If a fun park wants to reduce ride waiting time, should the length of the rides be increased or decreased?

Question

At a recent trip to a fun park I was struck by the long waiting time for rides (eg roller coasters). So I was wondering: if you changed the duration of a ride (eg let a merry-go-round go around ten times instead of five times or let a roller coaster go around twice), would the ride queues get longer or shorter?

My first guess was that to reduce the waiting times you should reduce the ride time because then the queues would move faster. But this is only taking one ride at a time into account.
So then my second guess was that if you increased the ride time the waiting times would reduce. This is because at a (simplified) fun park people can only do one of three things: 1. be on a ride, 2. queue for a ride, 3. walk between rides. If you increase the time on a ride the time, would the time for queuing decrease?

Total time at park = Ride time + Queue time + Travel time

So if time on a ride is the only variable you change, should you increase it or decrease it to reduce ride waiting time?

Can you more explicitly describe the model? I can't see any way of thinking about the problem where longer ride times would reduce queuing times. — Jack M, Nov 14 '17 at 14:10
@JackM: I'm not sure what else to add - the formula summarises my thinking. The longer times would mean people have less time to queue (since the total time at park is fixed). This assumes travel time is the same in both scenarios (which I'm beginning to think might be a false assumption and part of the answer). — Reinstate Monica - Goodbye SE, Nov 14 '17 at 14:19
So let's take total time at park to be fixed. I decide to spend 8 hours at the park. Your equation is indeed correct, but the problem is you're using "ride time" in two different ways. The "ride time" that the park designer gets to choose is the length of the ride. The "ride time" in your equation is the amount of time an actual visitor spends on a ride. If you make the ride 8 hours long, then by your logic I should spend the whole time riding, but that's not true if I arrive just as the ride starts and end up queuing for the full 8 hours. — Jack M, Nov 14 '17 at 15:13
@JackM: where am I using "ride time" as the park designer's ride length? If the ride is 8 hours long, then the queue will be zero since no one can get on it the same day. (I was trying the same approach BTW, take it to an extreme to try to prove it one way or the other.) — Reinstate Monica - Goodbye SE, Nov 14 '17 at 15:18
You say that "ride time" refers to the designers choice of ride length in the very statement of the question - for instance in the first paragraph when you talk about changing the number of loops on a merry go round. — Jack M, Nov 14 '17 at 15:20
@JackM: right, the default time of the ride, you are correct. So how is that a problem, since I am stating in the question that this default can be changed? — Reinstate Monica - Goodbye SE, Nov 14 '17 at 15:24

score 2 · Answer 1 · 2017-11-14T14:49:48.953

2

It is unclear how the traveling time works (will people stay for several rides or travel every time ?), so I absorb it in the waiting time.

Assume there are $n$ times more people than the coasters can accomodate. At any time, there are $n-1$ batches of people travelling/waiting and one batch riding. Thus on average people will be riding $1/n^{th}$ of the time and travelling/waiting $(n-1)/n^{th}$ of the time.

Whatever the ride duration. (What changes is the number of rides, but not the total time spent on or off the coasters. If there is a fixed travel time after a ride, the pure waiting time will indeed be shorter with shorter rides.)

edited Nov 14 '17 at 14:49

answered Nov 14 '17 at 14:46

Do you mean n times more people or n more people? I think the latter. – Reinstate Monica - Goodbye SE Nov 14 '17 at 14:47
@Wikis: $n$ times more. – Nov 14 '17 at 14:50
I don't understand. $n$ is normally an integer. If the coaster can accommodate 100 people and there are 2 people queuing then $n = 0.02$. That's what you mean, right? (I don't understand your answer yet and I'm parsing it line by line.) – Reinstate Monica - Goodbye SE Nov 14 '17 at 14:54
@Wikis: $n$ is not restricted to be an integer. If there are $250$ people, $n=2.5$ and you'll be riding $40%$ of your time. – Nov 14 '17 at 14:55
I see. Of course, $n$ actually depends on the duration of the ride because it is related to the coaster accommodation. – Reinstate Monica - Goodbye SE Nov 14 '17 at 15:00
@Wikis: not at all. The accomodation is the number of seats. But you are not obliged to believe me. – Nov 14 '17 at 15:24
It's not whether or not I believe you, I'm trying to understand. I thought $n$ was a factor of both time (ride duration) and space (number of seats). Now I understand it is specifically the latter. – Reinstate Monica - Goodbye SE Nov 14 '17 at 15:27

Bram28 · Answer 2 · 2017-11-14T14:41:15.977

1

My guess would be that the queue time will probably stay pretty much the same: Queueing for 20 minutes is worth it, queueing for 40 minutes is a drag, and no one likes to queue for an hour. So, increasing the time of a particular ride will probably decrease the length of the queue in such a way that the queue time will remain pretty much what it was.

Also, I would say there are various other ways in which people spend their time, e.g eating some food, shopping at a gift shop, sitting somewhere on a bench, watch some exhibit, etc. So, if there is any time gained or lost, people will mostly change how much time is spent on those 'other' things: spending 20 minutes more in the food court is a lot more enjoyable than spending 20 minutes more queueing.

edited Nov 14 '17 at 14:41

answered Nov 14 '17 at 14:30

Bram28

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You are proposing a more complicated model which might be a good follow up question! Many of those things you mention also involve queuing. If you replace "ride" with "activity" then you get a more general model which includes the things you mention (in my experience even shopping and eating involved a lot of queuing!). The other activities (sitting on a bench) have a queue time of zero. But the customer can choose the activity time eg how much time shopping! Like I said, a good follow up! – Reinstate Monica - Goodbye SE Nov 14 '17 at 14:41

score 1 · Answer 3 · answered Nov 14 '17 at 15:06

Think of the queuing time as a cost or as the price of a ride. If the ride is shorter the benefit is presumably less so people will pay less for this and thus not be prepared to queue for so long. As they now have spare time they will distribute their time amongst the other rides in the park including the one in question by riding it twice. So the queue for this ride will get shorter.

score 0 · Answer 4 · answered Nov 14 '17 at 14:30

The waiting times would be shorter if the rides were longer.

My argument:

The time it takes to buy tickets for a ride is unchanged by the duration of the ride, thus if the rides were longer, the time it takes to buy tickets would impact the whole waiting time a lot less than if the rides were shorter. For example it takes 2 minutes to sell tickets for a ride that's 5min long as well as a ride that's 15 min long. Now this argument's validity depends on the way tickets are being sold at a park.

Now if we completely ignore this argument, reducing or increasing the duration of the rides would not impact the waiting times as the queue time is proportional to the ride duration.

Incidentally, the park in question did not charge for individual rides but just one entrance price. But if the park did charge for individual rides, isn't the buying a ticket simply part of the queuing time in our simplified model? — Reinstate Monica - Goodbye SE, Nov 14 '17 at 14:43

If a fun park wants to reduce ride waiting time, should the length of the rides be increased or decreased?

4 Answers4