1. What if, instead of walking up to their desired floor if the elevator doesn’t stop there, people decide to walk down from the lowest floor the elevator stops on? The MC of skipping the 2nd floor is 1 (a single person walks down one flight); the MC of skipping the 3rd floor is 3 (one person walks down two flights, one person walks down one); the MC of skipping the 4th floor is 6 (one person walks down three, one person walks down two, one person walks down one). Notice that the sequence of numbers is just the same as when we assumed walking up instead of down: 1, 3, 6, 10, .... We should also include the time cost of riding the elevator up to the higher floor, but again, the number sequence is the same. But while the sequence is identical, the significance is different: it’s easier to walk down a flight than to walk up one. Recognizing this factor will therefore tend to increase the number of floors it makes sense to skip, because the cost of skipping floors is smaller. Maybe we’ll skip the 4th floor after all.
To complicate things further, in reality some people will walk up while others will walk down, depending on how far up their floor is. Say the elevator’s first stop is the 4th floor. Someone who works on the 2nd floor is more likely to walk up than someone who works on the 3rd floor. The 2nd floor person compares one-flight-up to two-flights-down, while the 3rd floor person compares two-flights-up to one-flight-down. I'm ignoring time spent in the elevator. There’s probably a whole post to be written on just this optimization problem. But regardless of how it works out, the result has to be a MC that is less than (or possibly equal to) the simple assumptions I’ve considered so far. Why? Take the assumption that everyone will ride up and walk down. If anyone deviates from that strategy, it will be because it makes them better off, and that means a lower cost of skipping a floor. Thus, taking into account passengers' different walk-up-versus-walk-down choices will only tend to increase the optimal number of floors for the elevator to skip.
2. What if, as seems likely, the marginal disutility of walking up (or down) a flight is increasing in the number of flights? In other words, what if walking up three flights is more than twice as annoying as walking up one? If so, then the MC will rise at a faster rate, and this could result in a smaller number of floors that should be skipped. Suppose, for instance, the disutility of walking up a flight equals the number of flights you’ve walked up total. E.g., if you walk up three flights, the first flight has disutility of 1, the second flight disutility of 2, the third disutility of 3, for a total disutility of 6. This changes the calculus a bit:
skipped fl. MB (time) MC (effort)
2nd 6 1
3rd 5 4
4th 4 10
5th 3 20
6th 2 35
7th 1 56
Assuming (as in the previous post) that effort and time costs are comparable with a one-to-one trade-off, it is now optimal to skip only the 2nd floor, whereas under the simpler assumption it made sense to skip the 3rd floor as well.