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 16, Problem 44P
To determine

(a)

The capital expenditure to build the water project.

Expert Solution
Check Mark

Answer to Problem 44P

The capital expenditure to build the water project is, $879629.11.

Explanation of Solution

Given:

Decades of Focus
First Second Third Fourth Fifth
Annual benefits (In thousands of dollars) $60 $64 $66 $66 $70

Calculation:

Calculate the present worth of project.

Presentworth=[$60000(PA,0.05,10)+$64000(PA,0.05,10)(PF,0.05,10)+$66000(PA,0.05,20)(PF,0.05,20)+$70000(PA,0.05,10)(PF,0.05,40)] ...... (I)

Calculate the factor (PA,0.05,10).

(PA,0.05,10)=[(1+i)n1i(1+i)n] ....... (II)

Here, the rate of interest is I and the time period is n.

Substitute 5% for i, and 10 years for n in Equation (II).

(PA,0.05,10)=[(1+0.05)1010.05(1+0.05)10]=7.722

Calculate the factor (PA,0.05,20).

Substitute 5% for i, and 20 years for n in Equation (II).

(PA,0.05,10)=[(1+0.05)2010.05(1+0.05)20]=12.462

Calculate the factor (PF,0.05,10).

(PF,0.05,10)=1(1+i)n ...... (III)

Substitute 5% for i, and 10 years for n in Equation (III).

(PF,0.05,10)=1(1+0.05)10=0.6139

Calculate the factor (PF,0.05,20).

Substitute 5% for i, and 20 years for n in Equation (III).

(PF,0.05,20)=1(1+0.05)20=0.3769

Calculate the factor (PF,0.05,40).

Substitute 5% for i, and 40 years for n in Equation (III).

(PF,0.05,40)=1(1+0.05)40=0.1420

Substitute 7.722 for (PA,0.05,10), 12.462 for (PA,0.05,20), 0.6139 for (PF,0.05,10), 0.3769 for (PF,0.05,20), and 0.1420 for (PF,0.05,40) in Equation (I)

Presentworth=[$60000×7.722+$64000×7.722×0.6139+$66000×12.462×0.3769+$70000×7.722×01420]=$463200+$303394.29+$309997.24+$76756.68=$1153468

The annual cost are $15000.

Calculate the justified capital expenditure.

JCE=PWBenefitsA(PA,i,n) ...... (IV)

Calculate the value of A(PA,i,n).

A(PA,i,n)=A[(1+i)n1i(1+i)n] ...... (V)

Substitute 5% for i, 50 years for n, and $15000 for A in Equation (V).

A(PA,i,n)=$15000[(1+0.05)5010.05(1+0.05)50]=$15000×18.256=$273838.89

Substitute $1153468 for PWBenefits, and $273838.89 for A(PA,i,n) in equation (IV).

JCE=$1153468$273838.89=$879629.11.

Conclusion:

Therefore, the capital expenditure to build the water project is, $879629.11.

To determine

(b)

The capital expenditure to build the water project.

Expert Solution
Check Mark

Answer to Problem 44P

The capital expenditure to build the water project is, $558614.

Explanation of Solution

Calculate the present worth of project.

Presentworth=[$60000(PA,0.08,10)+$64000(PA,0.08,10)(PF,0.08,10)+$66000(PA,0.08,20)(PF,0.08,20)+$70000(PA,0.08,10)(PF,0.08,40)] ...... (VI)

Calculate the factor (PA,0.08,10).

(PA,0.08,10)=[(1+i)n1i(1+i)n] ....... (VII)

Substitute 8% for i, and 10 years for n in Equation (VII).

(PA,0.08,10)=[(1+0.08)1010.08(1+0.08)10]=6.710

Calculate the factor (PA,0.08,20).

Substitute 8% for i, and 20 years for n in Equation (VII).

(PA,0.08,10)=[(1+0.08)2010.08(1+0.08)20]=9.818

Calculate the factor (PF,0.08,10).

(PF,0.08,10)=1(1+i)n ...... (VIII)

Substitute 8% for i, and 10 years for n in Equation (VIII).

(PF,0.08,10)=1(1+0.08)10=0.4632

Calculate the factor (PF,0.08,20).

Substitute 8% for i, and 20 years for n in Equation (VIII).

(PF,0.08,20)=1(1+0.08)20=0.2145

Calculate the factor (PF,0.05,40).

Substitute 8% for i, and 40 years for n in Equation (VIII).

(PF,0.08,40)=1(1+0.08)40=0.0460

Substitute 6.710 for (PA,0.08,10), 9.818 for (PA,0.08,20), 0.4632 for (PF,0.08,10), 0.2145 for (PF,0.08,20), and 0.0460 for (PF,0.08,40) in Equation (VI)

Presentworth=[$60000×6.710+$64000×6.710×0.4632+$66000×9.818×0.2145+$70000×6.710×0.0460]=$742116.23

The annual cost are $15000.

Calculate the justified capital expenditure.

JCE=PWBenefitsA(PA,i,n) ...... (IX)

Calculate the value of A(PA,i,n).

A(PA,i,n)=A[(1+i)n1i(1+i)n] ...... (V)

Substitute 8% for i, 50 years for n, and $15000 for A in Equation (V).

A(PA,i,n)=$15000[(1+0.08)5010.08(1+0.08)50]=$15000×12.233=$183502.27

Substitute $742116.23 for PWBenefits, and $183502.27 for A(PA,i,n) in equation (IX).

JCE=$742116.23$183502.27=$558614.

Conclusion:

Therefore, the capital expenditure to build the water project is, $558614.

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