Competition and coordination strategies of shared electric vehicles and public transportation considering customer travel utility

Abstract

With the rise of the sharing economy and the concept of “green environmental protection and low-carbon travel,” the emerging project of shared electric vehicles is booming. However, the accompanying coordination problem between shared electric vehicles and public transportation system needs to be urgently solved. In reality, customers’ choice of travel mode is influenced by their own travel perceived utility. Thus, this paper will discuss the competition and coordination problem between shared electric vehicles and public transportation system from the perspective of customer travel utility. Considering the travel cost and comfort in the customer travel utility, the game models of shared electric vehicle and public transportation system in different scenarios are established by using competitive game and cooperative game. Then, the equilibrium solutions under different scenarios are obtained by solving the models. The analysis results show that shared electric vehicles would bring some beneficial improvements to the transportation system under certain circumstances. Furthermore, public transportation system should adopt a coordinative strategy with the shared electric vehicles to promote the total customer travel utility for the entire system. It is worth considering the improvement of the service quality of shared electric vehicle and public transportation, which would affect the rate of increasing in the total customer travel utility.

Appendix

The proof of Proposition 1 .

Comparing the optimal number of customers using PT in Scenario 2 with Scenario 1, it can be found that $\frac{a-k}{2b}<\frac{a}{2b}$ owing to a > 0, b > 0, and k > 0. And the number of customers reduced is $\Delta {Q}_P=\frac{k}{2b}$.

The proof of Proposition 2 .

The change in the total utility is $\Delta U=U\left({Q}_P^{\ast },{Q}_S^{\ast}\right)-U\left({Q_P}^{\ast}\right)=\frac{k\left(k+4{bQ}_{all}-2a\right)}{4b}$. When k + 4bQ_all − 2a > 0, it can be obtained ΔU > 0. Therefore, when ${Q}_{all}>\frac{a}{2b}-\frac{k}{4b}$, $U\left({Q}_P^{\ast },{Q}_S^{\ast}\right)>U\left({Q_P}^{\ast}\right)$. In other words, after SEV enters the system and only competes with PT and ${Q}_{all}>\frac{a}{2b}-\frac{k}{4b}$, the total customer travel utility increases. Similarly, when ${Q}_{all}=\frac{a}{2b}-\frac{k}{4b}$, $U\left({Q}_P^{\ast },{Q}_S^{\ast}\right)={U}_{all}\left({Q}_P^{\ast}\right)$. When ${Q}_{all}<\frac{a}{2b}-\frac{k}{4b}$, $U\left({Q}_P^{\ast },{Q}_S^{\ast}\right)<{U}_{all}\left({Q}_P^{\ast}\right)$. In summary, the magnitude of the total utility of Scenario 1 and Scenario 2 depends on the relationship between Q_all and $\frac{a}{2b}-\frac{k}{4b}$ in the transport system.

The proof of Proposition 3 .

In Scenario 2, it can be known that the optimal numbers of customers using PT and SEV are ${Q}_P^{\ast }=\frac{a-k}{2b}$ and ${Q}_S^{\ast }={Q}_{all}-\frac{a-k}{2b}$, respectively. In addition, the maximum total utility of the system is $U\left({Q}_P^{\ast },{Q}_S^{\ast}\right)=\frac{a^2}{4b}+\frac{k\left(k+4{bQ}_{all}-2a\right)}{4b}$.

Since Q_all is a determined value in the transport system, when the number of customers taking SEV does not reach the equilibrium result, i.e., ${Q}_S<{Q}_{all}-\frac{a-k}{2b}$, it could be obtained that ${Q}_P>\frac{a-k}{2b}$. In this case, the total utility is less than that in the equilibrium, $U\left({Q}_P,{Q}_S\right)<U\left({Q}_P^{\ast },{Q}_S^{\ast}\right)=\frac{a^2}{4b}+\frac{k\left(k+4{bQ}_{all}-2a\right)}{4b}$. Thus, SEV needs to be subsidized to lower the price of taking SEV to attract more customers and ultimately promote the total customer travel utility of the whole system. Similarly, when ${Q}_S={Q}_{all}-\frac{a-k}{2b}$, it could be obtained ${Q}_P=\frac{a-k}{2b}$ and $U\left({Q}_P,{Q}_S\right)=U\left({Q}_P^{\ast },{Q}_S^{\ast}\right)$. SEV does not need to be subsidized to upset the optimal equilibrium of the whole system. When ${Q}_S>{Q}_{all}-\frac{a-k}{2b}$, it can have${Q}_P<\frac{a-k}{2b}$ and $U\left({Q}_P,{Q}_S\right)<U\left({Q}_P^{\ast },{Q}_S^{\ast}\right)$. Therefore, SEV does not need to be subsidized, but PT needs to be subsidized to improve the service quality to attract more customers and ultimately promote the total customer travel utility of the whole system.

In summary, the total customer travel utility of the system varies with the relationship between Q_S and ${Q}_{all}-\frac{a-k}{2b}$. When ${Q}_S<{Q}_{all}-\frac{a-k}{2b}$, the total customer travel utility of the whole system is decreasing. Therefore, it is recommended to subsidize SEV to promote the total customer travel utility of the whole system in this case. On the contrary, the result is reversed.

The proof of Proposition 4 .

As the travel utility function coefficients are determined in Scenario 3, it can be easily found that the size relationship is affected by u and η. For example, when μ < η, ${Q}_P^{\ast }-{Q^{\prime}}_P^{\ast }=\frac{k\left(\eta -\mu \right)}{2b\left(1+\mu \right)}>0$ and ${p}_S-{p}_A=\frac{\left(1+\mu +\eta \right)k-\left(1+2\mu \right)a}{2 bc}>0$. Therefore, it could be obtained that the optimal number of customers served by PT is reduced and the optimal price of using SEV at station A is reduced compared to that in Scenario 2. Meanwhile, the amount of change in the optimal number of customers taking PT is $\frac{k\left(\eta -\mu \right)}{2b\left(1+\mu \right)}$. Similarly, when μ = η, ${Q}_P^{\ast }={Q^{\prime}}_P^{\ast }$, and p_S = p_A. When μ > η, ${Q}_P^{\ast }<{Q^{\prime}}_P^{\ast }$, and p_S < p_A.

Further conclusions can be drawn from the above. The values of μ and η are influenced by the distance from station A to station B and the service quality of the travel modes (Zhang and Li 2014; Ma et al. 2007; Páez and Whalen 2010). Therefore, when station B in the transport system is settled, the values of μ and η are influenced only by the service quality. As μ < η, PT should promptly improve its service quality to get more customers. In addition, the price of taking SEV from station A in Scenario 3 is lower than that in Scenario 2, p_A < p_S, and the reduction is $\frac{\left(1+2\mu \right)a-\left(1+\mu +\eta \right)k}{2 bc}$. Based on the demand function of SEV, when the price of taking SEV decreases, it could result in the increasing number of customer taking SEV and the decreasing number of customer taking PT at station A. It explains why the optimal number of customers traveling by PT decreases in Scenario 3.

The proof of Proposition 5 .

The difference between ${U}_{all}\left({Q}_P,{Q}_S,{Q}_P^{\prime },{Q}_S^{\prime}\right)$ and U_all(Q_P, Q_S) is $\Delta U={U}_{all}\left({Q}_P,{Q}_S,{Q}_P^{\prime },{Q}_S^{\prime}\right)-{U}_{all}\left({Q}_P,{Q}_S\right)=\eta {kQ}_{all}+\frac{{\left[a\left(1+\mu \right)-k\left(1+\eta \right)\right]}^2}{4b\left(1+\mu \right)}$. Due to the positive parameter values, ΔU > 0 and it is always increasing regardless of the size relationship between μ and η.

The proof of Proposition 6 .

Taking the derivations of ΔU, it can be obtained that $\frac{\mathrm{\partial \Delta }U}{\partial \eta }=k\left[Q-\frac{a}{2b}+\frac{k\left(1+\eta \right)}{2b\left(1+\mu \right)}\right]=k\left(Q-{Q}_3\right)$ and $\frac{\mathrm{\partial \Delta }U}{\partial \mu }=\frac{2a\left[a\left(1+\mu \right)-k\left(1+\eta \right)\right]-4b{\left[a\left(1+\mu \right)-k\left(1+\eta \right)\right]}^2}{16{b}^2{\left(1+\mu \right)}^2}$. Owing to k > 0 and Q > Q₃ > 0, then$\frac{\mathrm{\partial \Delta }U}{\partial \eta }>0$. Therefore, ΔU is strictly monotonically increasing as the value of η increases. Let $\frac{\mathrm{\partial \Delta }U}{\partial \mu }=0$, we can get the following result, 2a[a(1 + μ) − k(1 + η)] − 4b[a(1 + μ) − k(1 + η)]² = 0. Thus, it can be found that there are two different cases between μ and $\frac{k\left(1+\eta \right)}{a}-1$ as follows.

Case 1. If a(1 + μ) − k(1 + η) = 0, $\mu =\frac{k\left(1+\eta \right)}{a}-1$.

Since $\mu =\frac{k\left(1+\eta \right)}{a}-1$, it can be obtained that $\frac{\mathrm{\partial \Delta }U}{\partial \mu}\equiv 0$. Therefore, ΔU is not changed as the value of μ changes. At this time, ΔU is a constant value, and this constant value is ΔU = ηkQ.

Case 2. If a(1 + μ) − k(1 + η) ≠ 0, $\mu \ne \frac{k\left(1+\eta \right)}{a}-1$.

Let $\frac{\mathrm{\partial \Delta }U}{\partial \mu }=0$, we can get $\mu =\frac{1}{2b}+\frac{k\left(1+\eta \right)}{a}-1$, and the following conclusions can be drawn. When $\mu >\frac{1}{2b}+\frac{k\left(1+\eta \right)}{a}-1$, $\frac{\mathrm{\partial \Delta }U}{\partial \mu }<0$, ΔU decreases as the value of μ increases at this time. When $\mu <\frac{1}{2b}+\frac{k\left(1+\eta \right)}{a}-1$, $\frac{\mathrm{\partial \Delta }U}{\partial \mu }>0$, ΔU increases as the value of μ increases at this time. Thus, $\mu =\frac{1}{2b}+\frac{k\left(1+\eta \right)}{a}-1$ is the only maximum point of ΔU. Therefore, $\mu =\frac{1}{2b}+\frac{k\left(1+\eta \right)}{a}-1$ is the maximum point of ΔU.

For μ > 0, we need to discuss the above calculations by different situations.

• (1) When $\mu =\frac{1}{2b}+\frac{k\left(1+\eta \right)}{a}-1\le 0$, it means that μ = 0 is the only maximum point of ΔU, and $\Delta {U}_{\mathrm{max}}<\eta kQ+\frac{{\left[a-k\left(1+\eta \right)\right]}^2}{4b}$. ΔU increases as the value of μ increases at this time.

• (2) When $\mu =\frac{1}{2b}+\frac{k\left(1+\eta \right)}{a}-1>0$, ΔU decreases as the value of μ increases. It means that $\mu =\frac{1}{2b}+\frac{k\left(1+\eta \right)}{a}-1$ is the only maximum point of ΔU, and $\Delta {U}_{\mathrm{max}}=\eta kQ+\frac{a^3}{8b\left[a+2 bk\left(1+\eta \right)\right]}$. At this time, when $0<\mu <\frac{1}{2b}+\frac{k\left(1+\eta \right)}{a}-1$, ΔU increases as the value of μ increases. Thus, it means that $\mu =\frac{1}{2b}+\frac{k\left(1+\eta \right)}{a}-1$ is the only maximum point of ΔU, and $\Delta {U}_{\mathrm{max}}=\eta kQ+\frac{a^3}{8b\left[a+2 bk\left(1+\eta \right)\right]}$.

Therefore, the increase or decrease of ΔU is different depending on the value of μ. There are three different relationships between ΔU and μ. When $\mu =\frac{k\left(1+\eta \right)}{a}-1$, ΔU is a fixed value and the fixed value is ηkQ. When $\mu \ne \frac{k\left(1+\eta \right)}{a}-1$, $\mu =\frac{1}{2b}+\frac{k\left(1+\eta \right)}{a}-1$ is the only maximum point of ΔU. If $\frac{1}{2b}+\frac{k\left(1+\eta \right)}{a}-1\le 0$, ΔU is strictly increased as μ increases at this time and $\Delta {U}_{\mathrm{max}}<\underset{\mu \to 0}{\lim}\Delta U=\eta kQ+\frac{{\left[a-k\left(1+\eta \right)\right]}^2}{4b}$. If $\frac{1}{2b}+\frac{k\left(1+\eta \right)}{a}-1>0$, $\mu =\frac{1}{2b}+\frac{k\left(1+\eta \right)}{a}-1$ is the maximum point of ΔU at this time and $\Delta {U}_{\mathrm{max}}=\eta kQ+\frac{a^3}{8b\left[a+2 bk\left(1+\eta \right)\right]}$.