Alternative 1 has a useful life of 5years. The initial investment is $15,000 and

Question

Alternative 1 has a useful life of 5years. The initial investment is $15,000 and the annual cost of operating the system is $6,000. The salvage value at the end of five years is $3,000. Alternative 2 has a useful life of three years. The initial investment is $20,000 and the annual operating expense is $2,000. It has no salvage value at the end of 3 years. MARR = 20% What is the AW2 - AW1 assuming repeatability?

So I understand this as you find PW and then find AW. Please choose from the following options...

-$-478

-$8,882

-$882

+$478

-$-478

-$8,882

-$882

+$478

Explanation / Answer

Therfore Answer is B

Years Discounting Factor @20% Alternative 1 Discounted Values Alternative 2 Discounted Values 0 OutFlows -15000 -15000 -20000 -20000 1 0.833333333 Annual Expense -6000 -5000 -2000 -1666.666667 2 0.694444444 Annual Expense -6000 -4166.666667 -2000 -1388.888889 3 0.578703704 Annual Expense -6000 -3472.222222 -2000 -1157.407407 4 0.482253086 Annual Expense -6000 -2893.518519 5 0.401877572 Annual Expense -6000 -2411.265432 5 0.401877572 Salvage Value 3000 1205.632716 Total -31738.04012 -24212.96296 Sum of Discounting Factor After 5 Years 2.9906 After 3 Years 2.10648 Annualized Values -10612.59952 -11494.51358 AW2-AW1 -881.9140589

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