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Average time waiting for bus
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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.
probability
asked Jul 30 at 11:34
March
134
 |Â
show 1 more comment
up vote
2
down vote
favorite
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.
probability
asked Jul 30 at 11:34
March
134
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 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 at 12:04
Buses are definitely not synced up.
– March
Jul 30 at 12:07
What does that mean? Please edit your post to explain all your assumptions.
– lulu
Jul 30 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 at 14:23
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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.
probability
asked Jul 30 at 11:34
March
134
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.
probability
asked Jul 30 at 11:34
March
134
edited Jul 30 at 12:23
asked Jul 30 at 11:34
March
134
asked Jul 30 at 11:34
March
134
asked Jul 30 at 11:34
March
134
134
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 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 at 12:04
Buses are definitely not synced up.
– March
Jul 30 at 12:07
What does that mean? Please edit your post to explain all your assumptions.
– lulu
Jul 30 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 at 14:23
 |Â
show 1 more comment
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 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 at 12:04
Buses are definitely not synced up.
– March
Jul 30 at 12:07
What does that mean? Please edit your post to explain all your assumptions.
– lulu
Jul 30 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 at 14:23
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 at 11:37
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 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 at 12:04
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 at 12:04
Buses are definitely not synced up.
– March
Jul 30 at 12:07
Buses are definitely not synced up.
– March
Jul 30 at 12:07
What does that mean? Please edit your post to explain all your assumptions.
– lulu
Jul 30 at 12:20
What does that mean? Please edit your post to explain all your assumptions.
– lulu
Jul 30 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 at 14:23
@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 at 14:23
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
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 $$mathbb E(T) = m, quad textrmwhere ; frac1m = frac1m_1+frac1m_2$$.
That is, $frac1m = (1/8+1/12) =frac524$. 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.
$$
mathbb E(T)=m_1-fracm_1^22left(frac1m_1+frac1m_2right)+fracm_1^33left(frac1m_1m_2right),
$$
thus, with $m_1 =8, m_2 = 12$, we get
$$m=8-frac642left(frac18+frac112right)+frac5123left(frac196right) = 28/9$$
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.
answered Jul 30 at 14:16
Martin Roberts
1,204318
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
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 $$mathbb E(T) = m, quad textrmwhere ; frac1m = frac1m_1+frac1m_2$$.
That is, $frac1m = (1/8+1/12) =frac524$. 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.
$$
mathbb E(T)=m_1-fracm_1^22left(frac1m_1+frac1m_2right)+fracm_1^33left(frac1m_1m_2right),
$$
thus, with $m_1 =8, m_2 = 12$, we get
$$m=8-frac642left(frac18+frac112right)+frac5123left(frac196right) = 28/9$$
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.
answered Jul 30 at 14:16
Martin Roberts
1,204318
add a comment |Â
up vote
2
down vote
accepted
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 $$mathbb E(T) = m, quad textrmwhere ; frac1m = frac1m_1+frac1m_2$$.
That is, $frac1m = (1/8+1/12) =frac524$. 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.
$$
mathbb E(T)=m_1-fracm_1^22left(frac1m_1+frac1m_2right)+fracm_1^33left(frac1m_1m_2right),
$$
thus, with $m_1 =8, m_2 = 12$, we get
$$m=8-frac642left(frac18+frac112right)+frac5123left(frac196right) = 28/9$$
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.
answered Jul 30 at 14:16
Martin Roberts
1,204318
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
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 $$mathbb E(T) = m, quad textrmwhere ; frac1m = frac1m_1+frac1m_2$$.
That is, $frac1m = (1/8+1/12) =frac524$. 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.
$$
mathbb E(T)=m_1-fracm_1^22left(frac1m_1+frac1m_2right)+fracm_1^33left(frac1m_1m_2right),
$$
thus, with $m_1 =8, m_2 = 12$, we get
$$m=8-frac642left(frac18+frac112right)+frac5123left(frac196right) = 28/9$$
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.
answered Jul 30 at 14:16
Martin Roberts
1,204318
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 $$mathbb E(T) = m, quad textrmwhere ; frac1m = frac1m_1+frac1m_2$$.
That is, $frac1m = (1/8+1/12) =frac524$. 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.
$$
mathbb E(T)=m_1-fracm_1^22left(frac1m_1+frac1m_2right)+fracm_1^33left(frac1m_1m_2right),
$$
thus, with $m_1 =8, m_2 = 12$, we get
$$m=8-frac642left(frac18+frac112right)+frac5123left(frac196right) = 28/9$$
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.
answered Jul 30 at 14:16
Martin Roberts
1,204318
answered Jul 30 at 14:16
Martin Roberts
1,204318
answered Jul 30 at 14:16
Martin Roberts
1,204318
answered Jul 30 at 14:16
Martin Roberts
1,204318
1,204318
add a comment |Â
add a comment |Â
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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 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 at 12:04
Buses are definitely not synced up.
– March
Jul 30 at 12:07
What does that mean? Please edit your post to explain all your assumptions.
– lulu
Jul 30 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 at 14:23