Critical level rationing in inventory systems with continuously distributed demand

Abstract

This paper analyzes the use of a constant critical level policy for fast-moving items, where rationing is used to provide differentiated service levels to two demand classes (high priority and low priority). The previous work on critical level models, with either a continuous or periodic review policy, has only considered slow-moving items with Poisson demand. In this work, we consider a continuous review (Q, r, C) policy with two demand classes that are modeled through continuous distributions, and the service levels are measured by the probability of satisfying the entire demand of each class during the lead time. We formulate a service level problem as an non-linear problem with chance constraints for which we optimally solve a relaxation obtaining a closed-form solution that can be computed easily. For instances, we tested, computational results show that our solution approach provides good-quality solutions that are on average $0.3~\%$ from the optimal solution.

Appendices

Appendix 1: Proof of proposition 2

Lemma 1

Let X and Y be two univariate continuous random variables, where Y has positive support. Then, for any C, we have

$$\begin{aligned} \mathbb {P}(X + Y> C )\ge \mathbb {P}(X > C). \end{aligned}$$

Proof

We note that the set of realizations $\{\omega ~|~ X(\omega )> C\}\subset \{\omega ~|~ X(\omega )+Y(\omega ) > C\}$, which gives the inequality $\mathbb {P}(X+Y> C) \ge \mathbb {P}(X> C)$. $\square$

Given Lemma 1, the demonstration of Proposition 2 is:

Proof

Let $\tau _{R,D}^{r-C}=\min \{\tau _{H,D}^{r-C},L\}$ be the time to rationing, which corresponds to the amount of time that elapses from the moment an order is placed until the critical level C is reached if this event occurs during the lead time. If the hitting time $\tau _{H,D}^{r-C}$ does not occur during lead time, then the time to rationing is defined as $\tau _{R,D}^{r-C}=L$. In this case, rationing coincides with the reception of the replenishment batch, and therefore, to be precise, rationing is not produced.

Given a $k>0$ we have that, for every demand realization $\omega$, the hitting time satisfies $\tau _{H,D}^{r-C}(\omega ) < \tau _{H,D}^{r+k-C}(\omega )$. This is because exactly k additional units of demand are necessary to reach C, and the demand is a strictly increasing non-negative demand. This implies that for any $k>0$ we have that

$$\begin{aligned} \tau _{R,D}^{r-C}(\omega )< \tau _{R,D}^{r+k-C}(\omega )&\forall \omega \mathrm{~s.t.~} \tau _{H,D}^{r-C}(\omega ) < L, \\ L = \tau _{R,D}^{r-C}(\omega ) = \tau _{R,D}^{r+k-C}(\omega )&\forall \omega \mathrm{~s.t.~} \tau _{H,D}^{r-C}(\omega ) \ge L \ . \end{aligned}$$

From these relations, we have that $\tau _{R,D}^{r+k-C}-\tau _{R,D}^{r-C}\ge 0$ with probability 1, which combined with the assumptions on the demand gives us $D_1(L-\tau _{R,D}^{r-C}) = D_1(L-\tau _{R,D}^{r+k-C})+ D_1(\tau _{R,D}^{r+k-C}-\tau _{R,D}^{r-C})$, where the last term is a positive support random variable when $\tau _{H,D}^{r-C} < L$. We, therefore, have

$$\begin{aligned} \alpha _1(r,C)= & {} \mathbb {P}(D_1(L-\tau _{R,D}^{r-C}) \le C ) \\= & {} \mathbb {P}(D_1(L-\tau _{R,D}^{r-C}) \le C ~|~ \tau _{H,D}^{r-C} \ge L)\mathbb {P}( \tau _{H,D}^{r-C} \ge L) \\&+\, \mathbb {P}(D_1(L-\tau _{R,D}^{r-C}) \le C ~|~ \tau _{H,D}^{r-C}< L)\mathbb {P}( \tau _{H,D}^{r-C}< L) \\= & {} \mathbb {P}(D_1(L-\tau _{R,D}^{r+k-C}) \le C ~|~ \tau _{H,D}^{r-C} \ge L)\mathbb {P}( \tau _{H,D}^{r-C} \ge L) \\&+\, \mathbb {P}(D_1(L-\tau _{R,D}^{r+k-C})\\&+\, D_1(\tau _{R,D}^{r+k-C}-\tau _{R,D}^{r-C}) \le C ~|~ \tau _{H,D}^{r-C}< L)\mathbb {P}( \tau _{H,D}^{r-C}< L) \\\le & {} \mathbb {P}(D_1(L-\tau _{R,D}^{r+k-C}) \le C ~|~ \tau _{H,D}^{r-C} \ge L)\mathbb {P}( \tau _{H,D}^{r-C} \ge L) \\&+\, \mathbb {P}(D_1(L-\tau _{R,D}^{r+k-C})\le C ~|~ \tau _{H,D}^{r-C}< L)\mathbb {P}( \tau _{H,D}^{r-C} < L) \\= & {} \mathbb {P}(D_1(L-\tau _{R,D}^{r+k-C}) \le C ) = \alpha _1(r+k,C). \end{aligned}$$

Here, the inequality uses Lemma 1 with $X= D_1(L-\tau _{R,D}^{r+k-C})$ and $Y=D_1(\tau _{R,D}^{r+k-C}-\tau _{R,D}^{r-C})$.

We repeat the argument to show the tendency of $\alpha _1(r,C)$ with respect to C. Given any $k>0$, we have that $\tau _{H,D}^{r-(C+k)}(\omega ) < \tau _{H,D}^{r-C}(\omega )$ for any demand realization $\omega$. Similarly, for any $k>0$, we now have

$$\begin{aligned} \tau _{R,D}^{r-(C+k)}(\omega )< \tau _{R,D}^{r-C}(\omega )&\forall \omega \mathrm{~s.t.~} \tau _{H,D}^{r-(C+k)}(\omega ) < L \\ L = \tau _{R,D}^{r-(C+k)}(\omega ) = \tau _{R,D}^{r-C}(\omega )&\forall \omega \mathrm{~s.t.~} \tau _{H,D}^{r-(C+k)}(\omega ) \ge L \ . \end{aligned}$$

The demand can be now separated $D_1(L-\tau _{R,D}^{r-(C+k)}) = D_1(L-\tau _{R,D}^{r-C})+D_1(\tau _{R,D}^{r-C}-\tau _{R,D}^{r-(C+k)})$, where for every demand realization $\omega$, this last term satisfies $D_1(\tau _{R,D}^{r-C}-\tau _{R,D}^{r-(C+k)})(\omega ) \le k$. This because $\tau _{R,D}^{r-C}(\omega )-\tau _{R,D}^{r-(C+k)}(\omega ) \le \tau _{H,D}^{r-C}(\omega )-\tau _{H,D}^{r-(C+k)}(\omega )$ and $D_1(\tau _{H,D}^{r-C}-\tau _{H,D}^{r-(C+k)})\le D(\tau _{H,D}^{r-C}-\tau _{H,D}^{r-(C+k)})=k$ by definition of hitting time. This gives

$$\begin{aligned} \alpha _1(r,C)= & {} \mathbb {P}(D_1(L-\tau _{R,D}^{r-C}) \le C ) \\= & {} \mathbb {P}(D_1(L-\tau _{R,D}^{r-C}) \le C ~|~ \tau _{H,D}^{r-(C+k)} \ge L)\mathbb {P}( \tau _{H,D}^{r-(C+k)} \ge L) \\&~~+ \ \mathbb {P}(D_1(L-\tau _{R,D}^{r-C}) \le C ~|~ \tau _{H,D}^{r-(C+k)}< L)\mathbb {P}( \tau _{H,D}^{r-(C+k)}< L) \\= & {} \mathbb {P}(D_1(L-\tau _{R,D}^{r-(C+k)}) \le C+k ~|~ \tau _{H,D}^{r-(C+k)} \ge L)\mathbb {P}( \tau _{H,D}^{r-(C+k)} \ge L) \\&~~+\ \mathbb {P}(D_1(L-\tau _{R,D}^{r-(C+k)})\\\le & {} C + D_1(\tau _{R,D}^{r-C}-\tau _{R,D}^{r-(C+k)}) ~|~ \tau _{H,D}^{r-(C+k)}< L)\mathbb {P}(\tau _{H,D}^{r-(C+k)}< L) \\\le & {} \mathbb {P}(D_1(L-\tau _{R,D}^{r-(C+k)}) \\\le & {} C+k ~|~ \tau _{H,D}^{r-(C+k)} \ge L)\mathbb {P}(\tau _{H,D}^{r-(C+k)} \ge L) \\&~~+\ \mathbb {P}(D_1(L-\tau _{R,D}^{r-(C+k)})\le C+k ~|~ \tau _{H,D}^{r-(C+k)}< L)\mathbb {P}(\tau _{H,D}^{r-(C+k)} < L) \\= & {} \mathbb {P}(D_1(L-\tau _{R,D}^{r-(C+k)}) \le C ) = \alpha _1(r,C+k) \ . \end{aligned}$$

Here, we add a k in the first term of the second equality, because $D_1(L-\tau _{R,D}^{r-(C+k)}) = D_1(0)$ when $\tau _{H,D}^{r-(C+k)} \ge L$, so that the first probability equals 1. The inequality comes from the fact that $\mathbb {P}(D_1(\tau _{R,D}^{r-C}-\tau _{R,D}^{r-(C+k)})\le k) = 1$. $\square$

Appendix 2: Partial derivative of $\alpha _1(r,C)$ with respect to C

Here, we give the expression of $\frac{\partial {\alpha _1(r,C)}}{\partial {C}}$ in the case when the demands for both classes are normally distributed and the density function of the hitting time $\tau _{H,D}^{r-C}$ is given by Eq. (19). We denote by $\bar{\varphi }_{\mu ,\sigma ^2}(x)$ the density function of a normal random variable with mean $\mu$ and variance $\sigma ^2$. The partial derivative then can be expressed as

$$\begin{aligned}&\frac{\partial {\alpha _1(r,C)}}{\partial {C}} \nonumber \\&= \int _0^L \left( \frac{r-C+\mu \tau }{2\tau } - \frac{C+\mu _1(L-\tau )}{2(L-\tau )}\right) \bar{\varphi }_{\mu _1(L-\tau ),\sigma _1^2(L-\tau )}(C) \bar{\varphi }_{\mu \tau ,\sigma ^2\tau }(r-C)\mathrm{d}\tau .\nonumber \\ \end{aligned}$$

(32)