Chapter 1. Engineering Economy is a collection of techniques that simplify comparisons of

Transcription

1 Chapter Engineering Economy Engineering Economy is a collection of techniques that simplify comparisons of alternatives on an economic basis. Engineering Economy involves formulating, estimating and evaluating the economic outcomes when alternatives are available to accomplish a defined purpose. What Engineering Economy is not? Engineering Economy is not a method or process for determining what the alternatives are. Therefore, Engineering Economy begins only after the alternatives have been identified Basic Concepts Alternatives An alternative is a stand-alone option (solution) for a given situation. Alternatives (economically) involve: - First cost (including purchase price, development and installation) - Useful life - Estimated annual income and expenses - Salvage value (resale or trade-in value) - Interest rate (rate of return - ROR) - Possibly inflation and income tax effects 1

2 Cash flow Cash flow is the estimated inflows (revenues) and outflows (costs) of money. These estimates are the heart of Engineering Economy analysis. Without cash flow estimates over a stated time period, no Engineering Economy study can be conducted Alternative selection The measure-of-worth values are compared and an alternative is selected. This is the result of the Engineering Economy analysis. If only one feasible alternative is defined, a second is often present in the form of the donothing alternative Evaluation criteria The best alternative will be selected according to: - Highest income (benefit) - Lowest cost - On tangible basis of financial units ($, TL) Intangible factors are difficult to quantify like goodwill, convenience, friendship Time value of money Time value of money is the change in the amount of money over a given time period. This is one of the most important concepts in Engineering Economy (borrowing money and owing more after a while). 2

3 1.3. Interest rate or Rate of Return (ROR) Computationally, the interest is the difference between an ending amount of money and the beginning amount. IIIIIIIIIIIIIIII = aaaaaaaaaa oooooooo nnnnnn oooooooooooooooo aaaaaaaaaaaa Interest rate is expressed as a percentage of the original amount (principal money) over a specific time unit. IIIIIIIIIIIIIIII rrrrrrrr (%) = iiiiiiiiiiiiiiii aaaaaaaaaaaaaa pppppp tttttttt uuuuuuuu oooooooooooooooo aaaaaaaaaaaa 100% The time unit of interest rate is called interest period. There are two perspectives to an amount of interest: Interest paid: when a person or organization borrowed money (obtained a loan) and repays a larger amount Interest earned: when a person or organization saved, invested or lent money and obtains a return of a larger amount. Example 1.1 An employee at LaserKinetics.com borrows $10,000 on May 1 and must repay a total of $10,700 exactly 1 year later. Determine the interest amount and the interest rate paid. 3

4 Example 1.2 A company plans to borrow $20,000 from a bank for 1 year at 9% interest for new recording equipment. (a) Compute the interest and the total amount due after 1 year. (b) Construct a column graph that shows the original loan amount and total amount due after 1 year used to compute the loan interest rate of 9% per year. Comment: Example 1.3 (a) Calculate the amount deposited 1 year ago to have $1000 now at an interest rate of 5% per year. (b) Calculate the amount of interest earned during this time period. 4

5 1.4. Terminology and Symbols The equations and procedures of engineering economy utilize the following terms and symbols. P = value or amount of money at the present or time 0. Also P is referred to as present worth (PW) or present value (PV). F = value or amount of money at some future time. Also F is called future worth (FW) and future value (FV). A = series of consecutive, equal, end-of-period amounts of money. Also A is called the annual worth (AW) n = number of interest periods; years, months, days i = interest rate per time period; percent per year, percent per month t = time, stated in periods; years, months, days Example 1.4 Today, we borrowed $5000 to purchase furniture for our new house. We can repay the loan in either of the two ways described below. Determine the engineering economy symbols and their value for each option. (a) Five equal annual installments with interest based on 5% per year. 5

6 (b) One payment 3 years from now with interest based on 7% per year. (a) The repayment schedule requires an equivalent annual amount A, which is unknown. P = $5000, i = 5% per year, n = 5 years, A =? (b) Repayment requires a single future amount F, which is unknown. P = $5000, i = 7% per year, n = 3 years, F =? Example 1.5 You plan to make a lump-sum deposit of $5000 now into an investment account that pays 6% per year, and you plan to withdraw an equal end-of-year amount of $1000 for 5 years, starting next year. At the end of the sixth year, you plan to close your account by withdrawing the remaining money. Define the engineering economy symbols involved. All five symbols are present, but the future value in year 6 is the unknown. P = $5000 A = $1000 per year for 5 years F =? at end of year 6 i = 6% per year n = 5 years for the A series and 6 for the F value 6

7 Example 1.6 Last year Jane s grandmother offered to put enough money into a savings account to generate $5000 in interest this year to help pay Jane s expenses at college. (a) Identify the symbols, and (b) calculate the amount that had to be deposited exactly 1 year ago to earn $5000 in interest now, if the rate of return is 6% per year. (a) Symbols P (last year is -1) and F (this year) are needed. P =? i = 6% per year n = 1 year F = P + interest =? + $5000 (b) Let F = total amount now and P = original amount. We know that F P = $5000 is accrued interest. Now we can determine P. F = P + Pi The $5000 interest can be expressed as Interest = F P = ( P + Pi ) P = Pi $5000 = P (0.06) therefore, P = $83, Cash Flows: Estimation and Diagramming Cash inflows are the receipts, revenues, incomes, and savings generated by project and business activity. A plus sign indicates a cash inflow. 7

8 Cash outflows are costs, disbursements, expenses, and taxes caused by projects and business Cash flow activity. A negative or minus sign indicates a cash outflow. The end-of-period convention means that all cash inflows and all cash outflows are assumed to take place at the end of the interest period in which they actually occur. When several inflows and outflows occur within the same period, the net cash flow is assumed to occur at the end of the period. Example 1.7 Each year Exxon-Mobil expends large amounts of funds for mechanical safety features throughout its worldwide operations. Carla Ramos, a lead engineer for Mexico and Central American operations, plans expenditures of $1 million now and each of the next 4 years just for the improvement of field-based pressure-release valves. Construct the cash flow diagram to find the equivalent value of these expenditures at the end of year 4, using a cost of capital estimate for safety-related funds of 12% per year. Example 1.8 An electrical engineer wants to deposit an amount P now such that she can withdraw an equal annual amount of A1 = $2000 per year for the first 5 years, starting 1 year after the deposit, and a different annual withdrawal of A2 = $3000 per year for the following 3 years. How would the cash flow diagram appear if i = 8.5% per year? 8

9 Example 1.9 A rental company spent $2500 on a new air compressor 7 years ago. The annual rental income from the compressor has been $750. The $100 spent on maintenance the first year has increased each year by $25. The company plans to sell the compressor at the end of next year for $150. Construct the cash flow diagram from the company s perspective and indicate where the present worth now is located. 9

10 Example 1.10 Manufacturers make backup batteries for computer systems available to Batteries+ dealers through privately owned distributorships. In general, batteries are stored throughout the year, and a 5% cost increase is added each year to cover the inventory carrying charge for the distributorship owner. Assume you own the City Center Batteries+ outlet. Make the calculations necessary to show which of the following statements are true and which are false about battery costs. (a) The amount of $98 now is equivalent to a cost of $ one year from now. (b) A truck battery cost of $200 one year ago is equivalent to $205 now. (c) A $38 cost now is equivalent to $39.90 one year from now. (d) A $3000 cost now is equivalent to $ one year earlier. (e) The carrying charge accumulated in 1 year on an investment of $20,000 worth of batteries is $1000. (a) Total amount accrued = 98(1.05) = $ $105.60; therefore, it is false. Another way to solve this is as follows: Required original cost is /1.05 = $ $98. 10

11 (b) Equivalent cost 1 year ago is /1.05 = $ $200; therefore, it is false. (c) The cost 1 year from now is $38(1.05) = $39.90; true. (d) Cost now is (1.05) = $ $3000; false. (e) The charge is 5% per year interest, or $20,000(0.05) = $1000; true Simple and Compound Interest Simple interest is calculated using the principal only. Simple interest = (principal)(number of periods)(interest rate) I = P*n*i Example 1.11 GreenTree Financing lent an engineering company $100,000 to retrofit an environmentally unfriendly building. The loan is for 3 years at 10% per year simple interest. How much money will the firm repay at the end of 3 years? The interest for each of the 3 years is Interest per year = $100,000(0.10) = $10,000. Total interest for 3 years = $30,000. The amount due after 3 years is $130, For compound interest, the interest accrued for each interest period is calculated on the principal plus the total amount of interest accumulated in all previous periods. Thus, compound interest means interest on top of interest. Compound interest = (principal + all accrued interest)(interest rate) Example

12 Assume an engineering company borrows $100,000 at 10% per year compound interest and will pay the principal and all the interest after 3 years. Compute the annual interest and total amount due after 3 years. Interest, year 1: 100,000(0.10) = $10,000 Total due, year 1: 100, ,000 = $110,000 Interest, year 2: 110,000(0.10) = $11,000 Total due, year 2: 110, ,000 = $121,000 Interest, year 3: 121,000(0.10) = $12,100 Total due, year 3: 121, ,100 = $133,100 Another solution Year 1: $100,000(1.10) 1 = $110,000 Year 2: $100,000(1.10) 2 = $121,000 Year 3: $100,000(1.10) 3 = $133,100 Total due after n years = principal(1 + interest rate) n years = P (1 + i ) n 1.7. Minimum Attractive Rate of Return The Minimum Attractive Rate of Return (MARR) is a reasonable rate of return that must be higher than the ROR of a bank or a safe investment. 12

13 MARR is also referred to as the hurdle rate, cutoff rate, benchmark rate, and minimum acceptable rate of return. Figure: Size of MAAR relative to other rate of return values Rule of 72: Estimating Doubling Time and Interest Rate (in 6 th edition of the text book) The rule of 72 for compound interest rates can be used to estimate I or n, to double the amount of deposited money in size. EEssssssssssssssss nn = 72 ii aannnn EEEEEEEEEEEEEEEEEE ii = 72 nn If the interest rate is simple, the rule of 100 may be used in the same way. 13

14 Example 1.13 What is the interest rate if we want to make our deposited money double in 10 years. EEEEEEEEEEEEEEEEEE ii = = 7.2% Example 1.14 After how many years our deposited money will be doubled if the interest rate is 5%. EEEEEEEEEEEEEEEEEE nn = 72 5 = 14.4 yyeeeeeeee Example 1.15 Solve examples 1.13 and 1.14 for the simple interest rate. EEEEEEEEEEEEEEEEEE ii = = 10% EEEEEEEEEEEEEEEEEE nn = = 20 yyeeeeeeee 14