Does car sharing help reduce the total number of vehicles?

Abstract

Car sharing has become a non-negligible complement to the existing transportation and is viewed as an environmentally beneficial alternative. In this paper, we develop a game-theoretic model to investigate how the introduction of car sharing has an impact on the market. Our results indicate that car sharing does not always reduce vehicle quantity. Specifically, only when the producing cost and transportation need are below some thresholds and the market size is greater than a threshold, can car sharing decrease the total number of vehicles. Otherwise, the manufacturer can generate a larger profit by increasing its production volume. We also find that the introduction of car sharing always makes the retailer incur loss, but it benefits the manufacturer if the market size is sufficiently large. Finally, our results show that the total number of vehicles does not always increase in sharing service level. In such cases, there is a positive spillover effect: by providing the car-sharing service, the manufacturer generates a larger profit with producing less vehicles, which helps alleviate some urban problems, such as air pollution and congestion.

Appendix

1.1 Proof of Proposition 4

To compare $Q^\mathrm{S}$ and $d^\mathrm{N}$, we let

$$\begin{aligned} G(t)&=Q^\mathrm{S}(t)-d^\mathrm{N} (t)\nonumber \\&=\frac{n}{4} \left( \left( (1-e)a-\frac{a(2-a)c}{1-a}\right) t - \frac{ac}{(1-a)t} \right. \nonumber \\&\left. \quad +\,\frac{2ac}{1-a}\right) +\frac{a}{1-a} ,\quad t \in (0,1). \end{aligned}$$

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Since $\partial G(t) / \partial t =\frac{n}{4}[ (1-e)a - \frac{a(2-a)c}{1-a} + \frac{ac}{(1-a)t^2}]$, G(t) increases in t, or first increases in t, then decreases in t. However, $\lim _{t \rightarrow 0^+} G(t) \rightarrow - \infty$, $G(1) =\frac{na(1-a(1-e-c))}{4(1-a)}+\frac{a}{1-a} > 0$ , hence G(t) has only one root in (0, 1). We denote this root as $\bar{t}$. Note that in our model $t > t_0$, otherwise no one owns a car in the car-sharing market, so we only care about the interval $(t_0 , 1)$. If $t_0 < \bar{t}$, $Q^\mathrm{S} (t) < d^\mathrm{N} (t)$ in $(t_0, \bar{t})$, and $Q^\mathrm{S} (t) > d^\mathrm{N} (t)$ in $(\bar{t}, 1)$. If $t_0 \ge \bar{t}$, $Q^\mathrm{S} (t) \ge d^\mathrm{N} (t)$ in $(t_0, 1)$. By checking the value $G(t_0)$, we have the following conditions:

When $G(t_0) <0$, we have

$$\begin{aligned}&c^2 - \left( \frac{(3-2a)(1-e)}{2-a} + \frac{4a}{n(1-a)(2-a)}\right) c\nonumber \\&\quad +\,\frac{1-e}{2-a}\left( (1-e)(1-a)-\frac{4}{n}\right) > 0 . \end{aligned}$$

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When $G(t_0) \ge 0$, we have

$$\begin{aligned}&c^2 - \left( \frac{(3-2a)(1-e)}{2-a} + \frac{4a}{n(1-a)(2-a)}\right) c\nonumber \\&\quad +\,\frac{1-e}{2-a}\left( (1-e)(1-a)-\frac{4}{n}\right) \le 0 , \end{aligned}$$

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in which $Q^\mathrm{S}$ is always larger than or equal to $d^\mathrm{N}$.

Let

$$\begin{aligned} K(c)&=c^2 - \left( \frac{(3-2a)(1-e)}{2-a} + \frac{4a}{n(1-a)(2-a)}\right) c\nonumber \\&\quad +\,\frac{1-e}{2-a}\left( (1-e)(1-a)-\frac{4}{n}\right) . \end{aligned}$$

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K(c) is a quadratic function which first decreases then increases in c. The discriminant $\varDelta = [\frac{(3-2a)(1-e)}{2-a} + \frac{4a}{n(1-a)(2-a)}]^2 - \frac{4(1-e)}{2-a}[(1-e)(1-a)-\frac{4}{n}]= \frac{(1-e)^2}{(2-a)^2}+\frac{16a^2}{n^2(1-a)^2(2-a)^2}+\frac{8a(3-2a)(1-e)}{(2-a)^2(1-a)n} + \frac{16(1-e)}{(2-a)n} > 0$, so K(c) has two roots $c_1$ and $c_2$ (we assume $c_1 < c_2$). Since $0< c< 1-e$, and $K(1-e)=- \frac{4(1-e)}{n(2-a)(1-a)} <0$. We have $c_2 > 1-e$.

Then we check $c_1$. First we find $K(0)= \frac{1-e}{2-a}[(1-e)(1-a)-\frac{4}{n}]$. If $n \le \frac{4}{(1-e)(1-a)}$, $K(c) \le 0$ in $(0,1-e)$, which means in this case $Q^\mathrm{S}$ is always bigger than $d^\mathrm{N}$. Let $n_0=\frac{4}{(1-e)(1-a)}$, if $n > n_0$, then $K(0) > 0$, in which case $K(c) > 0$ if $0< c < c_1$, and $K(c) < 0$ if $c_1< c < 1-e$. By applying the square root formula, we derive

$$\begin{aligned} c_1&= \frac{1}{2}\left( \frac{(3-2a)(1-e)}{2-a}+\frac{4a}{n(1-a)(2-a)} \right. \nonumber \\&\quad \left. -\,\sqrt{\frac{(1-e)^2}{(2-a)^2}+\frac{16a^2}{n^2(1-a)^2(2-a)^2}+\frac{16(1-e)}{n(2-a)}}\right) .\nonumber \\ \end{aligned}$$

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1.2 Proof of Proposition 5

One can easily show that $\partial d^\mathrm{S}_\mathrm{R} / \partial a < 0$, and $\partial d^\mathrm{S}_\mathrm{S} / \partial a > 0$. Then we investigate the relationship between $Q^\mathrm{S}$ and a. Firstly by checking the condition $t>t_0$, we have $0< a < \frac{1-e-c/t}{1-e-c}$. Next we solve the inequality $\partial Q^\mathrm{S} / \partial a > 0$, which is equal to $a^2-2a+A>0$, where $A= \frac{(1-e-2c)t^2+(2c+\frac{4}{n})t - c}{(1-e-c)t^2}$. Let $L(a)=a^2-2a+A$, then we can show that L(a) decreases in $(-\infty ,1)$. Since $0<a< \frac{1-e-c/t}{1-e-c}$, we have L(a) decreases in $(0, \frac{1-e-c/t}{1-e-c})$.

We divide this problem into three cases:

If $L(0) \le 0$, $L(a) < 0$ in $(0, \frac{1-e-c/t}{1-e-c})$. $Q^\mathrm{S}$ always decreases in a.

If $L(\frac{1-e-c/t}{1-e-c}) \ge 0$, $L(a) > 0$ in $(0, \frac{1-e-c/t}{1-e-c})$. $Q^\mathrm{S}$ always increases in a.

If $L(0) > 0$, and $L(\frac{1-e-c/t}{1-e-c}) < 0$. $Q^\mathrm{S}$ increases in $(0,a_1)$, and decreases in $(a_1, \frac{1-e-c/t}{1-e-c})$, where $a_1=1-\sqrt{1-A}$.

1.3 Proof of Proposition 6

${{\pi r}^\mathrm{S}} = (f^S-w^\mathrm{S})d^\mathrm{S}_\mathrm{R}$, and ${{\pi r}^\mathrm{N}} = (f^N-w^\mathrm{N})d^\mathrm{N}$. Since $f^\mathrm{S} <f^\mathrm{N}$, $w^\mathrm{S} = w^\mathrm{N}$, and $d^\mathrm{S}_\mathrm{R} < d^\mathrm{N}$ (i.e., $v_2 > v_0$), ${{\pi r}^\mathrm{S}}$ is always less than ${{\pi r}^\mathrm{N}}$. Then we consider the manufacturer’s profit:

Let

$$\begin{aligned} H(t)= & {} {{\pi m}^\mathrm{S}(t)} - {{\pi m}^\mathrm{N}(t)}\nonumber \\= & {} \frac{n}{8}\left( \left( a(1-e)^2 - 2ac(1-e) + \frac{a(2-a)c^2}{1-a}\right) t \right. \nonumber \\&\left. +\,\frac{ac^2}{(1-a)t} - \frac{2ac^2}{1-a}\right) - \frac{ca}{1-a} . \end{aligned}$$

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$H^{\prime }(t) = \frac{n}{8} [ a(1-e)^2 -2ac(1-e)+\frac{a(2-a)c^2}{1-a} - \frac{ac^2}{(1-a)t^2}]$, where $H^{\prime }(t)$ is the differential of H(t) . One can easily show that $H^{\prime }(t)$ is an increasing function in t.

$\lim _{t \rightarrow 0} H^{\prime }(t) \rightarrow - \infty$, and $H^{\prime }(1) = \frac{na}{8}(1-e-c)^2 > 0$. So H(t) first decreases in $(0, t^*)$, then increases in $(t^*, 1)$, where $t^* = \frac{c}{\sqrt{(1-a)(1-e)^2-2(1-a)(1-e)c + (2-a)c^2}} \in (0,1)$ is the root of $H^{\prime }(t)$.

Remember that in our model the condition $t>t_0$ must be satisfied, so we compare the values of $t_0$ and $t^*$. Let $t_0 > t^*$, which is the same as to solve the following inequality:

$$\begin{aligned} (2+a)c^2- 2(1-e)(1+a)c+a(1-e)^2 > 0 . \end{aligned}$$

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By solving the inequality we derive that when $c \in (0, \frac{a(1-e)}{2+a})$, $t_0 > t^*$; and $c \in (\frac{a(1-e)}{2+a}, 1-e)$, $t_0 < t^*$.

When $t_0 > t^*$, $H(t_0)$ is the minimum of H(t) . If $H(t_0)>0$, we have $n> \frac{8[ac+(1-e)(1-a)]}{(1-a)(2-a)(1-e-c)^2} =n_1$.

When $t_0 < t^*$, $H(t^*)$ is the minimum of H(t) . If $H(t^*)>0$, we have $n>\frac{4}{\sqrt{(1-a)(1-e)^2-2(1-a)(1-e)c+(2-a)c^2}-c} =n_2$.