Engineering Economy
Engineering Economy
8th Edition
ISBN: 9780073523439
Author: Leland T Blank Professor Emeritus, Anthony Tarquin
Publisher: McGraw-Hill Education
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Chapter 7, Problem 18P

(a)

To determine

Calculate the rate of return.

(a)

Expert Solution
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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:

S((1+i)n1i(1+i)n)=IN1,300,000((1+i)101i(1+i)10)=4,970,000((1+i)101i(1+i)10)=4,970,0001,300,000((1+i)101i(1+i)10)=3.8231

Substitute the rate of return as 22% by trial and error method in the above Equation.

((1+0.22)1010.22(1+0.22)10)=3.8231(7.30463110.22(7.304631))=3.8231(6.3046311.607019)=3.82313.9231>3.8231

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%.

((1+0.228)1010.228(1+0.228)10)=3.8231(7.79800810.228(7.798008))=3.8231(6.7980081.777946)=3.82313.82353.8231

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)

To determine

Calculate the rate of return.

(b)

Expert Solution
Check Mark

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:

S((1+i)n1i(1+i)n)=G×PG1,100,000((1+i)101i(1+i)10)=113×72,000((1+i)101i(1+i)10)=8,136,0001,100,000((1+i)101i(1+i)10)=7.3964

Substitute the rate of return as 6% by trial and error method in the above Equation.

((1+0.06)1010.06(1+0.06)10)=7.3964(1.79084810.06(1.790848))=7.3964(0.7908480.107451)=7.39647.3601<7.3964

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%.

((1+0.05896)1010.05896(1+0.05896)10)=7.3964(1.77335410.05896(1.773354))=7.3964(0.7733540.104557)=7.39647.3965<7.3964

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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