### Sample Chapter

**INSTANT DOWNLOAD COMPLETE TEST BANK WITH ANSWERS**

**Engineering Economics Financial Decision Making for Engineers 5th Edition by Niall M. Fraser – Test Bank**

Sample Questions

*Engineering Economics***, 5e (Fraser)**

**Chapter 3 Cash Flow Analysis**

** **

3.1 Multiple Choice Questions

1) What is an annuity?

- A) a series of payments that changes by the same proportion from one period to the next
- B) a series of equal payments over a sequence of equal periods
- C) a series of payments that changes by a constant amount from one period to the next
- D) a single payment
- E) present worth of a series of equal payments

Answer: B

Diff: 1 Type: MC Page Ref: 51

Topic: 3.5. Compound interest factors for annuities

Skill: Recall

User1: Qualitative

2) The present worth factor

- A) converts an annuity into the equivalent present amount.
- B) gives the future value equivalent to a series of equal payments.
- C) converts a series of repeated equal payments into the equivalent future amount.
- D) gives the present amount that is equivalent to some future amount.
- E) gives the future amount that is equivalent to a present amount.

Answer: D

Diff: 1 Type: MC Page Ref: 49

Topic: 3.4. Compound interest factors for single disbursements and receipts

Skill: Recall

User1: Qualitative

3) If the growth rate of a series is equal to 5% and annual interest rate is equal to 10%, what is the growth adjusted interest rate?

- A) +5.00%
- B) -5.00%
- C) +4.76%
- D) -4.76%
- E) +4.16%

Answer: C

Diff: 1 Type: MC Page Ref: 63

Topic: 3.7. Conversion factor for geometric gradient series

Skill: Applied

User1: Quantitative

4) Five years ago John invested $10 000 at 5% nominal interest rate compounded daily. What is his investment worth today?

- A) $10 513
- B) $11 763
- C) $12 763
- D) $12 840
- E) $13 763

Answer: D

Diff: 2 Type: MC Page Ref: 48

Topic: 3.4. Compound interest factors for single disbursements and receipts

Skill: Applied

User1: Quantitative

5) An arithmetic gradient series

- A) starts at zero at the end of the first period and then increases by a constant amount each period.
- B) starts at zero at the beginning of the first period and then increases by a constant amount each period.
- C) starts at zero at the end of the second period and then increases by a constant amount each period.
- D) starts at zero at the beginning of the second period and then increases by a constant amount each period.
- E) starts at zero at the end of the first period and then increases by an increasing amount each period.

Answer: A

Diff: 2 Type: MC Page Ref: 58

Topic: 3.6. Conversion factor for arithmetic gradient series

Skill: Recall

User1: Qualitative

6) Natalie received a gift of $1 000 from her grandmother. She decides to invest the money into a trip she wants to take when she graduates from college three years from now. What annual rate of return does she have to have to accumulate $1 250 by the time of her graduation?

- A) 7.7%
- B) 7.9%
- C) 8.4%
- D) 9.2%
- E) 12.5%

Answer: A

Diff: 2 Type: MC Page Ref: 49

Topic: 3.4. Compound interest factors for single disbursements and receipts

Skill: Applied

User1: Quantitative

7) How much money will you accumulate in a bank account by the end of a 5-year period if you deposit $1 200 today at an interest rate of 2% per year, compounded quarterly?

- A) $1 230
- B) $1 326
- C) $1 514
- D) $1 783
- E) $1 849

Answer: B

Diff: 3 Type: MC Page Ref: 49

Topic: 3.4. Compound interest factors for single disbursements and receipts

Skill: Applied

User1: Quantitative

8) One standard assumption for annuities and gradients is

- A) each payment occurs at the beginning of the period.
- B) annuities and gradients coincide with the beginning of sequential periods.
- C) annuities and gradients coincide with the end of preceding periods.
- D) payment period and compounding period differ.
- E) payment period and compounding period are the same.

Answer: E

Diff: 2 Type: MC Page Ref: 51, 58, 61

Topic: 3.5. Compound interest factors for annuities

Skill: Recall

User1: Qualitative

9) The compound amount factor produces

- A) the present amount, P, that is equivalent to a future amount, F.
- B) the future amount, F, that is equivalent to a present amount, P.
- C) the annuity, A, that is equivalent to a future amount, F.
- D) the annuity, A, that is equivalent to a present amount, P.
- E) the future amount of arithmetic gradient series.

Answer: B

Diff: 1 Type: MC Page Ref: 49

Topic: 3.4. Compound interest factors for single disbursements and receipts

Skill: Recall

User1: Qualitative

10) If Emily deposits $500 every other year into her bank account that pays 1.5% annual interest, compounded yearly, how much will she accumulate over a 10-year period?

- A) $2 500
- B) $2 568
- C) $2 576
- D) $2 656
- E) $5 738

Answer: D

Diff: 3 Type: MC Page Ref: 51-52

Topic: 3.8. Non-standard annuities and gradients

Skill: Applied

User1: Quantitative

11) The present worth of an infinitely long uniform series of cash flows is called

- A) capitalized value.
- B) salvage value.
- C) sinking value.
- D) compound value.
- E) continuous value.

Answer: A

Diff: 2 Type: MC Page Ref: 67

Topic: 3.9. Present worth of infinite annuity

Skill: Recall

User1: Qualitative

12) When is the growth-adjusted interest rate for a geometric series equal to zero?

- A) Growth is positive, but less than the interest rate.
- B) Growth is positive and greater than the interest rate.
- C) Growth is positive and equal to the interest rate.
- D) Growth is negative.
- E) Growth is positive.

Answer: C

Diff: 2 Type: MC Page Ref: 63

Topic: 3.7. Conversion factor for geometric gradient series

Skill: Recall

User1: Quantitative

13) Suppose that you want to evaluate the following non-standard cash flow: $1 000 paid at the end of every third year in a 12-year period with annual interest rate of 10%. What is the best method?

- A) Convert the non-standard cash flow into standard annuity by changing the interest rate.
- B) Convert the non-standard cash flow into standard annuity by changing the compounding period.
- C) Convert the non-standard cash flow into arithmetic gradient series.
- D) Treat each payment as a separate payment.
- E) Convert the non-standard cash flow into a geometric gradient series.

Answer: B

Diff: 3 Type: MC Page Ref: 65

Topic: 3.8. Non-standard annuities and gradients

Skill: Recall

User1: Qualitative

14) A municipality has just completed the construction of a bridge. It was calculated that operating and maintenance (O&M) costs of this bridge will be $20 000 in the first year with a 5% increase each year thereafter for the next 4 years. The interest rate used in calculations was 7.5% per year. What interest rate should be used to calculate the present worth of O&M costs over 5 years if we use the geometric series to present worth conversion factor?

- A) 1.9%
- B) 2.4%
- C) 5.0%
- D) 7.5%
- E) 8.2%

Answer: B

Diff: 2 Type: MC Page Ref: 63

Topic: 3.7. Conversion factor for geometric gradient series

Skill: Applied

User1: Quantitative

15) Calculating the growth-adjusted interest rate requires:

- A) the base amount and the rate of growth.
- B) the base amount and the interest rate.
- C) the number of periods and the rate of growth.
- D) the number of periods and the interest rate.
- E) the rate of growth and the interest rate.

Answer: E

Diff: 1 Type: MC Page Ref: 63

Topic: 3.7. Conversion factor for geometric gradient series

Skill: Recall

User1: Qualitative

16) A geometric gradient series

- A) starts with zero and from period to period increases by a constant amount.
- B) starts with zero and from period to period increases by a constant rate.
- C) starts with a certain amount and from period to period increases by a constant amount.
- D) starts with a certain amount and from period to period increases by a constant rate.
- E) starts with a certain amount and from period to period decreases by a constant percentage.

Answer: D

Diff: 1 Type: MC Page Ref: 61-63

Topic: 3.7. Conversion factor for geometric gradient series

Skill: Recall

User1: Qualitative

17) Calculate the uniform annuity equivalent to an arithmetic gradient series with a basic payment of $500 per year for 10 years that increases by $50 per year beginning in year 2, under 10% annual interest rate?

- A) $500
- B) $550
- C) $686
- D) $936
- E) $1 186

Answer: C

Diff: 2 Type: MC Page Ref: 60

Topic: 3.6. Conversion factor for arithmetic gradient series

Skill: Applied

User1: Quantitative

18) How many years will it take for an investment to triple itself if the interest rate is 12% compounded annually?

- A) 10.0
- B) 9.70
- C) 9.40
- D) 9.10
- E) 8.80

Answer: B

Diff: 1 Type: MC Page Ref: 48

Topic: 3.4. Compound interest factors for single disbursements and receipts

Skill: Applied

User1: Quantitative

19) If *i* stands for the interest rate, *g* stands for the growth rate, and* i*o stands for the growth-adjusted interest rate, which of the following is associated with deflation?

- A)
*i > g > 0*(*i*o*> 0*) - B)
*i > g > 0*(*i*o*< 0*) - C)
*i = g > 0*(*i*o*= 0*) - D)
*g < 0*(*i*o*> 0*) - E)
*i = g = 0*

Answer: D

Diff: 2 Type: MC Page Ref: 63

Topic: 3.7. Conversion factor for geometric gradient series

Skill: Recall

User1: Quantitative

20) The present worth of an infinitely long uniform series of cash flows is equal to

- A) A * [1/
*i*– N/((1 +*i*)N – 1)] - B) A*[(1 +
*i*)N – 1] - C) A/(
*i +*1) - D) A *
*i* - E) A/
*i*

Answer: E

Diff: 2 Type: MC Page Ref: 67

Topic: 3.9. Present worth of infinite annuity

Skill: Recall

User1: Qualitative

21) A company undertakes a 5-year project that requires annual payments. Payment for the first year is $2 000. It will then increase by 5% each subsequent year. The interest is fixed at 5% a year. What is the present worth of this cash flow?

- A) $9 112
- B) $9 328
- C) $9 442
- D) $9 524
- E) $10 226

Answer: D

Diff: 2 Type: MC Page Ref: 63-64

Topic: 3.7. Conversion factor for geometric gradient series

Skill: Applied

User1: Quantitative

22) How much Jim can accumulate in a private pension fund over 20 years if the fund offers 5% interest compounded annually, and he can afford to deposit $2 000 at the end of every second year?

- A) $28 946
- B) $32 259
- C) $66 132
- D) $94 256
- E) $117 853

Answer: B

Diff: 3 Type: MC Page Ref: 64-65

Topic: 3.8. Non-standard annuities and gradients

Skill: Applied

User1: Quantitative

23) How much should be set aside each month to accumulate $10 000 at the end of year 3 under 12% annual interest rate compounded monthly?

- A) $222.14
- B) $232.14
- C) $242.14
- D) $252.14
- E) $277.78

Answer: B

Diff: 2 Type: MC Page Ref: 52

Topic: 3.5. Compound interest factors for annuities

Skill: Applied

User1: Quantitative

24) A person deposits $100 to his savings account biweekly. The savings account pays a nominal interest rate of 5% per year, compounded every six months. What is the effective interest rate for a 6-month period?

- A) 2.1%
- B) 2.5%
- C) 3.2%
- D) 4.2%
- E) 5.1%

Answer: B

Diff: 2 Type: MC Page Ref: 64

Topic: 3.5. Compound interest factors for annuities

Skill: Applied

User1: Quantitative

25) A factor that relates a single cash flow in one period to another single cash flow in a later period is

- A) the uniform series compound amount factor.
- B) the capital recovery factor.
- C) the sinking fund factor.
- D) the annuity conversion factor.
- E) the compound amount factor.

Answer: E

Diff: 1 Type: MC Page Ref: 48

Topic: 3.4. Compound interest factors for single disbursements and receipts

Skill: Recall

User1: Qualitative

26) Capital recovery factor converts

- A) A into P.
- B) P into A.
- C) A into F.
- D) F into A.
- E) P into F.

Answer: B

Diff: 1 Type: MC Page Ref: 52

Topic: 3.5. Compound interest factors for annuities

Skill: Recall

User1: Qualitative

27) An *annuity due* is

- A) a series that starts at the end of the first period and increases by constant amount thereafter.
- B) a series that starts at the end of the first period and increases by constant percentage thereafter.
- C) a series that starts at the end of the first period and remains constant thereafter.
- D) a series that starts now and increases by constant amount thereafter.
- E) a series that starts now and remains constant thereafter.

Answer: E

Diff: 2 Type: MC Page Ref: 54

Topic: 3.5. Compound interest factors for annuities

Skill: Recall

User1: Qualitative

28) Maria wants to save up for a car. How much should she put in her bank account monthly to save $10 000 in two years if the bank pays 6% interest compounded monthly?

- A) $293.21
- B) $316.67
- C) $393.20
- D) $401.13
- E) $416.67

Answer: C

Diff: 2 Type: MC Page Ref: 51-53

Topic: 3.5. Compound interest factors for annuities

Skill: Applied

User1: Quantitative

29) Suppose that you have a series of payments: $100 in year 1, $150 in year 2 and $200 in year 3. If annual interest rate is 10%, what is the equivalent annuity for this series?

- A) $150.00
- B) $146.82
- C) $142.33
- D) $140.11
- E) $135.68

Answer: B

Diff: 2 Type: MC Page Ref: 51-53

Topic: 3.6. Conversion factor for arithmetic gradient series

Skill: Applied

User1: Quantitative

30) Suppose that your salary will increase by 2% per year over the next 4 years. If annual interest rate is also 2% over this period, what is the geometric series to present worth conversion factor in this case?

- A) 4.08
- B) 4.00
- C) 3.92
- D) 3.56
- E) 3.22

Answer: C

Diff: 3 Type: MC Page Ref: 61-64

Topic: 3.7. Conversion factor for geometric gradient series

Skill: Applied

User1: Quantitative

31) You will need to buy a replacement computer, costing $3 000, in five years time. If you have a bank account which earns 8% annual interest, how much must you put in the bank every year in order to have enough money for the replacement, assuming you make your first deposit in a year’s time?

- A) $565
- B) $597
- C) $666
- D) $675
- E) $712

Answer: C

Diff: 3 Type: MC Page Ref: 51-54

Topic: 3.5. Compound interest factors for annuities

Skill: Applied

User1: Quantitative

32) You want to have a million dollars in the bank when you retire. You think you can save $5 000 a year in a bank that offers you 5% interest. If you make your first deposit in a year’s time, how many years will it be from now before you can retire?

- A) 30
- B) 40
- C) 50
- D) 60
- E) 70

Answer: C

Diff: 3 Type: MC Page Ref: 51-54

Topic: 3.5. Compound interest factors for annuities

Skill: Applied

User1: Quantitative

33) You want to have a million dollars in the bank when you retire. You think you can save $5 000 this year, and increase that by $100 every subsequent year, in a bank that offers you 5% interest. If you make your first deposit in a year’s time, how many years will it be from now before you can retire?

- A) 30
- B) 35
- C) 40
- D) 45
- E) 50

Answer: D

Diff: 2 Type: MC Page Ref: 59-60

Topic: 3.6. Conversion factor for arithmetic gradient series

Skill: Applied

User1: Quantitative

34) You want to have a million dollars in the bank when you retire. You think you can save $5 000 this year, and increase that by 2% every subsequent year, in a bank that offers you 5% interest. If you make your first deposit in a year’s time, how many years will it be from now before you can retire?

- A) 41
- B) 42
- C) 43
- D) 44
- E) 45

Answer: D

Diff: 2 Type: MC Page Ref: 61-64

Topic: 3.7. Conversion factor for geometric gradient series

Skill: Applied

User1: Quantitative

35) Every leap year you get a bonus of $20 000, which you put into a retirement account at 5% interest. If your first payment into the account is made in four years time, and you put no other money into the account, how long will it be before you can retire with a million dollars?

- A) 36
- B) 40
- C) 44
- D) 48
- E) 52

Answer: E

Diff: 2 Type: MC Page Ref: 65

Topic: 3.8. Non-standard annuities and gradients

Skill: Applied

User1: Quantitative

36) You are offered a series of monthly payments of $10, continuing forever. If you deposit these at a nominal interest rate of 12%, compounded monthly, what is the present worth of the series?

- A) $120
- B) $1000
- C) $1200
- D) $1500
- E) Infinite

Answer: B

Diff: 2 Type: MC Page Ref: 67

Topic: 3.9. Present worth of infinite annuity

Skill: Applied

User1: Quantitative

37) You are promised that when you retire from your current job, in 40 years time, you will receive a gold watch valued at $1 000. If you can invest money at 5% annual interest, what is the present worth of this promise?

- A) $142
- B) $156
- C) $167
- D) $211
- E) $231

Answer: A

Diff: 2 Type: MC Page Ref: 48-49

Topic: 3.4. Compound interest factors for single disbursements and receipts

Skill: Applied

User1: Quantitative

3.2 Short Answers

1) A wholesale company has decided to build a new warehouse 3 years from now. It must accumulate $600 000 by the end of the 3-year period by putting aside an equal amount from its revenue at the end of each year. If the annual interest rate is 9%, how much the company should put aside each year?

Answer: The problem requires calculation of the annuity given the future worth of the entire cash-flow. The solution is obtained via the sinking factor:

A = $600 000 x (A/F, 9%, 3)

= $600 000 x 0.30505

= $183 032.85

Diff: 1 Type: SA Page Ref: 51

Topic: 3.5. Compound interest factors for annuities

Skill: Applied

User1: Quantitative

2) A trucking firm purchased two B-train trucks for $325 000. It paid 20% as a down payment and obtained a bank loan for the rest. The loan has a nominal interest rate of 10.5% compounded monthly with a 10-year amortization period. The loan term is 10 years. What are the firm’s monthly payments to the bank?

Answer: Down-payment is $65 000 (20% of $325 000). Subtract down-payment from the total truck costs to get the amount that should be borrowed from a bank: $325 000 – $65 000 = $260 000. The amortization period is the duration over which the bank loan should be paid back to the bank. Therefore: A = $260 000 x (A/P, 10.5/12%, [10 x 12]) = $3 508.31.

Diff: 3 Type: SA Page Ref: 52

Topic: 3.5. Compound interest factors for annuities

Skill: Applied

User1: Quantitative

3) An oil-extracting company expects to produce 1 000 barrels of oil a day over the next 5 years. If the oil price remains at $80 per barrel for the duration of the project, what will be the company’s accumulated total revenue at the end of the fifth year under 5% annual interest rate?

Answer:

Annual production = 1 000 barrels/day x 365 days = 365 000 barrels/year

Annual revenue = $80/barrel x 365 000 barrels/year = $29 200 000/year = 29.2 million/year

FW(Annual Revenue) = $29.2 x (F/A, 5%, 5) = $161.348 million

Diff: 2 Type: SA Page Ref: 52

Topic: 3.5. Compound interest factors for annuities

Skill: Applied

User1: Quantitative

4) John wants to buy a laptop computer in one year. He is working part time earning $600 per month. The computer John wants to buy costs $2 000. He decides to invest money in securities that pay a monthly rate of 2%. How much should John put aside every month to accumulate the required amount?

Answer: The following equation reflects the problem: 2 000 = A(F/A, 2%, 12) ;

hence A = $149.12/month.

Diff: 1 Type: SA Page Ref: 52

Topic: 3.5. Compound interest factors for annuities

Skill: Applied

User1: Quantitative

5) You want to buy a car that costs $20 000. You have to pay $2 000 upfront as a down-payment, and you are considering a financing option with a bank. If the bank charges 1.9% annual interest rate compounded daily, what would be your monthly payment for the 48-month financing period?

Answer: The effective monthly interest rate is *i*e = (1+0.019/365)30 – 1 = 0.0015628 or 0.15628% per month. The amount you want to finance is $18 000. Therefore, your monthly payment A is given by A(P/A, 0.15628%, 48) = 18 000. Hence A = $389.53.

Diff: 3 Type: SA Page Ref: 51-58

Topic: 3.5. Compound interest factors for annuities

Skill: Applied

User1: Quantitative

6) Stan saved $1 000 working part-time at a store. He wants to use this money for a trip to Europe. He is a smart guy and therefore decides to invest the money into an investment fund with a 24% annual interest rate compounded monthly. The trip to Europe costs $2 500. How long (in months) should Stan keep his money in the investment fund to accumulate the required amount?

Answer: The problem can be represented by the equation: $1 000 x (F/P, *i*e, N) = $2 500 where *i*e is the monthly rate of return and N is the number of months. The effective monthly rate of return is 24%/12 = 2% and therefore, the equation for N becomes 1 000 x (1 + 0.02)N = $2 500. N = 46.3 month is the solution.

Diff: 2 Type: SA Page Ref: 51-58

Topic: 3.4. Compound interest factors for single disbursements and receipts

Skill: Applied

User1: Quantitative

7) A manufacturing company expects a steady 2% annual growth in profits over the next three years. The company wants to invest the profits at a 10% interest rate. Expected annual profit of the company in year one is $1 000. How much will be accumulated by the company by the end of the third year?

Answer: The growth adjusted interest rate can be calculated as *i* = (1 + 0.1) x (1+0.02) – 1 = 0.122 or 12.2%. Therefore, the accumulated amount is:

1 000 x (F/A, 12.2%, 3)

= 1 000 x 3.38088

= $3 380.88.

Diff: 1 Type: SA Page Ref: 61-64

Topic: 3.7. Conversion factor for geometric gradient series

Skill: Applied

User1: Quantitative

8) The maintenance costs of a car are approximately $400 per year. With age the costs increase by $100 a year. What is the future worth of the maintenance costs in five years’ time if the interest rate is 5% compounded annually?

Answer: The maintenance costs are an arithmetic gradient series. The arithmetic gradient to annuity conversion factor is (A/G, 5%, 5) = 1/0.05 – 5/[(1+0.05)5 – 1]. This means that the geometrically-increasing annuity of $400 per year is equivalent to a uniform annuity of $400 + $100 x 1.90252 = $590.25 per year over the five-year period. So the future worth of the maintenance costs is $590.25 x (F/A, 5%,5) = $3 261.5.

Diff: 2 Type: SA Page Ref: 58-61

Topic: 3.6. Conversion factor for arithmetic gradient series

Skill: Applied

User1: Quantitative

9) Suppose you are evaluating a project that has various annual payments. These payments do not fall into any of the standard categories you have learned before. How would you approach this project?

Answer: There are three ways to deal with the problem: (i) convert non-standard annuities or gradients into standard by changing the compounding period; (ii) convert non-standard annuities or gradients into standard by re-defining them over the existing compounding period, (iii) if (i) and (ii) cannot be used, treat each payment as a single payment.

Diff: 2 Type: SA Page Ref: 65

Topic: 3.8. Non-standard annuities and gradients

Skill: Recall

User1: Qualitative

10) The Croesus Trust Fund currently has $1 million. It was established 10 years ago. The rate of return on the Fund’s money has been 5% compounded daily. How much money was originally invested into the Fund?

Answer: We are given future worth, and we need to find its present worth. However, we have to find the effective annual interest rate first. It is (1 + 0.05/365)365 – 1 = 0.05127 or 5.127%. Now we can use the present worth factor: $1 000 000 * (P/F, 5.127%, 10) = $606 552

Diff: 2 Type: SA Page Ref: 49

Topic: 3.4. Compound interest factors for single disbursements and receipts

Skill: Applied

User1: Quantitative

11) Explain why the growth-adjusted interest rate in the geometric gradient series to present worth conversion factor is less than the original interest rate when the growth factor is positive.

Answer: The interest rate reduces the present value of future cash flows. If the growth factor is positive, this reduction is offset by the increase in the value of each cash flow.

Diff: 2 Type: SA Page Ref: 64

Topic: 3.7. Conversion factor for geometric gradient series

Skill: Recall

User1: Qualitative

12) If the growth adjusted interest rate is negative, how should we proceed with calculation of the present worth of a geometric gradient series?

Answer: In such a case, we can still use the original formula for the geometric gradient series to present worth conversion factor.

Diff: 2 Type: SA Page Ref: 63

Topic: 3.7. Conversion factor for geometric gradient series

Skill: Recall

User1: Qualitative

13) It is known that a bridge costs $10 million to build, and it will be in operation forever. In order to recover initial investment, what should the annual revenue be in this case under 10% interest rate?

Answer: $10 million is the capitalized value of a series, and therefore, the required annual revenue can be defined as $10 000 000 ∗ 0.01 = $100 000/year.

Diff: 2 Type: SA Page Ref: 67

Topic: 3.9. Present worth of infinite annuity

Skill: Applied

User1: Quantitative

14) It is expected that a company is going to invest $8 million in a project. It is also expected that the annual revenue from this project will be $2 million dollars. How long would it take for the company to recover its investment under a 20% interest rate? Under 10%?

Answer: It is possible to set up the following equation: 8 = 2 ∗ [[(1 +* i*)x – 1]/[*i* ∗ (1 + *i*)x]] and solve it for x under 20% and 10% interest rate. Alternatively, we can set up a spreadsheet calculating the present worth of the annuity under each interest rate. Under 20% interest rate, it takes 8 years and 10 months to recover the investment. However, under 10% interest rate it takes only 5 years and 5 months to recover the investment. Therefore, with an increase in the interest rate the recovery period increases.

Diff: 2 Type: SA Page Ref: 51-58

Topic: 3.5. Compound interest factors for annuities

Skill: Applied

User1: Quantitative

15) Amy got $5 000 from her grandma to pay her tuition fee. She invested this amount at a 10% rate of return in an investment fund and simultaneously took a student loan for the same amount at a 6% interest rate compounded daily. Was it a smart move?

Answer: At the end of the first year, $5 000 in the investment fund becomes 5 000 ∗ (1 + 0.1) = $5 500. The accumulated debt of the student loan can be calculated using the effective annual interest rate *i*e:

*i*e = (1 + 0.06/365)365 = 0.06183 or 6.183%. Therefore, the accumulated debt will be

5 000 ∗ (1 + 0.06183) = $5 309.15. Since the amount in the investment fund exceeds the accumulated debt, Amy made a smart move.

Diff: 2 Type: SA Page Ref: 48

Topic: 3.4. Compound interest factors for single disbursements and receipts

Skill: Applied

User1: Quantitative

16) In this question, you are to write down the conversion factors that are required to transform the solid arrows into the equivalent dotted arrow. You can either indicate your answer by marking labelled arrows on the diagram, or by writing down a formula, as in the example given below:

Answer: P = F1(F/P,i,1) + F2(P/F,i,1)

Diff: 1 Type: SA Page Ref: 48-49

Topic: 3.4. Compound interest factors for single disbursements and receipts

Skill: Applied

User1: Quantitative

17) In this question, you are to write down the conversion factors that are required to transform the solid arrows into the equivalent dotted arrow. You can either indicate your answer by marking labelled arrows on the diagram, or by writing down a formula, as in the example given below:

Answer: P=A(P/A,i, 4)

Diff: 1 Type: SA Page Ref: 51-53

Topic: 3.5. Compound interest factors for annuities

Skill: Applied

User1: Quantitative

18) In this question, you are to write down the conversion factors that are required to transform the solid arrows into the equivalent dotted arrow. You can either indicate your answer by marking labelled arrows on the diagram, or by writing down a formula, as in the example given below:

Answer: P=A(F/A,i,4)(F/P,i,1) or P=A(P/A,i,4)(F/P,i,5)

Diff: 2 Type: SA Page Ref: 51-53

Topic: 3.5. Compound interest factors for annuities

Skill: Applied

User1: Quantitative

19) In this question, you are to write down the conversion factors that are required to transform the solid arrows into the equivalent dotted arrow. You can either indicate your answer by marking labelled arrows on the diagram, or by writing down a formula, as in the example given below:

Answer: P=(B+A(P/A,i,4))(F/P,i,5)-53

Diff: 2 Type: SA Page Ref: 48

Topic: 3.5. Compound interest factors for annuities

Skill: Applied

User1: Quantitative

20) In this question, you are to write down the conversion factors that are required to transform the solid arrows into the equivalent dotted arrow. You can either indicate your answer by marking labelled arrows on the diagram, or by writing down a formula, as in the example given below:

Answer: P = A((F/P,i,1)+(F/P,i,3)+(F/P,i,5)) (Treating each cashflow separately)

or

P = A(A/F,i,2)(F/A,i,6)(F/P,i,1) (Converting each cashflow to an annual annuity over two years)

Diff: 3 Type: SA Page Ref: 65-66

Topic: 3.8. Non-standard annuities and gradients

Skill: Applied

User1: Quantitative

21) In this question, you are to write down the conversion factors that are required to transform the solid arrows into the equivalent dotted arrow. You can either indicate your answer by marking labelled arrows on the diagram, or by writing down a formula, as in the example given below:

Answer: A=P(A/P,i,5)

Diff: 1 Type: SA Page Ref: 51-53

Topic: 3.5. Compound interest factors for annuities

Skill: Applied

User1: Quantitative

22) In this question, you are to write down the conversion factors that are required to transform the solid arrows into the equivalent dotted arrow. You can either indicate your answer by marking labelled arrows on the diagram, or by writing down a formula, as in the example given below:

Answer: A=P(F/P,i,1)(A/P,i,5)

Diff: 2 Type: SA Page Ref: 48-53

Topic: 3.5. Compound interest factors for annuities

Skill: Applied

User1: Quantitative

23) In this question, you are to write down the conversion factors that are required to transform the solid arrows into the equivalent dotted arrow. You can either indicate your answer by marking labelled arrows on the diagram, or by writing down a formula, as in the example given below:

Answer: A=G(A/G,i,6)

Diff: 2 Type: SA Page Ref: 58-61

Topic: 3.6. Conversion factor for arithmetic gradient series

Skill: Applied

User1: Quantitative

24) In this question, you are to write down the conversion factors that are required to transform the solid arrows into the equivalent dotted arrow. You can either indicate your answer by marking labelled arrows on the diagram, or by writing down a formula, as in the example given below:

Answer: P=G(A/G,i,6)(P/A,i,6) (We convert the arithmetic series to a uniform annuity, then find the present worth of the uniform annuity.)

Diff: 2 Type: SA Page Ref: 58-61

Topic: 3.6. Conversion factor for arithmetic gradient series

Skill: Applied

User1: Quantitative

25) In this question, you are to write down the conversion factors that are required to transform the solid arrows into the equivalent dotted arrow. You can either indicate your answer by marking labelled arrows on the diagram, or by writing down a formula, as in the example given below:

Answer: P=G(A/G,i,6)(F/A,i,6)(F/P,i,1)

(We convert the arithmetic progression to a uniform annuity, find the future worth of that annuity, then move the result one year further into the future.)

Diff: 2 Type: SA Page Ref: 59-61

Topic: 3.6. Conversion factor for arithmetic gradient series

Skill: Applied

User1: Quantitative

26) In this question, you are to write down the conversion factors that are required to transform the solid arrows into the equivalent dotted arrow. You can either indicate your answer by marking labelled arrows on the diagram, or by writing down a formula, as in the example given below:

Answer: P=(A+G(A/G,i,6))(P/A,i,6)

We convert the arithmetic series to an annuity, add this to the existing annuity, then find the annual worth of the sum of the annuities.

Diff: 2 Type: SA Page Ref: 59-61

Topic: 3.6. Conversion factor for arithmetic gradient series

Skill: Applied

User1: Quantitative

27) In this question, you are to write down the conversion factors that are required to transform the solid arrows into the equivalent dotted arrow. You can either indicate your answer by marking labelled arrows on the diagram, or by writing down a formula, as in the example given below:

Answer: P = (A+G(A/G,i,6))(P/A,i,6) – F(P/F,i,7)

Diff: 2 Type: SA Page Ref: 59-61

Topic: 3.6. Conversion factor for arithmetic gradient series

Skill: Applied

User1: Quantitative

*Engineering Economics***, 5e (Fraser)**

**Chapter 7 Replacement Decisions**

** **

7.1 Multiple Choice Questions

1) What is the economic life of an asset ?

- A) the service life that accounts for the costs of an asset replacement
- B) the service life that minimizes the marginal cost of an asset
- C) the service life that maximizes revenues generated by an asset
- D) the service life that minimizes the average cost of an asset over time
- E) the service life that minimizes the cost of replacing the asset

Answer: D

Diff: 1 Type: MC Page Ref: 223

Topic: 7.5. Defender and challenger are identical

Skill: Recall

User1: Qualitative

2) What is the equivalent annual cost (EAC)?

- A) annual cost of the initial investment in an asset spread over its economic life
- B) average annual cost of owning and operating an asset over its economic life
- C) the market cost of replacing the service provided by an asset
- D) the sum of the asset’s depreciation and its operating costs in any given year
- E) the difference between the profits generated by an asset and the cost of operating it

Answer: B

Diff: 1 Type: MC Page Ref: 218

Topic: 7.2. A replacement example

Skill: Recall

User1: Qualitative

3) TRINITY Ltd. produces different pieces of furniture. A set of electric drills used in the production of furniture wears out rapidly, after which the firm scraps them. Calculate the equivalent annual cost (capital costs) of a set of electric drills if the firm buys the set for $4 500 and uses it for 5 years. Assume an annual interest rate of 8%.

- A) $900
- B) $1 082
- C) $1 127
- D) $1 206
- E) $1 284

Answer: C

Diff: 2 Type: MC Page Ref: 218

Topic: 7.2. A replacement example

Skill: Applied

User1: Quantitative

4) In order to make a replacement decision, a firm calculated the equivalent annual cost of owning an asset as follows:

Replacement Period |
Salvage Value |
EAC Capital Costs |
Annual Repair Costs |
EAC Repair Costs |

1 | $1420 | $1,287 | – | – |

2 | $1,102 | $1,082 | $400 | $189 |

3 | $910 | $976 | $600 | $298 |

4 | $795 | $812 | $800 | $437 |

5 | $580 | $673 | $1,200 | $592 |

When should the firm replace its equipment?

- A) in 1 year
- B) in 2 years
- C) in 3 years
- D) in 4 years
- E) in 5 years

Answer: D

Diff: 2 Type: MC Page Ref: 222-224

Topic: 7.5. Defender and challenger are identical

Skill: Applied

User1: Quantitative

5) Currently a firm replaces its equipment every year. It has calculated the equivalent annual cost of replacement as follows:

Replacement Period |
Salvage Value |
EAC Capital Costs |
Annual Repair Costs |
EAC Repair Costs |

1 | $1420 | $1,287 | – | – |

2 | $1,102 | $1,082 | $400 | $189 |

3 | $910 | $976 | $600 | $298 |

4 | $795 | $812 | $800 | $437 |

5 | $580 | $673 | $1,200 | $592 |

How much can the firm save by making a replacement decision based on the economic life criterion?

- A) 16
- B) 22
- C) 25
- D) 38
- E) 44

Answer: D

Diff: 2 Type: MC Page Ref: 222-224

Topic: 7.5. Defender and challenger are identical

Skill: Applied

User1: Quantitative

6) Which of the following can be called the defender?

- A) a building that a company owns
- B) a bank account that a company owns
- C) a loan that a company has just taken from a bank
- D) a fleet of trucks that a company is currently planning to purchase
- E) a share held by a company

Answer: A

Diff: 1 Type: MC Page Ref: 219

Topic: 7.2. A replacement example

Skill: Recall

User1: Qualitative

7) Which of the following can be called the challenger?

- A) a revenue generated by a company
- B) machinery that a company wants to acquire
- C) an assembly line that a company wants to replace
- D) a piece of land that a company owns
- E) payments that a company receives from its clients

Answer: B

Diff: 1 Type: MC Page Ref: 219

Topic: 7.2. A replacement example

Skill: Recall

User1: Qualitative

8) What are the two reasons that a large portion of capacity cost is usually incurred early in the life of the capacity?

- A) (i) maintenance costs are usually higher in the early part of the capacity’s life, and (ii) assets lose their value most quickly early in their lives
- B) (i) installation costs are often large, and are incurred up-front, and (ii) maintenance costs are usually higher in the early part of the capacity’s life
- C) (i) assets lose their value most quickly early in their lives and (ii) installation costs are often large, and are incurred up-front
- D) (i) installation costs are often large, and are incurred up-front and (ii) there is a substantial cost associated with getting an asset ready for sale as salvage
- E) (i) there is a substantial cost associated with getting an asset ready for sale as salvage (ii) operating costs tend to fall as staff get more expert in using the asset

Answer: C

Diff: 2 Type: MC Page Ref: 221

Topic: 7.3. Reasons for replacement or retirement

Skill: Recall

User1: Qualitative

9) Which of the following can be treated as a part of installation costs?

- A) the costs of purchasing new equipment
- B) the costs of personnel retraining
- C) the costs of the loss in capacity value
- D) the difference between the purchase price and the salvage value
- E) the initial investment

Answer: B

Diff: 1 Type: MC Page Ref: 221

Topic: 7.2. A replacement example

Skill: Recall

User1: Qualitative

10) If it is said that the economic life of a challenger is four years it means that

- A) the payback period is four years.
- B) four years are required for the challenger to recover its initial investment.
- C) four years are required for the challenger to recover its operating and maintenance costs.
- D) the equivalent annual cost in year four is at its minimum.
- E) the equivalent annual cost is decreasing over four years.

Answer: D

Diff: 2 Type: MC Page Ref: 223

Topic: 7.5. Defender and challenger are identical

Skill: Recall

User1: Qualitative

11) In order to make a replacement decision when the defender and the challenger are identical, one should assume that

- A) the economic lives of the assets are very long relative to the planning time horizon.
- B) the technology can be changed during the economic life of an asset.
- C) the relative price and interest rate vary over time.
- D) the replacement decision is repeated an indefinitely large number of times.
- E) the defender and the challenger are made using different technologies.

Answer: D

Diff: 2 Type: MC Page Ref: 222-223

Topic: 7.5. Defender and challenger are identical

Skill: Recall

User1: Qualitative

12) A replacement study showed that the components of the total costs of an asset are as follows:

What is the economic life of an asset?

- A) 7 years
- B) 7.5 years
- C) 8 years
- D) 9 years
- E) 12 years

Answer: C

Diff: 1 Type: MC Page Ref: 223

Topic: 7.5. Defender and challenger are identical

Skill: Recall

User1: Quantitative

13) A company bought and installed an assembly line for $5 million and $0.5 million respectively. The company plans to use this assembly line for 8 years and then sell it for $100 000. Calculate the equivalent annual capital costs of the assembly line if the capital recovery factor is 0.1874, and a MARR is 10% for the capital investment.

- A) $0.997 mln.
- B) $1.012 mln.
- C) $1.022 mln.
- D) $1.027 mln.
- E) $1.046 mln.

Answer: C

Diff: 3 Type: MC Page Ref: 218

Topic: 7.2. A replacement example

Skill: Applied

User1: Quantitative

14) Capital cost per period

- A) mostly falls with an increase in the service life of an asset.
- B) always rises with an increase in the service life of an asset.
- C) can decline at some point in the service life of an asset if it requires a major overhaul.
- D) usually increases over time.
- E) remains constant over time.

Answer: A

Diff: 2 Type: MC Page Ref: 223

Topic: 7.4. Capital costs and other costs

Skill: Recall

User1: Qualitative

15) When making a replacement decision

- A) capital costs of the challenger are irrelevant.
- B) capital costs of the defender are irrelevant.
- C) operating costs of the defender are irrelevant.
- D) capital costs of the defender incurred in the past are irrelevant.
- E) operating costs of the challenger are irrelevant.

Answer: D

Diff: 2 Type: MC Page Ref: 230-231

Topic: 7.4. Capital costs and other costs

Skill: Recall

User1: Qualitative

16) What are the two conditions behind the One Year Principle?

- A) Capital costs for a defender are small compared to the operating costs, and the annual operating costs are gradually increasing.
- B) Capital costs for a defender are higher than the operating costs, and the annual operating costs are gradually increasing.
- C) Capital costs for a defender are less than the operating costs, and the increase in annual operating costs is not monotonic.
- D) Capital costs for a defender are higher than the operating costs, and the increase in annual operating costs is not monotonic.
- E) Capital costs for a defender are less than the operating costs, and the annual operating costs are monotonically declining.

Answer: A

Diff: 2 Type: MC Page Ref: 227-228

Topic: 7.6. Challenger is different from defender; challenger repeats indefinitely

Skill: Recall

User1: Qualitative

17) Shultz Ltd. produces portable electric band saws. Currently Shultz Ltd. pays its subcontractor $6.14 per power unit excluding material costs. It is expected that sales of the electric band saws will be as high as 7 500 per year. Should Shultz Ltd. produce power units itself if the equivalent annual cost for the company to install and run the production equipment would be $49 568?

- A) Yes, because own production would be cheaper by $0.22 per unit.
- B) Yes, because own production would be cheaper by $0.47 per unit.
- C) No, because own production would be $0.57 per unit more expensive.
- D) No, because own production would be $0.22 per unit more expensive.
- E) No, because own production would be $0.47 per unit more expensive.

Answer: E

Diff: 2 Type: MC Page Ref: 218-219

Topic: 7.2. A replacement example

Skill: Applied

User1: Quantitative

18) What asset will most likely contain sunk cost?

- A) refrigerator
- B) airplane
- C) administrative building
- D) rail tracks
- E) auto repair equipment

Answer: D

Diff: 2 Type: MC Page Ref: 230

Topic: 7.6. Challenger is different from defender; challenger repeats indefinitely

Skill: Recall

User1: Qualitative

19) Suppose that operating costs associated with the 5-year service life of an asset start with $1 000 in the first year increasing by $500 thereafter. Calculate the EAC(Operating) if annual rate is 10%.

- A) $1 505
- B) $1 905
- C) $2 005
- D) $2 105
- E) $2 205

Answer: B

Diff: 3 Type: MC Page Ref: 219

Topic: 7.2. A replacement example

Skill: Applied

User1: Quantitative

20) This table gives the data relevant to a replacement decision:

Year | Salvage Value | Maintenance Costs | EAC (Capital Costs) | EAC (Maintenance Costs) | Total |

1 | $32 000 | $4 000 | $38 416 | $4 000 | $42 416 |

2 | 24 500 | 8 000 | 24 783 | 5 923 | 30 706 |

3 | 16 350 | 16 000 | 20 263 | 9 027 | 29 291 |

4 | 11 280 | 32 000 | 17 182 | 14 125 | 31 307 |

5 | 8 420 | 64 000 | 14 895 | 22 627 | 37 521 |

If the MARR is 8.0%, what is the purchase price of this equipment?

- A) $65 200
- B) $66 700
- C) $67 400
- D) $68 500
- E) $69 300

Answer: A

Diff: 3 Type: MC Page Ref: 218-219

Topic: 7.5. Defender and challenger are identical

Skill: Applied

User1: Quantitative

21) This table gives the data relevant to a replacement decision:

Year | Salvage Value | Maintenance Costs | EAC (Capital Costs) | EAC (Maintenance Costs) | Total |

1 | $28 900 | $3 000 | $33 558 | $3 000 | $36 558 |

2 | 22 670 | 6 000 | 21 921 | 4 429 | 26 349 |

3 | 15 300 | 12 000 | 6 716 | ||

4 | 11 280 | 24 000 | 15 482 | 10 440 | 25 922 |

5 | 7 800 | 48 000 | 13 701 | 16 592 | 30 293 |

What is the EAC (Capital Costs) for year 3 at the MARR of 10%?

- A) $16 840
- B) $17 622
- C) $18 181
- D) $18 210
- E) $19 460

Answer: D

Diff: 3 Type: MC Page Ref: 218

Topic: 7.5. Defender and challenger are identical

Skill: Applied

User1: Quantitative

22) This table gives the data relevant to a replacement decision:

Year | Salvage Value | Maintenance Costs | EAC (Capital Costs) | EAC (Maintenance Costs) | Total |

1 | $35 260 | $4 500 | $30 597 | $4 500 | $35 097 |

2 | 18 256 | 7 000 | 25 803 | ||

3 | 14 520 | 14 000 | 19 688 | 8 201 | 27 889 |

4 | 12 569 | 27 500 | 16 179 | 12 359 | 28 538 |

5 | 8 598 | 44 900 | 14 385 | 17 689 | 32 075 |

What is the EAC (Maintenance Costs )for year 2 at the MARR of 10%?

- A) $5 220
- B) $5 690
- C) $5 980
- D) $6 210
- E) $6 740

Answer: B

Diff: 3 Type: MC Page Ref: 218-219

Topic: 7.5. Defender and challenger are identical

Skill: Applied

User1: Quantitative

23) This table gives the data relevant to a replacement decision:

Year | Salvage Value | Maintenance Costs | EAC (Capital Costs) | EAC (Maintenance Costs) | Total |

1 | $38 450 | $4 200 | $38 022 | $4 200 | $42 222 |

2 | 19 580 | 8 100 | 30 733 | 6 057 | 36 790 |

3 | 16 520 | 15 260 | 22 964 | 8 837 | 31 802 |

4 | 27 560 | 19 456 | 12 872 | 32 327 | |

5 | 9 210 | 49 500 | 16 831 | 18 871 | 35 702 |

What is the Salvage Value in year 4 at the MARR of 10%?

- A) $10 640
- B) $10 850
- C) $11 030
- D) $11 210
- E) $11 490

Answer: E

Diff: 3 Type: MC Page Ref: 218-219

Topic: 7.5. Defender and challenger are identical

Skill: Applied

User1: Quantitative

24) A printing and duplicating company owns ten photocopying machines. Which of the following associated expenses is a variable cost?

- A) the rental of the office space where the photocopiers are kept
- B) supplies of paper and toner
- C) the salary of the operator who runs and maintains the machines
- D) the interest on the loan the company took out to buy the copiers
- E) the lighting and heating of the office in which the photocopiers are kept

Answer: B

Diff: 2 Type: MC Page Ref: 221-222

Topic: 7.1. Introduction

Skill: Recall

User1: Qualitative

25) If the challenger is different from the defender, and the challenger does not repeat, what should one do to make an educated decision about replacement?

- A) Compute the economic life and the associated EAC of all the challengers.
- B) Compute the EAC of the remaining economic life of the defender.
- C) Keep the defender.
- D) Replace the defender by the challenger immediately.
- E) Structure the problem as a set of mutually exclusive alternatives.

Answer: E

Diff: 2 Type: MC Page Ref: 234-235

Topic: 7.7. Challenger is different from defender; challenger does not repeat

Skill: Recall

User1: Qualitative

26) Maintenance costs are mostly a part of

- A) fixed costs and as such should be included in the evaluation of EAC (Capital).
- B) fixed costs and as such should be included in the evaluation of EAC (Operating).
- C) variable costs and as such should be included in the evaluation of EAC (Capital).
- D) variable costs and as such should be included in the evaluation of EAC (Operating).
- E) Maintenance costs are neither fixed or variable costs and therefore, should be excluded from EAC calculations.

Answer: D

Diff: 2 Type: MC Page Ref: 223

Topic: 7.4. Capital costs and other costs

Skill: Recall

User1: Qualitative

27) If a challenger is different from the defender, and the challenger does not repeat, replacement decision should be based on

- A) economic life of the first challenger only.
- B) comparison of the defender with the last challenger.
- C) marginal costs of the sequence of all challengers.
- D) all possible combinations of the defender and all challengers over some study period.
- E) installation costs of all challengers.

Answer: D

Diff: 2 Type: MC Page Ref: 234-235

Topic: 7.7. Challenger is different from defender; challenger does not repeat

Skill: Recall

User1: Qualitative

28) Capital costs are usually:

- A) variable costs that directly depend on the level of production.
- B) fixed costs that do not directly depend on the level of production.
- C) negligible compared with operating costs.
- D) the most important factor in determining the economic life of an asset.
- E) costs of administrative overhead.

Answer: B

Diff: 1 Type: MC Page Ref: 221

Topic: 7.4. Capital costs and other costs

Skill: Recall

User1: Qualitative

29) What is the economic life of the following asset, given that its purchase price is $60 000 and the MARR = 10% ?

YEAR |
SALVAGE VALUE |
OPERATING COST |

1 | $40 000 | $20 000 |

2 | $20 000 | $20 000 |

3 | $10 000 | $40 000 |

4 | $5 000 | $50 000 |

- A) more than 4 years since EAC is increasing over 4 years
- B) 1 year and EAC = $46 000
- C) 2 years and EAC = $45 048
- D) 3 years and EAC = $47 147
- E) 4 years and EAC = $45 000

Answer: C

Diff: 3 Type: MC Page Ref: 222-223

Topic: 7.5. Defender and challenger are identical

Skill: Applied

User1: Quantitative

30) If the interest rate is 10%, the depreciation rate of an asset is 10%, its service life is 10 years and its first cost is $10 million then EAC(Capital) for a ten-year life is

- A) $1.000 million.
- B) $1.060 million.
- C) $1.349 million.
- D) $1.409 million.
- E) $1.977 million.

Answer: D

Diff: 3 Type: MC Page Ref: 224

Topic: 7.2. A replacement example

Skill: Applied

User1: Quantitative

31) You are considering purchase of a new furnace. Its initial cost, including installation, is $3 000, and it will cost $200 a year in fuel over its 10-year life. You expect that it can then be sold for $300. If your MARR is 10%, what is the equivalent annual cost of owning the furnace?

- A) $469
- B) $569
- C) $669
- D) $769
- E) $869

Answer: C

Diff: 2 Type: MC Page Ref: 218

Topic: 7.2. A replacement example

Skill: Applied

User1: Quantitative

32) An asset has an initial cost of $10 000. Its maintenance costs are $300 in the first year, and go up by 20% per year thereafter. Its salvage value declines by straight-line depreciation over ten years. If your MARR is 10%, what is its economic life?

- A) 2 years
- B) 3 years
- C) 4 years
- D) 5 years
- E) 6 years

Answer: E

Diff: 2 Type: MC Page Ref: 222-227

Topic: 7.5. Defender and challenger are identical

Skill: Applied

User1: Quantitative

33) You buy a new car for $20 000. Its salvage value declines by declining-balance depreciation of 10% per year, while its maintenance costs are $500 in the first year and go up by $400 a year. If your MARR is 5%, what is the EAC to you of keeping it until it reaches the end of its economic life?

- A) $4 358
- B) $4 335
- C) $4 327
- D) $4 333
- E) $4 352

Answer: C

Diff: 2 Type: MC Page Ref: 222-227

Topic: 7.5. Defender and challenger are identical

Skill: Applied

User1: Quantitative

34) A BMW costs $35 000, its salvage value declines by declining-balance depreciation of 10% per year, and its maintenance costs are $100 in the first year and go up by $500 a year. Mary and Tom both like to drive BMWs. Mary trades her car in for a new model every year, whereas Tom keeps his until it reaches its economic life, then trades it in. Mary and Tom both have MARRs of 5%. How much more does Mary pay for her car per year, on average, than Tom?

- A) They each pay the same.
- B) $569
- C) $615
- D) $769
- E) $869

Answer: C

Diff: 2 Type: MC Page Ref: 222-227

Topic: 7.5. Defender and challenger are identical

Skill: Applied

User1: Quantitative

35) A stamping machine currently has a salvage value of $10 000, and this will drop by 20% per year from now on. Its expected maintenance costs are $1 000 for this year, but in the following year it is expected to need a major overhaul, costing $3 500. In the year following the overhaul, its maintenance costs will be $500, and these will then go up by 30% per year. Your MARR is 10%. There is a challenger available that will do the same job for an EAC of $3 400. When should you replace the old machine?

- A) right now
- B) in two years time
- C) in four years time
- D) in six years time
- E) in eight years time

Answer: E

Diff: 2 Type: MC Page Ref: 231-233

Topic: 7.6. Challenger is different from defender; challenger repeats indefinitely

Skill: Applied

User1: Quantitative

36) A new laptop has just come onto the market that you estimate will have an equivalent annual cost of $500. You could sell your current laptop right away for $1 000, or you could spend $500 on an upgrade. At the end of this year, after the upgrade, it will have a resale value of $800, going down by $200 a year. Once its salvage value reaches zero, its physical life is over. If both laptops provide equivalent functionality, when should you replace your current laptop? Your MARR is 5%.

- A) Replace it now.
- B) Replace it in 2 years.
- C) Replace it in 3 years.
- D) Replace it in 4 years.
- E) Replace it in 5 years.

Answer: E

Diff: 2 Type: MC Page Ref: 231-233

Topic: 7.6. Challenger is different from defender; challenger repeats indefinitely

Skill: Applied

User1: Quantitative

37) A Tata costs $10 000, its salvage value declines by straight-line depreciation of $1 250 per year, and its maintenance costs are $100 in the first year and go up by 50% a year. Raj and Rashid both like to drive Tata’s. Raj trades her car in for a new model when it reaches its economic life, whereas Rashid keeps his until it reaches the end of its physical life. Raj and Rashid both have MARRs of 5%. How much more does Rashid pay for his Tata per year, on average, than Raj?

- A) They both pay the same.
- B) $53
- C) $64
- D) $77
- E) $112

Answer: B

Diff: 2 Type: MC Page Ref: 218

Topic: 7.5. Defender and challenger are identical

Skill: Applied

User1: Quantitative

7.2 Short Answer Questions

1) What is the difference between replacement of an existing physical asset and its retirement?

Answer: Retirement is removal of an existing physical asset from use without being replaced. It happens if the service that the asset provides is no longer needed. However, if there is an ongoing need for the service the asset provides, then it should be replaced whenever a cheaper way to provide the service appears.

Diff: 1 Type: SA Page Ref: 220

Topic: 7.3. Reasons for replacement or retirement

Skill: Recall

User1: Qualitative

2) Define the concept of the economic life of an asset and describe the circumstances under which the concept is applied to replacement decisions.

Answer: The economic life of a physical asset is the service life that minimizes the average cost of owning and using the asset. We base replacement decisions on a comparison of the annual costs of defender and challenger, assuming both are calculated over the economic life of the assets.

Diff: 2 Type: SA Page Ref: 223

Topic: 7.5. Defender and challenger are identical

Skill: Recall

User1: Qualitative

3) A trucking company evaluates its fleet of vehicles. According to its balance sheet, a three-year-old van has the book value of $21 870. Currently its maintenance costs are $1 000 per year. They have increased by 20% annually and are expected to increase by the same percentage in the future. Calculate the equivalent annual costs for the first six years knowing that the van’s purchase price was $30 000 and the minimum acceptable rate of return is 15%.

Answer: First it is necessary to calculate the depreciation rate d as follows:

d = 1 – = 0.1 or 10%. The economic life is associated with the minimum Equivalent Annual Cost (EAC). EAC is calculated in the following table:

Year |
Salvage Value |
EAC (Maintenance) |
EAC (Capital) |
EAC |

0 | 30 000 | – | – | – |

1 | 27 000 | 694 | 7 500 | 8 194 |

2 | 24 300 | 759 | 7 151 | 7 910 |

3 | 21 870 | 828 | 6 842 | 7 670 |

4 | 19 683 | 902 | 6 567 | 7 469 |

5 | 17 715 | 982 | 6 321 | 7 303 |

Salvage value in year N = BV(N) = P(1 – d)N

Annual maintenance cost AMC in year N = AMC(N) = AMC(1) x (1 + 0.2)N. And we know that the maintenance cost in year 1 was 1000/1.22 = $694

Equivalent Annual Cost (Operating) for a lifetime of N years is

EAC(operating, N) = ( * (P/F, *i*, *t*)) x (A/P, *i*, N)

Equivalent Annual Cost (Capital) in any given year N is

EAC(capital, N) = (P – S) x (A/P, *i*, N) + S*i*

The Equivalent Annual Cost is

EAC (N) = EAC(operating, N) + EAC(capital, N)

where BV is the book value, P is the purchase price, S is the salvage value and *i* is the MARR.

From the table, the economic life of the van is at least 5 years.

Diff: 3 Type: SA Page Ref: 218-219

Topic: 7.2. A replacement example

Skill: Applied

User1: Quantitative

4) A transportation company has bought a truck 5 years ago for $60 000. Currently its market value is $35 000. Maintenance costs of the truck are $2 000 per year, increasing by $500 with each year. A new truck of the same capacity costs $65 000 with constant maintenance costs of $500 for the duration of its service life of 10 years. Assuming 10% depreciation rate for both trucks and 5% annual interest rate, should the company buy the new truck now?

Answer: Let’s calculate the EAC of keeping the old truck one more year:

Salvage value at the end of that year is $35 000 * (1 – 0.1) = $31 500

EAC(Capital) = (35 000 – 31 500)*(A/P, 5%, 1) + 0.05 * 31 500 = $5 250

EAC = EAC(Capital) + EAC(operating) = 5 250 + 2 000 = $7 250

We expect that the EAC of keeping the old truck for longer than this will be higher, because of the rapid increase in maintenance costs.

Now let’s calculate the EAC of the new truck over its 10-year service life:

Salvage value = 65 000 * (1 – 0.1)10 = $22 664.10

EAC(Capital) = (65 000 – 22 664.10) * (A/P, 5%, 10) + 0.05 * 22 664.10 = $6 615.74

EAC = EAC(Capital) + EAC(Operating) = 6 615.74 + 500 = $7 115.74

Since the EAC of the new truck is lower, the company is better off buying it right now.

Diff: 3 Type: SA Page Ref: 218-219

Topic: 7.6. Challenger is different from defender; challenger repeats indefinitely

Skill: Applied

User1: Quantitative

5) What is sunk cost ? Explain how it affects replacement decisions.

Answer: Sunk cost is the unrecoverable portion of an investment. Once an asset has been installed and has been in operation for some time, all costs incurred up to that time are no longer relevant to replacement decisions. Only costs that will be incurred in keeping and operating the asset from this time on are relevant.

Diff: 2 Type: SA Page Ref: 230-231

Topic: 7.6. Challenger is different from defender; challenger repeats indefinitely

Skill: Recall

User1: Qualitative

6) Stan bought a car three years ago for $20 000. Recently he got a promotion and is deciding whether to keep his old car or to buy a new one. His dealer told him that the current market price of his old car is $15 000. The car maintenance costs are $1 000 now, and they are going to increase each year by at least $500. Stan compares his old car with a new one that, he calculates, would have an equivalent annual cost of $4 100. What is Stan’s optimal decision if his current interest rate is 7%?

Answer: First it is necessary to derive the implicit depreciation rate given the dealer’s information

d = 1 – = 0.1 or 10%

Equivalent Annual Cost (Capital, N) in any year N can be calculated as follows:

EAC(capital, N) = (P – S) x (A/P, *i*, N) + S*i*

where P is the current market price, S is the salvage value and *i* is the interest rate.

Equivalent Annual Cost (Operating, N) in any given year N is:

EAC(operating, N) = ( * (P/F, *i*, *t*)) * (A/P, *i*, N)

where AMCt is the annual maintenance cost in period *t*.

If Stan keeps his old car for one more year, then:

– EAC(capital, 1) = [15 000 – 15 000 x (1 – 0.1)] x 1.07 + 15 000 x (1 – 0.1) x 0.07 = $2 550

– EAC(operating, 1) = 1 500 = $1 500

– EAC = 2 550 + 1 500 = $4 050

Since the EAC is less than the EAC of the new car, Stan should keep his old car for at least one more year.

If Stan keeps his old car for two more years:

– EAC(capital, 2) = [15 000 – 15 000 * (1 – 0.1)2] * (0.5531) + 15 000 * (1 – 0.1)2 * 0.07 = $2 427

– EAC(operating, 2) = (1 500 * 0.934 + 2 000 * 0.873) * 0.5531 = $1 741

– EAC = 2 427 + 1 741 = $4 168

Since now the EAC of the old car exceeds the EAC of the new car, Stan should keep his old car for one more year and then consider buying a new one (taking into account any changes that may have occurred during the year, such as a new model coming out.)

Diff: 3 Type: SA Page Ref: 227

Topic: 7.6. Challenger is different from defender; challenger repeats indefinitely

Skill: Applied

User1: Quantitative

7) A new computer costs $4 000. It depreciates at 15% annually. The computer maintenance costs are $200 in the first year increasing by 50% per year thereafter. Calculate the EAC over the first five years of the computer’s life, assuming a 20% interest rate. What is the economic life of the computer?

Answer: The following table summarizes all calculations:

Year |
Salvage Value |
EAC (Capital) |
EAC (Maintenance) |
EAC Total |

0 | 4 000.00 | – | – | – |

1 | 3 400.00 | 1 400.00 | 200.00 | 1 600.00 |

2 | 2 890.00 | 1 304.42 | 245.43 | 1 549.85 |

3 | 2 456.50 | 1 223.93 | 301.62 | 1 525.55 |

4 | 2 088.03 | 1 156.25 | 371.22 | 1 527.47 |

5 | 1 774.82 | 1 099.19 | 457.45 | 1 556.64 |

Salvage value in any given year N is

S(N) = P x (1 – d)N

Equivalent Annual Cost (Capital, N) in any given year N is

EAC(capital, N) = (P – S) x (A/P, *i*, N) + S*i*

Annual operation cost in any given year N is

AOC(N) = 200 x (1 + 0.5)N-1

Equivalent Annual Cost (Operating, N) in any given year N is

EAC(operating, N) = ( x (P/F, *i*, *t*)) x (A/P, *i*, N)

Since the EAC of $1 525.55 in year 3 is the least, the economic life of the computer is 3 years.

Diff: 2 Type: SA Page Ref: 222-227

Topic: 7.2. A replacement example

Skill: Applied

User1: Quantitative

8) OMON Consulting is evaluating different scenarios to replace its existing technological line to produce compact computer discs. A four-year time horizon is used in this evaluation. One challenger is available at present, and it is expected that a new technology will be available in two years. List all potential scenarios associated with this evaluation.

Answer: The following table summarizes all potential scenarios:

Scenario | Defender Life | Challenger I Life | Challenger II Life |

1 | 4 | 0 | 0 |

2 | 3 | 1 | 0 |

3 | 3 | 0 | 1 |

4 | 2 | 2 | 0 |

5 | 2 | 1 | 1 |

6 | 2 | 0 | 2 |

7 | 1 | 3 | 0 |

8 | 1 | 2 | 1 |

9 | 1 | 1 | 2 |

10 | 0 | 4 | 0 |

11 | 0 | 3 | 1 |

12 | 0 | 2 | 2 |

Diff: 2 Type: SA Page Ref: 234-235

Topic: 7.7. Challenger is different from defender; challenger does not repeat

Skill: Applied

User1: Qualitative

9) NB Power is considering replacing its existing electricity generator. The current market value of the generator is $15 000. It depreciates at 25% per year. The generator produces 153 megawatt-hours of electricity annually and requires $3 000 in maintenance costs per year. Moreover, the maintenance costs are expected to increase by $1 200 per year in the future. On the other hand, there is another generator in the market that provides electricity at $0.059 per kilowatt-hour. What should NB Power do if the MARR is 8%?

Answer: If the old generator is kept for one more year then:

EAC(capital, 1) = (P-S) * (A/P, 8%, 1) + S*i*

= [15 000 – 15 000(1 – 0.25)] * (1 + 0.08) + 15 000(1 – 0.25) * 0.08

= $4 950

EAC(maintenance, 1) = 3 000 + 1 200 = $4 200

EAC(Total) = 4 950 + 4 200 = $9 150

Unit cost of electricity = $9 150/153 000 kilowatt-hours = $0.0598/kilowatt-hour

It is time to replace the old generator, since the new one provides a cheaper way to produce electricity.

Diff: 2 Type: SA Page Ref: 217-219

Topic: 7.2. A replacement example

Skill: Applied

User1: Quantitative

10) What are the three basic replacement schemes discussed in Chapter 7?

Answer: The three are:

- Identical defender and challenger.
- Different defender and challenger but all succeeding challengers are identical to the first.
- Different defender and challengers, and challengers do not repeat.

Diff: 1 Type: SA Page Ref: 222-235

Topic: 7.5. Defender and challenger are identical

Skill: Applied

User1: Qualitative

11) The salvage value of an electric generator is $10 000, while its purchase price five years ago was $20 000. It produces 45.5 megawatt-hours of electricity per year and requires $800 in annual maintenance costs. A new generator can produce electricity at $0.05 per kilowatt-hour. Should the new generator be purchased if the annual interest rate is 7%?

Answer: Since information on the new generator is given in terms of per unit cost, it is necessary to calculate the unit cost of electricity produced by the old generator. Unit costs can be defined as

Therefore, we have to calculate the EAC of the defender—old generator—if we keep it for one more year. Given information about the defender, we can calculate the depreciation rate as follows:

d = 1 – = 0.12945

Therefore, the salvage value of the defender at the end of year 6 will be

$10 000*(1 – 0.12945) = $8 705

EAC(Capital) = (10 000 – 8 705) * (A/P, 7%, 1) + 0.07 * 8 705 = $1 994

EAC(Total) = EAC(Capital) + EAC(Maintenance) = 1 994 + 800 = $2 794

Now we can calculate unit cost of electricity produced by the defender:

= $0.06/kilowatt-hour

Since the new generator produces cheaper electricity, it should be purchased

Diff: 2 Type: SA Page Ref: 217-219

Topic: 7.2. A replacement example

Skill: Applied

User1: Quantitative

12) A three-year-old small crane is being considered for early replacement. Its current market value is $13 000. Estimated future market values and annual operating costs for the next five years are given in the following table:

Year |
Market Value, P |
Annual Operating Cost, AOC |

0 | 13 000 | – |

1 | 9 000 | 2 500 |

2 | 8 000 | 2 700 |

3 | 6 000 | 3 000 |

4 | 2 000 | 3 500 |

5 | 0 | 4 500 |

What is the economic life for this crane if the interest rate is 10% per year?

Answer: In order to find the economic life, it is necessary to calculate the Equivalent Annual Costs (EAC) as the sum of the EAC (capital) and EAC (operating).

In any given year N:

EAC(operating, N) = ( * (P/F, *i*, *t*)) * (A/P, *i*, N)

where AOC*t* is the annual operating cost in period *t*.

In any given year N:

EAC(capital, N) = (P – S) * (A/P, *i*, N) + S*i*

where P is the market value, S is the salvage value, and *i* is the annual interest rate.

In this case, the salvage value is the market value of the crane next year. All the required calculations are presented in the following table:

Year |
Salvage Value |
EAC (Maintenance) |
EAC (Capital) |
EAC Total |

0 | 13 000 | – | ||

1 | 9 000 | 2 500 | 5 300 | 7 800 |

2 | 8 000 | 2 595 | 3 681 | 6 276 |

3 | 6 000 | 2 717 | 3 415 | 6 132 |

4 | 2 000 | 2 887 | 3 671 | 6 558 |

5 | 0 | 3 150 | 3 429 | 6 579 |

The economic life of the crane is three years, since this is when the EAC is minimized.

Diff: 2 Type: SA Page Ref: 222-223

Topic: 7.5. Defender and challenger are identical

Skill: Applied

User1: Quantitative

13) A communication system cost $65 000 to buy and $4 000 to install ten years ago. Currently the ten-year-old system can be sold for $7 000, but it will cost $1 000 to remove it. If it’s sold in a year’s time, the net income, after paying removal costs, will be $1 000. A new communication system costs $50 000 plus $500 to install it. Assuming 5% depreciation rate for the new system, 7% annual interest rate, by how much would the operating costs of the old system have to exceed operating costs of the new one for the old system to be replaced immediately?

Answer: First of all, the purchase price and the installation costs of the old system are sunk costs. They are irrelevant for the replacement decision.

Immediate replacement is needed only if annual equivalent cost of keeping the old system one more year exceeds the equivalent annual cost of the new system.

Current value of the old system is $7 000 – $1 000 (removal cost) = $6 000 and its salvage value at the end of the year is equal to $1 000. Therefore, equivalent annual cost (capital) to keep the system one more year is:

EAC(Capital, 1) = (6 000 – 1 000) * (A/P, 7%, 1) + 1 000 * 0.07

= $5 420

In turn, given the information on the new system, its equivalent annual cost (capital) is:

EAC(Capital, 1) = [50 000 + 500 – 50 000 * (1 – 0.05)] *(A/P, 7%, 1) + 50 000 * (1 – 0.05) * 0.07

= $6 535

So if the money saved on operating costs is to offset the difference in capital costs, operating the new system must be cheaper than operating the old system by $6 535 – $5 420 = $1 150.

Diff: 3 Type: SA Page Ref: 227-228

Topic: 7.6. Challenger is different from defender; challenger repeats indefinitely

Skill: Applied

User1: Quantitative

14) An old asset—a piece of equipment—has a current market value of $10 000. Its purchase price 5 years ago was $15 000. Installation costs were $2 000 in year 1 with some adjustment costs in year 2. If you are to calculate the EAC (Capital) to keep this asset in operation, what value should be assigned to P in the formula EAC(Capital) = (P – S)*(A/P, i, N) + iS?

Answer: This asset is a defender in terms of replacement decisions. Therefore, P should be assigned the value of $10 000, which is the current market value of the asset. All other costs are sunk costs, and are irrelevant for replacement decision

Diff: 1 Type: SA Page Ref: 230-231

Topic: 7.6. Challenger is different from defender; challenger repeats indefinitely

Skill: Applied

User1: Quantitative

15) The University has just invested $9 000 in a new desktop publishing system. From past experience, annual cash returns are estimated asA(t) = $8000 – $4000(1 + 0.15)t-1S(t) = $6000(1 – 0.3)twhere A(t) stands for the net cash flow in period t and S(t) stands for the salvage value at the end of year t, and t ≥ 1.If the MARR is 12%, compute the annual equivalent cost in year 2

Answer: We calculate salvage value in year 2 as

S(2) = 6 000 * (1 – 0.3)2 = $2 940

Now we can calculate EAC(Capital) as follows:

EAC(Capital) = (P – S) * (A/P, 12%, 2) +i S

= (9 000 – 2 940) * 0.5917 + 0.12 * 2 040

= $3 938.50

In terms of annual cash flows, in year 1 and year 2:

A(1) = 8 000 – 4 000*(1 + 0.15)0

= $4 000

A(2) = 8 000 – 4 000*(1 + 0.15)

= $3 400

Both are cash inflows or benefits (not costs!) Therefore, EAC (Operating) in this case is

EAC(Operating) = [- – ] * (A/P, 12%, 2)

= [-3 571.43 – 2 710.46] * 0.5917

= -$3 716.99

EAC = EAC(Capital) + EAC(Operating) = 3 938.50 – 3 716.99 = $221.51

Diff: 3 Type: SA Page Ref: 217-219

Topic: 7.2. A replacement example

Skill: Applied

User1: Quantitative

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