Transcript No Slide Title

1
Choosing the Appropriate Control Chart
Attribute (counts
or classification)
Defect
(count)
Defective
(classification)
Variable (measurable)
The Lean Six Sigma
Pocket Toolbook, p. 123.
(MJ II, p. 37)
2
Advantages of Statistical Control
1. Can predict its behavior.
2. Process has an identity.
3. Operates with less variability.
4. A process having special causes is unstable.
5. Tells workers when adjustments should not be made.
6. Provides direction for reducing variation.
7. Plotting of data allows identifying trends over time.
8. Identifies process conditions that can result in an
acceptable product.
3
source: Juran and Gryna, Quality Planning and Analysis, p. 380-381.
Different types of control charts
Attribute (or count) data
Situation
Number of
defects,
accidents or
flaws
Chart
Control Limits
C
# of accidents/week
# of
breakdowns/week
# of flaws on a
product
U
Lean Six Sigma Pocket Toolbook, p. 132.
4
source: Brian Joiner, Fourth Generation Management, p. 266-267.
Different types of control charts
Attribute (or classification) data
Situation
Fraction of
defectives
Chart
Control Limits
p
fraction of orders not
processed perfectly
on first trial (first pass
yield)
fraction of requests
not processed within
15 minutes
np
Lean Six Sigma Pocket Toolbook, p. 132.
5
source: Brian Joiner, Fourth Generation Management, p. 266-267.
Different types of control charts
Variables (or measurement ) data
Situation
Variables data,
sets of
measurements
Chart
Control Limits
Xbar and R
Charts
X-”BAR” CHART
X  A2 R
R CHART
See MJ II p. 42 for constants
A2, D3 and D4.
UCL  D4 R
Lean Six Sigma Pocket Toolbook, p. 127.
LCL  D3 R
6
source: Brian Joiner, Fourth Generation Management, p. 266-267.
PARAMETERS FOR CREATING X-BAR
CHARTS
Number of
Observations
in Subgroup
(n)
2
3
4
5
6
7
8
9
10
Factor for Xbar Chart
(A2)
1.88
1.02
0.73
0.58
0.48
0.42
0.37
0.34
0.31
Factor for
Lower
control Limit
in R chart
(D3)
0
0
0
0
0
0.08
0.14
0.18
0.22
Factor for
Factor to
Upper
estimate
control limit
Standard
in R chart
deviation, (d2)
(D4)
3.27
1.128
2.57
1.693
2.28
2.059
2.11
2.326
2.00
2.534
1.92
2.704
1.86
2.847
1.82
2.970
1.78
3.078
Lean Six Sigma Pocket Toolbook, p. 128.
7
PROCESS CAPABILITY
 USLx x  LSL 
USL LSL

Cp 
or C pk  min
,
6 * sigma
 3 * sigma 3 * sigma
EXCEL: =NORMDIST(x, mean, std dev,1) to calculate percent non-conforming material.
8
THE STATISTICAL MEANING OF SIX SIGMA
Process capability measure
Upper
Specification
Limit (USL)
Lower
Specification
Limit (LSL)
Process A
(with st. dev sA)
X-3sA
X-2sA
X-1sA
X
X+1sA X+2s
X+3sA
3s
Process B
(with st. dev sB)
X-6sB
X
Cp 
USL  LSL
6sˆ
xs
Cp
P{defect}
ppm
1s
0.33
0.317
317,000
2s
0.67
0.0455
45,500
3s
1.00
0.0027
2,700
4s
1.33
0.0001
63
5s
1.67
0.0000006 0,6
6s
2.00
2x10-9
0,00
X+6sB
sˆ =R/d2
• Estimate standard deviation:
• Look at standard deviation relative to specification limits
• Don’t confuse control limits with specification limits: a process can be out of
control, yet be incapable
9
Exercise An automatic filling machine is used to fill 16
ounce cans of a certain product. Samples of size 5 are
taken from the assembly line each hour and measured.
The results of the first 25 subgroups are shown in the
attached file with selected rows shown below.
Does the process appear to be in statistical control?
Source: Shirland, Statistical Quality Control, problem 5.2.
Filling Weights
subgroup
1
2
3
4
5
1
16.09
15.95
16.07
16.13
16.16
2
16.16
16.00
16.07
16.15
16.11
Sample
3
16.08
15.90
16.08
16.19
16.40
4
16.02
16.17
15.89
16.13
16.14
5
16.11
16.01
16.28
16.19
15.86
Average
16.09
16.01
16.08
16.16
16.13
Range
0.14
0.27
0.39
0.06
0.54
If the specification limits are USL = 16.539 and LSL = 15.829 is the
process capable?
10
STATISTICAL PROCESS CONTROL
Capability
Analysis
Eliminate
Assignable Cause
Conformance
Analysis
Investigate for
Assignable Cause
Capability analysis
• What is the currently "inherent" capability of my process when it is "in control"?
Conformance analysis
• SPC charts identify when control has likely been lost and assignable cause
variation has occurred
Investigate for assignable cause
• Find “Root Cause(s)” of Potential Loss of Statistical Control
Eliminate or replicate assignable cause
• Need Corrective Action To Move Forward
11
Identifying Special Causes of Variation
source: Brian Joiner, Fourth Generation Management, pp. 260.
See also Lean Six Sigma
Pocket Toolbook, p. 133-135.
12
Strategies for Reducing Special Causes of Variation
• Get timely data so special causes are signaled
quickly.
• Put in place an immediate remedy to contain any
damage.
• Search for the cause -- see what was different.
• Develop a longer term remedy.
13
source: Brian Joiner, Fourth Generation Management, pp. 138-139.
“In a common cause
situation, there is no such
thing as THE cause.”
Brian Joiner
14
Improving a Stable Process
• Stratify -- sort into groups or categories; look for
patterns. (e.g., type of job, day of week, time, weather,
region, employee, product, etc.)
• Experiment -- make planned changes and learn from
the effects. (e.g., need to be able to assess and learn
from the results -- use PDCA .)
• Disaggregate -- divide the process into component
pieces and manage the pieces. (e.g., making the
elements of a process visible through measurements
and data.)
15
source: Brian Joiner, Fourth Generation Management, pp. 140-146.
DESIGN OF EXPERIMENTS
Purposeful changes of the inputs (factors) to a
process in order to observe corresponding changes in
the output (response).
Inputs
Process
Outputs
16
Douglas Montgomery, Design and Analysis of Experiments
WHY USE DOE ?
• A basis of action -- allows purposeful changes.
• An analytic study -- one in which action will be taken
on a cause-and-effect system to improve
performance of a product or process in the future.
• Follows the scientific approach to problem solving.
• Provides a way to measure natural variation.
• Permits the clear analysis of complex effects.
• Most efficient way to derive the required information
at the least expenditure of resources.
Moen, Nolan and Provost, Improving Quality Through Planned Experimentation
17
INTERACTIONS
Varying factors together vs. one at a time.
B
+
U
C
K
--
DOE
+
18
George Box, Do Interactions Really Matter, Quality Engineering, 1990.
Voila!
B
+
U
C
K
- -
DOE
+
19
George Box, Do Interactions Really Matter, Quality Engineering, 1990.
• Experiment run at SKF -- largest producer of rolling
bearing in the world.
• Looked at three factors: heat treatment, outer ring
osculation and cage design.
• Results:
•choice of cage design did not matter (contrary to
previously accepted folklore -- considerable savings)
•life of bearing increased five fold if osculation and
heat treatment are increased together -- saved
millions of dollars !
20
George Box, Do Interactions Really Matter, Quality Engineering, 1990.
• Bearings like this have been made for decades. Why did
it take so long to discover this improvement ? One factor
vs. interaction effects !
16
128
26
85
Heat
19
21
Cage
17
Osculation
25
21
George Box, Do Interactions Really Matter, Quality Engineering, 1990.
The Power of Interactions !
21
106
18
23
Heat
Osculation
22
George Box, Do Interactions Really Matter, Quality Engineering, 1990.
2
2 Design Example
Consider an investigation into the effect of the
concentration of the reactant and the amount of
catalyst on the reaction time of a chemical process.
L
reactant (factor A)
catalyst (factor B)
15%
1 bag
H
25%
2 bags
23
Douglas Montgomery, Design and Analysis of Experiments
Design Matrix for 22
A
B
AB
-
-
+
+
-
-
-
+
-
+
+
+
Main effects
Total Average
Interaction
24
I
Replicates
II
III
Total
Factor
A-B-
28
25
27
80
Settings
A+B-
36
32
32
100
A-B+
18
19
23
60
A+B+
31
30
29
90
25
Douglas Montgomery, Design and Analysis of Experiments
An effect is the difference in the average response at
one level of the factor versus the other level of the
factor.
60
90
A effect =
( [90 + 100] - [60 + 80] )
/ 2(3) = 8.33
100
80
-
+
A
26
Douglas Montgomery, Design and Analysis of Experiments
Use a matrix to find the effects of each factor,
including the interaction effect between the two
factors.
A
B
AB
Total Average
-
-
+
80
26.7
+
-
-
100
33.3
-
+
-
60
20
+
+
+
90
30
Avg +
31.7
Avg -
23.3
Effect
8.4
27
Douglas Montgomery, Design and Analysis of Experiments
Completing the matrix with the effect calculations:
A
B
AB
Total Average
-
-
+
80
26.7
+
-
-
100
33.3
-
+
-
60
20
+
+
+
90
30
Avg +
31.7
25
28.3
Avg -
23.3
30
26.7
Effect
8.4
-5
1.7
28
Douglas Montgomery, Design and Analysis of Experiments
Dot Diagram
B
-10
-5
AB
0
A
5
10
29
Douglas Montgomery, Design and Analysis of Experiments
Response Plots
35
35
30
30
25
25
20
20
-
+
A
-
+
B
30
Douglas Montgomery, Design and Analysis of Experiments
Interaction
35
B-
30
B+
25
Response
Plot
20
B-
-
B+
+
A
A-
26.7
20
A+
33.3
30
31
Douglas Montgomery, Design and Analysis of Experiments
Prediction Equation
The ‘intercept’ in the equation is the overall average of
all observations.
The coefficients of the factors in the model are 1/2 the effect.
Y=
27.5 + 8.33/2 A - 5/2 B + 1.7/2 AB
or
Y=
27.5 + 4.165 A - 2.5 B + 0.85 AB
note: A and B will be values between -1 and +1.
32
Analysis of Variance
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square
F
A
B
AB
Error
208.33
75.00
8.33
31.34
1
1
1
8
208.33
75.33
8.33
3.92
53.15 *
19.13 *
2.13
Total
323.00
11
* = significant at 1% (see F table)
33
Calculating SS, df and MS for Effects and Interactions
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square
F
A
208.33
1
208.33
53.15 *
always 1 for
this type design
SS = Effect 2 x n
= 8.332 x 3
where n = replicates
Use this same process for A, B and AB
SS / df
34
Calculating total sum of squares and total degrees of freedom
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square
F
Total
323.00
11 Total df = n - 1 = 12 - 1 = 11
This is found by adding up every squared observation and then
subtracting what is called a correction factor (sum of all observations,
square this amount, then divide by the number of observations).
SST = 282 + 252 + 272 + ... + 292 - (3302 / 12)
= 9398.0 - 9075.0 = 323.0
35
Calculating error sum of squares, df and mean square
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square
Error
31.34
8
3.92
Found by subtraction:
Total SS - SSA - SSB - SSAB
= 323 - 208.33 - 75.0 - 8.33
= 31.34
F
SS / df
Found by subtraction:
= Total df - A df - B df - AB df
= 11 - 1 - 1 - 1 = 8
36
Calculating F ratios
Source of
Variation
Sum of
Squares
A
B
AB
Error
F ratios:
F = MS (A or B or AB)
------------------------MS (error)
Degrees of
Freedom
Mean
Square
F
208.33
75.33
8.33
3.92
53.15 *
19.13 *
2.13
208.33 / 3.92
75.33 / 3.92
8.33 / 3.92
Compare to F table
37
Interpreting F ratios
F table
at num df = 1
and denom df = 8
F .25
F .10
F .05
F .025
F .01
1.54
3.46
5.32
7.57
11.26
• F ratios confirm that
factors A and B are
significant at the 1%
level.
• F ratio shows there is
not a significant
interaction.
38
WHY USE 2K DESIGNS ?
• Easy to use and data analysis can be performed
using graphical methods.
• Relatively few runs required.
• 2k designs have been found to meet the majority of
the experimental needs of those involved in the
improvement of quality.
• 2k designs are easy to use in sequential
experimentation.
• Fractions of the 2k (fractional factorials) can be used
to further reduce the experiment size.
39
Moen, Nolan and Provost, Improving Quality Through Planned Experimentation