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. Arrival rates have been found to follow the Poisson distribution (i.e., exponentially distributed inter-arrival times), and the service times follow the exponential distribution. The average arrival rate is 10 patients per hour, and the average service rate is 18 patients per hour. On arrival, a patient needs to stay in the waiting room until the optometrist becomes available. Once the optometrist is available, the next patient goes to the examination room for the eye exam. How many patients, on the average, will be in the

Answer 1

Using queuing theory and given an arrival rate of 10 patients per hour and a service rate of 18 patients per hour, the average number of patients in the system at any given time is calculated to be 1.25. This is based on the M/M/1 queue model.

Step-by-step explanation:

When calculating the number of patients in a system like an eye care clinic where arrivals and services both follow exponential distributions, we use queuing theory. Given the arrival rate (λ) is 10 patients per hour and the service rate (μ) is 18 patients per hour, we can apply the formula L = λ / (μ - λ) for the average number of patients in the system (L). For our data, L = 10 / (18 - 10) = 10 / 8 = 1.25. Therefore, on average, 1.25 patients will be in the system at any given time, which includes both the patients being served and those waiting. This scenario is a classic example of an M/M/1 queue, a basic model in queuing theory in which arrivals are described by a Poisson process, service times are exponentially distributed, and there is one server.