Time years 005 inventory level units 100 the order

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· · · · · · · · · · · · · · · · · · · · · · · · · · Time (years) 0.05 Inventory level (units) 100 The order quantity of the warehouse should be twice larger than Q R , i.e., Q 0 W = Q W 200 units . Another way to see this is Q W = λT . As long as the order cycle time of the warehouse does not change, its order quantity cannot change. The inventory level of the warehouse is the thick line depicted below. Note that it is never 200: Whenever the warehouse get replenished, it ships half of it receives to the retail store immediately. - 6 Time (years) 0.05 Inventory level (units) 100 (d) The annual setup cost of the warehouse is still K W T = $500. The annual setup cost of the retail store becomes K R T 0 R = $200. The average inventory level of the warehouse throughout a year is 100 2 = 50. This implies that the annual holding cost of the warehouse is 50 h W = $50. The annual holding cost of the retail store is h R Q 0 R 2 = $500. The annual total cost of the system is thus $1250 . The policy with k = 2 is better. Though it incurs a higher setup cost of the retail store due to the more frequent ordering, it reduces the holding cost at the same time. Since the holding cost at the retail store is quite expensive, this reduces the total cost. 2
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3. (a) The basic EOQ formulas T * = r 2 K λh and G * = 2 Kλh gives us T * 1 = 0 . 447 year , T * 2 = 0 . 037 year , G * 1 = $894 . 43 , and G * 2 = $1095 . 45 . (b) Since log 2 ˆ T 1 T ! = 2 . 16 and log 2 ˆ T 2 T ! = - 1 . 45 , we know the candidates for k 1 are k F 1 = 2 and k C 1 = 3 and the candidates for k 2 are k F 2 = - 2 and k C 2 = - 1. With each of these four values, we may find the corresponding order cycle time and
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