- #1
IntegrateMe
- 217
- 1
I've been searching everywhere for an answer to this question but haven't been able to get one.
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?
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?