Transcript Session 3 and 4 Powerpoint Presentation File

Continuous Improvement
& Six Sigma Green Belt
Certification
Virtual Session 3 (Keisen v.2016.03.12m)
Review
•
•
•
•
•
•
Standard Deviation
Statistical 6 Sigma
DPU, DPO, DPMO, Defective%, PPM
Sigma Level
Descriptive Statistics
Quincuncx “artifact”
Brain Flex
Standard Deviation and its meaning
• What are your conclusions of the
following graph where the mean value
of the process is 28 minutes and the
standard deviation is 0.3 minutes
R:
•
Most of the times (99.7%), the process is
executed between 27.1 and 28.9 minutes.
•
We can estimate that for the next 100 services,
95.4% of them are expected to execute between
27.4 and 28.6 minutes.
Brain Flex
σ= 0.3
30
N=100
25
20
f 15
Mean = 28.0
10
Std Deviation = 0.3
5
27.4
27.1
27.7
28.0
68.3%
95.4 %
99.7%
29
.1
5
28
.9
5
28
.7
5
28
.5
5
28
.3
5
28
.1
5
27
.9
5
27
.7
5
27
.5
5
27
.3
5
27
.1
5
26
.9
5
26
.7
5
0
min.
28.3
28.6
28.9
Process Variation vs Specifications
• Different problems = Different Solutions
1
BIAS
Deviation vs Center
2
DISPERSION
3
ABNORMALITY
(Assignable cause)
Video
How to reduce variation?
Video: Reducing variation & dealing with
abnormalities
• Have you seen a Gauss “bell-shaped” curve?
– Use of Quincuncx to explain:
• Bell-shaped curve of a normal process.
• How to reduce variation? = Control of more variables.
• How to mess up a process? = Non-evidence based actions and
decisions.
• How to understand an abnormality?
– Use of Quincuncx to explain:
• Abnormalities and the need to investigate
Video
Benefits of understanding variation
Multistep process measurement
Video: Importance of measuring variation
• We usually know the average time of a 3 stage process can be
estimated adding the average values of each stage.
• However, how can we estimate the variation of the total time of
this 3 stage process?
– Use of Quincuncx and White board to explain:
• 3 different Stages or process steps with same mean but different
standard deviation.
– Total time mean estimation by “ADDITIVITY OF THE MEAN”.
– How to estimate the Final Standard Deviation as the “ADDITIVITY
OF THE STANDARD DEVIATION” will NOT give an accurate
value=
Video: Additivity of mean and variance
P1
Average
time 1
µ1=20
Whole
Process
(minutes)
P2
Average
time 2
µ2=30
Total average
time
µ=20+30+10=60
P3
Average
time 3
µ3=10
Video: Additivity of mean and variance
P1
• µ1=20
• Var1=4
• σ1=2
Whole
Process
(minutes)
P2
•
•
•
•
• µ2=30
• Var2=25
• σ2=5
µ=20+30+10= 60
Var = σ12+σ22+σ32
Var=4+25+1=30
σ= 30= 5.48
P3
• µ3=10
• Var3=1
• σ3=1
Variance = σ2
Standard Deviation = 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = σ2
Video: Additivity of mean and variance
P1
• µ1=20
• Var1=4
• σ1=2
P2
• µ2=30
• Var2=25
• σ2=5
P3
• µ3=10
• Var3=1
• σ3=1
σtotal= 5.48
Whole
Process
(minutes)
•
•
•
•
µ=20+30+10= 60
Var = σ12+σ22+σ32
Var=4+25+1=30
σ= 30= 5.48
σ2
Variance =
Standard Deviation = 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = σ2
43.56 min
60 min
Total spread estimation
76.44 min
Brain Re-Flex
Variances Additivity Property
• We need to estimate total average time and variation
of a 4 stage process.
• P1: Mean1= 5 min StandDev1= 2 min.
• P2: Mean2= 4 min. StandDev2= 2 min.
• P3: Mean2= 6 min. StandDev2= 2 min.
• P4: Mean2= 5 min. StandDev2= 2 min.
R:
• Means (averages) additivity property exists so total mean is
estimated by adding each stage´s mean (20 min).
• Also, Variances additivity property exists so in order to
estimate the whole process standard deviation, we must add
each stage´s variance (16 squared minutes).
• DO NOT add independent standard deviations to calculate
total processes SD.
• Total Standard Deviation is 4 minutes.
• What does this mean? How consistent is our process?
Probability Distribution
Means and Standard Deviations
Probability distribution is simply a
theoretical frequency distribution
characterized by Mean and Standard
Deviation.
σ
Normal Distribution is
represented by µ and σ.
Calculation of z-value (z score)
𝑥−𝜇
𝑧=
𝜎
Where:
x = Value of the data point of interest
µ = Mean of the population or data points
σ = Standard Deviation of the data points
Z = Number of standard deviations between x & the
mean (µ). z-value is negative when the data
point of interest is below the mean.
The z-value?
Assume the mean of service process A is 28.0
min. with a standard deviation of 0.3 min.
Calculate the z-value for process A service times:
a)
27.7 minutes
b)
28.3 minutes
c)
28.6 minutes
d)
29.2 minutes
µ=28.0 minutes and σ=0.3 minutes
σ
+0.08%
R:
+0.08%
27.7−28.0
a) Za=
= -1
b) zb=
=1
0.3
28.3−28.0
c) Zc=
d) Zc=
4σ
0.3
28.6−28.0
0.3
29.2−28.0
0.3
=2
=4
0.15%
0.15%
27.1
27.4
27.7
28.0
28.3
28.6
28.9
28.9
The z-value & Probability?
Assume the mean of service process A is 28.0
minutes with a standard deviation of 0.3 minutes.
Given the z-value for process A services times,
estimate the probability of finishing:
a)
Before 27.7 minutes (z= -1)
b)
Before 28.3 minutes (z= 1)
c)
After 28.6 minutes (z=2)
d)
After 29.2 minutes (z=4)
R:
a) 15.87% of the services are finishing before
27.7 minutes.
b) 84.13% of the services are delivered before
28.3 minutes.
c) As 97.72% of the services are finished by 28.6
minutes, 2.28% of the services might be
delivered after 28.6 minutes.
d) 30 services per million opportunities (0.003%)
of the services might deliver after 29.2 mins.
µ=28.0 minutes and σ=0.3 minutes
a)
σ
Probability of finishing
before 27.7 min or z= 1?
R: 15.87%
+0.08%
+0.08%
4σ
+0.003%
0.15%
0.15%
27.1
27.4
27.7
28.0
28.3
28.6
28.9
29.2
Using Normal Distribution Tables
•
•
If z-value=-1.00 search the first column for the
number -1.0 (one decimal) then each one of the
following columns define the second and third
decimal of the z-value.
In our case, z=-1 so the probability of services
finishing before z=-1 is equal to 15.87%
The z-value & Probability?
Assume the mean of service process A is 28.0
minutes with a standard deviation of 0.3 minutes.
Given the z-value for process A services times,
estimate the probability of finishing:
a)
Before 27.7 minutes (z= -1)
b)
Before 28.3 minutes (z= 1)
c)
After 28.6 minutes (z=2)
d)
After 29.2 minutes (z=4)
R:
a) 15.87% of the services are finishing before 27.7
minutes.
b) 84.13% of the services are delivered before
28.3 minutes.
c) As 97.72% of the services are finished by 28.6
minutes, 2.28% of the services might be
delivered after 28.6 minutes.
d) 30 services per million opportunities (0.003%)
of the services might deliver after 29.2 mins.
µ=28.0 minutes and σ=0.3 minutes
b) Before 28.3
min or z= 1?
R: 84.13%
σ
+0.08%
+0.08%
4σ
+0.003%
0.15%
0.15%
27.1
27.4
27.7
28.0
28.3
28.6
28.9
29.2
The z-value & Probability?
Assume the mean of service process A is 28.0
minutes with a standard deviation of 0.3 minutes.
Given the z-value for process A services times,
estimate the probability of finishing:
a)
Before 27.7 minutes (z= -1)
b)
Before 28.3 minutes (z= 1)
c)
After 28.6 minutes (z=2)
d)
After 29.2 minutes (z=4)
R:
a) 15.87% of the services are finishing before 27.7
minutes.
b) 84.13% of the services are delivered before
28.3 minutes.
c) As 97.72% of the services are finished by
28.6 minutes, 2.28% of the services might be
delivered after 28.6 minutes.
d) 30 services per million opportunities (0.003%)
of the services might deliver after 29.2 mins.
µ=28.0 minutes and σ=0.3 minutes
c) Over 28.6 min or z=2?
R: 1-97.72% = 2.28%
σ
+0.08%
+0.08%
4σ
+0.003%
0.15%
0.15%
27.1
27.4
27.7
28.0
28.3
28.6
28.9
29.2
The z-value & Probability?
Assume the mean of service process A is 28.0
minutes with a standard deviation of 0.3 minutes.
Given the z-value for process A services times,
estimate the probability of finishing:
a)
Before 27.7 minutes (z= -1)
b)
Before 28.3 minutes (z= 1)
c)
After 28.6 minutes (z=2)
d)
After 29.2 minutes (z=4)
R:
a) 15.87% of the services are finishing before 27.7
minutes.
b) 84.13% of the services are delivered before
28.3 minutes.
c) As 97.72% of the services are finished by 28.6
minutes, 2.28% of the services might be
delivered after 28.6 minutes.
d) 30 services per million opportunities
(0.003%) of the services might deliver after
29.2 mins.
µ=28.0 minutes and σ=0.3 minutes
d) Over 29.2 min or z=4?
R: 100-99.997= 0.003%
σ
+0.08%
+0.08%
4σ
+0.003%
0.15%
0.15%
27.1
27.4
27.7
28.0
28.3
28.6
28.9
29.2
How to calculate with MS Excel?
Function
Excel English
Excel Spanish
NORMAL DISTRIBUTION
FOR SPECIFIED MEAN AND
STD. DEVIATION
NORMDIST (x, mean,
standard deviation, TRUE)
DIST.NORM (x, media,
desviación estandar,
VERDADERO)
Function that estimates the
probability (to the left) given a
mean and a standard deviation.
X: The value for which you want the distribution
Mean: Arithmetic mean or average of the distribution
Standard deviation: Standard deviation of the distribution (a positive number)
TRUE: A logical value for the cumulative distribution function.
If you want the probability density function, then use “false” (i.e. use false when you want to calculate the “y” value
of the “x” value in a Normal Distribution Curve).
Brain Flex
How to estimate probable service
process output?
Assume the mean of service process A is
28.0 minutes with a standard deviation of
0.3 minutes. Estimate the % of process A
services finishing:
a)
b)
c)
Before 27.4 minutes
Between 27.4 and 28.3 minutes
Over 28.9 minutes
What is the probability?
Assume the mean of service process A is
28.0 minutes with a standard deviation of 0.3
minutes. Estimate the % of process A
services finishing:
a)
Before 27.4 minutes
b)
Between 27.4 and 28.3 minutes
c)
Over 28.9 minutes
µ=28.0 minutes and σ=0.3 minutes
a) Before 27.4 min?
σ
0.05%
0.05%
R:
a) 0.05+0.1+2.1 = 2.25% (in graph due to
rounding). In table: 2.28%
b) 13.6+34.1+34.1 = 81.8%
c) 1-(0.15+99.7) = 0.15%
0.15%
0.15%
27.1
27.4
27.7
28.0
28.3
28.6
28.9
What is the probability?
Assume the mean of service process A is
28.0 minutes with a standard deviation of 0.3
minutes. Estimate the % of process A
services finishing:
a)
Before 27.7 minutes
b)
Between 27.4 and 28.3 minutes
c)
Over 28.9 minutes
µ=28.0 minutes and σ=0.3 minutes
b) Between
27.4 and
28.3 min?
σ
0.05%
0.05%
R:
a) 0.05+0.1+2.1+13.6 = 15.85%
b) 13.6+34.1+34.1 = 81.8%
c) 1-(0.15+99.7) = 0.15%
0.15%
0.15%
27.1
27.4
27.7
28.0
28.3
28.6
28.9
What is the probability?
Assume the mean of service process A is
28.0 minutes with a standard deviation of 0.3
minutes. Estimate the % of process A
services finishing:
a)
Before 27.7 minutes
b)
Between 27.4 and 28.3 minutes
c)
Over 28.9 minutes
µ=28.0 minutes and σ=0.3 minutes
c) Over 28.9 min?
σ
0.05%
0.05%
R:
a) 0.05+0.1+2.1+13.6 = 15.85%
b) 13.6+34.1+34.1 = 81.8%
c) 1-(0.15+99.7) = 0.15%
0.15%
0.15%
27.1
27.4
27.7
28.0
28.3
28.6
28.9
Focus of CI & Six Sigma: Y=f(X)
Measuring and improving “Y”
Current Situation?
THE RESULTS
How well did it go?
“Y”
FOCUS ON THE
CAUSE (X),
Not the Result (Y).
How are the processes?
THE FACTORS THAT GENERATE
THE RESULTS
How well were resources allocated?
“X”
Pareto Graph
Type of Claim
Frequency
Cumulative
frequency
%
Cumulative
%
A:
44
44
36.7 %
36.7 %
B:
28
72
23.3 %
60.0 %
C:
25
97
20.8 %
80.8 %
D:
10
107
8.3 %
89.1 %
E:
5
112
4.2 %
93.3 %
Others
8
120
6.7%
100.0 %
TOTAL
120
100.0 %
PARETO GRAPH
100.0%
120
110
93.3%
89.1%
n=120,
Dec. 2015
120
80.8%
100
107
90
80
70
60.0%
70
50 %
36.7%
50
30
40
44
30
44
20
28
25
10
20
10
5
8
0
10
0
A
B
Pareto Graph Objectives:
1. Stratify the whole into its parts for
better understanding of the
situation.
2. Be able to Prioritize important in
order to Focus on key issues.
60
72
60
40
90
97
80
Frequency
112
100
C
D
E
Others
Interpretation:
•
Claim type A represents 36.7% of
all claims.
•
Claims A, B &C represent 80.8%
of all claims.
•
From the frequency standpoint,
our priority is to reduce claims
type A,B&C.
PARETO GRAPH (Stratified by A)
40
100.0%
93.2%
nA=44,
Dec. 2015
44
84.1%
35
100.0%
41
80.0%
37
68.2%
Frequency
30
30
60.0%
25
45.5%
20
15
Pareto Graph Objectives:
1. Stratify the whole into its parts for
better understanding of the
situation.
2. Be able to Prioritize important in
order to Focus on key issues.
20
%
20
40.0%
10
10
5
20.0%
7
4
3
0
0.0%
A1
A2
A3
A4
Others
Interpretation:
•
Claim type A represents 36.7% of
all claims.
•
Claims A, B &C represent 80.8%
of all claims.
•
From the frequency standpoint,
our priority is to reduce claims
type A,B&C.
PARETO GRAPH (Stratified by A)
90%
nA1=20,
Dec. 2015
100.0%
20
19
18
80.0%
75%
15
Frequency
100.0%
95%
20
15
60.0%
50%
10
%
10
10
40.0%
5
20.0%
5
3
1
1
0
0.0%
A1.1
Pareto Graph Objectives:
1. Stratify the whole into its parts for
better understanding of the
situation.
2. Be able to Prioritize important in
order to Focus on key issues.
A1.2
A1.3
A1.5
A1.6
Interpretation:
•
Claim type A represents 36.7% of
all claims.
•
Claims A, B &C represent 80.8%
of all claims.
•
From the frequency standpoint,
our priority is to reduce claims
type A,B&C.
Brain Flex
Analyze the following Pareto graph
and write your most important issue
to improve.
Why Why Analysis
A Simple and powerful Root Cause Analysis technique
Why the user complains about the service received?
Because of the errors in the bill account.
Why are they complaining about errors in the bill account?
Promised discount in an ad is not being applied
Why are promised discounts not being applied?
Online web based discount is applied, however “face to face” operations in Region 3 are
charging without applying the discount.
Why are Region 3 sales executives did not apply the discount?
Not aware of the existence of such a promotion and as their do a manual process without web
services, our “automated sales system” was not able to prevent these cases.
Why did Region 3 sales executives did not know about this particular promotion?
50% of Region 3 agents are commission agents (contractors) and were not considered in the
mailing list and two way communication protocols.
“Errors in the bill account occur because commission agents (contractors) only have
currently serving distant populations are not in the mailing list”
Why Why Analysis (wrong way)
Promote evidence based approach vs assumptions
Why the user complains about the service received?
Because of the errors in the bill account.
Why are they complaining about errors in the bill account?
Because customers never want to pay and lie seeking for any opportunity to extend
their payment period.
Why do customers do not want to pay?
Customers do not have money. They just like to buy things and then return them
hoping you will never remember them.
Why do customers do not have money?
Because they like to go on vacations.
Why do customers like to go on vacations?
In order to relax and prevent stress.
“So the user complains about the bill account errors, because users like to relax and
prevent stress”.
Activity
Practice Why Why Analysis
In search of causes (origin)
Why_________________________________________________
How can I validate the cause?
Cause & Effect Diagram
The Cause & Effect diagram was first used by
Dr. Kaoru Ishikawa of the University of Tokyo to
organize a discussion with Kawasaki Steel Co.
managers in 1953. This diagram is used to
identify all of the contributing possible causes
(factors) likely to be causing an effect (problem
or deviation).
Outside Japan, is also known as Ishikawa
Diagram and/or Fishbone Diagram.
This tool offers several benefits:
– Visual tool
– Straightforward and easy to learn
– Promotes participation of the whole team
– Organizes discussion & maintains focus
– Promotes systems thinking
– Supports further analysis and corrective
actions
Ishikawa Diagram Deployment
Video
Cause and Effect Diagram tips
Video: Ishikawa Diagram
• How to make an Ishikawa Diagram from
scratch.
–
–
–
–
Brainstorming with sticky notes.
Categorization (grouping).
Categories´ naming.
Organization of causes (primary, secondary, ….)
Video: Ishikawa Diagram
Category
Category
Category
Problem
Statement
Category
Category
Video: The Traditional 4M + New Categories
Man Machine Materials Method + Management Environment
MANAGEMENT
MACHINE / EQUIPMENT
MAN / PEOPLE
PROBLEM
STATEMENT
ENVIRONMENT
METHOD / PROCEDURE
MATERIALS / INPUT
Video: Ishikawa: Original Approach.
Process 3
Process 2
Process 1
PROBLEM
STATEMENT
Process 4
Process 5
Process “N”
Key Recommendations for an Effective Ishikawa Diagram
•
•
•
•
•
•
•
•
•
•
Correct definition of the problem and its statement.
Invite key stakeholders who live, know and understand the process where the
Improvement will be implemented.
Brainstorm all possible causes of the problem and write them in sticky notes (one
cause, one note).
Avoid “One Word” sticky notes. Try to write full sentences.
Avoid writing explanations: “look for causes, do not explain the problem”.
Avoid writing justifications: “look for the origin of the problem, not justify your actions”.
Avoid judging the current situation or causes: “Bad procedure”, “Inadequate operation”
Avoid writing causes as a “lack of your solution”: “Lack of training”, “not enough time”
Check that all participants understand all written causes in the same way.
You can decide to use the 4M rule or create your own categories for your diagram.
Pie Chart
Sales (million USD)
D; 5; 4%
C; 15; 11%
B; 30; 23%
A; 80; 62%
A
B
C
D
Bar graphs
Bar or column
100% Stacked column
%
100%
Jan Feb Mar Apr May Jun Jul Aug Sep
Jan Feb Mar Apr May Jun
Jul
Aug Sep
Radar Chart (Spider chart)
Skill 1
10
Test result Z
100
Skill 2
10
5
5
Test result Y
50
100
Skill 3
5
10
50
0
5
C
C-
10
Performance
evaluation
Skill 4
A
A+
Personality test 1
Boxplot (Box & Whisker)
Graphical representation of a data distribution
(center, width and outliers) based on:
• Minimum or smallest value of data set
• First or Lower Quartile (Q1)
• Second Quartile or Median (Q2)
• Third or Upper Quartile (Q3)
• Maximum or largest value of data set
• “Outliers” (value that lies at an abnormal distance
from other values or statistically outside current data
set distribution).
• Note: For symetric distributions, statistical limits of
the distribution (the whiskers) are calculated as
follows:
• Q3 + 1.5 IQR or largest value
• Q1 – 1.5 IQR or smallest value
Where IQR = Q3 – Q1 (the Box)
Boxplot
An example
Provided is an example of a call
center process where Average
Handle Time (AHT) of the calls is
compared between the
Supervisors of the process.
What does the Boxplot tell
you?
Scatter Plot Graph
n=50
Analyses possible
relationship between two
variables:
• Dependent (Y)
• Independent (x)
Y
X
n=50
Y
Y
Y
X
n=50
n=50
X
IMPORTANT:
Linear correlation between two variables does not necessarily imply a causal relationship
X
Brain Flex:
Demonstration: Big Feet mean higher intelligence.
Dr. Imaizumi, Masumasa, introduces this example in a
class to demonstrate a Japanese custom of praising babies
for the size of their feet.
Dr. Ricardo Hirata, disciple of Dr. Imaizumi applies to
graduate school through this innovative model of
admission.
Due to the size of his feet, he entered graduate school
without presenting any exam.
Math exam grade
Please discuss.
Size of feet
Brain Flex
Scatter Plot
What can be inferred from data presented below?
52
Homework
How to estimate probable service
process output using Z Score?
Assume the mean of service process A is
28.0 minutes with a standard deviation of
0.3 minutes. Estimate the % of process A
services finishing:
a)
b)
c)
Before 27.7 minutes
Between 27.4 and 28.3 minutes
Over 28.9 minutes
Using Excel functions and compare to
previous results of a Brain Flex.
Homework
Build a Pareto Graph
Collect data of one of your daily
processes that you would like to
understand and maybe improve.
1. Stratify your data and build a Pareto
Graph.
2. Write all your conclusions from your
interpretation of the graphs.
3. Select your priority: Your most
important issue to work on (you can
use this issue or topic for building a
Ishikawa Diagram).
Homework
Build a Box&Whisker Graph
Collect data of one of your daily processes you
would like to understand and maybe improve.
1.
2.
Build a Box&Whisker Plot Graph of one
process. Try to make more than one
Boxplot, for example
1. Different Boxplot of the process vs
time (one per month or week)
2. Same process, same month and
different Supervisors.
Write all your conclusions from your
interpretation of the graph.
Homework
Ishikawa / Cause and Effect Diagram
Define a small team of colleagues
(supervisors only or your natural team
you work with daily):
a)
b)
c)
Define a problem (define its
magnitude).
Build a Cause and Effect Diagram.
Find potential causes of the
problem.
Thank you