Statistical Process Control Introduction to Taguchi Methods

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

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

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

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

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

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

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

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

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 L(x) 400 10.05 10 400 = k(10.05 - 10.00)2 = k(.0025)

Estimating Loss Function L(x) 400 10.05 10 400 = 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 L(x) f(x) 25 1 1.5 25 = k(1 - 1.5)2 k = 100

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

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

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

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

Expected Loss

Expected Loss

Expected Loss

Expected Loss

Expected Loss

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 + 22 - 2T + 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 + 22 - 2T + T2 } =k {2 + ( - T)2 }

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 )

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

Example (Piston Diam.)

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

Tolerance (Pistons) Recall, 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

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

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

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