Inventory Record Inaccuracy, RFID Technology Adoption and Supply Chain Coordination

Abstract

Most retailers suffer from substantial discrepancies between inventory quantities recorded in the system and stocks truly available to customers. Promising full inventory transparency, RFID technology has often been suggested as a remedy to this problem. We consider inventory record inaccuracy in a supply chain model, where a Stackelberg manufacturer sets the wholesale price and a retailer determines how much to stock for sale to customers. We first analyze the impact of inventory record inaccuracy on optimal stocking decisions and profits. Contrasting optimal decisions in a decentralized supply chain with those in an integrated supply chain, we find that inventory record inaccuracy exacerbates the inefficiencies resulting from double marginalization in decentralized supply chains. Assuming that RFID technology can eliminate the problem of inventory record inaccuracy, we determine the cost thresholds at which RFID adoption becomes profitable. We show that a decentralized supply chain benefits more from RFID technology, such that RFID adoption improves supply chain coordination.

This chapter is based on the article “Inventory record inaccuracy, double marginalization and RFID adoption” by H.S. Heese, published 2007 in Production and Operations Management (volume 16, issue 5, pages 542–553).

Appendices

Appendix 1: Additional Results and Proofs

Lemma 1

 

• (a) ${Q_{1A}}$ is increasing in ${\mu_Y}$ for $0 < \alpha \le 2/3$, if $s > r/4$, and increasing-decreasing otherwise.

• (b) ${Q_{1B}}$ is increasing in ${\mu_Y}$ (for $2/3 < \alpha < 1$), if $s > r/4$, decreasing if $s < c/2$, and decreasing-increasing otherwise.

• (c) If ${\mu_Y} = 1$, then ${Q_{1A}} < \hat{Q}$, and ${Q_{1B}} > \hat{Q} \Leftrightarrow \alpha > \bar{\alpha }$, where $2/3 < \bar{\alpha } < 1$.

• (d) If $\alpha > 2/3$, then ${Q_{1B}} > {Q_{1A}}$.

Proof of Lemma 1

Part (a): $\frac{{\partial {Q_{1A}}}}{{\partial {\mu_Y}}} > 0 \Leftrightarrow {\mu_Y} < \frac{{2c}}{r}$; $\alpha \le \frac{2}{3} \Leftrightarrow {\mu_Y} \le \frac{{3c}}{{r + 2s}}$. ${Q_{1A}}$ is increasing in ${\mu_Y}$ for $0 < \alpha \le 2/3$, if $\frac{{3c}}{{r + 2s}} <\frac{{2c}}{r}\Leftrightarrow s> \frac{r}{4}$, and increasing-decreasing otherwise. Part (b): $\frac{{\partial {Q_{1B}}}}{{\partial {\mu_Y}}} > 0 \Leftrightarrow {\mu_Y} > \frac{c}{{2s}}$; $\alpha > \frac{2}{3} \Leftrightarrow {\mu_Y} > \frac{{3c}}{{r + 2s}}$. ${Q_{1B}}$ is increasing in ${\mu_Y}$ on $2/3 < \alpha < 1$, if $\frac{c}{{2s}} < \frac{{3c}}{{r + 2s}} \Leftrightarrow s > \frac{r}{4}$, decreasing if $\frac{c}{{2s}} > 1 \Leftrightarrow s < \frac{c}{2}$, and decreasing-increasing otherwise. Part (c) ${Q_{1A}} < \hat{Q}$ by inspection; ${Q_{1B}} < \hat{Q} \Leftrightarrow \frac{1}{{2\sqrt {{3(1 - \alpha )}} }} < \alpha \Leftrightarrow 12{\alpha^2}(1 - \alpha ) > 1$. The left side of this inequality is decreasing in $\alpha$(for $\alpha > 2/3$). At $\alpha = 2/3$, it is equal to $16/9 > 1$, and it is equal to zero at $\alpha = 1$. Hence there exists $\bar{\alpha } \in \left( {\frac{2}{3},1} \right)$, where the inequality is satisfied with equality. Part (d) ${Q_{1B}} > {Q_{1A}} \Leftrightarrow \frac{{{\mu_D}}}{{{\mu_Y}\sqrt {{3(1 - \alpha )}} }} > \frac{{3{\mu_D}\alpha }}{{2{\mu_Y}}} \Leftrightarrow \frac{1}{{3(1 - \alpha )}} > \frac{{9{\alpha^2}}}{4}$ (since both sides of the inequality are positive) $\Leftrightarrow \frac{1}{4}(3\alpha + 1){(2 - 3\alpha )^2} > 0 \Leftrightarrow \alpha \ne \frac{2}{3}$.□

Proof of Proposition 1

The optimal order quantities are derived in Inderfurth (2004). The resulting expected profits can be obtained by substituting these quantities into (3) and (4).□

Proof of Proposition 2

The optimal order quantity ${Q_2}$ is the well-known critical fractile solution to the classic newsvendor model. The expected profits follow from substituting ${Q_2}$ into the expected cost function (5) and simplification.□

Proof of Proposition 3

The retailer’s optimal order quantities are as given in Proposition 1, but with a critical fractile of ${\alpha_3} = \frac{{r - {w_3}/{\mu_Y}}}{{r - s}}$ instead of $\alpha$. The manufacturer’s profit equals $\pi_3^M = ({w_3} - c){Q_3}$. Two cases need to be distinguished depending on the resulting value of ${\alpha_3}$. Use ${\bar{w}_3} = \left( {r + 2s} \right)\frac{{{\mu_Y}}}{3}$ to denote the wholesale price at which ${\alpha_3} = 2/3$. For ${w_3} \ge {\bar{w}_3}$ (i.e., ${\alpha_3} \le 2/3$), $\pi_{3A}^M = ({w_{3A}} - c){Q_{3A}} = ({w_{3A}} - c)\frac{{3{\mu_D}}}{{2{\mu_Y}}}\left( {\frac{{r - {w_{3A}}/{\mu_Y}}}{{r - s}}} \right)$; $\frac{{{\partial^2}\pi_{3A}^M({w_{3A}})}}{{\partial {w_{3A}}^2}} = - \frac{{3{\mu_D}}}{{(r - s){\mu_Y}}} < 0$; $\frac{{\partial \pi_{3A}^M({w_{3A}})}}{{\partial {w_{3A}}}} = 0 \to {w_{3A}} = \frac{{c + r{\mu_Y}}}{2}$; ${w_{3A}} > {\bar{w}_3} \Leftrightarrow {\left. {{\alpha_3}} \right|_{{w_{3A}}}} < 2/3 \Leftrightarrow 3(c - {\mu_Y}s) + (r - s){\mu_Y} > 0$. For ${w_3} < {\bar{w}_3}$ (i.e., ${\alpha_3} > 2/3$), $\pi_{3B}^M = ({w_{3B}} - c){Q_{3B}} = ({w_{3B}} - c)\frac{{{\mu_D}}}{{{\mu_Y}}}\sqrt {{\frac{{r - s}}{{3({w_{3B}}/{\mu_Y} - s)}}}}$; $\frac{{\partial \pi_{3B}^M({w_{3B}})}}{{\partial {w_{3B}}}} = \frac{{{\mu_D}}}{{{\mu_Y}}}\sqrt {{\frac{1}{{3(1 - {\alpha_3})}}}} \left( {\frac{{c + {w_{3B}} - 2s{\mu_Y}}}{{2{w_{3B}} - 2s{\mu_Y}}}} \right) > 0 \to {w_{3B}} = {\bar{w}_3}$ (both cases are equivalent at ${\bar{w}_3}$). Since ${\left. {\pi_{3A}^M} \right|_{{\alpha_3} = 2/3}} = {\left. {\pi_{3B}^M} \right|_{{\alpha_3} = 2/3}}$ and ${w_{3A}} > {\bar{w}_3}$, the optimal wholesale price is ${w_3} = {w_{3A}}$. Substituting this optimal wholesale price gives the expressions in the Proposition.□

Proof of Proposition 4

The retailers optimal order quantity ${Q_4}$ is the standard critical fractile solution to the newsvendor problem in (5) – the critical fractile is $\frac{{r - {w_4}}}{{r - s}}$. The manufacturer’s profit equals $\pi_4^M = ({w_4} - (c + t)){Q_4} = ({w_4} - (c + t))2{\mu_D}\left( {\frac{{r - {w_4}}}{{r - s}}} \right)$; $\frac{{{\partial^2}\pi_4^M({w_4})}}{{\partial {w_4}^2}} = - \frac{{4{\mu_D}}}{{(p + r - s)}} < 0$; $\frac{{\partial \pi_4^M({w_4})}}{{\partial {w_4}}} = 0 \to {w_4} = \frac{{r + c + t}}{2}$. Substitution of the optimal wholesale price and simplification gives the expressions in the Proposition.□

Proof of Proposition 5

If $\alpha \le \frac{2}{3}:\pi_2^{SC} > \pi_{1A}^{SC} \Leftrightarrow \frac{{{{\left( {r - c - t} \right)}^2}}}{{r - s}}{\mu_D} > \frac{{3{{\left( {r - c/{\mu_Y}} \right)}^2}}}{{4(r - s)}}{\mu_D}$

$\Leftrightarrow 4{\left( {r - c - t} \right)^2} > 3{\left( {r - c/{\mu_Y}} \right)^2} \Leftrightarrow 2\left( {r - c - t} \right) > \sqrt {3} \left( {r - c/{\mu_Y}} \right)$ (both sides are positive) $\Leftrightarrow t < \left( {r - c} \right) - \frac{{\sqrt {3} }}{2}\left( {r - c/{\mu_Y}} \right)$; If $\alpha > \frac{2}{3}:\pi_2^{SC} > \pi_{1B}^{SC}$ $\Leftrightarrow \frac{{{{\left( {r - c - t} \right)}^2}}}{{r - s}}{\mu_D} > (r - s){\mu_D} - 2{\mu_D}\sqrt {{\frac{{\left( {r - s} \right)(c/{\mu_Y} - s)}}{3}}}$

$\Leftrightarrow {\left( {\frac{{r - c - t}}{{r - s}}} \right)^2} > 1 - 2\sqrt {{\frac{{1 - \alpha }}{3}}} \left( {1 - 2\sqrt {{\frac{{1 - \alpha }}{3}}} > 1 - 2\sqrt {{\frac{{1 - 2/3}}{3}}} = 1/3 > 0} \right)$

$\Leftrightarrow \frac{{r - c - t}}{{r - s}} > \sqrt {{1 - 2\sqrt {{\frac{{1 - \alpha }}{3}}} }} \Leftrightarrow t < (r - c) - (r - s)\sqrt {{1 - 2\sqrt {{\frac{{1 - \alpha }}{3}}} }}$.□

Proof of Proposition 6

The inequalities $\pi_4^R > \pi_3^R$ and $\pi_4^M > \pi_3^M$ (hence by definition also $\pi_4^{SC} > \pi_3^{SC}$) can both easily be transformed to $4{\left( {r - c - t} \right)^2} > 3{\left( {r - c/{\mu_Y}} \right)^2}$, which is the same inequality as in the proof of Proposition 5 for case A.□

Proof of Proposition 7

${t_A} > {t_B}$ $\Leftrightarrow (r - c) - \frac{{\sqrt {3} }}{2}(r - c/{\mu_Y}) > (r - c) - (r - s)\sqrt {{1 - 2\sqrt {{\frac{{1 - \alpha }}{3}}} }}$ $\Leftrightarrow \sqrt {{1 - 2\sqrt {{\frac{{1 - \alpha }}{3}}} }} > \frac{{\sqrt {3} }}{2}\alpha \Leftrightarrow 1 - 2\sqrt {{\frac{{1 - \alpha }}{3}}} > \frac{3}{4}{\alpha^2}$ (both sides of the inequality are positive for $\alpha > 2/3$) $\Leftrightarrow 1 - \frac{3}{4}{\alpha^2} > 2\sqrt {{\frac{{1 - \alpha }}{3}}}$ (both sides of the inequality are positive for $\alpha > 2/3$) $\Leftrightarrow 1 - \frac{3}{2}{\alpha^2} + \frac{9}{{16}}{\alpha^4} > \frac{4}{3} - \frac{4}{3}\alpha \Leftrightarrow \frac{9}{{16}}\left( {2 + \alpha } \right){\left( {\alpha - \frac{2}{3}} \right)^3} > 0 \Leftrightarrow \alpha > \frac{2}{3}.$□

Appendix 2: Overview of Notation

Symbol	Description
$c$	Unit cost
$r$	Unit revenue
$s$	Unit salvage value
$t$	RFID tag cost (per unit)
$Y\sim U[0,\bar{Y}]$	Yield (random variable)
${\mu_Y}$	Mean yield
$D\sim U[0,\bar{D}]$	Demand (random variable)
${\mu_D}$	Mean demand
$w$	Unit wholesale price (decision variable)
$Q$	Order quantity (decision variable)
$\pi$	Expected profit  Superscript (M = manufacturer; R = retailer; SC = supply chain)
$\alpha$	Critical fractile for the supply chain