An investment project will req

An investment project will require development costs of \$150million at time zero and \$80 million at the end of second year fromtime zero with incomes of \$25 million per year at the end of years1, 2 and 3 and incomes of \$60 million per year at the end of years4 through 10 with zero salvage value predicted at the end of year10. Calculate the rate of return for this project.

development cost at time 0 . d1 = \$ 150 million

development cost at time 2 , d2 = \$80 million

let the rate of return for this project = r

cash flow at the end of year 1 , c1 = \$25 million

cash flow at the end of year 2 , c2 = \$25 million

cash flow at the end of year 3 , c3 = \$25 million

cash flow at the end of year 4 , c4 = \$60 million

cash flow at the end of year 5 , c5 = \$60 million

cash flow at the end of year 6 , c6 = \$60 million

cash flow at the end of year 7 , c7 = \$60 million

cash flow at the end of year 8 , c8 = \$60 million

cash flow at the end of year 9 , c9 = \$60 million

cash flow at the end of year 10, c10= \$60 million

we can find r by solving the following equation

d1 =[c1/(1+r)]+[(c2-d2)/(1+r)2] +[c3/(1+r)3] + [c4/(1+r)4] +[c5/(1+r)5] +[c6/(1+r)6] +[c7/(1+r)7] +[c8/(1+r)8] +[c9/(1+r)9] + [c10/(1+r)10]

150 = [25/(1+r)] + [(25-80)/(1+r)2] +[25/(1+r)3] + [60/(1+r)4] +[60/(1+r)5] + [60/(1+r)6] +[60/(1+r)7]+ [60/(1+r)8] +[60/(1+r)9] + [60/(1+r)10]

By trial and error , we find that the value of r that satisfiesthe above equation is 0.1623

thus the rate of return , r = 0.1623 or16.23%

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