# Chapter 13: Replacement Analysis

Economic Life And Replacement Analysis

Video titled: Chapter 13: Replacement Analysis

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In this Excel tutorial, I'm going to be talking to you about Optimum economic life for an item. Under our information let's say that you purchased a refrigerator for \$16,000. It's going to cost you \$500 to sell. The salvage drops 30..the salvage value drops 30% in year 1, 20% in years 2 through 5 and 10% in a year after that. It's gonna cost you hundred dollars to repair in year 1 and year 2 and starting in year 3 it's going to increase by \$500 every year. The interest rate you're going to be evaluating is 6%. The information that you need to do in Optimum economic life is your cost, your salvage, your present worth of those items and then your equivalent uniform annual cost of those.

What you're looking for is over the life where was the minimum EUAC? And that is going to be your Optimum economic life. I've chosen to look at this for 15 years. I don't know if that'll be enough. Let's see. So first we're going to look at cost. In Year 0 we spent \$16,000 dollars to purchase our item. So thatâ€™s equals C2. In year 1 it costs us a hundred dollars to repair your \$200. And remember starting in year 3 it's going to increase by \$500 every year. Our salvage value for year 1, remember it dropped 30%, so this is going to equal C2, which is a cost of our fridge, times 1 minus C4 which was the 30%. Notice this stays In that for just year 1. Notice that in year 2 I had to change this to C5 because it changed to 20%. So year 2 through 5 is going to keep that 20%.

At year 6 something else is going to happen. So now to figure out our present worth of these items you're going to have to do a present worth of your Cost Plus a present worth of your Salvage. So let's take a look at the formula. Remember the cost of the fridge is already at your present worth and it's a negative value so you can see our formula we have a negative C2. I've locked it because that's always going to stay the same. Then I'm going to take the net present value of our cost. Notice we have C9 is our interest, then our values for cost are year z..is basically just year 1, which is the \$100. Notice I've locked C13 here but not after the colon because what I have..so when I drag the formula down the columns this will..this one will increment, this one won't.

Then I need to add in the present value of our salvage which is the interest rate of..of C9, we've only got it for one year at the moment and then it's minus D13 plus the \$500 It's going to cost to sell. And that's how we get the present worth there. Then we're simply going to find the equ..EUAC for that time period. Which is equal payment, C9 was our interest rate, then it's B3 for our number of years. We're currently just at year 1. And then E13 is our present worth. Now I locked the C9 so I can drag this formula down. So we've got that information. And that information the formula stayed pretty close for year 1 and 2 because this was a \$100. Notice here that our salvage did change but everything else going across should have hung out. Now what happens is we..we've established this formula so we can grab it all the way and I can take it down to year 5 because nothing has changed at that point.

Notice that my cost have incremented by the \$500. Notice that it's maintained my salvage. I've got my present worth and I've got my equivalent annual. Now what happens here is I'm just going to grab this once and I'm going to pull it down. Here I need to change the formula because my salvage has changed. I need to lock this to C6. Notice how it's switched down and then I should be okay for taking this and grabbing it and pulling it all the way down to year 15. Now what I'm looking for here is where is my minimum equivalent uniform annual cost? And it turns out it is right here. So the optimum economic life for this refrigerator is seven years.