Practice questions from Chapter 3, 7, 10

QUALITY MANAGEMENT AND SIX SIGMA (Chapter 10) A manufacturing company has been inspecting units of output from a process. Each product inspected is evaluated on five criteria. If the unit does not meet standards for the criteria it counts as a defect for the unit. Each unit could have as few as zero defects, and as many as five. After inspecting 2,000 units, they discovered 33 defects. What is the DPMO measure for this process? Defects per Million Opportunities (D P M O) DPMO = 33 5 ∗ 2,000 ∗ 1,000,000 = 3,300 21.Design specifications require that a key dimension on a product measure 100 ± 10 units. A process being considered for producing this product has a standard deviation of four units. a. What can you say (quantitatively) regarding the process capability? C pk = min { UTL − X 3 σ , X − LTL 3 σ } = min { 110 − 100 3 ( 4 ) , 100 − 90 3 ( 4 ) } = min { .8333 , .8333 } = 0.8333 b. Suppose the process average shifts to 92. Calculate the new process capability. C pk = min { UTL − X 3 σ , X − LTL 3 σ } = min { 110 − 92 3 ( 4 ) , 92 − 90 3 ( 4 ) } = min { 1.5000 , .1667 } = 0.1667 c. What can you say about the process after the shift? Approximately what percentage of the items produced will be defective? Many defects will be produced. Assuming a normal distribution, the left tail (Prob{x<90}) is z = (90 - 92)/4 = -0.50, which corresponds to a probability of . 3085. The right tail (Prob{x>110} = 1-Prob{x<110})is z = (110-92)/4 = 4.5,

which approximately corresponds to a probability of .1-1 = 0. Therefore 0.3085+0=0.3085 or approximately 31% are outside the specification limits. 22.C-Spec, Inc., is attempting to determine whether an existing machine is capable of milling an engine part that has a key specification of 4 ± .003 inches. After a trial run on this machine, C-Spec has determined that the machine has a sample mean of 4.001 inches with a standard deviation of . 002 inch. a. Calculate the C pk for this machine. Process Capability Index( C pk larger than one indicates process is capable) C pk = min { UTL − X 3 σ , X − LTL 3 σ } = min { 4.003 − 4.001 3 ( .002 ) , 4.001 − 3.997 3 ( .002 ) } = min { .333 , .667 } = 0.333 b. Should C-Spec use this machine to produce this part? Why? No, the machine is not capable of producing the part at the desired quality level.

b. The chart indicates that the process is out of control. The administrator should investigate the quality of the patient meals.

29.The following table contains the measurements of the key length dimension from a fuel injector. These samples of size five were taken at one-hour intervals. Construct a three-sigma ´ X -chart and R -chart (use Exhibit 10.13) for the length of the fuel injector. What can you say about this process? ´ X = .499, ´ R = .037 n = 5 → A 2 = .58, D 3 = 0, D 4 = 2.11 Control limits for X-bar chart: UCL ´ X , LCL ´ X = ´ X ± A 2 ´ R = .499 ± .58 ( .037 ) .520,.478 Control limits for R chart: UCL = D 4 ´ R = 2.11 ( .037 ) .078 LCL = D 3 ´ R = 0 ( .037 )= 0

The process is in statistical control. Students may note samples 4-8 on the X-bar chart, but strictly speaking there are only 4 decreases in a row there.

30.In the past, Alpha Corporation has not performed incoming quality control inspections but has taken the word of its vendors. However, Alpha has been having some unsatisfactory experience recently with the quality of purchased items and wants to set up sampling plans for the receiving department to use. For a particular component, X, Alpha has a lot tolerance percentage defective of 10 percent. Zenon Corporation, from which Alpha purchases this component, has an acceptable quality level in its production facility of 3 percent for component X. Alpha has a consumer’s risk of 10 percent and Zenon has a producer’s risk of 5 percent. d. When a shipment of product X is received from Zenon Corporation, what sample size should the receiving department test? AQL = .03, LTPD = .10 LTPD/AQL = .10/.03 = 3.333 From Exhibit 10.16, c = 5. Also from this Exhibit, n (AQL) = 2.613 n = 2.613 AQL = 2.613 .03 = 87.1 , round up to 88 e. What is the allowable number of defects in order to accept the shipment? Allow up to 5 defective components

17.The Bijou Theater 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.) Use Model 1. λ = 100 per hour μ = 120 per hour a. What is the average customer time in the system? L s = λ μ − λ = 100 120 − 100 = 5 customers W s = L s λ = 5 100 = .05 hours or 3 minutes b. 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? Now, μ = 180 per hour L s = λ μ − λ = 100 180 − 100 = 1.25 customers W s = L s λ = 1.25 100 = .0125 hours or .75 minutes or 45 seconds c. Would system waiting time be less than you found in ( b) if a second window was opened with each server doing all three tasks? Using model 3, λ = 100 per hour μ = 120 per hour S =2 , and ρ = λ μ = 100 120 = .8333 , L q = .1756 (exact value calculated using Ch7_Queue_Models spreadsheet)

L s = L q + λ μ = .1756 + 100 120 = 1.01 customers W s = L s λ = 1.01 100 = .0101 hours or .605 minutes or 36.3 seconds ---------------------------------------------------------------------------------------------------- 19. A cafeteria serving line has a coffee urn from which customers serve themselves. Arrivals at the urn follow a Poisson distribution at the rate of three per minute. In serving themselves, customers take about 15 seconds, exponentially distributed. Use model 1 λ = 3 per minute μ = 4 per minute a. How many customers would you expect to see on the average at the coffee urn? L s = λ μ − λ = 3 4 − 3 = 3 customers b. How long would you expect it to take to get a cup of coffee? W s = L s λ = 3 3 = 1 minute c. What percentage of time is the urn being used? ρ = λ μ = 3 4 = .75 or 75% d. What is the probability that three or more people are in the cafeteria? Probability of 3 or more is equal to 1 – probability of 0, 1, 2 P 0 = ( 1 − 3 4 )( 3 4 ) 0 = .2500, P 1 = ( 1 − 3 4 )( 3 4 ) 1 = .1875,

FORECASTING (Chapter 3)

22. Your manager is trying to determine what forecasting method to use. Based upon the following historical data, calculate the following forecast and specify what procedure you would utilize. a. Calculate the simple three-month moving average forecast for periods 4–12. b. Calculate the weighted three-month moving average using weights of 0.50, 0.30, and 0.20 for periods 4–12. c. Calculate the single exponential smoothing forecast for periods 2–12 using an initial forecast ( F 1 ) of 61 and an  of 0.30. d. Calculate the exponential smoothing with trend component forecast for periods 2–12 using an initial trend forecast ( T 1 ) of 1.8, an initial exponential smoothing forecast ( F 1 ) of 60, an  of 0.30, and a  of 0.30. e. Calculate the mean absolute deviation (MAD) for the forecasts made by each technique in periods 4–12. Which forecasting method do you prefer? Month (t) Demand 3-mo. MA Absolute deviation 3-mo WMA Absolute deviation F t Absolute deviation T t F t FIT t Absolute deviation 1 62 61.00 1.80 60.00 61.80 2 65 61.30 1.82 61.86 63.68 3 67 62.41 1.94 64.07 66.01 4 68 64.67 3.33 65.40 2.60 63.79 4.21 2.03 66.31 68.33 0.33 5 71 66.67 4.33 67.10 3.90 65.05 5.95 2.00 68.23 70.23 0.77 6 73 68.67 4.33 69.30 3.70 66.84 6.16 2.07 70.46 72.53 0.47 7 76 70.67 5.33 71.40 4.60 68.68 7.32 2.11 72.67 74.78 1.22 8 78 73.33 4.67 74.10 3.90 70.88 7.12 2.22 75.14 77.36 0.64 9 78 75.67 2.33 76.40 1.60 73.02 4.98 2.28 77.55 79.83 1.83 10 80 77.33 2.67 77.60 2.40 74.51 5.49 2.11 79.28 81.39 1.39 11 84 78.67 5.33 79.00 5.00 76.16 7.84 1.99 80.98 82.96 1.04 12 85 80.67 4.33 81.60 3.40 78.51 6.49 2.08 83.27 85.35 0.35 MAD 4.07 3.46 6.17 0.89 Based upon MAD, the exponential smoothing with trend component appears to be the best method. This should not be a surprise given the apparent upward trend.

25.Zeus Computer Chips, Inc., used to have major contracts to produce the Centrino-type chips. The market has been declining during the past three years because of the quad-core chips, which it cannot produce, so Zeus has the unpleasant task of forecasting next year. The task is unpleasant because the firm has not been able to find replacement chips for its product lines. Here is demand over the past 12 quarters: Use the regression and seasonal indexes to forecast demand for the next four quarters.