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# fm28 3 - c Transfers per year = \$4,500,000/\$45,000 = 100 or...

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SOLUTIONS TO END-OF-CHAPTER PROBLEMS 28-1 a. EOQ = ) P )( C ( ) S )( F ( 2 = ) 5 . 1 )( 2 . 0 ( ) 000 , 90 )( 15 (\$ 2 = 000 , 000 , 9 = 3,000 bags per order. b. The maximum inventory, which is on hand immediately after a new order is received, is 4,000 bags (3,000 + 1,000 safety stock). At \$1.50 per bag the dollar cost is \$6,000. c. = inventory Average 2 000 , 3 + 1,000 = 1,500 + 1,000 = 2,500 bags or \$3,750. d. 000 , 3 000 , 90 = 30 orders per year. 30 days 365 = 12.1 12 days. The company must place an order every 12 days. 28-2 a. C* = r ) T )( F ( 2 = . F = \$27; T = \$4,500,000; r = 12%. size n transactio Optimal C* = 12 . 0 ) 000 , 500 , 4 )(\$ 27 (\$ 2 = \$45,000. b. Average cash balance = \$45,000/2 = \$22,500.
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Unformatted text preview: c. Transfers per year = \$4,500,000/\$45,000 = 100, or one approximately every 3.6 ≈ 4 days. d. Total cost = 2 * C (r) + * C T (F) = 2 000 , 45 \$ (0.12) + 000 , 45 \$ 000 , 500 , 4 \$ (\$27) = \$2,700 + \$2,700 = \$5,400. If it maintained an average balance of \$50,000, this would mean transfers of \$100,000. There would be \$4,500,000/\$100,000 = 45 transfers per year. The cost would be 0.12(\$50,000) + 45(\$27) = \$7,215. If it maintained a zero balance, it would have to make 360 transfers per year, so its cost would be 360(\$27) = \$9,720. Answers and Solutions: 28 - 3...
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