Control Chart
Selection
3
Quality Characteristic
variable
attribute
defective
n>1?
no
x and MR
yes
n>=10 or no
computer?
yes
x and s
defect
x and R
constant
sample
size?
yes
no
p-chart with
variable sample
size
constant
sampling
unit?
p or
np
yes
no
c
u
Comparison of Variables v. Attributes
 Variables




Fit certain cases.
Both mean and variation
information.
More expensive?
Identify mean shifts
sooner before large
number nonconforming.
 Attributes




Fit certain cases – taste,
color, etc.
Larger sample sizes.
Provides summary level
performance.
Must define
nonconformity.
Types of Control
Control Chart
Attribute control charts
Monitors
p chart
Process fraction defective
c chart
number of defects
u chart
defects per unit
Variables control charts
X-bar chart
Process mean
R chart (Range Chart)
Process variability
When are Shifts Detected ?
Lower
Upper
Specification
Limit
Process Target
Specification
Limit
LCL
UCL
Nonconformity
Control Chart Identifies
Mean Shift Here
Attribute Chart Identifies
Mean Shift Here
Variables v. Attributes
 Both have advantages.
 At High levels - Attribute charts, identify problem
areas.
 At Lower levels – Variables charts, quantitative
problem solving tools.
Ch 12- Control Charts for Attributes
 p chart – fraction defective
 np chart – number defective
 c, u charts – number of defects
Defect vs. Defective
 ‘Defect’ – a single nonconforming quality
characteristic.
 ‘Defective’ – items having one or more defects.
Legal Concerns with Term ‘Defect’
 Often called ‘nonconformity’.
 Possible Legal Dialog
 Does your company make a lot of ‘defects’?
 Enough to track them on a chart ?
 If they are not ‘bad’, why do you call them
‘defects’, sounds bad to me.
 So you knowingly track and ship products with
‘defects’?
Summary of Control Chart Types and Limits
Table 12.3
These are again ‘3 sigma’ control limits
p, np - Chart
 P is fraction nonconforming.
 np is total nonconforming.
 Charts based on Binomial distribution.
 Sample size must be large enough (example p=2%)
 Definition of a nonconformity.
 Probability the same from item to item.
c, u - Charts
 c and u charts deal with nonconformities.
 c Chart – total number of nonconformities.
 u Chart – nonconformities per unit.
 Charts based on Poisson distribution.
 Sample size, constant probabilities.
How to Interpret Attribute Charts
 Points beyond limits- primary test.

Below lower limits means process has improved.
 Zone rules do not apply.
 Rules for trends, shifts do apply.
Only get One Chart !!
Examples of When to Use
 p,np charts–


Number of
nonconforming cables is
found for 20 samples of
size 100.
Number of
nonconforming floppy
disks is found for samples
of 200 for 25 trials.
 c,u charts

Number of paint
blemishes on auto body
observed for 30 samples.
Number of imperfections
in bond paper – by area
inspected and number of
imperfections.
p-chart
 Jika bohlam tidak menyala, lampu yang rusak
 Perusahaan ingin
 Perkirakan persentase bohlam rusak dan
 Tentukan apakah persentase bohlam rusak terus meningkat
dari waktu ke waktu..
 Sebuah peta kendali p adalah alat yang sesuai untuk
menyediakan perusahaan dengan informasi ini.
Notation
 Sample size = n = 100
 Number of samples (subgroups) = k = 5
 X = number of defective bulbs in a sample
 p = sample fraction defective = ???
 p-bar = estimated process fraction
defective

P = process fraction defective (unknown)
 p-bar is an estimate of P
Inspection Results
Day
1
2
3
4
5
n
100
100
100
100
100
X
20
5
30
35
24
Compute p and p-bar
Day
1
2
3
4
5
Sum
p-bar
n
100
100
100
100
100
X
20
5
30
35
24
p=X/n
0.20
0.05
0.30
0.35
0.24
1.14
0.23
p-bar
(Estimated Process Fraction Defective)
p

p - bar  p 
k
p 1.14

p

 0.23
k
5
p-Chart Control Limits
p (1  p )
UCL  p  3
n
p (1  p )
LCL  p  3
n
p-Chart - Control Limits
.23(1  .23)
UCL  .23  3
 .356
100
.23(1  .23)
LCL  .23  3
 .104
100
p-Chart for Bulbs
0.4
p
0.3
LCL
UCL
0.2
p-bar
p
0.1
0
1
2
3
Day
4
5
Interpretation
 The estimated fraction of defective bulbs produced is




.23.
On Day 2, p was below the LCL.
This means that a special cause occurred on that day
to cause the process to go out of control.
The special cause shifted the process fraction
defective downward.
This special cause was therefore favorable and should
be ???
Interpretation
 After Day 2, the special cause lost its impact
because on Day 4, the process appears to be back
in control and at old fraction defective of .23.
 Until the special cause is identified and made part of
the process, the process will be unstable and
unpredictable.
 It is therefore impossible to obtain a statistical valid
estimate of the process fraction defective because it
can change from day to day.
Trend Within Control Limits
Process fractions defective
is shifting (trending) upward
P
Sampling
Distribution
P
UCL
p-Chart
LC
L
P
P
P = process
fraction
defective
Applications
 Think of an application of a p-chart in:
 Sales
 Shipping
 Law
department
Use of c-Charts
 When we are interested in monitoring number of
defects on a given unit of product or service.






Scratches, chips, dents on an airplane wing
Errors on an invoice
Pot holes on a 5-mile section of highway
Complaints received per day
Opportunity for a defect must be infinite.
Probability of a defect on any one location or any
one point in time must be small.
c-Chart
c-chart
c = number of defects
notation:
k = number of samples
c - bar  c  estimated mean number of defects
c-Chart
 A car company wants to monitor the number of paint
defects on a certain new model of one of its cars.
• Each day one car in inspected.
• The results after 5 days are shown on the next slide.
c-Chart
Day
1
2
3
4
5
Sum c
c-bar
c
5
2
8
7
11
33
6.60
c-bar
6.6
6.6
6.6
6.6
6.6
LCL
0
0
0
0
0
UCL
14.307
14.307
14.307
14.307
14.307
c-Chart - Mean
c

c
k
c 33

c
  6.6
k
5
c-Chart – Control Limits
LCL  c  3 c
UCL  c  3 c
c-Chart – Control Limits
LCL  c  3 c
 6 .6  3 6 .6
 1.107 or 0
UCL  c  3 c
 6 .6  3 6 .6
 14.307
c-Chart for Number of Paint Defects
c, number of defects
16
14
12
c
10
LCL
8
UCL
6
c-bar
4
2
0
1
2
3
Car
4
5
Conclusion
 Process shows upward trend.
 Even though trend is within the control limits, the




process is out of control.
Mean is shifting upward
This is due to an unfavorable special cause.
Must identify special cause and eliminate it from
process.
Who is responsible for finding and eliminating
special cause?
Mini Case
 Think of an application of a c-chart bank.
u-Chart
 With a c chart, the sample size is one unit.
 A u-chart is like a c-chart, except that the
sample size is greater than one unit.
 As a result, a u-chart tracks the number of
defects per unit.
 A c-chart monitors the number of defects
on one unit.
u-Chart
 A car company monitors the number of
paint defects per car by taking a
sample of 5 cars each day over the
next 6 days.
 The results are shown on next side.
u-Chart
Day
1
2
3
4
5
6
Sum u
u-bar
n
5
5
5
5
5
5
c
45
58
48
53
68
44
u=c/n
9.0
11.6
9.6
10.6
13.6
8.8
63.2
10.5
u-bar
10.5
10.5
10.5
10.5
10.5
10.5
LCL
6.18
6.18
6.18
6.18
6.18
6.18
UCL
14.888
14.888
14.888
14.888
14.888
14.888
u-Chart
u

u 
k
u 63.2

u 

 10.5
k
6
u-Chart
u
LCL  u  3
n
u
UCL  u  3
n
u-Chart
u
LCL  u  3
n
10.5
 10.5  3
 6.18
5
u
UCL  u  3
n
10.5
 10.5  3
 14.89
5
u-Chart
u-Chart
c, number of defects
Number of Paint Defects Per Car
16
14
12
LCL
10
UCL
8
6
u
u-bar
4
2
0
1
2
3
4
Car
5
6
Conclusion
 The process appears stable.
 We can therefore get a statistically valid estimate the
process mean number of defects per car.
 Our estimate of the mean number of paint defects per
car is 10.5, the center line on the control chart.
 Thus, we expect each car to have, on average, 10.5
paint defects.
Conclusion
 Although the process is stable, the number of
defects per car is too high.
 Deming calls this a stable process for the production
of defective product.
 Important take away:


A stable process (process in control) is not necessarily a good
process because it can be in control at the wrong level.
A stable process is predictable, but this doesn’t mean that
what is being predicted is favorable.
Mini Case
 Who is responsible for improving this process?
 What is required to improve the process?
u-Chart vs. c-Chart
c
u
n
If n = 1, u = c and u  c .
Control limits of the two chart will therefore
be the same.
Sample Size
Control Chart
When To Use
Sample Size
p-Chart
Monitor the
proportion of
defectives in a process
At least 50
c-Chart
Monitor the number
of defects
1
u-chart
Monitor the number
of defects per unit
>1
In Practice
 You need 25 to 30 samples before computing initial
control limits.
 When a special cause occurs, you should eliminate
that sample and re-compute control limits if
 Special
cause is identified
 Eliminated or made part of process
 To identify special causes, workers must keep log
sheet, where they record any changes they make to
the process.
Tracking Improvements
UCL
UCL
LCL
LCL
Process centered
Process not centered and stable
and not stable
UCL
LCL
Additional improvements
made to the process
Sampel SAMA…p chart
• Proporsi diketahui
p  p  3 p
• Garis TengahUCL
= p¯
LCL p  p  3 p
p
p( 1  p )
n
• ProporsiSampel
TIDAK SAMA…p
diketahui chart
m nomer sampel (vertikal)
 n ukuran sampel (horisontal)
 D bagian tidak sesuai
p¯
= ∑Di/(mn)
Garis Tengah = p¯
UCL p  p  3 p
LCL p  p  3 p
p
p( 1  p )
n
Sampel BEDA …
a.
Metode INDIVIDU  Batas Kendali tergantung
ukuran sample tertentu shg BKA/BKB tidak
berupa garis LURUS
b.
Metode RATA_RATA  Ukuran sampel RATA RATA dg perbedaan tidak terlalu besar
( n¯ = ∑n/observasi)
c.
Peta Kendali TERSTANDAR dg GT=0 dan BK ± 3
np Chart
UCL = np  3 np( 1  p)
LCL = np  3 np(1  p)
assuming:
np > 5
n(1-p) > 5
Note: If computed LCL is negative, set LCL = 0
c-chart dan u-chart
 Mengetahui banyaknya kesalahan unit produk sbg
sampel
 Sampel konstan  c-chart
 Sampel bervariasi  u-chart
 Aplikasi : bercak pd tembok, gelembung udara pd
gelas, kesalahan pemasangan sekrup pd mobil
C - chart
Number of defects per unit:
c¯ = ∑ ci / n
c  c
UCLc  c  3 c
LCLc  c  3 c
U-chart
 u¯ = ∑ ci/n
 n ¯ = ∑ ni/g
g = banyaknya observasi
Model Individu
 BPA-u = u¯ + 3 √ (u¯ /ni)
 BPB-u = u¯ - 3 √ (u¯ /ni)
Model Rata-rata
 BPA-u = u¯ + 3 √ (u¯ /n¯)
 BPB-u = u¯ - 3 √ (u¯ /n¯)
Warning Conditions…..
Western Electric :
1. 1 titik diluar batas kendali (
3σ)
2. 2 dr 3 titik berurutan diluar
batas kend64li (2σ)
3. 4 dr 5 titik berurutan jauh dari
GT (1σ)
4. 8 titik berurutan di satu sisi
GT
5. Giliran panjang 7-8 titik
6. 1/beberapa titik dekat satu
batas kendali
7. Pola data TAK RANDOM
Patterns to Look for in Control Charts
Example………p-np chart
Twenty samples, each consisting of
250 checks, The number of defective
checks found in the 20 samples are
listed below.
(proporsi tidak diketahui)
4
2
1
8
5
5
3
3
2
6
7
4
4
2
5
5
2
3
3
6
Control Limits For a p Chart
Estimated p = 80/((20)(250)) = 80/5000 = .016
p(1  p)
.016(1  .016)
.015744
p 


 .007936
n
250
250
UCL = p  3 p  .016  3(.007936)  .039808
LCL = p  3 p  .016  3(.007936)  -.007808  0
Note that the
computed LCL
is negative.
Tdk sesuai
Proporsi
Tdk sesuai
Proporsi
4
(4/250) = 0,016
2
(2/250) = 0,008
1
(1/250) =0,004
8
(8/250) = 0,032
5
5
3
3
2
6
7
4
4
2
5
5
2
3
3
6
Control Limits For a p Chart
p Chart for Norwest Bank
0.045
Sample Proportion p
0.040
UCL
0.035
0.030
0.025
0.020
0.015
0.010
0.005
LCL
0.000
0
5
10
Sample Number
15
20
Ukuran sampel sama = 50 ( p-chart)
no
Banyak produk cacat
no
Banyak produk cacat
1
4
11
3
2
2
12
5
3
5
13
5
4
3
14
2
5
2
15
3
6
1
16
2
7
3
17
4
8
2
18
10
9
5
19
4
10
4
20
3
n






=
m
=
D
=
p¯
=
BKA
=
BKB
=
Tabel proporsi untuk plot ke grafik
n







= 50
m
= 20
D
= 72
p¯
= 72 / (20.50) = .072
p
= √ (0,072)(0,928)/50 = .037
BKA
= 0,072 + 3(0,037)
= 0,183
BKB
= 0,072 - 3(0,037)
= -0,039 = 0
Tabel proporsi untuk plot ke grafik
Ukuran sampel sama = 50 ( p-chart)
cacat
proporsi
cacat
4
(4/50 ) = 0,08
3
2
(2/50) = 0,04
5
proporsi
(5/50) = 0,01
5
5
3
2
2
3
1
2
3
4
(10/50) = 0,20 (out) revisi
2
10
(4/50) = 0,08
5
4
(3/50) = 0,06
4
3
Revisi
 p¯ = (72-10) / (1000-50) = 62/950 = 0,065

p = √ (0,065)(0,935)/50 = 0,035
 BKA = 0,065 + 3 (0,035)
= 0.17
 BKB = 0,065 - 3 (0,035)
= -0,04 = 0
 Grafiknya juga berubah
Ukuran sampel beda (p chart)
no
sampel
Produk cacat
no
sampel
Produk cacat
1
200
14
11
190
15
2
180
10
12
380
26
3
200
17
13
200
10
4
120
8
14
210
14
5
300
20
15
390
24
6
250
18
16
120
15
7
400
25
17
190
18
8
180
20
18
380
19
9
210
30
19
200
11
10
380
15
20
180
12
Jml
sampel
4860
Jml
Cacat
341
Metode Rata-rata
 Sampel rata-rata
n¯
= total sampel /observasi
= 4860/20 = 243
p¯ = D/(n¯m)
= 341 / (243.20) = 0,07 (CL)
p = √ (0,07(0,93))/243 = 0,0164
BPAp = 0,07 + 3 (0,0164) = 0,119
BPBp = 0,07 - 3 (0,0164) = 0,021
Metode Individu
 Sampel rata-rata
n¯
= total sampel /observasi
= 4860/20 = 243
p ¯ = D/(n¯m)
= 341 / (243.20) = 0,07 (CL) semua titik sama
 BP (obs-1)
p
= √ (0,07(0,93))/200 = 0,018
BPA = 0,07 + 3 (0,018) = 0,124
BPB = 0,07 - 3 (0,018) = 0,016……………….dst
Tabel Proporsi untuk Grafik
No observasi
sampel
cacat
proporsi
1
200
14
0,070
2
180
10
0,055
3
200
17
0,085
4
120
8
0,067
5
300
20
…
6
250
18
…
7
400
25
…
8
180
20
…
9
210
30
…
10
380
15
…
11
190
15
…
12
380
26
…
13
200
10
…
14
210
14
…
15
390
24
…
16
120
15
…
17
190
18
0,095
18
380
19
0,050
19
200
11
0,055
20
180
12
0,067
Example…c-chart
no
Byknya kesalahan
no
Byknya kesalahan
1
5
11
9
2
4
12
7
3
7
13
8
4
6
14
11
5
8
15
9
6
5
16
5
7
6
17
7
8
5
18
6
9
16
19
10
10
10
20
8
 c¯ = ∑c/n = 152/20 = 7,6
 BPA c = (7, 6) + 3 (√7,6) = 15,87
 BPB c = (7, 6) - 3 (√7,6)
= -0,67 = 0
Example…u-chart
no
Sampel
cacat
no sampel
cacat
1
20
5
11 30
9
2
30
14
12 25
16
3
25
8
13 25
12
4
15
8
14 25
10
5
25
12
15 10
6
6
10
6
16 20
8
7
20
20
17 20
5
8
15
10
18 10
5
9
15
6
19 30
14
10
25
10
20 20
8
Metode Rata-rata
 Sampel Rata-rata
u¯ = 192/415 = 0,462 (CL)
n¯ = 415/20 = 20,75
BPAu = (0,462) + 3 √ (0,462/20,75) = 0,906
BPBu = (0,462) - 3 √ (0,462/20,75) = 0,018
Metode Individu
 Sampel Rata-rata
u¯ = 192/415 = 0,462 (CL)
n¯ = 415/20 = 20,75
 Batas Kendali
 Observasi -1
BPA-1 = (0,462) + 3 √ (0,462/20) = 0,916
BPB-1 = (0,462) - 3 √ (0,462/20) = 0,008…….dst