Golab Roofing is considering the purchase of a crane that would cost $69,846, would have a useful life of 6 years, and would have no salvage value. The use of the crane would result in labor savings of $21,000 per year. The internal rate of return on the investment in the crane is closest to (Ignore income taxes.): Refer to Exhibit 12B-1 and Exhibit 12B-2, to determine the appropriate discount factor(s) using the tables provided.

20%

Explanation:

Purchase of Crane cost = $69,846

Salvage value = 0

Labor savings = $21,000 per year

Let us assume Internal rate of return = X %. At IRR, Present value of Inflows = Present value of outflows.

69,846 = 21,000/1.0X + 21,000/ 1.0X^2 + 21,000/1.0X^3 + 21,000/1.0X^4 + 21,000/1.0X^5 + 21,000/1.0X^6.

69,846 = 21,000 [1/1.0X + 1/1.0X^2 + 1/1.0X^3 + 1/1.0X^4 + 1/1.0X^5 + 1/1.0X^6]

69,846/21,000 = [1/1.0X + 1/1.0X^2 + 1/1.0X^3 + 1/1.0X^4 + 1/1.0X^5 + 1/1.0X^6]

69,846 / 21,000 = 3.326

From the given table (as attached below), period (6 of 20%) Indicates 3.326. So, the Internal rate of return = 20%.

a.

1) the stages a new product goes through from beginning to end, i.e., the stages that a product passes through from introduction through growth, maturity, and decline. 2) the time from initial research and development to the time at which sales and support of the product to customers are withdrawn. 3) the period of time during which a product can be produced and marketed profitably.

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