o.1 0.07N85 10.36761 0.07212 07 2733o 4. (a) A piece of equipment costs 100,000.

Question

o.1 0.07N85 10.36761 0.07212 07 2733o 4. (a) A piece of equipment costs 100,000. The salvage value slowly decreases linearly with time until at 100 years it is 10,000. It generates a revenue stream of 30,000 per year for the planning horizon period. Costs start at 2000 per year and increase each year by 100, The interest rate for the situation is 3% per year, what is the future worth of the system at the end of year 80? Show all work. (b) What is the annualized worth of the system? Show work. (Use back if needed). Us

Explanation / Answer

PV = 100,000

Terminal value = $10000 at the end of year 100.

Linear decrement in value = (100000-10000 )/100 = $900

Terminal value at the end of year 80 = 100000- 900*80 = $28000

FV of revenues = 30000* (F/A, 3%, 80) = 30000* 321.363 = $9640,890

FV of costs = 2000* (F/A, 3%, 80) + 100* (p/G, 3%, 80)* (F/P,3%,80)

= 2000* 321.363 + 100* 756.086* 10.641

= 1447,277

Future value = FV of revenues - FV of costs + FV of terminal value

= $9640,890 - 1447,277+ 28000=

8221613

B: Annualized worth = 8221613 * (A/F, 3%, 80) = 8221613 * 0.00311

= $25569.22

8221613

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