Question

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

Answer 1

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

Step-by-step explanation:

Step1 :- Given λ = 4.2 veh/min

and μ= 5 veh/min

a) Average number of customers in the queue

E(m) = $L_{q} = \frac{λ^{2} }{μ(μ-λ)}$

$L_{q} = \frac{(4.2)^{2} }{5(5-4.2)}$

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

$L_{q} = \frac{λ }{μ(μ-λ)}$

$L_{q} = \frac{4.2 }{5(5-4.2)}$

on simplification, we get

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