ECO 550 Week 3 Chapter 5 and Chapter 6 Problems

# Quarter 1 2010 t 32 d 11 1 d 21 0 d 31 0 y 1 825

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Quarter 1, 2010: t =32: D 11 =1 D 21 = 0 D 31 =0 Y' 1 = 8.25 + .125(32)  2.75(1) + .25(0) + 3.50(0) = \$9.5 (million) Quarter 2, 2010: t = 33: D 12 =0 D 22 =1 D 32 =0 Y' 2 = 8.25 + .125(33)  2.75(0) + .25(1) + 3.50(0) = \$12.625 (million) Quarter 3, 2010: t = 34: D 13 =0 D 23 =0 D 33 =1 Y' 3 = 8.25 + .125(34)  2.75(0) + 2.25(0) + 3.50(1) = \$16 (million) Quarter 4, 2010: t = 35 D 14 =0 D 24 =0 D 34 =0

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Y' 4 = 8.25 + .125(35) 2.75(0) + 2.25(0) + 3.50(0) = \$12.625 (million)
Chapter 6 2. Cummins could have established internal hedges by setting up offsetting payables in foreign currencies that become cheaper as the dollar appreciates. As the dollar appreciated, rising prices overseas in the export country currencies and the resulting decline in sales and revenue would be offset by falling dollar costs for the payables owed in those currencies. In addition, Cummins could have established a short position in currency futures or option markets to hedge the declining cash flow from their export sales receipts. An appreciating dollar implies a lower value of Euro or Yen that implies a profit on the short position that can purchase foreign currency at declining prices for delivery on fixed strike price contracts. Cummins could also have entered into a currency swap contract that makes money as the dollar appreciates and export sales receipts decline. 7.

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Quarter 1 2010 t 32 D 11 1 D 21 0 D 31 0 Y 1 825 12532 2751...

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