At work we have four lifts at a particular block, going from basement 2 through to floor 8. Every day I wait for a lift at ground floor level to go up to the seventh floor. Will I wait less, and if so by how much, if I take the stairs to the second floor and wait there?
The way I have formulated the problem: I want to determine the expected value of waiting time (expressed in units of "floors traversed" - I don't really care at this point whether the lift actually stops at another floor or not - so it's really a distance at this point) on the ground floor, versus expected waiting time on the second floor. I am also only considering time spent waiting for the lift to arrive - not concerned about time of arrival at the seventh floor.
So I set up the rules as follow:
At the time when I push the button to go up, each lift is on a path from a start position to end position.
For each lift I sample a pair (start position, end position) where each position is sampled from a uniform discrete distribution over the integers from 0 (basement 2) to 10 (floor 8). We need four pairs like this to represent the four lifts.
If the start position of a lift is below my floor, and the lift is moving upwards (end position greater or equal to start) then the wait time/distance is (wait position - start position).
For any other configuration the wait time/distance is |end position - start position| + |wait position - end position|.
Denote the wait distance for each lift by $d_1, \ldots, d_4$, then my waiting distance $d$ is $\min\{d_1, \ldots, d_4\}$. So $d$ is a random variable that depends on the random start and end points of each lift as well as the fixed floor position where I am waiting.
What is the expected value of $d$ at the ground floor (2) versus second floor (4)?
final note: My intuition was that one would wait less on a floor closer to the middle, BUT I ran some simulations where I took a million sets $\{(s_1,e_1), \ldots, (s_4,e_4)\}$, calculated $\{d_1, \ldots, d_4\}$ according to the rules as mentioned above and then took the minimum to get $d$ for each set...for the ground floor I got an average $d$ of 1.823865, and for the second floor 1.854256. So it seems I was wrong? It's better to just wait at ground floor level? I would love to see an exact solution or hear about any other insights you can provide...maybe there is some error in my formulation?
I haven't read your specifications carefully but have several comments too long for comments.
Are you sure the discrepancy between $1.82$ and $1.85$ is significant? What happens if you simulate the waiting times at each floor, perhaps repeatedly or with longer runs?
Your uniform distribution is not at all how real elevators behave. They usually move in the same direction as long as there is a task for them in that direction. If none, they reverse to do work, or remain where they are until called, or retreat to the ground floor. Morover, the distribution of requests is far from uniform, particularly in the morning when most people arrive at the first floor to go up.
See https://en.wikipedia.org/wiki/Elevator_algorithm and other links from a search for elevator algorithm.