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The continuous Elevator question

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Mr. Smith works on the 13th floor of a 15 floor building. The only elevator moves continuously through floors 1, 2, . . . , 15, 14, . . . , 2, 1, 2, . . . , except that it stops on a floor on which the button has been pressed. Assume that time spent loading and unloading passengers is very small compared to the travelling time.
Mr. Smith complains that at 5 pm, when he wants to go home, the elevator almost always goes up when it stops on his floor. What is the explanation?
Now assume that the building has n elevators, which move independently. Compute the proportion of time the first elevator on Mr. Smith’s floor moves up.
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Posted · Report post

It took a little to understand the problem, but the key point is that the elevator in essence follows a sinusoidal curve that ranges continuously from 1 to 15. Mr. Smith is trying to stop the elevator on the 13th floor, which is equivalent to finding the intersection of the curve with y=13. I don't know of an easy way to draw this out, but intuitively the only time the elevator will be going down when it reaches the 13th floor is when it is in one of 4 positions:


14th floor going up (at which it will reach the 15th floor and then turn to go back down and reach Mr. Smith
15th floor - direction is automatically downward, same as above
14th floor going down - also same as above
13th floor going down - right where Mr. Smith wants it

Meanwhile, there are 24 positions the elevator could be in when it will reach Mr. Smith going up: any time the elevator is at floor 12 or lower, or on the 13th floor going up, it will reach Mr. Smith going up. Thus, the proportion of times when the elevator is going up when it reaches Mr. Smith is 24/28, a quite large ratio.

I'm not sure at the moment how exactly to tackle the generalization, although I have the feeling it is likely the same answer as for the first part.

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Posted · Report post

good start.

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Posted (edited) · Report post

It took a little to understand the problem, but the key point is that the elevator in essence follows a sinusoidal curve that ranges continuously from 1 to 15. Mr. Smith is trying to stop the elevator on the 13th floor, which is equivalent to finding the intersection of the curve with y=13. I don't know of an easy way to draw this out, but intuitively the only time the elevator will be going down when it reaches the 13th floor is when it is in one of 4 positions:

14th floor going up (at which it will reach the 15th floor and then turn to go back down and reach Mr. Smith

15th floor - direction is automatically downward, same as above

14th floor going down - also same as above

13th floor going down - right where Mr. Smith wants it

Meanwhile, there are 24 positions the elevator could be in when it will reach Mr. Smith going up: any time the elevator is at floor 12 or lower, or on the 13th floor going up, it will reach Mr. Smith going up. Thus, the proportion of times when the elevator is going up when it reaches Mr. Smith is 24/28, a quite large ratio.

I'm not sure at the moment how exactly to tackle the generalization, although I have the feeling it is likely the same answer as for the first part.

Edited by dgreening
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Posted · Report post

A few observations

the 5 PM reference is just a red herring! the odds are the same all day

Going to n elevators operating independently will NOT be a straight forward expression. There will be some range of probabilities.

Let's start with the easiest case where n=2. Let's examine various possibilities:

[let's use the same circular notation Sairakan

Think of 15 at the top of the circle, elevators move clockwise with stops at 14 thru 2, then floor 1 is at the bottom.

continuing around with floors 2 thru 14 until you get to the top again. There are 28 positions around the circle

  • If the cars are 4+ floor apart: the cars each have a 4/28 chance of opening going down and [i think] the total likelihood of catching an elevator going down has increased to 8/28
  • If the cars are very close [less than half of a floor]. then the look like a single car [for the purposes of this exercise and the odds go back to about [slightly over] 4/28
  • If the cars are some distance apart, but less than 4 car lengths, then the odds should increase linearly from 4/28 [at 0 car length separation] to 8/28 at 4 car lengths.
  • Note: This is the right idea, but I don't think my probability calculations are quite correct. [see the case for n=7 below]

...

Thus at n=7 if all cars are equally spaced, you could have even odds of getting an upward going or a downward going elevator

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Posted · Report post

Suppose it takes 1 unit of time for an elevator to move 1 floor.

Generate a representative set of elevators as follows.

Start an elevator e1 from floor 1.

When e1 gets to floor 2 start another elevator e2 from floor 1.

When e2 gets to floor 2 start another elevator e3 from floor 1.

...

Repeat until e1 returns to floor 1.

That will take 28 units of time, and 28 elevators {ei} will comprise the group, which we call G28.

They travel in lock step, arriving (and departing) floors at the same instants of time.

At any instant when the elevators leave their floors,

two each will be leaving floors 2-14 (one up, one down) and

one each will be leaving floors 1 (up) and 15 (down.)

Of those, only four (13 up, 14 up, 15 down and 14 down) will next arrive at floor 13 going down.

We can now answer the two questions.

(1) Explain why Mr. Smith's elevator almost always hits floor 13 going up.

Mr. Smith's elevator is a randomly chosen member of G28.

With probability 24/28 = 0.857 his floor-13 elevator will be upward bound.

(2) With n elevators, what is the proportion of time a floor-13 elevator will be upward bound?

To make n arbitrarily large, we simply make many copies of G28. They need not all be in sync.

The result is the same.

The the proportion of time (instances) that Mr. Smith's next floor-13 elevator is upward bound is still 24/28.

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Posted · Report post

The reason he's more likely to catch an upgoing elevator instead of a downgoing elevator is because there are 4 positions an elevator could be in that would lead to him catching it while it's going down, and 24 positions where he would catch it while it'd going up, as has been pointed out. A key thing to notice: for most of those 24 positions where he catches an upgoing elevator, Mr Smith will be waiting quite a while as the elevator moves through several floors to reach him, while in the 4 positions where he catches a downgoing elevator he will only be waiting for a very short time.

If there are a large number of elevators in play, then the odds are slim that any elevator more than a couple of floors away will be the one that Mr Smith catches. As you reach a ridiculously large number of elevators and he will probably catch an elevator within a millisecond, the probability that an elevator is one millisecond from reaching the 13th floor heading up is equal to the probability that it's one millisecond from reaching the 13th floor heading down. So if Mr Smith wants to build enough elevators, he will have a 50/50 chance of heading down in the evening.

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Posted (edited) · Report post

The reason he's more likely to catch an upgoing elevator instead of a downgoing elevator is because there are 4 positions an elevator could be in that would lead to him catching it while it's going down, and 24 positions where he would catch it while it'd going up, as has been pointed out. A key thing to notice: for most of those 24 positions where he catches an upgoing elevator, Mr Smith will be waiting quite a while as the elevator moves through several floors to reach him, while in the 4 positions where he catches a downgoing elevator he will only be waiting for a very short time.

If there are a large number of elevators in play, then the odds are slim that any elevator more than a couple of floors away will be the one that Mr Smith catches. As you reach a ridiculously large number of elevators and he will probably catch an elevator within a millisecond, the probability that an elevator is one millisecond from reaching the 13th floor heading up is equal to the probability that it's one millisecond from reaching the 13th floor heading down. So if Mr Smith wants to build enough elevators, he will have a 50/50 chance of heading down in the evening.

Nice. If there are an infinite number of 12 ups and 14 downs, none of the others matter.

Edit:

Actually, if there are simply 28 elevators, the ones mentioned in my previous post, every floor will be served, in any desired direction, with a wait time no longer than the time it takes an elevator to go one floor.

That is the n-elevator solution I think.

Edited by bonanova
Include the n-elevator solution
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