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Assume the arrival in a parking facility follows a Poisson process with mean arrival of 4.2 veh/min. The processing time is exponentially distributed with a mean processing rate of 5 veh/min. Assume there is only 1 processing booth. Calculate the average length in queue, average time spent

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Answer:

Average number of vehicles in the queue is 4.4 ≅ 4 veh/min

Average waiting time of a vehicles in the queue = 1.05 veh/min

Explanation:

Step1 :- Given λ = 4.2 veh/min

and μ= 5 veh/min

a) Average number of customers in the queue

E(m) =
L_(q) = (λ^(2) )/(μ(μ-λ))


L_(q) = ((4.2)^(2) )/(5(5-4.2))

on simplification, we get

Average number of vehicles in the queue is 4.4 ≅ 4 veh/min

b) Average waiting time of a customer in the queue


L_(q) = (λ )/(μ(μ-λ))


L_(q) = (4.2 )/(5(5-4.2))

on simplification, we get

Average waiting time of a vehicles in the queue = 1.05 veh/min

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