Engineering Economics Financial Decision Making for Engineers, Fifth Edition with Companion Website,

Engineering Economics Financial Decision Making for Engineers, 5th E with Companion Website By Niall M. Fraser

Email: richard@qwconsultancy.com

Engineering Economics, 5e (Fraser) Chapter 1 Engineering Decision Making 1.1 Multiple Choice Questions 1) Evaluation of an engineering project involves the following constraints: A) financial, environmental, social and political constraints.

B) technical and financial constraints.

C) technological and social constraints.

D) time and money constraints.

E) budget and time constraints.

Answer: A Diff: 1 Type: MC Page Ref: 3 Topic: 1.1. Engineering decision making Skill: Recall User1: Qualitative 2) Engineering economics is A) a body of knowledge to determine which of several alternative is technically best.

B) a collection of techniques for quantitative analysis to select a preferable alternative from several technically viable ones.

C) a set of tools to calculate an engineering project's costs.

D) a set of rules to evaluate an engineering project.

E) a set of methods to determine an engineering project's costs in terms of time value of money.

Answer: B Diff: 1 Type: MC Page Ref: 3 Topic: 1.2. What is engineering economics Skill: Recall User1: Qualitative 3) From an economic standpoint, any engineering project can be thought of in terms of A) its costs and benefits over time.

B) its constraints.

C) its receipts and disbursements.

D) its revenues and profits over time.

E) its investment and costs over time.

Answer: A Diff: 1 Type: MC Page Ref: 3 Topic: 1.2. What is engineering economics Skill: Recall User1: Qualitative

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4) In the context of your textbook, abstraction means A) a study.

B) a method C) a methodology D) a model.

E) an approach Answer: D Diff: 1 Type: MC Page Ref: 6 Topic: 1.4. Dealing with abstractions Skill: Recall User1: Qualitative 5) When an engineer prepares a feasibility study, what economic information must she possess in order to do it correctly?

A) engineering specification B) expert opinion about this project C) macroeconomic situation in the world D) potential future costs and benefits of the project E) the level of uncertainty Answer: D Diff: 2 Type: MC Page Ref: 6 Topic: 1.4. Dealing with abstractions Skill: Applied User1: Qualitative 6) In which of the following problems do you need engineering economics?

A) to choose a course in your program B) to chose the right textbook C) to decide whether to borrow money from parents or from a bank D) to decide whether or not to buy a car E) to decide where to spend summer vacation Answer: C Diff: 2 Type: MC Page Ref: 6-8 Topic: 1.4. Dealing with abstractions Skill: Applied User1: Qualitative 7) Which of the following statements best describes the moral issues faced by engineers while evaluating projects?

A) moral issues are clearly defined in engineering design B) moral issues do not matter as long as technical goals are achieved C) your employer is the only person who can define what is ethical or unethical D) there are no general answers to moral questions E) since moral issues are value judgements they must not be taken into consideration Answer: D Diff: 3 Type: MC Page Ref: 9-10 Topic: 1.5. The moral question: Three true stories Skill: Applied 2 .

User1: Qualitative 8) The best way to find some answers to difficult moral questions in engineering design and project evaluation is A) to ask your friends.

B) to consult professional engineering associations.

C) to search the web.

D) to read newspapers.

E) to read professional textbooks.

Answer: B Diff: 2 Type: MC Page Ref: 9-10 Topic: 1.5. The moral question: Three true stories Skill: Recall User1: Qualitative 9) Stan borrowed $5 000 one year ago. Now he has to repay $5 100. The interest Stan pays is A) $5 100 B) 102% C) $100 D) 2% E) $5 000 Answer: C Diff: 2 Type: MC Page Ref: 6-8 Topic: 1.4. Dealing with abstractions Skill: Applied User1: Quantitative 10) Joan borrowed $1 000 one year ago. Now she has to repay $1 100. Therefore, the interest rate she pays is A) $1 000 B) $1 100 C) $100 D) 110% E) 10% Answer: E Diff: 2 Type: MC Page Ref: 6-8 Topic: 1.4. Dealing with abstractions Skill: Applied User1: Quantitative 11) All but one of the following are economic criteria to evaluate an engineering project. Which one is the non-economic criterion?

A) maximum profit B) minimum cost C) minimum pollution D) maximum rate of return E) minimum payback period Answer: C 3 .

Diff: 2 Type: MC Page Ref: 2-6 Topic: 1.2. What is engineering economics Skill: Applied User1: Qualitative 12) Analyze the following statement: "There is 50% probability of raining". This statement concerns A) extrapolation.

B) risk.

C) precise description.

D) abstraction.

E) expected scenario.

Answer: B Diff: 2 Type: MC Page Ref: 11 Topic: 1.6. Uncertainty, sensitivity analysis, and currencies Skill: Applied User1: Qualitative 13) To resolve uncertainty, engineers apply A) marginal analysis.

B) cash-flow analysis.

C) sensitivity analysis.

D) risk analysis.

E) economic impact analysis.

Answer: C Diff: 2 Type: MC Page Ref: 11 Topic: 1.6. Uncertainty, sensitivity analysis, and currencies Skill: Recall User1: Qualitative 14) If you are asked to choose between $100 today and $150 one year from now, you are being asked to A) make a guess.

B) reveal your implied interest rate.

C) compare two values which are not comparable in principle.

D) reveal your private financial information.

E) make a choice under uncertainty.

Answer: B Diff: 3 Type: MC Page Ref: 6-8 Topic: 1.4. Dealing with abstractions Skill: Applied User1: Qualitative 15) Which of the following items has negligible maintenance costs?

A) computer B) building C) ruler D) shoes E) fur hat Answer: C 4 .

Diff: 2 Type: MC Page Ref: 6-8 Topic: 1.4. Dealing with abstractions Skill: Applied User1: Qualitative

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16) You decide to buy a car. The following are some questions you have to answer with respect to your purchase. Which question is directly associated with engineering economics?

A) what colour?

B) what size?

C) CD player or tape player?

D) to lease or to own?

E) Ford or Toyota?

Answer: D Diff: 2 Type: MC Page Ref: 3 Topic: 1.3. Making decisions Skill: Applied User1: Qualitative 17) Susan is evaluating an engineering project. She assumes zero inflation for the duration of the project.

With respect to this situation, which statement is consistent with the subject matter of engineering economics?

A) A zero inflation assumption is always a good one.

B) A zero inflation assumption is not realistic but it is better than assuming some uncertain inflation rate.

C) A zero assumption is not a bad one for the base case, but then the project should be evaluated under different values in some range to see how inflation affects the project.

D) Since in Canada inflation is low, it is possible to neglect it.

E) The highest historical inflation rate must be chosen instead.

Answer: C Diff: 3 Type: MC Page Ref: 3-6 Topic: 1.4. Dealing with abstractions Skill: Applied User1: Qualitative 18) An owner of a small company is deciding to sell her business. She received several options specified bellow. Which one should the company's owner accept?

A) $600 000 in cash B) $500 000 in government bonds earning a 5% rate of return over 10 years C) $400 000 in securities earning an 8% rate of return over 8 years D) 15% of the future profits earned by the next owner during first 10 years E) it is impossible to compare these offers due to uncertainty about basic economic variables over time Answer: E Diff: 3 Type: MC Page Ref: 3-6 Topic: 1.4. Dealing with abstractions Skill: Applied User1: Quantitative

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19) Suppose that you are asked to evaluate a project of building a new bridge. Which of the following factors is the least important in terms of engineering economics?

A) service life of the bridge B) value added by the bridge C) cost structure D) vehicle stock in your city E) current interest rate Answer: D Diff: 3 Type: MC Page Ref: 3-6 Topic: 1.4. Dealing with abstractions Skill: Applied User1: Qualitative 20) Based on your intuition and experience, which of the following options has the highest value?

A) $100 government bond earning a 5% annual rate of return B) $100 invested in a fund with a 10% annual rate of return C) $200 government bond earning a 5% annual rate of return D) $200 invested in a fund earning a 10% annual rate of return E) $150 invested in a fund earning a 7% annual rate of return Answer: D Diff: 2 Type: MC Page Ref: 6-8 Topic: 1.4. Dealing with abstractions Skill: Applied User1: Quantitative 21) Engineering economics is important because A) it is a natural science.

B) it provides answers to all financial questions.

C) it facilitates the establishment of verifiable facts about decisions.

D) it uses mathematical models to address engineering issues.

E) it is based on objective laws.

Answer: C Diff: 2 Type: MC Page Ref: 4 Topic: 1.2. What is engineering economics Skill: Recall User1: Qualitative 22) What is the best way to describe a mathematical model?

A) It is an exact copy of the real world.

B) It is a simplification to describe the real world in a reasonable way.

C) It is a set of mathematical relationships with assumptions based on natural laws.

D) It is a relationship that includes all aspects of a modelling situation.

E) It describes all complex processes that occur in the real world.

Answer: B Diff: 2 Type: MC Page Ref: 6 Topic: 1.4. Dealing with abstractions Skill: Recall User1: Qualitative 7 .

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23) In engineering economics, sensitivity analysis A) addresses risk in engineering projects.

B) addresses inflation in a country.

C) involves random variables with their probabilities.

D) assesses the effect of uncertainty on a decision.

E) involves cash flows that are known with certainty.

Answer: D Diff: 3 Type: MC Page Ref: 11-12 Topic: 1.6. Uncertainty, sensitivity analysis, and currencies Skill: Recall User1: Qualitative 24) Most of the economic values we use in our evaluation of engineering projects are A) precise and very descriptive.

B) approximate.

C) based on robust statistical estimation.

D) derived from natural experiments.

E) results of precise statistical surveys.

Answer: B Diff: 2 Type: MC Page Ref: 11 Topic: 1.6. Uncertainty, sensitivity analysis, and currencies Skill: Recall User1: Qualitative 25) When we say "let us assume that economic agents are rational", in fact we A) precisely describe behaviour of economic agents.

B) want economic agents to behave that way.

C) make an assumption to predict the agents' behaviour.

D) impose restrictions on our model.

E) introduce uncertainty into our modelling process.

Answer: C Diff: 3 Type: MC Page Ref: 6 Topic: 1.3. Making decisions Skill: Applied User1: Qualitative 1.2 Short Answer Questions 1) What is the subject matter of Engineering Economics?

Answer: Engineering economics deals with techniques of quantitative analysis useful for selecting a preferable alternative from several technically viable ones. Its major objective is to allow an engineer to determine which of several alternatives is economically best.

Diff: 1 Type: SA Page Ref: 3 Topic: 1.2. What is engineering economics Skill: Recall User1: Qualitative

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2) Briefly describe the structure of a decision-making process as discussed in your textbook.

Answer: The textbook presents this structure in a form of a pyramid. At the top of the pyramid are preferences, which directly control the choices made. The next tier is composed of people and politics.

Politics means the use of power in organizations. The next tier is a collection of facts. The facts, which may not be verifiable, contribute to the politics and people, and indirectly to the preferences. At the bottom of the pyramid are the activities that contribute to the facts. These include the history of previous similar decisions, statistics of various sorts, and a determination of costs.

Diff: 2 Type: SA Page Ref: 4-5 Topic: 1.3. Making decisions Skill: Recall User1: Qualitative 3) Explain the role of a mathematical model in engineering economics.

Answer: When one describes something, one does so for a purpose. In the description, one selects aspects of the real world that are relevant to that purpose. The process of simplifying the complexities of the real world is necessary for any engineering analysis. This process of simplification is called abstraction or a model. Once a model is developed, it is used to analyze a situation, and perhaps make some predictions about the real world. The analysis and the predictions are then related back to the real world to make sure that the model is valid. The role of abstractions is to develop a viable mode of the real world.

Diff: 2 Type: SA Page Ref: 6 Topic: 1.4. Dealing with abstractions Skill: Recall User1: Qualitative 4) How should engineers address moral questions associated with economic evaluation of engineering projects?

Answer: Engineers have a responsibility to society to behave ethically and responsibly in all ways.

When many different issues must be taken into account in engineering decision making, it is often difficult to determine what course of action is ethical. There are no general answers to difficult moral questions. Practicing engineers often have to make choices with an ethical component, and can sometimes rely on no stronger foundation than their own sense of right and wrong. More information about ethical issues for engineers can be obtained from provincial professional engineering organizations.

Diff: 1 Type: SA Page Ref: 8-10 Topic: 1.5. The moral question: Three true stories Skill: Recall User1: Qualitative

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5) Suppose that you are about to start a new business: You will provide consulting services for those who want to launch their own web-sites. What major decisions based on the subject matter of engineering economics should be made before you set up your business activities?

Answer: You have to evaluate the expected future costs and benefits of your enterprise. In doing so, you have to understand your final product and identify a market for that product. As a result, you will be able to evaluate your expected revenues. This is known as demand-side analysis. On the supply side, you have to understand all materials, efforts and services involved in producing and/or providing your product. This will give you a cost structure. Then you have to define your planning horizon in order to use economic forecasts of major macroeconomic variables such as interest rate, inflation and others.

You have to understand the Canadian tax system as well. If you are uncertain about values of some variables involved, you have to use principles of sensitivity analysis. All this is the subject matter of engineering economics.

Diff: 3 Type: SA Page Ref: 3-5, 11 Topic: 1.4. Dealing with abstractions Skill: Applied User1: Qualitative 6) As a University student, when you look for a new apartment, what economic criteria consistent with the subject matter of engineering economics do you apply?

Answer: You have to understand your budget in terms of cash inflows and outflows. Cash inflows may include: a fellowship, a grant, a loan, an allowance provided by parents and others. Cash outflows to consider are: costs of utilities, transportation cost, cost of textbooks, tuition fee, food expenses, and others. As well, time should be explicitly introduced into your economic analysis. Therefore, for example, a more expensive apartment closer to the University can be preferred to a cheaper one far away, based on transportation costs.

Diff: 2 Type: SA Page Ref: 3-5 Topic: 1.3. Making decisions Skill: Applied User1: Qualitative 7) Suppose that in order to finance your studies at the University, three options are available to you: (i) a student loan, (ii) a student line of credit, and (iii) a part-time job at minimum wage. In making your decision, what factors should be taken into account and why?

Answer: Both a student loan and a student line of credit represent borrowing. There are costs associated with borrowing and it is necessary to compare them. In general, a student loan is interest-free as long as you stay in school, while a student line of credit is not interest-free. It means that the costs of borrowing associated with a student loan are lower. Of course, earning your own money is a good option since you avoid paying costs of borrowing. On the other hand, if you work you forgo opportunity to spend this time on your studies. There are some costs associated with that opportunity, called opportunity costs. If the opportunity costs of working exceed the costs of borrowing under student loan, then the latter is the best option. Otherwise, working is a better option.

Diff: 3 Type: SA Page Ref: 3-7 Topic: 1.3. Making decisions Skill: Applied User1: Qualitative

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8) List at least three non-economic factors (attributes) that may be used as evaluation criteria in the decision-making process.

Answer: Non-economic factors are mainly associated with social, environmental and political constraints. Examples are: - Results of new elections; - Introduction of a new welfare program; - Introduction of new environmental standards.

Diff: 1 Type: SA Page Ref: 4-6 Topic: 1.3. Making decisions Skill: Applied User1: Qualitative 9) When evaluating an engineering project, what types of constraints does a decision maker face?

Answer: A project might be technically feasible and the cheapest solution to a problem, but if it doesn't earn money at the minimum rate required by the company, it should not be done. The decision maker therefore has to be aware of the financial constraints on the problem. In addition, an engineering project can meet all other criteria, but may cause detrimental environmental effects, so we must also consider environmental constraints. Finally, any project can be affected by social and political constraints.

Diff: 2 Type: SA Page Ref: 3-6 Topic: 1.3. Making decisions Skill: Recall User1: Qualitative 10) Every year Stan takes a Canada student loan of $5 000 to pay his tuition fees at the University. The loan is interest-free for as long as Stan is a full-time student. After graduation, he will have to start repaying the loan within 6 months including interest. Is this a good decision? Why? Answer: Yes, this is a good decision. Stan can borrow $20 000 for four years or $25 000 for five years for free. If this money were invested in a mutual fund or in financial securities, it would earn some return. Or if Stan borrowed the money from a commercial bank, he would pay interest payments to service his debt. Student loan has no borrowing costs.

Diff: 2 Type: SA Page Ref: 3-6 Topic: 1.3. Making decisions Skill: Applied User1: Quantitative 11) If you decide to buy a home, list basic economic and non-economic factors you have to take into account and rank them Answer: You need to know the price of a home, current mortgage rate,potential duration of the mortgage and your own current and expected income. Those are fundamental economic factors. Noneconomic factors may include: location, environmental quality, and safety.

Diff: 2 Type: SA Page Ref: 3-6, 11 Topic: 1.6. Uncertainty, sensitivity analysis, and currencies Skill: Applied User1: Qualitative

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12) Suppose that a local government decides to build a bridge. In order to justify this project, economists want to ask people directly about their willingness to pay for the bridge. Using the discussion in your textbook, list at least three problems with this economic measure.

Answer: The following three problems can be mentioned in this regard: (i) willingness to pay for a nonmarket good is very difficult to define; (ii) willingness to pay for non-market good is very subjective measure; (ii) it will require a lot of data and comprehensive statistical analysis to evaluate total willingness to pay for the bridge Diff: 3 Type: SA Page Ref: 6-7 Topic: 1.4. Dealing with abstractions Skill: Applied User1: Qualitative 13) You were notified by an economics expert that the total costs of a project you are about to start are approximately $50 000. In order to justify this project, what additional economic information do you need?

Answer: First of all, it is necessary to understand what this value includes. Usually costs are divided into capital costs and operating costs and those are two different types of costs. Second, it is necessary to understand whether or not the value of total costs is given as of today or at some other time. Finally, the benefits of the project should be evaluated to compare them with total costs.

Diff: 1 Type: SA Page Ref: 3-4 Topic: 1.3. Making decisions Skill: Recall User1: Qualitative 14) Maintenance costs are an important component of the total costs of many engineering projects.

Normally in engineering projects these costs are added to the purchase price of a piece of engineering equipment. For the following items, describe how important the maintenance costs are compared to the item's price: (i) A tractor (ii) A desk (iii) A computer (iv) An oil pump Answer: The maintenance costs of a tractor, computer and oil pump are important. Particularly maintenance costs of a tractor are very important since the quantity and quality of the services, produced by the tractor, depend on these costs. To lesser extent maintenance costs are important in case of oil pump and computer. Maintenance costs of a desk are not important because they are very low, and have little effect on the services provided by the desk.

Diff: 1 Type: SA Page Ref: 3-6 Topic: 1.3. Making decisions Skill: Applied User1: Qualitative

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15) Suppose that you want to buy a used car for $3 000, but you don't have enough cash. List some of the (legal) strategies that you might consider.

Answer: Possible alternatives for financing the purchase of a used car are: - Not to buy (a "do nothing" alternative); - To borrow money from your parents; - To take a loan from a commercial bank; - To finance through a car dealer; - To postpone the purchase and earn money working extra hours.

Diff: 2 Type: SA Page Ref: 3-6 Topic: 1.3. Making decisions Skill: Applied User1: Qualitative

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Engineering Economics, 5e (Fraser) Chapter 2 Time Value of Money 2.1 Multiple Choice Questions 1) The price of money can be captured through A) the difference between benefits and costs that occur at different times.

B) the future worth of an investment.

C) the present worth of an investment.

D) the interest rate.

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?

A) variable interest rate B) a forgone opportunity of investment C) a diminishing purchasing power of money over time D) a growing inflation 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 A) the present value of money.

B) the future value of money.

C) the amount of money invested at the prime interest rate.

D) the annual equivalent value of money.

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

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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?

A) $11 629 B) $12 104 C) $12 958 D) $13 465 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?

A) 2.0 B) 2.5 C) 3.0 D) 3.5 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?

A) $3 000 B) $4 000 C) $5 000 D) $6 000 E) $7 000 Answer: D Diff: 3 Type: MC Page Ref: 26-27 Topic: 2.3 Compound and Simple Interest Skill: Applied User1: Quantitative

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7) Nominal interest rate is calculated by A) summing up all interest rates for all compounding periods.

B) converting a given interest rate with a compounding period to an equivalent interest rate with a oneyear compounding period.

C) dividing the interest rate per compounding period by the number of compounding periods per year.

D) multiplying the simple interest rate by the number of years.

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?

A) 11.6% B) 14.5% C) 18.3% D) 20.1% 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?

A) 3.8% B) 4.5% C) 4.8% D) 6.2% E) 8.6% Answer: B Diff: 2 Type: MC Page Ref: 25 Topic: 2.4 Effective and Nominal Interest Rates Skill: Applied User1: Quantitative

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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%?

A) 13 B) 12 C) 11 D) 10 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?

A) summary of benefits and costs of a project B) summary of the timing and magnitude of payments and receipts as they occur over time C) magnitude of cash flows at a given period of time D) summary of present, future, and annual worths of a project 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?

A) receipts = $45 000; disbursements = $25 000; project life = 6 years B) receipts = $35 000; disbursements = $45 000; project life = 6 years C) receipts = $45 000; disbursements = $35 000; project life = 6 years D) receipts = $45 000; disbursements = $35 000; project life = 7 years 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

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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?

A)

B)

C)

D)

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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?

A) indifference on the part of a decision maker among available choices B) the existence of a mathematical relationship between time and money C) the ability to exchange one cash flow for another at minimum cost D) the ability to exchange one cash flow for another at no cost 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?

A) $511.62 B) $537.79 C) $1 025.00 D) $1 049.41 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

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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?

A) $2 071 B) $7 362 C) $8 102 D) $8 399 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?

A) 5.74% B) 5.84% C) 5.94% D) 6.04% 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 A) gain $37.6 by the end of the year.

B) lose $37.6 by the end of the year.

C) gain $57.6 by the end of the year.

D) lose $57.6 by the end of the year.

E) make zero economic profit.

Answer: A Diff: 3 Type: MC Page Ref: 32-34 Topic: 2.1 Introduction Skill: Applied User1: Quantitative

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

A) $2 000 B) $4 000 C) $4 976.64 D) $976.64 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?

A) 14.2% B) 14.8% C) 15.2% D) 15.8% 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 A) the value of a cost at one time is numerically equal to the value of the related benefits received at a different time.

B) the present worth of a cost equals the future worth of a cost at any point in time.

C) the present worth of all costs and benefits equals the future worth of these costs and benefits at any point in time.

D) the project breaks even, meaning costs equal benefits at a certain point in time.

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

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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?

A) Year 1 ends at point A and year 2 begins at point B. B) Year 1 ends at point 2 and year 2 begins at point B. C) Year 1 ends at point 2 and year 2 begins at point 2.

D) A project has four periods.

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 A) an arbitrary compounding period.

B) a compounding period that is normally less than a year.

C) an exogenously given compounding period.

D) a one-year compounding period.

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?

A) $823 B) $1 284 C) $1 892 D) $2 324 E) $2 823 Answer: A Diff: 2 Type: MC Page Ref: 24 Topic: 2.3 Compound and Simple Interest Skill: Applied User1: Quantitative

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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?

A) $3 000 B) $3 033 C) $3 066 D) $3 100 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 A) the difference between the amount of money lent and the amount of money repaid later.

B) a percentage change in the time value of money.

C) the ratio of the amount of money lent to the amount of money repaid later.

D) the future worth of the money.

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 A) the actual but not usually stated interest rate.

B) the actual and usually stated interest rate.

C) the conventional method of stating the annual interest rate.

D) the key interest rate in an economy.

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

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

A) 11% B) 13% C) 15% D) 17% 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?

A) $1 000 B) $1 100 C) $1 610.51 D) $610.51 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?

A) 9% B) 10% C) 11% D) 12% E) 13% Answer: D Diff: 3 Type: MC Page Ref: 26 Topic: 2.4 Effective and Nominal Interest Rates Skill: Applied User1: Quantitative

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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.

A) 4.88% B) 9.54% C) 10.25% D) 21% 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?

A) 9% B) 9.52% C) 10.52% D) 11% 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?

A) 9.00% B) 9.53% C) 10.53% D) 11.53% E) 12.53% Answer: B Diff: 3 Type: MC Page Ref: 28 Topic: 2.5 Continuous Compounding Skill: Applied User1: Quantitative

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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? A) Bank B is always better.

B) 4 years C) 5 years D) 6 years 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?

A) Bank B is always better.

B) 4 years C) 5 years D) 6 years 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? A) $74.59 B) $92.50 C) $113.30 D) $137.39 E) $148.12 Answer: B Diff: 2 Type: MC Page Ref: 22-24 Topic: 2.4 Effective and Nominal Interest Rates Skill: Applied User1: Quantitative

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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?

A) 10 years B) 15 years C) 20 Years D) 25 years 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

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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

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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

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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

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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

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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

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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

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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

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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

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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 3 .

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

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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

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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 io stands for the growth-adjusted interest rate, which of the following is associated with deflation?

A) i > g > 0 (io > 0) B) i > g > 0 (io < 0) C) i = g > 0 (io = 0) D) g < 0 (io > 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

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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

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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 6month 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 8 .

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

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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

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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

11 .

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

12 .

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 ie = (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

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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, ie, N) = $2 500 where ie 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 geometricallyincreasing 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 14 .

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

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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 ie: ie = (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

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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 17 .

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 18 .

User1: Quantitative

19 .

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 20 .

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 21 .

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) 22 .

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:

23 .

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:

24 .

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

25 .

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 26 .

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 27 .

28 .

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 29 .

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) 30 .

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

31 .

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) 32 .

Diff: 2 Type: SA Page Ref: 59-61 Topic: 3.6. Conversion factor for arithmetic gradient series Skill: Applied User1: Quantitative

33 .

Engineering Economics, 5e (Fraser) Chapter 4 Comparison Methods: Part I 4.1 Multiple Choice Questions 1) A contingent project is an example of A) independent projects.

B) mutually exclusive projects.

C) related but not mutually exclusive projects.

D) public projects.

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?

A) The second one because it has higher annual savings.

B) The first one because its present worth is higher.

C) The second one because its present worth is higher.

D) The first one because incremental present worth is positive.

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 A) they have the same service life.

B) they have the same costs.

C) they have the same benefits.

D) they have the same rate of return.

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

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4) Two projects are mutually exclusive if A) the expected costs and benefits of one project do not depend on whether or not the other project is chosen.

B) the expected costs and benefits of one project depend on whether the other project is chosen.

C) if it is impossible to do both at the same time D) in the process of choosing one project at least one other project is chosen for comparison.

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 A) independent from the second project.

B) related but not mutually exclusive to the second project.

C) mutually exclusive to the second project.

D) contingent on the second project.

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 A) an interest rate, which is equal to a current bank interest rate.

B) the least interest rate among all alternative projects.

C) a highest interest rate among all alternative projects.

D) an interest rate that allows an investor to recoup the investment.

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

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

A) do nothing alternative B) other projects, earning MARR or higher interest rate C) putting money into a bank to earn a bank interest rate D) investing money in an alternative project that can recoup the investment 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?

A) to maximize the present worth of a project at the minimum acceptable rate of return B) to maximize the benefits of a project at the minimum acceptable rate of return C) to minimize the costs of a project at the minimum acceptable rate of return D) to maximize the difference between the project's benefits and costs at the minimum acceptable rate of return 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 A) dollars.

B) dollars per unit.

C) dollars per year.

D) dollars per unit per year.

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

3 .

10) A project is marginally acceptable if A) it earns positive present worth at the minimum acceptable rate of return.

B) it earns negative present worth at the minimum acceptable rate of return.

C) the present worth of a project is zero at the minimum acceptable rate of return.

D) it earns more than the minimum acceptable rate of return.

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?

A) $400 000 B) $406 475 C) $454 750 D) $464 750 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?

A) $1 573 B) $2 000 C) $8 681 D) $10 000 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

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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?

A) $2 846 B) $6 348 C) $7 154 D) $10 000 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 A) payback period.

B) present worth comparison method.

C) annual worth comparison method.

D) study period method.

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 A) similar to the present worth method but transforms all annuities to a uniform series at the minimum acceptable rate of return.

B) similar to the present worth method but transform all annuities to arithmetic gradient series at the minimum acceptable rate of return.

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.

D) similar to the present worth method since it transforms all annuities to the present worth as a single payment 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

5 .

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?

A) -$4 000 B) -$32 444 C) -$54 000 D) -$82 444 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 A) project B should be accepted on the basis of maximum benefits B) project A should be accepted because it has larger absolute value.

C) project B should be accepted on the basis of minimum costs.

D) the situation requires additional assessment since some of the costs or benefits might not be considered.

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 A) repeat service life of each alternative at least three times.

B) assume some certain study period for evaluation by introducing salvage values for each alternative.

C) assume that the service life of each project is indefinite.

D) apply only annual worth comparison method for each alternative.

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

6 .

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?

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

A) a period of time over which an initial investment can be recovered fully assuming market interest rate to be equal to the MARR B) a period of time after which a project starts generating profit C) a period of time over which an initial investment cannot be recovered D) a period of time over which an initial investment can be recovered fully 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? A) 7.5 years B) 7.9 years C) 8.4 years D) 9.2 years E) 10 years Answer: A Diff: 1 Type: MC Page Ref: 103 Topic: 4.5 Payback period Skill: Recall User1: Quantitative

7 .

22) By using the payback period method a decision-maker A) equally treats short- and long-term projects.

B) ignores benefits accumulated by a project C) accounts for time value of money.

D) disregards the need for a fast capital recovery.

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 # 1 2 3 4 5

Modification cost $6 000 $9 000 $7 000 $11 000 $7 000

Present Worth $54 000 $56 000 $55 000 $59 000 $55 000

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

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

8 .

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.

A) $201 500 B) -$215 000 C) $280 000 D) $350 000 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 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.

B) can be preferable to the payback period method because the latter discriminates over the short-term projects.

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.

D) can provide more accurate estimates of a project worth for the related but not mutually exclusive projects then other methods.

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 A) the PW of the project evaluated at MARR should be positive.

B) the PW of the project evaluated at MARR should be equal to zero.

C) the PW of the project evaluated at the MARR should be equal or greater than zero.

D) the PW of the project should be positive at the market interest rate.

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

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27) For two mutually exclusive projects with equal lives, the one with A) a lower PW at the MARR should be chosen.

B) a higher AW at the MARR should be chosen.

C) a lower FW at the MARR should be chosen.

D) a higher incremental PW at the MARR should be chosen.

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 0 1 2 3

A -$100 $200 $300 $400

B -$100 $470 $720 $360

C -$100 -$200 $200 $250

D -$100 0 0 $500

E -$100 $300 $250 -$40

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

A) A B) B C) C D) D 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 A) the repeated lives method.

B) the study period method.

C) Both repeated lives method and study period method produce the same result.

D) to choose the alternative with the highest FW. 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

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30) Suppose that cash flows of a project are given as follows: Year 0 1 2 3 4

Costs, $$ 100 000 20 000 20 000 20 000 20 000

Savings, $$ 0 60 000 70 000 80 000 90 000

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

A) one year B) two years C) three years D) four years 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?

A) $1 500 B) $1 600 C) $1 600 D) $1 700 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

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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?

A) $2 457 B) $3 246 C) $3 863 D) $4 127 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?

A) $333 B) $444 C) $555 D) $666 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?

A) 12 years B) 20 years C) 30 years D) 45 years 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

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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?

A) two years five months B) two years seven months C) two years eight months D) two years nine months 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?

A) two years five months B) two years seven months C) two years eight months D) two years nine months 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?

A) $300 B) $315 C) $363 D) $383 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

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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

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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:

Initial price, $ Productivity, pages/year

SUMY 1 000

ROVNO 1 200

PSEL 1 500

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:

Investment Annual operating and maintenance costs Annual income

Project 1 $3.0 million

Project 2 $3.3 million

Project 3 $4.6 million

$750 thousand $1.2 million

$200 thousand $750 thousand

$400 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 15 .

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

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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

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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 Equipment and tools: - solar cell - batteries - wiring - power control Operating costs

Value $2 600 $1 00 $800 $100 $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

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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

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14) Suppose that cash flows of a project are given as follows: Year 0 1 2 3 4

Costs $10 000 $2 000 $2 000 $2 000 $2 000

Savings 0 $2 000 $3 000 $4 000 $5 000

What is the project's payback period?

Answer: It is possible to summarize all calculations in the following table: Year 1 2 3 4 5

Net savings 2 000 - 2 000 = 0 3 000 - 2 000 = 1 000 4 000 - 2 000 = 2 000 5 000 - 2 000 = 3 000 6 000 - 2 000 = 4 000

Remaining costs 10 000 9 000 7 000 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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Engineering Economics, 5e (Fraser) Chapter 5 Comparison Methods: Part II 5.1 Multiple Choice Questions 1) The internal rate of return (IRR) is A) the interest rate that allows an investor to recoup the initial investment.

B) the interest rate that ensures the positive cash flow of a project.

C) the interest rate that breaks even a project's costs and benefits.

D) the interest rate that measures the return from operating costs E) the interest rate that is set up by an investor to guarantee that the return on investment will be higher than from a bank interest rate.

Answer: C Diff: 1 Type: MC Page Ref: 127 Topic: 5.2. Internal rate of return (IRR) Skill: Recall User1: Qualitative 2) A project pays $20 million right away, requires $10 million in investment in year 1, and pays $3 million in year 2. What is the approximate ERR of this project if the MARR is 10%??

A) 1.72% B) 172% C) 1.50% D) 150% E) 50% Answer: B Diff: 1 Type: MC Page Ref: 141 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative 3) The fundamental idea behind comparison of mutually exclusive projects on the basis of incremental IRR is A) different fractions of investments are associated with different levels of productivity.

B) all investments are non-simple.

C) some investments are not productive.

D) multiple IRRs. E) high uncertainty.

Answer: A Diff: 2 Type: MC Page Ref: 134-135 Topic: 5.3. Internal rate of return comparisons Skill: Recall User1: Qualitative

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4) The internal rate of return (IRR) is negative if A) a project is losing money.

B) a cash inflow exceeds a cash outflow.

C) a project just breaks even.

D) a project is a simple investment.

E) IRR cannot be negative.

Answer: A Diff: 1 Type: MC Page Ref: 128 Topic: 5.2. Internal rate of return (IRR) Skill: Recall User1: Qualitative 5) A project is represented by the following graphs:

What is the internal rate of return for this project?

A) 0% B) 4% C) 10% D) 14% E) 15% Answer: D Diff: 1 Type: MC Page Ref: 130-131 Topic: 5.2. Internal rate of return (IRR) Skill: Recall User1: Quantitative 6) Two mutually exclusive alternatives are being compared. We should choose the alternative that: A) has a higher minimum acceptable rate of return assuming the lives of the alternatives are equal.

B) has a higher internal rate of return assuming the lives of the alternatives are equal.

C) has a higher internal rate of return regardless of the lives of the alternatives.

D) has an incremental investment with a rate of return equal to the minimum acceptable rate of return.

E) has an incremental investment with the rate of return exceeding minimum acceptable rate of return.

Answer: E Diff: 2 Type: MC Page Ref: 132-136 Topic: 5.3. Internal rate of return comparisons 2 .

Skill: Recall User1: Qualitative 7) A project requires an initial investment of $100 000 and immediately pays $25 000. The next year this project requires an additional investment of $50 000 and does not pay anything. In the following year the project pays $150 000. The internal rate of return (i) for this project can be obtained by A) using a present worth factor.

B) solving a quadratic equation.

C) calculating an external rate of return.

D) solving a system of two equations.

E) solving an ith degree polynomial equation.

Answer: B Diff: 3 Type: MC Page Ref: 127 Topic: 5.3. Internal rate of return comparisons Skill: Recall User1: Quantitative 8) What is the difference between the internal rate of return (IRR) and the external rate of return (ERR)?

A) The IRR is earned by a project whereas the ERR is earned outside of it B) The ERR equals the difference between the IRR and the MARR C) The IRR usually equals the MARR whereas the ERR is always higher than the MARR D) The ERR is earned when a project's return is used for the purpose of further reinvestment while the IRR is typically used for calculating the net return of a project E) For a given explicit rate of return, a project can have more than one value for its ERR but can have only one value for its IRR Answer: A Diff: 2 Type: MC Page Ref: 140-142 Topic: 5.3. Internal rate of return comparisons Skill: Recall User1: Qualitative 9) What economic concept is used as background for the external rate of return?

A) a present worth B) a net cash flow C) an internal rate of return D) a payback period E) an opportunity cost Answer: E Diff: 3 Type: MC Page Ref: 140 Topic: 5.1. Introduction Skill: Applied User1: Qualitative

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10) A project requires no initial investment. It costs $4 000 a year from now and earns $8 000 two years from now. What is its internal rate of return?

A) 24% B) 50% C) 75% D) 100% E) 141% Answer: D Diff: 1 Type: MC Page Ref: 127 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative 11) What assumption makes the external rate of return a proxy?

A) All receipts of a project are assumed to be invested at the minimum acceptable rate of return B) Receipts that occur when a project balance is positive are assumed to be invested at the minimum acceptable rate of return C) All receipts and disbursements of a project are assumed to be invested at the minimum acceptable rate of return D) Receipts and disbursements that occur when a project balance is positive are assumed to be invested at the minimum acceptable rate of return E) A project's balance should be positive all the time Answer: A Diff: 2 Type: MC Page Ref: 141-142 Topic: 5.3. Internal rate of return comparisons Skill: Recall User1: Qualitative 12) The external rate of return must be used if A) there are multiple internal rates of return B) a project involves only simple investments.

C) a project starts with cash outflow D) a project requires multiple investments and generates multiple benefits E) it is impossible to calculate the minimum acceptable rate of return.

Answer: A Diff: 2 Type: MC Page Ref: 141-142 Topic: 5.3. Internal rate of return comparisons Skill: Recall User1: Quantitative

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13) A project is subject to the following cash flow diagram:

What rate of return would it be appropriate to use in this case?

A) ERR B) IRR C) MARR D) Market rate of return E) Explicit rate of return Answer: B Diff: 2 Type: MC Page Ref: 140-142 Topic: 5.3. Internal rate of return comparisons Skill: Recall User1: Qualitative 14) What is the major advantage of the internal rate of return comparison method?

A) It gives an explicit measure of profit.

B) It has familiar meanings to decision makers.

C) It is very easy to calculate and therefore it is commonly used.

D) It takes into account the need to have capital recovered quickly.

E) It facilitates the comparison of projects of different size.

Answer: E Diff: 3 Type: MC Page Ref: 147 Topic: 5.4. Rate of return and present/annual worth methods compared Skill: Recall User1: Qualitative 15) Which of the following statements is true with regard to the ERR? A) The approximate ERR is just a proxy for the IRR. B) The precise ERR is a proxy for the IRR. C) For a simple investment, the precise ERR is greater than the IRR. D) For a simple investment, the approximate ERR is greater than the IRR. E) An acceptable project will earn at least the approximate ERR. Answer: E Diff: 2 Type: MC Page Ref: 141 Topic: 5.3. Internal rate of return comparisons Skill: Recall User1: Qualitative

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16) What is the major disadvantage of the internal rate of return method?

A) It complicates the comparison of projects of the different sizes.

B) It can produce more than one internal rate of return.

C) It discriminates against long-term projects.

D) It ignores the time value of money.

E) It ignores the expected service life.

Answer: B Diff: 3 Type: MC Page Ref: 147 Topic: 5.4. Rate of return and present/annual worth methods compared Skill: Recall User1: Qualitative 17) A project is subject to the following cash flow diagram:

What is its rate of return?

A) 22% B) 36% C) 56% D) 60% E) 74% Answer: C Diff: 2 Type: MC Page Ref: 141 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative 18) Unlike the internal rate of return method, the present and annual worth computations A) are more complicated.

B) do not account for the time value of money.

C) give a direct measure of the profit provided by a project.

D) discriminate against long-term projects.

E) do not take into account expected service life of a project.

Answer: C Diff: 2 Type: MC Page Ref: 147 Topic: 5.4. Rate of return and present/annual worth methods compared Skill: Recall 6 .

User1: Qualitative 19) The following table summarizes information for five projects:

The data can be interpreted in the following way: The IRR on the incremental investment between project 5 and project 4 is 16%.

If all projects are independent and the company has at least $1 025 000 to invest, which projects should be undertaken if the MARR is 16%?

A) only 2 B) 2 and 4 C) 1, 3 and 5 D) 1, 3, 4 and 5 E) 1, 2, 3, 4 and 5 Answer: D Diff: 2 Type: MC Page Ref: 129-131 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative 20) The following table summarizes information for five projects:

The data can be interpreted in the following way: The IRR on the incremental investment between project 5 and project 4 is 16%.

If the projects are mutually exclusive, which projects should be undertaken if the MARR is 15%?

A) 3 only B) 5 only C) 3 and 5 D) 3, 4 and 5 E) 1, 3, 4, and 5 Answer: B Diff: 3 Type: MC Page Ref: 132-136 7 .

Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative 21) Dealing with mutually exclusive projects, we start with some current alternative. How do we chose it?

A) the alternative with the highest internal rate of return B) the alternative with a zero net benefit C) the alternative with the benefit-cost ratio of one D) the "do nothing" alternative E) the alternative with the smallest first cost Answer: E Diff: 1 Type: MC Page Ref: 132-136 Topic: 5.3. Internal rate of return comparisons Skill: Recall User1: Qualitative 22) What projects are called simple investments?

A) projects that consist of one inflow at the start followed by only one outflow B) projects that have only two internal rates of return C) projects that have a stream of revenues at the start followed by a stream of investments D) projects that have one or more periods of outflows at the start followed by one or more periods of inflows E) projects that require calculation of the external rate of return Answer: D Diff: 1 Type: MC Page Ref: 136-139 Topic: 5.3. Internal rate of return comparisons Skill: Recall User1: Qualitative 23) Computation of the precise external rate of return can be a complex procedure because A) positive balance earned by a project is to be invested outside this project at an unknown minimum acceptable rate of return.

B) of the difficulty in determining exactly when the explicit interest rate should be applied.

C) a project's balance can be negative for the trial external rate of return.

D) the calculation requires a lot of experimenting with the trial external rate of return.

E) a project that requires this calculation is typically a simple investment project.

Answer: B Diff: 3 Type: MC Page Ref: 140-142 Topic: 5.3. Internal rate of return comparisons Skill: Recall User1: Qualitative

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24) A two-year project has $100 million as initial investment and generates net savings of $60 million per year. What is the project's IRR? A) 10.1% B) 11.1% C) 12.1% D) 13.1% E) 14.1% Answer: D Diff: 2 Type: MC Page Ref: 127 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative 25) The approximate ERR will always be A) higher than the precise ERR. B) lower than the MARR. C) could be equal to precise ERR or MARR. D) between the precise ERR and the MARR. E) higher than both precise ERR and MARR. Answer: D Diff: 2 Type: MC Page Ref: 141 Topic: 5.3. Internal rate of return comparisons Skill: Recall User1: Qualitative 26) For two mutually exclusive projects with equal lives, the one with A) a higher IRR should be chosen.

B) a higher incremental PW at the MARR should be chosen.

C) a higher incremental FW at the MARR should be chosen.

D) a higher incremental IRR than the MARR should be chosen.

E) a higher incremental AW at the MARR should be chosen.

Answer: D Diff: 2 Type: MC Page Ref: 132-136 Topic: 5.3. Internal rate of return comparisons Skill: Recall User1: Qualitative 27) The problem of multiple IRRs arises whenever A) IRR > ERR. B) IRR = ERR. C) a project is a simple investment.

D) a project starts with net savings followed by investments.

E) a project starts with first cost followed by net annual savings for the rest of the project's service life.

Answer: D Diff: 2 Type: MC Page Ref: 136-139 Topic: 5.3. Internal rate of return comparisons Skill: Recall User1: Qualitative 9 .

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28) In general, the IRR comparison method and the PW comparison method A) produce the same results for independent projects but not for mutually exclusive projects.

B) produce the same results for mutually exclusive projects but not for independent projects.

C) produce different results for both independent projects and mutually exclusive projects.

D) produce the same results for independent projects and mutually exclusive projects with unequal lives.

E) produce the same results for independent projects and mutually exclusive projects with equal lives.

Answer: E Diff: 3 Type: MC Page Ref: 143-147 Topic: 5.4. Rate of return and present/annual worth methods compared Skill: Recall User1: Qualitative 29) External rate of return is used for A) projects with unequal lives.

B) projects which are non-simple investments.

C) projects with large capital costs but small operating costs.

D) projects with small capital costs but large operating costs.

E) projects that start with cash outflows followed by cash inflows.

Answer: B Diff: 1 Type: MC Page Ref: 140 Topic: 5.3. Internal rate of return comparisons Skill: Recall User1: Qualitative 30) You are going to buy a new car worth $14 500. A car dealer computes your monthly payment to be $267 for 7-year financing. What is the dealer's effective annual rate of return on this loan?

A) 9.33% B) 9.77% C) 10.23% D) 10.67% E) 11.33% Answer: B Diff: 2 Type: MC Page Ref: 126-127 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative 31) A project involves an immediate expenditure of $1 000, and will require additional expenditures of $100 a year for the next ten years, starting one year from now. After ten years it yields an income of $3 000. What is its rate of return?

A) 3.5% B) 4.5% C) 5.5% D) 6.5% E) 7.5% Answer: C Diff: 2 Type: MC Page Ref: 129-142 Topic: 5.3. Internal rate of return comparisons 11 .

Skill: Applied User1: Quantitative 32) A project involves an immediate expenditure of $2 000, and will require additional expenditures of $150 a year for the next ten years, starting one year from now. After ten years it yields an income of $8 000, but a year later a further expenditure of $1 000 will be required to close down the project. What is its rate of return?

A) 9% B) 10% C) 11% D) 12% E) 13% Answer: A Diff: 2 Type: MC Page Ref: 129-142 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative 33) I sign a contract that guarantees me an immediate payment of $2 000, but then I have to invest $800 a year for ten years. At the end of the tenth year I get a further payment of $8 000. What is my internal rate of return on the contract?

A) There are two solutions: 8% and 16%.

B) There are two solutions: 13% and 25%.

C) There are two solutions, 15% and 30%.

D) There are two solutions: 18% and 24%.

E) There is a single solution: about 22%.

Answer: B Diff: 2 Type: MC Page Ref: 129-142 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative 34) I sign a contract that guarantees me an immediate payment of $2 000, but then I have to invest $800 a year for ten years. At the end of the tenth year I get a further payment of $8 000. If I can invest money at 10% interest, what is my approximate external rate of return on the contract?

A) about 9% B) about 10% C) about 11% D) about 12% E) There are two solutions, about 8% and about 12%.

Answer: C Diff: 2 Type: MC Page Ref: 140-142 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative

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35) What is the IRR corresponding to this cashflow diagram?

A) about 9% B) about 12% C) about 19% D) about 22% E) about 25% Answer: C Diff: 2 Type: MC Page Ref: 129-139 Topic: 5.2. Internal rate of return (IRR) Skill: Applied User1: Quantitative 36) What is the IRR corresponding to this cashflow diagram?

A) About 10.5% B) About 11.5% C) About 12.5% D) About 13.5% E) About 14.5% Answer: E Diff: 2 Type: MC Page Ref: 129-139 Topic: 5.2. Internal rate of return (IRR) Skill: Applied User1: Quantitative

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37) If you can invest money at 10% interest, what is the approximate ERR corresponding to this cashflow diagram?

A) 15% B) 20% C) 25% D) 30% E) 35% Answer: C Diff: 2 Type: MC Page Ref: 140-142 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative 5.2 Short Answer Questions 1) What is the difference between the internal rate of return and the minimum acceptable rate of return?

Answer: The minimum acceptable rate of return (MARR) is a lower limit for investment acceptability.

The MARR is usually set by a business evaluating an investment project. Frequently market interest is used as the MARR. The internal rate of return (IRR) is that interest rate at which the present worth of all costs is equal to the present worth of all benefits associated with an investment project. For a project to be accepted, the IRR of the project must exceed the MARR. Diff: 1 Type: SA Page Ref: 127-128 Topic: 5.2. Internal rate of return (IRR) Skill: Recall User1: Qualitative

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2) You can buy a car for $40 000 paying cash now or you can finance it through a bank loan paying $700 per month for 5 years. What is the IRR of the financing option in annual terms?

Answer: By definition, the IRR equates present worth of costs and present worth of benefits. If you choose financing then $40 000 is your benefit while $700 per month for 60 months is your cost.

Equating the two values with the help of the present worth of equal payments factor : 40 000 = 700 * [(1+IRR)60 - 1)/IRR/(1 + IRR)60] and solving for IRR, we obtain 0.001615 by trial and error in a spreadsheet. This is the effective monthly IRR, so the effective annual IRR is: 1.00161512 - 1 = 0.0196 or 1.96% Diff: 2 Type: SA Page Ref: 127-128 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative 3) When does the problem of multiple IRRs arise and what are possible solutions to the problem?

Answer: If a project is not a simple investment, there may be more than one value of IRR satisfying the condition PW(Costs) = PW (Benefits). Such projects are usually characterized by one or more cash inflows at the start, followed by cash outflows. In such a case, application of the precise or approximate External Rate of Return (ERR) can resolve the problem.

Diff: 1 Type: SA Page Ref: 136-138 Topic: 5.3. Internal rate of return comparisons Skill: Recall User1: Qualitative 4) You have two ways to invest $1 000 000 you won in a lottery: (i) to buy government bonds that earn 4% annually, or (ii) to buy an apartment building that brings you $100 000 per year in revenues.

Compare the two options in terms of their internal rates of return.

Answer: The IRR of the first option is 4%. The IRR of the second option can be calculated as follows: Initial Investment = Present Worth of infinite equal annual revenues at an unknown IRR or 1 000 000 = 100 000/IRR. IRR = 0.1 or 10%. It turns out that the second option is better since it results in a higher rate of return.

Diff: 2 Type: SA Page Ref: 127-128 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative

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5) You are considering the following project: It pays you $5 000 at the end of the first year, costs $10 000 by the end of the second year and brings $4 000 next year. What is the project's internal rate of rate, external rate of return and the approximate external rate of return if current market interest rate is 25%?

Comment on the project acceptability on the basis of the given information.

Answer: The IRR is calculated from PW(Costs) = PW (Benefits) evaluated at the unknown IRR: 10 000/(1 + IRR)2 = 5 000/(1 + IRR) + 4,000/(1 + IRR)3 from which we can deduce that IRR = 0.447 or 44.7%.

The exact ERR can be obtained by calculating net benefits each year, and moving benefits at the current interest rate. For example, at the end of year 2, benefit of $5 000 becomes 5 000 x (1+0.25) = $6 250.

Since cost that year is $10 000, net benefit at the end of year 2 is 6 250 - 10 000 = -$3 750 which is actually net cost. Now it is possible to set the following equation: 3 750 x (1+ERR) = 4 000 from which we can deduce that ERR = 0.067 or 6.7%.

In order to calculate the ERRapp it is necessary to take all benefits at the current interest rate to the end of year 3 and to equate the result to all costs taken to the same year at the unknown ERRapp: 5 000 x (1 + 0.25)2 + 4 000 = 10 000 x (1 + ERRapp) and ERRapp = 0.181 or 18.1% The project would be accepted on the basis of the IRR, but rejected on the basis of the ERR and ERRapp.

Diff: 3 Type: SA Page Ref: 140-142 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative

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6) Suppose that a project pays $5 000 on odd years and costs $5 000 on even years. If the life of the project is 6 years, what is its approximate external rate of return if the MARR is 10%? Would you accept or reject such a project?

Answer: PW of all benefits at the 10% interest rate at the end of year 6 is: 5 000(1.1)5 + 5 000(1.1)3 + 5 000(1.1) = $20 207.55 PW of all costs at the unknown ERR at the end of year 6 is: 5 000(1+ERR)4 + 5 000(1 + ERR)2 + 5 000 Therefore, the following equation arises: 20 207.55 = 5 000(1 + ERR)4 + 5 000(1 + ERR)2 + 5 000 or (1 + ERR)4 + (1 + ERR)2 - 3.04151 = 0 Solution to this equation is ERR = 0.146 or 14.6%. Since the ERR is greater than the market interest rate, the project should be accepted.

Diff: 3 Type: SA Page Ref: 141-142 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative 7) If you invest $1 000 now, you will receive this amount back in two years plus dividends of $200 each year. What is the IRR of this investment?

Answer: We set up the following equation: 1 000 = 1 000/(1 + IRR)2 + 200/(1 + IRR) + 200/(1 + IRR)2 Solving for IRR results in 20% internal rate of return.

Diff: 2 Type: SA Page Ref: 127-128 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative

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8) Suppose that Maria borrowed $5 000 from a bank to be repaid in five equal annual payments of $1 250. What is the IRR of this transaction?

Answer: The underlying equation for the transaction is 5 000 = 1 250(P/A, IRR, 5). It can be simplified to: (P/A, IRR, 5) = 4 From tables, 9% < IRR < 10%. Interpolating: IRR = 0.09 + 0.01(4.0264 - 4.0)/(4.0264 - 3.9347) = 0.00929 or 9.29% where (P/A, 9%, 5) = 4.0264 and (P/A, 10%, 5) = 3.9347.

Diff: 1 Type: SA Page Ref: 129-131 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative 9) What are the advantages and disadvantages of the comparison method based on the IRR against the present worth comparison method?

Answer: IRR's advantages: facilitates comparison of projects of different sizes; commonly used.

IRR's disadvantages: relatively difficult to calculate; multiple IRRs. PW's advantages: gives explicit measure of profit contribution; difficult to compare projects of different sizes.

Diff: 1 Type: SA Page Ref: 147 Topic: 5.4. Rate of return and present/annual worth methods compared Skill: Recall User1: Qualitative 10) Assume that Stan takes a student loan of $5 000 at the beginning of his first year at Miskatonic University and graduates at the end of year 5. After graduation, a 7% interest rate on the debt is applied.

If Stan makes four equal annual payments to re-pay the debt in four years after graduation, what is the IRR on his loan?

Answer: The first step is to define the size of the annual payment after graduation A = 5 000 x (A/P, 7%, 4) = $1 476.14/year Now the following equation with respect to year 5 can be introduced: 5 000 x (F/P, IRR, 5) = 1 476.14 x (P/A, IRR, 4) Solution to the equation is IRR = 0.0225 or 2.25%.

Diff: 3 Type: SA Page Ref: 127-129 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative

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11) Suppose that you can buy a car now for $20 000. On the other hand, you can lease it at $350 per month for 60 months. If you buy a car now, then you will be able to sell it at the end of the fifth year for $8 000. If you choose to lease, what is the monthly and annual IRR of the lease compared to the buying a car now?

Answer: If you choose to lease, you forgo $20 000 of costs now (which is your benefit) and $8 000 of benefits in five years time (which is your cost) plus you pay $350 per month for 60 months, which is also your cost.

Let IRRm be the monthly internal rate of return. Then: 20 000 = 8 000 x (P/F, IRRm, 60) + 350 x (P/A, IRRm, 60) Solution to this equation is IRRm = 0.0102 or 1.02% per month.

So the effective annual IRR = 1.0112 - 1 = 12.68% Diff: 3 Type: SA Page Ref: 127-129 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative 12) Explain how you would compare two mutually exclusive projects on the basis of the internal rate of return.

Answer: First calculate the IRR for both alternatives. Reject any alternative for which IRR < MARR. Next calculate the incremental IRR of upgrading from the cheaper to the more expensive project. An alternative with a higher investment (first cost) is accepted only if it meets the criterion: IRR of incremental investment > MARR Otherwise, the alternative with lower investment is checked against MARR. It is accepted if its IRR > MARR. Diff: 2 Type: SA Page Ref: 132-136 Topic: 5.3. Internal rate of return comparisons Skill: Recall User1: Qualitative 13) There are two options to buy a plot of land for construction: (i) upfront payment of $250 000, or (ii) four equal payments of $50 000 in years 1-4 and the fifth payment of $150 000 in year 5. What is the implied IRR of choosing the second option over the first?

Answer: If the second option is chosen, then $250 000 is a benefit since it is payment avoided. Since the IRR equates the present worth of benefits and costs, the following equation arises: 250 000 = 50 x (P/A, IRR, 4) + 150 x (P/F, IRR, 5) and IRR = 10.21%.

Diff: 2 Type: SA Page Ref: 127-129 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative 19 .

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14) Steve is a professional web-site designer. He just bought a powerful computer for $5 000. According to the existing market, he will be able to sell this computer for $1 000 three years from now. In order for Steve to get 10% internal rate of return on his computer, what annual revenue should he generate over the three-year period?

Answer: It is possible to set up the following equation based on the definition of the IRR: 5 000 = 1 000 * (P/F, 10%, 3) + X * (P/A, 10%, 3) Solving this equation for X produces $1 708.46 per year.

Diff: 2 Type: SA Page Ref: 127-129 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative 15) YVY Consulting group received a contract to evaluate a new technological line to produce shoes.

The project is subject to the following cash flows: Cash flow Purchase price Annual operating costs Annual production Service life

Value $100 000 $20 000/year 800 pairs/year 5 years

If the market price of a pair of shoes is $70 and the MARR is 15%, is this a good investment on the basis of the IRR? Answer: It is necessary to calculate the IRR of this investment and compare it with the MARR = 15%.

However, first it is necessary to find annual revenue as follows: Annual Revenue = $70 * 800 = $56 000/year The IRR of this investment can be calculated from: 100 000 = (56 000 -20 000) x (P/A, IRR, 5) IRR comes to 23.4% by trial and error in a spreadsheet. Since the IRR is greater than the MARR, this is a good investment.

Diff: 2 Type: SA Page Ref: 127-129 Topic: 5.3. Internal rate of return comparisons Skill: Applied User1: Quantitative

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Engineering Economics, 5e (Fraser) Chapter 6 Depreciation and Financial Accounting 6.1 Multiple Choice Questions 1) What is depreciation?

A) The decline in value of a future good due to the time we have to wait to receive that good.

B) The decline in value of expected future income, due to inflation.

C) The tendency of money in a bank account to lose value over time.

D) The loss in an asset's value over time.

E) The accumulation of money in an interest-bearing account over time.

Answer: D Diff: 1 Type: MC Page Ref: 165 Topic: 6.2. Depreciation and depreciation accounting Skill: Recall User1: Qualitative 2) What is functional loss?

A) Functional loss occurs when the asset can still perform its function despite a loss in its market value.

B) Functional loss occurs when the asset becomes unusable to to misuse.

C) Functional loss occurs when the asset wears out due to being used for its intended function.

D) Functional loss occurs when the asset wears out due to the passage of time, whether or not it is used.

E) Functional loss occurs when the asset can still perform its original function, but that function is no longer valued.

Answer: E Diff: 1 Type: MC Page Ref: 165 Topic: 6.2. Depreciation and depreciation accounting Skill: Recall User1: Qualitative 3) What is the difference between market value and book value?

A) The market value represents the price for which an asset could be sold at the end of its physical life whereas the book value represents the depreciated value of the asset for accounting purposes.

B) The market value represents the price for which an asset could be sold at any point in its life, whereas the book value represents the depreciated value of the asset for accounting purposes.

C) The market value represents the price for which an asset could be sold at any point in its life whereas the book value represents the actual value of an asset at the end of its useful life.

D) The market value represents the price for which an asset could be sold at the end of its physical life whereas the book value represents the price for which an asset could be sold at the end of its useful life E) The market value represents the price for which an asset could be sold at the end of its useful life whereas the book value represents the depreciated value of an asset for accounting purposes.

Answer: B Diff: 2 Type: MC Page Ref: 165 Topic: 6.2. Depreciation and depreciation accounting Skill: Recall User1: Qualitative 1 .

4) The only depreciation models needed for corporate tax calculations in Canada are: A) straight-line and declining balance.

B) declining balance and double-declining balance.

C) straight-line and Sum-of-year's digits.

D) straight-line and units-of-production.

E) declining-balance and 150%-declining-balance.

Answer: A Diff: 2 Type: MC Page Ref: 166 Topic: 6.2. Depreciation and depreciation accounting Skill: Recall User1: Qualitative 5) Your accounting records show that an asset in use has book value of $7 119.14. The asset cost $30 000 when it was purchased and has been depreciated under the declining balance depreciation method with a 25% depreciation rate. How many years has the asset been in service?

A) 1 year B) 2 years C) 3 years D) 4 years E) 5 years Answer: E Diff: 2 Type: MC Page Ref: 168-171 Topic: 6.2. Depreciation and depreciation accounting Skill: Applied User1: Quantitative 6) Calculate the salvage value of equipment with a service life of 15 years if it was purchased 5 years ago for $120 000 and depreciates at the rate of 10% per year.

A) $5 400 B) $12 000 C) $24 707 D) $60 000 E) $70 859 Answer: C Diff: 2 Type: MC Page Ref: 168-171 Topic: 6.2. Depreciation and depreciation accounting Skill: Applied User1: Quantitative

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7) What are the two basic summary financial statements that provide information about a firm's revenues, expenses, assets and liabilities?

A) the cash flow statement and the statement of retained earning B) the income statement and the cash flow statement C) the statement of retained earning and the balance sheet D) the balance sheet and the cash flow statement E) the income statement and the balance sheet Answer: E Diff: 2 Type: MC Page Ref: 172 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative 8) For a given firm, the balance sheet A) summarizes firm's revenues and expenses over a period of time.

B) gives the firm's assets, liabilities and owners' equity at a moment in time.

C) provides the firm's net income.

D) lists the individuals and institutions that own the company assets.

E) gives the firm's assets, liabilities and owners' equity over a period of time.

Answer: B Diff: 2 Type: MC Page Ref: 172-177 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative 9) Owner's equity A) is the difference between a firm's assets and liabilities.

B) is cash and other assets that could be converted to cash within a relatively short period of time, usually a year or less.

C) is the cumulative sum of earnings from all transactions that can be reinvested in the business.

D) is the price per share set at the time the shares are originally issued.

E) is the sum of the firm's long-term assets and current assets.

Answer: A Diff: 3 Type: MC Page Ref: 175 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative

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10) What is the distinguishing feature of a limited partnership?

A) Some partners are involved as investors while other partners manage everyday operations.

B) The partners have liabilities limited to their personal assets only.

C) The partners can elect the board of directors that selects the managers to run the business in the interest of the partners D) The partners run the business together and share all profits and losses according to the partnership agreement.

E) The partners have their investments protected from the creditors; however, they are liable for up to the amount of their assets.

Answer: A Diff: 3 Type: MC Page Ref: 175 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative 11) You are one of five partners in a general partnership. You have invested $10 000 of your own money in the business. The business is unsuccessful and is left with no assets and liabilities of $100 000.

Your partners all leave the country leaving no forwarding addresses. How much of the debt are you legally responsible for?

A) $2 000 B) $10 000 C) $20 000 D) $30 000 E) $100 000 Answer: E Diff: 3 Type: MC Page Ref: 176 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative 12) You are one of five shareholders in a corporation. You have invested $10 000 of your own money in the corporation. The business is unsuccessful and is left with no assets and liabilities of $100 000. Your partners all leave the country leaving no forwarding addresses. How much of the debt are you legally responsible for?

A) $0 B) $2 000 C) $10 000 D) $20 000 E) $100 000 Answer: E Diff: 3 Type: MC Page Ref: 176 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative

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13) Your supervisor, Ms. Applebee, is asking you to provide her with the cost of equipment the firm owns. Where will you look for it?

A) in a balance sheet under the Current Assets B) in a balance sheet under the Long-Term Assets C) in an income statement under the Operating Expenses D) in an income statement under the Revenues E) in an income statement under the Expenses Answer: B Diff: 2 Type: MC Page Ref: 172-176 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative 14) In a balance sheet Current Assets are $4 000; Long-term Assets are $6 000; Total Owners' Equity is $6 000; Current Liabilities are $3 000. What are Long-term Liabilities?

A) $1 000 B) $5 000 C) $9 000 D) $13 000 E) $15 000 Answer: A Diff: 2 Type: MC Page Ref: 172-176 Topic: 6.3. Elements of financial accounting Skill: Applied User1: Quantitative 15) Which of the following are not quick assets?

A) Notes receivable B) Temporary investments C) Cash and deposits D) Capital assets E) Accounts receivable Answer: D Diff: 2 Type: MC Page Ref: 172-176 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative 16) In a balance sheet Current Assets are $9 000; Long-term Assets are $11 000; Current Liabilities are $7 000; Long-term Liabilities are $8 000; and Total Owners' Equity is $5 000. What is the equity ratio?

A) 0.25 B) 0.33 C) 0.50 D) 0.75 E) 1.0 Answer: A Diff: 2 Type: MC Page Ref: 183 Topic: 6.3. Elements of financial accounting 5 .

Skill: Applied User1: Quantitative 17) Calculate the depreciation rate of a vehicle if it was bought 5 years ago for $25 000 and can be sold now for $8 200.

A) 20% B) 32% C) 48% D) 68% E) 80% Answer: A Diff: 2 Type: MC Page Ref: 168-169 Topic: 6.2. Depreciation and depreciation accounting Skill: Applied User1: Quantitative 18) The net-profit ratio is defined as: A) net income divided by total equity.

B) revenue divided by net income.

C) net income divided by total assets.

D) net income divided by revenue.

E) revenue divided by total assets.

Answer: C Diff: 1 Type: MC Page Ref: 186 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative 19) Charles has just purchased a car for $9 520. He expects that the value of this car will decline by 5% each year. Eventually Charles wants to sell this car for at least $6 000 and buy a new one. How many years should Charles use this car before he can sell it?

A) 12 B) 10 C) 8 D) 6 E) 4 Answer: C Diff: 2 Type: MC Page Ref: 168-169 Topic: 6.2. Depreciation and depreciation accounting Skill: Applied User1: Quantitative

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20) The Balance Sheet of YYY Ltd. is as follows: ASSETS Current Assets Cash Accounts receivable Raw materials inventory Finished goods inventory Total current Assets

$30 000 $20 000 $50 000 $600 000 $700 000

Long-Term Assets Equipment Buildings Land Total Long-Term Assets

$2 000 000 $1 100 000 $400 000 $3 500 000

TOTAL ASSETS

$4 200 000

LIABILITIES & OWNER'S EQUITY Current Liabilities Accounts payable $10 000 Current loans $60 000 Total current liabilities $70 000 Long-Term Liabilities Long-term loans

$900 000

TOTAL LIABILITIES

$970 000

Owners' equity Common stock Retained Earnings TOTAL OWNERS' EQUITY

$2 130 000 $1 100 000 $3 230 000

TOTAL LIABILITIES AND OWNERS' EQUITY

$4 200 000

What is the current ratio of this company?

A) 0.43 B) 0.50 C) 2.00 D) 6.00 E) 10.00 Answer: E Diff: 1 Type: MC Page Ref: 181 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Quantitative 7 .

21) The Balance Sheet of YYY Ltd. is as follows: ASSETS Current Assets Cash Accounts receivable Raw materials inventory Finished goods inventory Total current Assets

$30 000 $20 000 $50 000 $600 000 $700 000

Long-Term Assets Equipment Buildings Land Total Long-Term Assets

$2 000 000 $1 100 000 $400 000 $3 500 000

TOTAL ASSETS

$4 200 000

LIABILITIES & OWNER'S EQUITY Current Liabilities Accounts payable $10 000 Current loans $60 000 Total current liabilities $70 000 Long-Term Liabilities Long-term loans

$900 000

TOTAL LIABILITIES

$970 000

Owners' equity Common stock Retained Earnings TOTAL OWNERS' EQUITY

$2 130 000 $1 100 000 $3 230 000

TOTAL LIABILITIES AND OWNERS' EQUITY

$4 200 000

If the sales of this company are $5 400 000, what is the company's inventory-turnover ratio?

A) 8.4 B) 9.0 C) 8.3 D) 12.0 E) 18.00 Answer: C Diff: 1 Type: MC Page Ref: 185 Topic: 6.3. Elements of financial accounting Skill: Recall 8 .

User1: Quantitative 22) What items should be treated as extraordinary items for a Canadian company that sells its goods to the US market?

A) loss of revenues caused by the Bank of Canada as it raises the bank rate B) loss of revenues caused by the US government as it imposes extra duties on the company's goods C) loss of revenues caused by the increased delays on the Canada-US border that occur due to new security measures D) loss of revenues caused by the changes in demand for the company's goods in the US E) loss of revenues caused by the longer haul of goods that occurs as a result of heavy snow on the roads Answer: E Diff: 2 Type: MC Page Ref: 186 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative 23) Working capital ratio is A) the ratio of total assets to total liabilities.

B) the ratio of total current assets to total current liabilities.

C) the ratio of total owners' equity to total assets.

D) the ratio of current assets to long-term assets.

E) the ratio of long-term assets to long-term liabilities.

Answer: B Diff: 1 Type: MC Page Ref: 187 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative 24) The extent to which a business relies on debt in its operations is reflected in A) the equity ratio.

B) the return-on-equity ratio.

C) the quick ratio.

D) the net-profit ratio.

E) the return-on-asset ratio.

Answer: A Diff: 1 Type: MC Page Ref: 187 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative 25) The return-on-assets ratio is a reflection of a business's A) marginal costs.

B) profitability.

C) marginal revenue.

D) marginal benefits.

E) net annual savings.

Answer: B Diff: 1 Type: MC Page Ref: 187 9 .

Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative 26) A company buys a corrugated-metal building to store fertiliser. Because of a change in plans, no fertiliser is ever stored in the building, but it rusts away due to exposure to the rain. What kind of depreciation is this?

A) a functional loss B) a time-related physical loss C) a use-related physical loss D) a technological loss E) a social loss Answer: B Diff: 1 Type: MC Page Ref: 165 Topic: 6.2. Depreciation and depreciation accounting Skill: Recall User1: Qualitative 27) From the standpoint of financial accounting, annual revenue is A) a component of the Balance Sheet.

B) a through variable.

C) an across variable.

D) an asset.

E) a liability.

Answer: B Diff: 2 Type: MC Page Ref: 172 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative 28) What is the depreciation rate of a physical asset with the purchase price of $150 000 and salvage value of $16 100 after 10 years of service?

A) 16.40% B) 16.66% C) 20.00% D) 25.00% E) 59.20% Answer: C Diff: 2 Type: MC Page Ref: 169 Topic: 6.2. Depreciation and depreciation accounting Skill: Applied User1: Quantitative

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29) A computer was bought for $2 000. After three years of service it can be sold for $500. If straight line depreciation is assumed, what was the computer's book value at the end of year 2?

A) $2 000 B) $1 500 C) $1 000 D) $500 E) $0 Answer: C Diff: 2 Type: MC Page Ref: 166-167 Topic: 6.2. Depreciation and depreciation accounting Skill: Applied User1: Quantitative 30) The most popular depreciation method for physical assets in Canada is A) 150%-declining-balance depreciation.

B) declining-balance depreciation.

C) sum-of-the-years'-digits depreciation.

D) double-declining-balance depreciation.

E) units-of-production depreciation.

Answer: B Diff: 1 Type: MC Page Ref: 169-170 Topic: 6.2. Depreciation and depreciation accounting Skill: Recall User1: Qualitative 31) One of the major drawbacks of the inventory-turnover ratio is that: A) Inventories are an across variable while sales are a through variable.

B) Inventories are a through variable while sales are an across variable.

C) Sales are measured at costs while inventories are measured in terms of market prices.

D) It includes only quick assets.

E) It includes only current liabilities.

Answer: A Diff: 2 Type: MC Page Ref: 186 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative

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32) Two assets, A and B, are purchased for the same price. Each loses 10% of its value in the first year.

Subsequently, the value of A continues to decline in the same way by declining-balance depreciation, while the value of B continues to decline in the same way by straight-line depreciation. Which will have the greater book value in five years time?

A) A B) B C) Their book values will always be equal.

D) Their book values will generally be different, but they are exactly equal at the end of the fifth year.

E) It is impossible to say without knowing the MARR. Answer: A Diff: 3 Type: MC Page Ref: 166-171 Topic: 6.2. Depreciation and depreciation accounting Skill: Recall User1: Quantitative 33) A factory produces two million hand-held computers a year. The annual fixed costs of the factory are $25 000 000, and the marginal cost of each computer is $7. What is the average cost per unit?

A) $12.25 B) $16.25 C) $17.30 D) $18.75 E) $19.50 Answer: E Diff: 2 Type: MC Page Ref: 208 Topic: 6.A. Cost Estimation Skill: Applied User1: Quantitative 34) You are trying to estimate the annual cost of setting up a branch of an engineering consulting business in Vancouver, based on your experience of running the original business in Halifax. The original business employs 80 engineers, whereas the Vancouver branch will initially employ 15. On the other hand, Vancouver salaries are typically 50% higher than those paid in Halifax. If the annual salary bill for the Halifax office is $9 000 000, give an order of magnitude estimate for the salary bill of the Vancouver office.

A) $844 000 B) $1 687 000 C) $2 400 000 D) $2 530 000 E) $2 953 000 Answer: D Diff: 2 Type: MC Page Ref: 209-210 Topic: 6.A. Cost Estimation Skill: Applied User1: Quantitative

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35) You are trying to estimate the annual cost of setting up a branch of a manufacturing business in Vancouver, based on your experience of running a similar business in Halifax. The original business produces 800 units per year, whereas the Vancouver branch will produce 1 700. If the cost of setting up the Halifax operation was $9 000 000 and the capacity factor is 0.6, give an order of magnitude estimate for the cost of setting up the Vancouver operation.

A) $5 400 000 B) $11 475 000$ C) $14 147 000 D) $19 125 000 E) $20 250 000 Answer: C Diff: 2 Type: MC Page Ref: 210 Topic: 6.A. Cost Estimation Skill: Applied User1: Quantitative 36) You run a software development business in Toronto. You are offered a chance to bid on a government contract that you estimate will require 600 000 lines of code, deliverable in one year.

You have developed a model: Effort = 3.0(KLOC)1.2 to predict the effort, in programmer-months, required to complete a job, where `KLOC' stands for `thousands of lines of code'. On the basis of this model, how many programmers should you assign to work on the project?

A) 150 B) 450 C) 540 D) 1 800 E) 6 500 Answer: C Diff: 2 Type: MC Page Ref: 210 Topic: 6.A. Cost Estimation Skill: Applied User1: Quantitative 37) Your company recently constructed a steel mill producing 600 000 tons of steel per year, for a construction cost of $12 000 000. You know that a rival company has constructed a similar mill, producing twice as much steel, at a cost of $20 900 000. Based on these figures, what is your estimate of the capacity factor for the steel industry?

A) 0.5 B) 0.6 C) 0.7 D) 0.8 E) 0.9 Answer: D Diff: 2 Type: MC Page Ref: 210 Topic: 6.A. Cost Estimation Skill: Applied User1: Quantitative 13 .

6.2 Short Answer Questions 1) A hydraulic press has just been purchased. It will have a book value of $6 465 in year 5. The present worth of its salvage value at the end of year 10 is $700, assuming a MARR of 7%. What is the purchase price of the press?

Answer: We can set the following equations: P * (1 - d)5 = 6 465 P * (1 - d)10/1.0710 = 700 The second equation can be re-written as P * (1 - d)5 * (1 - d)5/1.0710 = 700 Substituting the first one into the last expression: 6 465 * (1 - d)5/1.0710 = 700 and solving for d, we find that d = 0.266 or 26.6% P can be found from the first equation as follows: P * (1 - 0.266)5 = 6 465 So P = $30 353 Diff: 3 Type: SA Page Ref: 168-171 Topic: 6.2. Depreciation and depreciation accounting Skill: Applied User1: Quantitative 2) If a new technology comes into existence which partially replaces some old technology, how the loss in the value of the old technology can be classified?

Answer: This is an example of functional loss. The value of the old technology decreases because there is a better and cheaper way of producing services via the new technology.

Diff: 1 Type: SA Page Ref: 165 Topic: 6.2. Depreciation and depreciation accounting Skill: Recall User1: Qualitative 3) Explain the difference between scrap value and salvage value Answer: Scrap value is the value of an asset at the end of its physical life while salvage value is the value of an asset at the end of its useful life. Useful life can be shorter than or equal to, but never longer than an asset's physical life.

Diff: 1 Type: SA Page Ref: 165 Topic: 6.2. Depreciation and depreciation accounting Skill: Applied User1: Qualitative 14 .

4) It is known that the book value of an asset depreciated through declining balance depreciation is $3 000 in year 3. Purchase price of the asset was $3 800. What is the asset's salvage value at the end of its useful life, which is 10 years?

Answer: First we can find the depreciation rate from d=1-

= 0.0758 or 7.58%

Salvage value at the end of the asset's life is its book value in year 10 or BV(10) = P * (1 - d)10 = $3 800 * (1 - 0.0758)10 = $1 727.60 Diff: 2 Type: SA Page Ref: 169 Topic: 6.2. Depreciation and depreciation accounting Skill: Applied User1: Quantitative 5) Consider the following balance sheet: ASSETS Cash = $200 000 Accounts receivable = $200 000 Inventories = $100 000 Building and equipment = $2 000 000 Land = $3 000 000 TOTAL ASSETS = $5 500 000

LIABILITIES Accounts payable = $100 000 Taxes = $400 000 Long-term loans = $1 500 000

TOTAL LIABILITIES = $2 000 000

Define and calculate the acid-test ratio.

Answer: Acid-test ratio or quick ratio is defined as:

Quick assets are the most liquid assets. In this case they are cash of $200 000 and accounts receivable of $200 000. Current liabilities include accounts payable of $100 000 and taxes of $400 000. Therefore, the acid-test ratio is: = 0.8 Diff: 2 Type: SA Page Ref: 182 Topic: 6.3. Elements of financial accounting Skill: Applied User1: Quantitative

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6) According to a firm's balance sheet total assets account for $10 million while total liabilities account for $8 million. Calculate the equity ratio.

Answer: The equity ratio is defined as:

We know that Total Assets are $10 million. By definition, Total Owners' Equity is the difference between Total Assets and Total Liabilities which is $10 million - $8 million = $2 million. Therefore, the equity ratio is: = 0.2 Diff: 2 Type: SA Page Ref: 183 Topic: 6.3. Elements of financial accounting Skill: Applied User1: Quantitative 7) What is the major advantage and the major drawback of the straight line depreciation method?

Answer: The straight-line depreciation method has the great advantage of being easy to calculate. It also is easy to understand and is in common use. The main problem with the method is that its assumption of a constant rate of loss in asset value is often not valid.

Diff: 1 Type: SA Page Ref: 168 Topic: 6.2. Depreciation and depreciation accounting Skill: Recall User1: Qualitative

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8) Suppose that the purchase price of a piece of equipment is $1 million. Its salvage value at the end of its 5-year service life is $500 000. Depreciate this asset using straight line depreciation and declining balance depreciation Answer: For Straight Line depreciation (SLD), we need to find the depreciation amount per year: = $100 000 Under SLD, the asset's value loses this amount per year.

For Declining Balance Depreciation (DBD), we need to calculate the depreciation rate: d=1-

= 0.129

For DBD, the asset loses 12.9% of its value per year. Now we can calculate the book value in each year under both methods. The results are summarized in the following table:

BV(0) BV(1) BV(2) BV(3) BV(4) BV(5)

SLD 1 000 000 900 000 800 000 700 000 600 000 500 000

DBD 1 000 000 871 000 758 641 660 776 575 536 500 000

Diff: 2 Type: SA Page Ref: 166-171 Topic: 6.2. Depreciation and depreciation accounting Skill: Applied User1: Quantitative 9) A piece of equipment has just been purchased. It is expected to have a 5-year service life and to depreciate via the declining balance method. The present worth of its salvage value is $1 000. If the depreciation rate is 10%,and the MARR is also 10%, what was the purchase price of this equipment?

Answer: The following equation captures this P*(1 - 0.1 )5/1.15 = 1 000 where P is the purchase price of the equipment. Solving for P: P = 1 000 *

= $2 727.41

Diff: 3 Type: SA Page Ref: 169 Topic: 6.2. Depreciation and depreciation accounting Skill: Applied User1: Quantitative

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10) What is the difference between financial accounting and managerial accounting?

Answer: Financial accounting is concerned with recording and organizing the financial data of a business, which include revenues and expenses, and an enterprise's resources and the claims on those resources. Management accounting is concerned with the costs and benefits of the various activities of an enterprise. The goal of management accounting is to provide managers with information to help in decision making.

Diff: 1 Type: SA Page Ref: 171 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative 11) How are assets defined in a Balance Sheet? What types of assets are usually included in this financial document?

Answer: Assets are everything a business owns. Usually two types of assets are included in a Balance Sheet: (i) current assets and (ii) long-term assets. Current assets are cash and other assets that could be converted to cash within a relatively short period of time, usually a year or less.

Inventory and accounts receivable are examples of non-cash current assets. Long-term assets (also called fixed assets or non-current assets) are assets that are not expected to be converted to cash in the short term, usually taken to be one year Diff: 2 Type: SA Page Ref: 173 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative 12) Explain how owners' equity is defined Answer: Owners' equity is the interest of the owner or owners of a firm in its assets. Owners' equity usually appears as two components on a balance sheet of a corporation. The first is the par value of the owners' shares. When an enterprise is first organized, it is authorized to issue a certain number of shares.

Par value is the price per share set by the corporation at the time the shares are originally issued. At any time after the first sale, the shares may be traded at prices that are greater than or less than the par value, depending on investors' expectations of the return that will be earned by the business in the future.

The second part of owners’ equity usually shown on the balance sheet is retained earnings. Retained earnings includes the cumulative sum of earnings from normal operations in addition to gains (or losses) from transactions like the sale of plant assets or investments -the proceeds of which have been reinvested in the business Diff: 3 Type: SA Page Ref: 175-176 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative

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13) What is the fundamental difference between a Balance Sheet and an Income Statement?

Answer: A balance sheet is a snapshot of an enterprise’s financial position at a particular point in time, normally the last day of an accounting period. A firm’s financial position is summarized in a balance sheet by listing its assets, liabilities, and owners’ (or shareholders’) equity. An income statement summarizes an enterprise’s revenues and expenses over a specified accounting period. In accounting sense, a balance sheet includes across variables while an income statement includes through variables.

Diff: 2 Type: SA Page Ref: 172, 177 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative 14) What is the cost principle of accounting?

Answer: The cost principle of accounting states that assets are to be valued on the basis of their cost as opposed to market or other values. For example, the value of plant and equipment is based on cost. The value reported in the balance sheet is given by the initial cost minus accumulated depreciation Diff: 1 Type: SA Page Ref: 180 Topic: 6.3. Elements of financial accounting Skill: Recall User1: Qualitative 15) Suppose that a business has $100 million in long-term assets and $50 million in current assets. The business's liabilities are: current liabilities = $30 million and long-term liabilities = $80 million.

Calculate the equity ratio and comment on the business's performance on the basis of this ratio.

Answer: The equity ratio is defined as:

Total Assets = Current Assets + Long-Term Assets = $50 + $100 = $150 million Total Owners' Equity = Total Assets - Total Liabilities Total Liabilities = Current Liabilities + Long-Term Liabilities Therefore, Total Liabilities = $30 + $80 = $110 million Total Owners' Equity = $150 - $110 = $40 million Equity Ratio =

= 0.27. The ratio is worryingly low. It suggests that the business carries an excessive

amount of debt.

Diff: 3 Type: SA Page Ref: 183-184 Topic: 6.3. Elements of financial accounting Skill: Applied User1: Quantitative 19 .

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

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4) In order to make a replacement decision, a firm calculated the equivalent annual cost of owning an asset as follows: Replacement Period 1 2 3 4 5

Salvage Value $1420 $1,102 $910 $795 $580

EAC Capital Annual EAC Repair Costs Repair Costs Costs $1,287 $1,082 $400 $189 $976 $600 $298 $812 $800 $437 $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 1 2 3 4 5

Salvage Value $1420 $1,102 $910 $795 $580

EAC Capital Annual EAC Repair Costs Repair Costs Costs $1,287 $1,082 $400 $189 $976 $600 $298 $812 $800 $437 $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

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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

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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

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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

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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

6 .

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

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

Year 1 2 3 4 5

EAC Salvage Maintenance (Capital Value Costs Costs) $32 000 $4 000 $38 416 24 500 8 000 24 783 16 350 16 000 20 263 11 280 32 000 17 182 8 420 64 000 14 895

EAC (Maintenance Costs) $4 000 5 923 9 027 14 125 22 627

Total $42 416 30 706 29 291 31 307 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 1 2 3 4 5

EAC Salvage Maintenance (Capital Value Costs Costs) $28 900 $3 000 $33 558 22 670 6 000 21 921 15 300 12 000 11 280 24 000 15 482 7 800 48 000 13 701

EAC (Maintenance Costs) $3 000 4 429 6 716 10 440 16 592

Total $36 558 26 349 25 922 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

8 .

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

Year 1 2 3 4 5

EAC Salvage Maintenance (Capital Value Costs Costs) $35 260 $4 500 $30 597 18 256 7 000 25 803 14 520 14 000 19 688 12 569 27 500 16 179 8 598 44 900 14 385

EAC (Maintenance Costs) $4 500

Total $35 097

8 201 12 359 17 689

27 889 28 538 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 1 2 3 4 5

EAC Salvage Maintenance (Capital Value Costs Costs) $38 450 $4 200 $38 022 19 580 8 100 30 733 16 520 15 260 22 964 27 560 19 456 9 210 49 500 16 831

EAC (Maintenance Costs) $4 200 6 057 8 837 12 872 18 871

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

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Total $42 222 36 790 31 802 32 327 35 702

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

10 .

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 1 2 3 4

SALVAGE VALUE $40 000 $20 000 $10 000 $5 000

OPERATING COST $20 000 $20 000 $40 000 $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

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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

12 .

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

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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

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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 0 1 2 3 4 5

Salvage Value 30 000 27 000 24 300 21 870 19 683 17 715

EAC (Maintenance ) 694 759 828 902 982

EAC (Capital) 7 500 7 151 6 842 6 567 6 321

EAC 8 194 7 910 7 670 7 469 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) + Si

15 .

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 16 .

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) + Si 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

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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 0 1 2 3 4 5

Salvage Value 4 000.00 3 400.00 2 890.00 2 456.50 2 088.03 1 774.82

EAC (Capital) 1 400.00 1 304.42 1 223.93 1 156.25 1 099.19

EAC (Maintenance) 200.00 245.43 301.62 371.22 457.45

EAC Total 1 600.00 1 549.85 1 525.55 1 527.47 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) + Si 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

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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) + Si = [15 000 - 15 000(1 - 0.25)] * (1 + 0.08) + 15 000(1 - 0.25) * 0.08 = $4 950

EAC(maintenance, 1) EAC(Total)

= 3 000 + 1 200 = $4 200 = 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 19 .

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

Answer: The three are: 1. Identical defender and challenger.

2. Different defender and challenger but all succeeding challengers are identical to the first.

3. 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

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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 0 1 2 3 4 5

Market Value, P 13 000 9 000 8 000 6 000 2 000 0

Annual Operating Cost, AOC 2 500 2 700 3 000 3 500 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 AOCt is the annual operating cost in period t.

In any given year N: EAC(capital, N) = (P - S) * (A/P, i, N) + Si 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 0 1 2 3 4 5

Salvage Value 13 000 9 000 8 000 6 000 2 000 0

EAC (Maintenance) 2 500 2 595 2 717 2 887 3 150

EAC (Capital)

EAC Total

5 300 3 681 3 415 3 671 3 429

7 800 6 276 6 132 6 558 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 22 .

13) A communication system cost $65 000 to buy and $4 000 to install ten years ago. Currently the tenyear-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

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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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Engineering Economics, 5e (Fraser) Chapter 8 Taxes 8.1 Multiple Choice Questions 1) Why should businesses take into account tax impacts?

A) Taxes directly affect a business's cash flow.

B) Taxes increase the gross profit of a business.

C) Taxes allow the government to pay for public goods and services.

D) Taxes reduce the costs of a successful project.

E) Taxes are savings for a Canadian business.

Answer: A Diff: 1 Type: MC Page Ref: 264 Topic: 8.1. Introduction Skill: Recall User1: Qualitative 2) A tax is progressive if A) the rate of taxation increases by a constant amount with an increase in income.

B) the rate of taxation increases proportionally with an increase in income.

C) the rate of taxation increases by a constant percentage with an increase in income.

D) the rate of taxation increases faster than the income.

E) the rate of taxation increases slower than the income.

Answer: D Diff: 1 Type: MC Page Ref: 265 Topic: 8.2. Personal income taxes and corporate income taxes compared Skill: Recall User1: Qualitative 3) The before-tax MARR is A) higher than the after-tax MARR by a factor of (1 - t), where t is the corporate tax rate.

B) lower than the after-tax MARR by a factor of (1 - t), where t is the corporate tax rate.

C) higher than the after-tax MARR by a factor t, where t is the corporate tax rate.

D) lower than the after-tax MARR by a factor t, where t is the corporate tax rate.

E) approximately equal to the after-tax MARR. Answer: A Diff: 1 Type: MC Page Ref: 267 Topic: 8.4. Before- and after-tax MARR Skill: Recall User1: Quantitative

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4) SHMON Inc. wants to invest in future projects. The before-tax MARR is 14%. The after-tax MARR was found to be 7.4%. What is the corporate tax rate of the company?

A) 38% B) 42% C) 47% D) 53% E) 89% Answer: C Diff: 2 Type: MC Page Ref: 267 Topic: 8.4. Before- and after-tax MARR Skill: Applied User1: Quantitative 5) Which asset has the fastest depreciation rate?

A) land B) truck C) warehouse D) rental car E) office desk Answer: D Diff: 2 Type: MC Page Ref: 279 Topic: 8.3. Corporate tax rates Skill: Recall User1: Qualitative 6) What was the goal of the Canadian government in designing the Capital Cost Allowance system?

A) to calculate the total amount of taxes that the government collects from businesses' profits B) to define the after-tax minimum acceptable rate of return that businesses can use to account for the effects of taxation while making their investment decisions C) to define a corporate tax rate that businesses must use while calculating the amount of their taxes D) to define a specific amount of depreciation that businesses may claim in any year for any one depreciable asset E) to facilitate businesses in calculating their cash flows associated with the purchase of a long-term depreciable asset Answer: D Diff: 2 Type: MC Page Ref: 276-280 Topic: 8.8. Specific tax rules in Canada Skill: Recall User1: Qualitative

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7) Why do businesses want to depreciate their assets as soon as possible?

A) to balance their assets and liabilities B) to decrease the amount of taxes they must pay C) to release funds for further reinvestment D) to increase productivity of their operations E) to streamline their cash flows in the long-run Answer: B Diff: 2 Type: MC Page Ref: 277 Topic: 8.5. The effects of taxation on cash flows Skill: Recall User1: Qualitative 8) The effect of taxation on annual savings is captured by A) the capital tax factor.

B) the capital salvage factor.

C) the depreciation rate.

D) multiplying savings by the corporate tax rate.

E) multiplying savings by one minus corporate tax rate.

Answer: E Diff: 2 Type: MC Page Ref: 271-272 Topic: 8.5. The effects of taxation on cash flows Skill: Recall User1: Qualitative 9) What was the major reason for the government 's introduction of the Half-Year Rule?

A) to address the problem of delivery of public goods and services B) to address the problem of writing-off C) to reduce business activity D) to internalize externalities E) to follow international practice Answer: B Diff: 2 Type: MC Page Ref: 280 Topic: 8.8. Specific tax rules in Canada Skill: Recall User1: Qualitative 10) The undepreciated capital cost (UCC) is equal to A) the book value of an asset for the purpose of taxation under the CCA system.

B) the market value of an asset.

C) the salvage value of an asset.

D) the scrap value of an asset.

E) the non-taxable value of an asset.

Answer: A Diff: 2 Type: MC Page Ref: 280-283 Topic: 8.8. Specific tax rules in Canada Skill: Recall User1: Qualitative 3 .

11) In the CCA system "the half-year rule" implies A) that a firm can depreciate half the remaining capital cost of an asset in any given year after the asset is purchased.

B) that a firm can depreciate half the capital cost of an asset during the first half a year the asset is purchased.

C) that a firm can depreciate the capital cost of an asset during the first half a year the asset is purchased.

D) that a firm can depreciate half the capital cost of a new asset in the year the asset is purchased.

E) that a firm cannot depreciate the capital cost of an asset for at least half a year after the asset is purchased.

Answer: D Diff: 2 Type: MC Page Ref: 280 Topic: 8.8. Specific tax rules in Canada Skill: Recall User1: Qualitative 12) A firm has just purchased a vehicle for $100 000. The CCA rate for this vehicle is 40%. The firm's corporate tax rate is 50%. What would be the firm's tax savings due to the CCA by the end of the first year?

A) $5 000 B) $7 500 C) $10 000 D) $15 000 E) $20 000 Answer: C Diff: 2 Type: MC Page Ref: 280-283 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative

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13) Sample CCA Rates and Classes are presented as follows: CCA Rate 4 to 10 20 25 30 40 100

(%) 1, 3, 6 8 9 10 16 12

Class Description Buildings and additions Machinery, office furniture, and equipment Aircraft, aircraft furniture, and equipment Passenger vehicles, vans, trucks, and computers Taxis, rental cars, and freight trucks Dies, tools, and instruments that cost less than $200

What does the number "20" in the CCA Rate column mean?

A) A firm can claim a 20% depreciation rate per year for an asset such as machinery, office furniture, and equipment during its service life.

B) A firm can claim a 20% depreciation rate per year for an asset such as machinery, office furniture, and equipment beginning in the second half of the year after purchase of that asset.

C) A firm can fully depreciate an asset such as machinery, office furniture, and equipment in 5 years from the time it is purchased using the depreciation rate of 20% per year.

D) A firm can claim a 20% depreciation rate on half of the capital cost of a new asset such as machinery, office furniture and equipment in the year of purchase of that asset, while the other half is included in the following year.

E) A firm can only depreciate an asset such as machinery, office furniture, and equipment with the rate of 20% in the first year of purchase of that asset.

Answer: D Diff: 2 Type: MC Page Ref: 279-280 Topic: 8.8. Specific tax rules in Canada Skill: Recall User1: Qualitative

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14) SINCO Ltd. purchased a piece of equipment in 2000. The UCC amounts for this equipment are given in the following table:

Year 2000 2001 2002

Adjustments to UCC from Base UCC purchases and amounts for dispositions CCA $200 000 $100 000 0 $175 000

CCA $25 000 $43 750

Remaining UCC $175 000 $131 250

How much could SINCO Ltd. claim as CCA in 2002?

A) $26 250.00 B) $32 812.50 C) $40 250.50 D) $62 500.00 E) $131 250.00 Answer: B Diff: 3 Type: MC Page Ref: 280-281 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative 15) A company purchased a piece of equipment in 2000. The UCC amounts for this equipment are as follows : Adjustments to UCC from Base UCC purchases and amounts for Remaining Year dispositions CCA CCA UCC 2000 $200 000 $100 000 $25 000 $175 000 2001 0 $175 000 $43 750 $131 250 How much tax savings could the company accumulate due to the CCA by the end of 2001 if the corporate tax rate is 50%?

A) $12 465 B) $24 580 C) $30 925 D) $34 375 E) $68 750 Answer: D Diff: 3 Type: MC Page Ref: 280-282 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative

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16) GEMTECH Ltd. is an engineering construction company. It purchased an excavator for $200 000 in 1999 and a bulldozer for $180 000 in 2000. The UCC for the changes in asset holdings due to the purchases of an excavator are as follows:

Year 1999 2000

Adjustments to UCC from Base UCC purchases and amounts for dispositions CCA $200 000 $100 000

CCA $20 000

Remaining UCC $180 000

How much could the GEMTECH Ltd. claim as CCA in 2000?

A) $36 000 B) $54 000 C) $90 000 D) $130 000 E) $270 000 Answer: B Diff: 3 Type: MC Page Ref: 280-282 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative 17) An engineering construction company purchased an excavator for $200 000 in 1999 and a bulldozer for $180 000 in 2000. The UCC for the changes in asset holdings due to the purchases of an excavator are as follows:

Year 1999 2000

Adjustments to UCC from Base UCC purchases and amounts for dispositions CCA $200 000 $100 000

CCA $20 000

Remaining UCC $180 000

How much tax savings could the company accumulate due to the CCA by the end of 2000 if the corporate tax rate is 50%?

A) $20 000 B) $28 000 C) $32 000 D) $37 000 E) $54 000 Answer: D Diff: 3 Type: MC Page Ref: 280-282 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative 7 .

18) The capital tax factor (CTF) is a value that A) summarizes the effect of the net benefits due to CCA on the present worth of an asset.

B) summarizes the effect of the costs imposed on a firm through taxes due to CCA on the present worth of an asset.

C) summarizes the effect of the benefits from future tax savings due to CCA on the present worth of an asset.

D) summarizes the total effect of the taxes due to CCA on the present worth of an asset.

E) summarizes the effect of the benefits claimed as CCA on the future worth of an asset.

Answer: C Diff: 2 Type: MC Page Ref: 284 Topic: 8.8. Specific tax rules in Canada Skill: Recall User1: Qualitative 19) Sirius Ltd. purchased a piece of equipment at the very beginning of the 1999 fiscal year. The UCC amounts for this equipment are as follows:

Year 1999 2000

Adjustments to UCC from Base UCC purchases and amounts for dispositions CCA $40 000 $20 000 0 $36 000

CCA $4 000 $7 200

Remaining UCC $36 000 $28 800

What was the present worth at the beginning of fiscal 1999 of the company's savings due to CCA over the two-year period if the corporate tax rate was 25% and the interest rate was 10%?

A) $2 397 B) $2 488 C) $2 800 D) $3 214 E) $4 263 Answer: A Diff: 3 Type: MC Page Ref: 280-282 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative

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20) The salvage value of a ten-year-old truck is $10 000. If this truck is on the Balance Sheet of a transportation company as a Class 8 asset with a CCA rate of 20%, what was the present worth of this salvage value at the time of the truck's purchase if the corporate tax rate is 35% and annual after-tax interest rate is 5%?

A) $10 000 B) $8 200 C) $7 200 D) $6 240 E) $4 420 Answer: E Diff: 2 Type: MC Page Ref: 284 Topic: 8.6. Present worth and annual worth tax calculations Skill: Applied User1: Quantitative 21) In March of 1998 a company purchased a piece of equipment for $50 000. The corporate tax rate was and still is 30% and the after-tax MARR is 10%. What CTF should the company use if the present worth of the first cost of this equipment, taking into account all future tax savings due to the CCA, is $37 727?

A) 0.7018 B) 0.7213 C) 0.7425 D) 0.7545 E) cannot be determined Answer: D Diff: 2 Type: MC Page Ref: 284 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative 22) In 2002 a firm completed the present-worth tax calculations for the value of the salvage cost of an asset that had reached the end of its useful life. For this purpose, the firm A) multiplied the salvage value of the asset by d, where d is a depreciation rate.

B) multiplied the salvage value of the asset by (1 - d), where d is a depreciation rate.

C) multiplied the salvage value of the asset by (1 - t), where t is a corporate tax rate.

D) multiplied the salvage value of the asset by the CTF. E) multiplied the salvage value of the asset by the CSF. Answer: E Diff: 1 Type: MC Page Ref: 284 Topic: 8.8. Specific tax rules in Canada Skill: Recall User1: Quantitative

9 .

23) DON Corporation is making a decision about a project that has an after-tax internal rate of return of 18%. If the company pays 30% corporate income tax rate what is the before-tax internal rate of return?

A) 12.6% B) 18.0% C) 22.4% D) 25.7% E) 60.0% Answer: D Diff: 2 Type: MC Page Ref: 275 Topic: 8.7. IRR tax calculations Skill: Applied User1: Quantitative 24) Numerically the relationship between the CTF and the CSF is as follows: A) CTF is always equal to CSF. B) CTF is always smaller than CSF. C) CTF is always greater than CSF. D) CTF can be smaller or greater than CSF depending on circumstances.

E) CFS = 1 - CTF. Answer: C Diff: 3 Type: MC Page Ref: 284 Topic: 8.8. Specific tax rules in Canada Skill: Recall User1: Quantitative 25) The Capital Cost Allowance system uses a ________ depreciation rate to calculate the depreciation allowance for capital assets.

A) proportional B) fixed C) straight-line D) declining-balance E) arbitrary Answer: D Diff: 1 Type: MC Page Ref: 276 Topic: 8.8. Specific tax rules in Canada Skill: Recall User1: Qualitative 26) The Capital Salvage Factor A) captures the effect of taxation on an initial investment.

B) captures the effect on taxation of selling an asset for salvage.

C) captures the effect of taxation on operating costs.

D) captures the Half-Year Rule effect.

E) captures the effect of taxation on annual revenues.

Answer: B Diff: 1 Type: MC Page Ref: 284 Topic: 8.8. Specific tax rules in Canada Skill: Recall 10 .

User1: Qualitative 27) Suppose that the after-tax interest rate is 10%, and the corporate tax rate is 50%. What is the beforetax interest rate?

A) 40% B) 20% C) 15% D) 10% E) 5% Answer: B Diff: 1 Type: MC Page Ref: 267-268 Topic: 8.4. Before- and after-tax MARR Skill: Applied User1: Quantitative 28) Suppose that a Canadian company bought a car for $20 000. The CCA rate for the car is 20% and the corporate tax rate is 40%. How much money does the company save in the first year as a result of the CCA allowance?

A) $2 400 B) $1 600 C) $1 200 D) $800 E) It does not save at all since it still has to pay taxes.

Answer: D Diff: 2 Type: MC Page Ref: 276-280 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative 29) At the beginning of the year, a pie company has $P in the bank. During the year it takes in $A for pie sales. It pays $B for pie ingredients, and $C for wages and rent. It pays taxes at t%. At the end of the year, the amount the company has in the bank is: A) P + (A - B - C)t B) (P + A)t - B - C C) P + (A - B - C)(1 - t) D) P + (A - B - C)(1 - t) E) P + (A - B - C)t Answer: C Diff: 2 Type: MC Page Ref: 269-273 Topic: 8.5. The effects of taxation on cash flows Skill: Applied User1: Quantitative

11 .

30) At the beginning of the year, a pie company has $P in the bank. During the year it takes in $A for pie sales. It pays no wages or rent. It pays taxes at t%. It buys an oven for $F. If Revenue Canada allows ovens to depreciate at d% per year, the amount the company has in the bank at the end of the year is: A) P + A(1 - t) - F B) P + (A - F)(1 - t) C) P + A(1 - t) - F(CTF) D) P + (A - Fd/2)(1 - t) - F + Fd/2 E) P + At - F(CTF) Answer: D Diff: 2 Type: MC Page Ref: 280 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative 31) At the beginning of the year, a pie company has $P in the bank. During the year it takes in $A for pie sales. It pays $B for pie ingredients, and $C for wages and rent. It pays taxes at t%. It has an oven, purchased 10 years ago, which according to Revenue Canada rules depreciated by $D during the year.

At the end of the year, the amount the company has in the bank is: A) P + (A - B - C - D)t B) (P + A)t - B - C - D C) P + (A - B - C - D)(1 - t) + D D) P + A - (B + C + D)t E) P + A - (B + C + D)(1 - t) Answer: C Diff: 2 Type: MC Page Ref: 280-283 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative 32) At the beginning of the year, a pie company has $P in the bank. During the year it takes in $A for pie sales. It pays no wages or rent. It pays taxes at t%. It has an oven, whose undepreciated capital cost at the beginning of the year is $G. During the year it sells this oven for $G. At the end of the year, the amount the company has in the bank is: A) P + A(1 - t) + G B) P + A + G(CSF) C) P + A(1 - t) + G(CSF) D) P + (A + G)(1 - t) E) P + A + G(1 - t) Answer: A Diff: 2 Type: MC Page Ref: 266 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative

12 .

33) At the beginning of the year, a pie company has $P in the bank. During the year it takes in $A for pie sales. It pays no wages or rent. It pays taxes at t%. It has an oven, originally bought for $3G. whose undepreciated capital cost at the beginning of the year is $G. During the year it sells this oven for $2G. At the end of the year, the amount the company has in the bank is: A) P + A(1 - t) + 2G B) P + A + 2G(CSF) C) P + A(1 - t) + 2G(CSF) D) P + (A + G)(1 - t) + G E) P + (A + g)t + G Answer: D Diff: 2 Type: MC Page Ref: 281-282 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative 34) At the beginning of the year, a pie company has $P in the bank. During the year it takes in $A for pie sales. It pays no wages or rent. It pays taxes at t%. It has an oven, originally bought for $2G. whose undepreciated capital cost at the beginning of the year is $G. During the year it sells this oven for $3G. At the end of the year, the amount the company has in the bank is: A) P + A(1 - t) + 3G B) P + A + 3G(CSF) C) P + (A + 2G)(1 - t) + G D) P + (A + G)(1 - t) + 2G E) P + (A + 3G)(1 - t) Answer: D Diff: 2 Type: MC Page Ref: 266 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative 35) The after-tax IRR is A) always greater than the before-tax IRR. B) equal to the before-tax IRR. C) always smaller than the before-tax IRR. D) equal to the before-tax IRR minus the inflation rate.

E) equal to the before-tax IRR plus the inflation rate.

Answer: C Diff: 1 Type: MC Page Ref: 274-275 Topic: 8.7. IRR tax calculations Skill: Recall User1: Quantitative

13 .

36) A project involves an immediate expenditure of $10 000, and further expenditures of $10 000 every year for the next four years. It will yield an income of $8 000 at the end of the first year, and this will increase by $8 000 a year. This is the only project the company has; it is taxed at 50%, and its after-tax MARR is 10%. Assume that losses cannot be carried forward to offset future income. What is the present worth of the project to the company?

A) $7 452 B) $9 342 C) $12 296 D) $15 251 E) $27 552 Answer: B Diff: 2 Type: MC Page Ref: 273-274 Topic: 8.6. Present worth and annual worth tax calculations Skill: Applied User1: Quantitative 37) A Canadian company buys a capital asset for $100 000. The asset depreciates at 20% per year, and the half-year rule applies. Use of the asset brings in $5 000 a year in profit. If these are the company's only cash flows, what is the first year in which they have to pay taxes?

A) 1 B) 2 C) 3 D) 4 E) 5 Answer: E Diff: 2 Type: MC Page Ref: 276-283 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative 38) An asset is purchased for $100 000. It depreciates at 20% per year. After 10 years it will be sold for salvage for $10 700. If the company is taxed at 50% and has an after-tax MARR of 10%, what is the present worth of this sale in the year of purchase?

A) $1 374 B) $2 177 C) $2 747 D) $4 125 E) $7 126 Answer: A Diff: 2 Type: MC Page Ref: 283-284 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative

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39) A taxi company buys two new taxis at the beginning of every year. It keeps its taxis until they are worn out and have negligible value. Each taxi costs $20 000. The company is taxed at 50%, and the taxis depreciate at 40% per year. The company's after-tax MARR is 20%. What is the equivalent uniform annual cost to them of buying the taxis?

A) $14 668 B) $25 333 C) $30 400 D) $40 000 E) $48 000 Answer: C Diff: 2 Type: MC Page Ref: 283-284 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative 40) A company buys 10 new computers every year for $1 200 each. They depreciate at 30% per year.

After two years they are sold for their UCC value. The half-year rule applies. If the company pays taxes at 40% and has an after-tax MARR of 20%, what is the net present cost of buying and subsequently selling each computer?

A) $289 B) $393 C) $412 D) $478 E) $521 Answer: B Diff: 2 Type: MC Page Ref: 283-284 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative 8.2 Short Answer Questions 1) Explain when taxes are viewed as disbursements and when as savings Answer: When a firm makes an investment, the income from the project will affect the company’s cash flows. If the investment yields a profit, the profits will be taxed. Since the taxes are a direct consequence of the investment, they reduce the net profits associated with that investment. In this sense, taxes associated with a project are a disbursement. If the investment yields a loss, the company may be able to offset the loss from this project against the profits from another and end up paying less tax overall. As a result, when evaluating a lossgenerating project, the net savings in tax can be viewed as a negative disbursement. Income taxes thus reduce the benefits of a successful project, while at the same time reducing the costs of an unsuccessful project Diff: 2 Type: SA Page Ref: 264 Topic: 8.1. Introduction Skill: Recall User1: Qualitative

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2) A manufacturing company bought a truck for $35 000. After 10 years the truck's salvage value will be $5 000. The truck is expected to produce a revenue of $10 000 per year, while maintenance costs are expected to be $1 000 per year. If the annual interest rate is 10%, the corporate tax rate is 35% and straight-line depreciation is applied, what amount of taxes associated with the truck's operation will the company pay in the first year?

Answer: Tax is imposed on net income, which is the difference between annual revenue and total annual costs. Annual revenue is $10 000 and the total annual costs include depreciation costs plus maintenance costs. Assuming straight-line depreciation, the annual depreciation amount is: D=

=

= $3 000

Therefore, total annual costs are $3 000 + $1 000 = $4 000. The before-tax income is (revenue - total costs) = $10 000 - $4 000 = $6 000 and the amount of tax due in the first year is 0.35 x $6 000 = $2 100 Diff: 2 Type: SA Page Ref: 267-268 Topic: 8.5. The effects of taxation on cash flows Skill: Applied User1: Quantitative 3) An engineering project involves the purchase of a capital asset with first cost of $100 000, operating costs of $5 000 per year and a service life of 10 years. The expected revenue is $20 000 per year. If the CCA rate is 25%, interest rate is 10% and the corporate tax rate is 33%, what is the project's present worth?

Answer: The present worth of the project is: PW (Revenue) * (1 - t) - PW(Capital costs) - PW(operating costs) * (1 - t) where t is the corporate tax rate.

In turn, PW(Capital costs) is: PW = CTF * 100 000 = (1 -

) * 100 000 = 0.755 x 100 000 = $77 500

Therefore: PW(Project) = 20 000 * (P/A, 10%, 10) * ( 1 - 0.33) - 77 500 - 5 000 * (P/A, 10%, 10) * (1 - 0.33) = 20 000 * 6.1445 * 0.67 - 77 500 - 5 000 * 6.1445 * 0.67 = 82 337- 77 500 - 20 584 = -$15 747 Diff: 1 Type: SA Page Ref: 285 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative

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4) An engineering company just bought a bending machine for $50 000. Its service life is 10 years, and afterwards its salvage value will be $2 000. Services provided by the machine are expected to bring $10 000 in annual revenues, and operating costs are estimated at $2 000 per year. Assume that the CCA rate is equal to the depreciation rate, the after-tax interest rate is 5% and the corporate tax rate is 30%.

Calculate the present worth of this investment.

Answer: The depreciation rate is d = 1 -

= 0.275

The present worth of the investment is: PW = (10 000 - 2 000) x (P/A, 0.05, 10) x (1 - t) - CTF x 50 000 + CSF x 2 000 * (P/F, 0.05.10) where t is the corporate tax rate, CTF is the capital tax factor, and CSF is the capital salvage factor.

CTF = 1 -

) = 0.7522

CSF = 1 - 0.3 * 0.275/(0.05 + 0.275) = 0.746 And the present worth of the investment is therefore PW = 8 000 x 7.7217 x (1-0.3) - 0.7522 x 50 000 + 0.746 x 2 000 x 0.6139 = 43 241 - 37 610 + 916 = $6 547 Diff: 3 Type: SA Page Ref: 286 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative 5) Calculate the CSF and CTF given the following information: the before-tax interest rate is 10%, the corporate tax rate is 45%, the depreciation rate is 20% and the CCA rate is 25%. Which one is higher and why?

Answer: First, it is necessary to calculate the after-tax interest rate as follows: i = 0.1 * (1-0.45) = 0.055 The two capital cost tax factors are defined as follows: CSF = 1 CTF = 1 -

=1-

= 0.631 =1-

) = 0.641

In the above formulas, d is the CCA rate. The CTF is higher which means less tax savings under the current Canadian Capital Cost Allowance system. This is due to the Half-Year Rule introduced on November 13, 1981 to prevent fast depreciation of an asset in its first year.

Diff: 1 Type: SA Page Ref: 284 Topic: 8.8. Specific tax rules in Canada 17 .

Skill: Applied User1: Quantitative 6) Explain why the value of IRRafter-tax calculated as IRRafter-tax ≈ IRRbefore-tax * (1 - t) is only an approximation to the actual IRRafter-tax.

Answer: The calculated value is approximately correct because the IRR represents the percentage of the total investment that is net income. Since the tax rate is applied to net income, it correspondingly reduces the IRR by the same proportion. It is not exactly correct because it assumes that expenses offset receipts in the year that they occur. It does not take into account the CCA allowance, for example.

Consequently, if the after-tax IRR is close to the after-tax MARR, a more precise calculation is advisable.

Diff: 1 Type: SA Page Ref: 275 Topic: 8.7. IRR tax calculations Skill: Recall User1: Qualitative 7) What is the difference between the CCA rate and the depreciation rate, and how do the two affect financial analysis?

Answer: The depreciation rate is used in a company's balance sheet for internal purposes. It is based on the company's experience with capital assets of that type, and can be used to determine when the asset should be replaced. The CCA rate is the maximum rate a company can use to depreciate its assets for tax purposes under the Canadian Capital Cost Allowance system, and is used in calculating the company's tax burden.

Diff: 2 Type: SA Page Ref: 271, 272 Topic: 8.8. Specific tax rules in Canada Skill: Recall User1: Qualitative 8) What is the undepreciated capital cost and how it is related to an asset's book value?

Answer: The book value is the current value of an asset calculated on the basis of a prescribed depreciation method such as declining-balance depreciation. The Undepreciated Capital Cost (UCC) is the remaining book value for the asset subject to depreciation for taxation purposes, which may differ from its market or salvage value.

Diff: 2 Type: SA Page Ref: 280-282 Topic: 8.8. Specific tax rules in Canada Skill: Recall User1: Qualitative

18 .

9) Last year an engineering company earned $200 000 in annual revenues, and incurred $100 000 in total operating costs. It also bought a new lathe for $50 000. With a CCA rate of 30% and a corporate tax rate of 50%, how much income tax did the company pay?

Answer: Corporate tax is paid on net income (profit), which is the difference between total revenue and total costs including depreciation. According to the Half-Year Rule, only half of a new asset's value can be claimed in the year of the purchase. Therefore, the depreciation cost is: * 0.3 = $7 500 Taxable net income (profit) can be calculated as follows: Net Income = Revenue - Operating Costs - Depreciation = 200 000 - 100 000 - 7 500 = $92 500 Taxes paid at 50% rate are 0.5 * 92 500 = $46 250.

Diff: 2 Type: SA Page Ref: 280-282 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative 10) New Brunswick Metalworks is evaluating a $100 000 polishing machine (CCA class 8). Services provided by the machine are expected to bring $15 000 per year in value over the next five years.

Operating costs are expected to be $1 000 per year. Calculate the amount of taxes paid in each of the next five years of service life under the corporate tax rate of 40%.

Answer: All calculations are summarized in the following table:

Year 1 2 3 4 5

Operating Depreciation Revenue cost cost $20 000 $1 000 $10 000 $20 000 $1 000 $18 000 $20 000 $1 000 $14 400 $20 000 $1 000 $11 520 $20 000 $1 000 $9 216

Net income $9 000 $1 000 $4 600 $7 480 $9 784

Tax $3 600 $400 $1 840 $2 992 $3 913

In the first year, depreciation cost is calculated according to the Half-Year Rule. In all other consecutive years, the undepreciated capital cost is calculated first as UCC(t) = UCC(t - 1) - UCC(t - 1) * CCA rate and then depreciation cost is defined as UCC(t) * CCA rate where CCA rate is 20% (Table 8.4, page 279).

Diff: 3 Type: SA Page Ref: 280-282 Topic: 8.8. Specific tax rules in Canada Skill: Applied 19 .

User1: Quantitative 11) Explain the problem of "writing-off" and how the introduction of the Half-Year Rule addressed this problem.

Answer: Prior to November 12, 1981, a company was allowed to include in its base for the calculation of capital cost allowance the full purchase price of an asset purchased within the year, regardless of when in the year the asset was purchased. Consequently, there was considerable motivation for companies to purchase assets at the end of the fiscal year and claim their full value as depreciation amount. This was known as the problem of writing-off. The Canadian government, recognizing the problem, changed the rules effective November 13, 1981. Since that date, only half of the capital cost of acquiring an asset is considered in the Capital Cost Allowance system in the year of purchase of an asset, while the other half is then included in the second year. Since the change, there still remains an incentive to purchase equipment at the end of the fiscal year, however, the incentive has been reduced as the tax effects have been diminished.

Diff: 1 Type: SA Page Ref: 280 Topic: 8.8. Specific tax rules in Canada Skill: Recall User1: Qualitative 12) A manufacturing company just bought a new piece of equipment for $1 million. It is a class 8 asset.

The following information is given: - after-tax annual interest rate = 8% - corporate tax rate = 36% - service life = 10 years - historic depreciation rate = 25% Calculate the present worth of the equipment's salvage value with tax effects.

Answer: First of all, it is necessary to calculate salvage value of the asset using its historic depreciation rate: S = BV(10) = P * (1 - d)10 = 1 000 000 * (1 - 0.25)10 = $56 313.51 Present worth of the salvage value with tax effects can be defined as: PW(S) = S * CSF * (P/F, 8%, 10) where CSF is the capital salvage factor. Note that in calculating CSF, we don't use the historic depreciation rate, but the depreciation rate prescribed by Revenue Canada for Class 8 assets, which is 20%.

CSF = 1 -

=0.74286

Therefore: PW(S) = 56 313 * 0.74286 *

= $19 376 20 .

Diff: 2 Type: SA Page Ref: 285 Topic: 8.8. Specific tax rules in Canada Skill: Applied User1: Quantitative 13) An oil company uses a technology which it purchased for $15 million. Operating costs are $2 million per year, and output is 1 000 barrels per day. Calculate the after-tax IRR given the following information: The corporate tax rate is 25%, the price of oil is $20 per barrel, the service life of the technology is 5 years and salvage value is $2 million. If the after-tax MARR is 15%, is this a good investment?

Answer: The after tax IRR is IRRafter = IRRbefore x (1-t) where t is the corporate tax rate. Therefore, first it is necessary to calculate the before-tax IRR. By definition of the IRR: 20 * 1 000 * 365 * (P/A, IRRbefore, 5) + 2 000 000 * (F/A, IRRbefore, 5)=15 000 000 + 2 000 000 (P/A, IRRbefore, 5). and IRRbefore ≈ 0.247. Therefore the after-tax IRR is IRRafter = 0.247 x (1-0.25) = 0.185 or 18.5%.

Since the after-tax IRR exceeds the after-tax MARR, this is a good investment.

Diff: 3 Type: SA Page Ref: 275-276 Topic: 8.7. IRR tax calculations Skill: Applied User1: Quantitative 14) Explain the main objective of the Canadian Capital Cost Allowance system and its basic assumptions.

Answer: In order to limit the depreciation amount which companies use for tax purposes, the Canadian government established a maximum level of depreciation which a company can claim each year. This maximum amount is referred to as the firm's capital cost allowance (CCA). The CCA system specifies the amount and timing of depreciation expenses on capital assets. According to the CCA system, the declining-balance method of depreciation must be used for claiming capital costs associated with most tangible assets. Straight-line depreciation is used for intangible assets.

Diff: 1 Type: SA Page Ref: 276-280 Topic: 8.8. Specific tax rules in Canada Skill: Recall User1: Qualitative

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15) Explain why it is important to incorporate tax impacts into a business's cash flows.

Answer: Taxes can have a significant impact on the economic viability of a project because they change the actual cash flows experienced by a business. A vital component of a thorough economic analysis should therefore include the tax implications of an investment decision. When a business makes an investment, the income from the project will affect the business's cash flows. If the investment yields a profit, the profits will be taxed. Since the taxes result as a direct consequence of the investment, they reduce the net profits associated with that investment.

Diff: 1 Type: SA Page Ref: 264 Topic: 8.2. Personal income taxes and corporate income taxes compared Skill: Recall User1: Qualitative

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Engineering Economics, 5e (Fraser) Chapter 9 Inflation 9.1 Multiple Choice Questions 1) Suppose that the CPI was 105 on Jan 1, 2008 and 117 in Jan 1, 2011. What was the average annual inflation over the intervening period?

A) 4.00% B) 3.97% C) 3.87% D) 3.77% E) 3.67% Answer: E Diff: 2 Type: MC Page Ref: 306-309 Topic: 9.2. Measuring the inflation rate Skill: Applied User1: Quantitative

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2) Data from the International Monetary Fund are as follows:

where CPI = Consumer Price Index INF = Inflation Rate UEM = Unemployment Rate What is the base year in this case?

A) 1984 B) 1985 C) 1995 D) 1990 E) 1998 Answer: C Diff: 1 Type: MC Page Ref: 306-307 Topic: 9.2. Measuring the inflation rate Skill: Applied User1: Qualitative

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3) The price of a TV set was $600 in 1999. In 2003 it was priced at $520. If the CPI in 1999 was 123 and 2003 is the base year, has the TV set become cheaper or more expensive in real terms in 2003?

A) Real-dollar cost of the TV set has not changed.

B) Real-dollar cost of the TV set has increased by 6.6%.

C) Real-dollar cost of the TV set has decreased by 6.6%.

D) Real-dollar cost of the TV set has decreased by 15.8%.

E) Real-dollar cost of the TV set has increased by 15.8%.

Answer: B Diff: 2 Type: MC Page Ref: 309 Topic: 9.2. Measuring the inflation rate Skill: Applied User1: Quantitative 4) Mary graduated from a college in 2004 and she wanted to go on a sea cruise the same year. However, she could not afford it at that time because of the high price of $2 550. Today she can afford it, however, the cruise price is $2 870 now. If the price of the cruise in real dollars has not changed and the inflation rate has been 3.0% since 2004, what year is it now?

A) 2007 B) 2008 C) 2009 D) 2010 E) 2011 Answer: B Diff: 2 Type: MC Page Ref: 309 Topic: 9.3. Economic evaluation with inflation Skill: Applied User1: Quantitative 5) In general, inflation is defined as A) the increase in Gross Domestic Product over time.

B) a measure of the average person's preference for a present good compared to a future good.

C) an increase in prices of goods and services that are included in a consumption basket over time.

D) an increase in the purchasing power of money over time.

E) the difference between the real and nominal Gross Domestic Product (GDP).

Answer: C Diff: 1 Type: MC Page Ref: 305 Topic: 9.1. Introduction Skill: Recall User1: Qualitative

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6) The equation that converts actual dollars into real dollars can be written in terms of A) the present worth compound interest factor.

B) the series present worth factor.

C) the compound amount factor.

D) the capital recovery factor.

E) the capitalized value.

Answer: A Diff: 1 Type: MC Page Ref: 309 Topic: 9.3. Economic evaluation with inflation Skill: Recall User1: Qualitative 7) A valuable asset was priced at $1 million in 1995. What should be its nominal price in 2012 if the CPI was 95 in 1995 and 145 in 2012?

A) $0.95 million B) $1.25 million C) $1.45 million D) $1.53 million E) $1.65 million Answer: D Diff: 2 Type: MC Page Ref: 309 Topic: 9.2. Measuring the inflation rate Skill: Applied User1: Quantitative 8) The highest level of inflation in Canada in the post World War II period was observed in the A) 1950s.

B) 1960s.

C) 1970s.

D) 1980s.

E) 1990s.

Answer: C Diff: 2 Type: MC Page Ref: 313 Topic: 9.1. Introduction Skill: Recall User1: Qualitative

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9) A typical consumption basket for 2010 and 2011 is as follows: Item Eggs Bread Haircut

Quantity (2010) 400 100 200

Price (2010) $1.00 $2.50 $10.00

Price (2011) $0.90 $2.70 $11.00

Suppose that 2010 is the base year. What is the CPI for 2011?

A) 93.6 B) 106.8 C) 107.9 D) 108.2 E) 111.3 Answer: B Diff: 3 Type: MC Page Ref: 306-307 Topic: 9.2. Measuring the inflation rate Skill: Applied User1: Quantitative 10) The real MARR can be defined as A) the real interest rate minus an adjustment that reflects the inflation effect.

B) the actual interest rate plus an adjustment that reflects the inflation effect.

C) the real interest rate plus an adjustment that reflects the inflation effect.

D) the actual MARR plus an adjustment that reflects the inflation effect.

E) the actual MARR minus an adjustment that reflects the inflation effect.

Answer: E Diff: 2 Type: MC Page Ref: 312 Topic: 9.4. The effect of correctly anticipated inflation Skill: Recall User1: Qualitative 11) A municipality issued bonds to finance its infrastructure projects. The bonds are supposed to pay 6.5% per year. A city engineer calculated the real rate of interest to be 3.5% next year. What inflation rate did the engineer assume?

A) 2.7% B) 2.9% C) 4.8% D) 9.2% E) 22.2% Answer: B Diff: 2 Type: MC Page Ref: 312 Topic: 9.4. The effect of correctly anticipated inflation Skill: Applied User1: Quantitative

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12) A municipality issued bonds to finance its infrastructure projects. The bonds pay 6.5% per year. If expected annual inflation is 3.5%, what is the real rate of return on these bonds?

A) 2.90% B) 3.00% C) 3.50% D) 6.50% E) 10.23% Answer: A Diff: 1 Type: MC Page Ref: 312 Topic: 9.4. The effect of correctly anticipated inflation Skill: Applied User1: Quantitative 13) A company invested $3.5 million in a five-year project. It assessed that this project would generate $4.4 million revenue by the end of a five-year period in terms of today's dollars. If inflation is expected to be 3.0% per year over that period, what is the current IRR? A) 4.7% B) 5.7% C) 6.8% D) 7.8% E) 8.7% Answer: D Diff: 2 Type: MC Page Ref: 314-315 Topic: 9.5. Project evaluation methods with inflation Skill: Applied User1: Quantitative 14) A municipality decided to invest $500 000 in a four-year project. To decide whether the project should be undertaken, a city engineer was asked to run a financial analysis. Carrying out the analysis in actual dollars, the engineer found that by its very end the project would just break even. If inflation is expected to be 4.0% per year, what is the real IRR? A) -3.8% B) -4.0% C) -4.2% D) -4.4% E) -4.6% Answer: A Diff: 3 Type: MC Page Ref: 314-315 Topic: 9.4. The effect of correctly anticipated inflation Skill: Applied User1: Quantitative

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15) If Tonya as a municipal engineer evaluates a project which is subject to the effect of inflation, she should A) work with nominal values for cash flows and nominal interest rates.

B) work with real values for cash flows and nominal interest rates.

C) work with nominal values for cash flows and real interest rates.

D) work with constant values for cash flows and current interest rates.

E) rely on forecast of future interest rates and cash flows in real or nominal terms.

Answer: A Diff: 1 Type: MC Page Ref: 316-318 Topic: 9.5. Project evaluation methods with inflation Skill: Recall User1: Qualitative 16) If Trevor uses the real MARR and determines cash flows in actual dollars then he A) cannot estimate the profitability of investments.

B) estimates the profitability of investments incorrectly.

C) estimates the profitability of investments correctly.

D) overestimates the profitability of investments.

E) underestimates the profitability of investments.

Answer: D Diff: 2 Type: MC Page Ref: 311-312 Topic: 9.4. The effect of correctly anticipated inflation Skill: Recall User1: Qualitative 17) The cost of one of the following products has changed very differently from the others over time.

Which is the odd one out?

A) food B) clothing C) computers D) oil E) housing Answer: C Diff: 2 Type: MC Page Ref: 316-318 Topic: 9.3. Economic evaluation with inflation Skill: Recall User1: Qualitative 18) What should be done when the price of one good or service changes differently from the average price of other goods and services associated with a project?

A) Incorporate the price change directly into the cash flows of the project.

B) Incorporate the price change into the real MARR of the project.

C) Assume that the price change is the same for all goods and services of the project.

D) Use two different MARRs for different goods and services of the project.

E) Find the average price change for all goods and services of the project.

Answer: A Diff: 2 Type: MC Page Ref: 317 Topic: 9.5. Project evaluation methods with inflation 7 .

Skill: Recall User1: Qualitative 19) Jason inherited $6 000 from his grandfather. He decided to invest the money in a two-year project that promised a return of $2 000 by the end of this year and $4 500 at the end of the next year. Jason read an analytical report that inflation is going to be 3.0% over the next 5 years. He decided that he would only invest in this project if it could give him at least 10% real rate of return. What is the present worth of his investment under this rate of return?

A) $183 B) -$873 C) -$462 D) -$729 E) $5 289 Answer: D Diff: 3 Type: MC Page Ref: 316-318 Topic: 9.5. Project evaluation methods with inflation Skill: Applied User1: Quantitative 20) How should engineers deal with the fact that inflation rates are very difficult to predict?

A) Avoid any projects that involve cash flows occurring more than two years in the future.

B) Obtain estimates of inflation from expert economists.

C) Assume that inflation rates will average out to zero over the project life.

D) Base project analysis on the highest rates of inflation that have occurred in the last ten years.

E) Perform a sensitivity analysis on the profitability of the project for different inflation rates.

Answer: E Diff: 1 Type: MC Page Ref: 325 Topic: 9.5. Project evaluation methods with inflation Skill: Recall User1: Qualitative 21) XWave Ltd. has spent $150 000 for the retraining of its personnel this year, and decided to repeat the retraining every third year. How much will this retraining cost next time if annual inflation is going to be 4% over the next five years?

A) $162 450 B) $168 730 C) $169 780 D) $171 216 E) $174 549 Answer: B Diff: 2 Type: MC Page Ref: 309 Topic: 9.5. Project evaluation methods with inflation Skill: Applied User1: Quantitative

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22) Suppose John deposits some money in an investment account that offers 5% interest rate per year.

Annual inflation is expected to be 3.5% over the next 5 years. What is the inflation rate per half a year?

A) 1.50% B) 1.73% C) 1.75% D) 1.78% E) 1.80% Answer: B Diff: 2 Type: MC Page Ref: 309 Topic: 9.2. Measuring the inflation rate Skill: Applied User1: Quantitative 23) Annual inflation is expected to be 3.5% over the next 5 years. Heather's real MARR is 6%, compounded semiannually. What is the effective actual MARR per half a year?

A) 4.78% B) 5.67% C) 7.12% D) 8.26% E) 9.71% Answer: A Diff: 3 Type: MC Page Ref: 316, 317 Topic: 9.4. The effect of correctly anticipated inflation Skill: Applied User1: Quantitative 24) Heather deposits $8 000 in an investment account that offers 5% interest per year. The interest is due in 5 years. Annual inflation is expected to be 4% over the next 5 years. Heather's actual MARR per sixmonth period is 5%, compounded semiannually. What is the present worth of the interest?

A) $1 357 B) $1 471 C) $1 492 D) $1 504 E) $1 541 Answer: A Diff: 3 Type: MC Page Ref: 316, 317 Topic: 9.5. Project evaluation methods with inflation Skill: Applied User1: Quantitative

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25) A project involves a substantial expenditure right now, and is expected to yield a return of $X ten years in the future. We expect the annual inflation rate to be positive over the next ten years. Four analysts calculate the present worth of the project. Alice bases her calculation on the assumption that X is given in real dollars, using a real MARR. Bob also assumes that X is in real dollars, but uses the actual MARR to calculate present worth. Cecil assumes that X is in actual dollars, and calculates present worth using the real MARR. Donna also assumes that X is in actual dollars, and calculates present worh using the actual MARR. Who comes up with the highest figure for present worth?

A) Alice B) Bob C) Cecil D) Donna E) They will all come up with the same figure.

Answer: C Diff: 2 Type: MC Page Ref: 316-318 Topic: 9.5. Project evaluation methods with inflation Skill: Recall User1: Quantitative 26) If the CPI was 99.6% in January 2001 and 128% in January 2005, what was the approximate annual inflation over the intervening period?

A) 28.4% B) 7.1% C) 6.4% D) 5.0% E) 0.4% Answer: C Diff: 2 Type: MC Page Ref: 309 Topic: 9.2. Measuring the inflation rate Skill: Applied User1: Quantitative 27) Suppose that the CPI was 87.5% in 2000 and 132.2% in 2009. What is the value of $50 000 in 2009 in terms of 2000 dollars?

A) $66 100 B) 57 143 C) $43 750 D) $37 821 E) $33 094 Answer: E Diff: 2 Type: MC Page Ref: 309 Topic: 9.2. Measuring the inflation rate Skill: Applied User1: Quantitative

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28) A four-year project requires $1 000 000 as the first cost and brings $2 000 000 at the end of the fourth year. If expected annual inflation over this period is 5%, what is the real IRR of this project ?

A) 13.3% B) 18.9% C) 27.6% D) 47.6% E) 95.0% Answer: A Diff: 2 Type: MC Page Ref: 309 Topic: 9.4. The effect of correctly anticipated inflation Skill: Applied User1: Quantitative 29) It is known that the inflation rate over next three years is expected to be 5% in the first year, 10% in the second and 15% in the third. In such a case, average expected annual inflation rate is A) 5.00%.

B) 9.92%.

C) 10.00%.

D) 10.08%.

E) 15.00%.

Answer: B Diff: 2 Type: MC Page Ref: 309 Topic: 9.2. Measuring the inflation rate Skill: Applied User1: Quantitative 30) Suppose that you need to accumulate $10 000 in three years in today's dollars. What monthly amount in nominal (actual) dollars should you save if your real rate of return is 5% and expected annual inflation rate is 3%?

A) $246.14 B) $258.04 C) $268.96 D) $275.14 E) $281.96 Answer: C Diff: 3 Type: MC Page Ref: 316-317 Topic: 9.5. Project evaluation methods with inflation Skill: Applied User1: Quantitative

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31) I am promised a payment of $250 a year from now, repeating every year for a total of ten years. I can invest money at 10% interest, but the inflation rate over the next ten years is expected to be 5%. What is the smallest amount of money you could offer me right now that would persuade me to give up these payments?

A) 250(P/A,0.05,10) B) 250(P/A,0.15,10) C) 250(A/P,0.1,10) D) 250(P/A,0.155,10) E) 250(P/A,0.165,10) Answer: D Diff: 3 Type: MC Page Ref: 312 Topic: 9.4. The effect of correctly anticipated inflation Skill: Applied User1: Quantitative 32) The inflation rate is 10%. A project will yield $1 000 in real dollars at the end of two years. How many actual dollars will it yield?

A) $900 B) $909 C) $1 100 D) $1 110 E) Can't tell unless we know the MARR Answer: D Diff: 2 Type: MC Page Ref: 309 Topic: 9.3. Economic evaluation with inflation Skill: Applied User1: Quantitative 33) A bank offers me 10% annual interest on an investment, but the inflation rate is 5%. If I invest $500 in the bank, what will the value of my investment be, in real dollars, in five years time?

A) $396 B) $631 C) $638 D) $805 E) $1 028 Answer: B Diff: 2 Type: MC Page Ref: 309 Topic: 9.3. Economic evaluation with inflation Skill: Applied User1: Quantitative

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34) A bank offers me 10% annual interest on an investment, but the inflation rate is 5%. What is the real interest rate?

A) 2% B) 4.76% C) 5% D) 15% E) 15.5% Answer: B Diff: 2 Type: MC Page Ref: 312 Topic: 9.4. The effect of correctly anticipated inflation Skill: Applied User1: Quantitative 35) Jagdeep is considering a project that requires a large immediate investment and yields a return in five years. If he assumes that there will be be no inflation over the next five years, his calculated rate of return on the project is 25%. If he instead assumes a 10% annual rate of inflation, what will his expected real rate of return be?

A) It will still be 25%.

B) 13.64% C) 15% D) 35% E) 37.5% Answer: B Diff: 2 Type: MC Page Ref: 312 Topic: 9.4. The effect of correctly anticipated inflation Skill: Applied User1: Quantitative 36) I am considering a project which requires an immediate investment of $10 000 and yields a return in two years. If I expect the rate of inflation to be 15%, how big, in actual dollars, must the return be if my real rate of return on the project is to be 20%?

A) $13 500 B) $13 800 C) $14 400 D) $18 225 E) $19 044 Answer: E Diff: 2 Type: MC Page Ref: 316 Topic: 9.5. Project evaluation methods with inflation Skill: Applied User1: Quantitative

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37) I invested a sum of money in a project five years ago. The rate of inflation since then has been 5% per year. Now I receive a return of $5 000 on my original investment. I calculate that my real rate of return on the original investment has been 10%. What was the size of my original investment?

A) $2 435 B) $3 105 C) $3 918 D) $3 962 E) $6 309 Answer: A Diff: 2 Type: MC Page Ref: 316 Topic: 9.5. Project evaluation methods with inflation Skill: Applied User1: Quantitative 9.2 Short Answer Questions 1) John's annual income in 2002 was $52 000. Now in 2011 John earns $63 000 per year. The CPI was 96 in 2002 and is 115 now. In real terms, is John better off now than he was in 2002?

Answer: We can compare the two values of John's income with respect to the base year. In the base year, the CPI is always 100. Therefore, real incomes in 2002 and 2011 with respect to the base year are: Rb,2002 =

= $54 167

Rb,2011 =

= $54 783

Since real income in 2011 is higher, John is marginally better off now.

Diff: 1 Type: SA Page Ref: 309 Topic: 9.3. Economic evaluation with inflation Skill: Applied User1: Quantitative

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2) You want to invest $ 1 000 in a mutual fund that offers a 5% guaranteed rate of return. Current interest rate is 2%, and inflation is expected to be 1% per year. How much will you accumulate in your account in 5 years in terms of today's dollars?

Answer: First, let's calculate the accumulated amount in actual dollars: FW = 1 000 * (F/P, 5%, 5) = 1 000 * 1.27628 = $1 276.28 Now let's calculate the real dollar value of this amount: R0 =

=

= $1 214.33

Another approach is to calculate the real rate of return RRR as follows: RRR =

- 1 = 0.0396: the real rate of return is 3.96%. The future worth of $1 000 under the RRR

is: FW = 1 000 * (F/P, 3.96%, 5) = 1 000 * (1 + 0.0396)5 = $1 214.31 Diff: 2 Type: SA Page Ref: 309, 312 Topic: 9.3. Economic evaluation with inflation Skill: Applied User1: Quantitative 3) An engineering project requires $20 000 as first cost and has a planning horizon of five years.

Operating costs are $2 000 per year, and expected annual revenue is $8 000. Calculate the project's present worth if the real interest rate is 5%, the expected annual inflation is 2% and the salvage value is $5 000.

Answer: All dollar values are expressed in nominal (actual) terms while interest rate is the real interest.

Therefore, it is necessary to calculate the actual interest rate, as follows: iA = iR + f + iR * f = 0.05 + 0.02 + 0.05 * 0.02 = 0.071 or 7.1% PW(Project) = 8 000 * (P/A, 7.1%, 5) + 5 000 * (P/F, 7.1%, 5) - 20 000 - 2 000 * (P/A, 7.1%, 5) = 6 000 * (P/A, 7.1%, 5) + 5 000 * (P/F, 7.1%, 5) - 20 000 = 6 000 * 4.0892 + 5 000 * 0.70966 - 20 000 = 24 535 + 3 548 - 20 000 = $8 083 Diff: 2 Type: SA Page Ref: 316-318 Topic: 9.4. The effect of correctly anticipated inflation Skill: Applied User1: Quantitative

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4) Explain how Statistics Canada calculates the inflation rate.

Answer: Statistics Canada tracks the movement of average prices for a number of different sets of goods and services and calculates inflation rates based on the change in prices in these sets over time.

One set of prices monitored by Statistics Canada consists of goods and services bought by Canadian consumers. This set forms the basis for the consumer price index (CPI). The CPI for a given year relates the average price of a fixed "basket" of these goods and services in a given year to the average price of the same basket in some base year. The base year index is set at 100. Since the CPI is expressed as a percentage, the change in the CPI with respect to previous year is the inflation rate.

Mathematically: f=

x 100%

Diff: 1 Type: SA Page Ref: 306-308 Topic: 9.2. Measuring the inflation rate Skill: Recall User1: Qualitative 5) How does inflation affect the evaluation of engineering projects?

Answer: Inflation is an increase in average price level in an economy. It complicates the evaluation of engineering projects because it adds an additional factor to the evaluation of the present worth of future cash flows. Overall, there are two factors at work: - The time value of money captured by the interest rate; - The purchasing power of money captured by the inflation rate.

Both must be taken into account while evaluating an engineering project.

Diff: 2 Type: SA Page Ref: 316-318 Topic: 9.1. Introduction Skill: Recall User1: Qualitative 6) How is the consumer price index calculated and what does it show?

Answer: The consumer price index (CPI) for a given year relates the average price of a fixed basket of consumption goods and services to the average price of the same basket in a base year expressed in a percentage form. First, the cost of such a basket in the base year and in the current year are calculated, then the cost in the current year is divided by the cost in the base year, and then the ratio is multiplied by 100%.

Diff: 1 Type: SA Page Ref: 300 Topic: 9.2. Measuring the inflation rate Skill: Recall User1: Qualitative

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7) ROVNO Ltd. just invested $50 000 into a new local network. Services provided by the network will bring $15 000 per year in the next five years. The following information is given: - real MARR = 10% - expected annual inflation = 2% - the network will be sold for salvage after five years; its depreciation rate is 20% Is this a good investment in terms of the internal rate of return?

Answer: First, it is necessary to calculate the salvage value of the network at the end of its useful life: S = BVdb(5) = P * (1 - d)5 = 50 000 * (1 - 0.2)5 = $16 384 Since all cash flows are expressed in current (actual) dollars, it is necessary to calculate the actual MARR: MARRA = MARRR + f + MARRR x f = 0.1 + 0.03 + 0.1 * 0.03 = 0.133 or 13.3% This is the rate we have to compare the IRR with. The actual IRR can be found from the following equation: 15 000 * (P/A, IRR, 5) + 16 384 * (P/F, IRR, 5) = 50 000 Actual IRR = 0.211 or 21.1%. Since the actual IRR exceeds actual MARR, this is a good investment.

Diff: 3 Type: SA Page Ref: 316-318 Topic: 9.4. The effect of correctly anticipated inflation Skill: Applied User1: Quantitative

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8) Canada's government savings bonds pay 3% annual real rate of return compounded daily. Professor John Jones invests $40 bi-weekly as a part of his University payroll program. If the inflation rate is assumed to be 2% per year over the next 10 years, what is the purchasing power of the professor's bonds 10 years from now?

Answer: First, it is necessary to calculate actual annual rate of return (ARR) as follows ARRA = ARRR + f + ARRR x f = 0.03 + 0.02 + 0.03 * 0.02 = 0.0506 or 5.06% Now it is necessary to calculate the effective actual bi-weekly rate of return: )14 - 1 = 0.00194 or 0.194%

ARRA = (1 +

So the future worth of the professor's bonds in actual dollars is FW = 40 * (F/A, 0.2%, 260) = 40 * 340.577 = $13,623.09 This amount in real dollars is R0 =

=

= $11,175.68

which is the purchasing power of the professor's bonds 10 years from now Diff: 3 Type: SA Page Ref: 309, 316-318 Topic: 9.3. Economic evaluation with inflation Skill: Applied User1: Quantitative

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9) Suppose that an economy is producing only two goods—a consumption good C and an investment good I. Using the following data:

C-good I-good

Price 1990 4 4

Quantity 1990 50 100

Price 2000 5 4

Quantity 2000 60 110

Price 2004 6 6

Quantity 2004 70 120

Calculate the CPI in 2000 and 2004 with 1990 as the base year.

Answer: It is necessary to define the cost of basket (50 units of C and 100 units of I) in 1990, 2000 and 2004.

1990: $40 * 50 + $4 * 100 = $600 2000: $50 * 50 + $5 * 100 = $750 2004: $60 * 50 + $6 * 100 = $900 Now it is possible to calculate the consumer price index in 1990, 2000 and 2004: CPI1990 = 100 CPI2000 =

x 100 = 125

CPI2004 =

x 100 = 150

Diff: 2 Type: SA Page Ref: 306-308 Topic: 9.2. Measuring the inflation rate Skill: Applied User1: Quantitative 10) Explain the difference between an increase in average price level and an increase in relative prices in economic evaluation of engineering projects.

Answer: An increase in average price level is inflation, and it is expressed in terms of the inflation rate.

The inflation rate is then used in project evaluation to adjust the interest rate to calculate the project's PW, AW and FW. An increase in relative prices is an increase in the price of a specific good or service relative to the average price level. Changes in relative prices will generally not have a noticeable effect on the interest rate. Therefore, an analyst must incorporate expected increase in relative prices directly via growth rate. This can often be represented as a geometric gradient series.

Diff: 2 Type: SA Page Ref: 317 Topic: 9.5. Project evaluation methods with inflation Skill: Recall User1: Qualitative

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11) The purchase of a car requires a $25 000 loan to be repaid in monthly installments for four years at 12% annual real interest rate. If annual inflation rate is 4%, find the amount paid, in actual dollars, in the 20th month.

Answer: First, it is necessary to calculate the monthly payment in real dollars. It is: $25 000 * (A/P, 12%/12, 48) = 25 000 * 0.026333835 = $658.35 This amount in actual dollars in the 20th month is: $658.35 * (F/P, 4%/12, 20) = 658.35 * 1.0688 = $703.66 Diff: 3 Type: SA Page Ref: 309 Topic: 9.5. Project evaluation methods with inflation Skill: Applied User1: Quantitative 12) John borrowed $5 000 from a commercial bank for 4 years to pay his tuition fee. Calculate the interest owed on this loan after 4 years in real dollars if the bank charges 5% annual (actual) interest rate and annual inflation is 2%.

Answer: 5% charged by the bank is actual (nominal) interest rate which already includes future inflation. Real interest rate is -1 = 0.0294 or 2.94%

Therefore, the accumulated interest in real dollars will be: 5 000 * (1 + 0.0294)4 - 5 000 = $614.44 Diff: 2 Type: SA Page Ref: 309 Topic: 9.5. Project evaluation methods with inflation Skill: Applied User1: Quantitative 13) The price of oil is rising at 10% per year while inflation is just 2.5% per year. If the current price of oil is $45 per barrel, what would be the price of oil ten years from now in real dollars?

Answer: We assume that the 10% annual increase in the price of oil is an increase in its actual-dollar price. So in ten years, its actual price will be 45 * 1.110 = $116.71 / barrel The real-dollar equivalent is 116.71/(1.025)10 = $91.18 Diff: 2 Type: SA Page Ref: 317 Topic: 9.5. Project evaluation methods with inflation Skill: Applied User1: Quantitative

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14) A painting cost $1 million in 1980. Now in 2011 its price is $5 million. Assuming average annual inflation of 4% over this period, has the painting increased/decreased in value and by how much?

Answer: Current value of $5 million in 1980 dollars is: =$1.48 million

Therefore, this is a 48% increase in the value of the painting Diff: 1 Type: SA Page Ref: 309 Topic: 9.3. Economic evaluation with inflation Skill: Applied User1: Quantitative 15) A 10-year project is evaluated under two scenarios: (i) with inflation, and (ii) without inflation.

Without inflation, and with a 10% interest rate, its present worth is $24 254. If annual inflation of 2% is assumed, then the project's present worth becomes 20% lower. It is also known that the project's first cost involves the purchase of a capital asset which depreciates at 13% and will be sold for salvage at the end of the project, there are no operating costs, and there is a fixed annual revenue. What is the project's first cost?

Answer: If we define first cost as P and annual revenue as A then the present worth of the project without inflation can be written as follows: -P + A * (P/A, 10%, 10) + P * (1 - 0.13)10*(P/F, 10%, 10) = 24 254 When inflation is present, we need to calculate actual interest rate: 0.1 + 0.02 + 0.1 * 0.02 = 0.122 or 12.2% 20% reduction in the project's PW produces 0.8*24 254 = $19 403.20 Therefore, the PW of the project with inflation can be written as: -P + A * (P/A, 12.2%, 10) + P * (1 - 0.13)10 * (P/F, 12.2%, 10) = 19 403.20 As a result, we have two equations with two unknowns P and A. Solving them as a system, we can find that P = $28 103.30 Diff: 3 Type: SA Page Ref: 316-318 Topic: 9.5. Project evaluation methods with inflation Skill: Applied User1: Quantitative

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Engineering Economics, 5e (Fraser) Chapter 10 Public Sector Decision Making 10.1 Multiple Choice Questions 1) Engineering decisions associated with public projects should be based on A) market costs and benefits.

B) environmental costs and benefits.

C) marginal willingness to pay.

D) marginal costs and benefits.

E) social costs and benefits.

Answer: E Diff: 1 Type: MC Page Ref: 348-350 Topic: 10.1. Introduction Skill: Recall User1: Qualitative 2) In cost-benefit analysis associated with public projects, all costs and benefits should be evaluated in terms of A) market values.

B) social values.

C) marginal willingness to pay.

D) market supply and demand.

E) profitability.

Answer: B Diff: 2 Type: MC Page Ref: 348-350 Topic: 10.3. Decision making in the public sector Skill: Recall User1: Qualitative 3) Which of the following is an advantage of the modified benefit-cost ratio as compared with the conventional benefit-cost ratio?

A) It includes social costs.

B) It includes the effects of inflation.

C) It provides a measure of the net gain per dollar invested by the project sponsor.

D) It is easier to calculate.

E) It includes the effects of taxation.

Answer: D Diff: 2 Type: MC Page Ref: 357 Topic: 10.3. Decision making in the public sector Skill: Recall User1: Qualitative

1 .

4) What economic conditions underlie market failures?

A) External costs are greater than private costs.

B) There is environmental damage which is addressed by environmental tax.

C) There is a monopolist who sets the price equal to the marginal cost.

D) All costs are born by market participants.

E) Aggregate benefits to all persons are less than aggregate costs imposed.

Answer: E Diff: 2 Type: MC Page Ref: 347 Topic: 10.2. Market failures Skill: Recall User1: Qualitative 5) The hedonic price method is a way of setting a value on a social good based on: A) how much people say they value that good when surveyed.

B) how much an ideally rational individual would pay for that good.

C) the popularity of a political party that advocates that good.

D) how much it would cost the government to provide a substitute for that good.

E) how much people are prepared to pay to acquire that good.

Answer: E Diff: 3 Type: MC Page Ref: 355 Topic: 10.3. Decision making in the public sector Skill: Recall User1: Qualitative 6) Which of the following is an example of a remedy for a market failure?

A) building a refinery B) building a hospital C) building an assembly plant D) building an apartment complex E) building a movie theatre Answer: B Diff: 3 Type: MC Page Ref: 348-349 Topic: 10.2. Market failures Skill: Applied User1: Qualitative 7) How should market failures be addressed?

A) through the establishment of barriers to enter the market B) through the replacement of social equilibrium by market equilibrium C) through free market interactions between consumers and producers D) through government intervention and the courts E) through market deregulations Answer: D Diff: 1 Type: MC Page Ref: 348 Topic: 10.2. Market failures Skill: Recall User1: Qualitative 2 .

8) What is contingent valuation?

A) assigning a value to an intangible good based on the results of surveys B) assigning a value to an intangible good based on how much people will pay to obtain it C) assigning a value to an intangible good based on its historical price D) assigning a value to an intangible good based on its price in an ideal free market E) assigning a value to an intangible good based on what it would cost to provide a substitute for it Answer: A Diff: 2 Type: MC Page Ref: 354 Topic: 10.3. Decision making in the public sector Skill: Recall User1: Qualitative 9) An environmental tax is an example of A) a policy regulation.

B) a government takeover.

C) a financial incentive or deterrent.

D) a legal response.

E) a market failure.

Answer: C Diff: 1 Type: MC Page Ref: 348-349 Topic: 10.2. Market failures Skill: Recall User1: Qualitative 10) One of the following is a fundamental difference between financial analysis and BCA. Which one?

A) Inflation is not included in BCA. B) Tax effects are not included in BCA. C) Time value of money is not reflected in BCA. D) Operating costs are not included in BCA. E) Capital costs are not included in BCA. Answer: B Diff: 2 Type: MC Page Ref: 349-356 Topic: 10.3. Decision making in the public sector Skill: Recall User1: Qualitative 11) When a community puts pressure on a municipal government to preserve the municipal park, this is an example of A) regulation.

B) command and control policies.

C) monetary incentives.

D) litigation.

E) informal methods.

Answer: E Diff: 2 Type: MC Page Ref: 348-349 Topic: 10.2. Market failures Skill: Recall User1: Qualitative 3 .

4 .

12) What is the name for a market dominated by a single provider?

A) monopolistic competition B) perfect competition C) oligopoly D) monopoly E) market failure Answer: D Diff: 2 Type: MC Page Ref: 350 Topic: 10.2. Market failures Skill: Recall User1: Qualitative 13) All of the following except one are reasons for monopolies to be regulated. Which is the exception?

A) Monopolies are guaranteed to make economic profits in the long run.

B) Monopolies are inefficient.

C) The existence of monopolies may lead to a more unequal distribution of benefits.

D) Monopolists produce below efficient level of production.

E) Monopolies charge excessive prices.

Answer: A Diff: 2 Type: MC Page Ref: 350 Topic: 10.2. Market failures Skill: Recall User1: Qualitative 14) "Although it goes without saying that all relevant costs and benefits must be taken into account, the number of potential mistakes in applying the principle [of Account Comprehensively for Benefits and Costs] is surprisingly large. In the transit domain, for example, studies ... fail to assign economic value to ________." (HLB Decision Economics Inc., 2004).

A) operating costs B) reduction in risk of traffic accidents C) revenues collected D) capital costs E) taxes paid Answer: B Diff: 3 Type: MC Page Ref: 346-348 Topic: 10.2. Market failures Skill: Applied User1: Qualitative 15) One of the following is a fundamental limitation of BCA. Which one?

A) BCA does not take into account time value of money.

B) BCA ignores marginal principle.

C) BCA is a partial equilibrium analysis.

D) BCA uses market values.

E) BCA uses market interest rate.

Answer: B Diff: 2 Type: MC Page Ref: 349-356 Topic: 10.3. Decision making in the public sector 5 .

Skill: Recall User1: Qualitative 16) The point of view used in Benefit-Cost Analysis is associated with A) market participants.

B) social groups.

C) shareholders of a business.

D) the government.

E) consumers of a good or service.

Answer: B Diff: 1 Type: MC Page Ref: 350-352 Topic: 10.3. Decision making in the public sector Skill: Recall User1: Qualitative 17) Total benefits of a public project to a society A) include the value of all goods and services that result from a project, and effects of a project can be either positive or negative.

B) include the value of the goods and services directly produced by a project, and effects of a project can be either positive or negative.

C) include the value of all goods and services that result from a project, and effects of a project are generally positive.

D) include the value of the goods and services directly produced by a project, and effects of a project are generally positive.

E) include only feasible goods and services resulted from a project, and effects of a project can be either positive or negative.

Answer: A Diff: 2 Type: MC Page Ref: 353 Topic: 10.3. Decision making in the public sector Skill: Recall User1: Qualitative 18) Which of the following is viewed as direct costs?

A) depreciation B) advertising C) insurance D) costs of raw materials E) registration Answer: D Diff: 2 Type: MC Page Ref: 353 Topic: 10.3. Decision making in the public sector Skill: Recall User1: Qualitative

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19) Which of the following is an example of an intangible benefit?

A) improved technology B) improved custom service C) improved quality of a good D) improved safety E) improved design Answer: D Diff: 1 Type: MC Page Ref: 354 Topic: 10.3. Decision making in the public sector Skill: Recall User1: Qualitative 20) A study shows that widening a road shoulder by 0.3 metres results in a 10% decrease in fatal traffic accidents. Trying to improve the road safety of the Provincial part of a federal highway, a Provincial Department of Transportation widened the highway shoulders by 0.9 metres. Based on the data provided by the Transportation Safety Board (1997) in the table below, calculate the fatal accident cost savings that resulted from this project. Assume that the Provincial Department of Transportation used to record 10 fatalities on average per year on this part of the highway.

Severity of Accidents Fatalities Injuries (Road) Property damage (Road)

Occurrences per year 3 294 230 885 670 000

Compensation Cost per occurrence $1 560 000 $28 000 $5 600

A) $1.56 million B) $3.12 million C) $4.68 million D) $6.24 million E) $7.80 million Answer: C Diff: 2 Type: MC Page Ref: 353-356 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative 21) A 10-year public project that required $1.5 million generates annual revenue of $0.2 million beginning from year 1. What is the project's rate of return on its investment?

A) 5.6% B) 6.6% C) 7.6% D) 8.6% E) 9.6% Answer: A Diff: 2 Type: MC Page Ref: 353-355 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative 7 .

22) Nova Scotia government decided to reopen a gypsum mine site. A marketing study shows that if gypsum is sold locally, which requires a purchase of a fleet of trucks for hauling, it would generate $10.11 million in revenues over a 10-year period. If gypsum is hauled by rail to Ontario and sold there, it would generate $10.19 million in revenues over a 10-year period. It is calculated that the purchase of the truck fleet costs $7.1 million, operation and maintenance of the truck fleet require $2.4 million over a 10-year period (Alternative A). Rail haul option costs $9.6 million over a 10-year period (Alternative B). What is the benefit-cost ratio associated with incremental investment and what transportation option is preferred?

A) The benefit cost ratio in terms of incremental costs and benefits is 1.4, thus rail is preferred to trucking.

B) The benefit cost ratio in terms of incremental costs and benefits is 1.2, thus rail is preferred to trucking.

C) The benefit cost ratio in terms of incremental costs and benefits is 1, thus both options are identical.

D) The benefit cost ratio in terms of incremental costs and benefits is 0.7, thus trucking is preferred to rail.

E) The benefit cost ratio in terms of incremental costs and benefits is 0.8, thus trucking is preferred to rail.

Answer: E Diff: 3 Type: MC Page Ref: 359 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative 23) A municipality has just spent $8 million to build a second bridge across the city river to handle the increased traffic between two sides of the river. City engineer calculated that the bridge construction would allow a reduction in the travel time for the road users over the 25-year period of the bridge's service life with a present value of $15.5 million. The construction would lead to a decrease in the present value of surrounding property by $6.4 million over the bridge's service life. The present worth of the operation and maintenance of the bridge over its service life is $3.5 million. What is the benefit-cost ratio of the project?

A) 0.87 B) 0.96 C) 1.08 D) 1.32 E) 1.94 Answer: A Diff: 1 Type: MC Page Ref: 357 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative

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24) Which of the following effects should not be considered in BCA? A) direct benefits of the project B) changes in opportunities as a result of the project C) changes in income distribution D) changes in supply and demand E) changes in taxation Answer: E Diff: 3 Type: MC Page Ref: 352-356 Topic: 10.3. Decision making in the public sector Skill: Recall User1: Qualitative 25) Which of the following requires the use of BCA instead of financial analysis?

A) new sewage system B) new oil refinery C) new computer network D) new shopping mall E) new hockey arena Answer: A Diff: 2 Type: MC Page Ref: 350 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Qualitative 26) The conventional benefit-cost ratio of a public project reflects A) the future worth of net social benefits per $1 of the future worth of social costs.

B) the future worth of social benefits per $1 of the future worth of social costs.

C) the present worth of net benefits per $1 of the present worth of the project's costs.

D) the present worth of social benefits per $1 of the present worth of social costs.

E) the future worth of social benefits per $1 of the present worth of social costs.

Answer: C Diff: 1 Type: MC Page Ref: 357 Topic: 10.3. Decision making in the public sector Skill: Recall User1: Qualitative

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27) There are four mutually exclusive alternatives to build a rural road. Present worth of social benefits and costs as well as benefit-cost ratios are presented in the following table: ALTERNATIVE A B C D

PW(BENEFITS) 100 150 170 200

PW(COSTS) 80 85 100 150

BCR 1.25 1.76 1.70 1.33

Based on this information, what alternative should be chosen?

A) A B) B C) C D) D E) None Answer: C Diff: 3 Type: MC Page Ref: 359 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative 28) Suppose that as a result of a public project evaluation the following information was obtained:- PW of the project's capital costs = $9 million- PW of the project's operating costs = $5 million- PW of the project's benefits = $20 millionWhat is the BCRM for this project?

A) 0.42 B) 0.67 C) 1.07 D) 1.67 E) 2.20 Answer: D Diff: 2 Type: MC Page Ref: 358 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative 29) One of major drawbacks of the benefit-cost ratios is A) They are difficult to calculate.

B) Social benefits are not well defined.

C) Social costs are not well defined.

D) Ambiguity associated with a project's disbenefits.

E) Private benefits are not well defined.

Answer: D Diff: 2 Type: MC Page Ref: 361 Topic: 10.3. Decision making in the public sector Skill: Recall User1: Qualitative

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30) When choosing the appropriate interest rate to evaluate a public project, which of the following should be used?

A) current market interest rate B) the rate of return on private investment C) social discount rate D) private sector MARR E) external rate of return Answer: C Diff: 2 Type: MC Page Ref: 363 Topic: 10.3. Decision making in the public sector Skill: Recall User1: Qualitative 31) A city council is planning to build a new bridge across a creek. This will require an immediate expenditure of $50 000, but will save citizens an estimated $5 000 in travel costs every year over the next twenty years. If the council's MARR is 5%, what is the BCR for the project?

A) 0.1 B) 1.25 C) 1.44 D) 1.64 E) 2.00 Answer: B Diff: 2 Type: MC Page Ref: 357 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative 32) A city council is planning to build a new bridge across a creek. This will require an immediate expenditure of $50 000, and an estimated $2 000 a year in maintenance, but will save citizens an estimated $5 000 in travel costs every year over the next twenty years. If the council's MARR is 5%, what is the BCR for the project?

A) 0.83 B) 0.95 C) 1.20 D) 1.25 E) 2.5 Answer: A Diff: 2 Type: MC Page Ref: 357 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative

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33) A city council is planning to build a new bridge across a creek. This will require an immediate expenditure of $50 000, and an estimated $2 000 a year in maintenance, but will save citizens an estimated $5 000 in travel costs every year over the next twenty years. If the council's MARR is 5%, what is the BCRM for the project?

A) 0.75 B) 0.83 C) 1.25 D) 1.34 E) 2.00 Answer: A Diff: 2 Type: MC Page Ref: 357 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative 34) A city council is planning to build a new bridge across a creek. This will require an immediate expenditure of $50 000, but will save citizens an estimated $5 000 in travel costs every year over the next twenty years. Alternatively, they could renovate an existing bridge. This would only cost $20 000, but the saving in travel costs would only be $2 000 every year. If the council's MARR is 5%, what is the BCR for upgrading from the renovated bridge to the new bridge?

A) 0.75 B) 1.20 C) 1.25 D) 1.64 E) 2.00 Answer: C Diff: 2 Type: MC Page Ref: 359 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative

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35) The following mutually exclusive projects are available. Which should be chosen?

Project A B C D E

PW(Costs) 1 500 000 1 800 000 900 000 2 700 000 2 100 000

PW(Benefits) 2 400 000 1 900 000 1 800 000 4 200 000 1 900 000

A) A B) B C) C D) D E) E Answer: D Diff: 2 Type: MC Page Ref: 357 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative

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10.2 Short Answer Questions 1) The provincial department of transportation is considering investing in a provincial highway. Five mutually exclusive alternatives were identified. The present worths of their expected costs and benefits were calculated, and the data are summarized in the following table: Present worth of benefits, $M 3 10 13 18 25

Alternative A B C D E

Present worth of costs, $M 1 5 9 14 19

Calculate the BCR for each alternative, the incremental BCR and identify the best alternative.

Answer: By definition, BCR =

and therefore, the following table summarizes the BCR calculations

for these alternative:

Alternative A B C D E

Present worth of Present worth of benefits, $M costs, $M 3 1 10 5 13 9 18 14 25 19

BCR 3.00 2.00 1.44 1.29 1.32

Since our alternatives are already ranked in terms of costs, we can compare the alternatives by pairs: B against A: BCR =

=

= 1.75. Since incremental BCR > 1, alternative B becomes

=

= 0.75. Since incremental BCR < 1, alternative B remains the

=

= 0.89. Since incremental BCR < 1, alternative B is still the

=

= 1.07. Since incremental BCR > 1, alternative E is the best.

current best.

C against B: BCR = current best D against B: BCR = best.

E against B: BCR =

14 .

Although alternative A has the highest BCR ratio, nevertheless the most efficient use of public resources in this case is alternative E, which should be chosen.

Diff: 2 Type: SA Page Ref: 356-359 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative 2) The Department of Public Works has $1.1 million to allocate amongst several public projects. Data on these projects are presented in the following table:

Project A B C D E F

First cost, $K 400 300 250 500 350 200

Annual benefit, $K/year 100 100 80 70 90 60

Life, years 10 25 15 40 10 20

Using a MARR of 15%, rank the projects by their BCR, and comment on the most efficient way to allocate $1.1 million.

Answer: In this case, the benefit-cost ratio BCR of any project can be calculated as follows: BCR = where Ba is the annual benefit, and Cf is the first cost of a project. The results of calculations are presented below:

Project A B C D E F

First cost, $K 400 300 250 500 350 200

PW of benefits, $K 502 646 468 465 452 376

BCR 1.26 2.15 1.87 0.93 1.29 1.88

According to the BCR, the projects should be ranked as follows: 1.

2.

3.

4.

5.

B with first cost of $300K F with first cost of $200K C with first cost of $250K E with first cost of $350K A with first cost of $400K 15 .

6.

D with first cost of $500K

The first four projects—B, F, C and E—have a combined first cost of exactly $1.1 million. Therefore, the money should be assigned to these projects.

Diff: 2 Type: SA Page Ref: 356-359 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative 3) The City of Angels is considering two transportation projects to address its traffic congestion. The first one is a light-rail system while the second one is a new highway. Some economic data on both are presented below:

Project Light-rail system Highway

First cost, million $ 40 70

Operating cost, million$/year 8 6

In addition, the expected benefits from reduction in air pollution are $20 million per year in the case of the light-rail system and $21 million per year in the case of the highway. Service lives of the two alternatives are 50 years. Assuming a 10% MARR as suggested by the Canadian Treasury Board for public projects, which project is preferred? Does your conclusion change under 5% and 15% MARR? Answer: Let us calculate the present worth of costs and benefits associated with the two alternatives.

Light-rail system: PW(Benefits) = 20 * (P/A, 10%, 50) = 20 * 9.9148 = $198.3 M PW(Costs) = 8 * (P/A, 10%, 50) + 40 = 8 * 9.9148 + 40 = $119.3 M The benefit-cost ratio is BCRLR =

= 1.66

Highway: PW(Benefits) = 21 * (P/A, 10%, 50) = 21 * 9.9148 = $208.2 M PW(Costs) = 6 * (P/A, 10%, 50) + 70 = 6 * 9.9148 + 70 = $129.5 M The benefit-cost ratio is BCRHW =

= 1.61

The incremental cost of upgrading from the light-rail to the highway is $10.2M, and the incremental benefit is $9.9M, so the BCR for the upgrade is less than 1. Therefore we should choose the light rail system.

Now let us repeat the calculations under 5% MARR: Light-rail system: PW(Benefits) = 20 * (P/A, 5%, 50) = 20 * 18.2559 = $365.1M PW(Costs) = 8 * (P/A, 5%, 50) + 40 = 8 * 8.2559 + 40 = $186 M 16 .

The benefit-cost ratio is BCRLR =

= 1.96

17 .

Highway: PW(Benefits) = 21 * (P/A, 5%, 50) = 21 * 18.2559 = $383.4 M PW(Costs) = 6 * (P/A, 5%, 50) + 70 = 6 * 18.2559 + 70 = $179.5 M The benefit-cost ratio is BCRHW =

= 2.14

The incremental cost of upgrading from the highway to the light rail system is $6.5M, and the incremental benefit is -$18.3M, so the BCR for the upgrade is less than 1. Therefore we should choose the highway.

Finally let us repeat calculations under 15% MARR: Light-rail system: PW(Benefits) = 20 * (P/A, 15%, 50) = 20 * 6.6605 = $133.2 M PW(Costs) = 8 * (P/A, 15%, 50) + 40 = 8 * 6.6605 + 40 = $93.3 M The benefit-cost ratio is BCRLR =

= 1.43

Highway: PW(Benefits) = 21 * (P/A, 15%, 50) = 21 * 6.6605 = $139.9 M PW(Costs) = 6 * (P/A, 15%, 50) + 70 = 6 * 6.6605 + 70 = $110 M The benefit-cost ratio is BCRHW =

= 1.27

The incremental cost of upgrading from the light rail system to the highway is $16.7M, and the incremental benefit is $6.7M, so the BCR for the upgrade is less than 1. Therefore we should choose the light-rail system.

Again the light-rail system is preferred. It appears that the final decision is sensitive to the MARR. Diff: 2 Type: SA Page Ref: 349-356 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative

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4) A new public arena costs $60 million. The largest portion of this amount, $40 million, is for the land acquisition and major earthworks with infinite service life. The remaining $20 million is for the arena itself, which will be in operation for 20 years. Maintenance of the arena requires $2 million per year.

Assuming a 5% MARR, what annual benefit is required for the benefit cost ratio to be greater than 1?

Answer: In this case it is convenient to express the BCR in terms of annual worth as follows: BCR =

≥ 1 or AW(Benefits) ≥ AW(Costs).

The annual worth of costs is as follows: AW(Cost) = 40 *0.05 + 20 *(A/P, 5%, 20) + 2 = 2 + 20 *0.08 + 2 = $5.6 million So, $5.6 million of benefit per year is required for the project to be accepted.

Diff: 3 Type: SA Page Ref: 349-356 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative 5) A bridge costs $450 000 to build and $12 000 per year to maintain. The bridge is repaved every 10 years for $300 000. Assume that the bridge will last for 60 years and that the annual interest rate is 8%.

How many crossings are needed daily to justify the bridge if each user's willingness to pay for the crossing is $2?

Answer: In order to justify the bridge, the annual benefits should be equal or greater than annual costs.

Therefore, in terms of the annual worth, it is possible to set the following inequality: $2 * 365 * N ≥ AW(Costs) where N is the number of crossings per day.

The annual worth of the costs is: AW(Cost) = 450 000 * (A/P, 8%, 60) + 12 000 + 300 000 * A/F, 8%, 10) = 450 000 * 0.08 + 12 000 + 300 000 * 0.069 = $68 700 Therefore: N≥

= 94 or minimum 94 crossings per day are required to justify the bridge.

Diff: 2 Type: SA Page Ref: 349-356 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative

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6) What is a market failure and why is this concept so important in public decision-making?

Answer: A market failure occurs when the cost of a resource is not accurately represented by its market price. For example, a factory may produce pollution which destroys the ability of a river to support fish, but the factory owners may not have to pay for this cost; instead, the cost is born by fishers who can no longer fish in the river. One of the responsibilities of government is to correct market failures by legislation or by taxes and incentives.

Diff: 1 Type: SA Page Ref: 346-348 Topic: 10.2. Market failures Skill: Recall User1: Qualitative 7) A government and a large corporation might each consider building a nuclear reactor to provide power to a city. What similarities and differences might there be in how they evaluated the project?

Answer: The corporation is responsible to its shareholders, whereas the government is responsible to all the citizens. So although both the corporation and the government would consider factors such as the time value of money, the probable lifetime of the reactor, and the expected prices of electricity and uranium, the government would also need to take into account the effect on employment in the city, and the comparative health effects of nuclear power generation compared to the possible alternatives such as fossil fuels.

Diff: 2 Type: SA Page Ref: 353 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Qualitative

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8) A provincial program is subsidizing programs for alternative energy usage. Suppose there are two programs to consider: (i) a system that uses solar energy, and (ii) a system that uses wind energy. Each system lasts 25 years and has the same expected benefits. Economic data on both systems are presented below: Indicator First cost Provincial subsidy Annual maintenance cost

Solar System $6 000 40% $150

Wind system $2 000 0% $400

If the MARR is 5%, which system is preferred from the viewpoint of a private producer and from the viewpoint of a public decision-making?

Answer: Since benefits are the same, let us calculate the annual worth of costs for each of the two systems under two scenarios.

Private producer: 40% provincial subsidy for the solar system should be taken into account: AW(Solar) = 0.6 * 6 000 * (A/P, 5%, 25) + 40 = 3 600 * 0.071 + 150 = $405.6 AW(Wind) = 2 000 * (A/P, 5%, 25) = 2 000 * 0.071 + 400 = $542 Since AW(Solar) < AW(Wind), the solar system is preferred from a standpoint of the private producer.

Public decision-making: Subsidy should not be taken into account AW(Solar) = 6 000 * (A/P, 5%, 25) + 40 = 6 000 * 0.071 + 150 = $576 AW(Wind) = 2 000 * (A/P, 5%, 25) = 2 000 * 0.071 + 400 = $542 From a public decision-making perspective, the wind system is preferred.

Diff: 2 Type: SA Page Ref: 352-353 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative

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9) A provincial government's Department of Tourism has estimated that initial investment into a new permanent recreational park in the province is $2 million, and annual upkeep costs are $100 000. As well, a survey of potential visitors detected a $330 000 annual value of the park via willingness to pay measure. However, an increased inflow of tourists into the area will cause $40 000 per year of extra air pollution. Assuming a social discount rate of 10% per year, calculate (a) the conventional BCR and (b) the modified BCR. Answer: Let us identify the project's costs and benefits first: Costs of the project are: - initial investment of $2 million - upkeep costs of $100 000/year - environmental damage of $40 000/year Benefits of the project: - value of the park to visitors of $330 000/year Since three out of four are expressed as annual values, it is easier to calculate the BCRs via annual worth: AW(Costs) = 2 000 000 * 0.1 + 100 000 + 40 000 = 200 000 + 100 000 + 40 000 = $340 000 AW(Benefits) = $330 000 The benefit-cost ratios are: BCR = BCRM =

= 0.97 = 0.95

Both ratios show that the project is a bad investment.

Diff: 2 Type: SA Page Ref: 357 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative 10) What are the fundamental differences between financial analysis and social benefit-cost analysis?

Answer: There are four basic differences between the two. First, financial analysis is performed from a standpoint of a company's shareholders while social benefit-cost analysis (BCA) is performed from the viewpoint of the public. Second, financial analysis uses market values to calculate cash flows while social BCA uses social values. Third, financial analysis uses market interest rate to capture the time value of money while social BCA uses a social discount rate. Fourth, tax effects should be reflected in financial analysis and not in social BCA. Diff: 3 Type: SA Page Ref: 349-352 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Qualitative 22 .

11) Give at least two reasons for public decision making Answer: Public decision making is associated with goods and services that cannot be efficiently provided through free market. For example, national defence. In such a case there is no market to evaluate these services, and instead social values should be defined as the basis of public decision making. The other reason is market failures. Private markets cannot take them into account, and government intervention is required to correct this situation.

Diff: 2 Type: SA Page Ref: 341-346 Topic: 10.1. Introduction Skill: Recall User1: Qualitative 12) Suppose than a public project has the following profile: Years 0 1 2 3 4

Costs, million $$ 10 10 5 5 8

Benefits, million $$ 0 0 0 10 20

Assuming a social discount rate of 5%, calculate the project's BCR. Answer: The present worth of the costs and benefits can be calculated as follows: PW(Costs) = 10 + 10 * (P/F, 5%, 1) + 5 * (P/F, 5%, 2) + 5 * (P/F, 5%, 3) + 8 * (P/F, 5%, 4) = 10 + 9.524 + 4.535 + 4.319 + 6.582 = $34.960 million PW(Benefits) = 10 * (P/F, 5%, 3) + 20 * (P/F, 5%, 4) = 8.638 + 16.454 = $25.092 million The conventional benefit-cost ratio is BCR =

= 0.72

Diff: 1 Type: SA Page Ref: 357 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative

23 .

13) Suppose there are two mutually exclusive public alternatives A and B. They are characterized by the following data: Alternative A: First cost = 10 million Annual benefits = $2 million Annual operating costs =$1 million Alternative B: First cost = $15 million Annual benefits = $3.5 million Annual operating costs = $1 million Both alternatives have a 10-year service life. Social discount rate is 7%. Which alternative should be chosen on the basis of the conventional BCR? Answer: For mutually exclusive alternatives, we must evaluate the incremental benefit-cost ratio.

However, we first calculate the conventional benefit-cost ratios for each alternative and eliminate the alternatives with ratios less than one: Alternative A: PW(Benefits) = 2 * (P/A, 7%, 10) = 2 * 7.024 = $14.048 PW(Costs) = 10 + 1 * (P/A, 7%, 10) = 10 + 1 * 7.024 = $17.024 BCRA =

=0.825

Alternative B: PW(Benefits) = 3.5 * (P/A, 7%, 10) = 3 * 7.024 = $24.584 PW(Costs) = 15 + 1 * (P/A, 7%, 10) = 15 + 1 * 7.024 = $22.024 BCRB =

=1.116

Since Alternative A has BCR < 1 while Alternative B has BCR > 1 the latter should be chosen.

Diff: 1 Type: SA Page Ref: 357-359 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative 14) Give two examples of monetary incentives/disincentives Answer: Environmental tax on polluters is an example of a monetary disincentive. Once it is introduced, a polluter has to take it into account by adding the tax to the private costs. A subsidy for education is an example of a monetary incentive since it allows a private agent to acquire more education.

Diff: 2 Type: SA Page Ref: 348-349 Topic: 10.2. Market failures Skill: Applied 24 .

User1: Qualitative 15) A provincial government in a small city is considering a project of the city dump-site. Three alternative sites were identified with the following economic characteristics: Characteristic Value, million $$ Initial investment, million $$ Maintenance costs, million $$/year

Site A 10 5

Site B 12 4

Site C 14 3

0.5

0.6

0.4

Each site is expected to provide services valued at $3 million per year. Assume a 100-year study period, a 5% discount rate and a 10% sale value of the site after 100 years. Rank these sites in terms of their BCRs. Answer: The value of a site is its opportunity cost and therefore, it should be added as social cost.

Site A: PW(Costs) = 10 + 5 + 0.5 * (P/A, 5%, 100) = 10 + 5 + 0.5 * 19.848 = 10 + 5 + 9.924 = $24.924 million PW(Benefits) = 0.1 * 10 * (P/F, 5%, 100) + 3 * (P/A, 5%, 100) = 1 * 0.0076 + 3 * 19.848 = 0.008 + 59.544 = $59.552 million BCR =

= 2.389

Site B: PW(Costs) = 12 + 4 + 0.6 * (P/A, 5%, 100) = 12 + 4 + 0.6 * 19.848 = 12 + 4 + 11.909 = $27.909 million PW(Benefits) = 0.1 * 12 * (P/F, 5%, 100) + 3 * (P/A, 5%, 100) = 1.2*0.0076 + 3*19.848 = 0.009 + 59.544 = $59.553 million BCR =

= 2.134

Site C: PW(Costs) = 14 + 3 + 0.4 * (P/A, 5%, 100) = 14 + 3 + 0.4 * 19.848 = 14 + 3 + 7.939 = $24.939 million PW(Benefits) = 0.1 * 14 * (P/F, 5%, 100) + 3 * (P/A, 5%, 100) = 1.4 * 0.0076 + 3 * 19.848 = 0.011 + 59.544 = $59.555 million BCR =

= 2.388

In terms of the BCR, these sites should be ranked as follows: 1.

Site A 2.

Site C 3.

Site B Diff: 2 Type: SA Page Ref: 357 Topic: 10.3. Decision making in the public sector Skill: Applied User1: Quantitative 25 .

Engineering Economics, 5e (Fraser) Chapter 11 Project Management 11.1 Multiple Choice Questions 1) "CPM" stands for: A) Critical Path Method.

B) Company Profit Maximization.

C) Competitive Profit Methodology.

D) Complete Project Management.

E) Computer-Powered Management.

Answer: A Diff: 1 Type: MC Page Ref: 388 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative 2) The nature and scope of a project are defined in the A) planning stage.

B) execution stage.

C) monitoring and control stage.

D) initiation stage.

E) closure stage.

Answer: D Diff: 1 Type: MC Page Ref: 389 Topic: 11.2. Project management lifecycle Skill: Recall User1: Qualitative 3) The required resources for an engineering project are defined in the A) planning stage.

B) execution stage.

C) monitoring and control stage.

D) initiation stage.

E) closure stage.

Answer: A Diff: 1 Type: MC Page Ref: 389 Topic: 11.2. Project management lifecycle Skill: Recall User1: Qualitative

1 .

4) For an engineering project, the justification for the project is defined in the A) initiation stage.

B) planning stage.

C) execution stage.

D) monitoring and control stage.

E) closure stage.

Answer: A Diff: 1 Type: MC Page Ref: 390 Topic: 11.2. Project management lifecycle Skill: Recall User1: Qualitative 5) Who is responsible for the detailed project plan?

A) the project's sponsor B) the government C) the Canadian Engineering Association D) the project's manager E) the financial institution who finances the project Answer: D Diff: 2 Type: MC Page Ref: 390 Topic: 11.2. Project management lifecycle Skill: Recall User1: Qualitative 6) Project execution includes A) economic evaluation of the project's cash flows.

B) carrying out the processes defined in the project plan so that the project’s objectives are achieved.

C) estimation of the time, cost, and resources needed to complete the project.

D) identification of corrective actions.

E) review of the project's activities.

Answer: B Diff: 2 Type: MC Page Ref: 391 Topic: 11.2. Project management lifecycle Skill: Recall User1: Qualitative 7) The lifecycle of a project is given by the following sequence of stages: A) initiation—planning—execution—monitoring and control—closure.

B) planning—initiation—execution—monitoring and control—closure.

C) initiation—planning—monitoring and control—execution—closure.

D) initiation—evaluation—execution—monitoring and control—closure.

E) initiation—planning—execution—evaluation of cash flows—closure.

Answer: A Diff: 2 Type: MC Page Ref: 389 Topic: 11.2. Project management lifecycle Skill: Recall User1: Qualitative 2 .

3 .

8) The project management tools discussed in Section 11.3 of the textbook are associated with the A) initiation stage of a project.

B) planning stage of a project.

C) execution stage of a project.

D) monitoring and control stage of a project.

E) closure stage of a project.

Answer: B Diff: 2 Type: MC Page Ref: 392 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative 9) What does "WBS" stand for?

A) Weighted Brokerage System B) Work Bill of Sale C) Work Breakdown Structure D) World Banking System E) Whittaker's Bonding Scheme Answer: C Diff: 2 Type: MC Page Ref: 390 Topic: 11.2. Project management lifecycle Skill: Recall User1: Qualitative 10) Which of the following project management tools is used to organize the work to be done in a project by decomposing the project into successively smaller units?

A) WBS B) Gantt Chart C) CPM networks D) BCA E) CN networks Answer: A Diff: 2 Type: MC Page Ref: 392 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative

4 .

11) What is the difference between a diagram representation of the WBS and indented list representation?

A) a diagram is a 2-dimensional graph while a list is a sequence of actions B) a diagram is an organizational chart while a list is associated with evaluation of costs and benefits of a project C) a diagram is an organizational chart while a list is a numbered list of a project's layers D) a diagram is a three-dimensional graph of a project's layers while a list is a two-dimensional list of the project's costs and benefits E) a diagram is a cash-flow diagram while a list is a numbered list of a project's layers Answer: C Diff: 2 Type: MC Page Ref: 392-393 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative 12) A Gantt Chart is A) a chart that depicts the timing and magnitude of a project's cash flows.

B) a graph or bar chart that depicts the timing and sequence of project activities.

C) a numbered list of a project's activities.

D) an organizational chart of a project's activities.

E) a bar graph of a project's costs.

Answer: B Diff: 2 Type: MC Page Ref: 394 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative 13) Which of the following statements correctly describes precedence among project management tools?

A) A Gantt Chart precedes WBS. B) CPM networks precede WBS. C) CN networks precede WBS. D) WBS precedes a Gantt Chart.

E) WBS precedes BCA. Answer: D Diff: 2 Type: MC Page Ref: 395 Topic: 11.3. Project management tools Skill: Applied User1: Qualitative 14) A fundamental advantage of the CPM networks project management tool is that it provides A) a sequential representation of a project's activities.

B) a structural representation of a project's activities.

C) a representation that is able to capture the interdependence of a project's activities.

D) a graphical representation of a project's activities.

E) a three-dimensional representation of a project's activities.

Answer: C Diff: 1 Type: MC Page Ref: 396 Topic: 11.3. Project management tools 5 .

Skill: Recall User1: Qualitative 15) What does `PERT' stand for?

A) Probable Engineering Return Tool B) Projected Engineering Reaction Time C) Professional Engineering Resource Tool D) Profit Evaluation and Re-evaluation Tool E) Project Evaluation and Review Technique Answer: E Diff: 2 Type: MC Page Ref: 396 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative 16) Nodes and arcs are used in the following project management tool: A) CPM B) WBS C) BCA D) MARR E) ERR Answer: A Diff: 3 Type: MC Page Ref: 396 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative 17) AON stands for A) activity of a network.

B) activity on a node.

C) association of networks.

D) assembled and organized network.

E) activity of organized network.

Answer: B Diff: 2 Type: MC Page Ref: 397 Topic: 11.3. Project management tools Skill: Recall User1: Quantitative 18) In crashing a project, what should be the first activities you chose to crash?

A) the ones that can be sped up most cheaply B) the ones closest to the beginning of the project C) the ones on the critical path D) the ones closest to the end of the project E) the ones with the greatest slack Answer: C Diff: 3 Type: MC Page Ref: 403-408 Topic: 11.3. Project management tools 6 .

Skill: Applied User1: Qualitative

7 .

19) The critical path is A) a sequence of large cash outflows.

B) a sequence of large cash inflows.

C) the set of extremely large cash outflows at the very beginning of a project's lifecycle.

D) the set of activities that must be completed on time in order to keep the project on schedule.

E) the set of extremely large cash outflows at the very end of a project's lifecycle.

Answer: D Diff: 2 Type: MC Page Ref: 400 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative 20) Finding the critical path requires A) one pass through the network.

B) two passes through the network.

C) three passes though the network.

D) four passes through the network.

E) five passes through the network.

Answer: B Diff: 2 Type: MC Page Ref: 400-401 Topic: 11.3. Project management tools Skill: Applied User1: Quantitative 21) What is "crashing"?

A) time until a capital asset completely wears out B) reducing overall project completion time through the application of additional resources C) very low capitalized value of a business's cash flows D) time associated with the minimum of the equivalent annual costs E) when a project's costs exceed the project's benefits by more than 10 times Answer: B Diff: 3 Type: MC Page Ref: 403 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative 22) Suppose that the crash cost is $10 million, normal cost is $8 million, normal project time is 10 years and crash time is 8 years, what is the project's crash cost per year?

A) $1 million B) $2 million C) $3 million D) $4 million E) $5 million Answer: A Diff: 2 Type: MC Page Ref: 405 Topic: 11.3. Project management tools Skill: Applied User1: Quantitative 8 .

9 .

23) When resources such as overtime, extra equipment, and additional staff result in higher costs, what should a project manager do?

A) Hire additional staff and acquire additional resources.

B) Terminate the project.

C) Make a tradeoff between the extra costs associated with crashing and the benefits of a reduced project completion time.

D) Look for a new sponsor of the project.

E) Delay the project until economic conditions change.

Answer: C Diff: 3 Type: MC Page Ref: 403 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative 24) A construction company has just started building a new house. It was estimated that the project will take 12 months. However, the owner of the land asked the company to complete it in 10 months. What should the construction company do to accommodate the owner's request?

A) To deliver the project on time, one or several project activities should be crashed.

B) To deliver the project on time, the company should hire extra workers.

C) To deliver the project on time, the construction company should work overtime.

D) To deliver the project on time, the construction company should look for a new technology.

E) To deliver the project on time, the construction company should change the project.

Answer: A Diff: 3 Type: MC Page Ref: 403-405 Topic: 11.3. Project management tools Skill: Applied User1: Qualitative 25) A 5-year project requires $10 million in initial investment and will bring $0.5 million in net annual revenue under a 10% annual interest rate. If you were a project manager, would you proceed with project or would you terminate the project and at which stage?

A) Proceed with the project.

B) Terminate the project at initiation stage.

C) Terminate the project at planning stage.

D) Terminate the project at execution stage.

E) Terminate the project at monitoring and control stage.

Answer: B Diff: 2 Type: MC Page Ref: 390 Topic: 11.2. Project management lifecycle Skill: Recall User1: Qualitative

10 .

26) A feasibility study is associated with the ________ stage of a project lifecycle.

A) initiation B) planning C) monitoring and control D) execution E) closure Answer: A Diff: 1 Type: MC Page Ref: 390 Topic: 11.2. Project management lifecycle Skill: Recall User1: Qualitative 27) Suppose that a feasibility study done for a project by a consulting company brought the following results: PW (Benefits) = $13 million PW (Costs) = $10 million It was mentioned in the feasibility study that there is a risk of the project's loss due to potential high inflation. How should the project manager react to these results?

A) Terminate the project right away.

B) Terminate the project at initiation stage.

C) Proceed with the project to the planning stage during which the risk of inflation should be addressed.

D) Use Monte Carlo simulation to address the risk.

E) Look for a second opinion on feasibility study.

Answer: C Diff: 3 Type: MC Page Ref: 391 Topic: 11.2. Project management lifecycle Skill: Applied User1: Qualitative 28) A new bridge project passed the feasibility study test, and is now in the planning stage. The project manager thinks that evaluation of people's willingness to pay for the use of the bridge was done under the assumption of low inflation. In this regard, at this stage the project manager should A) initiate a public hearing.

B) perform sensitivity analysis with respect to variable levels of inflation.

C) terminate the project.

D) move on to the execution stage.

E) perform hedonic price evaluation.

Answer: B Diff: 2 Type: MC Page Ref: 390-391 Topic: 11.2. Project management lifecycle Skill: Applied User1: Qualitative

11 .

29) At what stage of a project's lifecycle can the project manager conclude that the project's goals have been met?

A) Initiation B) Planning C) Monitoring and control D) Execution E) Closure Answer: E Diff: 1 Type: MC Page Ref: 390 Topic: 11.2. Project management lifecycle Skill: Recall User1: Qualitative 30) In order to calculate the crash cost per unit of time, a project manager has to know A) the project's costs, benefits and service life.

B) the project's normal and crash costs as well as normal and crash time.

C) the project's equivalent annual costs and crash time.

D) the project's economic life and its crash costs.

E) the project's crash and normal time as well as the project's costs and benefits.

Answer: B Diff: 3 Type: MC Page Ref: 405 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative

12 .

31) The AON diagram shown represents a project. If the estimated times for each activity are as shown in the accompanying table, what is the estimated completion time for the project?

Activity 1 2 3 4 5 6 7 8 9

Time Start 2 days 3 days 10 days 12 days 4 days 6 days 5 days Finish

A) 16 days B) 20 days C) 26 days D) 30 days E) 32 days Answer: C Diff: 3 Type: MC Page Ref: 396-403 Topic: 11.3. Project management tools Skill: Applied User1: Quantitative

13 .

32) The AON diagram shown represents a project. If the estimated times for each activity are as shown in the accompanying table, how much slack is there in the completion of Activity 5?

Activity 1 2 3 4 5 6 7 8 9

Time Start 3 days 4 days 12 days 12 days 5 days 6 days 5 days Finish

A) There is no slack, because Activity 5 is on the critical path.

B) 1 day C) 2 days D) 3 days E) 4 days Answer: B Diff: 3 Type: MC Page Ref: 396-403 Topic: 11.3. Project management tools Skill: Applied User1: Quantitative

14 .

33) The AON diagram shown represents a project. Normal and crash times, and the associated costs, are shown for each activity in the accompanying table. If it is worth $5 000 a day to reduce the project duration, which activity or activities should be crashed?

Activity 1 2 3 4 5 6 7 8 9

Normal Time Crash Time Start Start 2 1 3 1 4 2 5 5 6 3 6 4 6 4 Finish Finish

Normal Cost 0 $10 000 $15 000 $20 000 $20 000 $6 000 $8 000 $10 000 0

A) There is no economical way to reduce completion time.

B) Crash Activity 3 by 2 days.

C) Crash Activity 7 by 2 days.

D) Crash Activities 3 and 7 by one day each.

E) Crash Activities 2 and 3 by one day each.

Answer: D Diff: 3 Type: MC Page Ref: 403-408 Topic: 11.3. Project management tools Skill: Applied User1: Quantitative

15 .

Crash Cost 0 $20 000 $20 000 $30 000 $20 000 $12 000 $16 000 $25 000 0

34) The AON diagram shown represents a project. Normal and crash times, and the associated costs, are shown for each activity in the accompanying table. The normal completion time for the project is 130 weeks. If it is worth $10 000 a week to reduce the project duration, what is the shortest economical crash time?

Activity 1 2 3 4 5 6 7 8 9

Normal Time Crash Time Start Start 20 weeks 10 weeks 30 weeks 10 weeks 40 weeks 20 weeks 40 weeks 20 weeks 60 weeks 60 weeks 10 weeks 5 weeks 20 weeks 10 weeks Finish Finish

Normal Cost 0 $100 000 $150 000 $200 000 $200 000 $60 000 $80 000 $100 000 0

A) There is no economical way to reduce completion time.

B) 10 weeks C) 15 weeks D) 20 weeks E) 25 weeks Answer: E Diff: 3 Type: MC Page Ref: 403-408 Topic: 11.3. Project management tools Skill: Applied User1: Quantitative

16 .

Crash Cost 0 $250 000 $200 000 $300 000 $400 000 $60 000 $160 000 $250 000 0

11.2 Short Answer Questions 1) Define project management Answer: Project management is concerned with planning a project, organizing the necessary resources, leading the individuals responsible for carrying out the project, monitoring its progress, and implementing its deliverables. The challenge is to achieve the goals and objectives of a project within the constraints of time, scope, and budget.

Diff: 1 Type: SA Page Ref: 388 Topic: 11.1. Introduction Skill: Recall User1: Qualitative 2) What is a project charter?

Answer: It is a document defining the terms of reference of a project, that is, what the project is intended to achieve, and the approximate timeline and resources available.

Diff: 1 Type: SA Page Ref: 389 Topic: 11.2. Project management lifecycle Skill: Recall User1: Qualitative 3) What are the five stages of a project's lifecycle?

Answer: The five stages are (i) initiation, (ii) planning, (iii) execution, (iv) monitoring and controlling, and (v) closure Diff: 2 Type: SA Page Ref: 389 Topic: 11.2. Project management lifecycle Skill: Recall User1: Qualitative 4) Explain what the closure stage of a project's lifecycle involves Answer: This is the stage at which the goals defined in the project charter have been accomplished. The project manager needs to get the project sponsor to sign off on the accomplishment of the goals. The project manager should also review and document lessons learned so that future projects can benefit from this information Diff: 2 Type: SA Page Ref: 389-392 Topic: 11.2. Project management lifecycle Skill: Recall User1: Qualitative 5) As a rule, each project includes a feasibility study. With reference to the five-stage project management lifecycle, where does this feasibility study belong?

Answer: It's part of the initiation stage. However, the project manager must be prepared to re-assess the feasibility of the project at any subsequent stage as new information becomes available.

Diff: 2 Type: SA Page Ref: 390 Topic: 11.2. Project management lifecycle Skill: Applied User1: Qualitative

17 .

6) Explain how and when a project's risk is addressed in the project management lifecycle Answer: The risks associated with the project are assessed at the planning stage. A risk plan will be developed, identifying the major risks, their potential impact, and methods for avoiding or minimizing them.

Diff: 3 Type: SA Page Ref: 390-391 Topic: 11.2. Project management lifecycle Skill: Recall User1: Qualitative 7) What is the Work Breakdown Structure (WBS)?

Answer: It is an analysis of the tasks needed to complete the project. Each task is recursively broken down in to sub-tasks, and each lowest level sub-task should be specified in sufficient detail to identify the exact material requirements. A useful rule of thumb is the `eighty-hour rule' —each lowest-level subtask should represent no more than 80 hours of labour. When the WSB is complete, it should be exhaustive—every activity needed to complete the project is listed—and non-overlapping—no activity is listed twice.

Diff: 3 Type: SA Page Ref: 384-393 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative 8) What is a Gantt Chart?

Answer: A Gantt chart is a graph or bar chart that depicts the timing and sequence of project activities.

It is one of the simplest and most fundamental project scheduling tools available. A Gantt chart consists of a list of the work items and a diagram indicating when each item starts and ends.

Diff: 1 Type: SA Page Ref: 394-395 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative 9) Define the critical path and explain its importance Answer: The critical path is determined by constructing a network representation of a project that captures all of the precedence relationships among activities. The critical path is the set of activities that must be completed exactly on time in order to keep the project on schedule. By definition, there is no slack for completing the activities on the critical path, as delaying any one of them will delay in the completion time of the entire project.

Diff: 2 Type: SA Page Ref: 400 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative

18 .

10) Identify some similarities and differences between the CPM and PERT project management tools.

Answer: Both tools are based on a network representation that captures the interdependence of activities. In both CPM and PERT, project activities are represented by a network of nodes and arcs.

CPM offers a method to make cost and time tradeoffs for a project. In contrast, PERT was developed to plan and control projects where task times are assumed to be uncertain.

Diff: 3 Type: SA Page Ref: 396 Topic: 11.3. Project management tools Skill: Applied User1: Qualitative 11) What is the crash cost of an activity and what information is needed to define it?

Answer: Crashing an activity means reducing the time taken to complete it through the application of additional resources. In CPM, we assume that this tradeoff is linear within certain limits. That is, we assume that if the fastest the activity can be completed is the crash time, and if the total cost of completing it within that time is the crash cost, then we can define: crash cost per unit time = Diff: 2 Type: SA Page Ref: 403-405 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative 12) Developing a new electronics product involves two activities that can be carried out independently: software development and hardware development. These activities would normally take 18 weeks and 20 weeks respectively, and the product development project is complete when both have been completed. If software development can be sped up at a cost of $1 million per week, and hardware development can be sped up at a cost of $2 million per week, what is the cost of reducing the project completion time to a) 19 weeks and b) 17 weeks?

Answer: a) $2 million (speed up hardware development one week) b) $7 million (speed up hardware development by 3 weeks, software development by 1 week) Diff: 2 Type: SA Page Ref: 405 Topic: 11.3. Project management tools Skill: Applied User1: Quantitative 13) What is meant by "crashing a project"?

Answer: Crashing a project means speeding it up. If it is necessary to shorten a project's lifecycle, the project manager will evaluate those project activities on the critical path to find the ones that can be speeded up given some extra resources such as additional staff or overtime. Given the costs of crashing each activity involved in the project, it is possible to calculate the cost of speeding up its completion.

Diff: 3 Type: SA Page Ref: 403-408 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative

19 .

14) How do we find a project's critical path?

Answer: The critical path is the set of activities that must be completed on time in order to keep the project on schedule. Finding the critical path involves making two passes through the CPM network diagram. In the first pass, the earliest possible start and finish times for each activity are determined.

This determines the earliest feasible completion time for the project. The second pass through the network goes in reverse, starting with the completion time established in the first pass and computing the latest time each activity can start and finish without delaying the project. The activities for which the earliest start time and the latest start time are identical are on the critical path. That is, there is no slack available in their start or finish times. They must be completed as scheduled or the project will be delayed Diff: 2 Type: SA Page Ref: 400 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative 15) What are the advantages and disadvantages of a Gantt Chart?

Answer: A Gantt chart is an excellent tool for determining precedence among tasks. When there are a large number of precedence relationships in a project, the Gantt chart is particularly useful. However, although tracking precedence relationships is reasonably straightforward for small projects, it may be difficult to manage and communicate for larger projects. This shortcoming reduces the effectiveness of Gantt charts as a planning and control tool for larger projects.

Diff: 2 Type: SA Page Ref: 394-396 Topic: 11.3. Project management tools Skill: Recall User1: Qualitative

20 .

Engineering Economics, 5e (Fraser) Chapter 12 Dealing with Uncertainty and Risk 12.1 Multiple Choice Questions 1) In general, sensitivity analysis can be described as A) establishing the range of variations in the key parameters of a project.

B) gaining a better understanding of how cash flows affect a project's performance.

C) predicting profitability of a project.

D) establishing the impacts of uncertainty in a project's inputs on its economic performance.

E) establishing the level of data errors.

Answer: D Diff: 1 Type: MC Page Ref: 425-426 Topic: 12.1. Introduction Skill: Recall User1: Qualitative 2) In a sensitivity graph, the slope of a line shows A) how sensitive a project's benefit-cost ratio is to changes in its inputs: the steeper, the more sensitive.

B) how sensitive a project's PW to changes in its inputs: the steeper, the more sensitive.

C) how sensitive a project's benefit-cost ratio to changes in its inputs: the flatter, the more sensitive.

D) how sensitive a project's PW to changes in its inputs: the flatter, the more sensitive.

E) how sensitive the project's IRR to changes in its inputs: the flatter, the more sensitive.

Answer: B Diff: 2 Type: MC Page Ref: 426-428 Topic: 12.2. Sensitivity graphs Skill: Applied User1: Quantitative 3) COOR Ltd. is buying a new sewing machine for $9 000. It can be sold at the end of its service life for $800. The firm's MARR is 10%. What is the minimum savings that the machine must yield in order to break even on the investment?

A) $2 012 B) $1 415 C) $1 815 D) $1 212 E) $1 692 Answer: B Diff: 2 Type: MC Page Ref: 429-432 Topic: 12.3. Break-even analysis Skill: Applied User1: Quantitative

1 .

4) An engineering project involves reinforcing a concrete beam. From previous experience it is thought that the probability of corrosion is 1%. A diagnostic test is performed on the reinforced beam. If corrosion is present, then the probability that the test will reveal it is 80% while the probability that the test will miss it is 20%. The diagnostic never yields false positives—that is, it will never indicate corrosion if corrosion is not present. What is the probability of discovering corrosion?

A) 0.001 B) 0.008 C) 0.09 D) 0.8 E) 0.002 Answer: B Diff: 2 Type: MC Page Ref: 436-442 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative 5) A company has developed a new product. Market price of the product is $10 per unit, variable costs are $6 per unit and total fixed costs are $100 000. Suppose that the company sets a target of $2 of profit per unit. How many units should be produced to meet the target?

A) 10 000 B) 20 000 C) 30 000 D) 40 000 E) 50 000 Answer: E Diff: 3 Type: MC Page Ref: 429-432 Topic: 12.3. Break-even analysis Skill: Applied User1: Quantitative

2 .

6) The sensitivity graph of a project is presented

below: What is the least sensitive input of the project?

A) MARR B) SV C) FC D) VC E) O&M Answer: E Diff: 2 Type: MC Page Ref: 429 Topic: 12.2. Sensitivity graphs Skill: Applied User1: Quantitative 7) The break-even point is defined as the point at which A) variable costs are equal to fixed costs.

B) the sum of variable and fixed costs is equal to total profit.

C) total income is just equal to the sum of fixed and variable costs.

D) profits are at a maximum.

E) profits are equal to variable costs.

Answer: C Diff: 2 Type: MC Page Ref: 429 Topic: 12.2. Sensitivity graphs Skill: Recall User1: Qualitative

3 .

8) A pie company has two employees who each earn $30 000 a year, and it pays $10 000 rent every year. The ingredients for a pie, including the gas to cook it, cost $10 per pie. If pies are sold at $20 each, the break-even point for the company (calculated pre-tax) is: A) 3 500 pies B) 4 000 pies C) 6 000 pies D) 7 000 pies E) 8 000 pies Answer: B Diff: 2 Type: MC Page Ref: 430-432 Topic: 12.3. Break-even analysis Skill: Applied User1: Quantitative 9) A graphical representation of the logical structure of a decision problem is usually A) an event tree.

B) a decision tree.

C) a probability tree.

D) a risk tree.

E) an uncertainty tree.

Answer: B Diff: 1 Type: MC Page Ref: 443 Topic: 12.5. Structuring decisions with decision trees Skill: Recall User1: Qualitative 10) A pie company has two employees who each earn $30 000 a year, and it pays $10 000 rent every year. The ingredients for a pie, including the gas to cook it, cost $10 per pie. If the company is taxed at 50% and pies are sold at $20 each, the break-even point for the company (calculated after-tax) is: A) 3 500 pies B) 4 000 pies C) 6 000 pies D) 7 000 pies E) 8 000 pies Answer: B Diff: 1 Type: MC Page Ref: 430-432 Topic: 12.3. Break-even analysis Skill: Applied User1: Quantitative

4 .

11) Below the results of a scenario analysis are presented:

Case Worst Base Best

Probability 25% 50% 25%

Sales, units 80 000 100 000 120 000

Profit per unit $30 $40 $60

What is the expected value of total profit according to this analysis?

A) $2.8M B) $3.2M C) $3.6M D) $4.2M E) $4.4M Answer: E Diff: 2 Type: MC Page Ref: 440 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative 12) A firm decides to invest $5 million in a new food processing plant. The plant has an expected life of 20 years with expected sales of 6 million cans of food per year. Fixed annual costs are $2 million and annual variable costs are $1 per can. The product will be priced at $2 per can. Corporate tax rate is 40%, and the MARR is 10%. Ignoring tax impacts on the first cost, what is the project's PW under the base case scenario?

A) $25.6 million B) $15.4 million C) $46.1 million D) $20.4 million E) $29.0 million Answer: B Diff: 2 Type: MC Page Ref: 425-426 Topic: 12.1. Introduction Skill: Applied User1: Quantitative

5 .

13) First cost of a construction project depends on weather. Historical weather data shows that in the fall weather is warm 20% of the time, it is cold 30% of the time and it is somewhat in between 50% of the time. During warm weather, first cost is expected to be $225 000, during cold weather $295 000 and $250 000 in between. What is the expected value of the first cost ?

A) $225 000 B) $250 000 C) $252 000 D) $258 500 E) $295 000 Answer: D Diff: 2 Type: MC Page Ref: 440 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative 14) Suppose a discrete random variable X can take on these values with the following probabilities: Value 0 1 2

Probability 0.25 0.50 0.25

What is the mean of the discrete random variable X? A) 0.25 B) 0.50 C) 0.75 D) 1.00 E) 1.25 Answer: D Diff: 2 Type: MC Page Ref: 440 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative 15) Suppose you are asked to do a sensitivity analysis of a project when an economy is in recession, what assumptions do you need to make to evaluate the project?

A) Market share is low, variable costs are high, product price is high.

B) Market share is low, fixed costs are high, product price is low.

C) Market share is high, variable costs are high, product price is low.

D) Market share is low, variable costs are high, product price is low.

E) Market share is low, variable costs are low, product price is low.

Answer: E Diff: 2 Type: MC Page Ref: 425-426 Topic: 12.1. Introduction Skill: Applied User1: Qualitative

6 .

16) The slope of a sensitivity graph of -5 with respect to operating costs means A) a 1% increase in operating costs leads to a 5% decrease in PW of the project.

B) a 1% increase in operating costs leads to a $5 decrease in PW of the project.

C) a $1 increase in operating costs leads to a 5% decrease in PW of the project.

D) a $1 increase in operating costs leads to a $5 decrease in PW of the project.

E) a 1% increase in operating costs leads to a 5% increase in PW of the project Answer: B Diff: 2 Type: MC Page Ref: 429 Topic: 12.2. Sensitivity graphs Skill: Applied User1: Quantitative 17) A decision tree of a project is given as follows:

What is the equivalent decision tree?

A)

7 .

B)

C)

D)

8 .

E)

Answer: E Diff: 3 Type: MC Page Ref: 442-448 Topic: 12.5. Structuring decisions with decision trees Skill: Applied User1: Qualitative 18) A decision tree of a project is given as follows:

What is the expected value of alternative B? A) $0 B) $4.5 C) $5.0 D) $5.2 E) $10.0 Answer: B Diff: 1 Type: MC Page Ref: 440, 442-448 9 .

Topic: 12.5. Structuring decisions with decision trees Skill: Applied User1: Quantitative 19) A decision tree of a project is given as follows:

What is the expected value of alternative A? A) $4.98 B) $5.08 C) $5.91 D) $8.83 E) $9.66 Answer: B Diff: 3 Type: MC Page Ref: 440, 442-448 Topic: 12.5. Structuring decisions with decision trees Skill: Applied User1: Quantitative 20) Graphically, the break-even value for any one of the input parameters to a project is captured by A) the slope of the project's sensitivity graph.

B) y-axis intercept of the graph of present worth as a function of that input parameter.

C) x-axis intercept of the graph of present worth as a function of that input parameter.

D) inverse of the slope of the graph of present worth as a function of that input parameter.

E) the ratio of the vertical intercept to horizontal intercept of the project's sensitivity graph.

Answer: C Diff: 2 Type: MC Page Ref: 429-432 Topic: 12.3. Break-even analysis Skill: Recall User1: Quantitative

10 .

21) In general, an increase in salvage value ________ and an increase in the interest rate ________ the present worth of a project A) increases, increases B) increases, decreases C) decreases, increases D) decreases, decreases E) does not change, decreases Answer: B Diff: 3 Type: MC Page Ref: 429-432 Topic: 12.2. Sensitivity graphs Skill: Applied User1: Quantitative 22) The decision tree of a project is represented by the following graph:

What is the outcome of an alternative A if the expected monetary value of a project is $4.56?

A) $2.0 B) $2.5 C) $5.0 D) $8.4 E) $10.3 Answer: E Diff: 3 Type: MC Page Ref: 442-448 Topic: 12.5. Structuring decisions with decision trees Skill: Applied User1: Quantitative 23) If a decision tree shows the possible profits associated with each of several alternative decisions, what is the optimal decision strategy?

A) the strategy that has the maximum profit as one of its possible outcomes B) the strategy that results in the highest expected value of profit C) the strategy for which the sum of leaf-node values is a maximum D) the strategy that minimizes the variance in profit E) the strategy that minimizes the risk of obtaining the lowest possible profit Answer: B Diff: 1 Type: MC Page Ref: 442-448 Topic: 12.5. Structuring decisions with decision trees Skill: Recall User1: Quantitative 11 .

24) A firm's evaluation of all possible alternatives with respect to production costs produced the following results:

Alternative 1 2 3 4 5

Mean Cost (in million $) 5.2 6.1 7.4 8.1 5.3

Standard deviation (in million $$) 1.7 1.7 1.9 2.0 1.6

Which alternatives should be eliminated from consideration?

A) 3, 4 B) 2, 3, 4 C) 1, 2 D) 2, 5 E) 1, 3, 4 Answer: B Diff: 3 Type: MC Page Ref: 440-442 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative 25) There are three alternatives for producing a needed resource. The table below presents mean costs and variance of costs for each of them.

ALTERNATIVE S MEAN, mln $ A 10 B 11 C 10

VARIANCE, mln $$ 5 4 6

If mean-variance dominance is applied to these alternatives then A) A dominates B. B) B dominates C. C) C dominates A. D) B dominates A. E) A dominates C. Answer: E Diff: 3 Type: MC Page Ref: 440-442 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative

12 .

26) As a rule, potential values of a random variable in a decision tree are represented by A) chance nodes.

B) decision nodes.

C) terminal positions.

D) branches.

E) links.

Answer: A Diff: 2 Type: MC Page Ref: 443 Topic: 12.5. Structuring decisions with decision trees Skill: Recall User1: Qualitative 27) Future annual revenue generated by a project is a random variable. Its probability distribution function is given as follows: Value, $/year 10 000 20 000 30 000 40 000 50 000

Probability 0.1 0.4 0.1 0.3 0.1

What is the probability of annual revenue being higher than $30 000/year?

A) 0.1 B) 0.3 C) 0.4 D) 0.5 E) 0.6 Answer: C Diff: 2 Type: MC Page Ref: 436-442 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative 28) The host of a TV show offers you a choice of five closed boxes. One contains $1 000, one contains $500, one contains $100, and two are empty. He offers, for a price, to remove one of the empty boxes and let you choose among the remaining four. How much is this offer worth?

A) It's not worth anything, because you still might choose the other empty box B) $80 C) $320 D) $400 E) $1 600 Answer: B Diff: 2 Type: MC Page Ref: 436-442 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative 13 .

14 .

29) The host of a TV show offers you a choice of ten closed boxes. Some of them contain money, some are empty. The total amount of money in them is $1 000. The host offers, for a price, to tell you how many are empty. How much is this offer worth?

A) It's not worth anything.

B) $10 C) $12.50 D) $100 E) $250 Answer: A Diff: 2 Type: MC Page Ref: 436-442 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative 30) The host of a TV show offers you a chance to pick one of four boxes. Three are empty, one contains $1 000. If you pick an empty box, you can pay $250 for a second pick from the remaining three boxes.

If you pick another empty box, you can pay $500 for a third pick from the two remaining boxes. If you pick wrong a third time, you are out of the game. What is your expected gain (or loss) from playing this game?

A) $250 B) $310 C) $330 D) $350 E) $500 Answer: B Diff: 2 Type: MC Page Ref: 442-448 Topic: 12.5. Structuring decisions with decision trees Skill: Applied User1: Quantitative 31) You have a contract with a company in the remote country of Placidia which guarantees you a profit of 100 000 Placidos in five years time. At the moment, one Placido is equal to one Canadian dollar.

However, there is about to be an election in Placidia, and if the Placidia First candidate gets in, it is expected that there will be 10% annual inflation in Placidia over the next five years. As far as you can tell, each of the two candidates has an equal chance of being elected. If your MARR is 10%, what is the expected present value to you of the contract, in Canadian dollars?

A) $38 554 B) $50 323 C) $62 091 D) $82 645 E) $86 777 Answer: A Diff: 2 Type: MC Page Ref: 436-439 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative

15 .

32) You are offered a chance to buy a ticket in a lottery. You have a one in ten chance of winning this lottery. If you win, your name will be entered along with 399 other winners as an entrant in a second lottery. All the entrants in the second lottery have an equal chance of winning the single prize, a sportscar valued at $400 000. What is the most it would be rational for you to pay for a ticket in the first lottery?

A) $10 B) $40 C) $100 D) $400 E) $1 000 Answer: C Diff: 2 Type: MC Page Ref: 436-439 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative 33) You are experimenting with a new process for manufacturing high-energy-density batteries. Your technical staff tell you that the process should only produce 25% defectives. However, the first three batteries produced by the process all turn out to be defective. What are the odds on this happening if your technical staff are telling the truth?

A) 1.6% B) 8.3% C) 12.5% D) 18.6% E) 22% Answer: A Diff: 2 Type: MC Page Ref: 436-439 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative 34) You are experimenting with a process for producing high-energy-density batteries. You have a contract with the government to deliver a batch of 20 batteries, due today. If you deliver them on time and they are all functional, you will get $100 000. Each battery has a 1% chance of being defective, and if any of them are defective, the government will not pay you the $100 000, but will fine you $10 000.

What is the expected value to you of your contract with the government?

A) -$1 821 B) $79 970 C) $81 791 D) $82 645 E) $86 777 Answer: B Diff: 2 Type: MC Page Ref: 436-439 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative

16 .

12.2 Short Answer Questions 1) The probability of heavy traffic on a provincial highway is 20%, while the probabilities of low and medium traffic are 50% and 30% respectively. It was estimated that annual maintenance costs of the highway are $20 000 in the case of low traffic, $30 000 in the case of medium traffic and $100 000 in the case of heavy traffic. Calculate the expected value of the annual maintenance costs.

Answer: The event under study is intensity of traffic. There are three possible outcomes of the event: low traffic, medium traffic and heavy traffic. Each is associated with specific maintenance costs of the highway. This is the case of a discrete event, and probability distribution function can be presented in the form of a table: Traffic intensity Low Medium Heavy

Probability, pi 0.5 0.3 0.2

Maintenance costs, $/year, xi 20 000 30 000 100 000

The expected value of annual maintenance costs is E(x) =

= 0.5 * 20 000 + 0.3 * 30 000 + 0.2 * 100 000 = $39 000

Diff: 1 Type: SA Page Ref: 440 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative

17 .

2) MMM Consulting is considering three scenarios of building a bridge. Data collected by the firm's experts are presented below:

Key Parameter Best Case First cost, million $ 4 Annual maintenance costs, million $/year 0.1 Annual value of crossings, million $/year 1 Service life 25

Expected Case 5

Worst Case 6

0.2

0.3

0.7 20

0.4 15

Probability of the best case scenario is 10% while probability of the expected and the worst case scenarios is 70% and 20% respectively. Assuming zero inflation and 10% annual interest rate, should the bridge be built?

Answer: Since the alternatives have different lives, it is useful to calculate the annual worth of each scenario AW(Best Case) = 1 - 4 * ( A/P, 10%, 25) - 0.1 = 1 - 0.441 - 0.1 = $0.459 million AW(Expected Case) = 0.5 - 5 * (A/P, 10%, 20) - 0.2 = 0.7 - 0.587 - 0.2 = -$0.087 million AW(Worst Case) = 0.4 - 6 * (A/P, 10%, 15) - 0.3 = 0.4 - 0.789 - 0.3 = -$0.689 million Now we calculate the expected value of the annual worth as follows: Expected value = 0.1 * 0.459 + 0.7 * (-0.087) + 0.2 * (-0.689) = -$0.1528 million Since the expected annual worth is negative, the project should be rejected.

Diff: 2 Type: SA Page Ref: 440 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative

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3) The expected value of the present worth of an engineering project is $103 015. It was evaluated on the basis of the following information on four potential outcomes:

Key parameter First cost, million $ Annual operating cost, million $/year Annual savings, million $/year Salvage value, million $ Service life, years

Alternative Alternative Alternative Alternative I II III IV 0.5

0.7

1.0

1.5

1

0.8

0.6

0.4

Unknown

1.0

0.8

0.7

0.2

0.5

0.7

0.8

4

4

4

4

If annual interest rate is 10%, and all the alternatives have equal probabilities of occurring, what is the annual savings under Alternative I? Answer: Since all the alternatives have the same service life, the present worth of each of them can be defined as follows PW = Annual Savings * (P/A, 10%, 4) + Salvage Value * (P/F, 10%, 4) - First Cost - Annual Operating Cost * (P/A, 10%, 4) = 3.17 * (Annual Savings - Annual Costs) + 0.683 * (Salvage Value) - First Cost Let the annual savings under Alternative I be X. Then, given definition of the expected value, the following equation arises: 0.25 * [3.17(X - 1) + 0.683 * 0.2 - 0.5] + 0.25 * [3.17 * (1 - 0.8) + 0.683 * 0.5 - 0.7] + 0.25 * [3.17 * (0.8 - 0.6) + 0.683 * 0.7 - 1.0] + 0.25 * [3.17 * (0.7 - 0.4) + 0.683 * 0.8 - 1.5 = 0.103015 X = $1.2 million is the solution.

Diff: 3 Type: SA Page Ref: 440 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative

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4) NB TECH is evaluating a dam-building project. The following estimates for probable cashflows were obtained: First cost, thousand $ 200 300 400 500

Probability of Net revenue, Probability of first cost thousand $/year net revenue 0.1 70 0.2 0.5 80 0.3 0.2 90 0.1 0.2 100 0.4

NB TECH uses a discount rate of 5%. If the life of a dam is 10 years and it has no salvage value thereafter, should it be built?

Answer: It is necessary to calculate the expected value of the present worth of the project. In this case it can be defined as: E(PW) = E(Net Revenue) * (P/A, 5%, 10) - E(First Cost) Expected values of the first cost and net revenue are: E(First Cost) = 0.1 * 200 000 + 0.5 * 300 000 + 0.2 * 400 000 + 0.2 * 500 000 = $350 000 E(Net Revenue) = 0.2 * 70 000 + 0.3 * 80 000+ 0.1 * 90 000 + 0.4 * 100 000 = $87 000/year Expected value of the present worth is: E(PW) = 87 000 * 7.722 - 350 000 = $321 814 Since the expected present worth is positive, the dam should be built.

Diff: 2 Type: SA Page Ref: 440 Topic: 12.4. Basic concepts of probability Skill: Applied User1: Quantitative

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5) The Kisu Company has just spent $100 000 to install a new technological line to produce kitchen knives. Estimates and probabilities for net annual savings, depreciation rates, CCA rates and useful life are given below: Key parameter Net annual savings Depreciation rate CCA rate Useful life

P = 0.3

P = 0.65

P = 0.05

$18 000 6% 10% 9

$22 000 4% 10% 10

$24 000 2% 10% 16

Kisu pays a 30% corporate tax rate. Calculate the expected value of the annual worth of the line if the after-tax MARR is 10%.

Answer: Under each outcome the annual worth can be defined as follows: AW = Net Annual Savings * (1 - 0.3) + Salvage Value * (A/F, 10%, T) * CSF - 100 000 * (A/P, 10%, T) * CCF First alternative (P = 0.3) Salvage value = 18 000 * (1- 0.06)9 = $10 313.91 CSF = 1 CCF = 1 -

= 0.85 = 0.8568

AW = 18 000 * 0.7 + 10 313.91 * 0.0736*0.85 - 100 000 * 0.1736 * 0.8568 = 12 600 + 645.24 - 14 874.05 = -$1 628.81/year Second alternative (P = 0.65) Salvage value = 22 000 * (1- 0.04)10 = $14 626.32 CSF and CCF are the same.

AW = 22 000 * 0.7 + 14 626.32 * 0.0627 * 0.85 - 100 000 * 0.1627 * 0.8568 = 15 400 + 779.51 - 13 940.14 = $2 239.37/year Third alternative (P = 0.05) Salvage value = 24 000 * (1- 0.02)16 = $17 371.15 CSF and CCF are the same.

AW = 24 000 * 0.7 + 17 371.15 * 0.0278 * 0.85 - 100 000 * 0.1278 * 0.8568 = 16 800 + 410.48 - 10 949.90 = $6 260.58/year Expected value of the annual worth of the technological line is: E(AW) = 0.3 * (-1 628.81) + 0.65 * 2 239.37 + 0.05 * 6 260.58 = -488.64 + 1 455.59 + 313.03 = $1 279.98/year.

Diff: 3 Type: SA Page Ref: 440 Topic: 12.4. Basic concepts of probability 21 .

Skill: Applied User1: Quantitative 6) New software has been developed, and $23 000 has been spent on its development. However, the implementation phase requires an additional $30 000. There is uncertainty about annual net revenue.

Three potential scenarios with their probabilities of occurring were identified., as shown in the following table: Parameter Probability Net revenue Life

Worst Case 0.4 -$3 000/year 5

Expected Case 0.5 $10 000/year 5

Best Case 0.1 $25 000/year 10

Assume a 10% MARR and use a decision tree to answer the following question: What is the expected PW for deciding whether or not to proceed?

Answer: The decision tree can be constructed like this:

The development cost of $23 000 is a sunk cost,. Only the implementation cost of $30 000 should be taken into account. It is necessary to calculate present worth associated with each chance node, depicted by a circle, as follows: PW(2) = -30 000 - 3 000 * (P/A, 10%, 5) = -30 000 - 3 000 * 3.7808 = -$41 372 PW(3) = -30 000 + 10 000 * (P/A, 10%, 5) = -30 000 + 10 000 * 3.7808 = $7 908 PW(4) = -30 000 + 25 000 * (P/A, 10%, 10) = -30 000 + 25 000 * 6.1446 = $123 615 Expected value of the present worth is E(PW) = 0.4 * (-41 372) + 0.5 * (7 908 )+ 0.1 * (123 615) = -$233 Since the expected value of the present worth is negative, it is inadvisable to proceed with implementation.

Diff: 2 Type: SA Page Ref: 442-448 Topic: 12.5. Structuring decisions with decision trees 22 .

Skill: Applied User1: Quantitative

23 .

7) A new technological line to produce shoes is being designed. It costs $100 000 and operating costs are expected to be $30 000 per year. Planned annual production is 800 pairs and the price of a pair of shoes is $80. The line's service life is five years, the depreciation rate is 20%. Assuming a 10% annual interest rate for the base case, perform a sensitivity analysis with respect to the interest rate in the interval [-20%, +20%] in terms of present worth. Comment on your results briefly.

Answer: Let us first evaluate the base case. In order to do so, it is necessary to calculate salvage value of the line as follows: S = BVDB(5) = P * (1 - d)5 = 100 000 * (1 - 0.2)5 = $32 768 Annual revenue is AR = $80/pair*800 pairs/year = $64 000 The present worth of the base case is: PW(Base Case) = 64 000 * (P/A, 10%, 5) + 32 768 * (P/F, 10%, 5) - 100 000 - 30 000 * (P/A, 10%, 5) = 64 000 * 3.7908 + 32 768 * 0.6209 - 100 000 - 30 000 * 3.7908 = 242 611 + 20 346 - 100 000 - 113 724 = $49 233 All calculations required for the sensitivity analysis are summarized in the table below: Key parameter First cost, $ Operating costs, $/year Annual revenue, $/year Salvage value, $ Interest rate, % PW, $

-20% 100 000

-10% 100 000

Base Case 100 000

+10% 100 000

+20% 100 000

30 000

30 000

30 000

30 000

30 000

64 000 32 768 8 58 053

64 000 32 768 9 53 547

64 000 32 768 10 49 233

64 000 32 768 11 45 107

64 000 32 768 12 41 156

With an increase in the interest rate by 1%, the PW decreases approximately by $4 000. However, in the interval [-20%, +20%] the PW stays positive and therefore, the new technological line is a good investment Diff: 2 Type: SA Page Ref: 426-429 Topic: 12.2. Sensitivity graphs Skill: Applied User1: Quantitative

24 .

8) A manager of a manufacturing enterprise is considering whether to produce an intermediate good in house or to buy it from another firm. If the good is produced in house, three potential outcomes are possible: (i) if production is done in one shift, then expected profit is -$30 000 per year, (ii) if production is done in two shifts, then expected profit is $10 000, and (iii) if production is done in three shifts, then expected profit is $55 000. Probabilities associated with these outcomes were estimated as 35%, 45% and 20% respectively. However, if the good is purchased, two potential outcomes arise: (i) on-time delivery of the good results in $20 000 of expected profit, (ii) delays in delivery result in an expected loss of -$17 000. Probabilities of these outcomes are 60% and 40% respectively. Draw a decision tree, and execute it to make a decision.

Answer: The tree looks like this:

The expected value of the profit associated with chance node 1 (in-house production) is calculated as follows: E(2) = 0.35 * (-30 000) + 0.45 * 10 000 + 0.2 * 55 000 = $5 000 The expected value of the profit associated with chance node 2 (purchase from another firm) is calculated as follows: E(3) = 0.6 * 20 000 + 0.4 * (-17 000) = $5 200 Based on the maximum expected value criterion, the purchase of the intermediate good from another firm is a better option.

Diff: 2 Type: SA Page Ref: 442-448 Topic: 12.5. Structuring decisions with decision trees 25 .

Skill: Applied User1: Quantitative 9) This decision tree represents the expected profits and the standard deviations associated with three decisions facing a mobile phone producer (All figures are in millions of dollars). The root node (the one on the left) represents the decision of whether to produce the phones in China or North America. The second pair of nodes represent the decision of whether to market the phones in China or North America; and the final nodes represent the choice of selling price: if the phones are sold in China, they will be sold for either $30 or $40, whereas if they are sold in North America, they will be sold for either $40 or $50.

Based on the calculated values, what is the company's best strategy? :

Answer: The two highest expected values, $1 250M, are those that correspond to making and selling in China for $30, and making and selling in North America for $40. But we note that the standard deviation of the China option is lower, indicating that this is the less risky of the two options. So this is the one we should pick.

Diff: 1 Type: SA Page Ref: 442-448 Topic: 12.5. Structuring decisions with decision trees Skill: Applied User1: Quantitative

26 .

10) The first cost of a highway project is $1.5 million. Operating and maintenance costs were estimated at $50 000 per year. There will be zero salvage value of the highway after 20 years. As a result of this project, a recreational facility worth $10 000 per year in benefits will be destroyed. Assuming a 4% social discount rate, what is the break-even value of time savings to the highway users? How would this value change if social discount rate increased to 10%?

Answer: Break-even value of annual time savings is exactly equal to the annual worth of the project's costs. Assuming a 20-year time horizon, the annual worth of costs is: AW(Costs) = 1 500 000 * (A/P, 4%, 20) + 50 000 + 10 000 = 1 500 000 * 0.07358 + 50 000 + 10 000 = $170 370 With a 10% social discount rate, the annual worth is AW(Costs) = 1 500 000 * (A/P, 10%, 20) + 50 000 + 10 000 = 1 500 000 * 0.11746 + 50 000 =10 000 = $236 190 The required time savings increase with an increase in the social discount rate.

Diff: 2 Type: SA Page Ref: 429-432 Topic: 12.3. Break-even analysis Skill: Applied User1: Quantitative 11) BC Power has just purchases a $0.5 million turbine. The turbine will allow it to produce 10 000 kilowatt hours (KWH) of electricity per day for twenty years. Afterwards the turbine's scrap value is expected to be negligible. Current price of electricity is $0.06/KWH. Operating costs of the turbine are $100 000 per year. Now BC Power pays a 40% corporate tax rate on its annual income. Depending on the outcome of the provincial elections, it is expected that the tax rate can either increase by 10% or decrease by 10%. Given this information and assuming a 5% annual interest rate and a 20% CCA rate, perform a sensitivity analysis on the company's after-tax profit in the turbine's first year of operation with respect to the expected changes in the corporate tax rate.

Answer: First, it is necessary to evaluate the base case before the elections. In the first year of operation, the company's profit is After-tax profit = Annual revenue - Annual operating costs - Taxes Taxes = t * (Annual revenue - Operating costs - Depreciation amount) Annual revenue = $0.06 * 10 000KWH/DAY * 365DAYS/YEAR = $219 000/YEAR Annual operating costs = $100 000/YEAR Depreciation amount should be defined according to the CCA system and the Half-Year Rule: Depreciation amount = CCA(1) = 0.2 * (500 000/2) = $50 000/YEAR Taxes = 0.4 * (219 000 - 100 000 - 50 000) = $27 600/YEAR After-tax profit = 219 000 - 100 000 - 27 600 = $91 400/YEAR A 10% reduction in taxes means a 35% corporate tax rate while a 10% increase means a 44% corporate tax rate.

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Under 36% tax rate: Taxes = 0.36 * (219 000 - 100 000 - 50 000) = $24 840/YEAR After -tax profit = 219 000 - 100 000 -24 840 = $94 160/YEAR Under 44% tax rate: Taxes = 0.44 * (219 - 100 000 - 50 000) = $30 360/YEAR After-tax profit = 219 000 - 100 000 - 30 360 = $88 640/YEAR The following table summarizes the sensitivity analysis with respect to the tax rate

After-tax profit

-10%

Base Case

+10%

$94 160

$91 400

$88 640

Diff: 1 Type: SA Page Ref: 426-429 Topic: 12.2. Sensitivity graphs Skill: Applied User1: Quantitative 12) A coal mine extracts 100 000 tons of coal annually using mining equipment that costs $10 million.

Operating and maintenance costs are $3 million per year. Define the break-even price of coal over planning horizon of six years under a 8% annual interest rate if the equipment's salvage value is $200 000.

Answer: The break-even price of the coal is that price which will make the PW of all cash inflows equal to the PW of all cash outflows. Annual revenue and salvage value are cash inflows while capital cost of equipment and operating costs are cash outflows: PW(Cash inflows) = 100 000C(P/A, 8%, 6) + 200 000 * (P/F, 8%, 6) = 462 290C + 126 034 where C is the price of one ton of coal.

PW (Cash Outflows) = 10 000 000 + 3 000 000 * (P/A, 8%, 6) = 10 000 000 + 13 868 700 = 23 868 700 We equate these two cashflows: 462 290C + 126 034 = 23 868 700 Solving for C: C = $51.36/ton. The mine is profitable if the price of coal is above $51.36 per ton.

Diff: 2 Type: SA Page Ref: 429-432 Topic: 12.3. Break-even analysis Skill: Applied User1: Quantitative

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13) New equipment was bought by an engineering firm for $5 million. For the base case, assume a 7% interest rate, a 20% depreciation rate and service life of four years. Construct a sensitivity graph for the PW of the equipment's salvage value with respect to a variation of 20% in its depreciation rate.

Answer: The relevant equation is PW(S) = BVDB * (P/F, 7%, 4) = 3.8145P(1 - d)4 where P is the purchase price The following table summarizes all calculations based on this model: Depreciation rate, % 16 18 20 22 24

PW of salvage value, million $ 1 899 1 725 1 562 1 412 1 273

Sensitivity graph in this case is the relationship between the PW of salvage value (dependent variable on the Y-axis) and percentage variation in the value of the depreciation rate from the base case (independent variable on the X-axis)

Diff: 2 Type: SA Page Ref: 426-429 Topic: 12.2. Sensitivity graphs Skill: Applied User1: Quantitative

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14) Describe the backward induction procedure used in decision tree analysis.

Answer: Backward induction, also known as the rollback procedure, involve the following steps: (i) At each chance node, compute the expected value of the possible outcomes. The resulting expected value becomes the value associated with the chance node and the branch on the left of that node. (ii) At each decision node, select the option with the best expected value. The best expected value becomes the value associated with the decision node and the branch on the left of that node. For the option(s) not selected at this time, indicate their termination by a double-slash (//) on the corresponding branch. (iii) Continue rolling back until the leftmost node is reached (iv) In some cases it may also be instructive to calculate the variance at each node. If two of the branches to the right of a given decision node have similar expected values, we will generally prefer the one with lower variance.

Diff: 1 Type: SA Page Ref: 446 Topic: 12.5. Structuring decisions with decision trees Skill: Recall User1: Qualitative

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