Interpreting Traffic on a Highway with On/Off Ramps in the Light of TASEP

Abstract

We raise the question whether a simple model for traffic, generic but based on microscopic rules, can provide an additional angle for interpreting flow through a road system. Using the totally simple exclusion process (TASEP) on a road segment with ramps, we show that measuring the flow directly at the road junctions may be a useful setup. We show that the presence of junctions affects the characterisation of traffic, suggesting that interpretations in terms of a 2-phase or a 3-phase description may be complementary, rather than contradictory. We furthermore argue that hysteresis-like features can appear in a system with junctions, which is intriguing as the TASEP dynamics as such do not lead to hysteresis. We discuss our findings in the light of boundary-driven phase transitions.

Appendices

Appendices

Appendix A: Phase Diagram in the $\tilde{\rho }-\beta$ Plane for the Symmetric Case ($\alpha _1=\alpha _2=\alpha$)

We begin by establishing/recalling the conditions for all possible phases:

a. Phase 2LD:LD We get four conditions from requiring segments A (i.e. A$_1$ and A$_2$) as well as segment C to be in the LD state:

$$\begin{aligned} \begin{array}{lllll} (A.i) &{} \alpha \le 1-\tilde{\rho }&{}\qquad &{}(A.ii) &{} \alpha \le 1/2 \\ (C.i) &{} \tilde{\rho }\le \beta &{}\qquad &{}(C.ii) &{} \tilde{\rho }\le 1/2 \end{array} \end{aligned}$$

(A1)

From which we can delimit the zone in the $\tilde{\rho }-\beta$-plane in which this phase may arise, if it exists.

In addition, we still need to impose the condition of current conservation at the junction, which leads to

$$\begin{aligned} 2 \times \alpha (1-\alpha ) = \tilde{\rho }(1-\tilde{\rho }) \end{aligned}$$

(A2)

and therefore the phase 2LD:LD requires that

$$\begin{aligned} \tilde{\rho }_{2LD:LD} = \frac{1}{2} \, \left( 1 - \sqrt{1 - 8 \alpha (1-\alpha )}\right) \qquad \left( \alpha \le \frac{1}{2} \, \left( 1-\sqrt{\frac{1}{2}}\right) \simeq 0.146\right) \end{aligned}$$

(A3)

where we have picked the negative branch, to ensure compatibility with condition C.ii, and restricted the values of $\alpha$ to guarantee that the expression under the square root remain positive.

b. Phase 2LD:HD The four conditions from the segments are

$$\begin{aligned} \begin{array}{lllll} (A.i) &{} \alpha \le 1-\tilde{\rho }&{}\qquad &{}(A.ii) &{} \alpha \le 1/2 \\ (C.i) &{} \beta \le \tilde{\rho }&{}\qquad &{}(C.ii) &{} \beta \le 1/2 \end{array} \end{aligned}$$

(A4)

Within this zone, current conservation imposes

$$\begin{aligned} 2 \times \alpha (1-\alpha ) = \beta (1-\beta ) \end{aligned}$$

(A5)

and therefore the phase 2LD:HD is restricted to a particular value of $\beta = \beta _{2LD:HD}$.

$$\begin{aligned} \beta _{2LD:HD} = \frac{1}{2} \, \left( 1 - \sqrt{1 - 8 \alpha (1-\alpha )}\right) \qquad \left( \alpha \le \frac{1}{2} \, \left( 1-\sqrt{\frac{1}{2}}\right) \simeq 0.146\right) \end{aligned}$$

(A6)

where we have picked the negative branch, to ensure compatibility with condition C.ii, and the condition on $\alpha$ ensures that there is a real solution to the square root.

c. Phase 2HD:LD This phase does not exist. The four conditions from the segments are

$$\begin{aligned} \begin{array}{lllll} (A.i) &{} 1-\tilde{\rho }\le \alpha &{}\qquad &{}(A.ii) &{} 1-\tilde{\rho }\le 1/2 \\ (C.i) &{} \tilde{\rho }\le \beta &{}\qquad &{}(C.ii) &{} \tilde{\rho }\le 1/2 \end{array} \end{aligned}$$

(A7)

Conditions A.ii and C.ii only allow for $\tilde{\rho }=1/2$, a special case, and from conditions A.i and C.i we require both $\alpha \ge 1/2$ and $\beta >1/2$.

However, current conservation furthermore requires

$$\begin{aligned} 2 \times \tilde{\rho }(1-\tilde{\rho }) = \tilde{\rho }(1-\tilde{\rho }) \end{aligned}$$

(A8)

which is impossible (except for the trivial cases $\tilde{\rho }=0$ or $\tilde{\rho }=1$). We must therefore exclude this phase.

d. Phase 2HD:MC The segment conditions are

$$\begin{aligned} \begin{array}{lllll} (A.i) &{} 1-\tilde{\rho }\le \alpha &{}\qquad &{}(A.ii) &{}1-\tilde{\rho }<1/2 \\ (C.i) &{} \tilde{\rho }\ge 1/2 &{}\qquad &{}(C.ii) &{} \beta \ge 1/2 \end{array} \end{aligned}$$

(A9)

Again this establishes a zone where this phase may arise, but current conservation furthermore yields

$$\begin{aligned} 2 \times \tilde{\rho }(1-\tilde{\rho }) = 1/4 \end{aligned}$$

(A10)

which translates to

$$\begin{aligned} \tilde{\rho }_{2HD:MC} = \frac{1}{2} \, \left( 1 + \sqrt{\frac{1}{2}}\right) \approx 0.85 \end{aligned}$$

(A11)

where the negative branch is excluded due to condition C.ii.

In this case we thus have two scenarios, due to this condition on top of condition A.i:

• if $\alpha \ge 1-\tilde{\rho }_{2HD:MC} = \frac{1}{2} \, \left( 1-\sqrt{\frac{1}{2}}\right) \approx 0.15$, then the solution $\tilde{\rho }= \tilde{\rho }_{2HD:MC}$ is valid anywhere in the above zone of the $\tilde{\rho }-\beta$-plane.

• in the opposite case, however, this condition is mutually exclusive with condition A.i, and therefore there is no phase 2LD : MC.

e. Phase 2LD:MC The conditions on segments A and C are

$$\begin{aligned} \begin{array}{lllll} (A.i) &{} \alpha \le 1-\tilde{\rho }&{}\qquad &{}(A.ii) &{}\alpha <1/2 \\ (C.i) &{} \tilde{\rho }\ge 1/2 &{}\qquad &{}(C.ii) &{} \beta \ge 1/2 \end{array} \end{aligned}$$

(A12)

from which we can identify a rectangle in the phase plane, which is at this stage bounded by $\tilde{\rho }< 1-\alpha$. However, current conservation shows that only one value is admissible for $\alpha$, as

$$\begin{aligned} 2 \, \alpha (1-\alpha ) = 1/4 \end{aligned}$$

(A13)

requires to have

$$\begin{aligned} \alpha = \frac{1}{2} \, \left( 1 - \sqrt{\frac{1}{2}}\right) \end{aligned}$$

(A14)

where the positive branch has been discarded due to condition A.ii. Thus the 2LD:MC phase arises only at this particular value for $\alpha$, but it covers the entire rectangle:

f. Phase 2HD:HD The segment conditions are

$$\begin{aligned} \begin{array}{lllll} (A.i) &{} 1-\tilde{\rho }\le \alpha &{}\qquad &{}(A.ii) &{}1-\tilde{\rho }<1/2 \\ (C.i) &{} \beta < \tilde{\rho }&{}\qquad &{}(C.ii) &{} \beta \le 1/2 \end{array} \end{aligned}$$

(A15)

This again identifies a zone in the phase plane, which is again dependent on the value of $\alpha$.

Requiring current conservation

$$\begin{aligned} 2 \times \tilde{\rho }(1-\tilde{\rho }) = \beta \, (1-\beta ) \end{aligned}$$

(A16)

links $\beta$ to t$\tilde{\rho }$ as

$$\begin{aligned} \beta _{2HD:HD} = \frac{1}{2} \, \left( 1 - \sqrt{1 - 8 \tilde{\rho }(1-\tilde{\rho })}\right) \end{aligned}$$

(A17)

choosing again the negative branch, in order to accommodate condition C.ii. Note that this only has real solutions in the required zone if

$$\begin{aligned} \tilde{\rho }\ge \tilde{\rho }_{2HD:HD}^* = \frac{1}{2} \, \left( 1+\sqrt{\frac{1}{2}}\right) \end{aligned}$$

(A18)

Here we must distinguish two cases:

• for large $\alpha$, $1-\alpha \ge \tilde{\rho }_{2HD:HD}^*$, the entire branch of the parabola $\tilde{\rho }_{2HD:HD}^*(\beta )$ is accessible

• for $\alpha$ below this value only the lower part of this branch, corresponding to small values of $\beta$, constitute a solution, the remainder being excluded by condition A.i

g. Overall phase diagram in the $(\tilde{\rho },\beta )$ plane as  $\alpha$ is varied We can thus recapitulate the location of the phases as $\alpha$ is progressively increased: