CVEN9701: Annual Equivalent - YouTube

Three designs are being considered

for a radio broadcast tower.

For each design, the capital cost, annual running cost

and salvage value are displayed here.

Determine the most economical design, assuming

a constant discount rate of 7%, to provide a facility

for an indefinite number of years.

So we've got three different designs we can use.

Project B is the design that costs the most up front.

But then in return, it's going to last the longest.

Project A costs a lot less to build in the first place,

but in 10 years, we'll have to rebuild it.

Project C has an in between cost and an in between life.

What we want to know is which one's the best.

So let's start by comparing the different projects.

Now, the important thing in this case

is the design lives are different.

Because they're different, we can't

use the net present value.

We need to use the annual equivalent.

So let's start and work out the annual equivalent

for Project A. First of all, we've got our capital cost.

Now this time, we need to change the capital cost using

our conversion factor, because we need to turn that

into a cost per year.

So it's going to be minus 20,000 times 0.07 times

1.07 to the power of 10, because it's

a 10 year project, all divided by 1.07 to the power of 10

minus 1.

Then we have to consider the annual cost.

So this is going to be minus $3,000.

However, this is already an annual cost, so that it's done.

Then we need to consider our salvage value.

So this time, we get the money.

So we're going to get $5,000 for selling our radio

tower as scrap.

So plus $5,000 times 0.7 divided by 1.07

to the power of 10 minus 1.

So this gives us minus $2,848 as our annual equivalent

for the capital cost, minus $3,000 for the operating

cost plus $362 for the salvage value,

which gives us a total of minus $5,486 per year.

OK, so that's Project A, which is the project that we

have to rebuild every 10 years.

So now, let's look at Project B. So the annual equivalent

for Project B is equal to--

well, this time our capital cost is $50,000,

so minus $50,000 times, we're using the same discount

rate, so 0.07 times 1.07.

But this time, the design life has changed.

This one can last for 30 years.

Divided by 1.07 to the power of 30 minus 1 minus,

the annual cost is lower this time, it's only $2,000.

That's already an annual amount, plus we have our salvage value,

which is $7,000, because this design is bigger and stronger,

so there will be more scrap metal

left at the end of the life of the tower.

OK, so this equals minus $4,029 minus $2,000

plus $74, which gives a total of minus $5,955.

So Project B as a cost per year costs more than Project A.

So we don't want to do Project B. Project A is better.

But we still have to consider Project C,

so let's look at it now.

Project C has a capital cost of $30,000,

so the annual equivalent for Project C

is equal to minus $30,000 times, same discount rate, 0.07,

times 1.07 to the power of, but this time the project life

is 20 years.

All divided by 1.07 to the power of 20 minus 1.

Our operating cost is the same, and it's already

an annual cost.

So it's minus $4,000.

And our salvage value, well, this is an in between project,

so we're going to have $6,000 worth of scrap metal at the end

of the life of the tower.

So this gives us minus $2,832 minus the $4,000 plus $146

as our present value of the scrap, which gives us

a grand total of minus $6,684.

So wow, that's even much more expensive than Project B.

So we're not going to be doing this one either,

and Project A ends up being our preferred option.

So we'll design it this way.