Control Chart Selection

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

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 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

When are Shifts Detected ?Lower

SpecificationLimit

LCL UCL

UpperSpecification

Limit

Process Target

Nonconformity

Control Chart IdentifiesMean Shift Here

Attribute Chart IdentifiesMean 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 defectivenp 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 LimitsTable 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 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.

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 n X1 100 202 100 53 100 304 100 355 100 24

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

p-bar (Estimated Process Fraction Defective)

kp

p bar -p

23.0514.1

k

pp

npppLCL

npppUCL

)1(3

)1(3

p-Chart Control Limits

104.100

)23.1(23.323.

356.100

)23.1(23.323.

LCL

UCL

p-Chart - Control Limits

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

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 ???

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.

UCL

LCL

p-Chart

Process fractions defective is shifting (trending) upward

SamplingDistribution

P = process fraction

defective PP

PP

Trend Within Control Limits

Applications

Think of an application of a p-chart in:SalesShipping departmentLaw

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 notation:c = number of defects

k = number of samples

defects ofnumber mean estimatedc bar -c

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.

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

kc

c

c-Chart - Mean

6.6533

k

cc

ccUCLccLCL

33

c-Chart – Control Limits

307.14

6.636.6

3

0or 107.1

6.636.6

3

ccUCL

ccLCL

c-Chart – Control Limits

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

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?

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.

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

ku

u

5.106

2.63

ku

u

u-Chart

nuuUCL

nuuLCL

3

3

u-Chart

89.145

5.1035.10

3

18.65

5.1035.10

3

nuuUCL

nuuLCL

u-Chart

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

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.

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

ncu

If n = 1, u = c and .cu

Control limits of the two chart will thereforebe 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 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.

Tracking Improvements

UCL

LCL

LCLLCL

UCLUCL

Process not centeredand not stable

Process centeredand stable

Additional improvementsmade to the process

Sampel SAMA…p chart

• Proporsi diketahui• Garis Tengah = p¯

pp p

n

( )1

UCL p

LCL pp p

p p

3

3

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

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 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

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

Number of defects per unit:

c¯ = ∑ ci / n

UCL cc c 3

LCL cc c 3

c c

C - chart

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 GT5. Giliran panjang 7-8 titik6. 1/beberapa titik dekat

satu batas kendali7. Pola data TAK RANDOM

Patterns to Look for in Control Charts

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

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

Tdk sesuai Proporsi Tdk sesuai Proporsi

4153274523

(4/250) = 0,016(1/250) =0,004

2853642536

(2/250) = 0,008(8/250) = 0,032

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

Ukuran sampel sama = 50 ( p-chart)

no Banyak produk cacat

no Banyak produk cacat

12345678910

4253213254

11121314151617181920

35523241043

n =m =D =p¯ =BKA =BKB =Tabel proporsi untuk plot ke grafik

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

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

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 sampe

lProduk cacat

12345678910

200180200120300250400180210380

1410178201825203015

11121314151617181920

190380200210390120190380200180

15261014241518191112

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 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

Tabel Proporsi untuk Grafik

No observasi sampel cacat proporsi1234567891011121314151617181920

200180200120300250400180210380190380200210390120190380200180

141017820182520301515261014241518191112

0,0700,0550,0850,067………………………………0,0950,0500,0550,067

Example…c-chartno Byknya

kesalahanno Byknya kesalahan

12345678910

547685651610

11121314151617181920

978119576108

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

Example…u-chartno Sampel cacat no sampel cacat12345678910

20302515251020151525

514881262010610

11121314151617181920

30252525102020103020

91612106855148

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 KendaliObservasi -1 BPA-1 = (0,462) + 3 √ (0,462/20) = 0,916 BPB-1 = (0,462) - 3 √ (0,462/20) =

0,008…….dst