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solution_quantity_discounts_problem_j

# solution_quantity_discounts_problem_j - With the highest...

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MGSC 487 QUANTITY DISCOUNTS EXAMPLE A firm estimates annual demand for an item at 1000 units. Setup costs are approximately \$200 per production run. Holding costs are estimated to be \$25 per unit per year. The price schedule the seller offers for this item is: Order Quantity Price per Unit 0 to 100 \$ 100.00 101 to 300 \$ 90.00 301 or more \$ 85.00 What should the order quantity of this item be? SOLUTION TO HOMEWORK (QUANTITY DISCOUNTS) First feasible EOQ, starting with the lowest price level: EOQ 85.00 = √2DS/H= √2(1000)(\$200.00)/0.25(\$85.00)= 137.19 ~ 138 units For the \$85.00 per unit price, minimum quantity is 301, so this EOQ is infeasible. Now with \$90.00 level: EOQ 90.00 = √2DS/H= √2(1000)(\$200.00)/0.25(\$90.00)= 133.33 units This quantity is feasible because a 133-unit order qualifies for the \$90.00 price.

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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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solution_quantity_discounts_problem_j - With the highest...

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