- #1

IntegrateMe

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Let's say we're coming two things at 12% compounded annually.

A's investment, salvage, life, and expense/year are given as:

$50,000 ; $10,000 ; 11 ; $5000

While B's are:

$40,000 ; $0 ; 10 ; $2000

If I do

**50000(A/P, 0.12, 11) - 10000(A/F, 0.12, 11) + 5000**for A and

**40000(A/P, 0.12, 10) + 2000**for B, can someone explain why this comparison "works?" For this problem in particular, we end up with the values $12000.94/year for A and $9000.08/year for B, so obviously B is better, but exactly what method is being used? I've tried researching "net annual cost" for a similar method, but haven't been able to find anything useful.

Any help?