Test Bank for Engineering Economic Analysis 14th Edition Newnan

Test Bank for Engineering Economic Analysis 14th Edition Newnan

Test Bank for Engineering Economic Analysis 14th Edition Newnan

Test Bank for Engineering Economic Analysis 14th Edition Newnan

CHAPTER 1 MAKING ECONOMIC DECISIONS Short Answer 1.1. Jim Watts, a senior engineer at a major global company, has been working in a foreign country all of 2016. He was paid in local currency, FC. One FC = 0.3$. The currency fluctuation in 2016 for each of the four quarters was 6.2, 6.9, 4.2, and 3.8. This contracted salary for the year was that he would be paid 75,000 in FC per quarter. What was his earnings in $ for 2016?

Answer: 31,507.50 Feedback: Earnings in local currency ($) = [(75,000)*(1.062) + (75,000)*(1.059) + (75,000)*(1.042) +(75,000)*(1.038)]*[0.1] = 31,507.50

Reference: Case Study 1 Ross Manufacturing in Atlanta is considering outsourcing a component from an offshore company. The company has collected the following data to make an economic decision.

Cost to make it in-house =T $E9S6TpB erAuNnK it.SELLER.COM Annual Demand = 500,000 units Price quoted by the Asian company = $90 per unit plus a shipping cost of $20,000 per shipment.

Short Answer 1.2A. Determine the number of shipments per year that the company can require the supplier to make in order to justify outsourcing.

Answer: 150 Refer to: Case Study 1 Feedback: Annual cost of making in house = 500,000(96) = $48,000,000 Let “X” be the number of deliveries that can be required for breakeven.

90 (500,000) + 20,000 X =48,000,000 X = {48,000,000 -90 (500,000)}/20,000 = 150.

1.2B. In addition to cost consideration, what other factors would weigh in decision making?

Answer: Reputation of overseas vendor, quality of parts, political climate of the country in which the vendor is from, labor regulations.

Refer to: Case Study 1

Test Bank for Engineering Economic Analysis 14th Edition Newnan

1.3. In decision making related to engineering projects, based on economics, the most important non-economic factor is: A) Operating cost B) Benefits C) Morale D) Taxes Answer: C

1.4. Decision making in engineering economy is concerned with choosing the best A) alternative with the longest life.

B) alternative with the smallest cost.

C) alternative with the largest annual benefit D) alternative that is the most cost-effective Answer: D

True/False 1.5. In an economic decision making, when the inputs and outputs are fixed, the criterion to use is minimize the input.

Answer: False Feedback: The criterion should be to maximize the difference between the input and output rather than maximizing the output.

1.6. Most engineering projects that have economic consequences have to be justified using economic decision making methods.

Answer: True Feedback: All economic cost and benefits have to be considered for justifying a project.

.

1.7. An example for fixed input in economic decision making is that the project manager has a budget of one million dollars for this project.

Answer: True Feedback: Since one million dollars is the input, it is fixed.

1.8. In engineering economic cost is a decision making tangible factor.

Answer: True

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CHAPTER 2 ESTIMATING ENGINEERING COSTS AND BENEFITS Reference: Table 1 Tech Engineering in TN is making a product for the overseas market.

The following cost data for the product has been compiled.

Item Cost Selling price $167 Materials and purchased parts $25/unit Direct Labor 2 hrs at $20 per hour Fixed Cost $1,400,000 2.1. If the overhead expenses are charged at 80 % of labor cost, determine the manufacturing cost per unit.

A) $72 B) $97 C) $65 D) None of these Answer: B Refer to: Table 1 Feedback: Total manufacturing cost = Materials cost + Direct labor cost + overhead cost = $25 + 2×20 + 0.80×2×20 = $97

2.2. The breakeven volume for this product is A) 14,433 B) 8,383 C) 20,000 D) None of these Answer: C Refer to: Table 1 Feedback: Total revenue = Total cost = Fixed cost + variable cost 167X = 1,400,000 + 97X X = 1,400,000 /(167-97) = 20,000 Units.

.

2.3. What is the profit per unit if 30,000 units are sold?

A) $23.33 B) $20.81 C) $24.35 D) None of these Answer: A Refer to: Table 1 Feedback: Total profit = Total revenue - Total cost = 30,000 × 167 – (1,400,000+ 97×30,000) = $700,000 Profit per unit = 700,000/30,000 = $23.33

2.4. To reduce the breakeven volume to 15,000 units, what should be the selling price?

A) $210.33 B) $190.33 C) $241.35 D) None of these Answer: B Refer to: Table 1 Feedback: Total revenue = Total cost = Fixed cost + variable cost Y × 15,000 = 1,400,000 + 97 × 15,000 Y = (1,400,000 + 97×15,000) /15,000 = $190.33

2.5. A 2000-gallon metal tank to store hazardous materials was bought 15 years ago at cost of $100,000. What will a 5,000-gallon tank cost today if the power–sizing exponent is 0.57 and the construction cost index for such facilities has increased from 180 to 600 over the last 15 years?

Choose the closest value.

A) $337,175 B) $666,667 C) $561,960 D) None of these.

Answer: C Feedback: Cost of a 5000-gallon tank without cost index = 100,000 (5,000/2,000)0.57 = $168,588 Cost of a 5000-gallon tank today with the cost index = (600/180)168,588 = $561,960

Reference: Case Study 2 A product that is very labor intensive assembled at Boyds Aero Structure in Memphis has an average labor cost of $20/hr. Overhead expenses are charged at 100% of labor at this company.

Time for the very first unit = 10 hours Time for the fourth unit = 8.1 hours.

2.6. The learning curve percentage for this operation is

.

A) 90% B) 85% C) 95% D) 75% Answer: A Solution:

Refer to: Case Study 2

y_4= 10(4)^n 8.1= 10(4)^n 4^n= 8.1/10=.81 n = -.0.1521 learning curve % = 90%

2.7. What is the cumulative time for 100 units to be made?

A) 869.12 hrs B) 581.33 hrs C) 761.84 hrs D) 621.68 hrs Answer: B Refer to: Case Study 2 Solution: Time for the first unit = 10 hrs Cumulative time for 100 units = {10 × [(100+0.5)(1-0.1521 - (0.5)(1-0.1521]/ (1-0.1521)} = 581.33 hrs.

2.8 What is the labor cost and overhead cost for the 500–th unit?

A) $162.30 B) $155.44 C) $139.88 D) $175.23 Answer: B Refer to: Case Study 2 Feedback: Time for the 500 units = 10 (500)-0.1521 = 3.8858 hrs Labor cost for the 500-th unit = 3.8858 (20) = $77.72 Labor and overhead costs = $77.72+77.72 = $155.44.88

True/False

2.9. The learning curve percentagTeEisS8T0B %AiN f tKheS lEe L a rL niEngR . c uCrvOeMexponent is given as -0.3214.

Answer: True Feedback: -0.3214 = log (learning curve expressed as a decimal) / log 2.0 Learning curve expressed as a decimal = 10^((log 2.0) * (-0.3214)) =0.80 or 80%

2.10. The breakeven volume is the quantity for which the unit cost is minimized.

Answer: False Feedback: The breakeven volume is the quantity for which the total revenue is exactly equal to the total cost, both fixed cost plus variable cost.

2.11. Sunk costs must be ignored in engineering economic decision making as sunk costs are money already spent and do not have any consequence on decision making.

Answer: True Solution: Costs incurred in the past do not make a difference.

2.12. An opportunity cost is associated with using a resource in one activity instead of another.

Answer: True Feedback: The amount of resources used in one activity cannot be used in another activity.

2.13. Life cycle costs (LCC) are costs incurred on a product from cradle to grave, (i.e) concept generation to retirement.

Answer: True Feedback: The cost increased from concept generation to retirement of a product is life cycle cost.

CHAPTER 3 INTEREST AND EQUIVALENCE True/False 3.1. Pam, when she turned 25, made an investment of $20,000 at an interest rate of 6.5% compounded semi-annually. Now that she is 50 years old, how much is the investment worth now.

A) $53,250 B) $44,491.96 C) $32,500 D) $98,976.71 Answer: D Feedback: Since the number of semi-annual periods in 25 years is 50, F can be determined as below.

F = P (F/P, 3.25%, 50) = 20,000 (1+ 0.0325) ^50 = $98,976.71

TESTBANKSELLER.COM Multiple Choice 3.2. Sam, a friend of Pam, made a similar investment of $20,000 at a much later date when he turned 35. Now that he is also 50, what is his investment worth if his investment also an earned an interest rate of 6.5% compounded semi-annually.

A) $52,207.37 B) $44,491.96 C) $32,500 D) None of these Answer: A Feedback: Since the number of semi-annual periods in 15 years is 30, F can be determined as below.

F = P (F/P, 3.25%, 30) = 20,000 (1 + 0.0325) ^30 = $52,207.37

3.3. Don Krump wants to triple his investment in 6 years. An investment firm offers him an

attractive interest rate. If the interest is compounded monthly, determine the nominal interest for this investment.

A) 20.09% B) 15.76% C) 18.45% D) 16.67% Answer: C Feedback: 3P = P (1 + i) ^ 72 (1 + i) = 1.015375509 i = 0.015375509 per month Nominal interest per year = 0.015375509 × 12 = 0.1845 or 18.45%

3.4. For the cash flows shown table below, evaluate the unknown value, X for an interest rate of 6% compounded annually.

Year 0 1 2 6 Cash Flow in $ 20,000 -5,000 -10,000 X A) $5,000 B) -$9,054 C) $6,377 D) $8,252 Answer: B Feedback: 20,000= -X (P/F, 6 %, 6) + 5,000 (P/F, 6%, 1) + 10,000 (F/P, 6%, 2) 20,000= -X (0.7050) + 5,000 (0.9434) + 10,000 (0.8900) X = -$9,053.91 = -$9,054

3.5. Tom Watson promises to deposit a sum of $100,000 for his granddaughter’s college education 18 years from now. If he invests $20,000 today, what should be return on investment he should get to be able keep up the promise? Assume interest is compounded yearly.

A) 9.35% B) 8.52% C) 10.04% D) 7.65% Answer: A

Feedback: 100,000 = 20,000(1 + i)18 (1 + i) = (100,000/20,000)1/18 = 1.09353 i = 0.0935 = 9.35%

3.6. Jane and Joe made two investments of $25,000 and $40,000 with different investors that yielded a combined rate of return of 10% compounded for 6 years. If the rate return on the first investment was 9%, what is the rate of return the couple obtained on the second investment?

A) 13.86% B) 11.18% C) 10.61% D) 10.00% Answer: C Feedback: The future worth of both the investments at the end of 6 years = 65,000(F/P, 10%, 6) = 65,000(1.772) = $115,180 115,180 = 25,000 (F/P, 9%, 6) + 40,000 (F/P, i%, 6) A ,N6K) SELLER.COM = 25,000(1.677) + 40,00T 0 E(FS/PT,Bi% (F/P, i%, 6) = {115,180 − 25,000(1.677)}/40,000 = 1.8314 (1 + i)6 = 1.8314 i = 0.1061 or 10.61%

3.7. A sum of $500,000 will be invested by a firm two years from now. If money is worth 12%, what will be the worth of this investment 10 years from now?

A) $1,553,000 B) $1,569,000 C) $1,463,100 D) $1,238,000 Answer: D Feedback: F = 500,000 (F/P,12%, 8) = 500,000 (2.476) = $1,238,000

3.8. Global Investment Corporation is exploring multiple investment opportunities to invest about $20 M. Four investment opportunities are being carefully evaluated. If the company is interested in maximizing the return on investment, which one is offers the best opportunity?

A) 12% compounded quarterly.

B) 11.85% compounded daily.

C) 11.95% compounded monthly.

D) 11.82% compounded continuously.

Answer: C Feedback: A comparison may be made by computing the effective interest rates. i. Effective interest rate = (1 + 0.03)^4 − 1= 0.1255 or 12.55% ii. Effective interest rate = (1 + 0.1185/365)^365− 1= 0.1258 or 12.58% i. Effective interest rate = (1 + 0.1195/12)^12 − 1= 0.1263or 12.63% i. Effective interest rate = e0.1182 − 1= 0.125469 or 12.5469%

True/False 3.9. To double an investment at an interest rate of 10% compounded annually, it will take exactly 5 years.

Answer: False Feedback: 2P = P (1 + 0.10)n (1 + 0.10)n = 2 n = ln 2/ ln 1.10 = 7.27 years

3.10. An amount $4,000 is invested today and another $10,000 invested four years from now. If both of them earn a simple interest of 10%, the interest accrued at the end of 5 years is $1,400 Answer: False Feedback: Accrued interest rate at the end of year 5 = 4,000*0.10*5 +10,000*0.10 = $3,000

3.11. One thousand dollars invested grew to be $3,000 six years hence. If the interest was compounded yearly, the interest rate on this investment was 20%.

Answer: True Feedback: 1,000(1+ i )6 = 3,000 (1+ i) = 3 1/6 = 1.2009

i = 1.2009 − 1 = 0.2009 or 20.09%

3.12. Sue borrowed $1,000 from Tom and paid him back very generously with a $2,000 check when she won a lottery after 8 years. The compound interest rate on this loan is 16.67%.

Answer: False Feedback: 2,000 = 1,000 (1 + i)8 (1 + i) = {2,000/ 1,000)(1/8) = 1.0905 i = 1.0905 − 1 = 0.0905 or 9.05%

3.13. If Sonja invested $10,000 in a good mutual fund that pays an average return of 10%, the investment will be worth $16,110 five years from now.

Answer: True Feedback: F = 10,000 (F/P, 10%, 5) = 10,000 (1.611) = $16,111

3.14. It is customary in engineering economic analysis to assume that the stated interest rate is for a one-year period.

Answer: True Feedback: It is required to state the time period if it is for any other period.

3.15. When there are different repayment plans available to repay a loan, the plans are all equivalent as long as the interest rates are different as each payment plan results in a different total repayment.

Answer: False Feedback: Since the equivalence of different repayment plans is based on the same interest rate.

3.16. Interest paid on a loan is analogous to rent paid on an apartment.

Answer: True Feedback: The interest paid is to use someone else’s capital or property.

3.17. Since cash flows like receipts and disbursements occur at different time periods, for evaluation of alternative, the cash flows can be added and subtracted ignoring the time vale of money.

Answer: False Feedback: Cash flows occurring in different times have to be accounted for equivalence using a TESTBANKSELLER.COM specific interest rate.

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CHAPTER 4 EQUIVALENCE FOR REPEATED CASH FLOWS

Multiple Choice 4.1. A set of cash flows are given in table below, using the principles of equivalence, determine the value “Y” for an interest rate of 8% compounded annually.

Year 0 1 2 3 4 5 5-10 Cash flow in $ -5,000 0 0 0 -1,000 -1,000 Y A) $1,402.34 B) $1,887.50 C) $1,107.78 D) $2,328.25 Answer: B Feedback: 5,000 (F/P, 8%, 4) +1,000 + 1,000(P/F,8%,1) = Y (P/A, 8%, 6) 5,000(1.360) +1,000 +1,000(0.9259) = Y (4.623) 4.623Y =8,725.90 A =8,725.90/4.623 = $1,887.50

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4.2. For an interest rate of 10% compounded annually, evaluate the value of “X” from the cash flows given in table below.

Year 0 1 2 3 4 5 Cash flows -10,000 + X 1,600 1,700 1,800 1,900 3,500 A) $2,316.85 B) $3,295.43 C) $1,064.74 D) $1,102.75 Answer: A Feedback: PW of negative cash flows = PW of positive cash flows 10,000 = X + 1,600 (P/A, 10%, 5) +100 (P/G, 10%, 5) + 1,500(P/F 10%, 5) 1

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10,000 = X +1,600(3.791) +100(6.862) +1,500 (0.6209) X = 10,000- 7,683.15 = $2,316.85

4.3. For the cash flow diagram below, determine the value of P

$10,000

0

10

20

$3,000

P A) $119.26 B) $195.49 C) $71 D) $147.20

i = 12%

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Answer: C Feedback: P = 10,000 (P/F, 12%, 20) -3,000(P/F, 12%, 10) = 10,000 (0.1037) -3,000(0.3220) =$71

4.4. Four different choices are given to determine the unknown value of “X” from the cash flow diagram given below. Use an interest rate of i =6%.

$20K

0

1

2

3

4

5

6

$1,600 2

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$1,700 $1,800

$2,000

X

$1,900

A) X= {20,000 – 2,000(P/A, 6%, 6) +100(P/G, 6%, 6)} (P/F, 6%, 6) B) X= {20,000 – 2,000(P/A, 6%, 5) +100(P/G, 6%, 5)} (F/P, 6%, 6) C) X= {20,000 – 2,000(P/A, 6%, 6) +100(P/G, 6%, 6)} (F/P, 6%, 6) D) X= {20,000 – 2,000(P/A, 6%, 6) +100(P/G, 6%, 5)} (P/F, 4%, 6) Answer: B Feedback: A uniform series and a negative gradient for 5 years can be recognized. “X” is a single cash flow occurring at the end of year 6.

4.5. What is the present worth of the cash flows in table below? Use an interest rate of 6% compounded annually.

Year 1 2 3 4 5 6-20 Cash flow 6,000 5,000 4,000 3,000 2,000 10,000

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A) $89,263 B) $85,496 C) $89,915.77 D) $91,611 Answer: C Feedback: P = 6,000 (P/A, 6%, 5) -1,000(P/G, 6%, 5) + 10,000(P/A, 6%, 15) (P/F, 6%, 5) = 6,000 (4.212) -1,000(7.934) + 10,000(9.712) (0.7473) = $89915.77

4.6. Given the cash flow diagram below, evaluate the value of “A” for an interest rate of 7%.

12,000 i = 7%

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0

1

2

3

4

2A

3A

4A

A

5

A =?

A) $585 B) $1045.72 C) $1200 D) $915 Answer: B Feedback: (12,000 +5A) (P/F, 7%, 5) = A (P/A, 7%, 5) + A (P/G, 7%, 5) (12,000 +5A) (0.7130) = A (4.100) + A (7.647) 8.182A = 8556 A = 8556/8.182 =1,045.72

TESTBANKSELLER.COM 4.7. Mary is planning to repay a debt of 50,000 with a quarterly payment $2,400 for the next 23 quarters and a final payment of “X” dollars at the end of 24-th quarter. If the interest rate is 12% per year, compounded quarterly, how much will be Mary’s final payment?

A) $21,426 B) $22,176 C) $23,010 D) $23,920 Answer: A Feedback: X = 50,000(F/P, 3%, 24) – 2,400 (F/A, 3%, 23) * (F/P, 3%, 1) = 50,000 (2.033) -2,400(32.453)*(1.030) =$21,426.18

4.8. Determine the value of P from the cash flows shown in table below. Interest rate = 10%.

Year Cash Flow

0 P

1 -4K

2 -3K

3 -2K

4 -1K

5 5K 4

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A) $8,102 B) $5,197.5 C) $9,211.2 D) $7,284.9 Answer: B Feedback: 𝑖 = 10% P + 5K (P/F, 10%, 5) = 4K (P/A, 10%, 4) – 1K (P/G, 10%, 4) P + 5K (0.6209) = 4K (3.170) – 1K (4.378) P = 5,197.5

4.9. Tom started investing, as soon as he started his first job, at the rate of $400 per month as soon as get paid into a savings account that earns an interest of 1% per month. Which of the following expression may be used to determine the account value 10 years from now?

A) F = [4,800(P/A, 12%, 10)] [(F/P, 12%, 10)] F = 400[(P/A, 1%, 120) (F/ P , 1 2 % , 5 )] B) T E ST B ANKSELLER.COM C) F= [400/0.01] (F/P, 1%, 60) D) F= 400(F/A, 1%, 120) Answer: D Feedback: Since the future worth of uniform series deposit of $400 is required, it can be evaluated from, F= 400(F/A, 1%, 120).

4.10. Choose the equation below that can be used to determine the value of “P” for a known interest rate, i.

Year Cash Flow, $

0 P

1 -0

2 -2,000

3 -4,000

4 -6,000

5 10,000

A) P = 10,000(P/F, i%, 5) – 2,000(P/G, i%, 5) B) P = 10,000- 1,000(P/G, i%, 5) (F/P, i%, 5) C) P= 10,000(P/F, i%, 5) -2,000(P/G, i%, 5) D) P= -10,000(P/F, i%, 5) + 2,000(P/G, i%, 4)

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Answer: D Feedback: It may be recognized that are three sets of cash flows occurring.

They are: a positive cash flow of P in year 0, a negative $2,000 gradient series from year 1 to 4, a $10,000 positive cash flow in year 5.

4.11. Ben invested $20,000 into a money market account and took out $5,000 at the end of year 5. He found out at the end of 10 years that he had as of $50,000 in the account.

What is the annual interest rate Ben had earned on this investment?

A) 8% B) 12.93% C) 15.23% D) 11.38% Answer: D Feedback: 20,000 = 5,000(P/F, i, 5) +50,000(P/F, i, 10) Try i =12% 20,000 = 5,000(0.5674) +50,000(T0.E3 S 22T0B) ANKSELLER.COM = 18,937? No Try i =10% 20,000 = 5,000(0.6209) +50,000(0.3855) = 22,379.5 Yes On linear interpolation between 10% and 12%, it is found that “i”=11.3

4.12. Given the cash flow diagram below, evaluate the “X” value using an interest rate of 8%.

Year Cash Flow, $ A. $7,201

0 10,000

1 -$1,000 B. $6,781

2 -$1,500

3 -$2,000

4 -$2,500

C. $3410.23

5 -(X+3,000) D. $6,525

Answer: C Feedback: 10,000 = X (P/F, 8%, 5) + 1,000(P/A, 8%, 5) + 500 (P/G, 8%, 5) = X (0.6806) + 1,000(3.993) + 500 (7.372) X = 3410.23 6

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4.13. IAN Tech is saving $10,000 a month in a money fund to pay for equipment that will cost $700,000 five years from now. If the interest rate is 1/4% per month, how much additional funds will be needed to pay for this equipment?

A) $48,239 B) $53,530 C) $59,750 D) $25,070 Answer: B Feedback: Additional money needed = 700,000 -10,000(F/A, 1/4%, 60) = 700,000 -10,000(64.647) =$53,530

4.14. Determine the unknown from the cash flow diagram shown below. i = 10%.

TESTBANKSELLER.COM 4000

3000 2,000 1,000 0

1

2

3

4

5 X

A) $12,160 B) $14,500 C) $10,108 D) $12,809 Answer: A Feedback: X+5,000=1000(F/A,10%,5)+1000(P/G,10%,5)(F/P,10%,5) =1000(6.105)+1000(6.862)(1.611) 7

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X=$12,159.68

4.15. Given a series of cash flows in the table below, determine the value of “y”. i=6% Year Cash flow

0 5000

5 -5000

10 10000

20 -Y

A) $21,025 B) $21,960 C) $23,821 D) $22,507 Answer: B Feedback: y+5000 (F/P,6%,15)=5,000(F/P,6%,20)+10,000(F/P,6%,10) y+5000 (2.397) =5,000(3.207)+10,000(1.791) = $21,960

4.16. Given the cash flow table below, determine the unknown value “X” for an interest rate of TESTBANKSELLER.COM 6%.

Year 0 1 2 3 4 5 6 7 8 9 10 Cash flow -X 400 500 -X 700 800 900 1,000 1,100 1,200 1,300 A) $2436.89 B) $3129.41 C) $2935.66 D) $1610.53 Answer: C Feedback: X = 400 (P/A, 6%, 10) + 100 (P/G, 6%, 10) − (X+600) (P/F, 6%, 3) X = 400 (7.360) + 100 (29.602) − (X+600) (0.8396) =5,400.44 – 0.8396 X X = 5400.44/1.8396 = $2935.66

Reference: Case Study 4.17

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Daniel borrowed $20,000 with a promise to repay the loan in 6 years with a uniform monthly payment and a single payment of $2,000 at the end of six years at a nominal interest rate of 12% per year.

4.17A. What is the amount of each payment?

Answer: A = [20,000(F/P,1%,72)-2,000](A/F,1%,72) = [20,000(2.047)-2,000](0.00955) = $371.87 Refer to: Case Study 4

4.17B. What is the amount of interest paid in the first payment?

Answer: Interest paid in the first payment = 0.01(20,000) = $200 Refer to: Case Study 4

4.17C. What will be the loan balance immediately after the 48th payment?

Answer: Loan balance immediately After the 48th payment = 371.87 (P/A,1%,24) = 371.87 (21.243) TESTBANKSELLER.CO=M $7,899.63 Refer to: Case Study 4.17

True/False 4.18. In order to use the gradient series factors to solve a set of given cash flows, the cash flows must increase or decrease gradually by the same amount every year, starting year 2 and must have zero cash flow in year 1.

Answer: True Feedback: Closely examining the gradient series factor formula, this is “True” as cash flow in year 1 must be “0” and must increase or decrease the same amount from year 2.

4.19. In developing cash flow diagrams the convention is to use a negative cash flow for receipts.

Answer: False

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Feedback: Anytime money is received by an individual or a company, it is definitely a positive cash flow.

4.20. The weekly payment for a $500 loan to be paid back in 104 payments at an interest rate of 0.25% per week is $6.82.

Answer: False Feedback: A = 500 (A/P, 0.25%, 104) = 500(0.0109) = $5.45

4.21. Interest compounding daily than continuous compounding for a known interest rate will provide a larger yield.

Answer: False Feedback: The continuous compounding will provide a higher yield than any daily compounding.

TESTBANKSELLER.COM 4.22. If the uniform series capital recovery factor is 0.1359 and the sinking fund factor is 0.0759, then the interest rate is 6%.

Answer: True Feedback: Since (A/P, i, n) = (A/F, i, n) + i, i can be evaluated as 0.1359 = 0.0759 + i i = 0.1359 − 0.0759 = 0.06 or 6%

10

CHAPTER 5 PRESENT WORTH ANALYSIS Multiple Choice 5.1. The annual maintenance cost of a monument in the state capital is estimated to be $4850. A perpetual i fund of $100,000 is set up to pay for this maintenance expenditure. Determine the interest rate this fund earns if the interest is compounded quarterly.

A) 4.58% B) 4.76% C) 4.39% D) None of these Answer: B Feedback: ia = 4850/100,000 = 0.0485 (1+i/4)4 – 1 = 0.0485 (1+i/4)4 = 1.0485 1+i/4 = (1.0485)1/4 1+i/4 = 1.0119 i = 0.0476 or 4.76%

TESTBANKSELLER.COM

Reference: Case Study 5 Dunn Manufacturing is considering the following two alternatives. The cost information for the two proposals for replacing an equipment are provided are in table below.

Machine X Machine Y Initial cost $120,000 $96,000 Benefits/year $20,000 for the first 10 years $12,000 per year for 20 and $9,000 for the next 10 years. years Life 20 years Salvage value $40,000 $20,000 MARR 8%

5.2. The NPW of machine X is

.

A) $35,158 B) $48,192 C) $50,752 D) $61,239 Answer: C Refer to: Case Study 5 Solution: Since both alternatives have the same life, the analysis period is 20 years.

NPW (Machine X) = 9,000 (P/A, 8%, 20) +11,000 (P/A, 8%, 10) +40,000 (P/F, 8%, 20) − 120,000 = 9,000 (9.818) +11,000 (6.710) +40,000 (0.2145) − 120,000 = $50,752

5.3. The NPW of machine Y is A. $42,196 B. $29,508

.

C. $26,106

D. $32,103

Answer: C Refer to: Case Study 5 Feedback: NPW (Machine Y) = 12,000 (P/A, 8%, 20) +20,000 (P/F, 8%, 20) − 96,000 = 12,000 (9.818)T+E 20S,T 00B0A(N 0.K 21S4E 5)L−L9E6R,0.0C 0 OM = $26,106

5.4. If machine “Y” has no salvage value, what would be the NPW of machine “Y”?

A) $19,891 B) $20,626 C) $23,478 D) $21,816 Answer: D Refer to: Case Study 5 Feedback: NPW (Machine Y) = 12,000 (P/A, 8%, 20) − 96,000 = 12,000 (9.818) − 96,000 = $21,816

5.5. How much money would have to be placed in a sinking fund each year to replace a machine costing $20,000 today at the end of 10 years if the fund yields 9% annual interest rate compounded yearly and if the cost of the machine is assumed to increase at 5% annually for the next 10 years? Assume the salvage value is zero.

A) $2,143.63 B) $2150 C) $2,340 D) $1923 Answer: A Feedback: The cost of the machine 10 years from now = (1 + 0.05) ^10 (20,000) = $32,577.89 A = $32,577.89 (A/F, 9%, 10) = $32,577.89 (0.0658) = $2,143.63

5.6. Given the cash flows in table below. Determine the value of P. i= 6% per year Year 0 1 2 3 4 5 Cash Flow -P 400 800 1200 T 16E0S 0T B 18A0N0K SELLER.COM A) $4709.14 B) $6125.15 C) $5125.63 D) $4978.23 Answer: A Feedback: Note, the cash flows appear to be increasing every year by $400 from year 1 to year 4. The cash flow in year 5 has increased only by 200.

P = 400(P/A, 6%, 4) + 400(P/G, 6%, 4) + 1800(P/F, 6%, 5) = 400(3.465) + 400(4.945) + 1800(0.7473) = $4709.14

5.7. Determine the net present worth (NPW) of the cash flows given in table below for an investment opportunity being presented to a company. MARR =12% Year 0 1-10 11-15 16-25 26-30 Cash Flow -$100K 10K 20K -5K 30K

A) $90,030 B) $80,914 C) $72,916 D) $112,200 Answer: B Feedback: P = 30K(P/A,10%,5)(P/F, 10%,25) − 5K(P/A,10%,10)(P/F,10%,15) + 20K (P/A, 10%, 5)(P/F,10%,10) + 10K(P/A,10%,10) − 100K =30K (3.605)(0.0588) − 5K(5.650)(0.1827) + 20K(3.605)(0.3220) + 10K(5.650) = $80,914.15

5.8. Kal Tech Engineering Systems is considering buying a CNC machining center for its operation in Tennessee. The net benefits in the first year is estimated to be $40,000 and increasing at the rate $5,000 for the next four years and stays at the same level as that of year 5 for the next 5 years. If MARR is 8%, determine the amount of money that the company can invest justifying on this machining center. A salvage value of 20% of the initial cost is reasonable to assume at the end of year 10.

A) $396,357 B) $416,182

TESTBANKSELLER.COM

C) $411,202 D) $399,500 Answer: A Feedback: The cash flows for the next 10 years for the CNC machining center tabulated as below.

Year 0 1 2 3 4 5 6-10 10 Cash flow -P $40K 45K 50K 55K 60K 60K 0.20P P = 40K (P/A,8%,5)+ 5K(P/G, 8%,5) +60K(P/A,8%,5)(P/F,8%,5) + 0.2P (P/F,8%,10) P = 40K (3.993) + 5K (7.372) +60K (3.993) (0.6806) +0.2P (0.4632) = $396,357

5.9. Sam is considering investing in a bond with a face value of $20,000. The bond pays an interest of 4% payable quarterly. If he expects to make a 1 ½ % return per quarter on this investment with a maturity of 20 years, determine the most he can pay for the bond .

A) $18,102.65 B) $14,923.86 C) $15,355.40 D) $16,000

Answer: C Feedback: The interest payment per quarter = 20,000 (0.04/4) = $200 P = 200 (P/A, 1 ½%, 80) + 20,000(P/F, 1 ½ %, 80) = 200 (46.407) + 20,000(0.3037) = $15,355.40

Reference: Table 5

Initial cost Annual benefit Salvage value Life in years MARR

A $15,000 $8,000 $5,000 2 years 10%

Problem 5-10 The NPW of alt. A is A) $13,420 B) $17,380 C) $11,000 D) $6,000

B C $9,000 $12,000 $2,000 $1,800 $9,000 0 T E S 3 Years InfTinBitA y N KSELLER.COM

.

Answer: B Refer to: Table 5 Feedback: The analysis period is not 6 years since the least common multiple of lives is NPW = PW of Benefits – PW of Costs = (8,000/0.10) – [{15,000(A/P, 10%, 2) – 5,000(A/F, 10%, 2)}/0.10] = (8,000/0.10) – [{15,000(0.5762) – 5,000(0.4762)}/0.10] = $17,380

5.11. The NPW of alt. B is

.

.

A) $13,420 B) $17,380 C) $11,000 D) $6,000 Answer: C Refer to: Table 5 Feedback: NPW = PW of Benefits – PW of Costs = (2,000/0.10) – [{9,000(A/P, 10%, 3) – 9,000(A/F, 10%, 3)}/0.10] = (2,000/0.10) – [{9,000(0.4021) – 9,000(0.3021)}/0.10] = $11,000

5.12. The NPW of alt. C is A) $13,420 B) $17,380 C) $11,000 D) $6,000

.

Answer: D Refer to: TabT leE5STBANKSELLER.COM Feedback: NPW = PW of Benefits – PW of Costs = (1,800/0.10) – 12,000 = $6,000

Problem 5-13 The best alternative among the three alternatives using the PW analysis is A) Alt. A B) Alt. B C) Alt. C D) Either alt. B or alt. C Answer: A Refer to: Table 5 Feedback: Alt. A has the largest net present worth and therefore, choose alt. A.

.

Short Response

5.14. Determine the better of the two alternatives using the present worth analysis. Use an interest rate of 10%.

Alt. X Initial cost $12,500 Annual benefit $6,800 Salvage value $5,000 Life in years 2 years MARR 10%

Alt.Y $8,900 $2,000 $8,900 3 Years

Solution: The analysis period is 6 years since the LCM of 2 and 3 is 6.

NPW (Alt. X) = 6,800(P/A,10%,6) + 5,000[(P/F, 10%,2) +(P/F,10%,4)+(P/F,10%,6)] − 12,500[1 + (P/F, 10%, 2) + (P/F, 10%, 4)] = 6,800(4.355) +5,000[0.8264+0.6830+0.5845] − 12,500[1+0.8264+0.6830] = $8,716 NPW (Alt. Y) = 2,000(P/A, 10%, 6) + 8,900[(P/F, 10%, 3) + (P/F, 10%, 6)] − 8,900[1+ (P/F, 10%, 3)] TESTBANKSELLER.COM = 2,000(4.355) +8,900 [0.7513 + 0.5645] − 8,900[1 + 0.7513] = $4,834.05 To maximize NPW, choose alt. X.

5.15. Solve the following problem using the present worth analysis for an interest rate of 8%.

Alt. A Alt. B Alt. C Initial cost $1,700 $2,100 $3,750 Benefit/year 1,000 1,000 1,000 Life in years 2 3 6 Solution: A six year analysis is required since the least common multiple of lives of 2, 3, and 6 is 6.

NPW (A)=1,000(P/A,8%,6)-[1,700+1,700(P/F,8%,2)+1,700(P/F,8%,4)] = 1,000 (4.623)–{1,700 + 1,700 (0.8573)+1,700(0.7250)}=$233.09 NFW(B)=1,000(P/A,8%,6)-[2,100+2,100(P/F,8%,3)] =1,000(4.623)-[2,100+2,100(0.7938)]= $856.02 NFW(C)=1,000(P/A,8%,6)-3,700 =1,000(4.623)-3,700 = $923

The best alternative is alt. C.

True/False 5.17. The capitalized cost of any investment may be determined using the equation P = A/i where P is the capitalized cost, A is the annual amount and i is the interest rate.

Answer: True Feedback: Since annuity, A = P i, P = A/i.

5.18. The NPW of a set of cash flows will decrease as the interest rate is increased.

Answer: True Feedback: For example (P/A, 12%, 10) = 5.00, and (P/A, 10%, 10) = 6.145, the NPW will decrease as TESTBANKSELLER.COM the interest is increased.

CHAPTER 6 ANNUAL CASH FLOW ANALYSIS Multiple Choice 6.1. Evaluate the value of A from the cash flows shown in table below using an interest rate of 7% compounded annually.

Year Cash Flow

0 -$15,000

1 5A

2 4A

3 3A

4 2A

5 5A

A) $734.63 B) $1,049.97 C) $955.11 D) $874.80 Answer: C Feedback: 5A (P/A, 7%, 5) - A (P/G, 7%, 5) +4A (P/F, 7%, 5) = 5,000 5A (4.100) - A (7.647) + 4A (0.7130) = 15,000 15.705 A = 15,000 A = 15,000/15.705 = $955.11 TESTBANKSELLER.COM

6.2. Determine the value of EUAW from the cash flow diagram shown below. i = 9%.

2000

0

7,500 A) $328.25 B) $377.75 C) $569.75 D) $451.06

1

2

3

4

5

6

Answer: A Feedback: EUAW = EUAB –EUAC = 2,000 – 7,500 (A/P, 9%, 6) = 2,000 – 7,500 (0.2229) = $ 328.25

6.3. Given: Cash flows for an investment.

Year 1 2 3 4 5 Cash flow $1400 1500 1600 1700 -2,000 Required: EUAC at 7% per year.

A) $674.63 B) $849.97 C) $925.68 D) $764.80

TESTBANKSELLER.COM

Answer: C Feedback:

$1700 $1500

$1800

$1600

$1400

0

$2000 Let A be the EUAC The equation to solve for “A” may be set up as below.

A = 1400+ 100(A/G, 7%, 5) − (3,800) (A/F, 7%, 5) = 1400 + 100 (1.865) − (3,800) (0.1739) =$925.68

$1800

6.4. Jason bought a car for $40,000 upon graduation from college with an engineering degree and a very good job offer. A down payment of $5,000 was paid by his dad as graduation gift. The rest of the amount was financed with Generous Motors at 6% nominal interest with 60 monthly payments, the first payment which is to start at the end of 13th month. Determine his monthly payment.

A) $583.33 B) $717.38 C) $652.68 D) $752.67 Answer: B Feedback: A = (40,000-5,000) (F/P, 1/2%, 12) (A/P, 1/2%, 60) = 35,000(1.062) (0.0193) = $717.38

TESTBANKSELLER.COM 6.5. An investment set up as a perpetual trust provides an annual disbursement of $25,000 for the first 20 years and an undetermined amount from year 21. If the trust is set up with $500,000, determine the disbursement that can be made from year 21 through infinity if the interest earned is 6% per year.

A) $41,031 B) $54,000 C) $60,090 D) $58,500 Answer: A Feedback: A = [500,000 (F/P, 6%, 20) − 25,000 (F/A, 6%, 20)] (0.06) = [500,000 (3.207) − 25,000 (36.786)] (0.06) = $41,031

6.6. A couple, both engineering alums from a reputable engineering school has decided to set up an endowment to help pay for 4 engineering scholarships at the rate of $X per year starting year

21 perpetually. If $200,000 is invested in the trust today and if it earns a very good rate of return of 12% per year, what will the amount of each scholarship starting year 21?

A) $49,756.2 B) $57,876 C) $62,110.5 D) $41,932 Answer: B Feedback: A = 200,000 (F/P, 12%, 20) (0.12) = 200,000 (9.646)(0.12) = $231,504 Amount for each scholarship = 231,504/4 = $57,876

6.7. Aaron invested on a corporate bond with a face value of $10,000 at a discounted price of $9,000. The bond pays a quarterly dividend. What is the quarterly dividend Aron should get if he hopes to get a 10% return on his investments? Assume the bond matures in 20 years.

A) $201 TESTBANKSELLER.COM B) $221 C) $190 D) $152 Answer: B Feedback: Assume that Aaron gets a dividend of X dollars per quarter.

X (F/A, 2 ½ %, 80) = 9,000(F/P, 2 ½%, 80) − 10,000 X (248.382) = 9,000(7.210) − 10,000 X = 54,890 / 248.382 = $220.99

6.8. For the cash flows given in the table below, evaluate the unknown value “A”. Use an interest rate of 6%.

Year

0

1

2

3

4

5

Cash flow

22,000

-2A

-2A

-2A

-2A

-5A

A) $1,982.57 B) $2,062.55 C) $1,862.55 D) $2,215.69 Answer: B Feedback: 22,000 1

3

2

4

5

2A 5A

TESTBANKSEL LER.COM

22,000(A/P, 6%, 5) = 2A + 3A (A/F, 6%, 5) 22,000(0.2374) = 2A + 3A (0.1774) 2.5322 A = 5222.8 A = 5222.8/2.5322 = $2062.55

6.9. Choose the best alternative among the three alternatives given in the table below. Use of annual cash flow analysis is required.

A B C Initial cost $4500 $1900 $1200 Annual benefit $2800 $700 $300 Salvage value $500 $900 0 Life in years 2 years 3 Years Infinity MARR 5% A) A B) B

C) C Answer: A Feedback: EUAW (A) = EUAB – EUAC = 2800 – [4,500(A/P, 5%, 2) − 500(A/F, 5%, 2)] = 2800 – [4,500(0.5378) − 500(0.4878)] = $623.80 EUAW (B) = 700 – [1,900(A/P, 5%, 3) − 900(A/F, 5%, 3)] = 700 – [1,900(0.3672) − 900(0.3172)] = $287.8 EUAW(C) = 300 – 1200(0.06) = $228 Decision: Choose A to maximize EUAW.

6.10. The cost data for two equipment alternatives are given below.

Determine the better alternative using the annual cash flow analysis.

TESTBANKSELLER.COM

Machine X Initial cost $200,000 Net Annual Benefit 30,000 Salvage value 45,000 Life in years 20 MARR 9%

Machine Y $180,000 24,000 20,000 15

A) Machine X B) Machine Y Feedback: [EUAB-EUAC]" of" ("Machine X " )=30,000-200,000(A/P,9%,20)+45,000(A/F,9%,20) =30,000-200,000(0.1095)+45,000(0.0195) =$8,977.5 ["EUAB" -EUAC] of (Machine Y)=24,000-180,000(A/P,9%, 15)+20,000(A/F,9%,15) =24,000-180,000(0.1241)+20,000(0.0341) =$2,344 Decision: Choose Machine X.

6.11. Given the financial data for three mutually exclusive alternatives in the table below, determine the best alternative using the annual cash flow analysis for a MARR of 8%.

A B First cost $45,000 $25,000 O &M Cost/ year 10,000 4,000 Benefit/year 18,000 13,000 Salvage value 3,000 6,000 Life in years 5

C $20,000 1,900 9,000 4,600

A) A B) B C) C Answer: B Feedback: EUAW(A) = EUAB – EUAC = (18,000 – 10,000) – [45,000(A/P, 8%, 5) – 3,000(A/F, 8%, 5)] = 8,000 – [45,000(0.2505) – 3,000(0.1705)] TESTBANKSELLER.COM = − $2,761 EUAW(B) = (13,000 – 4,000) – [25,000(0.2505) – 6,000(0.1705)] = $3,760.5 EUAW(C) = (9,000 − 1,900) – [20,000(0.2505) – 4,600 (0.1705)] = $2,874.30 Decision: Choose B to maximize EUAW.

Short Answer Reference: Case Study 6 Equipment is being financed for 10 years by XYZ Corp. with monthly payments. The arrangement with the vendor worked out is as below.

The interest rate is ¾ % per month for the first 48 payments and for the remainder of the loan the interest rate would be 1 % per month.

The equipment cost $500,000.

6.12. What is the monthly payment for the first 48 months?

Solution: 500,000 48

49

50

51

120

A1 A2 Answer: $6,350 A

Refer to: Case Study 6

= 500,000 (A/P, ¾ %, 120) = 500,000 (0.0127) = $6,350

TESTBANKSELLER.COM 6.13. Determine the amount of 49-th monthly payment.

Answer: 6,864.75 Refer to: Case Study 6 Feedback: A = [500,000 (F/P, ¾%, 48) − 6,350 (F/A, ¾%, 48)] (A/P, 1%, 72) = [500,000 (1.431) − 6,350 (57.521)] (0.0196) = $6,864.75

True/False 6.14. The equivalent uniform annual worth (EUAW) may be determined from net present worth from the equation, EUAW = PW (A/P, i, n).

Answer: True Feedback: Present worth can be multiplied by the capital recovery factor to get the EUAW.

6.15. The analysis period is the least common multiple of lives for evaluating mutually exclusive alternatives using the annual cash flow analysis.

Answer: False Feedback: For annual cash flow analysis, it is enough only one life cycle is considered to compute EUAW as it will be the same regardless of number of life cycles an asset may have.

6.16. An endowment has been setup to fund ten engineering scholarships of $10,000 each every year in a state university in Tennessee starting year 11. If the endowment is expected an earn 8% rate of return, the endowment is worth $1,000,000.

Answer: False Feedback: TESTBANKSELLER.COM Amount available for 10 scholarships from year 11 = [1,000,000 (F/P, 8%, 10) (0.08) = 1,000,000 (2.159) (0.08) = $172,720 Amount of one scholarship = 172,720/10 = $17,272.

6.17. The best alternative among the four mutually exclusive alternatives in the table below is alternative A. Alternative A EUAW

B

C

-$5,000 -$10,000 0

D (Do-nothing) -$15,000

Answer: False Feedback: The alternative that has the largest EUAW in the table above is alternative C.

CHAPTER 7 RATE OF RETURN ANALYSIS Multiple Choice

7.1. An investment of $100,000 today, will pay 10 annual payments of $15,000 each. Compute the rate of return on this investment to the second decimal place.

A) 8.15% B) 8.85% C) 8.52% D) 8.37% Answer: A Feedback: Equivalent Uniform Annual Cost = Equivalent Uniform Annual Benefits 100,000 = 15,000(P/A, i, 10) (P/A, i, 10) = 100,000/15,000 = 6.666% i (P/A,i,6) 9% 6.418 8% 6.315 RoR must be between 8% and 9 %T. ESTBANKSELLER.COM On interpolation, it is found to be 8.15%.

7.2. Determine the ROR for a project that has an initial cost of $82,000 and would provide positive cash flows of $12,000 the first year, $14,000 the second year, $16,000 the third year, $18,000 the fourth year, $20,000 the fifth year, and $15,000 the sixth year.

A0 4.12% B) 2.78% C) 4.95% D) 4.49% Answer: D Feedback: PW of Cost = PW of Benefits 82,000 = 12,000(P/A, i, 6) + 2,000 (P/G, i, 6) – 7000(P/F, i, 6) Try i = 4% 82,000 = 12,000(5.242) +2,000 (12.506) – 7000(.7903)

= 82383.9? No Try i = 4.5% 82,000 = 12,000(5.158) +2,000 (12.233) – 7000(.7679) = 80986.7? No RoR must be between 4.5% and 4%.

On interpolation, it is found to be 4.14%.

7.3. Given the cash flows described in table below. Determine the ROR. Year

0

Cash flow -$12,300

1-2

3-5

4,000

3,000

A) 13.85% B) 12.75% C) 14.21% D) 13.11% Answer: B Feedback: TESTBANKSELLER.COM PW of Cost = PW of Benefits 12,300 = 3,000 (P/A, i, 5) +1,000 (P/A, i, 2) Try i = 12% 12,300 = 3,000(3.605) +1,000 (1.690) = 12,505? No Try i =15% 12,300 = 3,000(3.352) +1,000 (1.626) = 11,682? No RoR must be between 12% and 15%.

On interpolation, RoR is found to be 12.75%.

7.4. Determine the IRR for an equipment that costs $250,000 and would provide positive cash flows of $100,000, $90,000, $80,000, and $20,000 at the end of each year for the next four years.

A) 9.75% B) 7.53% C) 10.90% D) 9.38%

Answer: B Feedback: Present Worth of Cost = Present Worth of Benefits Notice the series is gradient though year 3 and a single cash flow in year 4.

250,000 = 100,000(P/A, i, 3) -10,000 (P/G, i , 3) + 20,000((P/F,i,4) Try i = 7% 250,000 = 100,000(2.624) -10,000 (2.506) + 20,000(0.7629) = 252598? No Try i =8% 250,000 = 100,000(2.575) -10,000 (2.445) + 20,000(0.7350) = 247750? No RoR must be between 8% and 7%.

On interpolation, it is found to be 7.53%.

7.5. Tara purchases a 20-year, 5 percent savings bond with a face value of $10,000 at a premium price of $11,000. She gets the dividend paid yearly. Determine the rate of return on the bond if the bond is kept till maturity.

A) 3.79% TESTBANKSELLER.COM B) 4.62% C) 5.33% D) 2.38% Answer: A Feedback: Present Worth of Cost = Present Worth of Benefits Dividend per year = 0.05(10,000) =$500 11,000 = 10,000(P/F, i, 10) + 500 (P/A, i, 10) Try i = 3% 11,000 = 10,000(0.7441) + 500 (8.530)? = 11,706? No Try i =4% 11,000 = 10,000(0.6756) +500 (8.111)? = 10811.5? No RoR must be between 3.0% and 4%.

On interpolation, it is found to be 3.79%.

7.6. Pam promises to pay Sandy $4,000 in year 4 and another $6,000 in year 8 for a loan of $4,000 from Sandy today. What is the ROR that Sandy is getting? Assume interest is compounded

monthly.

A) 14.75%

B) 16.72%

C) 15.39%

D) 18.08%

Answer: C Feedback: Present Worth of Cost = Present Worth of Benefits 4,000 = 4,000(P/F, i, 48) + 6,000 (P/F, i, 96) Try i = 1% 4,000 = 4,000(0.6203) +6,000 (0.3847)? = 4789.4? No Try i =1.5% 4,000 = 4,000(0.4894) +6,000 (0.2395)? = 3394.6? No Interest rate per month must be between 1.0% and 1.50%.

On interpolation, it is found to be 1.2829*12%.

Annual interest rate = 12(1.2829) = 15.39%

7.7. An equipment costing $60,000 is being evaluated for a production process at Don Jones Co. The expected benefits per year is $4,500 and estimated salvage value is $20,000. Determine the rate of return the company can get in this equipment proposal. Equipment life = 20 years.

TESTBANKSELLER.COM A) 4.92% B) 5.96% C) 5.62% D) 4.18% Answer: C Feedback: 60,000=4,500(P/A,i,20)+20,000(P/F,i,20) Try i=5% 60,000=4,500(12.462)+20,000(0.3769) =$63,617.00 Try i=6% 60,000=4,500(11.740)+20,000(0.3118) =$57,851.00 Then, i = 5.62%

7.8. The table below provides cash flows for two mutually exclusive alternatives for developing a

“Pay for Recreation” facility being considered by a county government in Georgia. If money can be borrowed issuing bonds at the rate of 7% per year, using the RoR analysis, find the attractive alternative between the two given proposals.

Year 0 1-10 11-∞

Alt. A -10M 1M 1.5M

Alt. B -37M 4M 2.5M

A) Alt. A B) Alt. B Answer: B An incremental RoR analysis between the two alternatives is required. MARR = 7% Year Incremental Cash flow

0 -37M-(-10M) =-27M

1-10 4M-1M =3M

11-∞ 2.5M-1.5M= 1M

Need to prove that RoR is greater than MARR. Present Worth of cost = Present Worth of benefits 27M = (1M/i) + 2M (P/A, i, 10) TESTBANKSELLER.COM Try i = 8% 27M = (1M/0.08) + 2M (6.710) = 25.92M Try i = 7% 27M = (1M/0.07) + 2M (7.024) = 28.33M Since ∆ RoR is between 7% and 8%, it is greater than MARR; therefore, choose the higher cost alternative, Alt B.

7.9. Kruber company located Tennessee is considering two different makes of a blow molding machine for one of its automotive products. The cost data for the two alternatives are provided in table below. MARR =12%.

Machine X Initial cost $150K Annual operating cost 55K Benefits /Year 100K Salvage value 20K

Machine Y $275K 60K 120K 40K

Life

10 Years

A) Machine X B) Machine Y Answer: A Feedback: Year 0 1 thro ‘10 10 Incremental -125K 15K 20K Cash flow NPW = Present Worth of benefits - Present Worth of costs Try i = 12% 125K= 15K (P/A, 12%, 10) + 20K (P/F, 12%, 10) = 15K (5.650) + 20K (0.3220) = 99.885K? No Try i = 10% 125K= 15K (6.145) + 20K (0.3855) = 91.11K? No ∆RoR < 12% (MARR). Choose the lower cost alternative, alt. X

TESTBANKSELLER.COM True/False 7.10. MARR is minimum attractive rate of return.

Answer: True Feedback: MARR is the minimum attractive return an investor can expect to have in any investment.

7.11. In a capital equipment investment project, the smallest possible rate of return is 0%.

Answer: True Feedback: In any investment, at the worst case the investment made could all be lost.

Therefore, the lowest rate of return possible is “0%”.

7.12. If Carter expects a guaranteed ROR to triple his investment in 9 years, the nominal interest rate that he got is 30%.

Answer: False Feedback: 3P = P(1+i)9 (1+i)9 = 3P/P = 3 (1+i) = 31/9 = 1.1298 i = 0.1298 or 12.98%

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CHAPTER 8 CHOOSING THE BEST ALTERNATIVE Multiple Choice Reference: Case Study 8 The cash flows for three mutually exclusive alternatives are given in table below. MARR = 4%.

Alt. A Initial cost $15,000 Annual benefits $4,500 RoR 15% Life in years

Alt. B 27,000 7,600 13% 5

Alt. C 24,000 6,500 11%

8.1. ΔRoR for the first increment (Alt. C-Alt. A) is A) 3.36% B) 3.09% C) 2.85% D) 3.62%

.

Answer: D Refer to: Case Study 8 Feedback: TESTBANKSELLER.COM PW of incremental cost = PW of incremental benefits (24,000 - 15,000) = (6,500 -4,500) (P/A, i, 5) (P/A, i, 5) = 9,000/(2000) = 4.500 i% (P/A, i%, 5) 3.5% 4.515 4% 4.452 ΔRoR is between 3.5 %and 4%. On interpolation, ΔRoR = 3.62% Also, since MARR is 4%, alt. A is retained.

8.2. Determine the ΔRoR for the second increment (Alt. B – Alt.A) if A was retained during the first incremental analysis. .

A) 10.85% B) 8.12% C) 9.19% D) 5.78% Answer: C Feedback:

Refer to: Case Study 8

The second increment is (Alt.B-Alt.A).

PW of incremental cost = PW of incremental Benefits (27,000 -15,000) = (7,600 -4,500) (P/A, i, 5) (P/A, i, 5) = 12,000/(3,100) = 3.871 i% (P/A, i%, 5) 9% 3.890 10% 3.791 ΔRoR is between 9 %and 10%. On interpolation, ΔRoR = 9.19% For a MARR of 4%. Alt. B will be retained.

8.3. The best alternative for a MARR of 4.0% using the incremental rate of return analysis is .

A) Alt. C B) Alt. A C) Alt. B D) Do nothing Answer: C Refer to: Case Study 8 TESTBANKSELLER.COM Feedback: It is seen from solution to problem 8.1, for the first increment (C- A), the lower cost alternative, A will be chosen since ΔRoR is less tah MARR (4%).

From solution to problem 8-2, for the second increment (B-A), the higher cost alternative, B will be chosen since ΔRoR is greater than MARR(4%).

Therefore, the best alternative is, Alt. B

Reference: Table 8.1 Cash flows related to three mutually exclusive capital equipment projects are given in table below.

Alternative A B C

Initial Cost $100,000 $200,000 $300,000

8.4. ΔRoR for the first increment (alt.B - alt.A) is A) 10.2%

Computed ROR 18% 15% 13% .

B) 12% C) 14.67% D) 15.5% Answer: B Refer to: Table 8.1 Feedback: ΔRoR (B-A) = {(200,000) (0.15) - (100,000) (0.18)}/ {(200,000) - (100,000)} = 0.12 or 12 %.

8.5. ΔRoR for the second increment (C-B) is A) 10.2% B) 18.6% C) 14.67% D) 9.0%

.

Answer: D Refer to: Table 8.1 Solution: ΔRoR (C-B) = {(300,000) (0.13) - (200,000) (0.15)}/ {(300,000) - (200,000)} = 0.09 or 9%

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8.6. ΔRoR for the increment (C- A) is A) 12.3% B) 10.5% C) 14.67% D) 15.5%

Answer: B Refer to: Table 8.1 Solution: ΔRoR(C-A) = {(300,000) (0.13) - (100,000) (0.18)}/ {(300,000) - (100,000)} = < 0.105 or 10.5%

8.7. Based on ΔROR analysis, the best alternative , for a MARR of =10%, is .

A) Alt.A B) Alt.B C) Alt. C D) Do nothing

Answer: B Refer to: Table 8.1 Solution: The winner at the first increment between alt. B and alt. A is alt B as ΔROR = 12% When alt. B is compared to the next higher cost alternative C, alt. B is found to be better for a MARR of 10% since ΔROR(C-B) is 9%..

Therefore the ultimate winner is alt. B.

8.8. The cash flows for four mutually exclusive alternatives are provided in the table below.

If the useful life is 10 years for each alternative, find the best alternative using incremental rate of return analysis. MARR = 10%

Initial Cost EUAB A) Alt. P B) Alt. Q C) Alt. R D) Alt. S

P 19,000 3,500

Q 16,000 3,100

R 11,000 2,200

S 8,500 1,350

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Answer: B Solution: (R – S) Increment PW of cost = PW of benefits (11,000 – 8,500) = (2,200- 1,350) (P/A, i, 10) 2,500 = 850(P/A, i, 10) (P/A, i%, 10) = 2,500 / 850 = 2.941 i% (P/A, i%, 10) 30% 3.092 35% 2.715 ΔRoR > 30%: Retain R The next increment is (Q-R).

(16,000 - 11,000) = (3100 – 2200) (P/A, i%, 10) 5,000 = 900 (P/A, i%, 10) (P/A, i%, 10) = 5,000 /900 = 5.556 i% (P/A, i%, 10) 12% 5.650 15% 5.019 ΔRoR > 12%: Retain Q

The next increment is (P-Q).

(19,000 – 16,000) = (3,500 – 3,100) (P/A, i%, 10) 3,000 = 400 (P/A, i%, 10) (P/A, i%, 10) = 3,000/ 400 = 5.5556 i% (P/A, i%, 10) 6% 7.360 5% 7.722 ΔROR < 10%: Retain Q Winner: Alt. Q

8.9. Given the financial data for four mutually exclusive alternatives in the table below, determine the best alternative using the incremental rate of return (∆RoR) analysis. MARR =10%.

A First cost $15,000 O &M Cost/ year 1,600 Benefit/year 8,000 Salvage value 3,000 Life in years A) Alt. A B) Alt. B C) Alt. C D) Alt. D

B $36,000 400 13,000 6,000

C $21,200 900 9,000 4,600

D 45,000 1,000 15,000 10,000

4

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Answer: B Feedback: The first increment is (C-A).

PW of incremental cost = PW of incremental Benefits (21,200 -15,000) = [(9,000 -900) - (8,000 -1,600)] (P/A, i, 4)+ (4,600-3,000)( P/F, i, 4) 6,200 = 1,700(P/A, i, 4) – (1,600) ( P/F, i, 4) Try, i =12% 6,200 = 1,700(3.037) + 1,600 (0.6355)? = 6179.70? No Try, i =10% 6,200 = 1,700(3.170) + 1,600 (0.6830)? = 6,481.80? No ΔROR > 10%, Choose Alt C.

The second increment is (B-C).

PW of incremental cost = PW of incremental Benefits (36,000 -21,200) = [(13,000-400) - (9,000 -900)] (P/A, i, 4) + (6,000-4,600 )( P/F, i, 4) 14,800 = 4500 (P/A, i, 4) – 1400 (P/F, i, 4) Try, i =12% 14,800 = 4500 (3.037) + 1,400 (0.6355)? = 14,556.20? No Try, i =10% 14,800 = 4500(3.170) + 1,400 (0.6830)? = 15,221.20? No ΔROR > 10%, Choose Alt. B The third increment is (D – B).

PW of incremental cost = PW of incremental Benefits (45,000-36,000) = [(15,000-1000)-(13,000-400)][P/A, i, 4] + [10,000-6,000] [P/F, i, 4] 9,000 = 1400(P/A, i, 4) + 4000 (P/F, i, 4) Try, i =2% 9,000 = 1400(3.805) + 4000(.9238) = 9022.2 ≥ No Try, i =2.5% 9,000 = 1400(3.762) + 4000(.9060) TESTBANKSELLER.COM = 8890.80? No Since ΔROR is between 2% and 2.5%, Choose Alt. B The ultimate winner is Alt. B.

True/False Table 8.2 Given the financial data for three alternatives, find:

Initial cost EUAB Life RoR MARR = 8% ΔRoR (B-C) = 9.2%

A $600 82

B C $1,200 $400 196 70 10 Years 6.1% 10.1% 11.7%

8.10. The most attractive alternative for a MARR of 8% is Alt. C. Answer: False Refer to: Table 8.2 Solution: Since the incremental rate of return between Alt. B and Alt. C is 9.2%, one would choose the higher cost alternative, alt. B for a MARR of 8%.

8.11. For a MARR of 11 %, the most attractive is Alt. C. Answer: True

Refer to: Table 8.2

Solution: Since the incremental rate of return between Alt. B and Alt. C is 9.2%, one would choose the lower cost alternative for a MARR of 11%. Also, Alt. C has rate of return of over 11%.

TESTBANKSELLER.COM 8.12. For a MARR of over 11.7% the best alternative is “Do nothing.”

Answer: True

Refer to: Table 8.2

Solution: Since none of the alternatives provides a rate of return of better than 11.7%, the answer is “True”.

Chapter 9 OTHER ANALYSIS TECHNIQUES

Reference: Case Study 9 A major equipment purchase is being considered by Metro Atlanta. The initial cost is determined to be $1,000,000. It is estimated that this new equipment will save $100,000 the first year and increase gradually by $50,000 every year for the next 6 years. MARR=10% 9.1. The payback period for this equipment purchase.

A) 8 years B) 6 years C) 5 years D) 10 years.

Answer: C Feedback: Year 0 1 2 3 4 5 6

Refer to: Case Study 9 Cash flow -1,000K 100K 150K 200K 250K 300K 350K

Cum. Cash flow Remark −1,000K −1,000+100K = −900K −900K+150K = −750K −750TKE+S2T0B 0A KN=K−S5E 50LKL ER.COM −550K + 250K = −300K -300K+300K =0 Stop. Recovered the investment 0+350K

The payback is five years.

9.2. The B/C ratio for this investment is A) 1.48 B) 0.92 C) 0.51 D) 1.13

.

Answer: B Refer to: Case Study 9 Feedback: B/C = PW of benefits/PW of costs = {100K (P/A, 10%, 6) + 50K (P/G, 10%, 6)}/1,000K = {100K (4.355) + 50K (9.684)}/1,000K = 0.9287 The answer is “B.”

9.3. The NFW of this investment is A) -$72,678 B) -$66,756 C) -$116,129 D) -$142,398

.

Answer: D Refer to: Case Study 9 Feedback: NFW = FW of benefits - FW of costs = {100K (F/A, 10%, 6) + 50K (P/G, 10%, 6) (F/P, 10%, 6)} − 1,000K (F/P, 10%, 6) = {100K (7.716) + 50K (9.684) (1.772)} − 1,000K (1.772) = -$142,398 The answer is “D”.

9.4. The B/C ratio of an investment of $10,000 that provides a benefit of $1,500 at beginning every year for 10 years is , if money is worth 9%.

A) 2.50 TESTBANKSELLER.COM B) 1.38 C) 1.79 D) 1.05

the

Answer: D Feedback: B/C = Present worth of benefits/ PW of costs = 1500 [1+ (P/A, 9%, 9)] / 10,000 = 1500 [ 1+ 5.995]/10,000 = 1.05 The answer is “D”.

Table 9 Data for four mutually exclusive alternatives are given in the table below. Assume a life of 7 years and a MARR of 9%, Alt. A

Alt. B

Alt. C

Initial Cost $5,600 EUAB $1,400 Salvage Value $400

$3,400 $1,000 $0

$1,200 $400 $0

9.5. The ∆B /∆C ratio for the first increment, (C-D) is A) 1.19

Alt. D Do Nothing $0 $0 $0 .

B) 0.84 C) 1.68 D) 1.86 Answer: C Refer to: Table 9 Feedback: ΔB/ΔC = (400 − 0)/ [1200(A/P, 9%, 7)] = 400/ [1,200(0.1987)] = 1.68

9.6. The ∆B /∆C ratio for the second increment, (B-C) is A) 1.37 B) 0.84 C) 1.45 D) 1.86

.

Answer: A Refer to: Table 9 Feedback: ΔB/ΔC = (1,000 − 400)/ [(3,400 − 1,200) (A/P, 9%, 7)] = 600/ [2,200(0.1987)] = 1.37 The answer is “A”.

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9.7. The ∆B /∆C ratio for the third increment, (A-B) is A) 0.84 B) 0.92 C) 0.75 D) 0.69

.

Answer: B Refer to: Table 9 Feedback: ΔB/ΔC = (1,400 − 1,000)/ [(5,600 − 3,400) (A/P, 9%, 7)] = 400/ [2,200(0.1987)] = 0.92 The answer is “B”.

9.8. The best alternative using B/C ratio analysis is A) Alt. C

.

B) Alt. A C) Alt. B D) Alt. D Answer: B Refer to: Table 9 Feedback: The better alternative between” C” and “D” was found to be “C”.

On the next increment, it was found that “B” was better.

On the final increment, it was found that “B” was better. Therefore, “B” is the ultimate winner.

9.9. Tara invests $2,500 today and another $1,500 a year from now. Her investments starting year 2 keeps increasing by $100 every year for the next 10 years from today. She stops investing from year 11 until year 20. If she earns a rate of return of 7% on her investments, determine future worth of her investments 20 years from now?

A) $39,134 B) $61,175 C) $48,184 D) $59,290

TESTBANKSELLER.COM Answer: B Feedback: The cash flows are shown below.

Year 0 1 2 3 …. 10 11 …. 20 Cash flow -2,500 -1,500 -1,600 -2,400 0 F F is to be evaluated.

F = [2,500 + 1,500 (P/A, 7%, 10) + 100 (P/G, 7%, 10)] {F/P, 7%, 20} = [2,500 + 1,500 (7.024) + 100 (27.716)] {3.870} = $ 61,175.41 The answer is “B”.

9.10. Using the cost and benefits data for four different alternatives in table below, find the best alternative using the payback period analysis. MARR=7% Year A B C D 0 -$9,000 -$12,000 -$10,000 -$12,000 1 2,000 3,500 0 0

2 3 4 5 6

2,000 2,000 2,000 1,000 3,000

0 3,000 3,500 0 3,000

1,000 2,000 3,000 4,000 5,000

-2,000 0 18,000 0 0

A) A B) B C) C D) D Answer: D Feedback: Payback period (A) = 4 + (9,000 − 2,000 − 2,000 − 2,000 − 2,000)/1,000 = 5 Years Payback period (B) = 5 + (12,000 − 3,500 − 3,000 − 3,500)/3,000 = 5.67 Years Payback period (C) = 5 + (10,000 – 0 − 1,000 − 2,000 − 3,000 − 4,000)/5,000 = 5 Years Payback period (D) = 3 + (12,000 – (−2,000))/18,000 = 3.78 Years Alternative “D” has the smallest payback period.

9.11. Given the data for two a l t e rTn E a t iSv T e sB , A chN ooKsS eE thL e Lb E et tRer. aCl tOe rMnat i ve using the B/C ratio analysis.

MARR = 8%

Alternative First cost Operating costs/year Benefits /year Dis-benefits/year Life in years

X $100,000 50,000 100,000 30,000 5

Y $140,000 60,000 120,000 25,000 10

A) Alt. X B) Alt. Y Answer: Alt. Y Feedback: EUAC (Alt. X) = 100,000 (A/P, 8%, 5) + 50,000 = 100,000(0.2505) +50,000 =75,050 EUAB (Alt. X) = 100,000 − 30,000 = 70,000 EUAC (Alt. Y) = 140,000 (A/P, 8%, 10) + 60,000 = 140,000(0.1490) +60,000 = 80,860 EUAB (Alt. Y) = 120,000 − 25,000 = 95,000 (∆B) / ∆C = (95,000 – 70,000) / (80,860-75,050) = 3.44 Choose Alt. Y, the higher cost alternative.

Short Answer 9.12. Tech Engineering is considering whether to lease or buy earth-moving equipment.

MARR =8% The costs of buying an earth mover are as follows.

Purchase Price = $150,000 Life = 10 years Salvage value = $12,000 Maintenance = $1,800/year Insurance = $1,200 /year Operating cost day = $300/day If the equipment is leased, in addition to incurring an operating cost of $300 per day, the company has to pay a rental fee of $100 per day.

Determine the minimum number of days per year that the company has to use the earth mover in order justify buying the equipment.

Answer: 245.22 Feedback: Assume the “X” be the minimum # of days the equipment has to be operated per year.

N/KPS L,L Cost of owning equipment/year =T1E5S 0,T 00B0A(A , 8E% 10E)R–.

12C,0O0M 0(A/F,8%,10) +1,800+1,200+300X = 150,000(0.1490) − 12,000(0.0690) +1,800 +1,200 +300X = 24,522 + 300 X 24,522 + 300 X = 400 X X = 24,522 / (400-300) = 245.22 days.

9.13. Given the financial data in the table below for two mutually exclusive alternatives, determine the value “X” for the two alternatives to be equally attractive. Use an interest rate of 9% per year.

P Q Initial cost $2,500 $4,000 Annual benefit 400 X Salvage value 1,200 1,500 Life 5 years Answer: $835.52 Feedback: EUAW (P) = 500- [2,500(A/P, 9%, 5) – 1,200 (A/F, 9%, 5)]

= 500- [2,500(0.2571) – 1,200) (0.1671)] = $57.77 EUAW (B) = X – [4,000(0.2571) – 1,500(0.1671)] = X − 777.75 For the two alternatives to be equally attractive, X – 777.75 = 57.77 X = $835.52

True/False 9.14. The payback period analysis is the not the best decision making method for high tech industry as the method chooses the best choice based on how fast an investment can be recovered.

Answer: False Feedback: The answer is “False” as the payback is intended to figure out how fast an investment can be recovered rather than using an economic decision making method based on discount rates. In a high tech industry as technology is rapidly changing, it is important ti recover the investments as fast as possible. Therefore, this method is highly applicable for high tech industry.

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9.15. For an alternative to be considered as a viable alternative, the benefit /cost ratio must be greater than or equal to 1.

Answer: True Feedback: The answer is “True” as for an alternative to be viable the benefits must be greater than or equal to the cost. Therefore the B/C ratio must be greater than or equal to one.

9.16. If the B/C ratio for three alternatives A, B, and C are 2.8, 3.3 and 1.95 respectively, the best alternative is always “B” as it has the largest B/C ratio.

Answer: False Feedback: The answer is “False” as an incremental analysis is required to choose the best alternative.

9.17. Sensitivity analysis is concerned with determining how much variation in financial data, the decision maker can have to affect the economic decision Answer: True Feedback: The answer is “True” as sensitivity analyses are usually applied to determine the effect of changes in values for a chosen decision.

9.18. A continuous improvement team has helped to save $20,000 for the company on a process that will not be changed for the next 10 years. If the team has spent $50,000 on the improvement project, the net present worth (NPW) of savings on this improvement project is $150,000 at a MARR of 5%.

Answer: False Feedback: NPW = 20,000(P/A, 5%, 10) − 50,000 = 20,000(7.722) − 50,000 = $104,440

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CHAPTER 10 UNCERTAINITY IN FUTURE EVENTS Multiple Choice 10.1. MKBIC Systems is evaluating four projects A, B, C and D that have risks associated with the producing benefits. Based on the data given in the table below, which project is the best alternative?

Project A EUAW $2,500 $1,800 $3,200

Prob.

0.3 0.45 0.25

Project B EUAW Prob. $3,000 0.1 -$2,500 0.3 $4,000 0.6

Project C EUAW Prob.

-$5,000 0.25 $6,500 0.45 $2,000 0.3

Project D EUAW Prob. $4,000 0.35 $2,500 0.4 -$1,500 0.25

A) Project A B) Project B C) Project C D) Project D Answer: A Feedback: Expected EUAW (A) = (2,500) (0.3) + (1,800) (0.45) + (3,200) (0.25) = $2,360 Expected EUAW (B) = (3,000) (0T.1E)S+T(B −2A,N 50K0S )E (0L .3L) E+ R( 4., 0C0O0 )M( 0 . 6 ) = $1,950 Expected EUAW (C) = (−5,000) (0.25) + (6,500) (0.45) + (2,000) (0.3) = $2,275 Expected EUAW (D) = (4,000) (0.35) + (2,500) (0.4) + (−1,500) (0.25) = $2,025 Project A has the greatest expected value and should be selected.

The answer is “A”.

10.2. A new equipment is being considered at a local company at a cost of $200,000. The operation and maintenance costs of this equipment are estimated to be about $20,000 per year.

Salvage value is expected to be $40,000 at the end of its useful life. The life of this equipment is estimated to vary anywhere from 5 to 9 years with the associated probabilities as shown in the table below. If an interest rate of 10% is used, what is the expected EUAC for this equipment?

Life, Years Probability A) $66,460 B) $55,970 C) $62,960 D) $58,206.2

5 0.3

6 0.1

7 0.15

8 0.20

9 0.25

Answer: D Feedback: EUAC for a 5-year life = 200,000(A/P, 10%, 5) + 20,000 – 40,000(A/F, 10%, 5) = 200,000(0.2638) + 20,000 – 40,000(0.1638) = $66,208 EUAC for a 6-year life = 200,000(0.2296) + 20,000 – 40,000(0.1296) = $60,734 EUAC for a 7-year life = 200,000(0.2054) + 20,000 – 40,000(0.1054) = $56,864 EUAC for a 8-year life = 200,000(0.1874) + 20,000 – 40,000(0.0874) = $53,984 EUAC for a 9-year life = 200,000(0.1736) + 20,000 – 40,000(0.0736) = $51,776 Expected EUAC = 0.30(66,208) + 0.10(60,734) + 0.15(56,864) + 0.20(53,984) +0.25(51,776) = $58,206.2 The answer is “D”.

Reference: Table 10 First Cost, $ Probability

$60,000 0.25

$80,000 0.35

$100,000 0.30

$120,000 0.10

10.3. Find the expected EUAW from the financial data provided in the table for a new equipment. Because of the uncertainty of technology being used in this equipment, it has not EhLeLaEnnRu.aC been possible to get the initial cT osEt S acTcB urAatNelKyS .T l bOeMn e f i t , however, is estimated to be $25,000 with a possible equipment life of 5 years. The salvage value is expected to be 10% of the initial cost. MARR =8% A) $2977 B) $5157 C) $3957 D) $4628 Answer: B Refer to: Table 10 Feedback: Expected first cost = [60,000(0.25) +80,000(0.35) +100,000 (0.30) + 120,000(0.10) = $85,000 Expected salvage value = 85,000(0.10) = $8,500 Expected EUAW = 25,000 – [85,000 (A/P, 8%, 5) − 8,500 (A/F, 8%, 5)] = 25,000 – [85,000 (0.2505) − 8,500 (0.1705)] = $5156.75 The answer is “B”.

10.4. Determine the associated risk measure in this equipment investment in terms of standard deviation.

A) $4,923 B) $6,123 C) $4437.8 D) $8,523 Answer: C Refer to: Table 10 Feedback: EUAW for an initial cost of $60K = 25,000 – [60,000 (0.2505) − 6,000 (0.1705)] = $10,993 EUAW for an initial cost of $80K = 25,000 – [80,000 (0.2505) − 8,000 (0.1705)] = $6,324 EUAW for an initial cost of $100K = 25,000 – [100,000 (0.2505) − 10,000 (0.1705)] = $1,655 EUAW for an initial cost of $125K = 25,000 – [120,000 (0.2505) − 12,000 (0.1705)] = −$3,014 Standard deviation, σ = [EV (X2) – {EV(X)}2]1/2 = [0.25 (10,993)2 + 0.35 (6,324)2 + 0.30 (1,655)2 + 0.10 (−3,014)2 – {5157}2]1/2 = $4437.8 The answer is “C”.

10.5. The probabilities that the life of a machine will vary from 6 to 12 years are given in the table below. The expected life of the machine is .

Life , Years

5

Probability

0.10

TESTBANKSELLER.COM 6

7

8

9

10

11

12

0.15

0.2

0.15

0.20

0.10

0.05

0.05

A) 8.70 years B) 9 years C) 10.15 years D) 7.95 Answer: D Feedback: Expected life = 0.10(5) +0.15(6) + 0.2(7) + 0.15(8) + 0.20(9) + 0.10(10) +0.05(11) +0.05(12) = 7.95 years The answer is “D”.

10.6. What is standard deviation of the equipment life A) 1.73

.

B) 1.29 C) 1.88 D) 1.54 Answer: C Feedback: EV (X2) = 0.10(5)2 +0.15(6)2 + 0.2(7)2 + 0.15(8)2 + 0.20(9)2 + 0.10(10)2 +0.05(11)2 +0.05(12)2 = 66.75 Standard deviation of equipment life, σ = [EV (X2) – {EV(X)}2]1/2 = [66.75 − 7.952]1/2 = 1.88 The answer is “C”.

10.7. An equipment proposal from two vendors A and B have been received from a local company. The quality of these equipment have been varying and consequently the cost of manufacturing the product using these two different equipment has also varied. The varying costs based on the probabilities of defects are given in table below.

Determine the better vendor.

Vendor A

Prob.Defect.

0.10

0.25

0.10

0.12

0.13

0.16

0.14

Mfg. Cost

$20

25

19

21

23

22

20.80

Vendor B

Prob.Defect.

0.14

0.12

0.19

0.06

0.24

0.08

0.17

Mfg. Cost

$20

18

25

14

30

16

22

TESTBANKSELLER.COM

A) Vendor A B) Vendor B Answer: A Solution: Expected mfg. cost (Vendor A) = 0.10(20) +0.25(25) +0.10(19) +0.12(21) +0.13(23) +0.16(22) +0.14(20.80) = $22.09 Expected mfg. cost (Vendor B) = 0.14(20) +0.12(18) +0.19(25) +0.06(14) +0.24(30) +0.08(16) + 0.17(22) = $22.77 Choose vendor A to minimize manufacturing cost.

10.8. A continuous improvement has helped a company to raise savings with associated probabilities shown in the table below. The useful life is 5 years with a probability of0.6 and 3 years with probability of 0.4.

i. ii.

Determine the joint probability distribution for savings per year and useful life.

Determine the expected NPW if an investment of $50,000 is required. No salvage is expected. Use a MARR of 10%

Annual Benefit $15.000 $20,000 $25,000

Probability 0.25 0.45 0.3

Answer: D Feedback: Annual Benefit $15.000

Prob.

0.25

Life 5

$20,000

0.45

5

$25,000

0.3

5

$15.000

0.25

3

$20,000

0.45

3

$25,000

0.3

3

Prob.

0.6

Joint Prob.

NPW 0.15 {15,000 (P/A, 10%, 5) – 50,000} [0.15] = {15,000 (3.791) – 50,000} [0.15] = $1,029.75 0.6 0.27 = {20,000 (3.791) – 50,000} [0.27] = $6,971.4 0.6 0.18 = {25,000 (3.791) – 50,000}[0.18] = $8,059.5 0.4 0.10 {15,000 (P/A, 10%, 3) – 50,000} [0.10] = {15,000 (2.487) – 50,000} [0.10] = −$1,269.5 TES0T.4BANKS0E.1L8LER .=C{O2M 0,000 (2.487) – 50,000}[0.18] = −$46.8 0.4 0.12 = {25,000 (2.487) – 50,000}[0.12] = $1,461

Expected NPW = $1,029.75 + 6,971.4 + 8,059.5 -1,269.5-46.80+1,461= $16,205.25 The answer is “D”.

True/False 10.9. A simple rule of thumb for comparing a project’s risk and return by Joe Hardman is : If the expected worth is ≥ 2 standard deviation of the present worth, then the project is safe to invest.

Answer: True Feedback: Since a normal distribution has about 2.5% of its value less than 2 standard deviation below the mean. The answer is “True”.

10.10. At Tech Engineering, a new equipment is expected to cost $100,000. The annual savings are estimated to be $12,000, $15,000, and $20,000 respectively for 10 years of its life with probabilities of 0.4, 0.5, and 0.1 respectively. The annual savings for this equipment is $15,667.

Answer: False Feedback: The expected leasing cost = [12,000 (0.4) + 15,000 (0.5) + 20,000 (0.1)] = $14,300

10.11. An E-Commerce business has estimated the sales revenue for this Christmas season during a four-week period from Black Friday at $80M, $75M, $62M and $85M with probabilities of 0.5,0,2,0.1 and 0.2 respectively. The expected sales then may be estimated to be for this Christmas season is $76.5 M. Answer: False Feedback: The expected sales = [80M (0.5) + 75M (0.20) + 62M (0.1) + 85M (0.2)] = $78.2M. The answer is “False”.

TESTBANKSELLER.COM 10.12. If the annual cash flows of an investments are: $1,000, $1,500 and $2,000 with probabilities of 0.1, 0.7 and 0.2 respectively. Then expected annual cash flow is $1,800.

The expected sales then may be estimated to be for this Christmas season is $50.2 M Answer: False Feedback: Expected annual cash flow = [0.1 ($1,000) + 0.7 ($1,500) + 0.2 ($2,000)] = $1,550.

The answer is “False”.

10.13. One common measure of risk is the probability of a loss.

Answer: True Feedback: Risk may be thought of as the probability of getting an outcome other than the expected value such as a loss. The answer is “True”.

10.14. If the optimistic, most likely and pessimistic estimates of the life of an asset are 6, 8 and 10 respectively, then the estimates of the life is 8 years Answer: True Feedback: Estimate of life = [6 (1) + 8 (4) + 10 (1)] / 6 = 8. The answer is “True”.

CHAPTER 11 DEPRECIATION Multiple Choice 11.1. Jain Mart is to depreciate an asset bought for $500,000 using the SOYD method over a life of 8 years. If the depreciation charges in year 3 was $80,000, determine the salvage value used in computing the depreciation charges in year 3.

A) $50,000 B) $20,000 C) $100,000 D) $76,000 Answer: B Feedback: Sum of years digit for 8 years = 8 (8+1)/2 = 36 Depreciation in year 3 = 6/36(500,000 –X) = 80,000 X = 500,000 - 80,000 (36/6) = $20,000

11.2. An equipment at MNS Systems costing $600,000 was depreciated using the double Rp.aCnyOM declining balance (DDB) methodT. EInST yeBarAfNoK urS, E thL e LcE om decided switch to the straight-line depreciation method. Determine the depreciation charges in year 4. Assume a depreciable life of 10 years and a salvage value of $63,331.

A) $32,000 B) $50,000 C) $36,667 D) $40,000 Answer: C Feedback: DDB depreciation in year 1 = (2/10) (600,000) = $120,000 Book value at the end of year 1 = 600,000 -100,000 = $500,000 DDB depreciation in year 2 = (2/10) (500,000) = $100,000 Book value at the end of year 2 = 500,000 - 100,000 = $400,000 DDB depreciation in year 3 = (2/10) (400,000) = $80,000 Book value at the end of year 3 = 400,000 -80,000 = $320,000 SL depreciation in year 4 = (320,000 -63,331)/7 = $36,667 The answer is “C”.

11.3. An equipment bought at cost of $Y will be depreciated using either the SOYD (sum- of years - digit method) or SL (straight-line) method. This equipment will be depreciated over a period of 5 years with a salvage value of $X. Determine the percentage increase in depreciation charges in year 2 for the SOYD method over the SL method.

A) 10% B) 33.33% C) 0% D) 16.67% Answer: B Feedback: Sum-of-years digit for 5 years = 5(5+1)/2 = 15 Depreciation in year 2 using SOYD method = 4/15(Y-X) Depreciation in year 2 using SL method = 1/5(Y-X) % increase = [[4/15(Y-X) – 1/5(Y-X)]/(1/5)*(Y-X)] 100% = [[4/15 – 3/15]/[3/15]] *100 = 33.33% The answer is “B”.

11.4. An income producing asset costing $120,000 is being depreciated using the 150% Declining Balance method with TaEsSaT lvB agAeNvK alSuE eL ofL$E2R 0,.

00C0O, M determine the depreciation in year 3 assuming the equipment will be depreciated over a life of 5 years.

A) $37,500 B) $32,500 C) $17,640 D) $28,125 Answer: C Feedback: First year depreciation = (1.5/5) (120,000) = $36,000 Book value at the end of year 1 = 120,000 -36,000 = $84,000 Second year depreciation = (1.5/5) (84,000) = $25,200 Book value at the end of year 2 = 84,000 -25,200 = $58,800 Third year depreciation = (1.5/5) (58,800) = $17,640 The answer is “C”.

11.5. An automated inspection system purchased at a cost of $200,000 by Mega Tech Engineering was depreciated using the MACRS method. The system was sold after 4 years for $150,000. Determine the depreciation recapture on this equipment.

A) $50,000 B) $0 C) $ 37,488 D) $87,520 Answer: D Feedback: The automated inspection system is a 7-year property according to MACRS GDS Property Class.

The accumulated depreciation through year 4 = (0.1429 + 0.2449 +0.1749 +0.1249) 200,000 = $ 137,520 Book value at the end of year 4 = 200,000 – 137,520 = $62,480 Depreciation recapture = 150,000 – 62,480 = $87,520.

The answer is “D”.

11.6. Paragon Properties built a shopping center at a cost of $50M in year 2010. The company started leasing space in July of 2014. The land was purchased for $5M. Determine the depreciation charges through 2017 if the property was sold in November 2017.

A) $3,563,100 B) $5,002,200 TESTBANKSELLER.COM C) $3,991,050 D) $4,615,200 Answer: B Feedback: The percentage depreciation charges in year 2014 = 1.177% The percentage depreciation charges in year 2014, 2015 and 2016 = 2.564% The depreciation charges in year 2017 = 2.247% (Sold in November) Note that the cost of land is not depreciable.

Total depreciation charges = (0.01177 + 0.02564 + 0.02564 + 0.02564 + 0.02247) (50M- 5M) = $5,002,200 The answer is “B”.

Case Study 11.7 Given: Initial Cost, P = $50,000 Depreciable Life, N= 5 Years

Salvage Value at the end of 5 years, S = $10,000.

Year 1 2 3 4 5 Total

Projected Production , units 4,500 5,000 3,500 5,500 6,500 25,000

Actual Production , units 5,000 4,000 3,000 5,000 Not known

11.7. Determine the depreciation in year 4 using the UOP method.

A) $4,000 B) $5,600 C) $8,000 D) $3,200 Answer: C Feedback:

Refer to: Case Study 11.7 Year 1 2 3 4

UOP Depreciation (50,000 - 10,000)(5,000/25,000) = $8,000 (50,000 - 10,000)(4,000/25,000)= $6,400 (50,000 - 10,000)(3,000/25,000)= $4,800 (50,000 - 10T,0E0S 0)T(5B,0A0N0K /2S 5,E 00L0L)=ER $8.,0C0O 0M

The answer is “C”.

11.8. If the equipment in problem 11.7 is sold at the end of year 4 for $30,000, what is the depreciation recapture on this equipment?

A) $4,000 B) $5,600 C) $7,200 D) $8,400 Answer: C Refer to: Case Study 11.7 Feedback: Book Value of the equipment at the end of year 4 = $50,000 – (8,000 +6,400 + 4,800 +8,000) = $22,800 Depreciation recapture = 30,000 – 22,800 = $7,200.

The answer is “C”.

Short Answer 11.9. ABC Inc., a profitable corporation, purchased equipment for $20,000. The equipment may be depreciated using the following two depreciation methods for 5 years with a salvage value of $5,000.

The two methods are: i. DDB ii. SOYD Develop depreciation schedules for both the methods.

Answer: N = 5 years

B = $20,000

S = $5,000

i. DDB Year 1 2 3 4 5

Depreciation Book Value 2/5(20,000) = 8,000 20,000 – 8,000 = 12,000 2/5(12,000) = 4800 12,000 – 4800 = 7,200 2/5(7,200) = 2,200 7,200 – 2,200 = 5,000 T0 ESTBANKSELLER.COM 5,000 0 5,000

ii. SOYD SOYD = 5(5+1)/2 = 15 Year 1 2 3 4 5

Depreciation 5/15(20,000 - 5,000) = 5,000 4/15(15,000) = 4,000 3/15(15,000) = 3,000 2/15(15,000) = 2,000 1/15(15,000) = 1,000

Book Value 20,000 – 5,000 = 15,000 11,000 8,000 6,000 5,000

Case Study 11.10 The initial investment on a commercial building was $1M excluding the land. The building was occupied in March of the year in which it was purchased. After 10 years, the property was sold in the 11th year on September 2017 for $2M. The land cost was appraised at $100,000 for tax purposes.

11.10A. Determine the first year depreciation of the building.

Answer: $20,330 Refer to: Case Study 11.10 Feedback: Depreciation in year 1 = 0.02033*(1M) = $20,330

11.10B. What is the depreciation for the year in which the property is disposed?

Answer: 18,190 Refer to: Case Study 11.10 Feedback: Depreciation in year 11 = 0.01819*(1M) = 18,190

11.10C. What is the depreciation recapture for this asset?

Answer: $269,280 Refer to: Case Study 11.10 Feedback: Depreciation Recapture = [0.02033 + 0.01819 + 9*(0.02564)] * 1M = $269,280

11.10D. What is the capital gain on this asset?

Answer: $900,000 RefTeE r tS o:TCBaAseNSKtS udEyL1L 1.E1R 0 .COM Feedback: Capital Gain = 2M – 1M – 100,000 = $900,000.

11.11 Determine the depletion charges using the percentage depletion method for the first year only for a coal mine. The applicable rate for the percentage depletion method is 10%. Cost to acquire mine rights = $150,000 Estimated mine size = 40,000 Tons Selling price = $80 per Ton Amount sold in year 1 = 2,000 Tons Operating cost in year 1 = $50,000 Answer: $16,000 Feedback: Depletion charges in year 1 = (0.10) (2,000) (80) = $16,000 Taxable income in year 1 = 2,000(80) - 50,000 = $110,000 50% of taxable income = 0.50(110,000) = $55,000 Allowable depletion charges in year 1 = Minimum of (16,000; 55,000) = $16,000

True/False 11.12. Land is a depreciable asset as land appreciates overtime.

Answer: False Feedback: The answer is “False”. In spite of the fact that land appreciates in value over a time, it is not a depreciable asset as it does not wear out.

11.13. If an asset is sold before the end of its useful life for more than its salvage value, the difference between the two amounts is defined as capital gain for tax accounting.

Answer: False Feedback: The answer is “False” as depreciation recapture is the difference between selling price and book value.

11.14. An asset was sold for $50,000 at the end of its useful life of 7 years. The equipment was bought for $400,000. If it has been depreciated as a 7-year MACRS property, the depreciation recapture on this property is $50,000.

Answer: False Feedback: Book value at the end of year 7 = 0.0446(400,000) = $17,840 Depreciation recapture = Selling price- Book value. = 50,000 -17,840 = $32,160 The answer is “False”.

11.15. Depreciation may be caused by obsolescence.

Answer: True Feedback: The answer is “True” as some equipment may be still functional but lose their value because of newer models that have been introduced due to innovations in technology.

11.16. The straight – line (SL) method is often used for intangible property such as patents.

Answer: True Feedback: The answer is “True”. It is easier to compute depreciation charges using the SL method.

CHAPTER 12 INCOME TAXES FOR CORPORATIONS Multiple Choice 12.1. An equipment purchased at a cost $80,000 by a local company is being depreciated using MACRS method as a 5-year property. At the end of four years, the management decided to sell the equipment for a modest price of $20,000. The company is in the 34% tax bracket. Compute the tax consequence on the sale of this equipment.

A) $6,800 B) $926.40 C) $2,100 D) None of these Answer: C Feedback: Book value at the end year 4 = 80,000 – (0.20 + 0.32 + 0.192 + 0.1152) (80,000) = $13,824 Depreciation Recapture = $20,000 – $13,824 = $6,176 Income taxes on depreciation recapture = (0.34) 6,176 = $2,100.

The answer is “C”.

TESTBANKSELLER.COM 12.2. Eric Systems located in Georgia expects a 10% after-tax rate of return on equipment investments. The state income tax rate is 6%. If the company is in the 34% federal tax bracket, estimate the before-tax rate of return required.

A) 11.36% B) 13.80% C) 16.12% D) None of these Answer: C Feedback: Combined tax rate = (State tax rate) + (Federal tax rate) (1- State tax rate) = 0.06 + 0.34(1 - 0.06) = 0.3796 or 37.96% Before tax rate of return = (After-tax rate of return)/ (1- incremental tax rate) = (0.10) / (1 − 0.3796) = 0.1612 or 16.12% The answer is” C”.

Case Study 12 First cost of equipment = $150,000 Market value at the end of year 6 = $30,000

MACRS depreciation is used. The equipment is a 5-year property.

Incremental income-tax rate for the company = 35% Year BT-CF in $ O&M Expenses

0 -150K

1 60K 10K

2 63K 13K

3 66K 16K

12.3. The first-year after tax-cash flow is A) $33,000 B) $27,000 C) $53,000 D) $40,000

4 69K 19K

5 72K 22K

6 75K 25K

.

Answer: C Refer to: Case Study 12 Feedback: Year 1: Taxable income = BT-CF − Depreciation charge − O&M Expenses = 60,000 − 0.20(150,000) – 10,000 = $20,000 Income tax = 0.35(20,000) = $7,000 AT-CF = 60,000 – 7,000 = $53,000 The answer is “C”.

12.4. The fourth -year taxable income is equal to A) $500 B) $32,750 C) $50,000 D) $51,720

.

Answer: B Refer to: Case Study 12 Feedback: Year 2: Taxable income = BT-CF − O&M Expenses − Depreciation charges in year 4 = 69,000 – 19,000 − 0.1152 (150,000) = $32,720 The answer is “B”.

12.5. The tax on depreciation recapture in year 6 is equal to A) $6,000 B) $3,000

.

C) $10,500 D) $10,000 Answer: C Refer to: Case Study 12 Feedback: Depreciation recapture in year 6 = Market value – Book value = 30,000 - 0 = $30,000 Tax on depreciation recapture = 0.35(30,000) = $10,500 The answer is “C”.

12.6. A distribution center purchased an equipment for $100,000 and has depreciated the equipment using the MACRS depreciation schedule as a 7-year property. The operating income in year 2 was $200,000 and the expenses were $87,000. If the company is in the 40% income tax bracket, determine the after-tax cash flow in year 2 The tax on depreciation recapture in year 2 is equal to .

A) $40,00 B) $35,404 C) $24,490 D) None of these Answer: B Feedback: Depreciation in year 2 = 0.2449 (100,000) = $24,490 Taxable income = 200,000 – 87,000 – 24,490 = $88,510 Income tax in year 2 = 0.40 (88,510) = $35,404 The answer is “B”.

Short Answer 12.7. For a company with a taxable income of $10.5M, determine the: A) marginal and B) average tax rates Answer: A) 35% B) 34.04% Feedback: a) The marginal rate is 35% as the income is between $10 million and $15 million. b) Average income rate = [$3.4 M + (10.5 – 10) (0.35)]/10.5

= 0.3404 or 34.04% YES/NO 12.8. A production equipment at a cost of $500,000 has been purchased by a contract manufacturing company to meet the specific needs of customer that had awarded a 4-year contract with the possibility of extending the contract for another 4 years. The company plans to use the MACRS depreciation method for this equipment as a 7-year property for tax purposes.

The income tax rate for the company is 39%, and it expects to have an after -tax rate of return of 12% for all its investments.

The equipment generated an annual income of $100,000 for the first four years. The customer decided not to renew the contract after 4 years. Consequently, the company decided to sell the equipment for $180,000 at the end of 4 years. Determine if the company obtained the expected after-tax rate of return on this investment.

Answer: NO Feedback: Year 0 1 2 3 4 4

BT-CF, $ - 500,000 100,000 100,000 100,000 100,000 180,000

MACRS Depreciation Charges, $

Taxable Income, $

Income Tax @ 39%, $ AT-CF, $

0.1429(500T,0E0S0T ) B=A7N 1,K 45S0EL L2E8R ,5.5C 0 OM 11,251.50 0.2449(500,000) =122,450 0.1749(500,000) =87,450 0.1249(500,000) =62,450

-22,450 12,550 37,550 23,800†

−8,755.50 4,894.50 14,644.50 9,828.00

-500,000 88,748.50 108,755.50 95,105.50 85,355.50 170,718

Book value at the end of year 4 = $500,000 – (71,450+122,450+87,450+62,450) = $156,200 † Depreciation recapture = $180,000 − $156,200 = $23,800 (Note that this is taxed as ordinary income) NPW of AT-CF @ 12% = 88,748.50 (P/F, 12%, 1) + 108,755.50 (P/F, 12%, 2) + 95,105.50 (P/F, 12%, 3) + 85,355.50 (P/F, 12%, 4) + 170,718 (P/F, 12%, 4) − 500,000 =88,748.50 (0.8929) + 108,755.50 (0.7972) + 95,105.50 (0.7118) + (85,355.50) (0.6355) + 170,718 (0.6355) − 500,000 = −$103,626 The company did not get the 12% after-tax rate of return desired.

True/False 12.9. The income tax rates are the same for capital gains and depreciation recapture of an asset.

Answer: False Feedback: It is not uncommon for capital gains taxes to be different from those of ordinary income, whereas the depreciation recapture is treated as ordinary income and therefore the tax on depreciation recapture could be as high as 39%. The answer then is “False”.

12.10. A small business sold an equipment for $100,000 after depreciating the equipment using SOYD depreciation method. The federal tax liability on this depreciation recapture is $34,000 if the company also had other taxable income of $150,000 in that year.

Answer: False Feedback: The company will be paying a 39% tax on this depreciation recapture based on the taxable income. Income tax on depreciation recapture = 0.39 (100,000) = $39,000 The answer is “False”.

12.11. A firm operates in a state that has a corporate income tax rate of 5% and is deductible from the federal taxes. If the incremental federal tax is 34%, then the combined effective tax rate is 35.9% Answer: False Feedback: Combined tax rate = State income tax rate + (1 – State income tax rate) Federal income tax rate = 0.05 + (1 – 0.05)0.34 = 0.373 = 37.3% The answer is “False”.

12.12. Johnson’s Inc. made $250,000 in 2015 in consulting business in Tennessee. The company had other taxable income of $1,000,000 in year 2015. The federal income tax for the consulting work done in Tennessee is $85,500 Answer: True Feedback: Total taxable income for Johnson’s in year 2015 = 1,000,000 +250,000 = 1,250,000.

This amount of taxable income will put the company in the 34% tax bracket.

The income tax on this $250,000 consulting fee earned in Tennessee = 0.34(250,000) =$85,000 The answer is “True”.

12.13. A piece of property bought by XYZ Corporation a few years ago was sold for $5 M. The cost basis for this property was $2.75 M. The company had a taxable income of $12.15 million in the year the property was sold.

The capital gain tax on this property is $337,500.

Answer: True Feedback: Capital Gain = $5 M − $2.75 M = $2.25 M Capital gain taxes are computed at 15%.

Therefore, capital gain tax = 0.15 × 2,250,000 = $337,500 The answer is “True”.

TESTBANKSELLER.COM

CHAPTER 13 REPLACEMENT ANALYSIS Multiple Choice 13.1. An equipment cost $80,000 initially. The market value has been declining at the rate of $15,000 yearly. The O & M costs in year 1 were $10,000 and have been increasing by $2,000 from year 2. Determine the minimum cost life of this equipment for a MARR of 10 %.

A) 4 years B) 2 year C) 1 year D) 5 years Answer: C Feedback: Year, n 1 2 3 4 5

Loss in Interest in O&M cost in Marginal cost market value year, n year, n in year, n 15,000 15,000 (0.1) 10,000 11,500 = 1,500 15,000 1,500 12,000 13,500 15,000 1,500 14,000 15,500 1,5 0 0 1 6, 00 0 15,000 TESTBAN KSELLER.C O1M7,500 15,000 1,500 18,000 19,500

EUAC 11,500 12,452 13,842

EUAC (2) = [11,500 (P/F, 10%, 1) + 13,500 (P/F, 10%, 2)] [A/P, 10%, 2] = [11,500 (0.9091) + 13,500 (0.8264)] [0.5762] = $12,452 EUAC (3) = [12,452 (P/A, 10%, 2) + 15,500 (P/F, 10%, 3)] [A/P, 10%, 3] = [12,452 (1.736) + 15,500 (0.8264)] [0.4021] = $13,842 Since EUAC is increasing every year since year 1, the minimum cost life is one year.

The answer is “C”.

13.2. The O & M cost of an equipment is given in the table below. Determine the EUAC of this equipment in year 6.

Assume a MARR of 8%.

Year 1 2 3 4 5 6 O & M Cost $8,000 7,500 $6,000 $5,500 $4,500 8,000 A) $8,188.45

B) $9,149.87 C) 7,407.20 D) $7,168.52 Answer: C Feedback: EUAC (Year 6)

= 8,000 – 500(A/G, 8%, 6) + 4,000(A/F, 8%, 6) = 8,000 – 500(2.276) + 4,000(0.1363) = $7,407.20

The answer is “C”.

13.3. The EUAC of the challenger for an economic life of 5 years is $8,500. The marginal cost of the defender is provided in the table below. Determine when the defender should be replaced.

Year Marginal Cost, $

1 7,500

2 8,000

3 8,500

4 9,000

5 9,500

A) At the end of year 2 B) At the end of year 1 C) At the end of year 3 D) Never Answer: C Feedback: Since the marginal cost of the defender is greater than the EUAC of the challenger in year 4, it would economical to replace the defender at the end of year 3.

The answer is “C”.

13.4. Determine the minimum cost life of an equipment for which the marginal costs are given in table below. Use an interest rate of 10%.

Year Marginal Cost , $ A) 6 years B) 4 years C) 2 years D) 3 years

1 2 18,000 16,000

3 19,000

4 5 6 20,000 22,000 24,000

Answer: C Feedback: The computations are shown in table below.

Year 1 2 3 4

Equivalent Uniform Annual Cost (EUAC) $18,000 {18,000 (P/A, 10%, 1) + 16,000 (P/F, 10%, 2)}{A/P, 10%, 2} = {18,000 (0.9091) + 16,000 (0.8264)}{0.5762} = $17,047.57 {17,047.57 (P/A, 10%, 2) + 19,000 (P/F,10%,3)}{A/P,10%,3} = {17,047.57 (1.736) + 19,000(0.7513)}{0.4021} = $17,639.84 {17,639.84 (P/A, 10%, 3) + 20,000 (P/F,10%,4)}{A/P,10%,3} = {17,639.84 (2.487) + 20,000(0.6813)}{0.3155} = $18,140.08

Since the EUAC for the year 2 is the lowest, the economic life is 2 years.

The answer is “C”.

Case Study 13 A CNC machining center was puTrE chSasTeB d A6NyK eaSrsEaLgLoEfoRr.$C 10O0M,000. The current market value is $200,000, which will decline as follows over the next 5 years: $140,000, $130,000, $120,000, $100,000, and $80,000. The O & M costs are estimated to be $36,000 this year. These costs are expected to increase by $5,000 per year starting year 2. MARR = 10% 13.5. The marginal cost for defender in year 2 is A) $68,000 B) $65,000 C) $ 66,000 D) $56,000

.

Answer: B Refer to: Case Study 13 Feedback: Marginal cost of defender in year 2 = Loss in market value + Foregone interest + O&M cost = (140,000 − 130,000) + 140,000 (0.10) + 41,000 = $65,000 The answer is “B”.

13.6. The EUAC for defender in year 2 is A) $81,642 B) $91,715 C) $85,182 D) $105,240

.

Answer: B Refer to: Case Study 13 Feedback: Marginal cost in year 1 = (200,000 – 140,000) + 200,000 (0.10) + 36,000 = 116,000 EUAC for year 2 = {(Marginal cost in year 1)(P/F,10%,1) + (Marginal cost in year 2)(P/F, 10%, 2)}{A/P, 10%, 2} = {116,000 (P/F, 10%, 1) + 65,000 (P/F, 10%, 2)}{A/P, 10%, 2} = {116,000 (0.9091) + 65,000 (0.8264)}{0.5762} = $91,715 The answer is “B”.

13.7. The foregone interest in year 4 is A) $12,000 B) $8,000 C) $13,000 D) $14,000

.

Answer: A Refer to: Case Study 13 Feedback: Foregone interest in year 4 = (Book value at the beginning of year 4) (interest rate) = 120T ,0E 00ST (0B .1A) N = K$S 12E,0L0L0 ER.COM The answer is “A”

True/False 13.8. Obsolescence occurs when the technology used in an equipment has been surpassed by a better and newer technology.

Answer: True Feedback: If a newer laptop, for example, uses better and newer technology, the older laptop even tough is functional becomes obsolete. Therefore, the answer is “True”.

13.9. Marginal cost is the cost at which an asset has the minimum cost life.

Answer: False

Feedback: Marginal cost is the cost of ownership of an asset for the next year. Hence it is not the cost for minimum life. Therefore, the answer is “False”.

13.10. One of the replacement rule is when the defender’s marginal cost becomes greater than the challenger’s minimum EUAC, then replace the defender.

Answer: True Feedback: For example the defender’s marginal cost becomes 10,000 and the challenger’s EUAC is found to be 9,000, it will be economical to replace the defender.

Therefore, the answer is “True”.

13.11. Given the marginal cost data in the table below for a defender, it is economical to replace the defender in year 1 if EUAC of challenger is $12,500.

Year 1 2 3 4 5

Marginal cost of defender $13,000 $12,000 TEST BANKSEL$L1E1,R00.0COM $16,000 $15,000

Answer: False Feedback: Even though the marginal cost of the defender is less than the EUAC of the challenger, since the marginal cost decrease in years 2 and 3 and then increase further analysis is required before making a replacement decision.

Therefore, the answer is “False”.

CHAPTER 14 INFLATION AND PRICE CHANGE Multiple Choice 14.1. A production equipment at Dalton Manufacturing is priced at $40,000 in current $ – . The inflation rates for the next three years are expected to be 3%, 2%, and 4% respectively. What will be the cost of the equipment in R$ three years from now?

A) $39,156 B) $41,129 C) $36,609 D) $42,024 Answer: C Feedback: Cost in R$ three years from now = [40,000 / (1.03) (1.02) (1.04)] = $36,609 The answer is “C”.

14.2. The cost of a first-class postage in 1970 was $0.06 and in year 2016 was $0.47. Determine the average annual rate of increase in the price of postage stamps during the years 1970 to 2016.

A) 3.97% B) 4.58% C) 6.63% D) 4.84% Answer: B Feedback: Number of years between 1970 and 2016 = 46 (1 + f )46 = (.47/.06) = 7.8333 f = (7.8333)(1/46) -1 = .04576 or 4.58%.

The answer is “B”.

14.3. The tuition in a private university was $25,000 per academic year in 1999. If the inflation rate has been averaging at the rate of 3%, how much one should have invested in 1999 to send a kid to this private university in 2017. Assume the investment earned an 8% return.

A) 10,649 B) 12,156 C) 14,027 D) 13,252 Answer: A Feedback:

Tuition in the year 2017 in A$ = (25,000)(1 + 0.03)(18) = 42,560.83.

P = 42,560.83 (P/F, 8%, 18) = 42,560.83(0.2502) = $10,649 The answer is “A”.

14.4. Sam bought a rental property for 500,000 five years ago. He sold the property this year for $2,200,000 and spent $250,000 fixing it up before selling the property. If the inflation rate for the past 5 years has been steady at 4% annually, compute the after-tax real rate of return (i ′) on this investment. Assume a capital gain tax of 15%.

A) 33.17% B) 28.13% C) 22.11% D) 31.84% Answer: C Feedback: Capital gain = 2,200,000 - 250,000-500,000 = $1,450,000 Capital gain tax = 0.15(1,450,000) = 217,500 After-tax cash flow at the end of year 5 = 2,200,000-250,000-217,500 = $1,732,500.

1,732,500 = 500,000(1 + i)5 i = 0.2822 or 28.22% i = i ′+ f + i′f 0.2822 = i′ + 0.05 + 0.05 i′ i′ = 0.2211 or 22.11%.

The answer is “C”.

14.5. An equipment was purchased two years ago financing for 5 years at 8% interest compounded monthly. The inflation rate for the past two years has been 3% per year and is expected to remain the same for the next 3 years. Compute the real interest rate for this loan.

A) 5.45% B) 8.73% C) 6.86% D) 5.15% Answer: D Feedback: Effective interest rate = (1+0.08/12)12 -1 = 0.083 or = 8.3% i′ = (i - f) / (1 + f) = (0.083 - 0.03)/ (1 + 0.03) = 0.0515 or 5.15%.

The answer is “D”.

Short Answer 14.6. The university your daughter wants to go in the fall of 2107 requested that you pay the tuition for all 4 years upfront when she starts school. The current tuition bill is $50,000. You expect the inflation rate to average 2% per year, and that the university pays a 4 % interest per year for the four year period.

What is the present worth of your payment?

Answer: $190,571.38 Feedback: A =? f = 2% i = 4% n = 4 years.

Annual cost = $50,000 i′ = (i - f) / (1 + f) = (0.04 - 0.02) / (1 + 0.02) = 0.0196 = 1.96%.

PW of all payments = Annual cost (P/A, i′, n) = $50,000 [{1 - (1.0196)-4}/ 0.0196] = 50,000 (3.8114) = $190,571.38

True/False 14.7. If future cash flows are to be expressed in actual $ (A$), the interest rate that should be used in determining the present worth is the market interest rate.

Answer: True Feedback: Market interest rate Ti sE rSe T q uBi rAedNKt oSE b eL uLsEedR. i nCdOeMt erm i ni ng the present worth. The answer is “True”.

14.8. When the purchasing power of a currency ($) decreases over time, the result is inflation.

Answer: True Feedback: Due to inflation, future dollars can only buy less good and services then what can be bought now. The answer is “True”.

14.9. If the actual market interest rate is 15% and real interest rate is 9.2% then the inflation rate is 5.8%.

Answer: False Solution: i = i ′+ f + i′f i- i ′= f (1 + i ′) f = (i- i ′)/(1 + i ′) = (.15-.058)/(1 + 0.092) = .0842 or 8.42%

The answer is “False”.

CHAPTER 15 SELECTION OF A MINIMUM ATTRACTIVE RATE OF RETURN Multiple Choice Case Study 15.1 A Publically traded company has issued $50M worth of stocks. The stock holders expect a rate of return of 12%. The company has a bank loan at an interest rate of 10% worth $12M. The company has sold bonds worth $20M at a rate of 6% compounded yearly.

15.1. Determine the before tax cost of capital for the company.

A) 10.63% B) 9.32% C) 10.24% D) None of these Answer: C Refer to: Case Study 15.1 Feedback: Total outstanding loan capital = 50M + 12M + 20M = $82M Before tax cost of capital = (50/82) *(.12) + (12/82) * (.10) + (20/82) * (.06) = 0.1024 or 10.24%.

The answer is “C”.

15.2. What is the after tax cost of capital if the company is in the 34% income tax bracket.

A) 9.25% B) 6.2% C) 7.02% D) None of these Answer: A Refer to: Case Study 15.1 Feedback: Since interest payments are tax deductible, the after tax cost of capital may be computed as below.

After tax cost of capital = (50/82) *(.12) + (12/82)*(.10)*(1-.34) ˗ (20/82)*(.06)*(1-.34) = 0.0925 or 9.25%.

The answer is “A”.

Case Study 15.2 A distributing company has financed an expansion requiring $50 million as shown in table below even though historically the company has financed any capital requirement with 20% debt at 12% interest and 80% equity with a 8% rate of return.

Source of capital Stocks Retained earnings Bonds

Amount $16M $25M $9M

Interest Rate/ Dividend 8% 10% 6%

15.3. What is the weighted average cost of capital based on historical method?

A) 5.8% B) 7.9% C) 8.2% D) 8.80% Answer: D Refer to: Case Study 15.2 Feedback: Historical WACC = 0.2(0.12) +0.80(0.08) = 0.088 = 8.80% The answer is “D”.

15.4. What is the weighted average cost of capital based on current financing?

A) 6.4% B) 8.2% C) 8.64% D) 9.3% Answer: C Refer to: Case Study 15.2 Feedback: WACC on current financing = (16/50)(0.08) +(25/50)(0.10)+(9/50)(0.06) = 0.0864 = 8.64% The answer is “C”.

Multiple Choice (Multiple Answers) 15.5. A medium size manufacturing company has a budget of $200,000 to invest on five different capital projects. Each has a six year life. Additional financial data for the five project opportunities are given below.

A B C D E Initial Cost $50K $80K $30K $40K $60K

EUAB

12,160

18,368

7,932

8,652

13,374

Determine which projects should be funded.

A) A B) B C) C D) D E) E Answer: C, A, and B Solution: To find ROR, setup, EUAB = EUAC 12,160 = 50K (A/P, i, 6) (A/P, i, 6) = 0.2432, i = 12% 18,368 = 80K (A/P, i, 6) (A/P, i, 6) = 0.2296, i = 10% 7,932 = 30K (A/P, i, 6) (A/P, i, 6), i = 15% 8,652 = 40K (A/P, i, 6), i = 8% 13,374 = 60K (A/P, i, 6), i = 9% Projects Cum. cost C 30K A 30K + 50K = 80KTESTBANKSELLER.COM B 80K + 80K = 160K Choose projects C, A, and B.

True/False 15.6. The after-tax cost of capital for borrowed amount at a 10% interest rate is 7.5% if the corporate income-tax rate is 25% Answer: True Solution: After-tax interest rate = (1-0.25) 10% = 7.5%. The answer is “True”.

15.7. The cost of capital is average interest rate from all sources of funds generated by a company.

Answer: True

Solution: The answer is “True” as the cost of capital is also called as the weighted average cost of capital.

15.8. Treasury stocks are stocks sold by U.S Department of Treasury.

Answer: False Feedback: Treasury stocks are repurchased stocks of their own stocks of a company. The answer is “False”.

15.9. The opportunity cost is the rate of return on the best rejected project.

Answer: True Solution: The answer is “True” as the opportunity cost is the best opportunity foregone.

CHAPTER 16 ECONOMIC ANALYSIS IN THE PUBLIC SECTOR Multiple Choice 16.1. A public sector project will have the following cash flows.

Annual benefits = $900,000 Expected annual dis-benefits (quantifiable) = $50,000 Initial Cost = $10M Yearly maintained cost = $ 80,000.

Life = 50 years.

If MARR is 7%, what is the modified B/C ratio for this project?

A) 1.056 B) 1.053 C) 1.058 D) 1.062 Answer: D Solution: EAUW of net benefits = $500,000 EAUW of maintained cost = $80,000 EAUW of initial cost = 10M (A/P, 7%, 50) = 10(0.0725) = $0.725.

Modified B/C ratio = (900,000-80,000-50,000)/ (725,000) = 1.062 TESTBANKSELLER.COM The answer is “D”.

16.2. A small town is considering two potential alternatives for a garbage collection system. The following cost data has been compiled. If the interest rate is 6%, Compute the ΔB /ΔC ratio.

Year 0 1 - 30 30

System A -$250,000 20,000 40,000

System B -$375,000 32,000 50,000

A) 0.96 B) 1.09 C) 2.27 D) 1.34 Answer: D Solution: ΔB /ΔC = Incremental benefits/Incremental costs = [(32,000 – 20,000) + (50,000-40,000) (A/F, 6%, 30)]/ [375,000-250,000*(A/P,6%,30)]

= [(32,000 – 20,000) + (50,000 – 40,000) (.0126)]/ [(375,000 – 250,000) (.0726)] = 1.34 The answer is “D”.

Case Study 16 A project has the following cost and benefit data.

Initial Cost = $15M Annual O&M Cost = $500K Annual Benefit = $4M The life has the following Probability distribution.

Life Probability 7 0.20 8 0.50 9 0.30 MARR = 8% 16.3. Compute the conventional B/C ratio for a life of 7 years.

A) 1.290 B) 1.183 C) 1.286 D) 1.379 Answer: B Feedback: EUAC = 500K + 15M (A/P, 8%, 7) = 500K + 15M (.1921) = 3.3815M Conventional B/C ratio = 4M/3.3815M = 1.183 The answer is “B”.

16.4. Compute the conventional B/C ratio for a life of 8 years.

A) 1.290 B) 1.183 C) 1.286 D) 1.379 Answer: C Feedback: EUAC = 500K + 15M (A/P, 8%, 8) = 500K + 15M (.1740)

= 3.11M Conventional B/C ratio = 4M/3.11M = 1.286 The answer is “C”.

16.5. Compute the conventional B/C ratio for a life of 9 years.

A) 1.290 B) 1.183 C) 1.286 D) 1.379 Answer: D Feedback: EUAC = 500K + 15M (A/P, 8%, 9) = 500K + 15M (.1601) = 2.9015M Conventional B/C ratio = 4M/2.9015 = 1.379 The answer is “D”.

T/ECSraTtiBoAuNsiK 16.6. Compute the conventional B ngStEhL eL exEpR ec.teCdOEMUAC. A) 1.290 B) 1.183 C) 1.286 D) 1.379 Answer: A Feedback: Expected EUAC = 0.2(3.3815) + 0.5(3.11) +0.3(2.9015) = 3.10175M Conventional B/C ratio = 4M/3.10175M = 1.290 The answer is “A”.

Short Answer 16.7. For a project being considered by a city in Alabama, the following cost data is available.

Determine both the modified and conventional B/C ratios.

Initial cost = $10 million.

O & M cost / year = $450,000

Estimated annual benefit Life Salvage value MARR

= $850,000 = 25 Years = $1 million = 6%

Answer: Conventional B/C: 0.705 Modified B/C: 0.535 Feedback: EUAC of initial costs = $10M (A/P, 6%, 25) = 10M (0.0782) = $782,000 EUAB = $850,000 + 1M (A/F, 6%, 25) = 850,000 + 1M (0.0182) = $868,200 Conventional B/C = 868,200/ (782,000 + 450,000) = 0.705 Modified B/C = (868,200 - 450,000)/ (782,000) = 0.535

True/False 16.8. Modified B/C ratio may be T stE atS edTaBsA , NKSELLER.COM Modified B/C ratio = (E.W. of Benefits)/ (E.W. of initial costs + E.W. of operating and maintenance costs).

Answer: False Feedback: The answer is “False” as the ratio on the right will compute the conventional B/C ratio.

16.9. The office of management and Budget (OMB) stipulates that a 7% interest rate be used in economic analysis of federal projects.

Answer: True Feedback: The answer is “True” as OMB directive A94 (page 588 of the text book, 13th edition) stipulates this percentage interest rate.

CHAPTER 17 ACCOUNTING AND ENGINEERING ECONOMY Multiple Choice Case Study 17 Given: The financial data below for Tech Engineering in 2016.

Category Cash Accounts receivable Inventories Intangibles Accounts payable Accrued Expenses

Dollar Amount $800,000 1,000,000 600,000 100,000 650,000 180,000

17.1. Determine the working capital for the company.

A) $1,470,000 B) $1,570,000 C) $1,670,000 D) None of these Answer: D Feedback: Working capital

Refer to: Case Study 17 = Current assets – Current liabilities = (800,000 + 1,000,000 + 600,000+100,000) − (650,000 + 180,000) = 2,500,000 – 830,000 = $1,670,000

The answer is “C”.

17.2. Compute the current ratio from the financial data provided.

A) 1.57 B) 3.01 C) 2.72 D) 1.78 Answer: B Feedback: Current ratio

Refer to: Case Study 17

= Current assets/ Current liabilities = (2,500,000/830,000) = 3.01 The answer is “B”.

17.3. What is the acid-test ratio for the company in 2016?

A) 2.17 B) 1.61 C) 1.21 D) None of these Answer: A Refer to: Case Study 17 Feedback: Acid test ratio = Quick assets / Total current liabilities = (800,000 + 1,000,000)/ (830,000) = 2.17 The answer is “A”.

17.4. This is classified as current asset for accounting purposes.

A) Equipment B) Land C) Buildings D) Finished goods Inventories Answer: D TESTBANKSELLER.COM Feedback: The answer is “D” as the other three assets are non-current assets.

17.5. Given the data for two products being made at Putnam Manufacturing, determine: A) Overhead expenses based on labor cost for each of the two products.

B) 1) Manufacturing cost of each product. # of units produced / year Labor cost / unit Material cost / unit

Standard 500,000 $100 $250

Deluxe 40,000 $180 $320

Standard

Deluxe

Answer: A) $108.68M B) 1) Standard: $540 2) Deluxe: $842 Feedback:

# of units produced / year Labor cost / unit Total labor cost Allocated overhead $/$ of labor Allocation of overhead of each product Overhead expenses per unit Total unit cost

500,000 $100 100 × 500,000 = $50 M = 108.68 / (50 + 7.2) = 1.90 1.90 × 50 M = $95 M 1.90 × 100 = $190 $250 + 100 + 190 = $540

40,000 $180 180 × 40,000 = $7.2 M 1.90 × 7.2 M = $13.68 M 1.90 × 180 = $342 $320 + 180+ 342 = $842

True/False 17.6. The total overhead expenses for a company are $10 M. The total labor cost for the whole company is $18 M. If a particular product has a direct labor of $3.6 M then the overhead expenses for this product is $6.4 M if overhead is based on direct labor costs.

Answer: False Feedback: Allocation of overhead = 10/18 × 3.6 = $2 M.

17.7. An example of indirect costs in manufacturing engineered products is salaries of design engineers.

False: True Feedback: The answer is “True” since design engineers are involved with not one product but many products. Therefore, their salaries cannot be attributed to one product.

17.8. The total liabilities of a company from the balance sheet are found to be $30.78 M. If the equity is stated as $40.15 M, the asset of the company is $9.37.

Answer: False Feedback: Assets = Liabilities + Equity = 30.78 + 40.15 = $ 70.93 M Therefore, the answer is “False”.