© 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.

Geometric Gradient 7. Find i or n© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-3 Single Payment Factors (F/P and P/F) Single payment factors involve only P and F. Cash flow diagrams are as follows: F = P( 1 + i ) n P = F 1 / ( 1 + i ) n Formulas are as follows: Terms in parentheses or brackets are called factors. Values are in tables for i and n values Factors are represented in standard factor notation such as (F/P,i,n) , where letter to left of slash is what is sought; letter to right represents what is given© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-4 F/P and P/F for Spreadsheets Future value F is calculated using FV function: = FV(i%,n,,P) Present value P is calculated using PV function: = PV(i%,n,,F) Note the use of double commas in each function© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-5 Example: Finding F uture V alue A person deposits $ 5000 into an account which pays interest at a rate of 8% per year.

The amount in the account after 10 years is closest to: (A) $ 2,792 (B) $ 9,000 (C) $ 10 ,795 (D) $ 12 ,165 The cash flow diagram is: Solution: F = P(F/P,i,n ) = 5000 (F/P, 8%, 10 ) = $10 ,794.50 Answer is (C) = 5000 (2.1589 )© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-6 Example: Finding Present Value A small company wants to make a single deposit now so it will have enough money to purchase a backhoe costing $ 50 ,000 five years from now. If the account will earn interest of 10 % per year, the amount that must be deposited now is nearest to: (A) $ 10 ,000 (B) $ 31 ,050 (C) $ 33 ,250 (D) $ 319 ,160 The cash flow diagram is: Solution: P = F(P/F,i,n ) = 50 ,000 (P/F, 10 %, 5 ) = 50 ,000 (0.6209 ) = $31 ,045 Answer is (B)© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-7 U niform Series Involving P/A and A/P 0 1 2 3 4 5 A = ? P = Given The cash flow diagrams are: Standard Factor Notation P = A(P/A,i,n) A = P(A/P,i,n) Note: P is one period Ahead of first A value (1) Cash flow occurs in consecutive interest periods The uniform series factors that involve P and A are derived as follows: (2) Cash flow amount is same in each interest period 0 1 2 3 4 5 A = Given P = ?Example: Uniform Series Involving P/A © 2012 by McGraw -Hill, New York, N.

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.

7 , Page # 64 ? Please Work on Problem # 2.8 , Page # 65 © 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 1-11© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-12 Factor Values for Untabulated i or n 3 ways to find factor values for untabulated i or n values Use formula Use spreadsheet function with corresponding P, F, or A value set to 1 Linearly interpolate in interest tables Formula or spreadsheet function is fast and accurate Interpolation is only approximate© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-13 Example: Untabulated i Determine the value for (F/P, 8.3 %, 10 ) Formula: F = ( 1 + 0.083 )10 = 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.

0018 OK OK (Too high)Group Exercise # 2 ? Please Work on Problem # 2.21 , Page # 66 © 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 1-14© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-15 Arithmetic Gradients Arithmetic gradients change by the same amount e ach period The cash flow diagram for the P G of an arithmetic gradient is: 0 1 2 3 n G 2G 4 3G (n -1)G P G = ? G starts between periods 1 and 2 (not between 0 and 1) This is because cash flow in year 1 is usually not equal to G and is handled separately as a base amount (shown on next slide) Note that P G is located Two Periods Ahead of the first change that is equal to G Standard factor notation is P G = G(P/G,i,n)© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-16 Typical Arithmetic Gradient Cash Flow PT = ? i = 10 % 0 1 2 3 4 5 400 450 500 550 600 PA = ? i = 10 % 0 1 2 3 4 5 400 400 400 400 400 PG = ? i = 10 % 0 1 2 3 4 5 50 100 150 200 + This diagram = this base amount plus this gradient PA = 400 (P/A, 10 %, 5) PG = 50 (P/G, 10 %, 5) P T = P A + P G = 400 (P/A, 10 %, 5) + 50 (P/G, 10 %, 5) Amount in year 1 is base amount Amount in year 1 is base amountConverting Arithmetic Gradient to A © 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-17 i = 10 % 0 1 2 3 4 5 G 2G 3G 4G i = 10 % 0 1 2 3 4 5 A = ? Arithmetic gradient can be converted into equivalent A value using G(A/G,i,n) General equation when base amount is involved is A = base amount + G(A/G,i,n) 0 1 2 3 4 5 G 2G 3G 4G For decreasing gradients, change plus sign to minus A = base amount – G(A/G,i,n)© 2012 by McGraw -Hill, New York, N.

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.

25 , Page # 66 © 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 1-19Geometric Gradients © 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-20 Geometric gradients change by the same percentage each period 0 1 2 3 n A1 A 1(1+g) 1 4 A 1(1+g) 2 A 1(1+g) n-1 P g = ? There are no tables for geometric factors Use following equation for g ? i: P g = A 1{1- ( 1+g)/( 1+i) n}/(i -g) where: A 1 = cash flow in period 1 g = rate of increase If g = i, P g = A 1n/( 1+i) Note : If g is negative, change signs in front of both g values Cash flow diagram for present worth of geometric gradient Note : g starts between periods 1 and 2Example: Geometric Gradient © 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-21 Find the present worth of $ 1 ,000 in year 1 and amounts increasing by 7 % per year through year 10 . Use an interest rate of 12 % per year. (a) $ 5,670 (b) $ 7,333 (c) $ 12 ,670 (d) $ 13 ,550 0 1 2 3 10 1000 1070 4 1145 1838 P g = ? Solution: Pg = 1000 1-(1+0.07 /1+0.

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.

Y All Rights Reserved 2-24 Unknown Recovery Period n Unknown recovery period problems involve solving for n, given i and 2 other values (P, F, or A) (Like interest rate problems, they usually require a trial & error solution or interpolation in interest tables) Procedure: Set up equation with all symbols involved and solve for n A contractor purchased equipment for $ 60 ,000 that provided income of $ 8,000 per year. At an interest rate of 10 % per year, the length of time required to recover the investment was closest to: (a) 10 years (b) 12 years (c) 15 years (d) 18 years Can use either the P/A or A/P factor. Using A/P: Solution: 60 ,000 (A/P, 10 %,n) = 8,000 (A/P, 10 %,n) = 0.13333 From A/P column in i = 10 % interest tables, n is between 14 and 15 years Answer is (c)Group Exercise # 5 ? Please Work on Problem # 2.38 , Page # 67 ? Please Work on Problem # 2.45 , Page # 68 © 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 1-25Summary of Important Points © 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-26 In P/A and A/P factors, P is one period ahead of first A In F/A and A/F factors, F is in same period as last A To find untabulated factor values, best way is to use formula or spreadsheet For arithmetic gradients, gradient G starts between periods 1 and 2 Arithmetic gradients have 2 parts, base amount (year 1) and gradient amount For geometric gradients, gradient g starts been periods 1 and 2 In geometric gradient formula, A 1 is amount in period 1 To find unknown i or n, set up equation involving all terms and solve for i or n