- 120 Views
- Updated On :
- Presentation posted in: General

Quality Control

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Quality Control

Dr. Everette S. Gardner, Jr.

Correlation:

x

Strong positive

Positive

x

x

x

Negative

x

x

Strong negative

*

Competitive evaluation

Engineering characteristics

Source: Based on John R. Hauser and Don Clausing, “The House of Quality,” Harvard Business Review, May-June 1988.

Acoustic trans., window

Energy needed to open door

Check force on level ground

Energy needed to close door

x

Water resistance

= Us

Door seal resistance

Importance to customer

= Comp. A

A

= Comp. B

B

Customer requirements

(5 is best)

1 2 3 4 5

x

Easy to close

7

AB

Stays open on a hill

x

AB

5

Easy to open

3

AB

x

x

Doesn’t leak in rain

3

B

A

x

No road noise

2

B

A

Importance weighting

3

2

10

9

6

6

Relationships:

Strong = 9

Medium = 3

Target values

Reduce energy level to 7.5 ft/lb

Reduce energy to 7.5 ft/lb

Small = 1

Maintain current level

Maintain current level

Maintain current level

Reduce force to 9 lb.

5

BA

B

BA

x

x

A

4

B

B

B

x

Technical evaluation (5 is best)

A

x

3

A

x

2

A

x

1

Quality

Loss function

L(x) = k(x-T)2

where

x = any individual value of the quality characteristic

T = target quality value

k = constant = L(x) / (x-T)2

Average or expected loss, variance known

E[L(x)] = k(σ2 + D2)

where

σ2 = Variance of quality characteristic

D2 = ( x – T)2

Note: x is the mean quality characteristic. D2 is zero if the mean equals the target.

Quality

Average or expected loss, variance unkown

E[L(x)] = k[Σ ( x – T)2 / n]

When smaller is better (e.g., percent of impurities)

L(x) = kx2

When larger is better (e.g., product life)

L(x) = k (1/x2)

Quality

Definitions

- VariablesMeasurements on a continuous scale, such as length or weight
- AttributesInteger counts of quality characteristics, such as nbr. good or bad
- DefectA single non-conforming quality characteristic, such as a blemish
- DefectiveA physical unit that contains one or more defects
Types of control charts

Data monitored Chart name Sample size

- Mean, range of sample variables MR-CHART 2 to 5 units
- Individual variables I-CHART 1 unit
- % of defective units in a sample P-CHART at least 100 units
- Number of defects per unit C/U-CHART 1 or more units

Quality

Sample mean

value

0.13%

Upper control limit

Normal

tolerance

of

process

99.74%

Process mean

Lower control limit

0.13%

7

6

8

1

3

4

5

2

0

Sample number

Quality

n A A2 D3 D4 d2 d3

2 2.121 1.880 0 3.267 1.128 0.853

3 1.732 1.023 0 2.574 1.693 0.888

4 1.500 0.729 0 2.282 2.059 0.880

5 1.342 0.577 0 2.114 2.316 0.864

- Control factors are used to convert the mean of sample ranges
( R ) to:

(1) standard deviation estimates for individual observations, and

(2) standard error estimates for means and ranges of samples

For example, an estimate of the population standard deviation of individual observations (σx) is:

σx = R / d2

Quality

- Note that control factors depend on the sample size n.
- Relationships amongst control factors:
A2 = 3 / (d2 x n1/2)

D4 = 1 + 3 x d3/d2

D3 = 1 – 3 x d3/d2, unless the result is negative, then D3 = 0

A = 3 / n1/2

D2 = d2 + 3d3

D1 = d2 – 3d3, unless the result is negative, then D1 = 0

Quality

1. Compute the mean of sample means ( X ).

2. Compute the mean of sample ranges ( R ).

3. Estimate the population standard deviation (σx):

σx = R / d2

4. Estimate the natural tolerance of the process:

Natural tolerance = 6σx

5. Determine the specification limits:

USL = Upper specification limit

LSL = Lower specification limit

Quality

6. Compute capability indices:

Process capability potential

Cp = (USL – LSL) / 6σx

Upper capability index

CpU = (USL – X ) / 3σx

Lower capability index

CpL = ( X – LSL) / 3σx

Process capability index

Cpk = Minimum (CpU, CpL)

Quality

1. Compute the mean of sample means ( X ).

2. Compute the mean of sample ranges ( R ).

3. Set 3-std.-dev. control limits for the sample means:

UCL = X + A2R

LCL = X – A2R

4. Set 3-std.-dev. control limits for the sample ranges:

UCL = D4R

LCL = D3R

Quality

1. Compute the mean percentage defective ( P ) for all samples:

P = Total nbr. of units defective / Total nbr. of units sampled

2. Compute an individual standard error (SP ) for each sample:

SP = [( P (1-P ))/n]1/2

Note: n is the sample size, not the total units sampled.

If n is constant, each sample has the same standard error.

3. Set 3-std.-dev. control limits:

UCL = P + 3SP

LCL = P – 3SP

Quality

1. Compute the mean observation value ( X )

X = Sum of observation values / N

where N is the number of observations

2. Compute moving range absolute values, starting at obs. nbr. 2:

Moving range for obs. 2 = obs. 2 – obs. 1

Moving range for obs. 3 = obs. 3 – obs. 2

…

Moving range for obs. N = obs. N – obs. N – 1

3. Compute the mean of the moving ranges ( R ):

R = Sum of the moving ranges / N – 1

Quality

4. Estimate the population standard deviation (σX):

σX = R / d2

Note: Sample size is always 2, so d2 = 1.128.

5. Set 3-std.-dev. control limits:

UCL = X + 3σX

LCL = X – 3σX

Quality

1. Compute the mean nbr. of defects per unit ( C ) for all samples:

C = Total nbr. of defects observed / Total nbr. of units sampled

2. Compute an individual standard error for each sample:

SC = ( C / n)1/2

Note: n is the sample size, not the total units sampled.

If n is constant, each sample has the same standard error.

3. Set 3-std.-dev. control limits:

UCL = C + 3SC

LCL = C – 3SC

Notes:

● If the sample size is constant, the chart is a C-CHART.

● If the sample size varies, the chart is a U-CHART.

● Computations are the same in either case.

Quality

1.Compute a 4-quarter or 12-month moving average. Position the first average as follows:

a.Quarterly: Place the first average opposite the 3rd quarter. The first 2 quarters and the last quarter have no moving average.

b.Monthly: Place the first average opposite the 7th month. The first 6 months and the last 5 months have no moving average.

2. Divide each data observation by the corresponding moving average.

3.Compute a mean ratio for each quarter or month.

4.Compute a normalization factor to adjust the mean ratios so that they sum to 4 (quarterly) or 12 (monthly):

a.Quarterly: Normalization factor = 4 / Sum of mean ratios

b.Monthly: Normalization factor = 12 / Sum of mean ratios

Quality

5.Multiply each mean ratio by the normalization factor to get a set of final seasonal indices. Each quarter or month has an individual index.

6.Deseasonalize each data observation by dividing by the appropriate seasonal index.

7.Develop a control chart for the deseasonalized (seasonally-adjusted) data.

Quality

Step 1. Moving averages

t Qtr. Xt4-Qtr. moving average

1 1 53NA

2 2 83NA

3 3 95(53 + 83 + 95 + 72) / 4 = 75.75

4 4 72(83 + 95 + 72 + 50) / 4 = 75.00

5 1 50(95 + 72 + 50 + 75) / 4 = 73.00

6 2 75(72 + 50 + 75 + 102) / 4 = 74.75

7 3102(50 + 75 + 102 + 66) / 4 = 73.25

8 4 66(75 + 102 + 66 + 55) / 4 = 74.50

9 1 55(102 + 66 + 55 + 81) / 4 = 76.00

10 2 81(66 + 55 + 81 + 93) / 4 = 73.75

11 3 93(55 + 81 + 93 + 76) / 4 = 76.25

12 4 76NA

Quality

Step 2. Ratios

Ratio = Xt / Average

NA

NA

95 / 75.75 = 1.2541

72 / 75.00 = 0.9600

50 / 73.00 = 0.6849

75 / 74.75 = 1.0033

102 / 73.25 = 1.3925

66 / 74.50 = 0.8859

55 / 76.00 = 0.7237

81 / 73.75 = 1.0983

93 / 76.25 = 1.2197

NA

Quality

Step 3. Mean ratios

Qtr.Sum of ratios for each qtr. / Nbr.

1(0.6849 + 0.7237) / 2 = 0.7043

2(1.0033 + 1.0983) / 2 = 1.0508

3(1.2542 + 1.3925 + 1.2197) / 3 = 1.2888

4(0.9600 + 0.8859) / 2 = 0.9230

Sum of mean ratios = 3.9669

Step 4. Normalization Factor

Factor = 4 / (Sum of mean ratios)

Factor = 4 / 3.9669 = 1.0083

Quality

Step 5. Final seasonal indices

Qtr.Mean ratio x Factor = Index

10.7043 x 1.0083 = 0.7101

21.0508 x 1.0083 = 1.0595

31.2888 x 1.0083 = 1.2995

40.9230 x 1.0083 = 0.9307

Sum of indices = 3.9998

Quality

Step 6. Deseasonalize data

t Qtr. Xt / Index= Des. Xt

1 1 53 / 0.7101= 74.6

2 2 83 / 1.0595= 78.3

3 3 95 / 1.2995= 73.1

4 4 72 / 0.9307= 77.4

5 1 50 / 0.7101= 70.4

6 2 75 / 1.0595= 70.8

7 3102 / 1.2995= 78.5

8 4 66 / 0.9307= 70.9

9 1 55 / 0.7101= 77.5

10 2 81 / 1.0595= 76.5

11 3 93 / 1.2995= 71.6

12 4 76 / 0.9307= 81.7

Quality

1. Identify quality characteristics.

2. Choose a quality indicator.

3. Choose the type of chart.

4. Decide when to sample.

5. Choose a sample size.

6. Collect representative data.

7. If data are seasonal, perform seasonal adjustment.

8. Graph the data and adjust for outliers.

Quality

9. Compute control limits

10. Investigate and adjust special-cause variation.

11. Divide data into two samples and test stability of limits.

12. If data are variables, perform a process capability study:

a. Estimate the population standard deviation.

b. Estimate natural tolerance.

c. Compute process capability indices.

d. Check individual observations against specifications.

13. Return to step 1.

Quality

- Control factors
n A A2 D3 D4 d2 d3

2 2.121 1.880 0 3.267 1.128 0.853

3 1.732 1.023 0 2.574 1.693 0.888

4 1.500 0.729 0 2.282 2.059 0.880

5 1.342 0.577 0 2.114 2.316 0.864

- Process capability analysis
σx = R / d2

Cp = (USL – LSL) / 6σx CpU = (USL – X ) / 3σx

CpL = ( X – LSL) / 3σx Cpk = Minimum (CpU, CpL)

Quality

- Means and ranges
UCL = X + A2RUCL = D4R

LCL = X – A2RLCL = D3R

- Percentage defective in a sample
SP = [( P (1-P ))/n]1/2UCL = P + 3SP

LCL = P – 3SP

- Individual quality observations
σx = R / d2UCL = X + 3σX

LCL = X – 3σX

- Number of defects per unit
SC = ( C / n)1/2UCL = C + 3SC

LCL = C – 3SC

Quality