Interest Formulas For Single Cash Flows: Lecture No. 6 Contemporary Engineering Economics

Interest Formulas for Single Cash Flows Lecture No. 6 Chapter 3 Contemporary Engineering Economics Copyright © 2016

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Types of Common Cash Flows in Engineering Economics  Single cash flow  Equal (uniform) payment series at regular intervals  Linear gradient series  Geometric gradient series  Irregular (random) payment series

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Equivalence Relationship Between P and F Compounding Process Finding an equivalent future value of a current cash payment

 Discounting Process Finding an equivalent present value of a future cash payment Contemporary Engineering Economics, 6th edition Park

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Singe Cash Flow Formula Compound Amount Factor

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Example 3.7: Find F, Given i, N, and P Given: P = $2,000, i = 10%, N = 8 years

Excel Solution

Find: F F  $2,000(1  0.10)8  $2,000(F / P ,10%,8)  $4,287.18

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A Typical Compound Interest Table at 12%

To find the compound interest factor when the interest rate is 12% and the number of interest periods is 10, we could evaluate the following equation using the interest table.

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Single Cash Flow Formula Present Worth Amount Factor

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Example 3.8: Find P, Given i, N, and F Given: F = $1,000, i

Excel Solution

= 12%, N = 5 years

Find: P P  $1,000(1  0.12)5  $1,000(P / F ,12%,5)  $567.43

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Example 3.9: Find i, Given P, F, and N Given: F = $20, P = $10, N = 5 years Find: i

• Cash Flow Diagram

 Excel Solution

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Example 3.10: Find N, Given P, F, and i Given: P = $6,000, F = $12,000, i = 20% Find: N

• Solving for N

F  2P  P(1  0.20)N 2  1.2N log2  N log1.2 log2 N log1.2  3.80 years

 Excel Solution

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Rule of 72 Approximating how long it will take for a sum of money to double

• Number of Years Required to Double an Initial Investment at Various Interest Rates

72 interest rate (%) 72  20  3.6 years

N

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