Question

Vehicles leave an airport parking facility (arrive at parking fee collection booths) at a rate of 500 veh/h (the time between arrivals is exponentially distributed). The parking facility has a policy that the average time a patron spends in a queue waiting to pay for parking is not to exceed 5 seconds. If the time required to pay for parking is exponentially distributed with a mean of 15 seconds, what is the smallest number of payment processing booths that must be open to keep the average time spent in a queue below 5 seconds?

Answer 1

Answer:

At least 3 processing booths should be open.

Step-by-step explanation:

1 vehicle spends an average of 15 seconds in a booth

In N booths , there will be N/15 vehicles in 1 second

In 1 hr, there will be N/15 * 3600 vehicles

Therefore, there will be a total of 240N vehicles/hour

Actual arrival rate of vehicles at the airport is 500 veh/h

Probability of arrival, P = $\frac{Number of possible outcomes}{Total outcomes}$

P = 500/240N

P = 2.083/N

For the average time spends to be below 5 seconds

$\frac{(\frac{P^{2} }{1-p} )}{500} = 5$, since 500 is the average value.

$\frac{P^{2} }{1-p} } = 2500\\P^{2} = 2500 (1-P)\\P^{2} = 2500 - 2500P\\(\frac{2.083}{N}) ^{2} = 2500 - 2500(\frac{2.083}{N})$

$2.083^{2} = 2500N^{2} - 5208N\\2500N^{2} - 5208N - 4.34 = 0$

$N^{2} -2.0832N - 0.0017 = 0$

Solving for N using the quadratic formula with a = 1, b = -2.0832, c = -0.0017

$N = \frac{-b \pm \sqrt{b^{2} -4ac} }{2a} \\N = \frac{2.0832 \pm \sqrt{(-2.0832)^{2} -4*1*(-0.0017)} }{2}$

N = 1.0416 ± 1.0424

N = 2.084 or -0.0008

N = 2.084 is the only realistic value here

Therefore, the smallest number of processing booths that must be open should be 3