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 - Chart
 R - Chart
 S - Chart
VARIABLE CONTROL CHART :
Mean and Dispersion
1. To understand the Quality Characteristics
2. To understand the benefit of control chart
Targ
et
3. Able to develop the control chart
4. To know the control chart types
5. Able to evaluate the process using the
control chart
2
Introduction
The control chart can help to detect the change of process parameters.
Generally, there are two types of the control chart :
1. Variable control chart
2. Attribute control chart
3
The Change of Process Parameter
(Mean)
LSL
USL
m0
m1
4
The Change of Process Parameter
(Standard Deviation)
LSL
m0
USL
s0
s1
5
Quality Characteristics
Variable
Something that can be measured
and expressed by the numerical
scale.
Attribute
Something that can be classified
into conforming or non conforming.
6
To choose the quality characteristics
Develop
ing and
the
applicat
ion of
control
charts
Pareto analysis
Developing the control chart :
Preparation
Making the control chart
Implementation : Process evaluation
using the control chart.
7
To Choose the Quality Characteristic
 Product has many the quality characteristics.
 Choose the quality characteristics using the
Pareto analysis.
8
Pareto Analysis (1)
Defect
Code
1
2
3
4
5
6
7
8
Defect
Outside diameter of hub
Depth of keyway
Hub length
Inside diameter of hub
Width of keyway
Thickness of flange
Depth of slot
Hardness
Frequency
30
20
60
90
30
40
50
20
Percentage
8.82
5.88
17.65
26.47
8.82
11.77
14.71
5.88
9
Percentage of defects
Pareto Analysis (2)
30
25
20
15
10
5
0
4
3
7
6
1
5
2
8
Defect code
10
Preparation to use the control chart




Choose the sample
Sample size
Sampling Frequency
Choose the instrument for
measurement
 Design the form used to
collect the data
11
Non target based
X-bar and Range chart
Target based
X-bar and standard
deviation chart
Non target based
Target based
Making the Control Chart
12
X-bar and R Chart(1)
 Step 1
Write the measurement of the quality
characteristic in a Form.
 Step 2
Calculate Mean and Range for each sample.
n
X 
X
i 1
n
i
R  X max  X min
13
X-bar and R Chart(2)
 Step 3
Determine and draw a center line and trial control
limits for every chart.
X- bar chart
g
Center Line
Control Limit
X
X
i
i 1
g
X  3s X
3sˆ
3R
(UCLX , LCLX )  X 
X
 X  A2 R
n
nd 2
14
15
X-bar and R Chart(3)
R - Chart
g
R
Center Line
R
Control Limits
R  3s R
i 1
i
g
R
R
UCLR  R  3d3    D4 R LCLR  R  3d3    D3 R
d 
 d2 
 2
3d 3
D4  1 
d2

3d3 

D3  max  0,1 
d2 

16
X-bar and R Chart(4)
 Step 4
Plotting the range value at R-Chart.
Determine whether the point plotted in the
statistical control. If not, identify the assignable
causes that related to the out-of-control point and
then perform the improvement to eliminate the
assignable causes.
17
X-bar and R Chart(5)
 Step 5
Eliminate the out-of-control point after performing
the improvement.
Use the rest of sample to revise the center line and
the control limits.
 Step 6
Implement the control chart.
18
Example 1
Consider a process by which coils are
manufactured. Samples of size 5 are randomly
selected from the process, and the resistance
values (in ohms) of the coils are measured. The
data values are given in Table 7-2, as are the
sample mean X bar and the range R.
19
Example 1 (continue)
Table 7-2
Sample
Observations
X bar
R
1
3
.
.
.
22
23
25
20,22,21,23,22
25,18,20,17,22
.
.
.
21,18,18,17,19
21,24,24,23,23
19,20,21,21,22
21.6
20.40
.
.
.
18.6
23.00
20.6
3
8
.
.
.
4
3
3
Sum
521.00
87
Comments
New vendor
High Temp.
Wrong die
20
Example 1 (continue)
The initial of R-chart
87
R
 3.48
25
Center Line
UCLR  (2.114)(3.48)  7.357
Trial Control Limits
LCLR  (0)(3.48)  0
9
8
7
Range
6
5
4
3
2
1
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Sample
21
Example 1 (continue)
R-chart Revision 1
79
R
 3.29
24
Revised Center Line
Revised Control Limit
UCLR  (2.114)(3.29)  6.96
LCLR  (0)(3.29)  0
8
7
6
Range
5
4
3
2
1
0
1 22 43
54
65 76 87 98 109 11
25
10 12
11 13
12 14
13 15
14 16
15 17
16 18
17 19
18 20
19 21
20 22
21 23
22 24
23 24
Sample
22
Example 1 (Continue)
The initial of X-bar chart
500.6
X 
 20.858
24
Center Line
UCLX  20.858  (0.577)(3.29)  22.76
Trial Control Limits
LCLX  20.858  (0.577)(3.29)  18.96
25
X bar
20
15
10
5
0
11 22 43
54
65
76
10 12
11 13
121413 15
14 16
15 17
16 18
17 19
18 20
19 2120 22
21 23
22 24
23 25
24
87 98 109 11
Sample
23
Example 1 (Continue)
R-chart Revision 2
72
R
 3.27
22
Revised Center Line
UCLR  (2.114)(3.27)  6.92
Revised Control Limits
LCLR  (0)(3.27)  0
8
7
6
Range
5
4
3
2
1
0
1
22
43
54
65
76
87
8
9
9
10
10 12
11 13
12 14
13 15
14 16
15
11
Sample
16
17
17
18 18
19 19
20 2021 2124 22
25
24
Example 1 (Continue)
X-bar chart for revision 1)
459
X 
 20.86
22
Center Line
UCLX  20.86  (0.577)(3.27)  22.752
Control Limits
LCLX  20.86  (0.577)(3.27)  18.975
25
20
X bar
15
10
5
0
11
22
43
5
4
6
5
76
78
9
8
10
11 11
12
9 10
13 13
14 14
15 15
16 16
17 17
18 18
19
12
Sample
20 20
21 21
24 22
25
19
25
Standardized Control Chart (1)
 It is used when the sample size is
not the same.
 The Statistic is standardized by
subtraction the sample mean from
the grand mean and divide it by the
standard deviation.
 The standard value represents the
deviation from the mean with the
unit of standard deviation.
 The control limits for the
standardized control chart is ± 3.
26
Standardized Control Chart (2)
The mean control chart :
Grand mean
g
X
n X
i
i
i 1
g
n
i 1
i
g
The estimation of Standard
Deviation process
sˆ 
 (n
i 1
g
 1) si2
 (n
i 1
The standardized value
i
i
 1)
Xi  X
Zi 
ŝ / ni
The Zi values are plotted in the control chart with CL=0, UCL=3
dan LCL=-3.
27
Standardized Control Chart (3)
Range control chart
The value of ri
Ri
ri 
sˆ
ri  d 2
The value of ki k i 
d3
The ki values are plotted in the control chart with
CL=0, UCL=3 dan LCL=-3.
28
The Control Limits base on Target
X-bar control chart
CLX  X 0
3s 0
(UCLX , LCLX )  X 0 
 X 0  As 0
n
R control chart
CLR  d 2s 0
UCLR  R  3s R  d2s 0  3d3s 0  d2  3d3 s 0  D2s 0
LCLR  R  3s R  d2s 0  3d3s 0  d2  3d3 s 0  D1s 0
29
The Average and Standard Deviation Control
Chart


 Xi 
n
2
 i 1

X

 i
n
 X
n
s
i 1
i
X
n 1

2

i 1
2
n
n 1
E( s)  c4s
s s  s 1 c
2
4
30
The Average and Standard Deviation Control
Chart (No Standard)
Standard Deviation chart
g
Center Line
CLs  s 
s
i 1
i
g
UCLs  s  3s s  s  3s 1  c42
Control Limit
3s 1  c42
UCLs  s 
 B4 s
c4
3s 1  c42
LCLs  s 
 B3 s
c4
31
The Average and Standard Deviation Control
Chart (No Standard)
Average X-bar chart (grand mean)
g
Center Line
X
X
i
i 1
g
X  3s X  X  3
Control Limit
(UCLX , LCLX )  X 
3s
c4 n
s
n
 X  A3 s
32
The Average and Standard Deviation Control
Chart (There is a Standard)
Standard Deviation Chart
Center Line
CLs  c4s 0
UCLs  c4s 0  3s s  c4s 0  3s 0 1  c42
Control Limit


UCLs  c4  3 1  c42 s 0  B6s 0


LCLs  c4  3 1  c42 s 0  B5s 0
33
The Average and Standard Deviation Control
Chart (There is a Standard)
Average X-bar chart
Center Line
CLX  X 0
Control Limit
UCL
X
, LCLX   X 0  As 0
34
Control Chart Pattern
(Natural)
UCL
Sample Average
35
CL
20
LCL
5
1
2
3
4
5
6
7
8
9
10
11 12 13 14 15
Sample
35
Control Chart Pattern
(Sudden Shifts in the Level)
UCL
Sample Average
35
CL
20
LCL
5
1
2
3
4
5
6
7
8
9
10
11 12 13 14 15
Sample
36
Control Chart Pattern
(Sudden Shifts in the Level)
 Change in proportions of materials coming from
different sources.
 New worker or machine.
 Modification of production method or process.
 Change in inspection device or method.
37
Control Chart Pattern
(Gradual Shifts in the Level)
UCL
Sample Average
35
CL
20
LCL
5
1
2
3
4
5
6
7
8
9
10
11 12 13 14 15
Sample
38
Control Chart Pattern
(Gradual Shifts in the Level)
 The incoming quality of raw material
or components changed over time.
 The maintenance program changed.
 The style of supervision changed.
 New operator.
 A decrease in worker skill due to
fatigue.
 A gradual improvement in the
incoming quality of raw materials.
39
Control Chart Pattern
(Trending)
UCL
Sample Average
35
CL
20
LCL
5
1
2
3
4
5
6
7
8
9
10
11 12 13 14 15
Sample
40
Control Chart Pattern
(Trending)
 Gradual
deterioration
of
equipment.
 Worker fatigue.
 Deterioration of environmental
conditions.
 Improvement or deterioration
of operator skill.
 Gradual
change
in
homogeneity of incoming
material quality.
41
Control Chart Pattern
(Cyclic)
UCL
Sample Average
35
CL
20
LCL
5
1
2
3
4
5
6
7
8
9
10
11 12 13 14 15
Sample
42
Control Chart Pattern
(Cyclic)
 Temperature or other recurring changes in
physical environment.
 Worker fatigue.
 Differences in measuring or testing devices
which are used in order.
 Regular rotation of machines or operators.
43
Control Chart Pattern
(Freaks)
UCL
Sample Average
35
CL
20
LCL
5
1
2
3
4
5
6
7
8
9
10
11 12 13 14 15
Sample
44
Control Chart Pattern
(Freaks)
 The use of a new tool for a
brief test period.
 The failure of a component.
45
Control Chart Pattern
(Bunches)
UCL
Sample Average
35
CL
20
LCL
5
1
2
3
4
5
6
7
8
9
10
11 12 13 14 15
Sample
46
Control Chart Pattern
(Bunches)
 The use of a new vendor
for a short period of time.
 The use of different
machine for a brief time
period.
 A new operator used for a
short period.
47
Control Chart Pattern
(Mixture)
UCL
Sample Average
35
CL
20
LCL
5
1
2
3
4
5
6
7
8
9
10
11 12 13 14 15
Sample
48
Control Chart Pattern
(Mixture)
 The differences in the
incoming quality of
material from two
vendors.
 Overcontrol.
 Two or more machines
being represented on
the same control chart.
49
Control Chart Pattern
(Stratification)
UCL
Sample Average
35
CL
20
LCL
5
1
2
3
4
5
6
7
8
9
10
11 12 13 14 15
Sample
50
Control Chart Pattern
(Stratification)
 Incorrect calculation of
control limits.
 Incorrect subgrouping.
51
Process Capability
• Capability Process Estimation is performed
when the process is in control.
• Hitung standar deviasi proses.
• Proportion nonconforming item is performed
by viewing the average, standard deviasi
process, and specification limits (not the
control limits).
52
Example 2
The coil resistance specification is 21±3 ohms.
The sample with size 5 is taken with the result Rbar equal to 3.50 and the process average
estimation is 20.864. Determine the proportion
of nonconforming output with assumption that
the coil resistance data is normal distribution.
53
Example 2 (continue)
sˆ 
R
3.50

 1.505
d 2 2.326
0.0287
0.0188
LSL=18
X  20.864
sˆ  1.505
USL=24
X
54
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