We must decide whether existing (defender) equipment in an industrial plant should be replaced. A $4000 overhaul must be done now if the equipment is to be retained in service. Maintenance is $1800 in each of the next 2 years, after which it increases by $1000 each year. The defender has no present

Question

We must decide whether existing (defender) equipment in an industrial plant should be replaced. A $4000 overhaul must be done now if the equipment is to be retained in service. Maintenance is $1800 in each of the next 2 years, after which it increases by $1000 each year. The defender has no present or future salvage value. The equipment described in Example 13-1 is the challenger (EUAC =$4589). Should the defender be kept or replaced if the interest rate is 8%?

Accepted Answer

The first step is to determine the defender’s lowest EUAC. The pattern of overhaul and maintenance costs (Figure 13-4) suggests that if the overhaul is done, the equipment should be kept for several years. The computation is as follows:

Year, n	If Retired at End of Year n		Total EUAC
Year, n	EUAC of Overhaul $4000(A/P, 8%, n)	EUAC of Maintenance $1800+$1000 Gradient fromYear 3 on	Total EUAC
1	$4,320	$1,800	$6,120
2	2243	1800	4043
3	1552	1800 + 308*	3660←
4	1208	1800 + 683†	3691
5	1002	1800 + 1079	3881

[latex]{ }^{*} ext { For } t[/latex] the first 3 years, the maintenance is $1800, $1800, and $2800. Thus,

EUAC = 1800 + 1000(A/F, 8%, 3) = 1800 + 308.

[latex]{ }^{\dagger} EUAC =[/latex] 1800+1000(P/G, 8%, 3)(P/F, 8%, 1)(A/P, 8%, 4) = 1800+683.

The lowest EUAC of the overhauled defender is $3660. In Example 13-1, the challenger’s minimum cost life was 4 years with an EUAC of $4589. If we assume the equipment is needed for at least 4 years, the overhauled defender’s EUAC ($3660) is less than the challenger’s EUAC ($4589). Overhaul the defender.

If the defender’s and challenger’s cost data do not change, we can use replacement analysis technique 1 to determine when the overhauled defender should be replaced. We know from the minimum EUAC calculation that the defender should be kept at least 3 years. Is this the best life? The following table computes the marginal cost to answer this question.

Year, n	Overhaul Cost	Maintenance Cost	Marginal Cost to Extend Service
0	$4,000	$0
1	0	1800	$6120 = 4000(1.08) + 1800
2	0	1800	1800
3	0	2800	2800
4	0	3800	3800
5	0	4800	4800

Year 5 is the first year after Year 3, which has the overhauled defender’s lowest EUAC (in which the $4800 marginal cost exceeds the challenger’s $4589 minimum EUAC). Thus, the overhauled defender should be kept 4 more years if costs do not change. (Note that if the defender can be overhauled again after 3 or 4 years, that might be an even better choice.)

Question 13.7: We must decide whether existing (defender) equipment in an i...

Related Answered Questions

Verified Answer:

Refer to Example 13-2, where we calculated the before-tax marginal costs for a new piece of production machinery. Calculate the asset’s after-tax marginal costs considering this additional information. • Depreciation is by the straight-line method, with S = $0 and n = 5 years, so ...

Verified Answer:

Verified Answer:

Verified Answer:

A 5-year-old machine, whose current market value is $5000, is being analyzed to determine its minimum EUAC at a 10% interest rate. Salvage value and maintenance estimates and the corresponding marginal costs are given in the following table. ...

Verified Answer:

Let us look again at the defender and challenger assets in Example 13-4.This time let us arbitrarily change the defender’s marginal costs for its 5-year useful life. Now when, if at all, should the defender be replaced with the challenger? ...

Verified Answer:

Taking the machinery in Example 13-2 as the challenger and the machinery in Example 13-3 as the defender, use replacement analysis technique 1 to determine when, if at all, a replacement decision should be made. ...

Verified Answer:

Verified Answer:

A new piece of production machinery has the following costs. Investment cost =$25, 000 Annual operating and maintenance cost =$2000 in Year 1 and then increasing by $500 per year Annual cost for risk of breakdown =$5000 per year for 3 years, then increasing by $1500 per year Useful life =7 years ...

Verified Answer:

Verified Answer: