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Starting at time t = 0, a ferry service that carries cars across a river. The ferry holds c > 1 cars and waits at the dock until the ferry is full. At that time, the ferry departs and a new ferry immediately appears and begins waiting and loading newly arriving cars. Cars arrive as a Poisson process of rate lambda = 0.2 cars per minute. At time t, let K(t) denote the number of cars on the waiting ferry.

Q1: A car arrives at time t. At time t-, the instant before the arrival, let K = K(t-) denote the number of cars on the waiting ferry. What is the conditional expected time E [T|K = k] the car will wait?

I do not have any idea as to how to solve this question. It will be helpful if someone can give me some pointers to the way to solve this question.

Q2: D(t) is the number of ferries that have departed by time t. TRUE or FALSE: D(t) is a Poisson process. Explain.

According to me it is not a Poisson process because random variables generated for this process will not be independent as when the first ferry leaves then only second ferry arrives and cars starts loading. Correct me if I am wrong.

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  • $\begingroup$ If you need some hints, then look at Queueing theory; $K(t)$ is the arrival process of your queue, $c$ determines the departure process and then it is pretty standard there to calculate expected waiting time. $\endgroup$ – Arash Oct 20 '15 at 7:57
  • $\begingroup$ Okay. Is my second answer correct? $\endgroup$ – Shripad Oct 20 '15 at 8:00
  • $\begingroup$ It is not a Poisson; but I do not understand your argument. $\endgroup$ – Arash Oct 20 '15 at 8:12
  • $\begingroup$ Can you explain why it is not a poisson? $\endgroup$ – Shripad Oct 20 '15 at 8:19
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Q2: I would say that D(t) is a mean/expected value at time t. Given Central limit theorem, D(t) comes from Normal distribution and hence is not a Poisson process.

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