Continuous accounting of inventory costs with Brownian-motion and Poisson demand processes

Abstract

We investigate a periodic review inventory system with demand that is either a Brownian motion or a Poisson process and with costs that accrue in continuous time over the period (and not at the end of the period). We find explicit expressions for the expected inventory costs and for the optimality equation. On the basis of these expressions, a simple search can be designed to obtain the optimal order-up-to level. In addition, we develop a new distribution function with a closed-form quantile function to approximate this level. The approximation can be adjusted to consider lead time. Numerical examples show that the suggested approximation produces substantially lower deviations from the optimal solution compared with approximations recommended in the literature.

Appendices

Appendix 1: Calculating $\xi (S)$

When $\xi (S)=e^{2\mu S/\sigma ^{2}}\Phi [-(S+\mu )/\sigma ]$ is the product of a very large value multiplied by a very small one, it might be difficult to calculate its value directly. This function can be bounded by combining Laplace’s first-order lower and upper bounds to the Mills ratio of the standard normal distribution, $1/(z+1/z)<\Phi (-z)/\phi (z)<1/z$, $z>0$ (Patel and Read 1996, p. 56), with the relation $e^{2\mu S/\sigma ^{2}}\phi [(S+\mu )/\sigma ]=\phi [(S-\mu )/\sigma ]$ [Equation (18) in Appendix 4 of Hadley and Whitin 1963]:

$$\begin{aligned} \frac{1}{\frac{S+\mu }{\sigma }+\frac{\sigma }{S+\mu }}\phi \left( {\frac{S-\mu }{\sigma }} \right) \le \xi (S)\le \frac{\sigma }{S+\mu }\phi \left( {\frac{S-\mu }{\sigma }} \right) . \end{aligned}$$

(24)

When either $S$ or $\mu$ is large in comparison to $\sigma$, the lower and upper bounds are very close to each other, and their average value can be used to approximate the value of ${\xi }(S)$.

Claim When $\delta \equiv \mu /\sigma >3.5$ then $0\le \xi (S)\le 0.0576$ for every $S>0$.

Proof

Let $z\equiv (S-\mu )/\sigma$. Then, by the right hand-side of Eq. (24), $\xi (S)\le v_{\max } (\delta )$, where $v_{\max } (\delta )=\mathop {\max }\limits _{z>-\delta } \left\{ {v(z)=\phi (z)/(z+2\delta )} \right\}$. By the constraint $z>-\delta$ $(\Leftrightarrow S>0)$, the necessary condition $v {\prime }(z)=-\phi (z)(z^{2}+2\delta z+1)/(z+2\delta )^{2}=0$ has a unique solution, $z^{*}=\sqrt{\delta ^{2}-1}-\delta$. This solution also satisfies the sufficient condition for maximization, $v{''}(z^{*})=-\frac{2\phi (z^{*})(z^{*}+\delta )}{(z^{*}+2\delta )^{2}}<0$, so that $v_{\max } (\delta )\equiv v(z^{*})=\frac{\phi \left( {\sqrt{\delta ^{2}-1}-\delta } \right) }{\sqrt{\delta ^{2}-1}+\delta }$. Since $v_{\max } '(\delta )=-2\frac{\phi \left( {\sqrt{\delta ^{2}-1}-\delta } \right) }{\left( {\sqrt{\delta ^{2}-1}+\delta } \right) ^{ 2}}<0$, then, for $\delta >3.5, v_{\max } (\delta )<v_{\max } (3.5)=0.0576$. $\square$

Appendix 2: Proof of Theorem 1

Let $\bar{{\Phi }}(z)\equiv 1-\Phi (z)$ be the complementary CDF of the Normal distribution.

• (i) By modifying the integration boundaries in Eq. (1), we obtain

  $$\begin{aligned} C(S)=\int \limits _0^1 {\left( {(p+h)\int \limits _S^\infty {(x-S)f(x|t)} dx+h\int \limits _{-\infty }^\infty {(S-x)f(x|t)dx} } \right) } dt. \end{aligned}$$

  Substituting $f(x|t)=\frac{1}{\sigma \sqrt{t}}\phi \left( {\frac{x-\mu t}{\sigma \sqrt{t}}} \right)$ in the equation above yields

  $$\begin{aligned} C(S)=(p+h)\int \limits _0^1 {\sigma \sqrt{t}L\left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) } dt +h\left( S-\frac{1}{2}\mu \right) , \end{aligned}$$

  (25)

  where $L(z)\equiv \int \limits _z^\infty {(y-z)\phi (y)} dy=\phi (z)-z \bar{{\Phi }}(z)$ (Nahmias 2005, p. 261). In order to produce an explicit expression for $C(S)$, we decompose $\int \limits _0^1 {\sigma \sqrt{t}L\left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) } dt$ into three elements by the formula of $L(z)$:

  $$\begin{aligned} \int \limits _0^1 {\sigma \sqrt{t}L\left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) dt}&= \sigma ^{2}\int \limits _0^1 {\frac{\sqrt{t}}{\sigma }\phi \left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) dt} -S\int \limits _0^1 {\bar{{\Phi }}\left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) dt}\nonumber \\&+\,\mu \int \limits _{{0} }^1 { t \bar{{\Phi }}\left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) dt} . \end{aligned}$$

  (26)

  By Equations (10), (16) and (17) in Appendix 4 of Hadley and Whitin (1963), for $T_1 ,T_2 >0$:

  $$\begin{aligned} \int \limits _{T_1 }^{T_2 } {\frac{\sqrt{t}}{\sigma }\phi \left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) dt} =\left[ {\begin{array}{l} \frac{\sigma ^{2}}{\mu ^{3}}\left( {1+\frac{\mu S}{\sigma ^{2}}} \right) \Phi \left( {\frac{\mu t-S}{\sigma \sqrt{t}}} \right) -\frac{2\sigma \sqrt{t}}{\mu ^{2}}\phi \left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) \\ +\frac{1}{\mu ^{2}}\left( {S-\frac{\sigma ^{2}}{\mu }} \right) e^{2\mu S/\sigma ^{2}}\Phi \left( {-\frac{S+\mu t}{\sigma \sqrt{t}}} \right) \\ \end{array}} \right] {\begin{array}{l} {^{T_2 }} \\ \\ \\ {_{T_1 } } \\ \end{array} }. \end{aligned}$$

  (27)
  $$\begin{aligned} \int \limits _{T_1 }^{T_2 } {\bar{{\Phi }}\left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) dt} =\left[ {\begin{array}{l} \left( {t-\frac{S}{\mu }-\frac{\sigma ^{2}}{2\mu ^{2}}} \right) \Phi \left( {\frac{\mu t-S}{\sigma \sqrt{t}}} \right) +\frac{\sigma \sqrt{t}}{\mu }\phi \left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) \\ +\frac{\sigma ^{2}}{2\mu ^{2}}e^{2\mu S/\sigma ^{2}}\Phi \left( {-\frac{S+\mu t}{\sigma \sqrt{t}}} \right) \\ \end{array}} \right] {\begin{array}{l} {^{T_2 }} \\ \\ \\ {_{T_1 } } \\ \end{array} }. \end{aligned}$$

  (28)
  $$\begin{aligned} \int \limits _{T_1 }^{T_2 } {t \bar{{\Phi }}\left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) dt} =\left[ {\begin{array}{l} \frac{1}{2}\left( {t^{2}-\frac{S^{2}}{\mu ^{2}}-\frac{2\sigma ^{2}S}{\mu ^{3}}-\frac{3\sigma ^{4}}{2\mu ^{4}}} \right) \Phi \left( {\frac{\mu t-S}{\sigma \sqrt{t}}} \right) \\ +\frac{\sigma \sqrt{t}}{2\mu ^{2}}\left( {\mu t+\frac{3\sigma ^{2}}{\mu }+S} \right) \phi \left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) \\ -\frac{\sigma ^{2}}{2\mu ^{3}}\left( {S-\frac{3\sigma ^{2}}{2\mu }} \right) e^{2\mu S/\sigma ^{2}}\Phi \left( {-\frac{S+\mu t}{\sigma \sqrt{t}}} \right) \\ \end{array}} \right] {\begin{array}{l} {^{T_2 }} \\ \\ \\ \\ \\ {_{T_1 } } \\ \end{array} }. \end{aligned}$$

  (29)

  Hence, substituting Eqs. (27), (28) and (29) in Eq. (26) yields

  $$\begin{aligned}&\int \limits _{T_1 }^{T_2 } {\sigma \sqrt{t}L\left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) dt}\nonumber \\&\qquad =\left[ {\begin{array}{l} \left[ {\frac{\sigma ^{4}}{4\mu ^{3}}+\frac{S\sigma ^{2}}{2\mu ^{2}}+\frac{S^{2}}{2\mu }-St+\frac{\mu }{2}t^{2}} \right] \Phi \left( {\frac{\mu t-S}{\sigma \sqrt{t}}} \right) \\ +\frac{\sigma \sqrt{t}}{2}\left[ {t-\frac{S}{\mu }-\frac{\sigma ^{2}}{\mu ^{2}}} \right] \phi \left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) -\frac{\sigma ^{4}}{4\mu ^{3}}e^{2\mu S/\sigma ^{2}}\Phi \left( {-\frac{S+\mu t}{\sigma \sqrt{t}}} \right) \\ \end{array}} \right] {\begin{array}{l} {^{T_2 }} \\ \\ \\ {_{T_1 } } \\ \end{array} }. \end{aligned}$$

  (30)

  Using algebraic manipulations and substituting the expression of $L\left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right)$, we obtain

  $$\begin{aligned}&\int \limits _{T_1 }^{T_2 } {\sigma \sqrt{t}L\left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) dt}\nonumber \\&\qquad =\frac{\sigma }{2}\left[ {\begin{array}{l} \left( {t-\frac{S}{\mu }-\left( {\frac{\sigma }{\mu }} \right) ^{2}} \right) \sqrt{t}L\left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) +t\frac{\sigma }{\mu }\Phi \left( {\frac{\mu t-S}{\sigma \sqrt{t}}} \right) \\ +\frac{1}{2}\left( {\frac{\sigma }{\mu }} \right) ^{3}\left[ {\Phi \left( {\frac{\mu t-S}{\sigma \sqrt{t}}} \right) -e^{2\mu S/\sigma ^{2}}\Phi \left( {-\frac{S+\mu t}{\sigma \sqrt{t}}} \right) } \right] \\ \end{array}} \right] {\begin{array}{l} {^{T_2 }} \\ \\ \\ {_{T_1 } } \\ \end{array} }. \end{aligned}$$

  (31)

  Equation (2) is proved by substituting Eq. (31) in (25) with $T_2 =1$ and $T_1$ that approaches zero from above. In order to obtain the exact expression in Eq. (2), we show that the limit of each of the three elements in Eq. (31) is zero when $T_1$ approaches zero:

  • (a) $\mathop {\lim }\limits _{ t\rightarrow 0^{+}} \sqrt{t}L\left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) =0$ since $L(z)$ is bounded for $z\ge 0$;

  • (b) $\mathop {\lim }\limits _{ t\rightarrow 0^{+}} t \Phi \left( {\frac{\mu t-S}{\sigma \sqrt{t}}} \right) =0$ since $\Phi (z)$ is bounded;

  • (c) we distinguish between $S=0$ and $S>0$ in calculating the limit of the third element in Eq. (31) and show that the result is zero in both cases:

  $$\begin{aligned}&\mathop {\lim }\limits _{ t\rightarrow 0^{+}} \left( {\Phi \left( {\frac{\mu t-S}{\sigma \sqrt{t}}} \right) -e^{2\mu S/\sigma ^{2}}\Phi \left( {-\frac{S+\mu t}{\sigma \sqrt{t}}} \right) } \right) \\&\qquad =\left\{ {{\begin{array}{ll} {\Phi (0)-e^{0}\Phi (0)=1-1=0} &{} {S=0} \\ {\Phi (-\infty )-e^{2\mu S/\sigma ^{2}}\Phi (-\infty )=0-e^{2\mu S/\sigma ^{2}}0=0} &{} {S>0} \\ \end{array} }} \right. . \end{aligned}$$

• (ii) $F_X (S)=\int \limits _0^1 {\Phi \left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) dt} =\int \limits _0^1 {\left( {1-\bar{{\Phi }}\left( {\frac{S-\mu t}{\sigma \sqrt{t}}} \right) } \right) dt}$. Equation (3) is proved by substituting Eq. (28) with $T_2 =1$ and $T_1$ that approaches zero from above, and then using algebraic manipulations and limit calculations similar to those below Eq. (31).

• (iii) By differentiating Eq. (3) with respect to $S$, we obtain

  $$\begin{aligned} f_X (S)&= \frac{1}{\mu }\left( {\Phi \left( {\frac{\mu -S}{\sigma }} \right) -e^{2\mu S/\sigma ^{2}}\Phi \left( {-\frac{S+\mu }{\sigma }} \right) } \right) -\frac{1}{2\sigma }\left( {\frac{\sigma }{\mu }} \right) ^{2}\left( \phi \left( {\frac{S-\mu }{\sigma }} \right) \right. \nonumber \\&-\left. e^{2\mu S/\sigma ^{2}}\phi \left( {\frac{S+\mu }{\sigma }} \right) \right) . \end{aligned}$$

  (32)

  The claim is proved by substituting the relation $e^{2\mu S/\sigma ^{2}}\phi [(S+\mu )/\sigma ]=\phi [(S-\mu )/\sigma ]$ in Eq. (32).

Appendix 3: Proof of Corollary 1

In what follows, we let $\delta \equiv \mu /\sigma$ and use Laplace’s first-order lower and upper bounds to the inverse Mills ratio of the standard normal distribution (derived from Appendix 1):

$$\begin{aligned} \phi (z)/\Phi (-z)>z, z>0 , \end{aligned}$$

(33a)

and

$$\begin{aligned} \phi (z)/\Phi (-z)<z+1/z , z>0. \end{aligned}$$

(33b)

• (i) The claim is proved by substituting $S=0$ in Eq. (3) and using algebraic manipulations.

• (ii) By applying (i), $(\mu /\sigma )^{2}F_X (0)=y(\delta )$, where $y(\delta )\equiv 0.5\left[ {2\Phi (\delta )-1} \right] +\delta ^{2}\Phi (-\delta )-\delta \phi (\delta )$. By Eq. (33a), $0\le \delta ^{2}\Phi (-\delta )\le \delta \phi (\delta )$, and by using l’Hôpital’s rule, $\mathop {\lim }\limits _{\delta \rightarrow \infty } \delta \phi (\delta )=0$. Since $y{\prime }(\delta )=2\delta \Phi (-\delta )>0$, then for $\delta >3.5$, $y(3.5)=0.49956<y(\delta )\le \mathop {\lim }\limits _{\delta \rightarrow \infty } y(\delta )=0.5$.

• (iii) Differentiating Eq. (4) with respect to $S$ and substituting the relation $e^{2\mu S/\sigma ^{2}}\phi [(S+\mu )/\sigma ]=\phi [(S-\mu )/\sigma ]$ yields

  $$\begin{aligned} f'_X (S)=-\frac{2e^{2\mu S/\sigma ^{2}}}{\sigma ^{2}}\Phi \left( {-\frac{S+\mu }{\sigma }} \right) <0. \end{aligned}$$

  (34)

  Differentiating Eq. (34) with respect to $S$ and using algebraic manipulations yields

  $$\begin{aligned} f_X ^{\prime \prime } (S)=\frac{2e^{2\mu S/\sigma ^{2}}}{\sigma ^{3}}\Phi \left( {-\frac{S+\mu }{\sigma }} \right) \left[ {{\phi \left( {\frac{S+\mu }{\sigma }} \right) }\Bigg /{\Phi \left( {-\frac{S+\mu }{\sigma }} \right) }-2\frac{\mu }{\sigma }} \right] . \end{aligned}$$

  (35)

We prove now that $f_X (S)$, $S>0$, has a unique inflection point for $\delta >1$. Let $z\equiv (S+\mu )/\sigma$ and $u(z)\equiv \phi (z)/\Phi (-z)-2\delta$, so that ${u}^{\prime }(z)=\left( {\phi (z)/\Phi (-z)} \right) \left( {\phi (z)/\Phi (-z)-z} \right)$. Then $f''_X (S)=0$ has a unique solution for $S>0$ if and only if $u(z)=0$ has a unique solution for $z>\delta$. By Eq. (33a), $u{'}(z)>0$ and $u(2\delta )>0$, and by Eq. (33b), $u(\delta )<(1-\delta ^{2})/\delta$, so that $u(\delta )<0$ for $\delta >1$. According to the intermediate value theorem, there exists a unique value of $z$ that satisfies $u(z)=0$ for $\delta >1$.

Appendix 4: Proof of Theorem 2

Let $\bar{{F}}(n)\equiv 1-F(n-1)$ be the complementary CDF of the Poisson distribution.

• (i) By modifying the integration boundaries in Eq. (19), we obtain

  $$\begin{aligned} C_P (S)=(p+h)\sum _{n=S}^\infty {(n-S)} \int \limits _0^1 {P(n|t)dt} +h\left( S-\frac{1}{2}\mu \right) . \end{aligned}$$

  By Equation (16) in Appendix 3 of Hadley and Whitin (1963), in a Poisson process, $\int \limits _0^1 {P(n|t)} dt=\overline{F} (n+1)/\mu$. Thus,

  $$\begin{aligned} C_P (S)=\frac{p+h}{\mu }\left( {\sum _{n=S+1}^\infty {n\bar{{F}}(n)} -(S+1)\sum _{n=S+1}^\infty {\bar{{F}}(n)} } \right) +h\left( S-\frac{1}{2}\mu \right) . \end{aligned}$$

  (36)

  By Equation (8) in Appendix 3 of Hadley and Whitin (1963),

  $$\begin{aligned} \sum _{n=S+1}^\infty {n\bar{{F}}(n)} =\frac{1}{2}\mu ^{2}\bar{{F}}(S-1)+\mu \bar{{F}}(S)-\frac{1}{2}S(S+1)\bar{{F}}(S+1), \end{aligned}$$

  and by Equation (6) in Appendix 3 of Hadley and Whitin (1963),

  $$\begin{aligned} \sum _{n=S+1}^\infty {\bar{{F}}(n)} =\mu \bar{{F}}(S)-S\bar{{F}}(S+1) . \end{aligned}$$

  (37)

  Thus,

  $$\begin{aligned} C_P (S)&= \frac{p+h}{2\mu }\left( S+(S-\mu )^{2}-\mu ^{2}F(S-2)\right. \\&+\left. 2\mu S F(S-1)-S(S+1)F(S) \right) +h\left( S-\frac{1}{2}\mu \right) . \end{aligned}$$

• (ii) By Eq. (36), the difference function of the expected cost is

  $$\begin{aligned} \Lambda (S)\equiv C_P (S+1)-C_P (S) =h-\frac{p+h}{\mu }\sum _{n=S+1}^\infty {\bar{{F}}(n+1)} . \end{aligned}$$

Since the second-order difference function of the expected cost is $\Lambda (S+1)-\Lambda (S) =\frac{p+h}{\mu }\bar{{F}}(S+1)>0$, then $\Lambda (S)$ is an increasing sequence, implying that $C_P (S)$ is a discrete convex function. Hence, $C_P (S)$ is minimized for the smallest integer $S$ that satisfies $\Lambda (S) \ge 0$ [i.e., $C_P (S+1)\ge C_P (S)$]. By applying Eq. (37), we obtain

$$\begin{aligned} \sum _{n=S+1}^\infty {\bar{{F}}(n+1)}&= \sum _{n=S+2}^\infty {\bar{{F}}(n)} =\mu \bar{{F}}(S+1)-(S+1)\bar{{F}}(S+2)\\&= \mu (1-F(S))-(S+1)(1-F(S+1)), \end{aligned}$$

so that the optimal order-up-to level is found by $S^{*}=\min \left\{ {S\in \{0,1,2,\ldots \}|F_X (S)\ge \omega } \right\}$, where $F_X (S)\equiv F(S)+(S+1)(1-F(S+1))/\mu , S\in \{0, 1, 2,\ldots \}$.