## Replacement Example

We introduce some of the basic concepts involved in replacement decisions through an example.

### EXAMPLE 7.1

Sergio likes hiring engineering students to work in his landscaping business during the summer because they are such hard workers and have a lot of common sense. The students are always complaining about maintenance problems with the lawnmowers, which are subject to a lot of use and wear out fairly quickly. His routine has been to replace the machines every five years. Clarissa, one of the engineering student workers, has suggested that replacing them more often might make sense, since so much time is lost when there is a breakdown, in addition to the actual repair cost.

"I've checked the records, and have made some estimates that might help us figure out the best time to replace the machines," Clarissa reports. "Even' time there is a breakdown, the average cost to fix the machine is S60. In addition to that, there is an average loss of two hours of time at S20 per hour. As far as the number of repairs required goes, the pattern seems to be zero repairs the first season we use a new lawnmower. However, in the second season, you can expect about one repair, two repairs in the third season, four repairs in the fourth, and eight in the fifth season. I can see why you only keep each mower five years!"

"Given that the cost of a new lawnmower has averaged about S600 and that they decline in value at 40% per year, and assuming an interest rate of 5%, I think we have enough information to determine how often the machines should be replaced," Clarissa concludes authoritativelv.

How often should Sergio replace his lawnmowers? How much money will he save?

To keep things simple for now, let's assume that Sergio has to have lawns mowed every year, for an indefinite number of years into the future. If he keeps each lawnmower for, say, three years rather than five years, he will end up buying lawnmowers more

CHAPTER 7 Replacement Decisions

frequently, and his overall capital costs for owning the lawnmowers will increase. However, his repair costs should decrease at the same time since newer lawnmowers require fewer repairs. We want to find out which replacement period results in the lowest overall costs—this is the replacement period Sergio should adopt and is referred to as the economic life of the lawnmowers.

We could take any period of time as a study period to compare the various possible replacement patterns using a present worth approach. The least common multiple of the service lives method (see Chapter 3), suggests that we would need to use a 3 x 4 x 5 = 60-vear period if we were considering sendee lives between one and five vears. This is an awkward approach in this case, and even worse in situations where there are more possibilities to consider. It makes much more sense to use an annual worth approach. Furthermore, since we typically analyze the costs associated with owning and operating an asset, annual worth calculated in the context of a replacement study is commonly referred to as equivalent annual cost (EAC). In the balance of the chapter, we will therefore use EAC rather than annual worth. However, we should not lose sight of the fact that EAC computations are nothing more than the annual worth approach with a different name adopted for use in replacement studies.

Returning to Sergio's replacement problem, if we calculate the EAC for the possibility of a one-year, two-year, three-year (and so on) replacement period, the pattern that has the lowest EAC would indicate which is best for Sergio. It would be the best because he would be spending the least, keeping both the cost of purchase and the cost of repairs in mind.

This can be done in a spreadsheet, as illustrated in Table 7.1. The first column, "Replacement Period," lists the possible replacement periods being considered. In this case, Sergio is looking at periods ranging from one to five years. The second column lists the salvage value of a lawnmower at the end of the replacement period, in this case estimated using the declining-balance method with a rate of 40%. This is used to compute the entries of the third column. "EAC Capital Costs" is the annualized cost of purchasing and salvaging each lawnmower, assuming replacement periods ranging from one to five years. Using the capital recovery' formula (refer back to Close-Cp 3.1), this annualized cost is calculated as:

EAC(capital costs) = annualized capital costs for an TV-year replacement period

P = purchase price

S = salvage value of the asset at the end of TV years

 Replacement Salvage EAC Capital Annual EAC Repair EAC Period (Years) Value Costs Repair Costs Costs Total 1 S360.00 S270.00 S 0.00 S 0.00 S270.00 2 216.00 217.32 100.00 48.78 266.10 3 129.60 179.21 200.00 96.75 275.96 4 77.76 151.17 400.00 167.11 318.27 5 46.66 130.14 800.00 281.64 411.79

S can be calculated in turn as

BV„(n) = P(l-d)« For example, in the case of a three-year replacement period, the calculation is

EAC(capital costs) = [600 - 600(1 - 0.40)3](A/P,5%,3) + [600(1 - 0.40)3](0.05)

The "average" annual cost of repairs (under the heading "EAC Repair Costs"), assuming for simplicity that the cash flows occur at the end of the vear in each case, can be calculated as:

EAC(repairs) = [(60 + 40)(P/F,5%,2) + (60 + 40)(2)(P/F,5%,3)](a/P,5%,3)

= [100(0.90703) + 200(0.86584)](0.36721) = 96.75 The total EAC is then:

EAC(total) = EAC(capital costs) + EAC(repairs) = 179.22 + 96.75 = 275.96

Examining the EAC calculations in Table 7.1, it can be seen that the EAC is minimized when the replacement period is two years (and this is only slightly cheaper than replacing the lawnmowers every year). So this is saying that if Sergio keeps his lawnmowers for two years and then replaces them, his average annual costs over time will be minimized. We can also see how much money Clarissa's observation can save him, by subtracting the expected yearly costs from the estimate of what he is currently paying. This shows that on average Sergio saves S411.79 - S266.10 = S145.69 per lawnmower per year bv replacing his lawn-mowers on a two-year cycle over replacing them every five years.^

The situation illustrated in Example 7.1 is the simplest case possible in replacement studies. We have assumed that the physical asset to be replaced is identical to the one replacing it, and such a sequence continues indefinitely into the future. By asset, we mean any machine or resource of value to an enterprise. An existing physical asset is called the defender, because it is currently performing the value-generating activity. The potential replacement is called the challenger. However, it is not always the case that the challenger and the defender are the same. It is generally more common that the defender is outmoded or less adequate than the challenger, and also that new challengers can be expected in the future.

This gives rise to several cases to consider. Situations like Sergio's lawnmower problem are relatively uncommon—we live in a technological age where it is unlikely that a replacement for an asset will be identical to the asset it is replacing. Even lawnmowers improve in price, capability, or quality in the space of a few years. A second case, then, is that of a defender that is different from the challenger, with the assumption that the replacement will continue in a sequence of identical replacements indefinitely into the future. Finally, there is the case of a defender different from the challenger, which is itself different from another replacing it farther in the future. All three of these cases will be addressed in this chapter.

Before we look at these three cases in detail, let's look at why assets have to be replaced at all and how to incorporate various costs into the replacement decision. 