ENGM 620: Quality Management
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ENGM 620: Quality Management
26 November 2012
• Six Sigma
Problem Solving Quiz
• I am a good problem solver because:A. My organization has no problems, so I must
be good at solving them.B. I solve the same problems every day.C. I find the root cause and solve a problem
once.
Problem Solving Quiz• The people who work for me must be good
problem solvers because:A. I hear about no problems, so they must solve the
problems.B. They tell me they have no time for other things
because they spend all their time solving problems.C. Every member of my organization is trained in root-
cause problem solving techniques.
Fixing the symptoms, not the root cause!
Identify
Analyze
Plan
Implement
Evaluate
InstitutionalizeNew
Opportunity
Problem Solving Process
Select
Preven
t
Conta
in
Correct
People&
Teamwork
Six Sigma
• The purpose of Six Sigma is to reduce variation to achieve very small standard deviations so that almost all of your products or services meet or exceed customer requirements.
Reducing Variation
60 80 100 120 140
60 140
14060
Lower Spec Limit
Upper Spec Limit
Accuracy vs. Precision
• Accuracy - closeness of agreement between an observed value and a standard
• Precision - closeness of agreement between randomly selected individual measurements
Six Ingredients of Six Sigma
1. Genuine focus on the customer2. Data- and fact- driven management3. Process focus, management, and
improvement4. Proactive management5. Boundaryless collaboration6. Drive for perfection, tolerate failure
Key People in Six Sigma• Champion
– Work with black belts to identify possible projects• Master Black Belts
– Work with and train new black belts• Black Belts
– Committed full time to completing cost-reduction projects
• Green Belts– Trained in basic quality tools
Six Sigma Problem Solving Process
• Define the opportunity• Measure process performance • Analyze data and investigate causes• Improve the process • Control and process management
Define
• Four Phases (according to your text)– Develop the business case– Project evaluation– Pareto analysis– Project definition
• Project Charter
Some of the tools to “Define”
• Project Desirability Matrix • Problem/objective statement• Primary/secondary metric• Change Management • Process Map
– SIPOC, Flow chart, Value Stream, etc.• QFD Houses
Project Assessment
Return
Risk
DogsLow HangingFruit
Stars ???
Measure
• Two major steps: – Select process outcomes– Verifying measurements
Some of the tools to “Measure”• Magnificent 7• Basic Statistics • FMEA• Time Series analysis• Process capability
Analyze
• Three major steps: – Define your performance objectives– Identify independent variables– Analyze sources of variability
– Results of this step are potential improvements
Some of the tools to “Analyze”
• Graphic data analysis • Confidence intervals • Hypothesis tests • Regression/correlation• Process modeling / simulation
Improve
• Try your potential solutions– Off-line experiments– Pilot lines
• Assure true improvement
Some of the tools to “Improve”
• Hypothesis tests • Multi-variable regression • Taguchi methods • Design of experiments
Exercise: Anti-Solution• Objective: How do we best speed purchase
order preparation?• Anti-Objective: How do we slow purchase order
preparation down to a crawl?• Brainstorm the anti-objective• Examine each anti-objective for a positive idea• Record and add to the positive ideas
Control
• Sustain the improvements• Manage the process
Some of the tools to “Control”
• Implementation– Mistake proofing – Visible enterprise
• Control Plan– Documentation– Training
• Control Charts• Process Management Chart
Taguchi Methods
• The reduction of variability in processes and products
Equivalent definition:• The reduction of waste• Waste is any activity for which the
customer will not pay
Traditional Loss Function
x
LSL USL
T
TLSL USL
Example (Sony, 1979)
Comparing cost of two Sony television plants in Japan and San Diego. All units in San Diego fell within specifications. Japanese plant had units outside of specifications.
Loss per unit (Japan) = $0.44Loss per unit (San Diego) = $1.33
How can this be?
Sullivan, “Reducing Variability: A New Approach to Quality,” Quality Progress, 17, no.7, 15-21, 1984.
Example
x
TU.S. Plant (2 = 8.33)Japanese Plant (2 = 2.78)
LSL USL
Taguchi Loss Function
x
T
x
T
Taguchi Loss Function
T
L(x)
x
k(x - T)2
L(x) = k(x - T)2
Estimating Loss Function
Suppose we desire to make pistons with diameter D = 10 cm. Too big and they create too much friction. Too little and the engine will have lower gas mileage. Suppose tolerances are set at D = 10 + .05 cm. Studies show that if D > 10.05, the engine will likely fail during the warranty period. Average cost of a warranty repair is $400.
Estimating Loss Function
10
L(x)
10.05
400
400 = k(10.05 - 10.00)2
= k(.0025)
Estimating Loss Function
10
L(x)
10.05
400
400 = k(10.05 - 10.00)2
= k(.0025)
k = 160,000
Example 2
Suppose we have a 1 year warranty to a watch. Suppose also that the life of the watch is exponentially distributed with a mean of 1.5 years. The warranty costs to replace the watch if it fails within one year is $25. Estimate the loss function.
Example 2
1.5
L(x)
25
1
f(x)
25 = k(1 - 1.5)2
k = 100
Example 2
1.5
L(x)
25
1
f(x)
25 = k(1 - 1.5)2
k = 100
Single Sided Loss FunctionsSmaller is better
L(x) = kx2
Larger is better
L(x) = k(1/x2)
Example 2L(x)
25
1
f(x)
Example 2L(x)
25
1
f(x)
25 = k(1)2
k = 25
Expected Loss
Expected Loss: Piston DiameterProbability Probability
Diameter Process A Process B9.925 0.000 0.0259.950 0.200 0.0759.975 0.200 0.20010.000 0.200 0.40010.025 0.200 0.20010.050 0.200 0.07510.075 0.000 0.025
Expected Loss: Piston DiameterProbability Probability
Diameter Loss Process A Process B9.925 900 0.000 0.0259.950 400 0.200 0.0759.975 100 0.200 0.20010.000 0 0.200 0.40010.025 100 0.200 0.20010.050 400 0.200 0.07510.075 900 0.000 0.025
Expected Loss
Expected Loss: Piston DiameterProbability Weighted Probability
Diameter Loss Process A Loss Process B9.925 900 0.000 0.0 0.0259.950 400 0.200 80.0 0.0759.975 100 0.200 20.0 0.20010.000 0 0.200 0.0 0.40010.025 100 0.200 20.0 0.20010.050 400 0.200 80.0 0.07510.075 900 0.000 0.0 0.025
Expected Loss
Expected Loss: Piston DiameterProbability Weighted Probability Weighted
Diameter Loss Process A Loss Process B Loss9.925 900 0.000 0.0 0.025 22.59.950 400 0.200 80.0 0.075 30.09.975 100 0.200 20.0 0.200 20.010.000 0 0.200 0.0 0.400 0.010.025 100 0.200 20.0 0.200 20.010.050 400 0.200 80.0 0.075 30.010.075 900 0.000 0.0 0.025 22.5
Expected Loss
Expected Loss
Expected Loss: Piston DiameterProbability Weighted Probability Weighted
Diameter Loss Process A Loss Process B Loss9.925 900 0.000 0.0 0.025 22.59.950 400 0.200 80.0 0.075 30.09.975 100 0.200 20.0 0.200 20.010.000 0 0.200 0.0 0.400 0.010.025 100 0.200 20.0 0.200 20.010.050 400 0.200 80.0 0.075 30.010.075 900 0.000 0.0 0.025 22.5
Exp. Loss = 200.0 145.0
Expected Loss
Recall, X f(x) with finite mean and variance 2.
E[L(x)] = E[ k(x - T)2 ]
= k E[ x2 - 2xT + T2 ]
= k E[ x2 - 2xT + T2 - 2x + 2 + 2x - 2 ]
= k E[ (x2 - 2x+ 2) - 2 + 2x - 2xT + T2 ]
= k{ E[ (x - )2 ] + E[ - 2 + 2x - 2xT + T2 ] }
Expected Loss
E[L(x)] = k{ E[ (x - )2 ] + E[ - 2 + 2x - 2xT + T2 ] }Recall,
Expectation is a linear operator and E[ (x - )2 ] = 2
E[L(x)] = k{2 - E[ 2 ] + E[ 2x - E[ 2xT ] + E[ T2 ] }
Expected Loss
Recall, E[ax +b] = aE[x] + b = a + b
E[L(x)] = k{2 - 2 + 2 E[ x - 2T E[ x ] + T2 }
=k {2 - 2 + 22 - 2T + T2 }
Expected LossRecall,
E[ax +b] = aE[x] + b = a + b
E[L(x)] = k{2 - 2 + 2 E[ x - 2T E[ x ] + T2 }=k {2 - 2 + 22 - 2T + T2 }=k {2 + ( - T)2 }
Expected LossRecall,
E[ax +b] = aE[x] + b = a + b
E[L(x)] = k{2 - 2 + 2 E[ x - 2T E[ x ] + T2 }=k {2 - 2 + 22 - 2T + T2 }=k {2 + ( - T)2 }= k { 2 + ( x - T)2 } = k (2 +D2 )
Since for our piston example, x = T,
D2 = (x - T)2 = 0
L(x) = k2
Example
Example (Piston Diam.)
Expected Loss: Piston DiameterProbability Weighted Probability Weighted
Diameter (x-)2 Process A (x-)2 Process B (x-)29.925 0.0056 0.000 0.0000 0.025 0.00019.950 0.0025 0.200 0.0005 0.075 0.00029.975 0.0006 0.200 0.0001 0.200 0.000110.000 0.0000 0.200 0.0000 0.400 0.000010.025 0.0006 0.200 0.0001 0.200 0.000110.050 0.0025 0.200 0.0005 0.075 0.000210.075 0.0056 0.000 0.0000 0.025 0.0001
Var = 0.0013 0.0009E[LA(x)] = .0013*k E[LB(x)] = .0009*k
= 200 = 145
Example (Sony)
x
TU.S. Plant (2 = 8.33)Japanese Plant (2 = 2.78)
LSL USL
E[LUS(x)] = 0.16 * 8.33 = $1.33
E[LJ(x)] = 0.16 * 2.78 = $0.44
Tolerance (Pistons)
10
L(x)
10.05
400
400 = k(10.05 - 10.00)2
= k(.0025)
k = 160,000
Recall,
Next Class
• Homework– Ch. 13 #s: 1, 8, 10
• Preparation– Exam
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