Question

A university cafeteria line in the student center is a self-service facility in which students select the food items they want and then form a single line to pay the cashier. Students arrive at a rate of four per minute according to a Poisson distribution. The single cashier ringing up sales takes 12 seconds per customer on average, following an exponential distribution.

a. What is the probability that more than six students arrive in the line in the next minute?

b. What percentage of sales take more than 15 seconds to ring up?

c. Which of the models that we have studied in class is most appropriate for analyzing the cafeteria line?

Use the model you identified in part (c) to answer the following questions:

d. What is the average number of students in the line for the cashier?

e. On average how long will a student have to wait in line before reaching the cashier?

f. What is the probability that there are no students in line?

g. What is the probability that there are more than two students in line?

Answer 1

a) Probability that more than six students arrive in the line in the next minute = 1 - F(6,4) = 1 - $\sum_{k=0}^{6}\frac{e^{-4 }4^{k}}{k!}$ = 0.11067

This can also be calculated by Excel formula = 1 - POISSON(6,4,1)

b) Percentage of sales take more than 15 seconds to ring up = $\int \lambda e^{-\lambda x} dx$ , where λ = 1/12 = 0.0833 per sec and k = 1/15 = 0.067

Percentage of sales take more than 15 seconds to ring up = 0.00554

c) Accroding the characteristics of the queue system, M/M/1 model is the most appropriate for this with following parameters

Arrival rate, λ = 4 per minute

Sevice rate, μ = 60/12 = 5 per minute

d) Average number of students in line, Lq = λ²/(μ*(μ-λ)) = 4²/(5*(5-4)) = 3.2

e) Average waiting time in line before reaching the cashier, Wq = Lq/λ = 3.2/4 = 0.8 minute or 48 seconds

f) Probability that there are no students in line, P0 = 1-λ/μ = 1/4/5 = 0.20

g) Utilization, ρ = λ/μ = 4/5 = 0.80

Probability that there n students in line, P(n) =(1-ρ)*ρⁿ

Probability that there are more than two dstudents in line = 1-(P(0)+P(1)+P(2)) = 1-((1-0.8)*0.8⁰+(1-0.8)*0.8¹+(1-0.8)*0.8²) = 0.512