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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 nominalthebest quality loss function is proportional to the second moment of the quality characteristic about the target value. Solution: The nominalthebest 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  (ym ) 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.
 Fall '03
 lDavidMiller

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