How long will you wait for an elevator?

Question

So I am going to do my best to explain this problem, but feel free to add any variables I may have left out...

A hotel has n elevators and m floors. Each floor has a height of h.

Each elevator travels at some constant velocity v (for convenience let the units of v be ft/s). Assume the elevator does not stop until it reaches you.

Assume that the elevators are on random floors when you push the button, and the CLOSEST elevator travels downward when called from a floor below and upward when called from a floor above.

How many elevators are needed in order for the expected waiting time on the first floor to be 10 seconds?

If you can generalize the problem even more, show me how.

h has units of feet

If you think that you need an exact v, make a reasonable one up. I'm not looking for perfect numbers, I'm looking for a good method.

Answer 1

Note that I'm slightly amending my earlier answer.

If the nearest elevator is on the ith floor, then that elevator has to travel down i-1 floors to get to you.
That will take ti = (i-1)*h/v seconds

The probability that you'll have to wait ti seconds is the probability that the nearest elevator is on the ith floor and that no other elevator is nearer.
The prob that the nearest elevator is on the ith floor =
P(no elevators in 1st i-1 floors) - P(no elevators are in 1st i floors)
= P(i) = [((m+1-i)^n - (m-i)^n] / m^n

If T = waiting time
P(T = ti) = P(i) = [((m+1-i)^n - (m-i)^n] / m^n
Expected waiting time
E(T) = ∑ ti*P(T=ti)
= h/v * ∑ (i-1) * [((m+1-i)^n - (m-i)^n] / m^n
This summation goes from i=1 to m

In the example of n=4 elevators in an m=10 story building
E(T) = 1.5333 h/v

Note that this answer makes more sense than my earlier answer of 4.91 h/v. This means that the expected floor of the nearest elevator is 2.5333. Since there are 4 elevators and 10 floors, we expect something close to 2.5.
Note also that ∑P(i) = 1

Answer 2

Assuming when you say random you are implying equal probability that the elevator is on any particular floor, the elevator has a (1/m) chance of being on any particular floor, including the first. So, then I think I would set up the expected value equation:

In order of the floors
(1/m)(0) + (1/m)(h)*(s/ft) + 2(1/m)(h)*(s/ft)....+m(1/m)(h)(s/ft)

Then you could probably factor out all the matching stuff
(1/m)(h)(s/ft)* (0+1+...+m)

Summing numbers 1 to n gives you floor of (n*(n+1))/2

I hope this helps get you somewhere

Answer 3

You have almost everything needed to reasonably solve the problem. However, your expected wait time will depend upon v, the speed of the elevators.

In order to solve this problem we must know an exact value for v.