The Max-Pressure Controller for Arbitrary Networks of Signalized Intersections

Abstract

The control of an arbitrary network of signalized intersections is considered. At the beginning of each cycle, a controller selects the duration of every stage at each intersection as a function of all queues in the network. A stage is a set of permissible (non-conflicting) phases along which vehicles may move at pre-specified saturation rates. Demand is modeled by vehicles entering the network at a constant average rate with an arbitrary burst size and moving with pre-specified average turn ratios. The movement of vehicles is modeled as a “store and forward” queuing network. A controller is said to stabilize a demand if all queues remain bounded. The max-pressure controller is introduced. It differs from other network controllers analyzed in the literature in three respects. First, max-pressure requires only local information: the stage durations selected at any intersection depends only on queues adjacent to that intersection. Second, max-pressure is provably stable: it stabilizes a demand whenever there exists any stabilizing controller. Third, max-pressure requires no knowledge of the demand, although it needs turn ratios. The analysis is conducted within the framework of “network calculus,” which, for fixed-time controllers, gives guaranteed bounds on queue size, delay, and queue clearance times.

Appendices

Appendices

2.1.1 A Proof of Lemma 1

By induction. Since q(0) = 0, (2.2) holds for t = 0. Suppose (2.2) holds for t. Then

$$\displaystyle\begin{array}{rcl}q(t + 1)& =& \max \left \{0,\max _{0\leq s\leq t}[A(t,s) - C(t - 1,s)] - c(t)\right \} + a(t + 1) \\ & =& \max \left \{a(t + 1),\max _{0\leq s\leq t}[A(t + 1,s) - C(t,s)]\right \} \\ & =& \max _{0\leq s\leq t+1}[A(t + 1,s) - C(t,s)], \\ \end{array}$$

so (2.2) holds for t + 1. Since the queue size is the difference between arrivals and departures,

$$\displaystyle\begin{array}{rcl}B(t)& =& A(t) - [q(t) - q(0)] \\ & =& A(t) -\max _{0\leq s\leq t}[A(t,s) - C(t - 1,s)] \\ & =& \min _{0\leq s\leq t}[A(s) + C(t - 1,s)], \\ \end{array}$$

which proves (2.3).

2.1.2 B Proof of Theorem 1

Equation (2.4) follows from

$$q(t) =\max \limits_{0\leq s\leq t}[A(t,s)-C(t-1,s)] \leq \max \limits_{0\leq s\leq t}[f_{1}(t-s)-f_{2}(t-s)] =\max \limits_{0\leq \tau \leq t}[f_{1}(\tau )-f_{2}(\tau )].$$

Since always $B(t) \leq A(t)$,

$$\displaystyle\begin{array}{rcl} B(t,s)& \leq & A(t) - B(s) \\ & =& A(t) -\min \limits_{0\leq r\leq s}[A(r) + C(s - 1,r)] \\ & =& \max \limits_{0\leq r\leq s}[A(t,r) - C(s - 1,r)] \\ & \leq & \max \limits_{0\leq r\leq s}[f_{1}(t - r) - f_{2}(s - r)] \\ & =& \max \limits_{0\leq \tau \leq s}[f_{1}(t - s + \tau ) - f_{2}(\tau )] \\ & \leq & \max \limits_{0\leq \tau }[f_{1}(t - s + \tau ) - f_{2}(\tau )], \\ \end{array}$$

which proves (2.5). Next t + d(t) is the least time by which there are A(t) cumulative departures, so

$$d(t) =\min \{ d\ \vert \ B(t + d) \geq A(t)\}.$$

From (2.3),

$$\displaystyle\begin{array}{rcl} B(t + d) - A(t)& =& \min \limits_{s\leq t+d}\{A(s) + C(t + d - 1,s)\} - A(t) \\ & \geq & \min \{0,\min \limits_{s\leq t}\{ - A(t,s) + C(t + d - 1,s)\}\},\mbox{ as }A(s) - A(t) \\ & & +C(t + d - 1,s) \geq 0,s \geq t \\ & \geq & \min \{0,\min _{s\leq t}\{ - f_{1}(t - s) + f_{2}(t - s + d)\}\} \\ & =& \min \{0,\min _{0\leq \tau \leq t}\{ - f_{1}(\tau ) + f_{2}(\tau + d)\}\}.\end{array}$$

Hence $B(t + d) \geq A(t)$ if $f_{1}(\tau ) \leq f_{2}(\tau + d)$ for $\tau = 1,\cdots t$, which implies (2.6). Lastly, a busy period starting at s lasts until t if

$$A(s) = B(s),\;A(t + 1) = B(t + 1),\;\mbox{ and }A(s + \tau ) > B(s + \tau ),\;\tau = 1,\cdots \,,\ t - s,$$

and so

$$0 < A(s + \tau, s) - B(s + \tau, s) \leq f_{1}(\tau ) - f_{2}(\tau ),\;\tau = 1,\cdots \,,\ t - s,$$

from which (2.7) follows.

2.1.3 C Proof of Theorem 4

Proof.

According to (2.39) and (2.40) $\sum _{(l,m)}[q_{(l,m)}^{w}(t)/s(l,m)]$ is the same for all work-conserving controllers. So it is enough to exhibit one work-conserving controller for which (2.46) holds. For any controller $\sum \lambda _{(l,m)}(t)\delta _{(l,m)}$ write (2.37) in vector form ($Q(t) =\{ q_{(l,m)}(t)\}$)

$$Q(t + 1) = f(Q(t),t).$$

Because of (2.38) one can construct a work-conserving feedback controller $\sum \lambda _{(l,m)}^{w}(Q,t)\delta _{(l,m)}$ such that

$$[q_{(l,m)}^{w} - s(l,m)\lambda _{ (l,m)}^{w}({Q}^{{\prime}w},t)]_{ +} \leq [q_{(l,m)}(t) - s(l,m)\lambda _{(l,m)}(t)]_{+}$$

(2.90)

for all t, (l, m) and ${Q}^{w} \leq Q$ (the vector $\leq$ is interpreted component-wise). Write (2.37) for this work-conserving controller as

$${Q}^{w}(t + 1) = g({Q}^{w}(t),t).$$

It is not difficult to see that the functions f(Q, t) and g(Q, t) are both monotonic in Q, i.e.,

$${Q}^{w} \leq Q \Rightarrow f({Q}^{w},t) \leq f(Q,t)\mbox{ and }g({Q}^{w},t) \leq g(Q,t).$$

We claim that if Q ^w(0) = Q ^w(0) then

$${Q}^{w}(t) \leq Q(t),\;\;t \geq 0.$$

(2.91)

Equation (2.92) is clear for t = 0. Suppose it is true for t. Then

$${Q}^{w}(t + 1) = g({Q}^{w}(t),t) \leq g(Q(t),t) \leq f(Q(t),t) \leq Q(t + 1),$$

in which the first inequality is due to monotonicity of g and the second follows from (2.90). Thus this, and hence all, work-conserving controllers satisfy (2.46).

2.1.4 D Proof of (2.56)

We prove (2.56) in a few steps. For arrays $x =\{ x_{(l,m)}\}$ and $y =\{ y_{(l,m)}\}$ write $\langle x,y\rangle =\sum x_{(l,m)}y_{(l,m)}$, $\vert x{\vert }^{2} =\langle x,x\rangle$, $\min \{x,y\} =\{\min (x_{(l,m)},y_{(l,m)})\}$, $\max \{x,y\} =\{\max (x_{(l,m)},y_{(l,m)})\}$. Then (2.55) can be written as

$$q(t + 1) = [q(t) - {c}^{{\ast}}(t)]_{ +} + a(t + 1) =\max \{ q(t) - {c}^{{\ast}}(t),0\} + a(t + 1),$$

so

$$\delta = q(t + 1) - q(t) =\max \{ -{c}^{{\ast}}(t),-q(t)\} + a(t + 1) = -\min \{{c}^{{\ast}}(t),q(t)\} + a(t + 1).$$

Next,

$$\vert q(t + 1){\vert }^{2} -\vert q(t){\vert }^{2} = 2\langle \delta, q(t)\rangle + \vert \delta {\vert }^{2} = 2\alpha + \beta, \mbox{ say}.$$

(2.92)

We separately upper-bound $\alpha$ and $\beta$​.

2.1.4.1 Bound on α

$$\displaystyle\begin{array}{rcl} \alpha & =& \langle \delta, q(t)\rangle =\displaystyle\sum q_{(l,m)}(t)[a_{(l,m)}(t + 1) -\min \{ c{(l,m)}^{{\ast}}(t),q_{ (l,m)}(t)\}]\end{array}$$

(2.93)

$$\displaystyle\begin{array}{rcl} & =& \displaystyle\sum q_{(l,m)}(t)[a_{(l,m)}(t+1)-c{(l,m)}^{{\ast}}(t)+\max \{c{(l,m)}^{{\ast}}(t)-q_{ (l,m)}(t),0\}] \end{array}$$

(2.94)

$$\displaystyle\begin{array}{rcl}& =& \displaystyle\sum q_{(l,m)}(t)[a_{(l,m)}(t + 1) - c{(l,m)}^{{\ast}}(t)] \\ & & +\displaystyle\sum q_{(l,m)}(t)\max \{c{(l,m)}^{{\ast}}(t) - q_{ (l,m)}(t),0\} \end{array}$$

(2.95)

$$\displaystyle\begin{array}{rcl}& =& \alpha _{1} + \alpha _{2},\mbox{ say}.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \end{array}$$

(2.96)

Let $K =\max \{ a_{(l,m)}(t + 1),{c}^{{\ast}}(l,m)(t)\}$, the maximum taken over all (l, m), t. Then

$$\alpha _{2} \leq \displaystyle\sum q_{(l,m)}(t){c}^{{\ast}}(l,m)(t + 1)\mathbf{1}[q_{ (l,m)}(t) < {c}^{{\ast}}(l,m)(t)] \leq N{K}^{2},$$

(2.97)

in which N is the number of (l, m) pairs. Next

$$\displaystyle\begin{array}{rcl} \alpha _{1}& =& \displaystyle\sum q_{(l,m)}(t)[a_{(l,m)}(t + 1) - {c}^{{\ast}}(l,m)(t)] \\ & =& \displaystyle\sum q_{(l,m)}(t)[a_{(l,m)}(t + 1) - \rho (l,m)] +\displaystyle\sum q_{(l,m)}(t)[\rho (l,m) - c(l,m)] \\ & & +\displaystyle\sum q_{(l,m)}(t)[c(l,m) - {c}^{{\ast}}(l,m)(t)] \\ & =& \alpha _{11} + \alpha _{12} + \alpha _{13},\mbox{ say}.\end{array}$$

Let $\sigma _{(l,m)}(t + 1) = a_{(l,m)}(t + 1) - \rho (l,m)$. Since A _(l, m)(t) is $(\sigma (l,m),\rho (l,m))$ upper-bounded,

$$\alpha _{11} =\displaystyle\sum q_{(l,m)}(t)\sigma _{(l,m)}(t + 1),\mbox{ with }\displaystyle\sum _{t}\sigma _{(l,m)}(t) \leq \sigma (l,m).$$

By (2.54) $\rho (l,m) - c(l,m) < 0$ for all (l, m), so there exists $\eta > 0$ such that

$$\alpha _{12} \leq -\eta \displaystyle\sum q_{(l,m)}(t).$$

Lastly, since u ^∗(t) maximizes the pressure w(q(t), [U]), it follows that

$$\alpha _{13} =\displaystyle\sum q_{(l,m)}(t)[c(l,m) - {c}^{{\ast}}(l,m)(t)] = w(q(t),[U]) - w(q(t),{u}^{{\ast}}(t)) \leq 0.$$

Combining these three estimates gives

$$\alpha _{1} \leq \displaystyle\sum (-\eta + \sigma _{(l,m)}(t))q_{(l,m)}(t),\mbox{ with }\displaystyle\sum _{t}\sigma _{(l,m)}(t) \leq \sigma (l,m).$$

(2.98)

2.1.4.2 Bound on β

$$\displaystyle\begin{array}{rcl} \delta _{(l,m)}& =& a_{(l,m)}(t + 1) -\min \{ c{(l,m)}^{{\ast}}(t),q_{ (l,m)}(t)\} \\ & =& a_{(l,m)}(t + 1) - {c}^{{\ast}}(l,m)(t)\mathbf{1}[q_{ (l,m)}(t) > {c}^{{\ast}}(l,m)(t)] - q_{ (l,m)}(t)\mathbf{1}[q_{(l,m)}(t) \\ & \leq & {c}^{{\ast}}(l,m)(t)] \\ \vert \delta _{(l,m)}\vert & \leq & \vert a_{(l,m)}(t + 1) - {c}^{{\ast}}(l,m)(t)\vert + q_{ (l,m)}^{{\ast}}(t)\mathbf{1}[q_{ (l,m)}(t) \\ & \leq & {c}^{{\ast}}(l,m)(t)] \\ & \leq & \vert a_{(l,m)}(t + 1) - {c}^{{\ast}}(l,m)(t)\vert + {c}^{{\ast}}(l,m)(t) \leq 2K \\ \end{array}$$

So

$$\vert \delta {\vert }^{2} =\displaystyle\sum \delta _{ (l,m)}^{2} \leq 4N{K}^{2}.$$

(2.99)

Equation (2.56) follows from (2.92) to (2.99).

2.1.5 E Proof of (2.81)

The proof follows the same lines as in Appendix D. Write (2.74) in vector–matrix form as

$$q(t + 1) = [q(t) - {c}^{{\ast}}(t)]_{ +} +\tilde{ a}(t + 1).$$

Let

$$x = q(t + 1) - q(t) = -\min \{{c}^{{\ast}}(t),q(t)\} +\tilde{ a}(t + 1).$$

Then

$$\vert x{\vert }^{2} = 2\langle x,q(t)\rangle + \vert x{\vert }^{2} = 2\mu + \nu, \mbox{ say}.$$

(2.100)

We separately bound $\mu$, $\nu$.

2.1.5.1 Bound on μ

$$\displaystyle\begin{array}{rcl} \mu & =& \langle x,q(t)\rangle =\displaystyle\sum q_{(l,m)}(t)[\tilde{a}_{(l,m)}(t + 1) -\min \{ {c}^{{\ast}}(l,m)(t),q_{ (l,m)}(t)\}] \\ & =& \displaystyle\sum q_{(l,m)}(t)[\tilde{a}_{(l,m)}(t + 1) - {c}^{{\ast}}(l,m)(t) +\max \{ {c}^{{\ast}}(l,m)(t) - q_{ (l,m)}(t),0\}] \\ & =& \displaystyle\sum q_{(l,m)}(t)[\tilde{a}_{(l,m)}(t+1)-{c}^{{\ast}}(l,m)(t)]+\displaystyle\sum q_{ (l,m)}(t)\max \{{c}^{{\ast}}(l,m)(t)-q_{ (l,m)}(t),0\} \\ & =& \mu _{1} + \mu _{2},\mbox{ say}.\end{array}$$

Let $K =\max \{ {c}^{{\ast}}(l,m)(t)\}$ be the maximum over all t, (l, m). Then

$$\displaystyle\begin{array}{rcl} \mu _{2}& =& \displaystyle\sum q_{(l,m)}(t)\max \{{c}^{{\ast}}(l,m)(t) - q_{ (l,m)}(t),0\} \\ & \leq & \displaystyle\sum q_{(l,m)}(t){c}^{{\ast}}(l,m)(t)\mathbf{1}[{c}^{{\ast}}(l,m)(t) \geq q_{ (l,m)}(t)] \\ & \leq & N{K}^{2}, \end{array}$$

(2.101)

in which N is the number of (l, m) pairs in the network.

Using (2.74)–(2.77),

$$\displaystyle\begin{array}{rcl} \mu _{1}& =& \displaystyle\sum _{l,m}q_{(l,m)}(t)[\tilde{a}_{(l,m)}(t + 1) - {c}^{{\ast}}(l,m)(t)] \\ & =& \displaystyle\sum _{l,m}q_{(l,m)}(t)\left [e_{l}(t + 1)\gamma _{(l,m)} +\displaystyle\sum _{k}b_{(k,l)}(t)\gamma _{(l,m)} + \delta _{(l,m)}(t) - {c}^{{\ast}}(l,m)(t)\right ] \\ & =& \displaystyle\sum _{l,m}q_{(l,m)}(t)\left [e_{l}(t + 1)\gamma _{(l,m)} +\displaystyle\sum _{k}\min \{q_{(k,l)}(t),{c}^{{\ast}}(k,l)(t)\}\gamma _{ (l,m)}\right. \\ & & \left.+\delta _{(l,m)}(t) - {c}^{{\ast}}(l,m)(t)\right ] \\ & \leq & \displaystyle\sum _{l,m}q_{(l,m)}(t)\left [e_{l}(t + 1)\gamma _{(l,m)} +\displaystyle\sum _{k}{c}^{{\ast}}(k,l)(t)\gamma _{ (l,m)} - {c}^{{\ast}}(l,m)(t)\right ] \\ & & +\displaystyle\sum _{l,m}q_{(l,m)}(t)\delta _{(l,m)}(t) \\ & =& \displaystyle\sum _{l,m}q_{(l,m)}(t)\left [\beta _{l}\gamma _{(l,m)} +\displaystyle\sum _{k}{c}^{{\ast}}(k,l)(t)\gamma _{ (l,m)} - {c}^{{\ast}}(l,m)(t)\right ] \\ & & +\displaystyle\sum _{l,m}q_{(l,m)}(t)\left [\alpha _{l}(t + 1)\gamma _{(l,m)} + \delta _{(l,m)}(t)\right ] \\ & =& \mu _{11} + \mu _{12} + \mu _{13}.\end{array}$$

Above, $\alpha _{l}(t + 1) = e_{l}(t + 1) - \beta _{l}$, so $\sum _{t}\alpha _{l}(t) \leq \alpha _{l}$, since E _l is $(\alpha _{l},\beta _{l})$ upper-bounded;

$$\displaystyle\begin{array}{rcl} \mu _{11}& =& \displaystyle\sum _{l,m}q_{(l,m)}(t)\beta _{l}\gamma _{(l,m)}, \end{array}$$

(2.102)

$$\displaystyle\begin{array}{rcl}\mu _{12}& =& \displaystyle\sum _{l,m}q_{(l,m)}(t)\left [\displaystyle\sum _{k}{c}^{{\ast}}(k,l)(t)\gamma _{ (l,m)} - {c}^{{\ast}}(l,m)(t)\right ] \\ & =& \displaystyle\sum _{l,m}\left [\displaystyle\sum _{p}q_{(m,p)}(t)\gamma _{(m,p)} - q_{(l,m)}(t)\right ]{c}^{{\ast}}(l,m)(t) \\ & =& -w(q(t),{u}^{{\ast}}(t)), \end{array}$$

(2.103)

$$\displaystyle\begin{array}{rcl} \mu _{13}& =& \displaystyle\sum _{l,m}q_{(l,m)}(t)\left [\alpha _{l}(t + 1)\gamma _{(l,m)} + \delta _{(l,m)}(t)\right ].\end{array}$$

(2.104)

Substituting $\beta _{l} = \rho _{l} -\sum _{k}\rho _{k}\gamma _{(k,l)}$ from (2.59) into (2.102) gives

$$\displaystyle\begin{array}{rcl} \mu _{11}& =& \displaystyle\sum _{l,m}q_{(l,m)}(t)\left [\rho _{l} -\displaystyle\sum _{k}\rho _{k}\gamma _{(k,l)}\right ]\gamma _{(l,m)} \\ & =& \displaystyle\sum _{l,m}\rho _{l}\gamma _{(l,m)}q_{(l,m)}(t) -\displaystyle\sum _{l,m}q_{(l,m)}(t)\displaystyle\sum _{k}\rho _{k}\gamma _{(k,l)}\gamma _{(l,m)} \\ & =& \displaystyle\sum _{l,m}\rho _{l}\gamma _{(l,m)}q_{(l,m)}(t) -\displaystyle\sum _{m}\left [\displaystyle\sum _{l}\rho _{l}\gamma _{(l,m)}\right ]\displaystyle\sum _{p}q_{(m,p)}(t)\gamma _{(m,p)} \\ & =& \displaystyle\sum _{l,m}\rho _{l}\gamma _{(l,m)}\left [q_{(l,m)}(t) -\displaystyle\sum _{p}q_{(m,p)}(t)\gamma _{(m,p)}\right ]. \end{array}$$

(2.105)

By (2.80) there exists $[U] \in [\mathcal{U}]$ such that $S \circ [U] > [\rho ]\Gamma$. Since $0 \in [\mathcal{U}]$, this implies that $[\rho ]\Gamma$ is in the interior of $S \circ [\mathcal{U}]$. Hence there exist (possibly different) $[U]$ and $\eta > 0$ such that

$$S\circ [U](l,m) = \left \{\begin{array}{ll} \rho _{l}\gamma _{(l,m)} + \eta, \,\,\,\,&\mbox{ if $q_{(l,m)}(t) -\displaystyle\sum _{p}q_{(m,p)}(t)\gamma _{(m,p)} > 0$} \\ \rho _{l}\gamma _{(l,m)} - \eta, \,\,\,\,&\mbox{ if $q_{(l,m)}(t) -\displaystyle\sum _{p}q_{(m,p)}(t)\gamma _{(m,p)} \leq 0$} \end{array} \right.,$$

and so

$$w(q,[U]) \geq \mu _{11} + \eta \displaystyle\sum _{l,m}\vert q_{(l,m)}(t) -\displaystyle\sum _{p}q_{(m,p)}(t)\gamma _{(m,p)}\vert.$$

(2.106)

The linear transformation $\{q_{(l,m)}\}\mapsto \{q_{(l,m)} -\sum _{p}q_{(m,p)}\gamma _{(m,p)}\}$ is 1:1 from the conditions imposed on $\Gamma$. Hence (2.106) implies that there exists $\epsilon > 0$ so that

$$w(q(t),[U]) \geq \mu _{11} + \epsilon \vert q(t)\vert, $$

which together with (2.103) gives

$$\displaystyle\begin{array}{rcl} \mu _{11} + \mu _{12} \leq w(q(t),[U]) - w(q(t),{u}^{{\ast}}(t)) - \epsilon \vert q(t)\vert \leq -\epsilon \vert q(t)\vert, & &\end{array}$$

(2.107)

since the pressure w(q, [U] is maximized at u ^∗(t). Together with (2.104) we get the bound

$$\mu _{1} \leq -\epsilon \vert q(t)\vert + \sigma (t)\vert q(t)\vert, $$

(2.108)

for some $\sigma (t) \geq 0$, $\sum \sigma (t) < \infty$.

2.1.5.2 Bound on ν

From (2.100), $\nu =\sum _{l,m}\vert x_{(l,m)}{\vert }^{2}$, and

$$\displaystyle\begin{array}{rcl} x_{(l,m)}& =& \tilde{a}_{(l,m)}(t + 1) -\min \{ {c}^{{\ast}}(l,m)(t),q_{ (l,m)}(t)\} \\ & =& \tilde{a}_{(l,m)}(t + 1) - {c}^{{\ast}}(l,m)(t) -\min \{ q_{ (l,m)}(t) - {c}^{{\ast}}(l,m)(t),0\}, \\ \end{array}$$

so

$$\vert x_{(l,m)}\vert \leq \vert \tilde{a}_{(l,m)}(t+1)-{c}^{{\ast}}(l,m)(t)\vert +\vert {c}^{{\ast}}(l,m)(t)\vert \leq \vert \tilde{a}_{ (l,m)}(t+1)\vert +2\vert {c}^{{\ast}}(l,m)(t)\vert.$$

From (2.75) to (2.77) it follows that $\vert \tilde{a}_{(l,m)}(t + 1)\vert$ is bounded. Hence there is $k < \infty$ such that $\nu \leq k$, which together with (2.108) and (2.101) yield (2.81) as required.