In this screencast, we're going to look at comparing two different pieces of equipment,
and determining which is the better investment using an analysis known as capitalized costs.
Now, the idea of capitalized cost is that you create what's known as a perpetuity, and
we're going to say that a perpetuity is accounting for the fact that if you were to buy something,
like let's say you bought a car right now, and the other choice as to not buy a car.
It's not just the cost of buying the car, but it's the cost of replacing that car down
the road when it needs to be replaced and no longer has any value.
So you're accounting for the amount that you would need now to help you pay for that later
replacement cost.
So in analyzing the capital cost, we're going to take into consideration the amount it costs
to buy the piece of equipment, as well as this perpetuity that we would need to account
for our different alternatives.
So capital cost, which we'll designate as K, is going to be the initial cost of the
equipment, and we'll designate that CI, plus the present value of the perpetuity, which
we designate P, that is there for an infinite amount of replacements made every certain
amount of years.
So if we think of the replacement cost, CR, right now if I were to buy something new,
it would have an initial cost.
Later down the road, maybe if it's even a week later, I have to buy it again.
Maybe I could sell back the one I had and have some sort of salvage value associated
with it.
I would subtract that out from that initial cost, and that would be my replacement cost.
This happens all the time with cell phones, cars, so if we're trying to determine how
much money we need to cover the cost of the item plus the future replacement cost, we're
going to say that the future cost some years later, so we'll say ny standing for number
of years, is going to be equal to our replacement cost plus some amount of principle that we
need to put down that will grow in money to have enough to replace that item.
So not only are we replacing the item, and we have to pay for the cost of that, but we
need money to then account for the future replacement of the next item.
So if we go back to our investment factors, we can say some present worth, some principle
that we have now, we're going to multiply it by our compounding factor.
So this usually would be i, except that's assuming that it compounds on a yearly basis,
so we're going to use our nominal rate r, and divide it by the amount of times it's
compounded on a yearly basis, and we designate that m.
So then we take this to the exponent ny, for number of years, times the number of times
it's compounded per year, m.
So rearranging this, we can solve for our present value of the perpetuity as the following.
So as long as we know the nominal interest rate, the number of times it's compounded
per year, the amount of years we're looking at, and the replacement cost which again is
just the initial cost minus any salvage value, we could determine what our perpetuity is
going to be.
We could add this then to the initial cost of the equipment, and that gives us our capitalized
cost.
So let's go back to our example problem and use these tools to analyze the two pieces
of equipment.
So here are two pieces of equipment, reactor A and reactor B, and we have our initial cost,
where we do have end of year maintenance and annual costs that we need to account for.
So what's common to do with a capitalized cost analysis is to discount these annual
costs into present worth, and so we can use an annuity equation which we've seen before
to take this series of annual costs and bring it to a present value.
Now, we're also going to do that for our overhaul cost, since that's something that's in the
future that again we want to bring into a present value cost.
First we calculate the initial cost.
This is going to be the $25,000 that we see in the chart plus the present value of the
annuities, which is $2000 a year times the uniform series present worth factor, so this
is going to be 1.08^4, for 4 years, minus 1, and we divide this by 0.08, which again
is that rate of investment, times 1.08^4 and this brings our annuities into a present value.
You should get a value of $31,624.
So now our capitalized cost is going to be the $31,624 for our initial cost, plus the
second part of this equation, the replacement cost.
CR is $31,624 minus our salvage value, and we're going to divide this by 1 + r/m, since
we're using an annual compounding interest, this is just going to be 1.08 to the exponent
of 4 for 4 years, and then we subtract 1.
And this results in a capitalized cost of reactor A of roughly $111,000.
So before I get into what that actually means, let's do the same analysis for reactor B.
Our initial cost is $15,000 plus our maintenance cost of $4,000 per year, plus the overhaul
cost, and to bring that to a present value, we use the following compounding factor.
We do this at year 3.
Our initial cost of B comes out to roughly $36,000, so you can see that the initial cost
is more than reactor A. So now the capitalized cost for B is going to be the $36,270 plus
$36,270, since there's no salvage value we don't subtract anything, we divide this by
1.08^6, since it's a 6 year life span, and we solve and get $98,072.
So hopefully right away what you see is that reactor B has a lower capital cost.
So what does this mean?
First off, reactor B is the better investment because it has a lower capitalized cost.
Now, when we do a present worth analysis for comparing equipment purchases, we had to make
sure everything was on the same time scale.
In this case we don't, and the reason for that is what this value means, is that if
we had $98,000 right now to buy reactor B, then we would subtract out the cost of buying
B right at this moment.
The remaining amount of money could be invested at 8%, and that would allow us to create an
account that could constantly buy reactor B indefinitely every 6 years for as long as
we need to, because the interest we would be making on the difference would be enough
to buy that piece of equipment later down the road.
Now, this does not take into consideration inflation.
So, we would have to have more money to perpetually buy reactor A than we would reactor B, hence
why reactor B is the better investment.