# Mythbusters - Determine optimal boarding strategy based on time and satisfaction score

Most airlines board passengers starting from the back of the plane and then working their way towards the front (after boarding priority classes and passengers).

In an episode of Mythbusters, Adam and Jamie tested the myth that the boarding strategy favoured by most airlines, back to front, is the least efficient.

The myth was confirmed, and these were the results: The random no seats strategy is the fastest, followed by the WILMA straight strategy. However, random no seats strategy gives the lowest satisfaction scores.

The highest satisfaction score is given by the reverse pyramid strategy even though it is the fourth fastest.

How might one determine the optimal boarding strategy based solely on times and satisfaction scores given (not including advanced stuff like computing expected aisle or seat interferences)?

I can't seem to think of any kind of unit conversion except to convert the time to seconds and then multiply it with satisfaction score so it's like we're trying to maximise the product of time and satisfaction score:

$$f(t,s) = ts$$

What are some of the advantages or disadvantages of doing this?

One disadvantage seems to be that ranking by product of time and satisfaction score gives the same ranking by satisfaction score.

What else could be done? All that seem to come to mind are products so perhaps I might maximise anything like these:

$$f(t,s) = t^2s$$

$$f(t,s) = ts^{1/2} \text{(eliminating random no seats)}$$

$$f(t,s) = t(s-s_{ave})$$

I'm thinking we're going to have to relate time and satisfaction score to some unit such as money. So, one would have to find some relationship (for example, a linear relationship through linear regression) between boarding time and cost and then another between satisfaction score for boarding today and revenue from flight next month?

Does it have to be something like that?

I was suggested z-scores or something so I tried standardising, I think: Why did the sum of squares of z's turn out to be 6? Did I do something wrong? Is that the fourth moment or something?

• The first step would be to precisely define "optimal". Usually this would take the form of minimizing or maximizing some quantity under some constraints. This will give direction to your problem, which is lacking at the moment. Specifically, why would an optimal solution maximize t*s? This would mean that when two strategies provide equal amounts of satisfaction, the one that costs more time is preferable.
– user7935
Aug 23, 2016 at 13:25
• If this is truly for applied (real-life) purposes, then it is important to realize that there is probably no practical difference between 14:07 and 15:10. (Besides, if the mythbusters exercise were carried out scientifically with multiple repetitions, these numbers would probably average out to be about the same.) So, there are probably only 3 different times: 14:07 to 15:10 as one time; 17:15 and 24:29. Likewise, in real-life application, there are only 3 different satisfaction scores: -5; 12-19; and 102-113. Any applied model should take such a perspective if it hopes to be really useful. May 14, 2018 at 21:12

I would start with your generic $$f(t,s) = t \times s$$ and, instead of add weights or factors to this, I would add other variables related to time and satisfaction like:

• boarding time as function of luggage management ($TL$) times baggage number ($B$), and time to go to seat row ($T_g$) vs time to actually sit ($T_s$) for the different seat types (Window, Mid, Aisle) per number of passengers ($N$)
• satisfaction as function of easiness of boarding ($E$) (think about the WilMA and families with little kids), functionality of luggage positioning ($F_b$), amount of distraction required ($D$) (e.g. boarding while listen to loud music or boarding while talking at the phone distract from the boarding and leads to errors).

A proposal can be $$f(TL, Tg, Ts, W, M, A, N) = (TL\times B)+(TG\times N + \frac{N}{3}W + \frac{N}{3}M + \frac{N}{3}A)$$ $$f(E, Fb, D) = E \times Fb \times \frac{1}{D}$$

$$\mbox{boarding strategy score}=f(TL, Tg, Ts, W, M, A, N) \times f(E, Fb, D)$$

and I would start assigning weights while conducting some more simulations (I understood that Mythbusters example refers to single trials only for each strategy).

In my opinion, advantages / disadvantages are not coming from equations itself but from the methodology. Without more robust experimental data, all equations above, and even more factors, are arguable and contestable.

I also would not add "money" in the model, but rather added value for the airline vs added value for the passenger, and things will easily escalate: you may find that keep people stuffed in tunnels and in line waiting to enter the aircraft, or waiting in airports for delays or flight cancellations, can increase exposition time for advertisement boards, therefore potential revenue for airport services, therefore ...utility functions of delays.