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hw6solutions - IEOR 130 Methods of Manufacturing...

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IEOR 130 Methods of Manufacturing Improvement Solutions to HW #6 Spring 2007, Prof. Leachman 1. (a) M/C A: Avail = 0.896, Util = 0.875, Util of avail = 0.875 / 0.896 = 0.977 M/C B: Avail = 0.938, Util = 0.896, Util of avail = 0.896 / 0.938 = 0.955 M/C C: Avail = 0.979, Util = 0.833, Util of avail = 0.833 / 0.979 = 0.851 M/C D: Avail = 0.875, Util = 0.833, Util of avail = 0.833 / 0.875 = 0.952 (b) Machine A is the bottleneck; reducing its down time will increase fab throughput. Reducing down time at the other machines primarily serves to reduce cycle time. If other machines are upstream from machine A in the process flow, reducing their down time might help to reduce idle time of Machine A, but this is a weaker effect. (c) Machine A is the bottleneck; increasing its rate efficiency will increase fab throughput. Increasing rate efficiency at the other machines primarily serves to reduce cycle time. If other machines are upstream from machine A in the process flow, improving their rate efficiency might have a minor positive impact on reducing idle time of Machine A, but this is a very weak effect. 2. (a) OEE = 800*(0.5) / N*24 -> N = (800)(0.5) / (.667)(24) = 25 (b) OEE = U*RE*QE = U*RE = (0.85)*RE -> RE = OEE / U = (.667)/(.85) = .7882 (c) (.85)*RE*24*24 = 800*(0.5) -> RE = (800)(0.5) / (.85)(24)(24) = .817 3. (a) Avail A = 1 – 16/168 – 8/208 = 1 -.0952 - .0385 = 0.866 Avail B = 1 – 18/168 – 6/106 = 1 - .1071 - .0506 = 0.836
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