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