The optimal pricing decisions for e-tailers with different payment schemes

Abstract

Along with the quick development of e-commerce, different payment schemes are provided to online consumers to improve their shopping experience. Currently, the payment schemes can be divided into two categories, one is pay-to-order, and the other is pay-on-delivery. Payment scheme directly affects consumers’ behavior and e-tailer’s pricing decision in e-commerce. In this paper, we characterize the consumers’ purchase and returns behavior with consumer utility function, build the e-tailer’s profit functions and solve them to obtain the optimal pricing decisions, both in the condition of pay-to-order and dual scheme (including both pay-to-order and pay-on-delivery). We find that managers can affect consumers’ decision on choosing payment scheme by adjusting the e-tail price. We demonstrate the transfer of payment scheme, along with the transaction cost and return cost. Sensitivity analysis is provided to tell e-commerce managers that they should consider the influence of the e-purchase cost, together with the characteristic of both consumers and products when designing the payment scheme.

Appendix

1.1 Proof of Proposition 1

According to Figs. 2, 3 and 4, we can get, that ① when $\beta_{od} > 2\bar{\beta }$, we can always get $S_{1} (\beta ) > S_{2} (\beta )$, which means that all consumers will choose pay-to-order. Under this condition, $0 < p_{d} < \frac{{\alpha - 2\bar{\beta }}}{H}$. ② When $0 \le \beta_{od} \le 2\bar{\beta }$ and $S_{2} (\beta ) > 0$, we can get $S_{1} (\beta ) > S_{2} (\beta )$ if with $\beta \sim (0,\beta_{od} )$, indicating that consumers will choose pay-to-order; $S_{1} (\beta ) < S_{2} (\beta )$ if with $\beta \sim (\beta_{od} ,2\bar{\beta })$, indicating that consumers will choose pay-on-delivery. From $0 \le \beta_{od} \le 2\bar{\beta }$, we can get $\frac{{\alpha - 2\bar{\beta }}}{H} \le p_{d} \le \frac{\alpha }{H}$; from $S_{2} (\beta ) > 0$, we can get $p_{d} < \frac{\theta v - \alpha }{\theta (1 + \sigma + \eta )}$. ③ When $S_{1} (0) < S_{2} (\beta )$ and $S_{2} (\beta ) > 0$, the utility of the consumer choosing pay-on-delivery is always greater than the one choosing pay-to-order, that is, consumers will choose pay-on-delivery. From $S_{1} (0) = \theta (v - p_{d} ) - \sigma p_{d} - (1 - \theta )\gamma p_{d} < S_{2} (\beta ) = \theta (v - p_{d} - (\sigma + \eta )p_{d} ) - a$, we can get $p_{d} > \frac{\alpha }{H}$; from $S_{2} (\beta ) > 0$, we can get $p_{d} < \frac{\theta v - \alpha }{\theta (1 + \sigma + \eta )}$. If this situation is reasonable, then there should be $\frac{\alpha }{H} < \frac{\theta v - \alpha }{\theta (1 + \sigma + \eta )}$. Further simplification can be obtained, that is $\alpha < \frac{\theta vH}{H + \theta (1 + \sigma + \eta )}$.

1.2 Proof of Proposition 2

According to Proposition 1, ① when $p_{d} \in (0,\hbox{max} \{ 0,\frac{{\alpha - 2\bar{\beta }}}{H}\} )$, all consumers will choose pay-to-order, that is, $D = D_{do} = 1$; ② when $\frac{\alpha }{H} \le \frac{\theta v - \alpha }{\theta (1 + \sigma + \eta )}$, which means $\alpha \le \frac{\theta vH}{H + \theta (1 + \sigma + \eta )}$, we can get $\, p_{d} \in (\hbox{max} \{ 0,\frac{{\alpha - 2\bar{\beta }}}{H}\} ,\frac{\alpha }{H}]$, and both payment methods will be selected under this condition. Here, $D_{do} = F(\beta_{od} ) = \frac{{\alpha - Hp_{d} }}{{2\bar{\beta }}}$, $D_{dd} = 1 - D_{do} = \frac{{2\bar{\beta } - \alpha + Hp_{d} }}{{2\bar{\beta }}}$; when $\frac{\alpha }{H} > \frac{\theta v - \alpha }{\theta (1 + \sigma + \eta )}$, which is $\alpha > \frac{\theta vH}{H + \theta (1 + \sigma + \eta )}$, furthermore, we can get $\alpha < \frac{{\theta vH + 2\bar{\beta }\theta (1 + \sigma + \eta )}}{H + \theta (1 + \sigma + \eta )}$ from $\frac{{\alpha - 2\bar{\beta }}}{H} < \frac{\theta v - \alpha }{\theta (1 + \sigma + \eta )}$; then, we can see $\, p_{d} \in (\hbox{max} \{ 0,\frac{{\alpha - 2\bar{\beta }}}{H}\} ,\frac{\theta v - \alpha }{\theta (1 + \sigma + \eta )}]$. Under this situation, two payment methods coexist, among which $D_{do} = F(\beta_{od} ) = \frac{{\alpha - Hp_{d} }}{{2\bar{\beta }}}$ and $D_{dd} = 1 - D_{do} = \frac{{2\bar{\beta } - \alpha + Hp_{d} }}{{2\bar{\beta }}}$; if $\frac{\alpha }{H} \ge \frac{\theta v - \alpha }{\theta (1 + \sigma + \eta )}$ and $\frac{{\alpha - 2\bar{\beta }}}{H} \ge \frac{\theta v - \alpha }{\theta (1 + \sigma + \eta )}$, then $\alpha \ge \frac{{\theta vH + 2\bar{\beta }\theta (1 + \sigma + \eta )}}{H + \theta (1 + \sigma + \eta )}$; here, we can always get $S_{1} (2\bar{\beta }) > S_{2} (\beta )$, which is $\, D = D_{do} = 1$; ③ when $\alpha < \frac{\theta vH}{H + \theta (1 + \sigma + \eta )}$ and $\frac{\alpha }{H} < p_{d} < \frac{\theta v - \alpha }{\theta (1 + \sigma + \eta )}$, all consumers will choose pay-on-delivery, which is $D = D_{dd} = 1$. Taking the above situations into account, the distribution of market demands corresponding to different $\alpha$ and $p_{d}$ can be obtained.

1.3 Proof of Proposition 3

① In pay-to-order, according to the retailer’s profit function, when $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{p}_{o} < p_{o} < \bar{p}_{o}$, we can get $\frac{{\partial^{2} \pi_{1} }}{{\partial p_{o}^{2} }} = - \frac{AB}{{\bar{\beta }}} < 0$, that is, there is a unique optimal value $p_{o}^{*}$ that maximizes the profit function value, which satisfies $\frac{{\partial \pi_{1} }}{{\partial p_{o} }} = \frac{{A\theta v + Bc - 2ABp_{o} }}{{2\bar{\beta }}} = 0$. Hence, $p_{o}^{*} = (A\theta v + Bc)/(2AB)$. Substituting x into $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{p}_{o} < p_{o}^{*} < \bar{p}_{o}$, we can get $\frac{{\theta v - 2\bar{\beta }}}{B} < \frac{A\theta v + Bc}{2AB} < \frac{\theta v}{B}$ and further to simplify it to get $\frac{{A\theta v - 4A\bar{\beta }}}{B} < c < \frac{A\theta v}{B}$. Substituting $p_{o}^{*}$ into the market demand function, the profit function can obtain equilibrium market demand and equilibrium profits within this range. ② When $0 < p_{o} \le \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{p}_{o}$, the first derivative of the corporate profit function for retail prices is positive, that is $\frac{{\partial \pi_{1} }}{{\partial p_{o} }} = A > 0$, which means the profit is a monotonically increasing function of retail prices. Thus, the profit function gets the maximum value when $p_{o}$ takes the maximum $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{p}_{o}$, that is $p_{o}^{*\prime } = (\theta v - 2\bar{\beta })/B$. At this point, the market demand function is $D_{1}^{\prime } = 1$, and substituting it into the profit function can be obtained $\pi_{1}^{*\prime } = Ap_{o} - c$.

1.4 Proof of Corollary 1

According to Proposition 3, we can get: ① When $c \in ((A\theta v - 4A\bar{\beta })/B,A\theta v/B)$, $\frac{{\partial p_{o}^{*} }}{\partial c} = \frac{1}{2A} > 0$, which means $p_{o}^{*}$ is a monotonically increasing function of $c$; from $\frac{{\partial \pi_{1}^{*} }}{\partial c} = \frac{Bc - A\theta v}{{4A\bar{\beta }}}$, we see that $\frac{{\partial \pi_{1}^{*} }}{\partial c} < 0$ if with $c < \frac{A\theta v}{B}$, that is, $\pi_{1}^{*}$ is a monotonically decreasing function with respect to $c$ in this region. ② When $c \in [0,(A\theta v - 4A\bar{\beta })/B]$, $\frac{{\partial \pi_{1}^{*} }}{\partial c} < 0$, that is, $p_{o}^{*\prime }$ is not affected by $c$; the equation $\frac{{\partial \pi_{1}^{*'} }}{\partial c} = - 1 < 0$ indicates that $\pi_{1}^{*\prime }$ is a monotonically decreasing function about $c$.

1.5 Proof of Proposition 4

Name $G = \theta + (\xi - \lambda )(1 - \theta )$, $K = \lambda (1 - \theta ) + rt$, $G + K = A$, $H = (1 - \theta )(\sigma + \gamma ) - \theta \eta$ and $H + \theta (1 + \sigma + \eta ) = B$.

Then Eq. (11) can be simplified as $\pi_{2} = Ap_{d} D_{do} + Gp_{d} D_{dd} - c$.

Case (1) $0 < a \le \frac{\theta vH}{B}$

① When $\frac{a}{H} \le p_{d} \le \frac{\theta v - \alpha }{\theta (1 + \sigma + \eta )}$, the market demand function satisfies $D_{d} = D_{dd} = 1$, $D_{do} = 0$. Retailer profit function is $\pi_{2} = Gp_{d} - c$. Hence, the optimal retail price is $p_{d}^{*} = p_{dd} = \frac{\theta v - \alpha }{\theta (1 + \sigma + \eta )}$, and the optimal profit function is $\pi_{dd} = G\frac{\theta v - \alpha }{\theta (1 + \sigma + \eta )} - c$. ② When $a > 2\bar{\beta }$ and $\frac{{a - 2\bar{\beta }}}{H} < p_{d2} < \frac{\alpha }{H}$, retailer profit function is $\pi_{2} = Ap_{d} \frac{{\alpha - Hp_{d} }}{{2\bar{\beta }}} + Gp_{d} \frac{{2\bar{\beta } - \alpha + Hp_{d} }}{{2\bar{\beta }}} - c$; with further simplification, we can get $\pi_{2} = \frac{1}{{2\bar{\beta }}}[(G - A)Hp_{d}^{2} + (2\bar{\beta } - \alpha )Gp_{d} + Aap_{d} ] - c$ and the first derivative of the price $\frac{{\partial \pi_{2} }}{{\partial p_{d} }} = \frac{1}{{2\bar{\beta }}}[2(G - A)Hp_{d} + (2\bar{\beta } - \alpha )G + Aa]$. Since the second derivative of the profit function with respect to price is $\frac{{\partial \pi_{2}^{2} }}{{\partial^{2} p_{d} }} = \frac{1}{{\bar{\beta }}}(G - A)H < 0$, there is a unique optimal solution that maximizes the profit function value, which meets $\frac{{\partial \pi_{2} }}{{\partial p_{d} }} = \frac{1}{{2\bar{\beta }}}[2(G - A)Hp_{d} + (2\bar{\beta } - \alpha )G + Aa] = 0$, and we can get the optimal price $p_{d2} = \frac{\alpha }{2H} + \frac{{G\bar{\beta }}}{KH}$. Let $p_{d2} > \frac{{\alpha - 2\bar{\beta }}}{H}$, and substituting it into the above formula can achieve $a < 4\bar{\beta } + \frac{{2\bar{\beta }G}}{K}$; let $p_{d2} < \frac{\alpha }{H}$, and substituting it into the above formula can achieve $a > \frac{{2\bar{\beta }G}}{K}$. Therefore, we can get $\hbox{max} \{ 2\bar{\beta },\frac{{2\bar{\beta }G}}{K}\} < a \le \hbox{min} \{ \frac{\theta vH}{B},4\bar{\beta } + \frac{{2\bar{\beta }G}}{K}\}$. ③ When $0 < a < 2\bar{\beta }$ and $0 < p_{d2} < \frac{a}{H}$, the optimal retail price is $p_{d2} = \frac{\alpha }{2H} + \frac{{G\bar{\beta }}}{KH}$. Let $p_{d2} < \frac{\alpha }{H}$, we can get $a > \frac{{2\bar{\beta }G}}{K}$; then, $\frac{{2\bar{\beta }G}}{K} < a \le \hbox{min} \{ \frac{\theta vH}{B},2\bar{\beta }\}$. Combining ② and ③, there is $\frac{{2\bar{\beta }G}}{K} < a < \hbox{min} \{ \frac{\theta vH}{B},2\bar{\beta }\}$. ④ When $a > 2\bar{\beta }$ and $0 < p_{do} \le \frac{{a - 2\bar{\beta }}}{H}$, $D_{d} = D_{do} = 1$, retailer profit function is $\pi_{2} = Ap_{do} D_{do} - c$. Therefore, the optimal retail price is $p_{do} = \frac{{\alpha - 2\bar{\beta }}}{H}$, and the optimal profit function is $\pi_{2} = A\frac{{\alpha - 2\bar{\beta }}}{H} - c$.

Case (2) $\frac{\theta vH}{B} < \alpha < \frac{{\theta vH + 2\bar{\beta }\theta (1 + \sigma + \eta )}}{B}$

① When $a > 2\bar{\beta }$ and $\frac{{a - 2\bar{\beta }}}{H} < p_{d} < \frac{\theta v - \alpha }{\theta (1 + \sigma + \eta )}$, similar to the proof process of scenario (1), the retailer’s optimal retail price satisfies $p_{d2} = \frac{\alpha }{2H} + \frac{{G\bar{\beta }}}{KH}$. Let $p_{d2} < \frac{\theta v - \alpha }{\theta (1 + \sigma + \eta )}$, we can get $a < \frac{{2\theta KHv - 2\theta \bar{\beta }G(1 + \sigma + \eta ))}}{K(H + B)}$; combining the previous proof, we can get $a < 4\bar{\beta } + \frac{{2\bar{\beta }G}}{K}$ from $p_{d} > \frac{{a - 2\bar{\beta }}}{H}$. Hence, $\hbox{max} \{ 2\bar{\beta },\frac{\theta vH}{B}\} < a < \hbox{min} \{ \frac{{\theta vH + 2\bar{\beta }\theta (1 + \sigma + \eta )}}{B},4\bar{\beta } + \frac{{2\bar{\beta }G}}{K},\frac{{2\theta KHv - 2\theta \bar{\beta }G(1 + \sigma + \eta ))}}{K(H + B)}\}$.

② When $a < 2\bar{\beta }$ and $0 < p_{d} < \frac{\theta v - \alpha }{\theta (1 + \sigma + \eta )}$, the retailer’s optimal price is $p_{d2} = \frac{\alpha }{2H} + \frac{{G\bar{\beta }}}{KH}$, and we also get $a < \frac{{2\theta KHv - 2\theta \bar{\beta }G(1 + \sigma + \eta ))}}{K(H + B)}$. Therefore, $\frac{\theta vH}{B} < a < \hbox{min} \{ 2\bar{\beta },\frac{{2\theta KHv - 2\theta \bar{\beta }G(1 + \sigma + \eta ))}}{K(H + B)}\}$.

③ When $a \ge 2\bar{\beta }$ and $0 < p_{d} \le \frac{{a - 2\bar{\beta }}}{H}$, the retailer’s optimal retail price meets $p_{do} = \frac{{\alpha - 2\bar{\beta }}}{H}$ and the corresponding optimal profit function is $\pi_{do} = A\frac{{\alpha - 2\bar{\beta }}}{H} - c$.

Case (3) $\alpha \ge \frac{{\theta vH + 2\bar{\beta }\theta (1 + \sigma + \eta )}}{B}$

Combining with Proposition 2, we know that, under this condition, consumers choose pay-to-order to purchase online, so the optimal retail price is $p_{do} = \frac{{\theta v - 2\bar{\beta }}}{B}$, and the corresponding profit is $\pi_{do} = A\frac{{\theta v - 2\bar{\beta }}}{B} - c$.

Integrating above cases (1), (2) and (3), we can obtain the optimal retail prices and profits under different circumstances, as shown in Table 1.