Quality Management
“It costs a lot to produce a bad product.”
Norman Augustine
Cost of quality
1. Prevention costs
2. Appraisal costs
3. Internal failure costs
4. External failure costs
5. Opportunity costs
History: how did we get here…
• Deming and Juran outlined the principles of Quality
Management.
• Tai-ichi Ohno applies them in Toyota Motors Corp.
• Japan has its National Quality Award (1951).
• U.S. and European firms begin to implement Quality
Management programs (1980’s).
• U.S. establishes the Malcolm Baldridge National
Quality Award (1987).
• Today, quality is an imperative for any business.
What is quality management all about?
Try to manage all aspects of the organization
in order to excel in all dimensions that are
important to “customers”
Two aspects of quality:
features: more features that meet customer needs
= higher quality
freedom from trouble: fewer defects = higher
quality
What does Total Quality Management encompass?
TQM is a management philosophy:
• continuous improvement
• leadership development
• partnership development
Cultural
Alignment
Customer
Technical
Tools
(Process
Analysis, SPC,
QFD)
Developing quality specifications
Input
Design
Design quality
Process
Output
Dimensions of quality
Conformance quality
Continuous improvement philosophy
1. Kaizen: Japanese term for continuous improvement.
A step-by-step improvement of business processes.
2. PDCA: Plan-do-check-act as defined by Deming.
Plan
Do
Act
Check
3. Benchmarking : what do top performers do?
Tools used for continuous improvement
1. Process flowchart
Tools used for continuous improvement
2. Run Chart
Performance
Time
Tools used for continuous improvement
3. Control Charts
Performance Metric
Time
Tools used for continuous improvement
4. Cause and effect diagram (fishbone)
Machine
Man
Environment
Method
Material
Tools used for continuous improvement
5. Check sheet
Item
-------------------
A
B
C
√√
D
E
√
√√√
√√√
√√
√
√√
F
√
√
√
G
√
√√
Tools used for continuous improvement
6. Histogram
Frequency
Tools used for continuous improvement
7. Pareto Analysis
100%
75%
50
40
50%
30
20
10
25%
0%
A
B
C
D
E
F
Percentage
Frequency
60
Summary of Tools
1. Process flow chart
2. Run diagram
3. Control charts
4. Fishbone
5. Check sheet
6. Histogram
7. Pareto analysis
Case: shortening telephone waiting time…
• A bank is employing a call answering service
• The main goal in terms of quality is “zero waiting time”
- customers get a bad impression
- company vision to be friendly and easy access
• The question is how to analyze the situation and improve quality
The current process
Custome
rA
Custome
rB
Operator
Receiving
Party
How can we reduce
waiting time?
Fishbone diagram analysis
Absent receiving
party
Working system of
operators
Absent
Too many phone calls
Lunchtime
Out of office
Not at desk
Not giving receiving
party’s coordinates
Complaining
Absent
Lengthy talk
Does not know
organization well
Takes too much time to
explain
Leaving a
message
Customer
Does not
understand
customer
Operator
Makes
custome
r wait
Reasons why customers have to wait
(12-day analysis with check sheet)
Daily
average
Total
number
A
One operator (partner out of office)
14.3
172
B
Receiving party not present
6.1
73
C
No one present in the section receiving call
5.1
61
D
Section and name of the party not given
1.6
19
E
Inquiry about branch office locations
1.3
16
F
Other reasons
0.8
10
29.2
351
Pareto Analysis: reasons why customers have to wait
Frequency
Percentage
300
87.1%
250
71.2%
200
49%
150
100
0%
A
B
C
D
E
F
Ideas for improvement
1. Taking lunches on three different shifts
2. Ask all employees to leave messages when leaving desks
3. Compiling a directory where next to personnel’s name
appears her/his title
Results of implementing the recommendations
…After
Before…
Percentage
Frequency
Percentage
Frequency
100%
87.1%
300
300
71.2%
200
Improvement
200
49%
100
100
100%
0%
A
B
C
D
E
F
0%
B
C
A
D
E
F
In general, how can we monitor quality…?
By observing
variation in
output measures!
1. Assignable variation: we can assess the cause
2. Common variation: variation that may not be possible to
correct (random variation, random noise)
Statistical Process Control (SPC)
Every output measure has a target value and a level of
“acceptable” variation (upper and lower tolerance limits)
SPC uses samples from output measures to estimate the
mean and the variation (standard deviation)
Example
We want beer bottles to be filled with 12 FL OZ ± 0.05 FL OZ
Question:
How do we define the output measures?
In order to measure variation we need…
The average (mean) of the observations:
X 
1
N
N

xi
i 1
The standard deviation of the observations:
N
 

( xi  X )
i 1
N
2
What is the key assumption behind SPC?
LESS VARIABILITY implies BETTER PERFORMANCE !
High
Cost
Low
Lower spec
Target
Upper spec
Performance Measure
Capability Index (Cpk)
It shows how well the performance measure
fits the design specification based on a given
tolerance level
A process is k capable if
X  k   UTL
1
UTL  X
k
and
and
X  k   LTL
X  LTL
k
1
Capability Index (Cpk)
Another way of writing this is to calculate the capability index:
C pk
 X  LTL UTL  X 
 min 
,

k
k


Cpk < 1 means process is not capable at the k level
Cpk >= 1 means process is capable at the k level
Accuracy and Consistency
We say that a process is accurate if its mean X is close to
the target T.
We say that a process is consistent if its standard deviation
is low.
Example: Capability Index (Cpk)
X = 10 and σ = 0.5
LTL = 9
UTL = 11
LTL
C pk
 10  9
 min 
or
 3  0 .5
X
UTL
11  10 
  0 . 667
3  0 .5 
Example
Consider the capability of a process that puts
pressurized grease in an aerosol can. The design
specs call for an average of 60 pounds per square
inch (psi) of pressure in each can with an upper
tolerance limit of 65psi and a lower tolerance limit
of 55psi. A sample is taken from production and it
is found that the cans average 61psi with a standard
deviation of 2psi.
1. Is the process capable at the 3 level?
2. What is the probability of producing a defect?
Solution
LTL = 55 UTL = 65 X  61
=2
C pk
X  LTL UTL  X
 min(
,
)
3
3
C pk
61  55 65  61
 min(
,
)  min( 1, 0 . 6667 )  0 . 6667
6
6
No, the process is not capable at the 3 level.
Example (contd)
Suppose another process has a sample mean of 60.5 and
a standard deviation of 3.
Which process is more accurate? This one.
Which process is more consistent? The other one.
Solution
P(defect) = P(X<55) + P(X>65)
=P(X<55) + 1 – P(X<65)
=P(Z<(55-61)/2) + 1 – P(Z<(65-61)/2)
=P(Z<-3) + 1 – P(Z<2)
=G(-3)+1-G(2)
=0.00135 + 1 – 0.97725 (from standard normal table)
= 0.0241
2.4% of the cans are defective.
Control Charts
Upper Control Limit
Central Line
Lower Control Limit
Control charts tell you when a process measure is
exhibiting abnormal behavior.
Two Types of Control Charts
•
p Chart
This is a plot of proportions over time (used for
performance measures that are yes/no attributes)
•
X/R Chart
This is a plot of averages and ranges over time
(used for performance measures that are variables)
Statistical Process Control with p Charts
UCL = 0.117
p = 0.066
LCL = 0.015
Statistical Process Control with p Charts
When should we use p charts?
1. When decisions are simple “yes” or “no” by inspection
2. When the sample sizes are large enough (>50)
Sample (day)
Items
Defective
Percentage
1
200
10
0.050
2
200
8
0.040
3
200
9
0.045
4
200
13
0.065
5
200
15
0.075
6
200
25
0.125
7
200
16
0.080
Statistical Process Control with p Charts
Let’s assume that we take t samples of size n …
p
total number
( number
sp 
of " defects"
 (sample
of samples)
p (1  p )
n
UCL  p  zs
p
LCL  p  zs
p
size)
Statistical Process Control with p Charts
p
sp 
80
6  200

1
15
 0 . 066
0 . 066 (1  0 . 066 )
 0 . 017
200
UCL  0 . 066  3  0 . 017  0 . 117
LCL  0 . 066  3  0 . 017  0 . 015
Statistical Process Control with p Charts
UCL = 0.117
p = 0.066
LCL = 0.015
Statistical Process Control with X/R Charts
When should we use X/R charts?
1. It is not possible to label “good” or “bad”
2. If we have relatively smaller sample sizes (<20)
Statistical Process Control with X/R Charts
Take t samples of size n (sample size should be 5 or more)
X 
1
n

n
xi
i 1
X is the mean for each sample
R  max { x i }  min { x i }
R is the range between the highest and the lowest for each sample
Statistical Process Control with X/R Charts
X 
1
t

t
X
j
j 1
X is the average of the averages.
R 
1
t
R

t
j
j 1
R is the average of the ranges
Statistical Process Control with X/R Charts
define the upper and lower control limits…
UCL
X
 X  A2 R
LCL
X
 X  A2 R
UCL
R
 D4R
LCL
R
 D3 R
Read A2, D3, D4 from
Table TN 8.7
Example: SPC for bottle filling…
Sample
Observation (xi)
Average
1
11.90
11.92
12.09
11.91
12.01
2
12.03
12.03
11.92
11.97
12.07
3
11.92
12.02
11.93
12.01
12.07
4
11.96
12.06
12.00
11.91
11.98
5
11.95
12.10
12.03
12.07
12.00
6
11.99
11.98
11.94
12.06
12.06
7
12.00
12.04
11.92
12.00
12.07
8
12.02
12.06
11.94
12.07
12.00
9
12.01
12.06
11.94
11.91
11.94
10
11.92
12.05
11.92
12.09
12.07
Range (R)
Example: SPC for bottle filling…
Calculate the average and the range for each sample…
Sample
Observation (xi)
Average
Range (R)
1
11.90
11.92
12.09
11.91
12.01
11.97
0.19
2
12.03
12.03
11.92
11.97
12.07
12.00
0.15
3
11.92
12.02
11.93
12.01
12.07
11.99
0.15
4
11.96
12.06
12.00
11.91
11.98
11.98
0.15
5
11.95
12.10
12.03
12.07
12.00
12.03
0.15
6
11.99
11.98
11.94
12.06
12.06
12.01
0.12
7
12.00
12.04
11.92
12.00
12.07
12.01
0.15
8
12.02
12.06
11.94
12.07
12.00
12.02
0.13
9
12.01
12.06
11.94
11.91
11.94
11.97
0.15
10
11.92
12.05
11.92
12.09
12.07
12.01
0.17
Then…
X  12 . 00
is the average of the averages
R  0 . 15
is the average of the ranges
Finally…
Calculate the upper and lower control limits
UCL
X
 12 . 00  0 . 58  0 . 15  12 . 09
LCL
X
 12 . 00  0 . 58  0 . 15  11 . 91
UCL
R
 2 . 11  0 . 15  1 . 22
LCL
R
 0  0 . 15  0
The X Chart
UCL = 12.10
X = 12.00
LCL = 11.90
The R Chart
UCL = 0.32
R = 0.15
LCL = 0.00
The X/R Chart
UCL
X
LCL
What can you
conclude?
UCL
The process
is in control
R
LCL
Example
Sample
n
Defects
Sample
n
Defects
1
15
3
6
15
2
2
15
1
7
15
0
3
15
0
8
15
3
4
15
0
9
15
1
5
15
0
10
15
0
a. Develop a p chart for 95 percent confidence (z = 1.96)
b. Based on the plotted data points, what comments can you make?
Solution
Ten defectives were found in 10 samples of size 15.
10
P
 0 . 067
10 (15 )
Sp 
P (1  P )
n

. 067 (1  . 067 )
 0 . 0645
15
UCL = .067 + 1.96(.0645) = .194
LCL = .067 - 1.96(.0645) = -.060
zero
Defect proportion on Days 1 and 8 is 0.2, so process out of control.