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WELCOME TO SIX SIGMA TRAININGFOR
GREEN BELT CERTIFICATION
Trainer:Deepak Arora
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SIX SIGMA
THE GLOBAL CONCEPT
Six Sigma is a data-driven, high-performance approach to analyzing and solving root causes of businessproblems. It ties the outputs of a business directly to marketplace requirements.
As a result, Six Sigma projects lead to reduced costs, process improvement and reduced business cycletimes.
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Six sigma is about reducing variationSix sigma is about doing right thing the first timeSix sigma is about producing high quality with less moneySix sigma is not about working hard but working smarter .
Six Sigma is a highly disciplined process that helps us focus on developingand delivering near-perfect products and services.
Six Sigma is a business process that allows companies to dramatically improve
their Bottom line(Profitability) by designing and monitoring everyday businessactivities in way that minimizes waste and resources while increasing Customersatisfaction.
- - Mikel Harry (Father of Six Sigma)
- - GE
- - LG Electronics
What is Six Sigma ?
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Brief Role BB / GB guidance Drives Project
completion. Verification of results Education
6 Technical Leader Technical Background,
transmit skills
Full Time Project work toresolve neck issues.
Project execution Improvement TeamLeader
Team Member, tooleducation
Project execution Improvement Team
Leader
MasterB lack b elt
B lack b elt
G reen b elt
Belt
Part Time Project work toresolve neck issues.
Six Sigma Belt System
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Selecting CTQ to meet customer needs Deciding reasonable Tolerance Guarantee CTQs through capability analysis
MFG
Improve serious problem Real Time Monitoring system
CTQ Control system
SVC
Improve cycle time and accuracy Cost improvement
Guarantee for design completion
Quality assurance in manufacturing stage
Maximizing Sales and Service
R&D6
Manufacturing6
Transactional6
Six Sigma is a tool that is applied to all business systems, Design, Manufacturing, Salesand Service.
Application of Six Sigma
6
R&D
Sales&
SVCManufacturing
R&D
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4-6% of Sales
The Cos t of Poor
Qual i ty Typical lyFalls B etween 25 -35 % of Total Sales
Lost Opportunity
Tradit ionalQual i tyCos t sAdditional Costs of
Poor Quality
Rejects InspectionRework
Long cycle times
Increasing costs
Lost sales Late delivery
Lost customer loyalty
Excess inventory
Cost of Poor Quality Iceberg
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9+2 mm
MACHINE 1 MACHINE 2
9.1 8
9.2 8.5
9.3 7
9.4 109 10.5
9.1 9
9.2 7.5
Highvariation
Poorquality
Suppose two machines are making a rod of dia 9+2 mm.Following are the diameters of the rods made by both the machines.
The st.deviation can be found on by the formula:
Machine2
Lowvariation
Goodquality
Machine1
= Standard deviation= item or observation= population mean= total no. of items in the populationN
xx
Example
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Six Sigma Process
Is it OK ?
Define
Measure
Analyze
Improve
Control
Y
N
Clarifying the improvement target- Forecast the improvement effect
- CTQ selection for product and processes Understanding process capability for Y
- Clarifying measurement method for Y- Specific description of target
Clarifying target for improving Y- Clarifying factors which affect Y
Screening for the Vital Few- Understanding relationship of Vital Few- Process optimization and confirmation experiment
Determine control method for X- Build up process control system & audit Vital Few
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Practical Solution
for Positive
Business Impact
Statistical Solution
Statistical Problem
Long Term
Containment &
Control Plan
Practical Problem
Impacting the Business
The SixSigmaProject
Breaking down Six Sigma
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Six Sigma Reactive
DFSS Proactive
Design for Six Sigma
Solve Existing Problems
Fix Current Process
* Design & Develop Capable Product
* Design & Develop Capable Process
Design for
Six Sigma
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Tools used in various stages
Define
Measure
Analysis
Improve
Control
1. Process Mapping2. Logic Tree
3. Pareto Analysis4. Quality Function deployment5. 5 Y Analysis
5. Gage R&R6. Rational sub grouping7. Process Capability
8. Hypothesis Testing9. Regression
10. Graph Analysis
11. ANOVA12. Design of Experiments
13. SPC(statistical Process Control)
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DEFINED
M
A
I
C
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Statistical Terms
Mean Median Mode Standard Deviation
Statistical Concept
Type of data
Continuous MeasurableVariableEx. 1.20
5.00
Discrete AttributeCategoricalEg. Yes / no
High / low
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mInflection
Point
T USL
p(d)
3
The size or a standard deviationshows the distances betweenthe inflection point and the mean.We could say the process has 3
sigma capability, if 3 deviationsare fit able between the targetand the specification limit.
Mean
Standard Deviation
* Mean : Sample set or theaverage value of thepopulation
Average ofPopulation is
Sample averageis
* Standard Deviation is thesquare root of the variance
-Standard deviation ofpopulation
-Samples standard deviation
m=X i
i = 1
N
N
m= X =X i
i = 1
n
n
= S =( X i - m )
2
i = 1
N
N
= s =( X i - X )
2
i = 1
n
n - 1
1
3
Statistical Concept
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Define Stage
Pareto Analysis:This is a tool used to find the important Issues among the problem.Generally, 80%of the problem is caused by 20%of Issues/causes.(Vilfredo Pareto, Italian Economist in 1800s)
Type FrequencyIC 68Tuner 9Col 4Cable 3Connector 2Jack 2
Roller 2
Belt 1Filters 1Others 4
It refers to the process of identifying the independent factors or causes responsiblefor the chronic effect or result.
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Process Mapping
It is a diagrammatic representation of your process explaining all intrinsic and extrinsic parameters
involved in the process with there present level.
The Process Mapping Method
Define the Process boundary. (General area or specific process you intend to improve) Brainstorm and order process steps with your team. Code activities using symbols for easy analysis. Walk through the process to validate map. Add key process metrics
- Yield, costs, Rolled Throughput Yield, Scrap, Overtime $, Capacity, %Schedule, %OTD
Analyze map for key business issues -could be in the areas of :- Process loss or waste- Cycle time improvements- Quality improvements- Flow improvements
Define Stage
Intrinsic factors: Factors which are in control of operator.Extrinsic factors: Environmental parameters.
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Process Mapping Refrigerator - R1 Line Rolled Throughput Yield
D/Plate plate/paint
D/Liner extrusion/mold Door forming Door assembly
I/Case extrusion/mold Case forming
Front - CTQ, L painting O/Case, B/Plate
Cycle assembly
LQC & appearance
Door Assy 89.7%
Output
99.0%
99.7% 93.4% 97.3%
99.6% 81.0%
99.2%91.7%Case Assy 73.4%
97.7% 83.8%
96.5%
Rolled Though put Yield = 73.4% 89.7% 97.7% 83.8% 96.5% = 52.0%
Case Door Cycle assembly LQC& appearance
Define Stage -- Process mapping
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QFD is used to link key consumer requirements to technical specifications and
potential part CTQs. QFD is performed by a team of process experts.
QFD Process
Identify key consumer needs by reviewing market, reliability requirements, general requirementsand current quality issues.
Rank cues by importance and translate them into technical specifications required to meet customer cues. Rank technical specifications by impact on customer cues and translate them into potential partcharacteristics(CTQS).
Rank part characteristics by impact on meeting technical specifications(CTQS).
Define Stage QFD Process
QFD translates the Voice of the Consumer into the Voice of the Process owner.
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FMEA (Failure Mode & Effects Analysis)
FMEA is used to proactively identify and rank risks in a product design and assign appropriateactions to be taken to prevent the failure mode.
FMEA Process
Brainstorm potential failures of the product design.
Assign severity and probability (likelihood of occurrence) ratings to each potential failure mode. Determine existing control measures being taken to eliminate significant failure modes. Develop actions to be taken to eliminate or reduce risk on all remaining significant failure modes.
Define Stage FMEA Process
Failure Mode and Effect AnalysisAssyName
Key Date
ModelName
Supervisor
Purpose
Item/Function RecommendedActions
PotentialFailure Mode
Potential cause/Mechanism of failure
Potential Effectof Failure
Occur
PJT Name
Participant
Sev. Current DesignControls
Detection R.P.N operatorAction Results
Sev. R.P.NOcc. Det.Action taken
** RPN (Risk Priority number) = S x O x D
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Used to break down problem into manageable groups to identify root cause or area of focus. Breakdown the problem on the base of MECE
- MECE (Mutually Exclusive and Collective Exhaustive)
RPM
Rotor
Stator
Assembly
Lamination
End rings Area A
Area B
Electromagnetic
Mechanical
Losses
Inductance
OD
Core length
Why
WhyWhy
Why
Define Stage Logic tree
Brainstorming
Types of Brainstorming
A team approach to generate many ideas in a short time period.
Free Wheeling : Spontaneous flow of ideas by all team members Round Robin : Team members take turns suggesting ideas Card Method : Team members write ideas on cards with no discussion
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Write the Problem Statement
Problem Statement
It is critical to define the problem will. A problem statement should include bothAs is? And the Desired State? Of the issue, and be specific and measurable.
Example:
As is : The response time for 15% of our service calls is more than 2 hours.
Desired State : The response time for all service calls must be 2 hours or less.
As is ? Describes the problem as it is today Should not contain causes Should not imply solutions Should be as specific as possible and include measurement
Desired State ? What you want to achieve by solving the problem Be as objective as possible As specific as possible including measurement goal
Define Stage Theme registration
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MEASURED
M
A
I
C
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Rational Subgroup
It is the process of identifying all possible factors for variation in a system and giving them equal orweighted .
TIME
P R O C E S S R E S P O N S E
WHITE NOISE(Comm on Cause
variat ion)
BLA CK NOISE
(Signal)RATIONAL
SUBGROUPS
Measure Stage Rational Sub grouping
Rational sub grouping allows samples to be taken that include white noise which occurs, within theSamples and black noise which occurs between the samples.
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Common Cause Variation?
Assignable Cause Variation?
Common Cause Variation is the variation present in every process. Also known as white noise .
It is not controllable variation within the existing technology. Represents that best the process can be with the present technology (Inherent process capability).
Assignable Cause Variation represents the outside influences on a process that cause averageto shift and drift. Also known as black noise .
It is potentially controllable variation with the existing process technology. It represents how the process is actually performing over time(Sustained process capability).
Example:
Mixed lots of parts are currently loaded onto trailers at a supplier for shipment to the factory.Part number and count are entered into the factory computer system manually.Excessive variation exists between what shows in the factory computer and what is actually
unloaded at receiving because of errors in transcribing part numbers on the packing cartons.In order to reduce the variation in this process, minimize manual processing of shipping/receiving tickers.
Measure Stage Black / White noise
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Why should we do Rational Sub grouping?
Use the 5M of cause and Effect Analysis as a checklist to assist in identifyingthe potential causes of variation in the process.
Man ex ) Rotating operators, shift changes, new operators
Machine ex ) Setup changes, maintenance effects
Material ex ) Batch/lot/coil differences
Method ex ) Operator methodology, model line balance
Measurement ex ) Manual data entry, calibration effects
Aim is to capture as many of the influential factors on the data collection as possible.Use the Advocacy team to help define causes and plan the data collection period.
Measure Stage Rational Sub grouping
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Measurement : The foundation of Six Sigma.
Six Sigma is based on the measured data . There will be unfavorable consequences from analysisusing statistical tool if we have a problem with measuring system. Whats more, the process gets worse, then experiment will end up in failure. Therefore, we do better secure correct measurement systembefore the project
Observed processdata variation
Understanding Measurement Variation of system
Actual Partto Part
variation
Measurementvariation
Long termProcessvariation
WithinSamplevariation
Operatorvariation
(Reproducibility)
Gagevariation
Accuracy
Repeatability
Stability
Linearity
Six Sigma Project Y = f(X1....Xn)
ManpowerMethodMaterialMeasurementMachine &Environment
Short termProcessvariation
Measure Stage Gage R&R
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Repeatability ?Repeatability :
Getting consistent results
Measure/Re-measure variation
Variation observed with one measurementdevice when used several times by one operatorwhile measuring the identical characteristic onthe same parts.
Operator A
Operator B
Operator C
Reproducibility
Reproducibility ?
Variation obtained from different operatorsusing the same device when measuring theidentical characteristic on the same parts.
Measure Stage Gage R&R
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Linearity ?
Larger Bias Small Bias
LSL USL
Actualvalues
Reference values
Actualvalues
(No Bias)Reference value
Linearity is the difference inthe bias values throughout theexpected operating range of the
gage.(Gage is less accurate at the lowend of specification or operatingrange than at the high end).
Measure Stage Gage R&R
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Gage R&R Study ?
There are 3 kinds in Gage R&R Study as follows-. Repeatability-. Reproducibility-. Total measurement variation
Determines how much extent of their contribution to the total process variation or specification.
The importance of Gage R&R Study
2
Total = 2
Part-Part + 2
R&R
Total variation Variation due todifferences among the parts.
Measurement error variation
As study results, we can get information as follows : Gage resolution is adequate. The measurement system is statistically stable over time.. The measurement error is small enough . And acceptable relative to the process variation
or specification.(That is, Got ready to find X factor correctly caused by Y variation due to small variation
of measurement..) Tell you measurement system is good enough to gather process data.
Measure Stage Gage R&R
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What is Gage R&R Study ?
Sample of Gage R&R
* An improvement plan to lower R&R variationshould be implemented. If you decide not toimprove the measurement system, beware ofthe risk associated with a high Gage R&Rresult. Use of a gage with a conditional R&Rstudy result should be done cautiously.
20% : Acceptable
20% to 29% : Conditional
30% : Unacceptable
An acceptable value for a Gage R&R Study
Gage selection(Resolution)The Gage must have a resolution of less than or equal to 10% of the specification or process variation.* Resolution is the smallest unit of measure the gage is able to read.Ex) In case of part feature tolerance equals +/- 0.020, Gage must have resolution 0.002 and Gage R&R 20%
to be recommended.
Not selecting sample at random, the preparation must be proceeded by preliminary plan so that you cancover the total range of variation and specification.Ex) A pulley has a shaft I.D. = 0.500 +/-0.025 inch. For Gage RR Study, 10 parts should be selected that range
from 0.45 to 0.55 inch. For product acceptance? You must demonstrate that the gage can distinguish the goodpulleys from the bad ones.
Measure Stage Gage R&R
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The height of a CTQ component assembly Spec. = 2.000 0.015
Part Operator 1 Operator 2 |Range(1-2)|
Average range(R- Bar) = R /n = 0.015 / 5 = 0.003 Tolerance = 0.030 Gage Error = (5.15 / 1.19)*(R-bar) = 4.33 *0.003 = 0.013GRR as a % of Tolerance = (0.013 x 100) / 0.030 = 43.3%
d* values for distribution of the average range
12345678910
1.411.281.231.211.19 1.181.171.171.161.16
1.911.811.771.751.741.731.731.721.721.72
2.242.152.122.112.102.092.092.082.082.08
2.482.402.382.372.362.352.352.352.342.34
Gage error is calculated by multiplyingthe average range by a constant d*,where d* is determined from the followingtable. 5.15 is 99% confidence intervalby the gage.
12345
2.0031.9982.0072.0011.999
2.0012.0032.0061.9982.003
0.0020.0050.0010.0030.0040.015Range sum
Number of parts Number of operators2 3 4 5
Measure Stage Gage R&R
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Long study method (using Minitab)
File name : Session > gageaiag.mtw The paint thickness on a PC case was selected as the Six Sigma Theme. Spec. : 2.5 1.5
To confirmmeasurement system3 operators testedrepeatedly twice oneach 10 parts.
Selection : Gage R&R Study
Stat > Quality Tools > Gage R&R Study...(Minitab)
Measure Stage Gage R&R
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Long study method (using Minitab)
Select: ANOVA
Input : Parts, Operator& Measurement data
Measure Stage Gage R&R
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Long study method (using Minitab)
Sum of Squares ANOVA Table(Basis for the estimates)
Gage R&R itself is larger than20%, needs improvingmeasurement system process control applications as well.
If significant, P-value < 0.25 indicatesthat an operator is having a problemmeasuring some the parts.
Ok for product acceptance considering a productstolerance.
This value means the number,not duplicating confidence intervalof measuring part.
Two-Way ANOVA Table With Interaction
Source DF SS MS F P
Part 9 2.05871 0.228745 39.7178 0.00000Operator 2 0.04800 0.024000 4.1672 0.03256Operator*Part 18 0.10367 0.005759 4.4588 0.00016Repeatability 30 0.03875 0.001292Total 59 2.24913
Gage R R
Source VarComp StdDev 5.15*Sigma
Total Gage R&R 0.004438 0.066615 0.34306Repeatability 0.001292 0.035940 0.18509Reproducibility 0.003146 0.056088 0.28885Operator 0.000912 0.030200 0.15553Operator*Part 0.002234 0.047263 0.24340Part-To-Part 0.037164 0.192781 0.99282Total Variation 0.041602 0.203965 1.05042
Source %Contribution %Study Var %Tolerance
Total Gage R&R 10.67 32.66 11.44Repeatability 3.10 17.62 6.17Reproducibility 7.56 27.50 9.63
Operator 2.19 14.81 5.18Operator*Part 5.37 23.17 8.11Part-To-Part 89.33 94.52 33.09Total Variation 100.00 100.00 35.01
Number of Distinct Categories = 4
Number of Distinct Application Method of Categories
1) Number of Distinct Categories = 0 ~ 1 Not Acceptable
2) Number of Distinct Categories = 2 ~ 4 Conditional
3) Number of Distinct Categories 5 Acceptable
Measure Stage Gage R&R
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Attribute Gage R&R Study Each part is accepted if they meet the criteria of the attribute(Pass/Fail or Go/No go)
In case of VCR coating, it would be FAIL if fails to meet the attribute of the exterior appearance. Two checkers then inspect the 18 parts twice in a manner to prevent appraiser bias.
Appraiser "A" Appraiser "B"1 2 1 2
1 G G G G2 G G G G
3 NG G G G4 NG NG NG NG5 G G G G6 G G G G7 NG NG NG NG8 NG NG G G9 G G G G10 G G G G
11 G G G G12 G G G G13 G NG G G14 G G G G15 G G G G16 G G G G17 G G G G18 G G G G
Visual Inspection Gage Study
Different TestResults betweencheckers each other
The gage is acceptable if all the checkers(four per part) agree..
% Gage R&R = 3 / 18 x 100% = 17% If the results of checkers are different, the gage must be improved and re-evaluated. If the gage cannot be improved, it is unacceptable
and an alternate measurement system should befound.
Measure Stage Gage R&R
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Measure Stage PCA
Z BENCH is the reported baseline Z-value
Definition of Z. BENCH
10%9%
ZLSL = 1.34 Z USL = 1.22
19%
ZBENCH = .88
P USL is the probability of a defect relative to the USL.P LSL is the probability of a defect relative to the LSL.P TOT is the total probability of a defect. P TOT = P USL + P LSL Zbench is the Z value from the normal table which corresponds to the total number of defects.
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Executive Summary : Process Capability
Measure Stage PCA
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6 8 10 12 14 16 18
LSL USL
Process Capability Analysis for Quality
USL
Target
LSLMean
Sample N
StDev (ST)
StDev (LT)
Cp
CPUCPL
Cpk
Cpm
Pp
PPU
PPL
Ppk
PPM < LSL
PPM > USL
PPM Total
PPM < LSL
PPM > USL
PPM Total
PPM < LSL
PPM > USL
PPM Total
22.0000
*
8.000012.4368
38
1.90756
2.05922
1.22
1.670.78
0.78
*
1.13
1.55
0.72
0.72
26315.79
0.00
26315.79
10011.39
0.27
10011.66
15595.62
1.71
15597.33
Process Data
Potential (ST) Capability
Overall (LT) Capability Observed Performance Expected ST Performance Expected LT Performance
STLT
Executive Summary : Process Capability
Calculated value by ppm, counting defectnumber out of upper spec limit or lowerspec limit among actual measurement data.
In case of long term process data,Expected value of defect ratio for standardupper, lower limit.
Measure Stage PCA
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Zst = Xbar - mu Sigma(st)
6 SigmaCp = USL - LSL
Zlt = Xbar - mu Sigma(lt)
Zlt = Zst + 1.5
Cpk = (1-k) Cp K = M x barT/2
Determines the spreadof the data
Determines the shiftfrom the mean
Before proceeding for Process capability analysis, Normality test has to be done.And if p value is greater than 0.05 the data is normal.
Measure Stage PCA
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CTQ = Critical To Quality Looking at the customers view point, it means critical product, characterized service value of process.EX) Regarding the relation between the product and service, the customers require. The arrant centerwhich handles their complaints.CTQ is immediate, generous, understanding one, while service business in the center is very kind,informative for products and quick service as if they are likely to be in customers position.
D = Defect In order to satisfy customers requirements, Mishaps unfavorable, being waste, rework or failure.Ex) Error Information written wrongly for customers claim.
DO = Defect Opportunity Certain behavior or event cansing DefectEX) The number of articles should be written on one claim sheet.
U = Unit Certain item available for measurement.EX) Claim form.
DPU = Defect Per Unit The number of Defects existing in one unit.EX) DPU=2/1=2 in case of 2 Error articles out of 10 written on one claim form.
DPO = Defect Per Opportunity The number of defect which exists in one unit related to opportunity number.EX) Any 2 miswritten articles out of 10 which should be written in certain claim of one unit.
2 Defect / (1Unit x 10 Opportunity) = 0.2. That is, DPO=0.2.DPMO = Defect Per Million Opportunities
Converse into DPO number x 1,000,000. (Can be transferred as Sigma Scale)EX), 0.2 DPO x 1,000,000 = 200,000 DPMO or it comes to approximately 2.0 Sigma.
Terms
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DPU(Defect per Unit) / DPO(Defect per Opportunity)
dpu = d/u Here, d stands for the Number of observed defects or frequency. u stands for the Number of Units Produced. Ex : Out of the last 1,000 invoices issued, 1,000 inadequate items were defected. In this case,
What is the DPU for this sample?
dpu = 1,000 / 1,000 = 1.0 (100%)
This is average value, each invoices contained 1 defect.
dpo = d/(u*opp) Here, d : The Number of observed defects or frequency.u : The Number of units produced.
opp :The Number of opportunities
Ex : For each 1,000 invoices Issued, 1,000 defects were detected.There are 10 items per each Unit(Invoice). What is dpo for this sample?dpo = 1,000 / [(1,000)(10)] = 0.1 (10 %)
Is there possibility to have Defects more than one in each Unit ?
What will it be as a target for expectation ?
Measure Stage PCA (Discrete)
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Ex) Sigma level calculation against claim.Claim form - Example
Article 1 : Full in name Article 2 : address Article 3 : The name of city Article 4 : The name of province Article 5 : ZIP CODE
Article 6 : customer service No. Article 7 : The Number or claim. Article 8 : pay amount for claim. Article 9 : The deduction Amount. Article 10 :Actual Amount of Pay.
Claim form - Example
Article 1 : Hong Kil Dong Article 2 : Seongsan - Dong Article 3 : Changwon-City Article 4 : KyungSang South
Province Article 5 : 641 - 700
Article 6 : 000-11-2222 Article 7 : 5 Article 8 : 5,000 WON Article 9 : 40,000 WON Article 10 : 10,000 WON
Unit : 1
Defect Opp : 10
Defect : 2...
Define Problem
Unit = 1 (Claim form)Defect = 2 (Article 8&10)Defect Opp = 10 (10EA Article )
Calculation
Defects Per Unit = 2 Defect / 1 Unit = 2 DPUDefects Per Opportunity = 2 Defect / 10 Articles = 0.20 DPODefects Per Million Opp = 0.2 DPO x 1,000,000 = 200,000 DPMOSigma = 200,000 DPMO or approximately 2.0
Example PCA (Discrete)
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Sample Question
Q. If there are 34 Defects out of 750 units, Lets calculate sigma value of DPU / DPO / Yield /DPMO / Sigma.... (10 opportunities per each unit)
1) DPU = all Defect Number divided by all Unit Number, DPU is ( 34 ) ( 750 ) = ( 0.045 ) .
2) DPO = all Defect Number divided by (all Defect Number times opportunity) ,DPO = ( 34 ) ( 750 10 ) = ( 0.0045 )
3) Yield Value Zero Defect (r = 0), Poision Distribution is Y= e (-d/u) Y = 2.7183 -0.045 = ( 0.956 ) = ( 95.6 )%
OR Y = P(ND) 10 = (1-DPO) 10 = (1 - 0.0045 ) 10 = ( 0.956 ) = ( 95.6 )%
4) DPMO = DPO 1,000,000,DPMO = ( 0.0045 ) 1,000,000 = 4,500 (It has 4,500 ppm per one opportunity,
Thus, Defect has 45,000 ppm per 1 Unit.)
5) Sigma Value = Zinv( 0.956 ) + 1.5 shift = ( 1.71) + 1.5 = ( 3.21)
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Kinds of Yield
Three kinds of Yield.......
YFT = First Time Yield : The Yield without rework, or fix. Application : Deciding each Quality level individual process in use.
YRT = Rolled Throughput Yield : Passing Ratio of one item through the whole process withouteven one defect.(The Yield without rework or fix.)Presenting the possibility of zero defect.(100% Yield)
Application : Certain accumulating certain step of process in order, which is used forevaluating quality level.
YNA = Normalized Yield : Average Yield of successive process. Application : This is used for evaluating the level of quality in completed product.
Y (R ll d Th h Yi ld)
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YRT (Rolled Throughput Yield)
Receive from Supplier
45,000 ppm waste
51,876 ppmwaste
125,526 parts per millionwasted opportunities
28,650 ppm waste
97% Yield
94.4% Yield
YRT = 0.955*0.97*0.94.4 = 87.4%
95.5% Yield
RightFirstTime
YRT (Rolled Throughput Yield)
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Step 1
In order to calculate Y RT, every continues step- times(X) Y FT. YRT = 0.8 X 0.7 X 0.9 = 50.4%
( Theres no wasted opportunity due to Rework. )
Step 2 Step 3 A
If Product A consist of the 3 successive steps,
YFT = 80% YFT = 70% Y FT = 90% YRT = ???YND = ???
YRT (Rolled Throughput Yield)
In order to calculate Normalized Yield every step,- foundational average value is used based on the number of each step;YND(Normalized Yield) = nYRT , where n is the number of the number of the process.
- In this sample question , Y ND(Normalized Yield) = 30.504 = 79.6% Average Y FT in each step equals 79.6%.
Measure Stage PCA (Discrete)
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What is the parallel process ?Change your parallel process into a series process, when in Process Mapping
Ex:
99% 97% 98%
Operation 1 Operation 2 Operation 3 Operation 4
91% 99% 99%
2a 2b 2c
?
YRT = Y 1 x Y 2 xY3 x Y 4= .99 x (.91 x .99 x .99) 1/3 x .97 x .98= .99 x .96 x .97 x.98 = .9035
YNA = (YRT)1/4 = (.9035) 1/4p (defects) = 1 - .9749p (defects) = 0.0251 (Refer to the Normal table)
(2.51 E 2) = Z = 1.96
= 0.9749
Measure Stage PCA (Discrete)
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ANALYZE
D
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A
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C
l h l
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The procedure of Graph Analysis
1. Clarify What would you like to know?.
Clarify the output of what you want to know concerning the practical problem. Expected output is being focused on the content related to the next step Set up a related plan of data collection expecting output.
2. How do you want to know?
Decide whether the output is shown clearly when using any kind of graph for collected data.
3. What should be done about the results of graph analysis?
In regard to the result of graph analysis, confirm that you obtain the output youwanted and decide if there is the topic to be discussed more.
About the improvement of practical problem, take an urgent action when there is issue.
Analyze Stage Graph Analysis
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Example 1 : Graph Analysis
In the production line of air conditioner, the exposing duration in the moisture is very
important for compressor assembly . Set aside carefully assembly time by investigating3 workers on 3 production lines.
1.What do you want to know?2.How do you want to know?3.What kind of Tool is it adequate?4.What kind of data is necessary?5.Where can you get a data from?
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Example 1 : Understanding Graph
29. 0 29.5 30.0 30.5 31.0 31.5 32.0 32.5
0
1
2
3
4
5
6
7
8
time
F r e q u e n c y
The clue you got here?
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Example 1 : Understanding Graph
When we analyze the clue you obtained in different viewpoint ;
1 2 3
29
30
31
32
operator
t i m e
Ultimately what direction should it be taken actuallywith the found result ?
A l St H th i T ti
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Concept of Hypothesis TestingIt is the process of finding out the vital few factors by taking a practical problem and thenTranslating it to a statistical problem.
As we will be using samples (and relativity small ones at that) to estimate the populationparameters, there will always be a chance that we can select a weird sample for our experiment that may not represent a typical set of observations. Because of this, inferential statistics with some assumptions allows us to estimate theprobability of getting a weird sample.
Analyze Stage Hypothesis Testing
CorrectDecision
CorrectDecision
Ho Ha
Ho
Ha
True
Accept
The ratio which isbeing Ha even if its false.
Where is usuallyset up at 10%.
The ratio which isbeing rejected Ho even
though certain thing is truewhere is error.
(usually 5%)
Ho(Null Hypothesis) is assumed to be true.This is like the defendant being assumedto be innocent.
Ha(Alternative Hypothesis is alternativesthe Null Hypothesis.
Ha is the one that must be proved.
Type 1Error
producers
risk
Type 2Error
consumers
risk
A l St g H th i T ti g
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1- = Confidence
The probability that can be determined as a right thing when the Null Hypothesis
is correct.
1- = Power of the test
The rejecting probability when null Hypothesis you want to test is not right.
It is not possible to simultaneously commit a Type 1 and Type 2 decision error.
Analyze Stage Hypothesis Testing
A l St g H th i T ti g
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Test for means Test for VariancesTest for both
means &variances
When data is inform of OK / NG
When data hasequal Probability
of Occurance> Z Test > F Test > ANOVAWhen Population Testing of Variances Testing meansSigma is Known of 2 populations and variances for and data is normal or 2 samples two or more than
two population> T Test > Test for Equal variances1 T Test Testing of Variances Assuming Data2 T Test of more than 2 population to be normalWhen PopulationSigma is not known Data could be Normal or only population mean Non normalis knownData should be Normal
Tests for Discrete DataTest for Continous Data
Proportion test Chi Square Test
Analyze Stage Hypothesis Testing
Analyze Stage Hypothesis Testing
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1. If the Calculated Value is < or = toTable(Critical) value, no conclusions
can be drawn(fail to reject Ho).
If the Calculated Value is > toTable(Critical) value, thena difference exists( reject Ho, accept Ha).
2. If the p value is > or = to, no conclusions
can be drawn(fail to reject Ho) .
If the p value is > A1: 60 A2: 65 A3:70 A4:75
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One-way ANOVA: resp. versus A2
Analysis of Variance for resp.Source DF SS MS F PA2 3 1.9788 0.6596 31.19 0.000Error 8 0.1692 0.0211Total 11 2.1480
Individual 95% CIs For Mean
Based on Pooled StDevLevel N Mean StDev -------+---------+---------+---------a1 3 8.3600 0.0800 (---*---)a2 3 8.7000 0.1819 (---*---)a3 3 9.4800 0.1852 (---*--)a4 3 8.8600 0.1039 (---*---)
-------+---------+---------+---------Pooled StDev = 0.1454 8.50 9.00 9.50
As p value is ANOVA > One way
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Stat > ANOVA > Main effects
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Two way factorial DesignIt is the simplest of the designs it selects only one factor that affects the response.
Example : For Output Y , the input is A. We will study 4 levels of A.
>> A1: 60 A2: 65 A3:70 A4:75>> B1: M1 B2: M2 B3:M3 B4:M4
A1 A2 A3 A4
B1 97 .6 98 .6 99 98B2 97 .3 98 .2 98 97 .7B3 96 .7 96 .9 97 .9 96 .5
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Temp. Material Response
180 1 97.6
180 2 97.3
180 3 96.7
190 1 98.6
190 2 98.2
200 3 96.9
200 1 99.0
200 2 98.0
200 3 97.9
210 1 98.0
210 2 97.7
210 3 96.5Two-way ANOVA: Response versus Temp, Material
Analysis of Variance for Response
Source DF SS MS F P
Temp 3 2.2200 0.7400 7.93 0.016
Material 2 3.4400 1.7200 18.43 0.003
Error 6 0.5600 0.0933
Total 11 6.2200
As p value is
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Stat > ANOVA > Main Effect plot
From the main Effects plot we can judge which is the optimal condition for the response.
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What is a Factorial DesignIt is a experiment in which if there are n factors and the no. of levels at each factor is k, the experiment carried out at all levels of the combination.
Advantages of factorial DesignIt is implemented at combinations of all levels in all factors.We can estimate effects of all factors and their interactions.
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Example: Experiment of washing machine A new design has been developed for a washing machine, and several prototypes have been built.
We would like to run an experiment to quantify the effect of wash time and water volume
on the cleanliness of clothes. .
X1
Low ( - ) 10
High ( + ) 20Low ( - ) 10High ( + ) 20
X2
Low ( - ) 4Low ( - ) 4High ( + ) 8High ( + ) 8
X1: Wash time, in minutes
Level 1: 10 minutes Level 2: 20 minutes
X2: Amount of water, in gallons
Level 1: 4 gallons Level 2: 8 gallons
Y : Reflectance of clothes (a measure of brightness, or cleanliness)
Full factorial design: an experimental design in which all combinations offactors at all levels are tested.
We will repeat each possible combination of four times, for a total of 16 Runs of the experiment.This will give us more data at each run setting, and will give us more confidence in the results.
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Main Effects plot
Washing power
19
17
15
10 20 4 8time (minutes)
18.6
15.8
19.0
15.4
2.8
Time and Water both
appear to have an effect
on Wash Performance
(due to the steep slope
of the lines)
Experiment of washing machine
There are 3 graphs used for analysis of Factorial arrangements: Main Effects Plots - the effect of each individual X on Y In terac t ion Plo ts - the combined effects of two factors changing simultaneously Cube Plo ts - the Y response at each combination of factors
amount of water (gal)
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InteractionThe plot is to compare the changed effect together with 2 factor.
20 minutes20
10 minutes
18
16
14
4 8gallons
18.4
19.7
13.3
17.4Y=1.4
Interaction Plot
20 minutes wash time always givesbetter wash performance, regardless
of gallons of water used.
At 10 minutes wash time, 4 gallonsgives significantly better washperformance than 8 gallons.
Interpretation
change in reflectance
Use Interaction Plots to see if there are relationships between the factors thatcould change the response
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Cube PlotRepresents response value Y from each combination of the factors.
Cube plot
10 20
13.3X =8
gallons
4
time (minutes)
17.4X =
18.4X = 19.7X =
The optimized value of presentexperiment is 19.7 of washingpower on the condition of 20
minutes of wash time, 4 gallonof water.
The lowest washing powershows when in 10 minutes ofwash time, 8 gallon of water.
Interpretation
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Each 2 levels on time and water
Measuring 4 times repetition for each RUN
OK Click
Design of experiment washing power
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Input the factor name and level valueOK Click
Except the example of this case,practice randomly when DOE usually.
OK twice Click
Design of experiment : Wash Power
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Response value should beinput C7 column, after weproduce test equipmentalong with treatment of RUN.
Design of experiment : Wash Power
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What is it?
A DOE which allows for more factors (Xs) to be included in the experimentaldesign for the same number of runs.
Why use it?
To test a large number of potential Xs with a minimum of runs.
When is it used? When screening for the Vital Few Xs.
When economic considerations make it difficult to run a full factorial experiment.
Fractional Factorials are a powerful way toperform experiments with a large number of Xs
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A mathematical equation of describing a relationship between the Y and Xs Creating a Model of process
Y = b0 + b 1x + error where b 0 = constant b 1 = slope
The concept of Regression1. What is regression?
50 100 150
200
250
300
350
Floor Space
A n n u a
l S a
l e s
There appears to be a linear relationshipbetween floor space and annual sales
That is, Is the Annual sales reducing or
increasing according to change floor space.As floor area Increase the annual salesincreases.
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IF theres a relation, How could we make it as mathematical model ?
When the total measurement value become fitting to certain measurement model, Decide a and b so that sum of error which cant explain as a fitting model could be minimized(least sum of square method)?That is, if the data exists along the line above fitting line, Error variation become zero However, this case cant be happened .
80 130 180
200
250
300
350
Floor Space
A n n u a
l S a
l e s
Y = 155.083 + 0.855208XR-Sq = 75.9 %
Regression Plot
Y _TotalVariation
(Xi,Yi)
Can explain Variation
Cant explain Variation =Error
Total Variation = White Noise + Black Noise
( Y i - Y ) 2 = ( Y i - Yi ) 2 + ( Y i - Y ) 2 ^ ^
The concept of Regression
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Restart Minitab (Dont save anything!), and Enter the following data into C1 and C2:
Example:
You are trying to optimize the performance of an paint cure oven.
One theory says that blower fan velocity affects evaporation of solvent in the paint.You are trying to prove that such a relationship exists by analyzing the data below.
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One Variable Regression : Scatter Plot
0 100 200 300 400
0.0
0.5
1.0
1.5
Velocity
E v a p
Looks linear!!!
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One Variable Regression - Linear model
Independent
Graph Click
Storage Click
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One Variable Regression - Linear model
FITS are the predicted values of Ycalculated from the regression equation for each value of XC3 = 0.069 + 0.00383 C 1 (this is the Regression equation found in the Session Window)
or Predicted Response = 0.069 + 0.00383 (Velocity)
RESIDUALS are errors. The presence of residuals demonstrates that the model does not represent
the data without mistakes. (Actual Response minus Predicted Response (Fits) for each point).
Thus:C 4 = C2 - C3
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One Variable Regression - Linear model The average of the Residuals should always be 0.0
The Residuals should be randomly distributed. A pattern in the Residuals may indicatethat this model form is incorrect.
The Residuals should be normally distributed
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Regression Analysis
The regression equation is Evap = 0.069 + 0.00383 Velocity
Predictor Coef StDev T PConstant 0.0692 0.1010 0.69 0.512Velocity 0.0038288 0.0004378 8.75 0.000
S = 0.1591 R-Sq = 90.5% R-Sq(adj) = 89.3%
Analysis of Variance
Source DF SS MS F PRegression 1 1.9351 1.9351 76.49 0.000Residual Error 8 0.2024 0.0253Total 9 2.137
The Session Window contains the analysis results...
Fail to reject H o
Accept H a
For a good model, this number shouldbe close to the same value as R 2
p-value of the Constant
Ho: The line passes through the origin (0,0)...(0 velocity = 0 evaporation)
Ha: The line does not pass through the origin (0,0)...(0 velocity = 0 evaporation)
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Another Example:
Lot Size (X) 10 20 30 40 40 50 60 60 70 80Man-hour (Y) 20 29 50 60 70 85 90 95 109 120
C.I. = Confidence Interval:95% confidence that the means of all data will fall within this band
P.I. = Prediction Interval:95% confidence that the individual data points will fall within thisband
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CONTACTS
D
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A
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C
Control Stage
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What is the Control Phase?
Provides structured closure of projects and re-allocation of resources Provides systematic changes to ensure the process continues in a new path of optimization.
Transfers sustainability of the improvement to the appropriate members of the Advocacy Team Defines control plans specifying process monitoring and corrective actions Provides communication of new procedures and systems to process owners Ensures that the new process conditions are documented and monitored
Why is the Control Phase Important?
We do not want to re-fix the process later
We want to continue benefiting from the improvementsConfirmation of Improvement
In the Control phase, you need to ensure that you have indeed improved the process. This isaccomplished through re-base lining your process using rational subgroups.
You cannot confirm the improvement through initial DOE results only, or through sampling consecutiveparts. You must re-baseline the process to confirm the improvements are valid over the normal processvariation.
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What are the Control Phase Activities?
Confirmation of Improvement
Confirmation you solved the Practical Problem
Benefit Validation
Buy-in to the Control Plan
Quality Plan implementation
Procedural changes
System changes
Statistical Process Control implementation
Mistake-proofing the process (eliminate impact of X whenever possible)
Closure documentation
Audit process
Scoping next project
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Statistical Process Control (SPC)?
Statistical Statistical methods are used to monitor and analyze process variation from sample data
Process Any repetitive (manual or automatic) task or steps
Control Provides an early warning signal that a process has changed. The warning allows you tomake decisions about the process while there is still time to correct the problem before itcan be seen in the final output.
Six Sigma Quality focuses on moving control up stream in a process to leverage the inputcharacteristics for the Y response. If we can measure and control the vital few Xs, controlof the Y should be assured.
S tatistical P rocess C ontrol
Enables us to control our process using statistical methods to signal when processadjustments are needed.
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What makes SPC a good control tool?
Processes vary because they are influenced by common-cause variation (whitenoise) and special-cause variation
Common and special-caused variation can be seen in rational sub grouped samples: common-cause variation characterized by steady state stable process variation
(captured by the within subgroup variation) special-cause variation characterized by outside assignable causes on the
process variation (captured by the between: subgroup variation)
SPC signals when the steady state process variation has been influenced by outsideassignable causes
The Logic of SPC
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ProcessInput Output
Controller
Controllable factors- Assignable causes- Adjustable
- Special
Uncontrollable factors- Common causes- Noise
- Inherent causes SPC has traditionally been used to monitor and control the output of processes. In this application, we are
measuring the dimensions of finished parts or characteristics of finished assemblies.
Six Sigma Quality focuses on moving control upstream to the leverage input characteristic for Y. If we canmeasure and control the vital few Xs, control of Y should be assured.
A B C D E L M N O P
Samples
Process CapabilityDesiredOutput
X
Upper Control Limit
Lower Control Limit
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Types of Control Charts
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There are basically two types of control charts: Variables charts - these charts are used for
monitoring X variables that are continuous,such as, a diameter or consumer satisfactionrating.
Attribute charts - these charts are used for
monitoring discrete X variables, such as,good/bad counts, or inventory levels.
Refer to the diagram for a summary list of thespecific control chart types
Average & RangeXbar & Rn < 10,
typically 3-5
Average &
Std DeviationXbar &n 10
Median & RangeX & Rn
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File Open : S4 > Xbar - R
Calc > Random Data > Normal Distributions- Generate : 10- Store in column(s) : c1-c25- Mean : 4.0- Standard deviation : 0.6( Only c7 Mean : 2.8, Standard deviation : 1.6
Manip > Stack > Stack Columns- Stack the following columns : c1-c25- Store the stacked data in : c26
p
Example of a variable control chart : Customer Satisfaction Index
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p
0S u b g r o u p 5 1 0 1 5 2 0 2 52 . 5
3 . 5
4 . 5
S a m
p l e M
e a n
1
X = 3 . 9 3 8
3 . 0 S L = 4 . 5 4 6
- 3 . 0 S L = 3 . 3 2 9
0
1
2
3
4
S a m
p l e R a n g e
R = 1 . 9 7 5
3 . 0 S L = 3 . 5 0 9
- 3 . 0 S L = 0 . 4 4 0 6
Customer Satisfaction Index
The weekly evaluation averages 7 and 16 fell below 3.957. This change in consumer satisfaction score was driven by some assignable cause
(either system-related or region initiated). The appropriate action would be to investigate, identify and fix the assignablesource of the variation.
The variation among the regional centers for week 7 is larger than expected. An Out-of-control indication can come from either chart, independently.
Example of attribute control chart : appointment observation ratio.
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A local dental group wanted to know why a lot of their patients fail to keep their appointments. Aproblem solving team was assembled and decided to use a p Chart to track the percentages of noshows. The dental clinic began logging monthly percentages of no shows for each month. Of the
total appointments for each month, % no shows plus % shows equal 100%. Since a no showis a defective appointment, the average total fraction defective is called p.
Year 1996Month Jul Aug Sep Oct Nov Dec
% Failed 40 36 36 42 42 40Year 1997
Month Jan Feb Mar Apr May Jun% Failed 20 26 25 19 20 18
Month Jul Aug Sep Oct%Failed 16 10 12 12
monthly percentage defective
total average percentage defective
p = 236/600 = 0.39333, where np =40+36+36+42+42+40 = 236
the fraction is based on 600,total possible for 6 months
UCL = .39333+3(.39333*.60667)/100) ? =0.539LCL = .39333- 3(.39333*.60667)/100) ? =0.246
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Example of attribute control chart : appointment observation ratio.
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0 5 10 15
0.1
0.2
0.3
0.4
0.5
0.6
Sample Number
P r o p o r t
i o n
P Chart for Reject N
1 1 1 11
1
P=0.3933
3.0SL=0.5398
-3.0SL=0.2468
p pp
Control limits were established from the 1996 no show data. The control chart shows a dramatic reduction in the number of missed appointments
after the implementation of the flex-time policy.By adopting the new appointment policy the team was able to reduce the average
percentage of no shows from 39% to 18% (18% is the new average for 1997 data only).
Process Instability
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UCL
LCL
X
A
B
C
C
B
A
Rule 1 Rule 2 Rule 3 Rule 4 Rule 5
Note: A, B, and C representplus and minus one, two
and three sigma zonesfrom the overall process
average.
A lack of control (out of control is indicated when one or more of the following rules apply to yourchart data:
1. A single point above or below a control limit2. Two out of three consecutive points are on the same side of the central line, in Zone A or beyond.3. Four out of five consecutive points are on the same side of the central line, in Zone B or beyond.4 At least eight consecutive points are on the same side of the central line in Zone C