Exam II review - student version

P p u 2 18000 90 50 1837 q 1800 rounding within 5 d

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[ p /(p u)] = [2 * 18000 * 175 / 4.2] * 90 /(90-50) = 1837 Q* = 1800 (rounding within 5%) D = 18,000 units / year, S = $175, 1 year = 360 days H = $4.2 / unit / year, p = 90 units / day =1800*(90-50)/90=800 units Q” Q Inventory Run time = Q / p Cycle time = Q / u Q” = Q . (p – u)/ p (a) Find optimal lot size u = 18000 / 360 = 50 [units / day] (b) Calculate run time, cycle time and AIC. (c) What if machine cannot be run for more than 15 days? Run time = 15 days Q* = 15*90 = 1350 , Cycle time = 1350/50 = 27 days and AIC(1350) = $3593.33 = 1750 + 1680 = $3430
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