hw3_new_new_solutions - IEOR 130 Solutions to HW 3 Spring,...

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Solutions to HW 3 Spring, 2008, Prof. Leachman 1. Using a p-chart, the control limit is . . p n ) p ( p p UCL 04 0 1 3 = + = where Shift # of wafers # of wafers rework UCL Number processed reworked fraction ( p ) 1 500 20 .04 .0663 2 650 39 .06 .0631 3 550 36 .0654 .0651 4 600 18 .03 .0640 Rework was in statistical control during shifts 1, 2 and 4. 2. (a) The sample size is n = 100, and the average range, R-bar is 200.6. From the table in the back of the book, d 2 = 5.015. Hence σ = R-bar / d 2 = 200.6 / 5.015 = 40. Now μ + Z* σ = 150 for some Z, and we are given that 4% fall below the LSL, i.e., Prob. { Z < (150 - μ ) / σ } = 0.04. From the table in the back of the book, Z = -1.75. Hence μ = 150 + 1.75(40) = 220. (b) For bin 3, -1.75 <= Z <= (233-220) / 40 = 0.325. From the table in the back of the book, the probability Z lies in this range is 0.9599 – 0.372 = 0.588. Hence we expect 59 chips per wafer in bin 3. For bin 2, 0.325 <= Z <= (300 – 220) / 40 = 2.0. From the table in the back of the book, the probability Z lies in this range is 0.372 – 0.0228 = 0.349. Hence we expect 35 chips per wafer in bin 2. For bin 1, Z >= 2, which has probability 0.0228. Hence we expect 2 chips per wafer in bin 1. The expected revenue per wafer is thus 2(200) + 35(50) + 59(10) = 2740.
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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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hw3_new_new_solutions - IEOR 130 Solutions to HW 3 Spring,...

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