In parts 1, 3, 4, and 5 below, your answers will be algebraic expressions. Enter 'lambda' for and 'mu' for . Follow standard notation.
1. Shuttles bound for Boston depart from New York every hour on the hour (e.g., at exactly one o'clock, two o'clock, etc.). Passengers arrive at the departure gate in New York according to a Poisson process with rate per hour. What is the expected number of passengers on any given shuttle? (Assume that everyone who arrives between two successive shuttle departures boards the shuttle immediately following his/her arrival.)
2. Now, and for the remaining parts of this problem, suppose that the shuttles are not operating on a deterministic schedule. Rather, their interdeparture times are independent and exponentially distributed with common parameter per hour. Shuttle departures are independent of the process of passenger arrivals. Is the sequence of shuttle departures a Poisson process?
3. Let us say that an “event" occurs whenever a passenger arrives or a shuttle departs. What is the expected number of “events" that occur in any one-hour interval?
4. If a passenger arrives at the gate and sees people waiting (assume that is an integer), what is his/her expected waiting time until the next shuttle departs?
5. Find the PMF, pN(n) , of the number, N , of people on any given shuttle. Assume that lambda=20 and mu=2 .
For n>20 , pN(n)=
1) lambda
3) lambda + mu
4) 1/mu
Number 5 I don't have the answer
5. 2*20^n/(22^(n+1))
2 . Yes it is a poisson process !!
1) lambda
2) Yes it is a poisson process !!
3) lambda + mu
4) 1/mu
5) 2*20^n/(22^(n+1))
1. The expected number of passengers on any given shuttle is lambda/mu.
2. No, the sequence of shuttle departures is not a Poisson process because the interdeparture times are not fixed.
3. The expected number of "events" that occur in any one-hour interval is lambda + mu.
4. The expected waiting time until the next shuttle departs is 1/mu.
5. For n > 20, pN(n) = 0.
1. To find the expected number of passengers on any given shuttle, we need to calculate the mean of the Poisson distribution. The mean (or expected value) of a Poisson distribution is equal to the rate parameter. In this case, the rate parameter is λ. Therefore, the expected number of passengers on any given shuttle is λ.
2. No, the sequence of shuttle departures is not a Poisson process. For a process to be a Poisson process, the interarrival times need to be exponentially distributed, not the interdeparture times. In this case, the interdeparture times are exponentially distributed, but the arrivals of shuttles are not.
3. To find the expected number of events in any one-hour interval, we need to consider both passenger arrivals and shuttle departures. The rate of passenger arrivals is λ per hour, and the rate of shuttle departures is μ per hour. The total rate of events is the sum of these two rates. Therefore, the expected number of events in any one-hour interval is λ + μ.
4. To find the expected waiting time until the next shuttle departure for a passenger who sees m people waiting, we need to consider the rate of shuttle departures (μ) and the number of people waiting (m). The waiting time follows the exponential distribution with rate μ. Therefore, the expected waiting time is 1/μ. In this case, the expected waiting time is 1/μ hours.
5. To find the PMF (probability mass function) of the number of people on any given shuttle, we can use the M/M/1 queueing model. The PMF of the number of people on the queue (N) is given by the formula:
pN(n) = (1-ρ) * ρ^n
where ρ is the traffic intensity, given by λ/μ.
In this case, λ = 20 and μ = 2. Therefore, ρ = 20/2 = 10. For n > 20, pN(n) = 0, because the system cannot have more than 20 people.
Therefore, the PMF of the number of people on any given shuttle is:
pN(n) = (1-ρ) * ρ^n = (1-10) * 10^n = -9 * 10^n, for n <= 20
pN(n) = 0, for n > 20