Capital Cost Comparison: Present Worth Analysis

In this screencast, we’re going to go through
comparing the costs of capital equipment, you know it’s so that you’ve already designed
a plant, and you want to know whether or not later down the road, when you need to replace
a piece of equipment or do something else to the process, how are you going to compare
two different pieces of equipment based on initial cost, installation cost, maintenance,
salvage at a certain time, how long the lifespan of each piece of equipment might be, and when
you do this, it’s really important that you do compare them on the same time period so
that you can make a logical comparison between the two, and when we do this, we want to compare
them using what’s known as present worth analysis. So we’re going to take any future costs or
yearly costs and put them into a present value that we can make the comparison between the
types of equipment that we’re looking at. We’re going to disregard inflation, but we
are going to account for an annual interest rate. Let’s take the following example to
demonstrate how we would compare two pieces of equipment using two different methods.
One of those methods is going to be a present worth analysis, and the other method is going
to be one of capitalized cost. So we have two reactors that are being considered for
purchase, and you can see the information regarding these two different reactors, reactor
A and reactor B. Now, they have different service lifetimes. for an effective annual
interest of 8% a year, we want to determine which reactor is a better purchase, and explain
why. Let’s look at what’s known as a present worth analysis. So we’re going to start with
a cash flow diagram. I’ll go step by step through reactor A and then show you what it
looks like for reactor B. At the beginning of our timeline, time 0, we have an initial
cost of installation of $25,000. So we draw an arrow in the negative direction and label
this as -$25,000. Now, every year after that, we’re going to have maintenance charges of
$2000. This is going to be for up to 4 years, since that’s the lifetime of reactor A. Since
we’re comparing it on an equal timescale, we need to find a common multiple between
the two reactors. So we could compare this on a 12 year scale. So we’re going to draw
$2000, that’s negative, every year for the first 4 years, so that’s year 1, and the end
of year 2, year 3, year 4. Since it’s uniform, we’re taking into consideration the maintenance
costs over the entire year, but we’re doing our calculation at the end of the year. There’s
no overhaul, but there is a salvage value. So at the end of year 4, we can get $3000
back from this piece of equipment. Also at the end of year 4, we would have to buy a
new reactor. So this would start our new 4 year cycle, and then for the final 4 years
it would look again pretty similar, except we wouldn’t be buying a new piece of equipment,
since this would be the end of our cycle. And this just gives us a pictorial representation
of our cash flow on a yearly basis. So we can do the same thing for case B. Now, you
can see our initial cost of installation, $15,000, is shown at time 0, and every year
after that it’s $4000 a year for maintenance. Now, we’re told it’s a 6 year life span, so
here’s our 6 year mark, so we’d have to buy another pump, and this would get us through
the 12 years. At the end of year 3, we would have an overhaul, so that’s moreso than just
our maintenance, so we would have both our maintenance since again that’s accrued over
the course of the year. So you can see there’s no salvage value for case B. So just looking
at these two cash flow diagrams, the fact that we have some kind of salvage value in
case A and our maintenance is $2000 compared to case B where our maintenance is $4000,
it appears case A would be the better choice, but this is why it’s important to use present
worth analysis. So we’re going to take all of our future costs that we’ve written out,
since we’re really located at time 0 making this decision, and we’re going to sum them
up, accounting for the time value of money, into our present day cost, so that we can
make a decision as to which one is the better investment. For case A, we’re going to write
the present worth cost right off that bat is our initial cost, so -$25,000. Now we’re
going to sum up our maintenance cost for the 12 years. So the way I like to set this up
is to determine which equation we’re going to use for the time value of money calculations
for this cost, so because this is an annual repeating uniform series cost and we want
the present worth, we’re going to use a uniform series present worth factor calculation, and
we’re going to take into consideration the rate that we’re doing this at, which is 8%
as stated in the problem, and we’re going to do this for 12 years. I’ve put this in
parentheses to show us what information we would use in this calculation, and one thing
I forgot to write in here is that it’s helpful to write in the charge that we’re multiplying
by this factor, and then we need to account for our other charges on years 4, 8 and 12.
So we can take $25,000 at year 4 and subtract out the $3000, and that gives us $-22,000,
so this is a future cost that we’re now making a present cost, so we’re going to use a present
given future factor, again 8% but now this is at year 4. We do the same thing for year
8, and using 8 years now in our equation, and then lastly we’re adding $3000 as a future
value that we get from our salvage of the equipment. Again, 8%, and this is at year
12. So this is what our equation setup looks like, and now we’re going to fill in the appropriate
equations with these values. So I’ve written out the equations that we can use for these
factors, where i is the periodic interest rate, and n is the number of periods. So this
is assuming that our interest is compounded yearly. So we can rewrite our equation above
using these equations, and it should look like the following. So when we calculate this
out, the present value cost of reactor A comes out to -$66,937. For a 12 year period, it
would cost us $67,000 right now to have reactor A. Following those rules for case B, we would
have our negative $15,000 as our upfront cost. We would then be subtracting out $15,000 at
year 6, so this is going to be a present-future factor, 8%, 6 years. We would also be subtracting
out our maintenance cost, so that would be $3500, present-future, 8%, year 3, as well
as the same thing for year 9. And then we have to account for our maintenance charges,
our annual uniform series, and that’s going to be our $4,000, present over our annual
cost at 8% interest for 12 years. And I get a present cost of reactor B as $59,126. So
to have reactor B for 12 years, we would need $59,000 up front as our present worth calculation.
So examining these two reactors, reactor A cost us $67,000, reactor B cost us $59,000,
it makes more sense to invest in reactor B based on our present worth analysis.

Author Since: Mar 11, 2019

  1. Weren't you supposed to account for three salvages in case A? i.e. PW_A=-25000-2000(P/A,8,12)-22000[(P/F,8,4)+(P/F,8,8)]+3000[(P/F,8,4)+(P/F,8,8)+(P/F,8,12)]

  2. Very elaborative explanation. Than you very much. You are better than my professor. I wish I could take this course online

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