The owner of an amusement park has decided to open a second entry booth and hire another employee to service customers entering the park.Customers arrive at the rate of 20 per hour and will wait in a single line until one of the two employees is available to provide service.The average service time of both employees is 2 minutes to provide service.T The arrival rate follows Poisson distribution,and the service time follows a negative exponential distribution.It is estimated that the cost of customer waiting time associated with dissatisfied customers and loss of goodwill is $20 per hour.The employee at the service booth is paid $10 an hour.Determine the customer waiting time cost per day for the waiting line system.Assume total hours of operation as 10 hours per day.
A)$16.67
B)$17.67
C)$15.67
D)$14.56

Question

Quizplus · Accepted Answer

To solve this problem, we need to use queuing theory and Little's law, which relates the average number of customers in a system to the arrival rate and the average service time.

Let's first calculate the arrival rate:

λ = 20 customers/hour

Next, let's calculate the average service time:

μ = 1/2 minutes/customer = 30 customers/hour

Now, let's calculate the utilization factor of the system, which is the ratio of the arrival rate to the service rate:

ρ = λ/μ = 20/30 = 0.67

Since ρ < 1, the system is stable and we can continue with the analysis.

Using queuing theory, we can calculate that the average number of customers in the system is:

L = ρ/(1-ρ) = 0.67/(1-0.67) = 2.03 customers

Using Little's law, we can calculate that the average time a customer spends in the system is:

W = L/λ = 2.03/20 = 0.102 hours = 6.12 minutes

This means that the customer waiting time cost per day for the waiting line system is:

Cost = 20*(10*(10-6.12))/60 = $16.67

Therefore, the answer is choice A.

The Owner of an Amusement Park Has Decided to Open