Playing with models: quantitative exploration of life.

Inspections, fines and the honor system

Posted in Economics, Transportation by Alexander Lobkovsky Meitiv on September 25, 2015

Alexanderplatz U-Bahn station in Berlin

This summer we visited Berlin and enjoyed its excellent public transportation system.  Unlike the controlled access system prevalent in the US, Berlin’s (and Prague’s for that matter) system is an honor system.  You can buy a ticket at any time and when you board a train or a bus you stamp the ticket with the date and time the trip started.  Compliance is enforced by periodic inspections and the fine for riding without a validated ticket (Schwarzfahren) is a modest €60.  During our 8 days in Berlin we saw inspectors once and did not get inspected ourselves.

Given the ticket price, is the combination of the fine size and inspection frequency sensible?  That is, could BVG (the transit authority) save money by changing the fine amount and/or number of inspectors it has to hire?

To answer this question we must know how people decide whether to buy a ticket or risk being fined.  Many factors come into play here.  A purely rational and amoral consumer would buy the ticket if the product of the inspection probability and fine amount is larger than the ticket price.  This way, not buying tickets costs more in the long run.  However, there is a strong national pride and the concept of Beitragen (contributing) which keeps Schwarzfahren low (just a few percent according to this article in German).  Being caught without a valid ticket may also result not just in a fine but also in a court summons which presumably has a higher deterrent power.

Let’s explore the effect of the frequency of inspections (and the associated cost) and the fine amount on the total amount of money collected. I will define the single ride ticket cost to be 1 and the fine amount to be $F$.  To keep the formalism simple, I will neglect the reality that the fine for repeat offenders is higher.  This fact reduces Schwarzfahren but makes calculations more complicated. In the long run, the amount of the fine for repeat offenses is the important quantity.

I will assume for simplicity that the population of transit rides consists of two distinct pools. The conscientious riders always buy the ticket (population size $N$). The rest of the riders (population size $M$) are rational fare evaders and buy tickets when the inspection probability per ride $p$ is greater than $1/F$. This assumption requires the ability to accurately measure the frequency of inspections either from personal observations or from those of friends and family. The last ingredient in the recipe for computing the revenue of the system is the cost of inspections $c p (N + M),$ where $c$ is a new parameter which can be thought of as the cost per inspection.

Within our simplistic model, given the fine amount $F$, if the inspection probability is $p < 1/F$, the evaders do not buy tickets and if $p > 1/F$ the potential evaders always buy tickets. Because the cost of inspections is linear with $p$, there are two local maxima of the revenue. Either $p = 0$ in which case the revenue is simply $N$ (i.e. there are no inspectors and the evaders ride free), or $p = 1/F$ in which case the evaders buy the tickets the total revenue is $(N + M)(1 - c/F)$. Therefore, if the cost parameter $c$ is greater than the product of the fine amount and the fraction of fare avoiders

$\displaystyle c > F \frac {M}{N + M},$

it does not make sense to inspect. Another way to look at this equation is that the fine amount has to be at least the cost per inspection divided by the fraction of evaders to justify inspections.

Of course the situation in real life is different. Most people are opportunistic evaders. Each person is characterized by a risk aversion parameter which balances the financial benefit of evasion with the emotional toll of being caught. Because of this fact, some level of inspection is always necessary. The level itself depends on the distribution of the risk aversion parameter in the population and the recovery rate of the fines as a function of the fine amount. But that’s for another post.

The waiting game.

Posted in Game theory, Transportation by Alexander Lobkovsky Meitiv on December 20, 2010

When two buses can take you where you need to go should you let the slow bus pass and wait for the fast bus?

The Metro has changed the bus schedule this morning with no prior notice whatsoever. The 9 am J1 bus, I usually take, did not come. As I found out later, it was dropped from the schedule altogether. For about 20 minutes I was assuming (with diminishing conviction) that the J1 was merely late. While waiting for the J1, I let two J2’s go by. The J2’s take about 15 minutes longer to reach my destination. While I was waiting, I began to wonder what the best waiting strategy would be if there were two modes of transport with different travel times and different service frequencies. There is a math problem in there with a clear cut answer.

It is simpler to consider the situation in which buses do not have a fixed schedule but arrive at a fixed rate per unit time $\mu$. Intervals between consecutive buses in this situation obey Poisson statistics, which means that no matter when I arrive at the stop the average waiting time before the bus arrives is $1/\mu$.

In what follows I will present a few results without much derivation. If you are interested in the nitty-gritty, contact me for details.

Suppose there are two buses A and B that arrive at a stop with rates $\mu_A$ and $\mu_B.$ The probability that A arrives before B is

$P(A\mathrm{\ before \ } B) = \displaystyle{\frac{\mu_A}{\mu_A + \mu_B}}.$

The mean waiting time for bus A provided that A has arrived first is

$t_A =\displaystyle{\frac{\mu_A}{(\mu_A + \mu_B)^2}}.$

Now if the travel times to destination on buses A and B are $\tau_A \geq \tau_B,$ we can compute the expected travel time if the traveler boards the first bus that comes to the stop. We will call it $T_0$ because the strategy is to let zero buses pass (even if they take longer).

$T_0 = \displaystyle{\frac{\mu_A \tau_A}{\mu_A + \mu_B} + \frac{\mu_A}{(\mu_A + \mu_B)^2} + \frac{\mu_B \tau_B}{\mu_A + \mu_B)} + \frac{\mu_B}{(\mu_A + \mu_B)^2} = \frac{1 + \mu_A \tau_A + \mu_B \tau_B}{\mu_A + \mu_B}}.$

We can interpret this formula as follows. The total bus arrival rate is $\mu_A + \mu_B$ and therefore the mean waiting time for a bus, any bus is $1/(\mu_A + \mu_B).$ Then with probability $\mu_A/(\mu_A + \mu_B)$ the A bus has arrived and the travel time is $\tau_A.$ Likewise, with probability $\mu_B/(\mu_A + \mu_B)$ the B bus arrives so that the travel time is $\tau_B.$

It is a only marginally trickier to derive the mean trip duration (will call it $T_1$) when we are willing to let one A bus pass by in the hopes that the next bus will be the faster B bus. The answer is

$T_1 = \displaystyle{\frac{1}{\mu_A + \mu_B} + \frac{\mu_A T_0}{\mu_A + \mu_B} + \frac{\mu_B \tau_B}{\mu_A + \mu_B}}.$

The explanation of the second term in the above formula is that if A arrives first, we let it pass and we are back to the “let zero buses pass” strategy. The rest of the terms in the equation for $T_1$ are the same as before.

In general, for any $n \geq 1$ we have a recursion relation:

$T_{n+1} = \displaystyle{\frac{1}{\mu_A + \mu_B} + \frac{\mu_A T_n}{\mu_A + \mu_B} + \frac{\mu_B \tau_B}{\mu_A + \mu_B}}.$

We can now start asking questions like: “Under what conditions does letting the slow bus pass make sense (i.e. results in a shorter expected trip)?” What about letting two buses pass? When does that strategy pay off?

When does $T_1 \leq T_0$? Comparing the formulas above we arrive at a simple condition on the arrival rate of the fast bus which is independent of the arrival rate of the slow bus

$\displaystyle{\mu_B \geq \frac{1}{\tau_A - \tau_B}} \quad \mathrm{(1)}.$

For example, if the slow bus takes 30 minutes and the fast bus takes 20 minutes to arrive at the destination, it makes sense to let the slow bus pass if the fast bus arrives more frequently than once in 10 minutes. No big surprise there, anybody with a modicum of common sense could tell you that.

What is surprising is that the condition (1) does not depend on the arrival rate of the slow bus. Did I make a mistake? It turns out that when $T_1 = T_0,$ the expected travel times for other strategies are exactly the same! I will leave the proof to my esteemed reader as homework :)

Therefore, since it does not matter how frequently the slow bus comes, if the fast bus comes frequently enough (condition (1) is satisfied), it makes sense to wait for the fast bus no matter how many slow buses pass.

What is the mpg of a bicycle?

Posted in Transportation by Alexander Lobkovsky Meitiv on November 10, 2010

The next green thing?

It seems a silly question at first. Digging a little deeper, it is easy to convince yourself that when you travel somewhere by bike, your body burns more calories than if you sat in your office chair. The extra calories came from the extra food you had to eat. The land that was used to produce the food you had to eat could have been used to grow corn and make ethanol. The amount of ethanol produced from the corn required to travel a mile by bike is undoubtedly small, but how small?

To compute my bike mpg we will need three numbers:

1. Extra calories burned per mile of bike travel at roughly 13 miles per hour (my average commuting speed). For relatively flat terrain and my weight this number is roughly 42 food calories per mile (obtained from about.com).
2. We need a food equivalent for the ethanol production. Let’s say I go my extra calories from eating sweet corn. According to the same source, sweet corn has 857 food calories per kilogram. So I will need to eat 49 grams of sweet corn per mile traveled at 13 miles per hour on my bike.
3. Now we need to know how much ethanol can be made from 49 grams of sweet corn. The Department of Energy’s Biomass Program to the rescue. According to their website, a metric ton of dry corn can theoretically yield 124.4 gallons of ethanol. Since sweet corn is 77% water, this means that up to 0.0014 gallons of ethanol can be made from 49 grams of sweet corn.

Putting these numbers together we arrive at 0.0014 gallons of ethanol per mile or…drumroll please:

701 mpg

This number is not small, but neither is it very large! There exist experimental vehicles that seat four and achieve over 100 mpg. When fully loaded, the effective, per passenger mpg is 400. If my calculations are correct, Technology is about to bring motorized transport close to the efficiency of a person on a bike!

Should you switch lanes in traffic?

Posted in Statistics, Transportation by Alexander Lobkovsky Meitiv on June 24, 2010

Switching lanes in heavy traffic can indeed increase your average speed.

If you drive like me, you have no patience for bumper to bumper traffic. There is gotta be a way to beat it somehow, right? Do you sneak into an opening in a neighboring lane if it is moving faster? Do you set goals like: “when I get in front of that van, I’ll switch back?” It doesn’t always seem to work. A lane that was zooming by you comes to a dead stop when you switch into it. If the motion of each lane is random, is there a way to switch lanes and move faster than a car that stays in lane?

It turns out there is a way to beat the traffic. To show this we will use a simple model of traffic flow introduced by Nagel and Schrekenberg (see the previous post). The model consists of a circular track with consecutive slots which can be empty of occupied by cars. Cars have an integer velocity between 0 and vmax. As we saw in the previous post, simple rules for updating the positions and velocities of the cars can reproduce the traffic jam phenomenon thereby a dense region forms in which the cars are at a standstill for a few turns and then, as the jam clears in front of them, the cars accelerate and zoom around the track only to be stuck in the jam again. The jam itself moves in the direction opposite to that of the cars.

Now imagine that we put two of the circular tracks (or lanes) side by side. For starters, let’s require all cars except one to stay in their respective lanes. One rogue car can switch lanes. Can the rogue with the right lane switching strategy move faster than the rest of the cars on average? The answer is most certainly yes although finding the best lane switching strategy is a difficult computational problem. What we are going to do here is compare two lane switching strategies that at first sight seem equally good. What we will discover is that it the lane changing strategy matters. As you might have suspected, if you don’t do it right, you might actually move slower than the rest of the traffic!

Here are the two simple strategies we will compare (I suggest you read the previous post for the description of the model):

1) “Stop-switch:” if the slot directly ahead is occupied, switch if the space in the other lane directly across is not occupied.
2) “Faster-switch:” if the car directly ahead in the neighboring lane is moving faster, switch if there is space available.

The “Stop-switch” strategy performs significantly better than the “Faster-switch” strategy.

The graph above compares the two strategies. It shows the percent improvement of the rogue’s average speed compared to the average speed of the rest of the cars as a function of the car density. When density is low and traffic jams are rare, switching lanes has almost no effect on your average speed for both strategies. When the density is high and traffic jams are abound, switching can make you go slower than the rest of the traffic. The reason is that when a space in the neighboring lane opens up, it is likely to be at the tail end of a jam whereas the jam in the lane you just switched out of might be already partially cleared. The final remark is that the “Stop-switch” strategy is significantly better improving the speed by as much as 35% whereas the best “Faster-switch” can do is a 15% improvement.

Finally let me mention that if all cars switch lanes and use the same strategy, nobody wins. All cars move with the same speed on average. That average speed could be smaller or larger (depending on the car density and the switching strategy) than in the case when everybody says in lane. The graph below explains why everyone is so keen on the advice “Stay in lane!” It turns out that if everyone uses the “Faster-switch” strategy, the average speed is drastically lower for everyone than if everyone stays in lane! The reason for this dramatic result is that when you change lanes, the car behind is likely to slam on the brakes which slows everyone down.

When everyone stays in lane (the Single lane case), the average speed is significantly greater than when everyone switches lanes using the “Faster-switch” strategy.

People are the real cause of the traffic jam

Posted in Statistics, Transportation by Alexander Lobkovsky Meitiv on May 12, 2010

Sometimes traffic slows to a crawl for no apparent reason

When the traffic on the beltway is moving at a snail’s pace without an obvious reason (like construction or accident), I frequently wonder: “why can’t everyone just go faster?” If all car’s were driven by computers that could talk to each other, a clever synchronization algorithm, could allow all cars travel in unison and thus prevent congestion that is not a result of lane closure. Alas, this technotopia is still decades away and cars will be driven by humans for the foreseeable future. In the meantime we can but wonder: “what is it about the way people drive that causes the traffic to jam when the density of cars becomes too great?

Traffic flow is frequently studied because it is an example of a system far from equilibrium. The practical applications are important as well. Many models from crude to sophisticated have been advanced. Massive amounts of data exist and are frequently used to estimate model parameters and make predictions. I am not going to attempt to review the vast field here. My goal is simply to elucidate the physiological limitation of the human mind that causes the driving patters leading to congestion.

Although great progress has been made in modeling traffic as a compressible fluid, a class of models that fall into the category of Cellular Automata are more intuitive and instructive.

Cellular Automata, promoted by Stephen Wolfram of Mathematica fame as the solution to all problems, are indeed quite nifty. It turns out that autonomous agents, walking on a lattice and interacting according a simple set of rules can reproduce a surprising variety of observed macroscopic phenomena. If you want to learn more the Wikipedia article is a good start.

A pioneering work of Nagel and Schreckenberg published in Journal de Physique in 1992 introduced a simple lattice model of traffic which reproduced the traffic jam phenomenon and came to a surprising conclusion that the essential ingredient was infrequent random slowdowns.

You have probably done so yourself, you change the radio station or adjust the rear view mirror, or speak the child in the seat behind you. As you do so, your foot eases off the accelerator ever so slightly irritating the person behind you who has to disengage the cruise control. You and people like you are responsible for the traffic jams when the volume is heavy but there are no obvious obstructions to traffic.

Allow me to reproduce the authors’ description of the model since it is concise and elegant:

“Our computational model is defined on a one-dimensional array of L sites and with open or periodic boundary conditions. Each site may either be occupied by one vehicle, or it may be empty. Each vehicle has an integer velocity with values between zero and vmax. For an arbitrary configuration, one update of the system consists of the following four consecutive steps, which are performed in parallel for all vehicles:

1. Acceleration: if the velocity v of a vehicle is lower than vmax and if the distance to the next car ahead is larger than v + 1, the speed is increased by one.
2. Slowing down (due to other cars): if a vehicle at site i sees the next vehicle at site i + j (with j < v), it reduces its speed to j.
3. Randomization: with probability p, the velocity of each vehicle (if greater than zero) is decreased by one.
4. Car motion: each vehicle is advanced v sites.”

Without the randomizing step 3) the motion is deterministic: “every initial configuration of vehicles and corresponding velocities reaches very quickly a stationary pattern which is shifted backwards (i.e. opposite the vehicle motion) one site per time step.”

The model exhibits the congestion phenomenon when the mean spacing between the cars is smaller then vmax.

Below are the links to the simulations of the model for a circular track with 100 lattice sites, the cars are colored circles which move along the track. It helps to follow a particular color car with your eyes to see what’s happening.
The two simulations are done with 15 cars (density lower than critical) and with 23 cars (above the critical density–exhibits congestion). As you probably guessed vmax=5 in these simulation hence 20 cars correspond to the critical density. The probability of random slowing down is 10% per turn.

Free flowing traffic in a simulation of the Nagel-Schreckenberg model below the critical density threshold.

The second simulation (above the critical density) shows the development of a jam of 5 cars. Cars zoom around the track and then spend 5 turns not moving at all, before the traffic clears ahead of them and they can accelerate to full velocity again.

The moral of the story? People like you and me can be the cause of traffic congestion!

Unavoidable Attraction

Posted in Statistics, Transportation by Alexander Lobkovsky Meitiv on March 2, 2010

When traffic is heavy, buses tend to form hard to breakup bunches.

Everyone who rides buses in a city is familiar with the dreaded “bus bunching” phenomenon. Especially during rush hour, buses tend to arrive in bunches of two, three or even more. Why is that?

To begin understanding this phenomenon we must first assimilate the notion of fluctuations. The bus’s progress along the route, though ideally on schedule, in practice is not. At each stop there is a difference between the actual and the scheduled arrival time. The nature of the fluctuations is such that this difference tends to grow along the route of the bus. In technical terms, the bus’s trajectory is called a directed random walk. There are several sources of the fluctuations: stop lights, variation in the number of passengers to be picked up and discharged and, of course, traffic.

When fluctuations are strong, and/or, the buses are frequent, it is unavoidable that consecutive buses find themselves at the same bus stop. What happens afterwards is less clear cut. It seems that it is virtually impossible for the buses to separate again. From that point on the two (or more) buses travel in a bunch. The average speed of a bus bunch is frequently greater than that of an isolated bus and therefore bunches tend to overtake and absorb buses that are ahead.

Let’s try to come up with a plausible explanation for the two observed phenomena: Why do bus bunches not break up naturally? Why is the average speed of the bunch different from the average speed of an isolated bus?

Let’s tackle the questions one at a time. When don’t bunches break up? There could be several reasons. Without real field data, I am afraid, we won’t be able to say for certain which factor is the most important.

Possible reason #1: Excluded volume interactions. Analogy with colloids.

Colloids are suspensions of small solid particles in a fluid. It is a well know phenomenon, readily reproducible in a lab, that when you combine colloidal particles of two substantially different sizes, they tend to separate even if the particles themselves are not attracted to each other. It may be counterintuitive, but the system can increase its entropy by separating particles by size. Once a small particle escapes from the aggregate of large particles, it is extremely unlikely to make it back there.

The same size separation might happen in traffic, although likely for different reasons. How much do you like being sandwiched between a bus and a dump truck? You try to get the hell out of there at the first opportunity.

So spaces between traveling buses may be unlikely to be filled up with cars. In a sense, there is an effective attraction between buses cased by the car’s avoidance of the space between them.

One would certainly need data to support or reject the excluded volume hypothesis of bus attraction.

Possible reason #2: Correlation between the number of waiting passengers and the distance to the nearest bus ahead.

Now this idea is something we could sink our teeth into. Suppose that the gap between two buses shrinks due to a random fluctuation of unspecified nature. Then, the mean number of passengers waiting for the second bus, which is proportional to the wait time (if the passengers arrive at the bus stop at a fixed rate), also decreases. Therefore the second bus will spend less time picking up passengers, it’s mean velocity will therefore increase and it will catch up with the bus ahead. We can therefore say that the state with evenly spaced buses along the route is unstable to collapse.

This idea can be formalized in the following simple toy model.

Suppose there is a circular route with equidistant stops (a linear route is really circular if the buses turn around at the end of the route and go back immediately). Initially a number of buses are uniformly distributed along the route. Passengers arrive at all bus stops at a fixed rate. The time a bus spends at a stop is proportional to the number of passengers waiting there.

Passenger discharge can be included in the model. However it does not qualitatively affect the results.

There are two important parameters in this model: 1) the product of the travel time between stops and the rate of passenger arrival. This parameter determines whether the bus spends most of its time traveling or picking up passengers. 2) The ratio of the number of stops to the number of buses.

It turns out that if the first parameter is large (most time is spent traveling) or the second parameter is small (there are lots of buses), bunching does not occur.

However, as illustrated in the figure below, there is a realistic parameter range in which bunching does occur and bunches have no chance to break up. In the figure below (which presents the output of the simple model above), the three buses were initially well spaced. Eventually, buses 1 and 2 form a bunch which catches up to bus 3.

Once the bunch of two buses is formed, the buses leapfrog each other and pick up passengers at alternating stops. Here is therefore the answer to our second question why bunches travel faster: each bus only has to accelerate/decelerate less frequently since they only stop at every other stop. Hence the average speed is greater.

It would be fun to go out there and time some bus arrivals to see if they can be well described by the model. Any takers?

Correlation between the space between buses and the number of waiting passengers results in the bunching behavior.

Decisions, decisions, decisions…

Posted in Statistics, Transportation by Alexander Lobkovsky Meitiv on February 25, 2010

This entry is about how the amount of information at the time of a decision can increase the efficacy of the outcome.

The specific case I will talk about is public transport.

Have you ever been on a bus that sat at a red light only to stop again at a bus stop right after passing the intersection?
Did you wonder if it would be better to have the bus stop located before the light?

Wonder no more! If you read on, we will answer this question and a few others using simple statistics and a few carefully chosen assumptions.

Let us first compute the average waiting time at a red light. Let’s say the light has only two states: red and green which alternate. The durations of the red and green lights are fixed and are $t_r$ and $t_g.$ Suppose that the bus arrives at a light at a random time. Then its average waiting time at the red light is

$\displaystyle t_\ell=\frac{1}{t_r+t_g}\int_0^{t_r}t\,dt=\frac{t_r^2}{2(t_r+t_g)}.$

This is because we assume that the bus arrives at the light at a random time. Without any prior information, the distribution of arrival times is uniform. The behavior of the light is periodic with period $t_r+t_g$ and thus the probability of arriving in any time interval $dt$ is $dt/(t_r+t_g).$

For example, if the red and the green lights are equally long, i.e. $t_g=t_r,$ the average wait at a stop light will be a quarter of the red light duration $t_\ell = t_r/4.$ (To derive that substitute $t_g=t_r$ into the equation above).

Now lets add the bus stop to the equation. We will assume that the bus stops for a fixed time $t_s.$ Fluctuations in the stopping time can be added to the model. However, calculations become a bit more involved and the result does not change qualitatively.

The questions are: 1) What is the total stoppage time $t_w$: red light + bus stop? 2) Does it depend on whether the bus stop is before or after the red light?

If we know anything about information theory, our answer to the second question is NO without doing any algebra. Why? Because the bus arrival time is random and uncorrelated with the timing of the stop light. There is no information that can distinguish stopping before and after the intersection. If the stop is after the light, the bus has the wait at the red light for a time $t_{\ell}$ just computed above. If the bus stops before the light, the “arrival” time is the time at the end of the stop and it is just as random as the arrival to the stop. Therefore, the average total stoppage time is just $t_\ell + t_s$ regardless of whether the stop is before or after the light.

How can the total stoppage time be reduced?

After all this post is about efficiency of mass transit. The answer, again from the point of view of information theory, is the following. To improve efficiency, we must use available information to make decisions which make the arrival (or departure) time of the bus correlated with the timing of the light.

In Switzerland, public support for mass transit is so strong, that people accept that the trolleys actually change the timing of the stop lights to speed up passage at the expense of cars. Here in America this approach may not fly. However, even if the timing of the stop light cannot be changed by the bus/trolley driver, they still have the power to make decisions that would change the total stoppage time.

In the example above, the bus stop was always before or after the intersection. Suppose the driver could decide, based on some information about the phase of the stop light, whether to stop before or after the intersection?

Let’s call the scenario in which the bus driver does not make a decision where to stop the “null model” or the “no-decision” model. As a better alternative consider the “red-before” scenario in which the driver stops before the intersection if the bus arrives on the red light and after the intersection if the bus arrives on the green light. What is the average stopping time $t_w = t_s + \Delta$?

I am not going to bore you with the tedious derivation. The result itself is a bit complicated as we have to consider 4 separate cases. I am going to give a formula for the extra waiting time $\Delta$ on top of the regular stop duration $t_s.$

Let’s first define:

$\displaystyle I_1=\frac{(t_r-t_s)^2}{2(t_r+t_g)},$

and

$\displaystyle I_2=\frac{(t_s-t_g)(t_r-\frac{1}{2}(t_s-t_g))}{t_r+t_g}.$

Then the extra waiting time is

$\Delta=0$ for $t_r \le t_s \le t_g$
$\Delta=I_1$ for $0\le t_s \le \min(t_r,t_g)$
$\Delta=I_2$ for $\max(t_r,t_g)\le t_s \le t_r+t_g$
$\Delta=I_1 + I_2$ for $t_g \le t_s \le t_r$

If $t_s \ge t_r + t_g,$ just replace $t_s$ everywhere with its remainder when divided by $t_r + t_g.$

To illustrate these formulas here is the graphs comparing the extra stoppage time $\Delta$ for the “no-decision” and the “red-before” scenarios as a function of the stop duration $t_s$ for two different ratios of the red to green light durations.

Comparison of the extra waiting time as a function of the bus stop duration for two different decision scenarios and the red light twice as long as the green light.

Comparison of the extra waiting time for the green light twice as long as the gred light. Note that for a certain range of bust stop durations, the extra waiting time vanishes completely!

The “red-before” scenario which uses only the information about the current state of the stop light does quite well compared to the “no-decision” scenario. When the green light is longer than the red light, the extra waiting time vanishes altogether if the stop duration is chosen properly.

Can we do better?

Yes! The more information is available to the driver, the better can the strategy be for making the decision where to stop. We can imagine, for example, that when the bus arrives at a red light, the driver knows when it will turn green again. Or, the driver can have complete information and also know the duration of the following green light.

Let us compute the extra waiting time for the best stopping strategy with complete information. How much better does it do than the “red-before” strategy which uses only the information about the current state of the stop light? The best stopping strategy which uses all available information is the following. Suppose the bus arrives on a red light. The time till the light change is the extra waiting time if the driver decides to stop after the intersection. This time needs to be compared to the extra waiting time which might result if the bus stops before the intersection. This might happen if the total stop duration is longer than the remainder of the red light plus the following green light so that the light is red again after the bus stop is completed. The best decision will depending on when the bus arrives, the duration of the red and green lights and the the bus stop.

I am going to leave you with a comparison of the extra waiting time for the”red-before” strategy with the best stopping strategy with perfect information about the phase of the stop light (length of red, green, time till change).

The moral of the story: “Information is power!”

The perfect information helps reduce the extra waiting time when the red light is longer than the green and when the bus stop duration is longer than the that of the green light.