Real-Time Route Planning and Online Order Dispatch for Bus-Booking Platforms

Abstract

To cater to the high travel demands in Beijing Capital International Airport during 23:00–2:00, Beijing Traffic Management Bureau (BTMB) intends to develop a new service named bus-booking platforms. Compared to traditional airport shuttle buses, bus-booking platforms can conduct flexible route planning and online order dispatch while provide much lower price than the common car-hailing platform, e.g., Didi Chuxing. We conduct the real-time route planning by solving the standard Capacitated Vehicles Routing Problem based on the order prediction. In addition, we focus on the design of the online order dispatch algorithm for bus-booking platforms, which is novel and extremely different from the traditional taxi order dispatch in car-hailing platforms. When an order is dispatched, multiple influence factors will be considered simultaneously, such as the bus capacity, the balanced distribution of the accepted orders and the travel time of passengers, all of which aim to provide a better user experience. Moreover, we prove that our online algorithms can achieve the tight competitive ratio in this paper, where the competitive ratio is the ratio between the solution of an online algorithm and the offline optimal solution.

This work was supported by the National Key R&D Program of China [2018YFB1004703]; the National Natural Science Foundation of China [61872238, 61672353]; the Shanghai Science and Technology Fund [17510740200]; the Huawei Innovation Research Program [HO2018085286]; and the State Key Laboratory of Air Traffic Management System and Technology [SKLATM20180X].

Appendices

A Proof of Theorem 1

Proof

Consider a modified primal program as follows and let $\mathrm {OPT}, \mathrm {OPT}_\alpha$ be the optimal objective value for the primal program (6) and the modified program (10), respectively. Obviously, we have $\mathrm {OPT}_\alpha \ge (1 - \exp (-\alpha )) \mathrm {OPT}$.

$$\begin{aligned} \max&\quad \sum _{i \in \mathcal {I}} \sum _{j \in B_i: d_i \in S_j} x_{ij} p_i (1 - \exp (-\alpha )), \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \text {s.t.}&\quad \text {Constraint}~(1),(2),(3),(4),(5) \nonumber \\&\quad x_{ij} \ge 0, \ \forall i \in \mathcal {I}, j \in \mathcal {B} \nonumber \end{aligned}$$

(10)

Notice that Algorithm FPD exactly produces a feasible fractional solution for the dual of Program (10), where denote by $D_\alpha$ its objective value. Let P be the objective value computed by Algorithm FPD for the primal program (6). Therefore, we have $\dfrac{\partial P}{\partial x_{ij}} = p_i$ and

$$\dfrac{\partial D_\alpha }{\partial x_{ij}} = \dfrac{\partial y_i}{\partial x_{ij}} + K_j \dfrac{\partial z_j}{\partial x_{ij}} + T_j \dfrac{\partial u_j}{\partial x_{ij}} + R_v \dfrac{\partial r_v}{\partial x_{ij}} + T_v \dfrac{\partial q_v}{\partial x_{ij}} \ \ \ (\text {Note:} \ v = d_i)$$

From line 9 in Algorithm FPD, we get

$$\dfrac{\partial y_i}{\partial x_{ij}} = \alpha p_i \exp \alpha \left( \sum _{j \in B_i:d_i \in S_j} x_{ij} - 1\right) $$

Since $x_{ij}$ increases continuously and the condition in line 7 in Algorithm FPD is satisfied, $y_i \le p_i (1 - \exp (-\alpha ))$ is always guaranteed. Combining line 9, we have $\dfrac{\partial y_i}{\partial x_{ij}} \le \alpha p_i$. Similarly, we can get

$$\dfrac{\partial z_j}{\partial x_{ij}} \le \dfrac{\beta c_i }{2 K_j} \dfrac{p_{i_{*}}}{c_{i_{*}}} \exp \beta \left( \dfrac{\sum _{i \in I_j:d_i \in S_j} x_{ij} c_i}{2 K_j}\right) $$

Due to $c_i z_j \le p_i (1 - \exp (-\alpha )) \le p_i$, we have

$$\dfrac{\partial z_j}{\partial x_{ij}} \le \dfrac{\beta c_i }{2 K_j} \left( \dfrac{p_{i}}{c_{i}} + \dfrac{p_{i_{*}}}{c_{i_{*}}}\right) \le \dfrac{\beta c_i }{2 K_j} 2 \dfrac{p_{i}}{c_{i}} = \dfrac{\beta p_i}{K_j}$$

By the same way, we can get $\dfrac{\partial r_v}{\partial x_{ij}} \le \dfrac{\beta p_i}{R_v}, \ \ \dfrac{\partial u_j}{\partial x_{ij}} \le \dfrac{\beta p_i}{T_j}, \ \ \dfrac{\partial q_v}{\partial x_{ij}} \le \dfrac{\beta p_i}{T_v}$. Therefore, we have $\dfrac{\partial D_\alpha }{\partial x_{ij}} \le (\alpha + 4 \beta ) p_i = (\alpha + 4 \beta ) \dfrac{\partial P}{\partial x_{ij}}$. From the weak duality property, we have $\mathrm {OPT}_\alpha \le D_\alpha \le (\alpha + 4 \beta ) P$. Therefore, we further get

$$P \ge \dfrac{\mathrm {OPT}_\alpha }{\alpha + 4 \beta } \ge \dfrac{1 - \exp (-\alpha )}{\alpha + 4 \beta }\mathrm {OPT}$$

Let $\sigma = (1 - \exp (-\alpha ))/(\alpha + 4 \beta )$ and $\alpha = - \ln \sigma$. We can get

$$\beta = \dfrac{1 - \sigma + \sigma \ln \sigma }{4\sigma }$$

Let $g(\sigma ) = 1 - \sigma + \sigma \ln \sigma - \frac{1}{2}\sigma \ln ^2 \sigma$. Since $g'(\sigma ) = -\frac{1}{2} \ln ^2 \sigma \le 0$ and $\sigma \in (0, 1]$, we have $g(\sigma ) \ge g(1) = 0$. Therefore, we can get

$$\beta \ge \dfrac{\frac{1}{2}\sigma \ln ^2 \sigma }{4 \sigma } = \frac{1}{8}\ln ^2 \sigma $$

Now consider all the constraints in the primal program (6). First Constraint (1) is always satisfied because $y_i \le p_i(1 - \exp (-\alpha ))$ always holds. Let $i^* = \arg \max _{i \in I_j: d_i \in S_j} \dfrac{p_i}{c_i}$. Because $z_j$ is increased continuously and Algorithm FPD will not increase $z_j$ if the condition in line 7 is not satisfied, we have $z_j \le \dfrac{p_{i^*}}{c_{i^*}}$, i.e.,

$$\dfrac{p_{i_{*}}}{c_{i_{*}}} \left[ \exp \left( \beta \dfrac{\sum _{i \in I_j:d_i \in S_j} x_{ij} c_i}{2 K_j}\right) - 1 \right] \le \dfrac{p_{i^*}}{c_{i^*}}$$

Rearranging the above inequality, we have

$$\begin{aligned} \dfrac{\sum _{i \in I_j:d_i \in S_j} x_{ij} c_i}{K_j} \le \dfrac{2}{\beta } \log \left( \dfrac{p_{i^*}}{c_{i^*}}\dfrac{p_{i_{*}}}{c_{i_{*}}} + 1\right)&\le \dfrac{16}{\ln ^2 \sigma } \log \left( \dfrac{p_{i^*}}{c_{i^*}}\dfrac{p_{i_{*}}}{c_{i_{*}}} + 1\right) = O\left( \dfrac{\log \varGamma _j}{\log ^2 \sigma }\right) \end{aligned}$$

By the same way, we finish the proof.

B Proof of Theorem 2

Proof

Let x and (y, z, r, u, q) be the primal and dual solution computed by Algorithm IPD while P and D be the objective values, respectively. We finish the proof of this theorem from the following two properties:

• (1) The solution (y, z, r, u, q) is feasible for the dual program.

• (2) When order $(i,d_i, c_i, t_i)$ is accepted, the objective of the dual program increases $(1+\epsilon )p_i$.

For Property (1), if order $(i, d_i, c_i, t_i)$ is rejected, we have $c_i (z_j + r_v) + t_{ij} (u_j + q_v) \ge p_i$ for $\forall j \in B_i: d_i \in S_j$. Otherwise, we set $y_i \leftarrow p_i - c_i ( z_{j'} + r_v) - t_{ij'} (v_{j'} + q_v)$, which guarantees that $y_i + c_i (z_j + r_v) + t_{ij} (u_j + q_v) \ge p_i$ for $\forall j: d_i \in S_j$ because $j' \leftarrow \arg \min _{j \in B_i:d_i \in S_j} c_i z_j + t_{ij} (u_j + q_v)$.

For Property (2), when order $(i, d_i, c_i, t_i)$ is accepted, all the dual variables will be updated. Therefore, the objective of the dual program increases

$$\begin{aligned} \varDelta D&= \varDelta y_i + \varDelta K_{j'} z_{j'} + \varDelta R_v r_v + \varDelta T_{j'} u_{j'} + \varDelta T_v q_v \\&= p_i - c_i ( z_{j'} + r_v) - t_{ij'} (v_{j'} + q_v) + K_{j'} \left( \dfrac{z_{j'} c_i}{K_{j'}} + \dfrac{p_i \epsilon }{\ 4 K_{j'}} \right) \\&+\,R_v \left( \dfrac{r_v c_i}{R_v} + \dfrac{p_i \epsilon }{\ 4 R_v}\right) + T_{j'} \left( \dfrac{u_{j'} t_{ij'}}{T_{j'}} + \dfrac{p_i \epsilon }{\ 4 T_{j'}} \right) + T_{v} \left( \dfrac{q_{v} t_{ij'}}{T_{v}} + \dfrac{p_i \epsilon }{\ 4 T_{v}} \right) \\&= (1+\epsilon )p_i \end{aligned}$$

Based on Property (1) (2) and the weak duality, we get

$$P \ge \dfrac{D}{1 + \epsilon } \ge \dfrac{\mathrm {OPT}}{1 + \epsilon } \ge (1 - \epsilon ) \mathrm {OPT}$$

For the constraints, we consider the dual variables (z, r, u, q), respectively. Let $z_j^t$ be the value of $z_j$ after the t-th update. Let $i_* = \arg \min _{i: d_i \in S_j} \dfrac{p_i}{c_i}$. Therefore, we have

$$z_{j}^t = z_{j}^{t-1} \left( 1 + \dfrac{c_i}{K_{j}}\right) + \dfrac{p_i \epsilon }{\ 4 K_{j}} \ge z_{j}^{t-1} \left( 1 + \dfrac{c_i}{K_{j}}\right) + \dfrac{c_i \epsilon }{\ 4 K_{j}} \dfrac{p_{i_*}}{c_{i_*}}$$

Rearranging the above inequality, we can get

$$z_{j}^t + \dfrac{p_{i_*} \epsilon }{\ 4 c_{i_*}} \ge \left( z_{j}^{t-1} + \dfrac{p_{i_*} \epsilon }{\ 4 c_{i_*}} \right) \left( 1 + \dfrac{c_i}{K_{j}}\right) $$

Let $\gamma _j$ satisfy

$$\begin{aligned} \left( 1 + \dfrac{c_i}{K_j} \right) \ge \gamma _j^{\frac{c_i}{K_j}}, \ \forall i: d_i \in S_j \end{aligned}$$

(11)

Therefore, we have

$$\begin{aligned} z_{j}^t + \dfrac{p_{i_*} \epsilon }{\ 4 c_{i_*}}&\ge \left( z_{j}^{t-1} + \dfrac{p_{i_*} \epsilon }{\ 4 c_{i_*}} \right) \gamma _j^{\frac{c_i}{K_j}} \\&\ge \left( z_{j}^{0} + \dfrac{p_{i_*} \epsilon }{\ 4 c_{i_*}} \right) \gamma _j^{\frac{\sum _{i:d_i \in S_j} x_{ij}c_i}{K_j}} \nonumber \\&= \dfrac{p_{i_*} \epsilon }{\ 4 c_{i_*}} \gamma _j^{\frac{\sum _{i \in I_j:d_i \in S_j} x_{ij}c_i}{K_j}} \nonumber \end{aligned}$$

(12)

According to Algorithm IPD (lines 6–7), before the last update of $\widetilde{z_j}$, we have

$$c_i (z_j^{t-1} + q_v^{t-1})+ t_{ij} (u_j^{t-1} + q_v^{t-1}) < p_i $$

Hence, $z_j^{t-1} < \dfrac{p_i}{c_i}$. Let $i^{*} = \arg \max _{i \in B_j: d_i \in S_j} \dfrac{p_i}{c_i}$. After the last update, we can get

$$\begin{aligned} z_j^{t} < \dfrac{p_i}{c_i} \left( 1 + \dfrac{c_i}{K_{j}}\right) + \dfrac{p_i \epsilon }{\ 4 K_{j}} = \dfrac{p_i}{c_i} + \left( 1 + \dfrac{\epsilon }{4} \right) \dfrac{p_i}{K_j}&\le \dfrac{p_i}{c_i} + \left( 1 + \dfrac{\epsilon }{4} \right) \dfrac{p_i}{c_i} \\&\le \dfrac{p_{i^{*}}}{c_{i^{*}}} \left( 2 + \dfrac{\epsilon }{4} \right) \end{aligned}$$

Combining Inequality (12), we have

$$\dfrac{p_{i^{*}}}{c_{i^{*}}} \left( 2 + \dfrac{\epsilon }{4} \right) \ge z_{j}^t \ge \dfrac{p_{i_*} \epsilon }{\ 4 c_{i_*}} \left( \gamma _j^{\frac{\sum _{i \in I_j:d_i \in S_j} x_{ij}c_i}{K_j}} - 1\right) $$

Therefore,

$$\dfrac{\sum _{i \in I_j:d_i \in S_j} x_{ij}c_i}{K_j} \le \log _{\gamma _j} \left( \dfrac{p_{i^{*}}}{c_{i^{*}}} \dfrac{c_{i_*}}{p_{i_*}} \left( \dfrac{8}{\epsilon } + 1 \right) + 1\right) \le \log _{\gamma _j} \dfrac{p_{i^{'}}}{c_{i^{'}}} \dfrac{c_{i^*}}{p_{i^*}} \left( \dfrac{8}{\epsilon } + 2 \right) $$

Now consider $\gamma _j$. From Equality (11), we have

$$\ln \gamma _j \le \min _{i \in I_j: d_i \in S_j} \dfrac{\ln (1 + \frac{c_i}{K_j})}{\frac{c_i}{K_j}}$$

Because $0 \le \frac{c_i}{K_j} \le 1$ and $f(x) = \frac{\ln (1+x)}{x}$ is a monotone decreasing function, Equality (11) is satisfied when $\gamma _j = 2$. Therefore,

$$\begin{aligned} \dfrac{\sum _{i \in I_j:d_i \in S_j} x_{ij}c_i}{K_j} \le \left( \log _2 \dfrac{p_{i^{*}}}{c_{i^{*}}} \dfrac{c_{i_*}}{p_{i_*}} \left( \dfrac{8}{\epsilon } + 2 \right) \right) = O \left( \log \varGamma _j + \log \dfrac{1}{\epsilon } \right) \end{aligned}$$

By the same way, we finish this proof.