# Traffic flow Modelling

1. Jan 3, 2013

### ra_forever8

On a stretch ofsingle-lane road with no entrances or exits the traffic density r(x,t)is a continuous function of distance xand time t, for all t > 0, and the traffic velocity u(r) is a function of density alone.
Two alternative models are proposed to represent u:

(i) u= uSL (1 – r^n/ r^nmax)
where n is a positive constant
(ii) u= 1/3 uSL In (rmax/r)where uSLis the maximum speed limit on the road and rmax is themaximum density of traffic possible on the road (corresponding tobumper-to-bumper traffic).

a) Evaluate the maximum rate of traffic flow for cases (i)and (ii) above. Show that for both cases the maximum rate of traffic flow isless than uSLrmax and that for case (i)it increases towards this upper limit as nbecomes very large.
b) It is assumed that a model of the form given in case (i)is a reasonable representation of actual traffic behaviour. Apply the method ofcharacteristics to analyse the following situation. A queue of cars is stoppedat a red traffic light on a road for which the maximum speed limit is 40 m.p.h.It may be assumed that the queue is very long and that the road ahead of thelight is empty of traffic. The light turns green and remains green for only 45seconds. If a car which is initially a quarter of a mile behind the light getsthrough before the light changes back to red determine the smallest integervalue that n can have. (Hint: showinitially that the car will not start moving until a certain time after thelight has turned green, and then solve the appropriate differential equationfor the position of the car.)

=> The attempt at a solution

For part (a) you need to observe that the flow rate f(x,t) in vehicles per unit time is u(x,t)ρ(x,t) .

Mow,to show (i) and (ii) that u(x,t)≤u sl , then as ρ(x,t)≤ρ max I will have shown that the flow rate: f(x,t)≤u sl ρ max .

b)Consider a road element between x and x+Δx the traffic flow into the element at x per unit time is u(ρ(x,t))ρ(x,t) and out at x+Δx is u(ρ(x+Δx,t))ρ(x+Δx,t) Therefore the rate of change of car numbers in the element is:

∂N ∂t =u(ρ(x,t))ρ(x,t)−u(ρ(x+Δx,t))ρ(x+Δx,t)

and so the rate of change of density in the element is:

1 Δx ∂N ∂t =u(ρ(x,t))ρ(x,t)−u(ρ(x+Δx,t))ρ(x+Δx,t) Δx

Now take the limit as Δx→0 to get:

∂ρ ∂t =∂ ∂x u(ρ)ρ

2. Jan 3, 2013

### haruspex

It's not clear to me whether you have solved part (a).
For part (b), your differential eqn looks ok (apart from some missing division symbols and parentheses). You can substitute for u using the model and expand the derivative using the chain rule.

3. Jan 4, 2013

### ra_forever8

I think I have solved the qs part b). Please don't worry about part b now because i have done it. Would please help with the qs part a).

As i mentioned for part a)
To observe that the flow rate f(x,t) in vehicles per unit time is u(x,t)ρ(x,t) .
Now,to show (i) and (ii) that u(x,t)≤u sl , then as ρ(x,t)≤ρ max
I will have shown that the flow rate: f(x,t)≤u sl ρ max

4. Jan 4, 2013

### haruspex

For max flow rate, you can treat ρ (=r) as the same for all x and t. So it simply becomes a matter of maximising f wrt r, given f(r) = r*u(r). That's enough to find the max flow rate as a function of rmax and n in case (i). But I don't understand the equation for u(r) in case (ii). Is In() supposed to be ln()? But that doesn't make sense because as r tends to zero u will tend to infinity instead of levelling out at uSL. And is that really a factor of 1/3 out the front?

5. Jan 4, 2013

### ra_forever8

For case ii) Yes its correct, it is supposed to be In (r_max/r) and it is factor of 1/3 infront. Would please explain me clearly for case (i) again as i confused. Please try to solve the solution for case i) , it will easier for me to understand.

6. Jan 4, 2013

### haruspex

For (i), f(r) = r*u(r) = r*uSL (1 – (r/ rmax)n)
How do you find the r that maximises f? Find where df/dr = 0, right?
For (ii), I'm not familiar with any standard function In(). What is it?

7. Jan 6, 2013

### ra_forever8

On a stretch of single-lane road with no entrances or exits the traffic density ρ(x,t) is a continuous function of distance x and time t, for all t > 0, and the traffic velocity ) u( ρ) is a function of density alone.
Two alternative models are proposed to represent u:
i)u = u_(SL)*(1- ρ^n/ρ^n_max ), where n is a postive constant
ii) u = u_(SL)* In (ρ_max / ρ)
Where u_SL represents the maximum speed limit on the road and p_max represents maximum density of traffic possible on the road(meaning bumper-to-bumper traffic)

Compare the realism of the 2 models for u above. You should consider in particular the variations of velocity with density for each model, and the velocities for high and low densities in each case. State which model you prefer, giving reasons.
=>
I did for case i) which is u = u_(SL)*(1- ρ^n/ρ^n_max ),
u(ρ) = u_(SL)*(1- ρ^n/ρ^n_max ), for 0<ρ<ρ_max
Since ρ>= 0, cannot exceed u_SL
when ρ= ρ_max , u (ρ_max)= u_SL(1- ρ_max/ρ_max) =0
when ρ=0, u(0)= u_SL(1-0/ρ_max)= u_SL
Also, du/dρ= (- u_SL/ρ_max ) <0, so drivers reduce speed as density increase

8. Jan 6, 2013

### TSny

For case (ii), what do you get for u as ρ→0? (Is that the natural log function in case ii?)

9. Jan 6, 2013

### haruspex

This appears to be a continuation (different question, same scenario) of unfinished thread https://www.physicsforums.com/showthread.php?t=662238.
ra_forever8, I already asked you to explain this "In()" function on that thread. You said it was not ln() (and I agree that wouldn't make sense), but still haven't told me what it is. Do you know? If not, it's unlikely anyone will be able to help you.

10. Jan 6, 2013

### ra_forever8

Sorry,it is In() something there.
For case ii) u = u_(SL)* In(ρ_max /ρ)

11. Jan 6, 2013

### ra_forever8

For Tsny,
In case ii) u = u_(SL)* In (ρ_max / ρ) ,if ρ→ 0, then u = u_(SL)* In (ρ_max/0) which gives math error.

12. Jan 6, 2013

### haruspex

I have no idea what you mean by that.

13. Jan 6, 2013

### ra_forever8

it is natural logarithm like log

14. Jan 6, 2013

### haruspex

OK, I thought you denied that before.
But as you can see it doesn't make sense. It will exceed the speed limit whenever rmax > e*r.
Something like this would work: u = uSL - ln(1+(euSL-1)r/rmax)

15. Jan 7, 2013

### ra_forever8

what e there? is the natural logarithm like log?
rmax is same as ρ_max and r as ρ.
if it will exceed the speed limit whenever rmax > e*r then density will decrease right?

16. Jan 7, 2013

### haruspex

Yes, I mean e as in the base of natural logarithms. Btw, where you wrote In() in your posts, if you mean natural logarithm it's written ln() - that's a lowercase L not an uppercase I. Writing In() confuses everyone.
Before I go on, I want to correct what I wrote in may last post. In the second thread you started on this you wrote u= uSL ln (rmax/r). In that case, what I wrote about exceeding speed limit was correct: when rmax > e*r, u > uSL ln (e) = uSL. I.e. u will exceed the speed limit. But in this, the original thread you specified a factor one third: u= 1/3 uSL ln (rmax/r). That changes things, but not much. Now the speed limit will be exceeded if rmax > e3*r. Maybe this is just part of the answer to "is the model realistic?"
So let's just accept u= 1/3 uSL ln (rmax/r) and see where it takes us.
In the OP you had questions (a) and (b) to answer. In your repost (now a post within this thread) you asked a different question. Does that mean you have completed (a)(ii) and (b)(i)? I will assume so.
Do you mean, given a a running model in which the speed limit is being exceeded, will the speed decrease over time? Not necessarily. If the density is the same everywhere then it will stay that way. But that's not what this question is asking. It is asking whether increasing r decreases speed.
So, what do you get for du/dr in model (ii)?