Investment Option "South" This potential investment is less risky and shorter term, so it has a minimum rate of return of 8.90%. This investment would require an initial outlay of cash of $103,000, and at the end of the 5-year life of this investment IKK Corporation is expected to have to pay a wind-down cost of $10,300. For the first 4 years of this investment, net annual cash inflows are expected to be $36,250 and for the last year of the investment, the net annual cash inflow is expected to be $90,625. $ $ $ TA 6. How much is the present value of the purchase of equipment? 7. How much is the present value of the wind-down cost? 8. How much is the present value of the annual cash inflows for the first 4 years? 9. How much is the present value of the annual cash inflow for the final year of this investment? 10. How much is the Net Present Value? $ $ 11. What is the Internal Rate of Return? (round to the nearest hundredths place/input code: 0.00) % IKK Corporation is considering two possible investments and needs your help in determining which will be the better financial option. Perform the necessary calculations using Microsoft Excel so that maximum precision can be obtained. (Unless indicated otherwise, enter your answers rounded to the nearest whole dollar/input code: 0). Investment Option "North" This potential investment is a little more risky and longer term, so it has a minimum rate of return of 16.20%. This investment would require an initial outlay of cash to purchase a piece of equipment for $147,500, and at the end of the 7-year life of this investment is expected to have a salvage value of $44,250. For each year of this investment, net annual cash inflows are expected to be $29,500. $ 1. How much is the present value of the purchase of equipment? 2. How much is the present value of the salvage value? 3. How much is the present value of the annual cash inflows? $ $ 4. How much is the Net Present Value? $ 5. What is the value of the Present Value Index? (round to the nearest thousandths place/input code: 0.000)