saa report
TRANSCRIPT
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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
Cp =Upper Specification - Lower Specification
6s
Insurance claims process
Process mean x = 210.0 minutes
Process standard deviation s = .516 minutesDesign specification = 210 ± 3 minutes
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Process Capability Ratio
Cp =Upper Specification - Lower Specification
6s
Insurance claims process
Process mean x = 210.0 minutes
Process standard deviation s = .516 minutesDesign specification = 210 ± 3 minutes
= = 1.938213 - 207
6(.516)
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Process Capability Ratio
Cp =Upper Specification - Lower Specification
6s
Insurance claims process
Process mean x = 210.0 minutes
Process standard deviation s = .516 minutesDesign specification = 210 ± 3 minutes
= = 1.938213 - 207
6(.516)Process iscapable
SAA
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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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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
SAA
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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.
SAA
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SAA
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
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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