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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

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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.


Engm 620 quality management

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


Traditional loss function

Traditional Loss Function

LSL

USL

x

T

LSL

T

USL


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 function

Taguchi Loss Function

x

T

x

T


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 loss

Expected Loss


Expected loss1

Expected Loss


Expected loss2

Expected Loss


Expected loss3

Expected Loss


Expected loss4

Expected Loss


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 piston diam

Example (Piston Diam.)


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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