Exploring Net Annual Cost of Investments: A Comparison

[IntegrateMe]

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?

 

[HallsofIvy]

Not if you don't tell us what "50000(A/P, 0.12, 11)" means! I'm not even sure what "A/P" , means. It clearly does not have to do with "investment A" because you use it in the calculation for B also.

What I would do is this- A has a total cost of $50000, but we can recover $10000 so it really costs $40000, and lasts 11 years so we can prorate it at 40000/11= $3636.36 per year. It also has a "cost per year" of 5000 so it costs us $8636.36 per year.

B has a total cost of $40000 and we can recover nothing and lasts 10 years so we can prorate it at 40000/10= $4000 per year. It also has a "cost per year" of $2000 per year so it costs $6000 per year. It may be that you are doing a projected cost of not banking the money but that would require an assumed interest rate which is not given.

 

[RoshanBBQ]

Not if you don't tell us what "50000(A/P, 0.12, 11)" means! I'm not even sure what "A/P" , means. It clearly does not have to do with "investment A" because you use it in the calculation for B also.

What I would do is this- A has a total cost of $50000, but we can recover $10000 so it really costs $40000, and lasts 11 years so we can prorate it at 40000/11= $3636.36 per year. It also has a "cost per year" of 5000 so it costs us $8636.36 per year.

B has a total cost of $40000 and we can recover nothing and lasts 10 years so we can prorate it at 40000/10= $4000 per year. It also has a "cost per year" of $2000 per year so it costs $6000 per year. It may be that you are doing a projected cost of not banking the money but that would require an assumed interest rate which is not given.

He is using conversion factors that convert certain types of money to other types. The A/P converts a present value of money (i.e. $200 dollars today) to an equivalent annual rate at 12% per unit time over 11 units of time. The A/F does the same thing, except it does the conversion, using a future sum of money (you salvage the product at $10,000 in the future. Since there is an interest rate of 12%, you are effectively losing money by salvaging it then as opposed to now).

 

[Ray Vickson]

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?

Why do you add and subtract the way you do? Measuring in $000s, the cash outflow stream for A is (50,10,10,10,10,10,10,10,10,10,10,10-5) [we invest 50, then also spend 10 per year for 11 years, but get back 5 at the end of year 11]. The cash outflow stream for B is (40,0,0,0,0,0,0,0,0,0,0-2). You can compute the net present value (NPV) of each stream and compare them (assuming that in A the first payment, 50, occurs at time t = 0, the next payment, 10, at t = 1 year, etc.) I get NPVs of
[tex] \text{NPV}(A) = 50 + \sum_{n=1}^{11} \frac{10}{1.12^n} - \frac{5}{1.12^{11}} = 107.940\; (\$000),[/tex]
and
[tex] \text{NPV}(B) = 40 - \frac{2}{1.12^{10}} = 39.356 \; (\$000). [/tex]

You could, of course, convert these to equivalent annual amounts, but why bother?

RGV

 

[RoshanBBQ]

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?

It works by converting the present money and future money into an equivalent annual rate, given i = 12% per term and there are 11 terms. You then sum up all of the annuities to find the total annual loss (which is positive here) for each option.