Problem 3 Nanosys water engineering Inc plan to construct wa

Problem 3 Nanosys water engineering Inc. plan to construct water purification plant in Tulsa Oklahoma, to supply portable water to 40,000 Oklahoman homes. Nanosys will spend $230 million in constructing the plant and $135000 per year in operating it. If a salvage value of 15 % of the initial cost is assumed. Nanosys is intended to run the project for 9 years before they transfer the operation to the county government. A complete overhaul of the plant will be carrying out at the end of year 5 at an extra cost of $48200. Considering an interest rate of 8% per year, determine the engineering economy symbols and their value for each option.

Solution

P (initial investment) = $230 million

i (interest rate per year) = 8%

n (time) = 9 years

S (salvage value) = 15%*230 = $34.5 Million

A (Annual Operation cost) = $135000

Overhauling cost at the end of 5 years = $48200

NPW (net present worth) = 230000000 + 135000*(1-1/(1+i)^n)/I + 48200/(1+i)^5 – S/(1+i)^n

NPW (net present worth) = 230000000 + 135000*(1-1/1.08^9)/.08 + 48200/1.08^5 - 34500000/1.08^9

NPW (net present worth) = $213617544.6 or $213.62 Million approx.