The impact of the number of parallel warehouses on total inventory

Abstract

In the strategic design of a distribution system, the right number of stock points for the various products is an important question. In the past decade, a strong trend in the consumer goods industry led to centralizing the inventory in a single echelon consisting of a few parallel warehouses or even a single distribution center for a Europe-wide distribution system. Centralizing inventory is justified by the reduction in total stock which mostly overcompensates the increasing transportation cost. The effect of centralization is usually described by the “Square Root Law”, stating that the total stock increases with the square root of the number of stock points. However, in the usual case where the warehouses are replenished in full truck loads and where a given fill rate has to be satisfied, the Square Root Law is not valid. This paper explores that case. It establishes functional relationships between the demand to be served by a warehouse and the necessary safety and cycle stock for various demand settings and control policies, using an approximation of the normal loss function and its inverse. As a consequence, the impact of the number of parallel warehouses on the total stock can be derived. The results can be used as tools in network design models.

Appendix

The analysis of the stock functions $S^\mathrm{T}(N)$ is based on the following approximation to the normal loss integral $G(k) \approx R(k)$ which allows an analytic inversion $H(x)=G^{-1}(x) \approx R^{-1}(x)$: Let

$$\begin{aligned} G(k) = \frac{1}{\sqrt{2\pi }} \exp (-ak^2 - bk)\quad \text {with parameters } a, b > 0. \end{aligned}$$

(13)

The equation $x=G(k)$ or $k=G^{-1}(x)$ is equivalent to the quadratic equation in k $ak^2+bk+\ln (\sqrt{2\pi }x)=0$ with the solution

$$\begin{aligned} k = \frac{1}{2a}\left( -b + \sqrt{b^2-4a(\ln \sqrt{2\pi }+\ln x)}\right) . \end{aligned}$$

Hence the inverse to G(k) is

$$\begin{aligned} H(x) = G^{-1}(x) = -A + \sqrt{B-\frac{1}{a}\ln x}\quad \text {with } A = \frac{b}{2a}, B=A^2 - \frac{1}{a} \ln \sqrt{2 \pi }.\nonumber \\ \end{aligned}$$

(14)

The parameters a, b are fitted so that, for any given $k \in [0,4], k' = H(R(k))$ coincides with k as well as possible. This is done by minimizing the sum of squares $(\frac{k'}{k}-1)^2$ over $k=0.03, 0.1, 0.2, \ldots , 3.9, 4.0$ resulting in

$$\begin{aligned} a=0.36121504,\quad b=1.22377537,\quad A=1.69397068,\quad B=0.32551597 \end{aligned}$$

(15)

The resulting accuracy of the approximation is $G(0)=R(0)$ due to the definition (13) and $- 0.0016 \le G(k)-R(k) \le 0.001$ for $0 \le k \le 4$, and $- 0.0171 \le H(x) - R^{-1}(x) \le 0.00227$ for $R(4) \le x \le R(0)$.

The function H(x) has the following properties:

1.1 A.1 $H(x) > 0$

Proof

For $0 < x < \frac{1}{\sqrt{2 \pi }}$ we have, using (14), $H(x) = -A + \sqrt{B-\frac{1}{a}\ln x} > -A + \sqrt{B+\frac{1}{a}\ln \sqrt{2\pi }} = -A + \sqrt{A^2} = 0$.

1.2 A.2

The function $h_1(x) = H(c \, x^\theta )x^{1-\theta }$ with constants $c > 0$ and $0 < \theta < 1$ is positive for $0 < c \, x^{\theta } < \frac{1}{\sqrt{2 \pi }}$ and attains a maximum. It is concave if $\theta \le 0.5$.

Proof

$h_1(x) = (-A+w)x^{1-\theta }$ with $w = \sqrt{B-\frac{1}{a} \ln c - \frac{\theta }{a}\ln x} > A$ due to A.1 and $\frac{\mathrm d}{\mathrm d x}w=-\frac{\theta }{2axw}$. The derivatives are

$$\begin{aligned} h'_1(x) = \frac{1-\theta }{x^\theta }(-A+w)-\frac{\theta }{2ax^\theta w} = \frac{1}{x^\theta } \left( (1-\theta )(-A+w)-\frac{\theta }{2aw}\right) \end{aligned}$$

and

$$\begin{aligned} h''_1(x)= & {} -\frac{\theta }{x^{1+\theta }}\left( (1-\theta )(-A+w)-\frac{\theta }{2aw}\right) +\frac{1}{x^\theta }\left( \frac{-\theta (1-\theta )}{2axw}-\frac{\theta ^2}{4a^2xw^3}\right) \nonumber \\= & {} -\frac{\theta }{x^{1+\theta }}\left( (1-\theta )(-A+w)+\frac{1-2\theta }{2aw}+\frac{\theta }{4a^2w^3}\right) . \end{aligned}$$

(16)

Therefore, $h''_1(x) < 0$ for $\theta \le 0.5$ and hence the concavity of $h_1(x)$.

$h'_1(x)=0$ holds if $(1-\theta )(-A+w)-\frac{\theta }{2aw}=0$ or $w^2-Aw-\frac{\theta }{2a(1-\theta )}=0$.

From the solution of the last equation, $w_0 = \frac{1}{2}(A+\sqrt{A^2+\frac{2\theta }{a(1-\theta )}})$, the solution of $h'_1(x)=0$ is obtained, using the definition of w, as $x_0=[\frac{1}{c}\exp (a(B-w^2_0))]^{\frac{1}{\theta }}$.

This is a global maximum for $0 < \theta \le 0.5$. For $0.5<\theta <1$ it is at least a local maximum because for $w=w_0$

$$\begin{aligned} (1-\theta )(-A+w)+\frac{1-2\theta }{2aw}> (1-\theta )(-A+w)-\frac{\theta }{2aw}= 0 \end{aligned}$$

and hence, according to (16), $h''_1(x_0) < 0$.

1.3 A.3

For $\theta =0.5$, $h_1(x)$ attains the maximum approximately at $x_0 \approx \frac{1}{(6c)^2}$ with the value $h(x_0) \approx \frac{1}{10c}$. In the indifference area $[0.4x_0, 2 x_0]$, the value of $h_1(x)$ is less than $10~\%$ below the maximum.

Proof

For $\theta =0.5$, we have, using the parameter values (15), $w_0=\frac{1}{2}(A+\sqrt{A^2+\frac{2}{a}}) = 2.2967$ and $x_0 = [\frac{1}{c}\exp (a(B-5.2747))]^2 = (\frac{0.1673}{c})^2$, and $h_1(x_0)=\frac{0.1673}{c}H(0.1673) = \frac{0.1008}{c}$.

Let $y_1 = c \sqrt{0.4x_0} = \sqrt{0.4}\cdot 0.1673 = 0.1058$, $y_2=c\sqrt{2x_0}=\sqrt{2}\cdot 0.1673 = 0.2366$. Then, $h_1(0.4x_0)=H(y_1)\frac{y_1}{c}=\frac{0.09142}{c}=0.907h_1(x_0)$ and $h_1(2x_0)=H(y_2)\frac{y_2}{c}=\frac{0.09073}{c}=0.9h_1(x_o)$. Due to the concavity of $h_1(x)$, this proves the property of the indifference area.

1.4 A.4

The function $h_2(x) = H(\frac{c}{\sqrt{x}})\sqrt{x}$ is positive and concave for $x > 2 \pi c^2$ and increases asymptotically stronger than $\sqrt{x}$, but weaker than linearly.

Proof

$h_2(x)=(-A+w)\sqrt{x}$ with $w=\sqrt{B-\frac{1}{a}\ln c+\frac{1}{2a}\ln x}$. $h''_2(x)<0$ is proved analogously to $h''_1(x) < 0$ in A.2. For $x \rightarrow \infty$, $\frac{h_2(x)}{\sqrt{x}}=-A+w$ tends to infinity, but $\frac{h_2(x)}{x}=\frac{1}{\sqrt{x}}(-A+w)$ tends to zero.

1.5 A.5

The function $h_3(x)=H(\frac{c}{x})x$ is positive and convex for $x > c\sqrt{2\pi }$ and increases asymptotically stronger than linearly.

Proof

$h'_3(x)=-A+w+\frac{1}{2aw}$ and $h''_3(x) = \frac{1}{2axw}(1-\frac{1}{2aw^2}) > 0$, because $2aw^2 > 2a \, A^2=2.07$. This proves the convexity of $h_3(x)$. The asymptotic trend is obvious.