1 slides used in class may be different from slides in student pack chapter 9a process capability...

1Slides used in class may be different from slides in student pack

Chapter 9A

Process Capability and Statistical Quality Control Process Variation Process Capability Process Control Procedures

– Variable data– Attribute data

Acceptance Sampling– Operating Characteristic Curve


Basic Causes of Variation

Assignable causes are factors that can be clearly identified and possibly managed.

Common causes are inherent to the production process. In order to reduce variation due to common causes, the process must be changed.

Key: Determining which is which!


Types of Control Charts Attribute (Go or no-go information)

– Defectives refers to the acceptability of product across a range of characteristics.

– p-chart application

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

deviation.– X-bar and R chart applications


Types of Statistical Quality Control

StatisticalQuality Control

ProcessControl

AcceptanceSampling

VariablesCharts

AttributesCharts

Variables Attributes

StatisticalQuality Control

ProcessControl

AcceptanceSampling

VariablesCharts

AttributesCharts

Variables Attributes


UCL

LCL

Samples over time

1 2 3 4 5 6

Normal Behavior

UCL

LCL

Samples over time

1 2 3 4 5 6

Possible problem, investigate

UCL

LCL

Samples over time

1 2 3 4 5 6

Possible problem, investigate

Statistical Process Control (SPC) Charts

Excellent review in exhibit TN8.5.

Look for trends!


Control Limits

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

xLCL UCL

99.7%


Example of Constructing a p-Chart: Required Data

1 100 42 100 23 100 54 100 35 100 66 100 47 100 38 100 79 100 1

10 100 211 100 312 100 213 100 214 100 815 100 3

SampleSample size

Number of defectives


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

p =Total Number of Defectives

Total Number of Observations

ns

)p-(1 p = p

p

p

z - p = LCL

z + p = UCL

s

s

Given:

Compute control limits:


1. Calculate the sample proportions, p (these are what can be plotted on the p-chart) for each sample.

Sample n Defectives p1 100 4 0.042 100 2 0.023 100 5 0.054 100 3 0.035 100 6 0.066 100 4 0.047 100 3 0.038 100 7 0.079 100 1 0.01

10 100 2 0.0211 100 3 0.0312 100 2 0.0213 100 2 0.0214 100 8 0.0815 100 3 0.03

Example of Constructing a p-chart: Step 1


2. Calculate the average of the sample proportions.

0.037=1500

55 = p

3. Calculate the standard deviation of the sample proportion

.0188= 100

.037)-.037(1=

)p-(1 p = p n

s

Example of Constructing a p-chart: Steps 2&3


4. Calculate the control limits.

3(.0188) .037

UCL = 0.0930LCL = -0.0197 (0)

p

p

z - p = LCL

z + p = UCL

s

s

Example of Constructing a p-chart: Step 4


Example of Constructing a p-Chart: Step 55. Plot the individual sample proportions, the average

of the proportions, and the control limits

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Observation

p

p

UCL

LCL

p-bar


R Chart

Type of variables control chart– Interval or ratio scaled numerical data

Shows sample ranges over time– Difference between smallest & largest values in

inspection sample

Monitors variability in process Example: Weigh samples of coffee &

compute ranges of samples; Plot


R Chart Control Limits

k

RR

RDLCL

RDUCL

k

1ii

3R

4R

Sample Range Sample Range in samplein sample i i

# Samples# Samples

From Table From Table (function of sample (function of sample size)size)


R Chart Example

You’re manager of a 500-room hotel. You want to analyze the time it takes to deliver luggage to the room. For 7 days, you collect data on 5 deliveries per day. Is the process in control?


R Chart Hotel Data

SampleDay Delivery Time Mean Range

1 7.30 4.20 6.10 3.45 5.55 5.32

7.30 + 4.20 + 6.10 + 3.45 + 5.557.30 + 4.20 + 6.10 + 3.45 + 5.55 5 5

Sample Mean = Sample Mean =


R Chart Hotel Data


1 7.30 4.20 6.10 3.45 5.55 5.32 3.85

7.30 - 3.457.30 - 3.45Sample Range = Sample Range =

LargestLargest SmallestSmallest


R Chart Hotel Data


1 7.30 4.20 6.10 3.45 5.55 5.32 3.852 4.60 8.70 7.60 4.43 7.62 6.59 4.273 5.98 2.92 6.20 4.20 5.10 4.88 3.284 7.20 5.10 5.19 6.80 4.21 5.70 2.995 4.00 4.50 5.50 1.89 4.46 4.07 3.616 10.10 8.10 6.50 5.06 6.94 7.34 5.047 6.77 5.08 5.90 6.90 9.30 6.79 4.22


R Chart Control Limits Solution

0894.303

216.8894.311.2

89437

224274853

4

1

RDRLCL

RDUCL

....

k

RR

R

k

ii

From Table From Table ((nn = 5) = 5)


02468

1 2 3 4 5 6 7

R, Minutes

Day

R Chart Control Chart Solution

UCLUCL

R-barR-bar


X Chart

Type of variables control chart– Interval or ratio scaled numerical data

Shows sample means over time Monitors process average Example: Weigh samples of coffee &

compute means of samples; Plot


X Chart Control Limits

k

RR

k

XX

RAXLCL

RAXUCL

k

1ii

k

1ii

2X

2X

Range Range of of samplesample i i

# Samples# Samples

Mean of Mean of sample sample ii

From From TableTable


X Chart Example

You’re manager of a 500-room hotel. You want to analyze the time it takes to deliver luggage to the room. For 7 days, you collect data on 5 deliveries per day. Is the process in control?


X Chart Hotel Data


1 7.30 4.20 6.10 3.45 5.55 5.32 3.852 4.60 8.70 7.60 4.43 7.62 6.59 4.273 5.98 2.92 6.20 4.20 5.10 4.88 3.284 7.20 5.10 5.19 6.80 4.21 5.70 2.995 4.00 4.50 5.50 1.89 4.46 4.07 3.616 10.10 8.10 6.50 5.06 6.94 7.34 5.047 6.77 5.08 5.90 6.90 9.30 6.79 4.22


X Chart Control Limits Solution*

555.3894.358.0813.5

071.8894.358.0813.5

894.37

22.427.485.3

813.57

79.659.632.5

2

2

1

1

RAXLCL

RAXUCL

k

RR

k

XX

X

X

k

ii

k

ii

From TableFrom Table

((nn = 5) = 5)


X ChartControl Chart Solution*

02468

1 2 3 4 5 6 7

X, Minutes

Day

UCLUCL

LCLLCL

X-barX-bar


X AND R CHART EXAMPLEIN-CLASS EXERCISE

The following collection of data represents samples of the amount of force applied in a gluing process:

Determine if the process is in control

by calculating the appropriate upper and lower

control limits of the X-bar and R charts.


X AND R CHART EXAMPLEIN-CLASS EXERCISE

Sample Obs 1 Obs 2 Obs 3 Obs 4 Obs 51 10.68 10.689 10.776 10.798 10.7142 10.79 10.86 10.601 10.746 10.7793 10.78 10.667 10.838 10.785 10.7234 10.59 10.727 10.812 10.775 10.735 10.69 10.708 10.79 10.758 10.6716 10.75 10.714 10.738 10.719 10.6067 10.79 10.713 10.689 10.877 10.6038 10.74 10.779 10.11 10.737 10.759 10.77 10.773 10.641 10.644 10.72510 10.72 10.671 10.708 10.85 10.71211 10.79 10.821 10.764 10.658 10.70812 10.62 10.802 10.818 10.872 10.72713 10.66 10.822 10.893 10.544 10.7514 10.81 10.749 10.859 10.801 10.70115 10.66 10.681 10.644 10.747 10.728


Example of x-bar and R charts: Step 1. Calculate sample means, sample ranges, mean of means, and mean of ranges.

Sample Obs 1 Obs 2 Obs 3 Obs 4 Obs 5 Avg Range1 10.68 10.689 10.776 10.798 10.714 10.732 0.1162 10.79 10.86 10.601 10.746 10.779 10.755 0.2593 10.78 10.667 10.838 10.785 10.723 10.759 0.1714 10.59 10.727 10.812 10.775 10.73 10.727 0.2215 10.69 10.708 10.79 10.758 10.671 10.724 0.1196 10.75 10.714 10.738 10.719 10.606 10.705 0.1437 10.79 10.713 10.689 10.877 10.603 10.735 0.2748 10.74 10.779 10.11 10.737 10.75 10.624 0.6699 10.77 10.773 10.641 10.644 10.725 10.710 0.13210 10.72 10.671 10.708 10.85 10.712 10.732 0.17911 10.79 10.821 10.764 10.658 10.708 10.748 0.16312 10.62 10.802 10.818 10.872 10.727 10.768 0.25013 10.66 10.822 10.893 10.544 10.75 10.733 0.34914 10.81 10.749 10.859 10.801 10.701 10.783 0.15815 10.66 10.681 10.644 10.747 10.728 10.692 0.103

Averages 10.728 0.220400


Example of x-bar and R charts: Step 2. Determine Control Limit Formulas and Necessary Tabled Values

x Chart Control Limits

UCL = x + A R

LCL = x - A R

2

2

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


Example of x-bar and R charts: Steps 3&4. Calculate x-bar Chart and Plot Values

60110220405872810

85610220405872810

2

2

.)=.(-..R - AxLCL =

.)=.(..R + AxUCL =

10.550

10.600

10.650

10.700

10.750

10.800

10.850

10.900

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Sample

Mea

ns

SamplemeanUCL

LCL

grandmean of x


Example of x-bar and R charts: Steps 5&6: Calculate R-chart and Plot Values

0

0.46504

)2204.0)(0(R D= LCL

)2204.0)(11.2(R D= UCL

3

4

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

0.800

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Sample

R

RangeUCLLCLR-bar


SOLUTION:Example of x-bar and R charts:

1. Is the process in Control?

2. If not, what could be the cause for the process being out of control?


Process Capability

Process limits - actual capabilities of process based on historical data

Tolerance limits - what process design calls for- desired performance of process


Process Capability

How do the limits relate to one another?

You want: tolerance range > process range

Two methods of accomplishing this:

1. Make bigger 2. Make smaller

Bad idea Implies having greater control over processGood!


Process Capability Measurement

Cp index = Tolerance range / Process range

What value(s) would you like for Cp?

Larger Cp indicates a more reliable and predictable process (less variability)

The Cp index is based on the assumption that the process mean is centered at the midpoint of the tolerance range


-4

LTL UTL

6)( LTLUTL

C p

X

16

6

pC


26

12

pC

-4

LTL UTL

X


While the Cp index provides useful information on process variability, it does not give information on the process average relative to the tolerance limits. Note:

-4

9

UTLLTL

26

12

pC

X


Cpk Index

3

X-UTLor

3

LTLXmin=Cpk

Together, these process capability Indices show how well parts being produced conform to design specifications.

X = process mean (Unknown but can be estimated with the grand mean)

= standard deviation (Unknown but can be estimated with the average range)

Refers to the LTL Refers to the UTL


13

3,

3

9min

pkC

26

12

pC

-4

9

X

LTL UTLSince Cp and Cpk are different we can conclude that the process is not centered, however the Cp index tells us that the process variability is very low


An example of the use of process capability indicesThe design specifications for a machined slot is 0.5± .003 inches. Samples have been taken and the process mean is estimated to be .501. The process standard deviation is estimated to be .001.

What can you say about the capability of this process to produce this dimension?

6

LTLUTLC p

1001.6

497.503.

pC

3

X-UTL ,

3

LTLXmin=Cpk

667..0013

.501-.503 ,

001.3

497.501.min=Cpk


Process capability

0.497 inchesLTL

0.503 inchesUTL

Process mean0.501 inches

Machined slot(inches)

= 0.001inches


Sampling Distributions(The Central Limit Theorem)

Regardless of the underlying distribution, if the sample is large enough (>30), the distribution of sample means will be normally distributed around the population mean with a standard deviation of :

nx

/


Computing Process Capability Indexes Using Control Chart DataRecall the following info from our in class

exercise:

Since A2 is calculated on the assumption of three sigma limits:

728.10X

22.R

58.2 A

043.3)22(.58.

32 RA

x


From the Central Limit Theorem:

So,

Therefore,

nx

5043.

096.5043.


Suppose the Design Specs for the Gluing Process were 10.7 .2, Calculate the Cp and Cpk Indexes:

Answer:

694.096.6

5.109.10

6

LTLUTL

C p

3,

3min

XUTLLTLXC pk

597.597.,792.min096.3

728.109.10,

096.3

5.10728.10min

pkC


-4

1.792.8

Note, multiplying each component of the Cpk calculation by 3 yields a Z value. You can use this to predict the % of items outside the tolerance limits:

From Appendix E we would expect:

.008 + .036 = .044 or 4.4%

non-conforming product from this process

.792 * 3 = 2.38 .597 * 3 = 1.79

.008 or .8% of the curve .036 or 3.6% of the curve


Capability Index – In Class Exercise

You are a manufacturer of equipment. A drive shaft is purchased from a supplier close by. The blueprint for the shaft specs indicate a tolerance of 5.5 inches ± .003 inches. Your supplier is reporting a mean of 5.501 inches. And a standard deviation of .0015 inches.

What is the Cpk index for the supplier’s process?


3,

3min

XUTLLTLXC pk

444.444.,888.min0015.3

501.5503.5,

0015.3

497.5501.5min

pkC


Your engineering department is sent to the supplier’s site to help improve the capability on the shaft machining process. The result is that the process is now centered and the CP index is now .75. On a percentage basis, what is the improvement on the percentage of shafts which will be unusable (outside the tolerance limits)?


To answer this question we must determine the percentage of defective shafts before and after the intervention from our engineering department


Before:

-4

x.1.3x.88) =2.67

444.444.,888.min0015.3

501.5503.5,

0015.3

497.5501.5min

pkC

From Table

.089

From Table

.004

Total % outside

Tolerance = .089 + .004 = .093 or 9.3%


AfterSince the process is centered then Cpk = Cp; Cp = UTL-LTL / 6so the tolerance limits are .75 x 6 = 4.5 apart each 2.25 from the mean

-4

2.22.25

From Table

.012

So % outside of

Tolerance =

.012(2) = .024

Or 2.4%


So the percentage decrease in defective parts is 1 – (2.4/9.3) = 74%


Additional Questions A circular bar is produced on a draw bench by drawing the bar through a die. The bar has an UTL of 3.06 inches and a LTL of 2.94 inches. The process is in control and normally distributed. The following capability indices have been computed:

5.03

XUTL

5.13

LTLX

What is the CP index? What is the process mean (X bar bar)? Suppose 1000 bars were produced with this process. How many bars would you expect to be within the specification limits?


Basic Forms of Statistical Sampling for Quality Control Sampling to accept or reject the immediate

lot of product at hand (Acceptance Sampling).

Sampling to determine if the process is within acceptable limits (Statistical Process Control)


Acceptance Sampling Purposes

– Determine quality level– Ensure quality is within predetermined level

Advantages– Economy– Less handling damage– Fewer inspectors– Upgrading of the inspection job– Applicability to destructive testing– Entire lot rejection (motivation for improvement)


Acceptance Sampling

Disadvantages– Risks of accepting “bad” lots and rejecting

“good” lots– Added planning and documentation– Sample provides less information than 100-

percent inspection – No information is obtained on the process. Just

sorting “good” parts from “bad” parts


Risk

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

defined by producer. (Producer’s risk)

– The probability of rejecting a good lot. Lot Tolerance Percent Defective (LTPD)

– Percentage of defectives that defines consumer’s rejection point.

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

1 slides used in class may be different from slides in student pack chapter 9a process capability...

Documents

student pack control

student pack example

p chart

student pack chapter

compute control limits

control limits sl

student pack ucl lcl

upper control limits