SAA Report

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STATISTICALPROCESS

CONTROL

RUN TESTS

PROCESS CAPABILITY

OC CURVE

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Run Tests

An additional test for randomness

Even if all points are within the control limits the process may

not be random

Any sort of pattern in the data would suggest a non-random

process

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Nonrandom Patternsin Control Charts

Cycle

Trend

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UCL

LCL

UCL

LCL

UCL

LCL

Bias

Mean shift

Too muchdispersion

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U U D U D U D U U D

B A A B A B B B A A B

Counting Runs

Counting Above/Below Median

Counting Ups/Down

7

8SAA

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Run Tests Procedures

Ensure that process arein statistical control

based on the controlcharts

Make a run tests ofproducts and make

measurements ofdeviations from processspecifications or median

or mean

Count number ofobserved runs from thegraph, grouping similar consecutive patters as

one

Compute for z

Compute for thestandard deviation of

the run:

Ợmed= ( 1)/4

ỢU/D= (16 29)/90

Compute for theexpected run values:

E( r )med =

+ 1

E( r )u/d =−

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ILLUSTRATION:

Twenty sample means have beentaken from a process. The means are

shown in the following table. Usemedian and up/down run tests with z= 2 to determine if assignable causesof variation are present. Assume the

median is 11.

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Given:

Sample Mean

1 10.0

2 10.4

3 10.2

4 11.5

5 10.8

6 11.6

7 11.1

8 11.2

9 10.6

10 10.9

11 10.7

12 11.3

13 10.8

14 11.8

15 11.2

16 11.6

17 11.2

18 10.6

19 10.7

20 11.9

A/B

B

B

B

A

B

A

A

A

B

B

B

A

B

A

A

A

A

B

B

A

9.5

10

10.5

11

11.5

12

0 5 10 15 20 25

Mean

Mean

Median

N = 20z = 2

Median = 11

Observed Runmed = 10SAA

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Observed RunU/D = 17

9.5

10.0

10.5

11.0

11.5

12.0

0 5 10 15 20 25

M

e

a

n

Sample

Ups/Downs Chart

Mean

Sample Mean

1 10.0

2 10.4

3 10.2

4 11.5

5 10.8

6 11.6

7 11.18 11.2

9 10.6

10 10.9

11 10.7

12 11.3

13 10.8

14 11.815 11.2

16 11.6

17 11.2

18 10.6

19 10.7

20 11.9

Ups and DownsU/D

D

U

D

U

D

U D

U

D

U

D

U

D

U

D

D

U

U

D

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Computation of Expected Run (Er)

E( r )med =

+ 1 =

+ 1 = 11 10

E( r )u/d =−

2(20) 1

3 13 17

Er Or

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Chance Variability Computation

Ợmed= ( 1)/4 = (20 1)/4

ỢU/D= (16 29)/90 = (16(20) 29)/90

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Decision Point

Run Tests Computed Value ≤≥ ZDesired Findings

Zmed -0.46 < ±2 Random

ZU/D 2.22 > ±2 Not Random

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Common Terms For Variability Of Process

Tolerances or specifications

Range of acceptable values established by engineering

design or customer requirements

Process variability

Natural variability in a process

Process capability

Process variability relative to specification

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Capability Analysis

Lower

Specification

Upper

Specification

A. Process variability

matches specifications

LowerSpecification UpperSpecification

B. Process variability

well within specificationsLowerSpecification

UpperSpecification

C. Process variability

exceeds specificationsSAA

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(a) Acceptance sampling(Some bad units

accepted)

(b) Statistical processcontrol (Keep theprocess in control)

(c) Cpk >1 (Designa process thatis in control)

Lowerspecification

limit

Upperspecification

limit

Process mean, m SAA

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Actions In Case Of Tightness

redesign the process so that it can achieve thedesired output

use an alternative process that can achieve the

desired output retain the current process but attempt to

eliminate unacceptable output using 100 percentinspection; and,

examine the specifications to see whether they

are necessary or could be relaxed withoutadversely affecting customer satisfaction

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Process Capability

The natural variation of a process should be small enough to produce products that meet the standards required

A process in statistical control does not necessarily meet the design specifications

Process capability is a measure of the relationship between the naturalvariation of the process and the design specifications SAA

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Process Capability Ratio Cp

Cp =Upper Specification - Lower Specification

6s

A capable process must have a Cp of at least 1.0

Does not look at how well the process iscentered in the specification range

Often a target value of Cp = 1.33 isused to allow for off-center processes

Six Sigma quality requires a Cp = 2.0SAA

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Process Capability Ratio


6s

Insurance claims process

Process mean x = 210.0 minutes

Process standard deviation s = .516 minutesDesign specification = 210 ± 3 minutes

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6s




= = 1.938213 - 207

6(.516)

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6s




= = 1.938213 - 207

6(.516)Process iscapable

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ILLUSTRATION

Standard

Deviation

(mm)

Machine

Capability

(sd x 6)

Cp

(c/U-L)

0.13 0.78 1.03

0.08 0.48 1.67

0.16 0.96 0.83

0.78

0.48

0.96

1.03

1.67

0.83

B’s Cp of 1.67 > 1.33SAA

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Process

mean

Lowerspecification

Upper

specification

1350 ppm 1350 ppm

1.7 ppm 1.7 ppm

+/- 3 Sigma

+/- 6 Sigma

3 Sigma and 6 Sigma Quality

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Cpk

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A capable process must have a Cpk of at

least 1.0 A capable process is not necessarily in the

center of the specification, but it falls withinthe specification limit at both extremes

Cpk = minimum of ,

Upper Specification - x

Limit 3s

Lower x - Specification

Limit 3s

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New Cutting Machine

New process mean x = .250 inches Process standard deviation s = .0005 inches

Upper Specification Limit = .251 inchesLower Specification Limit = .249 inches

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New Cutting Machine

New process mean x = .250 inches Process standard deviation s = .0005 inches

Upper Specification Limit = .251 inchesLower Specification Limit = .249 inches

Cpk = = 0.67.001

.0015

New machine isNOT capable

Cpk = minimum of ,(.251) - .250

(3).0005

.250 - (.249)

(3).0005

Both calculations result in

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Cpk = negative number

Cpk = zero

Cpk = between 0 and 1

Cpk = 1

Cpk > 1SAA

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ILLUSTRATION

A process has a mean of 9.20 grams and a standard

deviation of .30 gram. The lower specification limit

is 7.50 grams and the upper specification limit is

10.50 grams. Compute Cpk·

Cpk =10.50−9.20

3(.30

)

(upper specs.)

=9.20−7.50

3(.30)(lower specs.)

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IMPROVING PROCESS

CAPABILITY

Simplify

StandardizeMistake-proof

Upgrade equipmentAutomate

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Acceptance Sampling

Form of quality testing used for incoming materials or finished goods

Take samples at random from a lot(shipment) of items

Inspect each of the items in the sample

Decide whether to reject the whole lot

based on the inspection results

Only screens lots; does not drivequality improvement efforts SAA

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Operating Characteristic Curve

Shows how well a sampling plandiscriminates between good andbad lots (shipments)

Shows the relationship betweenthe probability of accepting a lot

and its quality level

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Return wholeshipment

% Defective in Lot

P ( A c c

e p t W h o l e

S h i p m

e n t )

100 –

75 –

50 –

25 –

0 – | | | | | | | | | | |

0 10 20 30 40 50 60 70 80 90 100

Cut-Off

Keepwhole

shipment

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AQL and LTPD

Acceptable Quality Level (AQL)

Poorest level of quality we arewilling to accept

Lot Tolerance Percent Defective(LTPD)

Quality level we consider bad

Consumer (buyer) does not wantto accept lots with more defectsthan LTPD SAA

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Producer’s and Consumer’s Risks

Producer's risk ()

Probability of rejecting a good lot

Probability of rejecting a lot when thefraction defective is at or above theAQL

Consumer's risk (b)

Probability of accepting a bad lot

Probability of accepting a lot whenfraction defective is below the LTPDSAA

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Probability ofAcceptance

Percentdefective

| | | | | | | | |

0 1 2 3 4 5 6 7 8

100 – 95 –

75 –

50 –

25 –

10 –

0 –

= 0.05 producer’s risk for AQL

b = 0.10

Consumer’srisk for LTPD

LTPDAQL

Bad lotsIndifference

zoneGoodlots

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n = 50, c = 1

n = 100, c = 2

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A shipment of 2,000 portable battery units for microcomputers isabout to be inspected by a Malaysian importer. The Koreanmanufacturer and the importer have set up a sampling plan inwhich the risk is limited to 5% at an acceptable quality level (AQL)of 2% defective, and the ß risk is set to 10% at Lot Tolerance Percent

Defective (LTPD) = 7% defective. We want to construct the OCcurve for the plan of n = 120 sample size and an acceptance levelof c ≤ 3 defectives. Both firms want to know if this plan will satisfytheir quality and risk requirements. Use range values of 1 to 8percent as defectives.

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Selected Values

of % Defective

Mean of

Poisson

ʎ=np

Probability of

Acceptance

P(A)

0.01 1.2 0.966

0.02 2.4 0.779

0.03 3.6 0.515

0.04 4.8 0.294

0.05 6.0 0.151

0.06 7.2 0.072

0.07 8.4 0.032

0.08 9.6 0.014

1- at AQL = .221>.05

ß level at LTPD < .10

New calculation isnecessary with larger

sample size if the is to belowered

http://c/Users/Sonny/Desktop/POM/POM%20Report/Poisson%20Table.xlsx










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SAA

8.4%

Defective

B a d L

ot s

AQL = .221

GoodLots

-

0.200

0.400

0.600

0.800

1.000

1.200

0 2 4 6 8 10 12

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0.00

0.20

0.40

0.60

0.80

1.00

1.20

0 2 4 6 8 10 12

GOOD

LOTS ≤

2.4% D

= 0.22 PR for

AQL

ß = .03 CR

for LTPD

B

A

D

L

O

T

S

>

8.4%D

INDIFFERENCE

Percent Defective

P

r

o

b

a

b

i

l

i

t

y

o

f

A

c

c

e

p

t

a

n

c

e

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Average Outgoing Quality

1. If a sampling plan replaces alldefectives

2. If we know the incoming percent

defective for the lot, we can computethe average outgoing quality (AOQ) inpercent defective

The maximum AOQ is the highestpercent defective or the lowest averagequality and is called the averageoutgoing quality limit (AOQL) SAA

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Average Outgoing Quality

where

Pd = true percent defective of the lot

Pa = probability of accepting the lot

N = number of items in the lot

n = number of items in the sample

AOQ =(Pd)(Pa)(N - n)

N

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Table of Illustration 2

SAA

p pac AOQ

0 0 (P x Pac)

0.05 0.914 0.046

0.10 0.736 0.074

0.15 0.544 0.082

0.20 0.376 0.075

0.25 0.244 0.0610.30 0.149 0.045

0.35 0.086 0.030

0.40 0.046 0.019


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SAA

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0 0.1 0.2 0.3 0.4

The deliveries is 92.8%,(100-8.2)% good at 8.2AOQL.

A

O Q

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