Engineering Economic Analysis
Engineering Economic Analysis
13th Edition
ISBN: 9780190296902
Author: Donald G. Newnan, Ted G. Eschenbach, Jerome P. Lavelle
Publisher: Oxford University Press
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Chapter 5, Problem 69P
To determine

i.

Alternative to be selected if interest rate is 6%.

Expert Solution
Check Mark

Answer to Problem 69P

Alternative A is the correct answer.

Explanation of Solution

Given:

Interest Rate is 6%

A B C
Installed Cost $10,000 $15,000 $20,000
Annual Benefit 1,625 1,530 1,890
Useful Life(yrs) 10 20 20

Concept used:

RATE OF RETURN: It is a return which usually an investor gets on speculation. Investment can be done on shares, bonds etc. If rate of return is calculated for a fiscal year, then it is termed as annual rate of return.

Calculation:

Net Present worth analysis of Option- A

Formula used to calculate the present worth is given below:

PW=P+A(P/A,i,n)F(P/F,i,n) ........(1)

Here,

P is the initial cost of the alternative ($10000)

A is the uniform annual benefit ($1625)

F1is additional investment required in replacement ($10000) (in year- 10)

I is the interest rate (6%)

Substitute the values in above formula:

PW

=$10000+$1625(P/A,6%,20)$10000(P/F,6%,10)=$10000+$1625(11.470)$10000(0.5584)=$10000+$18638.75$5584=$3054.75

Thus, the present worth of the option A is $3054.75

Net present Worth Analysis of option B

Formula used to calculate the Present worth is given below:

PW=P+A(P/A,i,n)

Here,

P is the initial cost of the alternative ($15000)

A is the uniform annual benefit ($1530)

i is the interest rate (6%)

n is the time period (20 years)

Substitute the values in above formula

PW =

=$15000+$1530(P/A,6%,20)=$15000+$1530(11.470)=$15000+$17549.10=$2549.10

Thus, the present worth of the option B is $2549.10

Net Present Worth Analysis of option C

Formula used to calculate the present worth is given below:

PW=P+A(P/A,i,n)

Here,

P is the initial cost of the alternative ($20000)

A is the uniform annual benefit ($1890)

i is the interest rate (6%)

n is the time period (20 years)

Substitute the values in above formula

PW =

=$20000+$1890(P/A,6%,20)=$20000+$21678.30=$1678.30

Thus, the present worth of the analysis C is $1678.30.

Conclusion:

Using the present worth analysis, the best alternative is alternative A as the present worth of alternative A is highest among three options.

Thus alternative A is the correct answer.

To determine

ii.

Best alternative is selected.

Expert Solution
Check Mark

Answer to Problem 69P

Alternative A is selected.

Explanation of Solution

Given:

Interest Rate is 3%.

Concept used:

RATE OF RETURN: It is a return which usually an investor gets on speculation. Investment can be done on shares, bonds etc. If rate of return is calculated for a fiscal year, then it is termed as annual rate of return.

Calculation:

Net Present worth analysis of Option- A

Formula used to calculate the present worth is given below:

PW=P+A(P/A,i,n)F(P/F,i,n) ........(1)

Here,

P is the initial cost of the alternative ($10000)

A is the uniform annual benefit ($1625)

F1is additional investment required in replacement ($10000) (in year- 10)

I is the interest rate (6%)

Substitute the values in above formula:

PW

=$10000+$1625(P/A,6%,20)$10000(P/F,6%,10)=$10000+$1625(11.470)$10000(0.5584)=$10000+$18638.75$5584=$3054.75

Thus, the present worth of the option A is $3054.75

Net present Worth Analysis of option B

Formula used to calculate the Present worth is given below:

PW=P+A(P/A,i,n)

Here,

P is the initial cost of the alternative ($15000)

A is the uniform annual benefit ($1530)

i is the interest rate (6%)

n is the time period (20 years)

Substitute the values in above formula

PW =

=$15000+$1530(P/A,6%,20)=$15000+$1530(11.470)=$15000+$17549.10=$2549.10

Thus, the present worth of the option B is $2549.10

Net Present Worth Analysis of option C

Formula used to calculate the present worth is given below:

PW=P+A(P/A,i,n)

Here,

P is the initial cost of the alternative ($20000)

A is the uniform annual benefit ($1890)

i is the interest rate (6%)

n is the time period (20 years)

Substitute the values in above formula

PW =

=$20000+$1890(P/A,6%,20)=$20000+$21678.30=$1678.30

Thus, the present worth of the analysis C is $1678.30.

Conclusion:

Alternative A is selected.

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Chapter 5 Solutions

Engineering Economic Analysis

Ch. 5 - Prob. 8PCh. 5 - Prob. 9PCh. 5 - Prob. 10PCh. 5 - Prob. 11PCh. 5 - Prob. 12PCh. 5 - Prob. 13PCh. 5 - Prob. 14PCh. 5 - Prob. 15PCh. 5 - Prob. 16PCh. 5 - Prob. 17PCh. 5 - Prob. 18PCh. 5 - Prob. 19PCh. 5 - Prob. 20PCh. 5 - Prob. 21PCh. 5 - Prob. 22PCh. 5 - Prob. 23PCh. 5 - Prob. 24PCh. 5 - Prob. 25PCh. 5 - Prob. 26PCh. 5 - Prob. 27PCh. 5 - Prob. 28PCh. 5 - Prob. 29PCh. 5 - Prob. 30PCh. 5 - Prob. 31PCh. 5 - Prob. 32PCh. 5 - Prob. 33PCh. 5 - Prob. 34PCh. 5 - Prob. 35PCh. 5 - Prob. 36PCh. 5 - Prob. 37PCh. 5 - Prob. 38PCh. 5 - Prob. 39PCh. 5 - Prob. 40PCh. 5 - Prob. 41PCh. 5 - Prob. 42PCh. 5 - Prob. 43PCh. 5 - Prob. 44PCh. 5 - Prob. 45PCh. 5 - Prob. 46PCh. 5 - Prob. 47PCh. 5 - Prob. 48PCh. 5 - Prob. 49PCh. 5 - Prob. 50PCh. 5 - Prob. 51PCh. 5 - Prob. 52PCh. 5 - Prob. 53PCh. 5 - Prob. 54PCh. 5 - Prob. 55PCh. 5 - Prob. 56PCh. 5 - Prob. 57PCh. 5 - Prob. 58PCh. 5 - Prob. 59PCh. 5 - Prob. 60PCh. 5 - Prob. 61PCh. 5 - Prob. 62PCh. 5 - Prob. 63PCh. 5 - Prob. 64PCh. 5 - Prob. 65PCh. 5 - Prob. 66PCh. 5 - Prob. 67PCh. 5 - Prob. 68PCh. 5 - Prob. 69PCh. 5 - Prob. 70PCh. 5 - Prob. 71PCh. 5 - Prob. 72PCh. 5 - Prob. 73PCh. 5 - Prob. 74PCh. 5 - Prob. 75PCh. 5 - Prob. 76PCh. 5 - Prob. 77PCh. 5 - Prob. 78PCh. 5 - Prob. 79PCh. 5 - Prob. 80PCh. 5 - Prob. 81PCh. 5 - Prob. 82PCh. 5 - Prob. 83PCh. 5 - Prob. 84PCh. 5 - Prob. 85PCh. 5 - Prob. 86PCh. 5 - Prob. 87PCh. 5 - Prob. 88PCh. 5 - Prob. 89PCh. 5 - Prob. 90PCh. 5 - Prob. 91PCh. 5 - Prob. 92PCh. 5 - Prob. 93PCh. 5 - Prob. 94PCh. 5 - Prob. 95PCh. 5 - Prob. 96PCh. 5 - Prob. 97PCh. 5 - Prob. 98PCh. 5 - Prob. 99PCh. 5 - Prob. 100PCh. 5 - Prob. 101P
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