Question

Use Annualized Worth to determine best choice between these three.  I = 6% Show your work.

A     Planning horizon 10 years.   Initial Cost 50,000   annual maintenance costs 5000  annual revenue 40,000 salvage value 15000,

B   Planning horizon 8 years.    Initial Cost 75000

  annual maintenance cost   first year 5000  each year increases by 500.  

Annual revenue  first year 35000  each year increases 1000.  

Salvage value 18000.

C  Planning horizon 12 years   Initial cost 90,000

       Annual maintenance cost  first year 4000   each year increases 3000

       Annual Revenue   first year 65000   each year decreases 2000.

         Salvage value 16000

Answer 1

I = 6%

A.

Annual worth = (-50000 - 5000*(P/A, 6%, 10) + 40000*(P/A, 6%, 10) + 15000*(P/F, 6%, 10))*(A/P, 6%, 10)

Annual worth = (-50000 - 5000*7.3601 + 40000*7.3601 + 15000*.5584)*.1359

Annual worth = $29351.6 or $29352

==

B.

Annual worth = (-75000 - 5000*(P/A, 6%, 8) - 500*(P/G, 6%, 8) + 35000*(P/A, 6%, 8) + 1000*(P/G, 6%, 8) + 18000*(P/F, 6%, 8))*(A/P, 6%, 8)

Annual worth = (-75000 - 5000*6.2098 - 500*19.842 + 35000*6.2098 + 1000*19.842 + 18000*.6274)*.161

Annual worth = $21333.8 or $21334

==

C.

Annual worth = (-90000 - 4000*(P/A, 6%, 12) - 3000*(P/G, 6%, 12) + 65000*(P/A, 6%, 12) - 2000*(P/G, 6%, 12) + 16000*(P/F, 6%, 12))*(A/P, 6%, 12)

Annual worth = (-90000 - 4000*8.3838 - 3000*40.337 + 65000*8.3838  - 2000*40.337  + 16000*.4970)*.1193

Annual worth = $27162

The best choice is alternative A, because it has highest positive annual worth among all the three alternatives.