Optimal pricing and capacity sizing for online service systems with free trials

Abstract

This paper studies a revenue and social welfare optimization problem for an online service system which provides service-free trials to first-time users. There are two customer types: experienced customers and first-time customers, where type-1 (experienced) customers are fully rational, and they directly purchase service upon arrivals, while type-2 (first-time) customers are boundedly rational, so they may consider to purchase service or not to purchase service after experiencing the free trials. We investigate the optimal pricing strategy of service providers and social planners in two cases: service capacity is exogenous and endogenous. In the latter case, we jointly determine the optimal service price and capacity. We adopt a game theoretical approach: First, for a given price and service rate, we characterize the equilibrium behavior of post-trial customers; next, we solve optimization problems for the service provider and social planners to determine the optimal price and service rate. Our analysis reveals interesting insights and practical implications. For example, in contrast to the conventional wisdom, the service provider’s optimal revenue decreases as customers become less rational.

Appendix

In this appendix, we provide the proofs of all results in the main paper.

Proof of Theorem 3.1

Taking the first-order derivative of (10) with respect to $\phi$ gives that

$$\begin{aligned} \frac{d \varPi _{ex}}{d \phi }=\lambda q \left( R-\beta \ln \frac{\phi }{1-\phi }-\frac{C e^{\alpha \phi }}{\mu }\right) -\lambda y \left( \frac{\beta }{\phi (1-\phi )}+\frac{\alpha C e^{\alpha \phi }}{\mu }\right) , \end{aligned}$$

(44)

it is easily found that $d \varPi _{ex}/d \phi$ is decreasing in $\phi \in [1/2, 1)$ and $\lim _{\phi \rightarrow 1} d\varPi _{ex}/d \phi =- \infty$. We shall consider the value of $(d \varPi _{ex}/d \phi )|_{\phi =1/2}$ and define $g_1(\phi )=d \varPi _{ex}/d \phi$. Then, we consider the maximal value of $g_1(1/2)$:

• (1) if $g_1(1/2)<0$, $\varPi _{ex}$ is decreasing in $\phi \in [1/2,1)$, thus the maximal revenue was obtained at $\phi =1/2$;

• (2) if $g_1(1/2)\ge 0$, the function $\varPi _{ex}$ is concave in $\phi \in [1/2,1)$, the optimal joining probability of fresh customers $\phi _{ex}^{\varPi ^*}$ is the unique solution of $d \varPi _{ex}/d \phi =0$.

The results above are concluded in Theorem 3.1. $\square$

Proof of Theorem 3.3

By taking the first-order derivative of $SW_{ex}$ with respect to $\phi$, it gives us that

$$\begin{aligned} \frac{d SW_{ex}}{d \phi }=\lambda \left( q R-(\alpha y+q) \frac{C e^{\alpha \phi }}{\mu }\right) , \end{aligned}$$

(45)

where $d SW_{ex}/d \phi$ is decreasing in $\phi \in [1/2, 1)$, then we define $g_2(\phi )=q R-(\alpha y+q) C e^{\alpha \phi }/\mu$. Similar to Theorem 3.1,

• (1) if $g_2(1/2)<0$, the social welfare $SW_{ex}$ is decreasing in $\phi \in [1/2,1)$, thus the socially optimal joining probability is $\phi _{ex}^{SW^*}=1/2$;

• (2) if $g_2(1/2)\ge 0$, the function $SW_{ex}$ is concave in $\phi \in [1/2,1)$, then $\phi _{ex}^{SW^*}$ is the unique solution of $d SW_{ex}/d \phi =0$.

The results above are concluded in Theorem 3.3. $\square$

Proof of Proposition 3.4

By dividing the range of service reward R, we find:

• (1) if $R<\left( 1+\frac{\alpha (1-\frac{q}{2})}{q}\right) \frac{C e^{\frac{1}{2} \alpha }}{\mu }$, then $p_{ex}^{SW^*}=p_{ex}^{\varPi ^*}=R-\frac{C e^{\frac{1}{2} \alpha }}{\mu }$;

• (2) if $\left( 1+\frac{\alpha (1-\frac{q}{2})}{q}\right) \frac{C e^{\frac{1}{2} \alpha }}{\mu }\le R<\frac{2 \beta (2-q)}{q}+\left( 1+\frac{\alpha (1-\frac{q}{2})}{q}\right) \frac{C e^{\frac{1}{2} \alpha }}{\mu }$, then $p_{ex}^{SW^*}=R-\beta \ln \frac{\phi _{ex}^{SW^*}}{1-\phi _{ex}^{SW^*}}-\frac{C e^{\alpha \phi _{ex}^{SW^*}}}{\mu }\le R-\frac{C e^{\frac{1}{2} \alpha }}{\mu }=p_{ex}^{\varPi ^*}$ since $p(\phi )=R-\beta \ln \frac{\phi }{1-\phi }-\frac{C e^{\alpha \phi }}{\mu }$ is decreasing in $\phi \in [1/2,1)$;

• (3) if $R\ge \frac{2 \beta (2-q)}{q}+\left( 1+\frac{\alpha (1-\frac{q}{2})}{q}\right) \frac{C e^{\frac{1}{2} \alpha }}{\mu }$, we define

  $$\begin{aligned} f_1(\phi )=\frac{d SW_{ex}}{d \phi }-\frac{d \varPi _{ex}}{d \phi }=\lambda q \beta \ln \frac{\phi }{1-\phi }+\frac{\lambda y \beta }{\phi (1-\phi )}. \end{aligned}$$

  It is obvious that $f_1(\phi )$ is increasing in [1/2, 1) where $f_1(1/2)=4 \lambda y \beta >0$ and $\lim _{\phi \rightarrow 1} f_1(\phi )=+ \infty$, that is $d SW_{ex}/d \phi >d \varPi _{ex}/d \phi$ in the domain of $\phi$. Recall that $d SW_{ex}/d \phi$ and $d \varPi _{ex}/d \phi$ are decreasing in $\phi$, the order of optimal points is $\phi _{ex}^{SW^*}>\phi _{ex}^{\varPi ^*}$ and then $p_{ex}^{SW^*}<p_{ex}^{\varPi ^*}$. To sum up, the socially optimal price is no greater than the revenue-optimal one. $\square$

Proof of Corollary 3.6

In the revenue-optimal case,

• (1) if $R<2 \beta (2-q)/q+\left( 1+\alpha (1-q/2)/q\right) C e^{\frac{1}{2} \alpha }/\mu$, $\phi _{ex}^{\varPi ^*}=1/2$, and the equation $\partial \phi _{ex}^{\varPi ^*}/\partial \beta =0$ always holds;

• (2) otherwise, it is solved by equation (12). Differentiating it with respect to $\beta$ in both sides, we have

  $$\begin{aligned} \frac{\partial \phi _{ex}^{\varPi ^*}}{\partial \beta }=-\frac{\ln \frac{\phi _{ex}^{\varPi ^*}}{1-\phi _{ex}^{\varPi ^*}}+\frac{y_{ex}^{\varPi ^*}}{q \phi _{ex}^{\varPi ^*} (1-\phi _{ex}^{\varPi ^*})}}{\frac{(2 \phi _{ex}^{\varPi ^*}-y_{ex}^{\varPi ^*}) \beta }{q (\phi _{ex}^{\varPi ^*})^2 (1-\phi _{ex}^{\varPi ^*})^2}+\left( 2+\frac{\alpha y_{ex}^{\varPi ^*}}{q}\right) \frac{\alpha C e^{\alpha \phi _{ex}^{\varPi ^*}}}{\mu }}<0, \end{aligned}$$

  where $y_{ex}^{\varPi ^*}=1-q+q \phi _{ex}^{\varPi ^*}$. This shows that $\phi _{ex}^{\varPi ^*}$ is decreasing in $\beta$.

Therefore, the revenue-optimal joining probability of fresh customers is non-increasing with their level of bounded rationality.

In socially optimal case, we consider the value of service reward R,

• (1) if $R<\left( \alpha (1-q/2)/q+1\right) C e^{1/2\alpha }/\mu$, $\phi _{ex}^{SW^*}=1/2$, and the equation $\partial \phi _{ex}^{SW^*}/\partial \beta =0$ always holds;

• (2) otherwise, it is solved by equation (56). Differentiating it with respect to $\beta$ in both sides, we have

  $$\begin{aligned} \frac{\partial \phi _{ex}^{SW^*}}{\partial \beta }=0. \end{aligned}$$

  This shows that the socially optimal price is not dependent on $\beta$.

While considering the sensitivity of $\beta$ on price:

• (1) if $R<2 \beta (2-q)/q+\left( 1+\alpha (1-q/2)/q\right) C e^{1/2 \alpha }/\mu$, the optimal price $p_{ex}^{\varPi ^*}=R-C e^{1/2 \alpha }/\mu$, thus $\partial p_{ex}^{\varPi ^*}/\partial \beta =0$;

• (2) otherwise,

  $$\begin{aligned} \frac{\partial p_{ex}^{\varPi ^*}}{\partial \beta }=-\ln \frac{\phi _{ex}^{\varPi ^*}}{1-\phi _{ex}^{\varPi ^*}}-\left( \frac{\beta }{\phi _{ex}^{\varPi ^*} (1-\phi _{ex}^{\varPi ^*})}+\frac{\alpha C e^{\alpha \phi _{ex}^{\varPi ^*}}}{\mu }\right) \frac{\partial \phi _{ex}^{\varPi ^*}}{\partial \beta }, \end{aligned}$$

  the first item $-\ln (\phi _{ex}^{\varPi ^*}/(1-\phi _{ex}^{\varPi ^*}))\le 0$ and the second item is nonnegative. We do numerical examples to explore the monotonicity of $p_{ex}^{\varPi ^*}$ due to the complexity of it.

In the socially optimal case,

• (1) if $R<\left( \alpha (1-q/2)/q+1\right) C e^{1/2\alpha }/\mu$, the optimal price $p_{ex}^{SW^*} = R-C e^{\frac{1}{2} \alpha }/\mu$, thus $\partial p_{ex}^{SW^*}/\partial \beta =0$;

• (2) otherwise,

  $$\begin{aligned} \frac{\partial p_{ex}^{SW^*}}{\partial \beta }= & {} -\ln \frac{\phi _{ex}^{SW^*}}{1-\phi _{ex}^{SW^*}}-\left( \frac{\beta }{\phi _{ex}^{SW^*} (1-\phi _{ex}^{SW^*})}+\frac{\alpha C e^{\alpha \phi _{ex}^{SW^*}}}{\mu }\right) \frac{\partial \phi _{ex}^{SW^*}}{\partial \beta }\\= & {} -\ln \frac{\phi _{ex}^{SW^*}}{1-\phi _{ex}^{SW^*}}\le 0. \end{aligned}$$

The results above are concluded in Corollary 3.6. $\square$

Proof of Theorem 4.1

Taking the first-order derivative of both sides in equation (20) with respect to $\mu$, it yields

$$\begin{aligned} \frac{\partial \phi _{en}^{\varPi ^*}}{\partial \mu }=\frac{(\phi _{en}^{\varPi ^*})^2 (1-\phi _{en}^{\varPi ^*})^2 (q+\alpha y^{\varPi ^*}_{en}) C e^{\alpha \phi _{en}^{\varPi ^*}}}{(\phi _{en}^{\varPi ^*})^2 (1-\phi _{en}^{\varPi ^*})^2 \alpha \mu (2 q+\alpha y^{\varPi ^*}_{en}) C e^{\alpha \phi _{en}^{\varPi ^*}}+\mu ^2 (2 \phi _{en}^{\varPi ^*}-y^{\varPi ^*}_{en}) \beta }, \end{aligned}$$

(46)

where $y^{\varPi ^*}_{en}=1-q+q \phi _{en}^{\varPi ^*}$.

Based on (18), taking the first- and second-order derivatives with respect to $\mu$, and then combining them with (16), (20), and (46), we obtain

$$\begin{aligned}&\frac{d \varPi _{en}}{d \mu }=\frac{\lambda y^{\varPi ^*}_{en} C e^{\alpha \phi _{en}^{\varPi ^*}(\mu )}}{\mu ^2}-h, \end{aligned}$$

(47)

$$\begin{aligned}&\frac{d^2 \varPi _{en}}{d \mu ^2}= \frac{\lambda y^{\varPi ^*}_{en} C e^{\alpha \phi _{en}^{\varPi ^*}(\mu )}}{\mu ^3}\nonumber \\&\quad \times \left\{ \frac{T^{\varPi ^*}_{en} (q^2-2 \alpha q y^{\varPi ^*}_{en} -(\alpha y^{\varPi ^*}_{en})^2)-2 y^{\varPi ^*}_{en} \beta \mu (2 \phi _{en}^{\varPi ^*}(\mu )-y^{\varPi ^*}_{en})}{\alpha T^{\varPi ^*}_{en} (2 q+\alpha y^{\varPi ^*}_{en})+\mu (2 \phi _{en}^{\varPi ^*}(\mu )-y^{\varPi ^*}_{en}) \beta } \right\} , \end{aligned}$$

(48)

where $T^{\varPi ^*}_{en}=(\phi _{en}^{\varPi ^*}(\mu ))^2 (1-\phi _{en}^{\varPi ^*}(\mu ))^2 C e^{\alpha \phi _{en}^{\varPi ^*}(\mu )}$.

Since we have already assumed that $\alpha >0$ and $\phi \in [1/2,1)$, the denominator of $d^2 \varPi _{en}/d \mu ^2$ is positive, then we focus our research on the numerator. We find that $2 y^{\varPi ^*}_{en} \beta \mu (2 \phi _{en}^{\varPi ^*}-y^{\varPi ^*}_{en})\ge 0$, then we shall discuss the order between $(q^2-2 \alpha q y^{\varPi ^*}_{en}-(\alpha y^{\varPi ^*}_{en})^2)$ and zero in the following:

• (1) If $q^2-2 \alpha q y^{\varPi ^*}_{en}-(\alpha y^{\varPi ^*}_{en})^2\le 0$, the reasonable range of $\alpha$ is $\alpha \ge q/((\sqrt{2}+1) y^{\varPi ^*}_{en})$, then $d^2 \varPi _{en}/d \mu ^2\le 0$. Besides, the right-hand side of (25) is increasing in $\phi \in [1/2,1)$. Then, we define

  $$\begin{aligned} f_2(\phi )=\beta \left[ \ln \frac{\phi }{1-\phi }+\frac{y}{q \phi (1-\phi )}\right] + \left( 1+\frac{\alpha y}{q}\right) \sqrt{\frac{h C e^{\alpha \phi }}{\lambda y}}, \end{aligned}$$

  and $f_2(\phi )|_{\phi \rightarrow 1}=+\infty$. Therefore, there exists a unique $\phi$ which satisfies the equation $f_2(\phi )-R=0$. If $f_2(1/2)\le R$, the solution lies in [1/2, 1) and we obtain the corresponding $p_{en}^{\varPi ^*}$ and $\mu ^{\varPi ^*_{en}}$. But if $f_2(1/2)> R$, $d \varPi _{en}/d \mu < 0$ and the optimal service capacity is zero, the best response of service provider is stop operating the system.

• (2) If $q^2-2 \alpha q y^{\varPi ^*}_{en}-(\alpha y^{\varPi ^*}_{en})^2> 0$, the minuend of the numerator in (48) is positive, and we need a more detailed discussion of the numerator in $d^2 \varPi _{en}/d \mu ^2$.

$\square$

Proof of Theorem 4.2

While the denominator of (48) is positive, we only have to solve the numerator of (48) if $d^2 \varPi _{en}/d \mu ^2=0$. We obtain

$$\begin{aligned} {\overline{\mu }}=\frac{({\overline{\phi }}_{en})^2 (1-{\overline{\phi }}_{en})^2 C e^{\alpha {\overline{\phi }}_{en}} (q^2-2 q \alpha {\overline{y}}_{en}-(\alpha {\overline{y}}_{en})^2)}{2 y (2 {\overline{\phi }}_{en}-{\overline{y}}_{en}) \beta }, \end{aligned}$$

(49)

where ${\overline{y}}_{en}=1-q+q {\overline{\phi }}_{en}$ and ${\overline{\phi }}_{en}$ solves

$$\begin{aligned} \frac{R}{\beta }=\ln \frac{\phi }{1-\phi }+\frac{y}{q \phi (1-\phi )}+\frac{2 y (2 \phi -y) (q+\alpha y)}{q \phi ^2 (1-\phi )^2 (q^2-2 \alpha q y-(\alpha y)^2)}. \end{aligned}$$

(50)

With $\phi$ increasing, the left-hand side of (50) is a constant and the right-hand side of (50) is increasing. Then, there exists only one ${\overline{\phi }}_{en}$ satisfies $d^2 \varPi _{en}/d \mu ^2=0$. Based on (3.9), as $\mu$ grows to infinity, $d \varPi _{en}/d \mu$ decreases and converges to $-h$. Accordingly, $d \varPi _{en}/d \mu$ increases firstly, then decreases, and we denote the root of $d^2 \varPi _{en}/d \mu ^2$ as ${\overline{\mu }}$ and $d \varPi _{en}/d \mu$ is maximized at ${\overline{\phi }}_{en}$.

Consequently, there are at most two value of $\mu$ satisfies $d \varPi _{en}/d \mu =0$ and the number of roots depends on the order of h and $\lambda {\overline{\phi }}_{en} C e^{\alpha {\overline{\phi }}_{en}}/\mu ^2$.

Combining with (49), we set

$$\begin{aligned} {\bar{h}}=\frac{\lambda {\overline{y}}_{en}^3 [2 \beta (2 {\overline{\phi }}_{en}-{\overline{y}}_{en})]^2}{C e^{\alpha {\overline{\phi }}_{en}} {\overline{\phi }}_{en}^4 (1-{\overline{\phi }}_{en})^4 (q^2-2 \alpha q y-(\alpha {\overline{y}}_{en})^2)^2}. \end{aligned}$$

Thus, the order of h and $\lambda {\overline{\phi }}_{en} C e^{\alpha {\overline{\phi }}_{en}}/\mu ^2$ can be divided into three cases:

• $h>{\bar{h}}$, there is no $\mu$ satisfying $d \varPi _{en}/d \mu =0$, thus $d \varPi _{en}/d \mu <0$ and $\mu ^{\varPi ^*_{en}}=0$;

• $h={\bar{h}}$, there is only one $\mu$ satisfying $d \varPi _{en}/d \mu =0$, thus $d \varPi _{en}/d \mu \le 0$, and $\varPi _{en}$ is non-increasing function of $\mu$, then $\mu ^{\varPi ^*_{en}}=0$;

• $h<{\bar{h}}$, there are two roots of $d \varPi _{en}/d \mu =0$, and we denote them as $\mu ',\mu ''$ $(\mu '<\mu '')$, the revenue $\varPi _{en}$ is decreasing in $[0, \mu ')$ and $[\mu '', \infty )$. To find the maximum value of $\varPi _{en}$, we have to compare $\varPi _{en}(0)=0$ with $\varPi (\mu '')$. If $\varPi _{en}(\mu '')>0$ then $\mu ^{\varPi _{en}^*}=\mu ''$; if $\varPi _{en}(\mu '')<0$ then $\mu ^{\varPi _{en}^*}=0$ and if $\varPi _{en}(\mu '')=0$, the optimal service capacity can be zero or the local maximal point $\mu ''$.

Accordingly, the optimal solutions $p^{\varPi ^*}_{en}$ and $\phi ^{\varPi ^*}_{en}$ are derived as follows:

$$\begin{aligned} R= & {} \left[ \ln \frac{\phi }{1-\phi }+\frac{y}{q \phi (1-\phi )}\right] \beta +\frac{q+\alpha y}{q} \sqrt{\frac{h C e^{\alpha \phi }}{\lambda y}}, \end{aligned}$$

(51)

$$\begin{aligned} \mu= & {} \sqrt{\frac{\lambda y C e^{\alpha \phi }}{h}}, \end{aligned}$$

(52)

$$\begin{aligned} p= & {} \frac{y \beta }{q \phi (1-\phi )}+\frac{\alpha y}{q} \sqrt{\frac{h C e^{\alpha \phi }}{\lambda y}}. \end{aligned}$$

(53)

When $h<{\bar{h}}$, we consider $\mu ''$, the larger root of $d \varPi _{en}/d \mu =0$, and condition on $\varPi _{en}(\mu '')\ge 0$. Accordingly, there exists threshold which we denote it as ${\underline{h}}$, only if $h\le {\underline{h}}$ that $\varPi _{en}(\mu '')\ge 0$ and $\mu ^{\varPi _{en}^*}=\mu ''$. Hence, we solve ${\underline{h}}$ as follows.

Coupled with (18) and (52)-(53), we assume that $\varPi _{en}(\mu '')=0$ when $h={\underline{h}}$, and this equals to

$$\begin{aligned} \varPi _{en}(\mu '')=\lambda y \left( R-\beta \ln \frac{\phi }{1-\phi }\right) - 2 \sqrt{\lambda y {\underline{h}} C e^{\alpha \phi }}=0. \end{aligned}$$

Combining with (52), we have

$$\begin{aligned} {\underline{h}}=\frac{\lambda {\widetilde{y}}_{en}^3 \beta ^2}{(q-\alpha {\widetilde{y}}_{en})^2 C e^{\alpha {\widetilde{\phi }}_{en}} {\widetilde{\phi }}_{en}^2 (1-{\widetilde{\phi }}_{en})^2}, \end{aligned}$$

where ${\widetilde{y}}_{en}=1-q+q {\widetilde{\phi }}_{en}$ and ${\widetilde{\phi }}_{en}$ solves

$$\begin{aligned} \frac{R}{\beta }=\ln \frac{\phi }{1-\phi }+\frac{y}{q \phi (1-\phi )}+ \frac{(q+\alpha y) y}{q (q-\alpha y) \phi (1-\phi )}. \end{aligned}$$

In this case, we find that $h<{\bar{h}}$ holds, thus ${\underline{h}}<{\bar{h}}$. Therefore, $\varPi _{en}(\mu '')>0$ when $h<{\underline{h}}$. $\square$

Proof of Theorem 4.4

Taking the first-order derivative of the both sides in (34) with respect to $\mu$, it yields

$$\begin{aligned} \frac{\partial \phi ^{SW^*}_{en}}{\partial \mu }=\frac{q+\alpha y^{SW^*}_{en}}{ \alpha \mu (2 q+\alpha y^{SW^*}_{en})}, \end{aligned}$$

(54)

where $y^{SW^*}_{en}=1-q+q \phi ^{SW^*}_{en}$. Based on (59), taking the first- and second-order derivatives with respect to $\mu$, and then combining them with (34), (16), and (54), we obtain

$$\begin{aligned}&\frac{d SW_{en}}{d \mu }=\frac{\lambda y^{SW^*}_{en} C e^{\alpha \phi ^{SW^*}_{en}(\mu )}}{\mu ^2}-h, \end{aligned}$$

(55)

$$\begin{aligned}&\frac{d^2 SW_{en}}{d \mu ^2}=\frac{\lambda C e^{\alpha \phi ^{SW^*}_{en}}}{\mu ^3} \left\{ \frac{(q+\alpha y^{SW^*}_{en})^2}{\alpha (2 q+\alpha y^{SW^*}_{en})}-2 y^{SW^*}_{en}\right\} . \end{aligned}$$

(56)

• (1) If $(q+\alpha y^{SW^*}_{en})^2/(\alpha (2 q+\alpha y^{SW^*}_{en}))-2 y^{SW^*}_{en}\le 0$, the reasonable range of $\alpha$ is $\alpha \ge q/((\sqrt{2}+1) y^{SW^*}_{en})$, thus the second-order derivative of $SW_{en}$ is non-positive. Similar to Theorem 4.1, we research the rage right-hand of (38) in $\phi \in [1/2,1)$ and define it as function $m(\phi )$. By comparing R with m(1/2), we obtain the existence of the optimal solution and the value of it. If $m(1/2)>R$, $d SW_{en}/d \mu < 0$ and the optimal service capacity for social planner is zero, which is incompatible with his expectation.

• (2) If $(q+\alpha y^{SW^*}_{en})^2/(\alpha (2 q+\alpha y^{SW^*}_{en}))-2 y^{SW^*}_{en}> 0$, the first-order of social welfare is increasing, and we discuss next.

Proof of Theorem 4.5

According to Theorem 4.1, we have that

• (1) if $\alpha \ge q/(\sqrt{2}+1) y$,

  $$\begin{aligned} p_{en}^{\varPi ^*} = \left\{ \begin{array}{ll} \frac{y_{en}^{\varPi ^*} \beta }{q \phi _{en}^{\varPi ^*}(\mu ) (1-\phi _{en}^{\varPi ^*}(\mu ))}+\frac{\alpha y_{en}^{\varPi ^*}}{q} \sqrt{\frac{h C e^{\alpha \phi _{en}^{\varPi ^*}(\mu )}}{\lambda y_{en}^{\varPi ^*}}}, \quad &{} if \quad R\ge M,\\ 0,\quad &{} otherwise, \end{array} \right. \end{aligned}$$

  (57)

  where $M=\frac{2 \beta (2-q)}{q}+\frac{2+q}{2 q} \sqrt{\frac{h C e^{\frac{1}{2} \alpha }}{\lambda (1-\frac{q}{2})}}$

• (2) if $0<\alpha <q/((\sqrt{2}+1) y)$,

  $$\begin{aligned} p_{en}^{\varPi ^*} = \left\{ \begin{array}{ll} \frac{y_{en}^{\varPi ^*} \beta }{q \phi _{en}^{\varPi ^*}(1-\phi _{en}^*)} +\frac{\alpha y_{en}^{\varPi ^*}}{q} \sqrt{\frac{h C e^{\alpha \phi _{en}^{\varPi ^*}}}{\lambda y_{en}^{\varPi ^*}}},\quad &{} if \quad h<{\underline{h}}, \\ 0,\quad &{} otherwise.\end{array} \right. \end{aligned}$$

  (58)

where $y_{en}^{\varPi ^*}=1-q+q \phi _{en}^{\varPi ^*}$. In summary, $p_{en}^{\varPi ^*}\ge p_{en}^{SW^*}=0$.

When considering the optimal joining probability,

• (1) if $\alpha \ge q/((\sqrt{2}+1) y)$, $\phi _{en}^{SW^*}$ is solved by equation (39) and $\phi _{en}^{\varPi ^*}$ is solved by (38) thus $\phi _{en}^{SW^*}>\phi _{en}^{\varPi ^*}$;

• (2) if $0<\alpha <q/((\sqrt{2}+1) y)$ and $h<{\underline{h}}$, $\phi _{en}^{SW^*}$ is solved by (31), thus $\phi _{en}^{SW^*}>\phi _{en}^{\varPi ^*}$, otherwise, $\phi _{en}^{\varPi ^*}=0$.

In summary, $\phi _{en}^{SW^*}\ge \phi _{en}^{\varPi ^*}$. $\square$

Proof of Proposition 4.8

Based on (27) and (29), we obtain

$$\begin{aligned} {\underline{h}}=\frac{\lambda y}{C e^{\alpha \phi }} \left( \frac{(R-\beta [ln \frac{\phi }{1-\phi }+\frac{y}{q \phi (1-\phi )}]) q}{q+\alpha y}\right) ^2. \end{aligned}$$

(59)

Taking the first-order derivative of ${\underline{h}}$ with respect to $\beta$, and it includes $\partial {\widetilde{\phi }}_{en}/\partial \beta$, so we calculate it firstly. We take first-order derivative of the both sides of (29) with respect to $\beta$, and get

$$\begin{aligned} \frac{\partial {\widetilde{\phi }}_{en}}{\partial \beta }=\frac{-q {\widetilde{\phi }}_{en}^2 (1-{\widetilde{\phi }}_{en})^2 (q-\alpha {\widetilde{y}}_{en})^2 R}{\beta ^2 \{q G+(q-\alpha {\widetilde{y}}_{en}) [(2 {\widetilde{\phi }}_{en}-{\widetilde{y}}_{en}) (q-\alpha {\widetilde{y}}_{en})-{\widetilde{y}}_{en} (1-2 {\widetilde{\phi }}_{en}) (q+\alpha {\widetilde{y}}_{en})]\}}, \end{aligned}$$

(60)

where $G={\widetilde{\phi }}_{en} (1-{\widetilde{\phi }}_{en}) (q^2+2 \alpha q {\widetilde{y}}_{en}-(\alpha {\widetilde{y}}_{en})^2)$.

It is easy to find that $\partial {\widetilde{\phi }}_{en}/\partial \beta <0$.

Taking the first-order derivative of (59) and using (29), (60), we obtain

$$\begin{aligned} \frac{d {\underline{h}}}{d \beta }=-\frac{\lambda {\widetilde{y}}_{en}^2 \beta }{ C e^{\alpha {\widetilde{\phi }}_{en}}} \left\{ \frac{ln \frac{{\widetilde{\phi }}_{en}}{1-{\widetilde{\phi }}_{en}}}{{\widetilde{\phi }}_{en} (1-{\widetilde{\phi }}_{en}) (q-\alpha {\widetilde{y}}_{en})}\right\} , \end{aligned}$$

thus $d {\underline{h}}/d \beta \le 0$ with ${\widetilde{\phi }}_{en}\in [1/2, 1)$. Combining with $\partial {\widetilde{\phi }}_{en}/\partial \beta <0$, we find that ${\underline{h}}$ is non-decreasing in $\beta$. $\square$

Proof of Proposition 4.9

Based on (18) and Theorem 4.2, we get the maximal revenue

$$\begin{aligned} \varPi _{en}^*=\frac{\lambda (y_{en}^{\varPi ^*})^2 \beta }{q \phi _{en}^{\varPi ^*} (1-\phi _{en}^{\varPi ^*})}+\frac{\alpha y_{en}^{\varPi ^*}-q}{q} \sqrt{\lambda y_{en}^{\varPi ^*} h C e^{\alpha \phi _{en}^{\varPi ^*}}}, \end{aligned}$$

(61)

where $\phi _{en}^{\varPi ^*}$ solved by (31) and $y_{en}^{\varPi ^*}=1-q +q \phi _{en}^{\varPi ^*}$.

Based on (30), (31) and (61), we study the impacts of $\beta$ on $\phi _{en}^{\varPi ^*}, \mu ^{\varPi ^*}$ and $\varPi _{en}^*$, respectively.

By taking the first-order derivative of $\phi _{en}^{\varPi ^*}$ with respect to $\beta$, we have

$$\begin{aligned} \frac{\partial \phi _{en}^{\varPi ^*}}{\partial \beta }=-\frac{\ln \frac{\phi _{en}^{\varPi ^*}}{1-\phi _{en}^{\varPi ^*}}+\frac{y_{en}^{\varPi ^*}}{q \phi _{en}^{\varPi ^*} (1-\phi _{en}^{\varPi ^*})}}{\frac{(2 \phi _{en}^{\varPi ^*}-y_{en}^{\varPi ^*}) \beta }{q (\phi _{en}^{\varPi ^*})^2 (1-\phi _{en}^{\varPi ^*})^2}-\frac{(q^2-2 q\alpha y_{en}^{\varPi ^*}-(\alpha y_{en}^{\varPi ^*})^2)}{2 q y_{en}^{\varPi ^*}} \sqrt{\frac{h C e^{\alpha \phi _{en}^{\varPi ^*}}}{\lambda y_{en}^{\varPi ^*}}}}. \end{aligned}$$

(62)

Next, we identify the value of the denominator in (62). Based on (31), we define

$$\begin{aligned} s(\phi )=\beta \left[ \ln \frac{\phi }{1-\phi }+\frac{y}{q \phi (1-\phi )}\right] +\frac{q+\alpha y }{q} \sqrt{\frac{h C e^{\alpha \phi }}{\lambda y}}-R, \end{aligned}$$

where the first-order derivative of $s(\phi )$ is

$$\begin{aligned} \frac{\partial s(\phi )}{\partial \phi }=\frac{(2 \phi -y) \beta }{q \phi ^2 (1-\phi )^2}-\frac{q^2-2 \alpha q y-(\alpha y)^2}{2 q y} \sqrt{\frac{h C e^{\alpha \phi }}{\lambda y}}. \end{aligned}$$

According to Theorem 4.2, $\phi _{en}^{\varPi ^*}$ is the larger solution of $s(\phi )$ and $\mathop {\lim }\limits _{\phi \rightarrow 1} s(\phi )=+\infty$. If $\partial s(\phi _{en}^{\varPi ^*})/\partial \phi _{en}^{\varPi ^*}>0$, we have

$$\begin{aligned} \frac{(2 \phi _{en}^{\varPi ^*}-y_{en}^{\varPi ^*}) \beta }{q (\phi _{en}^{\varPi ^*}(1-\phi _{en}^{\varPi ^*}))^2}>\frac{q^2-2 \alpha q y_{en}^{\varPi ^*}-(\alpha y_{en}^{\varPi ^*})^2}{2 q y_{en}^{\varPi ^*}} \sqrt{\frac{h C e^{\alpha \phi _{en}^{\varPi ^*}}}{\lambda y_{en}^{\varPi ^*}}}. \end{aligned}$$

Besides, $\ln \frac{\phi }{1-\phi }+y/(q \phi (1-\phi ))>0$ when $\phi \in [1/2,1)$, thus $\partial \phi _{en}^{\varPi ^*}/\partial \beta <0$. Based on (30), as $\phi _{en}^{\varPi ^*}$ increasing, $\mu ^{\varPi _{en}^*}$ strictly increases and recalling that $\partial \phi _{en}^{\varPi ^*}/\partial \beta <0$, $\mu ^{\varPi _{en}^*}$ is decreasing in $\beta$ and converges to ${\overline{\mu }}=\sqrt{\lambda (1-\frac{q}{2}) C e^{1/2\alpha q }/h}$.

Using (61) and (62), we have

$$\begin{aligned} \frac{\partial \varPi _{en}^*}{\partial \beta }=-\lambda y_{en}^{\varPi ^*} \ln \frac{\phi _{en}^{\varPi ^*}}{1-\phi _{en}^{\varPi ^*}}, \end{aligned}$$

and $\ln \frac{\phi _{en}^{\varPi ^*}}{1-\phi _{en}^{\varPi ^*}}\ge 0$ when $\phi \in [1/2,1)$. Consequently, as $\beta$ increasing, $\varPi _{en}^*$ is decreasing. $\square$