[Guest]

On the route of x, busses are departing at exactly 10 minute intervals and run with constant speed. It's easy to deduce that people arriving to a stop have to wait for a bus for averagely 5 minutes.

Question: If the busses are departing and arriving to the stop irregularly, I mean, almost randomly. Averagely how many minutes will people have to wait for a bus? (The departure counts of busses still the same as the first case.)

I know the answer but didn't work on it, and don't know the solution way. I think some integral will be needed???

[Guest]

I'd say it's impossible to tell. There are too many unknowns:

1. It's unknown at which rate people arrive at the stop (it's unknown how many people know when the bus is departing and anticipate on that).

2. It's unknown how many people try to take the bus (you might need more than one bus to transport all the people or Mr and Mrs Jones are the only people taking the bus and they wait for an hour).

3. It's unknown how impatient people are and how quickly they decide to travel using an alternative (they might walk to their destination or take a taxi).

4. It's unknown whether you count people having a beer in the pub next to the stop while waiting for the bus as people waiting for the bus.

[Guest]

I'd say it's impossible to tell. There are too many unknowns:

1. It's unknown at which rate people arrive at the stop (it's unknown how many people know when the bus is departing and anticipate on that).

2. It's unknown how many people try to take the bus (you might need more than one bus to transport all the people or Mr and Mrs Jones are the only people taking the bus and they wait for an hour).

3. It's unknown how impatient people are and how quickly they decide to travel using an alternative (they might walk to their destination or take a taxi).

4. It's unknown whether you count people having a beer in the pub next to the stop while waiting for the bus as people waiting for the bus.

This is a math question, not a real life question. In real life the next bus might come a few minutes late or before. In this question you must assume that it will come maybe in seconds or hours.

Getting the average means you assume that infinite passengers may come to stop.

[Guest]

This is a math question, not a real life question. In real life the next bus might come a few minutes late or before. In this question you must assume that it will come maybe in seconds or hours.

Getting the average means you assume that infinite passengers may come to stop.

The best I can answer is that there's a variable amount of people waiting for a variable amount of time, which makes a total amount of x waiting time. This time should be divided by a y amount of busses that transports them. Assuming the same amount busses are in service, I suspect the average amount of waiting time is the same, thus 5 minutes.

However, I have no idea how to calculate this when there's an irregular stream of passengers and how to deal with the fact that busses might depart without passengers on board.

[Guest]

On the route of x, busses are departing at exactly 10 minute intervals and run with constant speed. It's easy to deduce that people arriving to a stop have to wait for a bus for averagely 5 minutes.

Question: If the busses are departing and arriving to the stop irregularly, I mean, almost randomly. Averagely how many minutes will people have to wait for a bus? (The departure counts of busses still the same as the first case.)

I know the answer but didn't work on it, and don't know the solution way. I think some integral will be needed???

So you're trying to solve this problem, mathematically supported without any practical variables? A realm of exact figures ?

I'm lost, bus spots are generally influence by human behavior ( obviously) therfore you'll need to take a statistical study.

But as you said, without anything to go on it is 50%. 5 minutes would be the mean. 20% will show up 0-2.5, 60% 2.5-7.5 , 20% 7.5-10

2 x 2.5

6 x 5

2x 7.5

n=10

Mean : 5

If Im wrong check out my screen name

[Prime]

I assume, the bus route makes a full circle (starts and ends at the same point).

Put enough buses on a circular route, so thtat they are at 10 min intervals.

Watch a bus leave from your favorite bus stop. Wait there until that bus makes a full circle and comes to that stop again. Since bus speed is constant, the wait time will be the same and you will count the same number of buses in the interim, no matter at what intervals they are arranged over the route. Thus the same average interval between the busses -- 10 minutes.

Whether it is the same average wait time for the people -- cannot say. Need additional data/clarification to solve that. How many people? At what times do they come to bus stop? Do you count waiting time for each person individually, or for the entire crowd at the bus stop? Does the problem refer to a single bus stop or all bus stops collectively?

As well as the proximity of the pub and all other good variables already noted by MarkTheGamer.

To make an illustration: imagine all buses are stacked one on top of the other, and it takes a bus 2 hours to make a full circle. Then, if you come to a bus stop at random time, your average wait is about 1 hour.

Either way, I see no need for integrals. The "qudrangle" problem is another matter...

[Guest]

I see that non of you admit that this is a soluble question. A few days ago Prof.Templeten referenced a math book here. Although it was hard for me, I tried to examine it, and found an interesting concept at 29. page of This old math book :

Suppose buses on a given route average 10-minute

intervals. If they run at exactly 10-minute intervals the

average time a passenger arriving randomly at a stop will

have to wait is 5 minutes. If the buses are irregular the

average time is greater ; if they are completely random

(Gaussian distribution) it is 10 minutes, and, what is more,

the average time since the previous bus is also 10 minutes.

If I hadn't read this on a math book, I couldn't challenge to ask it here. Maybe I misunderstood the concept or didn't word here thoroughly. Then, sorry; but else there still stands a question to solve its logic.

[Guest]

Well, I can see why this average is 5min when the busses are regularly timed.

People can arrive when a bus just left, when a bus just stopped (and they do not have to wait) or any time between these. This way, the average is the maximum (10min wait) plus the minimum (0min wait) divided by 2.

Now, if you need to consider a Gaussian Distribution, do not rely on integrals, because they're not integrable. Instead, you should use a Gaussian Cumulative Density Function Table.

[Guest]

It depends if a person waits exactly when the first bus leaves, he would have to w8 10 mins and if a person w8s exactly when the bus arrives, he wouldn't have to w8 at all.

The average should be 5.

[Guest]

It depends if a person waits exactly when the first bus leaves, he would have to w8 10 mins and if a person w8s exactly when the bus arrives, he wouldn't have to w8 at all.

The average should be 5.

Your and skywalkers answers are true for regularly departing busses. But the question is for busses that depart randomly, as Gaussian distribution.

I posted the answer

10 minutes

as written in a math book. And now I ask how did they calculated this???

[bonanova]

I see that non of you admit that this is a soluble question. A few days ago Prof.Templeten referenced a math book here. Although it was hard for me, I tried to examine it, and found an interesting concept at 29. page of This old math book :

Suppose buses on a given route average 10-minute

intervals. If they run at exactly 10-minute intervals the

average time a passenger arriving randomly at a stop will

have to wait is 5 minutes. If the buses are irregular the

average time is greater ; if they are completely random

(Gaussian distribution) it is 10 minutes, and, what is more,

the average time since the previous bus is also 10 minutes.

If I hadn't read this on a math book, I couldn't challenge to ask it here. Maybe I misunderstood the concept or didn't word here thoroughly. Then, sorry; but else there still stands a question to solve its logic.

Thank you -- for reminding me where I saw this problem.

I looked for it several other places [and of course did not find it.]

I read the book, after PT made reference to it and found most of it

interesting, especially the terse and clever writing style.

I took the "10-minute" answer as more or less obvious, it had a

certain conservation to it - 5 is average, so 0 and 10 must be

the extremes - but I didn't look [yet] at a proof. I might.

At least I can end my search for the original source.

[HoustonHokie]

It is an interesting problem. I wonder how the distribution of passengers' arrivals is assumed to be - regular or random? Does it matter?