EUAC Analysis A 5-year-old lathe has been analyzed as a defender and found to have an economic life of 5 years. Its current market value is $5,000, and maintenance cost is $500 per year. Its salvage value at the end of economic life is estimated to be $500. The challenger is a modern CNC lathe

Question

EUAC Analysis

A 5-year-old lathe has been analyzed as a defender and found to have an economic life of 5 years. Its current market value is $5,000, and maintenance cost is $500 per year. Its salvage value at the end of economic life is estimated to be $500.

The challenger is a modern CNC lathe whose analysis indicated its economic life to be 5 years. The CNC lathe is capable of operating as a turning cell due to its automatic tool changer, enhanced integratability, and automatic palletizing. Its first cost is $55,000, while the annual maintenance cost is $400. It is expected to save $14,000 per year through higher productivity. Its estimated salvage value at the end of 5 years is $5,000. Should the defender be replaced, if MARR = 12%?

Accepted Answer

Note that the defender’s economic life of 5 years is the same as that of the challenger. Since no criterion of comparison has been specified, we can choose any of those discussed
in Part II of the text.

Let us use EUAC as the basis of comparison. Our task then is to determine the EUAC of the defender and compare it with that of the challenger. The one with the lower EUAC should win. For both, [latex]i[/latex] = MARR = 12%, and n = 5 years.

Defender ( lathe ) : P = current market value [latex]= \$5,000[/latex]

S [latex]= \$500[/latex]

AM = annual maintenance [latex]= \$500[/latex]

From the defender’s cashflow diagram in Figure 12.6[latex]^{20}[/latex]

[latex]EUAC_{defender} =500+5,000({A}/{P},12\%,5)-500({A}/{F},12\%,5)[/latex]

[latex]=500+5,000 imes 0.2774-500 imes 0.1574[/latex]

[latex]=500+1,387-79[/latex]

[latex]=\$1,808[/latex]

Alternatively, we can use Equation (12.1), from which

EUAC[latex]_n[/latex] = P(A/P,i,n)-S(A/F,i,n)

= P(A/P,i,n)-S{(A/F,i,n)}

= (P-S)(A/P,i,n) - iS (12.1)

[latex]EUAC_{defender}[/latex] = Capital recovery cost +Maintenance cost

[latex]=\left\{(P-S)({A}/{P},12\%,5)+iS ight\} +AM[/latex]

[latex]=(5000-500) imes 0.2774+0.12 imes 500+500[/latex]

[latex]=4,500 imes 0.2774+60+500[/latex]

[latex]=\$1,808[/latex]

Challenger (CNC lathe): P=[latex]\$55,000[/latex]

S=[latex]\$5,000[/latex]

AM= annual maintenance = [latex]\$400[/latex]

AB= annual benefit = [latex] \$14,000[/latex]

Again, we can follow either the first-principles approach based on cashflow diagram, or use Equation (12.1). From the equation, we get

[latex]EUAC_{challenger}[/latex] = Capital recovery cost + Maintenance - Benefit

[latex]=\left[(P-S)({A}/{P},12\%,5)+iS ight]+AM-AB[/latex]

[latex]=(55,000-5,000)(0.2774)+0.12 imes 5,000+400-14,000[/latex]

[latex]=50,000 imes 0.2774+600-13,600[/latex]

[latex]=\$870[/latex]

Since the challenger’s EUAC is lower, replace the defender with the CNC lathe.

[latex]^{20}[/latex]Since the two cashflows at period 5 cancel each other out, you may be tempted to evaluate

EUAC[latex]_{defender}[/latex] = 500+ 5,000(A/P, 12%, 4),

But this is incorrect since this EUAC[latex]_{defender}[/latex] covers only 4 years. What we need to determine is the value of EUAC[latex]_{defender}[/latex] spread over the 5-year useful life.

Question 12.5: EUAC Analysis A 5-year-old lathe has been analyzed as a defe...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions