{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw_2_solution

hw_2_solution - Homework#2 Solution Waiting Lines...

This preview shows pages 1–3. Sign up to view the full content.

Homework #2 Solution Waiting Lines Management BUAD311- Operations Management Fall 2011 1. (10 pts.) A small credit union has a small branch with two tellers. The arrival rate of customers to the branch is 12 customers per hour and the average service time is 6 minutes. Both inter-arrival time and service time follow exponential distributions (that is, the coefficient of variation is 1). a. What is the utilization rate? The average service time is given by: p = 6 minutes. The average inter-arrival time is given by: a = 1/12 hr = 5 minutes. We have two servers ( m =2), so the utilization is u = p /( am ) = 6 minutes / (5 minutes * 2) = 0.6 = 60% b. What is the average waiting time in line? Using the formula for the average waiting time in the line with CV a = CV p = 1, T q = p m ÷× u 2( m + 1) - 1 1 - u × CV a 2 + CV p 2 2 = 6 2 ÷ 0.6 2(2 + 1) - 1 1 - 0.6 = 3.58min c. What is the average number of customers in system (waiting and in service)? T = T q + p = 9.58 min I = T / a = 9.68/ 5 = 1.916 customers Alternatively, I q = T q / a = 0.716 customers; I p = m(u) (or p/a) = 1.2 customers and hence I = I q + I p = 1.916 customers 2. (25 pts.) Question 3.2 on page 81. a. What is the average time a customer has to wait for the response to his/her e- mail, ignoring any transmission time? Note: This includes the time it takes the lawyer to start writing the e-mail and the actual writing time. Since we have 10 e-mails per hour, the average inter-arrival time is given by: a = 6 minutes, and we are given that CV a = 1. From the problem, we have m = 1 (since only one lawyer is “on call”), and p = 5 minutes, with CV p = 4 minutes / 5 minutes = 0.8. This means that the utilization is u = 5/6 = 0.8333. And, the average waiting time for a customer is given by: 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
T q = p m ÷× u 2( m + 1) - 1 1 - u × CV a 2 + CV p 2 2 = 5 1 ÷× 5 6 ÷ 2(1 + 1) - 1 1 - 5 6 × 1 2 + 0.8 2 2 = 5 × 5 × 1.64 2 = 20.5min Therefore, the total response time is given by: T = T q + p = 20.5 + 5 = 25.5 min b. How many e-mails will a lawyer have received at the end of a 10-hour day?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern