Interest Formulas – Equal Payment Series Lecture No.7 Chapter 3 Contemporary Engineering Economics Copyright © 2006 Contemporary Engineering Economics, 4th edition, © 2007

Equal Payment Series Equivalent Future Worth F P 1 2 N A A A N 1 2 N A A A P N 1 2 N Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Equal Payment Series – Compound Amount Factor F A A A 1 2 N N 1 2 F = 1 2 N A A A Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Process of Finding the Equivalent Future Worth, F F A A(1+i)N-2 A A A A(1+i)N-1 N 1 2 1 2 N Contemporary Engineering Economics, 4th edition, © 2007

Another Way to Look at the Compound Amount Factor Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Equal Payment Series Compound Amount Factor (Future Value of an Annuity) F 0 1 2 3 N A Example: Given: A = $5,000, N = 5 years, and i = 6% Find: F Solution: F = $5,000(F/A,6%,5) = $28,185.46 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Validation Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Finding an Annuity Value F 0 1 2 3 N A = ? Example: Given: F = $5,000, N = 5 years, and i = 7% Find: A Solution: A = $5,000(A/F,7%,5) = $869.50 Contemporary Engineering Economics, 4th edition, © 2007

Handling Time Shifts in a Uniform Series First deposit occurs at n = 0 i = 6% 0 1 2 3 4 5 $5,000 $5,000 $5,000 $5,000 $5,000 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Annuity Due Excel Solution Beginning period =FV(6%,5,5000,0,1) Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Sinking Fund Factor F 0 1 2 3 N A Example: College Savings Plan Given: F = $100,000, N = 8 years, and i = 7% Find: A Solution: A = $100,000(A/F,7%,8) = $9,746.78 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Excel Solution Given: F = $100,000 i = 7% N = 8 years Find: =PMT(i,N,pv,fv,type) =PMT(7%,8,0,100000,0) =$9,746.78 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Example 3.15 Combination of a Uniform Series and a Single Present and Future Amount Contemporary Engineering Economics, 4th edition, © 2007

Solution: A Two-Step Approach Step 1: Find the required savings at n = 5. Step 2: Find the required annual contribution (A) over 5 years. Contemporary Engineering Economics, 4th edition, © 2007

Comparison of Three Different Investment Plans – Example 3.16 Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solution: Investor A: Investor B: Investor C: Contemporary Engineering Economics, 4th edition, © 2007

How Long Would It Take to Save $1 Million? Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Loan Cash Flows Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Capital Recovery Factor P 1 2 3 N A = ? Example: Paying Off an Educational Loan Given: P = $21,061.82, N = 5 years, and i = 6% Find: A Solution: A = $21,061.82(A/P,6%,5) = $5,000 Contemporary Engineering Economics, 4th edition, © 2007

Example 3.17 Loan Repayment Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 Solution: Using Interest Factor: Using Excel: Contemporary Engineering Economics, 4th edition, © 2007

Example 3.18 – Deferred Loan Repayment Contemporary Engineering Economics, 4th edition, © 2007

Contemporary Engineering Economics, 4th edition, © 2007 A Two-Step Procedure Contemporary Engineering Economics, 4th edition, © 2007

Present Worth factor – Find P, Given A, i, and N 1 2 3 N A Contemporary Engineering Economics, 4th edition, © 2007

Example 3.19 Louise Outing’s Lottery Problem Given: A = $280,000 i = 8% N = 19 Find: P Using interest factor: P =$280,000(P/A,8%,19) = $2,689,008 Using Excel: =PV(8%,19,-280000) Contemporary Engineering Economics, 4th edition, © 2007