Calculating the probability of finishing service first
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I have a job where the service time follows an exponential distribution with mean $1/mu$. At the same time, there are new customers coming in with Poisson rate $lambda$. As soon as they arrive, they start a job whose service time follows an exponential distribution with mean $1/mu$. I would like to know the probability that I finish the job first (i.e., there are no incoming customers, if any, finishes their job before me).
Things I tried:
(1) Conditional on my own service time, then integrate over my service time. This looks very complicated.
(2) Assume $x_k$ is the probability that when $k$ people (other than me) are currently doing their job and I am still the first to finish. The original question asks for $x_0$.
I can write the system of equations according to the exponential race:
$$
x_n=frac{mu}{mu+lambda+nmu}+frac{nmu}{mu+lambda+nmu}x_{n-1}+frac{lambda}{mu+lambda+nmu}x_{n+1}
$$
where the first term stands for the case where I finish the first, the second term stands for the probability that a customer other than me finishes the first, and the third term stands for a new arrival happens first. Then I am not sure how to solve this.
Any hints will be extremely helpful. Thank you!
Edit:
$$
x_n=frac{mu}{mu+lambda+nmu}+frac{lambda}{mu+lambda+nmu}x_{n+1}
$$
probability stochastic-processes
asked Nov 25 at 0:17
hdiuwhciu
629
add a comment |
up vote
1
down vote
favorite
I have a job where the service time follows an exponential distribution with mean $1/mu$. At the same time, there are new customers coming in with Poisson rate $lambda$. As soon as they arrive, they start a job whose service time follows an exponential distribution with mean $1/mu$. I would like to know the probability that I finish the job first (i.e., there are no incoming customers, if any, finishes their job before me).
Things I tried:
(1) Conditional on my own service time, then integrate over my service time. This looks very complicated.
(2) Assume $x_k$ is the probability that when $k$ people (other than me) are currently doing their job and I am still the first to finish. The original question asks for $x_0$.
I can write the system of equations according to the exponential race:
$$
x_n=frac{mu}{mu+lambda+nmu}+frac{nmu}{mu+lambda+nmu}x_{n-1}+frac{lambda}{mu+lambda+nmu}x_{n+1}
$$
where the first term stands for the case where I finish the first, the second term stands for the probability that a customer other than me finishes the first, and the third term stands for a new arrival happens first. Then I am not sure how to solve this.
Any hints will be extremely helpful. Thank you!
Edit:
$$
x_n=frac{mu}{mu+lambda+nmu}+frac{lambda}{mu+lambda+nmu}x_{n+1}
$$
probability stochastic-processes
asked Nov 25 at 0:17
hdiuwhciu
629
1
I think you should delete the middle term, since if one of those other $n$ people win, we are done. So you can write $x_n$ in terms of $x_{n+1}$ only. You also know $lim_{nrightarrowinfty} x_n=0$.
– Michael
Nov 25 at 0:40
@Michael That's right! Thanks!
– hdiuwhciu
Nov 25 at 0:40
1
The sequence $x_n$ is interesting but looks hard to solve. Another (direct) way to solve is to do the first: If $X$ is your own service time and $N$ is the random number of arrivals during your service, condition on $X=t$ and $N=n$, then use the fact that, given $N=n$ and $X=t$, the $n$ arrival times are independently and uniformly distributed over the interval $[0,t]$.
– Michael
Nov 25 at 0:56
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a job where the service time follows an exponential distribution with mean $1/mu$. At the same time, there are new customers coming in with Poisson rate $lambda$. As soon as they arrive, they start a job whose service time follows an exponential distribution with mean $1/mu$. I would like to know the probability that I finish the job first (i.e., there are no incoming customers, if any, finishes their job before me).
Things I tried:
(1) Conditional on my own service time, then integrate over my service time. This looks very complicated.
(2) Assume $x_k$ is the probability that when $k$ people (other than me) are currently doing their job and I am still the first to finish. The original question asks for $x_0$.
I can write the system of equations according to the exponential race:
$$
x_n=frac{mu}{mu+lambda+nmu}+frac{nmu}{mu+lambda+nmu}x_{n-1}+frac{lambda}{mu+lambda+nmu}x_{n+1}
$$
where the first term stands for the case where I finish the first, the second term stands for the probability that a customer other than me finishes the first, and the third term stands for a new arrival happens first. Then I am not sure how to solve this.
Any hints will be extremely helpful. Thank you!
Edit:
$$
x_n=frac{mu}{mu+lambda+nmu}+frac{lambda}{mu+lambda+nmu}x_{n+1}
$$
probability stochastic-processes
asked Nov 25 at 0:17
hdiuwhciu
629
I have a job where the service time follows an exponential distribution with mean $1/mu$. At the same time, there are new customers coming in with Poisson rate $lambda$. As soon as they arrive, they start a job whose service time follows an exponential distribution with mean $1/mu$. I would like to know the probability that I finish the job first (i.e., there are no incoming customers, if any, finishes their job before me).
Things I tried:
(1) Conditional on my own service time, then integrate over my service time. This looks very complicated.
(2) Assume $x_k$ is the probability that when $k$ people (other than me) are currently doing their job and I am still the first to finish. The original question asks for $x_0$.
I can write the system of equations according to the exponential race:
$$
x_n=frac{mu}{mu+lambda+nmu}+frac{nmu}{mu+lambda+nmu}x_{n-1}+frac{lambda}{mu+lambda+nmu}x_{n+1}
$$
where the first term stands for the case where I finish the first, the second term stands for the probability that a customer other than me finishes the first, and the third term stands for a new arrival happens first. Then I am not sure how to solve this.
Any hints will be extremely helpful. Thank you!
Edit:
$$
x_n=frac{mu}{mu+lambda+nmu}+frac{lambda}{mu+lambda+nmu}x_{n+1}
$$
probability stochastic-processes
probability stochastic-processes
asked Nov 25 at 0:17
hdiuwhciu
629
asked Nov 25 at 0:17
hdiuwhciu
629
edited Nov 25 at 0:44
asked Nov 25 at 0:17
hdiuwhciu
629
asked Nov 25 at 0:17
hdiuwhciu
629
asked Nov 25 at 0:17
hdiuwhciu
629
629
1
I think you should delete the middle term, since if one of those other $n$ people win, we are done. So you can write $x_n$ in terms of $x_{n+1}$ only. You also know $lim_{nrightarrowinfty} x_n=0$.
– Michael
Nov 25 at 0:40
@Michael That's right! Thanks!
– hdiuwhciu
Nov 25 at 0:40
1
The sequence $x_n$ is interesting but looks hard to solve. Another (direct) way to solve is to do the first: If $X$ is your own service time and $N$ is the random number of arrivals during your service, condition on $X=t$ and $N=n$, then use the fact that, given $N=n$ and $X=t$, the $n$ arrival times are independently and uniformly distributed over the interval $[0,t]$.
– Michael
Nov 25 at 0:56
add a comment |
1
I think you should delete the middle term, since if one of those other $n$ people win, we are done. So you can write $x_n$ in terms of $x_{n+1}$ only. You also know $lim_{nrightarrowinfty} x_n=0$.
– Michael
Nov 25 at 0:40
@Michael That's right! Thanks!
– hdiuwhciu
Nov 25 at 0:40
1
The sequence $x_n$ is interesting but looks hard to solve. Another (direct) way to solve is to do the first: If $X$ is your own service time and $N$ is the random number of arrivals during your service, condition on $X=t$ and $N=n$, then use the fact that, given $N=n$ and $X=t$, the $n$ arrival times are independently and uniformly distributed over the interval $[0,t]$.
– Michael
Nov 25 at 0:56
1
1
I think you should delete the middle term, since if one of those other $n$ people win, we are done. So you can write $x_n$ in terms of $x_{n+1}$ only. You also know $lim_{nrightarrowinfty} x_n=0$.
– Michael
Nov 25 at 0:40
I think you should delete the middle term, since if one of those other $n$ people win, we are done. So you can write $x_n$ in terms of $x_{n+1}$ only. You also know $lim_{nrightarrowinfty} x_n=0$.
– Michael
Nov 25 at 0:40
@Michael That's right! Thanks!
– hdiuwhciu
Nov 25 at 0:40
@Michael That's right! Thanks!
– hdiuwhciu
Nov 25 at 0:40
1
1
The sequence $x_n$ is interesting but looks hard to solve. Another (direct) way to solve is to do the first: If $X$ is your own service time and $N$ is the random number of arrivals during your service, condition on $X=t$ and $N=n$, then use the fact that, given $N=n$ and $X=t$, the $n$ arrival times are independently and uniformly distributed over the interval $[0,t]$.
– Michael
Nov 25 at 0:56
The sequence $x_n$ is interesting but looks hard to solve. Another (direct) way to solve is to do the first: If $X$ is your own service time and $N$ is the random number of arrivals during your service, condition on $X=t$ and $N=n$, then use the fact that, given $N=n$ and $X=t$, the $n$ arrival times are independently and uniformly distributed over the interval $[0,t]$.
– Michael
Nov 25 at 0:56
add a comment |
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1
I think you should delete the middle term, since if one of those other $n$ people win, we are done. So you can write $x_n$ in terms of $x_{n+1}$ only. You also know $lim_{nrightarrowinfty} x_n=0$.
– Michael
Nov 25 at 0:40
@Michael That's right! Thanks!
– hdiuwhciu
Nov 25 at 0:40
1
The sequence $x_n$ is interesting but looks hard to solve. Another (direct) way to solve is to do the first: If $X$ is your own service time and $N$ is the random number of arrivals during your service, condition on $X=t$ and $N=n$, then use the fact that, given $N=n$ and $X=t$, the $n$ arrival times are independently and uniformly distributed over the interval $[0,t]$.
– Michael
Nov 25 at 0:56