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1 Power of DOE Some simple examples using 2- level factorials

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2 Newton’s Law of Motion Consider Newton’s Law of motion: F = m.a If nature knows this relationship but we don’t, how can we obtain the relationship without years of theoretical research? Let’s try to do it by experiment using the 2 factors mass (m) and acceleration (a) as input factors. The response is the Force (F).

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3 Experimental Design 2-level, 2 factors or 2 2 full factorial design Inputs: M: 5, 10; A: 100, 200 RunMAAvg. Force 15100 500 252001000 3101001000 4 102002000

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4 Analysis Using a sign table: Runmam x aForce 1-1-1 1500 2-1 1 -11000 3 1-1 -11000 4 1 1 12000 Overall average Force = 1125 [m] = (1500)-(750) = 750, [a] = (1500) – (750) = 750 [m.a] = (1250) – (1000) = 250

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5 Regression Equation F = 1125 + 375 m + 375 a + 125 m.a Now we need to convert the coded values into actual values M = (m + + m - )/2 + (m + - m - )/2 x m i.e. M = 7.5 + 2.5 x m Or m = (M -7.5)/2.5 Likewise for the other terms

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6 Substituting the actual transformed values into the regression equation, F = 1125 + 375 (M-7.5)/2.5 + 375 (A-150)/50 + 125 (M-7.5)/2.5 x (A-150)/50 Simplifying, F = M.A Imagine how quickly Newton would have closed the knowledge gap if he knew DOE!

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