Question

Patients arrive randomly at an eye care clinic for eye exam. Suppose that there is only one optometrist. The time required for the exam varies from patient to patient. Arrivals have been found to follow the Poisson process (i.e., exponentially distributed inter-arrival times), and the service times follow the exponential distribution. The average arrival rate is 12 patients per hour, and the average service rate is 20 patients per hour. A patient wait in a waiting room until the optometrist is ready to see them. How many patients, on the average, will be in the waiting room?

A. 3
B. 0.9
C. 36

Answer 1

Using the M/M/1 queue system calculations, the average number of patients in the waiting room at an eye care clinic, where arrivals and service times are both exponentially distributed, is found to be 0.5, which does not match any of the provided options (A, B, C).

Step-by-step explanation:

To calculate the average number of patients in the waiting room, we can use the properties of the M/M/1 queue system (a queue with a single server where arrivals are determined by a Poisson process and service times are exponentially distributed). The formula to calculate the average number of customers in the system (L) is given by:

L = λ / ( μ - λ ),
where λ is the average arrival rate and μ is the average service rate. In this scenario, λ = 12 patients per hour and μ = 20 patients per hour. Plugging these values into the formula we get:

L = 12 / (20 - 12) = 12 / 8 = 1.5.
However, this number represents the total number of patients in the system. Since there will always be one patient being served, we need to subtract one to find the number of patients in the waiting room:

Lqueue = L - 1 = 1.5 - 1 = 0.5.
Therefore, on average, there will be 0.5 patients in the waiting room, which is not one of the provided options. It seems there might be a misunderstanding in the options provided or we might need to review the full context of the given scenario for potential additional information.