# A Framework to Determine the Utilization of Vacant Taxis on HOV Lanes with the Optimal Deployment
## 3 Formulations
### Variational Inequality
Then the searching time for vacant taxis can be expressed as:  
\[w_{r}^{v}=\frac{\theta_{r}}{\lambda_{v}}+\frac{\Gamma}{\lambda_{v}}=\overline {\theta_{r}}+\overline{\varphi},\ r\in R \tag{29}\]  
we use \(E_{r}\) to represent \(\exp(-\sigma\theta_{r})\), so \(E_{r}\) can be denoted as:  
\[E_{r}=\exp(-\sigma\lambda_{v}\overline{\theta_{r}})=\left[\exp(-\sigma \overline{\theta_{r}})\right]^{\lambda_{v}} \tag{30}\]  
Let \(\overline{E_{r}}=(E_{r})^{\frac{1}{\lambda_{v}}}\), \(r\in R\), and \(g\overline{E_{r}}=\exp(-\sigma w_{r}^{v}),\ r\in R\), from Equation (17), so we obtain:  
\[g=\exp\left[\frac{-\sigma(N-\sum\limits_{r\in R}\sum\limits_{s\in S}d_{rs}^{v} h_{rs}-\sum\limits_{s\in S}\sum\limits_{r\in R}d_{sr}^{v}h_{sr})-\sum\limits_{r\in R }O_{r}^{o}\ln\overline{E_{r}}}{\sum\limits_{r\in R}O_{r}^{o}}\right] \tag{31}\]  
Then the passenger waiting time can be calculated by:  
\[w_{r}^{v}=-\frac{1}{\sigma}\ln(g\overline{E_{r}}),\ r\in R \tag{32}\]  
**Proposition 1**. : _Let vector \(\overline{\mathbf{d}}\) denote the set of the potential demand \(\overline{d_{rs}}\). If \(\overline{\mathbf{d}}\) is non-empty, there is at least one optimal solution of \(P1\). Namely, there is at least one optimal travel-demand pattern \(\mathbf{d}\) and flow pattern \(\mathbf{f}\)._  
**Proof of Proposition 1. ** The problem \(P1\) can be rewritten as:  
\[ \begin{array}{c}G(\mathbf{x})\\ s.t.\ J(\mathbf{x})\geq 0\end{array} \]  
where \(\mathbf{x}\) is the optimal solution vector. Specifically, \(\mathbf{x}=(f_{rs,p}^{m},f_{sr,p_{s}}^{s}d_{rs}^{m},d_{rs})^{\text{T}}\). Since \(\overline{\mathbf{d}}\) is non-empty and \(\exp(\cdot)>0\), we can always find an OD pair between which \(0<d_{rs}<\overline{d_{rs}}\). From Equation (8) it can be guaranteed that between this OD pair there is always \(0<d_{rs}^{m}\leq d_{rs}\). In addition, we can find at least one path flow \(f_{rs,p}^{m}\) satisfying \(0<f_{rs,p}^{m}\leq d_{rs}^{m}\leq d_{rs}<\overline{d_{rs}}\). Therefore, we know that \(\mathbf{x}\) is non-empty and bounded, which can guarantee the compactness of the solution space. The VI can be seen as \(\
abla G(\mathbf{x})(\mathbf{x}-\mathbf{x}^{*})\geq 0\); thus, \(G(\mathbf{x})\) is a continuous function. From the well-known Weierstrass theorem, the existence of the optimal solution can be guaranteed. \(\square\)  
**Proposition 2**. : _The optimal travel-demand pattern \(\mathbf{d}\) and the aggregate link flow pattern \(\mathbf{f}\) following the logit mode split and user equilibrium principle can be obtained by solving the above VI problem._  
**Proof of Proposition 2. ** Since P1 is a VI, the Karush-Kuhn-Tucker (KKT) points of P1 are identical to the optimal solution. Thus, the KKT conditions are given by:  
\[0\leq f_{rs,p}^{m}\bot(c_{rs,p}^{m}-\mu_{rs}^{m})\geq 0,\ r\in R,s\in S,p\in P,m\in M \tag{33}\]  
\[0\leq f_{sr,p}^{v}\bot(c_{sr,p_{v}}^{v}-\mu_{sr}^{v})\geq 0,\ r\in R,s\in S,p_{v} \in P_{v} \tag{34}\]  
\[0\leq d_{rs}^{m}\bot(\frac{1}{\beta}\ln\frac{d_{rs}^{m}}{d_{rs}}-k_{rs}^{m}+ \mu_{rs}^{m}-\varphi_{rs})\geq 0,\ r\in R,s\in S,m\in M \tag{35}\]  
\[0\leq d_{sr}^{v}\bot(\frac{1}{\beta}\ln d_{sr}^{v}+\mu_{sr}^{v}+\rho_{s}+\theta _{r})\geq 0,\ r\in R,s\in S \tag{36}\]  
\[0\leq d_{rs}\bot(\frac{1}{\pi_{rs}}\ln\frac{d_{rs}}{d_{rs}}+\varphi_{rs})\geq 0,\ r\in R,s\in S(8)\sim(16) \tag{37}\]  
By mathematical manipulations, the logit mode split and user equilibrium can be derived from the KKT conditions. Readers may refer to the similar process in [38] for the details. Therefore, the optimal solutions \(\mathbf{d}\) and \(\mathbf{f}\) from the KKT points follow the logit mode split and the user equilibrium. Since the KKT conditions are the necessary and sufficient conditions of the P1 problem, the optimal solution is also the solution of the VI above. \(\square\)  
According to the propositions, we are able to detect the network performances by solving the P1 problem.