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20/05/1436
University of Hail
College of Engineering
ISE 320 - Quality Control and Industrial Statistics
CHAPTER 04
CONTROL CHARTS FOR ATTRIBUTES
Professor Mohamed Aichouni
http://faculty.uoh.edu.sa/m.aichouni/ise230-quality/
Introduction
• Many quality characteristics cannot be
conveniently represented numerically.
• In such cases, each item inspected is classified
as either conforming or nonconforming to
the specifications on that quality characteristic.
• Quality characteristics of this type are called
attributes.
attributes
• Examples are nonfunctional semiconductor
chips, warped connecting rods, etc,.
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Types of Control Charts
Control Charts for Variables Data
X and R charts: for sample averages and ranges.
X and s charts: for sample means and standard deviations.
deviations
Md and R charts: for sample medians and ranges.
X charts: for individual measures; uses moving ranges.
Control Charts for Attributes Data
p charts: proportion of units nonconforming.
np charts: number of units nonconforming.
c charts: count of nonconformities.
u charts: count of nonconformities per unit.
Control Chart Selection
Quality Characteristic
Variable
Attribute
Defective
n>1?
no
x and MR
yes
no
n>=10?
Defect
x and R
constant
sample
size?
yes
no
yes
x and s
2
p-chart with
variable sample
size
constant
sampling
unit?
p or
np
yes
no
c
u
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Type of Attribute Charts
p charts
• This chart shows the fraction of nonconforming or defective product
produced by a manufacturing process.
• It is also called the control chart for fraction nonconforming.
np charts
• This chart shows the number of nonconforming. Almost the same as
the p chart.
c charts
• This shows the number of defects or nonconformities produced by a
manufacturing process.
u charts
• This chart shows the nonconformities per unit produced by a
manufacturing process.
p charts
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p charts
• In this chart, we plot the percent of
d f ti
defectives
((per b
batch,
t h per d
day, per machine,
hi
etc.).
• However, the control limits in this chart are
not based on the distribution of rate events
but rather on the binomial distribution (of
proportions).
Formula
• Fraction nonconforming:
p = (np)/n
• where p = proportion or fraction nc in the
sample or subgroup,
• n = number in the sample or subgroup,
• np = number nc in the sample or
subgroup.
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Example
• During the first shift, 450 inspection are made of bookof the month shipments and 5 nc units are found.
Production during the shift was 15,000 units. What is
the fraction nc?
p = (np)/n = 5/450 = 0.011
• The p, is usually small, say 0.10 or less.
• If p > 0.10, indicate that the organization is in serious
difficulty.
p-Chart contruction for constant
subgroup size
•
•
•
•
Select
S
l t the
th quality
lit characteristics.
h
t i ti
Determine the subgroup size and method
Collect the data.
Calculate the trial central line and control
limits.
• Establish
E tabli h the revised
e i ed ce
central
t al li
line
ea
and
d co
control
t ol
limits.
• Achieve the objective.
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Select the quality characteristics
The quality characteristic?
– A single quality characteristic
– A group of quality characteristics
– A part
– An entire product, or
– A number of products.
products
Determine the subgroup size and
method
• The size of subgroup is a function of the proportion
nonconforming.
g
• If p = 0.001, and n = 1000, then the average number
nc, np = 1. Not good, since a large number of values
would be zero.
• If p = 0.15, and n = 50, then np = 7.5, would make a
good chart.
• Therefore, the selection subgroup size requires some
preliminary observations to obtain a rough idea of the
proportion nonconforming.
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Collect the data
• The quality technician will need to collect
sufficient
ffi i t data
d t for
f att least
l
t 25 subgroups.
b
• The data can be plotted as a run chart.
• Since the run chart does not have limits,
its is not a control chart.
Calculate the trial central line and
control limits
• The formula:
p=
∑ np
∑n
UCL = p + 3
p (1 − p )
n
LCL = p − 3
p (1 − p )
n
•
= average of p for many subgroups
• n = number inspected in a subgroup
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∑ np = 138 = 0.018
p=
∑ n 7500
Subgroup
Number
Number
Inspected
n
np
p
1
300
12
0.040
0.018(1 − 0.018)
UCL = 0.018 + 3
300
= 0.041
0.018(1 − 0.018)
LCL = 0.018 − 3
300
= −0.005 = 0.0
2
300
3
0.010
3
300
9
0.030
4
300
4
0.013
5
300
0
0.0
6
300
6
0.020
7
300
6
0.020
8
300
1
0.003
19
300
16
0.053
300
2
0.007
7500
138
25
Total
Negative value of LCL is possible in a theoritical result, but
not in practical (proportion of nc never negative).
p Chart
0.053
p
UCL
0.04
0.03
0.02
p-bar
0.01
LCL
0
5
10
15
20
Subgroup
8
25
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Establish the revised central line and
control limits
• Determine the standard or reference value
f the
for
th proportion
ti nc, po.
pnew
np − np
∑
=
∑n − n
d
d
• where npd = number nc in the discarded subgroups
• nd = number inspected in the discarded subgroups
Revised control limits
po = pnew
pnew =
p (1 − po )
UCL = po + 3 o
n
LCL = po − 3
po (1 − po )
n
• where po is central line
9
138 − 16
= 0.017
7500 − 300
UCL = 0.017 + 3
0.017(1 − 0.017)
300
= 0.039
0.017(1 − 0.017)
300
= −0.005 = 0.0
LCL
C = 0.017
01 − 3
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Control Charts for Fraction
Nonconforming (p chart) Example 2
Example
p
• A process that produces bearing housings is
investigated. Ten samples of size 100 are
selected.
Sample #
# Nonconf.
1
5
2
2
3
3
4
8
5
4
6
1
7
2
8
6
9
3
10
4
• Is this process operating in statistical control?
Example
p
n = 100, m = 10
Sample #
# Nonconf.
Fraction
Nonconf.
1
5
2
2
3
3
4
8
5
4
6
1
7
2
8
6
9
3
10
4
0.05
0.02
0.03
0.08
0.04
0.01
0.02
0.06
0.03
0.04
m
p =
10
∑ pˆ i
i=1
m
= 0 . 038
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Example
p
Control Limits are:
0.038(1 − 0.038)
= 0.095
100
UCL = 0.038 + 3
CL = 0.038
0.038(1 − 0.038)
= −0.02 → 0
100
LCL = 0.038 + 3
C chart – Example 2
P Chart for C1
Proportion
0.10
3.0SL=0.09536
0.05
P=0.03800
0.00
- 3.0SL=0.000
0
1
2
3
4
5
6
7
Sampl e Number
11
8
9
10
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np Chart
• The np chart is almost the same as the p chart.
C t l liline = npo
Central
UCL = npo + 3 npo (1 − po )
LCL = npo − 3 npo (1 − po )
• If po is unknown, it must be determined by
collecting data, calculating UCL, LCL.
Example
12
Subgroup
n
np
UCL
np -bar
LCL
1
2
3
4
5
300
300
300
300
300
3
6
4
6
20
12.0
12 0
12.0
12.0
12.0
12.0
5.24
5 24
5.24
5.24
5.24
5.24
0.0
00
0.0
0.0
0.0
0.0
21
22
23
24
25
300
300
300
300
300
2
3
6
1
8
12.0
12.0
12.0
12.0
12.0
5.24
5.24
5.24
5.24
5.24
0.0
0.0
0.0
0.0
0.0
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c Chart
• The procedures for c chart are the same a s
those for the p chart.
chart
• If count of nonconformities, co, is unknown, it
must be found by collecting data, calculating
UCL & LCL.
UCL = c + 3 c
c=
13
LCL = c − 3 c
c = average count of nonconformities
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Example
ID Number Subgroup
MY102
1
MY113
MY121
MY125
MY132
MY143
MY150
MY152
MY164
MY166
MY172
MY267
MY278
MY281
MY288
c
UCL
2
3
4
5
6
7
8
9
10
11
7
6
6
3
20
8
6
1
0
5
14
12.76
12.76
12.76
12.76
12.76
12.76
12.76
12.76
12.76
12.76
12.76
c -bar
c 0141
5.64
c=
=
= 5.64
5.64 g 0 25
5.64
0
5.64
0
5.64
0 + 3 5.64
UCL = 5.64
5.64
0
5.64
0
LCL = 5.640− 3 5.64
5.64
5.64 = −1.48
0 =0
5.64
0
5.64
0
22
23
24
25
4
14
4
5
12.76
12.76
12.76
12.76
5.64
5.64
5.64
5.64
LCL
= 12 .76
0
0
0
0
c-Chart
25
20
Count of Nonconformities
c
UCL
c-bar
15
LCL
10
5
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Subgroup Num ber
14
16
17
18
19
20
21
22
23
24
25
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Revised
• Out-of-control: sample no. 5, 11, 23.
c new =
c − c d 141 − 20 − 14 − 14
=
= 4.23
g − gd
25 − 3
UCL = co + 3 co = 4.23 + 3 4.23 = 10.40
LCL = co − 3 co = 4.23 − 3 4.23 = −1.94 = 0
u Chart
• The u chart is mathematically equivalent to the c
chart.
chart
u=
c
n
u=
∑c
∑n
u
UCL = u + 3
n
15
LCL = u − 3
u
n
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Example
ID Number Subgroup
30-Jan
1
31-Jan
1-Feb
2-Feb
3-Feb
4-Feb
28 F b
28-Feb
1-Mar
2-Mar
3-Mar
4-Mar
u=
n
c
u
UCL
u -Bar
LCL
2
3
4
5
6
110
82
96
115
108
56
120
94
89
162
150
82
1.091
1.146
0.927
1.409
1.389
1.464
1.51
1.56
1.54
1.51
1.52
1.64
1.20
1.20
1.20
1.20
1.20
1.20
0.89
0.84
0.87
0.89
0.88
0.76
26
27
28
29
30
101
122
105
98
48
105
143
132
100
60
1.040
1
040
1.172
1.257
1.020
1.250
1.53
1
53
1.50
1.52
1.53
1.67
1.20
1
20
1.20
1.20
1.20
1.20
0.87
0
87
0.90
0.88
0.87
0.73
• For January 30:
u Jan 30 =
16
∑ c = 3389 = 1.20
∑ n 2823
c 120
=
= 1.09
n 110
UCLJan 30 =1.20 + 3
1.20
= 1.51
110
LCL Jan 30 =1.20 − 3
1.20
= 0.89
110
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Additional Example from Manufacturing
Surface defects have
b
been
counted
t d on 25
rectangular steel
plates, and the data
are shown below.
The control chart for
nonconformities is
set up using this
data.
Is the Process
under Control?
Comment.
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Additional Example from Manufacturing
Nonconformity Classification
• Critical nonconformities
– Indicate hazardous or unsafe conditions.
• Major nonconformities
– Failure
• Minor
Mi
nonconformities
f
iti
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Control Charts for Variables vs.
Charts for Attributes
• Sometimes, the quality control engineer
has a choice between variable control
charts and attribute control charts.
Advantages of attribute control
charts
• Allowing for quick summaries, that is, the engineer
may simply classify products as acceptable or
unacceptable, based on various quality criteria.
• Thus, attribute charts sometimes bypass the need
for expensive, precise devices and time-consuming
measurement procedures.
y managers
g
unfamiliar with
• More easilyy understood by
quality control procedures.
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Advantages of variable control
charts
• More sensitive than attribute control charts
charts.
• Therefore, variable control charts may alert us to
quality problems before any actual "unacceptables"
(as detected by the attribute chart) will occur.
• Montgomery (1985) calls the variable control charts
leading indicators of trouble that will sound an alarm
before the number of rejects (scrap) increases in the
production process.
Home Work 4.1
• Study the
process stability
using the data
shown on the
table on non
conforming units
in the
g
manufacturing
process.
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Conclusion
"Quality control truly
begins
and ends
g
with education",
K. Ishikawa (1990).
Lecture Finished
Any
Question?
No
Yes
Ask questions
Teachers answers
Train your self (Google, YouTube,
course webpage
End
(See you next lecture)
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