hw5_new_new_solutions - IEOR 130 Spring, 2009, Prof....

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1 IEOR 130 Spring, 2009, Prof. Leachman Solutions to HW 5 1. (a) We use the observed best yield to make a 3- σ estimate of Y R : 0.82 - Y R = 400 / ) 1 ( 3 R R Y Y ] )[ 400 / 9 ( 64 . 1 6724 . 0 2 2 R R R R Y Y Y Y = + 0 0225 . 1 6625 . 1 6724 . 0 2 = + R R Y Y 7555 . 0 ) 0225 . 1 ( 2 75012 . 2 7639 . 2 6625 . 1 = ± = R Y The fatal defect density equivalent to this random yield satisfies e -AD = Y R , where A = 0.5, i.e., D = 0.56. The in-line inspections have found the following fatal defect density: M1 (0.32)(0.38) = 0.1216 M2 (0.45)(0.50) = 0.225 Poly (0.25)(0.40) = 0.10 --------- 0.4466 in total Thus 0.56 – 0.4466 = 0.1134 in fatal defect density remains to be found. This corresponds to a yield of exp(- 0.5*0.1134) = 0.945, i.e., a yield loss of 5.5%. (b) The overall systematic mechanisms-limited yield is . 794 . 0 7555 . 0 / 6 . 0 / = = = R S Y Y Y The mechanisms identified to date account for (1 – 0.05 – 0.02)(0.98) = 0.9114 Thus about 0.117 in systematic yield loss remains to be explained. 2. c 1 = 4.5, c 2 = 0.5 t p t 1-F(t) denominator numerator G(t) Non-availability= G(t)/168 1 .15 .85 1.0 1.10 1.10 0.00655 2 .20 .65 1.85 1.90 1.027 0.00611 3 .25 .40 2.50 2.90 1.16 0.00690 4 .25 .15 2.90 3.90 1.34 0.00798 5 .15 0 3.05 4.50 1.48 0.00881 Replace O ring every 2 weeks. Note G(t) above is expressed as hours of down time per weeks of cycle length, so to express true non-availability we would have to convert units to hours of down time divided by hours of cycle length, i.e., non-availability equals G(t)/168.
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2 3. (a) c 1 = 200, c 2 = 50 t p t sum k*p k 1-F(t) denominator numerator G(t) 1 .08 .08 . 92 1.0 62 62.0 2 .12 .32 .80 1.92 80 41.7 3 .16 .80 .64 2.72 104 38.2 4 .24 1.76 .40 3.36 140 41.7 t=3 gives the lowest cost rate. (b) Let delta D denote the average defect density contributed by the machine in question. The production rate, λ = 10 wafers/day, A=0.5, DY = (0.5)*exp{-A*(delta D)} Avg. revenue per wafer = 2*100*(0.50)*exp{-A*(delta D)} Delta revenue = 100*[1 - exp{-A(delta D)}] per wafer If we PM every day, delta D = 0.1 (average of 0 and 0.2). Hence Delta revenue = 100*[1 - exp{-.05}] ~ 100*(.05) If we PM every 3 days, delta D = 0.3 (average of 0 and 0.6). Hence Delta revenue = 100*[1 - exp{-.15}] ~ 100*(.14) We should modify the G(t) function to be G*(t) = G(t) + { λ *{Delta revenue}*denominator} / denominator
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This note was uploaded on 03/16/2010 for the course ORMS 130 taught by Professor Leachman during the Spring '10 term at University of California, Berkeley.

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hw5_new_new_solutions - IEOR 130 Spring, 2009, Prof....

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