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.

Start

Produce Good
Provide Service
Take Sample

No
Assign.
Causes?
Yes

Inspect Sample

Stop Process

Create

Control Chart

Find Out Why

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

17 = UCL
16 = Mean

15 = LCL
| | | | | | | | | | | |
1 2 3 4 5 6 7 8 9 10 11 12

Sample number

Konsep Rata-rata dan Jarak

Rata-rata

X
Z

x

Menentukan Batas Diagram Rata-rata




Batas Kendali Atas (UCL) = X  Z x
Batas Kendali Bawah (LCL) = X  Z x
X = rata-rata dari sampel =

Z

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

x

= Standar deviasi rata-rata sampel

x
n

Cara Lain


Batas Kendali Atas =

Batas Kendali Bawah

Dimana :

R
A2
x

X  A2 R
X  A2 R

= rentangan rata-rata sampel
= Nilai batas kendali
= rata-rata dari sampel rata-rata

Batas Bagan Rentangan
UCLR D4 R
LCLR D3 R

Bagan Rata-rata

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

These
sampling
distributions
result in the
charts below
UCL

x-chart
LCL

(x-chart detects
shift in central
tendency)

UCL

R-chart
LCL

(R-chart does not
detect change in
mean)

Bagan
Jarak
(b)
(b)

These
sampling
distributions
result in the
charts below

(Sampling mean

is constant but
dispersion is
increasing)
UCL

x-chart
LCL

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

UCL

R-chart
LCL
Figure S6.5

(R-chart detects
increase in

dispersion)

Bagan Kendali Atribut
 Mengukur

persentase penolakan dalam
sebuah sampel, bagan-p
 Menghitung jumlah penolakan, bagan-c

Control Charts for Attributes

 For variables that are categorical
 Good/bad, yes/no,
acceptable/unacceptable
 Measurement is typically counting defectives
 Charts may measure
 Percent defective (p-chart)
 Number of defects (c-chart)

Control Limits for p-Charts

Population will be a binomial distribution, but
applying the Central Limit Theorem allows us to
assume a normal distribution for the sample
statistics

UCLp = p + z p^

p =
^

LCLp = p - z p^
where
^

p
z
p
n

=

=
=
=

p(1 - p)
n

mean fraction defective in the sample
number of standard deviations
standard deviation of the sampling dis
sample size

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 :

UCLx =

X  Z x

LCLx = X  Z x

1 

 9

= 16 + 3 

 1 
= 16 - 3  9 



Setting Control Limits
Process average x = 16.01 ounces
Average range R = .25
Sample size n = 5

Setting Control Limits
Process average x = 16.01 ounces
Average range R = .25
Sample size n = 5
UCLx

= x + A2R
= 16.01 + (.577)(.25)
= 16.01 + .144
= 16.154 ounces
From
Table S6.1

Setting Control Limits
Process average x = 16.01 ounces
Average range R = .25
Sample size n = 5
UCLx

LCLx

= x + A2 R
= 16.01 + (.577)(.25)
= 16.01 + .144
= 16.154 ounces
= x - A2 R
= 16.01 - .144
= 15.866 ounces

UCL = 16.154

Mean = 16.01

LCL = 15.866

Contoh Soal
Sample Number
Number of Errors

1
2
3
4
5
6
7
8
9
10

p=

Fraction
Defective

Sample Number
Number of Errors

Fraction
Defective

.06
.05
.00
.01
.04
.02
.05
.03
.03
.02

11
6
12
1
13
8
14
7
15
5
16
4
17
11
18
3
19
0
20
4
Total = 80

.06
.01
.08
.07
.05
.04
.11
.03
.00
.04

6
5
0
1
4
2
5
3
3
2
80
(100)(20)

= .04

 p^ =

(.04)(1 - .04)
100

= .02

p-Chart for Data Entry
UCLp = p + z p^ = .04 + 3(.02) = .10

Fraction defective

LCLp = p - z p^ = .04 - 3(.02) = 0
.11
.10
.09
.08
.07
.06
.05
.04
.03
.02
.01
.00

–
–
–
–
–
–
–
–
–
–
–
–

UCLp = 0.10

p = 0.04

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2

4

6

8

10

12

14

16

18

20

Sample number

LCLp = 0.00

p-Chart for Data Entry
UCLp = p + z p^ = .04 + 3(.02) = .10

Fraction defective

Possible
LCLp = p - z p^ = .04 - 3(.02) = assignable
0
causes present

.11
.10
.09
.08
.07
.06
.05
.04
.03
.02
.01
.00

–
–
–
–
–
–
–
–
–
–
–
–

UCLp = 0.10

p = 0.04

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2

4

6

8

10

12

14

16

18

20

LCLp = 0.00

Control Limits for c-Charts
Population will be a Poisson distribution,
but applying the Central Limit Theorem
allows us to assume a normal distribution
for the sample statistics
UCLc = c + 3 c
where

c

LCLc = c - 3 c
=

mean number defective in the sam

c-Chart for Cab Company

LCLc = c - 3 c
=3-3 6
=0

Number defective

c = 54 complaints/9 days = 6 complaints/day
UCLc = c + 3 c
UCL = 13.35
14 –
12 –
=6+3 6
10 –
= 13.35
c

8
6
4
2
0

–
–
–
–
– |

|
1 2

c= 6

| |
3 4

|
5

Day

|
6

|
7

LCLc = 0

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8 9