Pass-through decision analysis in a supply chain

Abstract

Firms have seen rising costs on one side of the supply chain, and frugal shoppers on the other end in today’s business environment. It is possible that increment of rising costs will be passed on to consumers. How will these organizations that are seeking trade promotion decide on passthrough rate in a supply chain? We build a model to analyze both the own-brand and cross-brand cost pass-through incentives of a supply chain selling two products. The key contributions of this paper are as follows. First, we formulate a general pass-through problem as a comparative static of the price equilibrium in a channel duopoly. Our model considers (1) competition intensity, (2) consumer marginal willingness to pay (MWTP), (3) consumer perceivable difference as a threshold level to compare utility from consuming different products, and (4) product quality difference. We show that whether retail cross-brand passthrough is positive, zero or negative depends on whether the hazard rate of consumer MWTP is increasing, affine or decreasing; whether the supplier own-brand and cross-brand pass-through is strictly increasing, affine or decreasing depends on whether the reciprocal of the consumer MWTP hazard rate is strictly increasing, affine or decreasing. Second, we use this model to compare differences in pricing decisions and in profits, resulting from using a centralized versus a decentralized chain in both monopoly and duopoly markets. Our approach allows us to derive pricing implications in both monopoly and duopoly settings.

Appendix

Proof of Proposition 1

Taking the second derivative in (9) and (10) with respect to $c_{i}^{{}}$($i\in \{1,2\}, j=3-i$) leads to:

$$\begin{aligned} \frac{\partial ^{2}p_{i}}{\partial c_{id}^{2}}= & {} \left( \frac{\partial ^{2}p_{i}}{ \partial c_{id}^{2}}-\frac{\partial ^{2}p_{j}}{\partial c_{id}^{2}} \right) m^{\prime }\left( \frac{p_{i}-p_{j}+\delta }{q_{i}{-}q_{j}}\right) +\frac{m^{\prime \prime }\left( \frac{p_{i}{-}p_{j}+\delta }{q_{i}{-}q_{j}}\right) }{q_{i}{-}q_{j}}\left( \frac{\partial p_{i} }{\partial c_{id}^{{}}}{-}\frac{\partial p_{j}}{\partial c_{id}^{{}}}\right) ^{2} \text { and} \end{aligned}$$

(27)

$$\begin{aligned} \frac{\partial ^{2}p_{j}}{\partial c_{id}^{2}}= & {} \left( \frac{\partial ^{2}p_{j}}{ \partial c_{id}^{2}}{-}\frac{\partial ^{2}p_{i}}{\partial c_{id}^{2}} \right) m^{\prime }\left( \frac{p_{i}{-}p_{j}{-}\delta }{q_{i}{-}q_{j}}\right) +\frac{\left( \frac{\partial p_{i}}{\partial c_{id}^{{}}}-\frac{\partial p_{j}}{\partial c_{id}^{{}}} \right) ^{2}m^{\prime \prime }\left( \frac{p_{i}-p_{j}{-}\delta }{q_{i}-q_{j}}\right) }{q_{i}-q_{j} }, \end{aligned}$$

(28)

where we use $m^{\prime }(\cdot )$ to represent $\frac{dm(\cdot )}{d(\cdot )}, m^{\prime \prime }(\cdot )$ to represent $\frac{d^{2}m(\cdot )}{d(\cdot )^{2}}$. Denote by $m^{{}}(\delta )=m(\frac{p_{i}-p_{j}+\delta }{q_{i}-q_{j}} )$ and $m^{{}}(-\delta )=m(\frac{p_{i}-p_{j}-\delta }{q_{i}-q_{j}})$ for simplicity, solving the system of equations in (12) and (13) with respect to $\frac{\partial ^{2}p_{i}}{\partial c_{id}^{2}}$ and $\frac{\partial ^{2}p_{j}}{\partial c_{id}^{2}}$ leads to:

$$\begin{aligned} \frac{\partial ^{2}p_{i}}{\partial c_{id}^{2}}= & {} \frac{\varkappa [m^{\prime }(\delta )m^{\prime \prime }(-\delta )+m^{\prime }(-\delta )m^{\prime \prime }(\delta )-m^{\prime \prime }(\delta )]}{m^{\prime }(\delta )+m^{\prime }(-\delta )-1}\text { and} \end{aligned}$$

(29)

$$\begin{aligned} \frac{\partial ^{2}p_{j}}{\partial c_{id}^{2}}= & {} \frac{\varkappa m^{\prime \prime }(\delta )}{m^{\prime }(\delta )}+\left[ 1-\frac{1}{m^{\prime }(\delta )}\right] \frac{\partial ^{2}p_{i}}{\partial c_{i}^{2}} \end{aligned}$$

(30)

$$\begin{aligned}= & {} \frac{\varkappa [m^{\prime }(\delta )m^{\prime \prime }(-\delta )+m^{\prime }(-\delta )m^{\prime \prime }(\delta )-m^{\prime \prime }(-\delta )]}{m^{\prime }(\delta )+m^{\prime }(-\delta )-1} \end{aligned}$$

(31)

where $\varkappa =\frac{1}{(q_{i}-q_{j})(m^{\prime }(\delta )+m^{\prime }(-\delta )-1)^{2}}>0$. For the convex Mills Ratio $m^{\prime \prime }\left( \cdot \right) \ge 0$ that follows from Assumption (iii), then $\frac{\partial ^{2}p_{i}}{\partial c_{id}^{2}}\ge 0$ and $\frac{\partial ^{2}p_{j} }{\partial c_{id}^{2}}\ge 0$ due to both $m^{\prime }(\delta )<0$ and $m^{\prime }(-\delta )<0$. Otherwise, for the concave Mills Ratio $m^{\prime \prime }\left( \cdot \right) <0$, then $\frac{\partial ^{2}p_{i}}{\partial c_{id}^{2}}\le 0$ and $\frac{\partial ^{2}p_{j}}{\partial c_{id}^{2}}\le 0$. This completes the proof. $\square$

Proof of Proposition 2

(a) Differentiating both sides in (24) with respect to $c_{i}$ yields

$$\begin{aligned} \frac{\partial w_{i}}{\partial c_{i}^{{}}}= & {} 1+\left( \frac{\partial p_{i}}{ \partial c_{i}^{{}}}-\frac{\partial p_{j}}{\partial c_{i}^{{}}}\right) T^{\prime } \end{aligned}$$

(32)

$$\begin{aligned}= & {} 1+\left[ \frac{\partial p_{i}}{\partial w_{i}}\frac{\partial w_{i}}{\partial c_{i}^{{}}}+\frac{\partial p_{i}}{\partial w_{j}^{{}}}\frac{\partial w_{j}}{ \partial c_{i}^{{}}}-\left( \frac{\partial p_{j}}{\partial w_{i}}\frac{\partial w_{i}}{\partial c_{i}^{{}}}+\frac{\partial p_{j}}{\partial w_{j}^{{}}}\frac{ \partial w_{j}}{\partial c_{i}^{{}}}\right) \right] T^{\prime } \end{aligned}$$

(33)

$$\begin{aligned}= & {} 1+\left( \frac{\partial p_{i}}{\partial w_{i}^{{}}}-\frac{\partial p_{j}}{ \partial w_{i}^{{}}}\right) \left( \frac{\partial w_{i}}{\partial c_{i}^{{}}}-\frac{ \partial w_{j}}{\partial c_{i}^{{}}}\right) T^{\prime } \end{aligned}$$

(34)

$$\begin{aligned}= & {} 1+\frac{T^{\prime }}{1-m^{\prime }\left( \delta \right) -m^{\prime }\left( -\delta \right) }\left( \frac{ \partial w_{i}}{\partial c_{i}^{{}}}-\frac{\partial w_{j}}{\partial c_{i}^{{}}}\right) , \end{aligned}$$

(35)

where (33) follows from the chain rule and (34) follows by the fact that the gap between retailer own-brand pass-through and cross-brand pass-through in both channels are the same, i.e., $\frac{ \partial p_{i}}{\partial w_{i}^{{}}}-\frac{\partial p_{j}}{\partial w_{i}^{{}}}=\frac{\partial p_{j}}{\partial w_{j}^{{}}}-\frac{\partial p_{i}}{ \partial w_{j}^{{}}}$, (35) follows from substituting (20) and (21) into (34).

Switching the position of $i$ and $j$ in (24) and differentiating both sides in the resulting equation with respect to $c_{i}$ yields

$$\begin{aligned} \frac{\partial w_{j}}{\partial c_{i}^{{}}}=-\frac{T^{\prime }}{1-m^{\prime }(\delta )-m^{\prime }(-\delta )}\left( \frac{\partial w_{i}}{\partial c_{i}^{{}}}- \frac{\partial w_{j}}{\partial c_{i}^{{}}}\right) \text {.} \end{aligned}$$

(36)

Solving the system of Eqs. (35) and (36) for $\frac{\partial w_{i}}{\partial c_{i}^{{}}}$ and $\frac{\partial w_{j}}{ \partial c_{i}^{{}}}$ leads to

$$\begin{aligned} \frac{\partial w_{i}}{\partial c_{i}^{{}}}= & {} \frac{m^{\prime }\left( -\delta \right) +T^{\prime }+m^{\prime }\left( \delta \right) -1}{m^{\prime }\left( -\delta \right) +2T^{\prime }+m^{\prime }\left( \delta \right) -1}=1- \frac{1}{2-\frac{1-m^{\prime }(\delta )-m^{\prime }(-\delta )}{T^{\prime }}} \in (\frac{1}{2},1]\text { and} \end{aligned}$$

(37)

$$\begin{aligned} \frac{\partial w_{j}}{\partial c_{i}^{{}}}= & {} \frac{-T^{\prime }}{ 1-m^{\prime }(\delta )-m^{\prime }(-\delta )-2T^{\prime }}=\frac{1}{2-\frac{ 1-m^{\prime }(\delta )-m^{\prime }(-\delta )}{T^{\prime }}}\in (0,\frac{1}{2} ], \end{aligned}$$

(38)

where $\frac{\partial w_{i}}{\partial c_{i}^{{}}}>\frac{1}{2}$ and $\frac{ \partial w_{j}}{\partial c_{i}^{{}}}<\frac{1}{2}$ holds due to the fact that $2-\frac{1-m^{\prime }(\delta )-m^{\prime }(-\delta )}{T^{\prime }}>2$ from (20) and $T^{\prime }(\cdot )<0$.

(b) (37) and (38) implies that supplier own-brand pass-through and cross-brand pass-through in both channels are the same, i.e., $\frac{\partial w_{i}}{\partial c_{i}^{{}}}=\frac{\partial w_{j}}{ \partial c_{j}^{{}}}$ and $\frac{\partial w_{j}}{\partial c_{i}^{{}}}=\frac{ \partial w_{i}}{\partial c_{j}^{{}}}$.

Substituting (20), (21), (37) and (38) into (34) yields the channel own-brand and cross-brand pass-through:

$$\begin{aligned} \frac{\partial p_{i}}{\partial c_{i}^{{}}}= & {} \frac{\partial p_{i}}{\partial w_{i}}\frac{\partial w_{i}}{\partial c_{i}^{{}}}+\frac{\partial p_{i}}{ \partial w_{j}^{{}}}\frac{\partial w_{j}}{\partial c_{i}^{{}}} \\= & {} \frac{1-m^{\prime }(-\delta )}{1-m^{\prime }(\delta )-m^{\prime }(-\delta )}\frac{1-m^{\prime }(\delta )-m^{\prime }(-\delta )-T^{\prime }}{ 1-m^{\prime }(\delta )-m^{\prime }(-\delta )-2T^{\prime }}\\&+\frac{-m^{\prime }(\delta )}{1-m^{\prime }(\delta )-m^{\prime }(-\delta )}\frac{-T^{\prime }}{ 1-m^{\prime }(\delta )-m^{\prime }(-\delta )-2T^{\prime }} \\= & {} \frac{1-m^{\prime }(-\delta )-T^{\prime }}{1-m^{\prime }(\delta )-m^{\prime }(-\delta )-2T^{\prime }}<\frac{\partial w_{i}}{\partial c_{i}^{{}}},\text { and} \\ \frac{\partial p_{j}}{\partial c_{i}^{{}}}= & {} \frac{\partial p_{j}}{\partial w_{i}}\frac{\partial w_{i}}{\partial c_{i}^{{}}}+\frac{\partial p_{j}}{ \partial w_{j}^{{}}}\frac{\partial w_{j}}{\partial c_{i}^{{}}} \\= & {} \frac{-m^{\prime }(-\delta )}{1-m^{\prime }(\delta )-m^{\prime }(-\delta ) }\frac{m^{\prime }\left( -\delta \right) +T^{\prime }+m^{\prime }\left( \delta \right) -1}{m^{\prime }\left( -\delta \right) +2T^{\prime }+m^{\prime }\left( \delta \right) -1}\\&+\frac{1-m^{\prime }(\delta )}{1-m^{\prime }(\delta )-m^{\prime }(-\delta )}\frac{-T^{\prime }}{1-m^{\prime }(\delta )-m^{\prime }(-\delta )-2T^{\prime }} \\= & {} \frac{-m^{\prime }(-\delta )-T^{\prime }}{1-m^{\prime }(\delta )-m^{\prime }(-\delta )-2T^{\prime }}>\frac{\partial w_{j}}{\partial c_{i}^{{}}}. \end{aligned}$$

This completes the proof. $\square$

Proof of Proposition 3

Using the threshold-based divisible Mills Ratio, we can rewrite (37) and (38) as

$$\begin{aligned} \frac{\partial w_{i}}{\partial c_{i}^{{}}}(2\widehat{T}-1)= & {} \widehat{T}-1 \text { and} \end{aligned}$$

(39)

$$\begin{aligned} \frac{\partial w_{j}}{\partial c_{i}^{{}}}(2\widehat{T}-1)= & {} \widehat{T} \end{aligned}$$

(40)

Taking the first derivative in both sides of (39) and (40) with respect to $c_{i}^{{}}$ and rearranging the result for $\frac{\partial ^{2}w_{i}}{\partial c_{i}^{2}}$ and $\frac{ \partial ^{2}w_{j}}{\partial c_{i}^{2}}$ yields the following

$$\begin{aligned} \frac{\partial ^{2}w_{i}}{\partial c_{i}^{2}}(2\widehat{T}-1)= & {} \frac{ \widehat{T}^{\prime }}{q_{i}-q_{j}}\left( \frac{\partial p_{i}}{\partial c_{i}^{{}} }-\frac{\partial p_{j}}{\partial c_{i}^{{}}}\right) -2\frac{\partial w_{i}}{ \partial c_{i}^{{}}}\frac{\widehat{T}^{\prime }}{q_{i}-q_{j}}\left( \frac{\partial p_{i}}{\partial c_{i}^{{}}}-\frac{\partial p_{j}}{\partial c_{i}^{{}}}\right) \end{aligned}$$

(41)

$$\begin{aligned}= & {} -\frac{\widehat{T}^{\prime }}{q_{i}-q_{j}}\left( \frac{\partial p_{i}}{\partial c_{i}^{{}}}-\frac{\partial p_{j}}{\partial c_{i}^{{}}}\right) \frac{1-m^{\prime }(\delta )-m^{\prime }(-\delta )}{1-m^{\prime }(\delta )-m^{\prime }(-\delta )-2T^{\prime }} \end{aligned}$$

(42)

$$\begin{aligned}= & {} -\frac{\widehat{T}^{\prime }}{q_{i}-q_{j}}\frac{1-m^{\prime }(\delta )-m^{\prime }(-\delta )}{[1-m^{\prime }(\delta )-m^{\prime }(-\delta )-2T^{\prime }]^{2}}, \end{aligned}$$

(43)

where (42) follows from substituting (37) into (41) and the last equality follows by use of the definition of $\widehat{T}$. From (43) we see that $\frac{\partial ^{2}w_{i} }{\partial c_{i}^{2}}>0$ if and only if $\widehat{T}^{\prime }>0, \frac{ \partial ^{2}w_{i}}{\partial c_{i}^{2}}=0$ if and only if $\widehat{T} ^{\prime }=0$ and $\frac{\partial ^{2}w_{i}}{\partial c_{i}^{2}}<0$ if and only if $\widehat{T}^{\prime }<0$ given that $2\widehat{T}-1<0$. The same results hold for $\frac{\partial ^{2}w_{j}}{\partial c_{i}^{2}}$ by use of exactly the same derivation process, which completes the proof. $\square$

Proof of Proposition 4

Solving the system of Eqs. (18) and (23) leads to analytical solutions

$$\begin{aligned} p_{i}^{*}= & {} \frac{A}{\left( k-2\right) \left( k+2\right) \left( k+2k^{2}-4\right) \left( -k+2k^{2}-4\right) }, \\ p_{j}^{*}= & {} \frac{B}{\left( k-2\right) \left( k+2\right) \left( -k+2k^{2}-4\right) \left( k+2k^{2}-4\right) }, \\ w_{i}^{*}= & {} \frac{C}{\left( k+2k^{2}-4\right) \left( -k+2k^{2}-4\right) }\text { and} \\ w_{j}^{*}= & {} \frac{D}{\left( -k+2k^{2}-4\right) \left( k+2k^{2}-4\right) }, \end{aligned}$$

where $A=48\delta -16c_{i}+36k\delta -12kc_{j}-48aq_{i}-34k^{2}\delta -24k^{3}\delta +6k^{4}\delta +4k^{5}\delta +14k^{2}c_{i}+10k^{3}c_{j}-3k^{4}c_{i}-2k^{5}c_{j}+12akq_{j}+70ak^{2}q_{i}-10ak^{3}q_{j}-30ak^{4}q_{i}+2ak^{5}q_{j}+4ak^{6}q_{i}, B=48\delta -16c_{j}+36k\delta -12kc_{i}-48aq_{j}-34k^{2}\delta -24k^{3}\delta +6k^{4}\delta +4k^{5}\delta +14k^{2}c_{j}+10k^{3}c_{i}-3k^{4}c_{j}-2k^{5}c_{i}+12akq_{i}+70ak^{2}q_{j}-10ak^{3}q_{i}-30ak^{4}q_{j}+2ak^{5}q_{i}+4ak^{6}q_{j}, C=-8\delta +8c_{i}-6k\delta +2kc_{j}+8aq_{i}+3k^{2}\delta +2k^{3}\delta -8k^{2}c_{i}-k^{3}c_{j}+2k^{4}c_{i}-2akq_{j}-9ak^{2}q_{i}+ak^{3}q_{j}+2ak^{4}q_{i},$ and $D=-8\delta +8c_{j}-6k\delta +2kc_{i}+8aq_{j}+3k^{2}\delta +2k^{3}\delta -8k^{2}c_{j}-k^{3}c_{i}+2k^{4}c_{j}-2akq_{i}-9ak^{2}q_{j}+ak^{3}q_{i}+2ak^{4}q_{j}$.

Taking the first derivative of $p_{i}^{*}$ and $w_{i}^{*}$ with respect to parameters leads to

$$\begin{aligned} \frac{dp_{i}^{*}}{d\delta }= & {} 2\frac{k^{2}-3}{\left( k-2\right) \left( k+2k^{2}-4\right) }<0, \\ \frac{dp_{i}^{*}}{dc_{i}}= & {} -\frac{\left( 3k^{2}-8\right) (k^{2}-2)}{ \left( k-2\right) \left( k+2\right) \left( k+2k^{2}-4\right) \left( -k+2k^{2}-4\right) }>0, \\ \frac{dp_{i}^{*}}{dc_{j}}= & {} -\frac{2k\left( k^{2}-2\right) (k^{2}-3)}{ \left( k-2\right) \left( k+2\right) \left( -k+2k^{2}-4\right) \left( k+2k^{2}-4\right) }>0, \\ \frac{dp_{i}^{*}}{dq_{i}}= & {} \frac{2a\left( k^{2}-3\right) (-9k^{2}+2k^{4}+8)}{\left( k-2\right) \left( k+2\right) \left( k+2k^{2}-4\right) \left( -k+2k^{2}-4\right) }>0, \\ \frac{dp_{i}^{*}}{dq_{j}}= & {} \frac{2ak\left( k^{2}-2\right) (k^{2}-3)}{ \left( k-2\right) \left( k+2\right) \left( k+2k^{2}-4\right) \left( -k+2k^{2}-4\right) } <0\text { and} \\ \frac{dw_{i}^{*}}{d\delta }= & {} \frac{k+2}{k+2k^{2}-4}<0, \\ \frac{dw_{i}^{*}}{dc_{i}}= & {} \frac{2\left( k^{2}-2\right) ^{2}}{\left( k+2k^{2}-4\right) \left( -k+2k^{2}-4\right) }>0, \\ \frac{dw_{i}^{*}}{dc_{j}}= & {} \frac{-k(k^{2}-2)}{\left( -k+2k^{2}-4\right) \left( k+2k^{2}-4\right) }>0, \\ \frac{dw_{i}^{*}}{dq_{i}}= & {} a\frac{-9k^{2}+2k^{4}+8}{\left( -k+2k^{2}-4\right) \left( k+2k^{2}-4\right) }>0, \\ \frac{dw_{i}^{*}}{dq_{j}}= & {} \frac{ ak(k^{2}-2)}{\left( k+2k^{2}-4\right) \left( -k+2k^{2}-4\right) }<0. \end{aligned}$$

This completes the proof. $\square$