Chapter Two - for practice.ppt statical quality

Chapter Two:
Statistical Process Control

In Grouped Frequency Distribution (GFD),
In Grouped Frequency Distribution (GFD),
 In order to estimate the number of classes, the ff
formula is used:
Number of classes=1+3.322(log N)
Number of classes=1+3.322(log N)
where:- N is the Number of observation.
The Class size =
The Class size = Range
Range (round up)
(round up)
(class width) 1+3,322(log N)
(class width) 1+3,322(log N)

Components of grouped frequency distribution
1. LCL: the smallest number that actually belongs to the class;
2. UCL: the largest number that actually belongs to the
respective classes.
3. Class boundaries: are used to separate adjoining classes
which should not coincide with the actual observations.
 LCB: is obtained by subtracting half a unit of measure
from the LCL;
 UCB: is obtained by adding half the unit of measure to the
upper class limits.
4. Class mark: is the midpoint of the class.
5. Class width/Class intervals/class size: the difference
between two consecutive LCLs or (UCLs, CMs, L/UCBs);
6. Unit of measure: the smallest possible positive difference
between any two measurements in the given data set.
4

CONSTRUCTION OF HISTOGRAMS
CONSTRUCTION OF HISTOGRAMS
Step I. Developing Group Frequency Distribution (GFD):
Step I. Developing Group Frequency Distribution (GFD):
Rules to construct GFD
Rules to construct GFD
i. Find the unit of measure(u)
unit of measure(u) of the given data;
ii. Find the range
range;
iii. Determine the number of classes(n)
number of classes(n) required;
iv. Find class width/size/interval (cw)
class width/size/interval (cw);
v. Determine a lowest class limit(LCL1)
lowest class limit(LCL1) and then find the successive lower
successive lower
(LCLi+1=LCLi+cw) and upper class limits ((UCL1=LCL1+cw-u), (UCLi+1=UCLi+cw))
(LCLi+1=LCLi+cw) and upper class limits ((UCL1=LCL1+cw-u), (UCLi+1=UCLi+cw))
forming non over lapping intervals
forming non over lapping intervals such that each observation falls into exactly
one of the class intervals;
vi. Determine the corresponding class boundaries(lcb, ucb)
vii.Find the number of observations
number of observations falling into each class boundaries that is taken as
the frequency of the class (class interval)
the frequency of the class (class interval) which is best done using a tally mark
tally mark.
.

Histogram: E.g.
Histogram: E.g.
0
5
10
15
20
20 - 3030 - 4040 - 5050 - 6060 - 70 70 -80
3-D Column 1
i. Mark the class boundaries on the horizontal axis (x- axis) and the class
frequencies along the vertical axis ( y- axis) according to a suitable scale.
ii. With each interval as a base draw a rectangle whose height equals the
frequency of the corresponding class interval. It describes the shape of
the data.
Step II to draw Histogram
Step II to draw Histogram

Example1 : Suppose the table below is the frequency
distribution of tensile test of 50 specimen.
Then the frequency table has 6 classes (no. of class).
What are the Unit of Measure, LCLs, UCLs, LCBs, UCBs, CW, & CM
Test Result 11-15 16-20 21-25 26-30 31-35 36-40
Frequency 7 8 10 12 9 4
 The unit of measure is 1
 LCLs: 11, 16, 21, 26, 31, 36 /// UCLs: 15, 20, 25, 30, 35, 40
 CMs: 13((11+15)/2), 18, 23, 28, 33, 38
 LCBs: 10.5(11-0.5), 15.5, …., 35,5 /// UCB: 15.5(15+0.5), ...35.5, 40.5
 Class width (size) = 5.

2. Check Sheet
A format in which items to be checked can be collected easily;
Main purpose: to arrange data automatically & systematically by
tallying each type of defect or problem;
It shows the types of defects and how many of each type
occurred
during that
period.
Defective item check sheet
Defective item check sheet

Example: The following table shows the different types of
defect and the total number of items that the defects
occurred on products in ABC company. Use the Pareto analysis
to determine the vital few cause, which results the majority of
the problem.
Table: Defects observed
Type of
defect
No. of
defects
Crack 10
Scratch 42
Stain 6
Strain 104
Gap 4
Pinhole 20
Others 14
Total 200
Type of
defect
No. of
defects
Cumm.
Tot.
Ind. % Cumm.
%
Strain 104 104 52 52
Scratch 42 146 21 73
Pinhole 20 166 10 83
Crack 10 176 5 88
Stain 6 182 3 91
Gap 4 186 2 93
Others 14 200 7 100
Total 200 - 100 -

 The value of r ranges between [ -1, +1], and
[ -1, +1], and
 r
r denotes the strength of the association as illustrated by
the following diagram.




















 
 
  
n
y)
(
y
.
n
x)
(
x
n
y
x
xy
2
2
2
2
r

7. Construction &
Interpretation of Control Charts

14
Types of Data
• Variable data
•Product characteristic
that can be measured
{Length, size, weight,
height, time, velocity}
• Attribute data
• Product characteristic
evaluated with a
discrete choice
{Good/bad, yes/no}
Types of Control
Charts
• CC for Variables
– Mean (x-Bar) Charts
– Range (R) Charts
• CC for Attributes
– P-Charts
– np - chart
– C-charts
– u- Chart

Control charts for variables
x-bar chart
• Two basic types: x-bar-chart and R-chart – one used
to monitor central tendency of the process, and the
other for variability.
• X-bar chart is used to plot the average measured
value of a certain quality variable taken from a
process. It indicates how the process mean is
varying over time.

• Usually these samples are small, with about four or five
observations. Each sample has its own mean. The center line
of the chart is then computed as the mean of all K sample
means, where K is the number of samples.
Center Line = Grand Average,

A2 value used for X bar chart
Samp
le
size n
A2
2 1.88
3 1.02
4 0.73
5 0.58
6 0.48
7 0.42
8 0.37
9 0.34
10 0.31
13 0.25
14 0.24
15 0.22
16 0.21
17 0.20
18 0.19
19 0.19
20 0.18
21 0.17
22 0.17
23 0.16
24 0.16

Range (R) charts
• Whereas x-bar charts measure shift in the central
tendency of the process, range charts monitor the
dispersion or variability of the process changes over
time.
• The center line of the control chart is the average
range, and the upper and lower control limits are
computed as follows:
n
R
R
n
i


 1

D3 and D4 values for constructing R bar charts
Samp
le
size n
D3 D4
2 0 3.27
3 0 2.57
4 0 2.28
5 0 2.11
6 0 2.00
7 0.08 1.92
8 0.14 1.86
9 0.18 1.82
10 0.22 1.78
13 0.31 1.69
14 0.33 1.67
15 0.35 1.65
16 0.36 1.64
17 0.38 1.62
18 0.39 1.61
19 0.40 1.60
20 0.41 1.59
21 0.43 1.58
22 0.43 1.57
23 0.44 1.56
24 0.45 1.55

Example: Control Charts for Variables
Ring Diameter (cm)
Sample 1 2 3 4 5 X R
1 5.02 5.01 4.94 4.99 4.96 4.98 0.08
2 5.01 5.03 5.07 4.95 4.96 5.00 0.12
3 4.99 5.00 4.93 4.92 4.99 4.97 0.08
4 5.03 4.91 5.01 4.98 4.89 4.96 0.14
5 4.95 4.92 5.03 5.05 5.01 4.99 0.13
6 4.97 5.06 5.06 4.96 5.03 5.01 0.10
7 5.05 5.01 5.10 4.96 4.99 5.02 0.14
8 5.09 5.10 5.00 4.99 5.08 5.05 0.11
9 5.14 5.10 4.99 5.08 5.09 5.08 0.15
10 5.01 4.98 5.08 5.07 4.99 5.03 0.10
50.09 1.15

Solution:-
Solution:-
From Table above:
– sum of X = 50.09…. For each sample
–Sum of R = 1.15…… for each sample
–n = 10
Thus;
–X-Double bar = X-D=X = 50.09/10 = 5.009
cm…. For the whole process
–R-bar=R = 1.15/10 = 0.115 cm…. For the
whole process

control limit
control limit
For X- chart
For X- chart
• UCLx-bar = X-D bar + A2 R-bar = 5.009 + (0.577)(0.115)
= 5.075 cm
• LCLx-bar = X-D bar - A2 R-bar = 5.009 - (0.577)(0.115)
= 4.943 cm
For R-chart
For R-chart
• UCLR = D4R-bar = (2.114)(0.115) = 0.243 cm
• LCLR = D3R-bar = (0)(0.115) = 0 cm
For A2, D3, D4: see Table
n = 5

X-bar Chart
X-bar Chart
4.94
4.96
4.98
5.00
5.02
5.04
5.06
5.08
5.10
0 1 2 3 4 5 6 7 8 9 10 11
Subgroup
X
bar
LCL
CL
UCL
outliers

R - Chart
R - Chart
0.00
0.05
0.10
0.15
0.20
0.25
0 1 2 3 4 5 6 7 8 9 10 11
Subgroup
Range
LCL
CL
UCL

Summary of attribute charts
• P chart – percent of occurrence – constant/varying
sample size;
• np chart – no. of occurrences – constant sample size;
• u chart – percent of occurrence - constant/varying
sample size - each sample can have more than one
occurrence;
• c chart – no. of occurrences – constant sample size –
each sample can have more than one occurrence;

Control Chart for Proportions – P-Chart
• Commonly known as fraction-defective control charts;
• Suppose D is the number of defective units in a random
sample of size n. We assume that D is a binomial random
variable with unknown parameter p. The fraction
defective
• of each sample is plotted on the chart.
• Suppose that m preliminary samples each of size n are
available, and let Di be the number of defectives in the ith
sample. The is the sample fraction defective in the
ith
sample. The average fraction defective is

• So,
• Example: suppose 20 preliminary samples, each of size 100 are
considered for analysis of a ceramic substrate production line. The
number of defectives in each sample is shown in table below.
Assuming that samples are numbered in the sequence of
production, determine the stability condition of the line.

Table: Number of Defectives in Samples of 100 Ceramic
Substrates

Solution
• Note that ; therefore, the trial
parameters for the control chart are

• NP charts can also be used in a similar approach.
This is just a control chart of , the number of
defectives in a sample. The points, center line, and
control limits for this chart are just multiples (times
n) of the corresponding elements of a P chart. The
use of an NP chart avoids the fractions in a P chart.

While the p-chart tracks the proportion of non-conformities per
sample, the np chart plots the number of non-conforming items
per sample.
The audit process of the samples follows a binomial distribution,
i.e. the expected outcome is “good” or “bad”, and therefore the
mean number of successes is np.
The control limits for an np chart are as follows:
UCL and LCL
np-Chart Calculations
Center Line,

Control Chart for Defects per Unit – U-Chart
The average count of occurrences of criteria of interest in sample
of items
• It is sometimes necessary to monitor the number of defects in
a unit of product rather than the fraction defective.
• Suppose that in the production of cloth it is necessary to
control the number of defects per yard or that in assembling an
aircraft wing the number of missing rivets must be controlled.
• Many defects-per-unit situations can be modeled by the
Poisson distribution.
• If each sample consists of n units and there are C total defects
in the sample,
• is the average number of defects per unit.

• If the number of defects in a unit is a Poisson random
variable with parameter , the mean and variance of
this distribution are both .
• Each point on the chart is U, the average number of
defects per unit from a sample of n units. Therefore,
the mean of U is and the variance of U is .
• If there are m preliminary samples, and the number of
defects per unit in these samples are U1, U2, ..., Um, the
estimator of the average number of defects per unit is

• The parameters of the U chart are defined as follows:
• Example: Printed circuit boards are assembled by a combination of manual
assembly and automation. A flow solder machine is used to make the
mechanical and electrical connections of the leaded components to the
board. The boards are run through the flow solder process almost
continuously, and every hour five boards are selected and inspected for
process-control purposes. The number of defects in each sample of five
boards is noted. Results for 20 samples are shown in table below:

Table: Number of Defects in Samples of Five Printed Circuit
Boards

Comment: because LCL is negative, it is set to 0.
From the chart, the process is in control. However, eight defects per
group of ﬁve circuit boards are too many (about 8/ 5 = 1.6
defects/board), and the process needs improvement.

Control Chart for Defects per Unit – U-Chart
The average count of occurrences of criteria of interest in sample of
items
Where nj
is the sample size (number of units)
of group j and u-bar is the Average percent.
Center
Line =
UCL,
LCL
• C-charts can also be produced using the total number
of defects in a sample.
• The points, center line, and control limits for this chart
are just multiples (times n) of the corresponding
elements of a U chart. The use of a C chart avoids the
fractions that can occur in a U chart.

The count of occurrences of a criterion of interest in a sample of
items
UCL, LCL
C Chart
Calculations
Center
Line =

• Specifying the control limits is one of the critical decisions that
must be made in designing a control chart.
• If the process is stable, then the distribution of subgroup
averages will be approximately normal.
• Analyze the patterns to see if there might be special causes of
variations by dividing the normal distribution into zones;
• Approximate %tage expected of each zone from a stable
process.
Control Chart Performance

• So far we have discussed ways of monitoring the production process
to ensure that it is in a state of control and isolate assignable causes
of variation, if there are;
• A critical aspect of SQC is evaluating the ability of a production
process to meet or exceed preset specifications. This is called
process capability;
• Specifications are often called tolerances and are preset ranges of
acceptable quality characteristics;
• For a product to be considered acceptable, its characteristics must
fall within this preset range. Otherwise, the product is not
acceptable. They are usually established by design engineers;
Process Capability

“Control” and “Capability”
A process can be in control and not capable.
• “Control” means that the process is stable.
• Capable means “within specification”.
• Control charting is a process improvement tool not a conformance
to specification tool.
• If a process is not in statistical control then capability has no
meaning. Therefore, a capable process involves only common
cause variations and not special cause variations.
• For example, the specification for the width of a machine part may be speciﬁed
as 15 inches  0.3. This means that the width of the part should be 15 inches,
though it is acceptable if it falls within the limits of 14.7 inches and 15.3 inches.

Measuring Process Capability
So, simply setting up control charts to monitor whether a
process is in control does not guarantee process capability.
To produce an acceptable product, the process must be
capable and in control before production begins.
• Process capability is measured by the process capability
index, Cp, which is computed as the ratio of the specification
width to the width of the process variability;
• The reason 6σ is used is that most of the process
measurement (99.74 percent) falls within  3σ, which is a
total of 6σ. So,
6σ
LSL
USL
width
Process
ion width
Specificat
Cp
or
PCR
or
PCI




• The definition of the PCR given above implicitly assumes that the
process is centered at the nominal dimension.
• If the process is running off-center, its actual capability will be less
than indicated by the PCR. If the process is not centered, a measure
of actual capability is often used. This ratio, called PCRk, is defined as:
• Because the process mean can shift in either direction, the direction
of shift and its distance from the design specification set the limit on
the process capability. In effect, PCRk is a one-sided process capability
ratio that is calculated relative to the speciﬁcation limit nearest to
the process mean.
Limitations of Cp





 


3σ
LSL
μ
,
3σ
μ
USL
min
Cp
or
PCR k
k

• The capability index (Cpk ) shows how well the parts being
produced ﬁt into the range specified by the design limits. If
the design limits are larger than the three sigma allowed in
the process, then the mean of the process can be allowed to
drift off-center before readjustment, and a high percentage of
good parts will still be produced.
• Hence, Capability index (Cpk) is the position of the mean and
tails of the process relative to design specifications; i.e. the
more off-center, the greater the chance to produce defective
parts.

Process fallout and
the process capability
ratio (PCR)
Many U.S. companies use
PCR = 1.33 as a minimum
acceptable target and PCR
= 1.66 as a minimum target
for strength, safety, or
critical characteristics.
Some companies require
that internal processes and
those at suppliers achieve a
PCRk = 2.0.

• A process with PCRk = 2.0 is referred to as a six-sigma process
because the distance from the process mean to the nearest
specification is six standard deviations. The reason that such a
large process capability is often required is that it is difficult to
maintain a process mean at the center of the specifications
for long periods of time.
• A common model that is used to justify the importance of a
six-sigma process is illustrated by referring to figure below. If
the process mean shifts off-center by 1.5 standard deviations,
the PCRk decreases to 4.5σ/3σ = 1.5.

• Assuming a normally distributed process, the fallout of the
shifted process is 3.4 parts per million. Consequently, the
mean of a 6-sigma process can shift 1.5 standard deviations
from the center of the speciﬁcations and still maintain a
fallout of 3.4 parts per million.

Guage Repeatability &
Reproducibility
(R & R) Studies evaluate the precision of a
measurement system.
It is important that the gauge be properly calibrated
before starting measurements;

Precision
Repeatability Problem:
The same person (or station)
can’t get the same result
twice on the same subject
Reproducibility Problem:
Different people (or stations)
can’t agree on the result
obtained on the same subject
Within Inspector Error between Inspector Error
Precision
Is the closeness of agreement between randomly selected
individual measurement or test results.

Gauge Repeatability & Reproducibility
• Gage R&R is a study to measure the
measurement error in measurement systems.
• The 10% rule tells us that if the percent of
tolerance consumed by the R&R does not
exceed 10%, the measurement system is
excellent.
• R&R = 15% means that 15% of the variation
in product measurements is due to the
measurement system.

• Each independent component of the model is variable:
– Product quality characteristic -
– Gauge, measurement error -
• Total observed variance in product quality characteristic
measurements:
2
( , )
product
X N  

 
2
0, gauge
N
 

2
2
2
gauge
product
total 

 

Total observed variance

Gauge variance
• If more than one operator used in study then
measurement (gauge) error has two components of
variance:
• Repeatability:
– - Variance due to measuring instrument
• Reproducibility:
– - Variance due to different operators
2
2
2
gauge
product
total 

 

2 2
repeatability reproducibiliy
 

2
repeatability

2
reproducibility


Experiments to estimate the components
&
are called:
gauge repeatability and reproducibility
studies (gauge R&R studies).
2
repeatability
 2
reproducibility


Repeatability & Reproducibility Report
n - number of parts r – number of operators
max min
i
R X X
 
For example :
2 operators, r=2 j=1…r
10 items , n=10 i=1…n
10
1
10
i
i
j
R
R 


2
2
1



j
j
R
R
10
10
2
1
10
1
10
1




 
 i
i
i
i
j
nd
X
st
X
X
2
1
2
j
j
X
X




Repeatability = Equipment Variation
1
2
0.9987 (3)
0.995 (2.576) 2 2.576 5.15
. . 5.15 5.15
EV
R
EV K R
d




   
 
     
 
 
 
Trials 2 3 4
K1
4.56 3.05 2.50
Average and Range method
Standard deviation for Repeatability . .
2
E V
R
d
 
Where d2 is a factor that depends on the number of trials (sample size):
n 2 3 4 5 6 7 8 9 10
d2 1.128 1.693 2.059 2.326 2.534 2.704 2.847 2.970 3.078
   
 
 
LSL
USL
Tolerance
V
E
V
E 

 /
.
.
100
.
.
%

Average and Range method
Standard deviation for Reproducibility
max min
. . * *
2 2
diff
AV
X
X X
d d


 
Reproducibility = Appraiser Variation
2 2
2
. . ( ) [( . .) /( )]
dif
AV X K E V n r n number of partes
r number of trails
    

Operators 2 3 4
K2
3.65 2.70 2.30
   
 
% . . 100 . . /
AV AV Tolerance USL LSL
 
  
 
n = number of parts
r = number of trials

Combined Gage Variation
2 2
&
( . .) ( . .) &
&
5.15
R R
CGV EV AV R R
R R

  

   
% & 100 & /
R R R R Tolerance
  
 

Assignment 2
• Based on the product each group selected in the 1st
assignment, collect pertinent data representing for some
quality characteristics of the product and answer the following
questions:
1) Identify & discuss the two main sources of variations in the process
through which the product you selected is manufactured.
2) Evaluate the process nature or quality using the application of the seven
SPC tools;
3) Determine the Process Capability indices and discuss whether the process
is stable and/or capable or not;
4) What is your conclusion regarding the existing quality level of the
underlying manufacturing/service process?

5) Which organizational entities (Men, Machine, Method, Material, or
Technology) do significantly contribute for the existing quality
problem, if any?
6) What is your judgment regarding the process’s quality in its near
future performance?
7) What do you recommend for the company regarding the potential
improvements that can be deployed to reduce the critical quality
problems resulted from process variations?
• Hint: you may consider one or more of the quality dimensions that
you identified in the 1st
assignment for deciding or determining the
quality characteristics that you will use for your analysis.

Specifications to Assignment 2:
• The assignment is to be submitted two days before the
mid exam;
• There will be a random question-&-answer session for
each member of each group when submitting the
assignments;
• Hence, every student should carefully understand each
question and be active participant in his/her group so
that when questioned randomly should give a proper
answer accordingly.

Chapter Two - for practice.ppt statical quality

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