ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 1 of 36. Introduction

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1 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 1 of 36 Introduction 1. Syllabus distributed: Dates of assignments, mid-terms, and final exams specified. (a) Let me know in writing about possible misses of midterm exam. (b) Reexamination policy is available in the Faculty Office. (c) University suggested course grade distribution will be used as a guideline when assigning the final grades. 2. Course Objective: Compare alternatives in monetary terms and make better engineering decisions. 3. The economic aspects of daily life and work: (a) Loans: student loans, car purchase loans, mortgages, credit card loans, etc. (b) Banking: checking account, savings account, term deposits, etc. (c) Job: starting salary, annual raise, inflation, etc. 4. Read Chapter 1. We now start Chapter 2. Equivalence and Interest Formulas 5. Cash flow diagram: From whose perspective is the diagram drawn? Positive cash flows: Negative cash flows: ; Time axis: horizontal line; Number of periods: days, weeks, months, years, etc; End-of-period convention Example 1.1: You borrowed $2000 from a bank to buy a used car. The bank requires you to make 24 equal monthly payments of $95 to pay off the loan (including principal and interest charges). Draw a cash flow diagram. 6. Interest: In a business environment, if you borrow money, you need to pay interest and if you lend money, you should earn interest. Money has earning power. There is a cost if you use someone else s money. Interest is defined to be the money paid for using borrowed money. 7. Interest rate per period, denoted by i: It is the ratio between the interest payable at the end of a period and the money borrowed at the beginning of the period. i = I 1 /P, or I 1 = P i 8. Method 1 of calculating Interest over duration of n periods: Simple Interest Interest is earned in each period only on the initial principal. Interest is earned every period, accumulated, and paid only at the end of a term of n periods. i --- Interest rate per period n --- Number of periods in the term P --- The initial borrowed amount F --- The total amount to be paid at the end of n periods F = P + P n i Total Interest Charges = P n i Example 1.2: Your checking account has a daily interest rate of 0.003%. Interest is calculated based on daily closing balance and paid monthly (every 30 days). (a) With an initial balance of $1000, what is your balance 30 days later? (b) If you deposit another $2000 on the 11 th day and withdraw $500 on the 26 th day, what is your balance at the end of the 30 th day?

2 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 2 of 36 Compound Interest and Equivalence 1. Method 2 of calculating interest over duration of n periods: Compound Interest Interest is calculated and paid every period. The interest you have earned in a period will earn you more interest in the following periods. i --- Interest rate per period n --- Number of periods in the term P --- The initial borrowed amount F --- The total amount to be paid at the end of n periods F = P (1 + i) n Example 2.1: You bought a 5-year term deposit certificate for $1000. The interest rate is 3% every 6 months and interest is paid every 6 months. How much will you have when the certificate matures? Two years after this purchase, you bought a 3-year term deposit for $1000 with an interest rate of 2% every 6 months. How much will you have when both these certificates mature? 2. Comparison of the two methods of calculating interest: The balance with the simple interest method grows linearly. The balance with the compound interest method grows exponentially. This exponential growth is often referred to as the power of compounding, as used by insurance salesmen and financial planners. The compound interest method is used much more widely than the simple interest method. In this book, we assume the compound interest method unless specified otherwise. For a checking account, simple interest calculation method is used within each month and compound interest calculation method is used from month to month. 3. The concept of equivalence: Money has time value. Money has earning power. It earns interest and the compound interest calculation method is usually used. $100 today should be worth more than $100, say, 10 years later. As a result, cash flows that are occurring at different times cannot be compared numerically directly. We often have to convert them into equivalent values at the same time point before we can compare them. Common comparison points are present point (we need to find the Present Worth or Present Equivalent of the cash flows), a future point (we need to find the Future Worth or Future Equivalent of the cash flows), and annual points with equivalent constant annual cash flows. Example 2.2: You have just signed a loan agreement with the following terms: Borrowed amount = $ , Monthly interest rate = 1% Payment = $100 per month for 24 months How to interpret this financial transaction using the concept of equivalence?

3 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 3 of 36 Interest Formulas for Equivalence Calculations 1. Single Cash Flows: F n-1 n P Example 3.1: If you invest $1000 today which can earn a rate of return of 10% per year, how much will you have in 20 years? F = P (1+i) n = P (F/P, i, n) P = F/(1+i) n = F (P/F, i, n) = F / (F/P, i, n), (F/P, i, n) = 1/(P/F, i, n) Example 3.2: What is the present worth of a lump sum of $1,000,000 to be received 50 years from today if your interest rate is 10%? What if your interest rate is 20%? 2. Equal Payment Series: F n-1 n Example 3.3: How much constant monthly deposit is needed for 5 years to accumulate $10,000 from an account earning 1%/month? A Example 3.4: A stereo store sells a stereo system for $1500 and offers the following financing: zero down payment and 24 monthly payments of $70 each. What is the monthly interest rate of such a loan? F = A + A(1+i) + A(1+i) A(1+i) n-1 = A [(1+i) n - 1] / i = A(F/A, i, n) A = F (A/F, i, n), (A/F, i, n) = 1/(F/A, i, n) P = A(P/A, i, n) (A/F, i, n) = (P/A, i, n) = i (1 i) n 1 n (1 i) 1 n i(1 i) Example 3.5: You just borrowed $1000 today at an interest rate of 1% per month. You don t need to make any payment in the next 5 months (however interest charges are accumulating). What is your monthly payment if you want to pay off the loan with 25 payments starting in month 6?

4 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 4 of 36 Interest Formulas for Equivalence Calculations (Continued) 3. Linear Gradient Series (Constant Amount Change): F F = G (F/G, i, n), P = G (P/G, i, n) A = G (A/G, i, n), n-1 n 0 G 2G 3G F = G (P/G, i, n) (F/P, i, n), F = G (A/G, i, n) (F/A, i, n) Example 4.1: You plan to deposit $1000 at the end of this month. Later monthly deposits will increase by $50 per month. If the deposits can earn 1% per month, how much will you have in 3 years? Example 4.2: You are now considering making a constant monthly deposit over the next 3 years instead of the pattern used in Example 4.1. What is the monthly deposit amount that will make you reach the same goal in savings in 3 years? 4. Geometric Gradient Series (Constant Rate Change): F n-1 n (P/A 1,g,i,n) = n /(1 i), if i g n 1 g 1 /(i g), if i g 1 i A 1 A 2 A 3 g A 2 = A 1 (1+g) A 3 = A 2 (1+g) = A 1 (1+g) 2 P = A 1 (P/A 1,g,i,n) F = A 1 (P/A 1,g,i,n) (F/P, i, n) A = A 1 (P/A 1,g,i,n) (A/P, i, n) The value of g may be positive or negative. Example 4.3: If you are to deposit $1000 in the first month and subsequent monthly deposits will decrease at a rate of 5% per month. How much will you save in 3 years if the savings account pays an interest rate of 2% per month? Example 4.4: You just borrowed $ from a bank. Your first payment will be made one month later. Subsequent monthly payments will increase at a rate of 2% per month. The loan is for 3 years with an interest rate of 1% per month. What is the amount of your first payment?

5 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 5 of 36 Applications of Interest Formulas Example 5.1: You are planning to make the following deposits to a savings account which pays 0.5% per month: $200/month in the first year, $300/month in the second year and $500/month in the third year. How much will you have in the savings account at the end of year three? Example 5.2: How to find the value of the interest factor (P/A, 5%, 42) using the interest tables? From the tables we have (P/A, 5%, 40) = , (P/A, 5%, 45) = Example 5.3: For the past three years you have been making monthly deposits to a savings account. The monthly deposits in the past three years were $100, $200 and $300, respectively. The monthly interest rates of the savings account for the past three years were 1%, 0.75%, and 0.5%. What is the balance of the account today? Example 5.4: Please calculate the present equivalent of the cash flows in the following cash flow diagram using i = 10%. Recall that the upward cash flows are positive and downward cash flows are negative. Also note that the present equivalent (or annual equivalent or future equivalent) of all the cash flows is equal to the sum of the present (or annual or future) equivalents of the individual cash flows. Solution for Example 5.4: PE A = + $600(P/A,10%,5) = $ PE B = + $200(P/G,10%,5) = $ PE C = [+$600(P/A,10%,5) - $100(P/G,10%,5)] (P/F,10%,5) = $ PE D = -$300(P/A1,10%,10%,10) = -$ PE = PE A +PE B +PE C +PE D = $ Limiting Behaviors of Some Interest Factors: F lim(, i, n), n P F lim(, i, n) 1, i0 P A lim(, i, n) i, n P A 1 lim(, i, n) i0 P n Example 5.5: Suppose that you have won a lottery of one million dollars. You have decided to invest the full amount earning 10% per year. How much can be withdrawn each year from this investment for an infinite number of years?

6 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 6 of 36 Nominal and Effective Interest Rates In Chapter 2, we assumed that we were given the interest rate for each period (be it a year, a month, a week, or a day). However, a bank often quotes the interest rates in term of per annum. This annual interest rate is called the nominal interest rate or annual percentage rate (APR). In addition, a compounding frequency is provided for each quoted nominal interest rate. For example: Credit card interest rate: 24.55% (per year) compounded daily; New car loan: 6.8% (per year) compounded monthly; Mortgage: 5% (per year) compounded semi-annually. Notation: r --- Nominal interest rate per year, usually given M --- The number of compounding periods in a year, usually given. i a --- Effective interest rate per year, to be computed The interest rate per compounding period = r/m F = P M r 1 = P (1 + ia ), i a = M M r 1 M - 1. Note that ia r unless M = 1. F M compounding periods in 1 year P The higher the compounding frequency, the higher the effective annual interest rate for a fixed nominal annual interest rate. Example 6.1: r = 12%, M = 1, i a = (1+12%) 1-1 = 12% r = 12%, M = 2, i a = (1+6%) 2-1 = 12.36% r = 12%, M = 12, i a = (1+1%) 12-1 = 12.68% r = 12%, M = 365, i a = ( %) = % r = 12%, M =, i a = e r - 1 = % A formula for finding the interest rate per payment period: i = number of compounding periods in a payment period. r 1 M C - 1, where C is the Example 6.2: Gerry Smith is planning to deposit $1500 every 3 years into an RRSP account that earns 9% compounded semi-annually. He will retire 39 years from today. The first deposit will occur three years from today. The last deposit will occur at the time of his retirement. What will be the balance of the account at the time of his retirement? If he plans to use up all the money in the account in 10 years after his retirement, how much will he be able to withdraw at the end of every month?

7 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 7 of 36 Amortized Loans If a loan is to be paid off with equal periodic installments, it is called an amortized loan. Examples of amortized loans include car loans, mortgages, loans for appliances, etc. The periodic payments may be made weekly, bi-weekly, or monthly. Most amortized loans cited in this book are based on monthly compounding. However, mortgages in Canada are based on semi-annual compounding. Issues to be addressed: (1) The principal portion and the interest portion in a payment (2) Two methods of finding the balance of a loan at any point of time (3) How to find the interest portion (and the principal portion) in any payment Example 7.1: Assume you have borrowed $10,000 for a new car. The financial institution charges you 12% compounded monthly for the loan. The term of the loan is 5 years. The calculated monthly payment is $ Find out the principal portions and the interest portions of your monthly payment for the first 5 months. What is the balance at the end of year 2? What is the interest charge for the first month in year 3? Month Total Interest Principal Remaining Number Payment Portion Portion Balance , , , , , , , , , , , , , , , , , , ,697.25

8 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 8 of 36 Mortgages and Bonds Mortgage refers to a long-term amortized loan used primarily for the purpose of purchasing a piece of real property. Amortization: The number of years it would take to repay a mortgage loan in full with a given interest rate and a specific repayment schedule. Term of the mortgage: The number of months or years during which the signed mortgage document is binding. At the end of the term, the balance of the loan is due and repayable. Interest rate: The nominal annual interest rate based on semi-annual compounding. Open mortgage: The balance may be paid off in full at any point of time. Closed mortgage: The mortgage is not allowed to be paid off in full until the end of the term. Fixed interest rate mortgage: The interest rate is fixed during the term of the mortgage. Variable rate mortgage: The interest rate of the mortgage fluctuates following the trend of the prime rate. Equity: Part of the value of the real property that belongs to the owner. Problem 3.70 in the textbook. Bonds terminology: Bonds are a way of borrowing money from the public. Companies and governments issue them. Par value or face value: It is printed on the bond certificate. This is the amount you get back when the bond matures. Maturity date: Most bonds have a maturity date. It is the date that the borrower returns the money to the lender. Coupon rate: The interest rate to be paid to the bond holders. Problem 3.87 in the textbook.

9 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 9 of 36 The Payback Method and the Discounted Payback Method Many investment projects have the pattern of a big initial investment followed by positive cash flows in future years. We are interested in assessing the profitability of these projects. Consider a new computer assembly plant as an example: (1) Initial investment: building, land, conveyor lines, assembly lines, etc. (2) Revenues: Sales of finished computer products (3) Operating costs: raw materials, wages and salaries, utilities, maintenance costs, etc (4) Estimated life of the facility: 5 years Is this a good investment? Assumptions: (1) Annual cash flows are used. (2) Taxes are ignored for now. The payback method and the discounted payback method are often used as initial project screening methods before more rigorous project analyses are performed. The payback method This method ignores the interest rate. It ranks projects according to their payback periods. The shorter the payback period, the better the project. This method ignores the time value of money. The payback period is the number of years to recover the initial investment. Example 9.1: The discounted payback method: This method considers interest rate in calculation of the payback period. It takes the time value of money into consideration in ranking projects. The project balance or the cumulative cash flow at the end of a year is equal to the future equivalent of all cash flows from time 0 to the end of that year. Example 9.2: Both versions of the payback methods discriminate against long-term projects. They also ignore the cash flows beyond the payback periods.

10 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 10 of 36 PE, AE, and FE Methods for Evaluating an Individual Project For a project to be evaluated, we need to estimate the investment required, the life of the project, the cash inflows (operating revenues), and cash outflows (operating costs). A cash flow diagram may be constructed based on these estimates. We also need to know the Minimum Attractive Rate of Return (MARR) of the company. The methods for determining MARR will be covered in Chapter 10. The decision rules that may be used for determining the acceptability of a project include: (a) If PE(MARR) 0, the project is acceptable; If PE(MARR) < 0, the project is not acceptable. (b) If AE(MARR) 0, the project is acceptable; If AE(MARR) < 0, the project is not acceptable. (c) If FE(MARR) 0, the project is acceptable; If FE(MARR) < 0, the project is not acceptable. Suggested procedure for solving problems: (1) List data given (2) List the objective (3) Draw cash flow diagram (4) Choose a decision rule (5) List the equation and do calculations (6) Draw conclusions Example 10.1: Consider a machine that costs $5000 and has a 5-year useful life. At the end of 5 years, it can be sold for $1000 after tax adjustment. This machine could generate after-tax revenues of $1100 per year. The company has a MARR value of 10%. Should this machine be purchased? Assume that all benefits and costs associated with this machine are accounted for in the given figures. Consider a cash flow diagram with a negative initial cash flow followed by a series of positive cash flows. Illustrate the trend of the PE as a function of the interest rate used. The higher the interest rate, the lower the PE value. For the above example, we have the following PE values at different interest rates: Interest Rate i: 2% 4% 6% 8% 10% 12% PE Value: $1, $ $ $72.56 ($209.21) ($467.32) The meaning of the equivalent values (PE, AE, and FE): The borrowed funds concept. Suppose the project borrows money from the company.

11 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 11 of 36 The ROR Method for Evaluating an Individual Project Investment Classification: We can classify investment projects into two categories: simple investments and non-simple investments. To classify a project, we count the number of sign changes in the cumulative cash flows of the project (the interest rate is ignored). A change from + to or from to + is counted as one sign change. (a) A simple investment has only one sign change during its life. (b) A non-simple investment has more than one sign changes during its life. Example 11.1: Two definitions of the rate of return (ROR): (a) The rate of return is the break-even interest rate (i * ) that makes the PE (or AE or FE) of a project's cash flows equal to zero, ie, PE(i * ) = 0. (b) The internal rate of return is the interest rate charged on the unrecovered project balance of the investment such that, when the project terminates, the unrecovered project balance is zero. For a simple investment, the equation PE(i) = 0 has only one real root. This root is equal to the internal rate of return of the project. For a non-simple investment, there may be two or more real roots that make PE(i) = 0. None of these roots is the internal rate of return of the project. Additional knowledge on the external rate of return of the company is needed. Feel free to read more on this topic in Appendix 4A. In this course, we will consider simple investments only. For a simple investment, we use the two terms, rate of return and internal rate of return, inter-changeably because they are equal to each other. The ROR method decision rule: If i * MARR, the project is acceptable; If i * < MARR, the project is not acceptable. Example 11.2: Hallmark Chemical Corporation is planning to expand one of its propylene manufacturing facilities. The land costs $300,000, the building costs $600,000, the equipment costs $250,000, and an additional $100,000 start-up cost is needed. It is expected that the product will result in sales of $625,000 per year for 12 years, at which time the land can be sold for $300,000, the building for $200,000 (net proceeds after tax adjustment), and the equipment for $50,000 (net proceeds after tax adjustment). The annual disbursements for labor, materials, and all other expenses (including income taxes) are estimated to be $425,000. What is the IRR? If the company requires a minimum rate of return of 15% on projects, determine if this is a good investment.

12 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 12 of 36 PE, AE, and FE Methods for Comparing Mutually Exclusive Projects Mutually exclusive alternatives: (a) Where to get a loan, a mortgage (b) Where to buy a car, what car to buy, what machine to buy (c) Where to build a plant, what design to adopt The individual analysis method for comparing mutually exclusive alternatives or projects: The best project is the one with the largest PE(MARR), or AE(MARR), or FE(MARR) The incremental analysis method for comparing mutually exclusive alternatives: Examine the so-called "incremental investment". This method is used to determine if the additional investment is justified or not. If P B > P A, we call (B A) the incremental investment. Under these assumptions, we can use the following decision rule: Project B is better than project A if PE BA (MARR) > 0, or AE BA (MARR) > 0, or FE BA (MARR) > 0. If PE, or AE, or FE of (B A) is 0, we say that A and B are equivalent. Example 12.1: The following two alternative machines are being considered for a certain manufacturing process: (a) Machine A has a first cost of $60,000. Its salvage value at the end of 6 years of estimated service life is $22,000. Operating costs of this machine are estimated to be $5000 per year. Extra income taxes with this machine are estimated to be $1400 per year. (b) Machine B has a first cost of $35,000. Its salvage value at the end of 6 years' service is estimated to be negligible. The annual operating costs will be $8000. Compare these two alternatives. Use an interest rate of 10%. What decision rule would you use if the operating costs of machine A increase at a pace of $500 per year and the operating costs of machine B increase at a pace of 5% per year?

13 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 13 of 36 The ROR Method for Comparing Mutually Exclusive Projects Mutually exclusive projects: A, B, C, D We should not always pick the project with the largest internal rate of return. This means that we cannot select the best project based on individual ROR calculations. Example 13.1: Assume that MARR = 10%. The following two mutually exclusive projects have equal lives. Which one is better? Project A: P A = $10,000, i A * = 15% Project B: P B = $50,000, i B * = 14% Now consider the following two mutually exclusive projects with equal lives. Which one is better? Project A: P A = $10,000, i A * = 14% Project B: P B = $50,000, i B * = 15% The incremental ROR method for comparing mutually exclusive alternatives: When the ROR method is used for comparing mutually exclusive projects, the incremental analysis method has to be used. Consider two projects, A and B, and assume that B requires a larger investment. The following decision rule should be used: Project B is better than project A if i BA * > MARR. If i BA * = MARR, we say that A and B are equivalent. Example 13.2: Assume that MARR = 10% and n = 8 years. Which of the following three projects is the best? A: P = $20000, OR = $8000/year, OC = $4000/year, S = $8000 B: P = $10000, OR = $6000/year, OC = $3000/year, S = $2000 C: P = $40000, OR = $25000/year, OC = $12000/year, S = $10000 Example 13.3: The following data are given for three mutually exclusive projects. Which one is the best if MARR = 25%? Project Cash flow at point 0 IRR A -$ % B -$ % C -$ % C - B 27% A - B 23% A - C 15%

14 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 14 of 36 Other Considerations when Comparing Mutually Exclusive Projects Different projects may have different useful lives. However, projects should be compared over an equal time span (this is called the analysis period). The analysis period should be selected properly to make the comparisons valid. When the analysis period is specified, we have to estimate all cash flows for the duration of this analysis period. When analysis period is not specified and the required service life is very long, we often use the lowest common multiple of the lives of the projects as our analysis period. When the analysis period is not specified, we often make the following assumptions: (1) The required service life is very long; (2) Like-for-like replacement (3) The lowest common multiple of project lives is used as the analysis period. In this case, the AE method is the easiest to use. Problem 5.31 in the textbook: When the analysis period is specified and it is shorter than all project lives, we need to consider terminating the projects early in the analysis. Use Example 5.7 in the textbook. When the analysis period is specified and it is longer than all project lives, we need to find other alternatives so that we have cash flow diagrams covering the duration of the analysis period. Use Example 5.8 in the textbook. The analysis period may also be dictated by the choice of project. Refer to Example 5.9 in the textbook.

15 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 15 of 36 Capital Recovery Cost and Unit Cost Calculation Capital Costs versus Operating Costs: Sometimes we need to compare only the costs of two projects because the revenues are irrelevant or they are the same in both projects. The costs of each project may be divided into two categories: capital costs and operating costs. To find the annual equivalent costs (AEC) of each project, we may find the annual equivalent capital costs (CR) and the annual equivalent operating costs (AEOC) separately and then add them together. The annual equivalent capital cost is given a special name called capital recovery cost, denoted by CR. CR = P (A/P, i, N) - S (A/F, i, N) = (P - S) (A/P, i, N) + i S AEC = CR + AEOC Example 15.1: Consider the purchase of a piece of equipment. i = 10%, P = $100,000, N = 5 years, S = $20,000, OC 1 = $20,000 and increases by $5000 per year. Find the annual equivalent cost of purchasing and operating this equipment. Unit cost calculation: Problem 6.22 in the textbook: Break-even point calculation: Problem 6.15 in the textbook:

16 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 16 of 36 Application Examples Comparing Mutually Exclusive Alternatives Problem 6.20 in the textbook: Danford Company, a farm-equipment manufacturer, currently produces 20,000 units of gas filters per year. The following costs are reported based on the previous year s production: Item Expense ($) Direct materials 60,000 Direct labor 180,000 Variable overhead (power and water) 135,000 Fixed overhead (light and heat) 70,000 Total cost 445,000 It is anticipated that gas-filters will be used for another 5 years. If the company continues to produce this product in-house, the annual direct material costs will increase at the rate of 5%. (For example, the annual material costs during the next production year will be $63,000.) The direct labor will increase at the rate of 6% per year. The variable overhead costs would increase at the rate of 3%, but the fixed overhead would remain at the current level over the next 5 years. Tompkins Company has offered to sell Danford 20,000 units of gas filters for $25 per unit. If Danford accepts the offer, some of the manufacturing facilities currently used to manufacture the gas filter could be rented to a third party at annual rental of $35,000. The firm s interest rate is known to be 15%. What is the unit cost of buying the gas filter from the outside source? What is the unit cost of producing the gas filter in-house? Which option is better? Problem 5.29 in the textbook: A chemical company is considering two types of incinerators to burn solid waste generated by a chemical operation. Both incinerators have a burning capacity of 20 tons per day. The following data have been compiled for comparison: Incinerator A Incinerator B Installed cost $1,200,000 $750,000 Annual O&M costs $50,000 $80,000 Service life 20 years 10 years Salvage value $60,000 $30,000 Income taxes $40,000 $30,000 If the firm's MARR is known to be 13%, Which is a better option. Assume that incinerator B will be available in the future at the same cost. Problem 5.19 in the textbook: An electric motor is rated at 10 horsepower (HP) and costs $800. Its full load efficiency is specified to be 85%. A newly designed, high-efficiency motor of the same size has an efficiency of 90%, but costs $1200. It is estimated that the motors will operate at the rated 10 HP output for 1500 hours a year, and the cost of energy will be $0.07 per kilowatt-hour. Each motor is expected to have a 15-year life. At the end of 15 years, the first motor will have a salvage value of $50, and the second motor will have a salvage value of $100. Consider the MARR to be 8%. (Note: 1 HP = kW.) Determine which motor should be installed. What if the motors operate 1000 hours a year instead of 1500 hours a year? What are the annual operating hours to make these two motors equivalent?

17 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 17 of 36 Replacement Analysis Up to this point, the mutually exclusive alternatives that we have compared are new investment opportunities. In replacement analysis, we compare an existing asset (an old asset) with a proposed asset (a new asset). These two options are mutually exclusive. The question to be answered is whether to replace the old asset with the new asset. Defender: an existing asset (an old asset) Challenger: a proposed asset (a new asset) Current Market Value (MV) of the Defender: the current selling price or current salvage value Sunk costs: A sunk cost is any past cost that cannot be changed by any new investment decisions. Sunk costs should never be used in any engineering economic analysis. Example 17.1: A machine was bought 6 months ago for $1.5M. Its current market value = $0.5M. Operating costs = $0.5M/year. A new machine is available today for $2M with annual operating costs of $0.2M. Both can be used for 10 more years. Revenues are the same. What data should be used to answer the question whether to replace the old one with the new one? Trade-off between defender and challenger: Defender does not require new investment while challenger does. Challenger usually has lower operating costs than defender. Two approaches for comparing defender and challenger: (1) Cash flow approach: (2) Opportunity cost approach: This is the preferred one. Sometimes this one has to be used. Example 17.2: A machine now in use was bought 5 years ago for $ It can be sold today for $3500, but could be used for 3 more years (remaining useful life), at the end of which time it would have a salvage value of $1000. The annual operating and maintenance costs for the old machine amount to $10,000. A new machine can be purchased at an invoice price of $14,000 to replace the present equipment. Because of the nature of the manufactured product, the new machine also has an expected life of 3 years, and it will have a salvage value of $2000 at the end of that time. The new machine s expected operating and maintenance costs amount to $4000 for the first year and $5000 for each of the next 2 years. The firm s interest rate is 15%. Should the defender be kept for three years or replaced by the challenger right away? Illustrate both approaches.

18 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 18 of 36 Economic Life and Finite Horizon Replacement Analysis The economic life of an asset is defined to be the number of years of its service, which will result in the lowest annual equivalent cost. We may need to find the remaining economic life of the defender and the economic life of the challenger. For the defender, we have to use the opportunity cost approach. To find the economic life of an asset (either a defender or a challenger), we need to consider the options of keeping it for 1 year, 2 years, 3 years, etc. The duration that results in the lowest AEC is its economic life. Example 18.1 (Part of Problem 6.37 in the textbook): An existing machine was bought four years ago at a price of $25,000. At the time it was purchased, its estimated life was 7 years with a salvage value of $5000. Now it has a market value of $7700. If it is retained, its updated market values and operating costs for the next four years are as follows: Year Market Value Operating Costs 0 $ $ Use an interest rate of 12%. What is the remaining economic life of this asset? Replacement analysis over a finite planning horizon Problem 6.56 in the textbook: A company currently has a machine for a certain type of service. This defender has a physical life of 4 more years. A challenger on the market has a physical life of 6 years. The AECs of using either the defender or the challenger for different durations are: Duration AEC for Defender AEC for Challenger 1 $3200 $ The company only needs such a machine for the next 10 years. MARR is 12%. Assume that there are no better challengers over the next 10 years. What is the best replacement strategy? Problem 6.39 in the textbook: Given the following data for a challenger: P = $20,000 S n = 12, n OC n = (n 1) What is the economic life of this asset if MARR = 10%?

19 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 19 of 36 Replacement Analysis under the Infinite Planning Horizon The required service for the type of equipment under consideration is very long or infinite. To answer the question whether the defender should be replaced by the challenger now: (1) Find the economic lives of both the defender and the challenger and their corresponding minimum annual equivalent cost (AEC) values. In other words, find n D * and AEC D * for the defender and n C * and AEC C * for the challenger. (2) Compare AEC D * and AEC C * and pick the smaller one. If AEC D * is smaller, the defender should be kept, that is, the defender should not be replaced by the challenger now. If AEC C * is smaller, the defender should be replaced by the challenger now. If the defender should not be replaced now, how to find when the defender should be replaced? (1) If the defender should not be replaced now, we should keep the defender for at least n D * years, i.e., we should continue to use the defender for at least its remaining economic life. (2) To find out whether the defender should be replaced at the end of its economic life, or 1 year, or 2 years, or 3 years after its economic life, we need to calculate the cost of keeping the defender for one additional year and compare this year-by-year cost with AEC C * of the challenger. The smaller value should be selected. Problem 6.7 in the textbook: The annual equivalent costs of retaining a defender over its 3-year remaining life and the annual equivalent costs for its challenger over its 4-year physical life are as follows: Holding Period AEC for Defender AEC for Challenger 1 $3000 $ Assume a MARR of 12%. Should the defender be replaced by the challenger now? Without further calculations, how long should the defender be kept? Problem 6.55 in the textbook: A 6-year-old CNC machine that originally cost $8000 has a current market value of $1500. If the machine is kept in service for the next 5 years, its O&M costs and salvage value are estimated as follows: Operating and Maintenance Costs Salvage Year Operation & Repair Costs Due to Breakdowns Value 1 $1300 $600 $ It is suggested that the machine be replaced by a new CNC machine of improved design at a cost of $6000. It is believed that this purchase will completely eliminate the costs due to break-downs and that operation and repair costs will be reduced by $200 a year from those incurred with the old machine. Assume a 5-year economic life for the challenger and a $1000 terminal salvage value. The firm's MARR is 12%. Should the old machine be replaced now? If not now, when?

20 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 20 of 36 Depreciation and Book Depreciation Methods Capital assets, or fixed assets or simply called assets, are equipment and real property that have lives longer than one year. Depreciation has been used to describe the phenomenon that the value of an asset decreases over time. For example, a computer becomes obsolete in three years, a mining shovel is worn out after being used for 2000 hours, and a device becomes useless because the company does not need it any more. Depreciation may occur through physical depreciation or functional depreciation. This phenomenon of depreciation is also called economic depreciation. Because assets lose their values as time goes, there is the concept of accounting depreciation, which is the systematic allocation of an asset's value in portions over its depreciable life. Accounting depreciation has two different purposes: (1) For financial reporting: The accounting depreciation methods used for this purpose are called book depreciation methods. (2) For filing tax returns: The accounting depreciation methods used for this purpose are called tax depreciation methods. Relevant concepts for accounting depreciation: (1) Depreciable property: used in business; has a life longer than one year; and generally has a decreasing value (2) Cost base P = purchase price + freight cost + installation cost + site preparation cost (3) Salvage value S = selling price - costs of selling the asset (4) Useful life N = the period over which the asset is useful to the company Book Depreciation Methods: The Straight Line (SL) Method: D n = D = (P - S) / N, for n = 1, 2, 3,..., N. It is a constant. B n = P - n D, for n = 1, 2, 3,..., N. The book value decreases linearly. B N = S The Sum of Years Digits (SOYD) Method: SOYD = N = 0.5 N (N + 1) D n = (P - S) (N - n + 1) /SOYD, for n = 1, 2, 3,..., N B n = B n-1 - D n, for n = 1, 2, 3,..., N B 0 = P B N = S Problem 7.23 in the textbook: Upjohn Company purchased new equipment with an estimated useful life of 5 years. The cost of the equipment was $20,000, and the salvage value was estimated to be $3,000 at the end of year 5. Compute the annual depreciation amounts and the book values of the equipment at the end of each year through its 5-year life under each of the following methods of book depreciation: (a) Straight-line, (b) Sum-of-the-years -digits method.

21 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 21 of 36 Book Depreciation: Declining Balance Methods The SOYD method and the declining balance methods consider the fact that the amount of loss in value in an asset is larger in the earlier years of its life. They are called the accelerated methods. A Declining Balance Method uses a constant declining balance rate, denoted by d. According to this method, the value of an asset declines by such a constant rate each year. d = N 1 (Multiplier), where N is the estimated useful life, the commonly used Multipliers are: 100%: The (100%) declining balance method, or DB in short, 150%: The 150% declining balance method, or 150% DB in short, 200%: The double declining balance method, or DDB in short. B 0 = P, D n = d B n-1, B n = B n-1 D n = B n-1 (1 d), for n = 1, 2,, N D n = d (1 d) n-1 P, B n = P (1 d) n, n = 1, 2, N These methods do not guarantee that B N = S. Either under-depreciation or over-depreciation may occur. To make sure that no over-depreciation occurs, the depreciation amounts in later years are simply reduced to zero. To prevent under-depreciation, accountants often choose the optimal time to switch to the Straight-Line method. We illustrate these provisions through an example. Problem 7.23 in the textbook (continued): Upjohn Company purchased new equipment with an estimated useful life of 5 years. The cost of the equipment was $20,000, and the salvage value was estimated to be $3,000 at the end of year 5. Compute the annual depreciation amounts and the book values of the equipment at the end of each year through its 5-year life under each of the following methods of book depreciation: (c) DDB with B N = S. (d) DB with B N = S. (c): DDB with d = 40%: Year Calculated Calculated Adjusted B n Adjusted D n n DDB B n DDB D n 0 $20,000 0 $20,000 1 $12,000 $8,000 $12,000 $8,000 2 $7,200 $4,800 $7,200 $4,800 3 $4,320 $2,880 $4,320 $2,880 4 $2,592 $1,728 $3,000 $1,320 5 $1,555 $1,037 $3,000 $0 (d): DB with d = 20%: Year n The D Value if switching to S-L method at the beginning of year n Calculated DB B-n Calculated DB D n Adjusted D n Adjusted B n 0 $20,000 $20,000 1 $16,000 $4,000 $3,400 $4,000 $16,000 2 $12,800 $3,200 $3,250 $3,250 $12,750 3 $10,240 $2,560 $3,267 $3,250 $9,500 4 $8,192 $2,048 $3,620 $3,250 $6,250 5 $6,554 $1,638 $5,192 $3,250 $3,000

22 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 22 of 36 Tax Depreciation Methods Tax depreciation methods are specified by Canada Customs and Revenue Agency (CCRA) for calculation of allowed deductions due to investments in capital assets when filing tax returns. Taxable income = Gross income Allowed deductions Income tax = Tax rate Taxable income Allowed deductions include operating costs (salaries, wages, materials, utilities, interest charges on debt, etc) and the so-called capital cost allowance due to investments in capital assets. Types of Expenditures in a Business: (1) Operating Expenditures: These expenditures are regular and on-going costs associated with providing products and/or services. They include salaries and wages of employees, raw material costs, utility costs, maintenance costs, utility costs, interest charges on borrowed money, etc. These costs are allowed deductions in annual income tax calculations. (2) Capital Expenditures: These expenditures are used to acquire assets with lives longer than one year. For example, purchasing land, building, equipment, etc. Capital expenditures are not tax deductible directly. Rather, they have to be spread over the useful lives of the assets following rules specified by CCRA. Tax Depreciation Methods: The allowed deduction in each year due to investment in a capital asset is called Capital Cost Allowance (CCA) by CCRA. The remaining portion of the total capital investment is called Undepreciated Capital Cost (UCC or simply U). (1) The Straight Line Method: This method applies to assets such as patents, franchise, concessions, and licenses. CCA = CCA n = (P S)/N, U n = P CCA n, 1 n N (2) The Declining Balance Method: This method is used for most assets. The declining balance rate is specified by CCRA for each class of assets. See Table 7.1 in the textbook. CCA n = P d (1 d) n-1, U n = P (1 d) n, 1 n N (3) The Available-for-use Rule: The asset must be available for use before you are allowed to deduct its CCA in income tax calculations. In this course, we assume that this rule is always satisfied. (4) The 50% rule: For newly acquired assets, only 50% of its cost base may be used in calculation of its first year s CCA amount. Because of this rule, we have the following revised equations for the declining balance method: CCA 1 = P d/2, CCA n = P d (1 d/2)(1 d) n-2, 2 n N U n = P (1 d/2)(1 d) n-1, 1 n N In this course, the 50% rule must be used unless otherwise specified. (5) Carry forward of CCA when filing tax returns. (6) The Asset Class Accounting Procedure: All assets in the same class have the same declining balance rate. Thus, only one account is needed for each class of assets for CCA calculations. As a result, the 50% rule applies to net acquisitions (= purchases disposals) in a year. See Figure 7.6 in the textbook (Schedule 8). Problem 7.34 in the textbook: Otto-Rentals Ltd is setting up a new car rental corporation. In the first year they purchase 12 vehicles for a total of $148,000. In year 2, they buy two more vehicles for $27,000. In year 3, they sell 4 vehicles for $29,000 and buy 5 new cars for $71,000. In year 4, they sell 3 vehicles for $16,000. These vehicles belong in CCA Class 16 (it has a CCA rate of 40%, see Table 7.1). What are the maximum CCA in each year and what are the UCC at the end of each year? Answers: CCA n : $29,600 $52,760 $45,456 $29,274 U n : $118,400 $92,640 $89,184 $43,910

23 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 23 of 36 Disposal Tax Effects Depreciable assets: Those assets that generally have declining values as they age. CCA may be claimed for these assets as they are used. Non-depreciable assets: Land, stocks, bonds, etc. No CCA can be claimed on these assets. When a capital asset is sold, its undepreciated capital cost is seldom equal to the salvage value. This difference has tax implications. We illustrate how to calculate these disposal tax effects. Notation: P: Initial Cost Based S: Salvage value at disposal U: Undepreciated capital cost at disposal t: Tax rate on regular income (1) When S > P, the amount of (S - P) is called Capital Gain. The tax rate for capital gains is approximately half of the tax rate for regular income, according to the current tax rules (since Oct 18, 2000). It was 75% before Feb 28, 2000 and 2/3 in between. We will use t CG to denote the tax rate on capital gains income. Capital Gains Tax = t CG (S - P), for S > P (2) When S < P for a non-depreciable asset, the amount of (P - S) is called Capital Loss. The capital loss may be used to offset the capital gain the company may have in the same year. As a result, tax savings can be generated by capital losses. Capital Loss Tax Savings = t CG (P - S), for S < P (3) When S < P for a depreciable asset, we need to compare the salvage value S with the undepreciated capital cost U. If S > U, we have a CCA Gain or Recaptured CCA in the amount of (S - U), which is taxable. If S < U, we have a CCA Loss in the amount of (U - S), which can be used to reduce tax payment. Disposal Tax Effect G = t (U - S), for S < P (4) The concept of Net Salvage Value, NS: NS = S + (Tax Savings due to disposal) - (Tax payments due to disposal) (5) When S > P for a depreciable asset, which does not happen very often, both capital gains and CCA gains are realized. These two types of gains are taxed at different rates. NS = S - t CG (S - P) - t (P - U) Example 23.1: A company purchased a piece of land for $300,000 in Downtown Edmonton 10 years ago. A big office building was built on the lot for $3 million in the second year (i.e., 9 years ago). Recently, the land and the building were sold for $2 million and $2.8 million, respectively. Assume that the company used a CCA rate of 10% during its 9 years of ownership of the building. The company's tax rate is 35% and the tax rate on capital gains is 17.5%. What are the net cash flows this year due to the selling of the building and the land? What if the building was sold for $3,200,000?

24 ENGM 310 Engineering Economy Lecture Notes (MJ Zuo) Page 24 of 36 Income Tax Calculations General Tax Framework: (1) Taxable Income = Gross Income Allowed Deductions (2) Income Tax = Tax Rate Taxable Income (3) Tax regulations are complicated. Our coverage in the textbook is simplified. (4) Different companies have different tax rates: Small business, Manufacturing, Non-manufacturing (5) Different types of income have different tax rates (6) Timing of tax payment: assumed to be December 31 Corporate Income Taxes: (1) We generally consider two types of income: operating income and disposal of capital assets (2) The tax rate for operating income is usually given. The tax rate for capital gains is approximately 50% of the tax rate for operating income (effective October 18, 2000). It was ¾ of the tax rate for operating income before October 18, 2000, as quoted in the textbook. (3) Major categories of allowed deductions are: Operating costs: salaries & wages, materials, utilities, etc. CCA: calculated based on capital expenditures and specified CCA rates. Interest charges on debt (4) Operating losses and capital losses can be carried backward for 3 years and forward for 7 years. (5) The tax rate for operating income is a sum of the federal tax rate and the provincial tax rate. (6) Corporate tax rates are considered non-progressive. Thus, we use marginal tax rate, incremental tax rate, and average tax rate interchangeably. (7) Tables 8.1 and 8.2 list the federal tax rates and the provincial tax rates separately (for 1999). (8) In this course, we assume that it is always possible to claim the maximum CCA in each year. Personal Income Taxes: (1) Personal income tax rates are progressive: The higher the income, the higher the tax rate. Federal tax rates for 1999 (Table 8.3): 0 29,590: 17%, 29,591 59,180: 26% 59,181 : 29% Alberta tax rates for 2001: 10% flat of federal tax. Other provinces tax rates are also progressive (Table 8.4). (2) Income categories: salaries/wages, interest income, pension income, dividends income, rental income, capital gains, etc. (3) Tax rates on salaries/wages, interest income, dividends income, and capital gains income are different (Table 8.5) (4) Allowed deductions: RRSP contributions, childcare expenses, tuitions, moving expenses, medical expenses, professional membership dues, etc. Problem 8.17 in the textbook: A small manufacturing company in Aylmer, Quebec has an estimated annual taxable income of $95,000. Owing to an increase in business, the company is considering purchasing a new machine that will generate additional (before-tax) net annual revenue of $50,000 over the next 5 years. The new machine requires an investment of $100,000. It has a CCA rate of 30%. (1) What is the increase in tax payment due to the purchase of the new machine? (2) What is the tax rate applicable to the investment in the machine? Problem 8.21 in the textbook: Anne Johnson completed her engineering degree in December She moved to Hinton, Alberta and started to work in January Assume that her taxable income for 1999 was $38,000, solely from her salary. What was her income tax amount for the year? What were her average tax rate and marginal tax rate?