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ENGM 620: Quality ManagementPowerPoint Presentation

ENGM 620: Quality Management

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ENGM 620: Quality Management

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26 November 2012

- Six Sigma

- I am a good problem solver because:
- My organization has no problems, so I must be good at solving them.
- I solve the same problems every day.
- I find the root cause and solve a problem once.

- The people who work for me must be good problem solvers because:
- I hear about no problems, so they must solve the problems.
- They tell me they have no time for other things because they spend all their time solving problems.
- Every member of my organization is trained in root-cause problem solving techniques.

Problem Solving

Fixing the symptoms, not the root cause!

- 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.

Lower Spec

Limit

Upper Spec

Limit

60

80

100

120

140

60

140

140

60

- Accuracy - closeness of agreement between an observed value and a standard
- Precision - closeness of agreement between randomly selected individual measurements

- Genuine focus on the customer
- Data- and fact- driven management
- Process focus, management, and improvement
- Proactive management
- Boundaryless collaboration
- Drive for perfection, tolerate failure

- 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

- Define the opportunity
- Measure process performance
- Analyze data and investigate causes
- Improve the process
- Control and process management

- Four Phases (according to your text)
- Develop the business case
- Project evaluation
- Pareto analysis
- Project definition

- Project Charter

- Project Desirability Matrix
- Problem/objective statement
- Primary/secondary metric
- Change Management
- Process Map
- SIPOC, Flow chart, Value Stream, etc.

- QFD Houses

Stars

???

Return

Low Hanging

Fruit

Dogs

Risk

- Two major steps:
- Select process outcomes
- Verifying measurements

- Magnificent 7
- Basic Statistics
- FMEA
- Time Series analysis
- Process capability

- Three major steps:
- Define your performance objectives
- Identify independent variables
- Analyze sources of variability
- Results of this step are potential improvements

- Graphic data analysis
- Confidence intervals
- Hypothesis tests
- Regression/correlation
- Process modeling / simulation

- Try your potential solutions
- Off-line experiments
- Pilot lines

- Assure true improvement

- Hypothesis tests
- Multi-variable regression
- Taguchi methods
- Design of experiments

- 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

- Sustain the improvements
- Manage the process

- Implementation
- Mistake proofing
- Visible enterprise

- Control Plan
- Documentation
- Training

- Control Charts
- Process Management Chart

- 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

LSL

USL

x

T

LSL

T

USL

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) = $0.44

Loss 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.

LSL

USL

x

T

U.S. Plant (2 = 8.33)

Japanese Plant (2 = 2.78)

x

T

x

T

L(x)

k(x - T)2

x

T

L(x) = k(x - T)2

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.

L(x)

400

10.05

10

400 = k(10.05 - 10.00)2

= k(.0025)

L(x)

400

10.05

10

400 = k(10.05 - 10.00)2

= k(.0025)

k= 160,000

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.

L(x)

f(x)

25

1

1.5

25= k(1 - 1.5)2

k= 100

L(x)

f(x)

25

1

1.5

25= k(1 - 1.5)2

k= 100

Smaller is better

L(x) = kx2

Larger is better

L(x) = k(1/x2)

L(x)

f(x)

25

1

L(x)

f(x)

25

1

25= k(1)2

k= 25

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 ] }

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 ] }

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 }

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 }

=k {2 + ( - T)2}

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 }

=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

LSL

USL

x

T

U.S. Plant (2 = 8.33)

Japanese Plant (2 = 2.78)

E[LUS(x)]= 0.16 * 8.33 = $1.33

E[LJ(x)] = 0.16 * 2.78 = $0.44

Recall,

L(x)

400

10.05

10

400 = k(10.05 - 10.00)2

= k(.0025)

k= 160,000

- Homework
- Ch. 13 #s: 1, 8, 10

- Preparation
- Exam