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 10, Problem 17P
To determine

(a)

The joint probability of saving and useful life.

Expert Solution
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Answer to Problem 17P

The joint probability of saving and useful life is shown in the following table.

Yearly saving Joint probability
$15000 0.18
$30000 0.30
$45000 0.12
$15000 0.12
$30000 0.20
$45000 0.08

Explanation of Solution

Given:

The first cost is $150000.

The salvage value is zero.

The interest rate is 20%.

Calculations:

The table for joint probability of saving and useful life is given below.

Yearly saving Probability of saving (A) Useful life Probability of useful life (B) Joint probability (A×B)
$15000 0.3 3 0.6 0.18
$30000 0.5 3 0.6 0.30
$45000 0.2 3 0.6 0.12
$15000 0.3 5 0.4 0.12
$30000 0.5 5 0.4 0.20
$45000 0.2 5 0.4 0.08

Table (1).

Conclusion:

The joint probability of saving and useful life is

Yearly saving Joint probability
$15000 0.18
$30000 0.30
$45000 0.12
$15000 0.12
$30000 0.20
$45000 0.08

Table (2).

To determine

(b)

The present worth of optimistic, most likely and pessimistic scenario.

Expert Solution
Check Mark

Answer to Problem 17P

The present worth for optimistic estimates is $15405.

The present worth for most likely estimates is $72330.

The present worth for optimistic estimates is $118440.

Explanation of Solution

Given:

The first cost is $150000 and zero salvage value. The interest rate is 20% and useful life is of 4 years.

Concept used:

The estimates which under any circumstance can give the best possible results are called as optimistic estimates. These estimates are estimated at the best future events.

The estimates which are most probable or likely to happen are called as most likely estimates. These estimates occur at normal circumstances.

The estimates which under any circumstance can give the worst possible results are called as pessimistic estimates. These estimates are estimated at the abnormal circumstances.

Write the expression for present worth.

PW=O+A[(1+i)n1i(1+i)n] ...... (I)

Here, the outflow is O, the annual inflow is A, the interest rate is i and useful life is n.

Calculations:

The optimistic estimates are saving of $45000 per year with project life of 5 years.

The most likely estimates are saving of $30000 per year with project life of 4 years.

The pessimistic estimates are saving of $15000 per year with project life of 3 years.

Calculate the present worth for optimistic estimates.

Substitute $150000 for O, $45000 for A, 20% for i and 5 for n.

PW=$150000+$45000[(1+0.2)510.2(1+0.2)5]=$150000+$45000[(1.2)510.2(1.2)5]=$150000+$45000(2.991)=$15405

The present worth for optimistic estimates is $15405.

Calculate the present worth for most likely estimates.

Substitute $150000 for O, $30000 for A, 20% for i and 4 for n.

PW=$150000+$30000[(1+0.2)410.2(1+0.2)4]=$150000+$30000[(1.2)410.2(1.2)4]=$150000+$30000(2.589)=$72330

The present worth for most likely estimates is $72330.

Calculate the present worth for optimistic estimates.

Substitute $150000 for O, $15000 for A, 20% for i and 3 for n.

PW=$150000+$15000[(1+0.2)310.2(1+0.2)3]=$150000+$15000[(1.2)310.2(1.2)3]=$150000+$15000(2.106)=$118440

The present worth for optimistic estimates is $118440.

Conclusion:

Thus, the present worth for optimistic estimates is $15405.

The present worth for most likely estimates is $72330.

The present worth for optimistic estimates is $118440.

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