Integrated Network Capacity Expansion and Traffic Signal Optimization Problem: Robust Bi-level Dynamic Formulation

Abstract

This paper presents a robust optimization formulation, with an exact solution method, that simultaneously solves continuous network capacity expansion, traffic signal optimization and dynamic traffic assignment when explicitly accounting for an appropriate robustness measure, the inherent bi-level nature of the problem and long-term O-D demand uncertainty. The adopted robustness measure is the weighted sum of expected total system travel time (TSTT) and squared up-side deviation from a fixed target. The model propagates traffic according to Daganzo’s cell transmission model. Furthermore, we formulate five additional, related models. We find that when evaluated in terms of robustness, the integrated robust model performs the best, and interestingly the sequential robust approach yields a worse solution compared to certain sequential and integrated approaches. Although the adopted objective of the integrated robust model does not directly optimize the variance of TSTT, our experimental results show that the robust solutions also yield the least-variance solutions.

Appendix

1.1 Reduction of MZC-QL-BLPP to QL-BLPP

The proof closely follows the work by Bard (1998) and Kalantari and Rosen (1982). Next, consider the standard optimization problem Eq. (12):

$$\left\{ \begin{aligned}& \mathop {\min }\limits_{x_1 ,x_2 } z\left( {x_1 ,x_2 ,y} \right) = c_{11} x_1 + c_{12} x_2 + d_1 y + \frac{1}{2}y^T Qy \\& {\text{subject to }}\left( {x_1 ,x_2 ,y} \right) \in P \\& {\text{ }}x_1 \in R_ + ^{n_1 } ,x_2 \in \left\{ {0,1} \right\}^{n_2 } ,y \in R_ + ^m \\ \end{aligned} \right\}\quad \quad \quad \left( {x_1^ * ,x_2^ * ,y^* } \right)$$

(12)

where $P = \cup ^{l}_{{i = 1}} P^{\prime }_{i}$ such that $P_i^\prime$ is a polyhedral set for i = 1,2,…,l and l is a positive integer. The optimal solution of Eq. (12) is denoted by $\left( {x_1^* ,x_2^* ,y^* } \right)$. Denote the feasible region of Eq. (12) by F:

$$F = \left\{ {\left. {\left( {x_1 ,x_2 ,y} \right)} \right|\left( {x_1 ,x_2 ,y} \right) \in P,x_1 \in R_ + ^{n_1 } ,x_2 \in \left\{ {0,1} \right\}^{n_2 } ,y \in R_ + ^m } \right\}$$

In this appendix, we will show that the mixed zero-one quadratic program Eq. (12) can be reduced to a quadratic program Eq. (13):

$$\left\{ \begin{aligned}& \mathop {\min }\limits_{x_1 ,x_2 } f\left( {x_1 ,x_2 ,y} \right) = c_{11} x_1 + c_{12} x_2 + d_1 y + \frac{1}{2}y^T Qy + M\theta \left( {x_2 } \right) \\& {\text{subject to }}\left( {x_1 ,x_2 ,y} \right) \in P \\& {\text{ }}x_1 \in R_ + ^{n_1 } ,0 \leqslant x_2 \leqslant e_{n_2 } ,y \in R_ + ^m \\ \end{aligned} \right\}\quad \quad \quad \left( {\tilde x_1 ,\tilde x_2 ,\tilde y} \right)$$

(13)

where $\theta :R^{n_2 } \to R$ is a continuous function such that θ(x ₂) ≥ 0 for all $0 \leqslant x_2 \leqslant e_{n_2 }$ and $\theta \left( {x_2 } \right) = 0$ if and only if $x_2 \in \left\{ {0,1} \right\}^{n_2 }$. $e_{n_2 }$ is a vector of ones with dimensions n ₂ . The optimal solution of Eq. (13) is denoted by $\left( {\widetilde{x_1}},{\widetilde{x_2}},{\widetilde{y}} \right)$. We denote the feasible region of Eq. (13) by $\tilde F$:

$$\tilde F = {\left\{ {\left. {\left( {x_1 ,x_2 ,y} \right)} \right|\left( {x_1 ,x_2 ,y} \right) \in P,x_1 \in R_ + ^{n_1 } ,0 \leqslant x_2 \leqslant e_{n_2 } ,y \in R_ + ^m } \right\}}$$

Thus, $F = {\left( {R_ + ^{n_1 } \times \left\{ {0,1} \right\}^{n_2 } \times R_ + ^m } \right)} \cap \tilde F$. It can be assumed without loss of generality that $c_{11} \leqslant 0$ _, $c_{12} \leqslant 0$, and $d_1 \leqslant 0$. If the ith component of c ₁₁ (respective c ₁₂ and d ₁) is positive, it is always possible to define a new variable $x_{1_i }^\prime = 1 - x_{1_i }$ ($x_{2_i }^\prime = 1 - x_{2_i }$, $y_i^\prime = 1 - y_i$) and make the appropriate substitutions to yield an equivalent program satisfying this requirement.

Theorem 1

If θ is a concave function, there exists a positive real number M such that the problem Eq. (12) and the problem Eq. (13) have the same optimal solutions.

Proof

If $\theta \left( {\tilde x_2 } \right) = 0$, then the result is true for any positive value of M. Suppose $\theta (\tilde x_2 ) >0$. Consider Eq. (12) when the integrality requirement is relaxed, giving problem Eq. (14):

$$\left\{ \begin{aligned} \mathop {\min }\limits_{x_1 ,x_2 } z\left( {x_1 ,x_2 ,y} \right) = c_{11} x_1 + c_{12} x_2 + d_1 y + \frac{1}{2}y^T Qy \\& {\text{subject to }}\left( {x_1 ,x_2 ,y} \right) \in \tilde F \\ \end{aligned} \right\}\quad \quad \quad \left( {x_1^0 ,x_2^0 ,y^0 } \right)$$

(14)

Let $\left( {x_1^0 ,x_2^0 ,y^0 } \right)$denote the optimal solution of Eq. (14).

Consider the following Eq. (15) with the optimal solution $\left( {\overline x _1 ,\overline x _2 ,\overline y } \right)$:

$$\left\{ \begin{aligned}& \min \theta \left( {x_2 } \right) \\& {\text{subject to }}\left( {x_1 ,x_2 ,y} \right) \in \widetilde F \\ \end{aligned} \right\}\quad \quad \quad \left( {\overline {x _1} ,\overline {x _2} ,\overline y } \right)$$

(15)

If $\theta \left( {\overline x _2 } \right) = 0$, then the result is true for any positive value of M. Suppose that $\theta \left( {\overline x _2 } \right) >0$.

Next, define M in Eq. (13) such that

$$M_0 = \frac{{ - c_{11} x_1^0 - c_{12} x_2^0 - d_1 y^0 - \frac{1}{2}y{^0} ^T Qy^0 + \frac{1}{2}y{^*} ^T Qy^* }}{{\theta \left( {\overline x _2 } \right)}} = \frac{{ - z\left( {x_1^0 ,x_2^0 ,y^0 } \right) + \frac{1}{2}y{^*} ^T Qy^* }}{{\theta \left( {\overline x _2 } \right)}}$$

Then, $f\left( {x_1^* ,x_2^* ,y^* } \right) - f\left( {\widetilde{x_1}} ,{\widetilde{x_2}} ,{\widetilde{y}} \right)$

$$\begin{array}{l} { = c_{11} x_1^* + c_{12} x_2^* + d_1 y^* + \frac{1}{2}y^{*T} Qy^* + M\theta \left( {x_2^* } \right) - \left( {c_{11} \widetilde{x}_1 + c_{12} \widetilde{x}_2 + d_1 \widetilde{y} + \frac{1}{2}\widetilde{y}^T Q\widetilde{y} + M\theta \left( {\widetilde{x}_2 } \right)} \right)} \\ { = c_{11} x_1^* + c_{12} x_2^* + d_1 y^* + \frac{1}{2}y^{*T} Qy^* - \left( {c_{11} \widetilde{x}_1 + c_{12} \widetilde{x}_2 + d_1 \widetilde{y} + \frac{1}{2}\widetilde{y}^T Q\widetilde{y} + M\theta \left( {\widetilde{x}_2 } \right)} \right)\,\left( {\text{as}\,x_2^ * \in \left\{ {0,1} \right\}^{n_2 } } \right)} \\ { \leqslant c_{11} x_1^* + c_{12} x_2^* + d_1 y^* + \frac{1}{2}y^{*T} Qy^* - \left( {c_{11} \widetilde{x}_1 + c_{12} \widetilde{x}_2 + d_1 \widetilde{y} + \frac{1}{2}\widetilde{y}^T Q\widetilde{y} + M\theta \left( {\overline x _2 } \right)} \right)\left( {\text{as }\theta \left( {\overline x _2 } \right) \leqslant \theta \left( {\widetilde{x}_2 } \right)} \right)} \\ { < c_{11} x_1^* + c_{12} x_2^* + d_1 y^* + \frac{1}{2}y^{*T} Qy^* - \left( {c_{11} \widetilde{x}_1 + c_{12} \widetilde{x}_2 + d_1 \widetilde{y} + \frac{1}{2}\widetilde{y}^T Q\widetilde{y} + M_0 \theta \left( {\overline x _2 } \right)} \right)\left( {\text{as}\,M > M_0 } \right)} \\ { \leqslant \frac{1}{2}y^{*T} Qy^* - \left( {c_{11} \widetilde{x}_1 + c_{12} \widetilde{x}_2 + d_1 \widetilde{y} + \frac{1}{2}\widetilde{y}^T Q\widetilde{y} + M_0 \theta \left( {\overline x _2 } \right)} \right)\quad \left( {\text{as}\,c_{11} ,c_{12} ,d_1 \leqslant 0\text{ and }x_1^* ,x_2^* ,y^* \geqslant 0} \right)} \\ { = \frac{1}{2}y^{*T} Qy^* - \left( {z\left( {\widetilde{x}_1 ,\widetilde{x}_2 ,\widetilde{y}} \right) + M_0 \theta \left( {\overline x _2 } \right)} \right)} \\ { = \frac{1}{2}y^{*T} Qy^* - \left( {z\left( {\widetilde{x}_1 ,\widetilde{x}_2 ,\widetilde{y}} \right) - z\left( {x_1^0 ,x_2^0 ,y^0 } \right) + \frac{1}{2}y^{*T} Qy^* } \right)\quad \left( {\text{as}\,M_0 = \frac{{ - z\left( {x_1^0 ,x_2^0 ,y^0 } \right) + \frac{1}{2}y^{*} TQy^* }}{{\theta \left( {\overline x _2 } \right)}}} \right)} \\ { = z\left( {x_1^0 ,x_2^0 ,y^0 } \right) - z\left( {\widetilde{x}_1 ,\widetilde{x}_2 ,\widetilde{y}} \right)} \\ { \leqslant 0} \\ \end{array}$$

(from definition of Eq. (14))

This establishes a contradiction since from the definition of Eq. (13), $f\left( {x_1^* ,x_2^* ,y^* } \right) - f\left( {\widetilde{x_1}} ,{\widetilde{x_2}} ,{\widetilde{y}} \right) \geqslant 0.$

Thus, $\theta \left( {\overline x _2 } \right) = 0$, $\theta \left( {\tilde x_2 } \right) = 0$ and $\left( {\widetilde{x_1}} ,{\widetilde{x_2}} ,{\widetilde{y}} \right) \in F$. From the definition of Eq. (13),

$$\begin{array}{l} {f\left( {x_1^* ,x_2^* ,y^* } \right) - f\left( {\widetilde{x}_1 ,\widetilde{x}_2 ,\widetilde{y}} \right)} \\ {= z\left( {x_1^* ,x_2^* ,y^* } \right) - z\left( {\widetilde{x}_1 ,\widetilde{x}_2 ,\widetilde{y}} \right)\quad \quad \quad \left( {\text{as}\,\theta \left( {x_2^* } \right) = 0\,\text{and}\,\theta \left( {\widetilde{x}_2 } \right) = 0} \right)} \\ { \geqslant 0.} \\ \end{array} $$

From the definition of Eq. (12), $z\left( {x_1^* ,x_2^* ,y^* } \right) - z\left( {\widetilde{x_1}} ,{\widetilde{x_2}} ,{\widetilde{y}} \right) \leqslant 0.$

Therefore, $z\left( {x_1^* ,x_2^* ,y^* } \right) = z\left( {\widetilde{x_1}} ,{\widetilde{x_2}} ,{\widetilde{y}} \right)$.

Theorem 2

There exists a positive real M′ such that MZC-QL-BLPP [Eqs. (10.1)–(10.6)] and QL-BLPP(M′) [Eqs. (11.1)–(11.8)], have the same optimal solutions.

Proof

From the theoretical properties of L-BLPP in Bard (1998), the inducible region IR can be stated as $IR = \left\{ {\left( {x_1 ,x_2 ,y} \right):A_{11} x_1 + A_{12} x_2 + B_1 y \leqslant b_1 ,y \in \arg \min \left( {d_2 y:B_2 y \leqslant b_2 - A_{21} x_1 - A_{22} x_2 ,y \geqslant 0} \right)} \right\}$.

The inducible region is a finite union of polyhedral sets. Thus, MZC-QL-BLPP is a special case of problem Eq. (12) mentioned in Theorem 1. If we define $\theta \left( {x_2 } \right) = \sum\limits_{j = 1}^{n_2 } {\min \left\{ {x_2}{ _j} ,1 - {x_2}{ _j } \right\}}$, which is a concave function in x ₂ , then we can apply Theorem 1 to assure the existence of a real positive M′ such that MZC-QL-BLPP and the problem Eqs. (16.1)–(16.6) have the same optimal solutions.

$$\mathop {\min }\limits_x F\left( {x,y} \right) = c_{11} x_1 + c_{12} x_2 + d_1 y + \frac{1}{2}y^T Qy + M\prime \sum\limits_{j = 1}^{n_2 } {\min \left\{ {x_2}{_j} ,1 - {x_2}{ _j } \right\}} $$

(16.1)

$${\text{subject}}\,{\text{to}}\quad A_{11} x_1 + A_{12} x_2 + B_1 y \leqslant b_1 $$

(16.2)

$$x_1 \geqslant 0,\;0 \leqslant x_2 \leqslant e_{n_2 } $$

(16.3)

$$\mathop {\min }\limits_y f\left( y \right) = d_2 y$$

(16.4)

$${\text{subject}}\,{\text{to}}\quad A_{21} x_1 + A_{22} x_2 + B_2 y \leqslant b_2 $$

(16.5)

$$y \geqslant 0$$

(16.6)

However, the problem Eqs. (16.1)–(16.6) is equivalent to QL-BLPP(M′) Eqs. (11.1)–(11.8), since the lower–level optimality in QL-BLPP(M′) enforces u _j to be equal to $\min \left\{ {x_2}{ _j} ,1 - {x_2}{ _j } \right\}$ for all $j \in \{ 1,...,n_2 \}$. We establish the equivalence between MZC-QL-BLPP and QL-BLPP(M′).