1. Statistical Process Control
Introduction to Taguchi Methods

2. Recall: What is Quality?
Juran – Quality is fitness for use
=> we should be able to determine a set of measurable characteristics which define quality

3. Recall: What is Quality?
Taguchi – Loss from quality is proportional to the amount of variability in the system
Why?
=> if we reduce variation, we reduce loss from quality
Less rework, reduction in wasted time, effort, and money

4. Quality Improvement
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
From Juran and Taguchi (and others);
Define the three categories of activities

5. Recall: Cost of Quality
Cost of Failure
Cost of
Control
Total Cost
Traditional View
Quality Level
Costs

6. Traditional Loss Function
LSL
USL x T
“goalpost” specifications – as long as you kick the ball between the posts, it doesn’t matter where you hit.
LSL T
USL

7. 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.

8. Example
LSL
USL x T
U.S. Plant (2 = 8.33)
Japanese Plant (2 = 2.78)

9. Taguchi Loss Function T x x T
The Taguchi loss function is a scientific approach to tolerance design.
Taguchi suggests that no strict cut-off point divides good quality from poor quality. Rather, Taguchi assumes that losses can be approximated by a quadratic function so that larger derivations from the target correspond to increasingly larger losses. x T

10. Taguchi Loss Function T L(x) = k(x - T)2 L(x) k(x - T)2 x
Nominal is best = quality deteriorates as the actual value moves away from the target on either side x T
L(x) = k(x - T)2

11. 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 = 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.

12. Estimating Loss Function
L(x)
400
10.05 10
400 = k( )2
= k(.0025)

13. Estimating Loss Function
L(x)
400
10.05 10
400 = k( )2
= k(.0025)
k = 160,000

14. 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.

15. Example 2
L(x)
f(x) 25 1
1.5
25 = k( )2
k = 100

16. Example 2
L(x)
f(x) 25 1
1.5
25 = k( )2
k = 100

17. Single Sided Loss Functions
Smaller is better
L(x) = kx2
Larger is better
L(x) = k(1/x2)

18. Example 2
L(x)
f(x) 25 1

19. Example 2
L(x)
f(x) 25 1
25 = k(1)2
k = 25

20. Expected Loss

21. Expected Loss

22. Expected Loss

23. Expected Loss

24. Expected Loss

25. 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 ] }

26. 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 ] }

27. 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 }

28. 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 }

29. 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 )

30. Example Since for our piston example, x = T, D2 = (x - T)2 = 0
L(x) = k2

31. Example (Piston Diam.)

32. Example (Sony) T x E[LUS(x)] = 0.16 * 8.33 = $1.33
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

33. Tolerance (Pistons) Recall, 10 400 = k(10.05 - 10.00)2 = k(.0025)
L(x)
400
10.05 10
400 = k( )2
= k(.0025)
k = 160,000

34. Tolerance
L(x)
Suppose repair for an engine which will fail during warranty can be made for only $200
400
200
10.05 10
LSL
USL

35. Tolerance
L(x)
Suppose repair for an engine which will fail during warranty can be made for only $200
200 =160,000(tolerance)2
400
200
10.05 10
LSL
USL

36. Tolerance
L(x)
Suppose repair for an engine which will fail during warranty can be made for only $200
200 = 160,000(tolerance)2
tolerance = (200/160,000)1/2
= .0354
400
200
10.05 10
LSL
USL