OIS FInal Exam Practice
-Q9. What is the average time a customer has to wait to rent an SUV? Using the wait time formula, Tq = (72/50) * (0.6^(sqrt(2*(50+1))-1) / (1-0.6) )*((1^2 + 0.33^2)/2) = 0.019 hours or 1.15 minutes.
-Q10. The company discovers that if it reduces its daily $80 rental price by $25, the average demand would increase to 12 rentals per day and the average rental duration will become 4 days. Should they go for it? If the average demand is increased to 12 rentals per day, then a = 2 hours. If the average rental duration increases to 4 days, then p = 96 hours. These values raise the utilization rate to p/(a x m) = 96/(2 x 50)= 96%. This means that m x u = 50 x 0.96 = 48 cars are rented on average. The initial average revenue per day = $80/day * 30 cars = $2400. New average revenue per day = $55*48 = $2640. Therefore, the company should make the proposed changes.
-Q11. How would the waiting time change if the company decides to limit all SUV rentals to exactly 4 days? Assume that if such a restriction is imposed, the average interarrival time will increase to 3 hours as will the std deviation. Stdev_p = 0 p = 4 days = 96 hours a = 3 hours stdev_a = 3 hours U = p/(m*a)=96/(50*3)=0.64 Using the wait time formula the average wait time is computed as Tq = (96/50)*(0.64^(sqrt(2*(50+1))-1) / (1-0.64) )*((1^2+0^2)/2) = 0.046 hours.
-Q12. With the above setting in Q11, assume the company provides snacks and sodas to customers waiting to rent an SUV in the waiting area. The snacks and sodas are known to cost $5.00 per hour of a customer wait. Then what is the average cost of snacks per day? Average number of customers waiting in line, I_q = R x Tq = 1/3 [cust/hour] x 0.046 [hour] = 0.01533 customers Daily cost = Iq x $5/hour x 24 hour/day = $1.84/day
Q4. Determine the annual ordering and holding cost for current order quantity. -We are given the following information: Annual demand: R = 1200 batteries per year Price: P = $28 per battery Holding cost (Annual): h =0.30*28 = $8.40 per battery per year Ordering cost: K = $20 per order Current order quantity: Q= 100 batteries -The current ordering and holding costs are: C(100) = (R/Q)*K+(Q/2)*h=1200/100*20+100/2*8.4 = 240 + 420 = $660.
-Q5. Determine the economic order quantity (EOQ). EOQ = sqrt(2KR/h) = sqrt(2*20*1200/8.4) = 75.6 -Q6. How many orders will be placed per year using the EOQ? R/EOQ = 1200/75.6 = 15.87 -Q7. Determine the total annual cost for the EOQ (including the ordering and the holding cost but not the purchasing cost). -C(75.6) = (R/Q)*K+(Q/2)*h=1200/75.6*20+75.6/2*8.4 = $634.98 (or you can use the other method below) Sqrt(2KRh) = sqrt(2 * 20 * 1200 * 8.4) = $634.98
-Car Rental Company (Q8 - Q12) The airport branch of a car rental company maintains a fleet of 50 SUVs. The interarrival time between requests for an SUV is 2.4 hours with a standard deviation of 2.4 hours. Assume that, if all SUVs are rented, customers are willing to wait until an SUV is available. An SUV is rented, on average for 3 days, with a standard deviation of 1 day.
-Q8. What is the average number of SUVs in the company lot? We approach this problem as though the rental car is the "server". We know that a = 2.4 hours, p = 72 hours, CVa= (2.4/2.4) = 1, CVp= (24/72) = 0.33, and m = 50 cars. To determine the number of cars on the lot, we can look at the utilization of our "servers" = p/(a x m) = 72/(2.4 x 50) = 60%. Therefore, on average 60% of the cars or 30 cars are in use, so on average 20 cars are in the lot.
Q1. Each year 1000 students are admitted to the undergraduate program of David Eccles School of business. The program takes, on average, 4 years to complete. What is the average number of students that are in the program? By Little's Law, 4[years] x 1000[students/year] = 4000 [students]
A Production Process (Q2 - Q3) 13 minutes/unit ->11 minutes/unit-> 8 minutes/unit -Q2. What is the capacity of the process in units per hour? --Bottleneck is Activity 1 (13 min because it is the longest). Capacity of Activity 1 is 60 minutes * 1 / (13 minutes/unit) = 4.6 units/hour
Q3. If the demand is 3 units per hour, what is the average labor utilization of the process? -When demand is 3 units per hour, the process is a demand-constrained. Flow rate = Demand = 3 units/hour Cycle time = 1/Flow rate = 1/3 hour = 1/3 hour x 60 min/hour = 20 min Idle time (1) = cycle time - proc. Time = 20 - 13 = 7 min Idle time (2) = cycle time - proc. Time = 20 - 11 = 9 min Idle time (3) = cycle time - proc. Time = 20 - 8 = 12 min Labor content = 13 + 11 + 8 = 32 min Average labor utilization = Labor content / (Labor content + Idle time) = 32/ (32 + 7 + 9 + 12) = 0.533
Auto Parts Supplier (Q4 - Q7) -An auto parts supplier sells Hardy-brand batteries to car dealers and auto mechanics. The annual demand is approximately 1,200 batteries. The supplier pays $28 for each battery and estimates that the annual holding cost is 30 percent of the battery's value. It costs approximately $20 to place an order (managerial and clerical costs). The supplier currently orders 100 batteries per month.
A process, while being "in control," is capable if it can produce output according to the design specifications. Ex bullseye target.
Design Specifications and Process Capability- Design specifications consist of a target value and a tolerance level- Tolerance level: Upper specification level, Lower specification level- Whether a process is capable of meeting the tolerance level established depends on two factors: (1) USL - LSL: The tightness of the design specification (2) σ: The amount of variation in the current process- Process capability index:- When Cp=1 : The process meets the quality requirements 99.7 percent of the time. - Six-sigma program requires Cp=2: The process meets the quality requirements 99.999999998 percent of the time
When we observe an entry for X-bar or R outside the control limits, we can say with 99.7 percent confidence that the process has gone "out of control", that is, that an assignable cause has occurred. We can look for patterns to see whether further investigation is needed
Limitations: Control charts only tell to what extent the process is "in control" A process being "in control" just means that there are only common causes but no assignable cause of variation in the process. However, a process that is in control might still fail to deliver the quality demanded from the customer or a downstream operation in the process The control limits are set according to how the process performed in the past We only measure to what extent the process is operating in line with its historical behavior Meeting the design specification can be another thing
The consistency requirement from the customer typically takes the form of a design specification A design specification include - A target value - A tolerance level, describing the range of values that are acceptable from the customer's perspective
Note that: Design specifications are driven by the needs of the downstream process or by the end customer, while control limits are driven by how the process step has been operating in the past Thus, it is very well possible that a process is "in control" yet incapable of providing sufficiently tight tolerances demanded by the customer
Utah Barber Shop (Q13 - Q15) Q13. Utah Barber Shop (UBS) has a demand of 3 customers per hour. There are 2 barbers who takes, on average, 45 minutes to cut a customer's hair. There is a competitor at the next door of UBS so that whenever a customer arrives and finds all barbers busy cutting other customer's hair, then the customer immediately goes to the competitor shop. What is the probability of a customer going to the competitor shop? Use the following Erlang Loss Table to answer the question. r=p/a = 45/20 = 2.25 m=2 Pm(r)=0.4378.
Q14. Market research shows that whenever a customer is lost to the competitor, Utah Barber Shop loses $30 per customer. Assuming that UBS opens 8 hours a day and 6 days a week, what is the weekly loss due to lost customer? Weekly demand = 6*8*3=144 customers/week. Average no of customers lost = 0.4378 *144 = 63.04 Average cost per week = 63.04*$30 = $1,891.30
Q15. Assuming that UBS pays its barbers $30 per hour, should UBS hire another barber? Additional weekly wage paid to the new barber = 30*8*6=$1440. Pm(r)=0.2472 when m=3. Weekly cost from lost customers when UBS hires another barber = 144*0.2472*30 = $1067.90. Thus, cost saving is $1,891.30 - $1067.90 = $823.4 UBS should not hire an additional worker.
Two causes of Variability -Common (chance) causes:Constant variation reflecting "pure randomness" in the process. ex) No two snowflakes look exactly the same -Assignable causes Effects that result in changes of the parameters of the underlying statistical distribution of the process. ex) mistake in programming machines, an operator error Control Charts Helps you determine whether your process is in control -In control: There is no assignable cause of variation in the process -Out of control: An assignable cause of variation has occurred. X-bar chart: plots the mean of individual samples R chart: plots the range of individual samples