1. Engineering Economics Contemporary Engineering Economics, 5th edition, © 2010

2. Understanding Money and Its Management – Main Focus 1. If payments occur more frequently than annual, how do you calculate economic equivalence? 2.If interest period is other than annual, how do you calculate economic equivalence? 3.How are commercial loans structured? 4.How would you manage your debt?

3. Nominal Versus Effective Interest Rates  Nominal Interest Rate: Interest rate quoted based on an annual period  Effective Interest Rate: Actual interest earned or paid in a year or some other time period Contemporary Engineering Economics, 5th edition, © 2010

4. Financial Jargon Contemporary Engineering Economics, 5th edition, © 2010 Nominal interest rate Annual percentage rate (APR) Interest period 18% Compounded Monthly

5. What It Really Means? Interest rate per month (i) = 18%/12 = 1.5% Number of interest periods per year (N) = 12 In words, Bank will charge 1.5% interest each month on your unpaid balance, if you borrowed money. You will earn 1.5% interest each month on your remaining balance, if you deposited money. Question: Suppose that you invest $1 for 1 year at 18% compounded monthly. How much interest would you earn? Contemporary Engineering Economics, 5th edition, © 2010

6. Effective Annual Interest Rate (Yield) Formula: r = nominal interest rate per year i a = effective annual interest rate M = number of interest periods per year Example: 18% compounded monthly What It really Means 1.5% per month for 12 months or 19.56% compounded once per year Contemporary Engineering Economics, 5th edition, © 2010

7. Practice Problem Suppose your savings account pays 9% interest compounded quarterly. (a) Interest rate per quarter (b) Annual effective interest rate (i a ) (c) If you deposit $10,000 for one year, how much would you have? Solution: Contemporary Engineering Economics, 5th edition, © 2010

8. Nominal and Effective Interest Rates with Different Compounding Periods Effective Rates Nominal Rate Compounding Annually Compounding Semi-annually Compounding Quarterly Compounding Monthly Compounding Daily 4%4.00%4.04%4.06%4.07%4.08% 55.005.065.095.125.13 66.006.096.146.176.18 77.007.127.197.237.25 88.008.168.248.308.33 99.009.209.319.389.42 1010.0010.2510.3810.4710.52 1111.0011.3011.4611.5711.62 1212.0012.3612.5512.6812.74

9. Why Do We Need an Effective Interest Rate per Payment Period? Contemporary Engineering Economics, 5th edition, © 2010 Payment period Interest period Payment period Interest period Whenever payment and compounding periods differ from each other, one or the other must be transformed so that both conform to the same unit of time.

10. Effective Interest Rate per Payment Period (i)  Formula:  C = number of interest periods per payment period  K = number of payment periods per year  CK = total number of interest periods per year, or M  r/K = nominal interest rate per payment period Functional Relationships among r, i, and i a, where interest is calculated based on 9% compounded monthly and payments occur quarterly Contemporary Engineering Economics, 5th edition, © 2010

11. Effective Interest Rate per Payment Period with Continuous Compounding  Formula: With continuous compounding Example: 12% compounded continuously (a) effective interest rate per quarter (b) effective annual interest rate Contemporary Engineering Economics, 5th edition, © 2010

12. Case 0: 8% compounded quarterly Payment Period = Quarter Interest Period = Quarterly 1 interest period Given r = 8%, K = 4 payments per year C = 1 interest period per quarter M = 4 interest periods per year 2 nd Q3 rd Q5th Q 1 st Q

13. Contemporary Engineering Economics, 5th edition, © 2010 Case 1: 8% compounded monthly Payment Period = Quarter Interest Period = Monthly 3 interest periods Given r = 8%, K = 4 payments per year C = 3 interest periods per quarter M = 12 interest periods per year 2 nd Q3 rd Q5th Q 1 st Q

14. Contemporary Engineering Economics, 5th edition, © 2010 Case 2: 8% compounded weekly Payment Period = Quarter Interest Period = Weekly 13 interest periods Given r = 8%, K = 4 payments per year C = 13 interest periods per quarter M = 52 interest periods per year 2 nd Q3 rd Q5th Q 1 st Q

15. Contemporary Engineering Economics, 5th edition, © 2010 Case 3: 8% compounded continuously Payment Period = Quarter Interest Period = Continuously  interest periods Given r = 8%, K = 4 payments per year 2 nd Q3 rd Q5th Q 1 st Q

16. Contemporary Engineering Economics, 5th edition, © 2010 Summary: Effective Interest Rates per Quarter at Varying Compounding Frequencies Case 0Case 1Case 2Case 3 8% compounded quarterly 8% compounded monthly 8% compounded weekly 8% compounded continuously Payments occur quarterly 2.000% per quarter 2.013% per quarter 2.0186% per quarter 2.0201% per quarter

17. Equivalence Calculations using Effective Interest Rates  Step 1: Identify the payment period (e.g., annual, quarter, month, week, etc)  Step 2: Identify the interest period (e.g., annually, quarterly, monthly, etc)  Step 3: Find the effective interest rate that covers the payment period. Contemporary Engineering Economics, 5th edition, © 2010

18. Case I: When Payment Period is Equal to Compounding Period  Step 1: Identify the number of compounding periods (M) per year  Step 2: Compute the effective interest rate per payment period (i)  Step 3: Determine the total number of payment periods (N) Contemporary Engineering Economics, 5th edition, © 2010

19. Example 4.4: Calculating Auto Loan Payments  Given:  MSRP = $20,870  Discounts & Rebates = $2,443  Net sale price = $18,427  Down payment = $3,427  Dealer’s interest rate = 6.25% APR  Length of financing = 72 months  Find: the monthly payment (A) Solution: Cont emporary Engineering Economics, 5th edition, © 2010

20. Dollars Down in the Drain  Suppose you drink a cup of coffee ($3.00 a cup) on the way to work every morning for 30 years. If you put the money in the bank for the same period, how much would you have, assuming your accounts earns a 5% interest compounded daily.  NOTE: Assume you drink a cup of coffee every day including weekends. Solution: Payment period = daily Compounding period = daily Contemporary Engineering Economics, 5th edition, © 2010

21. Case II: When Payment Periods Differ from Compounding Periods  Step 1: Identify the following parameters. M = No. of compounding periods K = No. of payment periods per year C = No. of interest periods per payment period  Step 2: Compute the effective interest rate per payment period.  For discrete compounding  For continuous compounding  Step 3: Find the total no. of payment periods.  N = K (no. of years)  Step 4: Use i and N in the appropriate equivalence formula. Contemporary Engineering Economics, 5th edition, © 2010

22. Example 4.5 Compounding Occurs More Frequently than Payments are Made (Discrete Case)  Given: A = $1,500 per quarter, r = 6% per year, M = 12 compounding periods per year, and N = 2 years  Find: F  Step 1:  M = 12 compounding periods/year  K = 4 payment periods/year  C = 3 interest periods per quarter  Step 2:  Step 3: N = 4(2) = 8 Solution: F = $1,500 (F/A, 1.5075%, 8) = $14,216.24 Contemporary Engineering Economics, 5th edition, © 2010

23. Example 4.6 Compounding is Less Frequent than Payments  Given: A = $500 per month, r = 10% per year, M = 4 quarterly compounding periods per year, and N = 10 years  Find: F  Step 1:  M = 4 compounding periods/year  K = 12 payment periods/year  C = 1/3 interest period per quarter  Step 2:  Step 3: N = 4(2) = 8 Solution: F = $500 (F/A, 0.826%, 120) = $101,907.89 Contemporary Engineering Economics, 5th edition, © 2010

24. A Decision Flow Chart on How to Compute the Effective Interest Rate per Payment Period Contemporary Engineering Economics, 5th edition, © 2010

25. Key Points Financial institutions often quote interest rate based on an APR. In all financial analysis, we need to convert the APR into an appropriate effective interest rate based on a payment period. When payment period and interest period differ, calculate an effective interest rate that covers the payment period. Then use the appropriate interest formulas to determine the equivalent values