Allocation planning in sales hierarchies with stochastic demand and service-level targets

Abstract

Matching supply with demand remains a challenging task for many companies, especially when purchasing and production must be planned with sufficient lead time, demand is uncertain, overall supply may not suffice to fulfill all of the projected demands, and customers differ in their level of importance. The particular structure of sales organizations often adds another layer of complexity: These organizations often have multi-level hierarchical structures that include multiple geographic sales regions, distribution channels, customer groups, and individual customers (e.g., key accounts). In this paper, we address the problem of “allocation planning” in such sales hierarchies when customer demand is stochastic, supply is scarce, and the company’s objective is to meet individual customer groups’ service-level targets. Our first objective is to determine when conventional allocation rules lead to optimal (or at least acceptable) results and to characterize their optimality gap relative to the theoretical optimum. We find that these popular rules lead to optimal results only under very restrictive conditions and that the loss in optimality is often substantial. This result leads us to pursue our second objective: to find alternative (decentral) allocation approaches that generate acceptable performance under conditions in which the conventional allocation rules lead to poor results. We develop two alternative (decentral) allocation approaches and derive conditions under which they lead to optimal allocations. Based on numerical analyses, we show that these alternative approaches outperform the conventional allocation rules, independent of the conditions under which they are used. Our results suggest that they lead to near-optimal solutions under most conditions.

Appendices

Appendix A: Proofs of the mathematical results

Proof of Proposition 1

Part 1 is straightforward from $\hat{x_l}(x_l)=\mathbb {E}[\min (D_l,x_l)]$. Clearly, $\hat{x_l}(x_l) + L_l (x_l) = \mathbb {E}[D_l]$. Hence, combining this with part 1 yields part 2. Part 3 holds by definition. $\square$

Proof of Lemma 1

As sums of convex functions are convex and $L_l$ is independent from $x_k$ for all $k \ne l$, we only show that $L_l(x_l)$ is convex in $x_l$, which is a straightforward consequence of the more general statement that $F(x):=\int _x^\infty f(t) \mathrm{d}t$ is convex if f is monotonously decreasing and uniformly bounded above and below, i.e., $|f| \le C$ for some $C<\infty$. In order to show the latter, note that F is continuous and fix some $-\infty< x< y < \infty .$ Now

$$\begin{aligned} \small F\left( \frac{x+y}{2}\right)&= \frac{1}{2}\left( \int _{\frac{x+y}{2}}^\infty f(t) \mathrm{d}t + \int _{\frac{x+y}{2}}^\infty f(t) \mathrm{d}t\right) \\&= \frac{1}{2}\left( \int _{x}^\infty f(t) \mathrm{d}t - \int _{x}^{\frac{x+y}{2}} f(t) \mathrm{d}t + \int _{\frac{x+y}{2}}^y f(t) \mathrm{d}t + \int _{y}^\infty f(t) \mathrm{d}t\right) \\&= \frac{1}{2}\left( \int _{x}^\infty f(t) \mathrm{d}t + \int _{x}^{\frac{x+y}{2}} \underset{\le 0}{\underbrace{\left[ f(t)- f\left( t + \frac{y-x}{2}\right) \right] }} \mathrm{d}t + \int _{y}^\infty f(t) \mathrm{d}t\right) \\&\le \frac{1}{2}\left( \int _{x}^\infty f(t) \mathrm{d}t + \int _{y}^\infty f(t) \mathrm{d}t\right) \\&= \frac{1}{2}F(x) + \frac{1}{2}F(y), \end{aligned}$$

which concludes the proof.$\square$

Proof of Theorem 1

(4) follows immediately from $x_0 < x_0^r$, which can easily be seen by contraposition: If $\epsilon :=x_0 - \sum _{l\in \mathcal {I}_K} x_l^*$ were strictly greater than 0, then for at least some $l'\in \mathcal {I}_K$ we would have $x_{l'}^* < x_{l '}^r$ and, thus, replacing $x_{l'}^*$ by $x_{l'}^*+\epsilon$ would further reduce L, which contradicts optimality.

Regarding the remaining proof, note that Problem 1c is a nonlinear optimization problem with linear constraints, namely, $\sum _{l \in \mathcal {I}_K} x_l \le x_0$ and $x_l \ge 0$ for all l. Since all constraints are linear and $\mathbf {x}\equiv 0$ is a feasible allocation, the refined version of Slater’s condition (cf. Boyd and Vandenberghe 2004) holds. In addition, by Lemma 1, W(x) is convex; hence, by standard nonlinear optimization theory (cf. Ruszczyński 2006), for any $\mathbf {x^*}$ there exist $\lambda ,\mu _l \in \mathbb {R}_0^+$ such that for all $l\in \mathcal {I}_K$

$$\begin{aligned} (\mu _l - \lambda ) / w_l&\in \partial L_l(x_l^*) \end{aligned}$$

(13)

$$\begin{aligned} \lambda (\sum _{l\in \mathcal {I}_K}x_l^* - x_0)&=0 \end{aligned}$$

(14)

$$\begin{aligned} \mu _l x_l^*&= 0 \end{aligned}$$

(15)

and any feasible point for which $\lambda ,\mu _l\in \mathbb {R}_0^+$ exist such that Eqs. (13)–(15) hold is a solution of Problem 1c.

Hence, (2) holds for all l with $\mu _l=0$ and (3) holds for all l with $\mu _l>0$. By definition of Problem 1c, $x_0 < x_0^r$, that is, the constraint $\sum _{l \in \mathcal {I}_K} x_l \le x_0$ is binding. Accordingly, $\lambda >0$. It remains to show that $\left\{ k~|~k\in \mathcal {I}_K, \mu _l>0 \right\} = A_\lambda .$

If $\mu _l>0$, then by (15) $x_l^*=0$. Furthermore, (13) together with the fact that $G_l$ is càdlàg and, therefore, $G_l(0)-1$ is an upper bound for $\partial L_l(0)$ yields $-\lambda / w_l < \mu _l/w_l - \lambda / w_l\le G_l(0) - 1.$ Accordingly, $\left\{ k~|~k\in \mathcal {I}_K, \mu _l>0 \right\} \subset A_\lambda .$

Conversely, if $\lambda \ge (1-G_l(0))w_l$, then for all $t\ge 0$ we have $- \lambda / w_l \le G_l(0) - 1 \le G_l(t) - 1$, where we used the monotonicity of $G_l$ and which immediately yields $\mu _l>0$ as (13) entails $\mu _l / w_l - \lambda /w_l \ge G_l(x_l^*) - 1$ (where we again used $G_l$’s being càdlàg). Hence, also $\left\{ k~|~k\in \mathcal {I}_K, \mu _l>0 \right\} \supset A_\lambda ,$ which concludes the proof.$\square$

Proof of Corollary 1

If $G_l$ is continuous, then by the first fundamental theorem of calculus L is differentiable with respect to $x_l$ and $\frac{\mathrm{d}}{\mathrm{d}x_l} L(x_l) = G_l(x_l) - 1$. Hence, (2) reduces to $\lambda / w_l = - L_l'(x_l^*) = (1-G_l(x_l^*))$, which—as $G_l$ is strictly increasing on $\{G_l< 1\}$ and, by Theorem 1, $\lambda >0$—implies the assertion.$\square$

Proof of Lemma 2

To avoid the use of obfuscating notation, we provide the proof for the case where $G_l$ is continuous and strictly increasing.

• 1. Fix l, k with $\alpha _l \ge \alpha _k$ and note that, in this case, $w_l \ge w_k$. Now, let $\mathbf {x^*}$ be the optimal allocation and let $\lambda ^*$ be the corresponding supply parameter. We distinguish three cases depending on the relation between $w_l, w_k$ and $\lambda ^*$:

  • $\lambda ^*>w_l\ge w_k$: In this case $x_k^*=x_l^*=0$ and the assertion holds.

  • $w_l\ge \lambda ^* \ge w_k$: In this case $x_l^*\ge 0 = x_k^*$ and the assertion holds.

  • $w_l \ge w_k \ge \lambda ^*$: In this case, ${\hat{\alpha }}_l = G_l(x_l^*)=1-\lambda /w_l=1-\lambda (1-\alpha _l)$$\ge 1-\lambda (1-\alpha _k) = G_k(x_k^*) = {\hat{\alpha }}_k.$

• 2. Without loss of generality, set $\lambda =1$, note that

  $$\begin{aligned} \lambda =1\le (\underset{\ge 1}{\underbrace{1/1- \alpha }})\cdot (\underset{=1}{\underbrace{1-G_l(0)}}) \end{aligned}$$

  and $x_l^r = G_l^{-1}(\alpha _l)=G_l^{-1}(1-1/w_l)=G_l^{-1}(1-\lambda /w_l)$. Hence, by Corollary 1$x_l^*=x_l^r$ is the unique solution of Problem 1c and, thus, the assertion holds.

$\square$

Proof of Lemma 3

Lemma 3 is a reformulation of Bellman’s (1957) principle of optimality. $\square$

Proof of Proposition 2

Proposition 2 is a straightforward consequence of Lemma 3. $\square$

Proof of Proposition 3

We first show that if (5) holds $x_{l'}^r/\mu _{l'} = x_0^r/\mu _0$ for all $l\in \mathcal {I}_K$. Set $\chi = x_{l'}^r / \mu _{l'}$; then, from (5) and the fact that $w_l = 1 / (1-\alpha _l)$ and $G_l(x_l^r) = \alpha _l$ follows

$$\begin{aligned} \frac{1-\alpha _{l''}}{1-\alpha _{l'}}&= \frac{1- G_{l''}(\mu _{l''}x_{l'}/\mu _{l'})}{1- \alpha _{l'}} \end{aligned}$$

and thus $\alpha _{l''} = G_{l''}(\mu _{l''}x_{l'}/\mu _{l'})$. From this, by definition of $x_{l''}^r$ and the fact that $G_{l''}$ is strictly increasing, follows $x_{l'}^r \mu _{l''} / \mu _{l'} = x_{l''}^r$. As $x_0^r = \sum _{l \in \mathcal {I}_K} x_l^r$, we derive that $x_0^r = x_l^r \mu _0 / \mu _l$. Hence, $x_{l'}^r/\mu _{l'}^r = x_0^r/\mu _0^r$.

With this $\chi \in \left( 0, x_{l'}^r/\mu _{l'}\right]$ is equivalent to $\chi \in \left( 0, x_0^r/\mu _0^r\right]$ and we can rearrange (5) to:

$$\begin{aligned} w_l'(1-G_{l'}(x_0\mu _{l'}/\mu _0)) = w_{l''}(1-G_{l''}(x_0\mu _{l''}/\mu _0)) \quad \text {for all } l',l''\in \mathcal {I}_K, x_0\in \left( 0, x_0^r\right] . \end{aligned}$$

As $x_{l'}^{\mathrm{PC}} = x_0\mu _{l'}/\mu _0$ it follows that $w_l'(1-G_{l'}(x_{l'}^{\mathrm{PC}})) = \lambda$ for all $l'\in \mathcal {I}_K$ and $x_0\in \left( 0, x_0^r\right]$. Replace $(1-G_l(x_l^*)) = - L_l'(x_l^*)$ to see that Theorem 1 applies and thus $x_{l'}^{\mathrm{PC}}$ is optimal.

Next, we proof the inverse implication: If (5) does not hold, then there exist at least one $\chi \in \left( 0, x_l^r/\mu _l \right]$ and a pair of customers $l,l'$ such that:

$$\begin{aligned} w_{l'}(1-G_{l'}(\mu _{l'}\chi )) \ne w_{l''}(1-G_{l''}(\mu _{l''}\chi )) \end{aligned}$$

(16)

For $x_0 = \chi \mu _{0}$ the per commit allocations are $x^{\mathrm{PC}}_{l}=\chi \mu _{l}$ for all $l\in \mathcal {I}_K$. As $\chi >0$, $x_{l}^{\mathrm{PC}} >0$ and hence, all customer groups receive the allocation $x_l^{\mathrm{PC}}$. By Theorem 1, this allocation is optimal if and only if

$$\begin{aligned} w_{l}(1-G_{l}(x^{\mathrm{PC}}_{l})) = \lambda \quad \text {for all }l\in \mathcal {I}_K \backslash A_\lambda \end{aligned}$$

This equivalent to

$$\begin{aligned} w_{l'}(1-G_{l'}(\mu _{l'}\chi )) = w_{l''}(1-G_{l''}(\mu _{l''}\chi )) \end{aligned}$$

which contradicts (16) and, thus, completes the proof. $\square$

Proof of Proposition 4

The proof of part 1 is straightforward from the definition of extended per commit.

For part 2, assume the requirements of Proposition 4 part 2 hold. Then, in analogy to the proof of Proposition 3, the identity $x_0^r/\mu _0 = x_l^r / \mu _l$ holds. Hence $x_0^r/x_l^r = \mu _0 / \mu _l$ with which it is easy to show that $x_l^{\mathrm{ePC}} = x_l^{\mathrm{PC}} = x_l^*$. $\square$

Proof of Proposition 5

Part 1 follows directly from the definition of rank-based allocations.

For part 2, we first prove that, if (8) holds, the rank-based allocation is optimal. Assume, without loss of generality, there are only two customers, $l'$ and $l''$, with $\alpha _{l'} > \alpha _{l''}$. Then, the allocation vector $\mathbf {x}^{\mathrm{RB}} = (x_{l'}, x_{l''})$ has two cases (cf. Definition 5):

$$\begin{aligned} \mathbf {x}^{\mathrm{RB}} = {\left\{ \begin{array}{ll} (x_0,0)&{}\text {if}\quad x_0 < x_{l'}^r\\ (x_{l'}^r,x_0 - x_{l'}^r)&{}\text {else.} \end{array}\right. } \end{aligned}$$

First, we regard the case that $x_0 < x_{l'}^r$. Set $\lambda = [1-\underset{x_l\nearrow x_0}{\lim }G_{l'}(x_l)]\cdot w_{l'}$; then, (2) holds for $l'$ and because $G_{l'}$ is increasing, $\lambda \ge [1-\underset{x_l\nearrow x_l^r}{\lim } G_{l'}(x_l)]\cdot w_{l'}$. By (8), it is straightforward that $l''\in A_\lambda$ and the allocation is optimal.

Now we are left to show that the allocation $\mathbf {x}^{RB} =(x_{l'}^r, x_0-x_{l'}^r)$ is optimal if $x_0 \ge x_{l'}^r$. Set $\lambda = [1-G_{l''}(x_0-x_{l'}^r)]$; then, (2) holds for $l''$. With (8) and as, by definition, $G_{l''}(x)\le \alpha _{l''}$ for all $x<x_{l''}^r$, we can bound $\lambda$ to

$$\begin{aligned}{}[1-\underset{x_l\nearrow x_0}{\lim }G_{l'}(x_l)]\cdot w_{l'} \ge \lambda > 1. \end{aligned}$$

By definition, $G_{l'}(x_{l'}^r)\ge \alpha _{l'}$ and $w_{l'}=1/(1-\alpha _{l'})$, therefore $w_{l'}(1-G_{l'}(x_{l'}^r))\le 1$. With this it is straightforward that (2) holds for $l'$ and the allocation is optimal.

Finally, we show that if (8) is violated, there is always at least one $x_0$ for which the rank-based allocation is not optimal. Assume that (8) does not hold for customers $l_k$ and $l_{k+1}$, and set $x_0 = \sum _{l\in \{l_1,\ldots ,l_k\}} x_l^r$. Then, the corresponding allocations are $x_{l_k}^{\mathrm{RB}} = x_{l_k}^r$ and $x_{l_{k+1}}^{\mathrm{RB}} = 0$ and the following holds by assumption

$$\begin{aligned} w_{l_k} (1- \underset{x_{l_k}\nearrow x_{l_k}^r}{\lim } G_{l_k}(x_{l_k})) < w_{l_{k+1}} (1 - G_{l_{k+1}}(0)). \end{aligned}$$

The conditions for an optimal allocation are (cf. Theorem 1):

$$\begin{aligned}&\text {for }l_k\text {:}&w_{l_k} \left( 1 - G_{l_k}(x_{l_k}^r) \right)&\le \lambda \le w_{l_k} (1- \underset{x_{l_k}\nearrow x_{l_k}^r}{\lim } G_{l_k}(x_{l_k})) \\&\text {for }l_{k+1}\text {:}&w_{l_{k+1}} \left( 1 - G_{l_{k+1}}(0) \right)&\le \lambda \end{aligned}$$

It follows that $\lambda \le w_{l_k} (1- \underset{x_{l_k}\nearrow x_{l_k}^r}{\lim } G_{l_k}(x_{l_k})) < w_{l_{k+1}} (1 - G_{l_{k+1}}(0)) \ge \lambda$ which leads to a contradiction. Hence, the allocation is not optimal which concludes the proof. $\square$

Proof of Proposition 6

As $x_m^H = x_m^{\mathrm{ePC}}$, part 1 follows directly from Proposition 4.

Part 2 follows directly from Corollary 1 and Lemma 3. $\square$

Proof of Proposition 7

Straightforward from Proposition 6 part 1 and the property that $x_n^H = x_n^{\mathrm{ePC}}$ for all $n \in \mathcal {I}_{K-1}$. $\square$

Proof of Proposition 8

Note that (11) mirrors the results of Corollary 1. Therefore, straightforward computations suffice to check optimality of the service level aggregation approach. $\square$

Appendix B: Formulae to determine customer parametrization from performance drivers

Forecast Heterogeneity Denote with $|\mathcal {I}_K|$ the number of customer classes, with ${\overline{\hbox {CV}}}$ the average CV (forecast accuracy) and with $H_{\mathrm{CV}}$ the targeted forecast heterogeneity. Then, with

$$\begin{aligned} r = \frac{2\sqrt{3} H_{\mathrm{CV}} \sqrt{|\mathcal {I}_K| - 1}{\overline{\hbox {CV}}}}{\sqrt{|\mathcal {I}_K| + 1}} \end{aligned}$$

the n’th customers CV is:

$$\begin{aligned} \hbox {CV}_n = {\overline{\hbox {CV}}} -\frac{r}{2} + (n-1)\cdot \frac{r}{|\mathcal {I}_K| - 1}. \end{aligned}$$

Service-level Heterogeneity Let $|\mathcal {I}_K|$ be the number of customer groups, $w_{\mathrm{max}}$ the maximum shortfall weight and $H_{\mathrm{SL}}$ the targeted service-level heterogeneity. Then, with

$$\begin{aligned} w_{\mathrm{min}} = \frac{2 \sqrt{3} \sqrt{(|\mathcal {I}_K|^2-1)H_{\mathrm{SL}}^2 } + 3 H_{\mathrm{SL}}^2 - |\mathcal {I}_K| - 3 H_{\mathrm{SL}}^2|\mathcal {I}_K| -1}{3H_{\mathrm{SL}}^2 |\mathcal {I}_K| - 3H_{\mathrm{SL}}^2 -1 }w_{\mathrm{max}} \end{aligned}$$

the n’th customer groups shortfall weight is

$$\begin{aligned} w_n = w_{\mathrm{min}} + (n-1)\frac{w_{\mathrm{max}}-w_{\mathrm{min}}}{|\mathcal {I}_K| - 1}. \end{aligned}$$