu t_1} t_1^S \mu ^{S+1}}{S!}dt_1&= \sum _{n=0}^{S} e^{-\mu L} \frac{(\mu L)^n}{n!} - \sum _{n=0}^{S} e^{-\mu T} \frac{(\mu T)^n}{n!}\\ \int \limits _{L}^T \frac{e^{-\mu t_1} t_1^{S-1}\mu ^{S}}{(S-1)!}dt_1&= \sum _{n=0}^{S-1} e^{-\mu L} \frac{(\mu L)^n}{n!} - \sum _{n=0}^{S-1} e^{-\mu T} \frac{(\mu T)^n}{n!} . \end{aligned}$$

To conclude, the average stock on hand is obtained as

$$\begin{aligned} E(IL)^+&= \mu \int \limits _{L}^T (t_1-L)f_{T_1}(t_1)dt_1 + (T-L)\pi \nonumber \\&= e^{\tau (\mu -\lambda )} \left[ \frac{C\cdot S}{\mu ^{S}} \left( \sum _{n=0}^S \frac{(\mu L)^n}{n!}e^{-\mu L}-\sum _{n=0}^S \frac{(\mu T)^n}{n!}e^{-\mu T}\right) \right. \nonumber \\&\quad - \left. \frac{C\cdot L}{\mu ^{S-1}} \left( \sum _{n=0}^{S-1} \frac{(\mu L)^n}{n!}e^{-\mu L}-\sum _{n=0}^{S-1} \frac{(\mu T)^n}{n!}e^{-\mu T}\right) \right] + (T-L)\pi . \end{aligned}$$

(33)

Let us proceed with the derivation of the average number of backorders when applying complete and partial backordering, respectively. Recall that for the complete backorder (CB) cost structures in (8) and (10), we have $\eta (t_1)=\lambda$ for $0\le t_1 < L$. Moreover, for the partial backorder (PB) cost structure in (9), we have $\eta (t_1)=0$ for $0\le t_1 < \tau$ (i.e., $\lambda =0$), and $\eta (t_1)=\mu$ for $\tau \le t_1 < L$. Hence, by using (4) and applying Little’s law we obtain

$$\begin{aligned} E(IL)_{CB}^-&= \lambda \int \limits _0^L (L-t_1)f_{T_1}(t_1)dt_1 = \lambda \int \limits _0^L (L-t_1) Ce^{-\lambda t_1} \frac{t_1^{S-1}}{(S-1)!}dt_1 ,\\ E(IL)_{PB}^-&= \mu \int \limits _\tau ^L (L-t_1)f_{T_1}(t_1)dt_1 = \mu \int \limits _\tau ^L (L-t_1) Ce^{- \mu (t_1-\tau )} \frac{t_1^{S-1}}{(S-1)!}dt_1 \end{aligned}$$

In view of the derivation of $E(IL)^+$ in (33), the rest of the proof is straightforward calculus and therefore omitted. $\square$

1.3 Proof of Proposition 3

It is clear that $k$ units in stock is equivalent to $0 \le T_{S} < T_{S-1} < \cdots < T_{k+1} < L \le T_k < \cdots < T_1 < T$. Hence, we have

$$\begin{aligned} P(IL=k)&= \int \limits _{L}^{\tau }\left( \int \limits _{L}^{t_1}\,\int \limits _{L}^{t_2}\, \cdots \, \int \limits _{L}^{t_{k-1}}\, \int \limits _{0}^{L}\,\int \limits _{0}^{t_{k+1}}\,\cdots \,\int \limits _{0}^{t_{S-1}}\,C \cdot e^{-\lambda t_1} dt_{S} dt_{S-1}\ldots dt_2 \right) dt_1 \nonumber \\&\quad + \int \limits _{\tau }^{T}\left( \int \limits _{L}^{t_1}\,\int \limits _{L}^{t_2}\,\cdots \,\int \limits _{L}^{t_{k-1}}\,\int \limits _{0}^{L}\,\int \limits _{0}^{t_{k+1}}\,\cdots \,\int \limits _{0}^{t_{S-1}}\,C \cdot \, e^{-(\lambda \tau + \mu (t_1-\tau ))} dt_{S} dt_{S-1}\ldots dt_2 \right) dt_1 \nonumber \\&= \frac{C\cdot L^{S-k}}{(S-k)!(k-1)!}\,\cdot \, \int \limits _L^{\tau } e^{-\lambda t_1} (t_1-L)^{k-1}dt_1 \nonumber \\&\quad +\, \frac{C\cdot L^{S-k}\, \cdot \, e^{(\mu -\lambda )\tau }}{(S-k)!(k-1)!}\,\cdot \, \int \limits _{\tau }^{T} e^{-\mu t_1} (t_1-L)^{k-1}dt_1 \end{aligned}$$

(34)

The rest of the proof is finished by simple calculus. The integral $\int _L^{\tau } e^{-\lambda t_1} (t_1-L)^{k-1}dt_1$ in (34) can be simplified as

$$\begin{aligned} \int \limits _L^{\tau } e^{-\lambda t_1} (t_1-L)^{k-1}dt_1&= e^{-\lambda L} \int \limits _0^{\tau -L} e^{-\lambda t_1} {t_1}^{k-1}dt_1 \nonumber \\&= \frac{e^{-\lambda L} (k-1)!}{\lambda ^k}\int \limits _0^{\tau -L} \frac{e^{-\lambda t_1}\lambda ^k {t_i}^{k-1}}{(k-1)!}dt_1 \nonumber \\&= \frac{e^{-\lambda L} (k-1)!}{\lambda ^k} \left( 1-e^{-\lambda (\tau -L)}\sum _{n=0}^{k-1} \frac{(\lambda (\tau -L))^n}{n!}\right) \end{aligned}$$

(35)

by recognizing $P(X\le \tau -L)=\int _0^{\tau -L} e^{-\lambda t_1}\lambda ^k {t_1}^{k-1}/(k-1)!dt_1$ as the distribution function of a r.v. $X\in \hbox {Erlang}(\lambda ,k)$. By similar calculations, the integral $\int _{\tau }^{T} e^{-\mu t_1} (t_1-L)^{k-1}dt_1$ in (34) can be stated as

$$\begin{aligned} \int \limits _{\tau }^{T} e^{-\mu t_1} (t_1-L)^{k-1}dt_1&= \frac{e^{-\mu L}(k-1)!}{\mu ^k}\cdot \nonumber \\&\left( e^{-\mu (\tau -L)}\sum _{n=0}^{k-1}\frac{(\mu (\tau -L))^n}{n!}-e^{-\mu (T-L)} \sum _{n=0}^{k-1} \frac{(\mu (T-L))^n}{n!}\right) .\nonumber \\ \end{aligned}$$

(36)

Inserting (35) and (36) in (34) yields the desired result.$\square$

1.4 Proof of Lemma 1

Proof of (i): By assuming that the inventory position is uniformly distributed over the integers $R+1,\ldots ,R+Q$ we have

$$\begin{aligned} \pi _{RQ}(R,Q)=\frac{1}{Q}\sum _{k=R+1}^{R+Q} \pi (k). \nonumber \end{aligned}$$

From Olsson and Tydesjö (2010) we know that $\pi (k)$ is increasing in $k$. Hence, keeping $R$ fixed we get

$$\begin{aligned} \pi _{RQ}(R,Q+1)-\pi _{RQ}(R,Q)&= \frac{1}{Q+1}\sum _{k=R+1}^{R+Q+1} \pi (k)-\frac{1}{Q}\sum _{k=R+1}^{R+Q} \pi (k) \\&= \frac{Q\sum _{k=R+1}^{R+Q+1}\pi (k) - (Q+1)\sum _{k=R+1}^{R+Q}\pi (k)}{Q(Q+1)} \\&= \frac{Q\pi (R+Q+1) - \sum _{k=R+1}^{R+Q}\pi (k)}{Q(Q+1)} \\&> \frac{Q[\pi (R+Q+1) - \pi (R+Q)]}{Q(Q+1)} > 0. \end{aligned}$$

Here we have used that $\pi (k)$ is increasing in $k$. Hence, $\pi _{RQ}(R,Q+1)>\pi _{RQ}(R,Q)$ for all positive integer values of $Q$. To show that $\pi _{RQ}$ is increasing in $R$ for a fixed $Q$ is straightforward and can be done in a similar way. $\square$

Proof of (ii): From (4) and (21), $\pi (k)$ can be stated as

$$\begin{aligned} \pi (k) = \frac{e^{-\lambda T} T^{k-1}}{\int _0^T t^{k-1} e^{-\lambda t} dt}. \end{aligned}$$

Since $\lambda >0$ we note that $e^{-\lambda t}< 1$ for $t>0$. This gives an upper bound on the integral,

$$\begin{aligned} \int \limits _0^T t^{k-1} e^{-\lambda t} dt \le \int \limits _0^T t^{k-1} dt = \frac{T^k}{k}. \end{aligned}$$

This means that

$$\begin{aligned} \pi (k) \ge e^{-\lambda T}\frac{k}{T} \end{aligned}$$

which results in

$$\begin{aligned} \pi _{RQ}(R,Q)&= \frac{1}{Q}\sum _{k=R+1}^{R+Q} \pi (k) \ge \frac{e^{-\lambda T}}{T}\cdot \frac{1}{Q} \sum _{k=R+1}^{R+Q} k = \frac{e^{-\lambda T}}{T}\cdot \frac{1}{Q} (2R+Q+1)Q \\&= \frac{e^{-\lambda T}}{T} (2R+Q+1) \rightarrow \infty \hbox { when } R \rightarrow \infty \hbox { or } Q \rightarrow \infty . \end{aligned}$$

$\square$