Question

solve this problem using engineering economy

Compare and pick the best alternative based on "Incremental ROR analysis for a MARR of 15%. All alternatives have the same lives with no salvage value. (6 + 1 pts) Alternatives Investment $5000 $4000 $2500 Operating Cost $240 $500 $1000

Answer 1

Please refer to below spreadsheet for calculation and answer. Cell reference also provided.

A C D E F G B Assumption: Lives of Alternative in years MARR 15% First: Second: Year А Year 0 А Т $5,000 $240 2 C $2,500 $1,000 $1,000 $1,000 $1,000 $1,000 B $4,000 $500 $500 $500 $500 $500 B-C $1,500 $500 -$500 -$500 -$500 -$500 В $4,000 $500 $500 $500 $500 $500 $240 -В $1,000 $260 -$260 -$260 -$260 $260 $2401 $240 $240 | Incremental ROR (B-C) 20% Incremental ROR (B-C) is greater than MARR thus we drop Alternative-C(Defender) to proceed with B for further analysis. Incremental ROR (A-B) 9.43% Incremental ROR (A-B) is less than MARR thus we drop Alternative- A(Challenger) and Select Alternative-B Preferred Alternative в Note: We analysis and compare alternatives in ascending order of their initial investments.

Cell reference -

C D E F G H I J B Assumption: Lives of Alternative in years 5 0.15 MARR First: Year C Second: Year 0 o 2500 1000 1000 1000 1000 1000 B B-C 4000 =D7-C7 500 =D8-C8 500 =D9-C9 500 =D10-C10 500 =D11-C11 500 =D12-C12 ONM BA A-B 4000 5000 =17-H7 500 240 =18-H8 500 240 =19-H9 500 240 =110-H10 500 240 =111-111 500 240 =112-H12 11 12 Incremental ROR (B-C) =IRR(E7:E12) Incremental ROR (B-C) is greater than MARR thus we drop Alternative-C(Defender) to proceed with B for further analysis. Incremental ROR (A-B) =IRR(J7:112) Incremental ROR (A-B) is less than MARR thus we drop Alternative-A(Challenger) and Select Alternative-B Preferred Alternative Note: We analysis and compare alternatives in ascending order of their initial investments.

Hope this will help, please do comment if you need any further explanation. Your feedback would be highly appreciated.