Management Chapter 8 Homework Answer See The Following Table And Figure

Chapter 8 – Instructor Reserve Problems

1. Reboard Electronics, Inc. is a reprocessor of electronic boards used in computers. The

quality manager is trying to determine the non-conformances per unit (NPU) and the

throughput yield. The quality manager has requested that 500 board be inspected and 25

non-conformances were found.

a. Calculate the nonconformances per unit (NPU) and the throughput yield (TY).

b. If the production process consists of four steps, with step 1 having a throughput yield

of 98 percent; step 2, 99 percent; step 3, 96 percent; and step 4, 97 percent, what is the

rolled throughput yield (RTY), and the proportion nonconforming?

Answer

1. a. To calculate the NPU, use:

NPU = Number of nonconformances found/number of units inspected = 48/500 = 0.096

2. Over the last year 752 drug doses were administered at the Goodhealth clinic. Quality is

measured by the proper dosage as well as the correct drug. In two instances, the incorrect

size of pill was given, and in one case, the wrong drug was given. At what sigma level is

Goodhealth’s process?

Answer

We use 3/752 to get the number of defects per unit (DPU’s). However, there are 2

opportunities per dose (wrong size, wrong drug) to make an error. They must be

3. A gauge repeatability and reproducibility study at TenGage, Inc., collected the data for

three operators, three trials, and ten parts, as found in the worksheet Ch08Inst-Rsv.xlsx in

the Instructor Reserve folder for this chapter. Analyze these data. The part specification is

1.3 ± 0.15 inches.

Answer

3. Detailed calculations for the first operator are as follows:

x

1 = (Mijk) /nr = 38.670 /30 = 1.289

R

1 = (Rij) / n = 1.060 /10 = 0.106

x

D = max {

x

i} – min {

x

i} = 1.289 – 1.246 = 0.043

R

= (

R

i) / m = (0.096 + 0.126 +0.049) / 3 = 0.091

D4 = 2.58 ; UCLR = D4

R

= (2.58) (0.091) = 0.235, all ranges below

K1 = 3.05; K2 = 2.70 (from Table 8.3)

R

% of TV

Equipment variation = 100 0.278

0.414 = 67.15%

Operator variation [AV] = 100 0.105

0.414= 25.36%

% of Tolerence

Equipment variation = 100 0.278

0.30 = 92.67%

Operator variation [AV] = 100 0.105

0.30 = 35.00%

Variance Ratios

EV% of Total Variance = 100 𝐸𝑉2

𝑇𝑉2 = 100 0.2782

0.4142 = 45.1%

AV% of Total Variance = = 100 𝐴𝑉2

𝑇𝑉2 = 100 0.1052

0.4142 = 6.43%

For detailed spreadsheet data, see Prob08-03IRR.xlsx. Spreadsheet results confirm prior

calculations, such as these that follow:

% of Total Variation

% of Tolerance

Variance Ratios

EV

66.94%

EV

92.04%

EV

44.81%

25.40%

34.93%

4. A gauge repeatability and reproducibility study was done at EngineBlader, Inc., which

makes and repairs compressor blades for jet engines. The quality analyst collected the

data for three operators, two trials, and ten parts, as found in the worksheet Ch08Inst-

Rsv.xlsx in the Instructor Reserve folder for this chapter. Analyze these data. The part

specification is 4.7 ± 0.1 inches. Calculate the process capability indexes for the parts.

What does this tell you about the relative importance of part variation versus equipment

variation and appraiser (operator) variation in assessing the gauging system?

Answer

4. a) Detailed calculations for the first operator are as follows:

x

1 = (Mijk) /nr = 93.960 /20 = 4.698

R

1 = (Rij) / n = 0.18 /10 = 0.018

x

R

% of TV

Equipment variation = 100 0.106

0.129 = 82.17%

% of Tolerance

Equipment variation = 100 0.106

0.20 = 53.00%

Operator variation [AV] = 100 0.057

0.20 = 28.50%

Variance Ratios

EV% of Total Variance = 100 𝐸𝑉2

𝑇𝑉2 = 100 0.1062

0.1292 = 67.52%

% of Total

Variation

% of Tolerance

Variance Ratios

EV

82.76%

EV

53.35%

EV

68.50%

Note that the calculator values, shown in the detailed calculations above, and computer

values do not match precisely, because a greater number of decimal places are used by

the computer to carry out calculations. All formulas are identical, however.

Process Capability Index Calculations

Average

4.6866

Standard deviation

0.0206

Cp

1.618

Cpl

1.401

Cpu

1.835

Cpk

1.401

10

12

14

16

18

Cell Upper Limit

Histogram

28.30%

19.27%

R&R

60.39%

87.77%

PV

22.55%

PV

12.23%

5. The cost data found in worksheet C08DataInsRsv.xlsx was taken from a paper

manufacturing process at Oakton Paper Company by cost category. Prepare a spreadsheet

and perform graphical analysis to show the “significant few” cost categories on which

Oakton should concentrate. Based on your analysis make a recommendation for

management on how they should act to improve quality.

Answer

See the following table and figure for Pareto analysis, based on the data in spreadsheet

C08DataInsRsv.xlsx.

OAKTON PAPER COMPANY

QUALITY COSTS AND PERCENTAGES

Percent

Cumulative %

Cost

Rejected paper

56.82

375000

Customer complaints

15.91

72.73

105000

Odd lot

10.61

83.33

Downtime

4.24

93.48

Excess inspection

3.18

96.67

Testing costs

2.12

98.79

Quality Imprv. Trng.

1.21

Total Costs

6. Data for Worldwide Measursys Repairs, Inc. cost of quality categories are found in the

spreadsheet C08DataInsRsv.xlsx. Determine which categories contribute the most to the

cost of quality at Worldwide. Show this, graphically, in a spreadsheet, and make a

recommendation to management.

Answer The Pareto chart for Worldwide Measursys Repairs, Inc. is shown below.

WORLDWIDE MEASURSYS REPAIRS, INC.

QUALITY COSTS AND PERCENTAGES

Percent

Cumulative %

Cost

Customer returns

40.00

$120,000

0.00%

20.00%

40.00%

60.00%

80.00%

100.00%

120.00%

Percent within Cost Categories

Quality Cost Categories

Pareto Chart for Oakton Paper Co.

Percent

Cumulative

Workstation downtime

Rework costs

Inspection – out

Training/improvement

Inspection – in

5.00

Total Costs

The data show that two categories of customer returns and workstation downtime total

56.7 percent of the defects. These two are possibly related, and may indicate “short

staffing,” and lack of training of setup personnel. Only 10% of total quality cost is

allocated to prevention (training/improvement). Steps should be taken to analyze root

causes for these problem areas in order to correct them as quickly as possible.

7. The temperature in a computer lab at Coyote University is very important for proper

functioning of the computer equipment. The data in the worksheet C08DataInsRsv.xlsx

show the results of 30 samples of 5 each, taken at random at different times of day over a

three month time period. Calculate the process capability statistics for the temperature in

the computer lab at Coyote University. The upper tolerance limit is 76 and the lower

tolerance limit is 68. What recommendation would you make to management concerning

the process, based on these findings?

Answer

The sample statistics are:

x

= 72.071; s = 1.268

0.00%

20.00%

40.00%

60.00%

80.00%

100.00%

120.00%

Percent Within Defect Category

Defect Categories

Pareto Chart for Worldwide Measursys Repairs

Percent

Cumulative %

Note that the spreadsheet uses an actual standard deviation of s = 1.268 calculated from

all of the sample values. This numeric value is the overall standard deviation for all

sample values, and is not the same average statistic, such would be obtained by summing

the standard deviations of each group of 10, then dividing by 15 to get

s

.

Process Capability

Upper specification

76.0

Cp

1.051

68.0

1.033

1.070

1.033

Management should be advised that, while the capability is minimally acceptable, they

8. Metropole Hospital is working on reducing waiting time in order to give customers better

service in their waiting rooms. Fifty samples of size 6 were taken at random times from

their main waiting room. These data can be found in the worksheet C08Data-InstRsv.xlsx

for this problem.

15

20

25

30

Frequency

Cell Upper Limit

Histogram

a. Compute the mean and range of each sample, and the control limits, and plot them on

x

and R control charts.

b. Does the process appear to be in statistical control? Why or why not?

Answer

Results from the 50 samples of 6 for Metropole show that both the

x

and R charts are

apparently out of control. (See spreadsheet Prob08-08InsRsvXR.xls for details).

For the Center Lines, CL

x

:

x

= 22.615; CLR:

R

= 2.192

For the R-chart:

R

21.00

21.50

22.00

23.50

24.00

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49

Averages

Sample number

Prob. 8-08 X-bar Chart

Averages

Lower control limit

Upper control limit

Center line

x

R

Point #6 on the

x

– chart and point #39 on the R chart are out of control. The analyst

9. The data in worksheet C08Data-Rsvprob.xlsx for Prob. 8-09 list electrical resistance

values (ohms) for 50 samples of size 5 that were taken from Babbage Chips, Inc.’s

computer chip-making process over a 25-hour period.

a. Compute the mean, standard deviation and other descriptive statistics for the data.

b. Calculate the control limits and construct the x– and R-charts, using the first 30

samples. Is the process under control at that point?

c. Specifications for the process are 9.2 ± 3.2 ohms. If the process is under control,

calculate the capability indexes, Cpu, Cpl, Cp, and Cpk using the part of the

x

and R-

charts Excel template that calculates the process capability. What do the indexes

indicate?

d. After calculating the control limits, the last 20 samples were collected. When plotted

using the control limits calculated earlier, does the process appear to be in statistical

control? Why or why not? What should be done if it is not under control?

0.00

3.00

3.50

4.00

4.50

5.00

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49

Sample number

Prob. 8-08 R-Chart

Ranges

Lower control limit

Upper control limit

Center line

Answer

a) Descriptive statistics for Babbage Chips, Inc., based on all 50 samples, are shown

below. The histogram shows the “classic” bell curve shape.

Descriptive Statistics – Prob. 8-09

Bin

Frequency

6.0

0

Mean

9.046

6.6

3

Standard Deviation

1.103

9.0

52

Sample Variance

1.218

9.6

48

Kurtosis

-0.323

10.2

44

Skewness

0.071

10.8

19

Range

5.716

11.4

12

Minimum

6.341

12.0

3

Maximum

12.057

12.6

1

Sum

2261.440

Count

250.000

Conf. Level(95.0%)

0.137

Standard Error

0.070

7.2

9

Median

9.011

7.8

23

Mode

9.215

8.4

36

Histogram

30

40

50

60

Bin

Frequency

b) Results from first 30 samples of 5 for Babbage show that both the

x

and R charts are

apparently in control. (See spreadsheet Prob08-09InsRsvAXR.xls for details).

For the Center Lines, CL

x

:

x

= 9.170; CLR:

R

= 2.543

Control limits for the

x

– chart are:

x

± A2

R

x

R

x

R

For the R-chart:

R

c) The computations for the process capability, taken from the spreadsheet template,

show that the process is not quite capable (see spreadsheet Prob08-

09InsRsvAXR.xlsx for details).

7

9.5

10

10.5

11

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49

Sample number

Prob. 8-09A X-bar Chart

Averages

Lower control limit

Upper control limit

Center line

0

5

6

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49

Sample number

Prob. 8-09A R-Chart Ranges

Lower control limit

Upper control limit

Center line