# Traffic Flow Modelling

#### grandy

##### Member
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#### CaptainBlack

##### Well-known member
Traffic Flow Modelling,
I really confuse and do not how to start .

View attachment 524
For part (a) you need to observe that the flow rate $$f(x,t)$$ in vehicles per unit time is $$u(x,t) \rho(x,t)$$.

Now you need to show that for (i) and (ii) that $$u(x,t)\le u_{sl}$$, then as $$\rho(x,t) \le \rho_{max}$$ you will have shown that the flow rate:
$f(x,t)\le u_{sl}\rho_{max}$

#### CaptainBlack

##### Well-known member
Traffic Flow Modelling,
I really confuse and do not how to start .

View attachment 524
The next step is to write down the partial differential equation satisfied by the traffic density. This is derivable from a conservation of mass (or vehicle numbers) argument that you will have seen innumerable times.

CB

#### grandy

##### Member
The next step is to write down the partial differential equation satisfied by the traffic density. This is derivable from a conservation of mass (or vehicle numbers) argument that you will have seen innumerable times.

CB

#### CaptainBlack

##### Well-known member
Consider a road element between $$x$$ and $$x+\Delta x$$ the traffic flow into the element at $$x$$ per unit time is $$u(\rho(x,t))\rho(x,t)$$ and out at $$x+\Delta x$$ is $$u(\rho(x+\Delta x,t))\rho(x+\Delta x,t)$$ Therefore the rate of change of car numbers in the element is:

$\frac{\partial N}{\partial t}=u(\rho(x,t))\rho(x,t)-u(\rho(x+\Delta x,t))\rho(x+\Delta x,t)$

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

$\frac{1}{\Delta x}\frac{\partial N}{\partial t}=\frac{u(\rho(x,t))\rho(x,t)-u(\rho(x+\Delta x,t))\rho(x+\Delta x,t)}{\Delta x}$

Now take the limit as $$\Delta x \to 0$$ to get:

$\frac{\partial \rho}{\partial t}=\frac{\partial}{\partial x}u(\rho)\rho$

CB

Last edited:

#### CaptainBlack

##### Well-known member
For part (a) you need to observe that the flow rate $$f(x,t)$$ in vehicles per unit time is $$u(x,t) \rho(x,t)$$.

Now you need to show that for (i) and (ii) that $$u(x,t)\le u_{sl}$$, then as $$\rho(x,t) \le \rho_{max}$$ you will have shown that the flow rate:
$f(x,t)\le u_{sl}\rho_{max}$
In small time interval $$\Delta t$$ all the vehicles less than a distance $$u(x,t)\Delta t$$ down stream of $$x$$ will pass $$x$$. The number of vehicles in this stretch of road is $$u(x,t)\rho(x,t)\Delta t$$, so $$u(x,t)\rho(x,t)\Delta t$$ vehicles pass $$x$$ in $$\Delta t$$ so the vehicle flow rate at $$x$$ is $$u(x,t)\rho(x,t)$$ vehicles per unit time.

CB

#### grandy

##### Member
Thank you very much for your lovely answer. Please explain me what do you mean by Triangle sign x represent. And also in next line below what does Triangle sign t mean?

#### grandy

##### Member
Traffic flow Modelling

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