7 quality control tool / qc

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1

By:Maruti Center for Excellence

(MACE)On 20th June’2008



2



3

ULTIMATE GOAL OF AN

ORGANIZATION

- Making Profits- Survival &- Growth



4

SURVIVAL AND GROWTH:

HOW?- Products & Services must be

preferable to Customers overCompetitors.

- Organization must continually meetNeeds & Expectations of Customersi.e. Voice of Customers also called

QUALITY



5

WHAT IS A PROBLEM?

A “problem” is the gap between the presentSituation And the ideal situation or objective

C o n t r o l

c h

a r a c t e r i s t i c

Good

Ideal situation or objective

Gap Problem (ideal situation or objective) -=(present level)

Present Level



6

PROBLEM DEFINED …

Deviation from the expectation- In maintaining the status quo.

- In improving the status quo

An opportunity for improvement



7



8

The Problem-Solving Process

Expose problem

Experience,intuition, nerve,

inspiration

Implementcountermeasures

Expose problem

Analyze causes

Implementcountermeasures

• Grasp problem• Set target• Identify gap betweenexisting situation and target

• Investigate cause

• Plan countermeasures• Implement countermeasure• Institutionalize

1. The conventional problemsolving approach

2. The QC problem-solvingapproach



9

TOOLS ARE USED TO IDENTIFY, ANALYSEAND RESOLVE PROBLEMS

TOOLS ARE SIMPLE, VERY POWERFUL

AND HELP TO IDENTIFY THE CAUSES FOR

WORK RELATED PROBLEMS AND TO FIND

SOLUTIONS FOR THE SAME IN A

SYSTEMATIC MANNER.

WHAT ARE SEVEN QC TOOLS :



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1 Cause and Effect Diagram 2 Graphs / Flowcharts

3 Pareto analysis 4 Checksheets

5 Control charts

6 Scatter diagram

7 Histogram



11

PARETO DIAGRAM

• Prioritisation Tool – It tells where to concentrate first.

• Vilfredo Pareto (1848-1923) Italianeconomist – 20% of the population has 80% of the

wealth

• Juran used the term “vital few, trivialmany”. He noted that 20% of the quality

problems caused 80% of the dollar loss.



12

PRINCIPLE OF PARETO

Isolate vital few from trivial many.

80/20 principle. 80% improvement can be achieved by

working on 20% of the causes.



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WHAT IS PARETO DIAGRAM :

It is a column graph. Differentiating major

factors contributing to problem from other

factors which have less contribution. Thus it

helps fixing priority to take first. OR

A technique to segregate vital few from

trivial many



14

DATA TALLY SHEET

DEFECT TALLY Total

A //// //// //// ....... //// 150

B //// //// //// …… //// 60

C //// //// …. //// 45D //// //// …. //// 30

E //// //// 9

F //// / 6TOTAL No.of defects

300

Total No. of defectives : //// //// //// ……. //// 150



15

What is a defective : A unit that contains at least one defect.

What is a defect :

An output of a process that does not meeta defined specification.



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DATA SHEET FOR PARETO DIAGRAM

S.No.

Defect No. ofdefects

Cumulativetotal

Percentageof overall

total

Cumulativepercentage

1) A 150 150 50 50

2) B 60 210 20 70

3) C 45 255 15 85

4) D 30 285 10 95

5) E 9 294 3 98

6) F 6 300 2 100

Total 300 - 100 -



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Pareto diagram

6930

4560

150

100989585

70

50

0

50100

150

200

A B C D E F

Type of defect

N o .

o f d e f e c t s

0

20

40

60

80

100

C u m m u

l a t i v e

%



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DATA TALLY SHEET

DEFECT TALLY TotalFlow mark //// //// //// ....... //// 200Sink Mark //// //// //// …… //// 100Warpage //// //// …. //// 50

Silver streak //// //// …. //// 25Crack //// //// //// 15Flash //// //// 10

TOTAL 400

Total No. of defectives : //// //// //// ……. //// 250



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DATA SHEET FOR PARETO DIAGRAM

S.No.

Defect No. ofdefects

Cumulativetotal

Percentageof overall

total

Cumulative

percentage

1) Flow

mark200 200 50 50

2) Sinkmark

100 300 25 75

3) Warpage

50 350 12.5 87.5

4) Silverstreak 25 375 6.25 93.755) Crack 15 390 3.75 97.5

6) Flash 10 400 2.5 100

Total 400 - 100 -



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Pareto diagram

200

100

5025 15 10

10097.593.75

87.575

50

050

100

150200250

F l o w m a r k

S i n k m

a r k

W a r p a

g e

S i l v e r s t r e a

k C r a

c k F l a s h

Type of defect

02040

6080100

C u m m u

l a t i v e %



21

• Shows the relationship between a problemand its possible causes.

• Developed by Kaoru Ishikawa (1953)

• Also known as … – Fishbone diagram – Ishikawa diagram

CAUSE AND EFFECTDIAGRAM



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WHAT IS CAUSE AND EFFECT DIAGRAM :

A systematic arrangement of all possible causes

which give rise to the effect are made. The

causes are first divided into major sources (4Ms)

i.e., MAN, MACHINE, METHOD & MATERIAL.

Then each source is divided into sub-sourcesand

so on. It helps to find out the root cause of the

roblem.



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METHODOLOGY : How to reach to rootcause of the problem

Step 1 : Make fish bone diagram and write all possiblecauses in 4 M’s after brainstorming session.

Step 2 : Find out suspect or potential causes from thepossible causes. Highlight it.

Step 3 : Validate all the suspect or potential causes andwrite the judgement. All the NG judgements are the

main causes of the problem.

Step 4 : Do the why why analysis of the main causes andfind out the root causes of the problem.



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METHODOLOGY : Step 1 : Possible

causes

Step 2 : Suspect orPotential causes

Step 3 : Maincauses

Step 4 : Root causes



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Cause & Effect diagram - Major and subsidiary causes

Material Methods Environment

Men MeasurementMachine

Quality

Assemblies

Components

Suppliers

Consumables

Procedures

Policies

Accounting

Noise level

Humidity

Temperature

Lighting

Training

Experience

Skill

Attitude

Variability

Tooling

Fixtures

Technology

Instruments

Gauging

Counting

Tests



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CAUSE & EFFECT DIAGRAM

POORWASHINGQUALITY

TRAINING

WORK LOAD

CAR SHAMPOORUBBING CLOTH

MATERIALMETHOD

MACHINE MAN

JOB SKILL

PUMP PRESSURE

PROCESSTIME

NOZZLE DIAEXPERIENCE

A P

DC

WATER

PROCESS SEQUENCE

POLISH QUALITYBRUSHES

VACUUM CLEANER

LAYOUT



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VALIDATION OF POTENTIAL CAUSES

S.No.

Potential Cause Validation Conclusion



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SAMPLE SIZE FOR VALIDATION

Formulae : np = 5 or n= 5 / p

Where n= number of sample sizep= no. of defects coming out of

100Example : Suppose 1% defect is there,

then sample size will ben= 5*100 / 1 = 500

5 WHY FOR ROOT CAUSE



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5 WHY FOR ROOT CAUSEASK “WHY” FIVE TIMES.

When a problem has just been discovered, think about it by asking the

question ―why?‖ five times. A haphazard idea of the cause cannot becounted on. Thinking it over repeatedly will discover the root cause.

For example , imagine that a bracket has broken from a pipe. 1. Why did the bracket broken?( It was not welded properly)

2. Why wasn’t it welded properly?( The bracket wasn’t set in the proper position when it was welded)

3. Why wasn’t the bracket placed in the proper position for welding?

( The welding jig was loose)

4. Why was the welding jig loose?( The welding jig is worn)

5. Why is the welding worn?( The material specification of the jig was not proper)



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WHAT IS A SCATTER DIAGRAM :

A graphical technique to show the dependency oftwo

variables.

It is used to study the variation of two

corresponding variables. For example, what

extent surface finish of a machined part be

varied by the change in speed of a lathe.



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SCATTER DIAGRAM - Exampleof Positive Correlation

0

5

10

15

0 5 10 15

X - Axis

Y -

A x i s



32

SCATTER DIAGRAM - Example

of Negative Correlation

0

5

10

15

0 5 10 15

X - Axis

Y -

A x i s

Scatter Diagram



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Scatter Diagram

variation of strength with varying air

pressure

1

1.2

1.4

1.6

1.8

3.5 4 4.5 5 5.5 6

Air pressure

S t r e n g

t h

02-Mar 03-Mar 04-Mar 05-Mar 06-Mar

EXAMPLE OF SCATTER DIAGRAM



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Two variables :

a) Speed of the car in kms / Hr.b) Petrol consumption in kms / Lt.

Speed ofthe car inkms / Hr.

Petrolconsumption in kms /Lt.

30 15, 15.5

35 16, 16.5

40 17, 17.5

45 18, 18.550 19, 19.5

55 20, 20.5

60 22, 22.5

65 21, 21.5

70 20, 20.4

75 19, 19.6

80 18, 18.6

85 18, 18.5

90 17, 17.4

95 16, 16.2

100 16, 16.1

CAUSE

E F F E C T

Here

Cause : Speed of the car in kms / Hr.

Effect : Petrol consumption in kms / Lt.

EXAMPLE OF SCATTER DIAGRAM :



35

16

16.1

16

16.2

17

17.418

18.5

18

18.6

19

19.6

20

20.421

21.5

22

22.5

20

20.5

19

19.5

18

18.5

17

17.5

16

16.5

15

15.5

1012

14

16

18

20

22

24

20 30 40 50 60 70 80 90 100 110

SPEED IN KMS / HR.

F

U E L A V E R A G E

I N K M S / L T

EXAMPLE OF SCATTER DIAGRAM :

Positive and negative co-relation



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OTHER EXAMPLES OF SCATTER DIAGRAM :

In machining process :1) Cause : Speed of the machine and

effect : surface finish of thecomponent.

2) Cause : Feed of the machine andeffect : surface finish of the component



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WHAT ARE CONTROL CHARTS :

Control charts serve to detectabnormal trends with the help of linegraphs.

Control charts differ from standardline graphs as they have control limitlines at the center, top and bottomlevels.





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WHAT IS X CHART

X chart shows the centering of theprocess,i.e. it shows the variation in theaverage of samples. It is the most

commonly used variable chart.



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WHAT IS R CHART

R chart shows the uniformity orconsistency of the process i.e. it

shows the variation in the range ofsamples.



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Diameter of Shaft: 23.75 + 0.1 mm

No. of samples : 6

The diameter of shafts are as given below :

1 st day 2 nd day 3 rd day 4 th day 5 th day 6 th day 7 th day 8 th day

23.77 23.80 23.77 23.79 23.75 23.78 23.76 23.76

23.80 23.78 23.78 23.76 23.78 23.76 23.78 23.79

23.78 23.76 23.77 23.79 23.78 23.73 23.75 23.77

23.73 23.70 23.77 23.74 23.77 23.76 23.76 23.72

23.76 23.81 23.80 23.82 23.76 23.74 23.81 23.78

23.75 23.77 23.74 23.76 23.79 23.78 23.80 23.78

Diameter of Shaft: 23.75 + 0.1 mm

No. of samples per day : 6

The diameter of shafts are as given below :



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Construct the X and R chart:

Average diameter for the first day

X1 = X 1+X 2+X 3+X 4+X 5+X 6

6

= 23.77+23.80+23.78+23.73+23.76+23.75

6= 23.765

Similarly, the averages for each day are calculated:

X1 X2 X3 X4 X5 X6 X7 X8

23.765 23.77 23.7716

23.7767

23.7717

23.7583

23.7767

23.7667



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Now X = X = 190.1567 = 23.7696

N 8

Ranges :

M

R1 R 2 R3 R 4 R5 R6 R7 R 8

.07 .11 .06 .08 .04 .05 .06 .07

R = R = 0.0675

N

For X chart :

UCL X =X + A 2R

=23.7696 + 0.48 x 0.0675

(A2= 0.48 for subgroup of from table )

M



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= 23.7696 + 0.0324 = 23.802LCLX =X - A2R

= 23.7696 – 0.0324 = 23.7322For R chart :

UCL R = D 4R

= 2 x 0.0675 = 0.1350LCR R = D 3R

= 0 (D 3 = 0 for subgroup of 6 or less)

Process capability : Sigma = R / d2 = 0.0675 / 2.534 = 0.0266

( for subgroup of 6, d2 = 2.534

Xmax = (USL) upper specification limit, Xmin= (LSL) lower

specification limit



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• Xmax – Xmin = 0.2 mm from data• Process capability Cp = (USL – LSL)/ 6

sigma• Cp= 0.2 / 6* 0.0266 = 0.2 / 0.15982 =

1.25

• Cpk = (USL – X ) / 3 sigma or (X – LSL) / 3 sigma• Cpk = (23.85 – 23.7696)/ 3*0.0266=1.0• Cpk = (23.7696 – 23.65)/ 3*0.0266 =

1.49• Cpk = 1.0 or 1.49 ( 1.0 is minimum )



46

X CHART

UCL —23.802

x—23.7696

LCL—23.7322



47

R chart

UCL – 0.1350 R1 R2 R3 R4 R5 R6 R7 R8

R – 0.0675

LCL – 0



48

No. I II III IV 1 1.38 1.42 1.42 1.42 1.38 1.39 1.42 1.41

3 1.41 1.41 1.39 1.384 1.36 1.45 1.41 1.395 1.36 1.42 1.46 1.376 1.38 1.45 1.39 1.407 1.40 1.40 1.41 1.398 1.40 1.41 1.38 1.399 1.40 1.40 1.40 1.39

10 1.41 1.41 1.39 1.38

11 1.41 1.40 1.39 1.4212 1.44 1.34 1.36 1.3813 1.37 1.43 1.41 1.3814 1.38 1.44 1.42 1.4015 1.39 1.46 1.40 1.4016 1.37 1.47 1.40 1.4017 1.38 1.42 1.45 1.3718 1.39 1.39 1.39 1.4519 1.38 1.44 1.46 1.3720 1.32 1.40 1.41 1.40

DIMENSION : 1.4 + / - 0.3

Draw X bar and R chart



49

No. I II III IV X 1 1.38 1.42 1.42 1.4 1.4052 1.38 1.39 1.42 1.41 1.4

3 1.41 1.41 1.39 1.38 1.39754 1.36 1.45 1.41 1.39 1.40255 1.36 1.42 1.46 1.37 1.40256 1.38 1.45 1.39 1.40 1.4057 1.40 1.40 1.41 1.39 1.48 1.40 1.41 1.38 1.39 1.3959 1.40 1.40 1.40 1.39 1.397510 1.41 1.41 1.39 1.38 1.397511 1.41 1.40 1.39 1.42 1.40512 1.44 1.34 1.36 1.38 1.3813 1.37 1.43 1.41 1.38 1.397514 1.38 1.44 1.42 1.40 1.4115 1.39 1.46 1.40 1.40 1.4125

16 1.37 1.47 1.40 1.40 1.4117 1.38 1.42 1.45 1.37 1.40518 1.39 1.39 1.39 1.45 1.40519 1.38 1.44 1.46 1.37 1.412520 1.32 1.40 1.41 1.40 1.3825

N l l t th R



50

Now we calculate the Range:-

Range:- Highest value of sample – Smallest value of sampleFor example:-

R1 1.42 – 1.38

0.04

Similarly we calculate for all the samples.



51

No. I II III IV X R 1 1.38 1.42 1.42 1.4 1.405 0.042 1.38 1.39 1.42 1.41 1.4 0.04

3 1.41 1.41 1.39 1.38 1.3975 0.034 1.36 1.45 1.41 1.39 1.4025 0.095 1.36 1.42 1.46 1.37 1.4025 0.16 1.38 1.45 1.39 1.40 1.405 0.077 1.40 1.40 1.41 1.39 1.4 0.028 1.40 1.41 1.38 1.39 1.395 0.039 1.40 1.40 1.40 1.39 1.3975 0.0110 1.41 1.41 1.39 1.38 1.3975 0.0311 1.41 1.40 1.39 1.42 1.405 0.0312 1.44 1.34 1.36 1.38 1.38 0.0113 1.37 1.43 1.41 1.38 1.3975 0.0614 1.38 1.44 1.42 1.40 1.41 0.0615 1.39 1.46 1.40 1.40 1.4125 0.07

16 1.37 1.47 1.40 1.40 1.41 0.117 1.38 1.42 1.45 1.37 1.405 0,0818 1.39 1.39 1.39 1.45 1.405 0.0619 1.38 1.44 1.46 1.37 1.4125 0.0920 1.32 1.40 1.41 1.40 1.3825 0.09

M



52

Now Calculate X XN

X1 + X2 + X3+ X4 + X5 + X6……………………..X19 +X20 20

1.405 + 1.4 + 1.3975 + 1.4025 + 1.4025 + 1.4……………………….1.4125 + 1.3825 20

1.4011

Now Calculate R RN

R1+ R2+R3+R4………….R19+R20 20

0.04 + 0.04 + 0.03 + 0.09……………………………..0.06 + 0.09 + 0.09 20

0.06

M

M



53

Calculation For Average (X-bar) Chart

Upper Control Limit(UCL) = X +A 2 R

= 1.4011+0.73*0.06= 1.4449= 1.44

Lower Control Limit (LCL) =X - A 2 R

= 1.4011 - 0.73 *0.06= 1.3573= 1.36

For future control it is advise to set theprocess at target value i.e., (USL + LSL)/2.



54

LCL = D 3 X R= 0 * 0.06= 0

UCL = D 4 X R= 2.28 * 0.06

= 0.1368= 0.14

Calculation for Range(R) Chart



55

THE X-bar & R CHARTS 1.44

1.40

1.36

0.14

0.06

0

THE X-bar CHART

THE R CHART



56

1.44

1.36X-bar CHART

R CHART

1.40

0.14

0.06

0

The Control Chart

TWO KINDS OF VARIATIONS



57

WO DS O V O S

1) Variation due to chance causes

2) Variation due to assignable causes.1) Variation due to chance causes:Variations due to chance causes are inevitable in

any process or product. They are difficult to trace anddifficult to control even under best conditions ofproduction. Since these variations may be due tosome inherent characterstic of the process ormachine which functions at random. For example, alittle play between nut and screw at random maylead to back-lash error and may cause a change indimension of a machined part.

2) Variation due to assignable causes:



58

2) Variation due to assignable causes:These variations possess greater magnitude

as compared to those due to chancecauses and can be easily traced ordetected. The variations due to assignablecauses may be because of the followingfactors:

a) Differences among machines.b) Differences among workersc) Differences among materialsd) Change in working conditions

d



59

How to read CONTROL CHARTS :

Whether a process is in the controlled state or not is

judged by the following criteria from the controlchart.

Process not in control :

a) Point out of control limit.b) Seven points on one side of the average (run)c) Seven points in a row continously increasing

and decreasing (trend).d) Points very close to control limits and close to

average.e) When the curve repeatedly shows an up and

down trend for the same interval (periodicity).



60

• A diagram that graphically depicts the

variability in a population.

WHAT IS HISTOGRAM :

WHAT IS HISTOGRAM :



61

WHAT IS HISTOGRAM :

The frequency data obtained frommeasurements display a peak around acertain value. The variation of qualitycharacterstics is called distribution. The

figure that illustrates frequency in the forma pole is referred to as a Histogram.

POPULATION AND SAMPLE



62

POPULATION AND SAMPLE

• The entire set of items is called thePopulat ion.

• The small number of items taken from thepopulation to make a judgment of thepopulation is called a Sample .

• The numbers of samples taken to make this judgment is called Sam p le si ze.

SAMPLE OF

SIZE THREEPOPULATION

Histogram steps



63

Histogram – steps

1.Obtain a set of 50 ~ 100 observations as shownbelow:SampleNumber

Results of Measurement

1-10 2.510 2.517 2.522 2.522 2.510 2.511 2.519 2.532 2.543 2.525

11-20 2.527

2.536 2.506 2.541 2.512 2.521 2.521 2.536 2.529 2.524

21-30 2.529

2.523 2.523 2.523 2.519 2.538 2.543 2.538 2.518 2.534

31-40 2.520

2.514 2.512 2.534 2.526 2.532 2.532 2.526 2.523 2.520

41-50 2.535

2.523 2.526 2.525 2.532 2.530 2.502 2.530 2.522 2.514

51-60 2.533

2.510 2.542 2.524 2.530 2.535 2.522 2.535 2.540 2.528

61-70 2.52 2.515 2.520 2.519 2.526 2.542 2.522 2.542 2.540 2.528



64

Histogram – steps 2. Obtain the maximum value and minimum value:SampleNumber

Results of Measurement Maximumvalue of theline

Minimum value of theline

1-10 2.510 2.517

2.522

2.522

2.510

2.511

2.519

2.532

2.543

2.525

2.543 2.510

11-20 2.527 2.536

2.506

2.541

2.512

2.521

2.521

2.536

2.529

2.524

2.541 2.506

21-30 2.529 2.523

2.523

2.523

2.519

2.538

2.543

2.538

2.518

2.534

2.543 2.518

31-40 2.520 2.514

2.512

2.534

2.526

2.532

2.532

2.526

2.523

2.520

2.534 2.512

41-50 2.535 2.523

2.526

2.525

2.532

2.530

2.502

2.530

2.522

2.514

2.545 2.502

51-60 2.533 2.510

2.542

2.524

2.530

2.535

2.522

2.535

2.540

2.528

2.542 2.510

61-70 2.525 2.515

2.520

2.519

2.526

2.542

2.522

2.542

2.540

2.528

2.542 2.515

71-80 2.531 2.545

2.524

2.522

2.520

2.519

2.519

2.529

2.522

2.513

2.545 2.513

81-90 2.518 2.527

2.511

2.519

2.531

2.527

2.529

2.528

2.519

2.521

2.531 2.511

The largest value2.545

The smallest value 2.502



65

Histogram- Steps3. Determine the number of classes: Let these be between 5-12,

Say it is “K”. Generally to decide the number of classes, dividethe range by 1,2 or 5 (their fractions or multiples).

In this case, the range is 0.043 and if it divided by 0.002 or 0.005or 0.010, we will get

- 0.043 / 0.002 = 21.5 ~ 22- 0.043 / 0.005 = 8.6 ~ 9- 0.043 / 0.010 = 4.3 ~ 4Therefore, number of interval of classes be taken as 9.

4. Determine class width rounded off to a convenient figure. Sothat it covers maximum and minimum value both

C = Max- Min = 0.043 / 9 ~ 0.005K



66

Histogram- Steps

5. Calculate the class boundaries so that itcovers minimum and maximum value both.In the case, let the boundary be 2.5055.

Note: First class boundary should containssmallest value and boundary value falls onhalf of the unit of measurement .

6. Calculate the mid point of first class by sumof upper and lower boundaries of first classi.e. (2.5005 + 2.5055)/2 = 2.503. Mid point ofsecond class shall be (2.5055+2.5105)/2=2.508 and so on.

Histogram – Steps



67

g p7. Make a frequency table as given below:

Class Mid-Point of class x Frequency

1 2.5005-2.5055 2.503 1

2 2.5055-2.5105 2.508 4

3 2.5105-2.5155 2.513 9

4 2.5155-2.5205 2.518 14

5 2.5205-2.5255 2.523 22

6 2.5255-2.5305 2.528 19

7 2.5305-2.5355 2.533 10

8 2.5355-2.5405 2.538 5

9 2.5405-2.5455 2.543 6

Total ……… 90



68

Histogram-Steps

8. Mark the horizontal axis with the classboundary values.

9. Mark the vertical axis with a frequencyscale.

10. Erect the rectangles over the classinterval having area proportion to thefrequencies.

11 Draw a line on the Histogram to representMean, number of data points and standarddeviation.

Types of Histogram



69

Types of Histogram

0

5

10

15

20

25

N=90Mean=2.5247

S.D=0.00906

2.51 2.52 2.53 2.54

2.5247

Histogram of shaft Dia

Shaft Dia

F r e q u e n c y

Normal Distribution



70

ử 3o99.7%

ử 2o95.4%

ử o68.3%

ử -3o ử -2o ử -o ử ử+o ử+2o ử+3o

Normal Distribution

USLLSL

Out of

Spec.

Out of

Spec.

o



71

Histogram for grade wisedistribution in a class

10

23

35

25

15

5

C- C B B+ A A +

Grade

N o .

o f s

t u d

e n

t s

WHAT IS CHECK SHEET :



72

WHAT IS CHECK SHEET :A check sheet is a paper form on which items

to bechecked have been printed so that data can

be

collected easily and concisely. Its mainpurpose is

twofold.• To make data gathering easy• To arrange data automatically so that they

can be used easily later on.

EXAMPLE 1 OF CHECK SHEET



73

Shown a check sheet used in final inspectionprocess of a certain molded plastic product. Atthe end of the day we can immediately calculatethe total number and types of defects that haveoccurred.

EXAMPLE -1 OF CHECK SHEET



74

Product: Date

Manufacturing stage: final insp. SectionType of defect: sink mark, silver Inspector’s name Streak,flow mark, crack, flash Lot no:.Total no. inspected: 1525

Remarks: all items inspected Order no:.

Type Check Subtotal

Sink markSilver streak

Flow markCrackFlash

//// //// //// // //// //// /

//// //// //// //// //// / /// ////

1711

2635

Total defects 62

Total Rejects //// //// //// //// //// //// //// //// // 42

Defective Item Check Sheet

CHECK SHEET



75

It is necessary to decide clearly how to record thedefects. When these are found in a product. We

shouldgive proper instructions to the staff regarding the

formatin which the defects are to be gathered. In this

case, 42out of 1525 components were found defective.

Howeverthe total nos of defects was 62 because two or

more

defects were found on the same piece.

EXAMPLE -2



76

Check Sheet

Product : Date :

Manufacturing stage : Final Inspection Section :

Type of defect :scratch, dents, door gap uneven Inspector's name :

Total no. inspected : 1500

Remarks : all itemsinspected

Type of defect Tally Mark Frequency

Scratch IIII IIII IIII 15

Dents IIII IIII 10 Door gap uneven IIII 4

Others III 3

Total No. Of Defects 32

Total No. Of Defectives IIII IIII IIII IIII 20

h k h



77

Check Sheet

Purpose of data recording : Variation in the Quality Characteristics of a proces

• Applicable for Histogram, Run chart / Control Charts

Difference between Groups/ Batches/ Machines• Applicable for Pareto Diagram / Bar-chart.

Relationship between two characteristics• Applicable for Scatter Diagram

Flow Chart / Graphs



78

Flow Chart / Graphs

Flow Chart :

A Tool that graphically represents the stepsof a process



79

Flow Chart

Different icons/symbols to indicate thedifferent types of actions in the process.

Start / End :Process Activities :Decision points :Movement :Storage :

Fl Ch t f li



80

Flow Chart of annealing process:

Annealing Process :

Rodreceipt

Start

Storage

Testing

Inspection

Storage

Base & BatchPreparation

Heating

EndPurging

FurnaceLoading

Parameter

setting

Soaking

Controlledcooling

Unloading

Controlledcooling



81

GRAPH : It is one of the meansof data stratification.

• Bar Graph

• Line Graph• Pie Graph• Radar Graph

Purpose of Graph: A picture is worth morethan thousand words

BAR GRAPH : A graph to compare the



82

18.8

12.8

21.6

8.6

14.8 13.69.8

0

5

10

15

20

25

R E A R B O D Y

C T R P L L R

F R O N T

P A N E L

R O O F

F R

. D O O R

R R

. D O O R

B . D

O O R

g p pdifference in numeric quantity.

% O

F D E N T S

DENT ANALYSIS

Pie Chart) :

A graph for the proportion of the different classifications



83

AREA OMNI W/B PERIOD 05/01/2001 NO.OF.VEHICLE 150

Gun TouchDents 25%

5

20% Spot Dent

4

Others 10%

2

HandlingDents 45%

9



SYSTEM AUDIT REPORT

A dit Points % Achie ed



85

Audit Points % Achieved

1. Production preparation 40%

2. Initial supply control 60%

3. Initial control- Changed parts 40%

4. Preventive measures for defects 70%

5. Education and training 80%

6. Quality audit 80%

7. Supplier control 70%

8. Control of drawings and engg. Changes 100%

9. Inspection standard & PCS 100%

10. Operation standard 100%

11. Observance of operation standard 100%

12. Role of Manager/ Supervisor 100%

13. Quality improvement of process 0%

14. Control of Manufacturing machine/ jig 50%

15. Control of Inspection equipment and Jig 40%

16. Statistical method 100%

17. Prevention of missing process/ wrong assembling 75%

18. Control of non conforming product 100%

19. Storage of product 100%

20. First-in , First-out 100%

21. History management of A parts 100%

22. Implementation of inspection 100%



Radar Graph

7. Supplier control, 70.0%

8. Control of drawings and engg. Changes,

100.0%

6. Quality audit, 80.0%

9. Inspection standard & PCS, 100.0%

3. Initial control- Changed parts, 40.0%

4. Preventiv e measures for defects, 70.0%

5. Education and training, 80.0%

2. Initial supply control, 60.0%

10. Operation standard, 100.0%

11. Observ ance of operation standard, 100.0%12. Role of Manager/ Supervisor, 100.0%

21. History m anagement of A parts, 100.0%

22. Implementation of inspection, 100.0%1. Production preparation, 40%

20. First-in , First-out, 100.0%

19. Storage of product, 100.0%

18. Control of non conforming product, 100.0%

17. Prevention of missing process/ w rong

assembling, 75.0%

14. Control of Manufacturing machine/ jig,

50.0%

13. Quality improvement of process, 0.0%

15. Control of Inspection equipment and Jig,40.0%

16. Statistical method, 100.0%

0%

10%

20%

30%

40%50%

60%

70%

80%

90%

100%

Series1

Radar Graph : To see the total sharp image as a compositegraph.

Documents

7 quality control tool / qc