ENGINEERING ECO ANALYSIS W/STUDY GUIDE
ENGINEERING ECO ANALYSIS W/STUDY GUIDE
13th Edition
ISBN: 9780190693053
Author: NEWNAN
Publisher: Oxford University Press
Question
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Chapter 10, Problem 38P
To determine

i.

Expected savings per year, expected life and rate of return for expected values.

Expert Solution
Check Mark

Answer to Problem 38P

Expected savings per year is $1980, expected life is 6 years and rate of return for the expected values is 12.18%.

Explanation of Solution

Given:

Cost of robot = $81000

Probability for optimistic annual savings of $22000 is 0.1

Probability for pessimistic annual savings of $18000 is 0.2

Probability for most likely annual savings of $20000 is 0.7

Probability for optimistic useful life of 4 year is 1/6

Probability for pessimistic useful life of 12 years is 2/3

Probability for most likely useful life of 5 year is 1/6.

Calculation:

Expected Savings (EAS):

EAS=(( Saving Optimistic ×SavingProbability Optimistic )+( Saving Pessimistic ×Savin Probability Pessimistic )+( Saving MostLikely×Saving Probability Mostlikely))

= ((22000×0.1)+(18000×0.2)+(20000×0.7))

=(2200+3600+14000)

=19800

Expected Saving - $19800

Expected Useful Life:

Jointprobability(( Useful life Optimistic ×Usefullife probability Optimistic )+( Useful life Pessimistic ×Usefullife Probability Pessimistic )+( Usefullife probability Mostlikely ×Usefullife probability Mostlikely ))

= ((4×16)+(12×16)+(5×23))

=((4×0.17)+(12×0.17)+(5×0.67))

=(0.68+2.04+3.35)

=6.07

Hence, expected useful life is 6 years

Rate of return (RR) for optimistic:

Present value can be calculated as follows:

Cost=JointProbabilityofsavings((1+Interest)JointprobabilityInterest×(1+Interest)JointProbability)

81000=19800(( 1+Interest)61Interest×(1+Interest)6)

8100019800=(( 1+Interest)61Interest×(1+Interest)6)

4.091=(( 1+Interest)61Interest×(1+Interest)6)

Interest rate = 12%

4.091=(( 1+0.12)610.12×(1+0.12)6)

4.091=(1.992910.1218×(1.9929))

4.091=4.091

As calculated value is equal to actual Value, thus rate of return is 12.18%.

Conclusion:

Expected savings per year is $1980, expected life is 6 years and rate of return for the expected values is 12.18%.

To determine

ii.

Expected rate of return for combination of savings per year and life.

Expert Solution
Check Mark

Answer to Problem 38P

Expected rate of return for combination of savings per year and life is 10.48%.

Explanation of Solution

Given:

Cost of robot = $81000

Probability for optimistic annual savings of $22000 is 0.1

Probability for pessimistic annual savings of $18000 is 0.2

Probability for most likely annual savings of $20000 is 0.7

Probability for optimistic useful life of 4 year is 1/6

Probability for pessimistic useful life of 12 years is 2/3

Probability for most likely useful life of 5 year is 1/6.

Calculation:

Expected rate of return for combination of savings per year and life:

Cost=(ExpectedSavings×Probability4years( (1+Interest) 4 1 Interest (1+Interest) 4 )+ExpectedSaving×Probability5years( (1+Interest) 5 1 Interest (1+Interest) 5 )+ExpectedSaving×Probability12years( (1+Interest) 12 1 Interest (1+Interest) 12 ))

81000=(19800×16( ( 1+Interest ) 4 1 Interest ( 1+Interest ) 4 )+19800×23( ( 1+Interest ) 5 1 Interest ( 1+Interest ) 5 )+19800×16( ( 1+Interest ) 12 1 Interest ( 1+Interest ) 12 ))

Interest rate = 11%

81000=(19800×0.1667( 1+0.11 ) 4 1 0.11 (1+0.11) 4 )+19800×0.6667( 1+0.11 ) 5 1 0.11 (1+0.11) 5 )+19800×0.1667( 1+0.11 ) 12 1 0.11 (1+0.11) 12 ))

81000=(3300.66( 1.51811 0.11(1.5181)+1320.66( 1.68511 0.11(1.6851))+3300.66( (3.4985)1 0.11(3.4985)))

81000=(3300.66( 0.5181 0.167)+1320.66( 0.6851 0.1854)+3300.66( (2.4985) 0.3848))

81000=(3300.66(3.1024)+1320.66(3.6959)+3300.66(6.4924))

81000=10240.12+48788.28+21429.068100080457.46

When interest rate is substituted as 11%, then calculated value is less than the actual value. Thus decrease interest rate further. If 10.84% is substituted, then actual value is equivalent to calculated value.

Conclusion:

Thus, 10.84% is rate of return for combination of savings per year.

To determine

iii.

Answers for part a and b match or not.

Expert Solution
Check Mark

Answer to Problem 38P

Answers donot match.

Explanation of Solution

Given:

Cost of robot = $81000

Probability for optimistic annual savings of $22000 is 0.1

Probability for pessimistic annual savings of $18000 is 0.2

Probability for most likely annual savings of $20000 is 0.7

Probability for optimistic useful life of 4 year is 1/6

Probability for pessimistic useful life of 12 years is 2/3

Probability for most likely useful life of 5 year is 1/6.

Concept used:

Since time period varies in part a and part b, rate of return for part a and part b are different.

Conclusion:

Answers donot match.

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