Question

XYZ Manufacturing decides to build a new plant. The plant will cost $20 million today and is expected to have a useful life of 20 years. At the end of year 5, 10 and 15, there will be major renovation expenses of $K each time. The plant will produce level returns of $2.5 million at the end of each year for the first 10 years and $5 million at the end of each year for the second 10 years. Find the maximum value of K that XYZ could pay so that the internal rate of return on its investment is at least 12%. (Round your answer to integer)

Answer 1

Answer:

The answer is "$ 3,005,010.27".

Explanation:

Let those cash flows be in millions of dollars

$NPV = -20+(\fatc{2.5}{1.12}+\frac{2.5}{1.12^2}+...+\frac{2.5}{1.12^{10}} +\frac{5}{1.12^9}+\frac{5}{1.12^{10}}+...+\frac{5}{1.12^{20}} - \frac{K}{1.12^5}+\frac{K}{1.12^{10}}+ \frac{K}{1.12^{15}})$

$\ max \ value \ of \ K \ so, \ NPV = 0\\\\ \Rightarrow -20+(\frac{2.5}{1.12}+\frac{2.5}{1.12^2}+...+\frac{2.5}{1.12^{10}} +\frac{5}{1.12^9}+\frac{5}{1.12^{10}}+...+\frac{5}{1.12^{20}}) - (\frac{K}{1.12^5}+\frac{K}{1.12^{10}}+ \frac{K}{1.12^{15}}) = 0$

$\Rightarrow -20+\frac{2.5}{0.12}\times(1-\frac{1}{1.12^{10}})+ \frac{1}{1.12^{10}}\times \frac{5}{0.12}\times (1-\frac{1}{1.12^{10}})- 1.072096*K = 0 \\\\\Rightarrow 3.221661 = 1.072096*K \\\\\Rightarrow K = 3.00501$

The value of k = "$ 3,005,010.27".