HW2_solution

# HW2_solution - Solution to 16.881 Homework#2 Quality Loss...

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Unformatted text preview: Solution to 16.881 Homework#2 Quality Loss Functions Dan Frey, 6 JUN 98 ORIGIN := 1 Let 1 be the first index in any vector. 1. Prove that the expected value of the nominal-the-best quality loss function is proportional to the second moment of the quality characteristic about the target value. Solution: The nominal-the-best quality loss function is defined as L(y) = k (y - m) 2 (1) Expectation is defined as E(f(x)) = f(x) p(x) dx (2) Substituting (1) into (2) E(L(y)) = E k (y - m) 2 = k (y - m) 2 p(y) dy (3) From the session 1 notes, the nth moment about the target (m) is E (x - m) n = (x - m) n p(x) dx QED 1 2. (60%) A plant manufactures 100 Ohm resistors. The acceptable tolerances are +/- 5% of the nominal value. The cost to scrap the product is \$0.10. a) Assume a normally distributed variation... Solution: ORIGIN := 1 Let 1 be the first index in any vector. A o := 10 Cost to scrap the resistor [cents] m := 100 Nominal value of the resistance (in ohms) D o := 5% m Allowable variation in the diameter is +/- 5% A o k 1 := 2 D o k 2 A o D o 2 := L(y) := k 1 (y - m) 2 if y > m Define the quadratic loss function k 2 (y - m) 2 if y m D o y := m - 1.2 D o , m - 1.2 D o + .. m + 1.2 D o Define a range over y for the purpose 10 of plotting Create a Monte Carlo simulation of the manufacture of the resistors. n := 1000 Number of resistors to be manufactured m m 0 s + := s D o 3 := R := rnorm ( n ,m,s) Create a vector with all of the resistance values of the resistors we manufactured. 2 Set up the format for a histogram of the data. number_of_bins := 20 4 D o width_of_bins := number_of_bins j := 1 .. number_of_bins + 1 Define a vector with the start and end points of bin j := m - 2 D o + width_of_bins j the bins. rel_freq := hist( bin , R) Compute the relative frequency distribution n over interval. bin_center := bin + 0.5 width_of_bins 3 Simulation Results for the Resistors m D o - m D o + L (y) 1 A o rel_freq 0.5 90 95 100 105 110 115 y , bin_center 1 Average_quality_loss n := L ( R i ) = n i 1 Average_quality_loss = 1.081 in cents How does this compare to the theoretically derived figure? m + 2 D o - (y-m ) 2 k 1 (y - m) 2 1 e 2 s 2 dy = 1.111 in cents s 2 p m- 2 D o k 1 (m - m ) 2 + s 2 = 1.111 Or, according to Phadke 2.10. 4 b) Change to "Six Sigma" Create a Monte Carlo simulation of the manufacture of the resistors. Number of resistors to be manufactured n 1000 := s D o 6 := m m s + := R := rnorm ( n ,m,s) Create a vector with all of the resistance values of the resistors we manufactured....
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## This note was uploaded on 11/08/2011 for the course AERO 16.851 taught by Professor Ldavidmiller during the Fall '03 term at MIT.

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HW2_solution - Solution to 16.881 Homework#2 Quality Loss...

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