© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-1 Slightly Modified Lecture Slides to Accompany Engineering Economy 7th edition By: Prof. Ramzi Taha (Most Credit Goes to Leland Blank & Anthony Tarquin) Chapter 2 Factors: How Time and Interest Affect Money© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-2 LEARNING OUTCOMES 1.

F/P and P/F Factors 2. P/A and A/P Factors 3. F/A and A/F Factors 4. Factor Values 5. Arithmetic Gradient 6.

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Y All Rights Reserved 2-8 A chemical engineer believes that by modifying the structure of a certain water treatment polymer, his company would earn an extra \$ 5000 per year. At an interest rate of 10 % per year, how much could the company afford to spend now to just break even over a 5 year project period? (A) \$ 11 ,170 (B) 13 ,640 (C) \$ 15 ,300 (D) \$ 18 ,950 The cash flow diagram is as follows: P = 5000 (P/A, 10 %, 5 ) = 5000 (3.7908 ) = \$ 18 ,954 Answer is (D) 0 1 2 3 4 5 A = \$ 5000 P = ? i = 10 % Solution:Uniform Series Involving F/A and A/F © 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-9 (1) Cash flow occurs in consecutive interest periods The uniform series factors that involve F and A are derived as follows: (2) Last cash flow occurs in same period as F 0 1 2 3 4 5 F = ? A = Given 0 1 2 3 4 5 F = Given A = ? Note: F takes place in the same period as last A Cash flow diagrams are: Standard Factor Notation F = A(F/A,i,n) A = F(A/F,i,n)© 2012 by McGraw -Hill, New York, N.

Y All Rights Reserved 2-10 Example: Uniform Series Involving F/A An industrial engineer made a modification to a chip manufacturing process that will save her company \$ 10 ,000 per year. At an interest rate of 8% per year, how much will the savings amount to in 7 years? (A) \$ 45 ,300 (B) \$ 68 ,500 (C) \$ 89 ,228 (D) \$ 151 ,500 The cash flow diagram is: A = \$ 10 ,000 F = ? i = 8% 0 1 2 3 4 5 6 7 Solution: F = 10 ,000 (F/A, 8 %, 7 ) = 10 ,000 (8.9228 ) = \$ 89 ,228 Answer is (C)Group Exercise # 1 ? Please Work on Problem # 2.5 , Page # 64 ? Please Work on Problem # 2.

2197 Spreadsheet: = FV( 8.3 %, 10 ,,1) = 2.2197 Interpolation: 8% —— 2.

1589 8.3 % —— x 9% —— 2.3674 x = 2.1589 + ( 8.3 – 8.0 )/( 9.0 – 8.0 ) 2.

3674 – 2.1589 = 2.2215 Absolute Error = 2.2215 – 2.2197 = 0.

Y All Rights Reserved 2-18 Example: Arithmetic Gradient The present worth of \$ 400 in year 1 and amounts increasing by \$ 30 per year through year 5 at an interest rate of 12 % per year is closest to: (A) \$ 1532 (B) \$ 1,634 (C) \$ 1,744 (D) \$ 1,829 0 1 2 3 Year 430 460 4 490 520 P T = ? 5 400 i = 12 % G = \$ 30 = 400 (3.6048 ) + 30 (6.3970 ) = \$ 1,633.

83 Answer is (B) PT = 400 (P/A, 12 %, 5) + 30 (P/G, 12 %, 5) The cash flow could also be converted into an A value as follows: A = 400 + 30 (A/G, 12 %, 5) = 400 + 30 (1.7746 ) = \$ 453.24 Solution:Group Exercise # 3 ? Please Work on Problem # 2.

12 )10/( 0.12 -0.07 ) = \$ 7,333 Answer is (b) g = 7% i = 12 % To find A, multiply P g by (A/P, 12 %, 10 )Group Exercise # 4 ? Please Work on Problem # 2.33 , Page # 67 © 2012 by McGraw -Hill, New York, N.

Y All Rights Reserved 1-22© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-23 Unknown Interest Rate i Unknown interest rate problems involve solving for i, given n and 2 other values (P, F, or A ) (Usually requires a trial and error solution or interpolation in interest tables) A contractor purchased equipment for \$ 60 ,000 which provided income of \$ 16 ,000 per year for 10 years. The annual rate of return of the investment was closest to: (a) 15 % (b) 18 % (c) 20 % (d) 23 % Can use either the P/A or A/P factor. Using A/P: Solution: 60 ,000 (A/P,i%, 10 ) = 16 ,000 (A/P,i%, 10 ) = 0.26667 From A/P column at n = 10 in the interest tables, i is between 22 % and 24 % Answer is (d) Procedure: Set up equation with all symbols involved and solve for i© 2012 by McGraw -Hill, New York, N. 