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Math Help - Poisson Process - Number of cars that a petrol station can service

  1. #16
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    Re: Poisson Process - Number of cars that a petrol station can service

    Thank you romsek, your assistance is so appreciated. I will try do this question by hand, because apparently it does not need any kind of computer software package and the answer should be able to be derived through algebra. I am trying to think of any "tricks" or short cuts...
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  2. #17
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    Re: Poisson Process - Number of cars that a petrol station can service

    Quote Originally Posted by usagi_killer View Post
    Thank you romsek, your assistance is so appreciated. I will try do this question by hand, because apparently it does not need any kind of computer software package and the answer should be able to be derived through algebra. I am trying to think of any "tricks" or short cuts...
    One other thing I noticed. Is that as the ratio of the initial amount of gas the station has to the max amount each car can request (in this problem it's 2) gets large the average number of cars serviced tends towards $\dfrac {GS_{init}}{Req_{max}}$ as you'd expect. I'm a bit surprised that the average number of cars wasn't just 100/25=4 myself but that's the nature of the uniform distribution.

    Usually you'd be able to model the sum of uniforms as a Normal rv but the numbers are so low in this problem it's not a very good fit.

    Good luck. I hope all this has at least shed some light on the problem.
    Thanks from usagi_killer
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  3. #18
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    Re: Poisson Process - Number of cars that a petrol station can service

    Thank you romsek. You have definitely gave me lots to work on, I think I may have solved part (a) algebraically by modelling it as a renewal process. I just want to ask about part (b) quickly. Assume for now that $E[N] \approx 3.67$ (remember that $N$ is defined to be the number of cars that the petrol station FULLY service), and let $T$ denote the time until the station runs out of petrol. Since cars arrive according to a Poisson process, then the expected time between two arrivals is $\lambda^{-1}$, now is $E[T] = \lambda^{-1}E[N]$ or $E[T] = \lambda^{-1}(E[N]+1)$? I add the $+1$ in the latter expression because the station runs out of gas "one car later", if you understand what I mean. However, I am unsure because the question says "...before the station runs out of petrol (and before any refilling occurs)?" That is, since we have defined $N$ to be the number of cars that are FULLY serviced, then any car that arrives after the $N$-th car cannot be fully served, thus, the process ends at the $N$-th car, so $E[T] = \lambda^{-1}E[N]$.

    I am just unsure which interpretation to follow and just want to see what you think would be the more appropriate one.

    Thanks again!
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