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Ch_10_HW_A_Solutions

# Ch_10_HW_A_Solutions - Ch 10 HW A Solutions 1 Initial cash...

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Ch 10 HW A Solutions 1 Ch 10 HW A Solutions 1. Initial cash flow = \$6.51 million + \$9.80 million + \$0.90 million = \$17.21 million \$0.32 million is sunk cost and not taken into consideration. 2. Using stand alone project principle, the annual sales figure for evaluating the project is: (280 * 150,000) + (3,000 * 40,000) – (300 * 275,000) = 79.50 million 3. Initial Outlay = 5,550,000 Operating Cashflow = (R t – C t )(1 – T) + D t T OCF = (4,900,000 – 2,000,000)(1 – 0.34) + (5,550,000 / 3)(0.34) D t = 5,550,000 / 3 because it is straight line depreciation to zero over 3 years. OCF = 2,543,000.00 4. Terminal cash flow = 0 NPV = 161,297.10 5. Initial Outlay = 5,550,000 + 500,000 = 6,050,000.00 6. Operating Cashflow = (R t – C t )(1 – T) + D t T = ( 4,900,000 2,000,000 )(1 – 0.34 ) + ( 5,550,000 / 3 )( 0.34 ) D t = 5,550,000 / 3 because it is straight line depreciation to zero over 3 years. OCF = 2,543,000.00 Year 1 Net Cash Flow = 2,543,000.00 7. Year 2 Net Cash Flow = 2,543,000.00 8. Terminal Cash Flow = 500,000 + ( 1,000,000 – 0)(1 – 0.34 ) Year 3 Net Cash Flow = Terminal Cash Flow + OCF = 500,000 + ( 1,000,000 – 0)(1 – 0.34 ) + 2,543,000.00 = 3,703,000.00 9. NPV = 404,460.00 10. Operating Cashflow 1 = (R t – C t )(1 – T) + D t T = ( 4,900,000 2,000,000 )(1 – 0.34 ) + ( 5,550,000 )( 0.3333 )( 0.34 ) D t = ( 5,550,000 )( 0.3333 ) because it is MACRS depreciation over 3 years.

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