Engineering Economy
Engineering Economy
8th Edition
ISBN: 9781259683312
Author: Blank
Publisher: MCG
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Chapter 7, Problem 22P
To determine

Calculate the rate of return.

Expert Solution & Answer
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Explanation of Solution

Additional investment (C) is -$950,000. Additional revenue in the 11th year (S) is $450,000 and additional revenue increases (SI) by $50,000 thereafter till 15th year. Time period (n) is 5 years and time period 1(n1) is 10 years.

Rate of return (i) for infinitive time period can be calculated as follows:

C=(S((1+i)n1i(1+i)n)+SI1i((1+i)n1i(1+i)nn(1+i)n))(1(1+i)n1)950,000=(450,000((1+i)51i(1+i)5)+50,000×1i((1+i)51i(1+i)55(1+i)5))(1(1+i)10)

Substitute the rate of return as 8% by trial and error method in the above Equation.

950,000=(450,000((1+0.08)510.08(1+0.08)5)+50,000×10.08((1+0.08)510.08(1+0.08)55(1+.0.08)5))(1(1+0.08)10)950,000=(450,000(1.46932810.08(1.469328))+50,000×12.5(1.46932810.08(1.469328)51.469328))(12.158925)950,000=(450,000(0.4693280.117546)+50,000×12.5(0.4693280.1175463.402916))(0.463193)950,000=(450,000(3.992718)+50,000×12.5(3.9927183.402916))(0.463193)950,000=(1,976,273.1+50,000×12.5(0.589802))(0.463193)950,000=(1,976,273.1+368,626.25)(0.463193)950,000<1,086,140.96

When the rate of return is substituted as 8%, the calculated value is greater than the additional investment. Thus, the rate of return increases to 8.45%.

950,000=(450,000((1+0.0845)510.0845(1+0.0845)5)+50,000×10.0845((1+0.0845)510.0845(1+0.0845)55(1+.0.0845)5))(1(1+0.0845)10)950,000=(450,000(1.50019510.0845(1.500195))+50,000×11.83432(1.50019510.0845(1.500195)51.500195))(12.250586)950,000=(450,000(0.5001950.126766)+50,000×11.83432(0.5001950.12676651.500195))(0.444329)950,000=(450,000(3.945814)+50,000×11.83432(3.9458143.3329))(0.444329)950,000=(1,775,616.3+50,000×11.83432(0.612914))(0.444329)950,000=(1,775,616.3+363,590.39)(0.444329)950,000950,511.57

The calculated value is nearly equal to the additional investment with rate of return 8.45%. Thus, it is confirmed that the rate of return is 8.45%.

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