# Capital Cost Comparison: Capitalized Cost Analysis 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.