Chapter 9AProcess Capability and

Statistical Quality Control (SQC)

Flower PlatoonAbad.Imperial.Javate.Palma.Uy,R., Valencia

• Process Variation• Process Capability• Process Control Procedures

– Variable data– Attribute data

• Acceptance Sampling– Operating Characteristic Curve

OBJECTIVES

• Statistical quality control (SQC)

– Quantitative aspects of quality management– How well are we doing at meeting the specifications

that were set for the design?• Requires periodic sampling of processes and analysis

– Use statistically derived performance criteria• Applications

– Manufacturing• Ex. Manufacturing defects

– Service• Ex.

Introduction

• Assignable variation– caused by factors that can

be clearly identified and possibly managed

• Ex. 1pt, different anthropometric measurements

• Common variation– inherent in the production

process• Ex. Donuts

Basic Forms of Variation

• As variation is reduced, quality is improved• However, it is also impossible to have zero

variability

• Solution: Define the target + acceptable limits about the target– Upper and Lower specification limits– Ex. 10 inches + 0.02 inches

Variation

Traditional View VS Taguchi’s View

• Graduation of acceptability away from aim

• Costs increases as variability increases

• Seek to achieve zero defects minimize quality costs

• Within LS and US is good • Cost is 0 if w/in range• Quantum leap if limit violated

IncrementalCost of Variability

High

Zero

LowerSpec

TargetSpec

UpperSpec

Traditional View

IncrementalCost of Variability

High

Zero

LowerSpec

TargetSpec

UpperSpec

Taguchi’s View

• Evaluate the ability of a production process to meet or exceed preset specifications– Ex. Motorola Six-Sigma Limits

• When is a Process CAPABLE?– Mean and SD of process are operating such that

UCL and LCL are acceptable relative to specification limits

Process Capability

• Shows how well parts being produced fit into the range specified by design limits

• Position of the mean and tails of the process relative to design specifications– More off center, more defective parts produced

Process Capability Index, Cpk

• Design Specifications: acceptable volume of liquid is preset at 16 ounces + .2 ounces– 15.8 (Lower) and 16.2 ounces (Upper)

Example

• Cp=1– process variability just meets specifications – Minimally capable

• Cp <1– process variability outside range of specification – Not capable of producing w/in specification, NI

• Cp >1 – process variability tighter than specifications – Exceeds minimal capability

Capability Index

Cpk MinX LTL

3;UTL X

3

• The quality assurance manager is assessing the capability of a process that puts pressurized grease in an aerosol can

• Design specification call for an average of 60 pounds per square inch (psi) of pressure in each can– W/ ULT of 65 ps and LLT of 55 psi

• Sample taken from production, average 61 psi w/ SD 2 psi• What is the capability of the process?• What is probability of producing a defect?

Sample problem:

• Given– LTL 55, UTL 65, X = 61, Sigma = 2

• Calculate the Cp

– Cp = Min 61-55 x 65-61 3(2) 3(2)

Cp = min 1 x 0.6667 = 0.6667 (*Cp <1, not capable of producing w/in specification)

Sample problem

Cpk MinX LTL

3;UTL X

3

• Calculate probability of producing a defect– Probability of a can w/ <55psi

• Z = LTL – Mean/σ • Z = 55-61/2 = -3

– Use Appendix E, p 745• NORMDIST (-3) = 0.00135

– Probability of a can >65 psi• Z = UTL – Mean/σ • Z = 65-61/2 = 2• 1 – NORMDIST(2) = 1-.97725 = 0.02275

• Probability of a can <55 or >65 psi– Probability = 0.00135 + 0.02275 = 00.0241– Approximately 2.4% of cans will be defective

Attribute (Go or no-go information)◦ Defectives

◦ acceptability of product across a range of characteristics.◦ Defects

◦ number of defects per unit which may be higher than the number of defectives.

◦ p-chart application

Variable (Continuous)◦ Usually measured by the mean and the standard

deviation.◦ X-bar and R chart applications

Types of Statistical Sampling

Control Limits are based on the Normal Curve!

x

0 1 2 3-3 -2 -1z

Standard deviation units or “z” units.

Standard deviation units or “z” units.

We establish the Upper Control Limits (UCL) and the Lower Control Limits (LCL) with plus or minus 3 standard deviations from some x-bar or mean value. Based on this we can expect 99.7% of our sample observations to fall within these limits.

Control Limits

xLCL UCL

99.7%

Statistical Process Control (SPC) Charts

UCL

LCL

Samples over time

1 2 3 4 5 6

UCL

LCL

Samples over time

1 2 3 4 5 6

UCL

LCL

Samples over time

1 2 3 4 5 6

Normal BehaviorNormal Behavior

Possible problem, investigatePossible problem, investigate

Possible problem, investigatePossible problem, investigate

P-Chart Example

• Page 349, #3– Ten Samples of 15 parts each were taken from an

ongoing process to establish a p chart for control. The samples and the number of defectives in each are shown as follows:

Sample N Number of defects per sample

1 15 3

2 15 1

3 15 0

4 15 0

5 15 0

6 15 2

7 15 0

8 15 3

9 15 1

10 15 0

• A. Develop a p chart for 95% (1.96 SD)• B. Based on the planned data points, what

comments can you make?

What we need!Sample

No.

No. of

Samples

Number of defects found in each sample

Sample N Number of defects per sample

1 15 3

2 15 1

3 15 0

4 15 0

5 15 0

6 15 2

7 15 0

8 15 3

9 15 1

10 15 0

Statistical Process Control Formulas:Attribute Measurements (p-Chart)

p =Total Number of Defectives

Total Number of Observationsp =

Total Number of Defectives

Total Number of Observations

ns

)p-(1 p = p n

s)p-(1 p

= p

p

p

z - p = LCL

z + p = UCL

s

s

p

p

z - p = LCL

z + p = UCL

s

s

Given:

Compute control limits:

Observations: Number of samples x Sample size

Constructing a p-chart• Calculate the sample

proportions, p for each sample

Sample N Defects per

sample

p

1 15 3 0.2

2 15 1 0.067

3 15 0 0

4 15 0 0

5 15 0 0

6 15 2 0.133

7 15 0 0

8 15 3 0.2

9 15 1 0.067

10 15 0 0

0.067=150

10 = p 0.067=150

10 = p

0.065= 15

0.067)-0.067(1=

)p-(1 p = p n

s 0.065= 15

0.067)-0.067(1=

)p-(1 p = p n

s

Calculate the average of the sample proportions

Calculate the standard deviation of the sample proportion

Constructing a p-chart

) 1.96(0.065 0.067 ) 1.96(0.065 0.067

UCL = 0.194LCL = - 0.060UCL = 0.194LCL = - 0.060

p

p

z - p = LCL

z + p = UCL

s

s

p

p

z - p = LCL

z + p = UCL

s

s

• Calculate the control limits

Constructing a p-chart

• Plot the individual sample proportions, the average of the proportions, and the control limits

Constructing a p-chart

UCL

LCL

Conclusion

• Control limits were established as 95 percent• Of the 10 Samples, 2 were out of the control

limits• Process is out of control and warrants an

investigation

X-bar and R Charts Example

• P. 349 #6– Resistors for electronic circuits are manufactured

on a high speed automated machine. The machine is set up to produce a large run of resistors of 1,000 ohms each

– To set up the machine and to create a control chart to be used throughout the run, 15 samples were taken with four resistors in each samples

Sample Number

Readings

1 1010 991 985 986

2 995 996 1009 994

3 990 1003 1015 1008

4 1015 1020 1009 998

5 1013 1019 1005 993

6 994 1001 994 1005

7 989 992 982 1020

8 1001 986 996 996

9 1006 989 1005 1007

10 992 1007 1006 979

11 996 1006 997 989

12 1019 996 991 1011

13 981 991 989 1003

14 999 993 988 984

15 1013 1002 1005 992

• Develop an x chart and an R chart and plot the values.

• From the charts, what comments can you make about the process? (Use three-sigma control limits as in Exhibit 9A.6)

X-bar and R Charts: Required DataSample Number

Readings

1 1010 991 985 986

2 995 996 1009 994

3 990 1003 1015 1008

4 1015 1020 1009 998

5 1013 1019 1005 993

6 994 1001 994 1005

7 989 992 982 1020

8 1001 986 996 996

9 1006 989 1005 1007

10 992 1007 1006 979

11 996 1006 997 989

12 1019 996 991 1011

13 981 991 989 1003

14 999 993 988 984

15 1013 1002 1005 992

Calculate sample means, sample ranges,

mean of means, and

mean of ranges.

Sample Number

Readings Average Range

1 1010 991 985 986993.000 25.000

2 995 996 1009 994998.500 15.000

3 990 1003 1015 10081004.000 25.000

4 1015 1020 1009 9981010.500 22.000

5 1013 1019 1005 9931007.500 26.000

6 994 1001 994 1005998.500 11.000

7 989 992 982 1020995.750 38.000

8 1001 986 996 996994.750 15.000

9 1006 989 1005 10071001.750 18.000

10 992 1007 1006 979996.000 28.000

11 996 1006 997 989997.000 17.000

12 1019 996 991 10111004.250 28.000

13 981 991 989 1003991.000 22.000

14 999 993 988 984991.000 15.000

15 1013 1002 1005 9921003.000 21.000

Average999.100 21.733

Determine Control Limit Formulas and Necessary Tabled Values

RA - x = LCL

RA + x = UCL

2

2

Limits ControlChart x

RA - x = LCL

RA + x = UCL

2

2

Limits ControlChart x

R Chart Control Limits

UCL = D R

LCL = D R

4

3

R Chart Control Limits

UCL = D R

LCL = D R

4

3

n A2 D3 D42 1.88 0 3.273 1.02 0 2.574 0.73 0 2.285 0.58 0 2.116 0.48 0 2.007 0.42 0.08 1.928 0.37 0.14 1.869 0.34 0.18 1.82

10 0.31 0.22 1.7811 0.29 0.26 1.74

From Exhibit 9A.6 p.341

From Exhibit 9A.6 p.341

X Bar Chart

983.235=(21.733) 0.73-999.1RA - x = LCL

014.965=3)0.73(21.73999.1RA + x = UCL

2

2

1

983.235=(21.733) 0.73-999.1RA - x = LCL

014.965=3)0.73(21.73999.1RA + x = UCL

2

2

1

R Chart

0

49.551

)733.21)(0(RD = LCL

)733.21)(28.2(RD = UCL

3

4

0

49.551

)733.21)(0(RD = LCL

)733.21)(28.2(RD = UCL

3

4

UCL

LCL

Conclusion

• All the points are well within the control limits• The process is controlled

• Acceptance Sampling– sampling to accept or

reject the immediate lot of product at hand

– What percentage of products conform to specifications?

Acceptance Sampling

• Determine(1) n - how many units to sample from a lot

(2) c - the maximum number of defective items that can be found in the sample before the lot is rejected

Acceptance Sampling: Single Sampling Plan

• Acceptable Quality Level (AQL)– Max. acceptable percentage of defectives defined by

producer

• The α (Producer’s risk)– The probability of rejecting a good lot

• Lot Tolerance Percent Defective (LTPD)– Percentage of defectives that defines consumer’s

rejection point

• The (Consumer’s risk)– The probability of accepting a bad lot

4 factors influencing n and c

• Hi-Tech industries manufactures Z-band radar scanners to detect speed traps

• Circuit boards in scanners purchased from outside vendor– Vendor produces boards to an AQL of 2% defectives– Willing to run 5% risk (or less) (α) of having defective lots

rejected

• Hi-Tech considers – Lots of 8% (or more) (LTPD) defectives unacceptable– Wants to ensure that no more than 10% poor quality lot is

accepted

Example, p344

• Given– AQL = 0.02, α= 0.05, LTPD = 0.08, β = 0.10

• Get c, Divide LTPD by AQL– 0.08/0.02 = 4

Determine n and c

• c = 4, n = 99• Take a random sample of 99 units from a lot• Reject the lot if more than 4 units are defective

Sampling plan

Operating Characteristic Curve

n = 99c = 4

AQL LTPD

00.10.20.30.40.50.60.70.80.9

1

1 2 3 4 5 6 7 8 9 10 11 12

Percent defective

Pro

bab

ilit

y of

acc

epta

nce

=.10(consumer’s risk)

= .05 (producer’s risk)

The OCC brings the concepts of producer’s risk, consumer’s risk, sample size, and maximum defects allowed together

The OCC brings the concepts of producer’s risk, consumer’s risk, sample size, and maximum defects allowed together

The shape or slope of the curve is dependent on a particular combination of the four parameters

The shape or slope of the curve is dependent on a particular combination of the four parameters

Bonus Problem

• P. 350, #10– The state and local police departments are trying to

analyze crime rates so they can shift their patrols from decreasing rate areas to areas where rates are increasing. The city and county have been geographically segmented into areas containing 5000 residences. The police recognizes that not all crimes and offenses are reported.

– Every month, because of this, the police are contacting by phone a random sample of 1000 of the 5000 residences for date on crime.

Month Crime Incidence Sample Size Crime Rate

January 7 1000 0.007

February 9 1000 0.009

March 7 1000 0.007

April 7 1000 0.007

May 7 1000 0.007

June 9 1000 0.009

July 7 1000 0.007

August 10 1000 0.010

September 8 1000 0.008

October 11 1000 0.011

November 10 1000 0.010

December 8 1000 0.008

• Construct a p chart for 95% Confidence (1.96) and plot each of the months

• If the next three months show crime incidences in this area as:– January = 10– February = 12– March = 11

• Comment on the crime rate

0.008=12000

100 = p 0.008=12000

100 = p

0.003= 1000

0.008)-0.008(1=

)p-(1 p = p n

s 0.003= 1000

0.008)-0.008(1=

)p-(1 p = p n

s

Calculate the average of the sample proportions

Calculate the standard deviation of the sample proportion

) 1.96(0.003 0.008 ) 1.96(0.003 0.008

UCL = 0.014LCL = 0.002UCL = 0.014LCL = 0.002

p

p

z - p = LCL

z + p = UCL

s

s

p

p

z - p = LCL

z + p = UCL

s

s

• Calculate the control limits

Constructing a p-chart

• Plot the individual sample proportions, the average of the proportions, and the control limits

UCL

LCL

Conclusion

• Crime Rate is within control– Crime rate is going up– January, March, April, May, July has lowest crime

rate– October has highest followed by August and

November

• If the next three months show crime incidences in this area as:– January = 10– February = 12– March = 11

Month Crime Incidence Sample Size Crime Rate

January 8.5 1000 0.007

February 10.5 1000 0.009

March 9 1000 0.007

April 7 1000 0.007

May 7 1000 0.007

June 9 1000 0.009

July 7 1000 0.007

August 10 1000 0.010

September 8 1000 0.008

October 11 1000 0.011

November 10 1000 0.010

December 8 1000 0.008

0.009=12000

105 = p 0.009=12000

105 = p

0.003= 1000

0.009)-0.009(1=

)p-(1 p = p n

s 0.003= 1000

0.009)-0.009(1=

)p-(1 p = p n

s

Calculate the average of the sample proportions

Calculate the standard deviation of the sample proportion

) 1.96(0.003 0.009 ) 1.96(0.003 0.009

UCL = 0.015LCL = 0.003UCL = 0.015LCL = 0.003

p

p

z - p = LCL

z + p = UCL

s

s

p

p

z - p = LCL

z + p = UCL

s

s

• Calculate the control limits

Constructing a p-chart

UCL

LCL

Conclusion

• Crime Rate Within Control– Crime rate going up– January, February, March had increased crime

rate– April, May, July lowest crime rate– October highest, followed by February, followed

by August and November

Thank You