r
What Is the Basic Comparison? 413
If Retired at Endof Year n
Years of
Remaining
Life, n
Estimated Salvage
Value (5) at
End of Year n
Estimated
Maintenance
Cost for Year
EUACof EUACof
Capital Recovery Maintenance
(P S) x(AI P,10%, n) +Si 100(AI G, 10%, n)Total "EUAC
$ 0
48
94
138
181
222
262
300
337
372
406
$l5,QQ
1262
1198
1177
1172
1111
1078
1062
1058+-
1060
1068
SOLUTION
A mWimumEUAC of $1,058is computed at Year 9 ,forthe:existing machiI.le.
Looking again at Examples 13-6and 13-7,we find that the two cases represent the same
machine being examined at different points in its life. (In Year 6 and later in Example 13-7
the costs match those of Example 13-6.)Example 13-7indicates that the 5-year-old machine
has a minimum EUAC at a life of 9 years.
For older equipment with a negligible or stable salvage value, it is likely that the
operating and maintenance costs are increasing. Under these circumstances, the useful life
at which EUAC is a minimum is 1 year.
It is necessary to calculate the defender's minimum EUAC when marginal costs are
not consistently increasing. Having made this calculation, we can now usereplacement
analysis technique2 to compare the defender against the challenger.
Replacement Analysis Technique 2:
Defender Marginal Cost Can Be Computed and Is Not Increasing
Using our second method of analyzing the defender asset against the best available chal-
lenger the basic comparison involvesthelowest EUAC of the defender and the EUAC of the
challenger at its minimum cost life.
We calculate the minimum EUAC of the defender and compare this directly against
the minimum EUAC of the challenger. Remember that thereplacement repeatability as-
sumptionsallow us to do this. In this comparison we then choose the alternative with !he
lowest EUAC. In Example 13-5 the comparison involved the defender with the lowest
EUAC= $14,618at a life of 3 years and the challenger with an economic life of 5 years
and EUAC=$15,430. Here we would recommend that the defender be retained for 4 more
years, as its marginal costs for later years will be increasing above the challenger's EUAC.
Then,replacement analysis technique 1would apply. Consider Example 13-8. ~
J
0 p=$5000
(^14000) .$ (^0) $1100 + 400
(^23500100) 864 + 350
(^33000200) 804 + 300
(^42500300) 789 + 250
(^52000400) 791 + 200
(^62000500) 689 + 200
7.1 2000 600 616 +- 200
(^82000700) 562 + 200
9 2000 800 521 + 200
10 2000 900 488 + 200
11 2000 1000 462 + 200