Engineering Economic Analysis

410 REPLACEMENTANALYSIS

-t.
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j Year,n
1
2
3
4
5

Lowest EUAC of the Defender

Ownership H

KeptnYears

$16,000

15,070

14,618

14,754

15,162

Challenger Minimum EUAC ='$15,430

= 16,000(P IF, 15%,I)(AI P,15%, 1)

= [16,000(P I F,15%, 1) + 14,000(P I F,15%,2)](AI P, 15%,2)

= [16,000(P IF, 15%, 1) + ... + 13,500(P IF, 15%,3)](AI P,15%,3)

= [16,000(PI F,15%,1) +... +15,300(PI F,15%,4)](AI P,15%,4)

= [16,000(P I F,15%, 1) +... + 17,5OO(PI F,15%,5)](AI P,15%,5)

The minimum EUAC of the defender for 3 years is $14,618, which is less than that of

the challenger. But this comparison alone is not sufficient to indicate that we should keep the

defender for 3 years and then replace it. In looking at a study period of 5 years, we really have six

options.

Option 1:

Option 2:

Option 3:

Option4:

Option 5:

Option 6:

:Implementthe challenger today (at $15,430/year).

Keep the defender for 1 year (at $16,000) and the challenger for 4 years

(at $15,430/year).

Keep the defender for 2 years (at $15,070/year) and the challenger for 3 years

(at $15,430/year).

Keep the defender for 3 more years (at $14,618/year) and the challenger for

2 years (at $15,430/year).

Keep the defender for 4 years (at $14,754/year)and the challenger for 1 year

(at $15,430).

Keep the defender for 5 years (at $15,162/year).

The EUAC of each of these options at 15% is:

One can see that 0I?tion5 produces the nrinimum.EUAC.Thus we should keep the defen~er

for 4 more years and then reevaluate the defender-'Challengerdecision. Notice that if we replace

;;; ~~the defenderno\V~witlf tfie"'C1ffi:neii@r'1>eFteplaZ'1fm~ni7lect~8n rutel:ll, w~ wlfrlid ifdr adYieft::

a minimum EUAC over this 5-year period. Notice also that the lowest cost life of 3 years for

the defender is NOT how-long we sJiouldkeep the defender. In Year 5, the defender's marginal 13 ~ - ='!II!CI! II:; =-- I::c 11:I11 II =' ,.

Option 1:

Option 2:

.Option 3:

Option 4:

Option 5:

Option 6:

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EUAC is $15,430.

EUAC is [16,000 +15,430(P I A,15%,4)](P I F,15%,I)(AI P,15%,5) = $15,578.

EUAC is[15,070(P lA, 15%,2) + 15,430(P lA, 15%,3)(P I F,15%, 2)] x

(AI P,15%,~) = $15,255.

EUAC is[14,618(P lA, 15%,3) + 15,430(P I A,15%,2)(P IF, 15%,3)] x

(AI P,15%,5)= $14,877.

EUAC is [14,754(P IA, 15%,4) +15,430(PI F, 15%,5)](AI P, 15%, 5)=$14,855.

EUAC is $15,162.

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