MAT109 · Erich Prisner · Franklin College · 2007-2011

Analysis of AIRPORT SHUTTLE

Prerequisites: Chapters 1, 2, and 8.

In Lugano, the hometown of our College, there are two competing shuttle companies serving the airport in Milano-Malpenso, which is about 1 hour 15 minutes away. Of course they have different schedules, and nowadays they have slightly different starting points, but for years they were both starting in front of the railway station. Often, the shuttle from company A was supposed to leave just briefly before the shuttle from company B, and I couldn't help but wondering how many, or rather few, customers would really take this shuttle from company B. Probably only those few who arrive at the railway station in the short time interval after shuttle A leaves and before shuttle B leaves.

It is rather obvious that the schedule, in relation to the schedule of the other company, matters, and the more general question we address in this chapter is how this schedule should look like.

1. The simple model

In the first model there are two competing airbus shuttles to the airport. In order to simplify things, the license granting department forces each shuttle to start at full hours only, and to drive four times per day every four hours. Thus the only decision left to Ann and Beth, the owners of the shuttles, is about when to start in the morning, 6:00 am, 7:00 am, 8:00 am, or 9:00 am. This decision for the schedule of the next year is to be announced to the license granting department by letter at a certain deadline in November, so we may think about this to be a simultaneous game.

1.1 To the Airport

Let's concentrate on the part of driving from the city to the airport first. Of course, eventually these shuttles will also pick up passengers at the airport for the ride back to the city. This will be dealt with in section 1.2, whereas in section 1.3 we will combine both.

The payoff for each shuttle is (proportional to) the number of passengers transported. We assume that at every full hour between a:00 and b:00, 8 persons have to leave for the airport. If no shuttle leaves at that time, they take one leaving earlier (if there is one) provided it is not more than 3 hours earlier---otherwise they would look for other transportation, as asking a friend to drive them. If two shuttles leave at the same time, they split the passengers equally.

Note that this game is almost, but not totally zero-sum. There may be some passengers that don't find a shuttle, either since they have to leave for the airport earlier than the earliest shuttle, or since they have to leave four hours or more later than the latest shuttle leaves, and don't take a shuttle for that reason.

Take as example a=4 and b=23, in other words, passengers have to leave from 4:00 until 23:00 Assume further that Ann's shuttle leaves at 6:00, 10:00, 14:00, and 18:00, and Beth's shuttle leaves at 7:00, 11:00, 15:00, and 19:00. The following table displays the choices the 8 passengers at that time make---"A" for using Ann's shuttle, "B" for using Beth's, and "-" for using none.

45678 910111213 1415161718 1920212223
--ABBBABBBA BBBABBBB-

Then Ann's shuttle leaving at 6:00 will have 8 passengers on board, and the same with Ann's shuttles at 10:00, 14:00, and 18:00. On the other hand, each of Beth's four shuttle journeys will have 24 passengers except the last, which will even transport 32 passengers. For simplicity let's assume that both shuttles are large buses, so that a shuttle must never leave passengers behind. Therefore Ann will transport 32 passengers, and Beth 104 passengers.

If we do this analysis for all possible pairs of start times for Ann's and Beth's shuttles, we get the following Normal form: There is an Excel sheet computing these values automatically.

The bimatrix for a=4, b=23, passengers leaving between 4:00 and 23:00.
6,10,14,187,11,15,198,12,16,209,13,17,21
6,10,14,18 64, 64 32, 104 64, 80 96, 48
7,11,15,19 104, 32 64, 64 32, 104 64, 72
8,12,16,20 80, 64 104, 32 64, 64 32, 96
9,13,17,21 48, 96 72, 64 96, 32 60, 60

Let us abbreviate the move "6,10,14,18" by "M6", and so on. None of these moves is dominated. The Maximin move for both players is M9, expecting a minimum payoff of 48. If both players would play their Maximin move, both would get a payoff of 60. But this schedule would also minimize the number of passengers transported. All the 24 persons who have to leave at 6:00 or at 7:00 or at 8:00 would need other transportation.

The symmetric best response digraph is cyclic: The best response to M6 is M7, the best response to M7 is M8, the best response to M8 is M9, and the best response to M9 is M6. It is best to start one hour later than your opponent. That implies that there is no Nash equilibrium in pure strategies.

Therefore we should look for mixed Nash equilibria. Let us define three mixed strategies:

Mix2 means essentially a mix of even start times (with a small probability of starting at 9:00) and Mix3 is essentially a mix of odd start times (with a tiny probability of starting at 8:00).

We claim that Mix1 versus Mix1 is a symmetric Nash equilibrium, and that Mix2 versus Mix3 are also Nash equilibria. Although finding them is at the moment a little beyond what we are supposed to do, at least checking the "Nashness" can be done in a rather straightforward way. We add these three mixed strategies to our pure strategies matrix:
M6M7M8M9 Mix1Mix2Mix3
M6 64, 64 32, 104 64, 80 96, 48  66.7, 72 66.1, 70 65.9, 74.4
M7 104, 32 64, 64 32, 104 64, 72  66.7, 67 69.8, 66.2 63.8, 68.5
M8 80, 64 104, 32 64, 64 32, 96  66.7, 66.7 69.8, 66.1 65.9, 65.9
M9 48, 96 72, 64 96, 32 60, 60  66.7, 63.7 69.8, 65.6 65.9, 61.7
Mix1 72, 66.7 67, 66.7 66.7, 66.7 63.7, 66.7  66.7, 66.7 69.1, 66.7 65.3, 66.7
Mix2 70, 66.1 66.2, 69.8 66.1, 69.8 65.6, 69.8  66.7, 69.1 68, 68 65.9, 69.8
Mix3 74.4, 65.9 68.5, 63.8 65.9, 65.9 61.7, 65.9  66.7, 65.3 69.8, 65.9 64.9, 64.9
The symmetric best response digraph is shown to the right. Obviously Mix1 versus Mix1 and Mix2 versus Mix3 are Nash equilibria.

Both players get a payoff of 12864/193=65.66 in the Mix1 versus Mix1 case. In the other two Nash equilibria there is essentially a mix of M6 and M8 versus a mix of M7 and M9. The player playing Mix2 gets an expected payoff of 9024/137=65.87, and the other player gets an expected payoff of 768/11=69.82%.

The payoffs of the three Nash equilibria are shown in the graph to the right.

If no communication is allowed between the players, then this game is difficult for both players, since it requires a certain amount of coordination between them in order to get a high payoff. If communication is allowed before the game starts, then the two players could agree to play one of the three Nash equilibria, whose payoff is shown in the graph. Which one would they agree upon? Without enforceable contracts and side-payments, probably the most likely outcome of the negotiation, the one considered fair by both players, would be the symmetric Nash equilibrium, since it is, as a symmetric one, rather focal. Both players could also expect a higher payoff as the 60 they would get if both play their Maximin move.

focal equilibrium, mention it before, and explain it here as well.

Would they obey the pre-game agreement? The answer obviously is: Yes. Remember that Nash equilibria are self-enforcing. If one player deviates, this player would make less or equal to what he or she would make by sticking to the agreement provided the other player obeys the agreement. Now remember that all these mixed Nash equilibria require both players to mix strategies with given probabilities. Since you couldn't even check whether the other players kept their promises--- except in cases where they would play some move with probability 0 in their agreed mixed strategy, they could always explain the outcome by (bad) luck, wouldn't that invite cheating? No---check on the bimatrix to assure yourself that a player deviating and playing any pure strategy cannot improve expected payoffs.

Note also that no pure strategies pair could serve as outcome of pregame negotiations. One player would always be tempted to deviate from such an agreement.

1.2 From the Airport

For the ride from the airport to the city, we assume that at each full hour between c:00 and d:00, 8 persons need a ride. They take the next available shuttle, provided they don't have to wait more than 3 hours, and again shuttles leaving at the same time split the passengers equally.

If we take the same start time c=a=4 and end time d=b=23 for passengers as in the analysis to the airport, we arrive at the following bimatrix: Again the Excel sheet can be used to compute this matrix.
7,11,15,198,12,16,209,13,17,2110,14,18,22
7,11,15,19 64, 64 104, 32 80, 64 56, 96
8,12,16,20 32, 104 64, 64 104, 32 80, 64
9,13,17,21 64, 80 32, 104 64, 64 104, 32
10,14,18,22 96, 56 64, 80 32, 104 64, 64

This game has only one Nash equilibrium. It is symmetric and in mixed strategies. Both players use a mix of 17/33=52% of "7,11,15,19", 4/11=36% of "9,13,17,21", and 4/33=12% of "10,14,18,22", and get a payoff of 2272/33=68.85.

1.3 Combining both

When to go to and from the airport are not independent decisions. If only one shuttle bus is to be used by a shuttle company, then it can not use the same time to leave from the city and from the airport. If we assume that the tour takes more than 1 hour, then even the combination of having a difference of 1 hour between leaving city and airport is impossible. Then there are only three possibilities:

Let us abbreviate these three moves by "6|8", "7|9", and "8|10". Although it is advantageous to go to the airport exactly one hour after the other shuttle does (since you collect three rounds of passengers, and the other shuttle only one), it is exactly the opposite for going back. For going back, it would be best to leave one hour before the other shuttle. So if you are forced to keep a regular pattern, any advantage you get for starting one hour later than the other in the city will be balanced by a disadvantage when coming back of almost equal weight.

For instance, if customers need the shuttle from 4:00 to 23:00 both ways, then the payoffs for both shuttle companies are equal in all occurring outcomes:

6|87|98|10
6|8 128, 128 136, 136 144, 144
7|9 136, 136 128, 128 136, 136
8|10 144, 144 136, 136 128, 128

Still they are different, so it is still worthwhile to analyze the game. We have five Nash equilibria, two pure ones of "6|8" versus "8|10" with payoff of 144 for both, and three mixed ones: 50% "6|8" and 50% "8|10" versus the same strategy, or versus the pure strategy "7|9". In each of these three mixed Nash equilibira, the payoffs for both players are 136.

2. Impatient customers

In this variant we make the assumption that the number of passengers actually using a shuttle decreases if they would have to drive too early, (meaning they would have to wait at the airport) for the trip to the airport, as well as if they would have to wait too long for the shuttle at the airport for the trip back. More precisely, out of the 8 persons who have to leave the city at a given full hour, one chooses other transportation (like the expensive taxi, or letting a friend drive him or her) if he or she would have to leave one hour earlier, and three, respectively six choose other transportation if they would have to leave two respectively three hours earlier. Remember that nobody having to leave four or more hours earlier would use the shuttle. The same assumptions are made for the trip from the airport to the city. One, three, or six of the 8 passengers would seek other transportation if they would have to wait one hour, two hours, or three hours.

Just to discuss a random example, assume that a=6 and b=20, the earliest passengers want to go to the airport at 6:00 and the latest at 20:00, and assume c=7 and d=21, the earliest passengers want to come from the airport at 7, and the latest at 21:00. Again the bimatrix can be computed on the Excel sheet.

6|87|98|10
6|8 83.5, 83.5 107, 107 120, 105
7|9 107, 107 83.5, 83.5 112, 92
8|10 105, 120 92, 112 77, 77

Although for the customers, the combination "6|8"-"8|10" would be best, since with this combination 225 of 240 possible passengers would be transported, this combination is not a Nash equilibrium. The player playing "8|10" would want to change to "7|9". Actually there are five Nash equilibria, the two pure ones "6|8" versus "7|9" with a payoff of 107 for both, and three mixed ones. The two pure ones have also the highest number of passengers transported (214) of all Nash equilibria, so maybe the authority should suggest this to both companies to make it a focal point. Note also that although the payoffs are equal, the equilibria themselves are not symmetric, so again pregame negotiations are necessary to talk about who will start at 6:00 and who at 7:00.

The 9 times 9 case could later be added, at the moment the chapter seems to be rich enough.


Exercises

  1. Try to get data from your local airport, how many passengers are leaving each hour, and how many are arriving at each hour. You could look through the departure and arrival schedules, or ask the local shuttle service whether they have data, or count yourself. Use the Excel sheet to analyze the game.
  2. If your airport has two different shuttles, look up their schedules and find out whether they form a Nash equilibrium, either with an assumption on the passenger distribution, or with the data obtained in the previous question.