Question

"You plan to operate the same type of machine for 12 years. Machine A lasts 4 years and Machine B lasts 6 years. Machine A costs $8,000 and Machine B costs $12,000. The salvage value of Machine A is $4,000 and the salvage value of Machine B is $2,000. Annual operation and maintenance costs are $3,000 for Machine A and $3,500 for Machine B. Both machines can be purchased in the future at the same price as today, and their salvage values and annual costs will remain as they are now. Your MARR is 17%. Enter the Annual Equivalent Cost (AEC) as a POSITIVE number for the machine that should be selected."

Answer #1

Machine A | Machine B | |

Cost | 8000 | 12000 |

Maintenance and operation cost | 3000 | 3500 |

Slavage Value | 4000 | 2000 |

Life | 4 | 6 |

MARR = 17% = 0.17

(A/F,i,n) = i x [ (i+1)^{N} - 1 ] ^{-1}

(A/P,i,n) = i x [ (i+1)^{N} - 1 ] ^{-1} x
(i+1)^{N}

Now AEC of machine A

AEC of A = -8000 (A/P,17%,4) + 4000 (A/F,17%,4) - 3000

AEC of A = - 2916+ 778 - 3000

AEC of A = -5138

Now AEC of machine B

AEC of B = -12000 (A/P,17%,6) + 2000 (A/F,17%,6) - 3500

AEC of B = - 3343+ 217 - 3500

AEC of B = -6626

So

select machine A sine AEC A > AEC B

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