# Interest Formulas – Equal Payment Series

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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 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 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% \$5,000 \$5, \$5, \$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 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 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
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, ) Contemporary Engineering Economics, 4th edition, © 2007