Expected Waiting Time in a Queue with exponential distributionExponential Distribution - memorylessaverage waiting timeExpected time, exponential distributionExponential distribution - lack of memoryAverage amount of time spend by customerExample of finding the conditional expected amount of time a person spends waiting in line given that she is eventually servedGeneralized expression for waiting time.Exponentially distributed probability at the post officeExponential Probability of customersTotal waiting time of exponential distribution is less than the sum of each waiting time, how so?
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Expected Waiting Time in a Queue with exponential distribution
Exponential Distribution - memorylessaverage waiting timeExpected time, exponential distributionExponential distribution - lack of memoryAverage amount of time spend by customerExample of finding the conditional expected amount of time a person spends waiting in line given that she is eventually servedGeneralized expression for waiting time.Exponentially distributed probability at the post officeExponential Probability of customersTotal waiting time of exponential distribution is less than the sum of each waiting time, how so?
$begingroup$
I am trying to solve the following problem in my exercise sheet towards preparation of competitive exam:
A post office has 2 clerks. $A$ enters the post office while 2 other customers, $B$ and $C$, are being served by the 2 clerks. She is next in line. Assume that the time a clerk spends serving a customer has the Expo($lambda$) distribution.
What is the expected time spent by $A$ in the post office?
My take:
Total time that A spends in the office is the sum of waiting time + the time taken to serve A.
$$E(textTotal Time) = E(textWaiting Time) + E(textService Time)$$
$$E(textService Time) = frac1lambda$$
Now I computed that the Probability that A is served at any of the counter is $1/2$. Therefore,
$$E(textWaiting Time) = 1/2 times E(textService Time of B) + 1/2 times E(textService Time of C)$$
$$=frac1lambda$$
Which gives me
$$E(textTotal Time) = frac2lambda$$
But the answer given is $frac32lambda$.
Please point out the mistake in what I am thinking or if you think there's a better way to do it.
Thanks.
probability probability-theory conditional-expectation expected-value exponential-distribution
asked 4 hours ago
VizagVizag
920113
$endgroup$
add a comment |
$begingroup$
I am trying to solve the following problem in my exercise sheet towards preparation of competitive exam:
A post office has 2 clerks. $A$ enters the post office while 2 other customers, $B$ and $C$, are being served by the 2 clerks. She is next in line. Assume that the time a clerk spends serving a customer has the Expo($lambda$) distribution.
What is the expected time spent by $A$ in the post office?
My take:
Total time that A spends in the office is the sum of waiting time + the time taken to serve A.
$$E(textTotal Time) = E(textWaiting Time) + E(textService Time)$$
$$E(textService Time) = frac1lambda$$
Now I computed that the Probability that A is served at any of the counter is $1/2$. Therefore,
$$E(textWaiting Time) = 1/2 times E(textService Time of B) + 1/2 times E(textService Time of C)$$
$$=frac1lambda$$
Which gives me
$$E(textTotal Time) = frac2lambda$$
But the answer given is $frac32lambda$.
Please point out the mistake in what I am thinking or if you think there's a better way to do it.
Thanks.
probability probability-theory conditional-expectation expected-value exponential-distribution
asked 4 hours ago
VizagVizag
920113
$endgroup$
add a comment |
$begingroup$
I am trying to solve the following problem in my exercise sheet towards preparation of competitive exam:
A post office has 2 clerks. $A$ enters the post office while 2 other customers, $B$ and $C$, are being served by the 2 clerks. She is next in line. Assume that the time a clerk spends serving a customer has the Expo($lambda$) distribution.
What is the expected time spent by $A$ in the post office?
My take:
Total time that A spends in the office is the sum of waiting time + the time taken to serve A.
$$E(textTotal Time) = E(textWaiting Time) + E(textService Time)$$
$$E(textService Time) = frac1lambda$$
Now I computed that the Probability that A is served at any of the counter is $1/2$. Therefore,
$$E(textWaiting Time) = 1/2 times E(textService Time of B) + 1/2 times E(textService Time of C)$$
$$=frac1lambda$$
Which gives me
$$E(textTotal Time) = frac2lambda$$
But the answer given is $frac32lambda$.
Please point out the mistake in what I am thinking or if you think there's a better way to do it.
Thanks.
probability probability-theory conditional-expectation expected-value exponential-distribution
asked 4 hours ago
VizagVizag
920113
$endgroup$
I am trying to solve the following problem in my exercise sheet towards preparation of competitive exam:
A post office has 2 clerks. $A$ enters the post office while 2 other customers, $B$ and $C$, are being served by the 2 clerks. She is next in line. Assume that the time a clerk spends serving a customer has the Expo($lambda$) distribution.
What is the expected time spent by $A$ in the post office?
My take:
Total time that A spends in the office is the sum of waiting time + the time taken to serve A.
$$E(textTotal Time) = E(textWaiting Time) + E(textService Time)$$
$$E(textService Time) = frac1lambda$$
Now I computed that the Probability that A is served at any of the counter is $1/2$. Therefore,
$$E(textWaiting Time) = 1/2 times E(textService Time of B) + 1/2 times E(textService Time of C)$$
$$=frac1lambda$$
Which gives me
$$E(textTotal Time) = frac2lambda$$
But the answer given is $frac32lambda$.
Please point out the mistake in what I am thinking or if you think there's a better way to do it.
Thanks.
probability probability-theory conditional-expectation expected-value exponential-distribution
probability probability-theory conditional-expectation expected-value exponential-distribution
asked 4 hours ago
VizagVizag
920113
asked 4 hours ago
VizagVizag
920113
edited 4 hours ago
Vizag
asked 4 hours ago
VizagVizag
920113
asked 4 hours ago
VizagVizag
920113
asked 4 hours ago
VizagVizag
920113
920113
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The mistake is that the waiting time is not distributed in the way that you have proposed, because the waiting time is the lesser of the service times of $B$ and $C$. This is because $A$ takes the first available opening, so whoever finishes first, $A$ chooses that clerk. That is to say, if $W_A$ is the waiting time of $A$ and $S_B$, $S_C$ are the service times of $B$ and $C$ respectively, then $$W_A = min(S_B, S_C),$$ and it is not generally true that $operatornameE[W_A] = operatornameE[S_B]/2$.
Rather, you must compute the minimum order statistic of $B$ and $C$'s service times. Since these are iid exponentially distributed with rate $lambda$, it follows that $$Pr[W_A > t] = Pr[min(S_B, S_C) > t] = Pr[(S_B > t) cap (S_C > t)] oversettextind= Pr[S_B > t] Pr[S_C > t] oversettextid= (e^-lambda t)^2.$$ That is to say, the probability that $A$ waits more than $t$ to begin service is the product of the probabilities that both $B$ and $C$ take more than $t$ to be finished. And this makes perfect intuitive sense--because if $A$ must wait more than $t$, then that means both $B$ and $C$ are still occupied, which means their service times are also more than $t$.
Now that you know $Pr[W_A > t] = e^-2lambda t$, what can you say about $operatornameE[W_A]$?
answered 4 hours ago
heropupheropup
66.1k866104
$endgroup$
$begingroup$
Many thanks to you :)
$endgroup$
– Vizag
4 hours ago
1
$begingroup$
You say "it is not generally true that $operatornameE[W_A] = operatornameE[S_B]/2$". It does seem to be true here because you have two servers with exponential distributions, so twice the rate and thus half the service time.
$endgroup$
– Henry
4 hours ago
$begingroup$
@Henry Yes, but that assertion nevertheless needs justification, which the explicit calculation of the order statistic furnishes. Moreover, my statement remains correct.
$endgroup$
– heropup
1 hour ago
add a comment |
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$begingroup$
The mistake is that the waiting time is not distributed in the way that you have proposed, because the waiting time is the lesser of the service times of $B$ and $C$. This is because $A$ takes the first available opening, so whoever finishes first, $A$ chooses that clerk. That is to say, if $W_A$ is the waiting time of $A$ and $S_B$, $S_C$ are the service times of $B$ and $C$ respectively, then $$W_A = min(S_B, S_C),$$ and it is not generally true that $operatornameE[W_A] = operatornameE[S_B]/2$.
Rather, you must compute the minimum order statistic of $B$ and $C$'s service times. Since these are iid exponentially distributed with rate $lambda$, it follows that $$Pr[W_A > t] = Pr[min(S_B, S_C) > t] = Pr[(S_B > t) cap (S_C > t)] oversettextind= Pr[S_B > t] Pr[S_C > t] oversettextid= (e^-lambda t)^2.$$ That is to say, the probability that $A$ waits more than $t$ to begin service is the product of the probabilities that both $B$ and $C$ take more than $t$ to be finished. And this makes perfect intuitive sense--because if $A$ must wait more than $t$, then that means both $B$ and $C$ are still occupied, which means their service times are also more than $t$.
Now that you know $Pr[W_A > t] = e^-2lambda t$, what can you say about $operatornameE[W_A]$?
answered 4 hours ago
heropupheropup
66.1k866104
$endgroup$
$begingroup$
Many thanks to you :)
$endgroup$
– Vizag
4 hours ago
1
$begingroup$
You say "it is not generally true that $operatornameE[W_A] = operatornameE[S_B]/2$". It does seem to be true here because you have two servers with exponential distributions, so twice the rate and thus half the service time.
$endgroup$
– Henry
4 hours ago
$begingroup$
@Henry Yes, but that assertion nevertheless needs justification, which the explicit calculation of the order statistic furnishes. Moreover, my statement remains correct.
$endgroup$
– heropup
1 hour ago
add a comment |
$begingroup$
The mistake is that the waiting time is not distributed in the way that you have proposed, because the waiting time is the lesser of the service times of $B$ and $C$. This is because $A$ takes the first available opening, so whoever finishes first, $A$ chooses that clerk. That is to say, if $W_A$ is the waiting time of $A$ and $S_B$, $S_C$ are the service times of $B$ and $C$ respectively, then $$W_A = min(S_B, S_C),$$ and it is not generally true that $operatornameE[W_A] = operatornameE[S_B]/2$.
Rather, you must compute the minimum order statistic of $B$ and $C$'s service times. Since these are iid exponentially distributed with rate $lambda$, it follows that $$Pr[W_A > t] = Pr[min(S_B, S_C) > t] = Pr[(S_B > t) cap (S_C > t)] oversettextind= Pr[S_B > t] Pr[S_C > t] oversettextid= (e^-lambda t)^2.$$ That is to say, the probability that $A$ waits more than $t$ to begin service is the product of the probabilities that both $B$ and $C$ take more than $t$ to be finished. And this makes perfect intuitive sense--because if $A$ must wait more than $t$, then that means both $B$ and $C$ are still occupied, which means their service times are also more than $t$.
Now that you know $Pr[W_A > t] = e^-2lambda t$, what can you say about $operatornameE[W_A]$?
answered 4 hours ago
heropupheropup
66.1k866104
$endgroup$
$begingroup$
Many thanks to you :)
$endgroup$
– Vizag
4 hours ago
1
$begingroup$
You say "it is not generally true that $operatornameE[W_A] = operatornameE[S_B]/2$". It does seem to be true here because you have two servers with exponential distributions, so twice the rate and thus half the service time.
$endgroup$
– Henry
4 hours ago
$begingroup$
@Henry Yes, but that assertion nevertheless needs justification, which the explicit calculation of the order statistic furnishes. Moreover, my statement remains correct.
$endgroup$
– heropup
1 hour ago
add a comment |
$begingroup$
The mistake is that the waiting time is not distributed in the way that you have proposed, because the waiting time is the lesser of the service times of $B$ and $C$. This is because $A$ takes the first available opening, so whoever finishes first, $A$ chooses that clerk. That is to say, if $W_A$ is the waiting time of $A$ and $S_B$, $S_C$ are the service times of $B$ and $C$ respectively, then $$W_A = min(S_B, S_C),$$ and it is not generally true that $operatornameE[W_A] = operatornameE[S_B]/2$.
Rather, you must compute the minimum order statistic of $B$ and $C$'s service times. Since these are iid exponentially distributed with rate $lambda$, it follows that $$Pr[W_A > t] = Pr[min(S_B, S_C) > t] = Pr[(S_B > t) cap (S_C > t)] oversettextind= Pr[S_B > t] Pr[S_C > t] oversettextid= (e^-lambda t)^2.$$ That is to say, the probability that $A$ waits more than $t$ to begin service is the product of the probabilities that both $B$ and $C$ take more than $t$ to be finished. And this makes perfect intuitive sense--because if $A$ must wait more than $t$, then that means both $B$ and $C$ are still occupied, which means their service times are also more than $t$.
Now that you know $Pr[W_A > t] = e^-2lambda t$, what can you say about $operatornameE[W_A]$?
answered 4 hours ago
heropupheropup
66.1k866104
$endgroup$
The mistake is that the waiting time is not distributed in the way that you have proposed, because the waiting time is the lesser of the service times of $B$ and $C$. This is because $A$ takes the first available opening, so whoever finishes first, $A$ chooses that clerk. That is to say, if $W_A$ is the waiting time of $A$ and $S_B$, $S_C$ are the service times of $B$ and $C$ respectively, then $$W_A = min(S_B, S_C),$$ and it is not generally true that $operatornameE[W_A] = operatornameE[S_B]/2$.
Rather, you must compute the minimum order statistic of $B$ and $C$'s service times. Since these are iid exponentially distributed with rate $lambda$, it follows that $$Pr[W_A > t] = Pr[min(S_B, S_C) > t] = Pr[(S_B > t) cap (S_C > t)] oversettextind= Pr[S_B > t] Pr[S_C > t] oversettextid= (e^-lambda t)^2.$$ That is to say, the probability that $A$ waits more than $t$ to begin service is the product of the probabilities that both $B$ and $C$ take more than $t$ to be finished. And this makes perfect intuitive sense--because if $A$ must wait more than $t$, then that means both $B$ and $C$ are still occupied, which means their service times are also more than $t$.
Now that you know $Pr[W_A > t] = e^-2lambda t$, what can you say about $operatornameE[W_A]$?
answered 4 hours ago
heropupheropup
66.1k866104
answered 4 hours ago
heropupheropup
66.1k866104
answered 4 hours ago
heropupheropup
66.1k866104
answered 4 hours ago
heropupheropup
66.1k866104
66.1k866104
$begingroup$
Many thanks to you :)
$endgroup$
– Vizag
4 hours ago
1
$begingroup$
You say "it is not generally true that $operatornameE[W_A] = operatornameE[S_B]/2$". It does seem to be true here because you have two servers with exponential distributions, so twice the rate and thus half the service time.
$endgroup$
– Henry
4 hours ago
$begingroup$
@Henry Yes, but that assertion nevertheless needs justification, which the explicit calculation of the order statistic furnishes. Moreover, my statement remains correct.
$endgroup$
– heropup
1 hour ago
add a comment |
$begingroup$
Many thanks to you :)
$endgroup$
– Vizag
4 hours ago
1
$begingroup$
You say "it is not generally true that $operatornameE[W_A] = operatornameE[S_B]/2$". It does seem to be true here because you have two servers with exponential distributions, so twice the rate and thus half the service time.
$endgroup$
– Henry
4 hours ago
$begingroup$
@Henry Yes, but that assertion nevertheless needs justification, which the explicit calculation of the order statistic furnishes. Moreover, my statement remains correct.
$endgroup$
– heropup
1 hour ago
$begingroup$
Many thanks to you :)
$endgroup$
– Vizag
4 hours ago
$begingroup$
Many thanks to you :)
$endgroup$
– Vizag
4 hours ago
1
1
$begingroup$
You say "it is not generally true that $operatornameE[W_A] = operatornameE[S_B]/2$". It does seem to be true here because you have two servers with exponential distributions, so twice the rate and thus half the service time.
$endgroup$
– Henry
4 hours ago
$begingroup$
You say "it is not generally true that $operatornameE[W_A] = operatornameE[S_B]/2$". It does seem to be true here because you have two servers with exponential distributions, so twice the rate and thus half the service time.
$endgroup$
– Henry
4 hours ago
$begingroup$
@Henry Yes, but that assertion nevertheless needs justification, which the explicit calculation of the order statistic furnishes. Moreover, my statement remains correct.
$endgroup$
– heropup
1 hour ago
$begingroup$
@Henry Yes, but that assertion nevertheless needs justification, which the explicit calculation of the order statistic furnishes. Moreover, my statement remains correct.
$endgroup$
– heropup
1 hour ago
add a comment |
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