Control Chart Selection
description
Transcript of Control Chart Selection
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3Control Chart Selection
Quality Characteristic
variable attribute
n>1?
n>=10 or computer?
x and MRno
yes
x and s
x and Rno
yes
defective defect
constant sample size?
p-chart withvariable samplesize
no
p ornp
yes constantsampling unit?
c u
yes no
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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.
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Types of ControlControl Chart Monitors
Attribute control chartsp chart Process fraction defectivec chart number of defectsu chart defects per unit
Variables control charts
X-bar chart Process mean
R chart (Range Chart) Process variability
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When are Shifts Detected ?Lower
SpecificationLimit
LCL UCL
UpperSpecification
Limit
Process Target
Nonconformity
Control Chart IdentifiesMean Shift Here
Attribute Chart IdentifiesMean Shift Here
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Variables v. Attributes
Both have advantages.At High levels - Attribute charts, identify
problem areas.
At Lower levels – Variables charts, quantitative problem solving tools.
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Ch 12- Control Charts for Attributes
p chart – fraction defectivenp chart – number defective
c, u charts – number of defects
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Defect vs. Defective
‘Defect’ – a single nonconforming quality characteristic.
‘Defective’ – items having one or more defects.
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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’?
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Summary of Control Chart Types and LimitsTable 12.3
These are again ‘3 sigma’ control limits
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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.
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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.
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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 !!
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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.
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p-chart
Jika bohlam tidak menyala, lampu yang rusakPerusahaan 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.
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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
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Inspection Results
Day n X1 100 202 100 53 100 304 100 355 100 24
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Compute p and p-bar
Day n X p=X/n1 100 20 0.202 100 5 0.053 100 30 0.304 100 35 0.355 100 24 0.24
Sum 1.14p-bar 0.23
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p-bar (Estimated Process Fraction Defective)
kp
p bar -p
23.0514.1
k
pp
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npppLCL
npppUCL
)1(3
)1(3
p-Chart Control Limits
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104.100
)23.1(23.323.
356.100
)23.1(23.323.
LCL
UCL
p-Chart - Control Limits
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p-Chart for Bulbs
0
0.1
0.2
0.3
0.4
1 2 3 4 5
Day
p
LCL
UCL
p-bar
p
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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 ???
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InterpretationAfter 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.
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UCL
LCL
p-Chart
Process fractions defective is shifting (trending) upward
SamplingDistribution
P = process fraction
defective PP
PP
Trend Within Control Limits
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Applications
Think of an application of a p-chart in:SalesShipping departmentLaw
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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.
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c-Chart
c-chart notation:c = number of defects
k = number of samples
defects ofnumber mean estimatedc bar -c
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c-ChartA 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.
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c-Chart
Day c c-bar LCL UCL1 5 6.6 0 14.3072 2 6.6 0 14.3073 8 6.6 0 14.3074 7 6.6 0 14.3075 11 6.6 0 14.307
Sum c 33c-bar 6.60
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kc
c
c-Chart - Mean
6.6533
k
cc
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ccUCLccLCL
33
c-Chart – Control Limits
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307.14
6.636.6
3
0or 107.1
6.636.6
3
ccUCL
ccLCL
c-Chart – Control Limits
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c-Chart for Number of Paint Defects
0
2
4
6
8
10
12
14
16
1 2 3 4 5
Car
c, n
umbe
r of d
efec
ts
c
LCL
UCL
c-bar
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Conclusion
Process shows upward trend.Even though trend is within the control limits, the
process is out of control.Mean is shifting upwardThis 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?
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Mini Case
Think of an application of a c-chart bank.
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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.
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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.
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Day n c u=c/n u-bar LCL UCL1 5 45 9.0 10.5 6.18 14.8882 5 58 11.6 10.5 6.18 14.8883 5 48 9.6 10.5 6.18 14.8884 5 53 10.6 10.5 6.18 14.8885 5 68 13.6 10.5 6.18 14.8886 5 44 8.8 10.5 6.18 14.888
Sum u 63.2u-bar 10.5
u-Chart
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ku
u
5.106
2.63
ku
u
u-Chart
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nuuUCL
nuuLCL
3
3
u-Chart
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89.145
5.1035.10
3
18.65
5.1035.10
3
nuuUCL
nuuLCL
u-Chart
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u-Chart
u-Chart Number of Paint Defects Per Car
02468
10121416
1 2 3 4 5 6
Car
c, n
umbe
r of d
efec
ts
LCL
UCL
u
u-bar
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ConclusionThe 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.
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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.
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Mini Case
Who is responsible for improving this process?
What is required to improve the process?
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u-Chart vs. c-Chart
ncu
If n = 1, u = c and .cu
Control limits of the two chart will thereforebe the same.
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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
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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 ifSpecial 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.
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Tracking Improvements
UCL
LCL
LCLLCL
UCLUCL
Process not centeredand not stable
Process centeredand stable
Additional improvementsmade to the process
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Sampel SAMA…p chart
• Proporsi diketahui• Garis Tengah = p¯
pp p
n
( )1
UCL p
LCL pp p
p p
3
3
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Sampel SAMA…p chart• Proporsi TIDAK diketahuim nomer sampel (vertikal) n ukuran sampel (horisontal) D bagian tidak sesuai p¯ = ∑Di/(mn)
Garis Tengah = p¯
pp p
n
( )1UCL p
LCL pp p
p p
3
3
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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
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np Chart
UCL = np np p 3 1( )
LCL = np np p 3 1( )
Note: If computed LCL is negative, set LCL = 0
assuming: np > 5n(1-p) > 5
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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
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Number of defects per unit:
c¯ = ∑ ci / n
UCL cc c 3
LCL cc c 3
c c
C - chart
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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¯)
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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 GT5. Giliran panjang 7-8 titik6. 1/beberapa titik dekat
satu batas kendali7. Pola data TAK RANDOM
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Patterns to Look for in Control Charts
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Twenty samples, each consisting of
250 checks, The number of defective
checks found in the 20 samples are
listed below.(proporsi tidak diketahui)
4 1 5 3 2 7 4 5 2 32 8 5 3 6 4 2 5 3 6
Example………p-np chart
$115006529 25447581 1445
2655
Simon SaysAugusta, ME 01227
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LCL = 3 .016 3(.007936) -.007808 0pp
(1 ) .016(1 .016) .015744 .007936250 250pp p
n
UCL = 3 .016 3(.007936) .039808pp
Note that thecomputed
LCLis negative.
Estimated p = 80/((20)(250)) = 80/5000 = .016
Control Limits For a p Chart$
115006529 25447581 1445
2655
Simon SaysAugusta, ME 01227
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Tdk sesuai Proporsi Tdk sesuai Proporsi
4153274523
(4/250) = 0,016(1/250) =0,004
2853642536
(2/250) = 0,008(8/250) = 0,032
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p Chart for Norwest Bank
0.000
0.005
0.010
0.015
0.020
0.025
0.030
0.035
0.040
0.045
0 5 10 15 20Sample Number
Sam
ple
Prop
ortio
n p UCL
LCL
Control Limits For a p Chart$
115006529 25447581 1445
2655
Simon SaysAugusta, ME 01227
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Ukuran sampel sama = 50 ( p-chart)
no Banyak produk cacat
no Banyak produk cacat
12345678910
4253213254
11121314151617181920
35523241043
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n =m =D =p¯ =BKA =BKB =Tabel proporsi untuk plot ke grafik
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n = 50m = 20D = 72p¯ = 72 / (20.50) = .072p = √ (0,072)(0,928)/50 = .037BKA = 0,072 + 3(0,037) = 0,183BKB = 0,072 - 3(0,037) = -0,039
= 0Tabel proporsi untuk plot ke grafik
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Ukuran sampel sama = 50 ( p-chart)
cacat
proporsi cacat proporsi
4253213254
(4/50 ) = 0,08(2/50) = 0,04
35523241043
(5/50) = 0,01
(10/50) = 0,20 (out) revisi(4/50) = 0,08(3/50) = 0,06
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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
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Ukuran sampel beda (p chart)no sampel Produk cacat no sampe
lProduk cacat
12345678910
200180200120300250400180210380
1410178201825203015
11121314151617181920
190380200210390120190380200180
15261014241518191112
Jml sampel 4860 Jml Cacat 341
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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
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Metode Individu
Sampel rata-rata n¯ = total sampel /observasi = 4860/20 = 243 p ¯ = D/(n¯m) = 341 / (243.20) = 0,07 (CL) semua
titik samaBP (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
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Tabel Proporsi untuk Grafik
No observasi sampel cacat proporsi1234567891011121314151617181920
200180200120300250400180210380190380200210390120190380200180
141017820182520301515261014241518191112
0,0700,0550,0850,067………………………………0,0950,0500,0550,067
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Example…c-chartno Byknya
kesalahanno Byknya kesalahan
12345678910
547685651610
11121314151617181920
978119576108
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c¯ = ∑c/n = 152/20 = 7,6
BPA c = (7, 6) + 3 (√7,6) = 15,87BPB c = (7, 6) - 3 (√7,6) = -0,67 = 0
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Example…u-chartno Sampel cacat no sampel cacat12345678910
20302515251020151525
514881262010610
11121314151617181920
30252525102020103020
91612106855148
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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
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Metode Individu
Sampel Rata-rata u¯ = 192/415 = 0,462 (CL) n¯ = 415/20 = 20,75
Batas KendaliObservasi -1 BPA-1 = (0,462) + 3 √ (0,462/20) = 0,916 BPB-1 = (0,462) - 3 √ (0,462/20) =
0,008…….dst