+1 vote
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Let’s say an average office building is 30 floors and there are 200 employees per floor. That means a total of 200 x 30 = 6,000 employees work there. Assuming that 10% are out of office at anytime and the number of visitors are negligible, we can say total number of employees in the building is 6,000 x 0.90 = 5,400.

Given people in the office usually arrive to work within a short window of time, I’d like to design the elevator in a way that employees don’t wait for longer than 30 seconds. Let’s assume that the 5,400 people arrive uniformly within 60 minutes in the morning.

Let’s say that the elevators are designed in a way that in the building are going to stop at a max of 5 floors, the speed of the elevator is 3 meters per second, each floor is 3 meters long, and it takes an elevator 15 seconds to open, unload some people in each floor and close and it takes elevator 60 second to get filled in ground floor. Let’s also assume the stops are disrtibuted unifromly. In other words, distance between the stops is 30 / 5 = 6 floors or 6 x 3 = 18 meters.

This means that each round takes the elevator time to travel to the floors + time to open and close in each floor + time to fill up the elevator in ground + time for the elevator to come back to ground floor. Let’s calculate the number:
5 stops x (18 meter / 3 meter per second) + 5 x 15 seconds + 30 seconds + (30 floors x 3 meters / 3 meters per second) = 30 + 75 + 30 + 30 = 165 seconds or around 3 minutes. In other words, one elevator can do 60 minutes / 3 minutes = 20 rides in an hour in the peak time. Let’s say each ride can consist of 20 people. This means that one elevator can help move 20 x 20 = 400 within one hour in the morning. Given total number of employees in the building is 5,400, we would need 5,400 / 400 = 13 elevators to move everyone to their floor in the morning in a way that no one has to wait for longer than 30 seconds.

+6
I like the approach @Nimesh, but I didn’t quite understand how the 30 seconds wait time came into play in your solution.

Here’s mine. Please let me know what you think.

To answer this, we need to figure out the rate at which we can move people (supply) and how many people to move (demand). We also need to consider: elevator capacity, acceptable wait time, and peak times.

Demand assumptions:
– 15 floors
– 200 people per floor
– 8-10AM peak times, so 2 hours
– 2/3 of the people arrive within the peak time

Given assumptions, we need to move
15×200 =3000 people
3000/2 = 1500 people per hour
1500*2/3 = 1000 people per hour after adjusting for not everyone coming to work in peak time.

We want to use peak time as our basis for the demand because if we can handle the load in peak time, we can be sure we can handle the load during off peak.

To figure out the supply, we need to calculate how quickly one elevator can return.

Supply assumptions:
– 2 seconds per floor
– 15 seconds per door open
– 5 door opens per trip

2 seconds x 15 floors x 2 round trip + 15 seconds x 5 door opens = 60 + 75 = 135 seconds
this is approximately 2 minutes.

So it takes 2 minutes for an elevator to be re-used.

In that time, how many people will be waiting?
1000 people/60 min = aprx 17 people per minute, so 34 people every two minutes.

If we want 30 seconds as the max wait time, then we need to move 9 people per 30 second increment (17 people per minute /2).

If each elevator can carry 10 people, then in 2 minutes we’d need 4 elevators because there are 4 30-second intervals in 2 minutes.

If we want to have wait time be 1 minute max, then we’d need to carry 17 people every minute. If the elevator can only take 10 people, we’d need two sets. In 2 minutes, there’s 2 1-minute intervals, so that requires 4 elevators.

If the elevator can take 20 people and acceptable wait time is 1 minute, then we’d only need 2 elevators!

So the levers are elevator capacity and acceptable wait time. A sanity check is in my building there’s 4 elevators. It’s 15 floors. I don’t think I’ve ever waited for longer than 1 minute. So it is reasonable estimate.