A new municipal power plant is being built in Highfill, AR. The initial installa

Question

A new municipal power plant is being built in Highfill, AR. The initial installation will cost $25 million. Experience shows that major repair and renovation will have to occur every 5 years (beginning at year 5) at a cost of $7.5 million. Annual operating and maintenance costs are $2 million per year, indefinitely. Based on a 5% time value of money, what will be the capitalized cost for indefinite operation of the power plant?

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Explanation / Answer

Overall Capitalized Cost for Indefinite operation for this plant is $101.57 Million

Workings

The Cash FLow across the years till indefinite time would look like this for this project

The cash flow for the project is calculated

Cash Flow in year (0) + Cash Flow in Year (1) + Cash Flow in year (2) + Cash Flow in Year (3) + Cash Flow in year (4) + Cash Flow in Year (5) +………………………………………..

= 25 + 2/(1+0.05) + 2/(1+0.05)^2+2/(1+0.05)^3 +2/(1+0.05)^4+ 9.5/(1+0.05)^5 + 2/(1+0.05)^6 + 2/(1+0.05)^7+2/(1+0.05)^8 +2/(1+0.05)^9+ 9.5/(1+0.05)^10 …………………….

The Cash Flow in year 5, 10,15 can be written as 2+7.5 to get a series

i.e

= 25 + 2/(1+0.05) + 2/(1+0.05)^2+2/(1+0.05)^3 +2/(1+0.05)^4 + (2+7.5)/(1+0.05)^5 + 2/(1+0.05)^6 + 2/(1+0.05)^7+2/(1+0.05)^8 +2/(1+0.05)^9+ (2+7.5)/(1+0.05)^10

Now this function has two Infinite Geometric Progressions

= 25 +[2/(1+0.05) + 2/(1+0.05)^2+2/(1+0.05)^3 +2/(1+0.05)^4 + 2/(1+0.05)^5 + 2/(1+0.05)^6+2/(1+0.05)^7 +2/(1+0.05)^8……………………….] +[ 7.5/(1+0.05)^5 + 7.5/(1+0.05)^10 +7.5/(1+0.05)^15 …………]

The r in first series being = 1/(1+0.05) and in Second = 1/(1+0.05)^5

= 25 + 2/[1 – 1/(1+0.05)] +7.5/[1 – 1/(1+0.05)^5]

= 25 + 2/(1-0.9523) + 7.5/(1-0.7835)

= 25 + 2/0.0477 + 7.5/0.2165

= 25 + 41.928 + 34.64

= 101.57 Million $

Year 0 25 Million Year 1 2 Million Year 2 2 Million Year 3 2 Million Year 4 2 Million Year 5 9.5 Million 2 Million + 7.5 Million Year 6 2 Million Year 7 2 Million Year 8 2 Million Year 9 2 Million Year 10 9.5 Million 2 Million + 7.5 Million Year 11 2 Million Year 12 2 Million Year 13 2 Million Year 14 2 Million Year 15 9.5 Million Year 16 2 Million This patter would continue for every 5 Years Year 17 2 Million Year 18 2 Million Year 19 2 Million Year 20 9.5 Million Year . . Year . . Year . . Year . Indefinitely

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