Congestion on Multilane Highways

Introduction

All computations in this section were run with the following equilibrium relations:

$$v_1(\rho) = v^{\infty}_1(1- \rho/\rho_{\rm max})\ {\rm and}\ v_2 (\rho) = v^{\infty}_2 (1-\rho/\rho_{\rm max}).$$ (1)

These transform to

$$V_1 (\gamma) = v^{\infty}_1 \left(1 - \frac{L}{\gamma}\right) \ {\rm and}\ V_2 (\gamma) = v^{\infty}_2 \left(1- \frac{L}{\gamma}\right)$$ (2)

where $L = {\displaystyle \frac{1}{\rho_{\rm max}}}$. The specific parameter used were

$$v^{\infty}_1 = 100\ {\rm feet/sec}\ = \frac{100 \times 3600}{5280} = 68.1818 \ldots\ {\rm mph}$$ (3)

$$v^{\infty}_2 = 40\ {\rm feet/sec} = \frac{40 \times 3600}{5280} = 27.2727 \ldots \ {\rm mph}$$ (4)

and

$$L = 15\ {\rm feet}.$$ (5)

The latter number corresponds to a maximum car density of

$$\rho_{\rm max} = \frac{1}{15}\ {\rm cars/foot}\ = \frac{5280}{15} = 352\ {\rm cars/mile}.$$ (6)

We used the constant switch curve introduced by Sopasakis [1]:

$$\gamma (u) = \gamma_* \ \ \ , \ \ \ 0 \leq u$$ (7)

with $\gamma_* = 20$ feet. For initial data we chose 3 sets of data:

$$x^{(k)}_m (0) = 20m + .1 \sin \left(\frac{km \pi}{200}\right)$$ (8)

for $- \infty \leq m \leq \infty$ and $k=1,2$, and 3. The observation that

$$x^{(k)}_{400}(0) = 8000\ {\rm feet}\ = 1.5151 \ldots \ {\rm miles}$$ (9)

and

$$x^{(k)}_{m+400}(0) = x^{(k)}_{m}(0) + 8000$$ (10)

implies we may interpret the data as initial data for a ring-road with 400 cars which is of length 1.5151... miles. We chose constant initial velocities

$$u^{(k)}_m (0) = .5 (V_1(\gamma_*)+V_2(\gamma_*)),\ 1 \leq m \leq 400$$ (11)

or

$$u^{(k)} (0) = 17.5\ {\rm feet/sec}\ = 11.931818 \ldots\ {\rm mph},\ 1 \leq m \leq 400.$$ (12)

These data guarantee points on both sides of the switch curve. Simulations were run with relaxation times

$$\epsilon = 1,2,4,\ {\rm and}\ 8.$$ (13)

A word is inorder about the simulations which follow. The first two frames in each figure are self-explanatory. In the third frame of each figure we plot the curve $m \rightarrow (\gamma_m = x_{m+1} - x_m, u_m)$. This curve is shown in black. The blue curves are the equilibrium curves $\gamma \rightarrow (\gamma, V_1,(\gamma))$ and $\gamma \rightarrow (\gamma ,V_2(\gamma))$ and the red curve is the image of $u\rightarrow (20,u)$. The red dot - o - is the image of $(\gamma_1,u_1)$. The discontinuities in the profiles propagate at the speed

$$c \simeq 227.6 \pm .1\ \ \ \ {\rm cars/minute}.$$ (3.14)

Simulation

Lagrangian Representation

	e = 1	e = 2	e = 4	e = 8
k = 1	view	view	view	view
k = 2	view	view	view	view
k = 3	view	view	view	view

Eulerian Representation

	e = 1	e = 2	e = 4	e = 8
k = 1	view	view	view	view
k = 2	view	view	view	view
k = 3	view	view	view	view

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Lagrangian Representation source code

Eulerian Representation source code

Paper (ps)

Paper (pdf)

2001-08-22