1. ENGR 112 Economic Analysis

2. Engineering Economic Analysis Evaluates the monetary aspects of the products, projects, and processes that engineers design Aids in the decision making process when alternatives are compared Especially useful when resources are limited Replace vs. keep Build vs. buy

3. Time Value of Money $1 today is more valuable than $1 a year later Engineering economy adjusts for the time value of money to balance current and future revenues and cost 05 10 Years

4. Capital vs. Interest Capital Invested money and resources Whoever owns it should expect a return from whoever uses it Bank lends you money Buy a car Go to college Firm invests in a project Buy equipment Buy “knowledge”

5. Capital vs. Interest Interest Return on capital Typically expressed as an interest rate for a year Interest rates are needed to evaluate engineering projects!! Interest rate = Interest $ amount Capital $ amount

6. “Interest”-ing Example The engineering group of Baker Designs must decide whether to spend $90K (K for thousands) on a new project. This project will cost $5K per year for operations, and it will increase revenues by $20K annually. Both the costs and the revenues will continue for 10 years. Should the project be done? Eschenbach, T.G., Engineering Economy: Applying Theory to Practice, Irwin, 1995, p.22

7. Types of Interest Nomenclature P = Initial deposit (or principal) N = Number of years i = Interest rate per year F = Future value after N years

8. Types of Interest Simple Interest The interest calculated for N periods is based on the initial deposit F 0 = P F 1 = P + P*i = P*(1 + i) F 2 = P + P*i + P*i = P* (1 + i + i) = P* (1 + 2i) … F n = P + P*i + P*i + … = P* (1 + i + i + …) = P* (1 + Ni)

9. Types of Interest Compound Interest Interest type used in engineering economic evaluations Interest is computed on the current balance that has not yet been paid Includes accrued interest F 0 = P F 1 = P + P*i = P *(1 + i) F 2 = P 1 + P 1 * i = P 1 *(1+i) = P *(1+i) *(1+i) = P *(1+i) 2 … F n = P n-1 + P n-1 * i = P n-1 *(1+i) = P *(1+i) n-1 *(1+i) = P *(1+i) n

10. Simple vs. Compound Interest If $100 is deposited in a savings account, how much is in the account at the end of each year for 20 years, if interest is deposited in the account and no withdrawals are made? Assume a 5% interest per year.

11. Simple vs. Compound Interest Compound Simple

12. In Class Problem #1 If $500 is deposited in a savings account, would a 5% simple interest rate be better than a 3% compound interest rate if you were planning to keep the money for 3 years? For 35 years? Assume that all earned interest is deposited in the account and no withdrawals are made.

13. Example #1 Sam Bostro borrows $4000 from his parents for his final year of college. He agrees to repay it 3 years later in one payment to which a 7% compound interest rate will be applied. a) How much does he repay? F = (4000)(1 + 0.07) 3 = $4,900.17 b) How much of this is interest and how much is principal? F = P + I  P = $4,000; I = $900.17

14. Example #2 Susan Cardinal deposited $500 in her savings account and six years later the account has $600 in it. What compound rate of interest has Susan earned on her capital? F = P(1+i) n  i = [F/P] 1/n – 1 i = [600/500] 1/6 – 1 =.031 i = 3.1%

15. In Class Problem #2 Your friend is willing to loan you $1500 to buy a new computer, but you must agree to pay him back $2500 when you graduate in 4 years? What is the compound interest rate you will be paying your friend?

16. Cash Flow Diagrams Pictorial description of when and how much money is spent or received Summarizes the economic aspects of an engineering project 012345

17. Cash Flow Categories First Cost Expense to build or to buy and install Operation and maintenance (O&M) Annual expenses (electricity, labor, minor repairs, etc.) Salvage value Receipt at project termination for sale or transfer of equipment There can also be a salvage COST

18. Cash Flow Categories Revenues Annual receipts due to sale of products or services Overhauls Major capital expenditures that occurs part way through the life of the asset Prepaid expenses Annual expenses that must be paid in advance (e.g., leases, insurance)

19. Examples Ernie’s Earthmoving is considering the purchase of a piece of heavy equipment. What is the cash flow diagram if the following cash flows are anticipated? First cost$120K O&M cost$30k per year Overhaul cost$35K in year 3 Salvage value$40K after 5 years How would the cash flow diagram change if Ernie decides to lease the equipment (at $25K per year) instead of purchasing it? a) b)

20. Solution 012345 012345 a) b) 120K 30K 30K + 35K 30K 40K 25K 25 + 30K 25 + 30K 25K + 30K + 35K 25K + 30K

21. Frequency of Compounding Interest rates are typically specified on an annual basis However, interest is often compounded more often Semiannually Quarterly Monthly Daily

22. Frequency of Compounding How does that change our formula? m = # of compounding periods per year n = # of compounding periods n = m x N Quarterly?  Monthly?  m = 4 n = 4N  F = P(1+i/m) m*N = P(1+i/4) 4N m=12 n = 12N  F = P(1+i/12) 12N

23. Example Suppose you borrow $3000 at the beginning of your senior year to meet college expenses. If you make no payments for 10 years and then repay the entire amount of the loan, including accumulated interest, how much money will you owe? Assume interest is 6% per year, compounded a) Annuallym=1P=3,000 b) Quarterlym=4N=10 years c) Monthlym=12i=6% per year d) Dailym=365 How significant is the frequency of compounding?

24. Solution F A = P(1+i/ 1 ) 1N = 3000(1+0.06/ 1 ) 10 = $5,372.54 F Q = P(1+i/ 4 ) 4N = 3000(1+0.06/ 4 ) 40 = $5,442.05 F M = P(1+i/ 12 ) 12N = 3000(1+0.06/ 12 ) 120 = $5,458.19 F D = P(1+i/ 365 ) 1N = 3000(1+0.06/ 365 ) 3650 = $5,466.08

25. Solution

26. In Class Problem #3 You need to borrow $500 to pay for the text books you will need for your sophomore level courses. Which is the better loan assuming that you will pay the entire loan amount plus interest when you graduate in 3 years?  3.0% compounded annually  2.5% compounded quarterly