## Mathematical Recreations

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#### Crossing the Street ("dimensionless" equations and Poisson statistics)

I've been going through a new Calculus text and looking over the exercises. This book is consistently poor in the way they handle units of measurement, and they make equations dimensionless at every opportunity. Very sad. I think it does the students a disservice—especially since this is supposed to be Calculus with Applications. It is an excellent opportunity for them to learn how to manage units in real problems.

The exercise below stood out so much that I had to reverse engineer it. That process turned out to be revealing and fun. It certainly reinforced my sense that students are being deprived of a proper treatment of the material, when the authors (or instructors) go to great lengths to sweep the units of measurement under the carpet.

This premise of this exercise is reproduced directly from the text.1 It is quoted for purposes of literary criticism:

Consider a child waiting at a street corner for a gap in traffic that is large enough so that he can safely cross the street. A mathematical model for the traffic shows that if the expected waiting time for the child is to be at most 1 minute, then the maximum traffic flow, in cars per hour is given by

 f(x)  = 29,000(2.322 – log x) x ,

where x is the width of the street in feet.*

*Bender, Edward, An Introduction to Mathematical Modeling, John Wiley & Sons, 1978, p. 213.

The topic for this month is to reverse-engineer this problem. Except in the blockquote above, I will use x to represent the width of the street, and I will use X to represent the dimensionless quantity x/(1 foot), which appears as x in the textbook exercise. I will also use τ to represent the maximum expected waiting time, which the exercise sets at one minute.

The first step is to express the problem in its natural form, with no "unexplained" numerical constants. (You may find it helpful to recall that "log" normally implies "log10" in math textbooks.) This step requires only an understanding of logarithms.

The second step is to try to derive the formula to infer what assumptions were made. Presumably the assumption is that the traffic follows a Poisson distribution. Anticipating that, let's agree that the traffic flow is represented by λ. That is, "f(x)" in the textbook exercise is related to λ by the expression λ = f(x)/(1 hour). If you make the same derivation that I did, then you will need to use integration and an understanding of the Poisson distribution.

#### Units

There are several steps that need to be taken to "clean up" this problem. First and foremost, the quantities x and f(x) need to be dimensioned. x is a length, so we need a reference length for the logarithm, since we can't take the logarithm of a dimensioned quantity. We will also need units for the overall expression:

(1)
 λ = 29000 ft/hr (2.322 – log10 (x / 1 ft) ) x

Next, we recognize that there are no physical processes here that revolve around 10 as a base for the logarithm. With a strong suspicion that Poisson statistics are involved, we convert the base of the logarithm to e.

(2)
 λ = 29000 ft/hr (2.322 – ln (x / 1 ft) / ln 10 ) x

Now, 1 ft doesn't seem to be a likely reference length in this problem, and there is the 2.322 floating around in a difference with the logarithm. Let's gather those terms:

(3)
 λ = (29000 ft/hr / ln 10) (2.322 ln 10 – ln (x / 1 ft) ) x = (29000 ft/hr / ln 10) ln (102.322 ft /x) x = (12595 ft/hr) ln (209.9 ft /x) x

Finally, the units of hr–1 don't match up with the only time value that we have in the problem, which is 1 min, so let's change the units to ft/min instead of ft/hr.

(4)
 λ = (209.9 ft/min) ln (209.9 ft /x) x

Well, we just got the same 209.9 ft constant! In fact, we have the expression 209.9 ft / x appearing twice in the equation. All that is left is a floating unit of min–1, and we recognize that this must be the maximum waiting time τ. We can now rewrite the dimensioned equation in it's final form, using x0 for the quantity 209.9 ft:

(5)
 λ = x0xτ ln (x0 / x)

#### Statistics

We have now reduced the equation to one constant that was not in the original problem (and we got rid of the 29,000 and 2.322 that seemingly came from nowhere). Let's see if we can understand this equation based on some assumptions about how the equation was probably derived.

If the cars arrive at an average rate λ and their arrival times are completely uncorrelated, then the system will obey Poisson statistics. Starting with this assumption, we can ask what the average wait time should be. We have to decide whether the average wait time includes the time spent crossing the street or not. Recognizing that λ becomes negative if x is greater than x0 = 209.9 ft, we can guess that x0 is the width of the widest street that a child can cross in one minute (or more generally in a time interval τ). This isn't necessarly a valid assumption, but it's not a bad place to start. Let's define T as the average wait time (including the time spent crossing the street). This implies a walking speed of v = x0 / τ, and it takes the child a time x / v to cross the street. When the child arrives at the corner, the probability that there is a traffic gap of at least x / v is given by:

(6)
 P =  e–λx/v

In that case, the wait time is just x / v, the time to cross the street. Let t represent the length of the initial gap. Then the probability density function for t is given by:

(7)
 g(t) =  λe–λt

If the child waits a time t for the first car to go by, then the average total wait time will simply be t + T. What we need is the average value of t for t < x / v. We'll let t1 represent this quantity. We find:

(8)
 (1 – e–λx/v ) t1 =  ∫0x/v t λ e–λtdt =  (1/λ) ∫0λx/v u e–u du =  –(x / v) e–λx/v – (1/λ) (e–λx/v – 1)

The factor (1 – e–λx/v ) is the probability that the initial gap is less than x / v. We can write the average wait time as:

(9)
 T =  (x / v)e–λx/v + (1 – e–λx/v ) (t1 + T)

From here, we can solve for T to get:

(10)
 T =  (x / v) + eλx/v(1 - e–λx/v ) t1 =  (x / v) – (x / v) + (1/λ) (eλx/v – 1) =  (1/λ) (eλx/v – 1)

If we expand the exponential as a power series, we can see that T is a monotonically increasing function of λ, so it has a unique inverse function. Furthermore, T < τ if and only if λ is less than the value of this inverse function at T = τ.

#### Approximating the result

Although λ is a well-defined function of x, τ and v, there is no analytical expression for the function. The parameter λ appears twice in the formula. As λ gets very large (compared to v / x), the exponential factor will dominate. This would make the wait time be much greater than one minute. Recongizing this, we find that either λx / v << 1 or λx / v ≈ 1.

Let's assume that λx / v ≈ 1. If this is the case, then we might consider the approximation that 1 / λ ≈ x / v. Then we could write

(11)
 τ ≈ (x / v) (eλx/v – 1)

We can solve this expression for λ

(12)
 λ ≈ (v / x) ln (1 + v τ / x)

In fact, we can use this "better" approximation for λ to solve for λ

(13)
 λ ≈ (v / x) ln [1 + (v τ / x) ln (1 + v τ / x)]

This iterative process seems to converge for all values of x, τ and v, although I haven't worked through it rigorously. In particular, though, it does eventually converge for λx / v << 1. The formula in the math textbook reduces to

(14)
 λ ≈ (v / x) ln (v τ / x)

What we find is that the formula in the math textbook is an underestimate of the exact value from Poisson statistics. In the regions where decisions about traffic loading would be made, it's not a gross underestimate. A value of vτ/x = 6 (corresponding to a street width of about 35 feet) gives a maximum traffic load of about 17.51 per minute using the exact formula, or 10.75 per minute using the approximation in the text. Using formula 12 gives 11.68 per minute, and formula 13 gives 15.24 per minute. A value of vτ/x = 4 (corresponding to a street width of about 52 feet) gives a maximum traffic load of 9.35 per minute using the exact formula, or 5.55 per minute using the approximation in the text. Using formula 12 gives 6.44 per minute, and formula 13 gives 8.03 per minute.

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Thanks,
Steve