I was looking at movie times today and was struck by the oddly-spaced showing times. For example, at the local Loew's Theater "Tron: Legacy 3D" (127 min.) is playing on two screens at the following interlaced times: 1:00 pm, 1:45 pm, 4:00 pm, 4:45 pm, 7:00 pm, 7:45 pm, 10:00 pm and 10:45 pm. Why not space the times equally? Is there an algorithm at work here? Other than optimizing food sales by cleverly keeping a pool of waiters, the strange times might have to do with overbooking and accommodating johnnys-come-lately.
Consider the following idealized scenario. Suppose only $1$ movie is a playing at a theater with $n$ screens, and free popcorn and refreshments is given upon sitting in the theater, so no other factors are relevant for spacing movie times but ticket sales. Suppose each showing can accommodate at most $N$ people. Suppose $N \pm M$ arrive at the kiosk reasonably before any particular showing time, where $0 < M < N$, and $0 < L < M$ people show up just a little too late for any particular show -- the same number of latecomers come by each time. If any person has to wait for more than some fraction $0 < R < 1$ of the time $t$ of the movie in question to watch the next movie in the cue, then he/she returns the ticket and goes home. Suppose the $\pm$ sign above is governed by tossing a fair coin, $+$ for heads, $-$ for tails.
Question: Given the above data, what is the optimal spacing of $X$ movie times, each movie of the same length $t$, on $n$ different screens that maximizes the total number of ticket purchases and (happy) moviegoers?
If this question is too easy, then generalize the above scenario to multiple movies showing at the same theater. If this question is too hard, then simplify it.
(Of course, feel free to edit and improve.)
(Added Thoughts) The constraints above are in place to try to model the scenario as closely as possible while keeping the mathematics simple.
I'd like to account for a little randomness, and the simplest truly non-trivial random event is the tossing of a fair coin. If $N - M$ or $N + M$ people come every time, then the problem is trivial or cumulatively impossible, respectively. What makes this problem tractable is that there are some occasions when some people are left out of a showing. These people are either at the end of a long cue or literally late; either way they must wait but few, if any, will wait longer than the length of the movie. I believe the answer of spacing depends heavily on the amount of wait time. That is, if $R = 0$, $L + M > 0$ people go home every time (not optimal). If $R = 1$, then any reasonable spacing should suffice to accommodate the extremely patient moviegoers. I think this possibility oversimplifies the problem, unless I'm missing something crucial or obvious.
I suppose also that the condition $L < M$ could be relaxed to $L < N$, but my reasoning is that latecomers seem to be rarer than overbookers. Are these constraints reasonable?