The following data pertain to a defender:

Purchase Price (3 yrs ago) $=\$50,000$

Current Market Value $=\$10,000$

Decrease in Market Value next 6 yrs ${\$750}/{year}$

Current Operating Cost $= \$7,500$

Operating Cost Increase by: ${\$1,000}/{year}$

Considering the system of Example 12.7 to be its challenger, and assuming 15% MARR, should the defender be replaced?

Question

The following data pertain to a defender:

Purchase Price (3 yrs ago) [latex]=\$50,000[/latex]

Current Market Value [latex]=\$10,000[/latex]

Decrease in Market Value next 6 yrs [latex]{\$750}/{year}[/latex]

Current Operating Cost [latex]= \$7,500[/latex]

Operating Cost Increase by: [latex]{\$1,000}/{year}[/latex]

Considering the system of Example 12.7 to be its challenger, and assuming 15% MARR, should the defender be replaced?

Accepted Answer

We first analyze the defender by determining its marginal costs following the procedure of Example 12.7. The results are summarized in the following table. With their marginal costs known, the defender and the challenger are compared later.

Year	Loss in Market Value	Interest Foregone	Operating Cost	Marginal Cost
1	$750	0.15 × $10,000 = $1,500	$ 7,500	$ 9,750
2	$750	0.15 × $ 9,250 = $1,388	$ 8,500	$10,638
3	$750	0.15 × $ 8,500 = $1,275	$ 9,500	$11,525
4	$750	0.15 × $ 7,750 = $1,163	$10,500	$12,413
5	$750	0.15 × $ 7,000 = $1,050	$11,500	$13,300
6	$750	0.15 × $ 6,250 = $ 938	$12,500	$14,188

Since the defender’s marginal cost increases during the 6-year analysis period, we compare its marginal cost, year-by-year, with the corresponding EUAC of the challenger to decide when to replace it.

However, what we have in the last column of the final table in Example 12.7 are the marginal costs, not EUACs, of the challenger. From these marginal cost data, we determine the EUAC corresponding to the year, and then compare it with the defender’s marginal cost, as follows. The defender is replaced when its marginal cost exceeds the challenger’s EUAC .

Year 1:

The analysis period for determining the challenger’s EUAC is 1 year. From Example 12.7, its marginal cost for year 1 is $13,500. Since there is no other cashflow, the EUAC equals the marginal cost. With the defender’s $9,750 marginal cost for year 1 (see the previous table) being less than the challenger’s $13,500 EUAC, the defender is not replaced.

Year 2:

The analysis period for determining the challenger’s EUAC is 2 years. The marginal costs of the challenger (Example 12.7) for years 1 and 2 are $13,500 and $12,350 respectively. These costs at periods 1 and 2 are converted (sketch a cashflow diagram, if helpful) to the EUAC by first determining their P-value and then converting the P-value into A-value as

[latex]EUAC_{2} =\left\{13,500({P}/{F},15\%,1)+12,350({P}/{F},15\%,2) ight\} ({A}/{P},15\%,2)[/latex]

[latex]=(13,500 imes 0.8696+12,350 imes 0.7561) imes 0.6151[/latex]

[latex]=(11,740+9,338) imes 0.6151[/latex]

[latex]=\$12,965[/latex]

Since the defender’s $10,638 marginal cost for year 2 is less than the above [latex]EUAC_{2}[/latex] of the challenger, the defender is not replaced.

Year 3:

The marginal costs of the challenger (Example 12.7) for years 1, 2, and 3 are $13,500, $12,350, and $11,350 respectively. These costs at periods 1, 2, and 3 are converted to their EUAC by first determining their P-value and then converting the P-value into A-value as

[latex]EUAC_{3} =\left\{13,500({P}/{F},15\%,1)+12,350({P}/{F},15\%,2)+11,350({P}/{F},15\%,3) ight\} ({A}/{P},15\%,3)[/latex]

[latex]=(13,500 imes 0.8696+12,350 imes 0.7561+11,350 imes 0.6575) imes 0.4380[/latex]

[latex]=(11,738+9,338+7,463) imes 0.4380[/latex]

[latex]=\$12,500[/latex]

Since the defender’s marginal cost for year 3 ($11,525 from the previous table) is less than the above [latex]EUAC_{3}[/latex] of the challenger ($12,500), the defender is not replaced.

The calculations for the other years are carried out the same way. A table is used to summarize the results and make the replacement decision, as follows.

Year	Defender’s Marginal Cost	Challenger’s EUAC	Decision to Replace
1	$ 9,750	$13,500	No
2	$10,638	$12,965	No
3	$11,525	$12,500	No
4	$12,413	$12,100	Yes

Since the defender’s marginal cost at year 4 ($12,413) exceeds [latex]EUAC_{4}[/latex] of the challenger ($12,100), the defender is replaced at year 4. Thus, the defender is kept operational for 4 years, at the end of which it is replaced by its challenger (of Example 12.7).

Question 12.8: The following data pertain to a defender:Purchase Price (3 y...

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