Question 10.1: A machine has an initial cost of $10 000. Being a special-pu......

The cost curve for this type of problem is the net EUAC, evaluated at the MARR, as a function of time.

From Section 7.4, we know that the net EUAC will be made up of two components: the capital recovery cost,

CR(j)=(\$10,000-\$500)(A/P,10\%,j)+(0.10)(\$500)          (1)

and the equivalent annualized operating cost,

A(j)=\$1400+\$600(A/G,10\%,j)+\$600(P/F,10\%,1)(A/P,10\%,j)       (2)

In deriving (I), the constancy of the salvage value was used; in deriving (2), the gradient series was extended backwards by writing the first year’s cost as $1400+ $600. In both expressions, j is the time, in years; thus, to keep the machine for 4 years would cost the company CR(4) + A(4) per year.
Substituting j = 1,2, . . . , 10 in (1) and (2), we generate Table 10-1; the points are plotted in Fig. 10-1. The data show that the machine should be retired at the end of 7 years.
The time interval for which the EUAC of an asset is smallest (7 years, for the machine of Example 10.1) is called the economic life of the asset. As we have seen, the economic life is given analytically as that value j* at which

is minimized.
The concept of economic life may also furnish the basis for a replacement decision, particularly when service lives are not precisely known, when salvage values in each year are not known, or when the equipment obsolesces rapidly.