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ENGM 620: Quality Management. 26 November 2012 Six Sigma. Problem Solving Quiz. 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.

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engm 620 quality management
ENGM 620: Quality Management

26 November 2012

  • Six Sigma
problem solving quiz
Problem Solving Quiz
  • 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.
problem solving quiz1
Problem Solving Quiz
  • 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.
slide4

Problem Solving

Fixing the symptoms, not the root cause!

six sigma
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
Reducing Variation

Lower Spec

Limit

Upper Spec

Limit

60

80

100

120

140

60

140

140

60

accuracy vs precision
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
Six Ingredients of Six Sigma
  • Genuine focus on the customer
  • Data- and fact- driven management
  • Process focus, management, and improvement
  • Proactive management
  • Boundaryless collaboration
  • Drive for perfection, tolerate failure
key people in six sigma
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
Six Sigma Problem Solving Process
  • Define the opportunity
  • Measure process performance
  • Analyze data and investigate causes
  • Improve the process
  • Control and process management
define
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
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
Project Assessment

Stars

???

Return

Low Hanging

Fruit

Dogs

Risk

measure
Measure
  • Two major steps:
    • Select process outcomes
    • Verifying measurements
some of the tools to measure
Some of the tools to “Measure”
  • Magnificent 7
  • Basic Statistics
  • FMEA
  • Time Series analysis
  • Process capability
analyze
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
Some of the tools to “Analyze”
  • Graphic data analysis
  • Confidence intervals
  • Hypothesis tests
  • Regression/correlation
  • Process modeling / simulation
improve
Improve
  • Try your potential solutions
    • Off-line experiments
    • Pilot lines
  • Assure true improvement
some of the tools to improve
Some of the tools to “Improve”
  • Hypothesis tests
  • Multi-variable regression
  • Taguchi methods
  • Design of experiments
exercise anti solution
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
Control
  • Sustain the improvements
  • Manage the process
some of the tools to control
Some of the tools to “Control”
  • Implementation
    • Mistake proofing
    • Visible enterprise
  • Control Plan
    • Documentation
    • Training
  • Control Charts
  • Process Management Chart
taguchi methods
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
example sony 1979
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) = $0.44

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.

example
Example

LSL

USL

x

T

U.S. Plant (2 = 8.33)

Japanese Plant (2 = 2.78)

taguchi loss function1
Taguchi Loss Function

L(x)

k(x - T)2

x

T

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

estimating loss function
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 function1
Estimating Loss Function

L(x)

400

10.05

10

400 = k(10.05 - 10.00)2

= k(.0025)

estimating loss function2
Estimating Loss Function

L(x)

400

10.05

10

400 = k(10.05 - 10.00)2

= k(.0025)

k = 160,000

example 2
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 21
Example 2

L(x)

f(x)

25

1

1.5

25 = k(1 - 1.5)2

k = 100

example 22
Example 2

L(x)

f(x)

25

1

1.5

25 = k(1 - 1.5)2

k = 100

single sided loss functions
Single Sided Loss Functions

Smaller is better

L(x) = kx2

Larger is better

L(x) = k(1/x2)

example 23
Example 2

L(x)

f(x)

25

1

example 24
Example 2

L(x)

f(x)

25

1

25 = k(1)2

k = 25

expected loss5
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 loss6
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 loss7
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 + 22 - 2T + T2 }

expected loss8
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 + 22 - 2T + T2 }

=k {2 + ( - T)2}

expected loss9
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 + 22 - 2T + T2 }

=k {2 + ( - T)2}

= k {2 + ( x - T)2} = k (2 +D2 )

example1
Example

Since for our piston example, x = T,

D2 = (x - T)2 = 0

L(x) = k2

example sony
Example (Sony)

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

tolerance pistons
Tolerance (Pistons)

Recall,

L(x)

400

10.05

10

400 = k(10.05 - 10.00)2

= k(.0025)

k = 160,000

next class
Next Class
  • Homework
    • Ch. 13 #s: 1, 8, 10
  • Preparation
    • Exam
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