We compare two alternatives using Annual Worth. However, the Salvage Va ue needs

Question

We compare two alternatives using Annual Worth. However, the Salvage Va ue needs some preparation. I = 6% At t = 0 the Salvage Value is the full refund value of the Machine which 500,000. At year 20 the Salvage Value is 50.000. Between year 0 a y 20. the Salvage Value decreases linearly (using constant amount eat year At t = 0 the Salvage Value is 800.000. Between year t= 0 and any year the ' Salvage Value decreases at 3% per year (geometrically). Analyze the Annual Worth and compare the two choices over a 10 year penod.

Explanation / Answer

So for alternative A , its a linear decrease over a 20 year period from 500000 to 50000

Hence value decreased over the time 500000-50000

= 450000

Over 20 years , hence per year = 450000/20

= 22500

Hence in 10 years time , the decrease in value would be 22500*10 = 225000 and salvage value at end of 10 years is

275000

And annual worth of 22500

B) Alternative has geometric decrease of 3% every year

Starting value 800000

At end of 1 year 800000*(1-0.03)= 800000*0.97= 776000

Similarly at 2nd year 800000*(0.97^2) =752720

Similarly at end of 10 years 800000*(0.97^10) = 800000*0.737 = 589937.42

Hence value decreased is 210062.57 and salvage values is 589937.42

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