The escape from L.A.

Probability Level 5

You're driving from Caltech, in Pasadena, to San Diego after a long day in the lab. You're cruising along and everything is going great until you hit Rosecrans Avenue, a horrible street in horrible Los Angeles. Here, everything grinds to a halt and you sit in mind numbing gridlock for a long time. Bored to tears, you start thinking about your situation and come up with the following simple model.

Let's approximate Rosecrans Ave as a 1d lattice with $n$ locations for cars $x_1,x_2,\ldots,x_n$ each of which can hold only one car. At each moment in time, traffic flow on Rosecrans works as follows:

If there is no car on the first position on Rosecrans ( $x_0$ ), a new car will enter it at the rate $p_\text{injection}$ .
If there is a car at last position on the avenue ( $x_n$ ), the car will leave Rosecrans (and move on to the highway) at the rate $p_\text{removal}$ .
If there is a car at $x_i$ and no car at $x_{i+1}$ , the car on $x_i$ will move to $x_{i+1}$ at the rate $v_\text{car}$ .
If there is a car at $x_i$ and a car at $x_{i+1}$ , the car at $x_i$ will stay put.

What is the flow rate (on average) of cars leaving Rosecrans?

Assumptions and details

$p_\text{injection} = 0.6$
$p_\text{removal} = 0.3$
$v_\text{car} = 1$
Assume the street is very long ( $n\rightarrow\infty$ )
Give your answer in $\text{cars}/\text{unit time}$

The answer is 0.21.

1 solution

David Vaccaro
Jul 20, 2014

Let $p_{i}$ represent the long-run probability that a car is present in space $x_{i}$ . Considering the long run net flow in and and out of $x_{n}$ we have $p_{n-1}(1-p_{n})=0.3p_{n}$ .

Since the road is long we can assume $p_{n-1}=p_{n}$ and this value is clearly non-zero! Simple algebra gives $p_{n}=0.7$ , and the flow out is $0.3\times 0.7=0.21$ .

Notice that the injection rate does not influence the net flow out at Rosecrans which is intuitively obvious- but it will influence the values of $p_{i}$ when $i$ is small. If the injection rate were to satisfy $p_{injection}=1-p_{removal}$ then $p_{i}$ would be constant throughout the road (by symmetry).

Here because 0.6<0.7 ( $p_{injection}<p_{removal}$ ) the traffic will get denser as we approach Rosecrans. If $p_{injection}$ were bigger than 0.7 then the jam would be heaviest closest to the Caltech.

It isn't at all clear how you derived the first equation. Would you care to explain?

Furthermore, can you point me to a source that will help me understand this concept of "rate" and how to apply it? I've done a quick Google search but can't seem to find anything that's talking about this.

Stewart Gordon - 4 years, 6 months ago

I guess the first equation is false and the answer is 0.3/1.3=0.2307692. Indeed, under the given scenario, you will never observe the pattern 00 near to the end. Consider the last two positions. The pattern 10 changes to 01 with probability, say, p {10}. The pattern x1 changes to y0 with unconditional probability p {removal} p(x n=1) and to x1 with unconditional probability (1-p {removal})p(x n=1). But y0 must be 10. Hence, due to stationarity, p {10}=p(x n=1)p {removal}. Moreover p {10}=p(x n=0) because 00 does not occur. Equating these results leads to p(x n=1)=1/(1+p {removal}). Then the outflow is p {removal} p(x n=1)=p {removal}/(1+p {removal}) = 3/13 =0.231

Jochem König - 4 years, 6 months ago

The escape from L.A.

The answer is 0.21.

1 solution

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