Capital Cost Comparison: Present Worth Analysis - YouTube

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.