Engineering Economic Analysis

l'

What Is the Basic Comparison? 411

-.
co~t again expeeds t}lechall~nger's minimum EUAC. We emphasize that this analysis is valid
only 1?ecauseof thereplacement repeatabilityassumptions.These assumptionshold that after the
5-yeat;pepod, all altemativ~shave the exact same yearly cash flowof $15,430 (thus we can ignore
theIil)~and these identical cash flows contillue indefinitely.

Lr Example 13-5demonstratesthat when the marginalcosts of the defenderare not consistently

increasing from year to year, it is necessary to calculate the lowest EUAC of the defender.

In the next section we describe this calculation.

Lowest EUAC of the Defender

How long can a defender asset be kept operating? Anyone who has seen or heard old

machinery in operation, whether it is a 50-year-old automobileor 20-year-oldpiece of pro-

duction equipment, has realized that almost any machine can be kept operating indefinitely,

provided it receives proper maintenance and repair. However, even though one might be

able to keep a defender going indefinitely, the cost may prove excessive. So, rather than

asking what remainingoperating life,the defender may have, we really want to ask what

is theminimum cost lifeof the asset. The minimum cost life of the defender is defined as

the number of years of ownership in the future that results in a minimum EUAC for the

defender and the challenger combined, as in Example 13-5.

However, for decision-making now, the answer to a simpler calculation is enough. Is

the defender's lowest or minimum EUAC less than the challenger's minimum EUAC?

An l1-year-old piece of equipment is being considered for replacement. It can be sold for $2000

now,and it is believed that the same salvage value can also be obtained in future years. The current

maintenance cost is $500 per year and is expected to increase $100 per year in future years. If the

equipment is retained in service, compute the minimum EUAC, based on 10% interest.

SOLUTION

Here the salvage value is not expecied to decline from its present $2000. The annual cost of this

invested capital isSi =2000(0.10)=$200. The maintenance is represented by $500 + $100G.

A year-by-year computation of EUAC is as follows:

EUAC of Maintenance Total

[=500 +100(AIG, 10%, n)] EUAC

$500 $700

548 748

-.. --- :>~q- ------ ~- - - -=;::;I- ,9'4,--- -

638 8~8

:;; z ~8111' = II au z=~Jl

--------

il

IIjt'Ii

lIfill

!

II'

I

I

II:!,

I'

I II

"

=

Age of EUAC of Invested

Year,n Equipment (years) Capital (=Si)

(^111) $200
(^212200)
== ;;;;;:0.3= :: = :==;;:'1 ' ;:; ;;; : -II 3 - .- - -2()0- -_. -- -- - ---- --..
4 14 200
5 "" (^15) II: (^200) II = ill
.. = =' ;