The given figure is a security model at an airport facility A single queue for all random screenings departure 1 random screener 1 -p A single queue for all passengers departure Security checkpoint 4 lanes The passenger arrival to the checkpoint is a Poisson distribution function. It averages 7.5 arrivals every minute. During this time 4 lanes are open. It takes an average of 0.5 minutes to go through each lane of the security checkpoint. It is exponentially distributed When every passenger has gone through the checkpoint, he has a 10% chance(p=10% in the picture) of being selected for random screening. There is only one random security officer who does screening. Also, the selection of passengers for screening is completely random and independent of whether the random security screening officer is busy (i.e. It is very much possible that someone is selected even when the security officer is busy). The random screening time is exponential distribution. Average is 0.5 minute Assume that the checkpoint and the random screening are two independent queueing systems. We know that the arrival to the screening officer, which is a random subset of a Poisson process, is again a Poisson process, but with a smaller rate (a) What is the average time a passenger has.to wait before s/he is admitted to security checkpoint? Suppose that the airport wants to ensure that there is enough space to hold the average number lined up to enter the checkpoint, how many people should it plan for? (b) What is the average time for a passenger to go through the whole process? (c) Due to possible terrorist activities, the percentage pulled for random screenings will be doubled to p=20%. One proposal on the table is to close one of the four security checkpoint lanes and use that resource to double the number of screeners (for random screening) to two. What do you think of the idea and why?

The given figure is a security model at an airport facility A single queue for all random screenings departure 1 random screener 1 -p A single queue for all passengers departure Security checkpoint 4 lanes The passenger arrival to the checkpoint is a Poisson distribution function. It averages 7.5 arrivals every minute. During this time 4 lanes are open. It takes an average of 0.5 minutes to go through each lane of the security checkpoint. It is exponentially distributed When

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