1. Quality And Performance Problems and Exercises5
For Operations Management, 9e by Krajewski/Ritzman/Malhotra © 2010 Pearson Education
Adapted for BA 357 by Ken Shaw
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Solved Problem 1
The Watson Electric Company produces incandescent lightbulbs. The following data on the number of lumens for 40-watt lightbulbs were collected when the process was in control.
Observation
Sample 1 2 3 4
604
612
588
600
597
601
607
603
581
570
585
592
620
605
595 5
590
614
608
a. Calculate control limits for an R-chart and an x-chart.
b. Since these data were collected, some new employees were hired. A new sample obtained the following readings: 570, 603, 623, and 583. Is the process still in control?
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Solved Problem 1
SOLUTION
a. To calculate x, compute the mean for each sample. To calculate R, subtract the lowest value in the sample from the highest value in the sample. For example, for sample 1, 4
= 601
x =
R =
612 – 588 = 24
Sample R 1
601 24 2
602 10 3
582 22 4 32 5
604
Total
2,991
112
Average
x = 598.2
R = 22.4 x
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Solved Problem 1
The R-chart control limits are
UCLR = D4R =
2.282(22.4) = 51.12
LCLR = D3R =
0(22.4) = 0
The x-chart control limits are
UCLx = x + A2R =
(22.4) =
LCLx = x – A2R =
598.2 – 0.729(22.4) =
b. First check to see whether the variability is still in control based on the new data. The range is 53 (or 623 – 570), which is outside the UCL for the R-chart. Since the process variability is out of control, it is meaningless to test for the process average using the current estimate for R. A search for assignable causes inducing excessive variability must be conducted.
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Solved Problem 2
The data processing department of the Arizona Bank has five data entry clerks. Each working day their supervisor verifies the accuracy of a random sample of 250 records. A record containing one or more errors is considered defective and must be redone. The results of the last 30 samples are shown in the table. All were checked to make sure that none was out of control.
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Solved Problem 2
Sample
Number of Defective Records 1 7 16 8 2 5 17 12 3 19 18 4 10 6 11 20 21 22 9 23 24 13 25 26 27 14 28 29 15 30
Total 300
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7. Number of Defective Records
Solved Problem 2
a. Based on these historical data, set up a p-chart using z = 3.
b. Samples for the next four days showed the following:
Sample
Number of Defective Records
Tues 17
Wed 15
Thurs 22
Fri 21
What is the supervisor’s assessment of the data-entry process likely to be?
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Solved Problem 2
SOLUTION
a. From the table, the supervisor knows that the total number of defective records is 300 out of a total sample of 7,500 [or 30(250)]. Therefore, the central line of the chart is
= 0.04
300
7,500
p =
The control limits are
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9. Number of Defective Records
Solved Problem 2
b. Samples for the next four days showed the following:
Sample
Number of Defective Records
Proportion
Tues 17
0.068
Wed 15
0.060
Thurs 22
0.088
Fri 21
0.084
Samples for Thursday and Friday are out of control. The supervisor should look for the problem and, upon identifying it, take corrective action.
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Solved Problem 3
The Minnow County Highway Safety Department monitors accidents at the intersection of Routes 123 and 14. Accidents at the intersection have averaged three per month.
a. Which type of control chart should be used? Construct a control chart with three sigma control limits.
b. Last month, seven accidents occurred at the intersection. Is this sufficient evidence to justify a claim that something has changed at the intersection?
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Solved Problem 3
SOLUTION
a. The safety department cannot determine the number of accidents that did not occur, so it has no way to compute a proportion defective at the intersection. Therefore, the administrators must use a c-chart for which
UCLc = c + z c
= = 8.20
LCLc = c – z c
= 3 – = –2.196
There cannot be a negative number of accidents, so the LCL in this case is adjusted to zero.
b. The number of accidents last month falls within the UCL and LCL of the chart. We conclude that no assignable causes are present and that the increase in accidents was due to chance.
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Solved Problem 4
Pioneer Chicken advertises “lite” chicken with 30 percent fewer calories. (The pieces are 33 percent smaller.) The process average distribution for “lite” chicken breasts is 420 calories, with a standard deviation of the population of 25 calories. Pioneer randomly takes samples of six chicken breasts to measure calorie content.
a. Design an x-chart using the process standard deviation.
b. The product design calls for the average chicken breast to contain 400 ± 100 calories. Calculate the process capability index (target = 1.33) and the process capability ratio. Interpret the results.
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Solved Problem 4
SOLUTION
a. For the process standard deviation of 25 calories, the standard deviation of the sample mean is
UCLx = x + zσx =
(10.2) = calories
LCLx = x – zσx =
420 – 3(10.2) = calories
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14. Solved Problem 4 b. The process capability index is
Cpk = Minimum of ,
x – Lower specification 3σ
Upper specification – x
= Minimum of = 1.60, = 1.07
420 – 300
3(25)
500 – 420
The process capability ratio is
Cp =
Upper specification – Lower specification 6σ
= = 1.33
500 – 300
6(25)
Because the process capability ratio is 1.33, the process should be able to produce the product reliably within specifications. However, the process capability index is 1.07, so the current process is not centered properly for four-sigma performance. The mean of the process distribution is too close to the upper specification.
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Application 5.1
Webster Chemical Company produces mastics and caulking for the construction industry. The product is blended in large mixers and then pumped into tubes and capped.
Webster is concerned whether the filling process for tubes of caulking is in statistical control. The process should be centered on 8 ounces per tube. Several samples of eight tubes are taken and each tube is weighed in ounces.
Tube Number
Sample 1 2 3 4 5 6 7 8
Avg
Range
7.98
8.34
8.02
7.94
8.44
7.68
7.81
8.11
8.040
0.76
8.23
8.12
8.41
8.31
8.18
7.99
8.06
8.160
0.43
7.89
7.77
7.91
8.04
8.00
7.93
8.09
7.940
0.32
8.24
7.83
8.05
7.90
8.16
7.97
8.07
8.050
0.41
7.87
8.13
7.92
8.10
8.14
7.88
7.980
0.33
8.130
0.03
Avgs
0.38
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Application 5.1
Assuming that taking only 6 samples is sufficient, is the process in statistical control?
Conclusion on process variability given R = 0.38 and n = 8:
UCLR = D4R =
1.864(0.38) = 0.708
LCLR = D3R =
0.136(0.38) = 0.052
The range chart is out of control since sample 1 falls outside the UCL and sample 6 falls outside the LCL. This makes the x calculation moot.
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Application 5.1
Consider dropping sample 6 because of an inoperative scale, causing inaccurate measures.
Tube Number
Sample 1 2 3 4 5 6 7 8
Avg
Range
7.98
8.34
8.02
7.94
8.44
7.68
7.81
8.11
8.040
0.76
8.23
8.12
8.41
8.31
8.18
7.99
8.06
8.160
0.43
7.89
7.77
7.91
8.04
8.00
7.93
8.09
7.940
0.32
8.24
7.83
8.05
7.90
8.16
7.97
8.07
8.050
0.41
7.87
8.13
7.92
8.10
8.14
7.88
7.980
0.33
Avgs
8.034
0.45
What is the conclusion on process variability and process average?
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Application 5.1
Now R = 0.45, x = 8.034, and n = 8
UCLR = D4R =
1.864(0.45) = 0.839
LCLR = D3R =
0.136(0.45) = 0.061
UCLx = x + A2R =
(0.45) = 8.202
LCLx = x – A2R =
8.034 – 0.373(0.45) = 7.832
The resulting control charts indicate that the process is actually in control.
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Application 5.2
A sticky scale brings Webster’s attention to whether caulking tubes are being properly capped. If a significant proportion of the tubes aren’t being sealed, Webster is placing their customers in a messy situation. Tubes are packaged in large boxes of 144. Several boxes are inspected and the following numbers of leaking tubes are found:
Sample
Tubes 1 3 8 6 15 5 2 9 4 16 10 17 11 18 12 19 13 20 7 14
Total = 72
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Application 5.2
Calculate the p-chart three-sigma control limits to assess whether the capping process is in statistical control.
The process is in control as the p values for the samples all fall within the control limits.
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Application 5.3
At Webster Chemical, lumps in the caulking compound could cause difficulties in dispensing a smooth bead from the tube. Even when the process is in control, there will still be an average of 4 lumps per tube of caulk. Testing for the presence of lumps destroys the product, so Webster takes random samples. The following are results of the study:
Tube #
Lumps 1 6 5 9 2 4 10 3 7 11 8 12
Determine the c-chart two-sigma upper and lower control limits for this process.
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Application 5.3
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Application 5.4
Webster Chemical’s nominal weight for filling tubes of caulk is 8.00 ounces ± 0.60 ounces. The target process capability ratio is 1.33, signifying that management wants 4-sigma performance. The current distribution of the filling process is centered on ounces with a standard deviation of 0.192 ounces. Compute the process capability index and process capability ratio to assess whether the filling process is capable and set properly.
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24. x – Lower specification Upper specification – x
Application 5.4
a. Process capability index:
Cpk = Minimum of ,
x – Lower specification 3σ
Upper specification – x
= Minimum of = 1.135, = 0.948
8.054 – 7.400
3(0.192)
8.600 – 8.054
Recall that a capability index value of 1.0 implies that the firm is producing three-sigma quality (0.26% defects) and that the process is consistently producing outputs within specifications even though some defects are generated. The value of is far below the target of Therefore, we can conclude that the process is not capable. Furthermore, we do not know if the problem is centering or variability.
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25. Upper specification – Lower specification
Application 5.4
b. Process capability ratio:
Cp =
Upper specification – Lower specification 6σ
= =
8.60 – 7.40
6(0.192)
Recall that if the Cpk is greater than the critical value (1.33 for four-sigma quality) we can conclude that the process is capable. Since the Cpk is less than the critical value, either the process average is close to one of the tolerance limits and is generating defective output, or the process variability is too large. The value of Cp is less than the target for four-sigma quality. Therefore we conclude that the process variability must be addressed first, and then the process should be retested.
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