
(a)
Calculate the
(a)

Explanation of Solution
Installation cost is (IN) $4,970,000. Saving per year (A) is $1,300,000. Time period is 10.
Rate of return (i) can be calculated as follows:
Substitute the rate of return as 22% by trial and error method in the above Equation.
When the rate of return is substituted as 22%, the calculated value is greater than the 3.8231. Thus, the rate of return increases to 22.8%.
The calculated value is nearly equal to initial cost with rate of return 22.8%. Thus, it is confirmed that the rate of return is 22.8%.
The interest rate can be calculated through spreadsheet function that given below:
=RATE(10,1300000,-4970000)
The above functions gives the value of 22.8%.
(b)
Calculate the rate of return.
(b)

Explanation of Solution
Length of the G (G) is 113. Cost of G (PG) per unit is $72,000. Saving per year (A) is $1,100,000. Time period is 10.
Rate of return (i) can be calculated as follows:
Substitute the rate of return as 6% by trial and error method in the above Equation.
When the rate of return is substituted as 6%, the calculated value is less than the 7.3964. Thus, the rate of return decreases to 5.896%.
The calculated value is nearly equal to the value 7.3964t with rate of return 5.896%. Thus, it is confirmed that the rate of return is 5.896%.
The interest rate can be calculated through spreadsheet function that given below:
= RATE(10,1100000,-72000*113)
The above functions gives the value of 5.896%.
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Chapter 7 Solutions
Engineering Economy
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