Optimal replenishment rate for inventory systems with compound Poisson demands and lost sales: a direct treatment of time-average cost

Abstract

Supply contracts are designed to minimize inventory costs or to hedge against undesirable events (e.g., shortages) in the face of demand or supply uncertainty. In particular, replenishment terms stipulated by supply contracts need to be optimized with respect to overall costs, profits, service levels, etc. In this paper, we shall be primarily interested in minimizing an inventory cost function with respect to a constant replenishment rate. Consider a single-product inventory system under continuous review with constant replenishment and compound Poisson demands subject to lost-sales. The system incurs inventory carrying costs and lost-sales penalties, where the carrying cost is a linear function of on-hand inventory and a lost-sales penalty is incurred per lost sale occurrence as a function of lost-sale size. We first derive an integro-differential equation for the expected cumulative cost until and including the first lost-sale occurrence. From this equation, we obtain a closed form expression for the time-average inventory cost, and provide an algorithm for a numerical computation of the optimal replenishment rate that minimizes the aforementioned time-average cost function. In particular, we consider two special cases of lost-sales penalty functions: constant penalty and loss-proportional penalty. We further consider special demand size distributions, such as constant, uniform and Gamma, and take advantage of their functional form to further simplify the optimization algorithm. In particular, for the special case of exponential demand sizes, we exhibit a closed form expression for the optimal replenishment rate and its corresponding cost. Finally, a numerical study is carried out to illustrate the results.

Appendix

Proof of Proposition 1

The proof of the necessary condition is trivial. To prove the sufficient condition, we first write Eq. (4.13) as

$$\begin{aligned} \int _{{\varvec{x}}_0 }^{\varvec{\infty }} {{\varvec{f}}({\varvec{x}}) {\varvec{dx}}} =-\int _0^{{\varvec{x}}_0 } {{\varvec{f}}({\varvec{x}}) {\varvec{dx}}} . \end{aligned}$$

(8.1)

Next, for all ${\varvec{z}}>0$,

$$\begin{aligned} \tilde{{\varvec{f}}}({\varvec{z}})= & {} \int _0^{{\varvec{x}}_0 } {{\varvec{e}}^{-{\varvec{z x}}}{\varvec{f}}({\varvec{x}}) {\varvec{dx}}} +\int _{{\varvec{x}}_0 }^{\varvec{\infty }} {{\varvec{e}}^{-{\varvec{z x}}}{\varvec{f}}({\varvec{x}}) {\varvec{dx}}} \\\le & {} \int _0^{{\varvec{x}}_0 } {{\varvec{e}}^{-{\varvec{z x}}}{\varvec{f}}({\varvec{x}}) {\varvec{dx}}} +{\varvec{e}}^{-{\varvec{z x}}_0 }\int _{{\varvec{x}}_0 }^{\varvec{\infty }} {{\varvec{f}}({\varvec{x}}) {\varvec{dx}}} \\= & {} \int _0^{{\varvec{x}}_0 } {{\varvec{e}}^{-{\varvec{z x}}}{\varvec{f}}({\varvec{x}}) {\varvec{dx}}} -{\varvec{e}}^{-{\varvec{z x}}_0 }\int _0^{{\varvec{x}}_0 } {{\varvec{f}}({\varvec{x}}) {\varvec{dx}}} \\= & {} \int _0^{{\varvec{x}}_0 } {\left[ {{\varvec{e}}^{-{\varvec{z x}}}-{\varvec{e}}^{-{\varvec{z x}}_0 }} \right] {\varvec{f}}({\varvec{x}}) {\varvec{dx}}} <0 \end{aligned}$$

Here, the first inequality holds since $\int _{{\varvec{x}}_0 }^{\varvec{\infty }} {{\varvec{e}}^{-{\varvec{z x}}}{\varvec{f}}({\varvec{x}}) {\varvec{dx}}} \le {\varvec{e}}^{-{\varvec{z x}}_0 }\int _{{\varvec{x}}_0 }^{\varvec{\infty }} {{\varvec{f}}({\varvec{x}}) {\varvec{dx}}}$ since ${\varvec{e}}^{-{\varvec{z x}}}\le {\varvec{e}}^{-{\varvec{z x}}_0 }$ for ${\varvec{x}}\ge {\varvec{x}}_0$; the second equality holds by Eq. (8.1), and the last inequality holds by the relations ${\varvec{e}}^{-{\varvec{z x}}}>{\varvec{e}}^{-{\varvec{z x}}_0 }$, whence ${\varvec{e}}^{-{\varvec{z x}}}-{\varvec{e}}^{-{\varvec{z x}}_0 }>0$ for $0\le {\varvec{x}}<{\varvec{x}}_0$ and ${\varvec{f}}({\varvec{x}})\le 0$ but not identically zero for $0\le {\varvec{x}}<{\varvec{x}}_0$. This completes the proof. $\square$

Proofs of Table 1 Formulas

(1) Constant demand size

Consider the first distribution row of Table 1, where ${\varvec{D}}={\varvec{d}}>0$ is a constant, so that

$$\begin{aligned} {\tilde{\varvec{f}}}_{\varvec{D}} ({\varvec{z}})={{\varvec{\exp }}} \{-{\varvec{zd}}\}. \end{aligned}$$

(8.2)

The corresponding ${\varvec{\xi }}^{*}$ follows by substituting Eq. (8.2) into Eq. (5.6), the corresponding ${\varvec{\rho }}^{*}$ follows by substituting this ${\varvec{\xi }}^{*}$ and Eq. (8.2) into Eq. (3.28), and the corresponding $\bar{{{\varvec{c}}}}_{{\varvec{\rho }}^*}$ follows by substituting ${\varvec{\xi }}^{*}$ and ${\varvec{\rho }}^{*}$ into Eq. (5.5).

(2) Exponentially-distributed demand size

Consider the second distribution row of Table 1, where ${\varvec{D\sim }}\, \mathrm{Exp}({\varvec{\beta }} )$. Substituting Eq. (5.14) into Eq. (5.5) yields

$$\begin{aligned} \bar{{{\varvec{c}}}}_{\varvec{\rho }} ={\varvec{\lambda K}}_0 +\frac{{\varvec{h}}}{{\varvec{\xi }} } - \frac{{\varvec{\lambda K}}_0 {\varvec{\beta }} }{{\varvec{\beta }} +{\varvec{\xi }} }. \end{aligned}$$

(8.3)

The corresponding ${\varvec{\xi }}^{*}$ is obtained by straightforward minimization of Eq. (8.3) in ${\varvec{\xi }}$, the corresponding ${\varvec{\rho }}^{*}$ follows by substituting this ${\varvec{\xi }}^{*}$ into Eq. (5.16), and the corresponding $\bar{{{\varvec{c}}}}_{{\varvec{\rho }}_*}$ follows by substituting this ${\varvec{\xi }}^{*}$ into Eq. (8.3).

(3) Uniformly-distributed demand size

Consider the third distribution row of Table 1, where ${\varvec{D\sim }}\,{\varvec{U}}({\varvec{a,b}})$, so that

$$\begin{aligned} \tilde{{\varvec{f}}}_{\varvec{D}} ({\varvec{z}})=\frac{{\varvec{e}}^{-{\varvec{a z}}}-{\varvec{e}}^{-{\varvec{b z}}}}{({\varvec{b}}-{\varvec{a}}) {\varvec{z}}}. \end{aligned}$$

(8.4)

The corresponding ${\varvec{\xi }}^{*}$ follows by substituting Eq. (8.4) into Eq. (5.6), the corresponding ${\varvec{\rho }}^{*}$ follows by substituting this ${\varvec{\xi }}^{*}$ and Eq. (8.4) into Eq. (3.28), and the corresponding $\bar{{{\varvec{c}}}}_{{\varvec{\rho }}^*}$ follows by substituting ${\varvec{\xi }}^{*}$ and ${\varvec{\rho }}^{*}$ into Eq. (5.5).

(4) Gamma-distributed demand size

Consider the fourth distribution row of Table 1, where ${\varvec{D\sim \varGamma }} ({\varvec{\alpha ,\beta }} )$, so that

$$\begin{aligned} \tilde{{\varvec{f}}}_{\varvec{D}} ({\varvec{z}})=\left( {1+\frac{{\varvec{z}}}{{\varvec{\beta }} }} \right) ^{-{\varvec{\alpha }}}. \end{aligned}$$

(8.5)

The corresponding ${\varvec{\xi }}^{*}$ follows by substituting Eq. (8.5) into Eq. (5.6), the corresponding ${\varvec{\rho }}^{*}$ follows by substituting this ${\varvec{\xi }}^{*}$ and Eq. (8.5) into Eq. (3.28), and the corresponding $\bar{{{\varvec{c}}}}_{{\varvec{\rho }}^*}$ follows by substituting ${\varvec{\xi }}^{*}$ and ${\varvec{\rho }}^{*}$ into Eq. (5.5). $\square$

Proofs of Table 2 Formulas

(1) Constant demand size

Consider the first distribution row of Table 2, where ${\varvec{D}}={\varvec{d}}$. Then, the corresponding ${\varvec{\xi }}^{*}$ follows by substituting Eq. (8.2) into Eq. (5.11), the corresponding ${\varvec{\rho }}^{*}$ follows by substituting this ${\varvec{\xi }}^{*}$ and Eq. (8.2) into Eq. (3.28), and the corresponding $\bar{{{\varvec{c}}}}_{{\varvec{\rho }}^*}$ follows by substituting ${\varvec{\xi }}^{*}$ and ${\varvec{\rho }}^{*}$ into Eq. (5.10).

(2) Exponentially-distributed demand size

Consider the second distribution row of Table 2, where ${\varvec{D}}\,{\varvec{\sim }}\,\mathrm{Exp}({\varvec{\beta }})$. Substituting Eq. (5.14) into Eq. (5.9) yields,

$$\begin{aligned} \bar{{{\varvec{c}}}}_{\varvec{\rho }} =\frac{{\varvec{h}}}{{\varvec{\xi }} }-\frac{{\varvec{\lambda K}}_1 }{{\varvec{\beta }} +{\varvec{\xi }} }+{\varvec{\lambda K}}_1 {\varvec{\mu _D}} . \end{aligned}$$

(8.6)

The corresponding ${\varvec{\xi }}^{*}$ is obtained by straightforward minimization of Eq. (8.6), the corresponding ${\varvec{\rho }}^{*}$ follows by substituting this ${\varvec{\xi }}^{*}$ into Eq. (5.16), and the corresponding $\bar{{{\varvec{c}}}}_{{\varvec{\rho }}^*}$ follows by substituting this ${\varvec{\xi }}^{*}$ into Eq. (8.6).

(3) Uniformly-distributed demand size

Consider the third distribution row of Table 2, where ${\varvec{D\sim U}}({\varvec{a,b}})$. Then, the corresponding ${\varvec{\xi }}^{*}$ follows by substituting Eq. (8.4) into Eq. (5.11), the corresponding ${\varvec{\rho }}^{*}$ follows by substituting this ${\varvec{\xi }}^{*}$ and Eq. (8.4) into Eq. (3.28), and the corresponding $\bar{{{\varvec{c}}}}_{{\varvec{\rho }}^*}$ follows by substituting ${\varvec{\xi }}^{*}$ and ${\varvec{\rho }}^{*}$ into Eq. (5.10).

(4) Gamma-distributed demand size

Consider the fourth distribution row of Table 2, where ${\varvec{D\sim \varGamma }} ({\varvec{\alpha ,\beta }} )$. Then, the corresponding ${\varvec{\xi }}^{*}$ follows by substituting Eq. (8.5) into Eq. (5.11), the corresponding ${\varvec{\rho }}^{*}$ follows by substituting this ${\varvec{\xi }}^{*}$ and Eq. (8.5) into Eq. (3.28), and the corresponding $\bar{{{\varvec{c}}}}_{{\varvec{\rho }}^*}$ follows by substituting ${\varvec{\xi }}^{*}$ and ${\varvec{\rho }}^{*}$ into Eq. (5.10). $\square$

Proof of Proposition 2

Assume first that the lost-sales penalty is of the form ${\varvec{w}}({\varvec{x}})=1_{(\varvec{0},\infty )} ({\varvec{x}}) {\varvec{K}}_0$ (see Sect. 6.1). Substituting Eq. (5.16) into Eq. (5.5), we have

$$\begin{aligned} \bar{{{\varvec{c}}}}_{\varvec{\rho }} =\frac{{\varvec{\rho h}}}{{\varvec{\lambda }} -{\varvec{\rho \beta }} }+{{\varvec{K}}}_0 [{\varvec{\lambda }} -{\varvec{\rho \beta }} ] = \bar{{{\varvec{h}}}}_{\varvec{\rho }} +\bar{{{\varvec{w}}}}_{\varvec{\rho }} , \end{aligned}$$

(8.7)

where the time-average carrying cost is

$$\begin{aligned} \bar{{{\varvec{h}}}}_{\varvec{\rho }} =\frac{{\varvec{\rho h}}}{{\varvec{\lambda }} -{\varvec{\rho \beta }} }, \end{aligned}$$

(8.8)

and the time-average lost-sales penalty is

$$\begin{aligned} \bar{{{\varvec{w}}}}_{\varvec{\rho }} ={{\varvec{K}}}_0 [{\varvec{\lambda }} -{\varvec{\rho \beta }} ]. \end{aligned}$$

(8.9)

Next, equate Eqs. (8.8) and (8.9) and solve for $\hat{{{\varvec{\rho }} }}$, yielding

$$\begin{aligned} \hat{{{\varvec{\rho }} }}=\frac{{\varvec{\lambda }} }{{\varvec{\beta }} }-\sqrt{\left( {\frac{{\varvec{h}}}{2 {{\varvec{K}}}_0 {\varvec{\beta }}^{2}}} \right) ^{2}+ \frac{{\varvec{\lambda h}}}{{{\varvec{K}}}_0 {\varvec{\beta }}^{3}}}+\frac{{\varvec{h}}}{2 {\varvec{K}}_0 {\varvec{\beta }}^{2}}. \end{aligned}$$

(8.10)

Letting ${\varvec{a}}=\frac{{\varvec{h}}}{2 {{\varvec{K}}}_0 {\varvec{\beta }}^{2}}>0$ and ${\varvec{b}}=\sqrt{ \frac{{\varvec{\lambda h}}}{{{\varvec{K}}}_0 {\varvec{\beta }}^{3}}}>0$ above, and noting that $\sqrt{{\varvec{a}}^{2}+{\varvec{b}}^{2}}<{\varvec{a}}+{\varvec{b}}$, we get

$$\begin{aligned} \sqrt{\left( {\frac{{\varvec{h}}}{2 {{\varvec{K}}}_0 {\varvec{\beta }}^{2}}} \right) ^{2}+ \frac{{\varvec{\lambda h}}}{{{\varvec{K}}}_0 {\varvec{\beta }}^{3}}}-\frac{{\varvec{h}}}{2 {{\varvec{K}}}_0 {\varvec{\beta }}^{2}}<\sqrt{\frac{{\varvec{\lambda h}}}{{{\varvec{K}}}_0 {\varvec{\beta }}^{3}}}. \end{aligned}$$

(8.11)

Equations (8.10) and (8.11) readily imply

$$\begin{aligned} \hat{{{\varvec{\rho }} }}>\frac{{\varvec{\lambda }} }{{\varvec{\beta }}}-\sqrt{\frac{{\varvec{\lambda h}}}{{{\varvec{K}}}_0 {\varvec{\beta }}^{3}}}={\varvec{\rho }}^{*}, \end{aligned}$$

(8.12)

where the equality in Eq. (8.12) follows from the exponential case in Table 1. This completes the proof of Eq. (5.17).

To prove Eq. (5.18), note first that $\bar{{{\varvec{h}}}}_{\varvec{\rho }}$ is an increasing function of ${\varvec{\rho }}$ by part (a) of Lemma 6, while $\bar{{{\varvec{w}}}}_{\varvec{\rho }}$ is a decreasing function of ${\varvec{\rho }}$ by part (b) of Lemma 6. Second, the aforementioned monotonicities of $\bar{{{\varvec{h}}}}_{\varvec{\rho }}$ and $\bar{{{\varvec{w}}}}_{\varvec{\rho }}$ in conjunction with Eq. (8.12) imply $\bar{{{\varvec{h}}}}_{\hat{{{\varvec{\rho }} }}} \ge \bar{{{\varvec{h}}}}_{{\varvec{\rho }}^{*}}$ and $\bar{{{\varvec{w}}}}_{\hat{{{\varvec{\rho }} }}} \le \bar{{{\varvec{w}}}}_{{\varvec{\rho }}^{*}}$. Eq. (5.18) now follows from the last two inequalities together with Eq. (5.12).

Finally, the corresponding proofs for the case ${\varvec{w}}({\varvec{x}})={\varvec{K}}_1 {\varvec{x}}$ are readily seen to be analogous to the proofs above for ${\varvec{w}}({\varvec{x}})={\varvec{K}}_0$, but with ${{\varvec{K}}}_0$ replaced by $\frac{{{\varvec{K}}}_1 }{{\varvec{\beta }} }$. $\square$