Service differentiation in a single-period inventory model with numerous customer classes

Abstract

We study critical-level inventory-management policies as means to provide differentiated ($\alpha$ and $\beta$) service levels to more than two classes of customers. First, we derive closed-form expressions for the service levels of a single-period critical-level policy with an arbitrary number of customer classes (with Poisson demand). Based on the service-level expressions, we derive additional structural insights and provide an efficient algorithm with which to compute the essential system parameters, that is, the minimum required starting inventory and the associated critical levels. Based on these results, we conduct numerical experiments and develop structural insights into the system’s behavior.

Appendices

Appendix 1: Additional proofs

All proofs from this paper have been gathered in the present appendix to improve legibility. Additionally, even in this appendix we have restricted the proofs to the main arguments. Standard calculations and repetitive arguments have been excluded.

Proof of Equation (1)

Disregarding the truncation at $t=1$, ${\bar{S}}_1$ follows an $\mathrm{Erl}(x_0+1,\lambda _1)$ distribution, hence

$$\begin{aligned} \mathbb {P}[{\bar{S}}_1 \ge 1] = 1 - F_{\mathrm{Erl}(x_0+1,\lambda _1)}(1) = \sum _{i=0}^{x_0} \frac{\lambda _1^i}{i!}e^{-\lambda _1}. \end{aligned}$$

$\square$

Proof of Equation (2)

Disregarding the truncation at $t=1$, $S_1$ is $\mathrm{Erl}(x_0,\lambda _1)$ distributed, accordingly

$$\begin{aligned} \mathbb {E}[S_1]= & {} \int _0^1 t \cdot f_{\mathrm{Erl}(x_0,\lambda _1)}(t) \mathrm{d}t + 1 \cdot \int _1^\infty f_{\mathrm{Erl}(x_0,\lambda _1)}(t) \mathrm{d}t \\= & {} \int _0^1 \lambda _1^{x_0}\frac{t^{x_0}}{(x_0 - 1)!}e^{-\lambda _1 t} \mathrm{d}t + 1 - F_{\mathrm{Erl}(x_0,\lambda _1)}(1) \\= & {} \frac{x_0}{\lambda _1} F_{\mathrm{Erl}(x_0+1,\lambda _1)}(1) + \sum _{i=0}^{x_0-1} \frac{\lambda _1^i}{i!}e^{-\lambda _1} \\= & {} \frac{1}{\lambda _1} \left( x_0\cdot \sum _{i=x_0+1}^{\infty } \frac{\lambda _1^i}{i!}e^{-\lambda _1} + \sum _{i=0}^{x_0-1}(i+1)\cdot \frac{\lambda _1^{i+1}}{(i+1)!}e^{-\lambda _1}\right) \\= & {} \frac{1}{\lambda _1} \left( x_0\cdot \sum _{i=x_0+1}^{\infty } \frac{\lambda _1^i}{i!}e^{-\lambda _1} + \sum _{i=1}^{x_0}i\cdot \frac{\lambda _1^{i}}{i!}e^{-\lambda _1}\right) .\\ \end{aligned}$$

$\square$

Proof of Proposition 3.5

The first assertion is obvious from the definition of an $\alpha$ service level. All orders from class k are filled if and only if one of the following two (disjoint) events occurs: either inventory never reaches $r_{k-1}$, i.e., $\{X_1 > r_{k-1}\}$, or no orders from class k arrive after $r_{k-1}$ has been reached, i.e., $\{X_1 \le r_{k-1} \text { and } D_{1,k} = D_{S_k,k}\}$. The union of these events equals $\{{\bar{S}}_k \ge 1\}$.

We prove the second assertion by induction over k and employ Doob’s optional stopping theorem, Theorem 3.4. To permit the latter, recall that each Poisson process $D_{t,i}$ only deviates from a martingale by a linear drift process, i.e., $\hat{D}_{t,i} = D_{t,i} - t\lambda _i$ is a right-continuous martingale (cf. Karatzas and Shreve 1991, Chapter 1, Problem 3.4). Additionally, note that the hit times of the critical levels are stopping times of the filtration generated by the $D_{t,k}$.

Let $k=n$ be a starting point for the induction. Then using Theorem 3.4 and denoting $\Lambda _n = \sum _{i=1}^{n} \lambda _i,$ we have

$$\begin{aligned} 0 = \sum _{i=1}^n \mathbb {E} [\hat{D}_{0,i}] \overset{\text {Doob}}{=} \sum _{i=1}^n \mathbb {E} [\hat{D}_{S_n,i}] = \mathbb {E}[\sum _{i=1}^n D_{S_n,i}] - \Lambda _n \cdot \mathbb {E}[S_n]. \end{aligned}$$

(4)

As the $D_{t,i}$ are independent Poisson processes,

$$\begin{aligned} \mathbb {E}[\sum _{i=1}^n D_{S_n,i}]/\Lambda _n = \mathbb {E}[D_{S_n,n}]/\lambda _n \end{aligned}$$

(5)

and, thus,

$$\begin{aligned} {\bar{\beta }}_n ^\mathrm{cl} = \mathbb {E}[D_{S_n,n}]/\lambda _n \overset{(5)}{=} \mathbb {E}\left[ \sum _{i=1}^n D_{S_n,i}\right] /\Lambda _n \overset{(4)}{=} \mathbb {E}[S_n]. \end{aligned}$$

Now fix some $k<n$ and assume the assertion holds for all j with $n\ge j>k$. Consequently, we have $[D_{S_j,j}] - \lambda _j\cdot \mathbb {E}[S_j]=0$ for $n\ge j >k$. Again by Theorem 3.4, ${\tilde{D}}_{t,j} = D_{\min (t,S_j),j} - \min (t, S_j)\lambda _j$ is a martingale for all $n\ge j \ge k$. Then (argueing as above),

As above,

$$\begin{aligned} {\bar{\beta }}_k^\mathrm{cl} = \mathbb {E}[D_{S_k,k}]/\lambda _k = \left( \sum _{j=1}^k \mathbb {E} [D_{S_k,j}]\right) /\left( \sum _{j=1}^k\lambda _j \right) = \mathbb {E}[S_k]. \end{aligned}$$

$\square$

Proof of Theorem 3.8

We first prove part (i). By Proposition 3.5, we have ${\bar{\alpha }}_k ^\mathrm{cl} = \mathbb {P}[{\bar{S}}_k \ge 1]$. By Lemma, 3.6 ${\bar{S}}_k$ equals $\min ({\bar{E}}, 1)$ in distribution. Consequently, by Lemma 3.7 we have

The proof of part (ii) is completely analogous to the proof of part (i). However, contrary to the requirements of Lemma 3.7, $\Lambda _{n+1}$ would equal $\Lambda _1$. Accordingly, we remove $E_{n+1}$ and increase $l_1$ by 1.

Now, we turn to the proof of part (iii). $S_k$ equals $\min (E, 1)$ in distribution, which by Proposition 3.5 leads to

$$\begin{aligned} {\bar{\beta }}_k^\mathrm{cl} = \int _0^1 t \cdot f_E(t) \mathrm{d}t\,+\, 1\cdot \int _1^\infty f_E(t) \mathrm{d}t. \end{aligned}$$

(6)

We will now compute the two summands of the right hand side of (6) separately. In complete analogy to the calculation of ${\bar{\alpha }}_k^\mathrm{cl}$ above, we have

$$\begin{aligned} \int _1^\infty f_E(t) \mathrm{d}t = \pi ^{k,n} \sum _{i=k}^{n} \sum _{j=1}^{l_i}c^{k,n}_{ij} \frac{1}{\Lambda _i^{j}} \sum _{m=0}^{j-1} e^{-\Lambda _i}\frac{\Lambda _i^m}{m!}. \end{aligned}$$

(7)

Again by Lemma 3.7,

Applying the last equality and (7) to (6) concludes the proof. $\square$

Proof of Corollary 3.9

Note that $c^{n,n}_{nj} = \delta _{l_n,j}$—where we used Kroneckers $\delta$-notation—as for $j\ne l_n$ the sum is empty and hence 0 by convention, and for $j=l_n$ the product is empty and hence 1 by convention. This observation allows us to calculate

$$\begin{aligned} {\bar{\beta }}^\mathrm{cl}_n= & {} \Lambda _n^{l_n} \sum _{i=n}^n\sum _{j=1}^{l_n} \delta _{l_n,j} \left[ \frac{j}{\Lambda _i^{j+1}} \left( 1 - \sum _{m=0}^{j} e^{-\Lambda _i}\frac{\Lambda _i^m}{m!} \right) + \frac{1}{\Lambda _i^{j}} \sum _{m=0}^{j-1} e^{-\Lambda _i}\frac{\Lambda _i^m}{m!} \right] \\= & {} \Lambda _n^{l_n} \left[ \frac{l_n}{\Lambda _n^{l_n+1}} \left( 1 - \sum _{m=0}^{l_n} e^{-\Lambda _n}\frac{\Lambda _n^m}{m!} \right) + \frac{1}{\Lambda _n^{r_n - r_{n-1}}} \sum _{m=0}^{l_n-1} e^{-\Lambda _n}\frac{\Lambda _n^m}{m!} \right] \\= & {} \frac{l_n}{\Lambda _n} \left( 1 - \sum _{m=0}^{l_n} e^{-\Lambda _n}\frac{\Lambda _n^m}{m!} \right) + \frac{1}{\Lambda _n} \sum _{m=0}^{l_n-1} e^{-\Lambda _n}\frac{\Lambda _n^{m+1}}{m!}\\= & {} \frac{l_n}{\Lambda _n} \left( \sum _{m=l_n+1}^\infty e^{-\Lambda _n}\frac{\Lambda _n^m}{m!} \right) + \frac{1}{\Lambda _n} \sum _{m=1}^{l_n} m\cdot e^{-\Lambda _n}\frac{\Lambda _n^{m}}{m!}. \end{aligned}$$

$\square$

Proof of Proposition 3.13

We only provide a proof for part (i) with $i=1$. The proofs for different values of i and the proof of part (ii) are completely analogous. For ease of presentation we focus on $\gamma = \beta$. The corresponding proof for $\alpha$ service levels builds on the same principles, but it is more involved.

First, let $j\ge k+2$. From Theorem 3.8, we see that ${\bar{\beta }}_j^\mathrm{cl}$ is independent of all $l_i$ with $i<j$. Hence, ${\bar{\beta }}_j^\mathrm{cl}$ remains unchanged.

Now, let $j=k+1$. From Proposition 3.5, we know ${\bar{\beta }}_j ^\mathrm{cl} = \mathbb {E}[S_j]$. By changing the allocation vector as depicted, $S_j$ is decreased (or remains unchanged) pathwise. Additionally, there are paths where $S_j$ is indeed decreased, for instance any path that has more than $x_0 - r_j$ orders from class j. Consequently, ${\bar{\beta }}_j ^\mathrm{cl} = \mathbb {E}[S_j]$ is decreased.

Finally, let $j\le k$. Without loss of generality, $j=k$. For $i=1,\ldots ,n$ denote the class i demand process stopped at $S_i$ as $\hat{D}_{t,i}=D_{\min (t, S_i),i}$. Now consider $E_{t,k}=\sum _{i=k+1}^n \hat{D}_{t,i} + \sum _{i=1}^k D_{t,i}$. The random process $E_{t,k}$ can be used to describe the behaviour of $S_k$. In fact, $S_k = \inf \{\tau \in [0,1] | E_{t,k} \ge x_0 - r_{k-1} \text { or } \tau = 1\}$. We have already seen above that — by changing the allocation vector—$S_{k+1}$ decreases pathwise and $S_j$ remains unchanged for all $j\ge k+2$. Consequently, $E_{t,k}$ decreases pathwise as class-$(k+1)$ demand is cut off earlier. But then, $S_k$ increases pathwise and therefore ${\bar{\beta }}_k ^\mathrm{cl} = \mathbb {E}[S_k]$ increases. $\square$

Proof of Proposition 3.14

Denote $z = \min \{y\in \mathbb {N} | {\bar{\gamma }}^\mathrm{std}(y,\sum \lambda _i)\ge \gamma _1\}$. A feasible vector of critical levels is given by $x_0 = z$ and $r_i = 0$ for $i=1,\ldots ,n-1$. (With this configuration ${\bar{\gamma }}_k^\mathrm{cl} \ge \gamma _1\ge \gamma _k$ for all k.) This already proves part (ii). As the set of possible parameter configurations with starting inventory $\le z$ is finite, there is at least one configuration with minimal $x_0$. This concludes the proof of part (i).

To prove part (iii), without loss of generality, let $n=2$. Denote $l_i=x_0^\mathrm{std}(\lambda _i,\gamma _i)$ for $i=1,2$. By definition of $l_2$, a critical-level policy with the allocation vector $(0,l_2)$ fulfills ${\bar{\gamma }}_2^\mathrm{cl}\ge \gamma _2$. By Proposition 3.13 part (iii) also the allocation vector $(l_1,l_2)$ fulfills ${\bar{\gamma }}_2^\mathrm{cl}\ge \gamma _2$. Following the same pattern, $(l_1,0)$ fulfills ${\bar{\gamma }}_1^\mathrm{cl}\ge \gamma _1$. By Proposition 3.13 part iv) also the allocation vector $(l_1,l_2)$ fulfills ${\bar{\gamma }}_1^\mathrm{cl}\ge \gamma _1$. Accordingly, $(l_1,l_2)$ is a feasible allocation vector which concludes the proof. $\square$

Proof of Proposition 3.17

We start proving part (i). According to Proposition 3.14, at least one solution of Problem 3.12 exists and the number of solutions is finite. As lexicographical ordering constitutes a strict total order relation on $\mathbb {N}^n$, it does so also on the set of solutions. The greatest element with respect to lexicographical ordering is the unique priority optimal solution.

Turning to part (ii), we first prove the “if” part of the assertion. Let $(l^*_1,l^*_2,l^*_3,\ldots ,l^*_n)$ be a solution and assume $(0,l^*_2,l^*_3,\ldots ,l^*_n)$ is the priority optimal solution of the relaxed problem. We need to show that any $(\tilde{l_1},\tilde{l_2}, \ldots , \tilde{l_n}) \overset{\mathrm{l.o.}}{>} (l^*_1,l^*_2,l^*_3,\ldots ,l^*_n)$ with $\sum _{i=1}^n l^*_i = \sum _{i=1}^n \tilde{l_i}$ cannot be a feasible allocation vector, i.e., not all the conditions of Problem 3.12 would be fulfilled. We prove this by contraposition and assume there is such a $(\tilde{l_1},\tilde{l_2}, \ldots , \tilde{l_n})$. Now recall from Theorem 3.8 that ${\bar{\gamma }}_k^\mathrm{cl}$ is independent of all $l_i$ for $i<k$. Thus, $(0,\tilde{l_2},\tilde{l_3}, \ldots , \tilde{l_n})$ fulfills ${\bar{\gamma }}_k^\mathrm{cl} \ge \gamma _k$ for all $k\ge 2$; hence, $(0,\tilde{l_2},\tilde{l_3}, \ldots , \tilde{l_n})$ is a solution of the relaxed problem. But as $(\tilde{l_1},\tilde{l_2}, \ldots , \tilde{l_n}) \overset{\mathrm{l.o.}}{>} (l^*_1,l^*_2,\ldots ,l^*_n)$ and $\sum _{i=1}^n l^*_i = \sum _{i=1}^n \tilde{l_i},$ there are only two possible cases each of which contradicts that $(0,l^*_2,l^*_3,\ldots ,l^*_n)$ is the priority optimal solution of the relaxed problem: either $\tilde{l_1}>l^*_1$, which implies $\sum _{i=2}^n l^*_i > \sum _{i=2}^n \tilde{l_i}$ as $\sum _{i=1}^n l^*_i = \sum _{i=1}^n \tilde{l_i}$, or $\tilde{l_1}=l^*_1,$ which implies $(0,\tilde{l_2},\tilde{l_3}, \ldots , \tilde{l_n}) \overset{\mathrm{l.o.}}{>} (0,l^*_2,l^*_3,\ldots ,l^*_n)$.

Now, we prove the “only if” part. Assume $(l^*_1,l^*_2,l^*_3,\ldots ,l^*_n)$ is priority optimal. There are two things to prove: (a) $(0,l^*_2,l^*_3,\ldots ,l^*_n)$ is a solution of the relaxed problem and (b) $(0,l^*_2,l^*_3,\ldots ,l^*_n)$ is priority optimal. We prove (a) by contraposition. Assume that there is a solution of the relaxed problem with total inventory $< \sum _{i=2}^n l^*_i$. Without loss of generality, this is $(0,l^*_2-1,l^*_3,\ldots ,l^*_n)$. Combining Proposition 3.13 part (i) applied to $(l^*_1,l^*_2,l^*_3,\ldots ,l^*_n)$ and the fact that $(0,l^*_2-1,l^*_3,\ldots ,l^*_n)$ is a solution of the relaxed problem, $(l^*_1+1,l^*_2-1,l^*_3,\ldots ,l^*_n)$ is a solution for the unrelaxed problem. However, this contradicts the assumption that $(l^*_1,l^*_2,l^*_3,\ldots ,l^*_n)$ is priority optimal as $(l^*_1+1,l^*_2-1,l^*_3,\ldots ,l^*_n) \overset{\mathrm{l.o.}}{>}(l^*_1,l^*_2,l^*_3,\ldots ,l^*_n)$. We prove (b) in the same pattern. Assume that there is another solution of the relaxed problem that is priority optimal. Without loss of generality, this is $(0,l^*_2+1,l^*_3-1,\ldots ,l^*_n)$. Then, $(l^*_1,l^*_2+1,l^*_3-1,\ldots ,l^*_n)$ is a solution for the unrelaxed problem. (As $(0,l^*_2+1,l^*_3-1,\ldots ,l^*_n)$ is a solution of the unrelaxed problem, the conditions on the ${\bar{\gamma }}_k^\mathrm{cl}$ are fulfilled for all $k\ge 2$. One obtains that the condition on ${\bar{\gamma }}_1^\mathrm{cl}$ is fulfilled by applying Proposition 3.13 part (i) to $(l^*_1,l^*_2,l^*_3,\ldots ,l^*_n)$.) Again, this contradicts the assumption that $(l^*_1,l^*_2,l^*_3,\ldots ,l^*_n)$ is priority optimal as $(l^*_1,l^*_2+1,l^*_3-1,\ldots ,l^*_n) \overset{\mathrm{l.o.}}{>}(l^*_1,l^*_2,l^*_3,\ldots ,l^*_n)$. $\square$

Proof of Corollary 3.18

We prove that $(0,\ldots ,0,l_k^*,\ldots ,l_n^*)$ is the priority-optimal solution of the k-relaxed problem by backward induction over k.

As a starting point note that, clearly, $(0,\ldots ,0,l_n^*)$ is the priority-optimal solution of the n-relaxed problem.

Now, assume that $(0,\ldots ,0,l_{k+1}^*,\ldots ,l_n^*)$ is the priority-optimal solution of the (k+1)-relaxed problem and let $(0,\ldots ,0,{\tilde{l}}_k,\ldots ,{\tilde{l}}_n^*)$ be the priority-optimal solution of the k-relaxed problem. By Proposition 3.17 and the assumption, ${\tilde{l}}_i = l_i^*$ for $i=k+1,\ldots ,n$. Then by (3) also ${\tilde{l}}_k = l_k^*$, i.e., $(0,\ldots ,0,l_k^*,\ldots ,l_n^*)=(0,\ldots ,0,{\tilde{l}}_k,\ldots ,{\tilde{l}}_n^*)$. Hence $(0,\ldots ,0,l_k^*,\ldots ,l_n^*)$ is the priority-optimal solution of the k-relaxed problem.

The n-th step of the induction, i.e., $k=1$, yields the assertion of the Corollary. $\square$

Appendix 2: Stratification procedure

To ensure a dispersed distribution of the weighted average service-level guarantees across the entire range between 0.75 and 0.95, we use stratification. Each sample consists of 20 strata, with each stratum containing 250 instances with weighted average service-level guarantee $\in [\gamma ^l, \gamma ^l+0.01[$ for $\gamma ^l = 0.75, 0.76, \ldots , 0.94$.

To compile each of the strata, we randomly generated vectors of service-level guarantees and demand allocations and then applied Algorithm 1 to all of these instances. We draw the $\lambda _1,\ldots ,\lambda _n$ from a uniform distribution on [0, 1] and rescale the vector $(\lambda _1,\ldots ,\lambda _n)$ to meet the total demand (i.e., 10). Similarly, we draw the service-level guarantees from a uniform distribution on [0.70, 0.99] and subsequently sort the service-level guarantees in descending order.

The compilation of each stratum is concluded by filtering the instances according to three criteria:

• (i) $\left( \sum _{k=1}^n \lambda _k \gamma _k \right) /\left( \sum _{k=1}^n \lambda _k \right) \in [\gamma ^l, \gamma ^l+0.01[,$

• (ii) $\lambda _k\ge 1$ for $k=1, \ldots , n,$

• (iii) $\gamma _k - \gamma _{k+1} \ge 0.01$ for $k=1, \ldots , n-1.$

Any instance that does not meet these conditions is disregarded. The first criterion ensures that the instance belongs to the stratum being compiled, while the second and third criteria serve to avoid pathologies like irrelevantly small instances and approximately equal service-level guarantees. This filtering procedure replaces the joint probability distribution of the service-level guarantees and the demand allocation by the corresponding distribution conditioned on the criteria above. Note that this procedure preserves the random nature of the simulation approach. (Consider a random variable Z that follows a uniform distribution on [0, 10]. If we condition this distribution on the criterion $Z \ge 1$, the corresponding conditional distribution is a uniform distribution on [1, 10].)

Note that a stratification with respect to $\max (\gamma _1, \ldots , \gamma _n)$, similar to the one we have carried out with respect to the weighted average, is not required. The random draws automatically cover the entire possible range of maximum service levels.

Appendix 3: Additional numerical results

This appendix presents further numerical results, namely the average required inventory per layer (and per number of customer classes) for the case of upward differentiation (Table 2) and downward differentiation (Table 3).

Table 2 Upward differentiation—average required inventory vs. number of customer classes with minimum service level (SL) controlled

Table 3 Downward differentiation—average required inventory vs. number of customer classes with maximum service level (SL) controlled

With respect to Table 2, note that our sample does not contain instances with $n=1$ and minimum service level in the range [0.70, 0.75[, by construction. However, it is used to verify that any such instance requires a starting inventory of 8 units. Hence, the value 8.0 is displayed in the corresponding cell of Table 2.