# Elementary Calculus and traffic flow problem

Question:

What should the speed limit be for cars on the Lions Gate Bridge in
Vancouver, British Columbia, during rush hour traffic, in order to maximize
the flow of traffic?

Solution:

This is of course an “open-ended” problem. The answer depends on one’s
assumptions. Here we make some simple-minded assumptions:
(a) The source of cars to the bridge is excellent.
(b) All cars move at the same speed v and have the same length.
(c) The separation distance between each car is the same.

(Note: How to find such simplifying assumptions. There is no clear cut answer. You just have start with some assumptions, as is true in mathematical modelling/applied mathematics, and then refine your assumptions to get a tractable problem).

Let q be the traffic flow rate, i.e., the flux of cars, corresponding to the number of cars per hour passing a given point or the capacity of the bridge;

let $\rho$ be the density of cars, i.e., the number of cars per mile of road; let $l^{*}$ be the effective space occupied by each car.

$l^{*}$ equals l plus the spacing between each car.

Then, $\rho = 1/l^{*}$ and $q=\rho v$

For a given bridge one can determine $\rho$ by aerial photographs and q by traffic counts. The problem now is to determine the density $\rho = \rho (v)$ and find the optimal speed $v=v_{opt}$, maximizing

$q(v)$ [$q=q(v_{opt})$].

Model 1. This model is based on radio and television advertisements in British Columbia which recommended that drivers should space out one car length for every 10 mph of speed, that is, $l^{*}=l[1+v/10]$.

corresponding to $\rho = \frac {1}{l(1+v/10)}$. The ratio $l^{*}/v$ is called headway.

Hence, $q(v)=\frac {v}{l(1+v/10)}$

Hence, $q(v)$ is a monotonically increasing function of v. Now, $\lim_{v \rightarrow +\infty}q/v=10/l$

This implies that there should be no speed limit!

For a typical Vancouver car such as a 1983 Ford Fairmont, $l \approx 1/325$ miles. Hence the optimal flow rate is $q_{opt}=\lim_{v \rightarrow +\infty}=3250$ cars per hour.

Note that when $v=15$ mph, q is 1950 cars per hour.

Note that when $v=30$ mph, q is approximately 2450 cars per hour.

Since for a given situation one is able to determine q and p from simple measurements, a traffic engineer is interested in the flow vs. density curve. For Model I, $q=q(\rho)=\frac {10}{l}-10\rho$

Experimentally it appears that figure will show the shape of a typical curve representing flow vs. density on a throughway.
It is observed that the optimal flow rate during rush hours for each lane of traffic on the Lions Gate Bridge is about 1600 to 1800 cars/hour.

My suggestion — try two other models on your own. You are welcome to discuss them on the blog.

More later,

Nalin Pithwa

# Permutations and combinations — an application to Statistical Mechanics

In Statistical Mechanics, one  encounters the situation of putting k particles into r distinct energy levels. The particles can thus be considered as discrete objects and the different energy levels as distinct boxes or cells. Three different situations are obtained by making  three different assumptions. These are:

a) Maxwell-Boltzmann: Here the particles are all distinct and any number of particles can be put into any of the r boxes. The number of possibilities are $r^{k}$ as given by the following theorem:

Let M be a multi set consisting of r distinct objects, each with infinite multiplicity. Then, the total number of d-permutations of M is $r^{d}$.

b) Bose-Einstein: Here the particles are all identical and any number of particles can be put in any one of the r boxes. The number of possibilities is $k+r-1 \choose k$ as given by the following theorem:

The following sets are in bijective correspondence:

i) The set of all increasing sequences of  length  k on an ordered set with r elements.

ii) The set of all the ways of putting k identical objects into  r distinct boxes.

iii) The set of all the k-combinations of a multi-set with r distinct elements.

c) Fermi-Dirac: Here the particles are all identical  but no  box can hold more than one particle. The number of possibilities is $r \choose k$.

More later…

Nalin Pithwa

# A cute problem on permutations and combinations

Here, is a problem on counting. Read it and crack it before you compare your solution with the one I present !!!

In a tennis tournament, there are $2n$ participants. In the  first round of  the tournament, each participant plays just once, so there are n games, each occupying a pair of  players. Show that the pairing for the first round can be arranged in exactly

$1 x 3 x 5 x 7 x 9 \ldots(2n-1)$

different ways.

Solution.

Call the required number of pairings of $2n$ players $P_{n}$. If you are a participant, you can be matched with any one of  the other $(2n-1)$ players. Once your antagonist is chosen, there remain

$(2n-2)=2(n-1)$

players who  can be paired in $P_{n-1}$ ways. Hence,

$P_{n}=(2n-1)P_{n-1}$.

More later

Nalin Pithwa

# everywhere differentiable function — an interesting problem

Question: Determine all the functions f, which are everywhere differentiable and satisfy

$f(x)+f(y)=f((x+y)/(1-xy))$ for all real x and y with $xy \neq 1$. —- Equation I

Solution:

Let $f(x)$ satisfy Equation I above. Differentiating Equation I partially with respect to each of x and y, we obtain

$f^{'}(x)=((1+y^{2})/(1-xy)^{2})f^{'}((x+y)/(1-xy))$ Equation II

$f^{'}(y)=((1+x^{2})/(1-xy)^{2})f^{'}((x+y)/(1-xy))$ Equation III

Eliminating common  terms in Equation II and Equation III, we deduce that

$(1+x^{2})f^{'}(x)=(1+y^{2})f^{'}(y)$

As the left side of Equation III depends only on x and the  right side only on y, each side of Equation III must be equal to a constant c. Thus, we have

$f^{'}(x)=\frac {c}{1+x^{2}}$

and so, $f(x)=c\arctan x +d$,

for some constant d. However, taking $y=0$ in Equation III, we obtain $f(x)+f(0)=f(x)$, so that $f(0)=0$ and

$d=0$. Clearly, $f(x)=c\arctan x$ satisfies Equation I and so all solutions of Equation I are given by

$f(x)=c\arctan x$ where c is a constant.

Note: In Equation II, the $f^{'}$ is w.r.t. x and in Equation III, the $f^{'}$ is w.r.t. y, yet in equation II

$f^{'}(x+y)/(1-xy)$ is the same as the corresponding $f^{'}$ in Equation III because the argument $(x+y)/(1-xy)$ is symmetric w.r.t. x and y.

# a cute problem on infinite series

Evaluate the infinite series $S = \sum_{n=1}^{\infty} \arctan (2/n^{2})$

Solution:

For $n \geq 1$ we have

$\arctan (1/n)-\arctan (1/(n+2))=\arctan \frac{(1/n)-(1/(n+2)}{1+(1/(n(n+2)))}=\arctan (2/(n+1)^{2})$

so that for $N \geq 2$ we have

$\sum_{n=2}^{N} \arctan (2/n^{2}) = \sum_{n=1}^{N-1}\arctan(2/(n+1)^{2})$ which equals the following

$\sum_{n=1}^{N-1} (\arctan (1/n) - \arctan(1/(n+2)) = \arctan (1) + \arctan (1/2) - \arctan (1/N) - \arctan (1/(N+1))$

Letting $N \rightarrow \infty$ we get

$\sum_{n=2}^{\infty} \arctan (2/n^{2}) = \arctan (1) + \arctan (1/2) = (\pi/4) + \arctan (1/2)$ and so

$S= (\pi/4) + \arctan (2) + \arctan (1/2) = ((3\pi)/4)$

More later…

Nalin Pithwa