Feedback Control Algorithms for the Dissipation of Traffic Waves with Autonomous Vehicles

Abstract

This article considers the problem of traffic control in which an autonomous vehicle is used to regulate human-piloted traffic to dissipate stop-and-go traffic waves. We first investigated the controllability of well-known microscopic traffic flow models, namely, (i) the Bando model (also known as the optimal velocity model), (ii) the follow-the-leader model, and (iii) a combined optimal velocity follow-the-leader model. Based on the controllability results, we proposed three control strategies for an autonomous vehicle to stabilize the other, human-piloted traffics. We subsequently simulate the control effects on the microscopic models of human drivers in numerical experiments to quantify the potential benefits of the controllers. Based on the simulations, finally, we conduct a field experiment with 22 human drivers and a fully autonomous-capable vehicle, to assess the feasibility of autonomous vehicle-based traffic control on real human-piloted traffic. We show that both in simulation and in the field test that an autonomous vehicle is able to dampen waves generated by 22 cars, and that as a consequence, the total fuel consumption of all vehicles is reduced by up to 20%.

Appendix

The linearization of system  (12.8) at the equilibrium point $(Y^*,0)\in \mathrm{I}\!\mathrm{R}^{2n}$ is described by

$$\begin{aligned} \dot{Y} =A Y + Bu, \end{aligned}$$

(12.20)

where

(12.21)

For $n\in \mathrm {I\!N}$ define the matrices $(D_{n-1}, F_{n-1}) \in \mathscr {M}_{2n-1}(\mathrm{I}\!\mathrm{R})^2$ by:

(12.22)

with $\gamma =bV'(y^*+d)$ and $\alpha =\frac{a}{(y^*+d)^2}$. Moreover, denote by $0_{n-1}\in \mathscr {M}_{n-1}(\mathrm{I}\!\mathrm{R})$ and $I_{n-1} \in \mathscr {M}_{n-1}(\mathrm{I}\!\mathrm{R})$ the zero matrix and the identity matrix respectively. Let $k\geqslant 1$. From (12.21), there exist $(A_{i,k})_{i\in \{1,\ldots ,4\}}\in \mathscr {M}_{n-1}(\mathrm{I}\!\mathrm{R})$ and $(C_{i,k})_{i\in \{1,2\}}\in \mathscr {M}_{n-1,1}(\mathrm{I}\!\mathrm{R})$ such that,

$$ A^k=\begin{bmatrix} A_{1,k} &{} A_{2,k} &{} C_{1,k} \\ A_{3,k} &{} A_{4,k} &{} C_{2,k}\\ 0_{1,n-1} &{} 0_{1,n-1}&{} 0 \\ \end{bmatrix}, $$

and for every $k\geqslant 1$ we have

$$\begin{aligned} \left\{ \begin{array}{l} A_{4,k+1}=D_{n-1}A_{2,k}+F_{n-1}A_{4,k},\\ A_{2,k+1}=A_{4,k},\\ C_{2,k+1}=D_{n-1}C_{1,k}+F_{n-1}C_{2,k},\\ C_{1,k+1}=C_{2,k}.\\ \end{array}\right. \end{aligned}$$

(12.23)

In particular, for every $k\geqslant 2$,

$$\begin{aligned} \left\{ \begin{array}{l} A_{4,k+1}=D_{n-1}A_{4,k-1}+F_{n-1}A_{4,k},\\ A_{2,k+1}=D_{n-1}A_{2,k-1}+F_{n-1}A_{2,k}, \\ C_{2,k+1}=D_{n-1}C_{2,k-1}+F_{n-1}C_{2,k},\\ C_{2,k}=D_{n-1}C_{1,k-1}+F_{n-1}C_{1,k}. \end{array}\right. \end{aligned}$$

(12.24)

Combining (12.23) with (12.24) and using that, for every $k\geqslant 1$, $A^kB(2n-1)=0$, we conclude that, for every $k\geqslant 2$,

$$\begin{aligned} A^{k+1} B= \mathscr {D} A^{k-1}B+\mathscr {F}A^k B, \end{aligned}$$

(12.25)

with

$$\begin{aligned} \mathscr {D}=\begin{bmatrix} D_{n-1} &{} 0_{n-1} &{} 0 \\ 0_{n-1} &{} D_{n-1} &{} 0\\ 0_{1,n-1} &{} 0_{1,n-1}&{} 0 \\ \end{bmatrix} \quad \text {and} \quad \mathscr {F}=\begin{bmatrix} F_{n-1} &{} 0_{n-1} &{} 0 \\ 0_{n-1} &{} F_{n-1} &{} 0\\ 0_{1,n-1} &{} 0_{1,n-1}&{} 0 \\ \end{bmatrix}. \end{aligned}$$

(12.26)

Moreover, we have

$$\begin{aligned} A^2B=X+ \mathscr {F}_{2n-1} AB \quad \text {with} \quad AB=\left( \begin{array}{l} 0_{n-2,1}\\ 1\\ 0_{n-3,1}\\ \alpha \\ -\alpha \\ 0 \end{array}\right) \, \, \text {and} \, \, X=\left( \begin{array}{l} 0_{n-2,1}\\ b\\ 0_{n-3,1}\\ \gamma \\ -\gamma \\ 0 \end{array}\right) . \end{aligned}$$

(12.27)

Proof of Theorem 12.3 If $b=0$ then $\gamma =0$. From (12.25) and (12.27), for every $k\geqslant 2$,

$$ \left\{ \begin{array}{l} A^{k+1}=\mathscr {F} A^kB, \\ A^2B=\mathscr {F} AB.\\ \end{array}\right. $$

Thus, the Kalman controllability matrix (12.6) satisfies

$$ \text {rank}\left( K(A,B)\right) =\text {rank}\left( B,AB,\mathscr {F}AB,\ldots ,\mathscr {F}^{2n-4}AB\right) . $$

By Cayley–Hamilton Theorem, there exists $(\alpha _0,\ldots , \alpha _{n-2}) \in \mathrm{I}\!\mathrm{R}^{n-1}$ such that $F^{n-1}_{n-1}=\sum _{i=0}^{n-2} \alpha _i F_{n-1}^i$. From (12.26), we conclude that

$$ \text {rank}\left( K(A,B)\right) =\text {rank}\left( B,AB,\mathscr {F}AB,\ldots ,\mathscr {F}^{n-2}AB\right) . $$

Using the expression of AB given in (12.27) and the equality $\mathscr {F}^k=\begin{bmatrix} F^k_{n-1} &{} 0_{n-1} &{} 0 \\ 0_{n-1} &{} F^k_{n-1} &{} 0\\ 0_{1,n-1} &{} 0_{1,n-1}&{} 0 \\ \end{bmatrix}$, by straightforward computations, we have

$$ \text {rank}\left( B,AB,\mathscr {F}AB,\ldots ,\mathscr {F}^{n-2}AB\right) =n, $$

whence the conclusion.

Proof of Theorem 12.4 Let’s prove by induction that, for every $k\geqslant 1$, there exist $\left( \lambda _i\right) _{i=1,\ldots ,2k}$ and $\left( \mu _i\right) _{i=1,\ldots ,2k+1}$ such that

$$\begin{aligned} \left\{ \begin{array}{l} A^{2k+1}B=\sum _{i=1}^{2k} \lambda _i A^i B+ \mathscr {D}^k(AB) \\ A^{2k+2}B=\sum _{i=1}^{2k+1} \mu _i A^i B+ \mathscr {D}^k(A^2B) \end{array}\right. . \qquad \qquad \qquad {(P_k)} \end{aligned}$$

Since $a=0$, we have $F_{n-1}=-b I_{n-1}$. Using (12.25), $A^3B=\mathscr {D}AB-bA^2B$ and $A^4B=\mathscr {D}A^2B-bA^3B$. Thus, ($P_k$) holds for $k=1$. Assuming that $(P_k)$ holds for $k=p$. From (12.25), we have

$$\begin{aligned} \left\{ \begin{array}{l} A^{2p+3}=\mathscr {D}A^{2p+1}-bA^{2p+2}B \\ A^{2p+4}=\mathscr {D}A^{2p+2}-bA^{2p+3}B \\ \end{array}\right. \quad \text {and} \quad \mathscr {D}A^{i}B=A^{i+2}B+bA^{i+1}B, \, i\geqslant 1. \end{aligned}$$

(12.28)

Using (12.28) and $(P_k)$ for $k=p$, we conclude that $(P_k)$ holds for $k=p+1$.

The equality $(P_k)$ for $k=n-2$ gives

$$ \begin{array}{ll} \text {rank}(K(A,B))&{}=\left( B,AB,\ldots , A^{2n-3}, \mathscr {D}^{n-2} A^2B \right) \\ &{}=\left( B,AB,\ldots , A^{2n-4},\mathscr {D}^{n-2} AB, \mathscr {D}^{n-2} A^2B \right) \\ &{}=\left( B,AB,A^2B,\mathscr {D}AB,\mathscr {D}A^2B,\ldots ,\mathscr {D}^{n-2} AB, \mathscr {D}^{n-2} A^2B \right) . \end{array} $$

Since AB and $A^2B$ are linearly independent and $\mathscr {D}^k=\begin{bmatrix} D^k_{n-1} &{} 0_{n-1} &{} 0 \\ 0_{n-1} &{} D^k_{n-1} &{} 0\\ 0_{1,n-1} &{} 0_{1,n-1}&{} 0 \\ \end{bmatrix}$, by straightforward computations, we have

$$ \text {rank}\left( B,AB,A^2B,\mathscr {D}AB,\mathscr {D}A^2B,\ldots ,\mathscr {D}^{n-2} AB, \mathscr {D}^{n-2} A^2B \right) =2n-1. $$

Thus, the linearization of system  (12.8) at the equilibrium point $(Y^*,0)$ is controllable. Using [8, Theorem 3.8], Theorem 12.4 is proved.

Proof of Theorem 12.5 Using the symbolic mathematics software Maple [https://www.maplesoft.com/products/Maple/], we establish that for every $3\leqslant n\leqslant 9$,

$$ \text {Det}(K(A,B))=\gamma ^{\frac{n^2-3n+2}{2}}b^{\frac{n^2-n}{2}}\left( \alpha -\frac{\gamma }{b}\right) ^{\frac{n^2-n}{2}}. $$

Thus, for every $\alpha \ne \frac{\gamma }{b}$, $\gamma \ne 0$, $b \ne 0$, the linearization of system  (12.8) at the equilibrium point $(Y^*,0)$ is controllable, whence the conclusion of Theorem 12.5 by using [8, Theorem 3.8].