Good Good Good Good Good Good, Exercises of Business Strategy

Download Good Good Good Good Good Good and more Exercises Business Strategy in PDF only on Docsity! Part 1: Please choose TRUE or FALSE for the statement. 1. One significant difference between manufacturing process and service process is that service cannot be inventoried. (True/False) Answer: True. N The seller may reduce its queue length by using reservation for the customer arrival. (True/False) Answer: True. Part 2: Queuing 1, If the arrival rate for loading trucks is 15 trucks per hour, what is the mean time between arrivals (in minutes)? Answer: 1/1 = 1/15 hour = 4 minutes. N Assume the service rate for a queue in a truck-loading operation is 14 trucks per hour. What is the average service time (in minutes)? Answer: 1/u = 1/14 hour. 3. The average number in the waiting line divided by the average arrival rate yields what measure of waiting line performance? Answer: average time spent in the waiting line. Problem 2.1 Benny the Barber owns a one-chair shop. At Barber College, they told Benny that his customers would exhibit a Poisson arrival distribution and that he would provide an exponential service distribution. His market survey data indicate that customers arrive at a rate of 2 per hour. It will take Benny an average of 20 minutes to give a haircut. Based on these figures, find the following: 1, The average number of customers waiting. Answer: = 2/our, 1/1=20 minutes = 1/3 hour, p=3/hour. So p= }/p=2/3. Using the formula for M/M/1 2 queue, we can calculate Lg = ca = 4/3. N The average time a customer waits (in minutes). 4 a 2 Answer: use Little’s Law. W, = = ; hours=40 minutes. 3. The average time a customer is in the shop (in minutes). Answer: W, = W, + 1/" = 40 + 20 = 60minutes. 4. The average utilization of Benny’s time (in %). Answer: utilization= p=2/3=66.7%. Problem 2.2 1/3 Customers enter the camera department of a store at the average rate of 6 per hour. The department is staffed by one employee, who takes an average of 6 minutes to serve each arrival. Assume this is a simple Poisson arrival exponentially distributed service time situation. 1, Asacasual observer, how many customers would you expect to see in the camera department? Answer: 4= 6/hour, 1/}1=6 minutes = 1/10 hour, =10/hour. So p= }/p=3/5. Using the formula for M/M/1 queue, we can calculate Ly = oS = 0.9, so the average number of people in the waiting line is 0.9, while the average number of people in the server is p = 0.6 , so the average number of people in the department is L, = 0.940.6=1.5. N As a casual observer, what is the probability would you expect to see three customers in the camera department? Answer: p3 = (1 — p)p*® =0.0864. 3. How long would a customer expect to spend in the camera department (total time) (in minutes)? Answer: use Little’s Law. W, = L,/A = 1.5/6=1/4 hour =15 minutes. 4. What is the utilization of the clerk (in %)? Answer: p=60%. Problem 2.3. The Bijou Theater in Hermosa Beach, California, shows vintage movies. Customers arrive at the theater line at the rate of 100 per hour. The ticket seller averages 30 seconds per customer, which includes placing validation stamps on customers’ parking lot receipts and punching their frequent watcher cards. (Because of these added services, many customers don’t get in until after the feature has started.) Assume this is a simple Poisson arrival exponentially distributed service time situation. 1 What is the average customer time in the system? Answer: Arrive rate: A = 100 /hr, Service rate: p= 120 /hr. . . . 1 A 1 1 . -. Average customer time in the systemis W, = W, +-= +-=—hr=3 minutes. s Te MuHa) 20 N What would be the effect on customer time in the system of having a second ticket taker doing nothing but validations and card punching, thereby cutting the average service time to 20 seconds? Answer: Arrive rate: A = 100 /hr, Service rate: p= 180 /hr, + . : 1 A 1 1 -. Average customer time in the system is W, = W, + n naa aa = J hour = 45 seconds. 3 Would system waiting time be less than you found in 2 if a second window was opened with each server doing all the tasks? Answer: In this sub-problem we have an M/M/2 system. Arrive rate: 2 = 100 /hr, Service rate: y= 120 /hr. Using the approximation formula (or using the table in the textbook to get an approximate number), we could 1_ 1 pv20"+) cyg+cve 1 get W, = W, += = 2.8 ae — ~ 0.01 hour= 36 seconds. nA 1p 2 120