Peculiaridades del hispanismo en la República Checa

When making present worth comparisons, we must always use the same time period in order to take into account the full benefits and costs of each alternative. If the lives of the alternatives are not the same, we can transform them to equal lives with one of the following two methods:

1. Repeat the service life of each alternative to arrive at a common time period for all alternatives. Here we assume that each alternative can be repeated with the same costs and benefits in the future—an assumption known as repeated lives. Usually we use the least common multiple of the lives of the various alternatives. Sometimes it is convenient to assume that the lives of the various alternatives are repeated indefinitely. Note that the assumption of repeated lives may not be valid where it is reasonable to expect

technological improvements.

2. Adopt a specified study period—a time period that is given for the analysis. To set an appropriate study period, a company will usually take into account the time of required service or the length of time it can be relatively certain of its forecasts. The study period method necessitates an additional assumption about salvage value whenever the life of one of the alternatives exceeds that of the given study period. Arriving at a reliable estimate of salvage value may be difficult sometimes.

Because they rest on different assumptions, the repeated lives and the study period methods can lead to different conclusions when applied to a particular project choice.

EXAMPLE 4.6 (MODIFICATION OF EXAMPLE4.3)

A mechanical engineer has decided to introduce automated materials-handling equipment for a production line. She must choose between two alternatives: building the equipment or buying the equipment off the shelf. Each alternative has a different service life and a different set of costs.

Alternative 1: Build custom automated materials-handling equipment.

First cost $15 000

Labour $3300 per year

Power $400 per year

Maintenance $2400 per year Taxes and insurance $300 per year

Service life 10 years

Alternative 2: Buy off-the-shelf standard automated materials-handling equipment.

First cost $25 000

Labour $1450 per year

Power $600 per year

Maintenance $3075 per year Taxes and insurance $500 per year

Service life 15 years

If the MARR is 9 percent, which alternative is better?

The present worth of the custom system over its 10-year life is PW(1)  15 000  (3300  400  2400  300)(P/A,9%,10)

 15 000  6400(6.4176)  15 000  PV(0.09,10,6400)

⬵ 56 073

The present worth of the off-the-shelf system over its 15-year life is PW(2)  25 000  (1450  600  3075  500)(P/A,9%,15)

 25 000  5625(8.0606)  25 000  PV(0.09,15,5625)

⬵ 70 341

The custom system has a lower cost for its 10-year life than the off-the-shelf system for its 15-year life, but it would be wrong to conclude from these calculations that the custom system should be preferred. The custom system yields benefits for only 10 years, whereas the off-the-shelf system lasts 15 years. It would be surprising if the cost of 15 years of benefits were not higher than the cost of 10 years of benefits. A fair compari-son of the costs can be made only if equal lives are compared.

Let us apply the repeated lives method. If each alternative is repeated enough times, there will be a point in time where their service lives are simultaneously completed. This will happen first at the time equal to the least common multiple of the service lives. The least common multiple of 10 years and 15 years is 30 years. Alternative 1 will be repeated twice (after 10 years and after 20 years), while alternative 2 will be repeated once (after 15 years) during the 30-year period. At the end of 30 years, both alternatives will be completed simultaneously. See Figure 4.3.

0 5 10 15 20 25 30

Alternative 2 Alternative 2

Alternative 1

Years Alternative 1 Alternative 1

Figure 4.3 Least Common Multiple of the Service Lives

With the same time period of 30 years for both alternatives, we can now compare present worths.

Alternative 1: Build custom automated materials-handling equipment and repeat twice.

PW(1)  15 000  15 000(P/F,9%,10)  15 000(P/F,9%,20)

 (3300  400  2400  300)(P/A,9%,30)

 15 000PV(0.09,10,15 000)PV(0.09,20,15 000)PV(0.09,30,6400)

 15 000  15 000(0.42241)  15 000(0.17843)  6400 (10.273)

⬵ 89 760

Alternative 2: Buy off-the-shelf standard automated materials-handling equipment and repeat once.

PW(2)  25 000  25 000(P/F,9%,15)  25 000

 (1450  600  3075  500)(P/A,9%,30) PV(0.09,15,25 000)

 25 000  25 000(0.27454)  5625(10.273) PV(0.09,30,5625)

⬵ 89 649

Using the repeated lives method, we find little difference between the alternatives.

An annual worth comparison can also be done over a period of time equal to the least common multiple of the service lives by multiplying each of these present worths by the capital recovery factor for 30 years.

AW(1)  89 760(A/P,9%,30)  PMT(0.09,30,89 760)

 89 760(0.09734)

⬵ 8737

AW(2)  89 649(A/P,9%,30)

 89 649(0.09734)

⬵ 8726

As we would expect, there is again little difference in the annual cost between the alternatives. However, there is a more convenient approach for an annual worth compari-son if it can be assumed that the alternatives are repeated indefinitely. Since the annual costs of an alternative remain the same no matter how many times it is repeated, it is not necessary to determine the least common multiple of the service lives. The annual worth of each alternative can be assessed for whatever time period is most convenient for each alternative.

Alternative 1: Build custom automated materials-handling equipment.

AW(1)  15 000(A/P,9%,10)  6400 PMT(0.09,10,15 000)  6400

 15 000(0.15582)  6400

⬵ 8737

Alternative 2: Buy off-the-shelf standard automated materials-handling equipment.

AW(2)  25 000(A/P,9%,15)  5625 PMT(0.09,15,25 000)  5625

 25 000(0.12406)  5625

⬵ 8726

If it cannot be assumed that the alternatives can be repeated to permit a calculation over the least common multiple of their service lives, then it is necessary to use the study period method.

Suppose that the given study period is 10 years because the engineer is uncertain about costs past that time. The service life of the off-the-shelf system (15 years) is greater than the study period (10 years). Therefore, we have to make an assumption about the salvage value of the off-the-shelf system after 10 years. Suppose the engineer judges that its salvage value will be $5000. We can now proceed with the comparison.

Alternative 1: Build custom automated materials-handling equipment (10-year study period).

PW(1)  15 000  (3300  400  2400  300)(P/A,9%,10)

 15 000  6400(6.4176)  15 000  PV(0.09,10,6400)

⬵ 56 073

Alternative 2: Buy off-the-shelf standard automated materials-handling equipment (10-year study period).

PW(2)  25 000  (1450  600  3075  500)(P/A,9%,10)

 5000(P/F,9%,10)

 25 000  5625(6.4176)  5000(0.42241)

⬵ 58 987

Using the study period method of comparison, alternative 1 has the smaller present worth of costs at $56 073 and is, therefore, preferred.

Note that here the study period method gives a different answer than the repeated lives method gives. The study period method is often sensitive to the chosen salvage value. A larger salvage value tends to make an alternative with a life longer than the study period more attractive, and a smaller value tends to make it less attractive.

In some instances, it may be difficult to arrive at a reliable estimate of salvage value.

Given the sensitivity of the study period method to the salvage value estimate, the analyst may be uncertain about the validity of the results. One way of circumventing this problem is to avoid estimating the salvage value at the outset. Instead we calculate what salvage value would make the alternatives equal in value. Then we decide whether the actual salvage value will be above or below the break-even value found. Applying this approach to our example, we set PW(1)  PW(2) so that

PW(1)  PW(2)

56 073  25 000  5625(6.4176)  S(0.42241) where S is the salvage value.

Solving for S, we find S  11 834. Is a reasonable estimate of the salvage value above or below $11 834? If it is above $11 834, then we conclude that the off-the-shelf system is the preferred choice. If it is below $11 834, then we conclude that the custom system is preferable. ________________________________________________________쏋

The study period can also be used for the annual worth method if the assumption of being able to indefinitely repeat the choice of alternatives is not justified.

EXAMPLE 4.7

Joan is renting a flat while on a one-year assignment in England. The flat does not have a refrigerator. She can rent one for a £100 deposit (returned in a year) and £15 per month (paid at the end of each month). Alternatively, she can buy a refrigerator for

£300, which she would sell in a year when she leaves. For how much would Joan have to be able to sell the refrigerator in one year when she leaves in order to be better off buying the refrigerator than renting one? Interest is at 6 percent nominal, compounded monthly.

Let S stand for the unknown salvage value (i.e., the amount Joan will be able to sell the refrigerator for in a year). We then equate the present worth of the rental alternative with the present worth of the purchase alternative for the one-year study period:

PW(rental)  PW(purchase)

100  15(P/A,0.5%,12)  100(P/F,0.5%,12)  300  S(P/F,0.5%,12)

100  15(11.616)  100(0.94192)  300  S(0.94192) S 127.35

If Joan can sell the used refrigerator for more than about £127 after one year’s use, she is better off buying it rather than renting one. ______________________________쏋