contoh kasus six sigma dipublikasi oleh

MENURUNKAN CACAT NO POWER PADA PRODUKSI
TV
DENGAN METODA SIX SIGMA
DI PT LG ELECTRONICS INDONESIA

A R I P I N
2002 - 21 – 140
TEKNIK INDUSTRI
UNIVERSITAS INDONUSA ESA UNGGUL
2004

Konsep Six Sigma
 Six Sigma : Suatu sistem yang flexibel untuk memperbaiki kinerja
yang mendukung sukses secara terus-menerus.
 Six Sigma dibangun berdasarkan banyak ide manajemen yang penting dan
praktek terbaik dari metoda sebelumnya

 Sigma adalah ukuran satuan statistik yang menggambarkan
kemampuan suatu proses dan ukuran nilai sigma dinyatakan dalam
DPU (Defect Per Unit) atau PPM
 Dapat dikatakan bahwa proses dengan nilai sigma lebih tinggi

dalam proses akan mempunyai cacat yang lebih sedikit
 Makin bertambah nilai sigma,
makin berkurang biaya kualitas dan waktu siklus

Konsep Six Sigma
Menggunakan Z sebagai satuan ukuran kemampuan proses
•Z =3

Sample’s average :

LSL

n

USL

=

X=


 Xi

i=1

n

Sample’s standard deviation :
n

 ( Xi - X )





=s=



2


i =1

n-1

Nilai Z (sigma level)

Z =

Kemampuan 3s

•Z =6

x - 


USL

LSL


Bila keragaman / variasi turun, maka
kemungkinan cacat juga turun,
maka kemampuan (capability)
1σ 2σ 3σ 4σ 5σ 6σ

Kemampuan 6s

meningkat

Konsep Six Sigma

DEFINE

• Standardize
solution
• Establish standard
measure to
maintain
performance


CONTROL

• Develop ideas to cut
of root causes
• Optimation solution
• Measure solution

IMPROVE

• Identify Customer’s CTQ
• Develop a team charter
• Defne and build a process
map

• Validate problem
/ process
• Measure key
problem
• Current
MEASURE condition


• Screening potential
factor
• Identify vital few root
causes
• Validate hypothesis

ANALYZE

Konsep Six Sigma

Geser
menuju
Target

Tepat namun tidak akurat

Shifting/bergeser ke Target
& Mereduksi Variasi


T
USL

USL

T
USL

μ

USL

μ

Object
dari 6
adalah

Akurat namun tidak tepat


Shift(geser)
ke target

T

Menurunkan
Variasi

USL

USL

μ

Turunkan
Variasi

Aplikasi detail Six Sigma

Perbaikan dengan Metoda Six Sigma ( DMAIC )


PT LG ELECTRONICS INDONESIA

REDUCE SMPS TRANSFORMER
SHORT ( NO POWER )

2004 06 14

Incoming Quality Control

Project Registration Sheet
Theme
Name

Project
Period

Quantity
Target


REDUCE SMPS TRANSFORMER
SHORT ( NO POWER )

Theme

Measurement February 3 ~ 6, 2004

Outline

Analysis

February 9 ~ 21, 2004

Improvement

February 24 ~ 28, 2004

Control

March ~ May, 2004


Current

Target

Improvement
Ratio (%)

840 PPM

0 PPM

100 %



6

Anticipation Short Test as check point at line
Effect
production of Samwha and IQC LGEIN.

Status
&
Problem

• Material Checking method
• Quality System set up

WIRE HOLE
WIRING CROSS
CHECKING PROSSES
Name

Member

Role

Aripin

$29000
Customer’s
Needs
Definition

Good reliability part

Telp.

QA

212

100%

Amount
Of target
(US$/Year)

$0
Before

Great Company Great People

Dept

2 - 22

After

Fun Innovation By New Culture

Problem Situation
Management policy said we have reduce field claim.
Data field claim show No Power is cause biggest
than other.

No power is TV Set have no supply voltage. It show by not light of Indicator power and no display on tube.
No power can caused by parts, like power cord, SMPS Transformer, IC, FBT, Transistor and others.
SMPS Transformer have function as divider power voltage to several sub circuit or part like FBT, IC, and
others.
If SMPS not function, it can’t supply voltage to parts or sub circuit .

Defect Analysis by No Power

Data field claim 2003

No Power

5 WORST SYMPTOM 29%
IN 2003

No Sound

Knowledge
Picture NG
CPT NG

10%

15%
Sound
SoundNo
NG

15%

A/V NG

No Power
42%

8.4%
Screen
16%

8%

27%

IC

5.2%

Crystal

3.4%

FBT

2.0%

Others

840 PPM

100%

13%

Diode

Resistor

70.2%

6170VMCBO1L

15%

Transistor

12%

Target

29%

SMPS

Fuse

14%

Defect Analysis by Model

26.3%

6170VMCBO1K

0PPM
Others

2.0%
3.6%

3.5%
Before

3 - 22

After

Theme Selection
Big Y

Little y1

Little y2

Improve
Improve
TV
TV Quality
Quality
Attribute
Attribute
30%
30% SCR
SCR
Reducing
Reducing

Knowledge

Part

TYPE

PIC

Statu
s

Crack, Ball
Opt. Skill, Ability
Wire Connection
Handling
Handling
Opt. Skill, Ability
Wrong Connection
Handling

1P1P

LQC

Finished

1P1P

LQC

On Progress

1P1P
1P1P

IQC
On Progress
Production On Progress

IIC
1P1P

Production Finished
Production On Progress

IIC

IQC/Prod

Finished

AGC Circuit

1P1P

IQC

On Progress

Opt. Skill, Ability

1P1P

Production On Progress

Antenna
How to Use

No connection

1P1P

OQC

On progress

How to Use

OQC

Product Spec
Electricity

How to used
Plug Connection

1P1P
1P1P

On Progress
On Progress

Environment

Environment

IC
Transistor
CPT

Chassis

Soldering

Process

Little y3

Adjustment
Speaker Connection
PWB Crack
Loosen Part
Bad Wiring Connection
PWB Short
Front Cabinet Dented
AGC Circuit
Bad System Connector

FBT
Tuner
Speaker
SMPS Trans

Reliability
Screen Focus, DYC,PRT
ABL Short
AGC, No Ruster
No Sound Problem
No Power

Fuse
Capacitor

Un Connect
Reliability
Reliability

Resistor

4 - 22

1P1P

OQC
OQC

1P1P
1P1P
1P1P

OQC
IQC
IQC

On Progress
On Progress
Finished

1P1P

IQC

Finished

IIC
1P1P

IQC
IQC

Finished
On Progress

1P1P
IIC

IQC
IQC

Finished
Finished

1P1P

IQC

On Progress

On Progress

Process Mapping
SAMWHA
Process

LGEIN IQC

LGEIN LQC

LGEIN OQC

•R&D

• Appearance

• Patrol

• ELT

• Purchase

• Dimension

• Performance

• LTLT

•IQC

• Structure

Final
Inspect

• Reliability
Product

•Production

• Performance

• LQC

• Assembling

• OQC

• Reliable Part

High defect
SMPS no power
At LGEIN production
Find Out what is
the Main Cause of
Problem
Selecting Best
Condition
for reliable Part

• OS&D

5 - 22

What is SMPS Transformer ?

9

7

SMPS
8
10

5

1

Output of SMPS (Voltage)
• Pin 9 - 8 : 24 ~ 25
• Pin 10 - 11 : 10 ~ 12
• Pin 14 - 13 : 110 ~ 115
• Pin 13 - 12 : 13 ~ 14
• Pin 1 – 2 : 7 ~ 10

14
13

SMPS Component :
• Core
• Bobbin
• Wire
• Insulation
• Tape
• Shield

12

2

Potential Short
Area

Short can be happen if one of wire on potential area
get direct connect.

6 - 22

Clarification Of Y
Kind of defect by SMPS
MARKING NG
APPEARANCE

Master
Signal

RUST
ISOLATION NG

SMPS
DEFECT

DIMENSION

Test
Signal

Test Signal

P/P OVER

NG SIGNAL DISPLAY
WIRE SHORT
PERFORMANCE

Master
Signal=

OK SIGNAL DISPLAY

No
Power

BROKEN

SMPS Transformer Characteristic
Display of Surge Comparison Tester ( ST-215 ) will
show two signal both master and part test.
If form signal different that’s part NG
and if same is OK.

TURNS

Demonstration of Data gathering
Display

Indicator

TV Set Power OK

TV Set No Power

7 - 22

Measurement (Gage R&R)
Instrument : Surge Comparison Tester ( ST-215 )

Agreement Analysis

21
22
23
24
25

Attribute
Pass
Pass
Pass
Fail

Fail
Pass
Pass
Fail
Fail

Try 1
Pass
Pass
Pass
Fail

Try 2
Pass
Pass
Pass
Fail

Fail
Pass
Pass
Fail
Fail

Within Appraisers

Fail
Pass
Pass
Fail
Fail

OPR 2
Try 1
Pass
Pass
Pass
Fail

Try 2
Pass
Pass
Pass
Fail

Fail
Pass
Pass
Fail
Fail

Fail
Pass
Pass
Fail
Fail

Try 1
Pass
Pass
Pass
Fail

Fail
Pass
Pass
Fail
Fail

Try 2
Pass
Pass
Pass
Fail

Within Appraisers

Fail
Pass
Pass
Fail
Fail

96

96

:

# Matched

Percent

25

25

100.00

(86.09, 100.00)

OPR-2

25

25

100.00

(86.09, 100.00)

OPR-3

25

25

100.00

(86.09, 100.00)

88

88

OPR-1

25

25

100.00

(86.09, 100.00)

OPR-2

25

25

100.00

(86.09, 100.00)

OPR-3

25

25

100.00

(86.09, 100.00)

Assessment Agreement
# Matched

Percent

25

100.00

OPR-3

Measurement System are
Acceptable

All appraisers' assessments agree with each other.

25

OPR-2
Appraiser

95 % CI

Between Appraisers :

# Inspected

OPR-1

 Instrument is valid and reliable

Appraiser's assessment across trials agrees with the known standard

Percent

OPR-3

 Operator is good in inspection

Each Appraiser vs Standard :

# Matched

86
OPR-2
Appraiser

Assessment Agreement : 100 %

95 % CI

# Inspected

92
90

OPR-1

# Inspected

94

90

86

OPR-1

Appraiser

98

92

95.0%  C I
P ercent

100

98

94

JA NUA RY 15. 2004
A RIPIN
SHORT TEST

Appraiser vs Standard
95.0%  C I
P ercent

100

Appraiser agrees with himself across trial
Appraiser

Date of study : 
Reported by :
Name of product:
Misc:

Assessment Agreement

OPR 3

Percent

Sample
1
2
3
4

OPR 1

Percent

Known Population

95 % CI
(86.09, 100.00)

8 - 22

Capability Process (Current)
Based on data October ~ December ‘03, defect SMPS 124 ea from 147595 ea.
• Defect Op. = 1 ( Wire Short )
• DPU

= 124 : 147595

• DPO

= 124 : ( 147595 x 1 ) = 8.4E-04

0.04

• DPMO = DPO x 1.000.000 = 840 PPM

From Z Normal table, we get :

DPO

= 8.4E-04

Z Value

= 3.1 + 0.04
= 3.14

Our Condition now :
Sigma level
= 3.14+ 1.5
= 4.64 
3.1

9 - 22

8.45E-04

Estimate Cause Analysis
By Analyzing all potential problem that could have significant effect for SMPS Performance,
thru Brainstorming and screening, we succeed to collect some root cause for the problem

Potential Factor

Dielectric Breakdown

wire
Material

Short
(No Power)

Hole

Core
Varnish

Press

Turn wiring

Space

Home pin

Finishing

Method

10 - 22

Selecting Vital Factor
X1 = Dielectric Breakdown
Hypothesis Testing
Ho : P P 2 ( There are no different Proportion Between D/B below 2250 V and D/B above 2250 V)
Ho : P≠P ( There are different Proportion Between D/B below 2250 V and above 2250 V)

D/B : 2250 V
Sample

2250

Decision

Sample

2250

Decision

1

2235

FAIL

14

1850

FAIL

2

1990

FAIL

15

2230

FAIL

Sample

X

N

Sample p

3

2010

FAIL

16

1990

FAIL

4

2240

FAIL

17

2010

FAIL

1

0

25

0.000000

5

2125

FAIL

18

1890

FAIL

2

19

25

0.760000

6

2150

FAIL

19

2110

FAIL

Difference = p (1) - p (2)

7

2250

FAIL

20

2225

FAIL

8

1990

FAIL

21

2000

FAIL

Estimate for difference:

FAIL

22

2010

FAIL

95% CI for difference:

23

1995

FAIL

9

2000

10

2230

11

2000
FAIL
Sample
2250
1985
FAIL
1960 1
FAIL5430
2
4795

12
13

D/B : 2250
V
FAIL

24
Decision
25
PASS

1890
FAIL
Sample
2250
1885
FAIL
14
4300

Test and CI for Two Proportions

-0.76
(-0.927414, -0.592586)

Test for difference = 0 (vs not = 0):

Z = -8.90

Decision
PASS

PASS

15

5800

PASS

3

4500

FAIL

16

7800

FAIL

4

5350

PASS

17

7900

PASS

5

6000

FAIL

18

4600

FAIL

6

5100

PASS

19

5800

PASS

7

5700

PASS

20

5500

PASS

8

5700

PASS

21

7500

PASS

9

4500

PASS

22

7900

PASS

10

5000

PASS

23

5600

PASS

11

4100

PASS

24

4500

FAIL

12

4250

FAIL

25

6500

PASS

13

5600

PASS

P-Value = 0.000

- Since P-Value < 0.05, Reject Ho
- It’s mean that Dielectric breakdown
has an effect or SMPS reliability.
Dielectric Breakdown is Vital Few factor

11 - 22

Selecting Vital Factor
X2 = Pin Hole Wire
Hypothesis Testing
Ho : P P 2 ( There are no different Proportion Between Pin Hole Below 5 and Above 5)
Ho : P≠P ( There are different Proportion Between Pin Hole Below 5 and Above 5)

P/H : 5
Test and CI for Two Proportions

Sample

5

Decision

Sample

5

Decision

1

0

PASS

14

1

PASS

Sample

2

0

PASS

15

1

PASS

3

1

PASS

16

0

PASS

4

0

PASS

17

0

PASS

5

0

PASS

18

0

PASS

Difference = p (1) - p (2)

6

2

PASS

19

1

PASS

Estimate for difference:

7

1

1

PASS

8

3

9

P/H : 205
PASS

1

PASS
Sample
PASS 1

21
5
22
7

1
Decision
0
PASS

10

1

PASS 2

11

0

PASS 3

23
7
24
8

12

3

PASS 4

13

0

PASS 5
6

X

N

Sample p

1

25

25

1.000000

2

14

25

0.560000

95% CI for difference:

PASS
Sample
PASS14

5

Decision

6

PASS

0
PASS
0
FAIL

PASS15
PASS16

7

PASS

7

FAIL

25
7

1
FAIL

PASS17

6

PASS

6

PASS

18

7

PASS

6

FAIL

19

8

FAIL

7

8

PASS

20

8

FAIL

8

8

FAIL

21

6

PASS

9

6

PASS

22

7

PASS

10

9

PASS

23

9

FAIL

11

10

FAIL

24

7

PASS

12

11

FAIL

25

8

PASS

13

10

FAIL

0.44
(0.245420, 0.634580)

Test for difference = 0 (vs not = 0):
Z = 3.76

P-Value = 0.000

- Since P-Value < 0.05, Reject Ho
- It’s mean that Pin Hole Wire
has an effect or SMPS reliability.
Pin Hole Wire is Vital Few factor

12 - 22

Selecting Vital Factor
X3 = Pressure
Hypothesis Testing
H0 : P 1 = P 2 (There are no different Proportion Between varnish and no varnish )
H1 : P 1 ≠ P 2 (There are different Proportion Between varnish and no varnish )
Test and CI for Two Proportions
Sample

Varnish

X

N

Sample p

1

22

25

0.880000

2

21

25

0.840000

Sam ple

Varnish

Decision

Sam ple

Varnish

Decision

1

V

PASS

14

V

PASS

2

V

PASS

15

V

PASS

3

V

PASS

16

V

PASS

Difference = p (1) - p (2)

4

V

PASS

17

V

PASS

5

V

PASS

18

V

PASS

Estimate for difference:

6

V

FAIL

19

V

PASS

7

V

V

PASS

8

V

V
Non Vr.22 DecisionV
NV 23
PASS V

PASS
Sample
FAIL

9

V

10

V

11

V

12

V

13

V

PASS
20
No
Varnish
PASS
21
Sample
PASS
1
PASS

95% CI for difference:

(-0.152036, 0.232036)

Test for difference = 0 (vs not = 0):
Non Vr.

Decision

14
PASS
15
PASS

NV

PASS

NV

PASS

2
FAIL
3
PASS

NV 24
NV 25

PASS V

PASS

NV

PASS

16
PASS
17

NV

4
PASS
5

NV

PASS

NV

PASS

18

NV

PASS

6

NV

PASS

19

NV

PASS

7

NV

PASS

20

NV

FAIL

8

NV

PASS

21

NV

PASS

FAIL V

0.04

9

NV

PASS

22

NV

PASS

10

NV

FAIL

23

NV

FAIL

11

NV

PASS

24

NV

PASS

12

NV

PASS

25

NV

PASS

13

NV

PASS

Z = 0.41

P-Value = 0.683

P Value > 0.05, Receive H0,
Varnish

13 - 22

not significant as Vital Factor.

Selecting Vital Factor
X4 = Space
Hypothesis Testing
H0 : P 1 = P 2 (There are no different Proportion Between space and no space )
H1 : P 1 ≠ P 2 (There are different Proportion Between space and no space )
No Space

Test and CI for Two Proportions
X

N

Sample p

1

22

25

0.880000

PASS

2

24

25

0.960000

S

PASS

Difference = p (1) - p (2)

S

PASS

S

PASS

Sample

Space

Decision

Sample

Space

Decision

1

S

PASS

14

S

PASS

2

S

PASS

15

S

PASS

3

S

PASS

16

S

4

S

PASS

17

5

S

PASS

18

6

S

PASS

7

S

8

S

PASS
Sample
PASS 1

9

S

10

S

11

S

12

S

13

S

Space19
No20Space
21NS
22NS
23NS
24NS
25NS

S
Decision
S
PASS
S
PASS
S
PASS
S
PASS
S
PASS

6

NS

7
8

PASSSample
PASS 14

No Space

Decision

NS

PASS

NS

PASS

NS

PASS

PASS 17
PASS 18

NS

PASS

NS

PASS

PASS

19

NS

PASS

NS

PASS

20

NS

PASS

NS

PASS

21

NS

PASS

9

NS

PASS

22

NS

PASS

10

NS

PASS

23

NS

PASS

11

NS

PASS

24

NS

PASS

12

NS

PASS

25

NS

PASS

13

NS

PASS

PASS 4
PASS 5

No Space

Estimate for difference:

PASS 15
PASS 16

PASS 2
PASS 3

PASS

Sample

14 - 22

95% CI for difference:

Space

- 0.08
(-0.152036, 0.232036)

Test for difference = 0 (vs not = 0):
Z = 0.41

P-Value = 0.297

P Value > 0.05, Receive H0,

Space not significant as Vital Factor.

Selecting Vital Factor
X5 = Finishing Method (Winding Method)
Hypothesis Testing
Ho : P P 2 ( There are no different Proportion Between Parallel and Cross)
Ho : P≠P ( There are different Proportion Between Parallel and Cross)

Analyzing Finishing Method by Comparing whether Parallel winding and Cross Winding contribute a significant
effect for wire short.
Parallel

Test and CI for Two Proportions

Sample

Cross

Decision

Sample

Cross

Decision

1

C

PASS

14

C

PASS

Sample

2

C

PASS

15

C

FAIL

3

C

FAIL

16

C

FAIL

4

C

PASS

17

C

PASS

5

C

PASS

18

C

PASS

6

C

19

C

FAIL

7

C

FAIL
Cross

8

C

9

C

10

C

11

C

12
13

FAIL

PASS Parallel
Sam ple

C

PASS

C

SampleFAIL Parallel

22PASS
23PASS

C
C

P

C

3 PASS
4 FAIL

P

24PASS
25PASS

C

5 PASS

P

6

P

7
8

Decision

14 FAIL
15 FAIL

P

PASS

P

PASS

16 PASS
17 PASS

P

PASS

P

PASS

PASS

18

P

PASS

FAIL

19

P

PASS

P

PASS

20

P

PASS

P

PASS

21

P

PASS

9

P

PASS

22

P

PASS

10

P

PASS

23

P

PASS

11

P

PASS

24

P

PASS

12

P

PASS

25

P

PASS

13

P

PASS

P

C
C

Sample p

1

14

25

0.560000

2

24

25

0.960000

Cross

Estimate for difference:

20
P

N

Parallel

Difference = p (1) - p (2)

21
Decision

1 PASS
2 FAIL

X

95% CI for difference:
0.190807)

-0.4
(-0.609193, -

Test for difference =0(vs not = 0):
Z =-3.75

P-Value = 0.000

- Since P-Value < 0.05, Reject Ho
- It’s mean that Finishing Method
has an effect or SMPS reliability.
Finishing Method is Vital Few factor
15 - 22

Selecting Vital Factor
Result hypothesis test, get vital few factor

Dielectric
Breakdown

Vital View

Hole

Vital View

wire
Material

Short
(No Power)

Core
Varnish

Press

Turn wiring

Space

Home pin

Cross

Method

Need improved
Not need improved

16 - 22

Vital View

Selecting Optimum Condition
In this Step, we try to find the optimum combination of all factors which have a
significant effect on SMPS Transformer Reliability.
• Dielectric Breakdown ( ↑ 2250 V and ↓ 2250V)

Cube Plot (data means) for PASS ( % )

• Wire Hole (↑ 5 and ↓5 / on 5 Meters)

0

• Method Finishing Home Pin ( Cross and parallel)

50

0

40

H +

Opt.
Condition

DB
DB DB DB +
DB +
DB +
DB +
DB DB -

PH
PH +
PH +
PH +
PH +
PH PH PH PH -

MTD
P
C
P
C
C
P
P
C

Pass %
0
0
50
40
70
100
0
0

H

0

100
P

0

MTD

70

H -

C
DB -

DB

DB +

Optimum condition :
 Dielectric Breakdown : ↑ 2250 V
 Pin Hole : ↓ 5 hole / 5 m
 Method Finishing : Parallel

17 - 22

Improvement
Improvement on Finishing Method at Samwha Process
Before :

Change Wire Winding

Finishing wire of pin 12 is upper ( entered to home pin )
After :





4 M Change Report

8

Cross Method

Finishing wire of pin 10 is upper ( entered to home pin )
Finishing wire of pin 12 is lower then pin 10.

Parallel Method

18 - 22

Improvement
Improvement on Inspection Checking Method
IQC Samwha
Before
Before

After
After

Supplier

Supplier

Delivery

Delivery

Incoming

Sampling

Inspection
(Conventional)

OK
Storage Area

Incoming

IQC
• Appearance
• Size

NG

Sampling

Content
Content
IQC

• Appearance
Size
• 5 m each roll wire
( test hole & D
Breakdown )

• Involving new item inspection as an
new inspection check point :
• Check hole wire by Phenolphthalein

Inspection
( > 5 hole/5m, NG )
• Dielectric Breakdown test
Check hole
&D
Breakdown

NG

OK
Storage Area
19 - 22

Maintain CTQ
To keep good condition, we do :
 Dielectric breakdown and hole wire control at wire vendor ( monthly )
 Wire hole test before to production
 Item CTQ Control
 QC and Process Irregular Audit at Supplier
 Total Inspection Part (100% Short Test)

CTQ
D/B

Inductance

Resistance

Control

Control

Control

I Chart of D/B

I Chart of INDUCTANCE
UCL=549.08

4

155.0

_
X=529.2

530

520

Individual Value

_
X=5.316
5

UCL=158.14

157.5

540

6

Individual Value

Individual Value

I Chart of RESISTANCE

550

UCL=7.135

7

_
X=152.4

152.5

150.0

147.5
LCL=3.497

LCL=146.66
510

3
2

4

6

8

10
12
Observation

14

16

18

20

LCL=509.32
2

4

6

8

10
12
Observation

14

20 - 22

16

18

20

145.0
2

4

6

8

10
12
Observation

14

16

18

20

Saving Cost
Part production on March ~ May ‘04 are 245915 EA, SMPS Transformer defect: 1 ea
DPU

= 1 : 184259 = 5.427E-6



DPMO = 4 PPM , Sigma Level = 5.97

Condition part Oct ~ Dec’03

Condition part Mar ~ May ’04

No Power by SMPS
(70.2%
)

840 PPM

4.64 

4.64

Improv
e

423

BOIL

(26.3%
)

56

BOIK

OTHE
RS

Part No

Defect PPM

Defect PPM

1128

No Power by SMPS

4 PPM

5.97 
5.97
BOIL

Part No

SAVING COST

HISTORY :
October ~ December ‘03 :
 Analysis cost : $ 1000

$29000

 Hold, rework & sorter cost : $ 1950
 Material cost : $ 4000
 Loss time prod : $ 300

$0
Before

BOIK

After

Total : $ 7250 ( 3 Month )

SAVING COST : $ 29,000 / YEAR
21 - 22

OTHER
S

THANK YOU

22 - 22