Improvement strategies of battery driving range in an electric vehicle supply chain considering subsidy threshold and cost misreporting

Abstract

This paper investigates a two-stage supply chain consisting of a battery supplier (BS) and an electric vehicle manufacturer (EVM). Considering consumers’ sensitivity to the battery driving range, the BS can improve the driving range level by investing, and obtain subsidies if the driving range level exceeds the subsidy threshold set by the government. Meanwhile, the BS may misreport his private information of the investment cost to the EVM. We mainly examine the effect of cost information misreporting and how the BS determines the optimal improvement strategy when the subsidy threshold increases. Firstly, a low (high) subsidy threshold makes the BS raise the driving range level above (below) the subsidy threshold; when the subsidy threshold is moderately raised, the choice of the improvement strategy is controlled by both the degree of raising the subsidy threshold and the technical upper limit. Secondly, participants may show different preferences for the improvement strategy under certain conditions. Finally, cost information misreporting brings adverse effects to participants and reduces the driving range level.

Appendix

Proof of Proposition 1

Obviously, $m_{1} = \frac{{\left( {k + s} \right)\left( {a - c_{1} - c_{2} } \right) + 4\beta \lambda R_{0} }}{{\left[ {4\beta \lambda - \left( {k + s} \right)^{2} } \right]R_{0} }} > \frac{{k\left( {a - c_{2} - c_{1} } \right) + 4\beta \lambda R_{0} }}{{\left( {4\beta \lambda - k^{2} } \right)R_{0} }} = m_{0}$, $\varPi_{M}^{{AI_{2} }} = \frac{{\beta^{2} \lambda^{2} \left[ {a - c_{2} - c_{1} + \left( {k + s} \right)R_{0} } \right]^{2} }}{{\left[ {4\beta \lambda - \left( {k + s} \right)^{2} } \right]^{2} }} > \frac{{\beta^{2} \lambda^{2} \left( {a - c_{1} - c_{2} + kR_{0} } \right)^{2} }}{{\left( {4\beta \lambda - k^{2} } \right)^{2} }} = \varPi_{M}^{{AI_{1} }}$. For the comparison of $\varPi_{S}^{{AI_{1} }}$ and $\varPi_{S}^{{AI_{2} }}$, we first analyze the size relationship between $\frac{{4\beta^{2} \lambda - \left( {k + s} \right)^{2} }}{{\left[ {4\beta \lambda - \left( {k + s} \right)^{2} } \right]^{2} }}$ and $\frac{{4\beta^{2} \lambda - k^{2} }}{{\left( {4\beta \lambda - k^{2} } \right)^{2} }}$. Let $\frac{{4\beta^{2} \lambda - \left( {k + s} \right)^{2} }}{{\left[ {4\beta \lambda - \left( {k + s} \right)^{2} } \right]^{2} }} > \frac{{4\beta^{2} \lambda - k^{2} }}{{\left( {4\beta \lambda - k^{2} } \right)^{2} }}$, we have $32\lambda^{2} \beta^{3} \left( {A - B} \right) + 4\lambda \beta^{2} \left( {B^{2} - A^{2} } \right) + 16\lambda^{2} \beta^{2} \left( {B - A} \right) + A^{2} B - AB^{2} > 0$, where $A = \left( {k + s} \right)^{2}$ and $B = k^{2}$. Let $F = 32\lambda^{2} \beta^{3} \left( {A - B} \right) + 4\lambda \beta^{2} \left( {B^{2} - A^{2} } \right) + 16\lambda^{2} \beta^{2} \left( {B - A} \right) + A^{2} B - AB^{2}$, it is easy to find that $F\left( {\beta = 1} \right) > 0$, $\frac{dF}{d\beta } = 8\lambda \beta \left( {A - B} \right)\left[ {4\lambda \beta - A + 4\lambda \beta - B + 4\lambda \beta - 4\lambda } \right] > 0$. Thus, $\frac{{4\beta^{2} \lambda - \left( {k + s} \right)^{2} }}{{\left[ {4\beta \lambda - \left( {k + s} \right)^{2} } \right]^{2} }} > \frac{{4\beta^{2} \lambda - k^{2} }}{{\left( {4\beta \lambda - k^{2} } \right)^{2} }}$ holds. Also, because $\frac{{\lambda \left[ {a - c_{2} - c_{1} + \left( {k + s} \right)R_{0} } \right]^{2} }}{2} > \frac{{\lambda \left( {a - c_{2} - c_{1} + kR_{0} } \right)^{2} }}{2}$, $\varPi_{S}^{{AI_{2} }} = \frac{{\lambda \left[ {4\beta^{2} \lambda - \left( {k + s} \right)^{2} } \right]\left[ {a - c_{2} - c_{1} + \left( {k + s} \right)R_{0} } \right]^{2} }}{{2\left[ {4\beta \lambda - \left( {k + s} \right)^{2} } \right]^{2} }} > \frac{{\lambda \left( {4\beta^{2} \lambda - k^{2} } \right)\left( {a - c_{1} - c_{2} + kR_{0} } \right)^{2} }}{{2\left( {4\beta \lambda - k^{2} } \right)^{2} }} = \varPi_{S}^{{AI_{1} }}$. As a result, when $1 < m \le m_{1}$, $\varPi_{S}^{{AI_{1} }} \le \varPi_{S}^{{AI_{2} }}$, $\varPi_{M}^{{AI_{1} }} \le \varPi_{M}^{{AI_{2} }}$; when $m > m_{1}$,$\varPi_{S}^{{AI_{1} }} > \varPi_{S}^{{AI_{2} }}$, $\varPi_{M}^{{AI_{1} }} > \varPi_{M}^{{AI_{2} }}$. Therefore, Proposition 1 is proved.

Proof of Proposition 2

(1)$R_{{}}^{{AI_{1} }} = \frac{{k\left( {a - c_{2} - c_{1} } \right) + 4\lambda R_{0} }}{{4\lambda - k^{2} }} > \frac{{4\lambda - k^{2} }}{{4\lambda - k^{2} }}R_{0} = R_{0}$, $R_{{}}^{{AI_{2} }} = \frac{{\left( {k + s} \right)\left( {a - c_{2} - c_{1} } \right) + 4\lambda R_{0} }}{{4\lambda - \left( {k + s} \right)^{2} }} > \frac{{k\left( {a - c_{2} - c_{1} } \right) + 4\lambda R_{0} }}{{4\lambda - \left( {k + s} \right)^{2} }} > \frac{{k\left( {a - c_{2} - c_{1} } \right) + 4\lambda R_{0} }}{{4\lambda - k^{2} }} = R_{{}}^{{AI_{1} }}$. Thus, $R_{0} < R^{{AI_{1} }} < R^{{AI_{2} }}$.(2) $w^{{AI_{1} }} - w^{AN} = \frac{{k^{2} \left( {a - c_{1} - c_{2} + kR_{0} } \right)}}{{2\left( {4\lambda - k^{2} } \right)}} > 0$, $w^{AN} - w^{NN} = \frac{{sR_{0} }}{2} > 0$. Let $w^{{AI_{2} }} < w^{NN}$. we have $\left( {k^{2} - s^{2} } \right)\left( {a - c_{1} - c_{2} } \right) + \left( {k - s} \right)\left( {k + s} \right)^{2} R_{0} < 0$ and it always holds. Thus, $w^{{AI_{2} }} < w^{NN} < w^{AN} < w^{{AI_{1} }}$.(3) $p^{{AI_{1} }} - p^{AN} = \frac{{3k^{2} \left( {a - c_{1} - c_{2} + kR_{0} } \right)}}{{4\left( {4\lambda - k^{2} } \right)}} > 0$, $p^{AN} - p^{NN} = \frac{{sR_{0} }}{4} > 0$. Let $p^{{AI_{2} }} < p^{NN}$, we have $\left( {3k - s} \right)\left( {k + s} \right)\left( {a - c_{1} - c_{2} } \right) + \left( {3k - s} \right)\left( {k + s} \right)^{2} R_{0} < 0$ and it always holds. Thus, $p^{{AI_{2} }} < p^{NN} < p^{AN} < p^{{AI_{1} }}$.(4) $q^{{AI_{2} }} - q^{{AI_{1} }} = \frac{{\left( {2ks + s^{2} } \right)\left( {a - c_{1} - c_{2} } \right) + \left( {4\lambda s + k^{2} s + ks^{2} } \right)R_{0} }}{{\left[ {4\lambda - \left( {k + s} \right)^{2} } \right]\left( {4\lambda - k^{2} } \right)}} > 0$, $q^{NN} - q^{AN} = \frac{{sR_{0} }}{4} > 0$. Let $q^{{AI_{1} }} > q^{NN}$, we have $a - c_{1} - c_{2} + \left( {k + s - \frac{4\lambda s}{{k^{2} }}} \right)R_{0} > 0$ and it always holds. Thus, $q^{AN} < q^{NN} < q^{{AI_{1} }} < q^{{AI_{2} }}$.In summary, Proposition 2 is proved.

Proof of Proposition 3

$\frac{{\partial w_{{}}^{{AI_{1} }} }}{\partial k} = \frac{{4\lambda k\left( {a - c_{1} - c_{2} } \right) + 2\lambda \left( {4\lambda + k^{2} } \right)R_{0} }}{{\left( {4\lambda - k^{2} } \right)^{2} }} > 0$, $\frac{{\partial p_{{}}^{{AI_{1} }} }}{\partial k} = \frac{{6\lambda k\left( {a - c_{1} - c_{2} } \right) + 3\lambda \left( {4\lambda + k^{2} } \right)R_{0} }}{{\left( {4\lambda - k^{2} } \right)^{2} }} > 0$, $\frac{{\partial R_{{}}^{{AI_{1} }} }}{\partial k} = - \frac{{4k\left( {a - c_{1} - c_{2} + kR_{0} } \right)}}{{\left( {4\lambda - k^{2} } \right)^{2} }} < 0$, $\frac{{\partial q_{{}}^{{AI_{1} }} }}{\partial k} = - \frac{{k^{2} \left( {a - c_{1} - c_{2} + kR_{0} } \right)}}{{\left( {4\lambda - k^{2} } \right)^{2} }} < 0$. $\frac{{\partial w_{{}}^{{AI_{1} }} }}{\partial \lambda } = - \frac{{2k^{2} \left( {a - c_{1} - c_{2} - kR_{0} } \right)}}{{\left( {4\lambda - k^{2} } \right)^{2} }} < 0$, $\frac{{\partial p_{{}}^{{AI_{1} }} }}{\partial \lambda } = - \frac{{3k^{2} \left( {a - c_{1} - c_{2} - kR_{0} } \right)}}{{\left( {4\lambda - k^{2} } \right)^{2} }} < 0$, $\frac{{\partial R_{{}}^{{AI_{1} }} }}{\partial \lambda } = - \frac{{4k\left( {a - c_{1} - c_{2} - kR_{0} } \right)}}{{\left( {4\lambda - k^{2} } \right)^{2} }} < 0$, $\frac{{\partial q_{{}}^{{AI_{1} }} }}{\partial \lambda } = - \frac{{k^{2} \left( {a - c_{1} - c_{2} - kR_{0} } \right)}}{{\left( {4\lambda - k^{2} } \right)^{2} }} < 0$. $\frac{{\partial w_{{}}^{{AI_{1} }} }}{{\partial R_{0} }} = \frac{2\lambda k}{{4\lambda - k^{2} }} > 0$, $\frac{{\partial p_{{}}^{{AI_{1} }} }}{{\partial R_{0} }} = \frac{3\lambda k}{{4\lambda - k^{2} }} > 0$, $\frac{{\partial R_{{}}^{{AI_{1} }} }}{{\partial R_{0} }} = \frac{4\lambda }{{4\lambda - k^{2} }} > 0$, $\frac{{\partial q_{{}}^{{AI_{1} }} }}{{\partial R_{0} }} = \frac{\lambda k}{{4\lambda - k^{2} }} > 0$. Therefore, Proposition 3 is proved.

Proof of Proposition 4

(1) $\frac{{\partial w_{{}}^{{AI_{2} }} }}{\partial k} = \frac{{\left( {4\lambda k - k^{2} s - 2ks^{2} - s^{3} } \right)\left( {a - c_{1} - c_{2} } \right) + \left( {8\lambda^{2} + 2\lambda k^{2} - 4\lambda ks - 6\lambda s^{2} } \right)R_{0} }}{{\left[ {4\lambda - \left( {k + s} \right)^{2} } \right]^{2} }}$. Because $a$ is large enough, $\frac{{\partial w_{{}}^{{AI_{2} }} }}{\partial k} > 0$ always holds in the case of $4\lambda k - k^{2} s - 2ks^{2} - s^{3} > 0$, and we obtain $\lambda > \frac{{k^{2} s + 2ks^{2} + s^{3} }}{4k}$. Let $8\lambda^{2} + 2\lambda k^{2} - 4\lambda ks - 6\lambda s^{2} \ge 0$, we get $\lambda \ge \frac{{3s^{2} - k^{2} + 2ks}}{4}$. Because $s > 3k$, $\frac{{k^{2} s + 2ks^{2} + s^{3} }}{4k} - \frac{{3s^{2} - k^{2} + 2ks}}{4} = \frac{{s\left( {s^{2} - ks - k^{2} } \right) - k^{3} }}{4k} > \frac{{7k^{2} }}{2} > 0$. Then, when $\lambda = \frac{{k^{2} s + 2ks^{2} + s^{3} }}{4k}$, $\frac{{\partial w_{{}}^{{AI_{2} }} }}{\partial k} > 0$ still holds. Thus, when $\lambda \ge \lambda_{1} = \frac{{k^{2} s + 2ks^{2} + s^{3} }}{4k}$, $\frac{{\partial w_{{}}^{{AI_{2} }} }}{\partial k} > 0$.$\frac{{\partial p_{{}}^{{AI_{2} }} }}{\partial k} = \frac{{\left( {2\lambda s + 6\lambda k - k^{2} s - 2ks^{2} - s^{3} } \right)\left( {a - c_{1} - c_{2} } \right) + \left( {12\lambda^{2} + 3\lambda k^{2} - 2\lambda ks - 5\lambda s^{2} } \right)R_{0} }}{{\left[ {4\lambda - \left( {k + s} \right)^{2} } \right]^{2} }}$. Let $2\lambda s + 6\lambda k - k^{2} s$$- 2ks^{2} - s^{3} > 0$, we have $\lambda > \frac{{k^{2} s + 2ks^{2} + s^{3} }}{2s + 6k}$. Let $12\lambda^{2} + 3\lambda k^{2} - 2\lambda ks - 5\lambda s^{2} \ge 0$, we have $\lambda \ge \frac{{2ks + 5s^{2} - 3k^{2} }}{12}$. Because of $s > 3k$, it is straightforward to find $\frac{{k^{2} s + 2ks^{2} + s^{3} }}{2s + 6k} > \frac{{2ks + 5s^{2} - 3k^{2} }}{12}$. Thus, when $\lambda \ge \lambda_{2} = \frac{{k^{2} s + 2ks^{2} + s^{3} }}{2s + 6k}$, $\frac{{\partial p_{{}}^{{AI_{2} }} }}{\partial k} > 0$. $\frac{{\partial R_{{}}^{{AI_{2} }} }}{\partial k} = \frac{{\left[ {4\lambda + \left( {k + s} \right)^{2} } \right]\left( {a - c_{1} - c_{2} } \right) + 8\lambda \left( {s + k} \right)R_{0} }}{{\left[ {4\lambda - \left( {k + s} \right)^{2} } \right]^{2} }} > 0$, $\frac{{\partial q_{{}}^{{AI_{2} }} }}{\partial k} = \frac{1}{{\left[ {4\lambda - \left( {k + s} \right)^{2} } \right]^{2} }}\left\{ {2\lambda \left( {s + k} \right)\left( {a - c_{1} - c_{2} } \right)} \right.$$+ \left. {\left[ {4\lambda^{2} + \lambda \left( {k + s} \right)^{2} } \right]R_{0} } \right\} > 0$.(2) $\frac{{\partial w_{{}}^{{AI_{2} }} }}{\partial \lambda } = \frac{{2\left( {s^{2} - k^{2} } \right)\left( {a - c_{1} - c_{2} } \right) + 2\left( {s - k} \right)\left( {k + s} \right)^{2} R_{0} }}{{\left[ {4\lambda - \left( {k + s} \right)^{2} } \right]^{2} }} > 0$, $\frac{{\partial p_{{}}^{{AI_{2} }} }}{\partial \lambda } = \frac{1}{{\left[ {4\lambda - \left( {k + s} \right)^{2} } \right]^{2} }}\left\{ {\left( {s - 3k} \right)\left( {s + k} \right) \cdot \left. {\left( {a - c_{1} - c_{2} } \right) + \left( {s^{3} - 3k^{3} - 5k^{2} s - ks^{2} } \right)R_{0} } \right\} > 0} \right.$, $\frac{{\partial R_{{}}^{{AI_{2} }} }}{\partial \lambda } = - \frac{{4\left( {s + k} \right)\left( {a - c_{1} - c_{2} } \right) + 4\left( {s + k} \right)^{2} R_{0} }}{{\left[ {4\lambda - \left( {k + s} \right)^{2} } \right]^{2} }} < 0$, $\frac{{\partial q_{{}}^{{AI_{2} }} }}{\partial \lambda } = - \frac{{\left( {s + k} \right)^{2} \left( {a - c_{1} - c_{2} } \right) + \left( {s + k} \right)^{3} R_{0} }}{{\left[ {4\lambda - \left( {k + s} \right)^{2} } \right]^{2} }} < 0$.(3) $\frac{{\partial w_{{}}^{{AI_{2} }} }}{{\partial R_{0} }} = - \frac{{2\lambda \left( {s - k} \right)}}{{4\lambda - \left( {k + s} \right)^{2} }} < 0$, $\frac{{\partial p_{{}}^{{AI_{2} }} }}{{\partial R_{0} }} = - \frac{{\lambda \left( {s - 3k} \right)}}{{4\lambda - \left( {k + s} \right)^{2} }} < 0$, $\frac{{\partial R_{{}}^{{AI_{2} }} }}{{\partial R_{0} }} = \frac{4\lambda }{{4\lambda - \left( {k + s} \right)^{2} }} > 0$, $\frac{{\partial q_{{}}^{{AI_{2} }} }}{{\partial R_{0} }} = \frac{{\lambda \left( {s + k} \right)}}{{4\lambda - \left( {k + s} \right)^{2} }} > 0$.In summary, Proposition 4 is proved.

Proof of Proposition 5

By Table 4, we can find that when $\bar{R} > \frac{{\left( {k + s} \right)\left( {a - c_{2} - c_{1} } \right) + 4\lambda R_{0} }}{{4\lambda - \left( {k + s} \right)^{2} }} = R_{1}$, $m^{\prime}_{1} \left( {\beta = 1} \right) = \frac{{\left( {k + s} \right)\left( {a - c_{2} - c_{1} } \right) + 4\lambda R_{0} }}{{\left[ {4\lambda - \left( {k + s} \right)^{2} } \right]R_{0} }} > \frac{{k\left( {a - c_{2} - c_{1} } \right) + 4\lambda R_{0} }}{{\left( {4\lambda - k^{2} } \right)R_{0} }} = m^{\prime}_{0} \left( {\beta = 1} \right)$, $\varPi_{S}^{{AI_{2} }} = \frac{{\lambda \left[ {a - c_{2} - c_{1} + \left( {k + s} \right)R_{0} } \right]^{2} }}{{2\left[ {4\lambda - \left( {k + s} \right)^{2} } \right]}}$$> \frac{{\lambda \left( {a - c_{2} - c_{1} + kR_{0} } \right)^{2} }}{{2\left( {4\lambda - k^{2} } \right)}} = \varPi_{S}^{{AI_{1} }}$, $\varPi_{M}^{{AI_{2} }} = \frac{{\lambda^{2} \left[ {a - c_{2} - c_{1} + \left( {k + s} \right)R_{0} } \right]^{2} }}{{\left[ {4\lambda - \left( {k + s} \right)^{2} } \right]^{2} }} > \frac{{\lambda^{2} \left( {a - c_{2} - c_{1} + kR_{0} } \right)^{2} }}{{\left( {4\lambda - k^{2} } \right)}} = \varPi_{M}^{{AI_{1} }}$. Therefore, when $1 < m \le m^{\prime}_{1}$ or $m > m^{\prime}_{1}$, the optimal strategy is to raise the driving range level above or below the subsidy threshold.When $\bar{R} \le R_{1}$, the BS needs to make the driving range level reach the technical upper limit if the driving range level is raised above the subsidy threshold. When $1 < m \le m^{\prime}_{0}$, the optimal strategy is to raise the driving range level above the subsidy threshold; when $m > m^{\prime}_{1}$, raising it below the subsidy threshold is optimal. To get the optimal strategy when $m^{\prime}_{0} < m < m^{\prime}_{1}$, the profits of the BS and EVM are compared respectively in $ij = AI_{1}$ and $ij = AI_{2}$. Let $\varPi_{M}^{{AI_{2} }} = \frac{{\left[ {a + \left( {k + s} \right)\bar{R} - c_{1} - c_{2} } \right]^{2} }}{16} > \frac{{\lambda^{2} \left( {a - c_{2} - c_{1} + kR_{0} } \right)^{2} }}{{\left( {4\lambda - k^{2} } \right)}} = \varPi_{M}^{{AI_{1} }}$, we have $\bar{R} \ge \frac{{k\left( {a - c_{1} - c_{2} } \right) + 4\lambda R_{0} }}{{4\lambda - k^{2} }} \cdot \frac{k}{k + s} = R_{2} < \frac{{\left( {k + s} \right)\left( {a - c_{1} - c_{2} } \right) + 4\lambda R_{0} }}{{4\lambda - \left( {k + s} \right)^{2} }} = R_{1}$. Thus, if $m \le \frac{{R_{2} }}{{R_{0} }}$, when $R_{2} \le \bar{R} \le R_{1}$, $\varPi_{M}^{{AI_{2} }} \ge \varPi_{M}^{{AI_{1} }}$; when $mR_{0} \le \bar{R} < R_{2}$, $\varPi_{M}^{{AI_{2} }} < \varPi_{M}^{{AI_{1} }}$. If $m > \frac{{R_{2} }}{{R_{0} }}$, when $mR_{0} \le \bar{R} \le R_{1}$, $\varPi_{M}^{{AI_{2} }} \ge \varPi_{M}^{{AI_{1} }}$. $\varPi_{S}^{{AI_{2} }} - \varPi_{S}^{{AI_{1} }} = \frac{{\left[ {a + \left( {k + s} \right)\bar{R} - c_{1} - c_{2} } \right]^{2} }}{8} - \frac{{\lambda \left( {\bar{R} - R_{0} } \right)^{2} }}{2} - \frac{{\lambda \left( {a - c_{2} - c_{1} + kR_{0} } \right)^{2} }}{{2\left( {4\lambda - k^{2} } \right)}}$. Let $n = a - c_{1} - c_{2}$, $g = 4\lambda - k^{2}$, we have $\varPi_{S}^{{AI_{2} }} - \varPi_{S}^{{AI_{1} }} = \frac{1}{8g}\left\{ {g\left[ {n + \left( {k + s} \right)\bar{R}} \right]^{2} - 4g\lambda \left( {\bar{R} - R_{0} } \right)^{2} - 4\lambda \left( {n + kR_{0} } \right)^{2} } \right\}$. Let $F\left( {\bar{R}} \right) = g\left[ {n + \left( {k + s} \right)\bar{R}} \right]^{2} - 4g\lambda \left( {\bar{R} - R_{0} } \right)^{2} - 4\lambda \left( {n + kR_{0} } \right)^{2}$.$\frac{{\partial F\left( {\bar{R}} \right)}}{{\partial \bar{R}}} = 2g\left( {k + s} \right)\left[ {n + \left( {k + s} \right)\bar{R}} \right] - 8g\lambda \left( {\bar{R} - R_{0} } \right)$, $\frac{{\partial^{2} F\left( {\bar{R}} \right)}}{{\partial \bar{R}^{2} }} = 2g\left( {k + s} \right)^{2} - 8g\lambda = 2g\left[ {\left( {k + s} \right)^{2} - 4\lambda } \right] < 0$. Let $\frac{{\partial F\left( {\bar{R}} \right)}}{{\partial \bar{R}}} = 0$, we obtain $\bar{R}^{*} = \frac{{4\lambda R_{0} + n\left( {k + s} \right)}}{{4\lambda - \left( {k + s} \right)^{2} }} = \frac{{4\lambda R_{0} + \left( {a - c_{1} - c_{2} } \right)\left( {k + s} \right)}}{{4\lambda - \left( {k + s} \right)^{2} }} = R_{1}$. Therefore, when $\bar{R} \le R_{1}$, $\frac{{F\left( {\bar{R}} \right)}}{8g}$ is monotonically increasing with $\bar{R}$. $\frac{F\left( 0 \right)}{8g} = \frac{{\left( {4\lambda - k^{2} } \right)n^{2} - 4\lambda \left( {n + kR_{0} } \right)^{2} - 4g\lambda R_{0}^{2} }}{8g} < 0$, $\frac{{F\left( {\bar{R}^{ * } } \right)}}{8g} = \frac{{\lambda \left[ {a - c_{1} - c_{2} + \left( {k + s} \right)R_{0} } \right]^{2} }}{{2\left[ {4\lambda - \left( {k + s} \right)^{2} } \right]}} - \frac{{\lambda \left( {a - c_{1} - c_{2} + kR_{0} } \right)^{2} }}{{2\left( {4\lambda - k^{2} } \right)}} > 0$. Then, when $\bar{R} \le R_{1}$, $\frac{{F\left( {\bar{R}} \right)}}{8g}$ only has one real root. Let $\frac{{F\left( {\bar{R}} \right)}}{8g} = 0$, we have $\bar{R} = \frac{{8\lambda R_{0} \left( {4\lambda - k^{2} } \right) + 2\left( {k + s} \right)\left( {4\lambda - k^{2} } \right)\left( {a - c_{1} - c_{2} } \right) - \sqrt \Delta }}{{2\left( {4\lambda - k^{2} } \right)\left[ {4\lambda - \left( {k + s} \right)^{2} } \right]}} = R_{3}$ or $\frac{{8\lambda R_{0} \left( {4\lambda - k^{2} } \right) + 2\left( {k + s} \right)\left( {4\lambda - k^{2} } \right)\left( {a - c_{1} - c_{2} } \right) + \sqrt \Delta }}{{2\left( {4\lambda - k^{2} } \right)\left[ {4\lambda - \left( {k + s} \right)^{2} } \right]}}$ which is rejected. Thus, if $m \le \frac{{R_{3} }}{{R_{0} }}$, when $R_{3} \le \bar{R} \le R_{1}$, $\varPi_{S}^{{AI_{2} }} \ge \varPi_{S}^{{AI_{1} }}$; when $mR_{0} \le \bar{R} < R_{3}$, $\varPi_{S}^{{AI_{2} }} < \varPi_{S}^{{AI_{1} }}$. If $m > \frac{{R_{3} }}{{R_{0} }}$, when $mR_{0} \le \bar{R} \le R_{1}$, $\varPi_{S}^{{AI_{2} }} \ge \varPi_{S}^{{AI_{1} }}$. Within the range of $m^{\prime}_{0} < m < m^{\prime}_{1}$, there are the following results. If $m \ge \hbox{max} \left\{ {\bar{R}_{2} ,\bar{R}_{3} } \right\}/R_{0}$, $\varPi_{S}^{{AI_{2} }} \ge \varPi_{S}^{{AI_{1} }}$, $\varPi_{M}^{{AI_{2} }} \ge \varPi_{M}^{{AI_{1} }}$. If $\hbox{min} \left\{ {\bar{R}_{2} ,\bar{R}_{3} } \right\}/R_{0} < m < \hbox{max} \left\{ {\bar{R}_{2} ,\bar{R}_{3} } \right\}/R_{0}$, when $\bar{R} \le \hbox{max} \left\{ {\bar{R}_{2} ,\bar{R}_{3} } \right\}$, the preferences of participants are inconsistent; when $\hbox{max} \left\{ {\bar{R}_{2} ,\bar{R}_{3} } \right\} < \bar{R} \le R_{1}$, $\varPi_{S}^{{AI_{2} }} \ge \varPi_{S}^{{AI_{1} }}$, $\varPi_{M}^{{AI_{2} }} \ge \varPi_{M}^{{AI_{1} }}$. If $m \le \hbox{min} \left\{ {\bar{R}_{2} ,\bar{R}_{3} } \right\}/R_{0}$, when $\bar{R} < \hbox{min} \left\{ {\bar{R}_{2} ,\bar{R}_{3} } \right\}$, $\varPi_{S}^{{AI_{2} }} < \varPi_{S}^{{AI_{1} }}$, $\varPi_{M}^{{AI_{2} }} < \varPi_{M}^{{AI_{1} }}$; when $\hbox{min} \left\{ {\bar{R}_{2} ,\bar{R}_{3} } \right\} \le \bar{R} < \hbox{max} \left\{ {\bar{R}_{2} ,\bar{R}_{3} } \right\}$, the preferences of participants are inconsistent; when $\hbox{max} \left\{ {\bar{R}_{2} ,\bar{R}_{3} } \right\} \le \bar{R} \le R_{1}$, $\varPi_{S}^{{AI_{2} }} \ge \varPi_{S}^{{AI_{1} }}$, $\varPi_{M}^{{AI_{2} }} \ge \varPi_{M}^{{AI_{1} }}$.In summary, Proposition 5 is proved.