Jpm Tqm Course Mat-5 T-3 Imba 2013

7/30/2019 Jpm Tqm Course Mat-5 T-3 Imba 2013

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RUN Test_

A RUN is a consecutive increase or decrease in the observations plotted on the

X -Control Chart. It indicates statisticallynon-randombehaviourof the process.

Long Run-UP or Long Run-DOWN (ie increasing ordecreasing

cont inuous lyformo re than5 consecutive observations) shows that theprocess is OUT OF CONTROL, though the observations are within the

control limits.

Long Run-up (8)

CL

UCL

LCL

_

X

N (Sample No.)

_

X

N (Sample No.)

LCL

CL

UCL

Long Run-down (7)

Runs Test is to test whether

the process is out of control, using the

data from control chart. Here, the first case is out of control due to Long

Run-up and the second case is out of control due to Long Run-down.

# When the process is out o f con tro l , ident i fy the Assign able Causes behindit and immediately remove (rect i fy) them.


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Chart SENSITIVITY Test_

It is the tool applied on X-Chart to judge how far the process is under Control.

_Rule 1: Even if only one Point (X)

falls beyon d Zone-3, the process isSaid to be out of control.

Rule 2 : Iftwo consecutive Points

fall in same side Zone-3, the process issaid to be out of control.

Rule 3 : Iffour consecutive Pointsfall in same side Zone-2, the process issaid to be out of control.

Rule 4 : Ifeight consecutive pointsfall in same

side Zone-1,The process is said to be out of control.

# When the process is out of control , ident i fy the Assig nable Causes

behind it and immediately remove (rect i fy) them.

_

X

N

( 3 Zone)

( 2) Zone

( 1) Zone

(+ 1) Zone 8

(+ 2) Zone

4

(+ 3) Zone 2

=X

UNTL=X +

3 1

= LNTLX - 3


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RUN - SUM Test _It is the tool applied on X-Chart to judge how far the process is under Control.

Steps1) Divide the Control Chart into 6zones, each with width of one ieStandard Deviation of population,in both Upper & Lower direction of the

_Centre Line of the X-chart

_

X

N

( 3 Zone)

( 2) Zone

( 1) Zone

(+ 1) Zone +1

(+ 2) Zone

+2

(+ 3) Zone +3

=X

UNTL

=X + 3

= LNTLX - 3

_

2) Successive Observations (X) are

plotted.

3) Assign each Observation a SCOREaccording to the ZONE in which it falls.

On Centre Line score= 0

In +1 Zone score= + 1

In - 1 Zone score= 1

In +2 Zone score= + 2

In - 3 Zone score= 3

4) Sum-up the individual scores (ki)

earned by all the observations. K = ki

5) If the Sum-up Score, I K I > 4, the Process is not under Control.# When the proc ess is not und er contro l , ident i fy the Assignable Causes

behind it and immediately remove (rect i fy) them.


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SPECIFIED TOLERANCE : (example : 250 +3 4 )

Process Control Limits :

USL = Upper specified Limit (here 253) LSL = Lower specified Limit (here 246)

(USL LSL) = SPECIFIED SPREADor simply SPREAD (S) (here 253 246 = 7)CENTER LINE of Specified Spread = (USL + LSL) / 2 (here 249.5)

NATURAL TOLERANCE: =Process Central Line = = X

Upper Natural Tolerance Limit = UNTL = + 3

Lower Natural Tolerance Limit = LNTL = 3

NATURAL /PROCESS SPREADor simply BAND (B) = (UNTL LNTL) = 6

PROCESS CAPABILITY RATIO (PCR) = Specified Spread.Process Spread

PCR = Cp =Specified spread = (USL LSL) = Spread (S) = S

Process Spread (UNTL LNTL) Band (B) 6


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Quality-Control based on SPREAD and Process BAND1) If 6 < SThe processcan be easily be

kept under control by adjusting=

the position of X very near to Centre

Line (SC) of the

Spread.Position2 is shifted and keptat position 1.

2) If 6 = SStrict Centering care to be taken

=to keep the Control (Centre) Line Xperfectly at the Centre Line of Spread.

It is also better to relax the SpecifiedTolerance to make S> 6 . Position2is shifted and kept at position 1.

3) If6 > S

Even with Extreme care & control andPerfect Centering, the process is not underControl. The Process is said to beOut of Control. Quality non-conformance will be there and there by Quality RejectionWill take place.Care to be taken : (1) Fundamental change in process,(2) Revision of Specified Tolerance, (3) Perfect Centering,

(4) 100% inspection

21

2

SC

USL

LSL

6

6S

USL

LSL

SC S2

1

6 2

6

1

1

UNTL

LNTL

UNTL

UNTL

LNTL

(

SC)S/2

S/2


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PROCESS CAPABILITY

PROCESS CAPABILITY RATIO = Specified Spread Process Spread PCR or Cp = (USL LSL) = S .

(UNTL LNTL) 6 (i) PCR should be considerably greater than 1, ie

Specified Spread (USL LSL) >> Natural Spread 6 . and

(ii) USL should be greater than UNTL ie ( + 3 ), and

LSL should be lesser than LNTL ie ( 3 ).

Even though, (USL LSL) >> 6 , if upper Natural Tolerance Limit(UNTL), UNTL > USL or LNTL < LSL, there will be quality rejections.

Thus, PCR ie Cp can not truly indicate the process capability, iewhether there will be no

quality-rejections (defects).

Cp is corrected to Cpk which truly indicates the Process Capability.

Cplower

= LSL = LSL , Cpupper

= USL = USL LNTL 3 UNTL 3

Cpk = Min [Cplower , Cpupper]

Cpk = (1 K) Cp

Where K is

Correction

Factor

K = I (SC) IS / 2

= Process CL.

SC = Specified CL.When sc= Cpk = Cp.

Cpk can never be > Cp).


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PROCESS CAPABILITY (Cp & Cpk) is less due to two reasons,i) If Process Capability Ratio (PCR orCp) is low, then Process

Improvement is needed)

ii) IfCpk is low (then Process Improvement as well as setting ofProcessCenter lineclose to Specified Center line are needed), [because, Process

Center line ie Design Center line ( ) has moved away from the specifiedCenter line (SC), increasing the value of Correction Factor K].

Recommendations on Cpk

Minimum Recommended Cpk : 2

Ideal Cpk : 5

Cases with different Cpk & Cp Case 1 When Cpk = Cp < 1 : Process mean is centrally located, K

=0,

but the Process is not CAPABLE,

Quality rejection is inevitable. Case 2 When Cpk = Cp = 1 : Process mean is centrally located, K

= 0,

and Specified spread exactlymatches with Process Spread, so theProcess is marginally capable,

any slight variation will cause rejection.


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Cases with different Cpk & Cp

Case 3 When Cpk = Cp >> 1 : Process mean is centrally located, K= 0,

Process is capableand safe,

less chance of rejections/defectives.

Case 4 When Cpk < 1,but Cp > 1 : Process mean is not centrally located, though Process is

Capable,

the process is not under controlto

prevent quality rejection.

Case 5 When Cpk = 1, but Cp > 1 : Process mean is not centrally located, though Process is

Capable,

Process is marginally undercontrol& any slight variation will cause

quality rejection.

Case 6 When Cpk = 0 : Process Mean (Centerline) comes on USL or LSL

Case 7 When Cpk < 0 : The Process Mean ie Process CL

() comes either above USL or belowLSL.

C ith diff t C k & C


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Cases with different Cpk & Cp

= Explanation of Case-6 and Case-7

Case 6 When Cpk = 0 : The Process Mean ie Process CL(X)

comes either on USL or on LSL.

[ Cpk = Min [Cplower, Cpupper], so either LSL = 0 or USL = 0

3 3

ie =USL or LSL ]

[then ( sc) = S/2 , K = I ( sc) I =1 then Cpk= (1 K) Cp =0 ]

S /2

= Case 7 When Cpk < 0 : The Process Mean ie Process CL (X)

comes either above USL or below LSL.

[ Cpk = Min [Cplower, Cpupper], so either ( LSL) < 0 or (USL) < 0 ie Process Center-line is beyond USL or LSL ]

[when ( sc) > S/2 , K = I ( sc) I >1 then Cpk= (1 K) Cp 0,


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Control Chart

Problem-1 Page - 1

For the quality measurement under Inspection process, if the sub-group Sample size is

14, no. of sub-group is 20, sum of measurements of the sample is 357.50 and Sum of

the sample-ranges is 9.90, (i) Indicate Center Line, UCL & LCL for X-chart & R- Chart,

(ii) Comment on the Process Capability and Whether the process is under control, if

specified dimension is 17.5 0.5. (iii) Suggest possible ways for improving theSituation, (iv) Test the Process Capability with Cpk .

[Given A2 = 0.18, d2 = 3.735, D3 = 0.41, D4 = 1.59]

Solution :Process Mean of measurements, = N = 357.50 / 20 = 17.875

_Mean Range, R = R N = 9.90 / 20 = 0.495

(i) For X-chart,Central Line = =17.875 , UCL = + A2R= 17.875 + 0.18 x0.495 = 17.9641

LCL=17.875 0.18 x0.495 = 17.7859

For R-chart _ _Central Line = R =0.495 , UCL = D4 x R = 1.59 x 0.495 = 0.78705

_LCL = D3 x R = 0.41 x 0.495 = 0.20295

_(ii) Standard Deviation of Universe, =R / d2 = 0.495 / 3.735 = 0.13253

Natural Spread (Band) = 6 = 6 x 0.13253 = 0.79518

and 3 = 0.39759

X X

X X

S l ti 1( td )


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Solution -1(contd.) (Control Chart) Page-2Upper Natural Tolerance Limit, UNTL = + 3 = 17.875 + 0.39759 = 18.2726

Lower Natural Tolerance Limit, LNTL = - 3 = 17.875 - 0.39759 = 17.4774

PROCESS CENTER LINE (Center Line of natural Spread) , = = 17.875,Specified dimension =17.5 0.5Upper Specified Limit, USL = 17.5 + 0.5 = 18.00Lower Specified Limit, LSL = 17.5 - 0.5 = 17.00

Specified Spread , S= USL LSL = 18 17 =1.0

Center line of Specified Spread , SC = (USL + LSL )/2 = (18 + 17)/2 =17.5

Shifting of Process Center Line (from Specified Center Line) = SC= 17.875 17.5 = 0.375

Process Capability Ratio, Cp = S = 1.0 . =1.257577 >1 but


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Solution -1(contd.) (Control Chart) Page-3Now,

UNTL > USL , UNTL is not within the specified limit,so considerable

rejections are expected which indicates that the process is not

under control.

LNTL > LSL , LNTL is within the specified limit,so no rejections are

expected which indicates that the process is under control.

(iii) Suggestion :

1) Process is marginally capable, soprocess improvement is

recommended

2) Designed Process Center-line must be set very close to

Center Line of the specified spread

3) Specified Spread is also to be relaxed to some extent, if

possible, asCp value is very low (close to 1).

S ( ) (C C )


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Solution -1(contd.) (Control Chart) Page-4

(iv) Testing of Process Capability with CpK

CpK

Testing :

CpK = (1 K) Cp = (1 K) . S

6

where K = SC = 0.375 = 0.75

S / 2 1.0 / 2

CpK = (1 0.75) x 1.257577 = 0.314394 < < 1.0 ,

so the process is not under control.

Jpm Tqm Course Mat-5 T-3 Imba 2013

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