HW2Solutions for operation management class PDF

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Operations Management Homework 2 Solutions Customers arrive to the checkout area of New Navy such that the interarrival rate is exponential with rate equal to 100 customers per hour. At the checkout area, the service time of each customer is exponentially distributed, and on average each cashier can...

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Operations Management Homework 2 Solutions 1. Customers arrive to the checkout area of New Navy such that the interarrival rate is exponential with rate equal to 100 customers per hour. At the checkout area, the service time of each customer is exponentially distributed, and on average each cashier can serve 10 customers per hour. The store has hired 12 cashiers to checkout customers and the checkout area consists of a single pool of cashiers with one common line in front of them. a. What is the probability that a customer’s service time is less than 19 minutes? Service times are exponentially distributed with a mean of 1/µ = 6 minutes, and hence we have µ = 1/6 customer per minute. Therefore, the probability that a customer’s service time is less than 19 minutes is given by (please, refer to slide 14 of session 5): Pr {Service time < 19 min} = 1 − Pr {Service time > 19 min} = 1 − e−(1/6)×19 = 0.958. b. What is the average number of customers waiting in the queue to be checked out? This does not include the customers being checked out. There are s = 12 cashiers, an arrival rate of λ = 100 customers per hour, and a service rate of µ = 10 customers per hour. Hence, we have λ/µ = 10. By using the LQ table (available in the course website), we find the average number of customers in the queue to be LQ = 2.2465. c. What is the average service time? The average service time is given by: 1 µ

= WS − WQ = 6 min.

d. What is the average total number of customers in the checkout area? This includes the customers who are waiting to be checked out, and the customers being checked out. There is an arrival rate of λ = 100 customers per hour and a service rate of µ = 10 customers per hour. Note that the average total number of customers in the checkout area is composed of the customers who are waiting in the queue, and the customers who are being checked out (i.e., being serviced). The average number of customers waiting in the queue was computed in part b above, and is equal to LQ = 2.2465. The average number of customers in service is Lse = λ/µ = 10 (using Little’s Law, and the fact that the service time is 1/µ). Therefore, the average total number of customers in the checkout area is: LS

= LQ + Lse = 2.2465 + 10 = 12.2465.

e. [Extra Credit] What is the probability that there are 12 or more customers in the checkout area? 1

If there are 12 or more customers in the checkout area, then a customer will be delayed before entering service. In other words, the probability of delay is exactly the same as the probability that there are 12 or more customers in the checkout area. From the M/M/s Excel spreadsheet, we find the probability of delay to be 0.4494 (using λ = 100, µ = 10, and s = 12). f. Suppose now that New Navy decides to replace the single common line for its customers with 12 separate lines, one line in front of each of the 12 cashiers. What is now the average number of customers waiting to be checked out? This only includes the customers who are waiting to be checked out (and not the customers being served). If we have 12 separate lines, then each cashier will behave like an M/M/1 queue with an arrival rate of λ = 100/12 = 8.3 customers per hour. Using the formula for the M/M/1 queue, we find that the average number of customers in the queue at each cashier will be: LQ =

λ2 (8.3)2 = 4.1. = 10 × (10 − 8.3) µ(µ − λ)

Since there are 12 cashiers, the average number of customers waiting to be checked out will be 12 × 4.1 = 49.2.

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