i. Present Worth of first Cost and revenue Expected Worth of Present worth.

Question

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Answer 1

Question

Chapter 10, Problem 36P

To determine

i.

Present Worth of first Cost and revenue

Expected Worth of Present worth.

Expert Solution

Answer to Problem 36P

Present worth of first cost is $95500.

Expected Worth of Present worth is -$19295.

Explanation of Solution

Given:

Useful Life = 10 year

Interest Rate = 12%.

Calculation:

First Costs	Probability	Net Revenues	Probability
$300,000	0.2	$70,000	0.3
$400,000	0.5	$90,000	0.5
$600,000	0.3	$100,000	0.2

PW = −C+A(P/A,i,n) = −$300000+$70000(P/A,0.12,10) = −$300000+$70000×5.650 = −$300000+$395500 = $95500

Where,

PW is the present value.

C is the initial cost of the option examined.

A is the annual amount that is consistent in the cash flow series.

P is the present worth value of the time series.

i is the interest rate.

n is the number of terms that the money is for.

Present worth for pessimistic is $95500.

First Costs	Probability	Net Revenues	Probability	Present Worth
$300,000	0.2	$70,000	0.3	$95,500
$400,000	0.5	$90,000	0.5	($108,500)
$600,000	0.3	$100,000	0.2	$35,000

Expected worth of present worth

Useful Life = 10 year

Interest Rate = 12%

EV of Present worth= PWP× P (joint Probability)+ PWM× P (joint Probability)+ PWO × P (joint Probability)

= $95500×0.06+$108500×0.25+$35000×0.06= −$19295.

Conclusion:

Expected Worth of Present worth is -$19295

Present worth of first cost is $95500.

To determine

ii.

Expected first costs, net revenues and present worth for the expected values.

Expert Solution

Answer to Problem 36P

Present worth of expected value is -$45900.

Explanation of Solution

Calculation:

PW = − Expected(first cost)+Expected (net revenue)A(P/A,i,n) = −$440000+$86000×5.650 = −$440000 +$485900 = −$45900

Expected first cost = $44000

Expected net Revenue

E (Net Revenue) = 0.3×$70000+0.5×$90000+0.2×$100000= $21000+$45000+$20000= $86000

Expected Net Worth = $86000

PW = − Expected(first cost)+Expected (net revenue)A(P/A,i,n) = −$440000+$86000×5.650 = −$440000 +$485900 = −$45900.

Conclusion:

Present worth of expected value is -$45900.

To determine

iii.

Answers in part a and b match or not.

Expert Solution

Answer to Problem 36P

Both values in both parts don’t match.

Explanation of Solution

Given:

Useful Life = 10 year

Interest Rate = 12%.

Concept used:

Both values in both parts don’t match. As in the first part, joint probability is used to get the expected worth of present worth for the first value and revenue whereas in second part expected first cost and and expected net revenue is calculated separately and after that present worth is calculated.

Conclusion:

Both values in both parts don’t match.

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Answer 2

Question

Chapter 10, Problem 36P

To determine

i.

Present Worth of first Cost and revenue

Expected Worth of Present worth.

Expert Solution

Answer to Problem 36P

Present worth of first cost is $95500.

Expected Worth of Present worth is -$19295.

Explanation of Solution

Given:

Useful Life = 10 year

Interest Rate = 12%.

Calculation:

First Costs	Probability	Net Revenues	Probability
$300,000	0.2	$70,000	0.3
$400,000	0.5	$90,000	0.5
$600,000	0.3	$100,000	0.2

PW = −C+A(P/A,i,n) = −$300000+$70000(P/A,0.12,10) = −$300000+$70000×5.650 = −$300000+$395500 = $95500

Where,

PW is the present value.

C is the initial cost of the option examined.

A is the annual amount that is consistent in the cash flow series.

P is the present worth value of the time series.

i is the interest rate.

n is the number of terms that the money is for.

Present worth for pessimistic is $95500.

First Costs	Probability	Net Revenues	Probability	Present Worth
$300,000	0.2	$70,000	0.3	$95,500
$400,000	0.5	$90,000	0.5	($108,500)
$600,000	0.3	$100,000	0.2	$35,000

Expected worth of present worth

Useful Life = 10 year

Interest Rate = 12%

EV of Present worth= PWP× P (joint Probability)+ PWM× P (joint Probability)+ PWO × P (joint Probability)

= $95500×0.06+$108500×0.25+$35000×0.06= −$19295.

Conclusion:

Expected Worth of Present worth is -$19295

Present worth of first cost is $95500.

To determine

ii.

Expected first costs, net revenues and present worth for the expected values.

Expert Solution

Answer to Problem 36P

Present worth of expected value is -$45900.

Explanation of Solution

Calculation:

PW = − Expected(first cost)+Expected (net revenue)A(P/A,i,n) = −$440000+$86000×5.650 = −$440000 +$485900 = −$45900

Expected first cost = $44000

Expected net Revenue

E (Net Revenue) = 0.3×$70000+0.5×$90000+0.2×$100000= $21000+$45000+$20000= $86000

Expected Net Worth = $86000

PW = − Expected(first cost)+Expected (net revenue)A(P/A,i,n) = −$440000+$86000×5.650 = −$440000 +$485900 = −$45900.

Conclusion:

Present worth of expected value is -$45900.

To determine

iii.

Answers in part a and b match or not.

Expert Solution

Answer to Problem 36P

Both values in both parts don’t match.

Explanation of Solution

Given:

Useful Life = 10 year

Interest Rate = 12%.

Concept used:

Both values in both parts don’t match. As in the first part, joint probability is used to get the expected worth of present worth for the first value and revenue whereas in second part expected first cost and and expected net revenue is calculated separately and after that present worth is calculated.

Conclusion:

Both values in both parts don’t match.

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Answer 3

Question

Chapter 10, Problem 36P

To determine

i.

Present Worth of first Cost and revenue

Expected Worth of Present worth.

Expert Solution

Answer to Problem 36P

Present worth of first cost is $95500.

Expected Worth of Present worth is -$19295.

Explanation of Solution

Given:

Useful Life = 10 year

Interest Rate = 12%.

Calculation:

First Costs	Probability	Net Revenues	Probability
$300,000	0.2	$70,000	0.3
$400,000	0.5	$90,000	0.5
$600,000	0.3	$100,000	0.2

PW = −C+A(P/A,i,n) = −$300000+$70000(P/A,0.12,10) = −$300000+$70000×5.650 = −$300000+$395500 = $95500

Where,

PW is the present value.

C is the initial cost of the option examined.

A is the annual amount that is consistent in the cash flow series.

P is the present worth value of the time series.

i is the interest rate.

n is the number of terms that the money is for.

Present worth for pessimistic is $95500.

First Costs	Probability	Net Revenues	Probability	Present Worth
$300,000	0.2	$70,000	0.3	$95,500
$400,000	0.5	$90,000	0.5	($108,500)
$600,000	0.3	$100,000	0.2	$35,000

Expected worth of present worth

Useful Life = 10 year

Interest Rate = 12%

EV of Present worth= PWP× P (joint Probability)+ PWM× P (joint Probability)+ PWO × P (joint Probability)

= $95500×0.06+$108500×0.25+$35000×0.06= −$19295.

Conclusion:

Expected Worth of Present worth is -$19295

Present worth of first cost is $95500.

To determine

ii.

Expected first costs, net revenues and present worth for the expected values.

Expert Solution

Answer to Problem 36P

Present worth of expected value is -$45900.

Explanation of Solution

Calculation:

PW = − Expected(first cost)+Expected (net revenue)A(P/A,i,n) = −$440000+$86000×5.650 = −$440000 +$485900 = −$45900

Expected first cost = $44000

Expected net Revenue

E (Net Revenue) = 0.3×$70000+0.5×$90000+0.2×$100000= $21000+$45000+$20000= $86000

Expected Net Worth = $86000

PW = − Expected(first cost)+Expected (net revenue)A(P/A,i,n) = −$440000+$86000×5.650 = −$440000 +$485900 = −$45900.

Conclusion:

Present worth of expected value is -$45900.

To determine

iii.

Answers in part a and b match or not.

Expert Solution

Answer to Problem 36P

Both values in both parts don’t match.

Explanation of Solution

Given:

Useful Life = 10 year

Interest Rate = 12%.

Concept used:

Both values in both parts don’t match. As in the first part, joint probability is used to get the expected worth of present worth for the first value and revenue whereas in second part expected first cost and and expected net revenue is calculated separately and after that present worth is calculated.

Conclusion:

Both values in both parts don’t match.

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Answer 4

Question

Chapter 10, Problem 36P

To determine

i.

Present Worth of first Cost and revenue

Expected Worth of Present worth.

Expert Solution

Answer to Problem 36P

Present worth of first cost is $95500.

Expected Worth of Present worth is -$19295.

Explanation of Solution

Given:

Useful Life = 10 year

Interest Rate = 12%.

Calculation:

First Costs	Probability	Net Revenues	Probability
$300,000	0.2	$70,000	0.3
$400,000	0.5	$90,000	0.5
$600,000	0.3	$100,000	0.2

PW = −C+A(P/A,i,n) = −$300000+$70000(P/A,0.12,10) = −$300000+$70000×5.650 = −$300000+$395500 = $95500

Where,

PW is the present value.

C is the initial cost of the option examined.

A is the annual amount that is consistent in the cash flow series.

P is the present worth value of the time series.

i is the interest rate.

n is the number of terms that the money is for.

Present worth for pessimistic is $95500.

First Costs	Probability	Net Revenues	Probability	Present Worth
$300,000	0.2	$70,000	0.3	$95,500
$400,000	0.5	$90,000	0.5	($108,500)
$600,000	0.3	$100,000	0.2	$35,000

Expected worth of present worth

Useful Life = 10 year

Interest Rate = 12%

EV of Present worth= PWP× P (joint Probability)+ PWM× P (joint Probability)+ PWO × P (joint Probability)

= $95500×0.06+$108500×0.25+$35000×0.06= −$19295.

Conclusion:

Expected Worth of Present worth is -$19295

Present worth of first cost is $95500.

To determine

ii.

Expected first costs, net revenues and present worth for the expected values.

Expert Solution

Answer to Problem 36P

Present worth of expected value is -$45900.

Explanation of Solution

Calculation:

PW = − Expected(first cost)+Expected (net revenue)A(P/A,i,n) = −$440000+$86000×5.650 = −$440000 +$485900 = −$45900

Expected first cost = $44000

Expected net Revenue

E (Net Revenue) = 0.3×$70000+0.5×$90000+0.2×$100000= $21000+$45000+$20000= $86000

Expected Net Worth = $86000

PW = − Expected(first cost)+Expected (net revenue)A(P/A,i,n) = −$440000+$86000×5.650 = −$440000 +$485900 = −$45900.

Conclusion:

Present worth of expected value is -$45900.

To determine

iii.

Answers in part a and b match or not.

Expert Solution

Answer to Problem 36P

Both values in both parts don’t match.

Explanation of Solution

Given:

Useful Life = 10 year

Interest Rate = 12%.

Concept used:

Both values in both parts don’t match. As in the first part, joint probability is used to get the expected worth of present worth for the first value and revenue whereas in second part expected first cost and and expected net revenue is calculated separately and after that present worth is calculated.

Conclusion:

Both values in both parts don’t match.

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Answer 5

Chapter 10, Problem 36P

To determine

i.

Present Worth of first Cost and revenue

Expected Worth of Present worth.

Expert Solution

Answer to Problem 36P

Present worth of first cost is $95500.

Expected Worth of Present worth is -$19295.

Explanation of Solution

Given:

Useful Life = 10 year

Interest Rate = 12%.

Calculation:

First Costs	Probability	Net Revenues	Probability
$300,000	0.2	$70,000	0.3
$400,000	0.5	$90,000	0.5
$600,000	0.3	$100,000	0.2

PW = −C+A(P/A,i,n) = −$300000+$70000(P/A,0.12,10) = −$300000+$70000×5.650 = −$300000+$395500 = $95500

Where,

PW is the present value.

C is the initial cost of the option examined.

A is the annual amount that is consistent in the cash flow series.

P is the present worth value of the time series.

i is the interest rate.

n is the number of terms that the money is for.

Present worth for pessimistic is $95500.

First Costs	Probability	Net Revenues	Probability	Present Worth
$300,000	0.2	$70,000	0.3	$95,500
$400,000	0.5	$90,000	0.5	($108,500)
$600,000	0.3	$100,000	0.2	$35,000

Expected worth of present worth

Useful Life = 10 year

Interest Rate = 12%

EV of Present worth= PWP× P (joint Probability)+ PWM× P (joint Probability)+ PWO × P (joint Probability)

= $95500×0.06+$108500×0.25+$35000×0.06= −$19295.

Conclusion:

Expected Worth of Present worth is -$19295

Present worth of first cost is $95500.

To determine

ii.

Expected first costs, net revenues and present worth for the expected values.

Expert Solution

Answer to Problem 36P

Present worth of expected value is -$45900.

Explanation of Solution

Calculation:

PW = − Expected(first cost)+Expected (net revenue)A(P/A,i,n) = −$440000+$86000×5.650 = −$440000 +$485900 = −$45900

Expected first cost = $44000

Expected net Revenue

E (Net Revenue) = 0.3×$70000+0.5×$90000+0.2×$100000= $21000+$45000+$20000= $86000

Expected Net Worth = $86000

PW = − Expected(first cost)+Expected (net revenue)A(P/A,i,n) = −$440000+$86000×5.650 = −$440000 +$485900 = −$45900.

Conclusion:

Present worth of expected value is -$45900.

To determine

iii.

Answers in part a and b match or not.

Expert Solution

Answer to Problem 36P

Both values in both parts don’t match.

Explanation of Solution

Given:

Useful Life = 10 year

Interest Rate = 12%.

Concept used:

Both values in both parts don’t match. As in the first part, joint probability is used to get the expected worth of present worth for the first value and revenue whereas in second part expected first cost and and expected net revenue is calculated separately and after that present worth is calculated.

Conclusion:

Both values in both parts don’t match.

Answer 6

Chapter 10, Problem 36P

To determine

i.

Present Worth of first Cost and revenue

Expected Worth of Present worth.

Expert Solution

Answer to Problem 36P

Present worth of first cost is $95500.

Expected Worth of Present worth is -$19295.

Explanation of Solution

Given:

Useful Life = 10 year

Interest Rate = 12%.

Calculation:

First Costs	Probability	Net Revenues	Probability
$300,000	0.2	$70,000	0.3
$400,000	0.5	$90,000	0.5
$600,000	0.3	$100,000	0.2

PW = −C+A(P/A,i,n) = −$300000+$70000(P/A,0.12,10) = −$300000+$70000×5.650 = −$300000+$395500 = $95500

Where,

PW is the present value.

C is the initial cost of the option examined.

A is the annual amount that is consistent in the cash flow series.

P is the present worth value of the time series.

i is the interest rate.

n is the number of terms that the money is for.

Present worth for pessimistic is $95500.

First Costs	Probability	Net Revenues	Probability	Present Worth
$300,000	0.2	$70,000	0.3	$95,500
$400,000	0.5	$90,000	0.5	($108,500)
$600,000	0.3	$100,000	0.2	$35,000

Expected worth of present worth

Useful Life = 10 year

Interest Rate = 12%

EV of Present worth= PWP× P (joint Probability)+ PWM× P (joint Probability)+ PWO × P (joint Probability)

= $95500×0.06+$108500×0.25+$35000×0.06= −$19295.

Conclusion:

Expected Worth of Present worth is -$19295

Present worth of first cost is $95500.

Answer 7

Chapter 10, Problem 36P

To determine

i.

Present Worth of first Cost and revenue

Expected Worth of Present worth.

Answer 8

Expert Solution

Answer to Problem 36P

Present worth of first cost is $95500.

Expected Worth of Present worth is -$19295.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

Useful Life = 10 year

Interest Rate = 12%.

Calculation:

First Costs	Probability	Net Revenues	Probability
$300,000	0.2	$70,000	0.3
$400,000	0.5	$90,000	0.5
$600,000	0.3	$100,000	0.2

PW = −C+A(P/A,i,n) = −$300000+$70000(P/A,0.12,10) = −$300000+$70000×5.650 = −$300000+$395500 = $95500

Where,

PW is the present value.

C is the initial cost of the option examined.

A is the annual amount that is consistent in the cash flow series.

P is the present worth value of the time series.

i is the interest rate.

n is the number of terms that the money is for.

Present worth for pessimistic is $95500.

First Costs	Probability	Net Revenues	Probability	Present Worth
$300,000	0.2	$70,000	0.3	$95,500
$400,000	0.5	$90,000	0.5	($108,500)
$600,000	0.3	$100,000	0.2	$35,000

Expected worth of present worth

Useful Life = 10 year

Interest Rate = 12%

EV of Present worth= PWP× P (joint Probability)+ PWM× P (joint Probability)+ PWO × P (joint Probability)

= $95500×0.06+$108500×0.25+$35000×0.06= −$19295.

Conclusion:

Expected Worth of Present worth is -$19295

Present worth of first cost is $95500.

Answer 13

To determine

ii.

Expected first costs, net revenues and present worth for the expected values.

Expert Solution

Answer to Problem 36P

Present worth of expected value is -$45900.

Explanation of Solution

Calculation:

PW = − Expected(first cost)+Expected (net revenue)A(P/A,i,n) = −$440000+$86000×5.650 = −$440000 +$485900 = −$45900

Expected first cost = $44000

Expected net Revenue

E (Net Revenue) = 0.3×$70000+0.5×$90000+0.2×$100000= $21000+$45000+$20000= $86000

Expected Net Worth = $86000

PW = − Expected(first cost)+Expected (net revenue)A(P/A,i,n) = −$440000+$86000×5.650 = −$440000 +$485900 = −$45900.

Conclusion:

Present worth of expected value is -$45900.

Answer 14

To determine

ii.

Expected first costs, net revenues and present worth for the expected values.

Answer 15

Expert Solution

Answer to Problem 36P

Present worth of expected value is -$45900.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

PW = − Expected(first cost)+Expected (net revenue)A(P/A,i,n) = −$440000+$86000×5.650 = −$440000 +$485900 = −$45900

Expected first cost = $44000

Expected net Revenue

E (Net Revenue) = 0.3×$70000+0.5×$90000+0.2×$100000= $21000+$45000+$20000= $86000

Expected Net Worth = $86000

PW = − Expected(first cost)+Expected (net revenue)A(P/A,i,n) = −$440000+$86000×5.650 = −$440000 +$485900 = −$45900.

Conclusion:

Present worth of expected value is -$45900.

Answer 20

To determine

iii.

Answers in part a and b match or not.

Expert Solution

Answer to Problem 36P

Both values in both parts don’t match.

Explanation of Solution

Given:

Useful Life = 10 year

Interest Rate = 12%.

Concept used:

Both values in both parts don’t match. As in the first part, joint probability is used to get the expected worth of present worth for the first value and revenue whereas in second part expected first cost and and expected net revenue is calculated separately and after that present worth is calculated.

Conclusion:

Both values in both parts don’t match.

Answer 21

To determine

iii.

Answers in part a and b match or not.

Answer 22

Expert Solution

Answer to Problem 36P

Both values in both parts don’t match.

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

Useful Life = 10 year

Interest Rate = 12%.

Concept used:

Both values in both parts don’t match. As in the first part, joint probability is used to get the expected worth of present worth for the first value and revenue whereas in second part expected first cost and and expected net revenue is calculated separately and after that present worth is calculated.

Conclusion:

Both values in both parts don’t match.