statistic process control

Statistic Process Control

Week 3

Ananda Sabil Hussein, SE, MCom

Latar Belakang

Pertengahan tahun 80 an pangsa pasar pager Motorola di rebut oleh produk-produk Jepang seperti halnya NEC, TOSHIBA dan Hitachi.

Motorola melakukan perubahan radikal dengan memperbaiki mutu, pengembangan produk dan penurunan biaya yang berbasis statistik

Statistical Process Control

Teknik statistik yang secara luas digunakan untuk memastikan bahwa proses yang sedang berjalan telah memenuhi standar.

StartProduce Good

Provide Service

Stop Process

Yes

NoAssign.

Causes?Take Sample

Inspect Sample

Find Out WhyCreate

Control Chart

Variasi Alami dan Khusus

Variasi alami adalah sumber-sumber variasi dalam proses yang secara statistik berada dalam batas kendali

Variasi Khusus/dapat dihilangkan yaitu variasi yang muncul disebabkan karena peralatan yang tidak sesuai, karyawan yang lelah atau kurang terlatih serta bahan baku baru.

Diagram Pengendalian

Plot of Sample Data Over Time

0

20

40

60

80

1 5 9 13 17 21

Time

Sam

ple

Va

lue

SampleValueUCL

Average

LCL

17 = UCL17 = UCL

15 = LCL15 = LCL

16 = Mean16 = Mean

Sample numberSample number

|| || || || || || || || || || || ||11 22 33 44 55 66 77 88 99 1010 1111 1212

Konsep Rata-rata dan Jarak

Rata-rata

Menentukan Batas Diagram Rata-rata

Batas Kendali Atas (UCL) = Batas Kendali Bawah (LCL) = = rata-rata dari sampel =

xZX xZX

x

Z

X

x

Z

X

= Standar deviasi = 2 (95.5%) 3(99.7%)

= Standar deviasi rata-rata sampel

nx

Cara Lain

RAX 2

RAX 2Batas Kendali Atas =

Batas Kendali Bawah

Dimana :

x

A

R

2

= rentangan rata-rata sampel

= Nilai batas kendali

= rata-rata dari sampel rata-rata

Batas Bagan Rentangan

RDLCL

RDUCL

R

R

3

4

Bagan Rata-rata

(a)(a)

These These sampling sampling distributions distributions result in the result in the charts belowcharts below

(Sampling mean is (Sampling mean is shifting upward but shifting upward but range is consistent)range is consistent)

R-chartR-chart(R-chart does not (R-chart does not detect change in detect change in mean)mean)

UCLUCL

LCLLCL

x-chartx-chart(x-chart detects (x-chart detects shift in central shift in central tendency)tendency)

UCLUCL

LCLLCL

R-chartR-chart(R-chart detects (R-chart detects increase in increase in dispersion)dispersion)

UCLUCL

LCLLCL

Figure S6.5Figure S6.5

(b)(b)

These These sampling sampling distributions distributions result in the result in the charts belowcharts below

(Sampling mean (Sampling mean is constant but is constant but dispersion is dispersion is increasing)increasing)

x-chartx-chart(x-chart does not (x-chart does not detect the increase detect the increase in dispersion)in dispersion)

UCLUCL

LCLLCL

Bagan Jarak

Bagan Kendali Atribut

Mengukur persentase penolakan dalam sebuah sampel, bagan-p

Menghitung jumlah penolakan, bagan-c

For variables that are categoricalFor variables that are categorical

Good/bad, yes/no, Good/bad, yes/no, acceptable/unacceptableacceptable/unacceptable

Measurement is typically counting defectivesMeasurement is typically counting defectives

Charts may measureCharts may measure

Percent defective (p-chart)Percent defective (p-chart)

Number of defects (c-chart)Number of defects (c-chart)

Control Charts for Attributes

Control Limits for p-Charts

Population will be a binomial distribution, but Population will be a binomial distribution, but applying the Central Limit Theorem allows us to applying the Central Limit Theorem allows us to

assume a normal distribution for the sample assume a normal distribution for the sample statisticsstatistics

UCLUCLpp = p + z = p + zpp^̂

LCLLCLpp = p - z = p - zpp^̂

wherewhere pp ==mean fraction defective in the samplemean fraction defective in the samplezz ==number of standard deviationsnumber of standard deviationspp ==standard deviation of the sampling distributionstandard deviation of the sampling distribution

nn ==sample sizesample size

^̂

pp(1 -(1 - p p))nn

pp = =^̂

Contoh Soal

Jam Rata2 Jam Rata2 Jam Rata2

1 17.1 5 16.5 9 16.3

2 18.8 6 16.4 10 16.5

3 14.5 7 15.2 11 14.2

4 14.8 8 16.4 12 17.3

Ditanyakan : Batas kendali proses 9 boks yang mencakup 99.7%

Jawab :xZX UCLx = = 16 + 3

9

1

LCLx = xZX = 16 - 3

9

1

Setting Control Limits

Process average x Process average x = 16.01= 16.01 ounces ouncesAverage range R Average range R = .25= .25Sample size n Sample size n = 5= 5

Setting Control Limits

UCLUCLxx = x + A= x + A22RR

= 16.01 + (.577)(.25)= 16.01 + (.577)(.25)= 16.01 + .144= 16.01 + .144= 16.154 = 16.154 ouncesounces

Process average x Process average x = 16.01= 16.01 ounces ouncesAverage range R Average range R = .25= .25Sample size n Sample size n = 5= 5

From From Table S6.1Table S6.1

Setting Control Limits

UCLUCLxx = x + A= x + A22RR

= 16.01 + (.577)(.25)= 16.01 + (.577)(.25)= 16.01 + .144= 16.01 + .144= 16.154 = 16.154 ouncesounces

LCLLCLxx = x - A= x - A22RR

= 16.01 - .144= 16.01 - .144= 15.866 = 15.866 ouncesounces

Process average x Process average x = 16.01= 16.01 ounces ouncesAverage range R Average range R = .25= .25Sample size n Sample size n = 5= 5

UCL = 16.154UCL = 16.154

Mean = 16.01Mean = 16.01

LCL = 15.866LCL = 15.866

Contoh SoalSampleSample NumberNumber FractionFraction SampleSample NumberNumber FractionFractionNumberNumber of Errorsof Errors DefectiveDefective NumberNumber of Errorsof Errors DefectiveDefective

11 66 .06.06 1111 66 .06.0622 55 .05.05 1212 11 .01.0133 00 .00.00 1313 88 .08.0844 11 .01.01 1414 77 .07.0755 44 .04.04 1515 55 .05.0566 22 .02.02 1616 44 .04.0477 55 .05.05 1717 1111 .11.1188 33 .03.03 1818 33 .03.0399 33 .03.03 1919 00 .00.00

1010 22 .02.02 2020 44 .04.04

Total Total = 80= 80

(.04)(1 - .04)(.04)(1 - .04)

100100pp = = = .02= .02^̂p p = = .04= = .04

8080

(100)(20)(100)(20)

.11 .11 –

.10 .10 –

.09 .09 –

.08 .08 –

.07 .07 –

.06 .06 –

.05 .05 –

.04 .04 –

.03 .03 –

.02 .02 –

.01 .01 –

.00 .00 –

Sample numberSample number

Fra

ctio

n d

efec

tive

Fra

ctio

n d

efec

tive

| | | | | | | | | |

22 44 66 88 1010 1212 1414 1616 1818 2020

p-Chart for Data Entry

UCLUCLpp = p + z = p + zpp = .04 + 3(.02) = .10= .04 + 3(.02) = .10^̂

LCLLCLpp = p - z = p - zpp = .04 - 3(.02) = 0 = .04 - 3(.02) = 0^̂

UCLUCLpp = 0.10= 0.10

LCLLCLpp = 0.00= 0.00

p p = 0.04= 0.04

.11 .11 –

.10 .10 –

.09 .09 –

.08 .08 –

.07 .07 –

.06 .06 –

.05 .05 –

.04 .04 –

.03 .03 –

.02 .02 –

.01 .01 –

.00 .00 –

Sample numberSample number

Fra

ctio

n d

efec

tive

Fra

ctio

n d

efec

tive

| | | | | | | | | |

22 44 66 88 1010 1212 1414 1616 1818 2020

UCLUCLpp = p + z = p + zpp = .04 + 3(.02) = .10= .04 + 3(.02) = .10^̂

LCLLCLpp = p - z = p - zpp = .04 - 3(.02) = 0 = .04 - 3(.02) = 0^̂

UCLUCLpp = 0.10= 0.10

LCLLCLpp = 0.00= 0.00

p p = 0.04= 0.04

p-Chart for Data Entry

Possible assignable

causes present

Control Limits for c-Charts

Population will be a Poisson distribution, Population will be a Poisson distribution, but applying the Central Limit Theorem but applying the Central Limit Theorem

allows us to assume a normal distribution allows us to assume a normal distribution for the sample statisticsfor the sample statistics

wherewhere cc ==mean number defective in the samplemean number defective in the sample

UCLUCLcc = c + = c + 33 c c LCLLCLcc = c - = c - 33 c c

c-Chart for Cab Company

c c = 54= 54 complaints complaints/9/9 days days = 6 = 6 complaintscomplaints//dayday

|1

|2

|3

|4

|5

|6

|7

|8

|9

DayDay

Nu

mb

er d

efec

tive

Nu

mb

er d

efec

tive14 14 –

12 12 –

10 10 –

8 8 –

6 6 –

4 –

2 –

0 0 –

UCLUCLcc = c + = c + 33 c c

= 6 + 3 6= 6 + 3 6= 13.35= 13.35

LCLLCLcc = c - = c - 33 c c

= 3 - 3 6= 3 - 3 6= 0= 0

UCLUCLcc = 13.35= 13.35

LCLLCLcc = 0= 0

c c = 6= 6