Fundamentals of Structural Analysis

Fundamentals of Structural Analysis

Lecture slides to accompany Engineering Economy, 8thth edition Leland Blank, Anthony Tarquin McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. Chapter 2 Factors: How Time and Interest Affect Money McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. 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 McGraw-Hill Education. Single Payment Factors (F/P and P/F) Single payment factors involve only P and F. Cash flow diagrams are as follows: Formulas are as follows: = (1+) n = [1/(1+ ) ] 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 McGraw-Hill Education.

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 McGraw-Hill Education. Example: Finding Future Value 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) 5000(2.1589) $10,794.50 Answer is (C) McGraw-Hill Education. 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 The cash flow diagram is: (C) $ 33,250 (D) $319,160 Solution: P F(P/F,i,n) 50,000(P/F,10%,5) 50,000(0.6209) $31,045 Answer is (B) McGraw-Hill Education. Uniform Series Involving P/A and A/P The uniform series factors that involve P and A are derived as follows: (1) Cash flow occurs in consecutive interest periods (2) Cash flow amount is same in each interest period The cash flow diagrams are: A ? A Given 0 1 2 P=? 3

4 5 0 1 2 4 P = Given P A(P/A,i,n) Standard Factor Notation A P(A/P,i,n) Note: P is one period Ahead of first A value McGraw-Hill Education. 3 5 Example: Uniform Series Involving P/A 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: Solution: A $5000 0 P ? McGraw-Hill Education. 1 2 i 10% 3 4 5 P 5000(P/A,10%,5) 5000(3.7908) $18,954 Answer is (D) Uniform Series Involving F/A and A/F The uniform series factors that involve F and A are derived as follows: (1) Cash flow occurs in consecutive interest periods (2) Last cash flow occurs in same period as F Cash flow diagrams are: A=? A Given

0 1 2 3 4 0 5 1 2 3 Standard Factor Notation A F(A/F,i,n) Note: F takes place in the same period as last A McGraw-Hill Education. 5 F = Given F=? F A(F/A,i,n) 4

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: F ? i 8% 0 1 2 3 4 A $10,000 McGraw-Hill Education. 5 6 7 Solution:

F 10,000(F/A,8%,7) 10,000(8.9228) $89,228 Answer is (C) 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 McGraw-Hill Education. Example: Untabulated i Determine the value for (F/P, 8.3%,10) Formula: OK Spreadsheet: OK Interpolation: 8% ------ 2.1589 8.3% -----x 9% ------ 2.3674 (Too high) McGraw-Hill Education.

Absolute Error Arithmetic Gradients Arithmetic gradients change by the same amount each period The cash flow diagram for the PG of an arithmetic gradient is: PG ? 1 2 3 4 n 0 G 2G 3G (n-1)G Standard factor notation is ) McGraw-Hill Education. 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 PG is located Two Periods

Ahead of the first change that is equal to G Typical Arithmetic Gradient Cash Flow PT ? i 10% 0 1 400 Amount in year 1 is base amount 2 3 450 4 500 5 550 600 This diagram = this base amount plus this gradient PG ? PA ? 0 Amount in

year 1 is base amount McGraw-Hill Education. 1 i = 10% 2 3 400 400 400 PA 400(P/A,10%,5) i 10% 4 5 400 400 0 + 1 2 50

PG 50(P/G,10%,5) 3 100 4 150 5 200 =+=(/ , %,)+(/ , %,) Converting Arithmetic Gradient to A Arithmetic gradient can be converted into equivalent A value using G(A/G,i,n) 0 1 i 10% 2 3 G 2G i 10% 4 3G 5 0

1 2 3 4 5 A ? 4G General equation when base amount is involved is 0 1 2 G McGraw-Hill Education. 3 2G 4 3G 5 4G For decreasing gradients, change plus sign to minus

= ( /, ,) 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 PT = ? 0 1 400 (B) $1,634 430 G $30 3 (D) $1,829 Solution: PT 400(P/A,12%,5) 30(P/G,12%,5) i12% 2 (C) $1,744 4 5

Year 400(3.6048) 30(6.3970) $1,633.83 Answer is (B) 460 490 520 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 McGraw-Hill Education. Geometric Gradients Geometric gradients change by the same percentage each period Cash flow diagram for present worth of geometric gradient There are no tables for geometric factors Pg? Use following equation for g i: 1 0 A1 2

A 1(1g)1 3 4 n where: A1 cash flow in period 1 A 1(1g)2 g rate of increase Note: g starts between periods 1 and 2 Pg A1{1 [(1g)/(1i)]n}/(i g) A 1(1g)n-1 If g i, Pg A1n/(1i) Note: If g is negative, change signs in front of both g values McGraw-Hill Education. Example: Geometric Gradient 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 (d) $13,550 Solution:

Pg ? 1 (c) $12,670 2 i 12% 3 Pg 1000[1(10.07/10.12)10]/(0.12 0.07) 4 10 $7,333 0 1000 1070 Answer is (b) 1145 g 7% 1838 McGraw-Hill Education. To find A, multiply Pg by (A/P,12%,10) Unknown Interest Rate i Unknowninterest interestrate rateproblems problemsinvolve

involvesolving solvingfor fori,i, Unknown givennnand and22other othervalues values(P, (P,F,F,ororA) A) given (Usuallyrequires requiresaatrial trialand anderror errorsolution solutionororinterpolation interpolationinininterest interesttables) tables) (Usually Procedure: Set up equation with all symbols involved and solve for i contractorpurchased purchasedequipment equipmentfor for$60,000 $60,000which whichprovided providedincome incomeofof$16,000 $16,000per per AAcontractor yearfor for10 10years. years.The

Theannual annualrate rateofofreturn returnofofthe theinvestment investmentwas wasclosest closestto: to: year (a)15% 15% (a) Solution: (b)18% 18% (b) (c)20% 20% (c) (d)23% 23% (d) Can use either the P/A or A/P factor. Using A/P: From A/P column at n 10 in the interest tables, i is between 22% and 24% McGraw-Hill Education. Answer is (d) Unknown Recovery Period n

Unknownrecovery recoveryperiod period problems problemsinvolve involvesolving solvingfor forn,n, Unknown giveni iand and22other othervalues values(P, (P,F,F,ororA) A) given (Likeinterest interestrate rateproblems, problems,they theyusually usuallyrequire requireaatrial trial&&error errorsolution solutionororinterpolation interpolationinininterest interesttables) tables) (Like Procedure: Set up equation with all symbols involved and solve for n contractorpurchased purchasedequipment equipmentfor for$60,000 $60,000that thatprovided

providedincome incomeofof$8,000 $8,000per per AAcontractor year.AtAtan aninterest interestrate rateofof10% 10%per peryear, year,the thelength lengthofoftime timerequired requiredtotorecover recoverthe the year. investmentwas wasclosest closestto: to: investment (a)10 10years years (b) (b)12 12years years (c)15 15years years (d) (d)18 18years years (a) (c)

Solution: Can use either the P/A or A/P factor. Using A/P: From A/P column in i 10% interest tables, n is between 14 and 15 years McGraw-Hill Education. Answer is (c) Summary of Important Points 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, A1 is amount in period 1 To find unknown i or n, set up equation involving all terms and solve for i or n McGraw-Hill Education.

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