Question

6-52. Compare alternatives A and B with the equivalent worth method of your choice if the MARR is 15% per year. Which one would you recommend? State al assumptions. (6.5) Capital investment $50,000 Operating costs S5,000 at end of year 1 and increasing by S500 per year thereafter S20,000 S10,000 at end of year I and increasing by $1,000 per year thereafter Overhaul costs $5,000 every 5 years None Life Salvage value S10,000 if just 20 years 10 years negligible overhauled

Answer 1

The two alternatives can be analysed on the basis of equivalent worth method and here we are taking Present worth method to analysing these two alternatives given that MARR is 15%. Assumptions we need to do here is that the alternatives are mutually exclusive and it means we can take any one alternative from these. Now we will calculate the PW of these two alternatives.

The two alternatives can be analyzed on the basis of equivalent worth method and here we are taking Present worth method to analyzing these two alternatives given that MARR is 15%. Assumptions we need to do here is that the alternatives are mutually exclusive and it means we can take any one alternative from these. Now we will calculate the PW of these two alternatives. Here we need to calculate the present worth of alternatives - A. pw of A = -$50,000 - $ 5000 (p/A, 15%,20) - 500 (p/G, 15%, 20) -5000 (p/F, 15%, 5) - 5000 (p/F, 15%, 10) -5000 (p/F, 15%, 15) -5000 (p/F, 15%,20) + 10000 (p/F, 15%,20) (Note: As all are cost so we take negative sign except salvage value which is positive.) therefore pw of A = -50,000 -5000 times (1.15)^20 -1/ 0.15 (1.15)^20 - 500 times (1.15)^20 -1/ 0.15^2 (1.15)^20 - 20/ 0.15 (0.15)^2 -5000 (1.15)-5 - 5000 (1.15)^-10 -5000 (1.15)^15 - 5000 (1.15)^-20 + 10000 (1.15)^-20