In: Operations Management
Customers enter the camera department of a store at the average rate of eight per hour. The department is staffed by one employee, who takes an average of 3.0 minutes to serve each arrival. Assume this is a simple Poisson arrival, exponentially distributed service time situation. Use Exhibit 10.9.
a-1. As a casual observer, how many people would you expect to see in the camera department (excluding the clerk)? (Round your answer to 2 decimal places.)
a-2. How long would a customer expect to spend in the camera department (total time)? (Round your answer to 1 decimal place.)
c. What is the probability that there are more than two people in the camera department (excluding the clerk)? (Round your intermediate calculations to 3 decimal places and final answer to 1 decimal place.)
d. Another clerk has been hired for the camera department who also takes an average of 3 minutes to serve each arrival. How long would a customer expect to spend in the department now? (Round your answer to 1 decimal place. Use the closest value of λ/μ in Exhibit 10.9 when determining Lq (i.e., do not interpolate).)
Given,
= Mean number of arrivals per time period = 8 arrivals per hour
= Mean number of people served per time period = 3 minutes per customer = 20 customers per hour
a.
Average number of customers in the system = = 8 / 20 - 8 = 8 / 12 = 2/3 = 0.67
b.
Average time a customer spends in the system = Waiting Time + Service time = = 1 / (20 - 8)
= 1/12 hours = 5 minutes
c.
Probability that there are more than k customers in the system =
Probability that there are more than 2 customers in the system = = 0.064
d.
Probability that there are no students in the system : (In the below image)
Hence, average time that a customer spends in the system = 3.126 minutes