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There are two buses: first arrives to the bus top every 8 minutes, and second arrives every 12 minutes. What is the average waiting time on a bus stop if we take whatever came first? It is expected that buses arrive at regular intervals, so first arrives every 8 minutes and second arrives every 12 minutes, but it is unknown when did they started.

I was thinking in the following way: there is 1/3 probability that second bus arrives at 8-12 minutes, so it is after the first one. In this case probability is 1/3 * 4(average waiting time for the first bus). Then we have 2/3 probability that second bus came at 0-7 minutes, and basically we have two buses that arrive every 8 minutes. In this case I estimated average waiting time as 2 minutes(4 min average waiting time and 2 minutes because there are two 8 minutes buses now) and in this case answer is 2 * 2/3 + 4 * 1/3 = 8/3 = 2 + 2/3. But I'm not sure that in case of two buses there is actually 2 minutes waiting time and not some other number, and also I think that this is a "standard" problem that should have standard way of solving. Please guide me to the correct answer.

March
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  • Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. – José Carlos Santos Jul 30 '18 at 11:37
  • Is it clear that the problem is well posed? Doesn't it matter how the buses are synced up? Consider the case where each bus arrives every two minutes. If they always arrive at the same time then your expected wait is one minute. If one arrives at each even minute and the other arrives at each odd minute then your expected wait is thirty seconds. – lulu Jul 30 '18 at 12:04
  • Buses are definitely not synced up. – March Jul 30 '18 at 12:07
  • What does that mean? Please edit your post to explain all your assumptions. – lulu Jul 30 '18 at 12:20
  • @lulu In your example, could you average over the two busses being [0, 2) minutes out-of-sync with each other? And therefore do something similar with the OP's figures? – TripeHound Jul 30 '18 at 14:23
  • @TripeHound Well, sure. You could say that the schedules are set uniformly and independently...something like that. But the assumption should be explicit. – lulu Jul 30 '18 at 14:36

1 Answers1

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There is an excellent answer to the more general question, "Expected Waiting time if the are many buses, which each stop every $m_k$ minutes", can be found here.

As mentioned in the comments on this question and the linked one, the answer depends very much on the model used to describe the passage times of the buses. Here are two possible solutions, each based on slightly different assumptions, for two bus scenario, but which give quite different results.

Case 1. Fully random times.

Here the passage times of buses of type $k$ are a Poisson process of intensity $1/m_k$ and the passage times of buses of different types are independent.

Expected Time is $$\mathbb E(T) = m, \quad \textrm{where } \; \frac{1}{m} = \frac{1}{m_1}+\frac{1}{m_2}$$.

That is, $\frac{1}{m} = (1/8+1/12) =\frac{5}{24}$. So $m$ =4 minutes 48 seconds.

Case 2. Fully periodic passage times with random uniform initializations

Here, buses of type $k$ pass at times in $S_k+mkN$ where $S_k$ is uniform on $(0,m_k)$ and the random variables $(S_k)$ are independent.

$$ \mathbb E(T)=m_1-\frac{m_1^2}2\left(\frac1{m_1}+\frac1{m_2}\right)+\frac{m_1^3}{3}\left(\frac1{m_1m_2}\right), $$ thus, with $m_1 =8, m_2 = 12$, we get $$m=8-\frac{64}{2}\left(\frac{1}{8}+\frac{1}{12}\right)+\frac{512}{3}\left(\frac{1}{96}\right) = 28/9$$ So $m=$ 3 minutes 7 seconds.

Thus, we can see from the quite different results that it depends a lot on what model/assumptions you use.

Martin Roberts
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