Question

The followng data for X1 and Y7 are in format [X7,Y]]. First cost in \$: [−50,000,−90,000], Aппual cost in \$: [−70,000,−4,000], Salvage value in $:[13,000,75,000], $ Life in years [3,6]. The alternatives shown are to be compared on the basis of a perpetual (i.e.. forever) equivalent anпual worh. At an interest rate of 10% per year, the equation that represents the perpetual AW of XI is: Select oпe: a. AWX1=−50,000(0.10)−10,000+13,000(0.10) b. AWX1=−50,000(0.10)−10,000+13,000(A ′ /F1 0% r​

3 ) c. AWX1=−50,000(0.10)−10,000−37,000(P/F,10%,3)(0.10)+ 13,000(0.70) d. AWX1=−50,000(A ′ (Pr 10% r )−10,000+13,000(A ′ /F 1 10% r
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Answer 1

The correct equation for the perpetual AW of X1 considering an interest rate of 10% includes the initial cost, annual cost, and conversion of the salvage value into an annuity over the 3-year lifespan of the asset.

Step-by-step explanation:

The question involves calculating the perpetual equivalent annual worth (AW) for alternative X1 given different financial parameters and considering the interest rate.

To find the perpetual AW of X1, we would typically calculate the present worth of the initial cost, the annual cost, and the salvage value, and then convert this to an annual worth using the perpetual annuity formula.

Given the options, the correct formula for the perpetual AW of X1, considering an interest rate of 10% would be:

AWX1 = -50,000(0.10) - 70,000 + 13,000(A/F, 10%, 3)

This formula accounts for the annual cost of -70,000 dollars, the initial cost of -50,000 dollars multiplied by the 10% interest rate, and the conversion of the salvage value into an annuity format over the 3-year life of the asset using the annuity factor (A/F).