Pricing and consumption in the P2P product sharing era: How does the dual-channel manufacturer cooperate with third-party sharing platforms?

Abstract

Recently, mobile communication technologies and sharing platforms have made peer-to-peer product sharing among consumers a major trend in the sharing economy. Product sharing has changed not only the purchasing and use behavior of consumers but also the operating decisions of manufacturers. This paper employs a game-theoretic analytical model to study consumers’ best choices, their sustainable consumption behavior, and the manufacturer’s optimal pricing strategy. Our analysis shows that peer-to-peer product sharing will increase the optimal prices, decrease customers’ demand for the product and reduce the profits of the manufacturer.

Appendices

Appendix 1: Proof of Proposition 2

Plugging Eq. (8) into Eq. (7) and Eq. (4), we can obtain consumers’ purchasing demand with product sharing ${D^S} = {{d(\beta (1 - d) + d)} / {(2(\beta + (1 - \beta ){d^2}))}}$. While in the no product-sharing scenario ${D^N} = {1 / 2}$, for $0 < d \le 1$ and $0 < \beta \le 1$, we can prove ${D^N} - {D^S} = {{\beta (1 - d)} / {(2(\beta + (1 - \beta ){d^2}))}} > 0$, thus, the product sharing will reduce the purchasing demand. Due to ${{d(\Delta {D^P})} / {d\beta }} = {{(1 - d){d^2}} / {(2(\beta + (1 - \beta ){d^2}))}} > 0$, $\Delta {D^P}$ will increase when $\beta$ increases. For ${{d(\Delta {D^P})} / {d(d)}} = - {{\beta (\beta {{(1 - d)}^2} + d(2 - d))} / {(2(\beta + (1 - \beta ){d^2}))}} < 0$, $\Delta {D^P}$ will decrease when d increases. Plugging Eq.  (8) into Eq. (7) and Eq. (5), we have consumers’ rent in demand ${D^R}$. Then we can obtain ${D^S} + {D^R} - {D^N} = {{d(1 - d)(1 - \beta )} / {(2(\beta + (1 - \beta ){d^2})}}$, we can easily prove $\Delta {D^U} > 0$, which implies product sharing will increase the total amount of products users. For ${{d(\Delta {D^U})} / {d\beta }} = - {{\beta (1 - \beta )} / {(2{{(\beta + (1 - \beta ){d^2})}^2})}} < 0$, $\Delta {D^U}$ will decrease when $\beta$ increases. For ${{d(\Delta {D^U})} / {d(d)}} = {{(1 - \beta )(\beta {{(1 - d)}^2} - {d^2})} / {2{{(\beta + (1 - \beta ){d^2})}^2}}}$, if $0 < d \le {{\sqrt{\beta }} / {(1 + \sqrt{\beta })}}$, we have ${{d(\Delta {D^U})} / {d(d)}} > 0$, then $\Delta {D^U}$ will increase in d, if ${{\sqrt{\beta }} \ {(1 + \sqrt{\beta })}} < d \le 1$, we have ${{d(\Delta {D^U})} / {d(d)}} < 0$, then $\Delta {D^U}$ will decrease in d.

Appendix 2: Proof of Proposition 3

First $\Delta p = {p^S} - {p^N} = {{\beta (1 - d)} / 2} > 0$,, for $\beta > 0$ and $1 - d > 0$, we can easily prove ${p^S} > {p^N}$ and $\Delta p$ will increase when $\beta$ increases, and decrease when d increases. Second, $\Delta \pi = {\pi ^N} - {\pi ^S} = {{d\beta (1 - \beta ){{(1 - d)}^2}} / {(4(\beta (1 - {d^2}) + {d^2}))}}$, we can easily prove that ${\pi ^N} > {\pi ^S}$. If ${d / {(1 + d)}}< \beta < 1$, we have ${(1 - \beta )^2}{d^2} - {\beta ^2} < 0$ and ${{d(\Delta \pi )} / {d\beta }} < 0$, then $\Delta \pi$ will decrease with the increase of $\beta$. If $0< \beta < {d / {(1 + d)}}$, we have ${(1 - \beta )^2}{d^2} - {\beta ^2} > 0$ and ${{d(\Delta \pi )} / {d\beta }} > 0$, then $\Delta \pi$ will increase with the increase of $\beta$. Third, if $0< d < {d^*}$, we have ${{d(\Delta \pi )} / {d(d)}} > 0$, then $\Delta \pi$ will increase with the increase of d. If ${d^*}< d < 1$, we have ${{d(\Delta \pi )} / {d(d)}} < 0$, then $\Delta \pi$ will decrease with the increase of d. Note that $d* = \root 3 \of {{\root 2 \of {{({A^2} - {B^3})}} - A}} + {B / {\root 3 \of {{\root 2 \of {{({A^2} - {B^3})}} - A}}}} - {1 / 3}$, and $A = {1 / {27}} - {\beta / {(1 - \beta )}}$, $B = {1 / 9} - {\beta / {(1 - \beta )}}$.