Engineering Economic Analysis
Engineering Economic Analysis
13th Edition
ISBN: 9780190296902
Author: Donald G. Newnan, Ted G. Eschenbach, Jerome P. Lavelle
Publisher: Oxford University Press
Question
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Chapter 15, Problem 22P
To determine

The projects to be funded and the opportunity cost of capital.

Expert Solution & Answer
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Answer to Problem 22P

Projects E,C,H,B,AandG should be funded.

The opportunity cost of the capital is 13.4%.

Explanation of Solution

Given:

Project First cost Annual benefits Life (years)
A $150,000 $44,290 5
B $200,000 $61,730 5
C $300,000 $98,780 5
D $250,000 $62610 6
E $400,000 $119,330 6
F $500,000 $115,500 6
G $350,000 $67940 10
H $600,000 $126920 10
I $750,000 $140580 10

The budget of the company is $2million.

Concept used:

Write the formula to calculate the present worth factor.

(PA,i,n)=(1+i)n1i(1+i)n

Here, the rate is i and the time period is n.

Calculation:

Calculate the Internal rate of return (IRR).

To determine the IRR, equate the present worth of project to zero.

PWofproject=(Firstcost)+Benefit(PA,i,n) ...... (I)

For project A.

Substitute 0 for PW, $150,000 for First cost, $44,290 for Benefit and 5 for n in Equation (I).

0=$150,000+$44,290((1+i)51i(1+i)5)3.387=((1+i)51i(1+i)5)i=15.12%

For project B.

Substitute 0 for PW, $200,000 for First cost, $61,730 for Benefit and 5 for n in Equation (I).

0=$200,000+$61,730((1+i)51i(1+i)5)3.24=((1+i)51i(1+i)5)i=16.10%

For project C.

Substitute 0 for PW, $300,000 for First cost, $98,780 for Benefit and 5 for n in Equation (I).

0=$300,000+$98,780((1+i)51i(1+i)5)3.04=((1+i)51i(1+i)5)i=19.04%

For project D.

Substitute 0 for PW, $250,000 for First cost, $62610 for Benefit and 6 for n in Equation (I).

0=$250,000+$62,610((1+i)61i(1+i)6)3.99=((1+i)61i(1+i)6)i=13.4%

For project E.

Substitute 0 for PW, $400,000 for First cost, $119,330 for Benefit and 6 for n in Equation (I).

0=$400,000+$119,330((1+i)61i(1+i)6)3.352=((1+i)61i(1+i)6)i=20%

For project F.

Substitute 0 for PW, $500,000 for First cost, $115,500 for Benefit and 6 for n in Equation (I).

0=$500,000+$115,500((1+i)61i(1+i)6)4.33=((1+i)61i(1+i)6)i=10.16%

For project G.

Substitute 0 for PW, $350,000 for First cost, $67940 for Benefit and 10 for n in Equation (I).

0=$350,000+$67940((1+i)101i(1+i)10)5.15=((1+i)101i(1+i)10)i=14.27%

For project H.

Substitute 0 for PW, $600,000 for First cost, $126920 for Benefit and 10 for n in Equation (I).

0=$600,000+$126920((1+i)101i(1+i)10)4.73=((1+i)101i(1+i)10)i=17.14%

For project I.

Substitute 0 for PW, $750,000 for First cost, $140580 for Benefit and 10 for n in Equation (I).

0=$750,000+$140580((1+i)101i(1+i)10)5.335=((1+i)101i(1+i)10)i=13.37%

Tabulate the results.

Project First cost IRR
A $150,000 15.12%
B $200,000 16.10%
C $300,000 19.04%
D $250,000 13.4%
E $400,000 20%
F $500,000 10.16%
G $350,000 14.27%
H $600,000 17.14%
I $750,000 13.37%

Projects E,C,H,B,AandG should be chosen as they have the highest IRR and a combined initial cost of $2million.

Thus, the opportunity cost of the capital is 13.4%.

Conclusion:

Projects E,C,H,B,AandG should be funded.

The opportunity cost of the capital is 13.4%.

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