1. Contemporary Engineering Economics, 4 th edition, © 2007 Equivalence Analysis using Effective Interest Rates Lecture No.11 Chapter 4 Contemporary Engineering Economics Copyright © 2006

2. Contemporary Engineering Economics, 4 th edition, © 2007 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.

3. Contemporary Engineering Economics, 4 th edition, © 2007 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) i = r/M  Step 3: Determine the total number of payment periods (N) N = M ×(number of years)

4. Contemporary Engineering Economics, 4 th edition, © 2007 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)

5. Contemporary Engineering Economics, 4 th edition, © 2007 Solution: Step 1: M = 12 Step 2: i = r/M = 6.25%/12 = 0.5208% per month Step 3: N = (12)(6) = 72 months Step 4: A = $15,000(A/P, 0.5208%,72) = $250.37

6. Contemporary Engineering Economics, 4 th edition, © 2007 Dollars Up in Smoke What three levels of smokers who bought cigarettes every day for 30 years at $4.75 a pack would have if they had instead banked that money each week: Level of smokerWould have had 1 pack a day 2 packs a day 3 packs a day $132,110 $264,220 $396,331 Note: Assumes constant price per pack, the money banked weekly and an annual interest rate of 5.5%

7. Contemporary Engineering Economics, 4 th edition, © 2007 Sample Calculation: One Pack per Day Step 1: Determine the effective interest rate per payment period. Payment period = weekly “5.5% interest compounded weekly” i = 5.5%/52 = 0.10577% per week Step 2: Compute the equivalent value at 30 years from now. Weekly deposit amount A = $4.75 x 7 = $33.25 per week Total number of deposit periods N = (52 weeks/yr.)(30 years) = 1,560 weeks F = $33.25 (F/A, 0.10577%, 1560) = $132,110

8. Contemporary Engineering Economics, 4 th edition, © 2007 Practice Problem Suppose you drink a cup of Starbuck 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.

9. Contemporary Engineering Economics, 4 th edition, © 2007 Solution Payment period: Daily Compounding period: Daily

10. Contemporary Engineering Economics, 4 th edition, © 2007 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  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

11. Contemporary Engineering Economics, 4 th edition, © 2007 Discrete Case: Quarterly deposits with Monthly compounding  Step 1: M = 12 compounding periods/year K = 4 payment periods/year C = 3 interest periods per quarter  Step 2:  Step 3: N = 4(3) = 12  Step 4: F = $1,000 (F/A, 3.030%, 12) = $14,216.24 F = ? A = $1,000 0 1 2 3 4 5 6 7 8 9 10 11 12 Quarters Year 1Year 2Year 3

12. Contemporary Engineering Economics, 4 th edition, © 2007 Continuous Case: Quarterly deposits with Continuous compounding  Step 1: K = 4 payment periods/year C =  interest periods per quarter  Step 2:  Step 3: N = 4(3) = 12  Step 4: F = $1,000 (F/A, 3.045%, 12) = $14,228.37 F = ? A = $1,000 0 1 2 3 4 5 6 7 8 9 10 11 12 Quarters Year 2Year 1Year 3

13. Contemporary Engineering Economics, 4 th edition, © 2007 Practice Problem A series of equal quarterly payments of $5,000 for 10 years is equivalent to what present amount at an interest rate of 9% compounded (a) quarterly (b) monthly (c) continuously

14. Contemporary Engineering Economics, 4 th edition, © 2007 Solution A = $5,000 0 1 240 Quarters

15. Contemporary Engineering Economics, 4 th edition, © 2007 (a) Quarterly Payment period : Quarterly Interest Period: Quarterly A = $5,000 0 1 2 40 Quarters

16. Contemporary Engineering Economics, 4 th edition, © 2007 (b) Monthly Payment period : Quarterly Interest Period: Monthly A = $5,000 0 1 2 40 Quarters

17. Contemporary Engineering Economics, 4 th edition, © 2007 (c) Continuously Payment period : Quarterly Interest Period: Continuously A = $5,000 0 1 2 40 Quarters

18. Contemporary Engineering Economics, 4 th edition, © 2007 A Decision Flow Chart on How to Compute the Effective Interest Rate per Payment Period