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Hw1-Solutions

Hw1-Solutions - HW 1 Solution 1(20 pts D = 5000/yr C =...

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HW 1 Solution 1. (20 pts) D = 5000/yr, C = 600/unit, 1 year = 300 days, i = 0.06, A = 300 Current ordering amount Q = 200 (a) T * = (b) Total(Holding + Setup) cost would be (c) The optimum cost would be (d) T * is 12 days. The closest power of two is 16 days(16/300 yr). The power of two on the other side of 12 days is 8 days(8/300 yr). 2. (20 pts) D = 200/month = 2400/yr, A = (100+55)*1.5 = 232.5 P = 50/hr = 50*6*20*12/yr = 72000/yr, i = 0.22, C = 2.50 (a) (b) (c) 3. (20 pts) (a) EOQ of A : EOQ of B : EOQ of C :

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Therefore, optimal order quantity is 4000 with source C. (b) Holding + Setup cost = (c) Cycle Time = 4000/20000 = 0.2 year = 2.4 months. Replenishment lead time = 3 months. Reorder point = 3/2.4 X 4000 = 5000 1000 units is reorder point It is interesting to interpret the above result for part (c) in terms of the definition of the Inventory Position IP(t) introduced in class during the discussion of the Stochastic Inventory Control theory. So, remember that IP(t) = OHI(t) + O(t) – BO(t) (1) where OHI(t) denotes the on‐hand‐inventory at time t; O(t) denotes the “pipeline” inventory at time t (i.e., material ordered but not received yet); BO(t) denotes the backorders at time t. Also, let Q l = l D denote the demand experienced over a replenishment lead time interval l. In our case, this quantity is Q l = (3/12)x20000 = 5000. Since we want to have no shortages, BO(t) = 0 for all t (2) Consider also the OHI(t) at any time t, and notice that at time t+ l ,
OHI(t+ l ) – BO(t+ l ) = OHI(t) + O(t) – Q l (3)

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