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

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

 

Engineering Economics, 5e (Fraser)

Chapter 2   Time Value of Money

 

2.1   Multiple Choice Questions

 

1) The price of money can be captured through

  1. A) the difference between benefits and costs that occur at different times.
  2. B) the future worth of an investment.
  3. C) the present worth of an investment.
  4. D) the interest rate.
  5. E) the consumer price index.

Answer:  D

Diff: 1    Type: MC     Page Ref: 19

Topic:  2.2 Interest and Interest Rates

Skill:  Recall

User1:  Qualitative

 

2) What makes one dollar in the future less desirable than one dollar today?

  1. A) variable interest rate
  2. B) a forgone opportunity of investment
  3. C) a diminishing purchasing power of money over time
  4. D) a growing inflation
  5. E) accumulated welfare of people

Answer:  B

Diff: 1    Type: MC     Page Ref: 20

Topic:  2.2 Interest and Interest Rates

Skill:  Recall

User1:  Qualitative

 

3) The principal amount is

  1. A) the present value of money.
  2. B) the future value of money.
  3. C) the amount of money invested at the prime interest rate.
  4. D) the annual equivalent value of money.
  5. E) the difference between the amount of money lent and the amount of money later repaid.

Answer:  A

Diff: 1    Type: MC     Page Ref: 20

Topic:  2.3 Compound and Simple Interest

Skill:  Recall

User1:  Qualitative

 

 

4) Bill wants to buy a new car in three years from now. He expects that the price of a car will be $15 000 in three years. How much money should Bill put in his savings account now if a bank pays 5% interest rate on this account?

  1. A) $11 629
  2. B) $12 104
  3. C) $12 958
  4. D) $13 465
  5. E) $14 286

Answer:  C

Diff: 2    Type: MC     Page Ref: 21

Topic:  2.3 Compound and Simple Interest

Skill:  Applied

User1:  Quantitative

 

5) Milo has just inherited $6 500 and immediately spent the money purchasing an investment certificate. He decided to use the investment certificate to finance his return to the university that he left because of the financial problems at the time. Milo calculated that the interest rate the bank would pay on his investment certificate would allow him to accumulate the $7 600 he would need over 4 years. What interest rate does the bank pay?

  1. A) 2.0
  2. B) 2.5
  3. C) 3.0
  4. D) 3.5
  5. E) 4.0

Answer:  E

Diff: 2    Type: MC     Page Ref: 21

Topic:  2.3 Compound and Simple Interest

Skill:  Applied

User1:  Qualitative

 

6) It is known that the total interest paid over a 5-year period is $2 081.13. What was the principal amount borrowed at a 6% nominal interest rate compounded quarterly?

  1. A) $3 000
  2. B) $4 000
  3. C) $5 000
  4. D) $6 000
  5. E) $7 000

Answer:  D

Diff: 3    Type: MC     Page Ref: 26-27

Topic:  2.3 Compound and Simple Interest

Skill:  Applied

User1:  Quantitative

 

 

7) Nominal interest rate is calculated by

  1. A) summing up all interest rates for all compounding periods.
  2. B) converting a given interest rate with a compounding period to an equivalent interest rate with a one-year compounding period.
  3. C) dividing the interest rate per compounding period by the number of compounding periods per year.
  4. D) multiplying the simple interest rate by the number of years.
  5. E) multiplying the interest rate per compounding period by the number of compounding periods per year.

Answer:  E

Diff: 2    Type: MC     Page Ref: 26

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Recall

User1:  Qualitative

 

8) Your credit card statement says that your card charges 0.0562% interest per day. What is the actual interest rate per year?

  1. A) 11.6%
  2. B) 14.5%
  3. C) 18.3%
  4. D) 20.1%
  5. E) 22.8%

Answer:  E

Diff: 2    Type: MC     Page Ref: 26

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Applied

User1:  Quantitative

 

9) If an interest rate is 18% per year, what is the equivalent interest rate per quarter?

  1. A) 3.8%
  2. B) 4.5%
  3. C) 4.8%
  4. D) 6.2%
  5. E) 8.6%

Answer:  B

Diff: 2    Type: MC     Page Ref: 25

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Applied

User1:  Quantitative

 

 

10) How many compounding periods are needed to obtain an effective interest rate of 25% if the interest rate per sub-compounding period is 1.88%?

  1. A) 13
  2. B) 12
  3. C) 11
  4. D) 10
  5. E) 9

Answer:  B

Diff: 2    Type: MC     Page Ref: 26

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Applied

User1:  Quantitative

 

11) What does a cash flow diagram of a project represent?

  1. A) summary of benefits and costs of a project
  2. B) summary of the timing and magnitude of payments and receipts as they occur over time
  3. C) magnitude of cash flows at a given period of time
  4. D) summary of present, future, and annual worths of a project
  5. E) change in value of money at different interest rates at various compounding periods

Answer:  B

Diff: 2    Type: MC     Page Ref: 29-31

Topic:  2.6 Cash Flow Diagrams

Skill:  Recall

User1:  Qualitative

 

12) A project is represented by the following cash flow diagram:

What are the project’s cash flows?

  1. A) receipts = $45 000; disbursements = $25 000; project life = 6 years
  2. B) receipts = $35 000; disbursements = $45 000; project life = 6 years
  3. C) receipts = $45 000; disbursements = $35 000; project life = 6 years
  4. D) receipts = $45 000; disbursements = $35 000; project life = 7 years
  5. E) receipts = $35 000; disbursements = $45 000; project life = 7 years

Answer:  C

Diff: 2    Type: MC     Page Ref: 29-31

Topic:  2.6 Cash Flow Diagrams

Skill:  Applied

User1:  Quantitative

 

 

13) J.D.Irving Ltd. is considering a construction project with $2 million initial investment that will last for 10 years. The duration of the construction phase is one year. Once the construction is over, the project starts yielding a constant annual revenue of $1.0 million. By the end of the fifth year the project generates $0.5 million extra revenue. The annual operation and maintenance expenses of $0.5 million will start at year four and last till the end of the project’s life. At the very end of the 10-year project the used equipment can be sold for $1.5 million. What cash flow diagram represents this project?

  1. A)
  2. B)
  3. C)
  4. D)

 

 

  1. E)

Answer:  B

Diff: 3    Type: MC     Page Ref: 29-31

Topic:  2.6 Cash Flow Diagrams

Skill:  Applied

User1:  Quantitative

 

14) What does the term “market equivalence” imply?

  1. A) indifference on the part of a decision maker among available choices
  2. B) the existence of a mathematical relationship between time and money
  3. C) the ability to exchange one cash flow for another at minimum cost
  4. D) the ability to exchange one cash flow for another at no cost
  5. E) the ability to obtain a zero net cash flow

Answer:  A

Diff: 2    Type: MC     Page Ref: 32

Topic:  2.7 Equivalence

Skill:  Recall

User1:  Qualitative

 

15) You invest $10 000 at 5% interest rate compounded monthly, what is your accumulated interest at the end of year 2?

  1. A) $511.62
  2. B) $537.79
  3. C) $1 025.00
  4. D) $1 049.41
  5. E) $1 089.41

Answer:  D

Diff: 2    Type: MC     Page Ref: 26-27

Topic:  2.3 Compound and Simple Interest

Skill:  Applied

User1:  Quantitative

 

 

16) You would like to have $8 500 for future spending in three years from now. How much should you deposit in your bank account now if the account pays you 0.4% interest per month?

  1. A) $2 071
  2. B) $7 362
  3. C) $8 102
  4. D) $8 399
  5. E) $8 429

Answer:  B

Diff: 3    Type: MC     Page Ref: 35

Topic:  2.3 Compound and Simple Interest

Skill:  Applied

User1:  Quantitative

 

17) The nominal interest rate is 6% per year compounded quarterly. What is the effective annual rate?

  1. A) 5.74%
  2. B) 5.84%
  3. C) 5.94%
  4. D) 6.04%
  5. E) 6.14%

Answer:  E

Diff: 1    Type: MC     Page Ref: 27

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Applied

User1:  Quantitative

 

18) Emily is considering two mutually exclusive financial options: (i) to deposit $4 000 in her bank’s savings account that pays 4.6% annual interest, or (ii) to purchase a $4 000 one-year guaranteed investment certificate with a monthly interest rate of 0.3%. From an opportunity cost standpoint, by making the decision to deposit $4 000 in the bank account, Emily will

  1. A) gain $37.6 by the end of the year.
  2. B) lose $37.6 by the end of the year.
  3. C) gain $57.6 by the end of the year.
  4. D) lose $57.6 by the end of the year.
  5. E) make zero economic profit.

Answer:  A

Diff: 3    Type: MC     Page Ref: 32-34

Topic:  2.1 Introduction

Skill:  Applied

User1:  Quantitative

 

 

19) If you borrow $2 000 today at 20% interest rate for 5 years, what is your simple interest in this case?

  1. A) $2 000
  2. B) $4 000
  3. C) $4 976.64
  4. D) $976.64
  5. E) $2 976.64

Answer:  A

Diff: 1    Type: MC     Page Ref: 24

Topic:  2.3 Compound and Simple Interest

Skill:  Applied

User1:  Quantitative

 

20) COSCO invested $5.5 million in a project ten years ago. As of today the worth of this project is $24.9 million. What annual interest rate has the project been earning if interest is compounded monthly?

  1. A) 14.2%
  2. B) 14.8%
  3. C) 15.2%
  4. D) 15.8%
  5. E) 16.2%

Answer:  C

Diff: 3    Type: MC     Page Ref: 27

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Applied

User1:  Quantitative

 

21) Equivalence is a condition that exists when

  1. A) the value of a cost at one time is numerically equal to the value of the related benefits received at a different time.
  2. B) the present worth of a cost equals the future worth of a cost at any point in time.
  3. C) the present worth of all costs and benefits equals the future worth of these costs and benefits at any point in time.
  4. D) the project breaks even, meaning costs equal benefits at a certain point in time.
  5. E) a decision-maker assesses two sets of cashflows as equally attractive.

Answer:  E

Diff: 1    Type: MC     Page Ref: 32

Topic:  2.7 Equivalence

Skill:  Recall

User1:  Qualitative

 

 

22) A project has the timing illustrated by the following cash flow diagram:

Which of the following statements about this cash flow diagram is correct?

  1. A) Year 1 ends at point A and year 2 begins at point B.
  2. B) Year 1 ends at point 2 and year 2 begins at point B.
  3. C) Year 1 ends at point 2 and year 2 begins at point 2.
  4. D) A project has four periods.
  5. E) First cost should be put at point 1.

Answer:  C

Diff: 1    Type: MC     Page Ref: 29-31

Topic:  2.6 Cash Flow Diagrams

Skill:  Applied

User1:  Quantitative

 

23) An effective interest rate has

  1. A) an arbitrary compounding period.
  2. B) a compounding period that is normally less than a year.
  3. C) an exogenously given compounding period.
  4. D) a one-year compounding period.
  5. E) no compounding periods.

Answer:  A

Diff: 1    Type: MC     Page Ref: 26

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Recall

User1:  Qualitative

 

24) Jennifer lends $2 000 to her friend who is launching a small business. Her friend promises to pay her 9% per year compounding interest. How much interest would Jennifer get at the end of four years?

  1. A) $823
  2. B) $1 284
  3. C) $1 892
  4. D) $2 324
  5. E) $2 823

Answer:  A

Diff: 2    Type: MC     Page Ref: 24

Topic:  2.3 Compound and Simple Interest

Skill:  Applied

User1:  Quantitative

 

 

25) At some point in time Peter had $3 000 in spare cash. He deposited this money in his bank account that pays a 1.1% annual interest rate. After one year he was approached by his friend who said that he could offer Peter an investment deal for a two-year period. What would the market equivalence be of Peter’s money?

  1. A) $3 000
  2. B) $3 033
  3. C) $3 066
  4. D) $3 100
  5. E) $3 133

Answer:  D

Diff: 3    Type: MC     Page Ref: 32-33

Topic:  2.3 Compound and Simple Interest

Skill:  Applied

User1:  Quantitative

 

26) In general, an interest rate is

  1. A) the difference between the amount of money lent and the amount of money repaid later.
  2. B) a percentage change in the time value of money.
  3. C) the ratio of the amount of money lent to the amount of money repaid later.
  4. D) the future worth of the money.
  5. E) the rate of return on direct investment.

Answer:  A

Diff: 1    Type: MC     Page Ref: 20-21

Topic:  2.2 Interest and Interest Rates

Skill:  Recall

User1:  Qualitative

 

27) Nominal interest rate is

  1. A) the actual but not usually stated interest rate.
  2. B) the actual and usually stated interest rate.
  3. C) the conventional method of stating the annual interest rate.
  4. D) the key interest rate in an economy.
  5. E) the overnight interest rate.

Answer:  C

Diff: 1    Type: MC     Page Ref: 26

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Recall

User1:  Qualitative

 

 

28) If the effective equivalent annual interest rate is 16.2%, and interest is compounded daily, what is the corresponding nominal annual interest rate?

  1. A) 11%
  2. B) 13%
  3. C) 15%
  4. D) 17%
  5. E) 19%

Answer:  C

Diff: 3    Type: MC     Page Ref: 26

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Applied

User1:  Quantitative

 

29) If you borrow $1 000 now at 10% interest for 5 years, what is the compound interest owed at the end of the fifth year?

  1. A) $1 000
  2. B) $1 100
  3. C) $1 610.51
  4. D) $610.51
  5. E) $500

Answer:  D

Diff: 2    Type: MC     Page Ref: 24-25

Topic:  2.3 Compound and Simple Interest

Skill:  Applied

User1:  Quantitative

 

30) Suppose that you just paid $9.91 monthly interest compounded daily on an outstanding balance of $1 000 on your credit card. What is the nominal annual interest rate in this case?

  1. A) 9%
  2. B) 10%
  3. C) 11%
  4. D) 12%
  5. E) 13%

Answer:  D

Diff: 3    Type: MC     Page Ref: 26

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Applied

User1:  Quantitative

 

 

31) Suppose the nominal rate is 10% per year and interest is compounded every two years. Use Equation 2.4 to calculate the effective annual rate.

  1. A) 4.88%
  2. B) 9.54%
  3. C) 10.25%
  4. D) 21%
  5. E) 44%,

Answer:  B

Diff: 3    Type: MC     Page Ref: 26

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Applied

User1:  Quantitative

 

32) If the nominal annual interest rate is 10% and interest is continuously compounded, what is the effective annual interest rate?

  1. A) 9%
  2. B) 9.52%
  3. C) 10.52%
  4. D) 11%
  5. E) 11.52%

Answer:  C

Diff: 2    Type: MC     Page Ref: 28

Topic:  2.5 Continuous Compounding

Skill:  Applied

User1:  Quantitative

 

33) If the effective annual interest rate is 10% and interest is continuously compounded, what is the nominal annual interest rate?

  1. A) 9.00%
  2. B) 9.53%
  3. C) 10.53%
  4. D) 11.53%
  5. E) 12.53%

Answer:  B

Diff: 3    Type: MC     Page Ref: 28

Topic:  2.5 Continuous Compounding

Skill:  Applied

User1:  Quantitative

 

 

34) You have $100 to deposit. Bank A offers 20% simple interest, Bank B offers 15% interest compounded annually. How many years would you have to keep your money in the bank for Bank B to be a better choice than Bank A?

  1. A) Bank B is always better.
  2. B) 4 years
  3. C) 5 years
  4. D) 6 years
  5. E) Bank B will never be better.

Answer:  C

Diff: 2    Type: MC     Page Ref: 22

Topic:  2.3 Compound and Simple Interest

Skill:  Applied

User1:  Quantitative

 

35) You have $100 to deposit. Bank A offers 16% interest, compounded annually, Bank B offers 15% interest, compounded monthly. How many years would you have to keep your money in the bank for Bank B to be a better choice?

  1. A) Bank B is always better.
  2. B) 4 years
  3. C) 5 years
  4. D) 6 years
  5. E) Bank B is never better.

Answer:  A

Diff: 2    Type: MC     Page Ref: 27

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Applied

User1:  Quantitative

 

36) You need to borrow $1 000 for a period of 10 years. Bank A will lend you the money at 10% interest, compounded annually, whereas Bank B will lend you the money at 10% interest, compounded monthly. At the end of ten years, how much more interest will you owe if you borrow from Bank B instead of Bank A?

  1. A) $74.59
  2. B) $92.50
  3. C) $113.30
  4. D) $137.39
  5. E) $148.12

Answer:  C

Diff: 2    Type: MC     Page Ref: 22-24

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Applied

User1:  Quantitative

 

 

37) You need to borrow $1 000. Bank A will lend you the money at 5% interest, compounded annually, whereas Bank B will lend you the money at 5% interest, compounded monthly.  Bank B also offers you a free cell phone, valued at $100, if you do business with them.  What is the longest duration of the loan for which Bank B would be a better choice?

  1. A) 10 years
  2. B) 15 years
  3. C) 20 Years
  4. D) 25 years
  5. E) 30 years

Answer:  D

Diff: 2    Type: MC     Page Ref: 22-24

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Applied

User1:  Quantitative

 

2.2   Short Answer Questions

 

1) Michael is indifferent about paying $1 500 for a new computer now and $2 000 two years from now. Define Michael’s implied interest rate.

Answer:  The implied interest rate can be defined from the following mathematical equivalence:

1 500 x (1 + i)2 = 2 000 and i = 15.5%.

Diff: 1    Type: SA     Page Ref: 32-33

Topic:  2.7 Equivalence

Skill:  Applied

User1:  Quantitative

 

2) When you borrow money from your bank, you pay a higher interest rate on that money compared with the interest rate offered on money in your savings account. How is this circumstance consistent with principles of engineering economics?

Answer:  This situation means that the interest rate associated with borrowing is higher than the one associated with saving. When you borrow, you convert future worth into present worth. However, when you save you convert present worth into future worth. As this example shows, conversion from present to future worth and vice versa in real life is not the same. This is inconsistent with market equivalence in engineering economics. Market equivalence is based on the idea that there is a market for money that permits cash flows in the future to be exchanged for cash flows in the present and vice versa at the same interest rate.

Diff: 1    Type: SA     Page Ref: 32-33

Topic:  2.7 Equivalence

Skill:  Recall

User1:  Qualitative

 

 

3) Stan has invested $1 000 into mutual fund at a 5% annual rate of return, compounded daily. What are the nominal and effective interest rates in this case? Discuss how these two interest rates affect Stan’s investment?

Answer:  The 5% annual rate of return is a nominal interest rate. The effective interest rate is the actual rate used in financial calculations. In order to convert the 5% nominal interest rate into effective interest rate, we have to use the following formula:

ie =   = 0.05127 or 5.127%.

Therefore, when calculating the real return on his investment, Stan should use 5.127% interest rate instead of 5%.

Diff: 2    Type: SA     Page Ref: 26

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Applied

User1:  Qualitative

 

4) Mary just earned $1 000 and wants to invest the money in Canada Savings Bonds. These bonds pay a 3% annual interest rate and have a ten-year maturity period. How much interest will Mary receive at the bonds’ maturity? In addition, compute simple interest and compare with actual interest.

Answer:

By the end of tenth year, $1 000 at 3% annual interest rate will become

1 000 x (1 + 0.03)10 = $1 343.92 and therefore, the accumulated interest is

$1 343.92 – 1 000 = $343.92.

 

It is also possible to directly apply the formula (2.2) on page 23 for compound interest rate

 

Ic = P(1 + i)N – P = 1 000 x (1 + 0.03)10 – 1 000 = $343.92

 

Simple interest can be defined as

 

Is = PiN = 1 000 x 0.03 x 10 = $300 (formula on page 24)

 

Real interest is larger by $43.93

Diff: 2    Type: SA     Page Ref: 24

Topic:  2.3 Compound and Simple Interest

Skill:  Applied

User1:  Qualitative

 

 

5) Suppose that the effective interest rate associated with a VISA credit card is 20.9% while the nominal interest rate is 18.9%. What are the card’s terms with respect to compounding?

Answer:  The following relationship between effective interest rate ie and nominal interest rate r should be used in this case:

ie =  – 1  where m is the number of compounding periods per year.

Plugging in values of ie and r

0.213 =  – 1  and solving for m by trial and error:

Trying m = 2(semiannual), 4 (quarterly), 12 (monthly) and 365 (daily) compounding, it turns out that m = 365 or nominal interest rate of 18.9% is compounded daily.

Diff: 3    Type: SA     Page Ref: 25-27

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Applied

User1:  Quantitative

 

6) Paul just bought a car for $15 000 and paid in cash. Calculate Paul’s opportunity cost as “funds tied up in the car” if you know that otherwise it was possible to invest the money at a 5% annual interest rate compounded monthly for five years.

Answer:  Opportunity cost in this case is the money forgone as a result of the car purchase, which is forgone interest. If this sum of money was invested under the specified conditions, it would earn the following interest in five years

Ic = P(1 + ie)N – P

where ie is the effective interest rate. In this case, the effective interest rate is

ie = = 0.05116 or 5.116%.

Therefore, interest forgone is

Ic = 15 000 x (1+0.05116)5 – 15 000 = $4 250.21

Diff: 3    Type: SA     Page Ref: 23-24

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Applied

User1:  Quantitative

 

7) Suppose you invested $1 000 in a new savings account with an annual interest rate of 3% compounded daily. What is your accumulated interest at the end of the first year?

Answer:  First, calculate the effective interest rate since 3% is nominal interest rate:ie =(1 + 0.03/365)365 – 1 = 0.03045 or it is 3.045%. Interest is given by the difference between future worth of the investment and its present worth which is $1 000 * (1 + 0.003045) – $1 000 = $30.45.

Diff: 2    Type: SA     Page Ref: 25-27

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Applied

User1:  Quantitative

 

 

8) Maria borrowed $2 000 for five years at a simple interest rate of 6% per year. How much money will Maria repay at the end of five years?

Answer:  Simple interest rate is a method of computing interest where interest earned during an interest period is not added to the principal amount used to calculate interest in the next period (p. 23). Therefore, the interest for each of five years is

 

Interest per year = 2, 000 x 0.06 = $120/year

 

Total interest = 120 x 5 = $600

 

The amount due after five years = 2 000 + 600 = $2 600.

Diff: 1    Type: SA     Page Ref: 24

Topic:  2.3 Compound and Simple Interest

Skill:  Applied

User1:  Quantitative

 

9) Explain why equivalences are just convenient assumptions. Give examples of two real world financial situations in which these equivalences do not hold

Answer:  Equivalences are needed to calculate and compare different costs and benefits over time. They are simplifications which capture the basic properties of cash flows without over-complicating the problem. They may not hold precisely true in the real world. For example, we borrow at a higher interest rate compared to savings. It means that in real life when we move along a time line in a cash flow diagram we might see different rates moving in two different directions; however, equivalences assume that the rate is the same. Another example is the cost of information. We assume (until Chapter 12) that information is free, while in real life information is costly.

Diff: 3    Type: SA     Page Ref: 32-34

Topic:  2.7 Equivalence

Skill:  Applied

User1:  Qualitative

 

 

10) A transportation company just bought a new truck for $25 000. The service life of the truck is seven years. The company has to pay a $100 registration fee at the beginning of every year plus maintenance costs of $1 000 in the first year and$200 at the beginning of the second year. At the end of the truck’s service life, it will be sold at 10% of its purchase price. Construct a cash flow diagram from the company’s perspective.

Answer:

Diff: 2    Type: SA     Page Ref: 29-31

Topic:  2.6 Cash Flow Diagrams

Skill:  Applied

User1:  Quantitative

 

11) Consider the following statement: “Financial data are collected based on discrete time periods.  However, in real life time is continuous. The error when using discrete compounding instead of continuous compounding is smaller the briefer the discrete compounding period is”. Do you agree or disagree with this statement and why?

Answer:  This statement is correct. With an increase in the number of discrete time periods, the error decreases. This can be seen by comparing two effective interest rates—compounded daily and continuously compounded—using the same nominal interest rate. In this case, the error is negligible.

Diff: 2    Type: SA     Page Ref: 28-29

Topic:  2.5 Continuous Compounding

Skill:  Applied

User1:  Qualitative

 

12) Joan is deciding whether she should remodel her house now or one year from now. If she does it now, the cost will be $1 500. If she waits one year, the cost is expected to be $1 600. At current interest rate of 5.6%, should Joan remodel her house now or one year from now?

Answer:  To compare the two alternatives, the concept of mathematical equivalence must be applied. According to the concept, $1 500 now is equivalent to 1 500 x (1 + 0.056) = $1 584 one year from now. This is less than $1 600 and therefore Joan should remodel her house now.

Diff: 1    Type: SA     Page Ref: 32

Topic:  2.3 Compound and Simple Interest

Skill:  Applied

User1:  Quantitative

 

 

13) George wants to buy a car. In order to accumulate money for a down payment, he decides to save $200 per month at 5% annual interest rate compounded monthly. How much money will Paul have for his down payment at the end of the first year?

Answer:  First, it is necessary to define monthly interest rate. It is 0.05/12 = 0.0041666. Each of Georges’ $200 payments should be compounded to the end of the year using this interest rate:

$200 * (1 + 0.0041666)11 + $200 * (1 + 0.0041666)10 + … + $200 = $2 455.78

 

Diff: 3    Type: SA     Page Ref: 22

Topic:  2.3 Compound and Simple Interest

Skill:  Applied

User1:  Quantitative

 

14) Suppose that a power plant project requires $10 million in period zero, has operating costs of $1 million per year over 10 years, and brings revenue of $2 million per year over that period of time. Based on this information and the concept of time value of money, comment on whether this is a profit generating project or not?

Answer:  If we forget about time value of money, then each year we have $2 million – $1 million = $1 million in net savings. Over 10 years it comes to $10 million. So, we invest $10 million now, and we will get $10 million in net savings over 10 years. However, if we take into account time value of money, net savings each year should be divided by some discount factor which means that net savings are less than $10 million. This project is not a good investment

Diff: 3    Type: SA     Page Ref: 25-27

Topic:  2.1 Introduction

Skill:  Applied

User1:  Quantitative

 

 

15) Suppose that the nominal interest rate is 18%. Calculate the effective interest rate when interest is compounded:

(i)     Annually

(ii)    Semiannually

(iii)   Quarterly

(iv)   Monthly

(v)    Biweekly

(vi)   Weekly

(vii)  Daily

(viii) Continuously

 

Answer:

Basic formula for (i) – (vii) is ie = (1 + )m – 1  where r is nominal interest rate and m is the number of sub-periods in the whole compounding period.

 

Therefore:

 

 

Diff: 2    Type: SA     Page Ref: 25-27

Topic:  2.4 Effective and Nominal Interest Rates

Skill:  Applied

User1:  Quantitative

 

 

 

Engineering Economics, 5e (Fraser)

Chapter 4   Comparison Methods: Part I

 

4.1   Multiple Choice Questions

 

1) A contingent project is an example of

  1. A) independent projects.
  2. B) mutually exclusive projects.
  3. C) related but not mutually exclusive projects.
  4. D) public projects.
  5. E) incremental projects.

Answer:  C

Diff: 1    Type: MC     Page Ref: 89

Topic:  4.2. Relations among projects

Skill:  Recall

User1:  Qualitative

 

2) Two mutually exclusive projects with the same service lives of 2 years are characterized by first costs of $100 million and $120 million respectively and annual savings of $60 million and $70 million respectively. If the MARR is 10%, which one should be chosen on the basis of the present worth comparison method?

  1. A) The second one because it has higher annual savings.
  2. B) The first one because its present worth is higher.
  3. C) The second one because its present worth is higher.
  4. D) The first one because incremental present worth is positive.
  5. E) Neither because both produce negative present worth.

Answer:  B

Diff: 2    Type: MC     Page Ref: 94

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Applied

User1:  Quantitative

 

3) Mutually exclusive projects can be compared in terms of present worth if

  1. A) they have the same service life.
  2. B) they have the same costs.
  3. C) they have the same benefits.
  4. D) they have the same rate of return.
  5. E) they have the same depreciation rate.

Answer:  A

Diff: 1    Type: MC     Page Ref: 94

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Qualitative

 

 

4) Two projects are mutually exclusive if

  1. A) the expected costs and benefits of one project do not depend on whether or not the other project is chosen.
  2. B) the expected costs and benefits of one project depend on whether the other project is chosen.
  3. C) if it is impossible to do both at the same time
  4. D) in the process of choosing one project at least one other project is chosen for comparison.
  5. E) the possibility of the implementation of one project directly depends on whether or not the other one is implemented.

Answer:  C

Diff: 1    Type: MC     Page Ref: 88

Topic:  4.2. Relations among projects

Skill:  Recall

User1:  Qualitative

 

5) If one project cannot be done by itself, a second project can be done alone, and both of them can be done together then the first project is said to be

  1. A) independent from the second project.
  2. B) related but not mutually exclusive to the second project.
  3. C) mutually exclusive to the second project.
  4. D) contingent on the second project.
  5. E) interrelated with the second project.

Answer:  D

Diff: 1    Type: MC     Page Ref: 89

Topic:  4.2. Relations among projects

Skill:  Recall

User1:  Qualitative

 

6) The minimum acceptable rate of return (MARR) is

  1. A) an interest rate, which is equal to a current bank interest rate.
  2. B) the least interest rate among all alternative projects.
  3. C) a highest interest rate among all alternative projects.
  4. D) an interest rate that allows an investor to recoup the investment.
  5. E) an interest rate that must be earned for a project to be accepted.

Answer:  E

Diff: 2    Type: MC     Page Ref: 90-91

Topic:  4.3. Minimum acceptable rate of return (MARR)

Skill:  Recall

User1:  Qualitative

 

 

7) For the purpose of comparison, what alternative should be used to an independent project?

  1. A) do nothing alternative
  2. B) other projects, earning MARR or higher interest rate
  3. C) putting money into a bank to earn a bank interest rate
  4. D) investing money in an alternative project that can recoup the investment
  5. E) investing money in an alternative project, which can earn the same interest rate as an independent project

Answer:  A

Diff: 1    Type: MC     Page Ref: 91

Topic:  4.2. Relations among projects

Skill:  Recall

User1:  Qualitative

 

8) What is the basis for decision-making using the present worth comparison method?

  1. A) to maximize the present worth of a project at the minimum acceptable rate of return
  2. B) to maximize the benefits of a project at the minimum acceptable rate of return
  3. C) to minimize the costs of a project at the minimum acceptable rate of return
  4. D) to maximize the difference between the project’s benefits and costs at the minimum acceptable rate of return
  5. E) to minimize the difference between the project’s costs and benefits at the minimum acceptable rate of return

Answer:  A

Diff: 2    Type: MC     Page Ref: 91

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Qualitative

 

9) The annual worth of a project is measured in terms of

  1. A) dollars.
  2. B) dollars per unit.
  3. C) dollars per year.
  4. D) dollars per unit per year.
  5. E) percentage.

Answer:  C

Diff: 1    Type: MC     Page Ref: 94-95

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Qualitative

 

 

10) A project is marginally acceptable if

  1. A) it earns positive present worth at the minimum acceptable rate of return.
  2. B) it earns negative present worth at the minimum acceptable rate of return.
  3. C) the present worth of a project is zero at the minimum acceptable rate of return.
  4. D) it earns more than the minimum acceptable rate of return.
  5. E) the difference between project’s costs and benefits is positive at the minimum acceptable rate of return.

Answer:  C

Diff: 1    Type: MC     Page Ref: 92

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Qualitative

 

11) An investor wants to invest in the trucking business. To run this business, 10 trucks should be acquired. The market price of a truck is $50 000. It is also necessary to pay $390 000 per year as wages to the truck drivers. Vehicle repair and maintenance costs are $10 000 per year. What is the annual worth of the project’s costs at 5% MARR over a 10-year period?

  1. A) $400 000
  2. B) $406 475
  3. C) $454 750
  4. D) $464 750
  5. E) $551 800

Answer:  D

Diff: 2    Type: MC     Page Ref: 91-92

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Quantitative

 

12) A project requires $10 000 as initial investment and will earn a revenue of $2 000 per year over the next seven years. The interest rate is 5.0% per year. What is the present worth of the project’s costs?

  1. A) $1 573
  2. B) $2 000
  3. C) $8 681
  4. D) $10 000
  5. E) $11 573

Answer:  D

Diff: 1    Type: MC     Page Ref: 91-92

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Quantitative

 

 

13) A project requires $10 000 as initial investment and will earn a revenue of $3 000 per year over the next five years. Annual interest rate is 5.4%. What is the present worth of the project’s benefits?

  1. A) $2 846
  2. B) $6 348
  3. C) $7 154
  4. D) $10 000
  5. E) $12 846

Answer:  E

Diff: 1    Type: MC     Page Ref: 91-92

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Quantitative

 

14) The best way to compare two projects with unequal lives is

  1. A) payback period.
  2. B) present worth comparison method.
  3. C) annual worth comparison method.
  4. D) study period method.
  5. E) incremental present worth comparison method.

Answer:  D

Diff: 2    Type: MC     Page Ref: 97

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Qualitative

 

15) The annual worth method is

  1. A) similar to the present worth method but transforms all annuities to a uniform series at the minimum acceptable rate of return.
  2. B) similar to the present worth method but transform all annuities to arithmetic gradient series at the minimum acceptable rate of return.
  3. C) methodologically different from the present worth method since it does not convert all annuities to the present worth at the minimum acceptable rate of return.
  4. D) similar to the present worth method since it transforms all annuities to the present worth as a single payment
  5. E) similar to the present worth method except that all payments are converted into a geometric gradient series

Answer:  A

Diff: 2    Type: MC     Page Ref: 94-96

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Qualitative

 

 

16) What is the present worth of an independent project that requires initial investment of $50 000 and annual maintenance costs of $4 000 for 10 years at a 4% minimum acceptable rate of return?

  1. A) -$4 000
  2. B) -$32 444
  3. C) -$54 000
  4. D) -$82 444
  5. E) -$17 556

Answer:  D

Diff: 2    Type: MC     Page Ref: 91-92

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Quantitative

 

17) If project A has present worth of -$27 000 and project B has present worth of -$26 000, then

  1. A) project B should be accepted on the basis of maximum benefits
  2. B) project A should be accepted because it has larger absolute value.
  3. C) project B should be accepted on the basis of minimum costs.
  4. D) the situation requires additional assessment since some of the costs or benefits might not be considered.
  5. E) the decision cannot be made because of uncertainty.

Answer:  C

Diff: 1    Type: MC     Page Ref: 91-92

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Quantitative

 

18) For comparison purposes, in order to evaluate alternatives with unequal lives one should

  1. A) repeat service life of each alternative at least three times.
  2. B) assume some certain study period for evaluation by introducing salvage values for each alternative.
  3. C) assume that the service life of each project is indefinite.
  4. D) apply only annual worth comparison method for each alternative.
  5. E) apply only present worth comparison method for each alternative.

Answer:  B

Diff: 1    Type: MC     Page Ref: 97

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Qualitative

 

 

19) Christine Robichaud, an engineer at Opus Ltd., has a $70 000 budget for upgrading the company’s warehouse. She has calculated that the improvements  could be completed in three months from now and would consume the entire budget. Due to these improvements the company saves $9 000 by the end of the first year, $14 000 by the end of the second year, and $35 000 by the end of the third year, $54 000 by the end of the fourth year and $17 000 by the end of the fifth year. What is the payback period for this  project?

  1. A) 1
  2. B) 2
  3. C) 3
  4. D) 4
  5. E) 5

Answer:  D

Diff: 2    Type: MC     Page Ref: 101-102

Topic:  4.5 Payback period

Skill:  Recall

User1:  Quantitative

 

20) What is the payback period?

  1. A) a period of time over which an initial investment can be recovered fully assuming market interest rate to be equal to the MARR
  2. B) a period of time after which a project starts generating profit
  3. C) a period of time over which an initial investment cannot be recovered
  4. D) a period of time over which an initial investment can be recovered fully
  5. E) a period of time over which an initial investment can be recouped assuming zero interest rate

Answer:  E

Diff: 2    Type: MC     Page Ref: 101-102

Topic:  4.5 Payback period

Skill:  Recall

User1:  Qualitative

 

21) What is the exact payback period for a 10-year project that requires $12 000 in initial investment,  $1 000 in annual maintenance costs and generates annual revenue of $2 600 per year  under 5% MARR?

  1. A) 7.5 years
  2. B) 7.9 years
  3. C) 8.4 years
  4. D) 9.2 years
  5. E) 10 years

Answer:  A

Diff: 1    Type: MC     Page Ref: 103

Topic:  4.5 Payback period

Skill:  Recall

User1:  Quantitative

 

 

22) By using the payback period method a decision-maker

  1. A) equally treats short- and long-term projects.
  2. B) ignores benefits accumulated by a project
  3. C) accounts for time value of money.
  4. D) disregards the need for a fast capital recovery.
  5. E) favour long-term projects over the short-term projects

Answer:  B

Diff: 3    Type: MC     Page Ref: 101-102

Topic:  4.5 Payback period

Skill:  Recall

User1:  Qualitative

 

23) A flight school operates 5 different types of airplanes. All airplanes are outdated and require some modifications. The school has $33 000 available in the budget for airplane modification this year. An aviation engineer came up with this plan to update all airplanes. The plan’s costs are shown in the following table:

 

Airplane # Modification cost Present Worth
1 $6 000 $54 000
2 $9 000 $56 000
3 $7 000 $55 000
4 $11 000 $59 000
5 $7 000 $55 000

 

Which airplane modification does the school have to postpone till next year?

  1. A) 1
  2. B) 2
  3. C) 3
  4. D) 4
  5. E) 5

Answer:  B

Diff: 3    Type: MC     Page Ref: 91-93, 101

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Quantitative

 

 

24) A municipality decided to build a new bridge to accommodate the city’s growing traffic across the river. It has to pay $200 000 to construct the bridge plus $15 000 per year for its maintenance. Assuming that the bridge will have an infinite service life, calculate the present worth of the project at 10% interest rate.

  1. A) $201 500
  2. B) -$215 000
  3. C) $280 000
  4. D) $350 000
  5. E) -$350 000

Answer:  E

Diff: 2    Type: MC     Page Ref: 91-93

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Quantitative

 

25) Annual worth comparison method

  1. A) should be used widely because it does not have to assume a fixed minimum acceptable rate of return over the duration of a project.
  2. B) can be preferable to the payback period method because the latter discriminates over the short-term projects.
  3. C) should be used whenever it requires less calculations than the present worth method since the two methods lead to the same result for projects with equal lives.
  4. D) can provide more accurate estimates of a project worth for the related but not mutually exclusive projects then other methods.
  5. E) is always preferred to the present worth method for comparison of two projects with the same service lives.

Answer:  C

Diff: 3    Type: MC     Page Ref: 94-95

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Quantitative

 

26) For an independent project to be accepted

  1. A) the PW of the project evaluated at MARR should be positive.
  2. B) the PW of the project evaluated at MARR should be equal to zero.
  3. C) the PW of the project evaluated at the MARR should be equal or greater than zero.
  4. D) the PW of the project should be positive at the market interest rate.
  5. E) the PW of the project should be equal to zero at the market interest rate.

Answer:  C

Diff: 2    Type: MC     Page Ref: 91-92

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Qualitative

 

 

27) For two mutually exclusive projects with equal lives, the one with

  1. A) a lower PW at the MARR should be chosen.
  2. B) a higher AW at the MARR should be chosen.
  3. C) a lower FW at the MARR should be chosen.
  4. D) a higher incremental PW at the MARR should be chosen.
  5. E) a higher incremental AW at the MARR should be chosen.

Answer:  B

Diff: 2    Type: MC     Page Ref: 94-95

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Qualitative

 

28) Consider the following investment alternatives:

 

YEAR A B C D E
0 -$100 -$100 -$100 -$100 -$100
1   $200    $470 -$200         0   $300
2   $300    $720   $200         0   $250
3   $400    $360   $250   $500   -$40

 

If MARR is 20%, which one is the best based on PW comparison method?

  1. A) A
  2. B) B
  3. C) C
  4. D) D
  5. E) E

Answer:  B

Diff: 2    Type: MC     Page Ref: 91-92

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Applied

User1:  Quantitative

 

29) If you have reliable economic forecast for five years only, and you would like to evaluate two alternatives with five and six years service lives, the best method to do so is

  1. A) the repeated lives method.
  2. B) the study period method.
  3. C) Both repeated lives method and study period method produce the same result.
  4. D) to choose the alternative with the highest FW.
  5. E) to choose the alternative with the highest PW.

Answer:  B

Diff: 3    Type: MC     Page Ref: 97

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Qualitative

 

 

30) Suppose that cash flows of a project are given as follows:

 

Year Costs, $$ Savings, $$
0 100 000 0
1   20 000 60 000
2   20 000 70 000
3   20 000 80 000
4   20 000 90 000

 

It is known that MARR is 10%. What is the project’s payback period?

  1. A) one year
  2. B) two years
  3. C) three years
  4. D) four years
  5. E) The project does not pay back.

Answer:  C

Diff: 3    Type: MC     Page Ref: 101-102

Topic:  4.5 Payback period

Skill:  Recall

User1:  Quantitative

 

31) I can invest for a pension in either the Senex or the Geriatrix pension plan. Senex requires me to invest $1500 a year for the next 15 years, whereas Geriatrix requires an immediate deposit of $5 000 and a subsequent annual investment of $1 200 a year. Senex pays 10% on my investments, whereas Geriatrix pays 9%. How much more money will I have in my pension plan in 15 years time if I invest in Senex?

  1. A) $1 500
  2. B) $1 600
  3. C) $1 600
  4. D) $1 700
  5. E) $1 800

Answer:  A

Diff: 2    Type: MC     Page Ref: 95

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Applied

User1:  Quantitative

 

 

32) I can invest for a pension in either the Senex or the Geriatrix pension plan. Senex requires me to invest $1500 a year for the next 15 years, whereas Geriatrix requires an immediate deposit of $5 000 and a subsequent annual investment of $1 200 a year. If my MARR is 15%, how much greater is the present cost to me of the series of payments I would make to Geriatrix versus the series of payments I would make to Senex?

  1. A) $2 457
  2. B) $3 246
  3. C) $3 863
  4. D) $4 127
  5. E) $4 222

Answer:  B

Diff: 2    Type: MC     Page Ref: 91-94

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Applied

User1:  Quantitative

 

33) I can invest for a pension in either the Senex or the Geriatrix pension plan. Senex requires me to invest $1500 a year for the next 15 years, whereas Geriatrix requires an immediate deposit of $5 000 and a subsequent annual investment of $1 200 a year. If my MARR is 15%, how much greater is the equivalent uniform annual cost to me of the series of payments I would make to Geriatrix versus the series of payments I would make to Senex?

  1. A) $333
  2. B) $444
  3. C) $555
  4. D) $666
  5. E) $777

Answer:  C

Diff: 2    Type: MC     Page Ref: 94-96

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Applied

User1:  Quantitative

 

34) I have 3 possible choices for a lawnmower. They have expected working lives of 3, 4 and 5 years.  If I expect lawnmower technologies to be stable for the foreseeable future, over what period of time should I compare the equivalent uniform annual costs of the three choices?

  1. A) 12 years
  2. B) 20 years
  3. C) 30 years
  4. D) 45 years
  5. E) 60 years

Answer:  E

Diff: 2    Type: MC     Page Ref: 97

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Applied

User1:  Quantitative

 

 

35) I spend $15 per week on bus fares to work. One day I decide to buy a mountain bike for $2000 and to ride to work instead. Assuming I work 50 weeks in a year, what is the payback time for this investment?

  1. A) two years five months
  2. B) two years seven months
  3. C) two years eight months
  4. D) two years nine months
  5. E) two years ten months

Answer:  C

Diff: 2    Type: MC     Page Ref: 101-103

Topic:  4.5 Payback period

Skill:  Applied

User1:  Quantitative

 

36) I spend $60 per month on bus fares to work. One day I decide to buy a mountain bike for $1680 and to ride to work instead. Assuming I work 12 months in a year, and my MARR is 1% per month, compounded monthly, what is the discounted payback time for this investment?

  1. A) two years five months
  2. B) two years seven months
  3. C) two years eight months
  4. D) two years nine months
  5. E) two years ten months

Answer:  D

Diff: 2    Type: MC     Page Ref: 103

Topic:  4.5 Payback period

Skill:  Applied

User1:  Quantitative

 

37) I can buy a Grapefruit laptop computer for $3 000, or a Doors laptop for $2 500. The Grapefruit has an expected life of five years, whereas the Doors is only expected to last four years. Both provide equivalent service. A four-year-old Grapefruit has a salvage value of $200. If my MARR is 10%, what is the present cost of choosing the Grapefruit over the Doors?

  1. A) $300
  2. B) $315
  3. C) $363
  4. D) $383
  5. E) $636

Answer:  C

Diff: 2    Type: MC     Page Ref: 90-93

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Applied

User1:  Quantitative

 

 

4.2   Short Answer Questions

 

1) Explain how related but not mutually exclusive projects are compared in terms of the present worth and the annual worth comparison methods

Answer:  Evaluation of related but not mutually exclusive projects can be simplified by combining them into mutually exclusive sets including the “do nothing alternative”. If there are “n” related projects, it is possible to make 2n mutually exclusive sets. Then if the service life is the same across the sets, the alternative with the highest present worth should be chosen. Otherwise the alternative with the highest annual worth should be chosen and the two lead to the same conclusion

Diff: 1    Type: SA     Page Ref: 88- 90

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Qualitative

 

2) How is the payback period defined and what is its economic interpretation?

Answer:  The payback period is the ratio of the first cost (initial investment) to the expected annual savings (revenues). It is an approximate criterion, and it shows how long it takes for an investment to be recouped when the interest rate is assumed to be zero.

Diff: 1    Type: SA     Page Ref: 101-102

Topic:  4.5 Payback period

Skill:  Recall

User1:  Qualitative

 

3) A manager is considering two technological lines to produce candies. The first one requires $1 million in initial investment and produces 150 kilograms (kg) of candies per day. The second one requires $1.3 million of initial investment and produces 200 kg of candies per day. Assuming the same service life of six years for both lines, a 5% annual interest rate and the price of candies of $4/kg, which line should the manager acquire?

Answer:  These are two mutually exclusive projects with the same service lives. Therefore, the one with the highest present worth (PW) should be chosen.

 

PW(Line 1) = -$1 000 000 + $4/kg x 150 kg/day x 365 days/year x (P/A, 5%, 6) =

-$1 000 000 + $219 000 x 5.0757 = $111 578.3

 

PW(Line 2) = -$1 300 000 + $4/kg x 200 kg/day x 365 days/year x (P/A, 5%, 6) =

-$1 300 000 + $292 000 x 5.0757 = $182 104.4

 

Since the present worth of Line 2 is higher, it should be chosen.

Diff: 2    Type: SA     Page Ref: 94-95

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Applied

User1:  Quantitative

 

 

4) Jim is considering new software to design web pages.  Currently there are three brands on the market: SUMY, ROVNO and PSEL. Based on the software manuals, Jim evaluated his productivity in terms of pages designed per year and put this information in a table:

 

  SUMY ROVNO PSEL
Initial price, $ 1 000 1 200 1 500
Productivity, pages/year 50 70 80

 

If Jim receives $10 per page designed, evaluate these alternatives in terms of their payback period.

Answer:  The payback period is the ratio of initial investment to annual savings (revenues).  In this example, the price of software is the initial investment. In turn,

 

Annual revenue = $10/page x productivity.

 

Using these definitions, the payback period is

–  for SUMY: 1 000/(10 x 50) = 2 years

–  for ROVNO: 1 200/(10 x 70) = 1.7 years

–  for PSEL: 1 500/(10 x 80) = 1.9 years

 

Since ROVNO has the lowest payback period it should be chosen.

Diff: 1    Type: SA     Page Ref: 101-102

Topic:  4.5 Payback period

Skill:  Applied

User1:  Quantitative

 

5) REMAX Development owns a parcel of land in a city and wants to sell it to the highest bidder. One of the requirements is that REMAX will be involved in the construction and maintenance on that land. Three bidders submitted their proposals. From a standpoint of REMAX information is as follows:

 

  Project 1 Project 2 Project 3
Investment $3.0 million $3.3 million $4.6 million
Annual operating and maintenance costs $750 thousand $200 thousand $400 thousand
Annual income $1.2 million $750 thousand $1.1 million

 

Evaluate these projects on the basis of their present worth if the interest rate is 12% and the assessment horizon is 25 years.

Answer:

PW(Project 1) = -3.0 – 0.75 x (P/A, 12%, 25) + 1.2 x (P/A, 12%, 25) =

-3.0 + (1.2-0.75) x 7.843 = $0.529 million

 

PW(Project 2) = -3.3 – 0.2 x (P/A, 12%, 25) + 0.75 x (P/A, 12%, 25) =

-3.3 + (0.75-0.2) x 7.843 = $1.014 million

 

PW(Project 3) = -4.6 – 0.4 x (P/A, 12%, 25) + 1.1 x (P/A, 12%, 25) =

-4.6 + (1.1-0.4) x 7.843 = $0.890 million

 

The second bidder’s project is more attractive since it produces the highest present worth.

Diff: 2    Type: SA     Page Ref: 94-95

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Applied

User1:  Quantitative

 

6) MMM Consulting is evaluating two oil pumps with the same operating costs. The first pump lasts three years and costs $12 000. The second one lasts 4 years and costs $15 000. Current market interest rate is 8%. Compare the two using the repeated lives method.

Answer:  The least common multiple of two lives is 12 years. Therefore, the first cash flow of $12 000 should be replicated 4 times giving $12 000 in year 0, $12 000 in year 3, $12 000 in year 6 and $12 000 in year 9. We can choose a new compounding period of three years and re-define interest rate as effective interest rate:

 

PW of First cost (Pump 1) = $12 000 + $12 000 x (P/A, 25.9712%, 3)

= 12 000 + 12 000 x 1.924 = $35 088

 

The second cash flow of $15 000 should be replicated 3 times, giving $15 000 in year 0, $15 000 in year 4 and $15 000 in year 8. The associated periodic (4-year) interest rate is

i = (1 + 0.08)4 – 1 = 0.36048896.

 

PW of the First Cost (Pump 2) = $15 000 + $15 000 x (P/A, 36.048896%, 2

= $15 000 + 15 000 x 1.275 = $34 125

 

Since we don’t have information on productivity, the maximum present worth of the project criterion becomes the minimum present worth of the first costs (operating costs are the same!). Based on this criterion, the second oil pump should be chosen.

 

Alternatively, we might treat each payment of $12 000 and $15 000 as single payments in respective years and use discounting via the(P/F, 8%, year) factor.

Diff: 3    Type: SA     Page Ref: 97-99

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Applied

User1:  Quantitative

 

7) Morris paid $500 per month for 20 years to pay off a bank mortgage on his house. If his down payment was $5 000, and the interest rate was 6% compounded monthly, how much did his house cost?

Answer:  The house price can be found as follows:

 

5 000 + 500 * (P/A, 6%/12, 12 * 20) = 5 000 + 500 * (P/A, 0.5%, 240) = $74 790

Diff: 1    Type: SA     Page Ref: 91-92

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Applied

User1:  Quantitative

 

 

8) MJ Consulting is doing a feasibility study on a new technological line to produce shoes. Experts identified three mutually exclusive alternatives. Information on all of them is presented in the following table:

 

  Alternative 1 Alternative 2 Alternative 3
Initial investment $1.2 million $1.3 million $1.4 million
Operating costs $0.3 million/year $0.5 million/year $0.6 million/year
Production 5 000 pairs/year 8 000 pairs/year 10 000 pairs/year

 

If the price of a pair of shoes is $100, the annual interest rate is 7% and service life is 5 years for all alternatives, which one is the best?

Answer:  Since service life is the same, the Present Worth method is applied:

 

PW(Alternative 1) = -1.2 – 0.3 x (P/A, 7%, 5) + (100 x 5 000/1 000 000) x (P/A, 7%, 5) =

-1.2 – 0.3 x 4.1 +0.5 x 4.1 = -$0.38 million

 

PW(Alternative 2) = -1.3 – 0.5 x (P/A, 7%, 5) + (100 x 8 000/1 000 000) x (P/A, 7%, 5) =

-1.3 – 0.5 x 4.1 + 0.8 x 4.1 = -$0.07 million

 

PW(Alternative 3) = -1.4 – 0.6 x (P/A, 7%, 5) + (100 x 10 000/1 000 000) x (P/A, 7%, 5) =

-1.4 – 0.6 x 4.1 +1.0 x 4.1 = $0.24 million

 

Alternative 3 is the best since unlike two others it results in a positive present worth.

Diff: 1    Type: SA     Page Ref: 94

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Applied

User1:  Quantitative

 

 

9) NB Power wants to assess the opportunity to install and use a photovoltaic system to produce electricity from solar energy. The unit cost of electricity currently produced by the thermal power plant is $0.06 per kilowatt-hour. The photovoltaic system is characterized by the following:

 

Item Value
Equipment and tools:

–  solar cell

–  batteries

–  wiring

–  power control

 

$2 600

$1 00

$800

$100

Operating costs $100/year

 

Batteries must be replaced every 5 years while power control is replaced every 10 years.

 

Assuming a service life of 20 years, annual interest rate of 10%, capacity of the photovoltaic system of 30 kilowatt-hours per day and ignoring the installation cost, should NB Power implement the system?

 

Answer:  First define the present worth of the total costs:

 

PW(Costs) = ($2 600 + $1 000 + $800+$100) + $1 000 x (P/F, 10%, 5) + ($1 000+$100) x (P/F, 10%, 10) + $1 000 x (P/F, 10%, 15) + $100 x (P/A, 10%, 20) = $6 635.77

 

Calculate annual cost of electricity produced by the photovoltaic system:

 

AW(Costs) = PW x (A/P, 10%, 20) = $6 635.77 x 0.11746 = $779.44/year

 

Annual production is (30 kilowatt-hours/day x 365 days) = 10 950 kilowatt-hours/year.

 

Therefore, unit cost of electricity produced by the photovoltaic system is:

 

[$779.44/year]/(10 950 kilowatt-hours/year] = $0.0712 per kilowatt-hour, which is higher than the existing unit cost. The photovoltaic system should be rejected.

Diff: 3    Type: SA     Page Ref: 94-96

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Applied

User1:  Quantitative

 

 

10) It was discovered by geologists that an oil deposit can produce oil commercially for 5 years. In order to develop this deposit $100 million is required in initial investment with the sale value of the equipment of $20 million at the end of the deposit’s life. In addition, $4 million annually is required for operating and maintenance costs. Current price of oil is $90 per barrel. How much oil must this deposit produce per year in order to cover all the costs under 2% interest rate?

Answer:  Let us define annual production by Q. Then we can set the following equation:

-100 + 90Q * (P/A, 2%, 5) – 4 * (P/A, 2%, 5) + 20 * (P/F, 2%, 5) = 0

Solving for Q, we obtain 237 472 barrels per year

Diff: 2    Type: SA     Page Ref: 91-92

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Applied

User1:  Quantitative

 

11) Explain when the repeated lives comparison method should be applied and how it works.

Answer:  The repeated lives method is applied if the lives of the alternatives are not the same. It is assumed that each alternative can be repeated with the same costs and benefits in the future. The least common multiple of the existing lives is calculated and each alternative is repeated the required number of times.

Diff: 2    Type: SA     Page Ref: 97-98

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Quantitative

 

12) How is the study period comparison method different from the repeated lives comparison method?

Answer:  Both comparison methods should be applied to compare the alternatives with unequal lives. The repeated lives method just replicates an alternative’s costs and benefits the required number of times according to the least common multiple of lives without extra assumptions. The study period method assumes some specific time horizon for all alternatives according to the evaluator’s perception of the future, and requires some extra assumptions about salvage value whenever the life of the alternatives exceeds that of the given study period.

Diff: 2    Type: SA     Page Ref: 97-98

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Recall

User1:  Qualitative

 

13) The present worth of a 4-year project is $145 million. It has a first cost of $76.8 million, annual savings and operating costs, and salvage value of $44 million at the end of its service life. What is the project’s annual net savings if its MARR is 4%?

Answer:  Present worth of the project can be written as -76.8 + X * (P/A, 4%, 4) + 44 * (P/F, 4%, 4) = 145 where X is annual net savings equal to the difference between annual savings and annual operating costs. Solving for X, we find that annual net savings are $50.742 million

Diff: 2    Type: SA     Page Ref: 93-103

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Applied

User1:  Quantitative

 

 

14) Suppose that cash flows of a project are given as follows:

 

Year Costs Savings
0 $10 000 0
1  $2 000 $2 000
2  $2 000 $3 000
3  $2 000 $4 000
4  $2 000 $5 000

 

What is the project’s payback period?

Answer:  It is possible to summarize all calculations in the following table:

 

Year Net savings Remaining costs
1 2 000 – 2 000 = 0 10 000
2 3 000 – 2 000 = 1 000 9 000
3 4 000 – 2 000 = 2 000 7 000
4 5 000 – 2 000 = 3 000 4 000
5 6 000 – 2 000 = 4 000  0

 

The payback period is 5 years, when net savings offset the first cost.

Diff: 2    Type: SA     Page Ref: 101-102

Topic:  4.5 Payback period

Skill:  Applied

User1:  Quantitative

 

15) A new computer system costs $20 000 and saves $5 000 per year over a five year service life. Resale value of the computer is estimated at $5 000 at the end of its service life. If the MARR is 4%, what is the annual net benefits of the computer system?

Answer:  The present worth of the system is:

 

PW = -20 000 + 5 000 x (P/A, 4%, 5) + 5 000 x (P/F, 4, 5)

= -20 000 + 5 000 x 4.45182 + 5 000 x 0.82192

= -20 000 + 22 259.11 + 4 109.63

= $6 368.74

 

Since the present worth is positive, the annual worth provides net annual benefits:

 

AW = PW x (A/P, 4%, 5) = 6 368.74 x 0.22462 = $1 430.59/year.

Diff: 1    Type: SA     Page Ref: 94-96

Topic:  4.4 Present worth (PW) and annual worth (AW) comparisons

Skill:  Applied

User1:  Quantitative

 

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