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It's been a while since I had my probability course based on Sheldon Ross' book "Probability Models", and while I never went into econometrics, I was very interested in the queuing theory section. I thought about this problem, and figured it has a simple answer, but don't recall the formulations well enough to work it:

In the United States and Europe, I observe that when people ride lift/elevator, they push the button for the direction they want to go. If an elevator stops on the floor where they're waiting, but is headed in the opposite direction, they do not embark. This is sound, as a) you want to ride the shortest path and b) the risk that the elevator that does arrive is overburdened is offset by the possibility that another elevator may arrive earlier or soon after with fewer intermediate stops to your destination. Call this the "shortest ride" approach.

However, in some African and Arab countries I've traveled to, I observed that people tend to push both the up and down arrows and embark on the first elevator that arrives, staying with it even if traveling in the wrong direction, until eventually it lands on the floor they want. Call this the "soonest on" approach.

Can we show (with a minimal number of assumptions) that, given other people ride soonest on, riding "shortest path" is on average faster?

AdamO
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    Say I want to go up, and say the elevator stops on every floor someone called it and there is always room inside. Then any elevator that currently going down, will meet me again on it's way up. So how can I lose time by not getting on it when it's going down? – Blueyedaisy Jan 30 '20 at 16:28
  • @Daphna then you understand the premise of the problem. Or at least that even a winning strategy will lose sometimes. But I want the notation and method to find the winning strategy. – AdamO Jan 30 '20 at 16:34

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Not sure I fully understand the problem, but I just think the so-called "soonest on" strategy obviously wins (or at least draws) every time.

Let's say I am currently on the 3rd floor and want to go up to 7th floor. The elevator is on the way down to 1st. At the time when the elevator reaches the 3rd floor on the way down, if I choose to enter it immediately and stay there, I can secure a slot and avoid the possibility of the elevator getting full of capacity when picking up people lined up on the 1st/2nd floor. Whereas if I choose to wait, the elevator still goes the same way, and I may lost the opportunity to get in if the elevator becomes full.

fermented_bean
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    This argument is, at best, incomplete: while you are descending the elevator, others going up may arrive, for instance. What is needed to even begin answering the quetsion is a clear statement of a model governing (a) the rates at which all people arrive; (b) their button-pushing and floor-selection strategies; (c) the numbers of floors and elevators in operation; (d) the distributions of speeds and durations of times stopped on floors; and (e) the current floor (and this list is likely incomplete). – whuber Sep 27 '20 at 13:43