Assignment 3 How Long to Wait

Due Friday February 12

Two neo-luddites are trying to meet up with each other at 0 (zero) o’clock.  They have no watches or cellphones or any way of communicating or knowing what time it is once they leave the sundials at their house, so when one of them shows up at the agreed upon meeting spot they don’t know whether they are the first to arrive and should wait, or if they are the second to arrive and that their friend arrived earlier and gave up on waiting for them.  The strategy that each neo-luddite must choose is how long to wait, before giving up on their friend and leaving.

(ah ha you say how can they know how long they’ve waited for if they don’t have watches or clocks or whatever. Well having been a waiting neo-luddite myself developed a consistent pattern of aggitation based on time spent waiting so a neo-luddite can infer from their annoyance levels quite accurately how long they’ve waited for… flimsy I know but it keeps the problem interesting, otherwise the answer is “awhile”)

There are three versions of this problem, that you need to solve.  In all three we assume:

1. The neo-luddites get crankeir the longer they wait according to the function n*(exp(2(t-2))/(1+exp(2(t-2)))), where t is the amount of time they’ve waited.  So they start off being just a little bit annoyed, n*(0.017986…) annoyed to be precise, and then get more annoyed as time passes.  You can see the pretty shape of this curve by plugging it in here.

2. The neo-luddites enjoy meeting eachother and get satisfaction m from meeting.  And we assume m < (n/2) otherwise the problem is boring.

In each problem you are trying to determine the Nash equilibrium waiting times for each neo-luddite.  To do this I suggest you determine the expected payoff function for each neo-luddite (which is their joy at having met their friend, m, if they do meet, less their annoyance at having waited which is a function of how long they’ve waited, n*(exp(2(t-2))/(1+exp(2(t-2)))). This expected payoff function will be a payoff of the waiting times of both neo-luddites.  Once you have the payoff functions you should find the best response functions.  Once you have the best response functions you need to find where they intersect because that is what a Nash Equilibrium is… an intersection of best responses.

The thing that changes from problem to problem is the arrival distribution of one of the neo-luddites.

Problem 1. Both neo-luddites arrive according to the following distribution:

• 0 for t < -2
• 1/2 + t/4 for -2 < t < 0
• 1/2 – t/4 for 0 < t < 2
• 0 for 2 < t

(Note that this problem is symmetric and that helps alot)

Problem 2. One of the neo-luddites is habitually late, so one of the neo-luddites arrives according to the distribution outlined in Problem 1 and the habitually late neo-luddite arrives according to this distribution.

• 0 for t < -2
• 1/2 + (t-1)/6 for -2 < t < 1
• 1/2  – (t-1)/2 for 1 < t < 2
• 0 for 2 < t

Problem 3. One of the neo-luddites is just a little less precise in their arrival so while one of the  arrives according to the distribution outlined in Problem 1 the less precise neo-luddite arrives according to this distribution.

• 0 for t < -3
• 1/3 + t/9 for -3 < t < 0
• 1/3  – t/2 for 0 < t < 3
• 0 for 3 < t

Good luck,

I’ll be going over a simpler version of this problem in class, as well as posting a simple version of the problem up here soon.