Taking the machinery in Example 13–4 as a new asset, calculate the EUACt for each year from the marginal costs.

Question

Taking the machinery in Example 13–4 as a new asset, calculate the [latex]EUAC_{t}[/lateax] for each year from the marginal costs.

Accepted Answer

The EUAC of keeping this asset for each year of its useful life is worked out as follows.

EUAC if Kept Through Year t	Present Cost if Kept Through Year t (PWcos[latex]t_{t}[/latex])	New Asset Total Marginal Cost in Year t	Year, t
× (A/P, 15%, 1) =$17,750	[17, 750(P/F, 15%, 1)]	$17,750	1
× (A/P, 15%, 2) =16,560	P[latex]C_{1}[/latex] + 15, 200(P/F, 15%,2)	15,200	2
× (A/P, 15%, 3) =15,810	P[latex]C_{2}[/latex] + 13, 950(P/F, 15%,3)	13,950	3
× (A/P, 15%, 4) =15,520	P[latex]C_{3}[/latex] + 14, 350(P/F, 15%,4)	14,350	4
× (A/P, 15%, 5) =15,430 ←	P[latex]C_{4}[/latex] + 14, 900(P/F, 15%,5)	14,900	5
× (A/P, 15%, 6) =15,450	P[latex]C_{5}[/latex] + 15, 600(P/F, 15%,6)	15,600	6
× (A/P, 15%, 7) =15,580	P[latex]C_{6}[/latex] + 16, 950(P/F, 15%,7)	16,950	7

A minimum EUAC of $15,430 is attained for the new asset at Year 5, which is the asset’s economic or minimum cost life. As long as the marginal cost for another year is less than the EUAC, the EUAC continues to decline. When the marginal cost in year 6 exceeds the EUAC, then the EUAC begins to increase.

We can see that marginal costs increase in each subsequent year of ownership. As before the asset is replaced when its marginal cost exceeds the minimum EUAC of the best new alternative.

Unfortunately, the $13,250 marginal cost of keeping the 5-year old asset for one more year cannot be repeated. Even if used assets were available the costs to find one, install it, and remove it 1 year later would far exceed the $1000 loss in value for the existing asset.

Question 13.5: Taking the machinery in Example 13–4 as a new asset, calcula...

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