### Capital Cost Comparison: Present Worth Analysis

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**Chester McEachern** - 2 months ago
- 15 comments

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.

in case A, why did you put 12 yrs ? isn't 4 yrs ?

isn't p/f take positive value ? why -22,000 (p/f,8%,8( and 4) ) ?

thanks bro

Thank you very much for the video , which software do you use?

for case B why for overhaul, just only at 3&9 year only?

where does the 1.08 come from in this equation

THANK YOU!

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)]

what does this interest rate mean?

Thank you Sir!

Very helpful, thank you

If lifetimes are different, isn’t it better to do an annual worth analysis?

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

How to input (P/A, (interest rate), years) in calculator?

I see it now hehehe