Solved Questions

Present Value:

F(3) = P(1+i)= $1,000 (1+0.12)3 = $1,405

2. What is the present value of $1500 received 3 years from now if the interest rate is 10 percent.

P = F (1 + i)n

P = $1500 (1.1)3 = $1126

3. what is the present value of the salvage on a machine if the salvage price 10 years from is $10,000 and if the interest rate is 10 percent?

P = F (1 + i)n

P = $10000(1.10)10 = $3855

4. Small interior design firm has borrowed $65,000 from wells fargo bank at 12 percent interest per year. How much must that firm repay 2 years later?

This question asks about the future value of a borrowed money now.

F = P (1 + i)n

F = $65,000 (1 + 0.12)2

F = $81,536

19. The project steering committee is considering which project they should invest capital in one of three projects to choose from. Project A has a potential to be worth $200,000 in 5 years, Project B has a potential to be worth $350,000 in 7 years, and Project C has a potential to be worth $420,000 in 8 years. the interest rate is 10 percent. The project steering committee decided the present value selection criteria is to be considered for the selection. Which Project is to consider?

PV (A) = $200,000 / (1 + 0.08)5 = $136,117

PV (B) = $$350,000 / (1 + 0.08)7 = $204,222

PV C) = $420,000 / (1 +0.08)10 = 194,541

The Best choice is Project B

There are some more sophisticated examples and questions about the calculations of present value and future value such as calculating the present value of series payments made over n years. those kind of questions are very rare on the PMP Exam and you are not required to go deep in the subject of Engineering Economics and Analysis , which present value calculation is part of.

Depreciation

Straight line depreciation Formula is Initial investment - SalvageNumber of years

Depreciation = $5.600,000 - $900,00012 =$391,667

2. An investment of $12,000 in new machine expected to have a salvage value of $1500 after 6 years of the machine’s life. Find the Straight-line depreciation expense per year.

Depreciation = (I - S )/ n = ($13,500 - $1,500)/ 6 yr

= $2,000/ yr

3. Restaurants chain company has purchased a100 oven of $8,000 each. they are expected to have 10 years of productive life, then they can be sold for $500 each. what is depreciation of those ovens considering using the straight line depreciation method?

Depreciation = (I - S )/ n = 100 x ($8,000 - $500)10 yr

= $75000/ yr

5. A new Boiler burner will cost $35,000 installed and it is expected to have $5,000 salvage value at the end of an 8-years economic life. maintenance is expected to cost about $5,000 per year. What is deprecation of the Burner using the straight line method. ?

Depreciation = (I-S)/n

= ($35,000 - $5,000)/8

= $3750

Payback Period

A $25,000 machine will last 10 years and have $5000 salvage value. it is expected to generate $5,000 yearly income from which $1000 should be paid for taxes. Find the payoff period?

payback period = (Investment - Salvage) / (Income - Taxes)

= $(25,000 - 5,000) / $(5000 - 1000)

= 4 years.

Present Value:

- Find the Future value due at the end of 3 years if $1000 is borrowed at a rate of 12 percent.

F(3) = P(1+i)= $1,000 (1+0.12)3 = $1,405

2. What is the present value of $1500 received 3 years from now if the interest rate is 10 percent.

P = F (1 + i)n

P = $1500 (1.1)3 = $1126

3. what is the present value of the salvage on a machine if the salvage price 10 years from is $10,000 and if the interest rate is 10 percent?

P = F (1 + i)n

P = $10000(1.10)10 = $3855

4. Small interior design firm has borrowed $65,000 from wells fargo bank at 12 percent interest per year. How much must that firm repay 2 years later?

This question asks about the future value of a borrowed money now.

F = P (1 + i)n

F = $65,000 (1 + 0.12)2

F = $81,536

19. The project steering committee is considering which project they should invest capital in one of three projects to choose from. Project A has a potential to be worth $200,000 in 5 years, Project B has a potential to be worth $350,000 in 7 years, and Project C has a potential to be worth $420,000 in 8 years. the interest rate is 10 percent. The project steering committee decided the present value selection criteria is to be considered for the selection. Which Project is to consider?

PV (A) = $200,000 / (1 + 0.08)5 = $136,117

PV (B) = $$350,000 / (1 + 0.08)7 = $204,222

PV C) = $420,000 / (1 +0.08)10 = 194,541

The Best choice is Project B

There are some more sophisticated examples and questions about the calculations of present value and future value such as calculating the present value of series payments made over n years. those kind of questions are very rare on the PMP Exam and you are not required to go deep in the subject of Engineering Economics and Analysis , which present value calculation is part of.

Depreciation

- meat packing company has installed some new plant equipment at a cost of $5.6 million. The equipment is estimated to have a $900,000 salvage value after an estimated useful life of 12 years. if the straight line distribution method is to be used, how much depreciation per year should be deducted?

Straight line depreciation Formula is Initial investment - SalvageNumber of years

Depreciation = $5.600,000 - $900,00012 =$391,667

2. An investment of $12,000 in new machine expected to have a salvage value of $1500 after 6 years of the machine’s life. Find the Straight-line depreciation expense per year.

Depreciation = (I - S )/ n = ($13,500 - $1,500)/ 6 yr

= $2,000/ yr

3. Restaurants chain company has purchased a100 oven of $8,000 each. they are expected to have 10 years of productive life, then they can be sold for $500 each. what is depreciation of those ovens considering using the straight line depreciation method?

Depreciation = (I - S )/ n = 100 x ($8,000 - $500)10 yr

= $75000/ yr

5. A new Boiler burner will cost $35,000 installed and it is expected to have $5,000 salvage value at the end of an 8-years economic life. maintenance is expected to cost about $5,000 per year. What is deprecation of the Burner using the straight line method. ?

Depreciation = (I-S)/n

= ($35,000 - $5,000)/8

= $3750

Payback Period

A $25,000 machine will last 10 years and have $5000 salvage value. it is expected to generate $5,000 yearly income from which $1000 should be paid for taxes. Find the payoff period?

payback period = (Investment - Salvage) / (Income - Taxes)

= $(25,000 - 5,000) / $(5000 - 1000)

= 4 years.