About the strategy of reducing the total suffering in a queue

[extranjero]

This is my funny theory (may be I have found already known things...).Let us assume the following abstract situation. We have a special place where people can get some kind of service (for instance any bureaucratic office). There is only one service clerk who spend a fixed time (we will call it "service time") for each visitor. Everyday the office opens at known time. Let us imagine that there are only two possible visitors (A and B) per day and these visitors know this fact, but they can not speak each with other. The waiting for service causes the suffering of visitors. If one waits in queue, inside the office, the suffering equal to 1 per service time. If one waits outside the office then suffering equal to 1/2 per service time. It is not possible go out the office to reduce suffering until getting the service, so, the only way is coming later. Let us give some possible examples of visitors behavior:

• A and B came both at the opening: one of them get service at once and his(her) suffering is zero, and other visitor needs to wait in the office and his(her) suffering is 1. Total suffering is 0 + 1 = 1.
• A comes at the opening and B comes to the office later after one "service time". Suffering of A in this case is zero, suffering of B is 1/2, total suffering is 0 + 1/2 = 1/2.
• A and B decide to wait outside the office (do you remember, they can not communicate?) and after they came simultaneously one of them (for example B) has to wait inside the office. The suffering of A in this case is 1/2, the suffering of B is 1/2 + 1. The total suffering is 1/2 + 1/2 + 1 = 2.

The question is: how A and B can reduce their average total suffering?

The one way is using random generators by A and B. A and B independently use coins with the probability 1/2 to be "0" and 1/2 to be "1". They assume that "0" means go to the office at opening time and "1" meas wait outside the office "service time" and after that visit the office. Let us present a table with all possible outcomes:
$$
\begin{array}{|c|c|c|}
\hline Coin A & Coin B & T \\
\hline 0 & 0 & 1 \\
\hline 0 & 1 & 1/2 \\
\hline 1 & 0 & 1/2 \\
\hline 1 & 1 & 2 \\
\hline
\end{array}
$$
In this case, the average total suffering <T> is:
$$<T> = \frac{1}{4} 1 + \frac{1}{4} \frac{1}{2} + \frac{1}{4} \frac{1}{2} + \frac{1}{4} 2 = 1$$
However, we can decrease this value by using non equal probabilities of A and B coins. If the probability of getting "0" is p (probability of "1" is 1 - p) we can get:
$$ <T> = 2p^2 - 3p + 2 $$
By solving equation ##\frac{d<T>}{dp} = 0## we have found the optimal value of p:

p = 3/4

This value corresponds to <T> = 7/8 < 1

Discussion

This simple model is only the first step of the deep investigations in the game and reflection theory. The next generalization of this model is assuming of more than two visitors. If each of these visitors can be simulated by neural network we can try to investigate dynamics of changing the strategy of visitors according to their experience. The open question here is the relation between visitors intention to reduce their individual suffering and behavior of the total suffering.

PS. Sorry for my bad English.

 

[mfb]

An interesting question.

Can you show that this strategy is indeed optimal? I checked adding the option to arrive at t=2 and (separately) a small chance to arrive at t=0.01 and t=1.01 (to reduce the in-office waiting time if the other one arrives at t=0), both increased the average total suffering, but I don't have a mathematical proof that there is no better strategy.

 

[extranjero]

Oh, I didn't think about the possibility to split "service time". In my consideration coming time is discrete and takes values 0 (at the opening) or 1 (after one "service time" after opening).

PS. It seems, the solution in your general case must be a special random generator which outputs the time of coming (like t = 0.3624, or t = 0. 8031 ) for A nd B (they have two independent generators) with the special distribution probability function f(t). This f(t) is unknown, but, it seems, can be found by using some variatonal approach. In my example, this f(t) is rough simple two peaks function like ##f(t) = \frac{3}{4}\delta(t) + \frac{1}{4}\delta(t-1)## (Dirac delta function was used), and I used simple equation to find coefficients near delta-functions.

PS2: It seems we can write more general expression for average suffering <T>
$$ <T> = \int_0^\infty\int_0^\infty dt_1dt_2 f(t_1)f(t_2) \left\{ (\frac{1}{2}t_1 + \frac{1}{2}t_1 ) + \int_0^\infty dt [\theta(t-t_1) - \theta(t-t_1-1)][\theta(t-t_2) - \theta(t-t_2-1)] \right\}$$
with
$$ \int_0^\infty dt f(t) = 1$$