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Midtermsolns2001

# Midtermsolns2001 - Solutions to IEOR 130 Midterm...

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Solutions to IEOR 130 Midterm Examination Spring, 2001 Prof. Leachman 1. (a) The down time is 2 hours. A= 22 / 24 = 0.9167 35 lots were run, but 5 lots were run twice. Thus there 40 machine cycles yesterday. The average process time per lot is 0.5 hours. U/A = (40)(0.5) / 22 = 20 / 22 = 0.9091 Since the rate efficiency is 85% and the average process time is 0.5 hours, the theoretical process time is (0.85)(0.5). Hence OEE = S / T = (35)(0.5)(0.85) / 24 = 0.6198 (b) QE = Theoretical process time for good units / Theoretical process time for all units. If we assume that all steps have the same process time and that the rework rate is the same across all process steps, then we can estimate QE = 1 / (1 + 0.06) . Since RE = Theoretical process time / Average process time, and since average process time is increasing by 2%, we can estimate RE = (0.85) [ 1 / ( + 0.02)] . Hence OEE would have been OEE = A (U/A) (QE) (RE) = (22/24)(20/22)[ 1 / (1 + 0.06) ] (0.85) [ 1 / ( + 0.02)] = 0.6551 (c) We require OEE = (20/24)[1 / (1+RW)](0.85/1.02) >= 0.6198 or (20/24)(1/0.6198)(0.85/1.02) >= 1 + RW or 1.1204 >= RW. 2. (a) Cp = (USL – LSL) / 6*sigma. Since there is no LSL, Cp is undefined in this case. The correct answer is: It has no value. Full credit also is given if the value of Cpk is correctly determined, as follows.

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