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Unformatted text preview: With the highest price level: EOQ 100.00 = 2DS/H= 2(1000)($200.00)/0.25($100.00)= 126 units This quantity is infeasible because it does not lie in the range corresponding to its price, P= $100.00. Since we had found the feasible EOQ for the second lowest price level, this would be one candidate for our best lot size. Now we need to calculate the total costs for this EOQ and each level to find the lowest total cost that is optimal. C= QH/2+ DS/Q+ PD C 301 = 301/2[(0.25)($85.00)]+ 1000($200.00)/ 301+ $85.00(1000)= $88,862.58 C 101 = 101/2[(0.25)($90.00)]+ 1000($200.00)/101+ $90.00(1000)= $93,116.45 C 133 =133/2[(0.25)($90.00)]+ 1000($200.00)/133+ $90.00(1000)= $93,000.01 The best purchase quantity is 301 units that is feasible and gives the minimum cost....
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This note was uploaded on 07/02/2011 for the course MGSC 487 taught by Professor Jayaram during the Spring '10 term at South Carolina.
 Spring '10
 Jayaram

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